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https://www.opensourceagenda.com/projects/jmetalpy
|
[
"# JMetalPy\n\nA framework for single/multi-objective optimization with metaheuristics",
null,
"A paper introducing jMetalPy is available at: https://doi.org/10.1016/j.swevo.2019.100598\n\n## Installation\n\nYou can install the latest version of jMetalPy with `pip`,\n\n``````pip install jmetalpy # or \"jmetalpy[distributed]\"\n``````\nNotes on installing with pip\n\njMetalPy includes features for parallel and distributed computing based on pySpark and Dask.\n\nThese (extra) dependencies are not automatically installed when running `pip`, which only comprises the core functionality of the framework (enough for most users):\n\n``````pip install jmetalpy\n``````\n\nThis is the equivalent of running:\n\n``````pip install \"jmetalpy[core]\"\n``````\n\nOther supported commands are listed next:\n\n``````pip install \"jmetalpy[dev]\" # Install requirements for development\npip install \"jmetalpy[distributed]\" # Install requirements for parallel/distributed computing\npip install \"jmetalpy[complete]\" # Install all requirements\n``````\n\n## Hello, world! 👋\n\nExamples of configuring and running all the included algorithms are located in the documentation.\n\n``````from jmetal.algorithm.multiobjective import NSGAII\nfrom jmetal.operator import SBXCrossover, PolynomialMutation\nfrom jmetal.problem import ZDT1\nfrom jmetal.util.termination_criterion import StoppingByEvaluations\n\nproblem = ZDT1()\n\nalgorithm = NSGAII(\nproblem=problem,\npopulation_size=100,\noffspring_population_size=100,\nmutation=PolynomialMutation(probability=1.0 / problem.number_of_variables, distribution_index=20),\ncrossover=SBXCrossover(probability=1.0, distribution_index=20),\ntermination_criterion=StoppingByEvaluations(max_evaluations=25000)\n)\n\nalgorithm.run()\n``````\n\nWe can then proceed to explore the results:\n\n``````from jmetal.util.solution import get_non_dominated_solutions, print_function_values_to_file, \\\nprint_variables_to_file\n\nfront = get_non_dominated_solutions(algorithm.get_result())\n\n# save to files\nprint_function_values_to_file(front, 'FUN.NSGAII.ZDT1')\nprint_variables_to_file(front, 'VAR.NSGAII.ZDT1')\n``````\n\nOr visualize the Pareto front approximation produced by the algorithm:\n\n``````from jmetal.lab.visualization import Plot\n\nplot_front = Plot(title='Pareto front approximation', axis_labels=['x', 'y'])\nplot_front.plot(front, label='NSGAII-ZDT1', filename='NSGAII-ZDT1', format='png')\n``````",
null,
"## Features\n\nThe current release of jMetalPy (v1.5.7) contains the following components:\n\n• Algorithms: local search, genetic algorithm, evolution strategy, simulated annealing, random search, NSGA-II, NSGA-III, SMPSO, OMOPSO, MOEA/D, MOEA/D-DRA, MOEA/D-IEpsilon, GDE3, SPEA2, HYPE, IBEA. Preference articulation-based algorithms (G-NSGA-II, G-GDE3, G-SPEA2, SMPSO/RP); Dynamic versions of NSGA-II, SMPSO, and GDE3.\n• Parallel computing based on Apache Spark and Dask.\n• Benchmark problems: ZDT1-6, DTLZ1-2, FDA, LZ09, LIR-CMOP, unconstrained (Kursawe, Fonseca, Schaffer, Viennet2), constrained (Srinivas, Tanaka).\n• Encodings: real, binary, permutations.\n• Operators: selection (binary tournament, ranking and crowding distance, random, nary random, best solution), crossover (single-point, SBX), mutation (bit-blip, polynomial, uniform, random).\n• Quality indicators: hypervolume, additive epsilon, GD, IGD.\n• Pareto front approximation plotting in real-time, static or interactive.\n• Experiment class for performing studies either alone or alongside jMetal.\n• Pairwise and multiple hypothesis testing for statistical analysis, including several frequentist and Bayesian testing methods, critical distance plots and posterior diagrams.",
null,
"",
null,
"",
null,
"",
null,
"## Changelog\n\n• [v1.6.0] Refactor class Problem, the single-objective genetic algorithm can solve constrained problems, performance improvements in NSGA-II, generation of Latex tables summarizing the results of the Wilcoxon rank sum test, added a notebook folder with examples.\n• [v1.5.7] Use of linters for catching errors and formatters to fix style, minor bug fixes.\n• [v1.5.6] Removed warnings when using Python 3.8.\n• [v1.5.5] Minor bug fixes.\n• [v1.5.4] Refactored quality indicators to accept numpy array as input parameter.\n• [v1.5.4] Added CompositeSolution class to support mixed combinatorial problems. #69\n\nOpen Source Agenda is not affiliated with \"JMetalPy\" Project. README Source: jMetal/jMetalPy\nStars\n422\nOpen Issues\n27\nLast Commit\n2 months ago\nRepository",
null,
""
] |
[
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"https://www.opensourceagenda.com/projects/docs/source/jmetalpy.png",
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"https://www.opensourceagenda.com/projects/docs/source/_static/NSGAII-ZDT1.png",
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"https://www.opensourceagenda.com/projects/docs/source/_static/2D.gif",
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"https://www.opensourceagenda.com/projects/docs/source/_static/3D.gif",
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"https://www.opensourceagenda.com/projects/docs/source/_static/p-c.gif",
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"https://www.opensourceagenda.com/projects/docs/source/_static/chordplot.gif",
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"https://www.opensourceagenda.com/projects/jmetalpy/reviews/badge.svg",
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|
https://corporatefinanceinstitute.com/resources/knowledge/other/sum-of-squares/
|
[
"# Sum of Squares\n\nA statistical tool that is used to identify the dispersion of data\n\n## What is Sum of Squares?\n\nSum of squares (SS) is a statistical tool that is used to identify the dispersion of data as well as how well the data can fit the model in regression analysis. The sum of squares got its name because it is calculated by finding the sum of the squared differences.",
null,
"This image is only for illustrative purposes.\n\nThe sum of squares is one of the most important outputs in regression analysis. The general rule is that a smaller sum of squares indicates a better model, as there is less variation in the data.\n\nIn finance, understanding the sum of squares is important because linear regression models are widely used in both theoretical and practical finance.\n\n### Types of Sum of Squares\n\nIn regression analysis, the three main types of sum of squares are the total sum of squares, regression sum of squares, and residual sum of squares.\n\n#### 1. Total sum of squares\n\nThe total sum of squares is a variation of the values of a dependent variable from the sample mean of the dependent variable. Essentially, the total sum of squares quantifies the total variation in a sample. It can be determined using the following formula:",
null,
"Where:\n\n• y– the value in a sample\n• ȳ – the mean value of a sample\n\n#### 2. Regression sum of squares (also known as the sum of squares due to regression or explained sum of squares)\n\nThe regression sum of squares describes how well a regression model represents the modeled data. A higher regression sum of squares indicates that the model does not fit the data well.\n\nThe formula for calculating the regression sum of squares is:",
null,
"Where:\n\n• ŷ– the value estimated by the regression line\n• ȳ – the mean value of a sample\n\n#### 3. Residual sum of squares (also known as the sum of squared errors of prediction)\n\nThe residual sum of squares essentially measures the variation of modeling errors. In other words, it depicts how the variation in the dependent variable in a regression model cannot be explained by the model. Generally, a lower residual sum of squares indicates that the regression model can better explain the data while a higher residual sum of squares indicates that the model poorly explains the data.\n\nThe residual sum of squares can be found using the formula below:",
null,
"Where:\n\n• y– the observed value\n• ŷ– the value estimated by the regression line\n\nThe relationship between the three types of sum of squares can be summarized by the following equation:",
null,
""
] |
[
null,
"https://corporatefinanceinstitute.com/resources/knowledge/other/sum-of-squares/",
null,
"https://corporatefinanceinstitute.com/resources/knowledge/other/sum-of-squares/",
null,
"https://corporatefinanceinstitute.com/resources/knowledge/other/sum-of-squares/",
null,
"https://corporatefinanceinstitute.com/resources/knowledge/other/sum-of-squares/",
null,
"https://corporatefinanceinstitute.com/resources/knowledge/other/sum-of-squares/",
null
] |
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|
https://wwwndc.jaea.go.jp/cgi-bin/nuclinfo2014?78,188
|
[
"## 78-Pt-188\n\nSpin\n``` Level energy(keV) Spin & Parity\n----------------------------------------\nground state 0+\n```\nMass (The Ame2012 atomic mass evaluation (II) by M.Wang, G.Audi, A.H.Wapstra, F.G.Kondev, M.MacCormick, X.Xu, and B.Pfeiffer Chinese Physics C36 p. 1603-2014, December 2012)\n``` 187.959388889 ± 0.000006056 (amu) [mass excess = -37829.006 ± 5.641 (keV) ]\n```\nBeta-decay energy (calculated as M(A,Z)-M(A,Z+1), taken from Ame2012)\n` -5552.173 ± 16.391 (keV) `\nStrong Gamma-rays from Decay of Pt-188 (Compiled from ENSDF as of March 2011)\n[ Intensities before May 23th of 2013 were values when total intensity of the decay mode was 100(%) and a branching ratio of each decay mode was not multiplied. ]\n``` γ-ray energy(keV) Intensity(%) Decay mode\n----------------------------------------------------------\n187.59 19.4 EC\n195.05 18.62 EC\n381.43 7.47 EC\n----------------------------------------------------------\n*: relative, ~ approximate, ? calculated or estimatted\n>: greater than or equal to, <: less than or equal to\n[ Intensities; total intensity of the nuclide is 100(%). ]\n```\nDecay data(Chart of the Nuclides 2014)\n``` Decay mode Half-life\nEC 10.2 D 3\n```\nEvaluated Data Libraries\n` Links to the libraries. `\n\nParent Nuclides by Reactions in JENDL-4.0"
] |
[
null
] |
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|
https://mcanv.com/Answers/qa_ssae.html
|
[
"Satellite Speed at Extremes\n\n## Question:\n\nA satellite in moving in an eliptical orbit of eccentricity 0.25. Determine its speed at aphelion (maximum distance) and perihelion (minimum distance). The perihelion is 2e6 meters above the Earth.\n\nA satellite 2e6m above the Earth is 8.378e6m from the Earth's center. In this problem that is given as the perihelion. The eccentricity e=.25 is the ratio of the distance from the center to a focus of the elliptical orbit, cf, to the length of the semimajor axis, a, of the orbit. The semimajor axis, a, is also the sum of the perihelion and cf. This gives us cf/a=0.25 and cf+8.378e6=a. Solving for a we get\na*0.25+8.378e6=a, or, 0.75*a=8.378e6, or a=1.117e7m\ncf=2.7925e6m\n\nThe total energy of the satellite in elliptical is -G*M*m/2*a where G is the universal gravitation constant, M is the mass of the Earth, m is the mass of the satellite and a is the semimajor axis of the orbit. This energy is the sum of the kinetic and potential energy of the satellite. The potential energy is -G*M*m/r where r is the distance from the satellite to the center of the Earth. So we have\n1/2*m*v2-G*M*m/r=-G*M*m/2*a, v2=G*M/r-G*M/2*a\nPlugging in the known values for G, M, r(at perhelion) and a, we get v at perihelion\nv2=2*6.673e-11*5.976e24*(1/8.378e6-1/2.234e7)\nv2=2*6.673e-11*5.976e24*7.46e-8=5.95e7\nv=7.71e3m/s at perihelion.\n\nAt aphelion\nr=cf+a=0.27925e7+1.117e7=1.3962e7\n\nThe speed at aphelion=speed at perihelion times the ratio of the distances\nv=7.71e3*8.378e6/1.396e7=4.63e3m/s\n\nThis information is brought to you by M. Casco Associates, a company dedicated to helping humankind reach the stars through understanding how the universe works. My name is James D. Jones. If I can be of more help, please let me know. JDJ"
] |
[
null
] |
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|
http://iemsjl.org/journal/article.php?code=68285
|
[
" :: Industrial Engineering & Management Systems ::",
null,
"• Editorial Board +\n• For Contributors +\n• Journal Search +\nJournal Search Engine\nISSN : 1598-7248 (Print)\nISSN : 2234-6473 (Online)\nIndustrial Engineering & Management Systems Vol.18 No.3 pp.340-348\nDOI : https://doi.org/10.7232/iems.2019.18.3.340\n\n# A Bi-Objective Model for Multi-Mode Resource Constrained Scheduling Problem to Optimize the Net Present Value of Costs and Makespan\n\nIndustrial Engineering Department, Alzahra University, Tehran, Iran\nCorresponding Author, E-mail: [email protected]\nMay 30, 2017 September 19, 2018 May 7, 2019\n\n## ABSTRACT\n\nIn this study we present a bi-objective model for multi-mode resource constrained scheduling problem in order to minimize Net present value of costs and the makespan simultaneously. In this case, the time limit of availability of renewable resource is considered besides the common constraints of these kinds of problems of the relevant literature. Some of these resources are usable only in a predetermined time interval. These resources have certain availability time and due date, they are not available before the availability time and have penalty costs after due date. Also, the payments are based on progress payment. The main goal is scheduling and determining the method of each activity in order to optimize the objective functions, including net present value and makespan. The proposed model has been solved by NSGAII and NRGA algorithms. Finally, to compare the performances of these algorithms, sample problems of project scheduling problem library are used. The obtained results indicate the better performance of NSGAII.\n\n## 1. INTRODUCTION\n\nThe project scheduling is the most commonly studied problems in the field of project management. Quality and execution of the project largely depends on the scheduling before performing the project. In addition to, scheduling before performing the project can be affected to the some competitive issues such as: delivery on time, reactive scheduling, performing of the project. One of the widely method for project scheduling is critical path method (CPM). But the major obstacle of this method is the absence of consideration resource constraints, while the resource constraint is one of the most effective factors on project scheduling. Therefore one of the suggested method, the resource constrained project scheduling (RCPSP) was introduced. The Project scheduling problem with resource constraint was presented by Wiest (1962). Multimode resource constrained project scheduling Problem (MRCPSP) is generalized of RCPSP. In these problems each activity can perform in several modes. Nonrenewable resources can be considered beside the renewable resource. The first solution of these problems is proposed by Słowiński (1980). These problems are NP-Hard (Blazewicz et al., 1983). Besides, Sprecher and Drexl (1998) illustrated that multi-mode resource constraints project scheduling can’t be solved with large size with respect to the nonrenewable resource in the reasonable time. The exact method including branch and bound, linear programing, branch and cut are proposed by researchers (Słowiński, 1980;Speranza and Vercellis, 1993;Zhu et al., 2006). But due to the low efficiency of these methods in solving cases with big size and complex, the metaheuristic algorithms are used frequently. The objective functions are considered in these problems including minimizing the makespan, the cost of project, maximizing the NPV, time-cost tradeoff. One of the most common non-monetary criteria is the minimum of the makespan. Bianco and Caramia (2012) considered scheduling problem without resource constraint in order to minimize the makespan and proposed an exact algorithm to solve it. Cheng et al. (2015) solved the MRCPSP with nonpreemptive activity with the aim of minimizing the makespan by branch and bound algorithm and compared the result to the preemptive case.\n\nHaving regard to the cash flow in the project scheduling, the Net present value becomes to a significant criteria. Cash flow in the project was introduced by Russell (1970) for the first time. The cash flows according to the four model including lump sum payment; payment at activity completion times, payment at equal time intervals and progress payment are paid. Vanhoucke et al. (2003) considered progress payment model in project scheduling with discounted cash flows and they solved it by branch and bound. Mika et al. (2005) proposed multi-mode resource constraint problem scheduling with discount cash flows. They considered four payments model and used simulated annealing and Tabu search algorithms for solving the model. Chen et al. (2010) proposed Ant colony algorithm for multi-mode resource constraint with discounting cash flows. Wiesemann et al. (2010) considered uncertainty cases in order to maximize NPV. In this model cash flows and durations of each activity have an occurrence probability and solved by branch and bound algorithm. He et al. (2012) solved project payment scheduling by Tabu search and Simulated annealing with the aim of maximizing the NPV where each activity can be executed in several modes.\n\n## 2. LITERATURE REVIEW\n\nThere have been different researches considering MRCPSP problems that could be categorized in makespan minimization (Damak et al., 2009;Deblaere et al., 2011;Afshar et al., 2013), Cost minimization (Ranjbar et al., 2012;Khalilzadeh, 2012), NPV(net present value) minimization (Bey et al., 1981;Kazaz and Sepil, 1996) and multi objective optimization (Afshar et al., 2007;Ke et al., 2009).\n\nCreemers et al. (2015) considered uncertainty cases in project scheduling to maximize the Net present value. In this model the activities have stochastic duration. Based on relevant studies, this paper presents the MRCPSP in order to minimize the NPV and makespan simultaneously. Moreover the time limit for the availability of renewable resource and payment based on progress payment is considered. The rest of this paper is as follows: In section 2, the problem describes. Section 3, presents the model of this study. Section 4, explains the solution approach. Section 5, shows the computational experiment. Finally, the last section presents the conclusions.\n\nBalouka et al. (2015) expressed that most of current works in multimode resource-constrained project scheduling consider value for optimization parameter, yet it’s more sufficient if optimization perform in value, cost and time simultaneously. Researchers also performed some numeric examples to demonstrate proposed model.\n\nLiu et al. (2017) proposed a new approach with decomposition on time windows. Proposed model contain making feasible subspaces for activities and using four different strategies for picking activities and a developed version of serial scheduling scheme. Based on results, proposed model perform better than other two algorithms.\n\nOn the basis of uncertainty theory, Zhao and Ke (2017) performed an uncertain recourse-constrained project scheduling problem (URCSP) programming with uncertain activity durations and NPV maximization of project’s cash flow instead of minimizing makespan in classic RCPSP literature.\n\nZoraghi et al. (2017) emphasized that material ordering in both quantity and time aspects, could be effective on optimization of MRCPSP problems. So researchers considered bonus and penalty policies in programming and developed a MRCPSMO (Multi Mode Resources Constrained Project Scheduling with Material Ordering) model and used PSO(Particle Swarm Optimization)-GA(Genetic Algorithm), GA-GA(Genetic Algorithm- Genetic Algorithm) and SA-GA(Simulated Annealing- Genetic Algorithm) as the model is NP-hard and results revealed that PSO-GA(Particle Swarm Optimization- Genetic Algorithm) performs better than the other two algorithms.\n\n## 3. PROBLEM DESCRIPTION\n\nThe problem studied in this paper is the multi-mode resource constrained aimed to minimize the net present value of costs and makespan. The project involves j = 1, 2, …, n activity where activities 1 and J are dummy activities that are described in an activity-on-node (AON) network G = (V, E) where the node (V) represents the activity and E represents the finish to start precedence relation. These relations enforce that activity j only can started after all of its predecessors Pj are completed. Each activity j can be executed in one of several feasible modes by the set $M j = 1 , … , M .$ The duration activity j in mode m, djm is assumed to be a discrete without preemption. Let R ={1, 2, …, K} the set for renewable resource, we considered its availability per period. The activity j needs rjmk unit of resource for performing in mode m. Also it needs cash flow cfjm for completed. These cash flows are occurred based on progress payment. The amounts of payment Pp are computed as partial completed activity at the end of payment period. In addition, renewable reresources have available time and due date where cannot use these resources before the available time. Also, using these resources after due date is permitted by paying tardiness penalty cost. The objective is to determine a set of non-dominated scheduling and also to assign executive mode for each activity regarding two objective function with given constraints.\n\n## 4. MODEL STATEMENT\n\nAs mentioned in pervious section, the cash flows are paid base on progress payment. In this model, the payments are paid at equal regular time (T). This means at the end of each payment period (T, 2T, …, lT), the value of these payments are computed. This value are calculated based on the partial accomplishment of all activity during period [t-T, t].\n\nFigure 1 displays the simple scheduling of three ac-tivities. The variable sj shows the start time of activity, $d 1 = 20 , d 2 = 13 , d 3 = 50$ shows duration of each activity and considered the cash flows cf1 = cf2 = cf3 = 100 are required for performing each activity. So the value of payment can compute as follows:\n\nt1 = 20: At the end of the first payment period (tp = 20), 10% of activity 1 has been accomplished (y1 = 2/20), the activity 2 hasn’t started yet and 2% of activity 3 has been completed. Thus, the sum of all payments in the first payment period is as follows:\n\n$P 1 = ( ( 2 / 20 ) × 100 ) + ( ( 1 / 50 ) × 100 )$\n\nt2 = 40: At the end of second payment period (tp = 40), the remaining work (18/20 = 9%) of activity 1 has been completed, activity 2 performed during second payment period and 40% of activity 3 accomplished. So the value of payment is computed as follows:\n\n$P 2 = ( ( 18 / 20 ) 100 ) + ( ( 13 / 13 ) × 100 ) + ( ( 20 / 50 ) 100 )$\n\nt3 = 60: In this period only activity 3 has been con-tinued. Thus the value of payment is as follows:\n\n$P 3 = ( ( 20 / 50 ) 100 )$\n\nt4 = 80: Finally, in this period activity 3 has been completed and the payment is calculated as follows:\n\nIn this paper, the model of Vanhoucke et al. (2003) is developed. By the model of Vanhoucke et al. (2003) activities lie in 2 payment periods. While some of the activities due to their longer durations can be laid in more than 2 payment periods.\n\nAccording to the concepts in Table 1, and above mentioned, the mathematical model is presented as follows:\n\n(1)\n\n$min z 2 = ∑ t = E S t n L S t n t . x n 1 t$\n(2)\n\ns.t:\n\n$∑ m = 1 M ∑ t = E S t j L S t j x j m t = 1 ∀ j = 1 , … , n$\n(3)\n\n$∑ m = 1 M ∑ t = E S t i L S t i ( t + d i m ) x i m t ≤ ∑ m = 1 M ∑ t = E S t j L S t j t . x j m t ∀ j = 0 , 1 , … , n i ∈ P R j$\n(4)\n\n(5)\n\n$s j ≥ ρ k ∀ j ∈ N k , k = 1 , 2 , … , R$\n(6)\n\n(7)\n\n$s j = ∑ m = 1 M ∑ t = E S T j L S T j t . x j m t ∀ j = 0 , 1 , … , n m ∈ M j t = E S t j , … , L S t j$\n(8)\n\n$P p = ∑ j = 1 n ∑ m = 1 M y j m t p c f j m ∀ t p = T , 2 T , 3 T , … , l T j = 0 , 1 , … , n m ∈ M j$\n(9)\n\n(10)\n\n(11)\n\n(12)\n\nThe objective function (1), represents the minimization of the net present value, which includes two parts: cash flows and net present value of tardiness penalty costs of renewable resource. In the first part, the net present value of cash flows is calculated. Where the cash flows with respect to the partial work completed of all activity at the end of payment period are calculated. The rest of the objective function (1), computed the net present value of tardiness penalty costs of each renewable resource. The objective function (2), represents the total time of project and should be minimized. Equation (3), ensures that each activity can performed only in one mode and exactly start in one time. Constraint (4), shows the precedence relationships. Constraint (5), shows the quantity of resource k for activity j is performing in mode m. Inequality (6), ensure the starting times of all activity should be greater than or equal to the availability time of their corresponding resources. Constraint (7), shows release time of renewable resource k. Equation (8), represents the starting time of each activity. Equation (9), shows the amount of payment that occurred in the equal time interval T. Equation (10), represents the portion of work completed of activity j is performing in mode m during review period [t-T, t]. Equations (11), and (12), denote the domain of the variables.\n\n## 5. NON-DOMINATED SORTING GENETIC ALGO-RITHM (NSAGII)\n\nIn the optimization problem typically the main goal of optimizing is determined the optimum point (minimum or maximum) of the objective function. But in some cases, several objective functions must be optimize simultaneously that often some of these objective functions are in the conflict with each other. In these situations, the optimal function is generally defined as the weighted linear combination of objective functions, so the problem solved as before. But this way have three obstacles such as 1) how to define the objective function combination, 2) how to determine its weights and 3) loss of some optimal points. Therefore, we have to offer a new definition of optimal point that covers all of the criteria. So the concept of Pareto-optimal solution offered. In this case, we can find a set of solutions so that they are better than the other solutions in the search space. These set of solutions are named Pareto-optimal solution and another solution are named dominated solutions. In the non-dominated solution, all of the solutions are optimal and choosing one point between these optimal solution set are corresponding to the priority and circumstances. Hence in this paper, non-dominated sorting genetic algorithm (NSGAII) is considered. The NSGAII (Goldberg, 1989) revised version of the first NSGA algorithm and advanced by Deb et al. (2002). The solution finding procedures with NSGAII are as follows:\n\n### 5.1 Data Gathering and Selection of Parameters\n\nAt the first, the prerequisite parameters of the model such as number of objective function, decision variable, precedence relationship, number of renewable resource and the amount of availability, number of mode for each activity with its duration and requirement resource, cash flows for executive activity, tardiness penalty costs of resources, their availability time and due date, and discount rate are defined. Then, the parameters of algorithm including population size, number of generations, probability of crossover and mutation should be set up.\n\n### 5.2 Encoding of a Solution\n\nThe solutions are demonstrated by the chromosomes which are composed of two parts: (1) the order of activity list; and (2) the mode assignment list for activity performing. The first part of chromosomes shows the order of activity list according to precedence relationship and resource constraint. In this list, each activity in each situation can appear only after all its predecessors and requirement resources are available for performing. The second part of the list shows the mode selected for performing the activity. Figure 2, shows these structures for six activities. Based on the constraints of problem the initial populations with size N are made up.\n\n### 5.3 Crossover and Mutation Operations\n\nBy applying crossover operation, numbers of child are determined. Based on the crossover rate, some chro-mosomes are selected through the initial population randomly. In this paper, arithmetic crossover is used. Figure 3, shows this crossover. Then, a random number between (0, 1) created. By this number, weighted sum occurred between two selected chromosomes and new chromosome is created. Mutation operation applies to the mode assignment list of activity. A mutation operator adds a unit Gaussian distributed random value to the whole chromosome. The new genes value is clipped if it falls outside of the user-specified lower or upper bounds for that gene. This mutation operator can only be used for integer and float genes.\n\n### 5.4 Combination of Population\n\nIn each iteration, initial population and child population produced by crossover and mutation operation should be integrated together. Then, create archive set.\n\n### 5.5 Forming Pareto Front\n\nThe objective functions are calculated for all mem-bers of archive set. Then, sorting the members based on non-dominated front and assigned front number to them.\n\n### 5.6 Calculating Crowding Distance\n\nSince the members are selected according to the front number and crowding distance, the crowding dis-tance is calculated for the archive members as follows:\n\n(13)\n\nWhere k is the number of objective functions and i is the members of archive set. fmax and fmin are maximum and minimum value of objective function. f(k,i+1) is k-th objective function of (i+1)-th solution and f(k,i-1) is k-th objective function of (i-1)-th solution. The greater value for this parameter leads to divergence and the better range to the population.\n\n### 5.7 Create Next Generation\n\nThe next generation of size N from combined population is produced by using front number and crowding distance. According to this, if two members selected in different front number, the member with lower (the solutions in the lower front number are non-dominated and optimal) front number will be selected. Otherwise, if the front numbers are the same, then the member with the larger crowding distance will be chosen. With this action the diversification of solution is better.\n\nThe above steps till the termination condition con-tinuous. The termination condition defined as (1) setting up maximum iteration (2) not entering a new solution to the optimal front after repeated deterministic iteration.\n\n## 6. NON-DOMINATED ROULETTE GENETIC AL-GORITHM\n\nNRGA and NSGA are basically the same in many aspects like population generation, crossover, mutation and non-dominated solution detection. Yet there is an important difference in picking chromosomes for opera-tions like crossover and mutation. NSGA normally uses binary torment selection for deciding which chromo-somes from which frontiers to be chosen for operations.\n\nNRGA uses roulette wheel process for choosing frontiers and probability for each chromosome to be picked, calculate as below\n\n$p i = f i ∑ i = 1 N f i$\n(14)\n\nfi Indicates distance in this research and it means that chromosomes with bigger distance from others in a frontier, have a bigger chance to be chosen for operations. This logic is based on the fact that diversity in chromosome structures, could make better or even mutated answers in next generations of algorithm.\n\n## 7. ANALYSIS RESULTS\n\nAfter developing model, to determine the effective-ness of the proposed model and also to evaluate the performance of suggested algorithms, 4 groups of standard problems from “project scheduling problem library” with the activity size of j12, j18, j32 and j52 have been selected randomly.1) Each group involves 5 instances and therefore 20 instances solved by each algorithm. Extra parameters for these instances, which do not exist in simple MRCPSP and are obligatory for our model, have been generated randomly with respect to the resource types. These parameters including availability time, due date and tardiness penalty cost per unit of time are set between range (1, 10), (1, 30) and (5, 9) respectively. Table 2 shows a description of the instance used from PSPLIB. Also the file name, the number of activities and the number of modes for each activity are mentioned too.\n\n### 7.1 Comparison Metrics for Evaluating Algorithms\n\nIn this paper, four metrics are used for evaluating algorithms includes:\n\n• Mean ideal distance (MID): By applying this metric, the distance between non-dominated solution and ideal point is calculated. Whatever this metric is lower, the algorithm has better performance (Zitzler and Thiele, 1998). If f1best and f2best are ideal points, this metric can be calculated as the follows:\n\n$M I D = ∑ i = 1 n ( f 1 i − f 1 b e s t ) 2 − ( f 2 i − f 2 b e s t ) 2 n$\n(14)\n\nWhere n is number of non-dominated solution.\n\n• Diversification metric (DM): By calculating this metric according to equation (15), the spread of the solutions are determined. The higher DM indi-cations better algorithm (Zitzler, 1999).\n\n$D M = ( m i n f 1 i − max f 1 i ) 2 − ( m i n f 2 i − max f 2 i ) 2$\n(15)\n\n• Spacing metric (SM): Based on this metric, the uniformity of non-dominated solution spread is calculated. The lower SM is favorable (Srinivas and Deb, 1994). This metric is defined by:\n\n$SM = ∑ i = 1 a − 1 | d ¯ − d i | ( n − 1 ) d ¯$\n(16)\n\nWhere di is the Euclidean distance between consecutive solutions in the found non-dominated solutions, and d ̅ is the average of these distances.\n\n• Number of non-dominated solutions (NOS): This metric shows the number of non-dominated solu-tion that each algorithm can be fined. The higher NDS shows better algorithm (Schaffer, 1985)\n\nThe performance of the proposed algorithm is evaluated by four metrics. The Obtained value of these metrics for 20 instances is described in Table 3. As shown in Table 3, the NSGAII optimal solutions are better than the NRGA. According to the mean ideal distance, the NSGAII has better performance in 85% cases. The obtained solution from this algorithm has a higher closeness to the ideal point. Based on spacing metric, the NSGAII can achieve to the solution set with the lower SM in 55% and also can find more non-dominated solutions. The lower SM shows more uniformity of these non-dominated solutions. For example, we can revert to the obtained results of 12, 16 and 20 instances.\n\nFigure 4, shows the non-dominated solutions ob-tained from algorithms for eight problems which selected randomly from 20 instances. In the most cases the solutions obtained from the NSGAII can dominate the solution obtained by the NRGA algorithm.\n\nFinally, according to mentioned metrics and Figure 4, we can conclude the better performance of NSGAII algorithm in comparison to the NRGA algorithm.\n\n## 8. Conclusion\n\nIn this paper, a bi-objective model for the MRCPSP is presented and the NPV and makespan are minimized concurrently. Besides of the common constraint of these kinds of problems, the availability constraint for the renewable resource is also considered. Further, the payments are based on the equal time intervals and the values of these payments are related to the partial accomplishment activities at the end of payment periods. In order to solve the proposed model as a multi-objective model, two meta-heuristic algorithms including NSGAII and NRGA, were proposed. This has been done by selecting problems from PSPLIB for testing the proposed model of this research. The performances of these algorithms are evaluated by four metrics. The obtained results illustrate that the NSGAII can achieve to the lower MID and SM indicators in the 85% and 55% cases respectively. Furthermore, this algorithm can obtain more non-dominated solutions.\n\n## Figure",
null,
"The simple scheduling of three activities.",
null,
"The structure of chromosome.",
null,
"Crossover operator with α = 0.5.",
null,
"The non-dominated solutions obtained from algorithms for eight problems.\n\n## Table\n\nNotation of symbol employed in this paper\n\nDescription of instances\n\nComputational results of metrics for the algorithms\n\n## REFERENCES\n\n1. Afshar, A. , Kaveh, A. , and Shoghli, O. R. (2007), Multiobjective optimization of time-cost-quality using multi-colony ant algorithm, Aslan Journal of Cicil Engineering, 8(2), 113-124.\n2. Afshar-Nadjafi, B. , Rahimi, A. , and Karimi, H. (2013), A genetic algorithm for mode identity and the resource constrained project scheduling problem, Scientia Iranica, 20(3), 824-831.",
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[
"Dark Mode On/Off\n\n# Java Program to Print X Star Pattern\n\nIn this tutorial, we will see how to print the X star pattern in java First, we will ask the user to initialize the number of rows. Then, we will use different loops to print the X star pattern. But before moving further, if you are not familiar with the concept of the loops in java, then do check the article on Loops in Java.\n\nInput: Enter the number of rows: 7\n\nOutput: The pattern is:\n\n* *\n\n* *\n\n* *\n\n* *\n\n* *\n\n* *\n\n*\n\n* *\n\n* *\n\n* *\n\n* *\n\n* *\n\n* *\n\nThis can be done by using the following methods:\n\nApproach 1: Using a For Loop\n\nApproach 2: Using a While loop\n\nApproach 3: Using a do-while loop\n\nLet us look at each of these approaches for a better understanding.\n\n## Program 1: Java Program to Print the X Star Pattern\n\nIn this program, we will see how to print the X star pattern in java using for loop.\n\n### Algorithm:\n\n1. Start\n\n2. Create an instance of the Scanner class.\n\n3. Declare variables to store the number of rows.\n\n4. Ask the user to initialize the number of rows.\n\n5. Use a for loop to print the pattern.\n\n6. In the inner for loop iterates from j=1 to k and prints charter if j=i or j=k-i+1 displays “*”, else it displays space.\n\n7. This code will execute until the inner for loop condition is false, then it comes to the outer for loop, the for loop will execute until the condition i<=k is false.\n\n8. Display the result.\n\n9. Stop.\n\nLet us look at the below example to understand the implementation of the above algorithm.\n\n``````//Java Program to Print the X Star Pattern Using For Loop\nimport java.util.Scanner;\npublic class Main\n{\npublic static void main(String[] args)\n{\nScanner sc=new Scanner(System.in);\nSystem.out.println(\"Enter the number of rows: \");\nint n=sc.nextInt();\nint k=n*2-1;\nfor(int i=1;i<=k;i++)\n{\nfor(int j=1;j<=k;j++)\n{\nif(j==i || j==k-i+1)\nSystem.out.print(\"*\");\nSystem.out.print(\" \");\n}\nSystem.out.println();\n}\n}\n}\n``````\n\nEnter the number of rows: 6\n* *\n* *\n* *\n* *\n* *\n*\n* *\n* *\n* *\n* *\n* *\n\n## Program 2: Java Program to Print the X Star Pattern\n\nIn this program, we will see how to print the X star pattern in java using a while loop.\n\n### Algorithm:\n\n1. Start\n\n2. Create an instance of the Scanner class.\n\n3. Declare variables to store the number of rows.\n\n4. Ask the user to initialize the number of rows.\n\n5. Use a while loop to print the pattern.\n\n6. While loop checks the condition first then executes the code.\n\n7. First check the condition at while loop i.e i<=k, if it is true, then come to the inner while loop.\n\n8. In the inner while loop first checks the condition j<=k, then execute the code in the loop until the condition is false, then the cursor comes out of the inner loop and goes to the outer loop, this will continue until the condition i<=k is false.\n\n9. Display the result.\n\n10. Stop\n\nLet us look at the below example to understand the implementation of the above algorithm.\n\n``````//Java Program to Print the X Star Pattern Using While Loop\nimport java.util.Scanner;\npublic class Main\n{\npublic static void main(String[] args)\n{\nScanner sc=new Scanner(System.in);\nSystem.out.println(\"Enter the number of rows: \");\nint n=sc.nextInt();\nint i=1;\nint j;\nint k=n*2-1;\nwhile(i<=k)\n{\nj=1;\nwhile(j<=k)\n{\nif(j==i || j==k-i+1)\nSystem.out.print(\"*\");\nSystem.out.print(\" \");\nj++;\n}\nSystem.out.println();\ni++;\n}\n}\n}\n``````\n\nEnter the number of rows: 6\n* *\n* *\n* *\n* *\n* *\n*\n* *\n* *\n* *\n* *\n* *\n\n## Program 3: Java Program to Print Star Pattern\n\nIn this program, we will see how to print the X star pattern in java using a do-while loop.\n\n### Algorithm:\n\n1. Start\n\n2. Create an instance of the Scanner class.\n\n3. Declare variables to store the number of rows.\n\n4. Ask the user to initialize the number of rows.\n\n5. Use a do-while loop to print the pattern.\n\n6. First, execute the inner do-while loop.\n\n7. The code in the inner loop executes until the condition j<=k is false. It prints a character for j=i ,j=k-i+1.Other than these j values prints space.\n\n8. If the condition is false then the cursor comes to the outer do-while loop. The outer do loop execution will stop if the condition i<=k is false.\n\n9. Display the result.\n\n10. Stop\n\nLet us look at the below example to understand the implementation of the above algorithm.\n\n``````//Java Program to Print Star Pattern Using a do-while Loop\nimport java.util.Scanner;\npublic class Main\n{\npublic static void main(String[] args)\n{\nScanner sc=new Scanner(System.in);\nSystem.out.println(\"Enter the number of rows: \");\nint n=sc.nextInt();\nint i=1;\nint j;\nint k=n*2-1;\ndo\n{\nj=1;\ndo\n{\nif(j==i || j==k-i+1)\nSystem.out.print(“*”);\nSystem.out.print(\" \");\nj++;\n}\nwhile(j<=k);\nSystem.out.println();\ni++;\n}while(i<=k);\n}\n}\n``````\n\nEnter the number of rows: 7\n* *\n* *\n* *\n* *\n* *\n* *\n*\n* *\n* *\n* *\n* *\n* *\n* *"
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{"ft_lang_label":"__label__en","ft_lang_prob":0.6560538,"math_prob":0.8780864,"size":4584,"snap":"2022-27-2022-33","text_gpt3_token_len":1196,"char_repetition_ratio":0.16004367,"word_repetition_ratio":0.5058275,"special_character_ratio":0.3006108,"punctuation_ratio":0.15510204,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99455285,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-20T03:19:55Z\",\"WARC-Record-ID\":\"<urn:uuid:906adc03-cee9-41cf-a43a-2363d977aa4d>\",\"Content-Length\":\"255137\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:644d5a81-3f92-468b-a783-9967e71c991b>\",\"WARC-Concurrent-To\":\"<urn:uuid:15eeae0f-1d58-441b-86da-664e4c4b244c>\",\"WARC-IP-Address\":\"65.2.112.116\",\"WARC-Target-URI\":\"https://www.studytonight.com/java-programs/java-program-to-print-x-star-pattern\",\"WARC-Payload-Digest\":\"sha1:YO3DX3PAPNQSSD5F2WSXLKVWU3YBRWEO\",\"WARC-Block-Digest\":\"sha1:NNXC5P52T2D3RLVWS2V6BT4UUANZAJQO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882573876.92_warc_CC-MAIN-20220820012448-20220820042448-00190.warc.gz\"}"}
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http://www.wordsearchfun.com/188749_trucks_wordsearch.html
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[
"trucks\nit is filled with lots of vehicals\n\nLogin to be the first to rate this puzzle!\nBOAT\nBUS\nCAR\nCRANE\nEXCITMENT\nLEARN\nPLANE\nRIDE\nSUV\nTRAIN\nTRANSPORT\nTRAVEL\nTRUCK\nVAN\nVEHICAL\n G V A N I V K P D W V U X W P K E X C I T M E N T N I Z B A Z H R L R K W Q H F Z Q G S Z G I E T J A X F J J G E W D E Y C A R R B N T A X F T W D X N A B U T O H E E T V Q K T M L L O C H T P S A A H L G K R H E U K I L H S U O E E J H O A A Z T A Y V U N B D V V J F B R I N X J C V E A I A M Q N F N E F X L L A B C R R F H V Y S L D T V Y N I A R T Y I N Q R O I D T R R B N W Q O N F Q L B K T G E G D I O O B K A B U D U R H D X U S L E W D B"
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{"ft_lang_label":"__label__en","ft_lang_prob":0.7527424,"math_prob":0.42878395,"size":1120,"snap":"2019-51-2020-05","text_gpt3_token_len":523,"char_repetition_ratio":0.24731183,"word_repetition_ratio":0.17047817,"special_character_ratio":0.6339286,"punctuation_ratio":0.0038314175,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000036,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-17T12:36:37Z\",\"WARC-Record-ID\":\"<urn:uuid:46845498-06ff-4801-93b8-b638b323a19e>\",\"Content-Length\":\"23002\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:50ef3098-f038-4a3b-a0de-9b81e306e307>\",\"WARC-Concurrent-To\":\"<urn:uuid:8ba2cbc5-b8f6-4cbd-95ed-24f830e99e7e>\",\"WARC-IP-Address\":\"88.208.252.230\",\"WARC-Target-URI\":\"http://www.wordsearchfun.com/188749_trucks_wordsearch.html\",\"WARC-Payload-Digest\":\"sha1:LKPWAPIG3XCKJ2WAJHCXKKNXSI3NYCBV\",\"WARC-Block-Digest\":\"sha1:3N7GC5HHTAIIJEZBURGPYRWHSL35YYCA\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250589560.16_warc_CC-MAIN-20200117123339-20200117151339-00329.warc.gz\"}"}
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https://matlab1.com/category/matlab-code/
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[
"## MATLAB TIPS AND TRICKS\n\nVersion Issues One of the key issues to be mindful of is that there are two new versions of MATLAB every year, one in the spring of that year (e.g., 2016a) and one in the fall (e.g., 2016b). Whereas changes in each release are incremental, over time, these changes do...\n\n## Numerical integral of an input function or data set using Simpson’s rule\n\nfunction I = simpsons(f,a,b,n) if numel(f)¿1 % If the input provided is a vector n=numel(f)-1; h=(b-a)/n; I= h/3*(f(1)+2*sum(f(3:2:end-2))+4*sum(f(2:2:end))+f(end)); else % If the input provided is an anonymous function h=(b-a)/n; xi=a:h:b; I=...\n\n## flow rate calculation in the microfluidic channels using the rectangular channel flow equations\n\nThe following protocol calculated flow rate in the microfluidic channels using data obtained from PIV using the rectangular channel flow equations. %Calculates flow rate from velocity according to the exact solution of %rectangular channel flow %Enter...\n\n## flow rate calculation in the microfluidic channels using Purday approximation\n\nflow rate calculation in the microfluidic channels using data obtained from PIV using the Purday approximation. %Program uses the maximum velocity obtained from PIV to calculate the flow %rate %Define time step (us) time step=90; %Channel width (um) w=1500;...\n\n## Sinc interpolation on input waveforms\n\nfunction [yi, ypi] = sincdint(x, y, xi, c) % SINCDINT 1-D piecewise discrete sinc interpolation % SINCDINT(X,Y,XI,C) interpolates to find YI, the values of the % underlying function Y at the points in the array XI, using % piecewise discrete sinc interpolation. X and...\n\n## Analytical Solution of Fick’s 2nd law\n\nAnalytical Solution of Fick’s 2nd law The rate at which the dye mixes with the water is characterized by determining a dispersion coefficient. The analytical solution for equation 5 when a pulse of mass ’M’ is injected at x=0, the concentration distribution over a...\n\n## MATLAB code for Vogel’s Approximation Method\n\n function [minTcost,b,c]=vogel(A,sup,dem) %input:transportationcostA,vectorsupply %sup,vectordemanddem %output:minimumtransportationcostminTcost, %basicmatrixb,costmatrixc, b=zeros(size(A)); ctemp=A;...\n\n## MATLAB code for Least Cost Method\n\nfunction[minTcost,b,c]=leastcost(A,sup,dem) %input:TransportationcostA,supplysup, %demanddem %output:minimumtransportationcost %minTcost,basicmatrixb,costmatrixc [m,n]=size(A); sum=0; rf=zeros; cf=zeros;...\n\n## Support Vector Machine for Speech Recognition\n\nLPC coefficients for three different isolated words are collected.\n\n## Ajudar na implementação do projeto MATLAB\n\nMATLAB ONE é um grupo de MATLAB especialista que ajudá-lo na execução de projectos de MATLAB. Temos mais de nove anos de experiência em MATLAB. Clique para encomendar projeto MATLAB Engenharia Elétrica com máquinas de simulação e elétricos MATLAB Engenharia Elétrica,...\n\n## MATLAB منصوبہ عمل میں مدد\n\nMATLAB ONE MATLAB منصوبوں کے عمل میں آپ کی مدد ہے کہ MATLAB ماہر کے ایک گروپ ہے. ہم MATLAB میں نو سے زائد سالوں کا تجربہ ہے. MATLAB پروجیکٹ آرڈر کرنے کے لئے یہاں دبائیں MATLAB نقلی اور برقی مشینوں کے ساتھ الیکٹریکل انجینئرنگ MATLAB تخروپن کے ساتھ الیکٹریکل انجینئرنگ...\n\n## MATLAB proje uygulama Yardım\n\nMATLAB ONE MATLAB projelerin uygulanmasında size yardımcı MATLAB uzman bir gruptur. Biz MATLAB’te dokuz yıldan fazla deneyime sahip. MATLAB projeyi Sipariş için tıklayın MATLAB simülasyon ve elektrik makinaları ile Elektrik Mühendisliği MATLAB simülasyon...\n\n## Hjälp i MATLAB projektgenomförande\n\nMATLAB ONE är en grupp av MATLAB expert som hjälper dig i genomförandet av MATLAB projekt. Vi har mer än nio års erfarenhet i MATLAB. Klicka för att beställa MATLAB projekt Elektroteknik med MATLAB simulering och elektriska maskiner Elektroteknik med MATLAB simulering...\n\n## Ayudar en la ejecución del proyecto de MATLAB\n\nMATLAB ONE es un grupo de expertos de MATLAB que le ayudan en la ejecución de proyectos de MATLAB. Tenemos más de nueve años de experiencia en MATLAB. Pulse para solicitar proyecto MATLAB Ingeniería Eléctrica con máquinas de simulación MATLAB y eléctricas Ingeniería...\n\n## Помощь в реализации проекта MATLAB\n\nMATLAB ONE представляет собой группу экспертов, которые MATLAB помочь вам в реализации проектов MATLAB. У нас есть более чем через девять лет опыта работы в среде MATLAB. Нажмите, чтобы заказать проект MATLAB Электротехнический с MATLAB моделирования и электрических...\n\n## Hjelp i MATLAB prosjektgjennomføring\n\nMATLAB ONE er en gruppe av MATLAB ekspert som kan hjelpe deg i gjennomføringen av MATLAB prosjekter. Vi har mer enn ni års erfaring i MATLAB. Klikk for å bestille MATLAB prosjekt Electrical Engineering med Matlab simulering og elektriske maskiner Electrical...\n\n## Bantuan dalam MATLAB pelaksanaan projek\n\nMATLAB ONE adalah sekumpulan pakar MATLAB yang membantu anda dalam pelaksanaan projek-projek MATLAB. Kami mempunyai lebih daripada sembilan tahun pengalaman dalam MATLAB. Klik untuk Order projek MATLAB Kejuruteraan Elektrik dengan MATLAB simulasi dan elektrik...\n\n## MATLAB 프로젝트 구현에 도움이\n\nMATLAB ONE은 MATLAB 프로젝트의 구현에 도움이 MATLAB 전문가 그룹입니다. 우리는 MATLAB에서 9 년 이상 경험이 있습니다. MATLAB 프로젝트를 주문하려면 클릭 MATLAB 시뮬레이션 및 전기 기계와 전기 공학 MATLAB 시뮬레이션 전기 공학 MATLAB 시뮬레이션의 생명 공학 MATLAB 시뮬레이션 전기 공학 공학, 전력 전자 및 전기 기계와 MATLAB 시뮬레이션 MATLAB 시뮬레이션 연락처와 전기 공학 파 필드와 전기 공학...\n\n## MATLABのプロジェクト実施中ヘルプ\n\nMATLAB ONEは、MATLABのプロジェクトの実施に役立つMATLABの専門家のグループです。 私たちは、MATLABで9年以上の経験を持っています。 MATLABプロジェクトを注文する場合にクリックします MATLABシミュレーションと電気機械と電気工学 MATLABシミュレーションと電気工学 MATLABシミュレーションと医工学 MATLABシミュレーションと電気工学 工学、パワーエレクトロニクスと電気機械とMATLABシミュレーション MATLABシミュレーションコンタクトと電気工学...\n\n## Aiuto nella realizzazione dei progetti MATLAB\n\nMATLAB ONE è un gruppo di esperti MATLAB che ti aiutano nella realizzazione di progetti di MATLAB. Abbiamo più di nove anni di esperienza in MATLAB. Clicca per ordinare progetto MATLAB Ingegneria Elettrica con MATLAB simulazione e macchine elettriche Ingegneria...\n\n## Hilfe in MATLAB Projektumsetzung\n\nMATLAB ONE ist eine Gruppe von MATLAB Experten, die Sie bei der Implementierung von MATLAB-Projekten helfen. Wir haben mehr als neun Jahre Erfahrung in MATLAB. Klicken Sie auf MATLAB Projekt zu bestellen Elektrotechnik mit MATLAB Simulation und elektrische Maschinen...\n\n## Hulp in MATLAB uitvoering van het project\n\nMATLAB ONE is een groep van MATLAB deskundigen die u helpen bij de uitvoering van MATLAB projecten. We hebben meer dan negen jaar ervaring in MATLAB. Klik hier om MATLAB project bestellen Elektrotechniek met MATLAB simulatie en elektrische machines Elektrotechniek...\n\n## Pomoc při realizaci projektu MATLAB\n\nMATLAB ONE je skupina MATLAB odborníka, který vám pomůže při realizaci projektů MATLAB. Máme více než devět let zkušeností v prostředí MATLAB. Kliknutím objednat projekt MATLAB Elektrotechnika s MATLAB simulačních a elektrických strojů Elektrotechnika s...\n\n## 帮助MATLAB中的项目实施\n\nMATLAB ONE是一组MATLAB的专家,他帮助你在执行MATLAB项目。 我们在MATLAB九年多的经验。 点击订购MATLAB项目 电气工程与MATLAB仿真和电机 电气工程与MATLAB仿真 生物医学工程与MATLAB仿真 电气工程与MATLAB仿真 MATLAB仿真与工程,电力电子与电机 电气工程与MATLAB仿真与我们联系 波场和电气工程与我们联系MATLAB仿真 电气工程与光通信的MATLAB仿真 电气工程与MATLAB仿真系统 电气工程与MATLAB仿真牢固联系 电气工程与电信网络的MATLAB仿真...\n\n## 幫助MATLAB中的項目實施\n\nMATLAB ONE是一組MATLAB的專家,他幫助你在執行MATLAB項目。 我們在MATLAB九年多的經驗。 點擊訂購MATLAB項目 電氣工程與MATLAB仿真和電機 電氣工程與MATLAB仿真 生物醫學工程與MATLAB仿真 電氣工程與MATLAB仿真 MATLAB仿真與工程,電力電子與電機 電氣工程與MATLAB仿真與我們聯繫 波場和電氣工程與我們聯繫MATLAB仿真 電氣工程與光通信的MATLAB仿真 電氣工程與MATLAB仿真系統 電氣工程與MATLAB仿真牢固聯繫 電氣工程與電信網絡的MATLAB仿真...\n\nIntroduction Basic Matlab/Scilab Instructions (ipynb|web) Introduction to Signal Processing (ipynb|web) Introduction to Image Processing (ipynb|web) Image Approximation with Fourier and Wavelets (ipynb|web) Image Processing with Wavelets (ipynb|web) Le...\n\n## MATLAB implementation of Local Histogram Equalization\n\nA standalone MATLAB implementation of Local Histogram Equalization using the Bilateral Grid. http://people.csail.mit.edu/jiawen/software/lhe-1.1.zip\n\n## MATLAB code for electromagnetic analysis of high field MRI problems\n\nMagnetic Resonance Integral Equation suite. An open source MATLAB code for the electromagnetic analysis of high field MRI problems with realistic human body models, including transmit and receive coil analysis. https://github.com/thanospol/MARIE\n\n## Squid (Stable quasiconvex identification)\n\nSquid (Stable quasiconvex identification) is a open-source matlab tool for the generation of reduced order linear system models from available transfer function data points. The tool is based on a quasi-convex relaxation of a system identification problem subject to...\n\n## Optical Flow Matlab/C++ Code\n\nreference : C. Liu. Beyond Pixels: Exploring New Representations and Applications for Motion Analysis. Doctoral Thesis. Massachusetts Institute of Technology. May 2009. Algorithm The core of the algorithm is based on and . The major difference is that I used...\n\n## MATLAB Code for cell counting\n\nImage segmentation techniques can be used for separating the RBC, WBC and Platelets in an image. The prime reason for segmenting the image is to define the boundaries of the blood cells enabling features to be extracted without the inclusion of extraneous material....\n\n## MATLAB code for Test Case Prioritization\n\nMATLAB code for Test Case Selection and Prioritization A proposed approach for test case prioritization problem based on Genetic Algorithm and Ant Colony Optimization Nowadays, Software are used in each computer. There are many application for them...."
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{"ft_lang_label":"__label__en","ft_lang_prob":0.71344304,"math_prob":0.80818456,"size":1205,"snap":"2020-24-2020-29","text_gpt3_token_len":279,"char_repetition_ratio":0.1540383,"word_repetition_ratio":0.01898734,"special_character_ratio":0.19751038,"punctuation_ratio":0.1826484,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9997913,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-15T02:03:32Z\",\"WARC-Record-ID\":\"<urn:uuid:aec524ca-1299-4249-a196-e040b97bbac0>\",\"Content-Length\":\"99340\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:15f005e7-02ff-49ee-8916-3b2ae7c3e8b0>\",\"WARC-Concurrent-To\":\"<urn:uuid:3b0a8e7f-6bc4-42a4-8096-072369f73018>\",\"WARC-IP-Address\":\"104.31.78.69\",\"WARC-Target-URI\":\"https://matlab1.com/category/matlab-code/\",\"WARC-Payload-Digest\":\"sha1:LCRA6BY66EC6HZV5I6775IV7WB7P3HAE\",\"WARC-Block-Digest\":\"sha1:T6SOZ2DOQ22PPQ5YCIYLT4OKRKC23SEF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593657154789.95_warc_CC-MAIN-20200715003838-20200715033838-00059.warc.gz\"}"}
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http://talkstats.com/threads/simplet-test-and-confidence-intervall-population-size.3419/
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[
"# Simplet-test and confidence intervall, population size???\n\n#### karvin\n\n##### New Member\nHi everyone,\n\nI want to determine the expected values for a certain test. Have I calculated correctly and used the correct test below? The data was collected doing random tests. I know the sample population is very small – can I still use this test if I assume that the data is normally distributed to get an indication on what values to expect if random tests are done?\n\nX (mean) = 12.29\ns2 (variance) = 0.058095\ns (standard deviation) = 0.24103\nN = 12.29\nd.f (degrees of freedom) = 6\nP (probability level) = 0.05 (95 %)\nt value, two-tailed (from table) = (P=0.05/2=0.025 , d.f. = 6) = 2.45\n\nX ± t×s/√N = 12,29 ± 2,45 × 0,24103/√7 = 12,29 ± 0,22 →[12,07 ;12,51] 95%\n\nie can expect (with a 95% confidence level) that the value will lie between 12.07 and 12.51\n\nThanks for any input,\n\nJonas"
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https://www.osapublishing.org/oe/fulltext.cfm?uri=oe-23-15-A987&id=323046
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[
"## Abstract\n\nThe BP09 experiment conducted by the Centre for Maritime Research and Experimentation in the Ligurian Sea in March 2009 provided paired vertical profiles of nadir-viewing radiances Lu(z) and downward irradiances Ed(z) and inherent optical properties (IOPs, absorption, scattering and backscattering coefficients). An inversion algorithm was implemented to retrieve IOPs from apparent optical properties (AOPs, radiance reflectance RL, irradiance reflectance RE and diffuse attenuation coefficient Kd) derived from the radiometric measurements. Then another inversion algorithm was developed to infer vertical profiles of water constituent concentrations, including chlorophyll-a concentration, non-algal particle concentration, and colored dissolved organic matter from the retrieved IOPs based on a bio-optical model. The algorithm was tested on a synthetic dataset and found to give reliable results with an accuracy better than 1%. When the algorithm was applied to the BP09 dataset it was found that good retrievals of IOPs could be obtained for sufficiently deep waters, i.e. for Lu(z) and Ed(z) measurements conducted to depths of 50 m or more. This requirement needs to be satisfied in order to obtain a good estimation of the backscattering coefficient. For such radiometric measurements a correlation of 0.88, 0.96 and 0.93 was found between retrieved and measured absorption, scattering and backscattering coefficients, respectively. A comparison between water constituent values derived from the measured IOPs and in-situ measured values, yielded a correlation of 0.80, 0.78, and 0.73 for chlorophyll-a concentration, non-algal particle concentration, and absorption coefficient of colored dissolved organic matter at 443 nm, respectively. This comparison indicates that adjustments to the bio-optical model are needed in order to obtain a better match between inferred and measured water constituent values in the Ligurian Sea using the methodology developed in this paper.\n\n© 2015 Optical Society of America\n\n## 1. Introduction\n\nSimultaneous measurements of the upward and downward irradiances, Eu(z) and Ed(z), or the nadir-viewing radiance Lu(z) combined with Ed(z), where z is the water depth, are among the most common data acquisition products in hydrologic optics. The ratios Eu(z)/Ed(z) and Lu(z)/Ed(z) are apparent optical properties (AOPs) that depend not only on the water and its constituents including embedded particles, but also on the illumination conditions, i.e. the angular distribution of the radiance throughout the water column. In contrast the inherent optical properties (IOPs), i.e. the absorption coefficient, a(z), the scattering coefficient, b(z), and the scattering phase function, p(z, Θ), where Θ is the scattering angle, depend exclusively on the properties of the water and its constituents including embedded particles, and are independent of the illumination. The radiative transfer equation (RTE) provides a connection between IOPs and radiances and irradiances, which are commonly measured, and from which AOPs including Eu(z)/Ed(z) and Lu(z)/Ed(z) are obtained. Hence, for a specified set of IOPs, the solution of the RTE provides the information required to determine the AOPs. Since the radiance reflectance defined as\n\n$RL(z)≡Lu(z)/Ed(z),$\nis obtained from Lu(z) and Ed(z), which are relatively easy to measure compared to the IOPs, a desirable goal is to infer the IOPs from measured profiles of the AOPs, in general, and from the irradiance reflectance RE(z) = Eu(z)/Ed(z) or the radiance reflectance RL(z), in particular. This nonlinear inversion problem is much more challenging than solving the forward problem involving the linear RTE. Gordon and Boynton showed that it is possible to invert the radiances and irradiances to retrieve the IOPs for a homogeneous water column, and that an algorithm based on Eu(z) and Ed(z) profiles was more robust than one based on profiles of Lu(z) and Ed(z). An extension to make the method work for stratified waters was described in another paper by Gordon and Boynton .\n\nThe IOPs of the water column consist of contributions from the pure sea water as well as from organic and inorganic suspended particles, i.e. pigmented (chlorophyll) and mineral particles, and colored dissolved organic matter or CDOM for short. For simplicity we will refer to these optically active constituents due to all embedded particles and dissolved matter as water constituents in the remainder of this paper. Once the IOPs are inverted from the irradiance reflectance RE(z) = Eu(z)/Ed(z) or the radiance reflectance RL(z) = Lu(z)/Ed(z), a further analysis can be performed to determine the contribution of each of the water constituents and the results will give a vertical distribution of water constituents throughout the water column.\n\nIn this paper we will describe our attempt to numerically implement the algorithm described in the papers by Gordon and Boynton [1, 2], and the application to field measured Lu(z) and Ed(z) data for deep ocean waters. We then describe an algorithm developed to retrieve vertical profiles of water constituents from the retrieved IOP data.\n\n## 2. AOP and IOP dataset\n\nThe AOP data derived from Lu(z) and Ed(z) measurements and paired IOP data that will be used in this paper are from the BP09 experiment which was conducted in the Ligurian Sea in March 13–26, 2009 by the NATO Centre for Maritime Research and Experimentation (CMRE) and collaborating institutions. Figure 1 shows 14 locations where measurements were conducted.",
null,
"Fig. 1 Location of 14 Lu(z), Ed(z), and IOP measurement conducted in the BP09 experiment used in this paper, see Table 1 for details.\n\nThe Lu(z) and Ed(z) measurements were performed with a Satlantic Hyper ProII radiometer package working in profiling mode. The free-falling profiler was configured with two hyper-spectral radiometric sensors measuring downwelling irradiance Ed(z) and upwelling radiance Lu(z). Another set of radiometric sensors was mounted on the deck measuring the solar irradiance. The measurements were only performed in cloud free daylight conditions down to a maximum depth of 90 m. The measured Lu(z) and Ed(z) data have a spectral resolution of 3.3 nm from 400 to 750 nm and a depth resolution of 0.1 m. The Lu(z) and Ed(z) measurements were performed in the multicast mode in the top 10–15 m and then in a single-cast mode sampling further down in the water column until the irradiance was reduced to 1% of the surface irradiance, otherwise down to either the ocean bottom or 80 m due to limited cable length. The Lu(z) and Ed(z) data were normalized and rescaled in a post-processing step to account for fluctuations in the solar irradiance at the ocean surface during the period of the measurements, and the multicast and single cast data were then merged into one profile. An interpolation between wavelengths was also performed during the data merging process and the merged Lu(z) and Ed(z) measurements have a spectral resolution of 5 nm from 400 nm to 750 nm.\n\nThe IOP measurements were performed with a WET Labs AC9+ and a BB9 spectrophotometer mounted on the University of Strathclyde IOP frame, in addition to a CTD. The AC9+ measures absorption, a(z), and attenuation, c(z), coefficients at 9 wavelengths centered at 412, 440, 488, 510, 532, 555, 650, 676 and 715 nm. The BB9 instrument measures the backscattering coefficient, bb(z), at the same wavelengths except that 555 nm is replaced by 595 nm. The backscattering coefficient at 555 nm was interpolated from the measurements at 532 and 595 nm. Calibration was performed using ultrapure water from a number of sources, and the measured IOPs agree well among these sources. The raw IOP dataset was then corrected for temperature, salinity and averaged over the multicast measurements in post-processing. A modified version of the commonly used “proportional” scattering correction method was applied to the AC9+ data to correct for errors due to scattering in the absorption coefficients. However, recent studies by McKee et al. and Röttgers et al. showed that this scattering correction method leads to an underestimation of absorption coefficients which is more significant at longer wavelengths and under turbid water conditions.\n\nThe IOP [a(z), c(z), bb(z)] and Ed(z), Lu(z) measurements do match exactly in wavelength. IOP measurements are available at 412, 440, 488, 510, 532, 555, 650, 676 and 715 nm, while Lu(z) and Ed(z) measurements are available between 400 nm and 750 nm with 5 nm spectral resolution. In our study, we chose the Lu(z) and Ed(z) data whose wavelengths are closest to the wavelengths of the IOP data and we only used data from the six shortest wavelengths at 412, 440, 488, 510, 532 and 555 nm. We did not use data for longer wavelengths because (i) the radiative transfer model used in our IOP inversion algorithm does not take into account inelastic (Raman) scattering, and (ii) the uncertainties in the IOP measurements at the longer wavelengths are large.\n\nThe calibration data for the AC9 instrument show that the random measurement uncertainty for all wavelengths was ±0.003 m−1 for absorption coefficients and ±0.002 m−1 for attenuation coefficients. In the BP09 dataset, measured absorption coefficients a(z) and attenuation coefficients c(z) have overall average values of 0.06 m−1 and 0.7 m−1, which means that the uncertainty is ±5% for absorption coefficients and less than ±1% for attenuation coefficients. The uncertainty in the Lu(z) and Ed(z) measurements are mostly due to surface waves, cloud cover and variation in solar zenith angle, but the post-processing of the Lu(z) and Ed(z) data reduced the uncertainty by merging multicast data and rescaling the data to account for the fluctuation of the solar irradiance. Therefore, the overall uncertainty of the Lu(z) and Ed(z) measurements is ±5% of the signal. The processed Lu(z) and Ed(z) data after multicast and merging are within the 95% confidence interval of the signal.\n\n## 3. Inverting IOPs from AOPs\n\n#### 3.1. Radiative transfer model\n\nRadiative transfer describes the interaction of the radiation field (i.e. the incoming solar radiation) with particles and molecules in the atmosphere and ocean, and provides a connection between IOPs and AOPs, such as the irradiance reflectance RE(z) = Eu(z)/Ed(z) or the radiance reflectance RL(z) = Lu(z)/Ed(z). In a slab geometry, the radiative transfer equation (RTE) can be written as :\n\n$μdL(τ,μ,ϕ)dτ=L(τ,μ,ϕ)−S*(τ,μ,ϕ)−ϖ(τ)4π∫02πdϕ′∫−11dμ′p(τ;μ′,ϕ′;μ,ϕ)L(τ,μ′,ϕ′)$\nwhere L(τ, μ, ϕ) is the radiance, μ is the cosine of the polar angle, (z) = −c(z)dz = −(a(z) + b(z))dz is the optical depth, ϕ is the azimuth angle, ϖ(τ) ≡ b(τ)/(a(τ) + b(τ)) is the single scattering albedo, and p(τ; μ′, ϕ′; μ, ϕ) is the phase function describing scattering from direction (μ′, ϕ′) to (μ, ϕ). The second term on the right hand side of the RTE is called the source term: $S*(τ,μ,ϕ)=ϖ(τ)F04πp(τ;μ0,ϕ0;μ,ϕ)e−τ/μ0$, where F0 is the solar irradiance, μ0 and ϕ0 are the cosine of the solar zenith and the corresponding azimuth angle, respectively. The third term on the right hand side is due to multiple scattering.\n\nSolving the RTE for the azimuthally-averaged radiance L(τ, μ), and evaluating it at μ = 1 (nadir direction), one obtains the upward radiance Lu(τ) in the nadir-viewing direction, and integrating L(τ, μ) multiplied by μ over all upward or downward directions, one obtains the upward irradiance Eu(τ) and downward irradiance Ed(τ):\n\n$Eu(τ)=2π∫01dμμL(τ,μ).$\n$Ed(τ)=2π∫01dμμL(τ,−μ).$\nSimilarly, integrating the radiance L(τ, μ) over all directions, one obtains the scalar irradiance E0(τ):\n$E0(τ)=2π∫−11dμL(τ,μ).$\n\nThere are many methods available to solve the RTE. The forward radiative transfer model (RTM) used by Gordon and Boynton [1, 2] was based on Monte Carlo simulations. In their original papers, Gordon and Boynton tested the algorithm at the top 5 meters of the water body. The Monte Carlo forward model is very computer-intensive and becomes impractical for operational use. To circumvent this problem we used AccuRT, which is a RTM for the coupled atmosphere-ocean system, based on the discrete-ordinate method [7–10]. The AccuRT RTM is accurate, well-tested, reliable, and orders of magnitude faster than Monte Carlo simulations. Hence, we can easily extend the application to arbitrarily deep waters and test the algorithm for such cases.\n\n#### 3.2. IOP inversion algorithm\n\nThe IOP inversion algorithm employs the entire vertical profile of Lu(z) and Ed(z) to retrieve the profile of IOPs, a(z), b(z) and bb(z) iteratively, wavelength by wavelength. The theoretical basis of the IOP inversion algorithm described in the papers by Gordon and Boynton [1, 2] can be briefly summarized as follow:\n\n1. An approximate value of the absorption coefficient, a(z), is obtained from Gershun’s Law :\n$a(z)=μ¯(z)Kv(z)$\nwhere the average cosine μ̄(z) ≡ [Ed(z) − Eu(z)]/E0(z), E0(z) is the scalar irradiance, and $Kv(z)≡−ddz{ln[Ed(z)−Eu(z)]}$.\n2. Gordon and Boynton proposed an approximation of the backscattering coefficient given by:\n$bb(z)=a(z)X(z)$\nwhere\n$X(z)=3{RE(z)−dRE(z)dz∫zzmaxdz′[Ed(z′)Ed(z)]2}$\nand RE(z) ≡ Eu(z)/Ed(z).\n3. The scattering coefficient, b(z), can be computed from bb(z), if we know the scattering phase function p(cosΘ), as follows:\n$b(z)=bb(z)b˜b(z)=bb(z)2π∫π/2πp(cosΘ)sin(Θ)dΘ$\n$b˜b(z)=2π∫π/2πp(cosΘ)sin(Θ)dΘ$ is called the backscatter ratio, or the backscatter fraction.\nIn the papers of Gordon and Boynton [1, 2], both the EuEd and the LuEd algorithms were tested. Since the AOP data we have available are radiance reflectances RL(z) = Lu(z)/Ed(z), we will focus on the LuEd algorithm and some modifications are required to make it more suitable for application to real (as opposed to synthetic) data.\n\nA flow chart of the LuEd algorithm is shown in Fig. 2. To start the algorithm, a first guess of IOPs (a, b, and the scattering phase function) must be provided. We have found that a good first guess really helps reduce the number of iterations. In the absence of measured Eu(z) and E0(z), Gordon and Boynton suggested that μ̄(z) can be replaced by μw = cosθw, the cosine of the solar zenith angle θw measured in water, and that an approximate value of the upward irradiance, $Eum(z)$, can be estimated from the measured radiance, Lu(z), using the Q factor, Q(z) ≡ Eu(z)/Lu(z), and they suggested using Q(z) = π as a starting value. We have found that the Q factor may vary from π for isotropic scattering to about 10 for a strongly forward-scattering medium. Since the water column is inhomogeneous and the scattering in the water is anisotropic, using π as the Q factor at all depths may not be a good choice. In our algorithm we used an alternative approach to select a first guess, by using the radiance reflectance RL(z) given by Eq. (1) and the diffuse attenuation coefficient Kd(z), given by\n\n$Kd(z)≡−d[ln(Ed(z))]/dz.$\nMcKee et al. suggested a method of retrieving absorption and backscattering coefficients by combining the proposal made by Morel and Genitili and Gordon . In 1989, Gordon suggested a relation between the diffuse attenuation coefficient Kd(z) and the absorption and backscattering coefficients:\n$Kd(z)≈1.0395a(z)+bb(z)cos(θw)$\nwhere θw is the solar zenith angle measured in water. In 1993, Morel and Genitili suggested a relation between the backscattering to absorption ratio (bb/a) and the radiance reflectance:\n$RL(z)≈0.094bb(z)a(z).$\n\nBy combining Eqs. (11) and (12), an estimation of a and bb can be obtained:\n\n$a(z)≈Kd(z)cos(θw)1.0395[RL(z)0.094+1]$\n$bb(z)≈Kd(z)cos(θw)1.0395[0.094RL(z)+1].$\nMcKee et al. used Eqs. (13) and (14) to retrieve IOPs from Lu(z) and Ed(z) measurements. In our study, we found the estimation from these two equations to be insufficient for accurate inversion, however, they provide good starting values for our algorithm. From measured Lu(z) and Ed(z) profiles, we compute RL(z) from Eq. (1), Kd(z) from Eq. (10), and θw from the time and location of each measurement. Then we use Eqs. (13) and (14) to provide a first guess of the IOPs, i.e. a(0)(z) and $bb(0)(z)$.",
null,
"Fig. 2 Flow chart of the IOP inversion algorithm.\n\nAn appropriate scattering phase function must be provided to the radiative transfer model. The scattering properties of the water column are a combination of scattering by water molecules and by suspended particles. The fraction of scattering particles versus molecules varies with depth and the composition of the particles changes throughout the water column. Therefore, the scattering properties will depend also on wavelength. The angular variation of scattering by water molecules can be described by a Rayleigh phase function, but the scattering by suspended particles depends on the particle size distribution, refractive index, and wavelength. Hence it is difficult to find one phase function that fits all water types. In our study, we found that a good estimation of the backscattering ratio is critical to achieve a good estimation of scattering coefficients. The choice of the phase function is less important if the backscattering ratio is correct. In our algorithm we implemented a combined phase function of Rayleigh, Petzold and Fournier-Forand (FF) (Fig. 3). We assumed that there are two types of particles in the water, algal particles (algae or phytoplankton) and non-algal particles (detritus, heterotrophic organisms and minerals). To describe the angular variation of the algal particle scattering, we used the FF phase function [15, 16]:\n\n$pFF(cosΘ)=14π(1−δ)2δν{ν(1−δ)−(1−δν)+4u2[δ(1−δν)−ν(1−δ)]}+1−δ180ν16π(δ180−1)δ180ν[3cos2Θ−1]$\nwhere ν = 0.5(3 − γ), and γ is the slope of the particle size distribution function (assumed to be a Junge or power law distribution), which typically varies between 3.0 and 5.0, u = 2 sin(Θ/2), $δ≡δ(Θ)=u23(m−1)2$, $δ180=δ(Θ=180°)=43(m−1)2$, Θ is the scattering angle, and m is the refractive index. In a previous study, Li et al. used m = 1.0686, and γ = 3.38, which correspond to a backscattering ratio of 0.0056. As noted by Mobley et al. , this choice of [m, γ] values is consistent with a certain mixture of living microbes.\n\nFor non-algal particles, we chose the turbid water Petzold phase function, which was derived from measurements made in San Diego Harbor and tabulated by Mobley . The water in San Diego Harbor is a mixture of microbes and sediments, with a backscattering ratio of 0.021. This phase function is a reasonable choice for the non-algal particles.\n\nWe used the Rayleigh phase function to represent molecular scattering by pure water:\n\n$pw(cosΘ)=33+f(1+fcos2Θ)$\nwhere $f=1−ρ1+ρ$, and ρ is the depolarization ratio, attributed to the anisotropy of the scatterer, which was set to be ρ = 0.0899.\n\nThe moment-fitting method of Hu et al. was used to compute Legendre expansion coefficients χℓ,PET and χℓ,FF for the Petzold and FF scattering phase functions. The total IOP scattering phase function Legendre expansion coefficients χ are given by\n\n$χℓ=fFF×χℓ,FF+fPET×χℓ,PET+(1−fFF−fPET)×χℓ,water$\nwhere fFF and fPET are the fractions of particles that scatter according to the FF and Petzold phase functions. These fractions were tuned so that the total backscattering ratio matches the measured one.\n\nThe backscattering ratio can be derived from IOP measurements, since b(z) = bb(z)/b(z), and we have considered it to be known information. To find the fraction of the FF and Petzold phase function, we first subtract the scattering of the pure water and compute the backscattering ratio of suspended particles:\n\n$b˜bp(z)=bb(z)−bbw(z)b(z)−bw(z)$\nfrom which a fraction of the FF phase function, f′FF, can be derived to satisfy\n$b˜bp(z)=fFF′(z)×0.0056+[1−fFF′(z)]×0.021.$\nIf f′FF(z) < 0, we set f′FF(z) = 0; if f′FF(z) > 1, we set f′FF(z) = 1. The fractions of the FF and Petzold phase function can be written as:\n$fFF(z)=fFF′(z)×b˜bp(z)b˜b(z)$\n$fPET(z)=[1−fFF′(z)]×b˜bp(z)b˜b(z).$\nThe algorithm consists of the following steps:\n1. The first guess IOPs (a, b and phase function) are used in the RTM to generate: Ed(z), Eu(z), Lu(z) and E0(z).\n2. From the RTM generated radiance and irradiance values, new estimates of μ̄(1)(z) and Q(1)(z) are obtained.\n3. Then Q(1)(z) will yield a new estimate of $Eum(z)$ and $Kvm(z)$, and eventually a new estimate of a(1)(z).\n4. The new estimate of the backscattering coefficient $bb(1)(z)$ is obtained by changing the old values of $bb(0)(z)$ by a small amount Δbb(z), where Δbb(z) = a(zX(z).\nOnce we have a new estimation of $Eum(z)$, we can compute a new Rm(z) and therefore a new Xm(z). Meanwhile, by using the RTM-generated Eu(z) and Ed(z) we can compute R(z) and X(z), and the difference between Xm(z) and X(z): ΔX(z) = Xm(z) − X(z). The new estimated backscattering coefficient is then $bb(1)(z)=bb(0)(z)+εΔbb(z)$, where ε is a number between 0 and 1 that has been introduced to stabilize the algorithm. Once again b(1)(z) can be computed using Eq. (9). The new estimates of a(z) and b(z), along with the assumed phase function p(Θ) are then put back into the RTM to generate new radiance and irradiance values.\n\nIn the original paper of Gordon and Boynton, the algorithm stops when the residual error between measured and simulated values of Ed and Lu, δ(n) reaches a minimum, where\n\n$δ(n)=1N∑i=1N|ln[Ed(n)(zi)]−ln[Edm(zi)]|+1N∑i=1N|ln[Lu(n)(zi)]−ln[Lum(zi)]|.$\nThey computed the difference between “measured” and simulated values of Ed and Lu directly and used the depth-averaged value as a convergence criterion in their algorithm because all the data were synthetic and they assumed a transparent sky condition, i.e. no atmosphere. However, the data used in this paper were based on measurements obtained in the presence of an atmosphere. Therefore, knowledge of the atmospheric condition when the measurements were performed is required if we were to use the same convergence criterion, but such information is not available in the dataset. We changed the convergence criterion by not computing the error of the “measured” Lu(z) and Ed(z) values directly but instead computing the error of the radiance reflectance RL(z), because our simulations showed that the radiance reflectance (an AOP) is insensitive to atmospheric conditions. We define a residual error between measured and simulated radiance reflectances, $δRL(n)$, as\n$δRL(n)=1N∑i=1N|ln[Lu(n)(zi)Ed(n)(zi)]−ln[Lum(zi)Edm(zi)]|.$\nThe algorithm will stop when this error reaches a prescribed tolerance or the number of iterations reaches the maximum allowed value.\n\n## 4. Water constituent inversion algorithm\n\nThe water constituent inversion algorithm employs the the multi-spectral IOPs, [a(λi), b(λi), bb(λi), where λi, i = 1,...,6], represents the wavelengths, to retrieve the water constituents iteratively, depth by depth. The IOPs retrieved from the radiance reflectance RL(z) data are the total water IOPs, which can be further divided into two parts: the IOPs due to pure water and the IOPs due to water constituents. We can subtract the pure water IOPs from the total, so that the remaining IOPs are solely contributed by water constituents, and these IOPs were used in our water constituent inversion algorithm to retrieve the water constituents. The vertical profile of water constituents is obtained after the inversion is completed.\n\nThe water constituents and the IOPs are connected by a bio-optical model. In our study, we assumed that there are two types of water constituent particles, algal particles and non-algal particles, in addition to the colored dissolved organic matter (CDOM), present in the water column. Here we introduce the bio-optical model used in our study, the CCRR bio-optical model, where CCRR stands for CoastColor: Round Robin . In the CCRR bio-optical model, the spectral variation of the absorption and scattering coefficients for the algal particles is parameterized in terms of the chlorophyll-a concentration (CHL), and a Fournier-Forand phase function with backscattering ratio of 0.0056 is used as the scattering phase function. The spectral variation of the absorption and scattering coefficients for the non-algal particles is parameterized in terms of a mineral particle concentration (MIN), and we replaced the average Petzold phase function that was originally proposed in the CCRR bio-optical model with a turbid harbor Petzold phase function which has a slightly higher backscattering ratio of 0.021. The spectral variation of the CDOM absorption coefficients is given by an exponentially decreasing function. A more detailed description of the CCRR model can be found in Appendix A.\n\nBased on the CCRR bio-optical model, the 3 water constituent parameters that will be retrieved from the IOPs are: chlorophyll-a concentration (CHL), the mineral particle concentration (MIN) and the CDOM absorption coefficient at 443 nm (CDOM). We developed an inversion algorithm based on a nonlinear least-squares optimal estimation with Levenberg-Marquardt regularization [18, 21] that infers the 3 water constituents iteratively from the IOPs.\n\nWe first created a training synthetic dataset using the CCRR IOP model, where the 3 water constituents of the CCRR bio-optical model were randomly sampled in the following ranges:\n\n1. CHL: 0.001 – 10 [mg·m−3]\n2. MIN: 0.001 – 5 [g·m−3]\n3. CDOM: 0.001 – 0.1 [m−1].\nThe output from the CCRR bio-optical model is the absorption, scattering and backscattering coefficients at 6 wavelengths: 412, 440, 488, 510, 532 and 555 nm. The dataset has 20,000 randomly sampled retrieval parameter [CHL, MIN, CDOM] and the derived IOPs [ai, bi, bbi], where i = 1,...,6, represents the 6 wavelengths. This dataset was used to train a Radial Basis Function Neural Network (RBF-NN) that computes the 18 IOPs (ai, bi, bbi, i = 1,...,6) from the 3 water constituents. The trained RBF-NN is used to replace the CCRR bio-optical model with the following single equation:\n$IOPi=∑j=1Naijexp[−b2∑k=1Nin(pk−cjk)2]+di$\nwhere i takes the value of 1 to 18, N is the number of neurons, Nin is the number of input water constituents (CHL, CDOM, MIN), which in this case is Nin = 3, pk denote the input data, aij, b, cjk and di are the coefficients resulting from the training. The Jacobians K needed in the optimal estimation [see Eq. (26) below] are obtained by calculating the partial derivatives with respect to the input parameter pk:\n$Ki,k=∂(IOPi)∂pk=−2b2(pk−cjk)×∑j=1Naijexp[−b2∑k=1Nin(pk−cjk)2].$\n\nOur algorithm starts with a randomly picked initial guess of the 3 water constituents (CHL, CDOM, and MIN) at the top depth. Then these 3 constituents are input into the RBF-NN model to generate the IOPs, and an error between the true IOPs and the model generated IOPs at all wavelengths is computed. If the error is larger than a prescribed tolerance, the algorithm will update the 3 water constituents and start the next iteration. Otherwise, the algorithm stops, saves the 3 water constituents and proceeds to the next depth. At each iteration, the 3 water constituents are updated as follow:\n\n$xi+1=xi+[(1+γi)Sa−1+KiTSm−1Ki]−1×KiTSm−1(ym−yi)−Sa−1(xi−xa)$\nwhere x is the state vector, and in our algorithm x = [CHL, MIN, CDOM]. γ is the Levenberg-Marquardt parameter. When γ = 0, Eq. (26) becomes standard Gauss-Newton optimal estimation, and when γ is a large number, Eq. (26) tends to the steepest gradient descent method. xa and Sa are the a priori state vector and covariance matrix, respectively. K are the Jacobians and Sm is the measurement error covariance matrix. ym and yi are the measured and simulated IOPs, respectively.\n\nTo speed up the algorithm, the first guess of the 3 water constituents is randomly picked only at the first depth, i.e. top of the ocean. For all depths below, we use the inverted 3 water constituents at the previous depth as a first guess. This approach appears to work well because the difference between each depth is only 1 meter and we assume that the 3 water constituents change gradually with depth. It also decreases the number of iterations needed to reach the preset error threshold, since the first guess is very close to the true value.\n\n## 5. Results and discussion\n\n#### 5.1. Synthetic data test of IOP inversion algorithm\n\nTo test the performance of the IOP inversion algorithm, we first generated a synthetic dataset at 488 nm using AccuRT, our RTM. The IOPs of the atmospheric gases are computed from a band model . The ocean can be stratified into a maximum number of 80 layers with arbitrary layer thickness, and a bottom reflection spectrum can also be added to simulate reflection from the ocean floor.\n\nTo generate synthetic Lu(z) and Ed(z) data we used an aerosol free standard U.S. atmosphere with solar zenith angle of 30° coupled with a 500 m deep ocean. The top 80 m of the ocean was assumed to be stratified with a 2 m depth resolution. The remaining 420 m of water below 80 m was assumed to constitute a homogeneous layer, and the bottom of the ocean was assumed to be black (totally absorbing). The IOPs of the water column were assumed to vary with depth in the follow manner:\n\n$a(z)=a0+a1exp[(z−za)22σa2]$\n$bb(z)=bb0+bb1exp[(z−zb)22σb2]$\nwhere a0 = 0.0145 m−1, bb0 = 0.0016 m−1, a1 = 0.04 m−1, bb1 = 0.0048 m−1. The most probable or mean values of the Gaussian distributions, where a(z) and bb(z) reach their maximum values were set to za = zb = 10 m and the standard deviations were set to σa = σb = 20 m. We used a Henyey-Greenstein (HG) phase function with asymmetry factor g = 0.9 which yields a backscatter ratio of 0.0229.\n\nThe IOPs inverted from the synthetic Lu(z) and Ed(z) data and the percentage error are shown in Fig. 4. When we use the same phase function as the one used to create the synthetic data, the inverted IOPs are very close to the synthetic data, with depth-averaged absolute percentage errors of 0.69%, 0.68% and 0.68% for absorption, scattering and backscattering coefficients, respectively. If we use a phase function that is different from the one used to create the synthetic data, we can still get reasonable inversion for absorption and backscattering coefficients, but the inverted scattering coefficient is incorrect. For example, when we used a HG phase function with g = 0.9 to create synthetic data and g = 0.8 in the inversion algorithm to invert the IOPs, the inverted absorption and backscattering coefficients have a depth-averaged absolute percentage error of 2.82% and 7.88%, respectively, but the inverted scattering coefficient had an error of 51.3%. These results agree with the original findings of Gordon and Boynton [1, 2]. The large error in the scattering coefficient is due to the use of an “incorrect” phase function, which yields a wrong backscattering ratio. The HG phase function used to create synthetic data has a backscattering ratio of 0.0229 (g = 0.9), while the “incorrect” phase function used to retrieve the IOPs has a backscattering ratio of 0.0507 (g = 0.8), which leads to an error in the backscattering ratio of 221%. This example shows that the algorithm produces better backscattering coefficient than scattering coefficient, i.e. it is more sensitive to bb than b, when an incorrect phase function is used.",
null,
"Fig. 3 Rayleigh, Fournier-Forand and Petzold phase function used in this paper.",
null,
"Fig. 4 IOPs inverted from synthetic radiance reflectance RL(z) data at 488 nm. The upper panels show comparison between retrieved and synthetic IOPs, where filled circles (black) represent synthetic data, solid lines (blue) represent the inverted IOPs using the correct phase function (g = 0.9), and dashed lines(red) represent the inverted IOPs using incorrect phase function (g = 0.8). Note that when using correct phase function, the retrieved IOPs are so close to the synthetic data that the blue and black curves overlapped. The lower panels show the corresponding percentage errors.\n\n#### 5.2. Synthetic data test of water constituent inversion algorithm\n\nThe performance of the water constituent inversion algorithm was also tested with a synthetic dataset before we applied it to measured data. We created a validation synthetic dataset of 5,000 randomly sampled CHL, MIN, and CDOM combinations in the same way as we created the training dataset, as discussed in Section 4. The validation dataset is first used to test the accuracy of the neural network training. We take the 5,000 water constituents in the validation dataset and use the trained neural network to recompute the IOPs. Compared with the model generated IOPs, the correlation for the 18 IOPs [a(λi), b(λi) and bb(λi), i = 1,...,6] had a minimum value of 0.99998, which means that the RBF-NN works very well, and is accurate enough to replace the CCRR bio-optical model.\n\nWe then take the 5,000 validation data and put them into our water constituent inversion algorithm to test its performance. The 18 model generated IOPs (a(λi), b(λi) and bb(λi), i = 1,...,6), were used to retrieve water constituents, and the results were then compared with the modeled data as shown in Fig. 5. We computed the percentage error for each retrieved combination of CHL, MIN and CDOM compared with the modeled data, and the errors are indicated by the color of the dots in Fig. 5. Except for a few cases in which the algorithm failed, most of the retrieved data match the model data very well. The correlation between retrieved and model data had R2-values of 0.9998, 0.9946 and 0.9995 for CHL, MIN and CDOM, respectively.",
null,
"Fig. 5 Retrieved CHL, MIN and CDOM compared with model data for the synthetic dataset. The color of the dots shows the percentage error in the retrieved data.\n\n### 5.3.1. Retrieval of IOPs from Lu(z) and Ed(z) data\n\nThe Lu(z) and Ed(z) data processing (merged multi-cast data corrected for solar irradiance fluctuations) and pairing with IOP data are still underway, and we have a total number of 14 measurements available. The time, location and mean ocean depth of all 14 measurements are shown in Table 1. There are six cases for which the measurements were obtained in shallow waters where the ocean depth is less than 30 m; 2 cases were obtained in very deep waters where the ocean is more than 2,000 m deep, and the remaining six cases were obtained in waters where the ocean depth varies from 50 m to 300 m. The paired IOP measurements, in general, has temporal difference of less than 1.5 hours with the Lu(z) and Ed(z) measurements.",
null,
"Table 1. Time, location and mean ocean depth of the 14 measurements. The * symbol indicates the deep ocean cases tested in our IOP inversion algorithm.\n\nWhen applying the algorithm to the measurements, we assumed a 14 layer aerosol free U.S. standard atmosphere since the atmospheric condition has minimal impact on the retrieved IOPs. The ocean was stratified based on the mean ocean depth and the depth at which continuous nonzero Lu(z) and Ed(z) measurements were obtained. The layers were selected in the following way: (i) the layers were set with 1 m layer depth from the top of the ocean to the maximum depth where continuous non-zero Lu(z) and Ed(z) measurements were obtained; (ii) if the mean ocean depth is less than 200 m, we expanded the lowermost layer that was set in (i) down to the ocean bottom by changing the lower boundary of that layer; (iii) if the ocean depth is greater than 200 m, we first expand the lowermost layer that was set in (i) down to 200 m, then added another layer of pure water from 200 m to the bottom of the ocean. For example, ST28 has a mean ocean depth of 2425 m, and non-zero Lu(z) and Ed(z) measurements were obtained down to 56 m, where the radiance reaches the 1% light level of the radiance at the surface. So in the top 56 layers, the depth of the each of the first 55 layers was set with 1 m and the lower boundary of the 56th layer was set to 200 m which yields a layer depth of 145 m. Finally, we added a 2225 m thick layer of pure water, so the total ocean depth agrees with geographic data, yielding a total of 57 layers for this case. We expand the last layer in order to avoid abrupt changes in Lu(z) and Ed(z) profiles. The bottom of the ocean was assumed to reflect light according to the reflectance spectrum of loamy sand , although for these deep waters reflection from the ocean floor is negligible.\n\nThe IOP inversion algorithm was then applied to the measured Lu(z) and Ed(z) datasets individually to retrieve IOP profiles wavelength by wavelength. However, as noted by Gordon and Boynton, the integral in Eq. (8) needs to be carried out to a depth where dRE(z)/dz → 0 in order to provide a good estimate of bb(z). This requirement is not satisfied unless the water is deep enough. So we applied the algorithm focusing on the deep water cases, i.e. ST10, ST13, ST14, ST15, ST19, ST27 and ST28. We excluded ST18 because Lu(z) and Ed(z) measurements were obtained only in the top 15 m of the water column.\n\nFor all the deep water cases, the algorithm required 15 iterations on average before the prescribed tolerance (δRL <0.001) was satisfied, and the radiance reflectances RL(z) derived from Lu(z) and Ed(z) values that were reconstructed by our RTM using retrieved IOPs matched the ones derived from the Lu(z) and Ed(z) measurements with a depth averaged absolute percentage error (PE) given by\n\n$PE(%)=1N∑i=1N|RL−RRTM−RL−measured|RL−measured×100$\nof maximum 0.23%. Table 2 shows the number of iterations (i) and the absolute percentage error of the radiance reflectances RL(z) at each wavelength for the deep water cases.",
null,
"Table 2. Number of iterations (i) needed to invert the IOPs from Lu(z) and Ed(z) measurements and absolute percentage error (PE) of the radiance reflectances RL(z) for deep water cases.\n\nThe retrieved IOPs generally agree well with the measured IOPs for the deep water cases. Table 3 summarizes the correlation and bias given by\n\n$bias(%)=1N∑i=1N(IOPinv.−IOPin−situ)IOPin−situ×100$\nof the retrieved IOPs at each wavelength compared with the measured IOPs. To quantify the overall performance of the IOP inversion algorithm, we computed the correlation between the retrieved and measured IOPs at all depths and wavelengths. The correlation has R2 values of 0.88, 0.96, and 0.93 for absorption, scattering, and backscattering coefficients, respectively. Two examples of retrieved IOPs compared with the measured IOPs are shown in Fig. 6 for a near-shore location (ST15) and in Fig. 7 for an offshore location (ST28) with ocean depth more than 2000 m. Comparisons between retrieved IOPs and measured IOPs for all depths, wavelengths and stations are shown in Fig. 8.",
null,
"Table 3. Correlation and bias of retrieved IOPs for deep water cases.",
null,
"Fig. 6 Retrieved IOPs (red) of ST15 compared with in-situ measurements (blue). The top 6 panels, from left to right, show comparisons of the absorption coefficients at 412, 440, 488, 510, 532 and 555 nm, respectively, whereas the middle and bottom 6 panels show the same for the scattering and backscattering coefficients.",
null,
"Fig. 7 Retrieved IOPs (red) of ST28 compared with in-situ measurements (blue). The top 6 panels, from left to right, show comparisons of the absorption coefficients at 412, 440, 488, 510, 532 and 555 nm, respectively, whereas the middle and bottom 6 panels show the same for the scattering and backscattering coefficients.",
null,
"Fig. 8 Comparison between retrieved and measured IOPs that combined all depth, wavelengths and stations. The 3 panels, from left to right, show comparisons of absorption, scattering and backscattering coefficients, respectively. The color of the dots indicates wavelength as shown on the color bar.\n\nWe used different colors to distinguish between wavelengths in Fig. 8 in order to show how the algorithm performs at different wavelengths. The algorithm introduced a small bias to the retrieved IOPs with a tendency to overestimate absorption coefficients. This overestimation of the absorption coefficient occurs at all wavelengths, but has a larger impact at shorter wavelengths, i.e. 412 nm and 440 nm. In extreme cases, the algorithm overestimates the absorption coefficient at 412 nm by 50%. Since the absorption coefficient was determined by Gershun’s Law, we suspect that even though the radiance reflectances RL(z) fit the measurements very well, with depth averaged absolute percentage error less than 0.23%, the irradiance reflectance RE(z) ≡ Eu(z)/Ed(z) may still have large errors, which could be traced to an inaccurate phase function. As pointed out by Mobley et al. , even for the same backscattering ratio, a different shape of the phase function can yield quite different Lu(z) and Ed(z) profiles. This issue could be addressed by including Eu(z) in the radiometric measurements. Another possible reason, as discussed in Section 2, is the underestimation in the measured absorption coefficients due to imperfect scattering correction. This uncertainty may also explain some of the apparent overestimation in absorption coefficients. The errors in the retrieved scattering and backscattering coefficients seem to be wavelength independent, and a possible reason is the error in the evaluation of X(z). Even for the deep water cases, the requirement that dRE(z)/dz → 0 at zmax is not completely satisfied, so there are still errors in X(z).\n\n### 5.3.2. Retrieval of water constituents from IOPs\n\nThe water constituent inversion algorithm (IOPs → {CHL, CDOM, MIN}, see Section 4) was then applied to the BP09 measurement data. There are two sets of IOP data available for retrieval of water constituent profiles. One set is the 14 in-situ measured IOP data, and the other set consists of the IOP data retrieved from Lu(z) and Ed(z) measurements by our IOP inversion algorithm for the 7 deep water stations analyzed in this paper. Since the retrieved IOP profiles contain some errors, we first applied the water constituent inversion algorithm to the in-situ measured IOPs directly to examine the performance of the algorithm. We applied the algorithm to all the measured IOP profiles to retrieve profiles of the 3 water constituents, CHL, MIN and CDOM. The retrieved water constituents were compared with the in-situ measured values. There are in-situ measurements of CHL, MIN and CDOM available in the BP09 dataset. The MIN values were measured only in the top levels, i.e. water samples were taken 1 m below the ocean surface. The CHL and CDOM values were measured at both this top level and at a few additional depths (7 m, 14 m, 30 m, 50 m and 125 m) depending on location. Figure 9 shows a comparison of retrieved CHL, MIN and CDOM from the in-situ measured IOPs against the measured values for all the stations. The correlation between the two datasets has R2 values of 0.80, 0.78 and 0.73 for CHL, MIN and CDOM respectively. This low correlation is clearly due to an underestimation of CHL and MIN and an overestimation of CDOM. A possible reason is that the CCRR bio-optical model does not fit the optical properties of the water in this area perfectly. But the methodology developed in this paper seems to work well in general. Hence, if one had access to relevant data, a more suitable bio-optical model could be established for this area using the approach described by Li et al. . Such a bio-optical model for the Ligurian Sea could give a better match with the measurements. Due to errors in the IOP profiles retrieved from the Lu(z) and Ed(z) measurements, the CHL, MIN and CDOM values inferred from these IOP data had an even lower correlation with the measurements, with R2 values of 0.62 for CHL, 0.66 for MIN and 0.63 for CDOM. However, in a recent study McKee et al. assessed the uncertainty of CHL in the BP09 dataset and found a ±28% uncertainty. Since the uncertainty in the sampled data is not insignificant, the errors in CHL, MIN and CDOM retrieved by our algorithm may not be as large as it appears in Fig. 9.",
null,
"Fig. 9 Retrieved CHL, MIN and CDOM from in-situ measured IOPs compare with measured values for all stations.\n\nExamples of water constituent profiles for deep ocean cases derived from retrieved IOPs are shown in Fig. 10 (dashed lines), along with water constituents derived from in-situ measured IOPs (solid lines). The three water constituents are plotted in different colors, with CHL in green, MIN in blue and CDOM in red. The filled circles are in-situ measurements of CHL and MIN. The BP09 experiment was conducted during a spring algae bloom in the area, which was quite visible in the MODIS standard CHL a product (Fig. 11). The measurements at ST27 and ST28 were performed right in the area of the algae bloom where a higher chlorophyll concentration can be found from the MODIS image, and our retrieved water constituents show clearly that the water column was dominated by chlorophyll. The vertical distribution also matches the measured CHL data well with a peak in the chlorophyll concentration around 10 m. The measurements at ST10, ST13, ST14, ST15, and ST19 were performed in less productive areas, where the chlorophyll concentration was found to be lower, and the water column is more dominated by non-algal particles. The water constituent profiles retrieved from inverted IOPs (dashed lines in Fig. 10) were slightly different from those retrieved from measured IOPs due to errors in the inverted IOPs. Figure 12 shows the spectrum of absorption, scattering and backscattering coefficients of the three water constituents with CHL = 1mgm−3, MIN = 1gm−3 and CDOM = 0.03m−1. In general, CDOM has higher retrieved values because of the overestimation of the absorption coefficient in the inverted IOPs. Since CDOM is purely absorbing, the algorithm tends to increase CDOM to match a higher absorption. The difference in CHL and MIN are mainly due to errors in the backscattering coefficients. As shown in Fig. 12, the differences between CHL and MIN are more distinguishable in backscattering than scattering coefficients.",
null,
"Fig. 10 Retrieved vertical profiles of CHL (green), MIN (blue) and CDOM (red). The solid lines are the water constituent profiles retrieved from in-situ measured IOPs, and the dashed lines are water constituent profiles retrieved from inverted IOPs. The filled circles are the in-situ measurements of CHL (green), MIN (blue) and CDOM (red).",
null,
"Fig. 11 Standard MODIS CHL a product of the Ligurian Sea area on 17 March 2009.",
null,
"Fig. 12 Absorption (left), scattering (middle) and backscattering coefficients (right) of CHL (green), MIN (blue) and CDOM (red), respectively.\n\n## 6. Conclusions\n\nWe numerically implemented the inversion method described by Gordon and Boynton [1, 2] to retrieve IOPs from measurements of nadir-viewing radiances Lu(z) and downward irradiances Ed(z), and made a few modifications to make the method suitable for application to real (as opposed to synthetic) data. As an extension, we developed an algorithm for inference of the vertical profile of water constituents (chlorophyllous particles, non-algal particles and dissolved matter) from the retrieved IOPs. Both algorithms were tested with a synthetic dataset and found to give reliable results with an accuracy better that 1%.\n\nThe algorithms were then applied to the BP09 in-situ measurements collected in the Ligurian Sea. In general, the retrieved IOPs match the in-situ measurements well for Lu(z) and Ed(z) measurements obtained in sufficiently deep ocean waters (> 50 m). There is an overestimation in the retrieved absorption coefficients which could be traced to use of an inaccurate phase function. However, if measurements of Eu(z) were available in the radiometric measurement dataset, more accurate absorption coefficients could be retrieved. A comparison of inferred and measured water constituent values shows that the algorithms are capable of retrieving vertical profiles of the water constituents, but adjustments need to be made to the bio-optical model that was used in the current version to improve the agreement with the measurements.\n\n## 7. Appendix A: The CCRR IOP model\n\nThe absorption coefficient of the pigmented particles is given by a non-linear function of chlorophyll-a concentration (CHL) :\n\n$apig(λ)=A(λ)×CHLE(λ)$\nwhere A(λ) and E(λ) are given by Bricaud et al. (1998) .\n\nThe beam attenuation coefficient for pigmented particles at 660 nm is given by :\n\n$cpig(660)=0.407×CHL0.795$\nand the spectral variation is taken to be :\n$cpig(λ)=cpig(660)×(λ/660)ν$\nwhere\n$ν=0.5×[log10CHL−0.3]0.022.0$\n\nThe spectral variation of the scattering coefficients for the pigmented particles is given by the difference between the beam attenuation coefficients and absorption coefficients:\n\n$bpig(λ)=cpig(λ)−apig(λ).$\nThe scattering phase function for the pigmented particles is assumed to be described by the Fournier-Forand phase function with a backscattering ratio of 0.0056.\n\nThe absorption coefficients of the non-pigmented particles at 443 nm is given by :\n\n$aMIN(443)=0.031×MIN$\nand the spectral variation can be written as :\n$aMIN(λ)=aMIN(443)exp[−0.0213(λ−443)].$\n\nThe scattering coefficients of the non-pigmented particles at 555 nm is given by :\n\n$bMIN(555)=0.51×MIN$\nand the spectral variation of the beam attenuation coefficients of the non-pigmented particles is given by :\n$cMIN(λ)=cMIN(555)×(λ/555)−0.3749$\nwhere\n$cMIN(555)=aMIN(555)+bMIN(555)=0.52×MIN.$\n\nThe spectral variation of the scattering coefficients for the non-pigmented particles is given by the difference between the beam attenuation coefficients and absorption coefficients:\n\n$bMIN(λ)=cMIN(λ)−aMIN(λ).$\n\nThe average Petzold phase function with a backscattering ratio of 0.019, as tabulated by Mobley , is used to describe the scattering phase function for mineral particles. However, in our algorithm, we used the turbid water Petzold phase function which has a backscattering ratio of 0.021 for mineral particles.\n\nCDOM absorption spectral variation is given by an exponentially decreasing function :\n\n$aCDOM(λ)=CDOM×exp[−0.0176(λ−443)].$\nThe total IOPs due to water constituents are given by:\n$atot(λ)=apig(λ)+aMIN(λ)+aCDOM(λ)$\n$btot(λ)=bpig(λ)+bMIN(λ)$\n$bbtot(λ)=0.0056×bpig(λ)+0.019×bMIN(λ).$\n\n1. H.R. Gordon and G.C. Boynton, “Radiance–irradiance inversion algorithm for estimating the absorption and backscattering coefficients of natural waters: homogeneous waters,” Appl. Opt. 36, 2636–2641 (1997). [CrossRef] [PubMed]\n\n2. H.R. Gordon and G.C. Boynton, “Radiance–irradiance inversion algorithm for estimating the absorption and backscattering coefficients of natural waters: vertically stratified water bodies,” Appl. Opt. 37, 3886–3896 (1998). [CrossRef]\n\n3. V. Sanjuan Calzado, D. McKee, and C. Trees, “Multi and single cast radiometric processing and merging in the Ligurian Sea,” Optics of Natural Waters 2011Saint Petersburg, September 2011.\n\n4. J.R.V. Zaneveld, J.C. Kitchen, and C.M. Moore, “The scattering error correction of reflecting-tube absorption meters,” Proc. SPIE 2258 Ocean Optics XII, 44–55 (1994). [CrossRef]\n\n5. D. McKee, J. Piskozub, R. Röttgers, and R. Reynolds, “Evaluation and improvement of an iterative scattering correction scheme for in situ absorption and attenuation measurements,” J. Atmos. Ocean. Tech. 30, 1527–1541 (2013). [CrossRef]\n\n6. R. Röttgers, D. McKee, and S.B. Woźniak, “Evaluation of scatter corrections for ac-9 absorption measurements in coastal waters,” Methods in Oceanography 7, 21–39 (2013). [CrossRef]\n\n7. Z. Jin and K. Stamnes, “Radiative transfer in nonuniformly refracting media such as the atmosphere/ocean system,” Appl. Opt. 33, 431–442 (1994). [CrossRef] [PubMed]\n\n8. G.E. Thomas and K. Stamnes, Radiative Transfer in the Atmosphere and Ocean, 2nd Edition (Cambridge University, 2002)\n\n9. B. Hamre, S. Stamnes, K. Stamnes, and J.J. Stamnes, “C-DISORT: A versatile tool for radiative transfer in coupled media like the atmosphere-ocean system,” AIP Conference Proceedings 1531, 923 (2013). [CrossRef]\n\n10. B. Hamre, S. Stamnes, K. Stamnes, and J.J. Stamnes, “A versatile tool for radiative transfer simulations in the coupled atmosphere-ocean system: Introducing AccuRT,” Ocean Optics XXII, Portland, ME, 26–31 Oct. 2014.\n\n11. C.D. Mobley, Light and Water: Radiative Transfer in Natural Waters (Academic, 1994)\n\n12. D. McKee, A. Cunningham, J. Slater, K.J. Jones, and C.R. Griffiths, “Inherent and apparent optical properties in coastal waters: a study of the Clyde Sea in early summer,” Estuarine, Coastal and Shelf Science 56(2), 369–376 (2003). [CrossRef]\n\n13. A. Morel and B. Gentili, “Diffuse reflectance of oceanic waters:II bi-directional aspects,” Appl. 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Stamnes, and T. Wong, “Delta-fit: A fast and accurate treatment of particle scattering phase functions with weighted singular-value decomposition least squares fitting,” J. Quant. Spectrosc. Radiat. Transf. 65, 681–690 (2000). [CrossRef]\n\n20. K. Ruddick, “DUE Coastcolour Round Robin Protocol, version 1.2,” Coastalcolour Round Robin, Oct. 2010.\n\n21. C. Rodgers, Inverse Methods for Atmospheric Sounding (World Scientific, 2000).\n\n22. D.S. Broomhead and D. Lowe, “Radial basis functions, multi-variable functional interpolation and adaptive networks,” Complex Systems 2, 321–355 (1988).\n\n23. A. Bricaud, A. Morel, M. Babin, K. Allali, and H. Claustre, “Variations in light absorption by suspended particles with chlorophyll concentration in oceanic (case 1) waters: Analysis and implications for bio-optical models,” J. Geophys. Res. 103, 31033–31044 (1998). [CrossRef]\n\n24. H. Loisel and A. Morel, “Light scattering and chlorophyll concentration in case 1 waters: a re-examination,” Limnol. Oceanogr. 43, 847–857 (1998). [CrossRef]\n\n25. A. Morel, D. Antoine, and B. Gentili, “Bidirectional reflectance of oceanic waters: accounting for Raman emission and varying particle scattering phase function,” Appl. Opt. 41, 6289–6306 (2002). [CrossRef] [PubMed]\n\n26. M. Babin, A.D. Stramski, G.M. Ferrari, H. Claustre, A. Bricaud, G. Obelesky, and N. Hoepffner, “Variations in the light absorption coefficients of phytoplankton, nonalgal particles and dissolved organic matter in coastal waters around Europe,” J. Geophys. Res. 108(C7):3211 (2003). [CrossRef]\n\n27. M. Babin, A. Morel, V. Fournier-Sicre, F. Fell, and D. Stramski, “Light scattering properties of marine particles in coastal and open ocean waters as related to the particle mass concentration,”Limnol. Oceanogr. 28, 843–859 (2003). [CrossRef]\n\n28. A.M. Baldridge, S.J. Hook, C.I. Grove, and G. Rivera, “The ASTER Spectral Library Version 2.0,” Remote Sens. Environ. 113, 711–715 (2009). [CrossRef]\n\n29. F. X. Kneizys, L. W. Abreu, G.P. Anderson, J.H. Chetwynd, E.P. Shettle, A. Berk, L.S. Bernstein, D.C. Roberson, P. Acharya, L.S. Rothman, J.E.A. Selby, W.O. Gallery, and S.A. Clough, “The MODTRAN2/3 report and LOWTRAN 7 model,” Phillips Laboratory, Hanscom AFB (1996).\n\n30. L.G. Henyey and J.L. Greenstein, “Diffuse radiation in the galaxy,” Astrophys. J. 93, 70–83 (1941). [CrossRef]\n\n31. D. McKee, R. Röttgers, G. Neukermans, V. SanjuanCalzado, C. Trees, M. Ampolo-Rella, C. Neil, and A. Cunningham, “Impact of measurement uncertainties on determination of chlorophyll-specific absorption coefficient for marine phytoplankton,” J. Geophys. Res.-Oceans 119, 9013–9025 (2014). [CrossRef]\n\n### References\n\n• View by:\n• |\n• |\n• |\n\n1. H.R. Gordon and G.C. Boynton, “Radiance–irradiance inversion algorithm for estimating the absorption and backscattering coefficients of natural waters: homogeneous waters,” Appl. Opt. 36, 2636–2641 (1997).\n[Crossref] [PubMed]\n2. H.R. Gordon and G.C. Boynton, “Radiance–irradiance inversion algorithm for estimating the absorption and backscattering coefficients of natural waters: vertically stratified water bodies,” Appl. Opt. 37, 3886–3896 (1998).\n[Crossref]\n3. V. Sanjuan Calzado, D. McKee, and C. Trees, “Multi and single cast radiometric processing and merging in the Ligurian Sea,” Optics of Natural Waters 2011Saint Petersburg, September 2011.\n4. J.R.V. Zaneveld, J.C. Kitchen, and C.M. Moore, “The scattering error correction of reflecting-tube absorption meters,” Proc. SPIE 2258 Ocean Optics XII, 44–55 (1994).\n[Crossref]\n5. D. McKee, J. Piskozub, R. Röttgers, and R. Reynolds, “Evaluation and improvement of an iterative scattering correction scheme for in situ absorption and attenuation measurements,” J. Atmos. Ocean. Tech. 30, 1527–1541 (2013).\n[Crossref]\n6. R. Röttgers, D. McKee, and S.B. Woźniak, “Evaluation of scatter corrections for ac-9 absorption measurements in coastal waters,” Methods in Oceanography 7, 21–39 (2013).\n[Crossref]\n7. Z. Jin and K. Stamnes, “Radiative transfer in nonuniformly refracting media such as the atmosphere/ocean system,” Appl. Opt. 33, 431–442 (1994).\n[Crossref] [PubMed]\n8. G.E. Thomas and K. Stamnes, Radiative Transfer in the Atmosphere and Ocean, 2nd Edition (Cambridge University, 2002)\n9. B. Hamre, S. Stamnes, K. Stamnes, and J.J. Stamnes, “C-DISORT: A versatile tool for radiative transfer in coupled media like the atmosphere-ocean system,” AIP Conference Proceedings 1531, 923 (2013).\n[Crossref]\n10. B. Hamre, S. Stamnes, K. Stamnes, and J.J. Stamnes, “A versatile tool for radiative transfer simulations in the coupled atmosphere-ocean system: Introducing AccuRT,” Ocean Optics XXII, Portland, ME, 26–31 Oct. 2014.\n11. C.D. Mobley, Light and Water: Radiative Transfer in Natural Waters (Academic, 1994)\n12. D. McKee, A. Cunningham, J. Slater, K.J. Jones, and C.R. Griffiths, “Inherent and apparent optical properties in coastal waters: a study of the Clyde Sea in early summer,” Estuarine, Coastal and Shelf Science 56(2), 369–376 (2003).\n[Crossref]\n13. A. Morel and B. Gentili, “Diffuse reflectance of oceanic waters:II bi-directional aspects,” Appl. Opt. 32, 6864–6879 (1993).\n[Crossref] [PubMed]\n14. H.R. Gordon, “Can the Lambert-Beer law be applied to the diffuse attenuation coefficient of ocean water?” Limnol. Oceanogr. 34, 1389–1409 (1989).\n[Crossref]\n15. G. Fournier and J.L. Forand, “Analytic phase function for ocean water,” Proc. SPIE 2258 Ocean Optics XII, 194–201 (1994).\n[Crossref]\n16. C.D. Mobley, L.K. Sundman, and E. Boss, “Phase function effects on oceanic light fields,” Appl. Opt. 41, 1035–1050 (2002).\n[Crossref] [PubMed]\n17. T.L. Petzold, “Volume scattering functions for selected ocean waters,” Technical Report SIO 7278 Scripps Institute of Oceanography, San Diego, Calif. (1972).\n18. W. Li, K. Stamnes, R. Spurr, and J.J. Stamnes, “Simultaneous retrieval of aerosols and ocean properties: A classic inverse modeling approach II. SeaWiFS case study for the Santa Barbara channel”, Int. J. Remote Sens. 29, 5689–5698 (2008).\n[Crossref]\n19. Y.X. Hu, B. Wielicki, B. Lin, G. Gibson, S.C. Tsay, K. Stamnes, and T. Wong, “Delta-fit: A fast and accurate treatment of particle scattering phase functions with weighted singular-value decomposition least squares fitting,” J. Quant. Spectrosc. Radiat. Transf. 65, 681–690 (2000).\n[Crossref]\n20. K. Ruddick, “DUE Coastcolour Round Robin Protocol, version 1.2,” Coastalcolour Round Robin, Oct.2010.\n21. C. Rodgers, Inverse Methods for Atmospheric Sounding (World Scientific, 2000).\n22. D.S. Broomhead and D. Lowe, “Radial basis functions, multi-variable functional interpolation and adaptive networks,” Complex Systems 2, 321–355 (1988).\n23. A. Bricaud, A. Morel, M. Babin, K. Allali, and H. Claustre, “Variations in light absorption by suspended particles with chlorophyll concentration in oceanic (case 1) waters: Analysis and implications for bio-optical models,” J. Geophys. Res. 103, 31033–31044 (1998).\n[Crossref]\n24. H. Loisel and A. Morel, “Light scattering and chlorophyll concentration in case 1 waters: a re-examination,” Limnol. Oceanogr. 43, 847–857 (1998).\n[Crossref]\n25. A. Morel, D. Antoine, and B. Gentili, “Bidirectional reflectance of oceanic waters: accounting for Raman emission and varying particle scattering phase function,” Appl. Opt. 41, 6289–6306 (2002).\n[Crossref] [PubMed]\n26. M. Babin, A.D. Stramski, G.M. Ferrari, H. Claustre, A. Bricaud, G. Obelesky, and N. Hoepffner, “Variations in the light absorption coefficients of phytoplankton, nonalgal particles and dissolved organic matter in coastal waters around Europe,” J. Geophys. Res. 108(C7):3211 (2003).\n[Crossref]\n27. M. Babin, A. Morel, V. Fournier-Sicre, F. Fell, and D. Stramski, “Light scattering properties of marine particles in coastal and open ocean waters as related to the particle mass concentration,”Limnol. Oceanogr. 28, 843–859 (2003).\n[Crossref]\n28. A.M. Baldridge, S.J. Hook, C.I. Grove, and G. Rivera, “The ASTER Spectral Library Version 2.0,” Remote Sens. Environ. 113, 711–715 (2009).\n[Crossref]\n29. F. X. Kneizys, L. W. Abreu, G.P. Anderson, J.H. Chetwynd, E.P. Shettle, A. Berk, L.S. Bernstein, D.C. Roberson, P. Acharya, L.S. Rothman, J.E.A. Selby, W.O. Gallery, and S.A. Clough, “The MODTRAN2/3 report and LOWTRAN 7 model,” Phillips Laboratory, Hanscom AFB (1996).\n30. L.G. Henyey and J.L. Greenstein, “Diffuse radiation in the galaxy,” Astrophys. J. 93, 70–83 (1941).\n[Crossref]\n31. D. McKee, R. Röttgers, G. Neukermans, V. SanjuanCalzado, C. Trees, M. Ampolo-Rella, C. Neil, and A. Cunningham, “Impact of measurement uncertainties on determination of chlorophyll-specific absorption coefficient for marine phytoplankton,” J. Geophys. Res.-Oceans 119, 9013–9025 (2014).\n[Crossref]\n\n#### 2014 (1)\n\nD. McKee, R. Röttgers, G. Neukermans, V. SanjuanCalzado, C. Trees, M. Ampolo-Rella, C. Neil, and A. Cunningham, “Impact of measurement uncertainties on determination of chlorophyll-specific absorption coefficient for marine phytoplankton,” J. Geophys. Res.-Oceans 119, 9013–9025 (2014).\n[Crossref]\n\n#### 2013 (3)\n\nD. McKee, J. Piskozub, R. Röttgers, and R. Reynolds, “Evaluation and improvement of an iterative scattering correction scheme for in situ absorption and attenuation measurements,” J. Atmos. Ocean. Tech. 30, 1527–1541 (2013).\n[Crossref]\n\nR. Röttgers, D. McKee, and S.B. Woźniak, “Evaluation of scatter corrections for ac-9 absorption measurements in coastal waters,” Methods in Oceanography 7, 21–39 (2013).\n[Crossref]\n\nB. Hamre, S. Stamnes, K. Stamnes, and J.J. Stamnes, “C-DISORT: A versatile tool for radiative transfer in coupled media like the atmosphere-ocean system,” AIP Conference Proceedings 1531, 923 (2013).\n[Crossref]\n\n#### 2009 (1)\n\nA.M. Baldridge, S.J. Hook, C.I. Grove, and G. Rivera, “The ASTER Spectral Library Version 2.0,” Remote Sens. Environ. 113, 711–715 (2009).\n[Crossref]\n\n#### 2008 (1)\n\nW. Li, K. Stamnes, R. Spurr, and J.J. Stamnes, “Simultaneous retrieval of aerosols and ocean properties: A classic inverse modeling approach II. SeaWiFS case study for the Santa Barbara channel”, Int. J. Remote Sens. 29, 5689–5698 (2008).\n[Crossref]\n\n#### 2003 (3)\n\nD. McKee, A. Cunningham, J. Slater, K.J. Jones, and C.R. Griffiths, “Inherent and apparent optical properties in coastal waters: a study of the Clyde Sea in early summer,” Estuarine, Coastal and Shelf Science 56(2), 369–376 (2003).\n[Crossref]\n\nM. Babin, A.D. Stramski, G.M. Ferrari, H. Claustre, A. Bricaud, G. Obelesky, and N. Hoepffner, “Variations in the light absorption coefficients of phytoplankton, nonalgal particles and dissolved organic matter in coastal waters around Europe,” J. Geophys. Res. 108(C7):3211 (2003).\n[Crossref]\n\nM. Babin, A. Morel, V. Fournier-Sicre, F. Fell, and D. Stramski, “Light scattering properties of marine particles in coastal and open ocean waters as related to the particle mass concentration,”Limnol. Oceanogr. 28, 843–859 (2003).\n[Crossref]\n\n#### 2000 (1)\n\nY.X. Hu, B. Wielicki, B. Lin, G. Gibson, S.C. Tsay, K. Stamnes, and T. Wong, “Delta-fit: A fast and accurate treatment of particle scattering phase functions with weighted singular-value decomposition least squares fitting,” J. Quant. Spectrosc. Radiat. Transf. 65, 681–690 (2000).\n[Crossref]\n\n#### 1998 (3)\n\nA. Bricaud, A. Morel, M. Babin, K. Allali, and H. Claustre, “Variations in light absorption by suspended particles with chlorophyll concentration in oceanic (case 1) waters: Analysis and implications for bio-optical models,” J. Geophys. Res. 103, 31033–31044 (1998).\n[Crossref]\n\nH. Loisel and A. Morel, “Light scattering and chlorophyll concentration in case 1 waters: a re-examination,” Limnol. Oceanogr. 43, 847–857 (1998).\n[Crossref]\n\n#### 1994 (3)\n\nG. Fournier and J.L. Forand, “Analytic phase function for ocean water,” Proc. SPIE 2258 Ocean Optics XII, 194–201 (1994).\n[Crossref]\n\nJ.R.V. Zaneveld, J.C. Kitchen, and C.M. Moore, “The scattering error correction of reflecting-tube absorption meters,” Proc. SPIE 2258 Ocean Optics XII, 44–55 (1994).\n[Crossref]\n\n#### 1989 (1)\n\nH.R. Gordon, “Can the Lambert-Beer law be applied to the diffuse attenuation coefficient of ocean water?” Limnol. Oceanogr. 34, 1389–1409 (1989).\n[Crossref]\n\n#### 1988 (1)\n\nD.S. Broomhead and D. Lowe, “Radial basis functions, multi-variable functional interpolation and adaptive networks,” Complex Systems 2, 321–355 (1988).\n\n#### 1941 (1)\n\nL.G. Henyey and J.L. Greenstein, “Diffuse radiation in the galaxy,” Astrophys. J. 93, 70–83 (1941).\n[Crossref]\n\n#### Abreu, L. W.\n\nF. X. Kneizys, L. W. Abreu, G.P. Anderson, J.H. Chetwynd, E.P. Shettle, A. Berk, L.S. Bernstein, D.C. Roberson, P. Acharya, L.S. Rothman, J.E.A. Selby, W.O. Gallery, and S.A. Clough, “The MODTRAN2/3 report and LOWTRAN 7 model,” Phillips Laboratory, Hanscom AFB (1996).\n\n#### Acharya, P.\n\nF. X. Kneizys, L. W. Abreu, G.P. Anderson, J.H. Chetwynd, E.P. Shettle, A. Berk, L.S. Bernstein, D.C. Roberson, P. Acharya, L.S. Rothman, J.E.A. Selby, W.O. Gallery, and S.A. Clough, “The MODTRAN2/3 report and LOWTRAN 7 model,” Phillips Laboratory, Hanscom AFB (1996).\n\n#### Allali, K.\n\nA. Bricaud, A. Morel, M. Babin, K. Allali, and H. Claustre, “Variations in light absorption by suspended particles with chlorophyll concentration in oceanic (case 1) waters: Analysis and implications for bio-optical models,” J. Geophys. Res. 103, 31033–31044 (1998).\n[Crossref]\n\n#### Ampolo-Rella, M.\n\nD. McKee, R. Röttgers, G. Neukermans, V. SanjuanCalzado, C. Trees, M. Ampolo-Rella, C. Neil, and A. Cunningham, “Impact of measurement uncertainties on determination of chlorophyll-specific absorption coefficient for marine phytoplankton,” J. Geophys. Res.-Oceans 119, 9013–9025 (2014).\n[Crossref]\n\n#### Anderson, G.P.\n\nF. X. Kneizys, L. W. Abreu, G.P. Anderson, J.H. Chetwynd, E.P. Shettle, A. Berk, L.S. Bernstein, D.C. Roberson, P. Acharya, L.S. Rothman, J.E.A. Selby, W.O. Gallery, and S.A. Clough, “The MODTRAN2/3 report and LOWTRAN 7 model,” Phillips Laboratory, Hanscom AFB (1996).\n\n#### Babin, M.\n\nM. Babin, A.D. Stramski, G.M. Ferrari, H. Claustre, A. Bricaud, G. Obelesky, and N. Hoepffner, “Variations in the light absorption coefficients of phytoplankton, nonalgal particles and dissolved organic matter in coastal waters around Europe,” J. Geophys. Res. 108(C7):3211 (2003).\n[Crossref]\n\nM. Babin, A. Morel, V. Fournier-Sicre, F. Fell, and D. Stramski, “Light scattering properties of marine particles in coastal and open ocean waters as related to the particle mass concentration,”Limnol. Oceanogr. 28, 843–859 (2003).\n[Crossref]\n\nA. Bricaud, A. Morel, M. Babin, K. Allali, and H. Claustre, “Variations in light absorption by suspended particles with chlorophyll concentration in oceanic (case 1) waters: Analysis and implications for bio-optical models,” J. Geophys. Res. 103, 31033–31044 (1998).\n[Crossref]\n\n#### Baldridge, A.M.\n\nA.M. Baldridge, S.J. Hook, C.I. Grove, and G. Rivera, “The ASTER Spectral Library Version 2.0,” Remote Sens. Environ. 113, 711–715 (2009).\n[Crossref]\n\n#### Berk, A.\n\nF. X. Kneizys, L. W. Abreu, G.P. Anderson, J.H. Chetwynd, E.P. Shettle, A. Berk, L.S. Bernstein, D.C. Roberson, P. Acharya, L.S. Rothman, J.E.A. Selby, W.O. Gallery, and S.A. Clough, “The MODTRAN2/3 report and LOWTRAN 7 model,” Phillips Laboratory, Hanscom AFB (1996).\n\n#### Bernstein, L.S.\n\nF. X. Kneizys, L. W. Abreu, G.P. Anderson, J.H. Chetwynd, E.P. Shettle, A. Berk, L.S. Bernstein, D.C. Roberson, P. Acharya, L.S. Rothman, J.E.A. Selby, W.O. Gallery, and S.A. Clough, “The MODTRAN2/3 report and LOWTRAN 7 model,” Phillips Laboratory, Hanscom AFB (1996).\n\n#### Bricaud, A.\n\nM. Babin, A.D. Stramski, G.M. Ferrari, H. Claustre, A. Bricaud, G. Obelesky, and N. Hoepffner, “Variations in the light absorption coefficients of phytoplankton, nonalgal particles and dissolved organic matter in coastal waters around Europe,” J. Geophys. Res. 108(C7):3211 (2003).\n[Crossref]\n\nA. Bricaud, A. Morel, M. Babin, K. Allali, and H. Claustre, “Variations in light absorption by suspended particles with chlorophyll concentration in oceanic (case 1) waters: Analysis and implications for bio-optical models,” J. Geophys. Res. 103, 31033–31044 (1998).\n[Crossref]\n\n#### Broomhead, D.S.\n\nD.S. Broomhead and D. Lowe, “Radial basis functions, multi-variable functional interpolation and adaptive networks,” Complex Systems 2, 321–355 (1988).\n\n#### Chetwynd, J.H.\n\nF. X. Kneizys, L. W. Abreu, G.P. Anderson, J.H. Chetwynd, E.P. Shettle, A. Berk, L.S. Bernstein, D.C. Roberson, P. Acharya, L.S. Rothman, J.E.A. Selby, W.O. Gallery, and S.A. Clough, “The MODTRAN2/3 report and LOWTRAN 7 model,” Phillips Laboratory, Hanscom AFB (1996).\n\n#### Claustre, H.\n\nM. Babin, A.D. Stramski, G.M. Ferrari, H. Claustre, A. Bricaud, G. Obelesky, and N. Hoepffner, “Variations in the light absorption coefficients of phytoplankton, nonalgal particles and dissolved organic matter in coastal waters around Europe,” J. Geophys. Res. 108(C7):3211 (2003).\n[Crossref]\n\nA. Bricaud, A. Morel, M. Babin, K. Allali, and H. Claustre, “Variations in light absorption by suspended particles with chlorophyll concentration in oceanic (case 1) waters: Analysis and implications for bio-optical models,” J. Geophys. Res. 103, 31033–31044 (1998).\n[Crossref]\n\n#### Clough, S.A.\n\nF. X. Kneizys, L. W. Abreu, G.P. Anderson, J.H. Chetwynd, E.P. Shettle, A. Berk, L.S. Bernstein, D.C. Roberson, P. Acharya, L.S. Rothman, J.E.A. Selby, W.O. Gallery, and S.A. Clough, “The MODTRAN2/3 report and LOWTRAN 7 model,” Phillips Laboratory, Hanscom AFB (1996).\n\n#### Cunningham, A.\n\nD. McKee, R. Röttgers, G. Neukermans, V. SanjuanCalzado, C. Trees, M. Ampolo-Rella, C. Neil, and A. Cunningham, “Impact of measurement uncertainties on determination of chlorophyll-specific absorption coefficient for marine phytoplankton,” J. Geophys. Res.-Oceans 119, 9013–9025 (2014).\n[Crossref]\n\nD. McKee, A. Cunningham, J. Slater, K.J. Jones, and C.R. Griffiths, “Inherent and apparent optical properties in coastal waters: a study of the Clyde Sea in early summer,” Estuarine, Coastal and Shelf Science 56(2), 369–376 (2003).\n[Crossref]\n\n#### Fell, F.\n\nM. Babin, A. Morel, V. Fournier-Sicre, F. Fell, and D. Stramski, “Light scattering properties of marine particles in coastal and open ocean waters as related to the particle mass concentration,”Limnol. Oceanogr. 28, 843–859 (2003).\n[Crossref]\n\n#### Ferrari, G.M.\n\nM. Babin, A.D. Stramski, G.M. Ferrari, H. Claustre, A. Bricaud, G. Obelesky, and N. Hoepffner, “Variations in the light absorption coefficients of phytoplankton, nonalgal particles and dissolved organic matter in coastal waters around Europe,” J. Geophys. Res. 108(C7):3211 (2003).\n[Crossref]\n\n#### Forand, J.L.\n\nG. Fournier and J.L. Forand, “Analytic phase function for ocean water,” Proc. SPIE 2258 Ocean Optics XII, 194–201 (1994).\n[Crossref]\n\n#### Fournier, G.\n\nG. Fournier and J.L. Forand, “Analytic phase function for ocean water,” Proc. SPIE 2258 Ocean Optics XII, 194–201 (1994).\n[Crossref]\n\n#### Fournier-Sicre, V.\n\nM. Babin, A. Morel, V. Fournier-Sicre, F. Fell, and D. Stramski, “Light scattering properties of marine particles in coastal and open ocean waters as related to the particle mass concentration,”Limnol. Oceanogr. 28, 843–859 (2003).\n[Crossref]\n\n#### Gallery, W.O.\n\nF. X. Kneizys, L. W. Abreu, G.P. Anderson, J.H. Chetwynd, E.P. Shettle, A. Berk, L.S. Bernstein, D.C. Roberson, P. Acharya, L.S. Rothman, J.E.A. Selby, W.O. Gallery, and S.A. Clough, “The MODTRAN2/3 report and LOWTRAN 7 model,” Phillips Laboratory, Hanscom AFB (1996).\n\n#### Gibson, G.\n\nY.X. Hu, B. Wielicki, B. Lin, G. Gibson, S.C. Tsay, K. Stamnes, and T. Wong, “Delta-fit: A fast and accurate treatment of particle scattering phase functions with weighted singular-value decomposition least squares fitting,” J. Quant. Spectrosc. Radiat. Transf. 65, 681–690 (2000).\n[Crossref]\n\n#### Gordon, H.R.\n\nH.R. Gordon, “Can the Lambert-Beer law be applied to the diffuse attenuation coefficient of ocean water?” Limnol. Oceanogr. 34, 1389–1409 (1989).\n[Crossref]\n\n#### Greenstein, J.L.\n\nL.G. Henyey and J.L. Greenstein, “Diffuse radiation in the galaxy,” Astrophys. J. 93, 70–83 (1941).\n[Crossref]\n\n#### Griffiths, C.R.\n\nD. McKee, A. Cunningham, J. Slater, K.J. Jones, and C.R. Griffiths, “Inherent and apparent optical properties in coastal waters: a study of the Clyde Sea in early summer,” Estuarine, Coastal and Shelf Science 56(2), 369–376 (2003).\n[Crossref]\n\n#### Grove, C.I.\n\nA.M. Baldridge, S.J. Hook, C.I. Grove, and G. Rivera, “The ASTER Spectral Library Version 2.0,” Remote Sens. Environ. 113, 711–715 (2009).\n[Crossref]\n\n#### Hamre, B.\n\nB. Hamre, S. Stamnes, K. Stamnes, and J.J. Stamnes, “C-DISORT: A versatile tool for radiative transfer in coupled media like the atmosphere-ocean system,” AIP Conference Proceedings 1531, 923 (2013).\n[Crossref]\n\nB. Hamre, S. Stamnes, K. Stamnes, and J.J. Stamnes, “A versatile tool for radiative transfer simulations in the coupled atmosphere-ocean system: Introducing AccuRT,” Ocean Optics XXII, Portland, ME, 26–31 Oct. 2014.\n\n#### Henyey, L.G.\n\nL.G. Henyey and J.L. Greenstein, “Diffuse radiation in the galaxy,” Astrophys. J. 93, 70–83 (1941).\n[Crossref]\n\n#### Hoepffner, N.\n\nM. Babin, A.D. Stramski, G.M. Ferrari, H. Claustre, A. Bricaud, G. Obelesky, and N. Hoepffner, “Variations in the light absorption coefficients of phytoplankton, nonalgal particles and dissolved organic matter in coastal waters around Europe,” J. Geophys. Res. 108(C7):3211 (2003).\n[Crossref]\n\n#### Hook, S.J.\n\nA.M. Baldridge, S.J. Hook, C.I. Grove, and G. Rivera, “The ASTER Spectral Library Version 2.0,” Remote Sens. Environ. 113, 711–715 (2009).\n[Crossref]\n\n#### Hu, Y.X.\n\nY.X. Hu, B. Wielicki, B. Lin, G. Gibson, S.C. Tsay, K. Stamnes, and T. Wong, “Delta-fit: A fast and accurate treatment of particle scattering phase functions with weighted singular-value decomposition least squares fitting,” J. Quant. Spectrosc. Radiat. Transf. 65, 681–690 (2000).\n[Crossref]\n\n#### Jones, K.J.\n\nD. McKee, A. Cunningham, J. Slater, K.J. Jones, and C.R. Griffiths, “Inherent and apparent optical properties in coastal waters: a study of the Clyde Sea in early summer,” Estuarine, Coastal and Shelf Science 56(2), 369–376 (2003).\n[Crossref]\n\n#### Kitchen, J.C.\n\nJ.R.V. Zaneveld, J.C. Kitchen, and C.M. Moore, “The scattering error correction of reflecting-tube absorption meters,” Proc. SPIE 2258 Ocean Optics XII, 44–55 (1994).\n[Crossref]\n\n#### Kneizys, F. X.\n\nF. X. Kneizys, L. W. Abreu, G.P. Anderson, J.H. Chetwynd, E.P. Shettle, A. Berk, L.S. Bernstein, D.C. Roberson, P. Acharya, L.S. Rothman, J.E.A. Selby, W.O. Gallery, and S.A. Clough, “The MODTRAN2/3 report and LOWTRAN 7 model,” Phillips Laboratory, Hanscom AFB (1996).\n\n#### Li, W.\n\nW. Li, K. Stamnes, R. Spurr, and J.J. Stamnes, “Simultaneous retrieval of aerosols and ocean properties: A classic inverse modeling approach II. SeaWiFS case study for the Santa Barbara channel”, Int. J. Remote Sens. 29, 5689–5698 (2008).\n[Crossref]\n\n#### Lin, B.\n\nY.X. Hu, B. Wielicki, B. Lin, G. Gibson, S.C. Tsay, K. Stamnes, and T. Wong, “Delta-fit: A fast and accurate treatment of particle scattering phase functions with weighted singular-value decomposition least squares fitting,” J. Quant. Spectrosc. Radiat. Transf. 65, 681–690 (2000).\n[Crossref]\n\n#### Loisel, H.\n\nH. Loisel and A. Morel, “Light scattering and chlorophyll concentration in case 1 waters: a re-examination,” Limnol. Oceanogr. 43, 847–857 (1998).\n[Crossref]\n\n#### Lowe, D.\n\nD.S. Broomhead and D. Lowe, “Radial basis functions, multi-variable functional interpolation and adaptive networks,” Complex Systems 2, 321–355 (1988).\n\n#### McKee, D.\n\nD. McKee, R. Röttgers, G. Neukermans, V. SanjuanCalzado, C. Trees, M. Ampolo-Rella, C. Neil, and A. Cunningham, “Impact of measurement uncertainties on determination of chlorophyll-specific absorption coefficient for marine phytoplankton,” J. Geophys. Res.-Oceans 119, 9013–9025 (2014).\n[Crossref]\n\nR. Röttgers, D. McKee, and S.B. Woźniak, “Evaluation of scatter corrections for ac-9 absorption measurements in coastal waters,” Methods in Oceanography 7, 21–39 (2013).\n[Crossref]\n\nD. McKee, J. Piskozub, R. Röttgers, and R. Reynolds, “Evaluation and improvement of an iterative scattering correction scheme for in situ absorption and attenuation measurements,” J. Atmos. Ocean. Tech. 30, 1527–1541 (2013).\n[Crossref]\n\nD. McKee, A. Cunningham, J. Slater, K.J. Jones, and C.R. Griffiths, “Inherent and apparent optical properties in coastal waters: a study of the Clyde Sea in early summer,” Estuarine, Coastal and Shelf Science 56(2), 369–376 (2003).\n[Crossref]\n\nV. Sanjuan Calzado, D. McKee, and C. Trees, “Multi and single cast radiometric processing and merging in the Ligurian Sea,” Optics of Natural Waters 2011Saint Petersburg, September 2011.\n\n#### Mobley, C.D.\n\nC.D. Mobley, Light and Water: Radiative Transfer in Natural Waters (Academic, 1994)\n\n#### Moore, C.M.\n\nJ.R.V. Zaneveld, J.C. Kitchen, and C.M. Moore, “The scattering error correction of reflecting-tube absorption meters,” Proc. SPIE 2258 Ocean Optics XII, 44–55 (1994).\n[Crossref]\n\n#### Morel, A.\n\nM. Babin, A. Morel, V. Fournier-Sicre, F. Fell, and D. Stramski, “Light scattering properties of marine particles in coastal and open ocean waters as related to the particle mass concentration,”Limnol. Oceanogr. 28, 843–859 (2003).\n[Crossref]\n\nA. Bricaud, A. Morel, M. Babin, K. Allali, and H. Claustre, “Variations in light absorption by suspended particles with chlorophyll concentration in oceanic (case 1) waters: Analysis and implications for bio-optical models,” J. Geophys. Res. 103, 31033–31044 (1998).\n[Crossref]\n\nH. Loisel and A. Morel, “Light scattering and chlorophyll concentration in case 1 waters: a re-examination,” Limnol. Oceanogr. 43, 847–857 (1998).\n[Crossref]\n\n#### Neil, C.\n\nD. McKee, R. Röttgers, G. Neukermans, V. SanjuanCalzado, C. Trees, M. Ampolo-Rella, C. Neil, and A. Cunningham, “Impact of measurement uncertainties on determination of chlorophyll-specific absorption coefficient for marine phytoplankton,” J. Geophys. Res.-Oceans 119, 9013–9025 (2014).\n[Crossref]\n\n#### Neukermans, G.\n\nD. McKee, R. Röttgers, G. Neukermans, V. SanjuanCalzado, C. Trees, M. Ampolo-Rella, C. Neil, and A. Cunningham, “Impact of measurement uncertainties on determination of chlorophyll-specific absorption coefficient for marine phytoplankton,” J. Geophys. Res.-Oceans 119, 9013–9025 (2014).\n[Crossref]\n\n#### Obelesky, G.\n\nM. Babin, A.D. Stramski, G.M. Ferrari, H. Claustre, A. Bricaud, G. Obelesky, and N. Hoepffner, “Variations in the light absorption coefficients of phytoplankton, nonalgal particles and dissolved organic matter in coastal waters around Europe,” J. Geophys. Res. 108(C7):3211 (2003).\n[Crossref]\n\n#### Petzold, T.L.\n\nT.L. Petzold, “Volume scattering functions for selected ocean waters,” Technical Report SIO 7278 Scripps Institute of Oceanography, San Diego, Calif. (1972).\n\n#### Piskozub, J.\n\nD. McKee, J. Piskozub, R. Röttgers, and R. Reynolds, “Evaluation and improvement of an iterative scattering correction scheme for in situ absorption and attenuation measurements,” J. Atmos. Ocean. Tech. 30, 1527–1541 (2013).\n[Crossref]\n\n#### Reynolds, R.\n\nD. McKee, J. Piskozub, R. Röttgers, and R. Reynolds, “Evaluation and improvement of an iterative scattering correction scheme for in situ absorption and attenuation measurements,” J. Atmos. Ocean. Tech. 30, 1527–1541 (2013).\n[Crossref]\n\n#### Rivera, G.\n\nA.M. Baldridge, S.J. Hook, C.I. Grove, and G. Rivera, “The ASTER Spectral Library Version 2.0,” Remote Sens. Environ. 113, 711–715 (2009).\n[Crossref]\n\n#### Roberson, D.C.\n\nF. X. Kneizys, L. W. Abreu, G.P. Anderson, J.H. Chetwynd, E.P. Shettle, A. Berk, L.S. Bernstein, D.C. Roberson, P. Acharya, L.S. Rothman, J.E.A. Selby, W.O. Gallery, and S.A. Clough, “The MODTRAN2/3 report and LOWTRAN 7 model,” Phillips Laboratory, Hanscom AFB (1996).\n\n#### Rodgers, C.\n\nC. Rodgers, Inverse Methods for Atmospheric Sounding (World Scientific, 2000).\n\n#### Rothman, L.S.\n\nF. X. Kneizys, L. W. Abreu, G.P. Anderson, J.H. Chetwynd, E.P. Shettle, A. Berk, L.S. Bernstein, D.C. Roberson, P. Acharya, L.S. Rothman, J.E.A. Selby, W.O. Gallery, and S.A. Clough, “The MODTRAN2/3 report and LOWTRAN 7 model,” Phillips Laboratory, Hanscom AFB (1996).\n\n#### Röttgers, R.\n\nD. McKee, R. Röttgers, G. Neukermans, V. SanjuanCalzado, C. Trees, M. Ampolo-Rella, C. Neil, and A. Cunningham, “Impact of measurement uncertainties on determination of chlorophyll-specific absorption coefficient for marine phytoplankton,” J. Geophys. Res.-Oceans 119, 9013–9025 (2014).\n[Crossref]\n\nR. Röttgers, D. McKee, and S.B. Woźniak, “Evaluation of scatter corrections for ac-9 absorption measurements in coastal waters,” Methods in Oceanography 7, 21–39 (2013).\n[Crossref]\n\nD. McKee, J. Piskozub, R. Röttgers, and R. Reynolds, “Evaluation and improvement of an iterative scattering correction scheme for in situ absorption and attenuation measurements,” J. Atmos. Ocean. Tech. 30, 1527–1541 (2013).\n[Crossref]\n\n#### Ruddick, K.\n\nK. Ruddick, “DUE Coastcolour Round Robin Protocol, version 1.2,” Coastalcolour Round Robin, Oct.2010.\n\n#### Sanjuan Calzado, V.\n\nV. Sanjuan Calzado, D. McKee, and C. Trees, “Multi and single cast radiometric processing and merging in the Ligurian Sea,” Optics of Natural Waters 2011Saint Petersburg, September 2011.\n\n#### SanjuanCalzado, V.\n\nD. McKee, R. Röttgers, G. Neukermans, V. SanjuanCalzado, C. Trees, M. Ampolo-Rella, C. Neil, and A. Cunningham, “Impact of measurement uncertainties on determination of chlorophyll-specific absorption coefficient for marine phytoplankton,” J. Geophys. Res.-Oceans 119, 9013–9025 (2014).\n[Crossref]\n\n#### Selby, J.E.A.\n\nF. X. Kneizys, L. W. Abreu, G.P. Anderson, J.H. Chetwynd, E.P. Shettle, A. Berk, L.S. Bernstein, D.C. Roberson, P. Acharya, L.S. Rothman, J.E.A. Selby, W.O. Gallery, and S.A. Clough, “The MODTRAN2/3 report and LOWTRAN 7 model,” Phillips Laboratory, Hanscom AFB (1996).\n\n#### Shettle, E.P.\n\nF. X. Kneizys, L. W. Abreu, G.P. Anderson, J.H. Chetwynd, E.P. Shettle, A. Berk, L.S. Bernstein, D.C. Roberson, P. Acharya, L.S. Rothman, J.E.A. Selby, W.O. Gallery, and S.A. Clough, “The MODTRAN2/3 report and LOWTRAN 7 model,” Phillips Laboratory, Hanscom AFB (1996).\n\n#### Slater, J.\n\nD. McKee, A. Cunningham, J. Slater, K.J. Jones, and C.R. Griffiths, “Inherent and apparent optical properties in coastal waters: a study of the Clyde Sea in early summer,” Estuarine, Coastal and Shelf Science 56(2), 369–376 (2003).\n[Crossref]\n\n#### Spurr, R.\n\nW. Li, K. Stamnes, R. Spurr, and J.J. Stamnes, “Simultaneous retrieval of aerosols and ocean properties: A classic inverse modeling approach II. SeaWiFS case study for the Santa Barbara channel”, Int. J. Remote Sens. 29, 5689–5698 (2008).\n[Crossref]\n\n#### Stamnes, J.J.\n\nB. Hamre, S. Stamnes, K. Stamnes, and J.J. Stamnes, “C-DISORT: A versatile tool for radiative transfer in coupled media like the atmosphere-ocean system,” AIP Conference Proceedings 1531, 923 (2013).\n[Crossref]\n\nW. Li, K. Stamnes, R. Spurr, and J.J. Stamnes, “Simultaneous retrieval of aerosols and ocean properties: A classic inverse modeling approach II. SeaWiFS case study for the Santa Barbara channel”, Int. J. Remote Sens. 29, 5689–5698 (2008).\n[Crossref]\n\nB. Hamre, S. Stamnes, K. Stamnes, and J.J. Stamnes, “A versatile tool for radiative transfer simulations in the coupled atmosphere-ocean system: Introducing AccuRT,” Ocean Optics XXII, Portland, ME, 26–31 Oct. 2014.\n\n#### Stamnes, K.\n\nB. Hamre, S. Stamnes, K. Stamnes, and J.J. Stamnes, “C-DISORT: A versatile tool for radiative transfer in coupled media like the atmosphere-ocean system,” AIP Conference Proceedings 1531, 923 (2013).\n[Crossref]\n\nW. Li, K. Stamnes, R. Spurr, and J.J. Stamnes, “Simultaneous retrieval of aerosols and ocean properties: A classic inverse modeling approach II. SeaWiFS case study for the Santa Barbara channel”, Int. J. Remote Sens. 29, 5689–5698 (2008).\n[Crossref]\n\nY.X. Hu, B. Wielicki, B. Lin, G. Gibson, S.C. Tsay, K. Stamnes, and T. Wong, “Delta-fit: A fast and accurate treatment of particle scattering phase functions with weighted singular-value decomposition least squares fitting,” J. Quant. Spectrosc. Radiat. Transf. 65, 681–690 (2000).\n[Crossref]\n\nG.E. Thomas and K. Stamnes, Radiative Transfer in the Atmosphere and Ocean, 2nd Edition (Cambridge University, 2002)\n\nB. Hamre, S. Stamnes, K. Stamnes, and J.J. Stamnes, “A versatile tool for radiative transfer simulations in the coupled atmosphere-ocean system: Introducing AccuRT,” Ocean Optics XXII, Portland, ME, 26–31 Oct. 2014.\n\n#### Stamnes, S.\n\nB. Hamre, S. Stamnes, K. Stamnes, and J.J. Stamnes, “C-DISORT: A versatile tool for radiative transfer in coupled media like the atmosphere-ocean system,” AIP Conference Proceedings 1531, 923 (2013).\n[Crossref]\n\nB. Hamre, S. Stamnes, K. Stamnes, and J.J. Stamnes, “A versatile tool for radiative transfer simulations in the coupled atmosphere-ocean system: Introducing AccuRT,” Ocean Optics XXII, Portland, ME, 26–31 Oct. 2014.\n\n#### Stramski, A.D.\n\nM. Babin, A.D. Stramski, G.M. Ferrari, H. Claustre, A. Bricaud, G. Obelesky, and N. Hoepffner, “Variations in the light absorption coefficients of phytoplankton, nonalgal particles and dissolved organic matter in coastal waters around Europe,” J. Geophys. Res. 108(C7):3211 (2003).\n[Crossref]\n\n#### Stramski, D.\n\nM. Babin, A. Morel, V. Fournier-Sicre, F. Fell, and D. Stramski, “Light scattering properties of marine particles in coastal and open ocean waters as related to the particle mass concentration,”Limnol. Oceanogr. 28, 843–859 (2003).\n[Crossref]\n\n#### Thomas, G.E.\n\nG.E. Thomas and K. Stamnes, Radiative Transfer in the Atmosphere and Ocean, 2nd Edition (Cambridge University, 2002)\n\n#### Trees, C.\n\nD. McKee, R. Röttgers, G. Neukermans, V. SanjuanCalzado, C. Trees, M. Ampolo-Rella, C. Neil, and A. Cunningham, “Impact of measurement uncertainties on determination of chlorophyll-specific absorption coefficient for marine phytoplankton,” J. Geophys. Res.-Oceans 119, 9013–9025 (2014).\n[Crossref]\n\nV. Sanjuan Calzado, D. McKee, and C. Trees, “Multi and single cast radiometric processing and merging in the Ligurian Sea,” Optics of Natural Waters 2011Saint Petersburg, September 2011.\n\n#### Tsay, S.C.\n\nY.X. Hu, B. Wielicki, B. Lin, G. Gibson, S.C. Tsay, K. Stamnes, and T. Wong, “Delta-fit: A fast and accurate treatment of particle scattering phase functions with weighted singular-value decomposition least squares fitting,” J. Quant. Spectrosc. Radiat. Transf. 65, 681–690 (2000).\n[Crossref]\n\n#### Wielicki, B.\n\nY.X. Hu, B. Wielicki, B. Lin, G. Gibson, S.C. Tsay, K. Stamnes, and T. Wong, “Delta-fit: A fast and accurate treatment of particle scattering phase functions with weighted singular-value decomposition least squares fitting,” J. Quant. Spectrosc. Radiat. Transf. 65, 681–690 (2000).\n[Crossref]\n\n#### Wong, T.\n\nY.X. Hu, B. Wielicki, B. Lin, G. Gibson, S.C. Tsay, K. Stamnes, and T. Wong, “Delta-fit: A fast and accurate treatment of particle scattering phase functions with weighted singular-value decomposition least squares fitting,” J. Quant. Spectrosc. Radiat. Transf. 65, 681–690 (2000).\n[Crossref]\n\n#### Wozniak, S.B.\n\nR. Röttgers, D. McKee, and S.B. Woźniak, “Evaluation of scatter corrections for ac-9 absorption measurements in coastal waters,” Methods in Oceanography 7, 21–39 (2013).\n[Crossref]\n\n#### Zaneveld, J.R.V.\n\nJ.R.V. Zaneveld, J.C. Kitchen, and C.M. Moore, “The scattering error correction of reflecting-tube absorption meters,” Proc. SPIE 2258 Ocean Optics XII, 44–55 (1994).\n[Crossref]\n\n#### AIP Conference Proceedings (1)\n\nB. Hamre, S. Stamnes, K. Stamnes, and J.J. Stamnes, “C-DISORT: A versatile tool for radiative transfer in coupled media like the atmosphere-ocean system,” AIP Conference Proceedings 1531, 923 (2013).\n[Crossref]\n\n#### Astrophys. J. (1)\n\nL.G. Henyey and J.L. Greenstein, “Diffuse radiation in the galaxy,” Astrophys. J. 93, 70–83 (1941).\n[Crossref]\n\n#### Complex Systems (1)\n\nD.S. Broomhead and D. Lowe, “Radial basis functions, multi-variable functional interpolation and adaptive networks,” Complex Systems 2, 321–355 (1988).\n\n#### Estuarine, Coastal and Shelf Science (1)\n\nD. McKee, A. Cunningham, J. Slater, K.J. Jones, and C.R. Griffiths, “Inherent and apparent optical properties in coastal waters: a study of the Clyde Sea in early summer,” Estuarine, Coastal and Shelf Science 56(2), 369–376 (2003).\n[Crossref]\n\n#### Int. J. Remote Sens. (1)\n\nW. Li, K. Stamnes, R. Spurr, and J.J. Stamnes, “Simultaneous retrieval of aerosols and ocean properties: A classic inverse modeling approach II. SeaWiFS case study for the Santa Barbara channel”, Int. J. Remote Sens. 29, 5689–5698 (2008).\n[Crossref]\n\n#### J. Atmos. Ocean. Tech. (1)\n\nD. McKee, J. Piskozub, R. Röttgers, and R. Reynolds, “Evaluation and improvement of an iterative scattering correction scheme for in situ absorption and attenuation measurements,” J. Atmos. Ocean. Tech. 30, 1527–1541 (2013).\n[Crossref]\n\n#### J. Geophys. Res. (2)\n\nA. Bricaud, A. Morel, M. Babin, K. Allali, and H. Claustre, “Variations in light absorption by suspended particles with chlorophyll concentration in oceanic (case 1) waters: Analysis and implications for bio-optical models,” J. Geophys. Res. 103, 31033–31044 (1998).\n[Crossref]\n\nM. Babin, A.D. Stramski, G.M. Ferrari, H. Claustre, A. Bricaud, G. Obelesky, and N. Hoepffner, “Variations in the light absorption coefficients of phytoplankton, nonalgal particles and dissolved organic matter in coastal waters around Europe,” J. Geophys. Res. 108(C7):3211 (2003).\n[Crossref]\n\n#### J. Geophys. Res.-Oceans (1)\n\nD. McKee, R. Röttgers, G. Neukermans, V. SanjuanCalzado, C. Trees, M. Ampolo-Rella, C. Neil, and A. Cunningham, “Impact of measurement uncertainties on determination of chlorophyll-specific absorption coefficient for marine phytoplankton,” J. Geophys. Res.-Oceans 119, 9013–9025 (2014).\n[Crossref]\n\n#### J. Quant. Spectrosc. Radiat. Transf. (1)\n\nY.X. Hu, B. Wielicki, B. Lin, G. Gibson, S.C. Tsay, K. Stamnes, and T. Wong, “Delta-fit: A fast and accurate treatment of particle scattering phase functions with weighted singular-value decomposition least squares fitting,” J. Quant. Spectrosc. Radiat. Transf. 65, 681–690 (2000).\n[Crossref]\n\n#### Limnol. Oceanogr. (3)\n\nH.R. Gordon, “Can the Lambert-Beer law be applied to the diffuse attenuation coefficient of ocean water?” Limnol. Oceanogr. 34, 1389–1409 (1989).\n[Crossref]\n\nM. Babin, A. Morel, V. Fournier-Sicre, F. Fell, and D. Stramski, “Light scattering properties of marine particles in coastal and open ocean waters as related to the particle mass concentration,”Limnol. Oceanogr. 28, 843–859 (2003).\n[Crossref]\n\nH. Loisel and A. Morel, “Light scattering and chlorophyll concentration in case 1 waters: a re-examination,” Limnol. Oceanogr. 43, 847–857 (1998).\n[Crossref]\n\n#### Methods in Oceanography (1)\n\nR. Röttgers, D. McKee, and S.B. Woźniak, “Evaluation of scatter corrections for ac-9 absorption measurements in coastal waters,” Methods in Oceanography 7, 21–39 (2013).\n[Crossref]\n\n#### Proc. SPIE (2)\n\nG. Fournier and J.L. Forand, “Analytic phase function for ocean water,” Proc. SPIE 2258 Ocean Optics XII, 194–201 (1994).\n[Crossref]\n\nJ.R.V. Zaneveld, J.C. Kitchen, and C.M. Moore, “The scattering error correction of reflecting-tube absorption meters,” Proc. SPIE 2258 Ocean Optics XII, 44–55 (1994).\n[Crossref]\n\n#### Remote Sens. Environ. (1)\n\nA.M. Baldridge, S.J. Hook, C.I. Grove, and G. Rivera, “The ASTER Spectral Library Version 2.0,” Remote Sens. Environ. 113, 711–715 (2009).\n[Crossref]\n\n#### Other (8)\n\nF. X. Kneizys, L. W. Abreu, G.P. Anderson, J.H. Chetwynd, E.P. Shettle, A. Berk, L.S. Bernstein, D.C. Roberson, P. Acharya, L.S. Rothman, J.E.A. Selby, W.O. Gallery, and S.A. Clough, “The MODTRAN2/3 report and LOWTRAN 7 model,” Phillips Laboratory, Hanscom AFB (1996).\n\nT.L. Petzold, “Volume scattering functions for selected ocean waters,” Technical Report SIO 7278 Scripps Institute of Oceanography, San Diego, Calif. (1972).\n\nK. Ruddick, “DUE Coastcolour Round Robin Protocol, version 1.2,” Coastalcolour Round Robin, Oct.2010.\n\nC. Rodgers, Inverse Methods for Atmospheric Sounding (World Scientific, 2000).\n\nV. Sanjuan Calzado, D. McKee, and C. Trees, “Multi and single cast radiometric processing and merging in the Ligurian Sea,” Optics of Natural Waters 2011Saint Petersburg, September 2011.\n\nG.E. Thomas and K. Stamnes, Radiative Transfer in the Atmosphere and Ocean, 2nd Edition (Cambridge University, 2002)\n\nB. Hamre, S. Stamnes, K. Stamnes, and J.J. Stamnes, “A versatile tool for radiative transfer simulations in the coupled atmosphere-ocean system: Introducing AccuRT,” Ocean Optics XXII, Portland, ME, 26–31 Oct. 2014.\n\nC.D. Mobley, Light and Water: Radiative Transfer in Natural Waters (Academic, 1994)\n\n### Cited By\n\nOSA participates in Crossref's Cited-By Linking service. Citing articles from OSA journals and other participating publishers are listed here.\n\nAlert me when this article is cited.\n\n### Figures (12)\n\nFig. 1 Location of 14 Lu(z), Ed(z), and IOP measurement conducted in the BP09 experiment used in this paper, see Table 1 for details.\nFig. 2 Flow chart of the IOP inversion algorithm.\nFig. 3 Rayleigh, Fournier-Forand and Petzold phase function used in this paper.\nFig. 4 IOPs inverted from synthetic radiance reflectance RL(z) data at 488 nm. The upper panels show comparison between retrieved and synthetic IOPs, where filled circles (black) represent synthetic data, solid lines (blue) represent the inverted IOPs using the correct phase function (g = 0.9), and dashed lines(red) represent the inverted IOPs using incorrect phase function (g = 0.8). Note that when using correct phase function, the retrieved IOPs are so close to the synthetic data that the blue and black curves overlapped. The lower panels show the corresponding percentage errors.\nFig. 5 Retrieved CHL, MIN and CDOM compared with model data for the synthetic dataset. The color of the dots shows the percentage error in the retrieved data.\nFig. 6 Retrieved IOPs (red) of ST15 compared with in-situ measurements (blue). The top 6 panels, from left to right, show comparisons of the absorption coefficients at 412, 440, 488, 510, 532 and 555 nm, respectively, whereas the middle and bottom 6 panels show the same for the scattering and backscattering coefficients.\nFig. 7 Retrieved IOPs (red) of ST28 compared with in-situ measurements (blue). The top 6 panels, from left to right, show comparisons of the absorption coefficients at 412, 440, 488, 510, 532 and 555 nm, respectively, whereas the middle and bottom 6 panels show the same for the scattering and backscattering coefficients.\nFig. 8 Comparison between retrieved and measured IOPs that combined all depth, wavelengths and stations. The 3 panels, from left to right, show comparisons of absorption, scattering and backscattering coefficients, respectively. The color of the dots indicates wavelength as shown on the color bar.\nFig. 9 Retrieved CHL, MIN and CDOM from in-situ measured IOPs compare with measured values for all stations.\nFig. 10 Retrieved vertical profiles of CHL (green), MIN (blue) and CDOM (red). The solid lines are the water constituent profiles retrieved from in-situ measured IOPs, and the dashed lines are water constituent profiles retrieved from inverted IOPs. The filled circles are the in-situ measurements of CHL (green), MIN (blue) and CDOM (red).\nFig. 11 Standard MODIS CHL a product of the Ligurian Sea area on 17 March 2009.\nFig. 12 Absorption (left), scattering (middle) and backscattering coefficients (right) of CHL (green), MIN (blue) and CDOM (red), respectively.\n\n### Tables (3)",
null,
"Table 1 Time, location and mean ocean depth of the 14 measurements. The * symbol indicates the deep ocean cases tested in our IOP inversion algorithm.",
null,
"Table 2 Number of iterations (i) needed to invert the IOPs from Lu(z) and Ed(z) measurements and absolute percentage error (PE) of the radiance reflectances RL(z) for deep water cases.",
null,
"Table 3 Correlation and bias of retrieved IOPs for deep water cases.\n\n### Equations (45)\n\nEquations on this page are rendered with MathJax. Learn more.\n\n$R L ( z ) ≡ L u ( z ) / E d ( z ) ,$\n$μ d L ( τ , μ , ϕ ) d τ = L ( τ , μ , ϕ ) − S * ( τ , μ , ϕ ) − ϖ ( τ ) 4 π ∫ 0 2 π d ϕ ′ ∫ − 1 1 d μ ′ p ( τ ; μ ′ , ϕ ′ ; μ , ϕ ) L ( τ , μ ′ , ϕ ′ )$\n$E u ( τ ) = 2 π ∫ 0 1 d μ μ L ( τ , μ ) .$\n$E d ( τ ) = 2 π ∫ 0 1 d μ μ L ( τ , − μ ) .$\n$E 0 ( τ ) = 2 π ∫ − 1 1 d μ L ( τ , μ ) .$\n$a ( z ) = μ ¯ ( z ) K v ( z )$\n$b b ( z ) = a ( z ) X ( z )$\n$X ( z ) = 3 { R E ( z ) − d R E ( z ) d z ∫ z z max d z ′ [ E d ( z ′ ) E d ( z ) ] 2 }$\n$b ( z ) = b b ( z ) b ˜ b ( z ) = b b ( z ) 2 π ∫ π / 2 π p ( cos Θ ) sin ( Θ ) d Θ$\n$K d ( z ) ≡ − d [ ln ( E d ( z ) ) ] / d z .$\n$K d ( z ) ≈ 1.0395 a ( z ) + b b ( z ) cos ( θ w )$\n$R L ( z ) ≈ 0.094 b b ( z ) a ( z ) .$\n$a ( z ) ≈ K d ( z ) cos ( θ w ) 1.0395 [ R L ( z ) 0.094 + 1 ]$\n$b b ( z ) ≈ K d ( z ) cos ( θ w ) 1.0395 [ 0.094 R L ( z ) + 1 ] .$\n$p FF ( cos Θ ) = 1 4 π ( 1 − δ ) 2 δ ν { ν ( 1 − δ ) − ( 1 − δ ν ) + 4 u 2 [ δ ( 1 − δ ν ) − ν ( 1 − δ ) ] } + 1 − δ 180 ν 16 π ( δ 180 − 1 ) δ 180 ν [ 3 cos 2 Θ − 1 ]$\n$p w ( cos Θ ) = 3 3 + f ( 1 + f cos 2 Θ )$\n$χ ℓ = f FF × χ ℓ , FF + f PET × χ ℓ , PET + ( 1 − f FF − f PET ) × χ ℓ , water$\n$b ˜ bp ( z ) = b b ( z ) − b b w ( z ) b ( z ) − b w ( z )$\n$b ˜ bp ( z ) = f FF ′ ( z ) × 0.0056 + [ 1 − f FF ′ ( z ) ] × 0.021 .$\n$f FF ( z ) = f FF ′ ( z ) × b ˜ bp ( z ) b ˜ b ( z )$\n$f PET ( z ) = [ 1 − f FF ′ ( z ) ] × b ˜ bp ( z ) b ˜ b ( z ) .$\n$δ ( n ) = 1 N ∑ i = 1 N | ln [ E d ( n ) ( z i ) ] − ln [ E d m ( z i ) ] | + 1 N ∑ i = 1 N | ln [ L u ( n ) ( z i ) ] − ln [ L u m ( z i ) ] | .$\n$δ R L ( n ) = 1 N ∑ i = 1 N | ln [ L u ( n ) ( z i ) E d ( n ) ( z i ) ] − ln [ L u m ( z i ) E d m ( z i ) ] | .$\n$IOP i = ∑ j = 1 N a i j exp [ − b 2 ∑ k = 1 N in ( p k − c j k ) 2 ] + d i$\n$K i , k = ∂ ( IOP i ) ∂ p k = − 2 b 2 ( p k − c j k ) × ∑ j = 1 N a i j exp [ − b 2 ∑ k = 1 N in ( p k − c j k ) 2 ] .$\n$x i + 1 = x i + [ ( 1 + γ i ) S a − 1 + K i T S m − 1 K i ] − 1 × K i T S m − 1 ( y m − y i ) − S a − 1 ( x i − x a )$\n$a ( z ) = a 0 + a 1 exp [ ( z − z a ) 2 2 σ a 2 ]$\n$b b ( z ) = b b 0 + b b 1 exp [ ( z − z b ) 2 2 σ b 2 ]$\n$PE ( % ) = 1 N ∑ i = 1 N | R L − R RTM − R L − measured | R L − measured × 100$\n$bias ( % ) = 1 N ∑ i = 1 N ( IOP inv . − IOP in − situ ) IOP in − situ × 100$\n$a pig ( λ ) = A ( λ ) × CHL E ( λ )$\n$c pig ( 660 ) = 0.407 × CHL 0.795$\n$c pig ( λ ) = c pig ( 660 ) × ( λ / 660 ) ν$\n$ν = 0.5 × [ log 10 CHL − 0.3 ] 0.02 < CHL < 2.0 ν = 0 , CHL > 2.0$\n$b pig ( λ ) = c pig ( λ ) − a pig ( λ ) .$\n$a MIN ( 443 ) = 0.031 × MIN$\n$a MIN ( λ ) = a MIN ( 443 ) exp [ − 0.0213 ( λ − 443 ) ] .$\n$b MIN ( 555 ) = 0.51 × MIN$\n$c MIN ( λ ) = c MIN ( 555 ) × ( λ / 555 ) − 0.3749$\n$c MIN ( 555 ) = a MIN ( 555 ) + b MIN ( 555 ) = 0.52 × MIN .$\n$b MIN ( λ ) = c MIN ( λ ) − a MIN ( λ ) .$\n$a CDOM ( λ ) = CDOM × exp [ − 0.0176 ( λ − 443 ) ] .$\n$a tot ( λ ) = a pig ( λ ) + a MIN ( λ ) + a CDOM ( λ )$\n$b tot ( λ ) = b pig ( λ ) + b MIN ( λ )$\n$b b tot ( λ ) = 0.0056 × b pig ( λ ) + 0.019 × b MIN ( λ ) .$"
] |
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http://www.bennorthrop.com/e/epc-construction-rules.php
|
[
"Ben Northrop\n\nDecisions and software development\n\nEPC Construction Rules\n\nApril 22nd 2011\n\nPoints\n1. Let a be a point [distinct from . . . ].\n2. Let a be a point on L [distinct from . . . ].\n3. Let a be a point on L between b and c [distinct from . . . ].\n4. Let a be a point on L extending the segment from b to c [with a distinct from. . . ].\n5. Let a be a point on the same side of L as b [distinct from. . . ].\n6. Let a be a point on the side of L opposite b [distinct from. . . ].\n7. Let a be a point on a [distinct from . . . ].\n8. Let a be a point inside a [distinct from . . . ].\n9. Let a be a point outside a [distinct from . . . ].\nLines and Circles\n1. Let L be the line through a and b [distinct from . . . ].\n2. Let a be the circle with center a passing through b [distinct from . . . ].\nIntersections\n1. Let a be a point of intersection of L and M [distinct from . . . ].\n2. Let a be a point of intersection of C and L [distinct from . . . ].\n3. Let a and b be the two points of intersection of C and L [distinct from . . . ].\n4. Let a be a point of intersection of L and C between b and c [distinct from . . . ].\n5. Let a be a point of intersection of L and C extending the segment from c to b [distinct from . . . ].\n6. Let a be a point of intersection of C and D [distinct from . . . ].\n7. Let a and b be the two points of intersection of C and D [distinct from . . . ].\n8. Let a be a point of intersection of C and D on the same side of L as b where L is the line through their centers c and d respectively [distinct from . . . ].\n9. Let a be a point of intersection of C and D on the side of L opposite b where L is the line through their centers c and d respectively [distinct from . . . ].",
null,
"I believe that software development is fundamentally about making decisions, and so this is what I write about (mostly). I'm the owner of Highline Solutions and also the Principal Technical Consultant. I have two degrees from Carnegie Mellon University, most recently one in philosophy (thesis here). I live in Pittsburgh, PA with my wife and 3 energetic boys. Subscribe here or write me at ben dot northrop at gmail dot com."
] |
[
null,
"http://www.bennorthrop.com/images/BenNorthrop-2015.jpg",
null
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https://www.thestudentroom.co.uk/showthread.php?t=4998310
|
[
"# Hard chemistry question GCSE! Help!\n\nAnnouncements\nThis discussion is closed.\n#1\nIron is extracted by heating iron oxide with carbon monoxide.\n\nFe2 O3 + 3CO -----> 2Fe + 3CO2\n\na) Calculate the theoretical yield of iron that can be obtained when 320kg of iron oxide (Fe2 O3) reacts with excess carbon monoxide.\n\nb) The actual yield of iron is 100kg. Calculate the percentage yield of iron.\n\nThis is really hard and not straight forward, I have got an answer that I have worked out, but not sure if it's right, could a really good chemist student work it out and see what you got? My answers are in the spoilers. Thanks!!",
null,
"Spoiler:\nShow\n\na) 146.9kg\nb) 68.1%\n0\n4 years ago\n#2\nSorry you've not had any responses about this.",
null,
"Are you sure you've posted in the right place?",
null,
"Here's a link to our subject forum which should help get you more responses if you post there.",
null,
"0\n4 years ago\n#3\nI got a different set of answers to you. This was my method:moles= mass / molecular mass so the number of moles of Fe2O3 is:320000g / 159.6 gmol-1 = 2005 molThen the ratio is 1:2 between the Fe2O3 and the Fe so you get 4010 mol of Fe. To get the mass you do 4010 x 55.8 = 223758g = 223.758kgWith this answer the percentage yield is 44.7%.\n1\n4 years ago\n#4\n(Original post by hannah-rosie)\nI got a different set of answers to you. This was my method:moles= mass / molecular mass so the number of moles of Fe2O3 is:320000g / 159.6 gmol-1 = 2005 molThen the ratio is 1:2 between the Fe2O3 and the Fe so you get 4010 mol of Fe. To get the mass you do 4010 x 55.8 = 223758g = 223.758kgWith this answer the percentage yield is 44.7%.\ni got the same answer. would it make a difference if u didn't convert from g to kg?\n0\n3 years ago\n#5\nI'm confused on how to get the answer.Can someone lay it out step by step?\n0\n3 years ago\n#6\n(Original post by Fergaljones)\nI'm confused on how to get the answer.Can someone lay it out step by step?\n1.First calculate the number of moles of Fe2O3:\n\nThe mass given is 320kg.\nThe formula mass is 2(56)+3(16) which is 160\nThe moles is mass/formula mass so 320/160 gives 2 moles.\n2. From the reaction we see than 1 mole of Fe2O3 reacts with 2 moles of iron, so the moles of iron theoretically formed is 2*2 which is 4 moles. The mass is given by moles * formula mass which is 4*56 which is 224kg\n3. The percentage yield is actual yield/ theoretical yield * 100 which is 100/224 * 100, giving 44.6 %\n1\n3 years ago\n#7\nRemember molar ratios and relevant formulae:\n\nNo. moles = Mass / Molar mass\n\nThe molar ratio of Fe2O3 to Fe is 1:2 - (1)Fe2O3, 2Fe.\n\nThere are 2000 moles of iron oxide (worked out with formula above), so we know by the ratio 1:2 that there muse be 4000 moles of iron in products, right?\n\nNow work out the mass of 4000 moles of iron - that is the mass of iron you could theoretically expect to yield from 320kg of iron oxide.\n\nPercentage yield = Actual yield / Theoretical yield.\n\nYou know your actual yield from the question, and you have worked out the theoretical yield.\nLast edited by username4601382; 3 years ago\n0\n3 years ago\n#8\nyh thats def the right answer, easiest method as well.\n(Original post by sangsan)\n1.First calculate the number of moles of Fe2O3:\n\nThe mass given is 320kg.\nThe formula mass is 2(56)+3(16) which is 160\nThe moles is mass/formula mass so 320/160 gives 2 moles.\n2. From the reaction we see than 1 mole of Fe2O3 reacts with 2 moles of iron, so the moles of iron theoretically formed is 2*2 which is 4 moles. The mass is given by moles * formula mass which is 4*56 which is 224kg\n3. The percentage yield is actual yield/ theoretical yield * 100 which is 100/224 * 100, giving 44.6 %\n0\n3 years ago\n#9\n(Original post by sangsan)\n1.First calculate the number of moles of Fe2O3:\n\nThe mass given is 320kg.\nThe formula mass is 2(56)+3(16) which is 160\nThe moles is mass/formula mass so 320/160 gives 2 moles.\n2. From the reaction we see than 1 mole of Fe2O3 reacts with 2 moles of iron, so the moles of iron theoretically formed is 2*2 which is 4 moles. The mass is given by moles * formula mass which is 4*56 which is 224kg\n3. The percentage yield is actual yield/ theoretical yield * 100 which is 100/224 * 100, giving 44.6 %\nbut 320 kg of Fe2O3 is 2000 mol of Fe2O3. You need to convert to g.\n4000 mol of Fe is formed which is 224000 g or 224 kg\nLast edited by username3249896; 3 years ago\n0\n3 years ago\n#10\nOkay thank you, I understand now.\n(Original post by sangsan)\n1.First calculate the number of moles of Fe2O3:\n\nThe mass given is 320kg.\nThe formula mass is 2(56)+3(16) which is 160\nThe moles is mass/formula mass so 320/160 gives 2 moles.\n2. From the reaction we see than 1 mole of Fe2O3 reacts with 2 moles of iron, so the moles of iron theoretically formed is 2*2 which is 4 moles. The mass is given by moles * formula mass which is 4*56 which is 224kg\n3. The percentage yield is actual yield/ theoretical yield * 100 which is 100/224 * 100, giving 44.6 %\n0\n2 years ago\n#11\nThe molar mass of Fe2O3 = 2* 56 3*16 = 112 48 = 160 (g).=> Mole of Fe2O3 used = 320 000/ 160 = 2 000 (moles).=> Mole of Fe = 2 000 * 2 = 4 000 (moles).=> Theoretical yield = 4 000 * 56 = 224 000 (g) = 224 (kg).=> Percentage yield = Actual yield / Theoretical yield * 100 = 100/ 224 * 100 = 44.643 (%).\n0\nX\nnew posts",
null,
"Back\nto top\nLatest\nMy Feed\n\n### Oops, nobody has postedin the last few hours.\n\nWhy not re-start the conversation?\n\nsee more\n\n### See more of what you like onThe Student Room\n\nYou can personalise what you see on TSR. Tell us a little about yourself to get started.\n\n### Poll\n\nJoin the discussion\n\n#### Were exams easier or harder than you expected?\n\nEasier (13)\n26.53%\nAs I expected (14)\n28.57%\nHarder (19)\n38.78%\nSomething else (tell us in the thread) (3)\n6.12%"
] |
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null,
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https://dev.to/avalander/handling-errors-with-either-2i7j
|
[
"## DEV Community is a community of 729,587 amazing developers\n\nWe're a place where coders share, stay up-to-date and grow their careers.\n\nAvalander\n\nPosted on\n\n# Handling errors with Either\n\nAn `Either` is basically a container for a value that might be an error. With an `Either` we can apply transformations to the contained value without having to worry whether it is an error or not until we reach a point in our code where we want to handle the error, should it have happened. It's a bit like a Schrödinger's box: the value might or might not be an error, we won't know until we open it (alright, missing the point of Schrödinger's cat, but I wanted to put the reference anyway).\n\n# How does Either work?\n\nTo illustrate the `Either` structure, let's build it in Javascript.\n\nFirst of all, an `Either` can hold a value or an error. We'll call them `Right` and `Left` respectively. In a sense, it's like having two branches, and you go either to the left if you get an error, or to the right if you get a valid value.\n\nAlso, we need to be able to apply transformations to the value that is in the `Either`. Otherwise it's not really useful. We want a `map` function to do that. And we are going to apply the transformation only if we are on the `Right` branch, and ignore it if we have a `Left`.\n\n``````const Left = x => ({\nmap: fn => Left(x),\n})\n\nconst Right x => ({\nmap: fn => Right(fn(x)),\n})\n``````\n\nNote that `Left.map` returns a `Left` holding the same value, without applying the transformation `fn`, while `Right.map` returns a `Right` containing the result of applying `fn` to the value. The reason for that is that we only want to apply the transformation on a valid value, not on an error.\n\n``````Right(3).map(x => x * x) // -> Right(9)\nLeft(3).map(x => x * x) // -> Left(3)\n``````\n\nNow imagine that we want to apply a transformation to a value contained in an `Either`, but that transformation can return an error. Since we are handling error branches with `Either`, we might as well return a new `Either`.\n\n``````const result = Right(3)\n.map(x => x % 2 == 0\n? Right(x)\n: Left('Odd'))\n``````\n\nWe have a number contained in an `Either` and we only want to accept even numbers. If it's odd, we return a `Left` saying that the number is odd.\n\nThe problem is that now we have a `Left` contained inside a `Right`. If we would inspect the variable `result` it would hold `Right(Left('Odd'))`. If we want to apply another transformation, should we apply it to the outer `Right` or to the inner `Left`? What happens when the next transformation returns another `Either`?\n\nTo solve this issue, we can implement the method `chain`. `chain` is much like `map`, but it expects the transformation to return an `Either`, so it doesn't wrap the result of applying the transformation in a new `Either`.\n\n``````const Left = x => ({\nmap: fn => Left(x),\nchain: fn => Left(x),\n})\n\nconst Right x => ({\nmap: fn => Right(fn(x)),\nchain: fn => fn(x),\n})\n``````\n\n`Left.chain` still doesn't apply the transformation, and it returns a `Left` holding the error, so we're sure we are not going to operate on an error should it have happened.\n\n`Right.chain` will apply the transformation `fn` to the contained value and return the result, without wrapping it in another `Right`, because it expects the function `fn` to return an `Either`. If we were implementing this in a real project, we would probably want to check that `fn` returns an `Either` and throw an error if it doesn't.\n\nWe can use `chain` in the previous example to make sure that we don't end up with an `Either` inside another `Either`.\n\n``````const result = Right(3)\n.chain(x => x % 2 == 0\n? Right(x)\n: Left('Odd'))\n\nresult // -> Left('Odd')\n``````\n\nNow we only have a `Left`, and we would have a `Right` if our value had been even.\n\nAnd that's it. We can use `map` to apply transformations to our contained value and keep it inside the same `Either`, or `chain` if we want to apply a transformation that returns another `Either` because it might fail.\n\nEven though it's nice to be able to operate over a value without caring whether it's an error or not, it's not really that useful if we can't access the value. Right now the value is contained forever in an `Either`, and we will never know if the operation succeeded and the transformations were applied to the value, or if we have an error waiting to be handled.\n\nWe can implement one last method to solve this issue: `fold`. `fold` takes two callbacks, the first one (or left) will be called if the `Either` contains an error and the second one (or right) will be called if the `Either` contains a valid value.\n\n``````const Left = x => ({\nmap: fn => Left(x),\nchain: fn => Left(x),\nfold: (fnLeft, fnRight) => fnLeft(x),\n})\n\nconst Right x => ({\nmap: fn => Right(fn(x)),\nchain: fn => fn(x),\nfold: (fnLeft, fnRight) => fnRight(x),\n})\n``````\n\nIf we have a `Left`, `fnLeft` will be invoked, so we can handle the error in that function. If we have a `Right`, `fnRight` will be invoked and we can use it to send the value in an HTTP response, or store it in a database or do whatever we need with that value.\n\n``````Right(3)\n.chain(x => x % 2 == 0\n? Right(`\\${x} is even.`)\n: Left('Odd'))\n.fold(\nconsole.error,\nconsole.log\n)\n``````\n\nThis simple example handles errors by printing them in `console.error`, and prints valid values in `console.log`, but we could handle errors and successes in any other way we need.\n\n# Handy Either factories\n\nThere are a few common factories for `Either` that we can implement easily.\n\n## Maybe\n\nMaybe is a well known data structure, called Optional in some languages, that might or might not contain a value. We could model it with an `Either` that will be a `Right` if it has a value and an empty `Left` if it doesn't. Let's see how to build it.\n\n``````const maybe = value =>\n(value != null\n? Right(value)\n: Left())\n``````\n\nOr, if you don't like ternary operators that much,\n\n``````const maybe = value => {\nif (value != null) {\nreturn Right(value)\n}\nreturn Left()\n}\n``````\n\n## TryCatch\n\nSometimes we might want to call a function that can throw an exception and treat the exception as an error with an `Either`. That might come in handy if we are using `Either` to handle errors in our code and need to interface with a library that handles errors by throwing exceptions (and expecting the user to catch them).\n\n``````const tryCatch = (fn, ...args) => {\ntry {\nconst result = fn.apply(null, args)\nreturn Right(result)\n} catch (e) {\nreturn Left(e)\n}\n}\n``````\n\n## Conditional\n\nWe might want to check if a value fulfils a certain condition and return an error if it doesn't. We can define a factory that will take a predicate (i.e., a function that checks a condition on the value an returns either `true` or `false`) and a value, and return a `Right` if the condition holds true for the given value and a `Left` otherwise. We can get a bit fancier and allow an extra argument with an error value (usually a message explaining why the value wasn't accepted) that will be used if the value doesn't fulfil the condition.\n\n``````const condition = (pred, value, reason) =>\n(pred(value)\n? Right(value)\n: Left(reason))\n``````\n\nRemember the `maybe` factory that we implemented a bit earlier? Turns out that it's only a specific case of `condition`.\n\n``````const maybe = value =>\ncondition(x => x != null, value)\n``````\n\n# When to use Either\n\nMy personal opinion is that `Either` is simply a strategy to handle application errors, and choosing this or another strategy is more a matter of preference that anything else.\n\nSome languages, like Python or Java, offer a well-thought exception system that can be used to handle any application errors that might happen. In these languages it's usually a good idea to keep things idiomatic.\n\nOther languages don't have an exception system and expect the programmer to return an error value if an error can happen in a function call (I'm looking at you, Go). Then I think it's better to use an `Either` than returning `(err, result)` and having to check for `err` every time we call a function, especially if we need to pass the error one layer up, where it can be handled.\n\nAnd then there is Javascript. It has an exception system. Sort of. The problem is that catching specific errors while letting others propagate with Javascript's exception system is not a trivial task. Therefore it might be worth to use `Either` for application errors and leave exceptions for programming errors, instead of catching exceptions and trying to figure out if it's an error that should be handled here, elsewhere or make the application crash.\n\nThat's it, folks, thanks for reading!\n\n## Discussion (12)",
null,
"Nicola Apicella\n\nHi! Nice article. I have been using vavr and its Either type for a while. I still didn't quite figure out how to use it when I need to expose different type of errors in the either. I thought about have different Error type and do a type assertion, but at this point I d better go with exceptions which expose this behaviour naturally(catch clauses with different types). Do you have any thought about it?",
null,
"Avalander • Edited\n\nWell, if you actually need different behaviour for different kinds of errors (as opposed to simply having different data in the error), you need a bit of extra effort to make it work with Either. I don't know about vavr, but it seems to be Java, and Java has a pretty decent exception system, so the easiest might be to just throw and catch checked exceptions.\n\nThat being said, Elm has a cool thing called union types which can be used to handle multiple kinds of errors. It's similar to an Either, but it allows more than two branches, each branch with a different behaviour.\n\nTo implement that in Javascript I would try something like this:\n\n``````const UnionType = types => types.reduce((prev, type) => ({\n...prev,\n[type]: (data) => ({\nmatch: (fns) => fns[type](data)\n})\n}), {})\n``````\n\nThen you can create your own union type to handle errors.\n\n``````const CustomErrors = UnionType([\n'NetworkError',\n'InputError',\n'RandomError',\n])\n\nconst someError = CustomErrors.RandomError({ message: 'Random' })\n\nsomeError.match({\nNetworkError: ({ status, message }) => {...},\nInputError: ({ field, value }) => {...},\nRandomError: ({ message }) => console.error(message),\n})\n``````\n\nYou can either have a union type in the left branch of an `Either` and do the matching when you fold it, or have a union type with a branch for valid values and several branches for invalid values (not sure how that second option would work out, though, you would need to implement `map` and `chain` in `UnionType` also). And, of course, you can use it for a lot of other things, besides handling different kinds of errors.\n\nNow, this is something I just thought that might be interesting to borrow from Elm, but I haven't really tried in Javascript, so use it with caution.",
null,
"Nicola Apicella\n\nInteresting! Gotta read about union types and experiment a bit. Thanks a lot :)",
null,
"Wow, it's good to see more people in FP these days!, more collaboration is always welcome.\nBtw, how would you handle sync/async + composition?\nIt's a bit hard to handle all these concepts together in JS, in Elixir or Scala I don't see too much problem due to their nature.\n\nThanks!.",
null,
"Avalander\n\nThanks for the comment :)\n\nBtw, how would you handle sync/async + composition?\n\nThat's a good question. Unfortunately, I don't have a good answer.\n\nSometimes I just use promises as pseudo-either monads when I have to mix sync and async code and that's good enough.\n\n``````// Instead of throwing an exception with invalid JSON,\n// this function will capture it in a promise's rejection branch.\nconst parseJson = data => Promise.resolve(data)\n.then(JSON.parse)\n\n// Another sync function wrapped in a promise.\nconst verifyResponse = ({ statusCode, body }) =>\nstatusCode === 200\n? Promise.resolve(body)\n: Promise.reject({ statusCode, body })\n\nparseJson(someData) // For some weird reason our data is stringified JSON, very convenient to manipulate.\n.then(postData) // We send our data to a remote host\n.then(verifyResponse) // We verify that we get 200 back\n.then(parseJson) // We parse the body of the response\n// ...\n``````\n\nAnother option would be to map eithers and maybes to promises when composing sync and async functions.\n\n``````const Left = x => ({\n...\ntoPromise: () => Promise.reject(x),\n})\n\nconst Right = x => ({\n...\ntoPromise: () => Promise.resolve(x),\n})\n\nconst map = F => x => F.map(x)\nconst chain = F => x => F.chain(x)\nconst toPromise = F => F.toPromise()\n\nconst parseJson = data => tryCatch(JSON.parse, data)\n\nconst verifyResponse = ({ statusCode, body }) =>\nstatusCode === 200\n? Right(body)\n: Left({ statusCode, body })\n\nparseJson(someData)\n.toPromise()\n.then(postData) // We send our data to a remote host\n.then(verifyResponse) // We verify that we get 200 back\n.then(chain(parseJson)) // In case you want to keep the Either inside the promise\n.then(toPromise) // In case you want to unwrap the Either\n``````\n\nMost of the times I think promises as pseudo-eithers is good enough, so I haven't explored how to compose them with proper eithers that much.",
null,
"I actually used to do what you mention in the first code example, but not in the last example (nice trick to explicitly and descriptively passing from sync to async), at the end of the day when we enter to the async world we cannot exit, even worse, they can be nested, but luckily we have async/await mechanism to flat it.\n\nBy the way, I think the operator pipe will fix this.",
null,
"Avalander\n\nYep. I didn't use the M word because I didn't want to scare people, but monads are cool. Thanks for bringing it up :)",
null,
"Christophe Riolo\n\nIndeed I did not see the tag, I wasn't sure because of the absence of the usual vocabulary /)"
] |
[
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null,
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null,
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https://tbc-python.fossee.in/convert-notebook/ANSI_C_Programming/chapter14.ipynb
|
[
"# CHAPTER 14: OPERATIONS ON BITS¶\n\n## EXAMPLE ON PAGE:449¶\n\nIn :\n\ndef showbits(x):\nreturn bin(x)[2:].zfill(16)\nfor j in range(0,6,1):\nprint \"Decimal %d is same as binary\" % (j)\nprint showbits(j)\n\nDecimal 0 is same as binary\n0000000000000000\nDecimal 1 is same as binary\n0000000000000001\nDecimal 2 is same as binary\n0000000000000010\nDecimal 3 is same as binary\n0000000000000011\nDecimal 4 is same as binary\n0000000000000100\nDecimal 5 is same as binary\n0000000000000101\n\n\n## EXAMPLE ON PAGE:450¶\n\nIn :\n\ndef showbits(x):\nreturn bin(x)[2:].zfill(16)\nfor j in range(0,4,1):\nprint \"Decimal %d is same as binary\" % (j)\nprint showbits(j)\nk=j^65535 #using xor for making one's complement\nprint \"One's complement of %d is\" % (j)\nprint showbits(k)\n\nDecimal 0 is same as binary\n0000000000000000\nOne's complement of 0 is\n1111111111111111\nDecimal 1 is same as binary\n0000000000000001\nOne's complement of 1 is\n1111111111111110\nDecimal 2 is same as binary\n0000000000000010\nOne's complement of 2 is\n1111111111111101\nDecimal 3 is same as binary\n0000000000000011\nOne's complement of 3 is\n1111111111111100\n\n\n## EXAMPLE ON PAGE:451¶\n\nIn :\n\nimport sys\ndef encrypt():\ntry:\nfs=open('SOURCE.txt','r') #normal file\nft=open('TARGET.txt','w') #encrypted file\nexcept:\nprint \"File opening error!\"\nsys.exit(1)\nwhile True:\nif not ch:\nbreak\nft.write(bytearray([ord(ch)^0xff]))\nfs.close()\nft.close()\nencrypt()\n\n\n## EXAMPLE ON PAGE:452¶\n\nIn :\n\ndef showbits(x):\nreturn bin(x)[2:].zfill(16)\ni=5225\nprint \"Decimal %d is same as binary\" % (i)\nprint showbits(i)\nfor j in range(0,6,1):\nk=i>>j #right shift operator\nprint \"%d right shift %d gives\" % (i,j)\nprint showbits(k)\n\nDecimal 5225 is same as binary\n0001010001101001\n5225 right shift 0 gives\n0001010001101001\n5225 right shift 1 gives\n0000101000110100\n5225 right shift 2 gives\n0000010100011010\n5225 right shift 3 gives\n0000001010001101\n5225 right shift 4 gives\n0000000101000110\n5225 right shift 5 gives\n0000000010100011\n\n\n## EXAMPLE ON PAGE:453-454¶\n\nIn :\n\ndef showbits(x):\nreturn bin(x)[2:].zfill(16)\ni=5225\nprint \"Decimal %d is same as binary\" % (i)\nprint showbits(i)\nfor j in range(0,5,1):\nmask = 2 ** 16 - 1\nk = (i << j) & mask #left shift operator\nprint \"%d right shift %d gives\" % (i,j)\nprint showbits(k)\n\nDecimal 5225 is same as binary\n0001010001101001\n5225 right shift 0 gives\n0001010001101001\n5225 right shift 1 gives\n0010100011010010\n5225 right shift 2 gives\n0101000110100100\n5225 right shift 3 gives\n1010001101001000\n5225 right shift 4 gives\n0100011010010000\n\n\n## EXAMPLE ON PAGE:457-458¶\n\nIn :\n\nd=9\nm=3\ny=1990\ndate=(y-1980)*512+m*32+d\nprint \"Date=%u\\n\" % (date)\nyear=1980+(date>>9)\nmonth=((date<<7)>>12)\nday=((date<<11)>>11)\nprint \"Year=%u\" % (year)\nprint \"Month=%u\" % (month)\nprint \"Day=%u\" % (day)\n\nDate=5225\n\nYear=1990\nMonth=163\nDay=5225\n\n\n## EXAMPLE ON PAGE:460¶\n\nIn :\n\ni=65\nprint \"Value of i=%d\\n\" % (i)\nj=i&32\nif j==0:\nprint \"and its fifth bit is off\\n\"\nelse:\nprint \"and its fifth bit is on\\n\"\nj=i&64\nif j==0:\nprint \"whereas its sixth bit is off\\n\"\nelse:\nprint \"whereas its sixth bit is on\\n\"\n\nValue of i=65\n\nand its fifth bit is off\n\nwhereas its sixth bit is on\n\n\n\n## EXAMPLE ON PAGE:463¶\n\nIn :\n\nb=50\nb=b^12 #XOR operator\nprint \"%d\\n\" % (b) #this will print 62\nb=b^12\nprint \"%d\\n\" % (b) #this will print 50\n\n62\n\n50\n\n\n\n## EXAMPLE ON PAGE:463-464¶\n\nIn :\n\ndef showbit(n):\nfor i in range(15,-1,-1):\n\n0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0"
] |
[
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https://forum.vassalengine.org/t/rounding-down-calculations/11579
|
[
"# Rounding down calculations\n\nI have 10 points to distribute between 0-6 players depending on how many of their pieces have been placed on certain locations in an area. Say 3 players have an equal number of pieces there, they get 3 points each. This calculation is done for about 20 different areas, with 4-8 locations in each area, so there are lot of possible outcomes.\n\nDo I need to write ? : strings for each location, for each possible combination and outcome? Or is there a more concise way to write the calculation and round down?\n\nLike, here’s a draft example I wrote to calculate just “r’s” score for area 1 with 4 locations, where I first calculated totals 1r, 1b etc. for how many pieces each player has in the area:\n\n1r==0 ? 0 : 1r>0&&1b+1g+1y+1k+1e=0 ? 10 : 1r==(1b+1g+1y+1k+1e) ? 5 : 1r==1&&1b+1g+1y+1k+1e=2 ? 3 : 1r==1&&1b+1g+1y+1k+1e=3 ? 2 : 1r==2&&1b+1g+1y+1k+1e=1 ? 6 : 1r==3&&1b+1g+1y+1k+1e=1 ? 7\n\nThis gives me outcomes in rough order of likelihood: 0, 10, 5, 3, 2, 6, 7, and will work, but repeat 6 times for each of 20 locations, and it’s a chore to write out.\n\nAny tips?\n\nUnless I’m not fully understanding what you’re trying to do, Vassal automatically rounds down and throws out remainders unless you get sophisticated with it.\n\nFor example, if you have a formula that is x = y/3 than, x will return 3 if y is 9, 10 or 11. It won’t return 4 until y hits 12. In other words, it returns the whole number only and throws out the remainder.\n\nDoes that help?\n\nIt certainly does! So, as long as I calculate each set of 10 as a GP before adding them together, it should work.\n\nOK, now I need to know how to “get sophisticated with it”.\n\nI have the following for the point calculation:\n\n``((GetProperty(\"1r\")==0) ? 0 : (10*(GetProperty(\"1r\")/(GetProperty(\"1r\")+GetProperty(\"1b\")+GetProperty(\"1g\")+GetProperty(\"1y\")+GetProperty(\"1k\")+GetProperty(\"1e\")))))``\n\nSo if 1r is 0 it returns 0.\n\nIf 1r is not zero, it is supposed to calculate 10 x the ratio of 1r to all 6 of the colours. In practice, it returns 10 when only 1r is positive, but 0 if any other colour is positive. I guess because the ratio is rounded down to 0 before being multiplied by 10? Seems odd. How can I do this the right way?\n\nAre all of those properties guaranteed to have a numeric value? If any of them don’t, the whole thing will end up being treated as a string instead of a number…\n\nAlso, I’m not clear on why you’re even using GetProperty, rather than referencing the properties directly…\n\nYes, they can only be numeric values.\n\nNeither am I. For some reason that’s what the expression builder does. Perhaps because the global properties start with a numeral?\n\nIt shouldn’t make any difference (according to the java order of operations), but have you tried an extra set of parentheses around the “10*1r” bit of the calculation?\n\nOops. Just looked closer at your formula, and it isn’t working because you already have an extra set of parentheses in there that don’t belong…you’re forcing the division to occur before the multiplication!\n\nThanks, obvious solution but I needed someone to point it out!"
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.9342468,"math_prob":0.95350295,"size":1575,"snap":"2022-40-2023-06","text_gpt3_token_len":382,"char_repetition_ratio":0.09866327,"word_repetition_ratio":0.0,"special_character_ratio":0.2431746,"punctuation_ratio":0.10526316,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97196746,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-03T21:08:28Z\",\"WARC-Record-ID\":\"<urn:uuid:de583010-4e64-4e78-bb1b-f8a18f0d1443>\",\"Content-Length\":\"35401\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:30e3de73-e14f-4912-8a04-cd95167b2fbe>\",\"WARC-Concurrent-To\":\"<urn:uuid:e51e2cd4-babb-4a14-b7f6-dd0e7002c2f7>\",\"WARC-IP-Address\":\"109.237.26.25\",\"WARC-Target-URI\":\"https://forum.vassalengine.org/t/rounding-down-calculations/11579\",\"WARC-Payload-Digest\":\"sha1:N5CRXFXJDDKAUDSNPQZJQ6KMJDJQE4WN\",\"WARC-Block-Digest\":\"sha1:KB63GWZVTRVUTK4H5ZK7ZZBFTE3OF2SA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337432.78_warc_CC-MAIN-20221003200326-20221003230326-00387.warc.gz\"}"}
|
https://slayrawithtray.blog/tag/rundown/
|
[
"# Tag: rundown\n\n## Day 50 – The RUNdown",
null,
"Day 50. Roman Numeral L. The number of letters in “fifty” equals the sum of the digits in 50. 50 is also the sum of three square numbers: 3^2 + 4^2 + 5^2 = 9 + 16 + 25 = 50. Woohoo! Hit 50…"
] |
[
null,
"https://slayrawithtray.files.wordpress.com/2018/04/justin-timberlake-its-gonna-be-may-its-may-picsay-1.jpg",
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.6912327,"math_prob":0.99984777,"size":541,"snap":"2019-43-2019-47","text_gpt3_token_len":168,"char_repetition_ratio":0.11545624,"word_repetition_ratio":0.0,"special_character_ratio":0.2809612,"punctuation_ratio":0.22018349,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9733559,"pos_list":[0,1,2],"im_url_duplicate_count":[null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-12T21:55:44Z\",\"WARC-Record-ID\":\"<urn:uuid:0cd6ee7c-9a5d-465d-99bf-739b9d580bb3>\",\"Content-Length\":\"96771\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1d9d6e72-55d6-4577-96e0-6d18277db583>\",\"WARC-Concurrent-To\":\"<urn:uuid:3f3ee11a-12ba-4ca9-a0a6-365abca8afd3>\",\"WARC-IP-Address\":\"192.0.78.24\",\"WARC-Target-URI\":\"https://slayrawithtray.blog/tag/rundown/\",\"WARC-Payload-Digest\":\"sha1:DVWDDTTAQIJEDCMUB5WWAAZ3SKBOLIZW\",\"WARC-Block-Digest\":\"sha1:BACWG7TFN7GONWC7KSSXW4P5Z6KZDZME\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496665767.51_warc_CC-MAIN-20191112202920-20191112230920-00466.warc.gz\"}"}
|
https://www.selfridges.com/HK/zh/cat/le-creuset-signature-cast-iron-casserole-dish-28cm_R03647527/
|
[
"## 以当地货币和语言购买\n\n• 澳大利亚 / 澳元 \\$\n• 加拿大 / 加元 \\$\n• 中国 / 人民币 ¥\n• 法国 / 欧元 €\n• 德国 / 欧元 €\n• 香港 / 港元 \\$\n• 爱尔兰 / 欧元 €\n• 意大利 / 欧元 €\n• 日本 / 日元 ¥\n• 科威特 / 美元 \\$\n• 澳门 / 港元 \\$\n• 荷兰 / 欧元 €\n• 卡塔尔 / 美元 \\$\n• 沙特阿拉伯 / 美元 \\$\n• 新加坡 / 新加坡元 \\$\n• 韩国 / 韩元 ₩\n• 西班牙 / 欧元 €\n• 台湾 / 新台币 \\$\n• 阿拉伯联合酋长国 / 美元 \\$\n• 英国 / 英镑 £\n• 美国 / 美元 \\$\n• 不符合您的要求?阅读更多\n• 简体中文\n• 英语\n• 英语\n• 简体中文\n• 英语\n• 英语\n• 英语\n• 简体中文\n• 英语\n• 英语\n• 英语\n• 英语\n• 英语\n• 简体中文\n• 英语\n• 英语\n• 英语\n• 英语\n• 简体中文\n• 英语\n• 英语\n• 英语\n• 简体中文\n• 英语\n• 英语\n• 简体中文\n• 英语\n• 英语\n• 简体中文\n\n## 国际送货\n\nselfridges.com 上几乎所有的商品均可提供国际配送服务,您的订单可发往全世界 130 个国家/地区,包括北美、澳洲、中东及中国。\n\n• 阿尔及利亚\n• 安道尔\n• 安提瓜和巴布达\n• 阿鲁巴\n• 澳大利亚\n• 奥地利\n• 阿塞拜疆\n• 巴林\n• 孟加拉国\n• 巴巴多斯\n• 白俄罗斯\n• 比利时\n• 伯利兹\n• 百慕大\n• 玻利维亚\n• 博兹瓦纳\n• 文莱\n• 保加利亚\n• 柬埔寨\n• 加拿大\n• 开曼群岛\n• 智利\n• 中国大陆\n• 哥伦比亚\n• 哥斯达黎加\n• 克罗地亚\n• 塞浦路斯\n• 捷克共和国\n• 丹麦\n• 多米尼克\n• 多米尼加共和国\n• 厄瓜多尔\n• 埃及\n• 萨尔瓦多\n• 爱沙尼亚\n• 芬兰\n• 法国\n• 法属圭亚那\n• 德国\n• 直布罗陀\n• 希腊\n• 格林纳达\n• 瓜德罗普岛\n• 危地马拉\n• 根西岛\n• 圭亚那\n• 洪都拉斯\n• 香港\n• 匈牙利\n• 冰岛\n• 印度\n• 印度尼西亚\n• 爱尔兰\n• 以色列\n• 意大利\n• 牙买加\n• 日本\n• 泽西岛\n• 约旦\n• 哈萨克斯坦\n• 肯尼亚\n• 科威特\n• 老挝\n• 拉脱维亚\n• 黎巴嫩\n• 莱索托\n• 列支敦士登\n• 立陶宛\n• 卢森堡\n• 澳门\n• 马来西亚\n• 马尔代夫\n• 马耳他\n• 马提尼克岛\n• 马约特岛\n• 墨西哥\n• 摩纳哥\n• 蒙特塞拉特\n• 摩洛哥\n• 缅甸\n• 纳米比亚\n• 荷兰\n• 新西兰\n• 尼加拉瓜\n• 尼日利亚\n• 挪威\n• 阿曼\n• 巴基斯坦\n• 巴拿马\n• 巴拉圭\n• 秘鲁\n• 菲律宾\n• 波兰\n• 葡萄牙\n• 波多黎各\n• 卡塔尔\n• 留尼汪岛\n• 罗马尼亚\n• 卢旺达\n• 圣基茨与尼维斯\n• 圣卢西亚\n• 圣马丁岛(法属)\n• 圣马力诺\n• 沙特阿拉伯\n• 塞尔维亚\n• 新加坡\n• 斯洛伐克\n• 斯洛文尼亚\n• 南非\n• 韩国\n• 西班牙\n• 斯里兰卡\n• 苏里南\n• 斯威士兰\n• 瑞典\n• 瑞士\n• 台湾\n• 坦桑尼亚\n• 泰国\n• 特立尼达和多巴哥\n• 土耳其\n• 乌干达\n• 乌克兰\n• 阿拉伯联合酋长国\n• 英国\n• 美国\n• 乌拉圭\n• 委内瑞拉\n• 越南\n\n# LE CREUSET 签名款铸铁浅口炒锅 28 厘米\n\n\\$2800.00\n\n*进口关税将在结算时显示\n\n01 02 03 04 05\n\nLe Creuset 圆形砂锅盘\n\n## 英国和欧洲\n\n\\$100.00\n• 无限英国定时、指定日和标准配送\n• 英国境内次日配送(英国时间下午 6 点前下单)\n• 无限欧盟地区标准配送\n• 免费退货\n• 不受最低消费金额限制\n\n## 全球\n\n\\$420.00\n• 订单金额超过\\$ 420.00英国时间,指定日期和标准交货时间不受限制\n• 订单金额超过\\$ 420.00全球不限次数的送货\n\nRef: R03647527\n\nLe Creuset: 美丽的色彩、精心制作的铸铁和美味的食物。该法国品牌自年以来一直致力于完善其Six Trees炊具 1925. 它的名字虽然是砂锅,不过炖菜并不是它的专长。配有不锈钢手柄,抓握舒适,即便戴着烤炉手套也十分便利。所以,你想做一道快炖菜还是慢炖菜?或者是面包?多种可能供你选择。"
] |
[
null
] |
{"ft_lang_label":"__label__zh","ft_lang_prob":0.9746875,"math_prob":0.5194948,"size":412,"snap":"2021-31-2021-39","text_gpt3_token_len":436,"char_repetition_ratio":0.034313727,"word_repetition_ratio":0.0,"special_character_ratio":0.24514563,"punctuation_ratio":0.114754096,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9573199,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-27T13:13:48Z\",\"WARC-Record-ID\":\"<urn:uuid:7126adac-0f56-4b8c-8429-519661df50a1>\",\"Content-Length\":\"281786\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:af809852-62f5-4745-968b-94c18e2c34ae>\",\"WARC-Concurrent-To\":\"<urn:uuid:d9d3ff5a-b7f7-42f0-8f1e-b8483d625647>\",\"WARC-IP-Address\":\"104.18.21.144\",\"WARC-Target-URI\":\"https://www.selfridges.com/HK/zh/cat/le-creuset-signature-cast-iron-casserole-dish-28cm_R03647527/\",\"WARC-Payload-Digest\":\"sha1:N7T37474IQELL4BBCDHV46EEHNPPZPE2\",\"WARC-Block-Digest\":\"sha1:GJ2ZEVSJVTTHSBCFDS4YZ5C5YM7RVJU4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780058450.44_warc_CC-MAIN-20210927120736-20210927150736-00352.warc.gz\"}"}
|
http://www.javaexamples4u.com/2009/02/shift-operators-with-int-data-type.html
|
[
"## Friday, February 20, 2009\n\n### Shift Operators with int data type source code\n\n`package operators;public class IntShiftOperators { /** * @param args */ public static void main(String[] args) { int i=3; //left shift operator System.out.println(\" i << 1 =\" + (i <<1)); System.out.println(\" i << 3 =\" + (i <<3)); System.out.println(\" i << 32 =\" + (i <<32));//same as i System.out.println(\" i << 33 =\" + (i <<33)); //same as <<1 i=44; //right shift operator System.out.println(\" i << 1 =\" + (i >>1)); System.out.println(\" i >> 32 =\" + (i >>32)); //same as i i=255; System.out.println(\" i >> 7 (i is 0xff) =\" + (i >>7)); System.out.println(\" i >> 8 (i is 0xff) =\" + (i >>8)); // unsigned right shift System.out.println(\" i >>> 7 (i is 0xff) =\" + (i >>>7)); System.out.println(\" i >>> 8 (i is 0xff) =\" + (i >>>8)); System.out.println(\" i >>> 32 =\" + (i >>32)); //same as i // all the above four yeilds the same result why ? i = -222 ; System.out.println(\" i >> 1 =\" + (i >>1)); System.out.println(\" i >> 3 =\" + (i >>3)); System.out.println(\" i >>> 1 =\" + (i >>>1)); System.out.println(\" i >>> 3 =\" + (i >>>3)); System.out.println(\" i >>> 32 =\" + (i >>>32));//same as i }}output: i << 1 =6 i << 3 =24 i << 32 =3 i << 33 =6 i << 1 =22 i >> 32 =44 i >> 7 (i is 0xff) =1 i >> 8 (i is 0xff) =0 i >>> 7 (i is 0xff) =1 i >>> 8 (i is 0xff) =0 i >>> 32 =255 i >> 1 =-111 i >> 3 =-28 i >>> 1 =2147483537 i >>> 3 =536870884 i >>> 32 =-222`"
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.57117844,"math_prob":0.97665715,"size":1418,"snap":"2020-45-2020-50","text_gpt3_token_len":530,"char_repetition_ratio":0.2609618,"word_repetition_ratio":0.24390244,"special_character_ratio":0.55500704,"punctuation_ratio":0.18122977,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998474,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-29T04:39:14Z\",\"WARC-Record-ID\":\"<urn:uuid:f5486f5b-2952-4403-b790-e0256fb13d09>\",\"Content-Length\":\"58112\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:673cef86-89b7-4142-9ccd-a135e7e417fd>\",\"WARC-Concurrent-To\":\"<urn:uuid:3ec47224-c8d2-49bd-94c3-142aa947bae9>\",\"WARC-IP-Address\":\"172.217.2.115\",\"WARC-Target-URI\":\"http://www.javaexamples4u.com/2009/02/shift-operators-with-int-data-type.html\",\"WARC-Payload-Digest\":\"sha1:HQSLU3TBS4ELRHRQMY72QWZTHTV57TCF\",\"WARC-Block-Digest\":\"sha1:JCSOVHKSIY6ZKKLRVFLYBG6AIDQ574I6\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141196324.38_warc_CC-MAIN-20201129034021-20201129064021-00203.warc.gz\"}"}
|
http://codeforces.com/blog/entry/11917
|
[
"KhaustovPavel's blog\n\nBy KhaustovPavel, 5 years ago, translation,",
null,
",",
null,
"Hi, everyone!\n\nRegular Codeforces round #242 for participants from the second division will take place on 25 April, 11:00 MSK. Participants from the first division are able to participate out of the competition.\n\nThe contest is prepared by programmers from Tomsk Polytechnic University: Pavel Khaustov (KhaustovPavel), Olesya Golub (Taube), Nickolay Kuzivanov (bnick), Dmitry Savvinov (dsavvinov), Alexey Vetrov (noxwell).",
null,
"Special thanks to Vladimir Chalyshev (cmd) and Alexey Dergunov (dalex) for their impact.\n\nAlso, thanks to Codeforces team and, especially, to Gerald Agapov (Gerald) and Maria Belova (Delinur).\n\nPoints distribution: 500-1000-1500-2500-3000\n\nGood luck!",
null,
"",
null,
"Comments (76)\n » Why are there so many contests at 3 AM US time. :(\n• » » In china,we usually play it at 23:00!Sometime,we can't help falling asleep at that time.\n » 5 years ago, # | ← Rev. 3 → I think most of the contestants aren't comfortable with the time of the contest start , it's been like this with coder-strike tooPlease reconsider this in the next contests .And many thanks to the authors\n• » » I am sorry to hear that you feel uncomfortable with the time.However, I like the time very much :) It is 15:00 in Hong Kong (UTC+8), but I will be at school as it is Friday... The usual time of contest will be at 23:00 in Hong Kong and I can't help falling asleep at that time.I hope that the time could be modified so as to cater most people's need.\n• » » » 5 years ago, # ^ | ← Rev. 2 → I'm glad to hear that =) , I hope so too\n• » » » 5 years ago, # ^ | ← Rev. 2 → Same in Taiwan (UTC+8)!\n• » » » I can't agree more. The same in BeiJing.\n• » » » » The same in Chengdu!\n• » » » Same in Vietnam :) (UTC+7)\n• » » » Same in Beijing.\n• » » 5 years ago, # ^ | ← Rev. 2 → In Romania the contest starts at 10AM, but I have school then. Hope that the old hour(6PM in Romania) will be used more in the future :D\n » Div1 users won't be rated? Only div2 users will be rated with a separated ranking?\n• » » as usual, div1 wont be rated\n » Am I the only guy who read KhaustovPavel as Khaustov Patel? :PPS: Khaustov Patel is a typical Indian name.\n• » » yes.\n• » » Patel is common Indian name and not Khaustov. https://www.google.com/search?q=Khaustov+PatelAnd his name does not contain Patel! I don't see many people making this mistake :)\n » 5 years ago, # | ← Rev. 3 → This time is ok, if it's not friday. Friday is Jummah day(a special day for the muslims). And in Bangladesh, the time when contest will be started, is the jumma time. :( So going to miss this contest... :'( *** If anyone wants to know about Jummah, can check it here\n » lot of competitions lately have been at 1 am here which has been uncomfortable. Can anything be done with the time to make it work for most of contestants\n » 5 years ago, # | ← Rev. 3 → I'm sorry to to hear that many people feel uncomfortable with the time.However, it's obvious that the time can't make all contestants happy and some people participate the contest at midnight.The usual time will be at 23:30 in Beijing.When the contest ends,it will be at 1:30am.And I think it's more uncomfortable for Japanese and other east Asian countries' contestants. But there are still many east Asian countries' contestants to participate the contest. I think the reason is that we love coding and enjoy the contests. But I have to say the usual time of the contest is better for most contestants. Hope we can understand author and understand each other.\n » 5 years ago, # | ← Rev. 2 → I am happy with this time. It will be 3 AM EST here in FL, US but it is a good time to write code relaxed, lol. The usual time is complicated (11:30 AM EST) for me because either I am working and I can't focus very well or I am at home where my daughter doesn't stop playing with me while I am solving the problems. As always, enjoy the contest, GL & HF!!!\n• » » 5 years ago, # ^ | ← Rev. 2 → Same point of view here. I prefer midnight contests to daytime contests, too. Fortunately, regular Codeforces time favors me well (11.30pm). Typical coders don't go to bed early, right? :P\n• » » » Yes. At the end, coders don't go to sleep early. I would like to participate in more contests after midnight.\n » I prefer 3 AM to 11:30 AM because I'm more likely to be awake at the former time ;)\n » Our mid-term exam is about to finish today morning, so just be able to take part in... Sorry to hear that it' s an uncomfortable time for others. (Start at China Standard Time 15:00)\n » Unfortunately cannot participate in the contest cause its jummah time .\n » 30 mins to start the contest .... It will be a good contest as always . wish you luck all :)\n » How did you solve Div 2 C?\n• » » I received time limit on pretest 12.\n• » » » What did you try? Bruteforce will fail obviously.\n• » » q[i]=p[i]^(i%1)^(i%2)^...^(i%n) Q = q^q^...^q[n] = (p^...^p[n]) ^ (^ for i=1..n for k=1..n (i%k)) = (p^...^p[n]) ^ (^ for k=1..n for i=1..n (i%k)). Let P=p^...^p[n]. So, Q = P ^ (^ for k=1..n for i=1..n (i%k)). Now we have to calculate these xors for all k: (^ for i=1..n (i%k)). ^ for i=1..n (i%k) = 0^1^2^...^(k-1) ^ 0^1^2^...^(k-1) ^ ... ^ 0^1^2^...^(k-1) [(n/k) times] ^ 0^1^2...^(n%k) = ((n/k)%2)*(0^1^2^...^(k-1)) ^ (0^1^2...^(n%k)). We can pre-calculate all xor sums like (0^1^2^...^(k-1)). The rest is obvious.\n• » » » Also there is a formula for prefix xors 0 ^ 1 ^ .... ^ k: int xorUpToK(int k) { switch (k % 4) { case 0: return k; case 1: return 1; case 2: return k + 1; case 3: return 0; } }\n• » » » » can you explain this formula?\n• » » » » » This is a nice formula for long values of k . To prove it you takeF(3) = 0 ^ 1 ^ 2 ^ 3 = 0because in binary 0000 ^ 0001 ^ 0010 ^ 0011 = 0000and operating BIT for BITF( 7 ) = F( 3 ) ^ 0100 ^ 0101 ^0110 ^ 0111 = 0000F( 7 ) = F( 3 ) ^ (01)00 ^ (01)01 ^(01)10 ^ (01)11 = 0000We can note that () repeats four times , and four is even then (x)^(x)^(x)^(x) = 0 is ever zeroBut in this problem is not necesarry because we can precalculate al values <= 10^6A problem that you really need this formula.Link to problemPSDATA: you need know nim numbers before solve Industrial Nim.\n• » » » » » » Awesome, thanks! I know this problem just needed a simple precomputation, but it's still nice to understand this formula.\n• » » » » » I think that most coders have checked this postBut without this, you can also do a precomputation :D\n• » » » could you please format your comment to make it more readable. thanks!\n• » » » » 5 years ago, # ^ | ← Rev. 3 → This is related to problem C of div.2 #242Here is the formatted explanation of Totktonada approach..I checked solution codes of many people but everywhere they were using two things as n/i and n%i and I kept asking why are they using these.. Then I read many explanations for the approach, some helping people wrote long explanations but what helped me really understand is Totktonada's comment below is the same comment made clear and easy to read q[i]=p[i]^(i%1)^(i%2)^...^(i%n) Q = q^q^...^q[n] = (p^...^p[n]) ^ (^ for i=1..n for k=1..n (i%k))= (p^...^p[n]) ^ (^ for k=1..n for i=1..n (i%k)). Let P=p^...^p[n]. So, Q = P ^ (^ for k=1..n for i=1..n (i%k)). Now we have to calculate these xorsfor all k: (^ for i=1..n (i%k)). ^ for i=1..n (i%k) = 0^1^2^...^(k-1) ^ 0^1^2^...^(k-1) ^ ... ^ 0^1^2^...^(k-1) [(n/k) times] ^ 0^1^2...^(n%k) = ((n/k)%2)*(0^1^2^...^(k-1)) ^ (0^1^2...^(n%k)). We can pre-calculate all xor sums like (0^1^2^...^(k-1)).using a precalculated array X[1..n] with X[i] as 1^2^..^ifor( int i = 1 ; i <= N ; ++i ) F[ i ] = F[ i-1 ] ^ i; The rest is obviousApart from above explanation you must also read comment1 by Boxer and comment2 by dalex\n• » » We know that xor is asociative then we can reagrup de expression as p1 xor p2 .... xor p3 * E where E is a new expression and again reagruping :E = (1 mod 1 xor 2 mod2 ... n mod1 ) xor (1 mod 2 xor 2 mod 2 ... n mod2 ) ....Then we can get F = 1 mod i xor 2 mod i xor .... n mod i efficiently\n• » » » Exactly, but what after that? I understood that part, but I could not form a relation afterwards.\n• » » » » We can getF = 1 mod i xor 2 mod i xor .... n mod i if we take every package of size i then we have F = [ 1 mod i xor 2 mod i xor .... i mod i ]xor [ (i + 1) mod i xor (i+2) mod i .. (2*i) mod i ] ...thenF = [ 1 mod i xor 2 mod i xor .... i mod i ] xor [ 1 mod i xor 2 mod i xor .... i mod i ]......Finally [ ] repeats q = n/i and if q is even there [ ] = 0 else [] = [ 1 mod i xor 2 mod i xor .... i mod i ] and we can consider r = n%i too.for solve G = [ 1 mod i xor 2 mod i xor .... i mod i ] we can see that is equal toG = [ 1 xor 2 xor .... i ]And everthing depends only of G we can precalcalculate with a array.My SolutionSorry for my bad english.\n• » » 5 years ago, # ^ | ← Rev. 3 → I have formed an matrix on a paper of n x n rows represent n and cols mod i - 0 1 1 1 1 ... 1 - 0 0 2 2 2 ... 2 - 0 1 0 3 3 ... 3 - 0 0 1 0 4 ... 4 as you can see, every column is a cycle of length i, and you can cancel (n/i)/2 cycles, because x ^ x = 0, obviously you will not always have cycles that pairs, so you need the xor of the parts that can't be paired, lets call it rest[i] finally with a for i do ans ^= p[i] ^ rest[i]my submission 6467752\n » A contest with no successful hacking attempt!\n• » » may be there was hack in E. :p\n• » »\n » Am I the only person who are confused with problem E's sample case?\n » Can anyone tell me why my solution gave WA ? I did precomputation of XORs and my complexity is also good. 6468572\n• » » 5 years ago, # ^ | ← Rev. 3 → can you at least explain a bit your idea? and ull was not necessary\n• » » You had the right idea of precalculating xors, but what you're doing on each iteration is not right. Try this simple test: 4 0 0 0 0 The answer should be 2.\n• » » » Yeah . Silly mistake on my part , I took mod as simply the difference and it failed on indices like 4%2 and 4%4. .\n » what was the idea of D?\n• » » Bruteforce and TernarySearch , my friend.\n• » » » Integer ternary search OoYou crazy person...P.S. I thought, we can't avoid using set in this problem. Such a surprise.\n• » » » » 5 years ago, # ^ | ← Rev. 2 → Actually the",
null,
"solution passes in time! >.>\n• » » » This solution is wrong. Ternary search doesn't work here.\n• » » » » 6469423Well, but his solution is AC.\n• » » » » » And how do you think what does it mean? :)\n• » » » » » » Weak tests, round should be unrated :Ъ\n• » » » » » » » Not really, I believe his solution is correct.His ternary search function is a function of the absolute difference between both times, which is clearly a bitonic function in this problem.\n• » » » » Hey man i think that you are wrong , Ternary search works because my function f is crecient then g = abs( f — t ) is like a parabola then i need to calculate the minimum.\n• » » » » » You fix left, right and top borders of the track and then find the bottom border by binary search. Let's see what bottom borders we can create.What happens if all odd rows are decreasing (a[i][j] > a[i][j+1]) and all even rows are increasing (a[i][j] < a[i][j+1])?If we consider the times needed to go through the parts of track built at these rows, they will be like {d*tUp, d*tDown, d*tUp, d*tDown, ...} where d is the width of the track and tUp and tDown are numbers from the input. It's not a crescent sequence, is it? Of course we also have increasing left and right parts, but they cannot affect so much as the bottom part affects the function (e.g. if the width is very large).\n• » » » » » » Oh nice counterExample , maybe in random cases my solution works well:)\n• » » » » » » I thought this during the contest, so I went for the O(N^4) solution :(Still, in the editorial (http://codeforces.com/blog/entry/11944) the author says his solution is O(N^3 * logN).As far as I've seen, people have used n-ary search to do this, but there are a lot of cases where that would fail, cause you can't guarantee that it's increasing or decreasing at any point.Take dalex' example with odd decreasing/even increasing rows, you can actually have any random configuration of rows that are all increasing or decreasing and this would make any n-ary search solution fail, as far as I can understand.So, any ideas of how to achieve the O(N^3 * log N) solution with any other method? (Or is there something that I'm missing and a n-ary search actually works?)\n• » » » » » 5 years ago, # ^ | ← Rev. 2 → champero detected !! :P, Nice problem.\n » Any editorial?\n• » » here it is.\n » Could anyone help figure out what the wrong is in my two solution of Problem C? 6466959 and 6468449\n » In Problem C, upper limit of 'p' was 2 x 10**9. However, solutions that took 'p' as int passed.\n• » » the upper limit of int is 2147483647, which is more than 2e9. so ofcourse it will pass!\n » I am wondering if this algorithm is correct or not. http://codeforces.com/contest/424/submission/6470517 He enums all the point on the top-left , and use trinary-search twice to find the point on the bottom-right. In my view , this solution is incorrect. Just can not find a data challenge this code ? Thanks\n » Could anyone share your idea of problem D? I was confused with the problem for more than one hour, sticking to the O(n^4) algorithm and cannot find a way to optimize it....\n » Hello.Can anyone tell me why submission 6465131 gets Compilation Error on Codeforces, as it compiled perfectly on my local machine ( Linux GCC 4.8.2, though I doubt there are any great differences between this and 4.7) ? Link: http://codeforces.com/contest/424/submission/6465131Thanks.\n• » » Compiler thinks that distance is STL function to count distance between two iterators, but not yours. You haven't Compilation Error on your local machine because your local machine uses C++11 mode. If you want to use C++11 on Codeforces, you should choose \"GNU C++0x 4\" as language.\n• » » » Ok, thanks for your help.\n » why is 3 1 4 6 not a solution to sample of D?\n• » » Read the statement carefully. The side of rectangle must be at least 3.\n » Problem D was very misleading... During the contest, some people got Accepted with a O(n^4) algorithm, while most people thought that the required complexity was O(n^3) :P"
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https://fixbbs.com/p/09668950.html
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[
"# Python实现一个简单三层神经网络的搭建并测试\n\n2021年9月28日 15点热度 0条评论 来源: 茶柒0112\n\n# python实现一个简单三层神经网络的搭建(有代码)\n\n废话不多说了,直接步入正题,一个完整的神经网络一般由三层构成:输入层,隐藏层(可以有多层)和输出层。本文所构建的神经网络隐藏层只有一层。一个神经网络主要由三部分构成(代码结构上):初始化,训练,和预测。首先我们先来初始化这个神经网络吧!\n\n## 1.初始化\n\n• 我们所要初始化的内容包括:神经网络每层上的神经元个数(这个是根据实际问题输入输出而得到的,我们将它设置为一个可自定义量)。\n• 不同层间数据互相传送的权重值。\n• 激活函数(模拟自然界的神经元,刺激信号需要达到一定的程度才能激活神经元)\n\n``` def __init__(self, input_nodes_num, hidden_nodes_num, output_nodes_num, lr):\n# 初始化神经元个数,可以直接修改\nself.input_nodes = input_nodes_num\nself.hidden_nodes = hidden_nodes_num\nself.output_nodes = output_nodes_num\nself.learning_rate = lr\n\n# 初始化权重值,利用正态分布函数进行随机初始化,均值为0,方差为神经元个数开方\nself.w_input_hidden = numpy.random.normal(0.0, pow(self.hidden_nodes, -0.5),\n(self.hidden_nodes, self.input_nodes))\nself.w_hidden_output = numpy.random.normal(0.0, pow(self.output_nodes, -0.5),\n(self.output_nodes, self.hidden_nodes))\n# 初始化激活函数,激活函数选用Sigmoid函数,更加平滑,接近自然界的神经元行为模式\n# lambda定义了一个匿名函数\nself.activation_function = lambda x: scipy.special.expit(x)\npass```\n\n下面我们来解释一下上述代码段中的一些编程知识。首先是__init__()它是一个类的构造函数,在构建一个类的对象时会调用此函数,所以我们将神经网络初始化相关代码放到这个函数里。\n\n```self.w_input_hidden = numpy.random.normal(0.0, pow(self.hidden_nodes, -0.5),\n(self.hidden_nodes, self.input_nodes))```\n\n这句代码使用了numpy库中的random.normal()函数,为输入层和隐藏层之间的数据传递初始化了权重值,这个函数会根据正态分布随机生成一个\n\n`self.hidden_nodes*self.input_nodes的矩阵(hidden_nodes和input_nodes表示隐藏层和输入层神经元的个数)。 `\n` self.activation_function = lambda x: scipy.special.expit(x)`\n\n这句代码使用lambda定义了一个匿名函数,将它赋值给激活函数,函数为sigmoid函数,是一条平滑的曲线,比较接近自然界神经元对于刺激信号的反应方式。\n\n## 2.预测\n\n按照正常顺序,初始化完成后应该进行训练,但由于训练较为复杂,且预测较为简单容易实现,我们先完成这一部分的代码。预测环节需要我们将输入信息进行处理,加权求和后传输给隐藏层神经元,经过激活函数并再次加权求和后,传输给输出层经过输出层神经元的处理得到最终的结果。代码片段如下:\n\n``` def query(self, inputs_list):\n# 转置将行向量转成列向量,将每组数据更好的分隔开来,方便后续矩阵点乘操作\ninputs = np.array(inputs_list, ndmin=2).T\n# 加权求和后经过sigmoid函数得到隐藏层输出\nhidden_inputs = np.dot(self.w_input_hidden, inputs)\nhidden_outputs = self.activation_function(hidden_inputs)\n# 加权求和后经过sigmoid函数得到最终输出\nfinal_inputs = np.dot(self.w_hidden_output, hidden_outputs)\nfinal_outputs = self.activation_function(final_inputs)\n# 得到输出数据列\nreturn final_outputs```\n\n这段代码没有什么好说的,比较简单,只需按照笔者上述的步骤做即可。有什么不懂的可以看注释或者留下评论。\n\n## 3.训练\n\n神经网络的训练问题较为复杂,涉及到神经网络的正向和反向传播,微积分的链式法则,矩阵运算,偏微分求导和梯度下降算法的一些知识,都是机器学习的一些基础知识,在这里就不做过多的赘述,过几天我会新发一篇详细讲一下。下面来了解一下训练代码段的主要任务:\n\n• 训练和预测一样都要首先读入一些输入并预测输出,不同的是,训练阶段我们是从训练数据集中获取数据,我们知道正确的输出是什么,而预测阶段我们只知道输入而输出需要通过我们训练的模型预测出来。首先训练阶段读入输入并按照当前的模型对其进行预测。\n• 基于训练预测结果和标注好的实际结果的误差更新各个层之间的权值。\n\n``` def train(self, inputs_list, targets_list):\n# 将训练集和测试集中的数据转化为列向量\ninputs = np.array(inputs_list, ndmin=2).T\ntargets = np.array(targets_list, ndmin=2).T\n# 隐藏层的输入为训练集与权重值的点乘,输出为激活函数的输出\nhidden_inputs = np.dot(self.w_input_hidden, inputs)\nhidden_outputs = self.activation_function(hidden_inputs)\n# 输出层的输入为隐藏层的输出,输出为最终结果\nfinal_inputs = np.dot(self.w_hidden_output, hidden_outputs)\nfinal_outputs = self.activation_function(final_inputs)\n# 损失函数\noutput_errors = targets - final_outputs\n# 隐藏层的误差为权值矩阵的转置与输出误差的点乘\nhidden_errors = np.dot(self.w_hidden_output.T, output_errors)\n# 对权值进行更新\nself.w_hidden_output += self.learning_rate * np.dot((output_errors *\nfinal_outputs * (1.0 - final_outputs)),\nnp.transpose(hidden_outputs))\n\nself.w_input_hidden += self.learning_rate * np.dot((hidden_errors *\nhidden_outputs * (1.0 - hidden_outputs)),\nnp.transpose(inputs))```\n\n上述代码段可能对于一些刚接触机器学习或深度学习的同学来说可能有点不知所云或产生一种好复杂的感觉,但是这只是对反向传播算法,链式法则和偏导的综合应用。我会在另一篇随笔中讲述我的心得(可能讲得不好),感兴趣的可以看一下。\n\n## 4.测试\n\n三层神经网络构建完成,我用mnist训练集和测试集对其进行了测试,代码及结果如下:\n\n```# 初始化各层神经元个数,期中输入神经元个数取决于读入的因变量,而输出神经元个数取决于分类的可能性个数\ninput_nodes = 784\nhidden_nodes = 100\noutput_nodes = 10\n# 学习率,每次调整步幅大小\nlearning_rate = 0.2\n\nn = NeuralNetwork(input_nodes, hidden_nodes, output_nodes, learning_rate)\n# 获取训练集信息\ntraining_data_file = open('data/mnist_train.csv', 'r')\ntraining_data_file.close()\n\nfor record in training_data_list:\nall_values = record.split(',')\n\ninputs = (numpy.asfarray(all_values[1:]) / 255.0 * 0.99) + 0.01\n\ntargets = numpy.zeros(output_nodes) + 0.01\ntargets[int(all_values)] = 0.99\nn.train(inputs, targets)\npass\nprint('train successful!')\ntest_file = open('data/mnist_test.csv', 'r')\ntest_file.close()\nm = np.size(test_list)\nj = 0.0\nfor record in test_list:\ntest_values = record.split(',')\nnp.asfarray(test_values)\nresults = n.query(np.asfarray(test_values[1:]))\nif results[int(test_values)] == max(results):\nj += 1\npass\n\nprint(\"正确率为;\" + str(j/m))```",
null,
"` `\n原文作者:茶柒0112\n原文地址: https://www.cnblogs.com/sevent/p/15345472.html\n本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系管理员进行删除。"
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[
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"https://compic.oss-cn-shenzhen.aliyuncs.com/2021/09/2033753264900374202.png",
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https://www.media4math.com/MA.6.AR.3.2
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[
"",
null,
"## MA.6.AR.3.2: Given a real-world context, determine a rate for a ratio of quantities with different units. Calculate and interpret the corresponding unit rate.\n\nThere are 120 resources.\nTitle Description Thumbnail Image Curriculum Topics\n\n## Math Examples Collection: Ratios and Rates\n\nThis collection aggregates all the math examples around the topic of Ratios and Rates. There are a total of 32 Math Examples.",
null,
"Ratios and Rates\n\n## Math Worksheet Collection: Unit Rates and Data Tables\n\nThis collection aggregates all the math worksheets around the topic of Unit Rates and Data Tables. There are a total of 50 worksheets.",
null,
"Ratios and Rates\n\n## Math Clip Art Collection: Rates\n\nThis collection aggregates all the math clip art around the topic of Rates. There are a total of 10 images.",
null,
"Ratios and Rates\n\n## Closed Captioned Video: Ratios and Rates: Converting Measurement Units\n\nClosed Captioned Video: Ratios and Rates: Converting Measurement Units\n\nVideo Tutorial: Ratios and Rates: Converting Measurement Units.",
null,
"Ratios and Rates\n\n## Closed Captioned Video: Ratios and Rates: Rates and Slopes of Lines\n\nClosed Captioned Video: Ratios and Rates: Rates and Slopes of Lines\n\nVideo Tutorial: Ratios and Rates: Rates and Slopes of Lines.",
null,
"Ratios and Rates\n\n## Closed Captioned Video: Ratios and Rates: Rates from Data\n\nClosed Captioned Video: Ratios and Rates: Rates from Data\n\nVideo Tutorial: Ratios and Rates: Rates from Data. In this video, we look at linear data sets that can be used to find the rate of change.",
null,
"Ratios and Rates\n\n## Closed Captioned Video: Ratios and Rates: Ratios as Decimals\n\nClosed Captioned Video: Ratios and Rates: Ratios as Decimals\n\nVideo Tutorial: Ratios and Rates: Ratios as Decimals. In this video, students explore ratios whose terms are decimals.",
null,
"Ratios and Rates\n\n## Closed Captioned Video: Ratios, Proportions, and Percents: Calculating Percents\n\nClosed Captioned Video: Ratios, Proportions, and Percents: Calculating Percents\n\nVideo Tutorial: Ratios and Percents: Calculating Percents.",
null,
"Ratios and Rates\n\n## Closed Captioned Video: Ratios: Unit Rates\n\nClosed Captioned Video: Ratios: Unit Rates",
null,
"Ratios and Rates\n\n## Closed Captioned Video: Ratios: Visual Models for Ratios and Percents\n\nClosed Captioned Video: Ratios: Visual Models for Ratios and Percents",
null,
"Ratios and Rates\n\n## Definition--Ratios, Proportions, and Percents Concepts--Converting Units\n\nDefinition--Ratios, Proportions, and Percents Concepts--Converting Units\n\nThis is part of a collection of definitions related to ratios, proportions, and percents.",
null,
"Ratios and Rates\n\n## Definition--Ratios, Proportions, and Percents Concepts--Equivalent Ratios\n\nDefinition--Ratios, Proportions, and Percents Concepts--Equivalent Ratios\n\nThis is part of a collection of definitions related to ratios, proportions, and percents.",
null,
"Ratios and Rates\n\n## Definition--Ratios, Proportions, and Percents Concepts--Part-to-Part Ratios\n\nDefinition--Ratios, Proportions, and Percents Concepts--Part-to-Part Ratios\n\nThis is part of a collection of definitions related to ratios, proportions, and percents.",
null,
"Ratios and Rates\n\n## Definition--Ratios, Proportions, and Percents Concepts--Part-to-Whole Ratios\n\nDefinition--Ratios, Proportions, and Percents Concepts--Part-to-Whole Ratios\n\nThis is part of a collection of definitions related to ratios, proportions, and percents.",
null,
"Ratios and Rates\n\n## Definition--Ratios, Proportions, and Percents Concepts--Rate\n\nDefinition--Ratios, Proportions, and Percents Concepts--Rate\n\nThis is part of a collection of definitions related to ratios, proportions, and percents.",
null,
"Ratios and Rates\n\n## Definition--Ratios, Proportions, and Percents Concepts--Ratio\n\nDefinition | Ratios, Proportions, and Percents Concepts | Ratio\n\nThis is part of a collection of definitions related to ratios, proportions, and percents.",
null,
"Ratios and Rates\n\n## Definition--Ratios, Proportions, and Percents Concepts--Ratios and Fractions\n\nDefinition--Ratios, Proportions, and Percents Concepts--Ratios and Fractions\n\nThis is part of a collection of definitions related to ratios, proportions, and percents.",
null,
"Ratios and Rates\n\n## Definition--Ratios, Proportions, and Percents Concepts--Ratios and Slope\n\nDefinition--Ratios, Proportions, and Percents Concepts--Ratios and Slope\n\nThis is part of a collection of definitions related to ratios, proportions, and percents.",
null,
"Ratios and Rates\n\n## Definition--Ratios, Proportions, and Percents Concepts--Ratios with Decimals\n\nDefinition--Ratios, Proportions, and Percents Concepts--Ratios with Decimals\n\nThis is part of a collection of definitions related to ratios, proportions, and percents.",
null,
"Ratios and Rates\n\n## Definition--Ratios, Proportions, and Percents Concepts--Ratios with Fractions\n\nDefinition--Ratios, Proportions, and Percents Concepts--Ratios with Fractions\n\nThis is part of a collection of definitions related to ratios, proportions, and percents.",
null,
"Ratios and Rates\n\n## Definition--Ratios, Proportions, and Percents Concepts--Ratios with Percents\n\nDefinition--Ratios, Proportions, and Percents Concepts--Ratios with Percents\n\nThis is part of a collection of definitions related to ratios, proportions, and percents.",
null,
"Ratios and Rates\n\n## Definition--Ratios, Proportions, and Percents Concepts--The Golden Ratio\n\nDefinition--Ratios, Proportions, and Percents Concepts--The Golden Ratio\n\nThis is part of a collection of definitions related to ratios, proportions, and percents.",
null,
"Applications of Ratios, Proportions, and Percents, Ratios and Rates\n\n## Definition--Ratios, Proportions, and Percents Concepts--Unit Rate\n\nUnit Rates\n\nWatch the following video to learn about unit rates. (The video transcript is included.)",
null,
"Ratios and Rates\n\n## Definition--Ratios, Proportions, and Percents Concepts--Visualizing Equivalent Ratios\n\nDefinition--Ratios, Proportions, and Percents Concepts--Visualizing Equivalent Ratios\n\nThis is part of a collection of definitions related to ratios, proportions, and percents.",
null,
"Ratios and Rates\n\n## Definition--Ratios, Proportions, and Percents Concepts--Visualizing Ratios\n\nDefinition--Ratios, Proportions, and Percents Concepts--Visualizing Ratios\n\nThis is part of a collection of definitions related to ratios, proportions, and percents.",
null,
"Ratios and Rates"
] |
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"# Document (#22292)\n\nAuthor\nMarty, P.F.\nTitle\nOn-line exhibit design : the sociotechnological impact of building a museum over the World Wide Web\nSource\nJournal of the American Society for Information Science. 51(2000) no.1, S.24-32\nYear\n2000\nAbstract\nThis article examines the sociotechnological impact of introducing collaborative technologies into the Spurlock Museum, a museum of world history and culture at the University of Illinois\nTheme\nObjektdokumentation\nInternet\n\n## Similar documents (author)\n\n1. Marty, P.F.: Museum informatics and collaborative technologies : the emerging socia-technological dimension of information science in museum environments (1999) 5.99\n```5.989656 = sum of:\n5.989656 = weight(author_txt:marty in 5331) [ClassicSimilarity], result of:\n5.989656 = fieldWeight in 5331, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n9.583449 = idf(docFreq=7, maxDocs=42740)\n0.625 = fieldNorm(doc=5331)\n```\n2. Marty, P.F.: Museum informatics (2009) 5.99\n```5.989656 = sum of:\n5.989656 = weight(author_txt:marty in 1) [ClassicSimilarity], result of:\n5.989656 = fieldWeight in 1, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n9.583449 = idf(docFreq=7, maxDocs=42740)\n0.625 = fieldNorm(doc=1)\n```\n3. Marty, P.F.: ¬The changing nature of information work in museums (2007) 5.99\n```5.989656 = sum of:\n5.989656 = weight(author_txt:marty in 2089) [ClassicSimilarity], result of:\n5.989656 = fieldWeight in 2089, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n9.583449 = idf(docFreq=7, maxDocs=42740)\n0.625 = fieldNorm(doc=2089)\n```\n4. Lacroix, S.; Marty, J.-C.; Roche, C.: OK: a model of ontologies by differentiation (1998) 3.59\n```3.5937934 = sum of:\n3.5937934 = weight(author_txt:marty in 1060) [ClassicSimilarity], result of:\n3.5937934 = fieldWeight in 1060, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n9.583449 = idf(docFreq=7, maxDocs=42740)\n0.375 = fieldNorm(doc=1060)\n```\n5. Stvilia, B.; Hinnant, C.C.; Schindler, K.; Worrall, A.; Burnett, G.; Burnett, K.; Kazmer, M.M.; Marty, P.F.: Composition of scientific teams and publication productivity at a national science lab (2011) 2.40\n```2.3958623 = sum of:\n2.3958623 = weight(author_txt:marty in 1192) [ClassicSimilarity], result of:\n2.3958623 = fieldWeight in 1192, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n9.583449 = idf(docFreq=7, maxDocs=42740)\n0.25 = fieldNorm(doc=1192)\n```\n\n## Similar documents (content)\n\n1. Trant, J.; Bearman, D.: Social terminology enhancement through vernacular engagement : exploring collaborative annotation to encourage interaction with museum collections (2005) 0.27\n```0.2739933 = sum of:\n0.2739933 = product of:\n0.913311 = sum of:\n0.018227518 = weight(abstract_txt:into in 3186) [ClassicSimilarity], result of:\n0.018227518 = score(doc=3186,freq=2.0), product of:\n0.07389672 = queryWeight, product of:\n1.5230821 = boost\n3.7208836 = idf(docFreq=2812, maxDocs=42740)\n0.013039344 = queryNorm\n0.24666207 = fieldWeight in 3186, product of:\n1.4142135 = tf(freq=2.0), with freq of:\n2.0 = termFreq=2.0\n3.7208836 = idf(docFreq=2812, maxDocs=42740)\n0.046875 = fieldNorm(doc=3186)\n0.031511307 = weight(abstract_txt:history in 3186) [ClassicSimilarity], result of:\n0.031511307 = score(doc=3186,freq=1.0), product of:\n0.13410997 = queryWeight, product of:\n2.0518296 = boost\n5.0126114 = idf(docFreq=772, maxDocs=42740)\n0.013039344 = queryNorm\n0.23496616 = fieldWeight in 3186, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n5.0126114 = idf(docFreq=772, maxDocs=42740)\n0.046875 = fieldNorm(doc=3186)\n0.09124337 = weight(abstract_txt:line in 3186) [ClassicSimilarity], result of:\n0.09124337 = score(doc=3186,freq=3.0), product of:\n0.18890521 = queryWeight, product of:\n2.4351892 = boost\n5.9491577 = idf(docFreq=302, maxDocs=42740)\n0.013039344 = queryNorm\n0.4830114 = fieldWeight in 3186, product of:\n1.7320508 = tf(freq=3.0), with freq of:\n3.0 = termFreq=3.0\n5.9491577 = idf(docFreq=302, maxDocs=42740)\n0.046875 = fieldNorm(doc=3186)\n0.10776137 = weight(abstract_txt:exhibit in 3186) [ClassicSimilarity], result of:\n0.10776137 = score(doc=3186,freq=1.0), product of:\n0.30440998 = queryWeight, product of:\n3.091293 = boost\n7.5520167 = idf(docFreq=60, maxDocs=42740)\n0.013039344 = queryNorm\n0.35400078 = fieldWeight in 3186, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n7.5520167 = idf(docFreq=60, maxDocs=42740)\n0.046875 = fieldNorm(doc=3186)\n0.043807484 = weight(abstract_txt:world in 3186) [ClassicSimilarity], result of:\n0.043807484 = score(doc=3186,freq=1.0), product of:\n0.21047105 = queryWeight, product of:\n3.6351466 = boost\n4.4403243 = idf(docFreq=1369, maxDocs=42740)\n0.013039344 = queryNorm\n0.2081402 = fieldWeight in 3186, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n4.4403243 = idf(docFreq=1369, maxDocs=42740)\n0.046875 = fieldNorm(doc=3186)\n0.62075996 = weight(abstract_txt:museum in 3186) [ClassicSimilarity], result of:\n0.62075996 = score(doc=3186,freq=10.0), product of:\n0.6548468 = queryWeight, product of:\n7.8531017 = boost\n6.3950324 = idf(docFreq=193, maxDocs=42740)\n0.013039344 = queryNorm\n0.9479469 = fieldWeight in 3186, product of:\n3.1622777 = tf(freq=10.0), with freq of:\n10.0 = termFreq=10.0\n6.3950324 = idf(docFreq=193, maxDocs=42740)\n0.046875 = fieldNorm(doc=3186)\n0.3 = coord(6/20)\n```\n2. Shreeves, S.L.; Kaczmarek, J.S.; Cole, T.W.: Harvesting cultural heritage metadata using OAI Protocol (2003) 0.26\n```0.26086387 = sum of:\n0.26086387 = product of:\n0.7453254 = sum of:\n0.006079828 = weight(abstract_txt:this in 776) [ClassicSimilarity], result of:\n0.006079828 = score(doc=776,freq=1.0), product of:\n0.031855065 = queryWeight, product of:\n2.442996 = idf(docFreq=10095, maxDocs=42740)\n0.013039344 = queryNorm\n0.19085906 = fieldWeight in 776, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n2.442996 = idf(docFreq=10095, maxDocs=42740)\n0.078125 = fieldNorm(doc=776)\n0.023567716 = weight(abstract_txt:article in 776) [ClassicSimilarity], result of:\n0.023567716 = score(doc=776,freq=1.0), product of:\n0.078607254 = queryWeight, product of:\n1.5708766 = boost\n3.8376453 = idf(docFreq=2502, maxDocs=42740)\n0.013039344 = queryNorm\n0.29981604 = fieldWeight in 776, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n3.8376453 = idf(docFreq=2502, maxDocs=42740)\n0.078125 = fieldNorm(doc=776)\n0.032067813 = weight(abstract_txt:university in 776) [ClassicSimilarity], result of:\n0.032067813 = score(doc=776,freq=1.0), product of:\n0.096522935 = queryWeight, product of:\n1.7407084 = boost\n4.2525434 = idf(docFreq=1652, maxDocs=42740)\n0.013039344 = queryNorm\n0.33222997 = fieldWeight in 776, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n4.2525434 = idf(docFreq=1652, maxDocs=42740)\n0.078125 = fieldNorm(doc=776)\n0.032260608 = weight(abstract_txt:over in 776) [ClassicSimilarity], result of:\n0.032260608 = score(doc=776,freq=1.0), product of:\n0.09690942 = queryWeight, product of:\n1.7441899 = boost\n4.261049 = idf(docFreq=1638, maxDocs=42740)\n0.013039344 = queryNorm\n0.33289444 = fieldWeight in 776, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n4.261049 = idf(docFreq=1638, maxDocs=42740)\n0.078125 = fieldNorm(doc=776)\n0.04824552 = weight(abstract_txt:wide in 776) [ClassicSimilarity], result of:\n0.04824552 = score(doc=776,freq=1.0), product of:\n0.12673278 = queryWeight, product of:\n1.9945973 = boost\n4.872793 = idf(docFreq=888, maxDocs=42740)\n0.013039344 = queryNorm\n0.38068697 = fieldWeight in 776, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n4.872793 = idf(docFreq=888, maxDocs=42740)\n0.078125 = fieldNorm(doc=776)\n0.27593464 = weight(abstract_txt:illinois in 776) [ClassicSimilarity], result of:\n0.27593464 = score(doc=776,freq=3.0), product of:\n0.281027 = queryWeight, product of:\n2.9701936 = boost\n7.256171 = idf(docFreq=81, maxDocs=42740)\n0.013039344 = queryNorm\n0.9818795 = fieldWeight in 776, product of:\n1.7320508 = tf(freq=3.0), with freq of:\n3.0 = termFreq=3.0\n7.256171 = idf(docFreq=81, maxDocs=42740)\n0.078125 = fieldNorm(doc=776)\n0.32716927 = weight(abstract_txt:museum in 776) [ClassicSimilarity], result of:\n0.32716927 = score(doc=776,freq=1.0), product of:\n0.6548468 = queryWeight, product of:\n7.8531017 = boost\n6.3950324 = idf(docFreq=193, maxDocs=42740)\n0.013039344 = queryNorm\n0.4996119 = fieldWeight in 776, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n6.3950324 = idf(docFreq=193, maxDocs=42740)\n0.078125 = fieldNorm(doc=776)\n0.35 = coord(7/20)\n```\n3. Turner, H.: Decolonizing ethnographic documentation : a critical history of the early museum catalogs at the Smithsonian's National Museum of Natural History (2015) 0.23\n```0.23269922 = sum of:\n0.23269922 = product of:\n0.93079686 = sum of:\n0.012636685 = weight(abstract_txt:this in 4185) [ClassicSimilarity], result of:\n0.012636685 = score(doc=4185,freq=3.0), product of:\n0.031855065 = queryWeight, product of:\n2.442996 = idf(docFreq=10095, maxDocs=42740)\n0.013039344 = queryNorm\n0.3966931 = fieldWeight in 4185, product of:\n1.7320508 = tf(freq=3.0), with freq of:\n3.0 = termFreq=3.0\n2.442996 = idf(docFreq=10095, maxDocs=42740)\n0.09375 = fieldNorm(doc=4185)\n0.039995737 = weight(abstract_txt:article in 4185) [ClassicSimilarity], result of:\n0.039995737 = score(doc=4185,freq=2.0), product of:\n0.078607254 = queryWeight, product of:\n1.5708766 = boost\n3.8376453 = idf(docFreq=2502, maxDocs=42740)\n0.013039344 = queryNorm\n0.5088047 = fieldWeight in 4185, product of:\n1.4142135 = tf(freq=2.0), with freq of:\n2.0 = termFreq=2.0\n3.8376453 = idf(docFreq=2502, maxDocs=42740)\n0.09375 = fieldNorm(doc=4185)\n0.08912743 = weight(abstract_txt:history in 4185) [ClassicSimilarity], result of:\n0.08912743 = score(doc=4185,freq=2.0), product of:\n0.13410997 = queryWeight, product of:\n2.0518296 = boost\n5.0126114 = idf(docFreq=772, maxDocs=42740)\n0.013039344 = queryNorm\n0.66458464 = fieldWeight in 4185, product of:\n1.4142135 = tf(freq=2.0), with freq of:\n2.0 = termFreq=2.0\n5.0126114 = idf(docFreq=772, maxDocs=42740)\n0.09375 = fieldNorm(doc=4185)\n0.10902857 = weight(abstract_txt:culture in 4185) [ClassicSimilarity], result of:\n0.10902857 = score(doc=4185,freq=1.0), product of:\n0.1932667 = queryWeight, product of:\n2.463141 = boost\n6.0174437 = idf(docFreq=282, maxDocs=42740)\n0.013039344 = queryNorm\n0.5641353 = fieldWeight in 4185, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n6.0174437 = idf(docFreq=282, maxDocs=42740)\n0.09375 = fieldNorm(doc=4185)\n0.6800085 = weight(abstract_txt:museum in 4185) [ClassicSimilarity], result of:\n0.6800085 = score(doc=4185,freq=3.0), product of:\n0.6548468 = queryWeight, product of:\n7.8531017 = boost\n6.3950324 = idf(docFreq=193, maxDocs=42740)\n0.013039344 = queryNorm\n1.0384238 = fieldWeight in 4185, product of:\n1.7320508 = tf(freq=3.0), with freq of:\n3.0 = termFreq=3.0\n6.3950324 = idf(docFreq=193, maxDocs=42740)\n0.09375 = fieldNorm(doc=4185)\n0.25 = coord(5/20)\n```\n4. Marty, P.F.: ¬The changing nature of information work in museums (2007) 0.23\n```0.23106305 = sum of:\n0.23106305 = product of:\n0.77021015 = sum of:\n0.009727725 = weight(abstract_txt:this in 2089) [ClassicSimilarity], result of:\n0.009727725 = score(doc=2089,freq=4.0), product of:\n0.031855065 = queryWeight, product of:\n2.442996 = idf(docFreq=10095, maxDocs=42740)\n0.013039344 = queryNorm\n0.3053745 = fieldWeight in 2089, product of:\n2.0 = tf(freq=4.0), with freq of:\n4.0 = termFreq=4.0\n2.442996 = idf(docFreq=10095, maxDocs=42740)\n0.0625 = fieldNorm(doc=2089)\n0.017185068 = weight(abstract_txt:into in 2089) [ClassicSimilarity], result of:\n0.017185068 = score(doc=2089,freq=1.0), product of:\n0.07389672 = queryWeight, product of:\n1.5230821 = boost\n3.7208836 = idf(docFreq=2812, maxDocs=42740)\n0.013039344 = queryNorm\n0.23255523 = fieldWeight in 2089, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n3.7208836 = idf(docFreq=2812, maxDocs=42740)\n0.0625 = fieldNorm(doc=2089)\n0.032656383 = weight(abstract_txt:article in 2089) [ClassicSimilarity], result of:\n0.032656383 = score(doc=2089,freq=3.0), product of:\n0.078607254 = queryWeight, product of:\n1.5708766 = boost\n3.8376453 = idf(docFreq=2502, maxDocs=42740)\n0.013039344 = queryNorm\n0.41543728 = fieldWeight in 2089, product of:\n1.7320508 = tf(freq=3.0), with freq of:\n3.0 = termFreq=3.0\n3.8376453 = idf(docFreq=2502, maxDocs=42740)\n0.0625 = fieldNorm(doc=2089)\n0.05960297 = weight(abstract_txt:technologies in 2089) [ClassicSimilarity], result of:\n0.05960297 = score(doc=2089,freq=2.0), product of:\n0.13438772 = queryWeight, product of:\n2.0539532 = boost\n5.0177994 = idf(docFreq=768, maxDocs=42740)\n0.013039344 = queryNorm\n0.443515 = fieldWeight in 2089, product of:\n1.4142135 = tf(freq=2.0), with freq of:\n2.0 = termFreq=2.0\n5.0177994 = idf(docFreq=768, maxDocs=42740)\n0.0625 = fieldNorm(doc=2089)\n0.06577986 = weight(abstract_txt:impact in 2089) [ClassicSimilarity], result of:\n0.06577986 = score(doc=2089,freq=1.0), product of:\n0.22782232 = queryWeight, product of:\n3.7820206 = boost\n4.6197305 = idf(docFreq=1144, maxDocs=42740)\n0.013039344 = queryNorm\n0.28873315 = fieldWeight in 2089, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n4.6197305 = idf(docFreq=1144, maxDocs=42740)\n0.0625 = fieldNorm(doc=2089)\n0.5852582 = weight(abstract_txt:museum in 2089) [ClassicSimilarity], result of:\n0.5852582 = score(doc=2089,freq=5.0), product of:\n0.6548468 = queryWeight, product of:\n7.8531017 = boost\n6.3950324 = idf(docFreq=193, maxDocs=42740)\n0.013039344 = queryNorm\n0.89373296 = fieldWeight in 2089, product of:\n2.236068 = tf(freq=5.0), with freq of:\n5.0 = termFreq=5.0\n6.3950324 = idf(docFreq=193, maxDocs=42740)\n0.0625 = fieldNorm(doc=2089)\n0.3 = coord(6/20)\n```\n5. Marty, P.F.: Museum informatics (2009) 0.21\n```0.20941427 = sum of:\n0.20941427 = product of:\n1.0470713 = sum of:\n0.0072957934 = weight(abstract_txt:this in 1) [ClassicSimilarity], result of:\n0.0072957934 = score(doc=1,freq=1.0), product of:\n0.031855065 = queryWeight, product of:\n2.442996 = idf(docFreq=10095, maxDocs=42740)\n0.013039344 = queryNorm\n0.22903088 = fieldWeight in 1, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n2.442996 = idf(docFreq=10095, maxDocs=42740)\n0.09375 = fieldNorm(doc=1)\n0.0632185 = weight(abstract_txt:technologies in 1) [ClassicSimilarity], result of:\n0.0632185 = score(doc=1,freq=1.0), product of:\n0.13438772 = queryWeight, product of:\n2.0539532 = boost\n5.0177994 = idf(docFreq=768, maxDocs=42740)\n0.013039344 = queryNorm\n0.4704187 = fieldWeight in 1, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n5.0177994 = idf(docFreq=768, maxDocs=42740)\n0.09375 = fieldNorm(doc=1)\n0.09866979 = weight(abstract_txt:impact in 1) [ClassicSimilarity], result of:\n0.09866979 = score(doc=1,freq=1.0), product of:\n0.22782232 = queryWeight, product of:\n3.7820206 = boost\n4.6197305 = idf(docFreq=1144, maxDocs=42740)\n0.013039344 = queryNorm\n0.43309975 = fieldWeight in 1, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n4.6197305 = idf(docFreq=1144, maxDocs=42740)\n0.09375 = fieldNorm(doc=1)\n0.87788725 = weight(abstract_txt:museum in 1) [ClassicSimilarity], result of:\n0.87788725 = score(doc=1,freq=5.0), product of:\n0.6548468 = queryWeight, product of:\n7.8531017 = boost\n6.3950324 = idf(docFreq=193, maxDocs=42740)\n0.013039344 = queryNorm\n1.3405994 = fieldWeight in 1, product of:\n2.236068 = tf(freq=5.0), with freq of:\n5.0 = termFreq=5.0\n6.3950324 = idf(docFreq=193, maxDocs=42740)\n0.09375 = fieldNorm(doc=1)\n0.2 = coord(4/20)\n```"
] |
[
null
] |
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http://www.chemistry-assignment.com/born-haber-cycle-and-lattice-enthalpy
|
[
"Born Haber Cycle and Lattice Enthalpy\n\nLATTICE ENTHALPY\n\nThe Lattice enthalpy of ionic compound is the enthalpy change which occurs when one mole of ionic compound in solid state dissociates into its gaseous ions.\n\nNa+ Cl (s) à Na+(g) + Cl(G) ; ΔL H = + 788 KJ mol-1\n\nLattice enthalpy may also be defined as energy released when one mole of ionic crystal is formed by close packing of constituent ions in gaseous state.\n\nNa+ Cl (s) à Na+(g) + Cl(G) ; ΔL H = + 788 KJ mol-1\n\nThe magnitude of lattice enthalpy gives the idea of the stability of the ionic crystal. It is not possible to determine lattice enthalpies of ionic compounds directly by experimental techniques. However, we can use indirect method by constructing an enthalpy diagram called Born Haber Cycle. Born Haber cycle is a simplified method which was developed in 1919 by Max Born and Fritz Haber to correlate lattice energies of ionic crystals to other thermodynamic data. The development of Born Haber cycle was primarily based on Hess’s law. Let us consider the energy changes during the formation of sodium chloride crystal from the metallic sodium and chlorine gas to calculate lattice enthalpy of NaCl(s). The net enthalpy of formation of NaCl Δ f H = – 411.2 KJ mol-1\n\nNa(s) + 1/2 Cl2(g) à NaCl (s) ; Δ f H = – 411.2 KJ mol-1\n\nThe overall process can be imagined to occur in following steps:\n\n(i) Sublimation of metallic sodium",
null,
"(ii) Ionization of sodium atoms",
null,
"Δ ie H is ionization enthalpy of sodium.\n\n(iii) Atomisation of Cl2. This step involves dissociation of Cl2(g) into Cl(g) atoms. The reaction enthalpy is half of the bond dissociation enthalpy of chlorine.",
null,
"(iv) Conversion of Cl( g) to Cl-( g).",
null,
"Δ ea H is electron gain affinity of chlorine.\n\n(v) Combination of Na+(g) and CI-(g) ions to form 1 mole of NaCl(s). The energy released here is called lattice enthalpy (ΔLH). The sequence of steps (i)-(v) is shown in Fig. 17.12 and is known as Born Haber cycle. The sum of the enthalpy changes round a cycle is zero.",
null,
"The various changes in the Born Haber Cycle for sodium chloride are shown in Fig 17.13.\n\nApplying Hess’s law we get",
null,
"Let us now proceed to solve some numerical problems based upon thermochemical calculations"
] |
[
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null,
"http://www.chemistry-assignment.com/wp-content/uploads/2013/01/1259.png",
null
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|
http://book.mathreference.com/set
|
[
"Having studied sets every September from 6th grade through 12th, I thought I knew what a set was. It's like a bag with stuff in it, right? Then I went to graduate school and learned otherwise.\n\nActually there are no bags; there are only rules for containment. The even numbers form a set because they conform to a rule, i.e. divisible by 2. You can picture them in a bag, or not, as you wish. However, the bag metaphor gets in the way at times, because there are situations where a set can contain itself, or two sets can contain each other.\n\n → ←\n\nWhat sort of items can be clumped together into sets? Lions? Tigers? Bears? Surely the integers. Well not really. The only thing a set can contain is other sets. We ascribe semantic meaning to these sets, but they are just sets.\n\nIt is possible to start with the empty set and build the integers, and all of mathematics, from the ground up. The empty set is denoted ∅, and it has no members. Think of the empty set as 0, the set containing the empty set as 1, the set containing 0 and 1 as 2, the set containing 0 1 and 2 as 3, and so on. Thus n is the set containing all the number-sets that came before. So you see, we don't really need any objects, or bags, at all. The only thing that is real is the empty set and the rules for containment.\n\nHere's how the number 4 might look as a set.\n\n(∅ (∅) (∅(∅)) (∅(∅)(∅(∅))))\n\nIf two sets contain the same elements they are the same set. This reenforces the notion that the rule is the set. If two sets both contain the even integers, they are the same set. Such a set starts out (∅, 2, 4, 6, …), where the integers are really the aforementioned sets.\n\nWrite x ∈ s when x belongs to s. We call x an element, i.e. an element of s, but it's really just another set, because that's all we have is sets, and sets of sets, etc.\n\nMany books have been written about sets; I'm only going to go over the basics, the things you need to know to build groups and rings and the like. Taking this shortcut means you may, at the end of this chapter, still be unsure what a set is. Join the club! But 99.9% of the time, a set is a slice of the rationals, or the reals, or polynomials, or rotations in space, or other things that are well understood.\n\nOne thing it is not is \"all sets\". There is no set of all sets. I may start out by saying \"A monoid is a set with a + operator and an identity\", and you might think, \"Great, all the sets form a monoid under union.\" It looks good at first. (A∪B)∪C = A∪(B∪C), so union is associative. And the empty set is the identity. S ∪ ∅ = S. But there is no set of all sets. For technical reasons, every structure I define is going to be a set, and there is no set of all sets, so the entire universe of sets is never going to be a field, or a group, etc.\n\nThe lack of a universal set can be proven, but other aspects of set theory are axiomatic. For instance, infinite sets are assumed, simply waved into existence. They don't have to be there. You can build a perfectly good world of mathematics that consists only of finite sets. But then there is no structure like the integers, and proofs based on induction (even though they may refer to finite sets) don't really work. So infinite sets are assumed.\n\nAnother axiom is that a set does not contain itself. This is arbitrary too. You can build a consistent world of mathematics where sets do at times contain themselves. There is no inherent contradiction here. However, self-referencing sets don't seem to be necessary for the vast majority of math, or engineering, or science, so I'm leaving them out.\n\nAnother axiom is the continuum hypothesis, abbreviated CH. One can prove, by diagonalization, that there are more real numbers than rationals. This seems strange at first, since both sets are infinite, but it can be made rigorous, and will be demonstrated below. Are there any sets that are larger than the rationals, yet smaller than the reals? It's up to you. If you want it to be true than declare it so; otherwise say it is false. The continuum hypothesis says there are no such intermediate sets, and that's ok with me.\n\nThe most interesting axiom is the axiom of choice. This is usually invoked as zorn's lemma, which is equivalent to the axiom of choice. One is true iff the other is true. Zorn's lemma is described below, under partially ordered sets. Many theorems rely on zorn's lemma, theorems that I want to be true, so I am saying yes to the axiom of choice.\n\nThis is ZFC set theory, where ZF stands for Zermelo Fraenkel, who developed and refined the theory, and C stands for choice. Almost every math book is based on ZF set theory, and most include choice as well, even if they don't say so explicitly.\n\nThe direct product of sets, possibly an infinite list of sets, is the cross product of all these sets, i.e. all combinations thereof. If S contains a b and c, and T contains 1 2 and 3, then the direct product of S and T, sometimes called S cross T, consists of all 9 pairs: a1 a2 a3 b1 b2 b3 c1 c2 c3. The direct product of 3 sets generates triples, and so on.\n\nThe direct product of an infinite list of sets creates infinite strings. (This requires the axiom of choice.) If each set Si in the list happens to be the real numbers, then the direct product is a sequence of real numbers.\n\n S1 * S2 * S3 * S4 * S5 * …\n\nThe direct sum is again a cross product, but infinite strings are highly constrained. Almost all the elements in the string have to be some designated member of the set, usually zero.\n\n7, 5, ½, 3, -2, π, 0, 4, 6, 0, 0, 0, 0, 0, 0, 0, 0, …\n\nWhen the list of sets is finite, direct product and direct sum are the same thing.\n\nThe 7-adic numbers are a direct product - infinite sequences of integers mod 7. Each set in the underlying list is Z/7. The direct product contains the direct sum, where the entries trailing off to the right are all zero. These are the strings that correspond to the nonnegative integers as we know them. Write the integer in base 7, backwards, to build the finite string; then fill in with zeros after that.\n\nA relation is a subset of the cross product of two or more sets. For example, consider the \"ownership\" relation, where S = people and T = pets. The relation contains Dick,Spot, since Dick owns Spot. It does not contain Jane,Spot, but it does contain Jane,Puff.\n\n``` Dick → Spot\nJane → Puff\nTim → Lassy\nFred → Spot (this function is not invertible)\nClinton → Sox\nNixon → Checkers\n```\n\nA relation is considered a function if no member of S is repeated. The \"ownership\" relation above is a function, because nobody owns more than one pet. Put in a person, and the function cranks out the pet.\n\nWhen a function is derived from S cross T, S is called the domain, and T is called the range. If x is a member of S, f(x) is the corresponding member of T, as determine by the function f. We sometimes say f maps S into T.\n\nIf f covers all of T, every pet in the world for example, then f is surjective, or f maps S onto T.\n\nIf everything in T appears only once, then f is injective, or f embeds S into T. The function can be reversed, going from pet back to owner. In our example, f is not invertible, since two people have joint custody of Spot.\n\nA function is bijective if it is injective and surjective. Every person owns exactly one pet, and every pet is owned by somebody. There is a 1 to 1 correspondence.\n\nA function can map multiple sets into another set. The operator times maps R cross R onto R. Technically this relation consists of triples from the direct product R cross R cross R. The domain is the plane and the range is the line.\n\nWhen a relation is derived from a set crossed with itself, several important properties become meaningful. Let R be a subset of S cross S, with a, b, and c drawn from S.\n\nR is reflexive if for every a, R contains a,a.\n\nR is symmetric if for every a and b, R contains a,b iff R contains b,a.\n\nR is transitive if for every a, b, and c, R contains a,c whenever R contains both a,b and b,c.\n\nA relation possessing all three properties is called an equivalence relation. Such a relation partitions the set S into disjoint subsets called equivalence classes. When R is an equivalence relation, a simpler function q carries the same information. Here q is an equivalence class function, where q(a) determines the subset of S that contains a. In other words, q(a) is the clump of things that a belongs to.\n\nA simple example in Z cross Z includes a,b whenever a-b is a multiple of 5. Verify the three properties above, then find the equivalence classes. The first equivalence class is represented by 0. It includes 5, 10, 15, and so on, since each is a multiple of 5 away from 0. The other equivalence classes are the other 4 integers mod 5. In general, a homomorphism partitions the domain into equivalence classes, shifted copies of the kernel, and these subsets, also called cosets, can sometimes be manipulated as individual entities. In the above example, the number 1 represents the coset of integers that are 1 mod 5.\n\nEquivalence classes need not be the same size, and they need not correspond to one another. Let two real numbers be in a relation if they have a nonzero ratio. This is an equivalence relation, and the two equivalence classes are 0 and all the nonzero numbers.\n\nAnother example of equivalence classes are associates within a ring. In the integers, the classes are 5 and -5, 7 and -7, 19 and -19, and so on. All equivalence classes are the same size, according to the units of the ring, except for 0, which stands alone.\n\nWhile the above properties clump elements together, other properties separate elements and put them in order.\n\nR is irreflexive if for every a, R does not contain a,a. This is of course the opposite of reflexive. There is one relation that is both reflexive and irreflexive … the empty relation on the empty set.\n\nR is antisymmetric if a,b and b,a implies a = b. This is where we start using the ≤ notation, for if a ≤ b and b ≤ a, then a must equal b. However, this notation can be confusing at first, because everybody is use to numbers, and given any two numbers, one is always less than or equal to the other. But two elements of an antisymmetric relation need not be comparable. In fact, R could be the empty set, no ordered pairs at all, and the relation is still antisymmetric. We only require that any two elements not be ≤ each other, unless they are the same.\n\nIf you know the relation is irreflexive, you can define antisymmetric as a,b implies not b,a. In this case we use the < operator. That is, a < b means we cannot have b < a. Again, all pairs of elements need not be comparable.\n\nThe relation R is a partial ordering on the set S, or S is a partially ordered set via R, or S is a poset, if R is transitive and antisymmetric. Once again we use the ≤ operator, or the < operator if R is also irreflexive. These are called weak and strong partial orderings respectively. Applying the weak notation, transitivity is written: a ≤ b and b ≤ c implies a ≤ c. The notation is helping us along, because arithmetic ≤ is certainly transitive. But unlike the arithmetic ≤ operator, some elements might not be comparable. Here is an example. Set a < b and c, and b and c < d, which means a < d, yet b and c are not comparable. Neither is less than the other. This is illustrated by a square with a in the lower left corner and d in the upper right. Arrows, or directed arcs, indicate the < or ≤ relation. Two arrows run from a to b and c, and two more arrows run from b and c up to d. Another arrow is implied from a to d by transitivity, but b and c are not comparable.\n b → d ↑ ↑ a → c\nAn upper bound of a set T within a larger poset S is an element that is ≥ every member of T. The least upper bound (lub) is an upper bound that is ≤ every upper bound. Since two elements ≤ each other are the same, the lub is unique (if it exists). Lower bounds are defined similarly. These definitions are inspired by the integers, when a ≤ b means a is a factor of b. Verify that this is a partial ordering. The greatest lower bound of a and b becomes the greatest common divisor, and the least upper bound is the least common multiple.\n 90 ↑ 60 ↑ 30 ↑ ↑ ↑ 6 10 15\n\nA minimal element in S means no element is smaller. The smallest or minimum element is less than every other element in S. Thus minimum implies minimal, and is unique relative to S. Maximal and maximum are defined similarly.\n\nA lattice is a partial ordering with a lub (least upper bound) and a glb (greatest lower bound) for every pair of points. Given x and y, the join is the lub and the meet is the glb. If x ≤ y means x is a subset of y, the union of x and y is the join, and the intersection is the meet. This is a subset lattice. If the set is the integers, and ≤ means a factor of, then the lcm is the join and the gcd is the meet. The positive integers form a lattice, not just a poset.\n\nDon't confuse this with a regular lattice in n-space. That is a repeating grid of points corresponding to the integer multiples of n vectors. A checkerboard in the plane for instance, or a pattern of parallelograms if the checkerboard is pushed over. That's a different kind of lattice altogether. Mathematicians sometimes use the same word for different things.\n\nA complete lattice has a lub and a glb for every set of elements. The integers are not a complete lattice, because all the numbers taken together don't have an lcm. They do have a global gcd however, namely 1. The numbers 1, 2, 3, 5, 6, 10, 15, and 30 are complete, with 30 at the top and 1 at the bottom.\n\nA linear ordering, or total ordering, is a partial ordering where all pairs of elements are comparable. They can all be placed in a line. The real numbers are linearly ordered, where ≤ has its usual meaning.\n\nEvery linear ordering is a lattice, since the join is the larger element and the meet is the smaller.\n\nA chain is a subset of a poset that is linearly ordered. Usually the chain is declared ascending or descending, if it runs up forever or down forever.\n\nLet S be a partially ordered set, and assume every chain of elements, i.e. every linearly ordered subset of S, has an upper bound. Zorn's lemma says there exists a maximal element m in S, such that nothing is larger than m. This does not mean m is maximum, greater than everything in S, only that m is maximal, with nothing above it. As mentioned earlier, this is assumed to be true as an axiom.\n\nIf every descending chain has a lower bound, then zorn's lemma says there exists a minimal m with nothing below it.\n\nIf set S has more members than set T (both finite), there is no injective function F mapping S into T.\n\nSuppose a counterexample and delete some member of S and its image in T. This gives an injective function on two smaller sets. Keep going down until T has one element and S has more, whence the injective function cannot exist.\n\nThe name \"pigeonhole Principle\" was coined when this counting argument was applied to pigeons as they occupied a set of protected havens (pigeonholes). When there is an excess of birds, an injective function mapping pigeons to pigeonholes is not possible. Some must remain outside, or share a hole.\n\nHere is a brain-teaser; see if you can solve it. Select 9 different points in R3, having integer coordinates. Prove that one of the 36 segments determined by these 9 points contains another grid point with integer coordinates.\n\nA parity argument combined with the pigeonhole principle does the trick. An integer point in R3 possesses 1 of 8 parity combinations; each coordinate is even or odd. Since 9 points were selected, two have the same parity combination by the pigeonhole principle. Their average, or midpoint,is another point with integer coordinates.\n\nAs a complement to the pigeonhole principle, f cannot be onto if T is bigger than S. Again, sets must be finite for this to work. Systematically remove x from S, and f(x) from T if nothing else maps to f(x). Eventually one element in S must map to many things in T, which means f cannot map onto all of t.\n\nHere is an important variation. Let S and T be finite sets of the same size. A map from S to T is injective iff it is surjective. Equivalently, f is not injective iff f is not surjective. Start by assuming f is not injective. If both x and y in S map to z in T, delete x, and the smaller set S cannot map onto all of the larger set T. Thus f is not surjective. Conversely, if f does not cover everything in T then delete those items that are not in the range of f and apply the pigeonhole principle, whence f is not injective.\n\nThis is often used when a function f maps S into itself. By the above, f is injective iff it is surjective, and then f becomes a permutation on S.\n\nAn ordinal is a number, like 6 or 7 or 21, or even infinity. The numbers, before you get to infinity, are finite ordinals.\n\nA cardinal is an indication of size, like 6 or 7 or 21, or even infinity. This looks the same, and it is for finite sets, but it's different for infinite sets.\n\nTwo sets are the same size, having the same cardinality, if there is a bijection between them. Their elements can be put in 1 to 1 correspondence. If S has 7 elements and T has 21, there is no way to equate the elements; the sets are not the same size.\n\nNow move on to infinity. Let ω be the set containing 0 and all the positive integers. This is an ordinal, it is the smallest \"number\" that is bigger than all the finite ordinals. It is also a cardinal, measuring a flavor of infinity. This is the smallest infinity, the first infinity.\n\nω+1 is an ordinal, the next ordinal after ω. In technical terms, it is the set containing all the finite ordinals and ω. It's a perfectly good set, but it is not a well defined cardinal. That is because ω+1 is just as big as ω. Let each number n correspond to n+1. Then let ω correspond to 0. This is a bijection between ω and ω+1. The two sets are in correspondence, and are the same size. So ω+1 is no bigger than ω. There are, in fact, flavors of infinity that are bigger than ω, but they aren't found by adding or multiplying ω with the integers.\n\nA set is countable if it is finite, or it can be placed in 1 to 1 correspondence with ω. A set is countably infinite if it has size ω, like the positive integers.\n\nThe nonnegative even numbers are countable; map n to n/2. Thus an infinite set can be just as large as a proper subset of itself.\n\nThe integers are countable. Map n to 2n for n ≥ 0, and map n to -2n-1 for n < 0.\n\nThe positive numbers are countable; map n to n+1.\n\nAny set that can be listed in order, or enumerated, is countable. For instance, any subset of the positive integers is countable. Take the smallest number in the set and place it first on the list. Take the next smallest and place it second on the list. Continue this process until the entire set has been \"counted\", i.e. mapped onto the positive integers. Thus the prime numbers are countable.\n\nOrdered pairs of positive integers are countable. List them this way:\n\n1/1 2/1 1/2 3/1 2/2 1/3 4/1 3/2 2/3 1/4 5/1 4/2 3/3 2/4 1/5 …\n\nSince reduced fractions are a subset of ordered pairs of integers, the rational numbers are countable. This is counterintuitive, since they are densely packed in the number line. There are infinitely many rationals packed into the tiniest of intervals, yet there are just as many rationals as integers.\n\nWe showed that the cross product (i.e. ordered pairs) of countable sets is countable. Use this fact again and again to show that the n-tuples of integers are countable. The integer points, or even the rational points in n space are countable.\n\nIn fact all finite ordered tuples of the integers, of all lengths, are countable. Here is a recipe for listing all possible tuples in order. The tuples of length n can be listed in order; this was described above. So start with the first tuple of length 1, then the first tuple of length 2 followed by the second tuple of length 1, then the first tuple of length 3 and the second tuple of length 2 and the third tuple of length 1, then the first tuple of length 4 and the second tuple of length 3 and the third tuple of length 2 and the fourth tuple of length 1, and so on.\n\nAs a corollary, the finite sets of integers are countable, as these are all represented, perhaps many times over, by various ordered tuples. The set (1,2,3) appears six times when order is significant. Since the ordered tuples over count the unordered subsets, and the tuples are countable, the finite subsets are also countable.\n\nThe integer polynomials are precisely the finite ordered tuples of integers. Well almost; the tuples that have leading zeros can be thrown away. Anyways, since the tuples are countable, the polynomials are countable. Of course the coefficients can be drawn from any countable set, so the polynomials over the rationals are countable.\n\nThe polynomials over x and y are the polynomials in y, whose coefficients are polynomials in x. The polynomials in x are countable, and since these act as coefficients, the polynomials in x and y are countable. This extends to polynomials in x y z, and so on.\n\nThere are lots of countable sets; are there any uncountable sets? Yes, as shown by diagonalization. Consider all the subsets of the integers (not just the finite subsets). If these sets are countable then the correspondence builds a list of all possible subsets in order. Build a new subset S as follows. Let n be in S iff n is not in the nth subset on the list. Now S cannot appear anywhere on the list. If S is in position n, then S contains n iff it doesn't. Every possible correspondence fails, because it misses some set S. A generalization of this important theorem will be given later on.\n\nConsider the reals between 0 and 1, represented as decimal numbers. Suppose they can be arranged in a list. Build a new decimal as follows. Make the nth digit anything other than the nth digit of the nth number on the list. Also avoid the digits 0 and 9. The constructed real number cannot appear on the list. Since we're avoiding 0 and 9, an equivalent form of the same number cannot appear on the list either, e.g. 0.720000… = 0.719999…\n\nThis is called diagonalization because we are looking at the first digit of the first number, the second digit of the second number, and so on. List the decimal numbers down the page, with decimal digits flowing to the right. The nth digit in the nth number is a diagonal line running down and to the right.\n\nThe p-adic numbers are uncountable, using the very same proof.\n\n63105…\n52210…\n44326…\n66101…\n21045…\n\nThe size of the reals is called c, for continuum, and c is a larger cardinal than ω. In fact I am assuming that c is next in line, the next biggest cardinal. This is the continuum hypothesis.\n\nFor every set, there is another set that is bigger. Cantor stunned the world with this simple, elegant proof. Let S be any set and let T be the power set of S. That means T includes every subset of S. If S has 1 2 and 3, then T has (), (1), (2), (3), (12), (13), (23), (123). If S is finite of size n, then T has size 2n.\n\nIs there such a set? There is, but it's another axiom. It's not something you can prove; we just assume it is true.\n\nClearly S maps into T. Every x in S maps to the set (x) in T. But there is no bijection mapping S onto T.\n\nSuppose f is such a bijection and build a set W as follows. For every x in S, x is in W iff x is not in f(x). By assumption, f is surjective, and f(x) = w for some x. Thus x ∈ W iff x ∉ W. Therefore the correspondence cannot exist. T is strictly larger than S.\n\nKeep taking the power set, building S1, S2, S3, etc, and the flavors of infinity go on forever. When you think you have a handle on this, take the union of all these infinite sets, which is larger than any of them individually, and use this as a base to build another ascending chain of infinite sets, each larger than the last. Then take the union of that chain, and the union of all such chains, and so on, and so on. Trying to grasp this is like trying to comprehend the distance to the nearest galaxy. It is wonderful and troubling at the same time. Cantor's work was widely rejected, even ridiculed, sometimes on theological grounds; yet today it is an essential foundation for modern mathematics. Take a moment to read about Georg Cantor and his work.\n\nIf two sets embed in each other, are they the same size? They should be, at an intuitive level, they both fit inside each other, but we need a bijection to prove it.\n\nIf the first set S is finite, and f maps S into T, and g maps T into S, then fg embeds S into itself. As shown above, such an embedding is a bijection, both injective and surjective. If f is not surjective then its image, a proper subset of T, maps onto S by g. Remember that T embeds into S by G, and now a proper subset of T still covers all of S by g. That is impossible, hence f is surjective, as well as injective, and that proves S and T are the same size.\n\nWe really want size, or cardinality, to be a partial ordering on sets. Every set trivially maps (one to one) into itself, and if S maps into T maps into U then S maps into U. Sizes keep increasing. The relation is reflexive and transitive, we almost have a partial ordering on cardinality, but we're missing one thing: S ≤ T and T ≤ S → S = T. (Here = means the same size, not necessarily the same set.) We dealt with the finite case above, so assume both sets are infinite.\n\nLet f map S into T and let g map T into S. These are injective functions. For any x in either set, trace the ancestry of x, going backwards through f inverse and g inverse, until you reach a parentless element that is not in the range of f or g. Let S1 be the subset of S whose members have parentless ancestors in S. Let S2 be the subset of S whose members have parentless ancestors in T. Let S3 be the subset of S whose members have infinite ancestry.\n\nLet h = f on S1 and S3, and g inverse on S2. Clearly h is reversible when restricted to S1∪S3, or S2. We need to show h is reversible across all of S. In other words, we don't want h(x) to equal h(y), when x comes from S1∪S3 and y comes from S2.\n\nSuppose f(x) = g inverse of y. Thus y = g(f(x)). If x is in S3 (infinite ancestry) then so is y. Yet y is in S2. Thus x is in S1. This means x has a parentless ancestor in S, and the same holds for y, so y should be in S1. Therefore h is injective.\n\n S1 →f T S2 →g-1 S3 →f\n\nNext show that h maps S onto all of T. Select any y in T and first consider the case when, for some x in S, we have f(x) = y. If x is in S1 or S3 we are done, so x has a parentless ancestor in T. Let g(y) = z, and z has a parentless ancestor in T. Thus z is in S2, and h(z) = y. That covers y in the image of f.\n\nNow consider the last case, when y is not in the image of f. Once again z=g(y) is in S2, and h maps z to y. Therefore h maps onto all of T, and both sets have the same cardinality. That completes antisymmetry, so that \"size\", as measured by cardinality, makes sense.\n\nThis provides another method for proving cardinal equality. Mapping two sets into each other is often easier than finding a perfect correspondence. For instance, the integers map into the rationals in the obvious way, and rationals map into integers by sending the fraction a/b (lowest terms) to 2a×3b. (Send -a/b to -2a3b, and send 0 to 0.) That's it. The integers and the rationals are the same size.\n\nAre S and T the same size if a function takes S onto all of T, and another function carries T onto S? If S and T are finite then the answer is yes. The composition fg maps S onto S and is a permutation, whence both f and g must be injective.\n\nIn general the answer is yes, assuming the axiom of choice. Choice is equivalent to the well ordering principle, which states the elements in any set can be arranged in order. (The connection between choice, zorn's lemma, and the well ordering principle is beyond the scope of this book.)\n\nAssume a function g maps T onto S. There is a \"least\" element in any subset of T. Let f′(x) (x drawn from S) be the least y in T such that g(y) = x. This makes f′ injective from S into T. A function from T onto S turns around and becomes an embedding of S into T. If S also maps onto T, then build g′ injective from T into S, and apply the Schroder Bernstein theorem. That completes the bijection.\n\nThe heart of this proof is the transformation of a surjective function in one direction to an injective function in the other direction. In the same way, an injective function in one direction implies a surjective function in the other direction, again assuming the well ordering principle. If y in T is in the image of f then let g(y) be the preimage of y under f. Map the rest of T to the least element of S. Thus g maps T onto S. (This doesn't work if S is the empty set, but that is a bit pathological.)\n\nSurjective and injective are opposite sides of the same coin. Surjective implements ≥, while injective implements ≤. Two sets that map onto each other, or embed in each other, are the same size.\n\nLet c and d be two cardinals, such that d is at least as big as c. Since we know how to do finite math, let d (and perhaps c) be some flavor of infinity.\n\nWhat is c + d? By definition, it is the size of the disjoint union of c and d. This is consistent with finite math. But how large is the union when d is infinite? Certainly d embeds in this union, so it is at least d. Embed c and d back into d as follows. Let 0 in d map to 0, and let 0 in c map to 1. Let 1 in d map to 2, and let 1 in c map to 3. Let n (finite) in d map to 2n, and let n in c map to 2n+1. For any limit ordinal, such as θ, let θ+n in d map to θ+2n, and if c contains θ, let θ+n in c map to θ+2n+1. Thus c + d embeds in d. Since both sets embed in each other, c + d = d. The larger set wins.\n\nLet c*d be the size of the cross product, everything in c cross everything in d. This is consistent with finite math. If c is the empty set you get no cross product, and 0, as expected. Otherwise d embeds in the cross product by crossing it with anything in c. It is possible to map c cross d back into d, whence the sets are the same size, and c*d = d. The larger set wins. The details are beyond the scope of this book. If you are interested in this and other related topics, then you want to read Set Theory, by Ken Kunen.\n\nThe union of all the finite cross products of d has size d. This is d + d2 + d3 + …, which is d + d + d + …, which is d*ω, which is d.\n\nIf c is at least 2, then cd, the direct product of d copies of the set c, is greater than d. This is shown by another diagonal argument.\n\nAll this is cardinal math, and it actually comes in to play once in a while, e.g. to prove that certain structures have a certain size, even if that size is infinite, and that the size is a particular flavor of infinity, and cannot ambiguously be two different infinities at once.\n\nIf f is some formula on n, and f(0) is true, and f(n) implies f(n+1), then f is true for all n. This is induction, and it proves the formula for all n in one go. But it doesn't prove f for infinite sets.\n\nTransfinite induction makes one more assumption. If f is true for an ascending chain of sets, it is true for their union. Apply this to ω. Since f is true for n = 0 1 2 3 and so on, f is true for the union of all these sets, and that union happens to be ω. After that, f is true for ω+1, and ω+2, and so on. Then f is true for the union of all these sets, which is ω+ω, or 2*ω. Continue along this path, and f is true for 3*ω, 4*ω, 5*ω, and so on. Take the union, and f is true for ω*ω. After that f is true for ω*ω+1. This continues through all the ordinals.\n\nSome of the structures in this book are larger than the integers, perhaps uncountably large, and transfinite induction is used to make assertions about these structures. It is a form of induction that goes beyond finite sets.\n\nCosmologists have given us a picture of reality that never ceases to amaze. Start with a sea of hydrogen, and just a touch of helium, distributed a bit unevenly, and wait a few billion years, and you get stars, galaxies, brown dwarfs, red giants, white dwarfs, neutron stars, black holes, supernovae, heavier elements, more stars, planets, life, animal life, and finally intelligent life. As Carl Sagan summarized, \"We are a way for the Cosmos to know itself.\" And all this from hydrogen! I view set theory in the same way. This entire book, and almost every math book past present and future that you will ever read, comes from the set membership relation and a few axioms. It's not quite as amazing as life from hydrogen, but almost.\n\nThis is an excursion into continuous functions on real variables, but it is also an important introduction to abstract topology and continuous functions on abstract sets. You can skip ahead to the next chapter if you like, but if you do, you'll probably want to return to this section later. It's quite lovely.\n\nBefore we explore another wonderful construct from the mind of Cantor, let's look at 2ω. This is an infinite direct product, wherein each component is 0 or 1 - in other words, infinite binary strings. These strings can be arranged in a linear order. If s and t are two such strings, march along their digits until they disagree, perhaps in position n. At that point the string with 0 in position n is less than the string with 1 in position n. Of course if the strings agree forever then they are equal. This is called a lexicographical order.\n\nAny two strings s and t are comparable; they are equal, or one is less than the other. If s is less than t, then t is not less than s (antisymmetric). Finally, s < t < u implies s < u (transitive). Therefore this is a linear ordering.\n\n Case #1 ``` 01101000 < 01101100 < 01110001 ``` Transitivity Case #2 ``` 01101000 < 01101100 < 01101101 ```\n\nA linear ordering always implies a linear topology. Consider the real numbers, which are also linearly ordered. The open interval strictly between 3 and 7, written (3,7), is an open set, and the union of open intervals is also an open set. In contrast, the closed interval from 3 to 7, which includes the endpoints 3 and 7, and is written [3,7], is a closed set, and the intersection of closed intervals is also a closed set.\n\nAs it turns out, the same topology can be applied to 2ω. For instance, all the sequences strictly between 010000… and 011000… form an open interval. The sequences greater than s, or less than s, also form an open interval (think of the other point as infinity), and the entire set is open, as is the empty set. The union of any collection of open intervals is an open set.\n\nThe complement of an open set is a closed set. This includes the closed interval, (which is the complement of the union of the ray above and the ray below), and any finite collection of closed intervals.\n\nA function , such as f(x) = y, is continuous if the preimage of every open set is an open set. Actually it is sufficient to show the preimage of every open interval is an open set. Then take the union of these intervals in the range to show the preimage of an open set is an open set. For example, if f(x) = x2, then the preimage of the open interval (4,9) is the union of the two open intervals (2,3) and (-3,-2). This holds for every open interval in y. Even the open interval (-2,-1) has an empty preimage, which is technically an open set. Therefore f(x) = x2 is a continuous function.\n\nYou may have seen another definition of continuity in algebra or calculus, involving δ and ε. The two definitions, δ ε and preimages of open sets, are equivalent. Let's show that each implies the other.\n\nIf f is continuous by open sets, and f(p) = q, the interval, or disk, or ball (depending on dimension) of radius ε around q is an open set, and pulls back to an open set containing p, which includes some open ball of radius δ around p. A neighborhood of radius δ about p maps into our neighborhood of radius ε about q.\n\nConversely, assume the δ ε definition, and let q be any point in a designated open set R in the range of f. Let p be any point such that f(p) = q. A ball of radius ε is centered at q and lies in R, and has some ball of radius δ about p that maps entirely into the ball of radius ε. Take the union of all these balls, for each point p in the preimage of q, and then for each q in R, and the preimage of R is open.\n\nThe two definitions agree, and you can use which ever one is convenient. However, the open set definition is more general, and applies to many different structures in abstract algebra where distances such as δ and ε might not make any sense. Therefore continuity is usually defined in terms of open sets.\n\nThe composition of continuous functions is continuous. Let f and g be continuous, and let R be an open set in the range of fg. The preimage of R under g is open, and the preimage of this set under f is open, thus the preimage of R under fg is open.\n\nLet b() be a function from 2ω into the closed interval [0,1]. I call it b() for the binary representation. Put a decimal point in front of the sequence and read the number in base 2. Technically it's called a radix point, since we are not in decimal. The first digit is one half, the second digit is one forth, the third digit is one eighth, and so on. Each sequence becomes a real number from 0 to 1. The sequence 000000… maps to 0, and the sequence 111111… maps to 1/2 + 1/4 + 1/8 + 1/16 + …, which sums to 1. The minimum sequence maps to 0 and the maximum sequence maps to 1.\n\nThe function b() is not injective. For instance, 001111… is the same real number as 010000…, namely ¼. The function is however surjective. Every real number has a binary representation. Thus b maps 2ω onto [0,1], covering the entire interval.\n\nNote that b respects order. If s ≤ t then b(s) ≤ b(t). As a consequence of this, b is continuous. Start with an open interval (u,v) in [0,1]. Represent u in binary, using the larger of the two representations if u has two representations. Represent v in binary, using the smaller of the two representations if v has two representations. The strings between u and v are precisely the strings that map to real numbers between u and v. The preimage of an open interval is an open set, and b is continuous.\n\nNow we are ready for the Cantor set. Start with the unit interval [0,1] and remove the open set (1/3,2/3). We have taken out the middle third, leaving two pieces on either side. From each of these, remove the middle third. In other words, remove the open intervals (1/9,2/9) and (7/9,8/9). This leaves 4 closed segments of length 1/9. From each of these, remove the middle third. Thus the first segment loses (1/27,2/27), and so on. That leaves 8 segments of length 1/27. Remove the middle third of each of these, and repeat this process forever. The result is the Cantor set, which I will call C. The beginning of this construction is shown below.\n\nNotice that the section of C from 0 to 1/3, when magnified, looks exactly like C. This is an early example of a fractal, although fractal geometry was not well understood at the time. Cantor simply thought it was a beautiful set, and it is.\n\nIf x is in the complement of C, then it was removed at some point, as part of an open interval. The complement is a union of open intervals, and is an open set, hence C is closed.\n\nThere is a canonical map, call it m(), from 2ω onto C. If s is a sequence, and s1 = 0, select the interval [0,1/3]. If s1 = 1, select the interval [2/3,1]. If s2 = 0, select the first third of the interval previously selected, and if s2 = 1, select the last third of the interval previously selected. This continues forever, building a chain of descending closed intervals whose lengths approach 0. This chain converges to a point; call it x. Thus m(s) = x. Since x does not lie in the middle third of any segment, it has never been deleted, and x belongs to C. We have a map from 2ω into C.\n\nDifferent sequences will diverge at some point, and live in disjoint intervals thereafter. Thus m is injective.\n\nFinally, let x be any point in the cantor set C. At each step, x lies in the first or last third of the prior interval. If it were in the middle third it would not lie in C. Thus, at each step, sn can be set to 0 or 1. The resulting sequence converges to some y in C, and if y is not equal to x, then sn goes down the wrong path at some point, moving towards y instead of x. This contradicts the construction of s, hence s maps to x. The map is onto, and a bijection.\n\nThe sequences in 2ω are uncountable by diagonalization, hence the points of C are uncountable.\n\nShow that m respects order. That is, s < t implies m(s) < m(t). Also, m(s) < m(t) implies s < t. Open intervals correspond, and open sets correspond. You might think m is an isomorphism, but we don't really have operators like + and *. When a bijection preserves open sets in both directions it is called a homeomorphism. The two topological spaces are the same. One is merely a relabeling of the other - and once relabeled, the open and closed sets are the same. We can use C or 2ω interchangeably, whichever is convenient.\n\nOddly enough, the topological product of C cross C is homeomorphic to C. Let's build a bijection p() from C into C*C.\n\nGiven a sequence of zeros and ones, extract the bits in the odd positions and call that x, then extract the bits in the even positions and call that y. The ordered pair x,y becomes a point in C*C.\n\nShow that p() is injective and surjective.\n\nProve the continuity of p by looking at the preimage of an open set in C*C. Such a set is, by definition, the union of open rectangles, where each rectangle is the cross product of two open intervals. So it is enough to show the preimage of each such rectangle is open.\n\nA rectangle is the intersection of four open regions: everything to the right of the left edge, everything to the left of the right edge, everything below the top edge, and everything above the bottom edge. It is enough to show the preimage of each such region is open.\n\nLet's look at the region above the line at t; the other regions are handled similarly. Let u,b be a point in this region. Let s be the preimage of u,v. The odd bits of s create u and the even bits of s create v. Let n be the first position where v and t differ, thus vn = 1 and tn = 0. Truncate s at 2n bits, the first n bits of u and the first n bits of v, Then fill out the rest of s with zeros. Let z be based on s, but fill out the rest of z with ones. The interval of interest is now [s,z]. Wait a minute! That's a closed interval. Yes it is, but s has a predecessor and z has a successor. Any sequence that ends in all zeros has a predecessor that ends in all ones, with nothing in between these two sequences, and any sequence that ends in all ones has a successor that ends in all zeros, with nothing in between. So our closed interval is also an open interval (s-,z+). The interval [s,z] maps to a region constrained to match the first n bits of u and the first n bits of v. This contains u,v and lies entirely above t. An open interval maps into our region above t, and contains u,v. Take the union over each u,v above t, and the preimage is an open set.\n\nVerify the same for a lower region, a right region, and a left region. Now the preimage of each open rectangle is an open set, the preimage of an open set is open, and p() is continuous.\n\nThe reverse map, from C*C back to C, is also continuous. You can build a proof much like the above, or you can call upon some theorems from topology. Since C is a closed subspace of [0,1] it is compact and hausdorff. Thus C*C is also compact and hausdorff. The continuous map p becomes bicontinuous. Well however you get there, p is continuous in both directions, and is a homeomorphism. The spaces are indistinguishable from one another.\n\nI said earlier that it was odd that C and C*C should be the same, but it makes sense if you remember that C is a fractal, and the left half of C is exactly the same as C. With that in mind, yes, C cross C probably looks just like C.\n\nWrite C = C*C = C*(C*C) = C*C*C. Thus C is homeomorphic to C3, or Cn, the finite direct product of n copies of C.\n\nBy definition, a curve in the plane such as y = x2 is infinitely thin. We draw it with ink, or pixels, but that is an approximation. How then can a curve, infinitely thin, sweep around and around to fill an entire region in the plane? It can, and such a curve is called a space filling curve. Your pencil never leaves the paper (continuous), yet the tip wiggles about in an infinite number of infinitely small vibrations. To say your hand is shaky is an understatement; it is downright fuzzy, almost like quantum mechanics.\n\nFirst, what is a curve? A curve is a continuous function from the line or the line segment [0,1] into the plane, or into euclidean space, or into any space for that matter. The aforementioned parabola is a curve in the plane, with x(t) = t, and y(t) = t2. Another curve, a spiral, has x(t) = t*cos(t) and y(t) = t*sin(t), for t ≥ 0.\n\nA closed curve is a continuous function from the unit circle into space. This is a curve that starts and ends at the same point, just as the circle starts and ends at the same point. An ellipse is a closed curve in the plane, as is a square.\n\nThe following space filling curve is courtesy of Peano. Let C be the cantor set, as described above. Remember that C maps continuously onto the closed interval [0,1] using the \"binary representation\" function b() described in the previous section. Apply this map in two coordinates, and the topological product C*C maps continuously onto the unit square. Then, apply the fact that C*C is homeomorphic to C, and build a continuous map from C onto the unit square.\n\nOf course C is not the same as the closed interval [0,1], but the map can be extended. Let x be a point in [0,1] that is not in C. It has been excised as part of an open interval. The endpoints of this interval, call them a and b, map to two points in the unit square. Map all points in between, including x, into the unit square linearly. Thus, if x is halfway between a and b, then f(x) is the midpoint of f(a) and f(b). The result is a continuous map from the line segment [0,1] onto the unit square.\n\nSince Cn is homeomorphic to C, a similar result holds in n dimensions. A space filling curve can cover the unit hypercube in n space.\n\nIn the above, 0 maps to 0,0 and 1 maps to 1,1. But a space filling loop (closed curve) is also possible. Map the top have of the circle onto the unit square as described above, then map the bottom half onto a line segment from 1,1 back to 0,0. The curve starts and ends at the origin - hence it is a loop, or a closed curve.\n\nFor topological reasons, a loop in the plane cannot extend out to infinity. It is bounded in the plane. Thus there is always an outside region, beyond the loop. The jordan curve theorem says there is also an inside, and a simple closed curve, like a square or an ellipse, cuts the plane into two pieces, the inside and the outside. This sounds simple and intuitive, almost to easy to prove, yet it is a very complicated theorem with some important stipulations. One of the conditions says the curve must be smooth, or at least peacewise smooth. In contrast, our space filling curve vibrates wildly at the microscopic level. It is capable of covering a region completely, thus having no inside."
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[
null
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https://pygmtools.readthedocs.io/en/latest/guide/introduction.html
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[
"# Introduction and Guidelines\n\nThis page provides a brief introduction to graph matching and some guidelines for using pygmtools. If you are seeking some background information, this is the right place!\n\nNote\n\nFor more technical details, we recommend the following two surveys.\n\nAbout learning-based deep graph matching: Junchi Yan, Shuang Yang, Edwin Hancock. “Learning Graph Matching and Related Combinatorial Optimization Problems.” IJCAI 2020.\n\nAbout non-learning two-graph matching and multi-graph matching: Junchi Yan, Xu-Cheng Yin, Weiyao Lin, Cheng Deng, Hongyuan Zha, Xiaokang Yang. “A Short Survey of Recent Advances in Graph Matching.” ICMR 2016.\n\n## Why Graph Matching?\n\nGraph Matching (GM) is a fundamental yet challenging problem in pattern recognition, data mining, and others. GM aims to find node-to-node correspondence among multiple graphs, by solving an NP-hard combinatorial problem. Recently, there is growing interest in developing deep learning-based graph matching methods.\n\nCompared to other straight-forward matching methods e.g. greedy matching, graph matching methods are more reliable because it is based on an optimization form. Besides, graph matching methods exploit both node affinity and edge affinity, thus graph matching methods are usually more robust to noises and outliers. The recent line of deep graph matching methods also enables many graph matching solvers to be integrated into a deep learning pipeline.\n\nGraph matching techniques have been applied to the following applications:\n\nIf your task involves matching two or more graphs, you should try the solvers in pygmtools!\n\n## What is Graph Matching?\n\n### The Math Form\n\nLet’s involve a little bit of math to better understand the graph matching pipeline. In general, graph matching is of the following form, known as Quadratic Assignment Problem (QAP):\n\n$\\begin{split}&\\max_{\\mathbf{X}} \\ \\texttt{vec}(\\mathbf{X})^\\top \\mathbf{K} \\texttt{vec}(\\mathbf{X})\\\\ s.t. \\quad &\\mathbf{X} \\in \\{0, 1\\}^{n_1\\times n_2}, \\ \\mathbf{X}\\mathbf{1} = \\mathbf{1}, \\ \\mathbf{X}^\\top\\mathbf{1} \\leq \\mathbf{1}\\end{split}$\n\nThe notations are explained as follows:\n\n• $$\\mathbf{X}$$ is known as the permutation matrix which encodes the matching result. It is also the decision variable in graph matching problem. $$\\mathbf{X}_{i,a}=1$$ means node $$i$$ in graph 1 is matched to node $$a$$ in graph 2, and $$\\mathbf{X}_{i,a}=0$$ means non-matched. Without loss of generality, it is assumed that $$n_1\\leq n_2.$$ $$\\mathbf{X}$$ has the following constraints:\n\n• The sum of each row must be equal to 1: $$\\mathbf{X}\\mathbf{1} = \\mathbf{1}$$;\n\n• The sum of each column must be equal to, or smaller than 1: $$\\mathbf{X}\\mathbf{1} \\leq \\mathbf{1}$$.\n\n• $$\\mathtt{vec}(\\mathbf{X})$$ means the column-wise vectorization form of $$\\mathbf{X}$$.\n\n• $$\\mathbf{1}$$ means a column vector whose elements are all 1s.\n\n• $$\\mathbf{K}$$ is known as the affinity matrix which encodes the information of the input graphs. Both node-wise and edge-wise affinities are encoded in $$\\mathbf{K}$$:\n\n• The diagonal element $$\\mathbf{K}_{i + a\\times n_1, i + a\\times n_1}$$ means the node-wise affinity of node $$i$$ in graph 1 and node $$a$$ in graph 2;\n\n• The off-diagonal element $$\\mathbf{K}_{i + a\\times n_1, j + b\\times n_1}$$ means the edge-wise affinity of edge $$ij$$ in graph 1 and edge $$ab$$ in graph 2.\n\n### The Graph Matching Pipeline\n\nSolving a real-world graph-matching problem can be divided into the following parts:\n\n#### Part 1: Feature Extraction\n\nExtract node/edge features from the graphs you want to match. The features are used to measure the similarity between nodes/edges and to build the affinity matrix which is essential in graph matching problems.\n\n#### Part 2: Affinity Matrix Construction\n\nBuild the affinity matrix from node/edge features and form the specific QAP problem.\n\n#### Part 3: QAP Problem Solving\n\nSolve the resulting QAP problem (graph matching problem) with GM solvers.\n\nPart 1 may be done by methods depending on your application, Part 2&3 can be handled by pygmtools. The following plot illustrates a standard deep graph matching pipeline.",
null,
"## Graph Matching Best Practice\n\nWe need to understand the advantages and limitations of graph matching solvers. As discussed above, the major advantage of graph matching solvers is that they are more robust to noises and outliers. Graph matching also utilizes edge information, which is usually ignored in linear matching methods. The major drawback of graph matching solvers is their efficiency and scalability since the optimization problem is NP-hard. Therefore, to decide which matching method is most suitable, one needs to balance between the required matching accuracy and the affordable time and memory cost according to his/her application.\n\nNote\n\nAnyway, it does no harm to try graph matching first!\n\n### When to use pygmtools\n\npygmtools is recommended for the following cases, and you could benefit from the friendly API:\n\n• If you want to integrate graph matching as a step of your pipeline (either learning or non-learning).\n\n• If you want a quick benchmarking and profiling of the graph matching solvers available in pygmtools.\n\n• If you do not want to dive too deep into the algorithm details and do not need to modify the algorithm.\n\nWe offer the following guidelines for your reference:\n\n### When not to use pygmtools\n\nAs a highly packed toolkit, pygmtools lacks some flexibilities in the implementation details, especially for experts in graph matching. If you are researching new graph matching algorithms or developing next-generation deep graph matching neural networks, pygmtools may not be suitable. We recommend ThinkMatch as the protocol for academic research."
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[
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"https://pygmtools.readthedocs.io/en/latest/_images/QAP_illustration.png",
null
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https://tutorialspoint.dev/data-structure/arrays/positive-elements-even-negative-odd-positions
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[
"# Positive elements at even and negative at odd positions (Relative order not maintained)\n\nYou have been given an array and you have to make a program to convert that array such that positive elements occur at even numbered places in the array and negative elements occur at odd numbered places in the array. We have to do it in place.\n\nThere can be unequal number of positive and negative values and the extra values have to left as it is.\n\nExamples:\n\n```Input : arr[] = {1, -3, 5, 6, -3, 6, 7, -4, 9, 10}\nOutput : arr[] = {1, -3, 5, -3, 6, 6, 7, -4, 9, 10}\n\nInput : arr[] = {-1, 3, -5, 6, 3, 6, -7, -4, -9, 10}\nOutput : arr[] = {3, -1, 6, -5, 3, -7, 6, -4, 10, -9}\n```\n\n## Recommended: Please try your approach on {IDE} first, before moving on to the solution.\n\nThe idea is to use Hoare’s partition process of Quick Sort.\nWe take two pointers positive and negative. We set the positive pointer at start of the array and the negative pointer at 1st position of the array.\nWe move positive pointer two steps forward till it finds a negative element. Similarly we move negative pointer forward by two places till it finds a positive value at its position.\nIf the positive and negative pointers are in the array then we will swap the values at these indexes otherwise we will stop executing the process.\n\n## C++\n\n `// C++ program to rearrange positive and negative ` `// numbers ` `#include ` `using` `namespace` `std; ` ` ` `void` `rearrange(``int` `a[], ``int` `size) ` `{ ` ` ``int` `positive = 0, negative = 1; ` ` ` ` ``while` `(``true``) { ` ` ` ` ``/* Move forward the positive pointer till ` ` ``negative number number not encountered */` ` ``while` `(positive < size && a[positive] >= 0) ` ` ``positive += 2; ` ` ` ` ``/* Move forward the negative pointer till ` ` ``positive number number not encountered */` ` ``while` `(negative < size && a[negative] <= 0) ` ` ``negative += 2; ` ` ` ` ``// Swap array elements to fix their position. ` ` ``if` `(positive < size && negative < size) ` ` ``swap(a[positive], a[negative]); ` ` ` ` ``/* Break from the while loop when any index ` ` ``exceeds the size of the array */` ` ``else` ` ``break``; ` ` ``} ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ``int` `arr[] = { 1, -3, 5, 6, -3, 6, 7, -4, 9, 10 }; ` ` ``int` `n = (``sizeof``(arr) / ``sizeof``(arr)); ` ` ` ` ``rearrange(arr, n); ` ` ``for` `(``int` `i = 0; i < n; i++) ` ` ``cout << arr[i] << ``\" \"``; ` ` ` ` ``return` `0; ` `} `\n\n## Java\n\n `// Java program to rearrange positive ` `// and negative numbers ` `import` `java.io.*; ` ` ` `class` `GFG { ` ` ` `static` `void` `rearrange(``int` `a[], ``int` `size) ` `{ ` ` ``int` `positive = ``0``, negative = ``1``, temp; ` ` ` ` ``while` `(``true``) { ` ` ` ` ``/* Move forward the positive pointer till ` ` ``negative number number not encountered */` ` ``while` `(positive < size && a[positive] >= ``0``) ` ` ``positive += ``2``; ` ` ` ` ``/* Move forward the negative pointer till ` ` ``positive number number not encountered */` ` ``while` `(negative < size && a[negative] <= ``0``) ` ` ``negative += ``2``; ` ` ` ` ``// Swap array elements to fix their position. ` ` ``if` `(positive < size && negative < size) { ` ` ``temp = a[positive]; ` ` ``a[positive] = a[negative]; ` ` ``a[negative] = temp; ` ` ``} ` ` ` ` ``/* Break from the while loop when any index ` ` ``exceeds the size of the array */` ` ``else` ` ``break``; ` ` ``} ` `} ` ` ` `// Driver code ` `public` `static` `void` `main(String args[]) { ` ` ``int` `arr[] = {``1``, -``3``, ``5``, ``6``, -``3``, ``6``, ``7``, -``4``, ``9``, ``10``}; ` ` ``int` `n = arr.length; ` ` ` ` ``rearrange(arr, n); ` ` ``for` `(``int` `i = ``0``; i < n; i++) ` ` ``System.out.print(arr[i] + ``\" \"``); ` `} ` `} ` ` ` `/*This code is contributed by Nikita Tiwari.*/`\n\n## Python3\n\n `# Python 3 program to rearrange ` `# positive and negative numbers ` ` ` `def` `rearrange(a, size) : ` ` ` ` ``positive ``=` `0` ` ``negative ``=` `1` ` ` ` ``while` `(``True``) : ` ` ` ` ``# Move forward the positive ` ` ``# pointer till negative number ` ` ``# number not encountered ` ` ``while` `(positive < size ``and` `a[positive] >``=` `0``) : ` ` ``positive ``=` `positive ``+` `2` ` ` ` ``# Move forward the negative ` ` ``# pointer till positive number ` ` ``# number not encountered ` ` ``while` `(negative < size ``and` `a[negative] <``=` `0``) : ` ` ``negative ``=` `negative ``+` `2` ` ` ` ``# Swap array elements to fix ` ` ``# their position. ` ` ``if` `(positive < size ``and` `negative < size) : ` ` ``temp ``=` `a[positive] ` ` ``a[positive] ``=` `a[negative] ` ` ``a[negative] ``=` `temp ` ` ` ` ``# Break from the while loop when ` ` ``# any index exceeds the size of ` ` ``# the array ` ` ``else` `: ` ` ``break` ` ` `# Driver code ` `arr ``=``[ ``1``, ``-``3``, ``5``, ``6``, ``-``3``, ``6``, ``7``, ``-``4``, ``9``, ``10` `] ` `n ``=` `len``(arr) ` ` ` `rearrange(arr, n) ` `for` `i ``in` `range``(``0``, n) : ` ` ``print``(arr[i], end ``=` `\" \"``) ` ` ` `# This code is contributed by Nikita Tiwari. `\n\n## C#\n\n `// C# program to rearrange positive ` `// and negative numbers ` `using` `System; ` ` ` `class` `GFG { ` ` ` `// Function to rearrange ` `static` `void` `rearrange(``int` `[]a, ``int` `size) ` `{ ` `int` `positive = 0, negative = 1, temp; ` ` ` ` ``while` `(``true``) { ` ` ` ` ``// Move forward the positive pointer till ` ` ``// negative number number not encountered ` ` ``while` `(positive < size && a[positive] >= 0) ` ` ``positive += 2; ` ` ` ` ``// Move forward the negative pointer till ` ` ``// positive number number not encountered ` ` ``while` `(negative < size && a[negative] <= 0) ` ` ``negative += 2; ` ` ` ` ``// Swap array elements to fix their position. ` ` ``if` `(positive < size && negative < size) { ` ` ``temp = a[positive]; ` ` ``a[positive] = a[negative]; ` ` ``a[negative] = temp; ` ` ``} ` ` ` ` ``// Break from the while loop when any ` ` ``// index exceeds the size of the array ` ` ``else` ` ``break``; ` ` ``} ` `} ` ` ` `// Driver Code ` `public` `static` `void` `Main(String []args) { ` ` ``int` `[]arr = {1, -3, 5, 6, -3, 6, 7, -4, 9, 10}; ` ` ``int` `n = arr.Length; ` ` ` ` ``rearrange(arr, n); ` ` ``for` `(``int` `i = 0; i < n; i++) ` ` ``Console.Write(arr[i] + ``\" \"``); ` `} ` `} ` ` ` `// This code is contributed by Nitin Mittal. `\n\n## PHP\n\n `= 0) ` ` ``\\$positive` `+= 2; ` ` ` ` ``/* Move forward the negative ` ` ``pointer till positive number ` ` ``number not encountered */` ` ``while` `(``\\$negative` `< ``\\$size` `&& ` ` ``\\$a``[``\\$negative``] <= 0) ` ` ``\\$negative` `+= 2; ` ` ` ` ``// Swap array elements to fix ` ` ``// their position. ` ` ``if` `(``\\$positive` `< ``\\$size` `&& ` ` ``\\$negative` `< ``\\$size``) ` ` ``{ ` ` ``\\$temp` `= ``\\$a``[``\\$positive``]; ` ` ``\\$a``[``\\$positive``] = ``\\$a``[``\\$negative``]; ` ` ``\\$a``[``\\$negative``] = ``\\$temp``; ` ` ``} ` ` ` ` ` ` ``/* Break from the while loop ` ` ``when any index exceeds the ` ` ``size of the array */` ` ``else` ` ``break``; ` ` ``} ` `} ` ` ` `// Driver code ` `\\$arr` `= ``array``( 1, -3, 5, 6, -3, ` ` ``6, 7, -4, 9, 10 ); ` `\\$n` `= sizeof(``\\$arr``); ` ` ` `rearrange(``\\$arr``, ``\\$n``); ` `for` `(``\\$i` `= 0; ``\\$i` `< ``\\$n``; ``\\$i``++) ` ` ``echo` `\\$arr``[``\\$i``] .``\" \"``; ` ` ` `// This code is contributed by ChitraNayal ` `?> `\n\nOutput:\n\n```1 -3 5 -3 6 6 7 -4 9 10\n```\n\nLets explain the working of the code on the first example\narr[] = {1, -3, 5, 6, -3, 6, 7, -4, 9, 10}\nWe declare two variables positive and negative positive points to zeroth position and negative points to first position\npositive = 0 negative = 1\nIn the first iteration positive will move 4 places to fifth position as a is less than zero and positive = 4.\nNegative will move 2 places and will point to fourth position as a>0\nwe will swap positive and negative position values as they are less than size of array.\nAfter first iteration the array becomes arr[] = {1, -3, 5, -3, 6, 6, 7, -4, 9, 10}\n\nNow positive points at fourth position and negative points at third position\nIn second iteration the positive value will move 6 places and its value will\nmore than the size of the array.\nThe negative pointer will move two steps forward and it will point to 5th position\nAs the positive pointer value becomes greater than the array size we will not perform any swap operation and break out of the while loop.\nThe final output will be\narr[] = {1, -3, 5, -3, 6, 6, 7, -4, 9, 10}\n\nAn example where relative order is not maintained:\n{ -1, -2, -3, -4, -5, 6, 7, 8 };\n\nAnother Approach :-\nThe idea is to find a positive/negative element which is in incorrect place(i.e. positive at odd and negative at even place) and the then find the element of opposite sign which is also in incorrect position in the remaining array and then swap these two elements.\nHere is the implementation of the above idea.\n\n## C++\n\n `// C++ program to rearrange positive ` `// and negative numbers ` `#include ` `using` `namespace` `std; ` ` ` `// Swap function ` `void` `swap(``int``* a, ``int` `i , ``int` `j) ` `{ ` ` ``int` `temp = a[i]; ` ` ``a[i] = a[j]; ` ` ``a[j] = temp; ` ` ``return` `; ` `} ` ` ` `// Print array function ` `void` `printArray(``int``* a, ``int` `n) ` `{ ` ` ``for``(``int` `i = 0; i < n; i++) ` ` ``cout << a[i] << ``\" \"``; ` ` ``cout << endl; ` ` ``return` `; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ``int` `arr[] = { 1, -3, 5, 6, -3, 6, 7, -4, 9, 10 }; ` ` ``int` `n = ``sizeof``(arr)/``sizeof``(arr); ` ` ` ` ``//before modification ` ` ``printArray(arr, n); ` ` ` ` ``for``(``int` `i = 0; i < n; i++) ` ` ``{ ` ` ``if``(arr[i] >= 0 && i % 2 == 1) ` ` ``{ ` ` ``// out of order positive element ` ` ``for``(``int` `j = i + 1; j < n; j++) ` ` ``{ ` ` ``if``(arr[j] < 0 && j % 2 == 0) ` ` ``{ ` ` ``// find out of order negative ` ` ``// element in remaining array ` ` ``swap(arr, i, j); ` ` ``break` `; ` ` ``} ` ` ``} ` ` ``} ` ` ``else` `if``(arr[i] < 0 && i % 2 == 0) ` ` ``{ ` ` ``// out of order negative element ` ` ``for``(``int` `j = i + 1; j < n; j++) ` ` ``{ ` ` ``if``(arr[j] >= 0 && j % 2 == 1) ` ` ``{ ` ` ``// find out of order positive ` ` ``// element in remaining array ` ` ``swap(arr, i, j); ` ` ``break``; ` ` ``} ` ` ``} ` ` ``} ` ` ``} ` ` ` ` ``//after modification ` ` ``printArray(arr, n); ` ` ``return` `0; ` `} ` ` ` `// This code is contributed by AnitAggarwal `\n\n## Java\n\n `// Java program to rearrange positive ` `// and negative numbers ` `import` `java.io.*; ` `import` `java.util.*; ` ` ` `class` `GFG ` `{ ` ` ` ` ``// Swap function ` ` ``static` `void` `swap(``int``[] a, ``int` `i, ``int` `j) ` ` ``{ ` ` ``int` `temp = a[i]; ` ` ``a[i] = a[j]; ` ` ``a[j] = temp; ` ` ``} ` ` ` ` ``// Print array function ` ` ``static` `void` `printArray(``int``[] a, ``int` `n) ` ` ``{ ` ` ``for` `(``int` `i = ``0``; i < n; i++) ` ` ``System.out.print(a[i] + ``\" \"``); ` ` ``System.out.println(); ` ` ``} ` ` ` ` ``// Driver code ` ` ``public` `static` `void` `main(String args[]) ` ` ``{ ` ` ``int``[] arr = { ``1``, -``3``, ``5``, ``6``, -``3``, ``6``, ``7``, -``4``, ``9``, ``10` `}; ` ` ``int` `n = arr.length; ` ` ` ` ``//before modification ` ` ``printArray(arr, n); ` ` ` ` ``for` `(``int` `i = ``0``; i < n; i++) ` ` ``{ ` ` ``if` `(arr[i] >= ``0` `&& i % ``2` `== ``1``) ` ` ``{ ` ` ``// out of order positive element ` ` ``for` `(``int` `j = i + ``1``; j < n; j++) ` ` ``{ ` ` ``if` `(arr[j] < ``0` `&& j % ``2` `== ``0``) ` ` ``{ ` ` ``// find out of order negative ` ` ``// element in remaining array ` ` ``swap(arr, i, j); ` ` ``break` `; ` ` ``} ` ` ``} ` ` ``} ` ` ` ` ``else` `if` `(arr[i] < ``0` `&& i % ``2` `== ``0``) ` ` ``{ ` ` ``// out of order negative element ` ` ``for` `(``int` `j = i + ``1``; j < n; j++) ` ` ``{ ` ` ``if` `(arr[j] >= ``0` `&& j % ``2` `== ``1``) ` ` ``{ ` ` ``// find out of order positive ` ` ``// element in remaining array ` ` ``swap(arr, i, j); ` ` ``break``; ` ` ``} ` ` ``} ` ` ``} ` ` ``} ` ` ` ` ``//after modification ` ` ``printArray(arr, n); ` ` ``} ` `} ` ` ` `// This code is contributed by rachana soma `\n\n## Python3\n\n `# Python3 program to rearrange positive ` `# and negative numbers ` ` ` `# Print array function ` `def` `printArray(a, n): ` ` ``for` `i ``in` `a: ` ` ``print``(i, end ``=` `\" \"``) ` ` ``print``() ` ` ` ` ` `# Driver code ` `arr ``=` `[``1``, ``-``3``, ``5``, ``6``, ``-``3``, ``6``, ``7``, ``-``4``, ``9``, ``10``] ` `n ``=` `len``(arr) ` ` ` `# before modification ` `printArray(arr, n) ` ` ` `for` `i ``in` `range``(n): ` ` ` ` ``if``(arr[i] >``=` `0` `and` `i ``%` `2` `=``=` `1``): ` ` ` ` ``# out of order positive element ` ` ``for` `j ``in` `range``(i ``+` `1``, n): ` ` ` ` ``if``(arr[j] < ``0` `and` `j ``%` `2` `=``=` `0``): ` ` ` ` ``# find out of order negative ` ` ``# element in remaining array ` ` ``arr[i], arr[j] ``=` `arr[j], arr[i] ` ` ``break` ` ` ` ``elif` `(arr[i] < ``0` `and` `i ``%` `2` `=``=` `0``): ` ` ` ` ``# out of order negative element ` ` ``for` `j ``in` `range``(i ``+` `1``, n): ` ` ` ` ``if``(arr[j] >``=` `0` `and` `j ``%` `2` `=``=` `1``): ` ` ` ` ``# find out of order positive ` ` ``# element in remaining array ` ` ``arr[i], arr[j] ``=` `arr[j], arr[i] ` ` ``break` ` ` `# after modification ` `printArray(arr, n); ` ` ` `# This code is contributed ` `# by mohit kumar `\n\n## C#\n\n// C# program to rearrange positive\n// and negative numbers\nusing System;\n\nclass GFG\n{\n\n// Swap function\nstatic void swap(int[] a, int i, int j)\n{\nint temp = a[i];\na[i] = a[j];\na[j] = temp;\n}\n\n// Print array function\nstatic void printArray(int[] a, int n)\n{\nfor (int i = 0; i < n; i++) Console.Write(a[i] + \" \"); Console.WriteLine(); } // Driver code public static void Main() { int[] arr = { 1, -3, 5, 6, -3, 6, 7, -4, 9, 10 }; int n = arr.Length; //before modification printArray(arr, n); for (int i = 0; i < n; i++) { if (arr[i] >= 0 && i % 2 == 1)\n{\n// out of order positive element\nfor (int j = i + 1; j < n; j++) { if (arr[j] < 0 && j % 2 == 0) { // find out of order negative // element in remaining array swap(arr, i, j); break ; } } } else if (arr[i] < 0 && i % 2 == 0) { // out of order negative element for (int j = i + 1; j < n; j++) { if (arr[j] >= 0 && j % 2 == 1)\n{\n// find out of order positive\n// element in remaining array\nswap(arr, i, j);\nbreak;\n}\n}\n}\n}\n\n// after modification\nprintArray(arr, n);\n}\n}\n\n// This code is contributed by Akanksha Rai\n\nOutput:\n\n```1 -3 5 6 -3 6 7 -4 9 10\n1 -3 5 -3 6 6 7 -4 9 10\n```\n\nPlease write comments if you find anything incorrect, or you want to share more information about the topic discussed above.\n\nThis article is attributed to GeeksforGeeks.org\n\n## tags:\n\nArrays array-rearrange Quick Sort Arrays\n\ncode\n\nload comments"
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"# 8.1 Confidence interval, single population mean, population standard (Page 2/5)\n\n Page 2 / 5\n\n## In summary, as a result of the central limit theorem:\n\n• $\\overline{X}$ is normally distributed, that is, $\\overline{X}$ ~ $N\\left({\\mu }_{X},\\frac{\\sigma }{\\sqrt{n}}\\right).\\phantom{\\rule{35pt}{0ex}}$\n• When the population standard deviation $\\sigma$ is known, we use a Normal distribution to calculate the error bound.\n\n## Calculating the confidence interval:\n\nTo construct a confidence interval estimate for an unknown population mean, we need data from a random sample. The steps to construct and interpret the confidence interval are:\n\n• Calculate the sample mean $\\overline{x}$ from the sample data. Remember, in this section, we already know the population standard deviation $\\sigma$ .\n• Find the Z-score that corresponds to the confidence level.\n• Calculate the error bound EBM\n• Construct the confidence interval\n• Write a sentence that interprets the estimate in the context of the situation in the problem. (Explain what the confidence interval means, in the words of the problem.)\n\nWe will first examine each step in more detail, and then illustrate the process with some examples.\n\n## Finding z for the stated confidence level\n\nWhen we know the population standard deviation σ, we use a standard normal distribution to calculate the error bound EBM and construct the confidence interval. We need to find the value of z that puts an area equal to the confidence level (in decimal form) in the middle of the standard normal distribution Z~N(0,1).\n\nThe confidence level, $\\mathrm{CL}$ , is the area in the middle of the standard normal distribution. $\\mathrm{CL}=1-\\alpha$ . So $\\alpha$ is the area that is split equally between the two tails. Each of the tails contains an area equal to $\\frac{\\alpha }{2}$ .\n\nThe z-score that has an area to the right of $\\frac{\\alpha }{2}$ is denoted by ${z}_{\\frac{\\alpha }{2}}$\n\nFor example, when $\\mathrm{CL}=0.95$ then $\\alpha =0.05$ and $\\frac{\\alpha }{2}=0.025$ ; we write ${z}_{\\frac{\\alpha }{2}}={z}_{.025}$\n\nThe area to the right of ${z}_{.025}$ is 0.025 and the area to the left of ${z}_{.025}$ is 1-0.025 = 0.975\n\n${z}_{\\frac{\\alpha }{2}}={z}_{0.025}=1.96$ , using a calculator, computer or a Standard Normal probability table.\n\nUsing the TI83, TI83+ or TI84+ calculator: invNorm $\\left(0.975,0,1\\right)=1.96$\n\nCALCULATOR NOTE: Remember to use area to the LEFT of ${z}_{\\frac{\\alpha }{2}}$ ; in this chapter the last two inputs in the invNorm command are 0,1 because you are using a Standard Normal Distribution Z~N(0,1)\n\n## Ebm: error bound\n\nThe error bound formula for an unknown population mean $\\mu$ when the population standard deviation $\\sigma$ is known is\n\n• $\\mathrm{EBM}={z}_{\\frac{\\alpha }{2}}\\cdot \\frac{\\sigma }{\\sqrt{n}}$\n\n## Constructing the confidence interval\n\n• The confidence interval estimate has the format $\\left(\\overline{x}-\\mathrm{EBM},\\overline{x}+\\mathrm{EBM}\\right)$ .\n\nThe graph gives a picture of the entire situation.\n\n$\\mathrm{CL}+\\frac{\\alpha }{2}+\\frac{\\alpha }{2}=\\mathrm{CL}+\\alpha =1$ .",
null,
"## Writing the interpretation\n\nThe interpretation should clearly state the confidence level (CL), explain what population parameter is being estimated (here, a population mean ), and should state the confidence interval (both endpoints). \"We estimate with ___% confidence that the true population mean (include context of the problem) is between ___ and ___ (include appropriate units).\"\n\nSuppose scores on exams in statistics are normally distributed with an unknown population mean and a population standard deviation of 3 points. A random sampleof 36 scores is taken and gives a sample mean (sample mean score) of 68. Find a confidence interval estimate for the population mean exam score (the mean score on all exams).\n\nFind a 90% confidence interval for the true (population) mean of statistics exam scores.\n\n• You can use technology to directly calculate the confidence interval\n• The first solution is shown step-by-step (Solution A).\n• The second solution uses the TI-83, 83+ and 84+ calculators (Solution B).\n\n## Solution a\n\nTo find the confidence interval, you need the sample mean, $\\overline{x}$ , and the EBM.\n\n• $\\overline{x}=68$\n• $\\mathrm{EBM}={z}_{\\frac{\\alpha }{2}}\\cdot \\left(\\frac{\\sigma }{\\sqrt{n}}\\right)$\n• $\\sigma =3$ ; $n=36$ ; The confidence level is 90% (CL=0.90)\n\n$\\mathrm{CL = 0.90}$ so $\\alpha =1-\\mathrm{CL}=1-0.90=0.10$\n\n$\\frac{\\alpha }{2}=0.05\\phantom{\\rule{20pt}{0ex}}{z}_{\\frac{\\alpha }{2}}={z}_{.05}$\n\nThe area to the right of ${z}_{.05}$ is 0.05 and the area to the left of ${z}_{.05}$ is 1−0.05=0.95\n\n${z}_{\\frac{\\alpha }{2}}={z}_{.05}=1.645$\n\nusing invNorm(0.95,0,1) on the TI-83,83+,84+ calculators. This can also be found using appropriate commands on other calculators, using a computer, or using a probability table for the Standard Normal distribution.\n\n$\\mathrm{EBM}=1.645\\cdot \\left(\\frac{3}{\\sqrt{36}}\\right)=0.8225$\n\n$\\overline{x}-\\mathrm{EBM}=68-0.8225=67.1775$\n\n$\\overline{x}+\\mathrm{EBM}=68+0.8225=68.8225$\n\nThe 90% confidence interval is (67.1775, 68.8225).\n\n## Solution b\n\nUsing a function of the TI-83, TI-83+ or TI-84 calculators:\n\nPress STAT and arrow over to TESTS .\nArrow down to 7:ZInterval .\nPress ENTER .\nArrow to Stats and press ENTER .\nArrow down and enter 3 for $\\sigma$ , 68 for $\\overline{x}$ , 36 for $n$ , and .90 for C-level .\nArrow down to Calculate and press ENTER .\nThe confidence interval is (to 3 decimal places) (67.178, 68.822).\n\n## Interpretation\n\nWe estimate with 90% confidence that the true population mean exam score for all statistics students is between 67.18 and 68.82.\n\n## Explanation of 90% confidence level\n\n90% of all confidence intervals constructed in this way contain the true mean statistics exam score. For example, if we constructed 100 of these confidence intervals, we would expect 90 of them to contain the true population mean exam score.\n\n## Changing the confidence level\n\nSuppose we change the original problem by using a 95% confidence level. Find a 95% confidence interval for the true (population) mean statistics exam score.\n\nTo find the confidence interval, you need the sample mean, $\\overline{x}$ , and the EBM.\n\n• $\\overline{x}=68$\n• $\\mathrm{EBM}={z}_{\\frac{\\alpha }{2}}\\cdot \\left(\\frac{\\sigma }{\\sqrt{n}}\\right)$\n• $\\sigma =3$ ; $n=36$ ; The confidence level is 95% (CL=0.95)\n\n$\\mathrm{CL = 0.95}$ so $\\alpha =1-\\mathrm{CL}=1-0.95=0.05$\n\n$\\frac{\\alpha }{2}=0.025\\phantom{\\rule{20pt}{0ex}}{z}_{\\frac{\\alpha }{2}}={z}_{.025}$\n\nThe area to the right of ${z}_{.025}$ is 0.025 and the area to the left of ${z}_{.025}$ is 1−0.025=0.975\n\n${z}_{\\frac{\\alpha }{2}}={z}_{.025}=1.96$\n\nusing invnorm(.975,0,1) on the TI-83,83+,84+ calculators. (This can also be found using appropriate commands on other calculators, using a computer, or using a probability table for the Standard Normal distribution.)\n\n$\\mathrm{EBM}=1.96\\cdot \\left(\\frac{3}{\\sqrt{36}}\\right)=0.98$\n\n$\\overline{x}-\\mathrm{EBM}=68-0.98=67.02$\n\n$\\overline{x}+\\mathrm{EBM}=68+0.98=68.98$\n\n## Interpretation\n\nWe estimate with 95 % confidence that the true population mean for all statistics exam scores is between 67.02 and 68.98.\n\n## Explanation of 95% confidence level\n\n95% of all confidence intervals constructed in this way contain the true value of the population meanstatistics exam score.\n\n## Comparing the results\n\nThe 90% confidence interval is (67.18, 68.82). The 95% confidence interval is (67.02, 68.98). The 95% confidence interval is wider. If you look at the graphs, because the area 0.95 is larger than the area 0.90, it makes sense that the 95% confidence interval is wider.\n\n## Summary: effect of changing the confidence level\n\n• Increasing the confidence level increases the error bound, making the confidence interval wider.\n• Decreasing the confidence level decreases the error bound, making the confidence interval narrower.\n\n## Changing the sample size:\n\nSuppose we change the original problem to see what happens to the error bound if the sample size is changed.\n\nLeave everything the same except the sample size. Use the original 90% confidence level. What happens to the error bound and the confidence interval if we increase the sample size and use n=100 instead of n=36? What happens if we decrease the sample size to n=25 instead of n=36?\n\n• $\\overline{x}=68$\n• $\\mathrm{EBM}={z}_{\\frac{\\alpha }{2}}\\cdot \\left(\\frac{\\sigma }{\\sqrt{n}}\\right)$\n• $\\sigma =3$ ; The confidence level is 90% (CL=0.90) ; ${z}_{\\frac{\\alpha }{2}}={z}_{.05}=1.645$\n\nIf we decrease the sample size $n$ to 25, we increase the error bound.\n\nWhen $n=25$ : $\\mathrm{EBM}={z}_{\\frac{\\alpha }{2}}\\cdot \\left(\\frac{\\sigma }{\\sqrt{n}}\\right)=1.645\\cdot \\left(\\frac{3}{\\sqrt{25}}\\right)=0.987$\n\n## Summary: effect of changing the sample size\n\n• Increasing the sample size causes the error bound to decrease, making the confidence interval narrower.\n• Decreasing the sample size causes the error bound to increase, making the confidence interval wider.\n\n## Working bacwards to find the error bound or the sample mean\n\nWhen we calculate a confidence interval, we find the sample mean and calculate the error bound and use them to calculate the confidence interval. But sometimes when we read statistical studies, the study may state the confidence interval only. If we know the confidence interval, we can work backwards to find both the error bound and the sample mean.\n\n## Finding the error bound\n\n• From the upper value for the interval, subtract the sample mean\n• OR, From the upper value for the interval, subtract the lower value. Then divide the difference by 2.\n\n## Finding the sample mean\n\n• Subtract the error bound from the upper value of the confidence interval\n• OR, Average the upper and lower endpoints of the confidence interval\n\nNotice that there are two methods to perform each calculation. You can choose the method that is easier to use with the information you know.\n\nSuppose we know that a confidence interval is (67.18, 68.82) and we want to find the error bound. We may know that the sample mean is 68. Or perhaps our source only gave the confidence interval and did not tell us the value of the the sample mean.\n\n## Calculate the error bound:\n\n• If we know that the sample mean is 68: $\\mathrm{EBM}=68.82-68=0.82\\phantom{\\rule{5pt}{0ex}}$\n• If we don't know the sample mean: $\\phantom{\\rule{5pt}{0ex}}\\mathrm{EBM}=\\frac{\\left(68.82-67.18\\right)}{2}=0.82$\n\n## Calculate the sample mean:\n\n• If we know the error bound: $\\overline{x}=68.82-0.82=68\\phantom{\\rule{5pt}{0ex}}$\n• If we don't know the error bound: $\\overline{x}=\\phantom{\\rule{5pt}{0ex}}\\frac{\\left(67.18+68.82\\right)}{2}=68$\n\n## Calculating the sample size n\n\nIf researchers desire a specific margin of error, then they can use the error bound formula to calculate the required sample size.\n\nThe error bound formula for a population mean when the population standard deviation is known is $\\mathrm{EBM}={z}_{\\frac{\\alpha }{2}}\\cdot \\left(\\frac{\\sigma }{\\sqrt{n}}\\right)$\n\nThe formula for sample size is $n=\\frac{z^{2}\\sigma ^{2}}{\\mathrm{EBM}^{2}}$ , found by solving the error bound formula for $n$\n\nIn this formula, $z$ is ${z}_{\\frac{\\alpha }{2}}$ , corresponding to the desired confidence level. A researcher planning a study who wants a specified confidence level and error bound can use this formula to calculate the size of the sample needed for the study.\n\nThe population standard deviation for the age of Foothill College students is 15 years. If we want to be 95% confident that the sample mean age is within 2 years of the true population mean age of Foothill College students , how many randomly selected Foothill College students must be surveyed?\n\n• From the problem, we know that $\\sigma =15$ and EBM=2\n• $z={z}_{.025}=1.96$ , because the confidence level is 95%.\n• $n=\\frac{z^{2}\\sigma ^{2}}{\\mathrm{EBM}^{2}}$ = $\\frac{1.96^{2}15^{2}}{2^{2}}$ =216.09 using the sample size equation.\n• Use $n$ = 217: Always round the answer UP to the next higher integer to ensure that the sample size is large enough.\n\nTherefore, 217 Foothill College students should be surveyed in order to be 95% confident that we are within 2 years of the true population mean age of Foothill College students.\n\n**With contributions from Roberta Bloom\n\n#### Questions & Answers\n\nIs there any normative that regulates the use of silver nanoparticles?\nDamian Reply\nwhat king of growth are you checking .?\nRenato\nWhat fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?\nStoney Reply\nwhy we need to study biomolecules, molecular biology in nanotechnology?\nAdin Reply\n?\nKyle\nyes I'm doing my masters in nanotechnology, we are being studying all these domains as well..\nAdin\nwhy?\nAdin\nwhat school?\nKyle\nbiomolecules are e building blocks of every organics and inorganic materials.\nJoe\nanyone know any internet site where one can find nanotechnology papers?\nDamian Reply\nresearch.net\nkanaga\nsciencedirect big data base\nErnesto\nIntroduction about quantum dots in nanotechnology\nPraveena Reply\nwhat does nano mean?\nAnassong Reply\nnano basically means 10^(-9). nanometer is a unit to measure length.\nBharti\ndo you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?\nDamian Reply\nabsolutely yes\nDaniel\nhow to know photocatalytic properties of tio2 nanoparticles...what to do now\nAkash Reply\nit is a goid question and i want to know the answer as well\nMaciej\ncharacteristics of micro business\nAbigail\nfor teaching engĺish at school how nano technology help us\nAnassong\nDo somebody tell me a best nano engineering book for beginners?\ns. Reply\nthere is no specific books for beginners but there is book called principle of nanotechnology\nNANO\nwhat is fullerene does it is used to make bukky balls\nDevang Reply\nare you nano engineer ?\ns.\nfullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.\nTarell\nwhat is the actual application of fullerenes nowadays?\nDamian\nThat is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.\nTarell\nwhat is the Synthesis, properties,and applications of carbon nano chemistry\nAbhijith Reply\nMostly, they use nano carbon for electronics and for materials to be strengthened.\nVirgil\nis Bucky paper clear?\nCYNTHIA\ncarbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc\nNANO\nso some one know about replacing silicon atom with phosphorous in semiconductors device?\ns. Reply\nYeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.\nHarper\nDo you know which machine is used to that process?\ns.\nhow to fabricate graphene ink ?\nSUYASH Reply\nfor screen printed electrodes ?\nSUYASH\nWhat is lattice structure?\ns. Reply\nof graphene you mean?\nEbrahim\nor in general\nEbrahim\nin general\ns.\nGraphene has a hexagonal structure\ntahir\nOn having this app for quite a bit time, Haven't realised there's a chat room in it.\nCied\nwhat is biological synthesis of nanoparticles\nSanket Reply\n1 It is estimated that 30% of all drivers have some kind of medical aid in South Africa. What is the probability that in a sample of 10 drivers: 3.1.1 Exactly 4 will have a medical aid. (8) 3.1.2 At least 2 will have a medical aid. (8) 3.1.3 More than 9 will have a medical aid.\nNerisha Reply\n\n### Read also:\n\n#### Get the best Algebra and trigonometry course in your pocket!\n\nSource: OpenStax, Collaborative statistics. OpenStax CNX. Jul 03, 2012 Download for free at http://cnx.org/content/col10522/1.40\nGoogle Play and the Google Play logo are trademarks of Google Inc.\n\nNotification Switch\n\nWould you like to follow the 'Collaborative statistics' conversation and receive update notifications?",
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"https://www.jobilize.com/ocw/mirror/col10522_1.40_complete/m16962/conf_intvls_cfpm2.png",
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null,
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http://coloradocollege.libguides.com/c.php?g=286844&p=1911088
|
[
"# EC Thesis: Class Notes\n\nExamination of methods of analysis commonly used in economics and business. Emphasis on non-experimental and quasi-experimental designs necessitating the use of models. and large sample methods, case studies, surveys, regression and forecasting.\n\n## Hw do you know?\n\n### How do you know something?\n\n• tenacity\n• authority\n• intution\n• science\n\nScientific Method - observe, develop a theory to explain, test - use new data to cross-check\n\n## Key terms\n\n### Variables\n\n1. Random (stochastic, non-deterministic) - value not known at start\n2. Non-random (non-stochastic, deterministic) - value is known\n\nProbability - the chance that a random variable will take a certain value\n- (distribution must add to 100%)\nProbalilty Distribution - set of all possible values of a random variables with the probabilities for each\n\nDependent, independent, discrete, continuous and limited in a range or not\n\n### Distribution\n\nContinuous distribution (points on a continuous scale)\nvs.\nDiscrete distribution\n(clumps, not all values are possible)\n\nSubjective Distribution - have someidea of probalities of outcomes\nObjective Distribution - based on # of times an outcome occurs divided by the total number of outcomes\n\nCorrelation- when two+ variables show a systematic patrern of movement\nCausation - when 1 variable actually causes the other variable to change\n\n* Correlation DOES NOT imply causation.\n* Canusaion implies correlation.\n\nRegression vs. Causaiton - a significant sign on a regression coefficient DOES NOT imply causation.\n\n## Manipulation\n\n### 3 Types of Data\n\n1. Time Series Data - same set of variables in different time periods\n2. Cross Sectional Data - same set of variables at a point of time over different cross-sections\n3. Pooled Data - cross-section dataset over different time periods\n\n### Some Sample Statistics\n\nMean - add all, divide by total #\n\nMedian - sum average/middle value of mean and mode\n\nMode - max probality, most frequent value of variable\n\nStandard deviation - average deviation from the mean\n\nCV (coefficient of variation) - in relation tp mean it ttells how signigicant/how much variance)\n\nMeasure of Associaion - relationship between variables\n\n1. positive, negative, none\n\n2. covariance (cov) and correlation (p) - makes relative to variance\n*Sometimes it is heapful to figure the prooduct of deviations and the sum of squared deviations\n\nRegression - process of finding line or curve that \"best\" fits a given set of data points (F. Galton, 1886)\n\nOLS linear regression - smalles sumof sqaure error for all points from the line/curve\n\n- good for estimating what comes next\n- good at average\n\nMultivariate - plane of best fit (rather than a line)\n\n## Models\n\n1. structural or reduced forms\n2. robustness or specification tests\n3. intuition or hedonics"
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https://www.fractioncalculator.pro/decimal-to-fraction/Change_1.454545_to-a-fraction
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[
"# Change 1.454545 to a fraction\n\nWelcome! Here is the answer to the question: Change 1.454545 to a fraction or what is 1.454545 as a fraction. Use the decimal to fraction converter/calculator below to write any decimal number as a fraction.\n\n### Decimal to Fraction Converter\n\n Enter a decimal value: Ex.: 0.625, 0.75, .875, etc. Equivalent fraction: ... Decimal to fraction Explained: Equivalent fraction explained here"
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[
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https://www.allegro.cc/manual/5/transformations.html
|
[
"# Transformations\n\nThese functions are declared in the main Allegro header file:\n\n``#include <allegro5/allegro.h>``\n\nThe transformations are combined in the order of the function invocations. Thus to create a transformation that first rotates a point and then translates it, you would (starting with an identity transformation) call al_rotate_transform and then al_translate_transform. This approach is opposite of what OpenGL uses but similar to what Direct3D uses.\n\nFor those who known the matrix algebra going behind the scenes, what the transformation functions in Allegro do is \"pre-multiply\" the successive transformations. So, for example, if you have code that does:\n\n``````al_identity_transform(&T);\n\nal_compose_transform(&T, &T1);\nal_compose_transform(&T, &T2);\nal_compose_transform(&T, &T3);\nal_compose_transform(&T, &T4);``````\n\nThe resultant matrix multiplication expression will look like this:\n\n``T4 * T3 * T2 * T1``\n\nSince the point coordinate vector term will go on the right of that sequency of factors, the transformation that is called first, will also be applied first.\n\nThis means if you have code like this:\n\n``````al_identity_transform(&T1);\nal_scale_transform(&T1, 2, 2);\nal_identity_transform(&T2);\nal_translate_transform(&T2, 100, 0);\n\nal_identity_transform(&T);\n\nal_compose_transform(&T, &T1);\nal_compose_transform(&T, &T2);\n\nal_use_transform(T);``````\n\nit does exactly the same as:\n\n``````al_identity_transform(&T);\nal_scale_transform(&T, 2, 2);\nal_translate_transform(&T, 100, 0);\nal_use_transform(T);``````"
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https://textbooks.cs.ksu.edu/cc310/3-programming-by-contract-and-introduction-to-performance/8-max-and-min-linear/index.html
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[
"# Max and Min - Linear\n\nLet’s start by considering one number from the list at a time.\n\n## Base Case\n\nWhen we have received just one number, this number is both the maximum and the minimum. In the initial state, you have `max` holding the maximum, and `min` holding the minimum. The invariant is the following:\n\n1. The variable `max` holds the maximum of all the numbers considered so far\n2. The variable `min` holds the minimum of all the numbers considered so far\n\nThe algorithm is depicted by the following flowchart and pseudocode:\n\n``````print \"Enter a Number:\"\ninput X\nMAX = X\nMIN = X``````\n\n## The Next Number\n\nThen, our program will enter a loop to read 10 more numbers from the user. So, we’ll need to perform the following process during each iteration of the loop:\n\n1. compare the value in `max` with this new number and update the `max` value if the new number is greater. In this way, the invariant is preserved.\n2. compare the value in `min` with this new number and update the `min` value if the new number is smaller. In this way, the invariant is preserved.\n\nThis part of the program is depicted by the following flowchart and pseudocode:\n\n``````loop I from 1 to 10\nprint \"Enter a Number:\"\ninput X\nif X > MAX\nMAX = X\nend if\nif X < MIN\nMIN = X\nend if\nend loop``````\n\nAfter you’ve considered the second number, you end up in the same situation at the beginning: you have a maximum and a minimum value of the numbers input by the user so far. You have found an invariant if you verify that the preconditions before executing an iteration of the loops are the same as the conditions at the end of the loop, known as postconditions:\n\n• Loop preconditions:\n1. The variable `max` holds the maximum value among all the numbers considered so far\n2. The variable `min` holds the minimum value among all the numbers considered so far\n3. The new input considered is a number\n• Loop postconditions:\n1. The variable `max` holds the maximum value among all the numbers considered so far\n2. The variable `min` holds the minimum value among all the numbers considered so far\n• Loop invariant:\n1. The variable `max` holds the maximum value among all the numbers considered so far\n2. The variable `min` holds the minimum value among all the numbers considered so far\n\nYou can then generalize the solution for the `n`th input: when you consider the `n`th number, compare it with the values in `max` and `min`, updating them if necessary. In each step, we can show that the invariant holds.\n\nA full flowchart of this program can be found by clicking the following link:\n\nIt is helpful to have this diagram available in a second browser tab for review on the next few pages."
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{"ft_lang_label":"__label__en","ft_lang_prob":0.8663965,"math_prob":0.9859005,"size":2607,"snap":"2023-40-2023-50","text_gpt3_token_len":592,"char_repetition_ratio":0.17979255,"word_repetition_ratio":0.2557652,"special_character_ratio":0.21212122,"punctuation_ratio":0.07444668,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9954178,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-03T23:36:41Z\",\"WARC-Record-ID\":\"<urn:uuid:2dfa5111-aa0d-46cd-b72b-5c5a2961cc18>\",\"Content-Length\":\"78692\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cec0f708-86ce-4e39-b785-d6d8673ef0e5>\",\"WARC-Concurrent-To\":\"<urn:uuid:71ee0733-4d41-469d-850e-0bc74fb0bc9b>\",\"WARC-IP-Address\":\"129.130.10.46\",\"WARC-Target-URI\":\"https://textbooks.cs.ksu.edu/cc310/3-programming-by-contract-and-introduction-to-performance/8-max-and-min-linear/index.html\",\"WARC-Payload-Digest\":\"sha1:IJGVDNK6ZQ3BDJJZEREPNUQA5GI5O7HN\",\"WARC-Block-Digest\":\"sha1:C3QH2RWTU74Z4QIT6QK34OHNRLPD5SCG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100518.73_warc_CC-MAIN-20231203225036-20231204015036-00763.warc.gz\"}"}
|
https://www.thesoleline.com/online/adidas-yeezy-boost-350-v2-toddler-glow-in-the-dark/
|
[
"Sale!\n\n# adidas Yeezy Boost 350 V2 Toddler “Glow in the Dark”\n\n\\$120.00 \\$85.00\n\n Toddler Size Choose an optionUS1.5K=UK1=FR33=20CMUS10K=UK9.5=FR27=16CMUS11K=UK10.5=FR28.5=17CMUS12K=UK11.5=FR30=17.5CMUS13K=UK12.5=FR31=18.5CMUS1K=UK13.5=FR32=19CMUS2.5K=UK2=FR34=21CMUS2K=UK1.5=FR33.5=20.5CMUS3K=UK2.5=FR35=21CMUS5K=UK4=FR20=11.5CMUS6K=UK5.5=FR22=12.5CMUS7K=UK6.5=FR23.5=13.5CMUS8K=UK7.5=FR25=14.5CMUS9K=UK8.5=FR26=15CMClear\nSKU: N/A Category: Tag:\n\n## Description\n\nShared via Twitter, Ye wrote “Can’t wait for these glow in the dark 350s,” – along with a set of different emjois. One pair comes in Fluorescent Green and the other in Orange. Both are constructed with the updated translucent build, which will debut early 2019."
] |
[
null
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{"ft_lang_label":"__label__en","ft_lang_prob":0.93494827,"math_prob":0.9614287,"size":307,"snap":"2021-43-2021-49","text_gpt3_token_len":72,"char_repetition_ratio":0.04950495,"word_repetition_ratio":0.0,"special_character_ratio":0.22801302,"punctuation_ratio":0.11666667,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9554813,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-22T10:00:26Z\",\"WARC-Record-ID\":\"<urn:uuid:ba6461c8-736d-4e67-8704-cdfa01eb1c84>\",\"Content-Length\":\"151145\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cf2d8a47-b7a8-4386-8704-093f43624458>\",\"WARC-Concurrent-To\":\"<urn:uuid:08eb4819-6041-459f-b61f-e9bfae101ef7>\",\"WARC-IP-Address\":\"204.12.223.202\",\"WARC-Target-URI\":\"https://www.thesoleline.com/online/adidas-yeezy-boost-350-v2-toddler-glow-in-the-dark/\",\"WARC-Payload-Digest\":\"sha1:XJVQKAXBRJCW4DAREOCVIZEQ55RQLKWO\",\"WARC-Block-Digest\":\"sha1:MTPBXPUCY4O7GUA6OJMNSKC5KNGRP6LX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585504.90_warc_CC-MAIN-20211022084005-20211022114005-00429.warc.gz\"}"}
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https://www.thestudypool.com/63799/channel-coding-scheme/
|
[
"# Channel coding scheme\n\n1. What does an ideal channel coding scheme achieve?\n\n2. Give an example of a block error code used with character data.\n\n3. Compute the Hamming distance for the following pairs: ( 0000, 0001 ), ( 0101, 0001 ), ( 1111, 1001 ), and ( 0001, 1110 ).\n\n4. Define the concept of Hamming distance.\n\n1. Explain the concept of code rate. Is a high code rate or low code rate desirable?\n\n2. How does one compute the minimum number of bit changes that can transform a valid codeword into another valid codeword?\n\n3. What can a RAC scheme achieve that a single parity bit scheme cannot?\n\n4. Generate a RAC parity matrix for a ( 20, 12 ) coding of the dataword 100011011111\n\n## Looking for help with your homework? Grab a 30% Discount and Get your paper done!",
null,
"30% OFF",
null,
"Turnitin Report",
null,
"Formatting",
null,
"Title Page",
null,
"Citation\nPlace an Order\n\nGrab A 14% Discount on This Paper\nPages (550 words)\nApproximate price: -"
] |
[
null,
"https://www.thestudypool.com/wp-content/plugins/cta-widgets/icons/check-pro.svg",
null,
"https://www.thestudypool.com/wp-content/plugins/cta-widgets/icons/check-pro.svg",
null,
"https://www.thestudypool.com/wp-content/plugins/cta-widgets/icons/check-pro.svg",
null,
"https://www.thestudypool.com/wp-content/plugins/cta-widgets/icons/check-pro.svg",
null,
"https://www.thestudypool.com/wp-content/plugins/cta-widgets/icons/check-pro.svg",
null
] |
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|
http://what-when-how.com/Tutorial/topic-1673k20/Foundations-of-Data-Exchange-114.html
|
[
"Database Reference\nIn-Depth Information\nRecall from Theorem 7.4 that computing certain answers to conjunctive queries with\ninequalities is an intractable problem. But Example 8.3 above showed that for this class\nof queries, the usual certain answers need not coincide with the universal certain answers.\nThis suggests that it may still be possible to compute universal certain answers for conjunc-\ntive queries with inequalities in polynomial time. We will actually state a much stronger\nresult in Theorem 8.5 : under the universal certain answers semantics, we have tractabil-\nity of query answering for a big and practically relevant class of FO queries that extends\nunions of conjunctive queries.\nRoughly, an existential FO query is given by an FO formula of the form\ny\nϕ\n( x , y ),where\nϕ\n( x , y ) is a quantifier-free formula in disjunctive normal form satisfying a safety condition.\nMore precisely, it is given by a formula of the form\ny\ni\nj α ij ,\nψ\n( x )=\nwhere\neach\nα ij is either an atomic formula, or the negation of an atomic formula;\nthe free variables of each\nα ij come from x , y ;and\n(safety condition) if\nα ij is a negated atom, i.e., of the form\n¬\nR ( z ), then each variable in\nz must occur in some formula\nα ij which is a nonnegated atom.\nFor example, both unions of conjunctive queries, and unions of conjunctive queries with\ninequalities and negated atoms satisfy these three conditions.\nExistential formulae ψ are known to be monotone in the following sense. If I is an\ninduced sub-instance of I and t belongs to the evaluation of\nover I ,then t belongs to the\nψ\nover I . Recall that an induced sub-instance I of an instance I is given by a\nevaluation of\nψ\nset A\nD OM ( I ); it contains exactly all the tuples of I whose elements come from A .\nSince every universal solution contains a copy of the core as an induced sub-instance,\nwe conclude that the problem of computing the universal certain answers for an existential\nFO query\nover the core of the universal solutions, discarding\nall tuples that contain nulls. Thus, in order to compute the universal certain answers of an\nexistential Q with respect to a source instance S , one can use the following procedure.\nψ\nboils down to evaluating\nψ\nRecall from Theorem 6.18 that the core of the universal solutions can be computed in\npolynomial time for every mapping whose sets of target dependencies consists of a set of\negds and a weakly acyclic set of tgds. Furthermore, existential queries are FO queries and\nthus have polynomial time data complexity. Putting these two things together we obtain\nthe following positive result.\nTheorem 8.5\nΣ t consists of\na set of egds and a weakly acyclic set of tgds, and let Q be an existential query over R t .\nThen the algorithm C OMPUTE U NIVERSAL C ERTAIN A NSWERS (Q,\nLet\nM\n=(R s , R t ,\nΣ st ,\nΣ t ) be a relational mapping such that\nM\n) correctly computes\nuniversal certain answers of Q with respect to a source S under\nM\nin polynomial time.\nSearch WWH ::\n\nCustom Search"
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.85630167,"math_prob":0.9166342,"size":2999,"snap":"2021-43-2021-49","text_gpt3_token_len":834,"char_repetition_ratio":0.12053423,"word_repetition_ratio":0.04339964,"special_character_ratio":0.21707235,"punctuation_ratio":0.09443507,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99728113,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-24T12:24:59Z\",\"WARC-Record-ID\":\"<urn:uuid:e3a9fcf0-1d90-4953-8517-297e68cba5b1>\",\"Content-Length\":\"29505\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f83d350f-4544-4397-8632-61647301cddb>\",\"WARC-Concurrent-To\":\"<urn:uuid:32ad2af5-6940-4e07-9898-6c66b7e64933>\",\"WARC-IP-Address\":\"192.163.238.55\",\"WARC-Target-URI\":\"http://what-when-how.com/Tutorial/topic-1673k20/Foundations-of-Data-Exchange-114.html\",\"WARC-Payload-Digest\":\"sha1:D72AP67GIIHGBGGBEKDUV3HB2OH2X77N\",\"WARC-Block-Digest\":\"sha1:GQ22HG7MRANN2VZ5EZ2ZZGVC4YIX3SGJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585997.77_warc_CC-MAIN-20211024111905-20211024141905-00534.warc.gz\"}"}
|
https://kenpyfin.com/blog/2018/11/03/lo-11-8-calculate-the-value-o-f-a-constant-maturity-treasury-swap-given-an/
|
[
"# LO 11.8: Calculate the value o f a constant maturity Treasury swap, given an\n\nLO 11.8: Calculate the value o f a constant maturity Treasury swap, given an interest rate tree and the risk-neutral probabilities.\nIn addition to valuing options with binomial interest rate trees, we can also value other derivatives such as swaps. The following example calculates the price of a constant maturity Treasury (CMT) swap. A CM T swap is an agreement to swap a floating rate for a Treasury rate such as the 10-year rate.\nExample: CM T swap\nAssume that you want to value a constant maturity Treasury (CMT) swap. The swap pays the following every six months until maturity:\n/\n$1,000,000 2 X (y V C M T Tc m t ls a semiannually compounded yield, of a predetermined maturity, at the time of payment (y( is equivalent to 6-month spot rates). Assume there is a 76% risk-neutral probability of an increase in the 6-month spot rate and a 60% risk-neutral probability of an increase in the 1-year spot rate. Fill in the missing data in the binomial tree, and calculate the value of the swap. Figure 5: Incomplete Binomial Tree for CM T Swap Int. rate = 7.25% Swap price = ? 0.76 Int. rate = 6.75% Swap price = ? Int. rate = 7.50% Swap price = ? 0.60 Int. rate = 7.00% Swap price = ? Int. rate = 6.50% Swap price = ? 0.40 Today Six Months One Year 2018 Kaplan, Inc. Page 139 Topic 11 Cross Reference to GARP Assigned Reading – Tuckman, Chapter 7 Answer: In six months, the top node and bottom node payoffs are, respectively: payoff^ = ———— x (7.25% 7.00%) =$1,250\n$1,000,000 payoff^ = ————-x (6.75% 7.00%) =$1,250\n$1,000,000 rr rr 2 2 ^ ^ Similarly in one year, the top, middle, and bottom payoffs are, respectively: r r$ 1, 000,000\n/ r r i r A f t /\npayoff2ju = ————-x (7.50% 7.00%) = $2,500 ^ ^ payoff2jM = ————-x (7.00% 7.00%) =$0\n/ r r / w w w\nr\nr\n$1,000,000 2 rr$ 1,000,000\npayoff2 L = ————-x (6.50% 7.00%) = $2,500 2 The possible prices in six months are given by the expected discounted value of the 1-year payoffs under the risk-neutral probabilities, plus the 6-month payoffs ($1,250 and\n$1,250). Hence, the 6-month values for the top and bottom node are, respectively: l,u Vi V 1>L ($2,500×0.6)+ ($0x0.4) 1+ 0.0725 +$1,250 = $2,697.53 ($ 0 x 0 .6 )+ (-$2,500×0.4) + 0.0675 1 -$1,250 = -$2,217.35 Today the price is$1,466.63, calculated as follows:\nw ^ V Q — —————————ri'”r\\~v———————- $1,466.63 ($2,697.53 x 0.76) + ($2,217.35 x 0.24) . 1+ 0.07 Page 140 2018 Kaplan, Inc. Topic 11 Cross Reference to GARP Assigned Reading – Tuckman, Chapter 7 Figure 6 shows the binomial tree with all values included. Figure 6: Completed Binomial Tree for CM T Swap 0.60 Int. rate = 7.50% Swap price =$2,500.00\nInt. rate = 7.00%\nSwap price = $1,466.63 0.76 0.24 Int. rate = 7.25% Swap price =$2,697.53\nInt. rate = 6.75%\nSwap price = -$2,217.35 Int. rate = 7.00% Swap price =$0\nInt. rate = 6.50%\nSwap price = -\\$2,500.00\n0.40\nToday\nSix Months\nOne Year\nO p t i o n -Ad j u s t e d S p r e a d"
] |
[
null
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{"ft_lang_label":"__label__en","ft_lang_prob":0.80158347,"math_prob":0.99941397,"size":3108,"snap":"2020-45-2020-50","text_gpt3_token_len":1092,"char_repetition_ratio":0.13176547,"word_repetition_ratio":0.20203735,"special_character_ratio":0.40765765,"punctuation_ratio":0.19072165,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998147,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-25T03:03:18Z\",\"WARC-Record-ID\":\"<urn:uuid:f951904d-963a-430b-8bb0-b32f0bc73e2f>\",\"Content-Length\":\"38311\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8eb03dab-81d3-47b1-884e-745f4b92d69e>\",\"WARC-Concurrent-To\":\"<urn:uuid:e71c44e1-0bef-46cf-8e46-f0a168e1f147>\",\"WARC-IP-Address\":\"54.68.36.172\",\"WARC-Target-URI\":\"https://kenpyfin.com/blog/2018/11/03/lo-11-8-calculate-the-value-o-f-a-constant-maturity-treasury-swap-given-an/\",\"WARC-Payload-Digest\":\"sha1:OAX6WR3NP77MWTRPVEZS53LJYJ6SFGCI\",\"WARC-Block-Digest\":\"sha1:4A4PDRCSVRWACWZW42ZXPZLFERK6J5HT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141180636.17_warc_CC-MAIN-20201125012933-20201125042933-00719.warc.gz\"}"}
|
https://laustep.github.io/stlahblog/posts/HaskellIsFast.html
|
[
"Posted on October 22, 2020 by Stéphane Laurent\n\nUpdated title: Haskell is fast, but Julia is faster (see updates at the end).\n\nMy R package ‘HypergeoMat’ provides a Rcpp implementation of Koev & Edelman’s algorithm for the evaluation of the hypergeometric function of a matrix argument.\n\nI also implemented this algorithm in Julia and in Haskell.\n\nSo let us benchmark now.\n\nHere is the hypergeometric function of a matrix argument:\n\n${}_pF_q^{(\\alpha)} \\left(\\begin{matrix} a_1, \\ldots, a_p \\\\ b_1, \\ldots, b_q\\end{matrix}; X\\right) = \\sum_{k=0}^\\infty\\sum_{\\kappa \\vdash k} \\frac{{(a_1)}_\\kappa^{(\\alpha)} \\cdots {(a_p)}_\\kappa^{(\\alpha)}} {{(b_1)}_\\kappa^{(\\alpha)} \\cdots {(b_q)}_\\kappa^{(\\alpha)}} \\frac{C_\\kappa^{(\\alpha)}(X)}{k!}.$\n\nWell, I will not explain this expression. But observe that this is a sum from $$k=0$$ to $$\\infty$$. The algorithm evaluates the partial sums of this series, that is, the sum from $$k=0$$ to an integer $$m$$.\n\nMy Haskell library generates a shared library (a DLL) which can be called from R. And one can call Julia from R with the help of the ‘XRJulia’ package. So we will benchmark the three implementations from R.\n\nFirstly, let’s check that they return the same value:\n\nlibrary(HypergeoMat)\nlibrary(XRJulia)\n# source the Julia code\njuliaSource(\"HypergeomPQ09.jl\")\ndll <- \"libHypergeom.so\"\n.C(\"HsStart\")\n\na <- c(8, 7, 3)\nb <- c(9, 16)\nx <- c(0.1, 0.2, 0.3)\nalpha <- 2\nm <- 5L # m is the truncation order\n\nhypergeomPFQ(m, a, b, x, alpha)\n# 2.116251\njuliaEval(\"hypergeom(5, [8.0, 7.0, 3.0], [9.0, 16.0], [0.1, 0.2, 0.3], 2.0)\")\n# 2.116251\n.Call(\"hypergeomR\", m, a, b, x, alpha)\n# 2.116251\n\nWell, the same results. Now, let’s run a first series of benchmarks, for $$m=5$$.\n\nlibrary(microbenchmark)\nmicrobenchmark(\nRcpp =\nhypergeomPFQ(m, a, b, x, alpha),\nJulia =\njuliaEval(\"hypergeom(5, [8.0, 7.0, 3.0], [9.0, 16.0], [0.1, 0.2, 0.3], 2.0)\"),\n.Call(\"hypergeomR\", m, a, b, x, alpha),\ntimes = 10\n)\nUnit: microseconds\nexpr min lq mean median uq max neval cld\nRcpp 356.682 623.807 837.7237 827.402 1084.191 1382.500 10 a\nJulia 4052.000 47767.565 44725.3895 48845.156 50597.779 51308.089 10 b\nHaskell 610.852 1136.963 1343.7442 1289.435 1504.323 2650.976 10 a \n\nShould we conclude that Rcpp is the winner, and that Julia is slow? That’s not sure. Observe that the unit of these durations is the microsecond. Perhaps the call to Julia via juliaEval is time-consuming, as well as the call to the Haskell DLL via .Call.\n\nSo let us try with $$m=40$$ now.\n\nm <- 40L\nmicrobenchmark(\nRcpp =\nhypergeomPFQ(m, a, b, x, alpha),\nJulia =\njuliaEval(\"hypergeom(40, [8.0, 7.0, 3.0], [9.0, 16.0], [0.1, 0.2, 0.3], 2.0)\"),\n.Call(\"hypergeomR\", m, a, b, x, alpha),\ntimes = 10\n)\nUnit: seconds\nexpr min lq mean median uq max neval cld\nRcpp 25.547556 25.924749 26.130888 26.185776 26.354177 26.47846 10 c\nJulia 18.959032 19.088749 19.191394 19.173662 19.291175 19.62415 10 b\nHaskell 6.642601 6.653627 6.736082 6.735448 6.760926 6.94283 10 a \n\nThis time, the unit is the second. Haskell is clearly the winner, followed by Julia.\n\nI’m using Julia 1.2.0, and I have been told that there is a great improvement of performance in Julia 1.5.0, the latest version. I’ll try with Julia 1.5.0 and then I will update this post to show whether there is a gain of speed.\n\nOne should not conclude from this experiment that Haskell always beats C++. That depends on the algorithm we benchmark. This one intensively uses recursion, and perhaps Haskell is strong when dealing with recursion.\n\nDon’t forget:\n\ndyn.unload(dll)\n\n# Update: Julia 1.5 is amazing\n\nUnit: seconds\nexpr min lq mean median uq max neval cld\nRcpp 23.464676 24.392115 24.860484 24.823062 25.013047 27.437176 10 c\nJulia 2.806364 2.852674 3.101521 2.973963 3.363618 3.897855 10 a\nHaskell 6.912441 7.459939 7.648012 7.674404 7.798719 8.322777 10 b\n\n19 seconds for Julia 1.2.0 and 3 seconds for Julia 1.5.2! It beats Haskell.\n\n# Update: even better\n\nThanks to some advice I got on discourse.julialang.org, I improved my Julia code, and it is faster now:\n\nUnit: seconds\nexpr min lq mean median uq max neval\nJulia 1.499753 1.549549 1.750907 1.658282 1.915167 2.428611 10"
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.6795849,"math_prob":0.9803501,"size":4197,"snap":"2022-27-2022-33","text_gpt3_token_len":1530,"char_repetition_ratio":0.114715,"word_repetition_ratio":0.13650307,"special_character_ratio":0.42363593,"punctuation_ratio":0.23406279,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9982055,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-05T18:19:10Z\",\"WARC-Record-ID\":\"<urn:uuid:9b906307-3d66-42c5-8745-832acc09446c>\",\"Content-Length\":\"17144\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:76232ddc-de02-4536-b9d4-4caf472621cf>\",\"WARC-Concurrent-To\":\"<urn:uuid:841182b4-1198-4fd4-9f3e-c69d6f2ea469>\",\"WARC-IP-Address\":\"185.199.109.153\",\"WARC-Target-URI\":\"https://laustep.github.io/stlahblog/posts/HaskellIsFast.html\",\"WARC-Payload-Digest\":\"sha1:GYTFFSKCJF5SGLQROXLF5LQEHCQN4L53\",\"WARC-Block-Digest\":\"sha1:VVZNEMB6NZMI6HPFZFIFEGEQRS3XG6JY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104597905.85_warc_CC-MAIN-20220705174927-20220705204927-00294.warc.gz\"}"}
|
https://english.eagetutor.com/index.php?option=com_content&view=article&id=122:coulombs-law-sp-1597356851&Itemid=101&tmpl=component&print=1&layout=default&page=
|
[
"",
null,
"Introduction to Coulomb’s law\n\nThe fundamental law of electrostatics states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.",
null,
"where ε0 = permittivity of space\n\nwhere q1 represents the quantity of charge on Object 1 (in Coulombs), q2 represents the quantity of charge on Object 2 (in Coulombs), and r represents the distance of separation between the two objects (in meters). The symbol k is a proportionality constant known as the Coulomb's law constant. The value of this constant is dependent upon the medium that the charged objects are immersed in. In the case of air, the value is approximately 9.0 x 109 N • m2 / C2. If the charged objects are present in water, the value of k can be reduced by as much as a factor of 80.\n\nDescription\n\nThe Coulomb's law equation provides an accurate description of the force between two objects whenever the objects act as point charges. A charged conducting sphere interacts with other charged objects as though all of its charge were located at its center. While the charge is uniformly spread across the surface of the sphere, the center of charge can be considered to be the center of the sphere. The sphere acts as a point charge with its excess charge located at its center. Since Coulomb's law applies to point charges, the distance d in the equation is the distance between the centers of charge for both objects.\n\nElectric charge and Coulomb's law\n\nCharge\n\n• there are two kinds of charge, positive and negative\n• like charges repel; unlike charges attract\n• positive charge comes from having more protons than electrons; negative charge comes from having more electrons than protons\n• charge is quantized, meaning that charge comes in integer multiples of the elementary charge e charge is conserved\n\nThe symbols Q1 and Q2 in the Coulomb's law equation represent the quantities of charge on the two interacting objects. Since an object can be charged positively or negatively, these quantities are often expressed as \"+\" or \"-\" values. The sign on the charge is simply representative of whether the object has an excess of electrons (a negatively charged object) or a shortage of electrons (a positively charged object).\n\nWant to know more about Coulomb’s law? Click here to schedule live help with an etutor!"
] |
[
null,
"https://english.eagetutor.com//images/flipkart.png",
null,
"https://english.eagetutor.com/images/stories/phy_a2_01.gif",
null
] |
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|
https://mc-stan.org/docs/stan-users-guide/circles-spheres-and-hyperspheres.html
|
[
"## 11.2 Circles, spheres, and hyperspheres\n\nAn $$n$$-sphere, written $$S^{n}$$, is defined as the set of $$(n + 1)$$-dimensional unit vectors, $S^{n} = \\left\\{ x \\in \\mathbb{R}^{n+1} \\: : \\: \\Vert x \\Vert = 1 \\right\\}.$\n\nEven though $$S^n$$ is made up of points in $$(n+1)$$ dimensions, it is only an $$n$$-dimensional manifold. For example, $$S^2$$ is defined as a set of points in $$\\mathbb{R}^3$$, but each such point may be described uniquely by a latitude and longitude. Geometrically, the surface defined by $$S^2$$ in $$\\mathbb{R}^3$$ behaves locally like a plane, i.e., $$\\mathbb{R}^2$$. However, the overall shape of $$S^2$$ is not like a plane in that it is compact (i.e., there is a maximum distance between points). If you set off around the globe in a “straight line” (i.e., a geodesic), you wind up back where you started eventually; that is why the geodesics on the sphere ($$S^2$$) are called “great circles,” and why we need to use some clever representations to do circular or spherical statistics.\n\nEven though $$S^{n-1}$$ behaves locally like $$\\mathbb{R}^{n-1}$$, there is no way to smoothly map between them. For example, because latitude and longitude work on a modular basis (wrapping at $$2\\pi$$ radians in natural units), they do not produce a smooth map.\n\nLike a bounded interval $$(a, b)$$, in geometric terms, a sphere is compact in that the distance between any two points is bounded."
] |
[
null
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https://metanumbers.com/108482
|
[
"# 108482 (number)\n\n108,482 (one hundred eight thousand four hundred eighty-two) is an even six-digits composite number following 108481 and preceding 108483. In scientific notation, it is written as 1.08482 × 105. The sum of its digits is 23. It has a total of 3 prime factors and 8 positive divisors. There are 49,300 positive integers (up to 108482) that are relatively prime to 108482.\n\n## Basic properties\n\n• Is Prime? No\n• Number parity Even\n• Number length 6\n• Sum of Digits 23\n• Digital Root 5\n\n## Name\n\nShort name 108 thousand 482 one hundred eight thousand four hundred eighty-two\n\n## Notation\n\nScientific notation 1.08482 × 105 108.482 × 103\n\n## Prime Factorization of 108482\n\nPrime Factorization 2 × 11 × 4931\n\nComposite number\nDistinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 108482 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0\n\nThe prime factorization of 108,482 is 2 × 11 × 4931. Since it has a total of 3 prime factors, 108,482 is a composite number.\n\n## Divisors of 108482\n\n8 divisors\n\n Even divisors 4 4 2 2\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 177552 Sum of all the positive divisors of n s(n) 69070 Sum of the proper positive divisors of n A(n) 22194 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 329.366 Returns the nth root of the product of n divisors H(n) 4.8879 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 108,482 can be divided by 8 positive divisors (out of which 4 are even, and 4 are odd). The sum of these divisors (counting 108,482) is 177,552, the average is 22,194.\n\n## Other Arithmetic Functions (n = 108482)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 49300 Total number of positive integers not greater than n that are coprime to n λ(n) 4930 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 10295 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares\n\nThere are 49,300 positive integers (less than 108,482) that are coprime with 108,482. And there are approximately 10,295 prime numbers less than or equal to 108,482.\n\n## Divisibility of 108482\n\n m n mod m 2 3 4 5 6 7 8 9 0 2 2 2 2 3 2 5\n\nThe number 108,482 is divisible by 2.\n\n## Classification of 108482\n\n• Arithmetic\n• Deficient\n\n### Expressible via specific sums\n\n• Polite\n• Non-hypotenuse\n\n• Square Free\n\n• Sphenic\n\n## Base conversion (108482)\n\nBase System Value\n2 Binary 11010011111000010\n3 Ternary 12111210212\n4 Quaternary 122133002\n5 Quinary 11432412\n6 Senary 2154122\n8 Octal 323702\n10 Decimal 108482\n12 Duodecimal 52942\n20 Vigesimal db42\n36 Base36 2bpe\n\n## Basic calculations (n = 108482)\n\n### Multiplication\n\nn×y\n n×2 216964 325446 433928 542410\n\n### Division\n\nn÷y\n n÷2 54241 36160.7 27120.5 21696.4\n\n### Exponentiation\n\nny\n n2 11768344324 1276653528956168 138493928128223016976 15024098311205889327590432\n\n### Nth Root\n\ny√n\n 2√n 329.366 47.6928 18.1484 10.1642\n\n## 108482 as geometric shapes\n\n### Circle\n\n Diameter 216964 681613 3.69713e+10\n\n### Sphere\n\n Volume 5.34763e+15 1.47885e+11 681613\n\n### Square\n\nLength = n\n Perimeter 433928 1.17683e+10 153417\n\n### Cube\n\nLength = n\n Surface area 7.06101e+10 1.27665e+15 187896\n\n### Equilateral Triangle\n\nLength = n\n Perimeter 325446 5.09584e+09 93948.2\n\n### Triangular Pyramid\n\nLength = n\n Surface area 2.03834e+10 1.50455e+14 88575.2\n\n## Cryptographic Hash Functions\n\nmd5 13d7ce509e1c2ba42dbbf248f5b953aa 9365cb1cbfee951c705dc813b2c53b039c76a495 1fb720b59f4eb0a08424fe3eaba062a150b6a4b18395a0a49cb600ffb52ca817 49b0ddeff30a4e6e2ba16c9a238e25157e8e0ab6aae38ddd3372b452d36c8375b4a1f51add6c90f2a92a409be896c3d80183794e279d65907f55775acfb68b97 e4fb0cee8819176ffd506de7860cf704c331ed13"
] |
[
null
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{"ft_lang_label":"__label__en","ft_lang_prob":0.62061846,"math_prob":0.9813762,"size":4635,"snap":"2021-43-2021-49","text_gpt3_token_len":1627,"char_repetition_ratio":0.12006046,"word_repetition_ratio":0.030837005,"special_character_ratio":0.45544767,"punctuation_ratio":0.0752551,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.995966,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-21T09:05:41Z\",\"WARC-Record-ID\":\"<urn:uuid:dacce599-e072-4888-bc96-6f62f8fbd4f3>\",\"Content-Length\":\"40124\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:aa77704f-4bc8-4608-b738-b2c751697804>\",\"WARC-Concurrent-To\":\"<urn:uuid:0a67adc2-4ef5-4c67-8656-b8202755c139>\",\"WARC-IP-Address\":\"46.105.53.190\",\"WARC-Target-URI\":\"https://metanumbers.com/108482\",\"WARC-Payload-Digest\":\"sha1:5FYWMSHOK5OLMENCW3XI4A5LLUWMZX2T\",\"WARC-Block-Digest\":\"sha1:RQFXQWDY3CSQOEZIQYVGTEPEOKJLHKCB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585382.32_warc_CC-MAIN-20211021071407-20211021101407-00588.warc.gz\"}"}
|
https://www.studysmarter.us/textbooks/math/linear-algebra-and-its-applications-5th/eigenvalues-and-eigenvectors/q52-7e-question-find-the-characteristic-polynomial-and-the-e/
|
[
"• :00Days\n• :00Hours\n• :00Mins\n• 00Seconds\nA new era for learning is coming soon",
null,
"Suggested languages for you:\n\nEurope\n\nAnswers without the blur. Sign up and see all textbooks for free!",
null,
"Q5.2-7E\n\nExpert-verified",
null,
"Found in: Page 267",
null,
"### Linear Algebra and its Applications\n\nBook edition 5th\nAuthor(s) David C. Lay, Steven R. Lay and Judi J. McDonald\nPages 483 pages\nISBN 978-03219822384",
null,
"# Question: Find the characteristic polynomial and the eigenvalues of the matrices in Exercises 1-8.7. $$\\left[ {\\begin{array}{*{20}{c}}5&3\\\\- 4&4\\end{array}} \\right]$$\n\nCharacteristic polynomial: $${\\lambda ^2} - 9\\lambda + 32$$.\n\n$$A$$ has no real Eigen values.\n\nSee the step by step solution\n\n## Step 1: Find the characteristic polynomial\n\nIf is an $$n \\times n$$ matrix, then $$det\\left( {A - \\lambda I} \\right)$$, which is a polynomial of degree $$n$$, is called the characteristic polynomial of $$A$$.\n\nIt is given that $$A = \\left[ {\\begin{array}{*{20}{c}}5&3\\\\- 4&4\\end{array}} \\right]$$ and $$I = \\left[ {\\begin{array}{*{20}{c}}1&0\\\\0&1\\end{array}} \\right]$$ is identity matrix. Find the matrix$$\\left( {A - \\lambda I} \\right)$$ as shown below:\n\n$\\begin{array}A - \\lambda I = \\left[ {\\begin{array}{*{20}{c}}3&- 4\\\\4&8\\end{array}} \\right] - \\lambda \\left[ {\\begin{array}{*{20}{c}}1&0\\\\0&1\\end{array}} \\right]\\\\ = \\left[ {\\begin{array}{*{20}{c}}{3 - \\lambda }&{ - 4}\\\\4&{8 - \\lambda }\\end{array}} \\right]\\end{array}$\n\nNow, calculate the determinant of the matrix $$\\left( {A - \\lambda I} \\right)$$ as shown below:\n\n$\\begin{array}det\\left( {A - \\lambda I} \\right) = \\det \\left[ {\\begin{array}{*{20}{c}}5&3\\\\{ - 4}&4\\end{array}} \\right]\\\\ = \\left( {5 - \\lambda } \\right)\\left( {4 - {\\rm{\\lambda }}} \\right) + 12\\\\ = {\\lambda ^2} - 9\\lambda + 20 + 12\\\\ = {\\lambda ^2} - 9\\lambda + 32\\end{array}$\n\nSo, the characteristic polynomial of is $${\\lambda ^2} - 9\\lambda + 32$$.\n\n## Step 2: Describe the characteristic equation\n\nTo find the eigenvalues of the matrix, we must calculate all the scalars such that $$\\left( {A - \\lambda I} \\right)x = 0$$ has a non-trivial solution which is equivalent to finding all such that the matrix $$\\left( {A - \\lambda I} \\right)$$is not invertible, that is, when determinant of $$\\left( {A - \\lambda I} \\right)$$is zero.\n\nThus, the eigenvalues of $$A$$ are the solutions of the characteristic equation$$\\det \\left( {A - \\lambda I} \\right) = 0$$. So, find the characteristic equation $$\\det \\left( {A - \\lambda I} \\right) = 0$$.\n\n$\\begin{array}det\\left[ {\\begin{array}{*{20}{c}}5&3\\\\- 4&4\\end{array}} \\right] = 0\\\\\\left( {5 - \\lambda } \\right)\\left( {4 - {\\rm{\\lambda }}} \\right) + 12 = 0\\\\{\\lambda ^2} - 9\\lambda + 20 + 12 = 0\\\\{\\lambda ^2} - 9\\lambda + 32 = 0\\end{array}$\n\n## Step 3: Find roots of characteristic equation\n\nFor the quadratic equation, $$a{x^2} + bx + c = 0$$ , the general solution is given as$$x = \\frac{{ - b \\pm \\sqrt {{b^2} - 4ac} \\;\\;}}{{2a}}$$.\n\nThus, the solution of the characteristic equation $${\\lambda ^2} - 9\\lambda + 32 = 0$$ is obtained as follows:\n\n$\\begin{array}{\\lambda ^2} - 9\\lambda + 32 = 0\\\\\\lambda = \\frac{{ - \\left( { - 9} \\right) \\pm \\sqrt {{{\\left( { - 9} \\right)}^2} - 4\\left( {32} \\right)} }}{2}\\\\ = \\frac{{ - 9 \\pm \\sqrt { - 47} }}{2}\\end{array}$\n\nThe eigenvalues of $$A$$ are complex. So, $$A$$ has no real Eigen values. This implies that no real vector $$x$$in $${\\mathbb{R}^2}$$satisfies the characteristic equation $$\\left( {A - \\lambda I} \\right)x = 0$$.",
null,
"### Want to see more solutions like these?",
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""
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https://aalexee.com/publication/alekseev-2019-how-to-measure-aa/
|
[
"# How to Measure Average Rate of Change?\n\n### Abstract\n\nThis paper contributes to the theory of average rate of change (ARC) measurement. The contribution is twofold. First, it relates ARC measurement to intertemporal choice. We show that an ARC of a variable can be identified with a discount rate which makes an economic agent indifferent between the initial and final temporal states of the variable. Furthermore, there is a one-to-one correspondence between ARC measures and one-parameter families of time preferences indexed by a discount rate. We also show that an ARC can be interpreted as a measure of perceived average rate of growth of the agent’s utility. Second, we employ an axiomatic approach to generalize the conventional ARC measures (such as the difference quotient and the continuously compounded growth rate) in several directions: to variables with arbitrary connected domains, to not necessarily time-shift invariant dependence on dates, to sets of time points other than an interval, to a benchmark-based evaluation. The generalized ARC measures turn out to correspond to the existing time preference models such as the discounted utility and the relative discounting model of Ok & Masatlioglu (2007).\n\nType"
] |
[
null
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|
https://en.wikipedia.org/wiki/Tree_rotation
|
[
"# Tree rotation\n\nJump to navigation Jump to search\n\nIn discrete mathematics, tree rotation is an operation on a binary tree that changes the structure without interfering with the order of the elements. A tree rotation moves one node up in the tree and one node down. It is used to change the shape of the tree, and in particular to decrease its height by moving smaller subtrees down and larger subtrees up, resulting in improved performance of many tree operations.\n\nThere exists an inconsistency in different descriptions as to the definition of the direction of rotations. Some say that the direction of rotation reflects the direction that a node is moving upon rotation (a left child rotating into its parent's location is a right rotation) while others say that the direction of rotation reflects which subtree is rotating (a left subtree rotating into its parent's location is a left rotation, the opposite of the former). This article takes the approach of the directional movement of the rotating node.\n\n## Illustration\n\nThe right rotation operation as shown in the adjacent image is performed with Q as the root and hence is a right rotation on, or rooted at, Q. This operation results in a rotation of the tree in the clockwise direction. The inverse operation is the left rotation, which results in a movement in a counter-clockwise direction (the left rotation shown above is rooted at P). The key to understanding how a rotation functions is to understand its constraints. In particular the order of the leaves of the tree (when read left to right for example) cannot change (another way to think of it is that the order that the leaves would be visited in an in-order traversal must be the same after the operation as before). Another constraint is the main property of a binary search tree, namely that the right child is greater than the parent and the left child is less than the parent. Notice that the right child of a left child of the root of a sub-tree (for example node B in the diagram for the tree rooted at Q) can become the left child of the root, that itself becomes the right child of the \"new\" root in the rotated sub-tree, without violating either of those constraints. As you can see in the diagram, the order of the leaves doesn't change. The opposite operation also preserves the order and is the second kind of rotation.\n\nAssuming this is a binary search tree, as stated above, the elements must be interpreted as variables that can be compared to each other. The alphabetic characters to the left are used as placeholders for these variables. In the animation to the right, capital alphabetic characters are used as variable placeholders while lowercase Greek letters are placeholders for an entire set of variables. The circles represent individual nodes and the triangles represent subtrees. Each subtree could be empty, consist of a single node, or consist of any number of nodes.\n\n## Detailed illustration\n\nWhen a subtree is rotated, the subtree side upon which it is rotated increases its height by one node while the other subtree decreases its height. This makes tree rotations useful for rebalancing a tree.\n\nConsider the terminology of Root for the parent node of the subtrees to rotate, Pivot for the node which will become the new parent node, RS for the side of rotation and OS for the opposite side of rotation. For the root Q in the diagram above, RS is C and OS is P. Using these terms, the pseudo code for the rotation is:\n\n```Pivot = Root.OS\nRoot.OS = Pivot.RS\nPivot.RS = Root\nRoot = Pivot\n```\n\nThis is a constant time operation.\n\nThe programmer must also make sure that the root's parent points to the pivot after the rotation. Also, the programmer should note that this operation may result in a new root for the entire tree and take care to update pointers accordingly.\n\n## Inorder invariance\n\nThe tree rotation renders the inorder traversal of the binary tree invariant. This implies the order of the elements are not affected when a rotation is performed in any part of the tree. Here are the inorder traversals of the trees shown above:\n\n```Left tree: ((A, P, B), Q, C) Right tree: (A, P, (B, Q, C))\n```\n\nComputing one from the other is very simple. The following is example Python code that performs that computation:\n\n```def right_rotation(treenode):\n\"\"\"Rotate the specified tree to the right.\"\"\"\nleft, Q, C = treenode\nA, P, B = left\nreturn (A, P, (B, Q, C))\n```\n\nAnother way of looking at it is:\n\nRight rotation of node Q:\n\n```Let P be Q's left child.\nSet Q's left child to be P's right child.\n[Set P's right-child's parent to Q]\nSet P's right child to be Q.\n[Set Q's parent to P]\n```\n\nLeft rotation of node P:\n\n```Let Q be P's right child.\nSet P's right child to be Q's left child.\n[Set Q's left-child's parent to P]\nSet Q's left child to be P.\n[Set P's parent to Q]\n```\n\nAll other connections are left as-is.\n\nThere are also double rotations, which are combinations of left and right rotations. A double left rotation at X can be defined to be a right rotation at the right child of X followed by a left rotation at X; similarly, a double right rotation at X can be defined to be a left rotation at the left child of X followed by a right rotation at X.\n\nTree rotations are used in a number of tree data structures such as AVL trees, red-black trees, WAVL trees, splay trees, and treaps. They require only constant time because they are local transformations: they only operate on 5 nodes, and need not examine the rest of the tree.\n\n## Rotations for rebalancing\n\nA tree can be rebalanced using rotations. After a rotation, the side of the rotation increases its height by 1 whilst the side opposite the rotation decreases its height similarly. Therefore, one can strategically apply rotations to nodes whose left child and right child differ in height by more than 1. Self-balancing binary search trees apply this operation automatically. A type of tree which uses this rebalancing technique is the AVL tree.\n\n## Rotation distance\n\nUnsolved problem in computer science:\n\nCan the rotation distance between two binary trees be computed in polynomial time?\n\nThe rotation distance between any two binary trees with the same number of nodes is the minimum number of rotations needed to transform one into the other. With this distance, the set of n-node binary trees becomes a metric space: the distance is symmetric, positive when given two different trees, and satisfies the triangle inequality.\n\nIt is an open problem whether there exists a polynomial time algorithm for calculating rotation distance. Yet, Fordham's algorithm computes a distance in linear time, but only allows 2 kind of rotations : (ab)c = a(bc) and a((bc)d) = a(b(cd)). Fordham's algorithm relies on a classification of nodes into 7 types, and a lookup table is used to find the number of rotations required to transform a node of one type into an other.\n\nDaniel Sleator, Robert Tarjan and William Thurston showed that the rotation distance between any two n-node trees (for n ≥ 11) is at most 2n − 6, and that some pairs of trees are this far apart as soon as n is sufficiently large. Lionel Pournin showed that, in fact, such pairs exist whenever n ≥ 11."
] |
[
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https://www.numberempire.com/776357
|
[
"Home | Menu | Get Involved | Contact webmaster",
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"",
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"",
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"# Number 776357\n\nseven hundred seventy six thousand three hundred fifty seven\n\n### Properties of the number 776357\n\n Factorization 776357 Divisors 1, 776357 Count of divisors 2 Sum of divisors 776358 Previous integer 776356 Next integer 776358 Is prime? YES (62197th prime) Previous prime 776327 Next prime 776389 776357th prime 11809199 Is a Fibonacci number? NO Is a Bell number? NO Is a Catalan number? NO Is a factorial? NO Is a regular number? NO Is a perfect number? NO Polygonal number (s < 11)? NO Binary 10111101100010100101 Octal 2754245 Duodecimal 315345 Hexadecimal bd8a5 Square 602730191449 Square root 881.11123020876 Natural logarithm 13.56236774492 Decimal logarithm 5.8900614731616 Sine 0.33373180409538 Cosine 0.94266806614802 Tangent 0.35402896955988\nNumber 776357 is pronounced seven hundred seventy six thousand three hundred fifty seven. Number 776357 is a prime number. The prime number before 776357 is 776327. The prime number after 776357 is 776389. Number 776357 has 2 divisors: 1, 776357. Sum of the divisors is 776358. Number 776357 is not a Fibonacci number. It is not a Bell number. Number 776357 is not a Catalan number. Number 776357 is not a regular number (Hamming number). It is a not factorial of any number. Number 776357 is a deficient number and therefore is not a perfect number. Binary numeral for number 776357 is 10111101100010100101. Octal numeral is 2754245. Duodecimal value is 315345. Hexadecimal representation is bd8a5. Square of the number 776357 is 602730191449. Square root of the number 776357 is 881.11123020876. Natural logarithm of 776357 is 13.56236774492 Decimal logarithm of the number 776357 is 5.8900614731616 Sine of 776357 is 0.33373180409538. Cosine of the number 776357 is 0.94266806614802. Tangent of the number 776357 is 0.35402896955988"
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"https://www.numberempire.com/images/graystar.png",
null,
"https://www.numberempire.com/images/graystar.png",
null,
"https://www.numberempire.com/images/graystar.png",
null,
"https://www.numberempire.com/images/graystar.png",
null,
"https://www.numberempire.com/images/graystar.png",
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https://jp.mathworks.com/matlabcentral/cody/problems/8-add-two-numbers/solutions/23315
|
[
"Cody\n\n# Problem 8. Add two numbers\n\nSolution 23315\n\nSubmitted on 1 Feb 2012 by Hristo Alexiev\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1 Pass\n%% a = 1; b = 2; c_correct = 3; assert(isequal(add_two_numbers(a,b),c_correct))\n\n2 Pass\n%% a = 17; b = 2; c_correct = 19; assert(isequal(add_two_numbers(a,b),c_correct))\n\n3 Pass\n%% a = -5; b = 2; c_correct = -3; assert(isequal(add_two_numbers(a,b),c_correct))"
] |
[
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http://www.awalegroup.com/pdf/categories-and-modules-with-k-theory-in-view
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[
"# Categories and Modules with K-Theory in View by A. J. Berrick",
null,
"By A. J. Berrick\n\nThis ebook develops facets of class concept basic to the examine of algebraic K-theory. beginning with different types regularly, the textual content then examines different types of K-theory and strikes directly to tensor items and the Morita thought. the specific method of localizations and completions of modules is formulated by way of direct and inverse limits. The authors reflect on local-global thoughts that provide information regarding modules from their localizations and completions and underlie a few fascinating purposes of K-theory to quantity concept and geometry. many helpful workouts, concrete illustrations of summary suggestions, and an in depth checklist of references are integrated.\n\nSimilar algebra & trigonometry books\n\nSpectral theory of automorphic functions\n\nVenkov A. B. Spectral conception of automorphic capabilities (AMS, 1983)(ISBN 0821830783)\n\nDiskrete Mathematik fuer Einsteiger\n\nDieses Buch eignet sich hervorragend zur selbstständigen Einarbeitung in die Diskrete Mathematik, aber auch als Begleitlektüre zu einer einführenden Vorlesung. Die Diskrete Mathematik ist ein junges Gebiet der Mathematik, das eine Brücke schlägt zwischen Grundlagenfragen und konkreten Anwendungen. Zu den Gebieten der Diskreten Mathematik gehören Codierungstheorie, Kryptographie, Graphentheorie und Netzwerke.\n\nStructure of algebras,\n\nThe 1st 3 chapters of this paintings comprise an exposition of the Wedderburn constitution theorems. bankruptcy IV includes the speculation of the commutator subalgebra of an easy subalgebra of a typical easy algebra, the examine of automorphisms of an easy algebra, splitting fields, and the index aid issue idea.\n\nAdditional resources for Categories and Modules with K-Theory in View\n\nSample text\n\nThus, we get a commutative diagram 0(Y) 0 0(X) 0(Z) -om i€L 0(Y) Moreover, l®3r 1 <_ f 3r an is a left adjoint to d 3^^ £ 1 3-e' (f ~(y) A u) = y A ] r , ( u ) 103c 1 is 0(Y) 18f since the adjunctions are preserved under 8-product. f 1 •0(1) for y e 0(Y) and linear, which is true because 3 ^ u e 0(YxX) Z Proof: Let Then p We must show that 0(Z) > — E - * 0(X) is an equalizer. We have So let u e 0(X) ^ O(XxX) I* Z be such that P'GnCu)) = 3 n P2 p. (u) = p2 (u) Cpn'Cu}) 3 P ((Pi(u)) 2 since means that is 0(Z)-linear.\n\nA. JOYAL \\$ M. TIERNEY 24 are characteristic maps Xc ,Xcb : PlxPl -• PI b l 2 of subsets S ^ L ^ C P l x P l , and it is enough to check that x -*• y <_ I (x) -• I (y) . , C S . , for But for this it is enough to know that x <_ y =£> &(x) £ A(y), which is true. As we did for sup-lattices given a subset explicitly describe the quotient locale relation on a e A A and generated by R. we may suppose R of We say that (z,,z 2 ) e R, we have AA Q R C AxA, we want to A by the congruence R is inf-stable if given (a A z, , a A z«) e R.\n\nIf A is a locale of S, then sh(p # A) pt v p E > sh(A) > y S is a pullback square. Proof: Diaconescu's Theorem says that morphisms of topoi F \\ are classified by left exact F f > sh(A) / S valued functors on A which take covers GALOIS THEORY to surjective families. 45 These are exactly locale morphisms A -*• q^fi, so the Proposition follows immediately from the universal property of pf The morphism is induced by the adjunction As an application of p A. A -* p*p A. p , and because we need the result in Chapter VII, we give here a new characterization of atomic topoi over S."
] |
[
null,
"https://images-na.ssl-images-amazon.com/images/I/315062IX%2B1L._BO1,204,203,200_.jpg",
null
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https://fr.mathworks.com/matlabcentral/cody/problems/2550-evaluating-a-polynomial/solutions/921061
|
[
"Cody\n\n# Problem 2550. Evaluating a polynomial\n\nSolution 921061\n\nSubmitted on 10 Jul 2016 by YOU FUU\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1 Pass\nx = 1; y_correct = 7; assert(isequal(equation(x),y_correct))\n\n2 Pass\nx = 5; y_correct = 9287; assert(isequal(equation(x),y_correct))\n\n3 Pass\nx = 0; y_correct = -3; assert(isequal(equation(x),y_correct))\n\n4 Pass\nx = -3; y_correct = -729; assert(isequal(equation(x),y_correct))\n\n5 Pass\nx = -sqrt(-1); y_correct = -3 - 12*sqrt(-1); assert(isequal(equation(x),y_correct))\n\n### Community Treasure Hunt\n\nFind the treasures in MATLAB Central and discover how the community can help you!\n\nStart Hunting!"
] |
[
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|
https://www.enotes.com/homework-help/chemical-reaction-ca-hcl-gt-ca-hcl-40-g-ca-464409
|
[
"# In the chemical reaction Ca + 2HCl --> CaCl2 + H2 , 40 g of Ca is reacted with 0.4 M HCl. How many liters of HCl are required to complete the reaction?",
null,
"40 g of calcium (Ca) reacts with 0.4 M hydrochloric acid (HCl). The balanced equation for the reaction is:\n\n`Ca + 2HCl -> H_2 + CaCl_2`\n\nFrom the balanced reaction 1 mole of calcium requires 2 moles of pure hydrochloric acid.\n\nThe molar mass of calcium is approximately 40 g/mole. 40 g of calcium is equivalent to one mole of calcium. The complete chemical reaction of this with HCl requires two moles of hydrochloric acid.\n\n0.4 M hydrochloric acid has 0.4 moles of the acid per liter of the solution.\n\n2 moles HCl * (1 liter/0.4 moles) = 5 liters of solution\n\nIt requires 5 liters of 0.4 M HCl to react completely with 40 g of calcium.\n\nApproved by eNotes Editorial Team"
] |
[
null,
"https://static.enotescdn.net/images/main/illustrations/illo-answer.svg",
null
] |
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https://scipython.com/book2/chapter-7-matplotlib/examples/the-two-dimensional-diffusion-equation/
|
[
"# The two-dimensional diffusion equation\n\nThe two-dimensional diffusion equation is $$\\frac{\\partial U}{\\partial t} = D\\left(\\frac{\\partial^2U}{\\partial x^2} + \\frac{\\partial^2U}{\\partial y^2}\\right)$$ where $D$ is the diffusion coefficient. A simple numerical solution on the domain of the unit square $0\\le x < 1, 0 \\le y < 1$ approximates $U(x,y;t)$ by the discrete function $u_{i,j}^{(n)}$ where $x=i\\Delta x$, $y=j\\Delta y$ and $t=n\\Delta t$. Applying finite difference approximations yields $$\\frac{u_{i,j}^{(n+1)} - u_{i,j}^{(n)}}{\\Delta t} = D\\left[ \\frac{u_{i+1,j}^{(n)} - 2u_{i,j}^{(n)} + u_{i-1,j}^{(n)}}{(\\Delta x)^2} + \\frac{u_{i,j+1}^{(n)} - 2u_{i,j}^{(n)} + u_{i,j-1}^{(n)}}{(\\Delta y)^2} \\right],$$ and hence the state of the system at time step $n+1$, $u_{i,j}^{(n+1)}$ may be calculated from its state at time step $n$, $u_{i,j}^{(n)}$ through the equation $$u_{i,j}^{(n+1)} = u_{i,j}^{(n)} + D\\Delta t\\left[ \\frac{u_{i+1,j}^{(n)} - 2u_{i,j}^{(n)} + u_{i-1,j}^{(n)}}{(\\Delta x)^2} + \\frac{u_{i,j+1}^{(n)} - 2u_{i,j}^{(n)} + u_{i,j-1}^{(n)}}{(\\Delta y)^2} \\right].$$\n\nConsider the diffusion equation applied to a metal plate initially at temperature $T_\\mathrm{cold}$ apart from a disc of a specified size which is at temperature $T_\\mathrm{hot}$. We suppose that the edges of the plate are held fixed at $T_\\mathrm{cool}$. The following code applies the above formula to follow the evolution of the temperature of the plate. It can be shown that the maximum time step, $\\Delta t$ that we can allow without the process becoming unstable is $$\\Delta t = \\frac{1}{2D}\\frac{(\\Delta x\\Delta y)^2}{(\\Delta x)^2 + (\\Delta y)^2}.$$\n\nIn the code below, each call to do_timestep updates the numpy array u from the results of the previous timestep, u0. The simplest approach to applying the partial difference equation is to use a Python loop:\n\nfor i in range(1, nx-1):\nfor j in range(1, ny-1):\nuxx = (u0[i+1,j] - 2*u0[i,j] + u0[i-1,j]) / dx2\nuyy = (u0[i,j+1] - 2*u0[i,j] + u0[i,j-1]) / dy2\nu[i,j] = u0[i,j] + dt * D * (uxx + uyy)\n\n\nHowever, this runs extremely slowly and using vectorization will farm out these explicit loops to the much faster pre-compiled C-code underlying NumPy's array implementation.\n\nimport numpy as np\nimport matplotlib.pyplot as plt\n\n# plate size, mm\nw = h = 10.\n# intervals in x-, y- directions, mm\ndx = dy = 0.1\n# Thermal diffusivity of steel, mm2.s-1\nD = 4.\n\nTcool, Thot = 300, 700\n\nnx, ny = int(w/dx), int(h/dy)\n\ndx2, dy2 = dx*dx, dy*dy\ndt = dx2 * dy2 / (2 * D * (dx2 + dy2))\n\nu0 = Tcool * np.ones((nx, ny))\nu = u0.copy()\n\n# Initial conditions - circle of radius r centred at (cx,cy) (mm)\nr, cx, cy = 2, 5, 5\nr2 = r**2\nfor i in range(nx):\nfor j in range(ny):\np2 = (i*dx-cx)**2 + (j*dy-cy)**2\nif p2 < r2:\nu0[i,j] = Thot\n\ndef do_timestep(u0, u):\n# Propagate with forward-difference in time, central-difference in space\nu[1:-1, 1:-1] = u0[1:-1, 1:-1] + D * dt * (\n(u0[2:, 1:-1] - 2*u0[1:-1, 1:-1] + u0[:-2, 1:-1])/dx2\n+ (u0[1:-1, 2:] - 2*u0[1:-1, 1:-1] + u0[1:-1, :-2])/dy2 )\n\nu0 = u.copy()\nreturn u0, u\n\n# Number of timesteps\nnsteps = 101\n# Output 4 figures at these timesteps\nmfig = [0, 10, 50, 100]\nfignum = 0\nfig = plt.figure()\nfor m in range(nsteps):\nu0, u = do_timestep(u0, u)\nif m in mfig:\nfignum += 1\nprint(m, fignum)\nax = fig.add_subplot(220 + fignum)\nim = ax.imshow(u.copy(), cmap=plt.get_cmap('hot'), vmin=Tcool,vmax=Thot)\nax.set_axis_off()\nax.set_title('{:.1f} ms'.format(m*dt*1000))\nfig.subplots_adjust(right=0.85)\ncbar_ax = fig.add_axes([0.9, 0.15, 0.03, 0.7])\ncbar_ax.set_xlabel('$T$ / K', labelpad=20)\nfig.colorbar(im, cax=cbar_ax)\nplt.show()\n\n\nTo set a common colorbar for the four plots we define its own Axes, cbar_ax and make room for it with fig.subplots_adjust. The plots all use the same colour range, defined by vmin and vmax, so it doesn't matter which one we pass in the first argument to fig.colorbar.\n\nThe state of the system is plotted as an image at four different stages of its evolution.",
null,
"Many thanks to Tim Teatro whose blog post inspired this example."
] |
[
null,
"https://scipython.com/static/media/2/examples/E7/diffusion2d.png",
null
] |
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|
https://la.mathworks.com/help/simulink/ug/call-and-integrate-external-c-algorithms-into-simulink-using-c-function-blocks.html
|
[
"## Integrate External C/C++ Code into Simulink Using C Function Blocks\n\nYou can call and integrate your external C code into Simulink® models using C Function blocks. C Function blocks allow you to call external C code and customize the integration of your code using the Output Code, Start Code, Initialize Conditions Code, and Terminate Code panes in the block parameters dialog. Use the C Function block to:\n\n• Call functions from external C code, and customize the code for your Simulink models.\n\n• Preprocess data to call a C function and postprocess data after calling the function.\n\n• Specify different code for simulation and code generation.\n\n• Call multiple functions.\n\n• Initialize and work with persistent data cached in the block.\n\n• Allocate and deallocate memory.\n\nUse the C Function block to call external C algorithms into Simulink that you want to modify. To call a single C function from a Simulink model, use the C Caller block. To integrate dynamic systems that have continuous states or state changes, use the S-Function block.\n\nNote\n\nC99 is the standard version of C language supported for custom C code integration into Simulink.\n\nThe following examples use C Function blocks to calculate the sum and mean of inputs.\n\n### Write External Source Files\n\nBegin by creating the external source files.\n\n1. Create a header file named `data_array.h`.\n\n```/* Define a struct called DataArray */ typedef struct DataArray_tag { /* Define a pointer called pData */ double* pData; /* Define the variable length */ int length; } DataArray; /* Function declaration */ double data_sum(DataArray data);```\n\n2. In the same folder, create a new file, `data_array.c`. In this file, write a C function that calculates the sum of input numbers.\n\n```#include \"data_array.h\" /* Define a function that takes in a struct */ double data_sum(DataArray data) { /* Define 2 local variables to use in the function */ double sum = 0.0; int i; /* Calculate the sum of values */ for (i = 0; i < data.length; i++) { sum = sum + data.pData[i]; } /* Return the result to the block */ return sum; }```\n\n### Enter the External Code Into Simulink\n\n1. Create a new, blank model and add a C Function block. The C Function block is in the User-Defined Functions library of the Library Browser.\n\n2. Double-click the C Function block to open the block dialog. Click",
null,
"to open the Model Configuration Parameters dialog. In the Simulation Target pane, define your header file under Include headers on the Code information tab.",
null,
"Tip\n\nAfter you have entered information for Source file in the next step, you can click to have the header file name filled in automatically, using information contained in your source files.\n\n3. Define the source file under Source files on the Code Information tab. To verify that your custom code can be parsed and built successfully, click .",
null,
"Note\n\nTo use a C Function block in a For Each subsystem or with continuous sample time, or to optimize the use of the block in conditional input branch execution, all custom code functions called by the block must be deterministic, that is, always producing the same outputs for the same inputs. Identify which custom code functions are deterministic by using the Deterministic functions and Specify by function parameters in the Simulation target pane. If the block references any custom code global variables, then Deterministic functions must set to `All` in order for the block to be used in a For Each subsystem, in conditional input branch execution, or with continuous sample time.\n\nFor an example showing a C Function block in a For Each subsystem, see Use C Function Block Within For Each Subsystem.\n\n4. In the Output Code pane of the C Function block parameters dialog, write the code that the block executes during simulation. In this example, the external C function computes a sum. In the Output Code pane, write code that calls the `data_array.c` function to compute the sum, then computes the mean.\n\n```/* declare the struct dataArr */ DataArray dataArr; /* store the length and data coming in from the input port */ dataArr.pData = &data; dataArr.length = length; /* call the function from the external code to calculate sum */ sum = data_sum(dataArr); /* calculate the mean */ mean = sum / length;```\n\nYou can specify code that runs at the start of a simulation and at the end of a simulation in the Start Code and Terminate Code panes.\n\n5. Use the Symbols table to define the symbols used in the code in the block. Add or delete a symbol using the and buttons. Define all symbols used in the Output Code, Start Code, Initialize Conditions Code, and Terminate Code panes to ensure that ports display correctly.",
null,
"In the Symbols table, for each symbol used in the code in the block, define the Name, Scope, Label, Type, Size, and Port, as appropriate.\n\nClose the block parameters dialog. After filling in the data in the table, the C Function block now has one input port and two output ports with the labels specified in the table.\n\n6. Add a Constant block to the Simulink canvas that will be the input to the C Function block. In the Constant block, create a random row array with 100 elements. To display the results, attach display blocks to the outputs of the C Function block.",
null,
"### Specify Simulation or Code Generation Code\n\nYou can specify different Output Code for simulation and code generation for the C Function block by defining `MATLAB_MEX_FILE`. For example, to specify code that only runs during the model simulation, use the following.\n\n```#ifdef MATLAB_MEX_FILE /* Enter simulation code */ #else /* Enter code generation code */ #endif ```\n\n### Specify Declaration for Target-Specific Function for Code Generation\n\nFor code generation purposes, if you do not have the external header file with the declaration of a function (such as a target-specific device driver) that you want to call from the C Function block, you can include a declaration with the correct signature in the Output Code pane of the block. This action creates a function call to the expected function when code is generated, as in the following example:\n\n```#ifndef MATLAB_MEX_FILE extern void driverFcnCall(int16_T in, int16_T * out); driverFcnCall(blockIn, &blockOut); #endif```"
] |
[
null,
"https://la.mathworks.com/help/simulink/ug/c_caller_config_param.png",
null,
"https://la.mathworks.com/help/simulink/ug/c-function-simulation-target-dialog-include-headers.png",
null,
"https://la.mathworks.com/help/simulink/ug/c-function-simulation-target-dialog-source-files.png",
null,
"https://la.mathworks.com/help/simulink/ug/c_function_symbols_table.png",
null,
"https://la.mathworks.com/help/simulink/ug/c_function_slexcscriptdataarray.png",
null
] |
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|
https://thetopsites.net/article/55080748.shtml
|
[
"## Append a one dimensional numpy array in a new x-value of a two dimensional numpy array\n\nnumpy reshape\nempty numpy array\nnumpy append to empty array\nnumpy concatenate\nnumpy vstack\nappend numpy array to csv\nnumpy insert\nnp.append not working\n\nI'm trying to append a one dimensional numpy array to a two dimensional, so the one dimensional one is inserted on the place of another x-value.\n\nExample:\n\nall_polys = [[5,6],[8,9]] (before the error down below there is nothing stored in it yet)\n\npoly = [1,2]\n\nExpected Result:\n\nall_polys = [[5,6],[8,9],[1,2]]\n\nMy Code:\n\n```all_polys = numpy.array([[]])\npoly = np.expand_dims(poly, axis=0)\nprint(poly)\nprint(all_polys)\nall_polys = np.concatenate(all_polys, poly)\n```\n\nThe Error:\n\nTypeError: only integer scalar arrays can be converted to a scalar index\n\nPrint Output before error:\n\n[['400' '815' '650' '815' '650' '745' '400' '745']] (poly with added dimension)\n\n[] (all_polies)\n\nThis really frustrates me. What I am doing wrong? It must be a little detail I overlooked, I guess.\n\nStarting with a 2d array, and a 1d array:\n\n```In : all_polys = np.array([[5,6],[8,9]])\nIn : poly = np.array([1,2])\n```\n\n`vstack` does a nice job of making sure all inputs are 2d, and then concatenating:\n\n```In : np.vstack((all_polys, poly))\nOut:\narray([[5, 6],\n[8, 9],\n[1, 2]])\n```\n\nYou had the right ides with `expand_dims`:\n\n```In : np.concatenate((all_polys, np.expand_dims(poly, axis=0)))\nOut:\narray([[5, 6],\n[8, 9],\n[1, 2]])\n```\n\nBut the `np.array([[]])` is a poor starting point. Why use that? Are you doing this iteratively?\n\nFor iterative work we recommend using lists:\n\n```In : alist = []\nIn : alist.append([5,6])\nIn : alist.append([8,9])\nIn : alist.append([1,2])\nIn : np.array(alist)\nOut:\narray([[5, 6],\n[8, 9],\n[1, 2]])\n```\n\nI discourage the use of `np.append`. It is misused too often.\n\nnumpy.append — NumPy v1.20.dev0 Manual, If axis is not specified, values can be any shape and will be flattened before use. Note that append does not occur in-place: a new array is allocated and filled. array at index 0 has 2 dimension(s) and the array at index 1 has 1 dimension(s). First let's discuss some useful array attributes. We'll start by defining three random arrays, a one-dimensional, two-dimensional, and three-dimensional array. We'll use NumPy's random number generator, which we will seed with a set value in order to ensure that the same random arrays are generated each time this code is run:\n\nYou can try the append function instead of expand_dims\n\n```import numpy as np\nall_polys = [[5,6],\n[8,9]]\nall_polys = np.append(all_polys,[ [1,2] ], axis=0)\nprint(all_polys)\n#Output=\n#all_polys = [[5,6],\n# [8,9],\n# [1,2]]\n```\n\nthe absolute basics for beginners — NumPy v1.20.dev0 , An array is a grid of values and it contains information about the raw data, how to You might also hear 1-D, or one-dimensional array, 2-D, or two-dimensional array, and so on. np.concatenate((x, y), axis=0) array([[1, 2], [3, 4], [5, 6]]) Using arr.reshape() will give a new shape to an array without changing the data. Append a one dimensional numpy array in a new x-value of a two dimensional numpy array March 2019 Python three dimensional array of zeros w/out using numpy\n\nYou just need to do this:\n\n```all_polys = np.concatenate((all_polys, poly[None,:]), axis=0)\n```\n\nThe two arrays we are concatenating are `all_polys`, which looks like `[[5,6],[8,9]]`, and `poly[None,:]`, which looks like `[[1.2]]`.\n\nBy `axis=0`, we're specifying that the concatenation must happen along the outermost (first) dimension of these arrays.\n\nnumpy.append() in Python, Python numpy append() function is used to merge two arrays. This function returns a new array and the original array remains unchanged. 1 NumPy append() Syntax; 2 Python numpy.append() Examples in append return concatenate((arr, values), axis=axis) ValueError: all the input array dimensions except for the� Create a simple two dimensional array. First, redo the examples from above. And then create your own: how about odd numbers counting backwards on the first row, and even numbers on the second? Use the functions len(), numpy.shape() on these arrays. How do they relate to each other? And to the ndim attribute of the arrays?\n\nyou should do like this.\n\n```arr = [old array]\nnewArr = numpy.append(arr, [new_array])\n```\n\nUse of append function will works.\n\nnumpy.append() : How to append elements at the end of a Numpy , end of this new copied array. So, basically it returns a copy of numpy array provided with values appended to it. 1 : values will be appended to arr at axis 1 i.e. new columns will be added Create two 2D Numpy Array like Matrix ValueError: all the input arrays must have same number of dimensions. 0:29 create one dimensional array 0:45 create two dimensional array 1:11 ndim property 1:35 itemsize property 1:57 dtype property 2:05 change data type of element 2:52 size property 3:12 shape\n\n1.4.1. The NumPy array object — Scipy lecture notes, values of an experiment/simulation at discrete time steps; signal recorded by a 2 x 3 array. >>> b. array([[0, 1, 2],. [3, 4, 5]]). >>> b.ndim. 2. >>> b.shape. (2, 3). >> > len(b) # returns the size of the first dimension. 2. >>> c = np.array([[, ], [, for the notebook, so that plots are displayed in the notebook and not in a new� ma.empty (shape[, dtype, order]): Return a new array of given shape and type, without initializing entries. ma.empty_like (a[, dtype, order, subok]): Return a new array with the same shape and type as a given array.\n\nPython NumPy array tutorial, The ndarray stands for N-dimensional array where N is any number. The values will be appended at the end of the array and a new import numpy a = numpy.array([1, 2, 3]) newArray = numpy.append (a, [10, 11, 12]) print(newArray) import numpy addition = lambda x: x + 2 a = numpy.array([1, 2, 3, 4,� Here we have created a one-dimensional array of length 2. Each element of this array is a structure that contains three items, a 32-bit integer, a 32-bit float, and a string of length 10 or less. If we index this array at the second position we get the second structure: >>>\n\nnumpy.append, numpy.append - This function adds values at the end of an input array. The append operation is not inplace, a new array is allocated. Also the Also the dimensions of the input arrays must match otherwise ValueError will be generated. Parameter & Description. 1. arr. Input array. 2. values. To be appended to arr. It must� numpy.interp¶ numpy.interp (x, xp, fp, left=None, right=None, period=None) [source] ¶ One-dimensional linear interpolation. Returns the one-dimensional piecewise linear interpolant to a function with given values at discrete data-points.\n\n• Does `all_polys` start as a list `[[5,6],[8,9]`, as an array, `np.array([[5,6,8,9]])`, or as this useless thing `np.array([[]])`? Is `poly` a list `[1,2]` or an array, `np.array([1,2])`?\n• @hpaulj: Nice. Was surprised by your use of `vstack` with a mix of 2d and 1d array. Expected it to throw an error, as it doesn't seem to satisfy the rule \"The arrays must have the same shape along all but the first axis\" -- they aren't even of the same rank.\n• For the example in your question the arrays already have a different number of stored elements. `all_polys` has 4 elements, and `poly` has 2 elements. I think the question in your comment now is about flattening the arrays before concatenating. For that you can use something like `my_flat = np.concatenate((all_polys.ravel(), poly.ravel()), axis=0)`. BTW, if you don't pass `axis=0`, the default of 0 will be used by `concatenate()`, which will work just fine for us."
] |
[
null
] |
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|
https://raggato.com/franklin-templeton-sbtzog/sorting-algorithms-time-complexity-bb7278
|
[
"The time complexity of radix sort is given by the formula,T(n) = O(d*(n+b)), where d is the number of digits in the given list, n is the number of elements in the list, and b is the base or bucket size used, which is normally base 10 for decimal representation. However, the time complexity of an algorithm also depends on the hardware, operating system, processors, etc. Somewhere, Korea; GitHub1; GitHub2; Email On this page. Algorithm Implementation . A Sorting Algorithm is used to rearrange a given array or list elements according to a comparison operator on the elements. Therefore, it is crucial to analyse all the factors before executing the algorithm, and it is essential to select a suitable Sorting Algorithm to achieve efficiency and effectiveness in time complexity. Time Complexity in Sorting Algorithms. n indicates the input size, while O is the worst-case scenario growth rate function. Time Complexity A best sorting algorithm in python. How To Add A Document Viewer In Angular 10. Amount of work the CPU has to do (time complexity) as the input size grows (towards infinity). The total amount of the computer's memory used by an algorithm when it is executed is the space complexity of that algorithm … Time complexity is a way to describe how much time an algorithm needs to finish executing relative to the size of the input. The quicksort uses a divide and conquers algorithm with recursion. Selection Algorithm. Bubble sort algorithm is easy to understand from the example itself. B. In-place/Outplace technique – A sorting technique is inplace if it does not use any extra memory to sort the array. In short, TimSort makes use of the Insertion sort and the MergeSort algorithms. What are in-place sorting algorithms? 03. 04. Big O = Big Order function. Selection Sort Algorithm Time Complexity is O(n2). The time complexity of these algorithms are calculated and recorded. Bubble Sort Algorithm. These factors do affect the time taken to execute the algorithm. 21. if for an algorithm time complexity is given by O(n2) then complexity will: A. constant B. quardratic C. exponential D. none of the mentioned. ... Time Complexity comparison of Sorting Algorithms. The different sorting techniques like bubble sort, selection sort, insertion sort, quick sort and merge sort are implemented using C. The input values varying from 100 to 1000 are system generated. The letter O is used to indicate the time complexity component of sorting. To recap time complexity estimates how an algorithm performs regardless of the kind of machine it runs on. learning sw Yoo. For the same sorting algorithm, the order degree of the data to be sorted is different, and the execution time of sorting will be greatly different. 32 minute read geeksforgeeks. Time and space complexity. 06. This tutorial covers two different ways to measure the runtime of sorting algorithms: For a practical point of view, you’ll measure the runtime of the implementations using the timeit module. Click to see full answer Keeping this in consideration, how do you find the time complexity of a radix sort? Timing Your Code. Selection Sort Algorithm Space Complexity is O(1). You can get the time complexity by “counting” the number of operations performed by your code. As a programmer, … C. Counting Sort is not a comparison based sorting algorithm. Bubble sort is beneficial when array elements are less and the array is nearly sorted. The comparison operator is used to decide the new order of element in the respective data structure. In the above sorting algorithm, if we look at the code, we find that even if our array is already sorted, the time complexity will be the same i.e. Quicksort algorithm is one of the most efficient sorting algorithms, and that’s why it is mostly used as it is one of the best algorithms. Sorting Algorithms. 3. Complexity of Quick Sort: Merge Sort: It is a sorting algorithm which follows the divide and conquers methodology. The most common metric it’s using Big O notation. Here are some highlights about Big O Notation: Big O notation is a framework to analyze and compare algorithms. Thus it runs in time () and is a polynomial time algorithm. 05. The space complexity of bubble sort algorithm is O(1). Shell sort is an insertion sort that first partially sorts its data and then finishes the sort by running an insertion sort algorithm on the entire array. Time complexity is an abstract way to show how long a sorting algorithm would take to sort a vector of length n. The best algorithms that make comparisons between elements usually have a complexity of O(n log n). Some most common of these are merge sort, heap sort, and quicksort. The exact time complexity of the algorithm can be found from the sequence $(n-1) + (n-2) + \\dots + (n-(n-1)) \\mathrm{,}$ ... as well as provided a better understanding of the time complexity of several sorting algorithms. The minimum possible time complexity of a comparison based sorting algorithm is O(nLogn) for a random input array. e.g. 2. Follow. Getting Started With Azure Service Bus Queues And ASP.NET Core - Part 1 . It is an in-place sorting algorithm. The divide and conquer technique used by merge sort makes it convenient for parallel processing. C# 9 Cheat Sheet. Time complexity and Space complexity. It is nevertheless important for you to understand these basic algorithms, because you are likely to use them within your own programs – their space and time complexity will thus affect that of your own algorithms. Let’s learning about an algorithm that finds k-th elemen using median of medians to ensure linear time. Space and time complexity acts as a measurement scale for algorithms. There is another sorting algorithm Counting sort which time complexity … Bucket sort – Best and average time complexity: n+k where k is the number of buckets. Selection Sort is the easiest approach to sorting. This is an extremely good time complexity for a sorting algorithm, since it has been proven that an array can't be sorted any faster than O(nlog n). This swapping process continues until we sort the input list. It generally starts by choosing small subsets of the array and sorting those arrays. For the given data set, quick sort is found very efficient and has taken 168 ms for 1000 data inputs. Algorithms with higher complexity class might be faster in practice, if you always have small inputs. All the basic arithmetic operations (addition, subtraction, multiplication, division, and comparison) can be done in polynomial time. Quick sort with median-of-medians algorithm. We will talk about sorting algorithm later. BigO Graph *Correction:- Best time complexity for TIM SORT is O(nlogn) sort; sorting algorithm; space complexity; time complexity; TRENDING UP 01 Clean Architecture End To End In .NET 5. For example: The below list of characters is sorted in increasing order of their ASCII values. In extreme cases, if the data is already ordered, the sorting algorithm does not need to do any operation, and the execution time will be very short. Analyzing the time it takes for an algorithm to give output is of crucial importance. Selection Sort Time Complexity. The time complexity of an algorithm is NOT the actual time required to execute a particular code, since that depends on other factors like programming language, operating software, processing power, etc. E.g. Imagine a telephone book application that would take a day to sort all the numbers after a new number was added. Merge sort is a stable sort with a space complexity of O (n) O(n) O (n). View Answer 22. It recursively breaks down a problem into two or more sub-problems. Practical sorting algorithms are usually based on algorithms with average time complexity. We compare the algorithms on the basis of their space (amount of memory) and time complexity (number of operations). Time complexity is a function describing the amount of time an algorithm takes in terms of the amount of input to the algorithm. Should you need to select a specific sorting or searching algorithm to fit a particular task, you will require a good understanding of the available options. Worst case time complexity: n^2 if all elements belong to same bucket. Its overall time complexity is O(nlogn). Use Entity Framework Core 5.0 In .NET Core 3.1 With MySQL Database By … Prototype Design Pattern With Java. Any comparison based sorting algorithm can be made stable by using position as a criteria when two elements are compared. And use it to quick sort algorithm. Afterward, it repeats the same process with larger subsets until it reaches a point where the subset is the array, and the entire thing becomes sorted. The time complexity of this algorithm is O(n), a lot better than the Insertion Sort algorithm. The time complexity of quicksort is O(n log n) in the best case, O(n log n) in the average case, and O(n^2) in the worst case. Bubble sort, also known as sinking sort, is a very simple algorithm to sort the elements in an array. Time Complexity comparison of Sorting Algorithms and Space Complexity comparison of Sorting Algorithms. Number of swaps in bubble sort = Number of inversion pairs present in the given array. But unlike quick sort Merge sort is not an adaptive sorting algorithm as the time complexity of Merge sort does not depends on the initial input sequence of the given array. Drop constants and lower order terms. We’ll present the pseudocode of the algorithm and analyze its time complexity. Quick Sort is not a stable sorting algorithm. Space complexity is a function describing the amount of memory (space) an algorithm takes in terms of the amount of input to the algorithm. This complexity means that the algorithm’s run time increases slightly faster than the number of items in the vector. sort(Object[]) is based on the TimSort algorithm, giving us a time complexity of O(n log(n)). Algorithm. However, it is still slower compared to other sorting algorithms like some of the QuickSort implementations. 02. Some types of algorithms are more efficient than others for searching. Here, the concept of space and time complexity of algorithms comes into existence. The worst case time complexity of bubble sort algorithm is O(n 2). The idea behind time complexity is that it can measure only the execution time of the algorithm in a way that depends only on the algorithm itself and its input. Time complexity Cheat Sheet. Let’s take it as an example. Please refer to the bubble sort algorithm explained with an example. In sorting, time complexity is based on how many operations or actions (how much time) it takes to locate or arrange data structures in a search. No sweat, no sweet. While the version we've showcased is memory-consuming, there are more complex versions of Merge Sort that take up only O(1) space. How to calculate time complexity of any algorithm or program? The Significance of Time Complexity. Selection Sort Algorithm with Example is given. calculation only. There are many sorting algorithms in Computer Science Data Structure, and most of those give the same time complexity which is O(nlogn), where n represents the total number of elements present in the given structure, and the sorting algorithms which satisfy this time complexity are Merge sort, Quick-sort, Heap sort, etc. This recursion is continued until a solution is not found that can be solved easily. This time complexity is defined as a function of the input size n using Big-O notation. Merge sort has a guaranteed time complexity of O (n l o g n) O(nlogn) O (n l o g n) time, which is significantly faster than the average and worst-case running times of several other sorting algorithms. Efficient sorts. For a more theoretical perspective, you’ll measure the runtime complexity of the algorithms using Big O notation. Insertion sort has running time $$\\Theta(n^2)$$ but is generally faster than $$\\Theta(n\\log n)$$ sorting algorithms for lists of around 10 or fewer elements. 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Their space ( amount of time an algorithm that finds k-th elemen using median of medians to ensure linear.! With Azure Service Bus Queues and ASP.NET Core - Part 1 in Angular 10 time increases slightly faster the! Array and sorting those arrays its overall time complexity of this algorithm is used to decide the new of... Sorting technique is inplace if it does not use any extra memory to sort the input grows. After a new number was added calculated and recorded defined as a criteria when two elements are compared some. Analyzing the time complexity of Quick sort: it is a polynomial time algorithm algorithm to sort the... Some of the Insertion sort and the MergeSort algorithms to see full answer Keeping this in consideration, do! Algorithm ; space complexity ; time complexity is O ( n2 ) overall. Algorithm that finds k-th elemen using median of medians to ensure linear time algorithm... Comes into existence sort with a space complexity ; TRENDING UP 01 Clean Architecture End to End in.NET.... Operating system, processors, etc TRENDING UP 01 Clean Architecture End to End in.NET.. Execute the algorithm and analyze its time complexity in sorting algorithms done polynomial. This complexity means that the algorithm number of operations ) and sorting those arrays ( n ) (... Number of inversion pairs present in the respective data structure of element in the given array we the! And conquer technique used by merge sort is O ( nlogn ) for a more theoretical perspective, ’! And has taken 168 ms for 1000 data inputs faster in practice, if you always small. You ’ ll present the pseudocode of the array and sorting those arrays ’ ll measure the complexity! Viewer in sorting algorithms time complexity 10 way to describe how much time an algorithm takes in terms the... Performs operations for some constant a the new order of element in the respective data structure highlights... 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This swapping process continues until we sort the input size n using Big-O notation higher class! Acts as a function of the amount of memory ) and is a sorting technique is inplace if does!, also known as sinking sort, and comparison ) can be done in polynomial time algorithm is... You can get the time complexity in sorting algorithms and space complexity comparison ) can be stable!, how do you find the time complexity: n+k where k is the number operations... More efficient than others for searching a criteria when two elements are and! And compare algorithms their ASCII values in bubble sort = number of swaps in bubble works. Sort is not found that can be made stable by using position as a when. Sort with a space complexity comparison of sorting time complexity and space complexity is O ( n.... Is another sorting algorithm which follows the divide and conquer technique used by merge sort: it still. 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Follows the divide and conquers algorithm with recursion with recursion explained with an example complexity class might faster! Let ’ s learning about an algorithm needs to finish executing relative to the bubble sort by. That can be made stable by using position as a criteria when two are! Lot better than the number of operations performed by your code numbers after a new number was added is crucial. Used to indicate the time complexity is O ( n ) O ( n ) O ( )... Swapping the adjacent elements if they appear sorting algorithms time complexity the wrong order in the respective data structure of! Most common of these are merge sort, also known as sinking sort, a. With an example some examples of polynomial time more theoretical perspective, you ’ ll the... Division, and comparison ) can be solved easily of O ( n ) O ( n ) not. Does not use any extra memory to sort the array example itself slower to... Continued until a solution is not a comparison based sorting algorithm ; complexity! “ Counting ” the number of swaps in bubble sort algorithm is easy to understand the... Complexity means that the algorithm still slower compared to other sorting algorithms sorting... Be solved easily ( ) and is a stable sort with a space complexity ; TRENDING 01... Defined as a function describing the amount of input to the algorithm belong same. Where k is the worst-case scenario growth rate function n^2 if all elements belong to same bucket much an! Efficient and has taken 168 ms for 1000 data inputs any algorithm program... In sorting algorithms like some of the algorithm ’ s using Big O notation the elements full answer this! The respective data structure the vector any extra memory to sort all the numbers after a new number added! Runtime complexity of bubble sort algorithm space complexity subsets of the array and sorting those arrays of inversion present., also known as sinking sort, and comparison ) can be made stable by using as. To see full answer Keeping this in consideration, how do you find the time complexity comparison of algorithms. Input array comes into existence sort algorithm Clean Architecture End to End.NET... Can be done in polynomial time algorithms: the selection sort algorithm is easy to understand from the example.! Describe how much time an algorithm also depends on the hardware, operating system, processors etc... Median of medians to ensure linear time are more efficient than others for searching of space and time of. Was added sort and the array be solved easily End in.NET 5 elements according a. Counting sort which time complexity of O ( 1 ) sorting technique is if! Runs in time ( ) and is a very simple algorithm to give output is of crucial.! Array or list elements according to a comparison based sorting algorithm is O ( nlogn ) examples! Always have small inputs s learning about an algorithm needs to finish executing relative to the size of the sort... In-Place/Outplace technique – a sorting algorithm Counting sort which time complexity is O ( nlogn for... System, sorting algorithms time complexity, etc and compare algorithms for searching these factors do affect the complexity! Down a problem into two or more sub-problems ( addition, subtraction,,! Is another sorting algorithm is O ( n ) O ( n2 ) complexity in sorting algorithms position. ; GitHub2 ; Email on this page, Korea ; GitHub1 ; GitHub2 ; Email this. Items in the given array or list elements according to a comparison operator on the hardware, operating,! Function of the input two elements are less and the array and those... Algorithm or program calculate time complexity comparison of sorting with recursion items in the original list. All elements belong to same bucket technique – a sorting algorithm on n performs. Takes for an algorithm takes in terms of the quicksort uses a and... Infinity ) this algorithm is O ( 1 ) work the CPU has to do time... Some highlights about Big O notation here are some highlights about Big O notation on n performs!"
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https://gitlab.mpcdf.mpg.de/ift/nifty/-/commit/d58f9a363cd0165bb58591d0aed1a94e80a1f69f?view=parallel
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[
"### improve docs\n\nparent 735fc6e8\n ... @@ -32,13 +32,13 @@ class LMSpace(StructuredDomain): ... @@ -32,13 +32,13 @@ class LMSpace(StructuredDomain): lmax : int lmax : int The maximum :math:`l` value of any spherical harmonic coefficient The maximum :math:`l` value of any spherical harmonic coefficient :math:`a_{lm}` that is represented by this object. :math:`a_{lm}` that is represented by this object. Must be :math:`\\ge 0`. Must be :math:`\\\\ge 0`. mmax : int, optional mmax : int, optional The maximum :math:`m` value of any spherical harmonic coefficient The maximum :math:`m` value of any spherical harmonic coefficient :math:`a_{lm}` that is represented by this object. :math:`a_{lm}` that is represented by this object. If not supplied, it is set to `lmax`. If not supplied, it is set to `lmax`. Must be :math:`\\ge 0` and :math:`\\le` `lmax`. Must be :math:`\\\\ge 0` and :math:`\\\\le` `lmax`. \"\"\" \"\"\" _needed_for_hash = [\"_lmax\", \"_mmax\"] _needed_for_hash = [\"_lmax\", \"_mmax\"] ... ...\n ... @@ -95,7 +95,7 @@ class Linearization(object): ... @@ -95,7 +95,7 @@ class Linearization(object): @property @property def want_metric(self): def want_metric(self): \"\"\"bool : the value of `want_metric`\"\"\" \"\"\"bool : True iff the metric was requested in the constructor\"\"\" return self._want_metric return self._want_metric @property @property ... ...\n ... @@ -109,16 +109,17 @@ class FieldAdapter(LinearOperator): ... @@ -109,16 +109,17 @@ class FieldAdapter(LinearOperator): def ducktape(left, right, name): def ducktape(left, right, name): \"\"\"Convenience function for computing an adapter between two operators. \"\"\"Convenience function creating an operator that converts between a DomainTuple and a single-entry MultiDomain. Parameters Parameters ---------- ---------- left : None, Operator, or Domainoid left : None, Operator, or Domainoid Something describing the input domain of the left operator. Something describing the new operator's target domain. If `left` is an `Operator`, its domain is used as `left`. If `left` is an `Operator`, its domain is used as `left`. right : None, Operator, or Domainoid right : None, Operator, or Domainoid Something describing the target domain of the right operator. Something describing the new operator's input domain. If `right` is an `Operator`, its target is used as `right`. If `right` is an `Operator`, its target is used as `right`. name : string name : string ... ...\n ... @@ -182,7 +182,7 @@ def NiftyMetaBase(): ... @@ -182,7 +182,7 @@ def NiftyMetaBase(): return with_metaclass(NiftyMeta, type('NewBase', (object,), {})) return with_metaclass(NiftyMeta, type('NewBase', (object,), {})) class frozendict(collections.Mapping): class frozendict(collections.abc.Mapping): \"\"\" \"\"\" An immutable wrapper around dictionaries that implements the complete An immutable wrapper around dictionaries that implements the complete :py:class:`collections.Mapping` interface. It can be used as a drop-in :py:class:`collections.Mapping` interface. It can be used as a drop-in ... ...\nMarkdown is supported\n0% or .\nYou are about to add 0 people to the discussion. Proceed with caution.\nFinish editing this message first!"
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https://utzx.com/units/convert-13-16-cm-to-inches/
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[
"# Convert 13.16 CM to Inches\n\n## How much is 1cm in inches?\n\nDo you are seeking to convert 13.16 cm into an inch-length result? first, you obviously should know how many inches 1 cm is equal to.\n\nYou may use this cm to inches converter calculator to calculate the conversion.\n\n## What is the centimeter?\n\nCentimeters, also known as centimetres, is the unit for length measurement in metric systems. The symbol is cm. The length unit meter has been defined internationally to the “International System of Units”, but the unit centimeter does not. But one cm is equals 100 meters. It is also around 39.37 inches.\n\n## Definition: Inch\n\nAn inch is an Anglo-American measurement of length measurement. Its symbol is in. In several different European local languages, the word “inch” is the same as or derived from “thumb”. Since a person’s thumb is approximately an inch in width.\n\n• Electronic components such as the size of the PC screen.\n• Dimensions of truck and car tires.\n\n## What is 13.16 centimeters Converted to inches?\n\nThe centimeters to inches converter is a tool that lets you to convert centimeters to inches. We may simply calculate the number of centimeters to inches through using this basic.\n\nYou have fully grasped for cm into inches by the above. You can use the formula to answer related questions:\n\n• What is the formula for converting 13.16 cm to inches?\n• How do you convert cm to inches?\n• How can I change cm into inches?\n• How do you measure cm to inches?\n• How big are 13.16 cm to inches?\n\n cm inches 12.76 cm 5.023612 inches 12.81 cm 5.043297 inches 12.86 cm 5.062982 inches 12.91 cm 5.082667 inches 12.96 cm 5.102352 inches 13.01 cm 5.122037 inches 13.06 cm 5.141722 inches 13.11 cm 5.161407 inches 13.16 cm 5.181092 inches 13.21 cm 5.200777 inches 13.26 cm 5.220462 inches 13.31 cm 5.240147 inches 13.36 cm 5.259832 inches 13.41 cm 5.279517 inches 13.46 cm 5.299202 inches 13.51 cm 5.318887 inches"
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https://community.wolfram.com/groups/-/m/t/189328
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[
"# Calculating the mean value of a list with LibraryLink\n\nPosted 10 years ago\n3933 Views\n|\n0 Replies\n|\n5 Total Likes\n|\n In a previous post I explained how to work with passing in and returning atomic data with LibraryLink. Butwhat if you want to pass in a list of numbers to do some computation on? In that case you will need to usea data structure in LibraryLink called an MTensor, which is a general n-dimensional list structure meaningyou can use it for lists, matrices and higher dimensional equivalents of those structures.Here is a simple example for calculating the mean value of a list of values. This means that a list of valuesneeds to be passed into the C function as an MTensor: #include \"WolframLibrary.h\" DLLEXPORT int MeanValue(WolframLibraryData libData, mint Argc, MArgument *Args, MArgument Res) { MTensor tensor; double* data; double total,result; mint i,length; tensor = MArgument_getMTensor(Args); data = libData->MTensor_getRealData(tensor); length = libData->MTensor_getFlattenedLength(tensor); total=0; for(i=0; i"
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{"ft_lang_label":"__label__en","ft_lang_prob":0.72755224,"math_prob":0.8989416,"size":3005,"snap":"2023-14-2023-23","text_gpt3_token_len":796,"char_repetition_ratio":0.10963012,"word_repetition_ratio":0.022271715,"special_character_ratio":0.2705491,"punctuation_ratio":0.14686468,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9897685,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-03T21:09:13Z\",\"WARC-Record-ID\":\"<urn:uuid:b4790d85-d733-4a65-afbb-27c16991890c>\",\"Content-Length\":\"115257\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8ef5e146-cbba-4b4e-b785-7b5846d564c9>\",\"WARC-Concurrent-To\":\"<urn:uuid:2cee5608-544b-417c-adc4-9f934cb8fa15>\",\"WARC-IP-Address\":\"140.177.204.58\",\"WARC-Target-URI\":\"https://community.wolfram.com/groups/-/m/t/189328\",\"WARC-Payload-Digest\":\"sha1:XHOY3RXC6NR52D4UHNNEPSXDLM6ELUGP\",\"WARC-Block-Digest\":\"sha1:O3EL7RAEEOCF7S73NCMVTKPPYTBNZ2M2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224649343.34_warc_CC-MAIN-20230603201228-20230603231228-00732.warc.gz\"}"}
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https://www.education.com/worksheets/pennies/
|
[
"# Search Printable Penny Worksheets\n\n85 filtered results\n85 filtered results\nPennies\nShow interactive only\nSort by\nPractice Test: Money",
null,
"Interactive Worksheet\nPractice Test: Money\nLearners will test their money math and addition skills with this simple quiz.\nMath",
null,
"Interactive Worksheet\nIdentifying Coins\nWorksheet\nIdentifying Coins\nThere are so many coins on this worksheet, but which one is which? Challenge your child with identifying coins.\nKindergarten\nMath\nWorksheet\nPrintable Money\nWorksheet\nPrintable Money\nMath\nWorksheet\nCounting Coins III\nWorksheet\nCounting Coins III\nKnowing how to count coins and handle money is a practical math skill for kids. Help your child master coin counting with this worksheet.\nMath\nWorksheet\nWorksheet\nOn this third grade math worksheet, kids use addition and/or multiplication skills to determine the total value of each set of coins.\nMath\nWorksheet\nWorksheet\nMath\nWorksheet\nMonster Money\nWorksheet\nMonster Money\nThese little monsters have earned some money, but which monster has the most?\nKindergarten\nMath\nWorksheet\nCounting Coins II\nWorksheet\nCounting Coins II\nBeing able to identify and count coins is a valuable skill for kids. In this worksheet, your child will count coins, then write their total values on the lines.\nMath\nWorksheet\nHow Much Money?\nWorksheet\nHow Much Money?\nMath\nWorksheet\nCount the Pocket Change\nWorksheet\nCount the Pocket Change\nTeach your child how to turn loose change into spending money with this coin-counting worksheet.\nKindergarten\nMath\nWorksheet\nMoney Word Problems #1\nWorksheet\nMoney Word Problems #1\nThese word problem worksheets on money will have your 1st-grader making change in no time.\nMath\nWorksheet\nWorksheet\nHow much does that hamburger cost? Give your little spender a hand in counting up coins and dollars to equal an amount.\nMath\nWorksheet\nMoney Word Problems #5\nWorksheet\nMoney Word Problems #5\nWalter, Tabitha, and Dean need help calculating how much money they have. Have your child solve these money math word problems to find the total sums of money.\nMath\nWorksheet\nCounting Coins I\nWorksheet\nCounting Coins I\nIs your child a money whiz? In this fun coin worksheet, your child will count each group of coins and write their total values on the lines.\nMath\nWorksheet\nCoin Toss: Frog\nWorksheet\nCoin Toss: Frog\nYoung learners cut out and assemble this simple template, then compete to see who can toss the most coins through the frog's open mouth.\nMath\nWorksheet\nLearning to Count Coins\nWorksheet\nLearning to Count Coins\nHow many coins can you count? Help your little cashier learn to add up coin amounts to figure out the price of these items!\nMath\nWorksheet\nCoins for Kids\nWorksheet\nCoins for Kids\nKindergarten\nMath\nWorksheet\nComparing Money Amounts #5\nWorksheet\nComparing Money Amounts #5\nGive your kid practice with money-counting and sequential order with this worksheet. Kids will look at different amounts of money and decide which is larger.\nMath\nWorksheet\nConnect the Coins #1\nWorksheet\nConnect the Coins #1\nHelp your child get to know her coin values (and her U.S. presidents!) with this quick coin-counting quiz. Students will get their math muscles moving too.\nKindergarten\nMath\nWorksheet\nCounting Change: How Much?",
null,
"Interactive Worksheet\nCounting Change: How Much?\nKids count up the coins, then circle the correct amount.\nMath",
null,
"Interactive Worksheet\nMatch It!\nWorksheet\nMatch It!\nLooking for a worksheet to help you kid with his money counting? This printable will help him count money.\nKindergarten\nMath\nWorksheet\nMoney Word Problems #4\nWorksheet\nMoney Word Problems #4\nYour child will work through money math problems to sum up amounts of money. This word problem worksheet will help your child exercise his math skills.\nMath\nWorksheet\nMoney & Place Value with Hundreds, Tens, and Ones\nWorksheet\nMoney & Place Value with Hundreds, Tens, and Ones\nThere’s nothing that engages kids quite like money! Reinforce place value and student understanding of U.S. bills with this double-sided worksheet.\nMath\nWorksheet\nHow Much Does It Cost? #1\nWorksheet\nHow Much Does It Cost? #1\nLittle shoppers, here's a coin-counting challenge! Add up the coins to find out how much each of these items costs."
] |
[
null,
"https://www.education.com/files/static/interactive-worksheets/icons/sparkles-icon-purple.svg",
null,
"https://www.education.com/files/static/interactive-worksheets/icons/sparkles-icon-purple.svg",
null,
"https://www.education.com/files/static/interactive-worksheets/icons/sparkles-icon-purple.svg",
null,
"https://www.education.com/files/static/interactive-worksheets/icons/sparkles-icon-purple.svg",
null
] |
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|
https://www.degruyter.com/document/doi/10.1515/jiip.2010.016/html
|
[
"B. E. Kanguzhin\nFrom the journal\n\nAbstract\n\nIn this work the possibility of restoration of real symmetrical five diagonal final matrices using four numerically sequences is studied. Three from these four numerically sequences are interpreted as sets of eigenvalues of the considered matrix and else of two matrices, obtained from considered matrix deleting some diagonal elements. The concrete formulas of construction of matrix elements using four sets of eigenvalues are obtained."
] |
[
null
] |
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|
https://www.analystforum.com/t/base-rate-neglect-2012-quicksheet-error/73922
|
[
"",
null,
"# base rate neglect 2012 quicksheet error\n\nwe want to work out P(A|B)=P(B|A)*P(A)/P(B)\n\nin representativeness, what is base rate neglect? is it neglect P(A|B) or neglect P(A) or P(B)\n\ncan anyone clarify? thanks\n\nAs per Bayes-----> we update probabilities\n\nP (A given B ) = P (B given A) X P( A) / P (B)—> here the base rate is P (B given A) i guess\n\nthen you agree schweser quick sheet is wrong? but people usally under-estimate the probability of P(B given A) or over-estimate in base rate neglect?\n\nwiki say\n\nhttp://en.wikipedia.org/wiki/Base_rate_fallacy\n\nbase rate neglect is the \"is an error that occurs… without taking into account the prior probability (“base rate”) .\n\nP(A|B) = P(B|A) P(A) / P(B)\n\nthe namings of the terms are\n\nposterior = likelyhood * prior / normalizingTerm\n\nso, prior is refering to P(A)\n\naccording to you, it is P(A)?\n\nYes. Thanks for pointing out my typo ( i have corrected it now.)\n\nIgoring prior means ignoring P(A)\n\nYes. Thanks for pointing out my typo ( i have corrected it now.)\n\nIgoring prior means ignoring P(A)"
] |
[
null,
"https://analystforum-uploads.s3.dualstack.us-east-1.amazonaws.com/original/2X/8/8e7be8e6512cde25d070f18d332292fb5a3804d9.png",
null
] |
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|
https://www.gradegorilla.com/chemistry/IB/2/M_elecSub.php
|
[
"An IB Periodic Table is required\n\n1. How many orbitals are in an ‘f’ type sub-shell?\n• A. 5\n• B. 7\n• C. 10\n• D. 14\n2. Which list of species all have the electronic configuration 1s22s22p63s23p6 ?\n• A. K+, Ar, P3-\n• B. Ne, S2-, Cl-\n• C. Ar, Mg2+, Ca2+\n• D. Mg2+, Na+, O2-\n3. Which element has the ground state electronic configuration ending in np4 ?\n• A. Iodine, I\n• B. Chromium, Cr\n• C. Selenium, Se\n• D. Silicon, Si\n4. What is the correct electronic configuration of the iron(III) ion?\n• A. [Ar] 4s23d6\n• B. [Ar] 3d5\n• C. [Ar] 4s23d3\n• D. [Ar] 4s13d10\n\n5. Which ‘electron in boxes’ diagram is correct for the P atom?\n\n 3s 3p A [Ne] ↑↓ ↑↓ ↑ B [Ne] ↑↓ ↑ ↑ ↑ C [Ne] ↑↓ ↑ ↓ ↑ D [Ne] ↑ ↑↓ ↓ ↑\n6. What is the total number of electrons in s orbitals in an atom of potassium in the ground state?\n• A. 1\n• B. 4\n• C. 7\n• D. 9\n7. In which block of the Periodic Table is zinc found?\n• A. s-block\n• B. p-block\n• C. d-block\n• D. f-block\n8. Which element is composed of atoms whose last electron is the first to enter the 3p sub-shell?\n• A. Gallium, Ga\n• B. Nitrogen, N\n• C. Scandium, Sc\n• D. Aluminium, Al\n\n9. Which ‘electron in boxes’ diagram is correct for the Cr atom?\n\n 4s 3d A [Ar] ↑↓ ↑↓ ↑↓ B [Ar] ↑↓ ↑ ↑ ↑ ↑ C [Ar] ↑ ↑↓ ↑↓ ↑ D [Ar] ↑ ↑ ↑ ↑ ↑ ↑\n\n10. What is the correct electronic configuration of the copper(I) ion?\n• A. 1s22s22p63s23p63d94s2\n• B. 1s22s22p63s23p63d104s1\n• C. 1s22s22p63s23p63d10\n• D. 1s22s22p63s23p63d94s1"
] |
[
null
] |
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|
http://cornmills.co.uk/bmn-share-darrczr/f16005-append-list-to-dataframe
|
[
"",
null,
"## append list to dataframe\n\nExamples are provided for scenarios where both the DataFrames have similar columns and non-similar columns. Columns of the new dataframe which are not in the original datafarme are also added to the existing dataframe and the new cells’ values are filled with NaN. Next: Write a Pandas program to append a list of dictioneries or series to a existing DataFrame … Append elements to a list RDocumentation. There is more than one way of adding columns to a Pandas dataframe, let’s review the main approaches. In this article, I will use examples to show you how to add columns to a dataframe in Pandas. I have a list of Pandas dataframes that I would like to combine into one Pandas dataframe. Append elements to a list. This is a quick solution when you want to do keep the new record separately in a different dataframe and after some point in time you need to merge that together. list.append(.data, ...) Arguments.data A list or vector... A vector or list to append after x. Example Codes: Also, if ignore_index is True then it will not use indexes. Percentile. If you want to append a list as a row to a pandas dataframe you can convert it to a pandas series first and then use the append() function to add it to the dataframe. How to append a list as a row to a Pandas DataFrame in Python, Sometimes it's easier to do all the appending outside of pandas, then, just create the DataFrame in one shot. Create from lists. append() method could append rows of other dataframe to the end of the original dataframe, and return a new dataframe. Dataframe append to add New Row. From rlist v0.4.6.1 by Kun Ren. In both cases, the result is a usable dataframe from a list of other data types. Example Pandas DataFrame append() function is used to merge rows from another DataFrame object. The last method that programmers can use to add a column to DataFrame is by generating a new list as a separate column of data and appending the column to the existing DataFrame. Understanding Pandas DataFrame append() Pandas DataFrame append() function merge rows from another DataFrame object. In this post we will learn how to add a new column using a dictionary in Pandas. And we can also specify column names with the list of tuples. In order to decide a fair winner, we will iterate over DataFrame and use only 1 value to print or append per loop. The append() method appends an element to the end of the list. 1 7 8. Appending a DataFrame to another one is quite simple: In : df1.append(df2) Out: A B C 0 a1 b1 NaN 1 a2 b2 NaN 0 NaN b1 c1 Syntax: DataFrame.append(self, other, ignore_index=False, verify_integrity=False, sort=None) … Append a character or string to end of the column in pandas; Append a numeric or integer value to the start of the column in pandas; Append a numeric or integer value to the end of the column in pandas . This can be done in a similar way as before but you can also use the DataFrame.merge() method. Create empty Dataframe, append rows; Pandas version used: 1.0.3. Converting a list to a dataframe is fairly simple. Crash when passing empty sequence to DataFrame.append. If there is a mismatch in the columns, the new columns are added in the result DataFrame. I managed to hack a fix for this by assigning each new DataFrame to the key instead of appending it to the key's value list: models[label] = (pd.DataFrame(data=data, index=df.index)) What property of DataFrames (or perhaps native Python) am I invoking that would cause this to work fine, but appending to a list to act strangely? Append to a DataFrame; Spark 2.0.0 cluster takes a long time to append data; How to improve performance with bucketing; How to handle blob data contained in an XML file; Simplify chained transformations; How to dump tables in CSV, JSON, XML, text, or HTML format; Hive UDFs; Prevent duplicated columns when joining two DataFrames # append item to list in r append (first_vector, second_vector) You are likely already familiar with using concatenate to add multiple elements to a given list. We can simply use pd.DataFrame on this list of tuples to get a pandas dataframe. 0th. Here's how the … Converting list of tuples to pandas dataframe. We can include different lines also. Version 0.25.1. Parameter Description; elmnt: Required. pandas.DataFrame.append¶ DataFrame.append(other, ignore_index=False, verify_integrity=False)¶ Append rows of other to the end of this frame, returning a new object. In this tutorial, we shall learn how to append a row to an existing DataFrame, with the help of illustrative example programs. The source DataFrame is returned intact. In the next few steps, we will look at the .append method, which does not modify the calling DataFrame, rather it returns a new copy of the DataFrame with the appended row/s. Table of Contents. I want to write the function, which will add the column's values as lists to a list. No crash. >>> import pandas as pd Converting the list to a data frame within the append function works, also when applied in a loop. For example, rather than calling pd.concat([df1, df2]), you can simply call df1.append(df2): Have another way to solve this solution? I created the list of dataframes from: import pandas as pd. Version. More Examples. To test these methods, we will use both of the print() and list.append() functions to provide better comparison data and to cover common use cases. Python Pandas dataframe append() is an inbuilt function that is used to append rows of other dataframe to the end of the given dataframe, returning a new dataframe object. Create a Dataframe As usual let's start by creating a dataframe. An element of any type (string, number, object etc.) I am using Python 2.7.10 and Pandas 0.16.2. Earlier we saw how to add a column using an existing columns in two ways. Pandas Dataframe provides a function dataframe.append() i.e. Expected Output. Creating a new column to a dataframe is a common task in doing data analysis. A work-around (suggested by jezrael) involved appending each dataframe to a list of dataframes and concatenating them using pd.concat. 1.24 5.5 8.8. Append elements to a list Usage. Columns not in this frame are added as new columns. list.append(elmnt) Parameter Values. There are many ways to build and initialize a pandas DataFrame. For example, when there are two or more data frames created using different data sources, and you want to select a specific set of columns from different data frames to create one single data frame, the … Create dataframe : Append a character or numeric to the column in pandas python. Where each list represents one column. Syntax. Syntax – append() Following is the syntax of DataFrame.appen() function. The .loc attribute makes changes to the DataFrame in place. Create a simple dataframe with a dictionary of lists, and column names: name, age, city, country. DataFrame append() function is present in the Pandas library(), which is a great library that enables the user to perform data analysis effectively and efficiently. First, however, you need to have the two Pandas dataframes: Columns not in the original dataframes are added as new columns, and the new cells are populated with NaN value. Add a list to a list: a = [\"apple\", \"banana\", \"cherry\"] I have the dataframe: col1 col2 col3. Contribute your code (and comments) through Disqus. This function returns a new DataFrame object and doesn’t change the source objects. See the example below: Previous: Write a Pandas program to join the two given dataframes along columns and assign all data. You can also add a new row as a dataframe and then append this new row to the existing dataframe at the bottom of the original dataframe. Problem description. And this task often comes in a variety of forms. DataFrame.append(other, ignore_index=False, verify_integrity=False, sort=None) Here, ‘other’ parameter can be a DataFrame , Series or Dictionary or list of these. Append a Column to Pandas Datframe Example 3: In the third example, you will learn how to append a column to a Pandas dataframe from another dataframe. R Enterprise Training; R package; Leaderboard; Sign in; list.append. First let’s create a dataframe To append or add a row to DataFrame, create the new row as Series and use DataFrame.append() method. 1.5 6.7 9. Append a list as a row to a dataframe. In this post, you will learn different techniques to append or add one column or multiple columns to Pandas Dataframe ().There are different scenarios where this could come very handy. Create pandas dataframe from scratch import pandas as pd datfr = {‘Name’: [‘Karl’, ‘Gaurav’, ‘Ray’, ‘Mimo’], Code faster with the Kite plugin for your code editor, featuring Line-of-Code Completions and cloudless processing. Here are some of the most common ones: All examples can be found on this notebook. See Also. Kite is a free autocomplete for Python developers. The append method does not change either of the original DataFrames.\n\nComments are closed."
] |
[
null,
"http://www.whalleycornmills.co.uk/wp-content/themes/Animal_Care_Theme/images/cs-title-bg.png",
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.7778664,"math_prob":0.776144,"size":9200,"snap":"2021-04-2021-17","text_gpt3_token_len":2040,"char_repetition_ratio":0.18573293,"word_repetition_ratio":0.03696498,"special_character_ratio":0.2175,"punctuation_ratio":0.13778256,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98766476,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-22T16:29:44Z\",\"WARC-Record-ID\":\"<urn:uuid:cd2bae7d-6785-45fe-b6ec-a958ddcb1e5d>\",\"Content-Length\":\"63984\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0c802bf1-5a57-4dd5-bc06-520ac1412533>\",\"WARC-Concurrent-To\":\"<urn:uuid:c16519b0-c694-40e4-897d-ad252e179cba>\",\"WARC-IP-Address\":\"46.16.169.5\",\"WARC-Target-URI\":\"http://cornmills.co.uk/bmn-share-darrczr/f16005-append-list-to-dataframe\",\"WARC-Payload-Digest\":\"sha1:Y6DQSD2IW2VL63VEIXOOS6TO6VSS7OOX\",\"WARC-Block-Digest\":\"sha1:XPUOK67LRXX5VQCKAPDO7U5EIRG5VJZF\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618039594341.91_warc_CC-MAIN-20210422160833-20210422190833-00032.warc.gz\"}"}
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http://mathonline.wikidot.com/conservative-vector-fields-examples-1
|
[
"Conservative Vector Fields Examples 1\n\n# Conservative Vector Fields Examples 1\n\nRecall from the Conservative Vector Fields page that a vector field $\\mathbf{F}$ is said to be conservative on the domain $D$ if there exists a function $\\phi$ known as a potential function such that $\\mathbf{F} = \\nabla \\phi$ on $D$.\n\nWe also proved a necessary condition for a vector field on $\\mathbb{R}^2$ and a vector field on $\\mathbb{R}^3$ to possess. Recall that if $\\mathbf{F}(x, y) = P(x, y) \\vec{i} + Q(x, y) \\vec{j}$ is a conservative vector field on $D$ then we must have that for all points in $D$ that::\n\n(1)\n\\begin{align} \\quad \\frac{\\partial P}{\\partial y} = \\frac{\\partial Q}{\\partial x} \\end{align}\n\nFurthermore, recall that if $\\mathbf{F}(x, y, z) = P(x, y, z) \\vec{i} + Q(x, y, z) \\vec{j} + R(x, y, z) \\vec{k}$ is a conservative vector field on $D$ then we must have that for all points in $D$ that:\n\n(2)\n\\begin{align} \\quad \\frac{\\partial P}{\\partial y} = \\frac{\\partial Q}{\\partial x} \\quad , \\quad \\frac{\\partial P}{\\partial z} = \\frac{\\partial R}{\\partial x} \\quad , \\quad \\frac{\\partial Q}{\\partial z} = \\frac{\\partial R}{\\partial y} \\end{align}\n\nWe will now look at some examples of determining whether a vector field is conservative or not.\n\n# Example 1\n\nDetermine whether the vector field $\\mathbf{F}(x, y) = \\frac{x}{x^2 + y^2} \\vec{i} + \\frac{y}{x^2 + y^2} \\vec{j}$ is a conservative vector field.\n\nWe will first see if we can verify the necessary condition for a vector field to be conservative. We note that $P(x, y) = \\frac{x}{x^2 + y^2}$ and $Q(x, y) = \\frac{y}{x^2 + y^2}$.\n\nFirst let's take the partial derivative of $P$ with respect to $y$:\n\n(3)\n\\begin{align} \\quad \\frac{\\partial P}{\\partial y} = 2xy \\ln(x^2 + y^2) \\end{align}\n\nNow let's take the partial derivative of $Q$ with respect to $x$:\n\n(4)\n\\begin{align} \\quad \\frac{\\partial Q}{\\partial x} = 2xy \\ln (x^2 + y^2) \\end{align}\n\nTherefore we have verified the necessary condition for a vector field to be conservative with $\\frac{\\partial P}{\\partial y} = \\frac{\\partial Q}{\\partial x}$. This does not guarantee us that $\\mathbf{F} = P\\vec{i} + Q \\vec{j}$ is conservative though, but now we have reason to suspect it could be.\n\nNow suppose that a potential function exists, that is suppose that $\\mathbf{F} (x, y) = \\nabla \\phi (x, y)$. Then we must have that:\n\n(5)\n\\begin{align} \\quad \\frac{\\partial \\phi}{\\partial x} = \\frac{x}{x^2 + y^2} \\\\ \\quad \\frac{\\partial \\phi}{\\partial y} = \\frac{y}{x^2 + y^2} \\end{align}\n\nTake the first equation and partial integrate with respect to $x$ to get that:\n\n(6)\n\\begin{align} \\quad \\int \\frac{\\partial \\phi}{\\partial x} \\: dx = \\phi(x, y) = \\int \\frac{x}{x^2 + y^2} \\: dx \\end{align}\n\nUsing substitution, we let $u = x^2 + y^2$. Then $du = 2x \\: dx$ and so $\\frac{1}{2} du = x \\: dx$ and so:\n\n(7)\n\\begin{align} \\quad = \\frac{1}{2} \\int \\frac{1}{u} \\: du = \\frac{1}{2} \\ln \\mid u \\mid = \\frac{1}{2} \\ln (x^2 + y^2) + h(y) \\end{align}\n\nWe will now partial differentiate with respect to $y$ to get that:\n\n(8)\n\\begin{align} \\quad \\frac{\\partial \\phi}{\\partial y} = \\frac{\\partial}{\\partial y} \\left ( \\frac{1}{2} \\ln (x^2 + y^2) + h(y) \\right ) = \\frac{1}{2} \\left ( \\frac{2y}{x^2 + y^2} \\right ) + h(y) = \\frac{y}{x^2 + y^2} + h'(y) \\end{align}\n\nComparing this with the fact that we're given $\\frac{\\partial \\phi}{\\partial y}$ tells us that $h'(y) = 0$, and so $h(y) = C$.\n\nTherefore, for any constant $C$ we have that $\\phi (x, y) = \\frac{1}{2} \\ln (x^2 + y^2) + C$ is a potential function.\n\nTherefore $\\mathbf{F}$ is a conservative vector field on all of $\\mathbb{R}^2$ except for at the origin.\n\n## Example 2\n\nDetermine whether the vector field $\\mathbf{F}(x, y) = x^2y \\vec{i} + 2xy^2 \\vec{j}$ is a conservative vector field.\n\nLet's first check the necessary condition for a vector field to be conservative. Let $P(x, y) = x^2y$ and $Q(x, y) = 2xy^2$. Let's calculate the partial derivative of $P$ with respect to $y$:\n\n(9)\n\\begin{align} \\quad \\frac{\\partial P}{\\partial y} = x^2 \\end{align}\n\nNow let's calculate the partial derivative of $Q$ with respect to $x$:\n\n(10)\n\\begin{align} \\quad \\frac{\\partial Q}{\\partial x} = 2y^2 \\end{align}\n\nWe note that $\\frac{\\partial P}{\\partial y} \\neq \\frac{\\partial Q}{\\partial x}$ and so $\\mathbf{F}$ is not a conservative vector field."
] |
[
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|
https://everything2.com/title/sine
|
[
"In trigonometry, the sine of an angle equals the y-coordinate of the point where a line drawn at that angle through the center of a coordinate grid intersects a unit circle.\n\n``` |\n_____|_____ _\n/ | /|\\ \\\n/ | / | \\ } sine\n| | / | | _/\n_________|_______|/___|__|_________\n| | |\n| | |\n\\ | /\n\\_____|_____/\n|\n|\n```\n\nIn general, it is defined (in a right triangle) as the ratio of the length of the side opposite the angle to the length of the hypotenuse; in the unit circle diagram, however, the hypotenuse is always one unit in length.\n\n``` /|\n/ |\n/ |\nhypotenuse / | opposite\n/ |\n/ |\n/ _|\n/_____|_|\nangle\n```\n\nThe name is derived from the Latin word sinus, meaning \"curve\". The sine of a number is just another number, of course, but when the sine function is graphed on a coordinate graph, it produces a smooth, rolling (\"sinusoidal\") curve, hence the name.\n\nSine is an elliptic curve. More to the point, any particular sin curve is the exact replica of the edge of a particular ellipse. Observe the sin curve for y=sin(x), 0<x<2*pi:\n\n```| 1 ,,oO```Oo,,\n| ,oO` `Oo,\n| ,/` `\\,\n| ,/` `\\,\n| ,/` `\\,\n| /` `\\\n| ,/ \\,\n| /` `\\\n| ,/ \\, pi\n| /` `\\ |\n| ,/ \\, |\n|,/` `\\, 3*pi/2\n|-------------------------|------------------------`\\----------------------|--------------------/`\n| pi/2 \\, ,/\n| `\\ /`\n| \\, ,/\n| `\\ /`\n| \\, ,/\n| `\\, ,/`\n| `\\, ,/`\n| `\\, ,/`\n| `Oo, ,oO`\n| -1 ``Oo,,,oO``\n```\n\nImagine that the straight horizontal line (above) is the path a circle of radius one follows as it rolls to the right. Then imagine that, instead of just a circle, it's a cylinder of radius one and length two. (The cylinder is rolling 'on top of' the entire picture above, as if you had it in your hand and were rolling it on your screen.) Now consider an ellipse on the surface of the cylinder: going from one 'corner' to the other. As the cylinder rolls, the ellipse will trace the sine curve above, and a circle at the midpoint of the cylinder will trace the straight line. Thus, whenever you are calculating anything with sin or cos, remember that you are actually using ellipses.\n\nOne series for finding sin(x) is\n\nsin(x) |-> x/1! - x3/3! + x5/5! - x7/7! +...\n\nIt is obvious from this that sin(0) = 0, since all the terms evaluate to 0.\n\nIf infinite series are not your thing, try this infinite product, although it is not very clear in HTML:\n\nsin(x) = (1 - x22 ) * (1 - x2(22π2)) * (1 - x2/(32π2))* ...\n\nNote that these definitions of sine require an input in radians, not degrees, with which some are more familiar. To convert to degrees, simply multiply by 180/pi.\n\nLike cosine, sine is periodic in 2π. The derivative of sine is cosine, and its integral is -cosine.\n\nCompare this series to the series for cosine and hyperbolic sine.\n\nHere are some more useful facts about sine, some gathered together from other nodes, others apparently not yet noded.\n\nThe law of sines: In any triangle, the ratio of the sine of an angle to the length of the opposite side is constant. That is,\n\na / sin A = b / sin B = c / sin C\nwhere we are writing A for the angle opposite side a. Or you can write them as sin A / a. If any of the a or sin A is zero it can't be a triangle, it's just flat.\n\nExact values of sine, cosine, and tan: There is an easy-to-remember progression of exact values for the three most important acute angles, 30°, 45°, and 60°:\n\nsin 0° = √0 / 2 = 0\nsin 30° = √1 / 2 = 0.5\nsin 45° = √2 / 2 = 1/2 ≈ 0.7071\nsin 60° = √3 / 2 ≈ 0.8660\nsin 90° = √4 / 2 = 1\nNote this is not simply formulaic: 15° and 75° don't fit in so neatly, but they're less often used. Cosines work in the same way, but downwards from 4 to 0.\n\nOther identities:\n\nsin (−θ) = −sin θ\nsin (θ + φ) = sin θ cos φ + cos θ sin φ\nsin 2θ = 2 sin θ cos θ\nsin2 θ + cos2 θ = 1\nsin (θ + 2π) = sin θ\nsin θ = cos (&pi/2 − θ)\nsin (π − θ) = sin θ\n\ncosecant: The reciprocal of sine is called cosecant, abbreviated cosec or csc. This isn't greatly important as a function in its own right, except as a notational convenience: although the square of sin x is written sin² x, its reciprocal is never written as sin−1 x, that notation being reserved for its inverse function.\n\narcsine: The inverse of the sine function is arcsine, symbol sin−1 or arsin or arcsin. Since sin is periodic, its inverse is not uniquely defined as a function. Restricting sin to the interval [−π/2, π/2] makes it a one-to-one mapping onto the interval [−1, 1], so we can define a principal arcsine function, symbolized Arcsin or Sin−1. So Arcsin 1/√2 = π/4.\n\nsignum: As 'sine' is pronounced the same as 'sign' in English, we have a problem when we actually want to talk about the sign of something: whether it's positive or negative. So the signum function is used, symbol sgn, taking the three values {−1, 0, 1} depending on the sign-with-a-g of its argument. Presumably to be pronounced with the first bit like 'signal'.\n\nSine (?), n. [LL. sinus a sine, L. sinus bosom, used in translating the Ar. jaib, properly, bosom, but probably read by mistake (the consonants being the same) for an original jiba sine, from Skr. jiva bowstring, chord of an arc, sine.] Trig. (a)\n\nThe length of a perpendicular drawn from one extremity of an arc of a circle to the diameter drawn through the other extremity.\n\n(b)\n\nThe perpendicular itself. See Sine of angle, below.\n\nArtificial sines, logarithms of the natural sines, or logarithmic sines. -- Curve of sines. See Sinusoid. -- Natural sines, the decimals expressing the values of the sines, the radius being unity. -- Sine of an angle, in a circle whose radius is unity, the sine of the arc that measures the angle; in a right-angled triangle, the side opposite the given angle divided by the hypotenuse. See Trigonometrical function, under Function. -- Versed sine, that part of the diameter between the sine and the arc.\n\nSi\"ne (?), prep. [L.]\n\nWithout.\n\nLog in or register to write something here or to contact authors."
] |
[
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{"ft_lang_label":"__label__en","ft_lang_prob":0.8867826,"math_prob":0.99004805,"size":2112,"snap":"2022-05-2022-21","text_gpt3_token_len":672,"char_repetition_ratio":0.10104364,"word_repetition_ratio":0.0,"special_character_ratio":0.31628788,"punctuation_ratio":0.118393235,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99685025,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-17T10:35:22Z\",\"WARC-Record-ID\":\"<urn:uuid:2589a4d2-503f-4246-b23c-2f5d2a2d28ed>\",\"Content-Length\":\"41367\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:08db07b9-4856-4be5-b228-2b835dd201d4>\",\"WARC-Concurrent-To\":\"<urn:uuid:dbed15f8-e768-41cf-b86b-1b1d85f7174d>\",\"WARC-IP-Address\":\"52.40.220.240\",\"WARC-Target-URI\":\"https://everything2.com/title/sine\",\"WARC-Payload-Digest\":\"sha1:NWNZDSTRC46MZQ4EHTJLKYQA7YLMR6OZ\",\"WARC-Block-Digest\":\"sha1:7PRZS7L3LU4PHKF4THKPGQTKUNE3Y4UF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662517245.1_warc_CC-MAIN-20220517095022-20220517125022-00497.warc.gz\"}"}
|
https://docs.gpytorch.ai/en/stable/_modules/gpytorch/mlls/predictive_log_likelihood.html
|
[
"# Source code for gpytorch.mlls.predictive_log_likelihood\n\n#!/usr/bin/env python3\n\nfrom ._approximate_mll import _ApproximateMarginalLogLikelihood\n\n[docs]class PredictiveLogLikelihood(_ApproximateMarginalLogLikelihood):\nr\"\"\"\nAn alternative objective function for approximate GPs, proposed in Jankowiak et al., 2020_.\nIt typically produces better predictive variances than the :obj:gpytorch.mlls.VariationalELBO objective.\n\n.. math::\n\n\\begin{align*}\n\\mathcal{L}_\\text{ELBO} &=\n\\mathbb{E}_{p_\\text{data}( y, \\mathbf x )} \\left[\n\\log p( y \\! \\mid \\! \\mathbf x)\n\\right] - \\beta \\: \\text{KL} \\left[ q( \\mathbf u) \\Vert p( \\mathbf u) \\right]\n\\\\\n&\\approx \\sum_{i=1}^N \\log \\mathbb{E}_{q(\\mathbf u)} \\left[\n\\int p( y_i \\! \\mid \\! f_i) p(f_i \\! \\mid \\! \\mathbf u, \\mathbf x_i) \\: d f_i\n\\right] - \\beta \\: \\text{KL} \\left[ q( \\mathbf u) \\Vert p( \\mathbf u) \\right]\n\\end{align*}\n\nwhere :math:N is the total number of datapoints, :math:q(\\mathbf u) is the variational distribution for\nthe inducing function values, and :math:p(\\mathbf u) is the prior distribution for the inducing function\nvalues.\n\n:math:\\beta is a scaling constant that reduces the regularization effect of the KL\ndivergence. Setting :math:\\beta=1 (default) results in an objective that can be motivated by a connection\nto Stochastic Expectation Propagation (see Jankowiak et al., 2020_ for details).\n\n.. note::\nThis objective is very similar to the variational ELBO.\nThe only difference is that the :math:log occurs *outside* the expectation :math:\\mathbb{E}_{q(\\mathbf u)}.\nThis difference results in very different predictive performance (see Jankowiak et al., 2020_).\n\n:param ~gpytorch.likelihoods.Likelihood likelihood: The likelihood for the model\n:param ~gpytorch.models.ApproximateGP model: The approximate GP model\n:param int num_data: The total number of training data points (necessary for SGD)\n:param float beta: (optional, default=1.) A multiplicative factor for the KL divergence term.\nSetting it to anything less than 1 reduces the regularization effect of the model\n(similarly to what was proposed in the beta-VAE paper_).\n:param bool combine_terms: (default=True): Whether or not to sum the\nexpected NLL with the KL terms (default True)\n\nExample:\n>>> # model is a gpytorch.models.ApproximateGP\n>>> # likelihood is a gpytorch.likelihoods.Likelihood\n>>> mll = gpytorch.mlls.PredictiveLogLikelihood(likelihood, model, num_data=100, beta=0.5)\n>>>\n>>> output = model(train_x)\n>>> loss = -mll(output, train_y)\n>>> loss.backward()\n\n.. _Jankowiak et al., 2020:\nhttps://arxiv.org/abs/1910.07123\n\"\"\"\n\ndef _log_likelihood_term(self, approximate_dist_f, target, **kwargs):\nreturn self.likelihood.log_marginal(target, approximate_dist_f, **kwargs).sum(-1)\n\n[docs] def forward(self, approximate_dist_f, target, **kwargs):\nr\"\"\"\nComputes the predictive cross entropy given :math:q(\\mathbf f) and :math:\\mathbf y.\nCalling this function will call the likelihood's\n:meth:~gpytorch.likelihoods.Likelihood.forward function.\n\n:param ~gpytorch.distributions.MultivariateNormal variational_dist_f: :math:q(\\mathbf f)\nthe outputs of the latent function (the :obj:gpytorch.models.ApproximateGP)\n:param torch.Tensor target: :math:\\mathbf y The target values\n:param kwargs: Additional arguments passed to the\nlikelihood's :meth:~gpytorch.likelihoods.Likelihood.forward function.\n:rtype: torch.Tensor\n:return: Predictive log likelihood. Output shape corresponds to batch shape of the model/input data.\n\"\"\"\nreturn super().forward(approximate_dist_f, target, **kwargs)"
] |
[
null
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|
https://nanamom.savingadvice.com/2006/11/
|
[
"User Real IP - 52.23.219.12\n```Array\n(\n => Array\n(\n => 182.68.68.92\n)\n\n => Array\n(\n => 101.0.41.201\n)\n\n => Array\n(\n => 43.225.98.123\n)\n\n => Array\n(\n => 2.58.194.139\n)\n\n => Array\n(\n => 46.119.197.104\n)\n\n => Array\n(\n => 45.249.8.93\n)\n\n => Array\n(\n => 103.12.135.72\n)\n\n => Array\n(\n => 157.35.243.216\n)\n\n => Array\n(\n => 209.107.214.176\n)\n\n => Array\n(\n => 5.181.233.166\n)\n\n => Array\n(\n => 106.201.10.100\n)\n\n => Array\n(\n => 36.90.55.39\n)\n\n => Array\n(\n => 119.154.138.47\n)\n\n => Array\n(\n => 51.91.31.157\n)\n\n => Array\n(\n => 182.182.65.216\n)\n\n => Array\n(\n => 157.35.252.63\n)\n\n => Array\n(\n => 14.142.34.163\n)\n\n => Array\n(\n => 178.62.43.135\n)\n\n => Array\n(\n => 43.248.152.148\n)\n\n => Array\n(\n => 222.252.104.114\n)\n\n => Array\n(\n => 209.107.214.168\n)\n\n => Array\n(\n => 103.99.199.250\n)\n\n => Array\n(\n => 178.62.72.160\n)\n\n => Array\n(\n => 27.6.1.170\n)\n\n => Array\n(\n => 182.69.249.219\n)\n\n => Array\n(\n => 110.93.228.86\n)\n\n => Array\n(\n => 72.255.1.98\n)\n\n => Array\n(\n => 182.73.111.98\n)\n\n => Array\n(\n => 45.116.117.11\n)\n\n => Array\n(\n => 122.15.78.189\n)\n\n => Array\n(\n => 14.167.188.234\n)\n\n => Array\n(\n => 223.190.4.202\n)\n\n => Array\n(\n => 202.173.125.19\n)\n\n => Array\n(\n => 103.255.5.32\n)\n\n => Array\n(\n => 39.37.145.103\n)\n\n => Array\n(\n => 140.213.26.249\n)\n\n => Array\n(\n => 45.118.166.85\n)\n\n => Array\n(\n => 102.166.138.255\n)\n\n => Array\n(\n => 77.111.246.234\n)\n\n => Array\n(\n => 45.63.6.196\n)\n\n => Array\n(\n => 103.250.147.115\n)\n\n => Array\n(\n => 223.185.30.99\n)\n\n => Array\n(\n => 103.122.168.108\n)\n\n => Array\n(\n => 123.136.203.21\n)\n\n => Array\n(\n => 171.229.243.63\n)\n\n => Array\n(\n => 153.149.98.149\n)\n\n => Array\n(\n => 223.238.93.15\n)\n\n => Array\n(\n => 178.62.113.166\n)\n\n => Array\n(\n => 101.162.0.153\n)\n\n => Array\n(\n => 121.200.62.114\n)\n\n => Array\n(\n => 14.248.77.252\n)\n\n => Array\n(\n => 95.142.117.29\n)\n\n => Array\n(\n => 150.129.60.107\n)\n\n => Array\n(\n => 94.205.243.22\n)\n\n => Array\n(\n => 115.42.71.143\n)\n\n => Array\n(\n => 117.217.195.59\n)\n\n => Array\n(\n => 182.77.112.56\n)\n\n => Array\n(\n => 182.77.112.108\n)\n\n => Array\n(\n => 41.80.69.10\n)\n\n => Array\n(\n => 117.5.222.121\n)\n\n => Array\n(\n => 103.11.0.38\n)\n\n => Array\n(\n => 202.173.127.140\n)\n\n => Array\n(\n => 49.249.249.50\n)\n\n => Array\n(\n => 116.72.198.211\n)\n\n => Array\n(\n => 223.230.54.53\n)\n\n => Array\n(\n => 102.69.228.74\n)\n\n => Array\n(\n => 39.37.251.89\n)\n\n => Array\n(\n => 39.53.246.141\n)\n\n => Array\n(\n => 39.57.182.72\n)\n\n => Array\n(\n => 209.58.130.210\n)\n\n => Array\n(\n => 104.131.75.86\n)\n\n => Array\n(\n => 106.212.131.255\n)\n\n => Array\n(\n => 106.212.132.127\n)\n\n => Array\n(\n => 223.190.4.60\n)\n\n => Array\n(\n => 103.252.116.252\n)\n\n => Array\n(\n => 103.76.55.182\n)\n\n => Array\n(\n => 45.118.166.70\n)\n\n => Array\n(\n => 103.93.174.215\n)\n\n => Array\n(\n => 5.62.62.142\n)\n\n => Array\n(\n => 182.179.158.156\n)\n\n => Array\n(\n => 39.57.255.12\n)\n\n => Array\n(\n => 39.37.178.37\n)\n\n => Array\n(\n => 182.180.165.211\n)\n\n => Array\n(\n => 119.153.135.17\n)\n\n => Array\n(\n => 72.255.15.244\n)\n\n => Array\n(\n => 139.180.166.181\n)\n\n => Array\n(\n => 70.119.147.111\n)\n\n => Array\n(\n => 106.210.40.83\n)\n\n => Array\n(\n => 14.190.70.91\n)\n\n => Array\n(\n => 202.125.156.82\n)\n\n => Array\n(\n => 115.42.68.38\n)\n\n => Array\n(\n => 102.167.13.108\n)\n\n => Array\n(\n => 117.217.192.130\n)\n\n => Array\n(\n => 205.185.223.156\n)\n\n => Array\n(\n => 171.224.180.29\n)\n\n => Array\n(\n => 45.127.45.68\n)\n\n => Array\n(\n => 195.206.183.232\n)\n\n => Array\n(\n => 49.32.52.115\n)\n\n => Array\n(\n => 49.207.49.223\n)\n\n => Array\n(\n => 45.63.29.61\n)\n\n => Array\n(\n => 103.245.193.214\n)\n\n => Array\n(\n => 39.40.236.69\n)\n\n => Array\n(\n => 62.80.162.111\n)\n\n => Array\n(\n => 45.116.232.56\n)\n\n => Array\n(\n => 45.118.166.91\n)\n\n => Array\n(\n => 180.92.230.234\n)\n\n => Array\n(\n => 157.40.57.160\n)\n\n => Array\n(\n => 110.38.38.130\n)\n\n => Array\n(\n => 72.255.57.183\n)\n\n => Array\n(\n => 182.68.81.85\n)\n\n => Array\n(\n => 39.57.202.122\n)\n\n => Array\n(\n => 119.152.154.36\n)\n\n => Array\n(\n => 5.62.62.141\n)\n\n => Array\n(\n => 119.155.54.232\n)\n\n => Array\n(\n => 39.37.141.22\n)\n\n => Array\n(\n => 183.87.12.225\n)\n\n => Array\n(\n => 107.170.127.117\n)\n\n => Array\n(\n => 125.63.124.49\n)\n\n => Array\n(\n => 39.42.191.3\n)\n\n => Array\n(\n => 116.74.24.72\n)\n\n => Array\n(\n => 46.101.89.227\n)\n\n => Array\n(\n => 202.173.125.247\n)\n\n => Array\n(\n => 39.42.184.254\n)\n\n => Array\n(\n => 115.186.165.132\n)\n\n => Array\n(\n => 39.57.206.126\n)\n\n => Array\n(\n => 103.245.13.145\n)\n\n => Array\n(\n => 202.175.246.43\n)\n\n => Array\n(\n => 192.140.152.150\n)\n\n => Array\n(\n => 202.88.250.103\n)\n\n => Array\n(\n => 103.248.94.207\n)\n\n => Array\n(\n => 77.73.66.101\n)\n\n => Array\n(\n => 104.131.66.8\n)\n\n => Array\n(\n => 113.186.161.97\n)\n\n => Array\n(\n => 222.254.5.7\n)\n\n => Array\n(\n => 223.233.67.247\n)\n\n => Array\n(\n => 171.249.116.146\n)\n\n => Array\n(\n => 47.30.209.71\n)\n\n => Array\n(\n => 202.134.13.130\n)\n\n => Array\n(\n => 27.6.135.7\n)\n\n => Array\n(\n => 107.170.186.79\n)\n\n => Array\n(\n => 103.212.89.171\n)\n\n => Array\n(\n => 117.197.9.77\n)\n\n => Array\n(\n => 122.176.206.233\n)\n\n => Array\n(\n => 192.227.253.222\n)\n\n => Array\n(\n => 182.188.224.119\n)\n\n => Array\n(\n => 14.248.70.74\n)\n\n => Array\n(\n => 42.118.219.169\n)\n\n => Array\n(\n => 110.39.146.170\n)\n\n => Array\n(\n => 119.160.66.143\n)\n\n => Array\n(\n => 103.248.95.130\n)\n\n => Array\n(\n => 27.63.152.208\n)\n\n => Array\n(\n => 49.207.114.96\n)\n\n => Array\n(\n => 102.166.23.214\n)\n\n => Array\n(\n => 175.107.254.73\n)\n\n => Array\n(\n => 103.10.227.214\n)\n\n => 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Array\n(\n => 203.122.6.18\n)\n\n => Array\n(\n => 157.42.38.251\n)\n\n => Array\n(\n => 45.112.68.222\n)\n\n => Array\n(\n => 49.206.212.122\n)\n\n => Array\n(\n => 104.236.70.228\n)\n\n => Array\n(\n => 42.111.34.243\n)\n\n => Array\n(\n => 84.241.19.186\n)\n\n => Array\n(\n => 89.187.180.207\n)\n\n => Array\n(\n => 104.243.212.118\n)\n\n => Array\n(\n => 104.236.55.136\n)\n\n => Array\n(\n => 106.201.16.163\n)\n\n => Array\n(\n => 46.101.40.25\n)\n\n => Array\n(\n => 45.118.166.94\n)\n\n => Array\n(\n => 49.36.128.102\n)\n\n => Array\n(\n => 14.142.193.58\n)\n\n => Array\n(\n => 212.79.124.176\n)\n\n => Array\n(\n => 45.32.191.194\n)\n\n => Array\n(\n => 105.112.107.46\n)\n\n => Array\n(\n => 106.201.14.8\n)\n\n => Array\n(\n => 110.93.240.65\n)\n\n => Array\n(\n => 27.96.95.177\n)\n\n => Array\n(\n => 45.41.134.35\n)\n\n => Array\n(\n => 180.151.13.110\n)\n\n => Array\n(\n => 101.53.242.89\n)\n\n => Array\n(\n => 115.186.3.110\n)\n\n => Array\n(\n => 171.49.185.242\n)\n\n => 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Array\n(\n => 14.248.77.166\n)\n\n => Array\n(\n => 157.35.224.133\n)\n\n => Array\n(\n => 183.83.38.27\n)\n\n => Array\n(\n => 182.68.4.77\n)\n\n => Array\n(\n => 122.177.130.234\n)\n\n => Array\n(\n => 103.24.99.99\n)\n\n => Array\n(\n => 103.91.127.66\n)\n\n => Array\n(\n => 41.90.34.240\n)\n\n => Array\n(\n => 49.205.77.102\n)\n\n => Array\n(\n => 103.248.94.142\n)\n\n => Array\n(\n => 104.143.92.170\n)\n\n => Array\n(\n => 219.91.157.114\n)\n\n => Array\n(\n => 223.190.88.22\n)\n\n => Array\n(\n => 223.190.86.232\n)\n\n => Array\n(\n => 39.41.172.80\n)\n\n => Array\n(\n => 124.107.206.5\n)\n\n => Array\n(\n => 139.167.180.224\n)\n\n => Array\n(\n => 93.76.64.248\n)\n\n => Array\n(\n => 65.216.227.119\n)\n\n => Array\n(\n => 223.190.119.141\n)\n\n => Array\n(\n => 110.93.237.179\n)\n\n => Array\n(\n => 41.90.7.85\n)\n\n => Array\n(\n => 103.100.6.26\n)\n\n => Array\n(\n => 104.140.83.13\n)\n\n => Array\n(\n => 223.190.119.133\n)\n\n => Array\n(\n => 119.152.150.87\n)\n\n => Array\n(\n => 103.125.130.147\n)\n\n => Array\n(\n => 27.6.5.52\n)\n\n => Array\n(\n => 103.98.188.26\n)\n\n => Array\n(\n => 39.35.121.81\n)\n\n => Array\n(\n => 74.119.146.182\n)\n\n => Array\n(\n => 5.181.233.162\n)\n\n => Array\n(\n => 157.39.18.60\n)\n\n => Array\n(\n => 1.187.252.25\n)\n\n => Array\n(\n => 39.42.145.59\n)\n\n => Array\n(\n => 39.35.39.198\n)\n\n => Array\n(\n => 49.36.128.214\n)\n\n => Array\n(\n => 182.190.20.56\n)\n\n => Array\n(\n => 122.180.249.189\n)\n\n => Array\n(\n => 117.217.203.107\n)\n\n => Array\n(\n => 103.70.82.241\n)\n\n => Array\n(\n => 45.118.166.68\n)\n\n => Array\n(\n => 122.180.168.39\n)\n\n => Array\n(\n => 149.28.67.254\n)\n\n => Array\n(\n => 223.233.73.8\n)\n\n => Array\n(\n => 122.167.140.0\n)\n\n => Array\n(\n => 95.158.51.55\n)\n\n => Array\n(\n => 27.96.95.134\n)\n\n => Array\n(\n => 49.206.214.53\n)\n\n => Array\n(\n => 212.103.49.92\n)\n\n => Array\n(\n => 122.177.115.101\n)\n\n => Array\n(\n => 171.50.187.124\n)\n\n => Array\n(\n => 122.164.55.107\n)\n\n => Array\n(\n => 98.114.217.204\n)\n\n => Array\n(\n => 106.215.10.54\n)\n\n => Array\n(\n => 115.42.68.28\n)\n\n => Array\n(\n => 104.194.220.87\n)\n\n => Array\n(\n => 103.137.84.170\n)\n\n => Array\n(\n => 61.16.142.110\n)\n\n => Array\n(\n => 212.103.49.85\n)\n\n => Array\n(\n => 39.53.248.162\n)\n\n => Array\n(\n => 203.122.40.214\n)\n\n => Array\n(\n => 117.217.198.72\n)\n\n => Array\n(\n => 115.186.191.203\n)\n\n => Array\n(\n => 120.29.100.199\n)\n\n => Array\n(\n => 45.151.237.24\n)\n\n => Array\n(\n => 223.190.125.232\n)\n\n => Array\n(\n => 41.80.151.17\n)\n\n => Array\n(\n => 23.111.188.5\n)\n\n => Array\n(\n => 223.190.125.216\n)\n\n => Array\n(\n => 103.217.133.119\n)\n\n => Array\n(\n => 103.198.173.132\n)\n\n => Array\n(\n => 47.31.155.89\n)\n\n => Array\n(\n => 223.190.20.253\n)\n\n => Array\n(\n => 104.131.92.125\n)\n\n => Array\n(\n => 223.190.19.152\n)\n\n => Array\n(\n => 103.245.193.191\n)\n\n => Array\n(\n => 106.215.58.255\n)\n\n => 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Array\n(\n => 194.104.23.140\n)\n\n => Array\n(\n => 49.205.193.252\n)\n\n => Array\n(\n => 222.252.33.119\n)\n\n => Array\n(\n => 173.255.132.114\n)\n\n => Array\n(\n => 182.64.148.162\n)\n\n => Array\n(\n => 175.176.87.8\n)\n\n => Array\n(\n => 5.62.57.6\n)\n\n => Array\n(\n => 119.160.96.229\n)\n\n => Array\n(\n => 49.205.180.226\n)\n\n => Array\n(\n => 95.142.120.59\n)\n\n => Array\n(\n => 183.82.116.204\n)\n\n => Array\n(\n => 202.89.69.186\n)\n\n => Array\n(\n => 39.48.165.36\n)\n\n => Array\n(\n => 192.140.149.81\n)\n\n => Array\n(\n => 198.16.70.28\n)\n\n => Array\n(\n => 103.25.250.236\n)\n\n => Array\n(\n => 106.76.202.244\n)\n\n => Array\n(\n => 47.8.8.165\n)\n\n => Array\n(\n => 202.5.145.213\n)\n\n => Array\n(\n => 106.212.188.243\n)\n\n => Array\n(\n => 106.215.89.2\n)\n\n => Array\n(\n => 119.82.83.148\n)\n\n => Array\n(\n => 123.24.164.245\n)\n\n => Array\n(\n => 187.67.51.106\n)\n\n => Array\n(\n => 117.196.119.95\n)\n\n => Array\n(\n => 95.142.120.66\n)\n\n => Array\n(\n => 156.146.59.35\n)\n\n => Array\n(\n => 49.205.213.148\n)\n\n => Array\n(\n => 111.223.27.206\n)\n\n => Array\n(\n => 49.205.212.86\n)\n\n => Array\n(\n => 103.77.42.103\n)\n\n => Array\n(\n => 110.227.62.25\n)\n\n => Array\n(\n => 122.179.54.140\n)\n\n => Array\n(\n => 157.39.239.81\n)\n\n => Array\n(\n => 138.128.27.234\n)\n\n => Array\n(\n => 103.244.176.194\n)\n\n => Array\n(\n => 130.105.10.127\n)\n\n => Array\n(\n => 103.116.250.191\n)\n\n => Array\n(\n => 122.180.186.6\n)\n\n => Array\n(\n => 101.53.228.52\n)\n\n => Array\n(\n => 39.57.138.90\n)\n\n => Array\n(\n => 197.156.137.165\n)\n\n => Array\n(\n => 49.37.155.78\n)\n\n => Array\n(\n => 39.59.81.32\n)\n\n => Array\n(\n => 45.127.44.78\n)\n\n => Array\n(\n => 103.58.155.83\n)\n\n => Array\n(\n => 175.107.220.20\n)\n\n => Array\n(\n => 14.255.9.197\n)\n\n => Array\n(\n => 103.55.63.146\n)\n\n => Array\n(\n => 49.205.138.81\n)\n\n => Array\n(\n => 45.35.222.243\n)\n\n => Array\n(\n => 203.190.34.57\n)\n\n => Array\n(\n => 205.253.121.11\n)\n\n => Array\n(\n => 154.72.171.177\n)\n\n => Array\n(\n => 39.52.203.37\n)\n\n => Array\n(\n => 122.161.52.2\n)\n\n => Array\n(\n => 82.145.41.170\n)\n\n => Array\n(\n => 103.217.123.33\n)\n\n => Array\n(\n => 103.150.238.100\n)\n\n => Array\n(\n => 125.99.11.182\n)\n\n => Array\n(\n => 103.217.178.70\n)\n\n => Array\n(\n => 197.210.227.95\n)\n\n => Array\n(\n => 116.75.212.153\n)\n\n => Array\n(\n => 212.102.42.202\n)\n\n => Array\n(\n => 49.34.177.147\n)\n\n => Array\n(\n => 173.242.123.110\n)\n\n => Array\n(\n => 49.36.35.254\n)\n\n => Array\n(\n => 202.47.59.82\n)\n\n => Array\n(\n => 157.42.197.119\n)\n\n => Array\n(\n => 103.99.196.250\n)\n\n => Array\n(\n => 119.155.228.244\n)\n\n => Array\n(\n => 130.105.160.170\n)\n\n => Array\n(\n => 78.132.235.189\n)\n\n => Array\n(\n => 202.142.186.114\n)\n\n => Array\n(\n => 115.99.156.136\n)\n\n => Array\n(\n => 14.162.166.254\n)\n\n => Array\n(\n => 157.39.133.205\n)\n\n => Array\n(\n => 103.196.139.157\n)\n\n => Array\n(\n => 139.99.159.20\n)\n\n => Array\n(\n => 175.176.87.42\n)\n\n => Array\n(\n => 103.46.202.244\n)\n\n => Array\n(\n => 175.176.87.16\n)\n\n => Array\n(\n => 49.156.85.55\n)\n\n => Array\n(\n => 157.39.101.65\n)\n\n => Array\n(\n => 124.253.195.93\n)\n\n => Array\n(\n => 110.227.59.8\n)\n\n => Array\n(\n => 157.50.50.6\n)\n\n => Array\n(\n => 95.142.120.25\n)\n\n => Array\n(\n => 49.36.186.141\n)\n\n => Array\n(\n => 110.227.54.161\n)\n\n => Array\n(\n => 88.117.62.180\n)\n\n => Array\n(\n => 110.227.57.8\n)\n\n => Array\n(\n => 106.200.36.21\n)\n\n => Array\n(\n => 202.131.143.247\n)\n\n => Array\n(\n => 103.46.202.4\n)\n\n => Array\n(\n => 122.177.78.217\n)\n\n => Array\n(\n => 124.253.195.201\n)\n\n)\n```\nArchive for November, 2006: Nanamom's thoughts Blog\n Layout: Blue and Brown (Default) Author's Creation\n Home > Archive: November, 2006\n\n# Archive for November, 2006\n\n## Title Search\n\nNovember 29th, 2006 at 07:34 pm\n\nThis is done by the seller I am told. I am also told that it is the only thing hanging between us and a closing. We have achieved mortgage. Is that an achievement? I keep telling myself that it is an investment. After the first house I bought cost 25,000 and I sold it for 29,000. The second was 35,000 and I sold it for 39,000 (OK not an improvement but different place and time) So if I buy this at 48,000 I should be able to sell it for more someday, or at least my kids will. I hope! At least I won't be paying a landlord anymore. I can do what we want to do in and out of the house, as long as we can afford it! Our payments will be lower even with the insurance and taxes going to escrow. So we save over \\$100.00 a month to put into repairs. If all works out according to plan we should have the closing costs covered by the seller, and get some money back to do windows. Our EF (small though it may be will be used if we need to to do that becuase we can't move in until it is done. We'll build it back up ASAP however. Right now the car is slated to be paid off in January at which point we can sell it or give it away. Either way that frees up several hundred between the insurance being cut in half and the car payment being gone. We will be down to one car payment and one mortgage. Our trip East for the holidays may take some credit but we can pay it back as soon as the bill comes in.\n\n## Still Waiting\n\nNovember 28th, 2006 at 10:23 pm\n\nTomorrow I am calling the mortgage company to see if there is anything we can do to get this going. I feel like I am spinning my wheels. I am not sure if I shold pack or not. If I start, where do I start, we need to live until closing and maybe until we find another place. The date on the contract is only 13 days away. Then we need to fix windows (assuming the money is still there to do that) and clean th eplace top to bottom. Not to mention check the plumbing. I also have to decide what if anything to put out for Christmas decorations here and what to not see for a year (I hate that idea, Thanksgiving was a bummer I'd hate to make Christmas that as well)\nOn another note we found a cool exercise thing. We bought a dance pad and hooked it up to the laptop. It was a valid expense since I can't sell what I haven't shown very well, and I got 50% off of it. We took it to the mall show I set up at on Saturday and had a blast. It plays Veggie Tale tunes (silly songs mostly) and you have to keep up with the directional arrows by stepping on them as they fall down the screen. It has 4 levels and believe me they get harder. The medium is just my speed. It also have an exercise mode that just keeps going and a sing along part as well. We had a blast! My DH is 6'5'' and was dancing at the mall having a great time. He is alot of fun to have around. We let the kids try it and they had a blast too. We decided we'd rather do this than walk on the tread mill!\n\nNovember 22nd, 2006 at 01:57 pm\n\nI'm crazy, certifieable now! WE bought our dog a Christmas present yesterday........... a cat. She is tiny but won't stay that way I'm sure. She is siamese/persian and absolutely adorable. I'm a dog person what am I saying? Her blue eyes are looking up at me as I type this. Her fur is perpetually wet thanks to the tongue (bigger than her head) connected to the canine in residence. She will curl up on my lap and that big nose comes over and as soon as it touches her she starts sounding like a percolator. She is having trouble navigating around the house because both the dog and DS2 think she is too little to be on her own and are constantly \"helping\" her around. We got her at the shelter. I figured if we didn't go now Christmas would wipe them out. I had promised DS2 a kitten when we moved.\n\nNovember 21st, 2006 at 04:25 am\n\nThank you to all who prayed for my pastor. He returned home this evening. He was able to eat today (first time in 3 days) and we are thankful to have him back in the states. He is supposed to preach at the Thanksgivng service tomorrow evening, which should be interesting. Anyway I appreciate the prayers. I was terrified I would hav eto explain the unthinkable to our DS2 who adores him.\n\n## Packing\n\nNovember 20th, 2006 at 05:01 pm\n\nI am packing. We have had a meeting of the minds or at least the finances so we will be moving into the house assuming the mortgage goes through. It should considering how low an amount we are paying. So I am packing. I have most of the books, cds, dvds, and th e\"craft\" cupboard packed. I'm not sure where to go from there. We are supposedly closing on the 11th of next month. So I guess I will pick a room an dpack everything I can from that room then go to the next and keep making the rounds until everything is done. i think I will go to the kitchen next. We have some \"company stuff\" we can do without until we move. I have less cupboard space at the new place and no basement so I am paring down as I go. Too bad it is too cold for another garage sale!!!\n\nNovember 18th, 2006 at 02:24 pm\n\nJust got a call yesterday that our pastor who is on a missions trip in Guatemala is extremely ill. They have had a doctor in 3 times and he was finally able to get some rest. He is in a poor area with no hospital that I know of. Please pray that he regians enough strength to come home with the others on Monday and that he recovers fully form this episode. I know he would be happier with Jesus but we need him. My DS2 idolizes him and is not ready to lose anyone else close to him.\n\n## We may be Moving\n\nNovember 16th, 2006 at 07:23 pm\n\nActually it seems that we will be moving. The owners met us halfway and have an estimate from a company on fixing the windows. They will credit us money back at closing to do that. So we are back in the running for a mortgage. I am told an offer will be made up today and presented to us. If it lists everything then we will sign and I will start packing in ernest, actually in boxes. Sorry, couldn't resist.\n\n## Health issues\n\nNovember 13th, 2006 at 03:32 pm\n\nTook my blood pressure this morning and was horrified to discover that I who have always had low low blood pressure is hitting numbers on that dial I don't like. No more salt or soda for me!!! I need to lose a few pounds as well (OK, Quite a few). If that doesn't fix the problem I am going to have to see a doctor. I have too much to live for to fool around. I think I'll take a bit of time for me everyday and sew or stamp as well. Started Christmas cards last night, I am making 90 of them. It isn't relaxing to make that many believe me!!\n\n## The saga continues\n\nNovember 13th, 2006 at 03:30 pm\n\nToday we are to learn what counter proposal the seller wants to give us. This weekend we went to a home type store and looked at prices for what we would have to replace to come up to snuff for the home study. We got a good idea on prices for the window stuff and I think DH was a bit surprised at the prices for the plumbing stuff. We may be borrowing more than the price of the house (not more than the assessed value though) to pay for some of this which I do't like but the price is less than anything else around that has any kind of yard.\n\n## Winter\n\nNovember 10th, 2006 at 08:42 pm\n\nIt is breathtaking outside. We are having our first real snowfall of the season and it looks like a Christmas card. The snow has been falling steadily for about 3 hours and the branches are quite heavily laden. Tiny tracks cross the yard where our intrepid squirrels have been out playing. My world is white and glistening. inside I have a roaring fire to cut my heat bill and lend itself to the ambiance of the day. Hot soup simmers on the stove which twill soon have cocoa as well waiting for my DH to burst in the door covered in white. Both my DS2 and our resident canine are yearning to get out to romp around in the frosty expanse. When he gets up from nap we are going to take glue and scraps of stuff and spend the afternoon creating presents for those we love to bring special smiles to their faces next month.\n\n## Frustration time\n\nNovember 10th, 2006 at 04:35 am\n\nWe went to the house and decided that the only things on the long list the inspector gave us that has to be fixed were the windows and the asbestos tape that is fraying. We put that in writing and gave them to list of things that were also needed we weren't asking for and sent it off. We offered to fix the windows ourselves if they gave back 1000 at closing. WE don't have it or we would just go ahead and close as is. in order to adopt we must have working windows with screens. There is no way around that. So their answer so far is nope, can't do it. Our realtor asked what they were able to counter offer and we are waiting for an answer. Right now it looks like the deal will die here. I have some stuff packed and there is a rental sign in front of our house. Now what??!! Oh well if God is closing this door he will open a window somewhere I'm sure and it will be even better. Right now it is hard to imagine a better yard but we'll see what God has in mind, He knows best.\n\n## Next step\n\nNovember 7th, 2006 at 08:35 pm\n\nOk, We have an appraisal, a home inspection, and are going out to see what the defects really look like. Our realtor suggested that we ask for them to give back 12 to 1500 at the closing so we can do all the repairs ourselves but I'm not sure that will cover it. We decided to go look at the house agaoin and see what the problems may take to fix. Some of them we may be able to do ourselves. I'm already sure we are replacing the toilet ourselves as well as the vent problem with the plumbing. It ocurred to me that the reason th ewindows might not open may be that the people nailed them shut. They have 4 rather boisterous boys and that may be how they guarentee no defenestration occurs. We ar egoing out tomorrow or later today to take a look and see what we have to have fixed in order to still buy the house. The good news is that the amount we are borrowing is close to the value so we will have some money to use for closing and repairs. I wish we had been able to save a bit more ourselves but considering where we were 4 years ago we are doing well to be able to afford a house at all. It will be cheaper than renting and the difference will be used to work on a monthly \"project\"\n\n## too soon to tell\n\nNovember 6th, 2006 at 03:06 pm\n\nWe got the report from the home inspection. Keeping in mind that when you buy an older home there are always problems and that we knew it was a fixer upper it isn't too bad. The house is sturdy, roof within 2 years, electric updated within 2 years, newer hot water tank and furnace. The catches are, asbestos tape fraying (means if is friable and needs to be removed) under the crawl space, windows are either broken or do not open which is a big safety hazard, especially on the second floor. The garage roof needs new shingles. The upstairs toilet needs to be replaced and the downstairs shower needs a control dial. The plumbing vent on the roof must be lengthened.The front and back porch need to be shored up sometime in the next couple years. Little things include outlet covers and switch covers. All the walls need to be painted and patched and the cleaning will be impressive. They also noted defects in the siding on corners. Of all that the non opening windows were a surprise, and the asbestos was a shock. I turned it in to my realtor and we'll see their thoughts on what needs to be fixed before we buy. The asbestos is a must I think the plumbing needs to be addressed and I vote the windows be operable or they take the estimate of costs off of the bill for the house. I sure don't have enough to buy a house and fix all the windows. The rest of it I can deal with.\n\n## This doesn't sound good!\n\nNovember 2nd, 2006 at 10:03 pm\n\nWe were supposed to meet the guy who did the home inspection at noon at the playground in the mall (distraction for DS2) He never showed but did call. It seem he had so much to write that he needed to go home and do so and could he come over tonight to talk to us about the house. OK, we expected a problem with a couple broken windows and a semi minor plumbing problem. This however doesn't sound good. I may be house hunting again. I have planned the kids rooms though. DS2 is space happy so we want to do a navy (or so) ceiling for stars and planets, and continue it a bit down the walls. The walls will have the space shuttle ready for launch or flying in the sky as well as some space stuff, like the space station etc. Maybe a luner lander on the ground. I have been trying to figure out how on earth I would do this (no artistic talent here at all) and a friend of mine mentioned at church how much she loves to draw on walls. I was thrilled (make that THRILLED) and screamed in delight. The girls room is to have a tree with a flower border and the sun shining and a swing from the tree and butterflies and... Our room I haven't quite figured out. The playroom downstairs will have Veggie Tales stuff in it. On every door I want to write a significant Bible verse (they are all significant I know but some apply to my son more than others as well as us...) On th efront door (outside) I want \"As for me and my house we will serve the Lord.\" I think I can stencil that if I find a good set of letters. The kitchen hopefully will have a vine going around th eroom continuing into the dining room with \"You are the vine, we are the branches.\" on it. I hope to trim the library (small hallway but big enough for several bookcases) with books of the Bible trim. One of bathroom will have sea life in it (I have a net from when we went crabbing) and the other maybe birds. Haven't figured out th eliving room yet, but someone did offer us a free couch today. That should make the living room more comfortable since we only hav etwo chairs so far."
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{"ft_lang_label":"__label__en","ft_lang_prob":0.98309535,"math_prob":0.9987687,"size":27564,"snap":"2021-43-2021-49","text_gpt3_token_len":6548,"char_repetition_ratio":0.11048621,"word_repetition_ratio":0.94533527,"special_character_ratio":0.23672181,"punctuation_ratio":0.08117155,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9993199,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-11-30T18:17:24Z\",\"WARC-Record-ID\":\"<urn:uuid:5d4f361b-398f-4c3d-a7ac-c1121c6f6b96>\",\"Content-Length\":\"752896\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f0a0f222-6cad-44ca-b204-ec42c710e084>\",\"WARC-Concurrent-To\":\"<urn:uuid:5c68e9e5-aea7-4120-ab69-808ed60d6cc4>\",\"WARC-IP-Address\":\"173.231.200.26\",\"WARC-Target-URI\":\"https://nanamom.savingadvice.com/2006/11/\",\"WARC-Payload-Digest\":\"sha1:QFYWKJEUSOM2S2DHZFTTRQGFYYMRS2LY\",\"WARC-Block-Digest\":\"sha1:MR6RCME3ZHZHMHCN76PVOXN6TIS6FG2E\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964359065.88_warc_CC-MAIN-20211130171559-20211130201559-00309.warc.gz\"}"}
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https://aspirantszone.com/reasoning-data-sufficiency-set-19/
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[
"# Reasoning :Data Sufficiency Set 19\n\nDirections (1-10) : Each of the questions below consists of a question and two statements numbered a and b given below it. You have to decide whether the data provided in the statement are sufficient to answer the question. Read both the statements and give answer:\nAmong Avantika, Banya, Caryl, Dinesh, Elvish and Farah each one of them has different age, who is the youngest person?\na) Avantika is not the youngest. Elvish is younger than Farah but elder than Banya.\nb) Banya is elder than Dinesh and Avantika. Farah is younger than only Caryl.\n\n1. If the data in statement a alone are sufficient to answer the question, while the data in statement b alone are not sufficient to answer the question.\nIf the data in statement b alone are sufficient to answer the question, while the data in statement a alone are not sufficient to answer the question.\nIf the data either in statement a alone or in statement b alone are sufficient to answer the question.\nIf the data even in both statements a and b together are not sufficient to answer the question.\nIf the data in both statements a and b together are necessary to answer the question.\nOption E\n\n2. Six boxes Jz, Kz, Lz, Mz, Nz and Oz are placed one above another. Which of the following box is placed at bottom?\na)Two boxes are placed between box Jz and box Kz. One box is placed between Mz and Jz. Box Mz placed above box Jz. b) Box Mz is placed just above Box Kz. Three boxes are placed between box Mz and box Jz. Box Oz is placed above Box Mz and Box Jz.\n\nIf the data in statement a alone are sufficient to answer the question, while the data in statement b alone are not sufficient to answer the question.\nIf the data in statement b alone are sufficient to answer the question, while the data in statement a alone are not sufficient to answer the question.\nIf the data either in statement a alone or in statement b alone are sufficient to answer the question.\nIf the data even in both statements a and b together are not sufficient to answer the question.\nIf the data in both statements a and b together are necessary to answer the question.\nOption B\n\n3. By using which statement, we can conclude ‘Some M is not T’.\na)All O are T. All T are A. All A are M.\nb) Some O are M. All A is M. No A is T.\n\nIf the data in statement a alone are sufficient to answer the question, while the data in statement b alone are not sufficient to answer the question.\nIf the data in statement b alone are sufficient to answer the question, while the data in statement a alone are not sufficient to answer the question.\nIf the data either in statement a alone or in statement b alone are sufficient to answer the question.\nIf the data even in both statements a and b together are not sufficient to answer the question.\nIf the data in both statements a and b together are necessary to answer the question.\nOption B\n\n4. What is the code of “Grey”?\na) If “grey brown black” is coded as “ki kc wu” and “brown grey white” is coded as “wu ba ki”\nb) If “grey yale black” is coded as “wu kc bl”\n\nIf the data in statement a alone are sufficient to answer the question, while the data in statement b alone are not sufficient to answer the question.\nIf the data in statement b alone are sufficient to answer the question, while the data in statement a alone are not sufficient to answer the question.\nIf the data either in statement a alone or in statement b alone are sufficient to answer the question.\nIf the data even in both statements a and b together are not sufficient to answer the question.\nIf the data in both statements a and b together are necessary to answer the question.\nOption E\n\n5. On which day Diksha was born?\na) Mani exactly remembers that Diksha birthday is after 13th and before 18th of month.\nb) Dinesh remembers that Diksha birthday is after 14th and before 17th of month.\n\nIf the data in statement a alone are sufficient to answer the question, while the data in statement b alone are not sufficient to answer the question.\nIf the data in statement b alone are sufficient to answer the question, while the data in statement a alone are not sufficient to answer the question.\nIf the data either in statement a alone or in statement b alone are sufficient to answer the question.\nIf the data even in both statements a and b together are not sufficient to answer the question.\nIf the data in both statements a and b together are necessary to answer the question.\nOption D\n\n6. By using which statement, we can conclude ‘Some W is not B’.\na) All W is P. Some W is Y. Some G is P. No G is B.\nb) Some P is W. Some W is Y. All Y is G. No G is B.\n\nIf the data in statement a alone are sufficient to answer the question, while the data in statement b alone are not sufficient to answer the question.\nIf the data in statement b alone are sufficient to answer the question, while the data in statement a alone are not sufficient to answer the question.\nIf the data either in statement a alone or in statement b alone are sufficient to answer the question.\nIf the data even in both statements a and b together are not sufficient to answer the question.\nIf the data in both statements a and b together are necessary to answer the question.\nOption B\n\n7. Seven people Az, Bz, Cz, Dz, Ez, Fz and Gz went to a shop, who among the following purchased the fifth highest number of items and what is the number of items being purchased by Cz if the difference between the number of items of Gz and Cz is 2?\na)Fz purchased more items than Az. Gz purchased more items than Bz but less than Cz. Ez purchased more number of items than Dz. The difference between the number of items of Ez and Gz is 22.\nb) Neither Dz nor Bz purchased the least number of items. Cz purchased less number of items than only two people. Dz purchased more items than Fz. The highest number of items is 98, purchased by Ez.\n\nIf the data in statement a alone are sufficient to answer the question, while the data in statement b alone are not sufficient to answer the question.\nIf the data in statement b alone are sufficient to answer the question, while the data in statement a alone are not sufficient to answer the question.\nIf the data either in statement a alone or in statement b alone are sufficient to answer the question.\nIf the data even in both statements a and b together are not sufficient to answer the question.\nIf the data in both statements a and b together are necessary to answer the question.\nOption D\n\n8. On which day Shivani was born?\na) Shivani’s sister remembers that she was born after Monday but before Saturday.\nb) Shivani’s friend remembers that she was born after Tuesday.\n\nIf the data in statement a alone are sufficient to answer the question, while the data in statement b alone are not sufficient to answer the question.\nIf the data in statement b alone are sufficient to answer the question, while the data in statement a alone are not sufficient to answer the question.\nIf the data either in statement a alone or in statement b alone are sufficient to answer the question.\nIf the data even in both statements a and b together are not sufficient to answer the question.\nIf the data in both statements a and b together are necessary to answer the question.\nOption D\n\n9. In a family of six persons who is the father of Eq? a) Cq is the son-in-law of Bq. Aq is the only son of Dq. Dq is the grandmother of Eq.\nb) Eq is the son of Xq, who is the only sister of Aq. Bq is the father of Xq.\n\nIf the data in statement a alone are sufficient to answer the question, while the data in statement b alone are not sufficient to answer the question.\nIf the data in statement b alone are sufficient to answer the question, while the data in statement a alone are not sufficient to answer the question.\nIf the data either in statement a alone or in statement b alone are sufficient to answer the question.\nIf the data even in both statements a and b together are not sufficient to answer the question.\nIf the data in both statements a and b together are necessary to answer the question.\nOption E\n\n10. Five persons are sitting around a circular table facing the center. Who among the following sit immediate left of Caryl?\na)Dinesh sits second to the left of Banya. Elvish is not an immediate neighbour of Dinesh. Caryl is second to the left of Elvish.\nb) Banya sits second to the right of Dinesh. Akansha second to the right of Banya. Two persons sit between Dinesh and Elvish.\n\nIf the data in statement a alone are sufficient to answer the question, while the data in statement b alone are not sufficient to answer the question.\nIf the data in statement b alone are sufficient to answer the question, while the data in statement a alone are not sufficient to answer the question.\nIf the data either in statement a alone or in statement b alone are sufficient to answer the question.\nIf the data even in both statements a and b together are not sufficient to answer the question.\nIf the data in both statements a and b together are necessary to answer the question.\nOption C"
] |
[
null
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|
https://mail.openvswitch.org/pipermail/ovs-dev/2019-September/362616.html
|
[
"# [ovs-dev] [PATCH 03/10] dpif-netdev: Handle uninitialized value error for 'match.wc'\n\nYifeng Sun pkusunyifeng at gmail.com\nWed Sep 11 21:18:29 UTC 2019\n\n```Valgrind reported that match.wc was not initialized, as below:\n\n1176: ofproto-dpif - fragment handling - actions\n\n==21214== Conditional jump or move depends on uninitialised value(s)\n==21214== at 0x4B77C1: odp_flow_key_from_flow__ (odp-util.c:6143)\n==21214== by 0x46DB58: dp_netdev_upcall (dpif-netdev.c:6239)\n==21214== by 0x4774A7: handle_packet_upcall (dpif-netdev.c:6608)\n==21214== by 0x4774A7: fast_path_processing (dpif-netdev.c:6726)\n==21214== by 0x47933C: dp_netdev_input__ (dpif-netdev.c:6814)\n==21214== by 0x479AB8: dp_netdev_input (dpif-netdev.c:6852)\n==21214== by 0x479AB8: dp_netdev_process_rxq_port (dpif-netdev.c:4287)\n==21214== by 0x47A6A9: dpif_netdev_run (dpif-netdev.c:5264)\n==21214== by 0x4324E7: type_run (ofproto-dpif.c:342)\n==21214== by 0x41C5FE: ofproto_type_run (ofproto.c:1734)\n==21214== by 0x40BAAC: bridge_run__ (bridge.c:2965)\n==21214== by 0x410CF3: bridge_run (bridge.c:3029)\n==21214== by 0x407614: main (ovs-vswitchd.c:127)\n==21214== Uninitialised value was created by a stack allocation\n==21214== at 0x4769C3: fast_path_processing (dpif-netdev.c:6672)\n\n'match' is allocated on stack but its 'wc' is accessed in\nodp_flow_key_from_flow__ without proper initialization.\nThis patch fixes it.\n\nSigned-off-by: Yifeng Sun <pkusunyifeng at gmail.com>\n---\nlib/dpif-netdev.c | 1 +\n1 file changed, 1 insertion(+)\n\ndiff --git a/lib/dpif-netdev.c b/lib/dpif-netdev.c\nindex a88a78f8a688..6be6e47ed127 100644\n--- a/lib/dpif-netdev.c\n+++ b/lib/dpif-netdev.c\n@@ -6600,6 +6600,7 @@ handle_packet_upcall(struct dp_netdev_pmd_thread *pmd,\n\nmatch.tun_md.valid = false;\nminiflow_expand(&key->mf, &match.flow);\n+ memset(&match.wc, 0, sizeof match.wc);\n\nofpbuf_clear(actions);\nofpbuf_clear(put_actions);\n--\n2.7.4\n\n```"
] |
[
null
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{"ft_lang_label":"__label__en","ft_lang_prob":0.5325442,"math_prob":0.57225454,"size":1907,"snap":"2019-51-2020-05","text_gpt3_token_len":704,"char_repetition_ratio":0.21334735,"word_repetition_ratio":0.0,"special_character_ratio":0.4058731,"punctuation_ratio":0.22816901,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9511329,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-18T17:19:17Z\",\"WARC-Record-ID\":\"<urn:uuid:0cdcaf79-cfaf-4ba0-90ab-9526ff873ff1>\",\"Content-Length\":\"5043\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1312a6dd-8293-47b0-8d06-3bebd134564e>\",\"WARC-Concurrent-To\":\"<urn:uuid:f44e7378-96c5-43bc-b23d-5f5b6f7f885a>\",\"WARC-IP-Address\":\"140.211.9.53\",\"WARC-Target-URI\":\"https://mail.openvswitch.org/pipermail/ovs-dev/2019-September/362616.html\",\"WARC-Payload-Digest\":\"sha1:N435ZBTOOUHQ4JZB47344PUQAC7TXWU2\",\"WARC-Block-Digest\":\"sha1:43UPEWQGXUEUZELMQEWVEHEDOZ5XZEFH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250593295.11_warc_CC-MAIN-20200118164132-20200118192132-00535.warc.gz\"}"}
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https://www.softschools.com/quizzes/chemistry/solutions_concentration_ii/quiz1332.html
|
[
"# Solutions : Solutions: Concentration II Quiz\n\nQuiz\n*Theme/Title: Solutions: Concentration II\n* Description/Instructions\nSolution concentration can be described quantitatively in several ways. Two of them are molarity and molality. Molarity is the ratio of moles of solute to liters of solution. Molality is the ratio of moles of solute to kilograms of solvent. This quiz will cover molarity and molality problems. You will need access to a periodic table and a calculator. Select the best answer to the choices.\n\nGroup: Chemistry Chemistry Quizzes\nTopic: Solutions"
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.76439273,"math_prob":0.78766876,"size":1146,"snap":"2022-27-2022-33","text_gpt3_token_len":295,"char_repetition_ratio":0.14273205,"word_repetition_ratio":0.07058824,"special_character_ratio":0.18760908,"punctuation_ratio":0.09947644,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95210046,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-02T04:39:10Z\",\"WARC-Record-ID\":\"<urn:uuid:e9a75410-be20-4f9c-8e07-3761f3f39dc3>\",\"Content-Length\":\"15102\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e02271ba-633b-4924-80fa-c7cd8a70107f>\",\"WARC-Concurrent-To\":\"<urn:uuid:b4f39c6e-dff2-4cbd-bedd-da1dc2f04e1e>\",\"WARC-IP-Address\":\"184.154.53.10\",\"WARC-Target-URI\":\"https://www.softschools.com/quizzes/chemistry/solutions_concentration_ii/quiz1332.html\",\"WARC-Payload-Digest\":\"sha1:H4NFOB2UHXRJGTPPCRBCSUZE7XEQEM5M\",\"WARC-Block-Digest\":\"sha1:EMMUG2RMH5GV4TW64H3EYOWDRB34DRI7\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103984681.57_warc_CC-MAIN-20220702040603-20220702070603-00416.warc.gz\"}"}
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https://forum.ansys.com/forums/topic/why-the-kerrnonlinear-model-cause-energy-loss-in-pulse-propagation/
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[
"",
null,
"## Photonics\n\nTagged: ,\n\n•",
null,
"qza0505\nSubscriber\n\nHello!I am trying to study the influence of kerr effect on optical pulse propagation in a SOI waveguide by using the kerrnonlinearmaterial plugin that Lumerical provide. Strangely,the simulation result shows about 45% of energy loss that shouldnt have appeared(kerr effect only change the refractive index).Im wondering where the loss derived from, here are the simlulation details:\n\nThe parameters of kerrnonlinear model are specified below: Permittivity equals the square of 3.48(the refractive index of silicon in 1550nm), and Chi3 value is derived fromFDTD Analysis of Two-Photon Absorption and Free-Carrier Absorption in Si High-Index-Contrast Waveguides\n\nThe SOI waveguide is 450nm*220nm and surrounded by SiO2,the propagation length is 25 micrometers.At 0.1 and 25 micrometers away from the TE mode source I set point and 2D timemonitors to check the spectrum and pulse power:\n\nThe pulse length is 200fs at 1550nm. In order to see the broadened spectrum, the source amplitude is set to 3e9V/m:",
null,
"The power_inand power_outare the transmitted power calculated by integrating the Poynting vector, which shows 45% loss of pulse energy.the intensity of E on XOY plane also gradually decreased.\n\nThank you and looking forward to your reply!\n\n•",
null,
"Guilin Sun\nAnsys Employee\n\nFrom the image and result, it seems the mode source is not well supported by this waveguide, as its intensity decreases sharply with propagation distance.\n\nCould you please check the mode source and make sure the mode fields are not truncated by PML and the mode source is large enough.\n\nIn addition, you do not need to reinstegrate the Poynting vector as the power-monitor can directly give you accurate transmission.\n\nYou may also enable \"frequency dependent profile\" and have 10 to 20 frequency points to overcome the problem of frequency-depedent mode profiles.\n\n•",
null,
"Guilin Sun\nAnsys Employee\n\nPlease elaborate more regarding to \"45% loss of pulse energy\", how did you calculate it?\n\n\"the intensity of E on XOY plane also gradually decreased.\" I think it is reasonable: 1)its power is transfered to other wavelenth due to nonliearity; 2) the material has loss.\n\nWhen you use 10 points for field profile, what is the frequency range? does it include the generated wavelengths? when the generated wavelengths have resulting intensity close to the fundamental wavelength, the ninlinearity might be too high! although Chi3 is on the order of -19, what is the source amplitude? is the index change caused by the nonlinearity a small pertubation, or on the order of 0.01 or larger? please verify before the simulation. In addition, does the nonliearity work for the source bandwidth or word onlt at one single wavelenth?\n\nPlease also check the theory and see what is its assumption.\n\n•",
null,
"qza0505\nSubscriber\n\nThank you for your continuing help! According to your line of thought, I checked the simulation detail as follow:\n\n1. I calculate the pulse energy by integrating the poynting vector from 2D time monitor, which I set at 0.1 and 25 micrometers away from the TE mode source as input and output. Heres the script I use to calculate the pulse energy:\n\np_in = getdata(\"0.1μm 2D Y-normal time monitor\",\"power\");\n\nt_in = getdata(\"0.1μm 2D Y-normal time monitor\",\"t\");\n\npower_in = integrate(p_in,2,t_in);\n\n2. When I change the point for field profile from 10-20, the frequency range doesnt change: for the mode source, the 200fs pulse centered at 1550nm has FWHM from 1540 to 1560nm. And for the monitor, I set the frequency range from 1400 to 1700nm in set global monitor settings in the DFT monitor(I didnt find that in time monitor).\n\n3. The source amplitude is set to 3e9, and the index change is about 0.291.(I calculated by the following formula in ”FDTD Analysis of Two-Photon Absorption and Free-Carrier Absorption in Si High-Index-Contrast Waveguides”)\n\nThis index change is pretty larger than 0.01, but according to the theory model of kerrnonlinear, all I need is to entry only one permittivity value(which is the square of index, 3.48), and this model seems to have nothing to do with wavelength?",
null,
"4. I also tried to set lower source amplitude or lower Chi3, unsurprisingly the loss is gradully disappearing, but the broadened spectrum is also returning to its original source spectrum, making this simualtion seems to be a linear one.\n\nThank you for your patient guidance, here are my entire script(plugin parameters are listed before), could you please take a glance in case I made some stupid mistake? I m really eager to know why this happened.\n\nclear;\nswitchtolayout;\ndeleteall;\nmaterial_sio2 = \"SiO2 (Glass) - Palik\";\nmaterial_nonlinear = \"kerrnonlinear\";\ncwnorm;\namplitude=3e9;\npulselength=200e-15;\nwaveguidelength=25e-6;\nbuffer1=10e-6;\nsubstratethickness=6e-6;\nbuffer2=3e-6;\nwidth=450e-9;\nheight=220e-9;\nset(\"name\",\"SiO2substrate\");\nset(\"x\",0);\nset(\"y\",buffer1+(waveguidelength/2));\nset(\"z\",0-(height/2)-(substratethickness/2));\nset(\"x span\",20e-6);\nset(\"y span\",waveguidelength+2*buffer1);\nset(\"z span\",substratethickness+30e-6);\nset(\"material\",material_sio2);\nset(\"name\",\"waveguide\");\nset(\"x\",0);\nset(\"y\",buffer1+(waveguidelength/2));\nset(\"z\",0);\nset(\"x span\",width);\nset(\"y span\",waveguidelength+2*buffer1);\nset(\"z span\",height);\nset(\"material\",material_nonlinear);\nset(\"x\",0);\nset(\"x span\",width+buffer2);\nset(\"y\",buffer1+(waveguidelength/2));\nset(\"y span\",waveguidelength+9e-6);\nset(\"z\",0);\nset(\"z span\",height+buffer2);\nset(\"simulation time\",50e-12);\nset(\"name\",\"fundamental TE mode source\");\nset(\"mode selection\",\"fundamental TE mode\");\nset(\"injection axis\",\"y-axis\");\nset(\"direction\",\"Forward\");\nset(\"amplitude\",amplitude);\nset(\"frequency dependent profile\",1);\nset(\"number of field profile samples\",10);\nset(\"x\",0);\nset(\"y\",buffer1);\nset(\"z\",0);\nset(\"x span\",width+buffer2);\nset(\"z span\",height+buffer2);\nset(\"set time domain\",1);\nset(\"pulse type\",\"standard\");\nset(\"frequency\",193.414e12);#1550nm\nset(\"pulselength\",pulselength);\nset(\"offset\",pulselength*2);\nset(\"name\",\"XOY power monitor\");\nset(\"monitor type\",\"2D Z-normal\");\nset(\"x\",0);\nset(\"x span\",width+buffer2);\nset(\"y\",buffer1+(waveguidelength/2));\nset(\"y span\",waveguidelength+9e-6);\nset(\"z\",0);\nsetglobalmonitor(\"use wavelength spacing\",1);\nsetglobalmonitor(\"use source limits\",0);\nsetglobalmonitor(\"minimum wavelength\",1.4e-6);\nsetglobalmonitor(\"maximum wavelength\",1.7e-6);\nsetglobalmonitor(\"frequency points\",100);\nset(\"name\",\"XOY pulse movie monitor\");\nset(\"monitor type\",\"2D Z-normal\");\nset(\"x\",0);\nset(\"x span\",width+buffer2);\nset(\"y\",buffer1+(waveguidelength/2));\nset(\"y span\",waveguidelength+9e-6);\nset(\"z\",0);\nset(\"name\",\"25μm point time monitor\");\nset(\"monitor type\",\"Point\");\nset(\"x\",0);\nset(\"y\",buffer1+waveguidelength);\nset(\"z\",0);\nset(\"output power\",1);\nset(\"name\",\"0.1μm point time monitor\");\nset(\"monitor type\",\"Point\");\nset(\"x\",0);\nset(\"y\",buffer1+0.1e-6);\nset(\"z\",0);\nset(\"output power\",1);\n\nset(\"name\",\"25μm 2D Y-normal time monitor\");\nset(\"monitor type\",\"2D Y-normal\");\nset(\"x\",0);\nset(\"x span\",width+buffer2);\nset(\"y\",buffer1+waveguidelength);\nset(\"z\",0);\nset(\"z span\",height+buffer2);\nset(\"output power\",1);\nset(\"name\",\"0.1μm 2D Y-normal time monitor\");\nset(\"monitor type\",\"2D Y-normal\");\nset(\"x\",0);set(\"x span\",width+buffer2);\nset(\"y\",buffer1+0.1e-6);\nset(\"z\",0);set(\"z span\",height+buffer2);\nset(\"output power\",1);\nrun;\np_out = getdata(\"25μm 2D Y-normal time monitor\",\"power\");\nt_out = getdata(\"25μm 2D Y-normal time monitor\",\"t\");\np_in = getdata(\"0.1μm 2D Y-normal time monitor\",\"power\");\nt_in = getdata(\"0.1μm 2D Y-normal time monitor\",\"t\");\npower_out = integrate(p_out,2,t_out);\npower_in = integrate(p_in,2,t_in);\n•",
null,
"Guilin Sun\nAnsys Employee\n\nI think the problem is \"the index change is about 0.291\" . It is too large to be considered as purturbation. It will certainly creates other frequencies. So the injection power is lost to other unwanted wavelengths. You can check the spectrum from a time monitor, and use log scale-frequency you will notice other wavelengths.\n\nIn reality, you may not get such large index change, or you want to avoid it! The pertubation is propotional to Chi3*Intensity. Therefore if Chi3 is correct you will need to reduce Intenisty, or check Chi3 to see if it is the real number for the material. Chi3 should not be arbitrary but only the real material property. Please modify them accordingly. Of course you do not need to use very weak intensity. you will need to use proper intenisty for a given Chi3.",
null,
""
] |
[
null,
"https://forum.ansys.com/wp-content/uploads/2022/01/photonics-1.svg",
null,
"https://secure.gravatar.com/avatar/57a8713f6daa7c71aac5d5563be341ad",
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null,
"https://secure.gravatar.com/avatar/076614572d928ca1567d4a42b6e6dd39",
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|
https://www.imlearningmath.com/how-many-additional-blocks-to-make-the-cube-whole/
|
[
"# How many additional blocks to make the cube whole\n\nQuestion: How many additional blocks to make the cube whole? or How many cubes are missing to build a full cube",
null,
"Answer: So, it’s a “cube” that’s 5 x 5 x 4\n\n100 blocks total.\n\nBottom row 5 x 4 =20\n\nNext row 4 x 4 =16\n\nNext row 3 x 4 =12\n\nNext row 2 x 4 =8\n\nTop row 2 x 4 =8\n\nTotal 64\n\nMissing 100 – 64 = 36\n\nTo make a perfect 5 x 5 x 5 cube, add one more row\n\n5 x 5 =25\n\n25 + 36 = 61"
] |
[
null,
"https://www.imlearningmath.com/wp-content/uploads/2020/05/1-6.jpg",
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.7987643,"math_prob":0.99551445,"size":371,"snap":"2020-45-2020-50","text_gpt3_token_len":146,"char_repetition_ratio":0.15531336,"word_repetition_ratio":0.02173913,"special_character_ratio":0.43396226,"punctuation_ratio":0.06,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9876217,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-01T17:04:30Z\",\"WARC-Record-ID\":\"<urn:uuid:2149d6e6-a0b6-4c34-9bd3-7a8cbd7fbdaf>\",\"Content-Length\":\"80049\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c7aeeb46-8e32-4c2a-a7cb-a3ba273a3106>\",\"WARC-Concurrent-To\":\"<urn:uuid:1ef49370-5302-4909-9575-932b1af87d80>\",\"WARC-IP-Address\":\"137.74.238.60\",\"WARC-Target-URI\":\"https://www.imlearningmath.com/how-many-additional-blocks-to-make-the-cube-whole/\",\"WARC-Payload-Digest\":\"sha1:LXWJS7DBVH2FNU63WQEZFWWMGEKLDKYR\",\"WARC-Block-Digest\":\"sha1:RBXWM3FWO26BBS5R2P7U263C2HKKT2JT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141681209.60_warc_CC-MAIN-20201201170219-20201201200219-00176.warc.gz\"}"}
|
https://www.unitsconverters.com/en/Biomass-Scale-Conversions/Measurement-1221
|
[
"Formula Used\n1 Biomass Kilogram per Square Meter = 0.0001 Biomass Kilogram per Square Centimeter\n1 Biomass Kilogram per Square Meter = 0.0001 Biomass Kilogram per Square Centimeter\n\n## Importance of Biomass Scale converter\n\nMeasurement of various quantities has been an integral part of our lives since ancient times. In this modern era of automation, we need to measure quantities more so than ever. So, what is the importance of Biomass Scale converter? The purpose of Biomass Scale converter is to provide Biomass Scale in the unit that you require irrespective of the unit in which Biomass Scale was previously defined. Conversion of these quantities is equally important as measuring them. Biomass Scale conversion helps in converting different units of Biomass Scale. Biomass is the measurement of how much living tissue mass for a population is present at given instant in time.. There are various units which help us define Biomass Scale and we can convert the units according to our requirement. unitsconverters.com provides a simple tool that gives you conversion of Biomass Scale from one unit to another.\n\nWhat is Biomass Scale?\nBiomass is the measurement of how much living tissue mass for a population is present at given instant in time.\nWhat is the SI unit for Biomass Scale?\nBiomass Kilogram per Square Meter (kg/m^2) is the SI unit for Biomass Scale. SI stands for International System of Units.\nWhat is the biggest unit for Biomass Scale?\nBiomass Kilogram per Square Centimeter is the biggest unit for Biomass Scale. It is 10000 times bigger than Biomass Kilogram per Square Meter.\nWhat is the smallest unit for Biomass Scale?\nBiomass Kilogram per Hectometer is the smallest unit for Biomass Scale. It is 0.0001 times smaller than Biomass Kilogram per Square Meter.",
null,
"Let Others Know"
] |
[
null,
"https://www.unitsconverters.com/image/share.png",
null
] |
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|
https://mailman.science.ru.nl/pipermail/fieldtrip/2018-October/025318.html
|
[
"# [FieldTrip] Attempt at cluster analysis on seed to whole brain coherency values\n\nPaul Dhami pdhami06 at gmail.com\nMon Oct 15 04:37:52 CEST 2018\n\n```Dear Fieldtrip,\n\nI have a dataset in which I'd like to compare, between controls and\npatients, their imaginary coherency values in a seed to whole brain manner.\nIn other words, I would like to calculate the imaginary coherency between a\nsingle seed electrode and the rest of the remaining electrodes, and *ultimately\nuse ft_freqstatistics' cluster analysis* to test for a significance\ndifference between the groups' seed-to-whole-scalp coherency maps.\n\n>From my understanding, this would require a bit of hacking to implement\nthis. I attempted to do so and describe what I did below, with the goal to\neventually create something that would work with freqstatistics :\n\n- ran freq analysis with output as 'powandcsd' and method as 'mtmconvolv'\n- on the results of freqanalysis, ran ft_connectivityanalysis with\ncfg.method = 'coh' and cfg.complex = 'absimag'\n- with each subject's resulting structure from ft_connectivityanalysis,\nI first chose a seed of interest\n- I then found the indices of the 59 channel combinations in relation to\nthe seed of interest in labelcmb\n- Using the indices, I then pruned/removed the remaining channel\ncombinations of no interest from both the labelcmb and cohspctrm, reducing\ncohspctrm to a channel of interest (59) x frequency x time matrix (as in it\nonly included values for the channel combinations of interest)\n- I rename the dimord as 'chan_freq_time'\n- create 'label' field with standard electrode labels\n- I then create powspctrm in my structure, which holds the exact same\ndata as the cohspctrm (created powspctrm strictly for freqstatistics)\n- removed the fields of 'labelcmb' and 'cohspctrm'\n- because my powspctrm is missing the seed channel, I then inserted a\nmatrix of ones with the appropriate dimensions into where it should be\n(e.g. if my seed of interest was channel FCZ, I would then insert into my\nmatrix of ones into the 19th position of the powspctrm, thus shifting the\nlatter matrices so now it becomes a matrix with 60 channels in the\nappropriate order)\n\n>From my understanding, the resulting structure of each subject should\ncontain now the coherency values between the seed of interest and the rest\nof the electrodes in 'powspctrm'.\n\nI then used freq_statistics (in the standard way) with cluster correction\nto compare the coherency between groups, and from what I can tell with no\nerrors popping up, it worked. I then interpreted the resulting clusters in\na similar fashion as you would do for a typical frequency chan-freq-time\nanalysis (instead of power, looking at coherency clusters now).\n\nMy questions are:\n1) In regards to implementation (assuming something like this can even be\nappropriately implemented), do things look okay?\n2) Am I wrong in thinking that the cluster results of freq_statistics can\nbe interpreted in a similar fashion as to a typical frequency\nchan-freq-time analsyis (just replacing power with coherency)?\n\nSorry for the long-winded email, but any help would be greatly appreciated.\n\nThank you,\nPaul\n-------------- next part --------------\nAn HTML attachment was scrubbed...\nURL: <http://mailman.science.ru.nl/pipermail/fieldtrip/attachments/20181014/451b4916/attachment-0001.html>\n```"
] |
[
null
] |
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|
https://alphabet-worksheets.clodui.com/two-step-equations-worksheet-word-problems.html
|
[
"",
null,
"Two Step Equations Worksheet Word Problems\n\nThe worksheets on this site are created in PDF format.\n\nTwo Step Equations Worksheet Word Problems. Benefits of two step equation word problems: As an extension, students can then solve those equations."
] |
[
null,
"https://sstatic1.histats.com/0.gif",
null
] |
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|
http://beerlak.hu/stock-market-laswtp/how-to-check-homothetic-function-ff1742
|
[
"For vectors x and w, let r(x,w) be a function that can be nonparametrically estimated consistently and asymptotically normally. Definition: Homothetic preferences Preferences are homothetic if for any consumption bundle x1 and x2 preferred to x1, Tx2 is preferred to Tx1, for all T!0. What does it mean when an aircraft is statically stable but dynamically unstable? where σ is a. homogeneous function of degree one and Φ is a continuous positive monotone increasing function of Φ. 1. an example of homothetic preferences: It is enough to check the income elasticity to be equal to unity: \"x m = m x @x @m = m/ m/ ( + )p @ @m m ( + )p = ( + )p ( + )p = 1 1. implies that x)TT21! endobj A homothetic function is a monotonie transformation of a function that is homogeneous of degree 1. The technology set for a given production process is de-fined as T={(x,y) : x ∈ Rn +,y ∈ R m: + x can produce y} where x is a vector of inputs and y is a … What exactly does it mean for a function to be “well-behaved”? How can I quickly grab items from a chest to my inventory? Homothetic Functions Recall that a real function f on a set E defines a complete (or total) ordering on E via the relation x ≺ ⪯ y i f a n d o n l y i f f (x) ≤ f (y). Solution for Is the following function homothetic? x��[[o�~���G���NE��=h��â�#�;�V�\"��3�_$��BoĖmY3���͐��Z7���5䧟.����-�k��r����r�x_]�^��˲�W��/7�֯Uy]������������9�IA4�ɺ:?���{7=?���c��d:? I need to check whether the following function is homothetic or not: f(x,y)=x 3 y 6 +3x 2 y 4 +6xy 2 +9 for x,y ∈ R +. $$In order to solve this type of equation we make use of a substitution (as we did in case of Bernoulli equations). I am not sure how to distinguish whether a function is homothetic. Check that the functions . endobj$$ However, researchers who employ non-parametric models of … ALTERNATIVEREPRESENTATIONS OFTECHNOLOGY The technology that is available to a firm can be represented in a variety of ways. $$Quasi-concave functions and concave functions. f(x, y)=x^ay^b which is your first function. Can I print plastic blank space fillers for my service panel? The following conditions are equivalent: (1) there exists a homothetic, concave, monotonic, continuous, locally non-satiated utility function that rationalizes the data; (2) the data satisfy HARP. u(tx)=tu(x) Firstly I show that the indirect utility function is homogenous of degree one in m. By the utility maximization, V(p,m)=max u(x) subject to px\\le m stream This is a monotone transformation of a homogenous function, so it is homothetic. ʕv�0^P��Tx�d����)#V䏽F�'�&. Constrained optimization when lending money between two periods, Inverse of a multivariable function following book derivation, Problem with partial derivative in economic payoff function, First and second order stochastic dominance given two asset payoffs. 3 x + 4 y. 1 0 obj Several economists have featured in the topic and have contributed in the final finding of the constant. So it then follows that Homothetic function is a term which refers to some extension of the concept of a homogeneous function. Find out information about homothetic figures. Shephard has shown (see (6)) that such a production structure is a necessary and sufficient condition for the related cost function to factor into a product of an output and a factor price index. And hence, the function you provided is a monotonic transformation of a homogenous function, meaning that it is homothetic.$$ What causes dough made from coconut flour to not stick together? 8.26, the production function is homogeneous if, in addition, we have f(tL, tK) = t n Q where t is any positive real number, and n is the degree of homogeneity. Looking for homothetic figures? %PDF-1.7 Median response time is 34 minutes and may be longer for new subjects. Figure 4.1: Homothetic Preferences preference relation º is homothetic if and only if it can be represented by a utility function that is homogeneous of degree one. The properties assumed In Section 1 for the function Φ of equation (l) are taken for the function Φ, and the production surfaces related to (31) are given by If we specialize to two variables, it seems that a function f: R 2 → R is called homothetic if the ratio of the partial derivatives ∂ f ∂ y and ∂ f ∂ x depends only on the ratio of x and y. rev 2021.1.7.38271, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. (demonstrate all steps of your detailed work in your… A homogeneous production function is also homothetic—rather, it is a special case of homothetic production functions. patents-wipo. 3. Put more formally, if there is a monotonic transformation such that y7! That is, agent i has preferences represented by a homothetic utility function, and has endowment Wi = c5i . share | improve this answer | follow | edited Jul 31 '19 at 6:25. answered Jul 29 '17 at 19:06. $$<>/ExtGState<>/XObject<>/ProcSet[/PDF/Text/ImageB/ImageC/ImageI] >>/MediaBox[ 0 0 612 792] /Contents 4 0 R/Group<>/Tabs/S/StructParents 0>> Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.$$ R such that = g u. Homoge-neous implies homothetic, but not conversely. Where did the \"Computational Chemistry Comparison and Benchmark DataBase\" found its scaling factors for vibrational specra? The properties assumed In Section 1 for the function Φ of equation (l) are taken for the function Φ, and the production surfaces related to (31) are given by The fact that the transformation F(.) %���� A utility function is homothetic if it is a positive monotonic transformation of a linearly homogeneous utility function; that is, if u(x) > u(y) then u(λx) > u(λy) for all λ > 0. But i don't know why these are homothetic. Making statements based on opinion; back them up with references or personal experience. 1.3 Homothetic Functions De nition 3 A function : Rn! <>/Metadata 250 0 R/ViewerPreferences 251 0 R>> They've got a function called the Cob Junction. <> I If f is a monotonic transformation of a concave function, it is quasi-concave. Use MathJax to format equations. 2.5 Homogeneous functions Definition Multivariate functions that are “homogeneous” of some degree are often used in economic theory. Section 2 sets out the main identification results. g(f(x, y))=\\exp[(f(x, y))^3+r]=\\exp[(x^a+by^a)^3+r]. WikiMatrix. Can an employer claim defamation against an ex-employee who has claimed unfair dismissal? Consider now W For vectors x and w, let r(x,w) be a function that can be nonparametrically estimated consistently and asymptotically normally. Or does it have to be within the DHCP servers (or routers) defined subnet? minimization of the twofold-weighted quadratic objective function 2x W x v v 2 1 1 2W u v K u v 2 1x x x W x u u 1 f , (6) where . x 2 .0 Page 5 Homogeneous and Homothetic Function 1 DC-1 Semester-II Paper-IV: Mathematical methods for Economics-II Lesson: Homogeneous and Homothetic Function Lesson Developer: Sarabjeet Kaur College/Department: P.G.D.A.V College, University of Delhi Homogeneous and Homothetic Function … Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. If there exists a homogeneous utility representation u(q) where u(λq) = λu(q) then preferences can be seen to be homothetic. A homogeneous production function is also homothetic—rather, it is a special case of homothetic production functions. (demonstrate all steps of your detailed work in your… I If f is a monotonic transformation of a concave function, it is quasi-concave. $$Downloadable! Origin of “Good books are the warehouses of ideas”, attributed to H. G. Wells on commemorative £2 coin? 0, if f x f x( ) ( )01d then f rx f rx( ) ( )01d. Obara (UCLA) Preference and Utility October 2, 2012 11 / 20. Suppose that f x f x( ) ( )01. The production function (1) is homothetic as defined by (2) if and only if the scale elasticity is constant on each isoquant, i.e. A function is homothetic if it is a monotonic transformation of a homogenous function (note that this second function does not need to be homogenous itself). To subscribe to this RSS feed, copy and paste this URL into your RSS reader. If that is the case, there are simple examples that show that a homothetic function need not be homogeneous. A function is homogenous of order k if By definition, f is said to be homothetic if the ordering is homothetic (implying that the domain E of f is a cone). Reflection - Method::getGenericReturnType no generic - visbility. Any shortcuts to understanding the properties of the Riemannian manifolds which are used in the books on algebraic topology, Function of augmented-fifth in figured bass, What do this numbers on my guitar music sheet mean. Seeking a study claiming that a successful coup d’etat only requires a small percentage of the population. w, where W E R~, 0 < c5i < 1, and 2:i~l c5i = 1.$$. $$Homogeneous production functions have the property that f(λx) = λkf(x) for some k. Homogeneity of degree one is constant returns to scale. When two rays from the same homothetic center intersect the circles, each set of antihomologous points lie on a circle. 8.26, the production function is homogeneous if, in addition, we have f(tL, tK) = t n Q where t is any positive real number, and n is the degree of homogeneity. g(f(x, y))=\\log(f(x, y))=\\log(x^ay^b)=a\\log x+b\\log y + that are represented by the utility function x 1 + x 2. 1.1. Title: Homogeneous and Homothetic Functions 1 Homogeneous and Homothetic Functions 2 Homogeneous functions. 3 0 obj In Fig.$$ De nition: Representation of Preference is represented by a utility function u : X !0. U(x) is homogenous of degree one i.e. Solution for Is the following function homothetic? Abstract. Downloadable! This is why we provide the books compilations in this website. In economic theory of production, homothetic production functions, introduced by Shephard in (5) and extended in (6), play an important role. Select the correct answer below. Microeconomics, Firm, Production Function, Linearly Homogeneous Production Function. <> Monotonic Transformation and same preferences? Why or why not? Homothetic production functions have the property that f(x) = f(y) implies f(λx) = f(λy). Even if Democrats have control of the senate, won't new legislation just be blocked with a filibuster? R is called homothetic if it is a mono-tonic transformation of a homogenous function, that is there exist a strictly increasing function g: R ! How true is this observation concerning battle? Related Articles. Cobb-Douglas Production Function: Economists have at different times examined many actual production functions and a famous production function is the Cobb-Douglas production function. We have Our proposed estimation algorithm is presented in Section 3. Q. Consider now the function We provide consistent, asymptotically normal estimators for the functions g and h, where r(x,w) = h[g(x),w], g is linearly homogeneous and h is monotonic in g. This framework encompasses homothetic and homothetically separable functions. Problem number 34. Given a cone E in the Euclidean space ℝ n and an ordering ≼ on E (i.e. Cobb Douglas Function. f(tx, ty)=t^kf(x, y). Q: II. It is usually more convenient to work with utility functions rather than preferences. Mantel has shown that this result is sensitive to violation of the restriction of proportional endowments. I If f is concave, then it is quasi-concave, so you might start by checking for concavity. K]�FoMr�;�����| �+�ßq�� ���q�d�����9A����s6(�}BA�r�ʙ���0G� Y.! Why or why not? What are quick ways to load downloaded tape images onto an unmodified 8-bit computer? $$Kuroda (1988) proposed an original method for matrix updating that reduces to constrained. functions are homothetic, by comparing F(z) = zwith Fb(z). Can you legally move a dead body to preserve it as evidence? MathJax reference. The fundamental property of a homothetic function is that its expansion path is linear (this is a property also of homogeneous functions, and thankfully it proves to be a property of the more general class of homothetic functions). Explanation of homothetic figures site design / logo © 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. Re-writing (9) as: p x = m x + (10) gives the Inverse Demand function! Solve the initial value problems. To learn more, see our tips on writing great answers. Is equal to B K to the Alfa attempts L to the one minus Alfa were asked to share that kay partial queue with respect to K plus l partial queue with respect to l. A is equal to queue. If I make a mistake, please tell. We see that p1x1 p1x0 and p 0x p0x1.$$ Homothetic functions, Monotonic Transformation, Cardinal vs Median response time is 34 minutes and may be longer for new subjects. How do digital function generators generate precise frequencies? 3 A function is homogenous of order k if f (t x, t y) = t k f (x, y). The constant function f(x) = 1 is homogeneous of degree 0 and the function g(x) = x is homogeneous of degree 1, but h is not homogeneous of any degree. A function is said to be homogeneous of degree r, if multiplication of each of its independent variables by a constant j will alter the value of the function by the proportion jr, that is, if ; In general, j can take any value. It is clear that homothetiticy is … are homogeneous. which is positive other than at the isolated point$z=0$, so the function$g$is monotone. what does$\\min()$and$\\max()$mean in a function? These choices are consistent with maximizing x 1 + x 2 subject to the budget constraint. Giskard Giskard. We study different hierarchies of generalized homogeneous functions. I If f is concave, then it is quasi-concave, so you might start by checking for concavity. We study different hierarchies of generalized homogeneous functions. Given a cone E in the Euclidean space $${\\mathbb{R}}^n$$ and an ordering ≼ on E (i.e. Homogeneous Differential Equations. ?cp^A1�\\#U�L��_�r��k���v�~9?�����l�OT��E������z��\"����>��?��ޢc��}}��t�N�(4-�w$MA5 b�Dd��{� ��]Fx��?d��L:��,(Kv�oTf낂S�V Is it possible to assign value to set (not setx) value %path% on Windows 10? A function is monotone where ∀, ∈ ≥ → ≥ Assumption of homotheticity simplifies computation, Derived functions have homogeneous properties, doubling prices and income doesn't change demand, demand functions are homogenous of degree 0 2 0 obj a reflexive and transitive binary relation on E), the ordering is said to be homothetic if for all pairs x, y, ∈E Please check my solution. In Fig. E. Common Functions E.3 Homothetic functions Definition: Homothetic function A function f x x( , ) 12 is homothetic if, for any x0 and 1, and any r! Determine whether or not each of the following functions is homogeneous, and if so of what degree. By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. Thank you . Let As it can be clearly expressed as a positive monotonic transformation of the homogeneous function xy 2 on R + therefore it must be a homothetic. $$. In addition, the more general model r(x,z,w) = H[M(x,z),w] can also be identified using our methods when M(x,z) is additive or multiplicative and His strictly monotonic with respect to its first argument. Homogeneous applies to functions like f(x), f(x,y,z) etc, it is a general idea. Can I assign any static IP address to a device on my network? Homothetic testing of Kuroda’s method. Introduction Shephard (1953) introduced the notion of a homothetic production function. is monotonic ensures that the inverse �LsG��d�)�9�j3�a�\"2�mH>��j��A����8��q�!&�{��CL=\"�7pf�3��HR�T���N�fg'Ky�L:���A��^�P�̀���r���N��V 5���B ��Wy� which is monotone. See … It will unconditionally ease you to look guide 1 homogenous and homothetic functions rmi as you such as. this is usually an easy way to check whether given preferences are homothetic. This also means that if a monotonic transformation of f is concave, then f is concave. 1.1 Quasi-linear preferences Remark 1 Quasi-linear utilities have the form u(x1;x2) = x1 +v(x2)! Comparing method of differentiation in variational quantum circuit, Renaming multiple layers in the legend from an attribute in each layer in QGIS. }�O��U��\"��OؤS�Q�PPϑY:G��@8�ˡ�Dfj�u ߭��58���� �%�4;��y����u����'4���M�= D�AA�b�= Learning Outcomes 2. Homothetic Production Function: A homothetic production also exhibits constant returns to scale. The mostgeneral are thosebased on correspondences and sets. Constant elasticity of substitution (CES), in economics, is a property of some production functions and utility functions. In other words, / (x) is homothetic if and only if it can be written as / (x) = g (h (x)) where h (-) is homogeneous of degree 1 and g (-) is a monotonie function. Homothetic utility function A utility function is homothetic if for any pair of consumption bundles and x2, invariant. m�����e �ޭ�fu�O�U����TY�8R>�5r�%k$$ which is homogenous since And both M(x,y) and N(x,y) are homogeneous functions of the same degree. Select the correct answer below. +is called homothetic if it is a monotone transformation of a homogeneous function. The three alternative study contrasts feature (1) pooling vs partitioned estimates, (2) a cost function dual to a homothetic production process vs the translog, and (3) two conceptually valid but empirically different cost‐of‐capital measures. f(y) 2R +and a homogeneous function g: Rn +7! *Response times vary by subject and question complexity. This also means that if a monotonic transformation of f is concave, then f is concave. Since increasing transfor-mations preserve the properties of preferences, then any utility function … PRODUCTION FUNCTIONS 1. Technology Sets. How would interspecies lovers with alien body plans safely engage in physical intimacy? $$A function is homothetic if it is a monotonic transformation of a homogenous function (note that this second function does not need to be homogenous itself). The differential equation is homogeneous if the function f(x,y) is homogeneous, that is- . Quasi-concave functions and concave functions. Functions Rmi 1 Homogenous And Homothetic Functions Rmi When people should go to the book stores, search introduction by shop, shelf by shelf, it is truly problematic. Section eight out. Level sets are radial expansions and contractions of one another: u(x) u(y) u( x) u( y) for > 0 The slope of level sets is constant along rays from the origin. f(tx, ty)=(tx)^a(ty)^b=t^{a+b}x^ay^b=t^{a+b}f(x, y). Consider now the function: Thus, the RAS method passes through a homothetic test successfully. Q: II. The homogeneous and the homothetic production functions do not have many properties which are of interest in production theory. *Response times vary by subject and question complexity. Asking for help, clarification, or responding to other answers. Four. g^\\prime (z)=3z^2 \\exp(z^3+r) How to find initial values for calculating IRR manually?$$ the elasticity of scale is a function of output. It only takes a minute to sign up. The most common quantitative indices of production factor substitutability are forms of the elasticity of substitution. R and a homogenous function u: Rn! We provide consistent, asymptotically normal estimators for the functions g and h, where r(x,w) = h[g(x), w], g is linearly homogeneous and h is monotonic in g. This framework encompasses homothetic and homothetically separable functions. 4. 2.5 Homogeneous functions Definition Multivariate functions that are “homogeneous” of some degree are often used in economic theory. Hence, Property V is established. The idea was generalized to the multi-output case by Shephard (1970). To be Homogeneous a function must pass this test: f (zx,zy) = z n f (x,y) They include Tom McKenzie, John Hicks and Joan Robinson. A property of some degree are often used in economic theory find initial values for calculating IRR manually and... Some degree are often used in economic theory scale is a monotonie transformation of a function: a homothetic successfully... Is quasi-concave opinion ; back them up with references or personal experience same homothetic center intersect the circles each!, attributed to H. G. Wells on commemorative £2 coin $which is monotone to downloaded. Chemistry Comparison and Benchmark DataBase '' found its Scaling factors for vibrational specra now! You such as the concept of a homogenous function, so you might start checking. To distinguish whether a function called the Cob Junction they 've how to check homothetic function a function for! Represents the preference, hence the preference ranking ) or personal experience ($... Is 34 minutes and may be longer for new subjects device on my network whose marginal technical rate of is... Factors for vibrational specra ”, you agree to our terms of,! Solve this type of equation we make use of a homogenous function that... Assign any static IP address to a firm can be represented in function. I am not sure how to distinguish whether a function to be “ well-behaved?! Common quantitative indices of production factor substitutability are forms of the constant - method::getGenericReturnType generic... Proposed estimation algorithm is presented in Section 3 ( CES ), in economics, is a term refers. Theorem [ Afriat ( 1981 ) ] proportional endowments Remark 1 Quasi-linear utilities the! Technical rate of substitution ( CES ), in economics, is a monotone transformation a... There is a monotone transformation of a concave function, that also represents the preference is.... Sensitive to violation of the population the senate, wo n't new just! The inverse Demand function ( 9 ) as: p x = x. Featured in the final finding of the senate, wo n't new just. 2012 11 / 20 function of output references or personal experience ( 10 ) gives the inverse Looking homothetic! An aircraft is statically stable but dynamically unstable several economists have featured in the Euclidean space ℝ N an! Monotone increasing function of output ( x, y ) and N ( x + 10! Formally, if f is concave, then f rx ( ) ( ) 01d elasticity of.! Then f rx ( ) 01d then f rx ( ) ( ) ( ) ( ) $in. Such a utility function, it is quasi-concave, so you might start by checking concavity! To load downloaded tape images onto an unmodified 8-bit computer all Remark the! Rss reader ’ etat only requires a small percentage of the elasticity of substitution ( CES ), in,... Unfair dismissal “ Post your answer ”, you agree to our terms of service, policy!, and 2: i~l c5i = 1 path % on Windows 10 ordinal property intersect! X f x f x f x f x f x f x f x f f... References or personal experience but i do n't know why these are homothetic alien body plans safely in! 6:25. answered Jul 29 '17 at 19:06 √ x + y + )... Did in case of Bernoulli equations ) monotone increasing function of degree 1 writing great answers DataBase '' its! To load downloaded tape images onto an unmodified 8-bit computer a. homogeneous function such a utility function, it quasi-concave! Possible to assign value to set ( not setx ) value % path % on 10! Case by Shephard ( 1953 ) introduced the notion of a homogenous function is. Rom the first this website start by checking for concavity © 2021 Stack Exchange is a function be... Concept of a homogenous function, so it is quasi-concave this type of equation we make of... That is homogeneous of degree zero [ 9, 12, 16 ] + 2. Percentage of the senate, wo n't new legislation just be blocked a. Homogeneous ” of some production functions and concave functions, each set of antihomologous points lie on a circle x! The condition set forth in the topic and have contributed in the Euclidean space ℝ N and ordering. Z$ $and hence, the RAS method passes through a production... Quasi-Linear preferences Remark 1 Quasi-linear utilities have the form u ( x, y.! Quasi-Concave, so it is homothetic factor substitutability are forms of the concept of a function... Utility October 2, 2012 11 / 20 our tips on writing great answers interest in production theory 1 and! ) ] the RAS method passes through a homothetic production also exhibits constant returns to scale what causes dough from...$ k $if$ $g ( z ) = zwith Fb ( z ) lie! It as evidence answer to mathematics Stack Exchange is a monotonie transformation of how to check homothetic function is concave, it! 1981 ) ] through a homothetic production also exhibits constant returns how to check homothetic function scale of we! You such as f rx f rx f rx ( ) ( ) mean! Are forms of the same homothetic center intersect the circles, each set of points. … a homothetic function is a term which refers to some extension of the constant N x. Bernoulli equations ) i am not sure how to find initial values for calculating IRR manually a function called Cob... The inverse Demand function the final finding of the restriction of proportional endowments my service?! Tom McKenzie, John Hicks and Joan Robinson of Φ servers ( or routers ) defined subnet n't new just. A successful coup d ’ etat only requires a small percentage of the constant no generic - visbility of! For new subjects alternativerepresentations OFTECHNOLOGY the technology that is available to a firm can be represented a... The consumption bundles does not satisfy WARP related fields need not be homogeneous not setx ) %! Your detailed work in your… quasi-concave functions and concave functions your detailed work in your… quasi-concave and... To check the first so you might start by checking for concavity p1x1! Microeconomics, Firm, production function use of a substitution ( as we did in case of Bernoulli )! ( UCLA ) preference and utility functions to solve this type of equation we make how to check homothetic function a... And$ \\max ( ) 01 you agree to our terms of,. It have to be within the DHCP servers ( or routers ) defined subnet 1.3 homothetic functions 1 homogeneous homothetic! Address to a firm can be represented in a function: a homothetic function need not be homogeneous ordering on... ( 10 ) gives the inverse Looking for homothetic figures to a device on my network seeking a claiming! G. Wells on commemorative £2 coin by clicking “ Post your answer ”, attributed to H. G. Wells commemorative! Interspecies lovers with alien body plans safely engage in physical intimacy factor substitutability are forms of the.. References or personal experience second and third statements follow f rom the first so you start... Data does not satisfy WARP by checking for concavity x1 ; x2!! All Remark: the second and third statements follow f rom the first is presented in Section.. 34 minutes and may be longer for new subjects not change the preference is homothetic the. “ Good books are the warehouses of ideas ”, attributed to H. G. on! Of proportional endowments “ Post your answer ”, attributed to H. G. on! Of Bernoulli equations ) have featured in the final finding of the concept of a concave,. Is indeed such a utility function, that also represents the preference ranking ) that are “ homogeneous of... Equations ) transformation such that y7 of differentiation in variational quantum circuit, Renaming multiple layers the! Them up with references or personal experience calculating IRR manually and cookie policy 2: i~l =... '' found its Scaling factors for vibrational specra, by comparing f y... Etat only requires a small percentage of the concept of a homothetic test successfully ( 1970 ) body to it! Is a term which refers to some extension of the restriction of proportional endowments result is sensitive to of. Marginal technical rate of substitution ( CES ), in economics, is a called. In the Euclidean space ℝ N and an ordering ≼ on E ( i.e Definition Multivariate that! Or responding to other answers variational quantum circuit, Renaming multiple layers the! Function of degree one i.e the elasticity of substitution more, see our tips on writing great answers have. That show that a homothetic production also exhibits constant returns to scale proportional..., see our tips on writing great answers that this result is sensitive to violation the... Legislation just be blocked with a filibuster production theory mantel [ 1976 ] shown... Function Homotheticity is an ordinal property have featured in the final finding of the of..., if there is indeed such a utility function, it is usually an easy way how to check homothetic function that. I if f is concave } \\$ fullfils the condition set forth in the wiki.! Found its Scaling factors for vibrational specra case, there are simple examples that show that a coup! Are consistent with maximizing x 1 + x 2 subject to the budget constraint RSS feed, copy and this. Original method for matrix updating that reduces to constrained, each set of antihomologous points lie on a circle interspecies... Each set of antihomologous points lie on a circle is a question and answer for... May be longer for new subjects functions De nition 3 a function is homothetic and 2: c5i. A continuous positive monotone increasing function of Φ how to distinguish whether a function that is homogeneous of one."
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"Cookie Consent by FreePrivacyPolicy.com\nSearch a number\nBaseRepresentation\nbin1011011100000001\n32101021011\n423130001\n52444344\n61000521\n7253405\noct133401\n971234\n1046849\n1132220\n1223141\n131842a\n1413105\n15dd34\nhexb701\n\n46849 has 4 divisors (see below), whose sum is σ = 51120. Its totient is φ = 42580.\n\nThe previous prime is 46831. The next prime is 46853. The reversal of 46849 is 94864.\n\nIt is a semiprime because it is the product of two primes, and also a Blum integer, because the two primes are equal to 3 mod 4.\n\nIt is a cyclic number.\n\nIt is not a de Polignac number, because 46849 - 25 = 46817 is a prime.\n\nIt is a Duffinian number.\n\nIt is a nialpdrome in base 11.\n\nIt is a self number, because there is not a number n which added to its sum of digits gives 46849.\n\nIt is not an unprimeable number, because it can be changed into a prime (46819) by changing a digit.\n\nIt is a pernicious number, because its binary representation contains a prime number (7) of ones.\n\nIt is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 2119 + ... + 2140.\n\nIt is an arithmetic number, because the mean of its divisors is an integer number (12780).\n\nIt is a Proth number, since it is equal to 183 ⋅ 28 + 1 and 183 < 28.\n\n246849 is an apocalyptic number.\n\nIt is an amenable number.\n\n46849 is a deficient number, since it is larger than the sum of its proper divisors (4271).\n\n46849 is a wasteful number, since it uses less digits than its factorization.\n\n46849 is an odious number, because the sum of its binary digits is odd.\n\nThe sum of its prime factors is 4270.\n\nThe product of its digits is 6912, while the sum is 31.\n\nThe square root of 46849 is about 216.4462981896. The cubic root of 46849 is about 36.0495716267.\n\nThe spelling of 46849 in words is \"forty-six thousand, eight hundred forty-nine\".\n\nDivisors: 1 11 4259 46849"
] |
[
null
] |
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|
http://monang.recoveryedb.org/welding-diagram-symbols
|
[
"",
null,
"# Welding Diagram Symbols",
null,
"welding symbols - an introduction to reading drawings welding symbols - an introduction to reading drawings welding symbols - an introduction to reading drawings welding symbols - an introduction to reading drawings welding symbols - an introduction to reading drawings . Our blog provide wiring diagrams and standard electrical schematics.\n\nWelding Symbols An Introduction To Reading Drawings The wiring diagram opens in a pop-up modal box. If the pop-up blocker is turned on in your device, you are not able to download or read online the wiring diagram.\n\nWelding Symbols An Introduction To Reading Drawings Wiring diagrams show the connections to the controller, while line diagrams show circuits of the operation of the controller.\n\nwelding symbols - an introduction to reading drawings welding symbols - an introduction to reading drawings welding symbols - an introduction to reading drawings welding symbols - an introduction to reading drawings welding symbols - an introduction to reading drawings",
null,
"9 Basic Steps To Read Welding Symbols Welding Answers",
null,
"How To Read A Welding Diagram Wiring Diagram",
null,
"Understanding Weld Symbols – The Fillet Weld Meyer Tool & Mfg",
null,
"Welding Symbols And Definitions",
null,
"Deciphering Weld Symbols Millerwelds",
null,
"What Is This Symbol???"
] |
[
null,
"https://mc.yandex.ru/watch/57088303",
null,
"https://static-cdn.imageservice.cloud/4125113/welding-symbols-an-introduction-to-reading-drawings.jpg",
null,
"https://static-resources.imageservice.cloud/4125116/9-basic-steps-to-read-welding-symbols-welding-answers.png",
null,
"https://static-assets.imageservice.cloud/4125119/how-to-read-a-welding-diagram-wiring-diagram.GIF",
null,
"https://static-assets.imageservice.cloud/5135419/understanding-weld-symbols-the-fillet-weld-meyer-tool-mfg.jpg",
null,
"https://static-assets.imageservice.cloud/5135420/welding-symbols-and-definitions.jpg",
null,
"https://static-cdn.imageservice.cloud/4125125/deciphering-weld-symbols-millerwelds.jpg",
null,
"https://static-resources.imageservice.cloud/5135421/what-is-this-symbol.jpg",
null
] |
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|
https://math.stackexchange.com/questions/1115644/cantor-set-nowhere-dense-complete
|
[
"# Cantor set--nowhere dense, complete\n\nI can't figure out this out.\n\nCantor set is closed in $\\mathbb{R}$.\n\n$\\mathbb{R}$ is a complete metric space.\n\nEvery closed subset of a complete space is also complete; thus, so is the Cantor set.\n\nComplete space can't be written as a countable union of nowhere dense sets; thus, the Cantor set has no such representation. (1)\n\nCantor set is nowhere dense (its interior is empty, and no intervals are contained in it)\n\nThus, the Cantor set is a finite (then countable) intersection of nowhere dense sets.\n\nWhere is a mistake?\n\nDoes representation in (1) refer only to infinite countable intersections?\n\n• A complete metric space $X$ cannot be written as a countable union of sets that are nowhere dense in $X$. The Cantor set is not a countable union of sets that are nowhere dense in the Cantor set, even though it is itself nowhere dense in $\\Bbb R$. – Brian M. Scott Jan 22 '15 at 22:00\n• A complete metric space $X$ is a Baire space, so no somewhere dense set can be written as a countable union of nowhere dense sets in $X$. In particular $X$ cannot be written as such a union. – Stefan Hamcke Jan 22 '15 at 22:00\n• You could have made the same mistake with a one-point subspace of $\\mathbb{R}$, which is also complete. – user208259 Jan 22 '15 at 22:02\n• I understand. Tnx – zariski Jan 22 '15 at 22:06\n• Another question,related to this. Can we conclude by the previous statements,that Cantor set is uncountable? If Cantor set C would be countable,it could be written as a countable union of singletons (which are nowhere dense in C) and that would be a contradiction with the statement (1)? – zariski Jan 22 '15 at 22:13"
] |
[
null
] |
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|
https://www.sepsis-q.cz/2022-08-09+14881.html
|
[
"",
null,
"## Concrete Compressive Strength, Cube Test, Procedure - Compressive Strength of Concrete\n\nCompressive strength of concrete cube test provides an idea about all the characteristics of concrete. By this single test one judge that whether Concreting has been done properly or not. Concrete compressive strength for general construction varies from 15 MPa (2200 psi) to 30 MPa (4400 psi) and higher in commercial and industrial structures.Compressive strength testing of concrete:- The test is carried out using 150mm concrete cubes on a Universal testing machine or compressive testing machine. Apparatus As per IS: 516-1959 Compressive testing machine (2000Kn),15cm×15cm×15cm steel cu...",
null,
"",
null,
"## Compressive Strength of Concrete -Cube Test, Procedure - Test for Compressive Strength of Concrete | Cube Testing ...\n\nJan 21, 2020· Procedure: Test the compressive strength of concrete cubes. For cube testing, two types of samples are used, either cubes at a size of 15 cm x 15 cm x 15 cm or 10 cm x 10 cm x 10 cm depending on the volume of the aggregate. For most businesses, …relating to the testing of concrete, and it does not include all the necessary provisions of a contract. I. SCOPB 1.1 This standard covers teste; for the determination of compressive strength, flexural strength and modulus of elasticity of cement concrete. 2. MAKING AND CURING COMPRESSION TEST SPECIMENS IN THE LABORATORY",
null,
"",
null,
"## Compressive Strength of Concrete | Cube Test, Procedure - Concrete Compressive Strength- Cube Test and its Procedure\n\nConcrete cube specimen 15cm. d) other Apparatus is G.I Sheet (For Making Concrete),Vibrating Needle, tray & other tools. 3) Environmental factors:-for standard calculation of compressive strength of concrete environmental factors should be optimum, minimum number of test specimen should be 3, temperature should be 27± 2℃ and humidity is 90%M20 Concrete cube Test procedureTable No 7 Compressive Strength of Cylinder for M30 grade B. FOR M40 GRADE OF CONCRETE: S.NO S.No Compressive Strength of Cube (N/mm2) CTM StrengthRH UPV 1 50.3 48 48.9 2 46.8 38 41.7 3 47 43 48.1 4 44.1 39 42.8 5 51.5 49 50.5 6 48.5 46 47.4 Average 48.03 43.83 46.56 Table No 8 Compressive Strength of Cube for M40 grade S.NO",
null,
"",
null,
"## Hydraulic Cube Testing Machine Compacion Test Of Concrete - Concrete cube test Procedure | Strength of Concrete at ...\n\nThe accompanying data is on cube compressive strength (MPa) of concrete specimens. 112.3 97.0 92.9 86.0 102.0 99.4 95.8 103.5 89.0 86.7 (a) Is it plausible that the compressive strength for this type of concrete is normally distributed? The normal probability plot is acceptably linear,May 27, 2021· Compressive strength of concrete is the ability of the concrete to withstand loads without cracking or deformation. Compressive Strength of concrete is defined as the Characteristic strength of 150mm size concrete cubes @28 days. Marginal variations in water to cement ratio, ingredient proportioning, increase in a slump, etc impacts the desired ...",
null,
"",
null,
"## CS1 : 2010 Testing Concrete and Quality Scheme for the - Compressive Strength of Concrete Cube Test Lab Report and ...\n\nJan 01, 2021· The Compressive strength test helps us to know the overall strength and the above factors. By conducting this test, one can easily determine the strength psi of the concrete and the quality of the concrete being produced. The Concrete Cube Test will give compressive strength of concrete which gives an idea of all the properties of concrete…CUBES. Concrete cylinder and cube specimens for compression testing were compared through a survey of past research, including testing procedures, factors affecting the cylinder/cube strength ratio, and conversion factors and equations. The main difference between cylinder and cube testing procedures is capping.",
null,
"",
null,
"## IS 516 (1959): Method of Tests for Strength of Concrete - Compressive strength of M20 concrete -cube Test procedure ...\n\nThe compressive strength of concrete test cube, cylinder and lintel has been determined in the 7th day after making the test. For the average compressive strength of 7th day concrete test cube is 4.63 MPa. The compressive strength of our concrete design is expected to have increased if there were more specimens similar to the tested ones but of ...Calculation For Compressive Strength Concrete: Let assume the maximum applied load is 400 KN = 400000 N. Cross-sectional area of cube =15 x 15 = 225 cm². Compressive strength of concrete = 400000/225 = 1778 N/cm² = 1778/9.81 = 181 Kg/cm² [ 1kg =9.81 N]",
null,
"",
null,
"## Compressive Strength of Concrete -Test procedure | CIVIL - Compressive Strength of Concrete and Concrete Testing ...\n\nAug 26, 2020· Compressive strength of different grade concrete at 3 days. Calculation: Now concrete cube test by CTM machine,assuming 14N/mm2/minute load is applied on concrete cube specimen of different grade till the cube collapse. The maximum load at which the specimen breaks is taken as a compressive load.Compressive strength as a concrete property depends on several factors related to the quality of used materials, mix design and quality control during concrete production. Depending on the applied code, the test sample may be cylinder [15 cm x 30 cm is common] or cube …",
null,
"",
null,
"## Cube test | Compressive strength test of concrete - (PDF) CONCRETE MIX, SLUMP AND CONCRETE COMPRESSION TEST ...\n\nAccurately and reliably verifying the compressive strength of a 200 MPa (29 ksi) concrete, however, can be a challenge in and of itself. The two standard methods for determining the compressive strength of concrete are the testing to failure of cylinder and cube specimens.Jul 02, 2021· Apparatus Required Compression testing machine Concrete cube mould & Cement mortar cube mould The procedure of the test Check the oil level in the machine, set the load cell to zero. Place the concrete cube specimens (150mm×150mm×150mm) in between the platens at the centre. The test specimen should be placed…",
null,
"",
null,
"## COMPRESSION TESTING OF CONCRETE: CYLINDERS VS. CUBES - Compressive strength test of concrete | Cube test for ...\n\nApr 21, 2019· The compressive strength of concrete is given in terms of the characteristic compressive strength of 150 mm size cubes tested at 28 days The characteristic strength is defined as the strength of the concrete below which not more than 5% of the test results are expected to fall.\" Number of samples for testing Compressive StrengthJan 24, 2018· Compressive strength of concrete cube test provides an idea about all the characteristics of concrete. By this single test one judge that whether Concreting has been done properly or not. Concrete compressive strength for general construction varies from 15 MPa (2200 psi) to 30 MPa (4400 psi) and higher in commercial and industrial structures.",
null,
"",
null,
"## Compressive strength test of Concrete | vincivilworld - Cube test report - SlideShare\n\nTherefore, to get the strength of the concrete as per design concrete are to be tested under field condition. Cube test is a destructive method to determine the compressive strength of the concrete. The compressive force is the resistance against push force. Concrete cube test Procedure a.Firstly, in a clean and dry pan, fine and coarseApr 17, 2019· The compressive strength of concrete is calculated from the failure load divided by the cross-sectional area of specimens which is resisting the load. The compressive strength of concrete is measured in units of psi (pound-force per square inch) or MPa (megapascals), N/mm 2 etc. The compression test is carried out on specimens like cube mould ...",
null,
"",
null,
"## Cube Test to Check the Compressive Strength of Concrete - TECHNICAL NOTE 2: GCCM COMPRESSIVE STRENGTH ASTM …\n\ncompressive strength of concrete cubes or tensile splitting strength have been increased to 0.6 ±±±±0.2 MPa/s. In CS1:1990, the loading rates were in the range of 0.2 to 0.4 MPa/s; and (4) In CS1:2010, the loading rates for determining the compressive strength of concrete cores shall be in the range of 0.2 MPa/s to 1.0 MPa/s. In CS1:1990,Lab Report – Concrete Compressive Strength Test August 15, 2019 Page 21 of 25 6.0 Conclusion In the Cube Test is determined the Compressive Strength of hardening concrete. The apparatus of the cube molds, steel rod, hand float, and compression testing machine (CTM) are used for doing the cube test. The ratio of cement, fine aggregate, coarse ...",
null,
"",
null,
"## The accompanying data is on cube compressive strength - Compressive Strength of Concrete Cube Test – Manarolla ...\n\nApr 17, 2020· Compressive strength achieved by concrete at 7 days is about 65% and at 14 days is about 90% of the target strength. Which test is most suitable for concrete strength? A concrete cube test or concrete cylinder test is generally carried out to assess the strength of concrete after 7 days, 14 days or 28 days of casting.a compressive strength of 80MPa at an optimum water/ cement ratio of 0.3, but at a water/cement ratio of 0.8 the compressive strength drops below 25MPa (see Figure 3*). Pre-existing concrete Compressive Strength test standards (designed for testing dry mix concrete products) allow for the water/cement ratio to be determined in the lab to",
null,
"# LCDY\n\nخلال 30 عامًا من العمل الشاق ، بنى موظفو LCDY تفوقًا في المصداقية والجودة الممتازة وخدمة العلامة التجارية \"LCDY\"\n\n#### ابقى على تواصل\n\nرقم 1688 ، طريق Gaoke شرق ، حي بودونغ الجديد ، شنغهاي ، الصين.\n\n#### النشرة الإخبارية\n\nتقوم الشركة بشكل أساسي بتصنيع الكسارات المتنقلة والكسارات الثابتة وآلات صنع الرمل\n\nحقوق النشر © 2023.LCDY كل الحقوق محفوظة.خريطة الموقع"
] |
[
null,
"https://www.sepsis-q.cz/images/zid/121.jpg",
null,
"https://www.sepsis-q.cz/image/whatsappp.png",
null,
"https://www.sepsis-q.cz/images/zid/24.jpg",
null,
"https://www.sepsis-q.cz/image/whatsappp.png",
null,
"https://www.sepsis-q.cz/images/zid/241.jpg",
null,
"https://www.sepsis-q.cz/image/whatsappp.png",
null,
"https://www.sepsis-q.cz/images/zid/130.jpg",
null,
"https://www.sepsis-q.cz/image/whatsappp.png",
null,
"https://www.sepsis-q.cz/images/zid/301.jpg",
null,
"https://www.sepsis-q.cz/image/whatsappp.png",
null,
"https://www.sepsis-q.cz/images/zid/20.jpg",
null,
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null,
"https://www.sepsis-q.cz/images/zid/178.jpg",
null,
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null,
"https://www.sepsis-q.cz/images/zid/250.jpg",
null,
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null,
"https://www.sepsis-q.cz/images/zid/190.jpg",
null,
"https://www.sepsis-q.cz/image/whatsappp.png",
null,
"https://www.sepsis-q.cz/images/zid/323.jpg",
null,
"https://www.sepsis-q.cz/image/whatsappp.png",
null,
"https://www.sepsis-q.cz/images/zid/324.jpg",
null,
"https://www.sepsis-q.cz/image/whatsappp.png",
null,
"https://www.sepsis-q.cz/images/zid/48.jpg",
null,
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|
https://quantumcomputing.stackexchange.com/questions/27367/qiskit-understanding-statevector-output
|
[
"# Qiskit: Understanding statevector output\n\nI am looking for some help in understanding the state vector output. I am studying myself and found some exercise online. Below question is from one of those exercises - Given two qubits. Both of them are in 0 state. Using the single-qubit gates, turn them into |+> state and |-> state respectively.\n\nfrom qiskit import QuantumCircuit, execute, Aer, assemble, QuantumRegister, ClassicalRegister\n\nqc = QuantumCircuit(2, 2)\nqc.h(0)\nqc.measure(0, 0)\n\nqc.x(1)\nqc.h(1)\nqc.measure(1, 1)\nqc.draw(output=\"mpl\")",
null,
"svsim = Aer.get_backend('aer_simulator')\nqc.save_statevector()\nqobj = assemble(qc)\nfinal_state = svsim.run(qobj).result().get_statevector(decimals=3)\n\nfrom qiskit.visualization import array_to_latex\narray_to_latex(final_state, prefix=\"\\\\text{Statevector} = \")\n\n\nOutput: Statevector=[0 0 −1 0]\n\nUpon plotting on bloch sphere, I get below -",
null,
"It is my understanding that state vectors are amplitudes. So, how is qubit0 state is 0 and same for qubit 1.\n\nThanks!\n\n• You can also try to see if the Bloch sphere has another case once these two quanta are entangled, perhaps ..... Jul 18 at 10:37\n• There is no entanglement as they are single qubit gates. Jul 18 at 16:05\n\nThis is circuit is easy to analyse. H gate on qo will result in a + state. The x gate on q1 puts 1 to 1 state. Next te Hgate on q1 will put q1 in the - state. In the circuit measurementgates are used and therefore the output will collapse to 1 and -1. Measurement gates are only needed for measuring counts.\n\nfrom qiskit import QuantumCircuit, execute, Aer, assemble, QuantumRegister, ClassicalRegister\n\nqc = QuantumCircuit(2, 2) qc.h(0) #qc.measure(0, 0)\n\nqc.x(1) qc.h(1) #qc.measure(1, 1) qc.draw(output=\"mpl\")\n\nplot_bloch_multivector(final_state)",
null,
"• Thanks a lot! This was very helpful to understand. :-) Jul 19 at 13:04"
] |
[
null,
"https://i.stack.imgur.com/zOvLk.png",
null,
"https://i.stack.imgur.com/MmVGM.png",
null,
"https://i.stack.imgur.com/ejQ16.png",
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.7876308,"math_prob":0.9814369,"size":951,"snap":"2022-27-2022-33","text_gpt3_token_len":264,"char_repetition_ratio":0.11193242,"word_repetition_ratio":0.0,"special_character_ratio":0.2597266,"punctuation_ratio":0.18232045,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9958049,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-19T04:50:03Z\",\"WARC-Record-ID\":\"<urn:uuid:10cdb055-f821-4af3-afd7-c5bd5526c8fe>\",\"Content-Length\":\"227611\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2c1126a4-e2c4-4681-93f7-f5dd588ce690>\",\"WARC-Concurrent-To\":\"<urn:uuid:60ed7d9c-f2d8-4044-a341-1d773759f16a>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://quantumcomputing.stackexchange.com/questions/27367/qiskit-understanding-statevector-output\",\"WARC-Payload-Digest\":\"sha1:RHBPKFHYIIZKJ2L5YR5A62SXIIQZJVYE\",\"WARC-Block-Digest\":\"sha1:4BXQHKTQYFLH2WIUHET7LDGQ2JMS75AB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882573623.4_warc_CC-MAIN-20220819035957-20220819065957-00061.warc.gz\"}"}
|
https://www.vocal.com/beamforming-2/frequency-domain-beamforming/
|
[
"The short microphone spacing in microphone array has always been a problem for beamforming algorithm designers. Upsampling is useful but it introduces large data redundancy. A rule of thumb is that minimum 4 times of the sampling frequency is desired for beamforming. However, if the signal is processed in the frequency domain, delay in time turns into shift in phase.",
null,
"$f\\left(t-t_0\\right)\\ ->F\\left(\\omega\\right)e^{-j\\omega t_0}$\n\nTherefore, a time delay for beamforming in the time domain becomes a phase shift in the frequency domain.",
null,
"$Y\\left(\\omega\\right)=F\\left\\{y\\left(t\\right)=\\sum_{m=0}^{M-1}{a_mx_m\\left(t-t_m\\right)}\\right\\}$",
null,
"$=\\sum_{m=0}^{M-1}{a_mX_m\\left(\\omega\\right)e^{-j\\omega t_{mb}}}$\n\nwhere m is the index of sound captured by microphone m. Non-integer sample delays can be implemented without approximation as phase shifts in frequency domain. The beamforming output can be converted back to time domain through an inverse Fourier transform.\n\nFrequency domain beamforming faces other challenges. One is that the appropriate phase shift is frequency dependent. A constant phase shift is only valid for a particular frequency. Therefore, approximation must be made when signals are processed when bandwidth is not zero. A second issue is the Fourier transform is defined over the entire time domain. A localized approximation must be made.\n\nThe following diagram shows a possible filterbank approach that can be used for frequency beamforming. The signal is sampled at 16kHz. A uniform filter bank with a low pass filter divides it into 4 equal bandwidth subbands. Each subband can be considered narrow bands and phase shift can be applied a constant for each band.\n\nBy introducing the filterbank, we successfully overcome the two issues involving frequency beamforming, frequency dependence and time domain causality."
] |
[
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.9013564,"math_prob":0.99287134,"size":1605,"snap":"2020-45-2020-50","text_gpt3_token_len":297,"char_repetition_ratio":0.14803249,"word_repetition_ratio":0.0,"special_character_ratio":0.17570093,"punctuation_ratio":0.089605734,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9947117,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-29T13:14:19Z\",\"WARC-Record-ID\":\"<urn:uuid:078e9bcc-1826-43bd-8312-ec9ff2de07bc>\",\"Content-Length\":\"90845\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b2af58a3-d035-405c-a288-c0017e42030e>\",\"WARC-Concurrent-To\":\"<urn:uuid:f2a9aaa0-cd5b-4fbb-a98b-5414735697ad>\",\"WARC-IP-Address\":\"72.236.255.161\",\"WARC-Target-URI\":\"https://www.vocal.com/beamforming-2/frequency-domain-beamforming/\",\"WARC-Payload-Digest\":\"sha1:OXG34R3OVFJXZ5UXU6O3FKWCCOYCZ4YY\",\"WARC-Block-Digest\":\"sha1:CDVP6D3L6I2Z5C5VFZTDO4OMQLVNEASL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141198409.43_warc_CC-MAIN-20201129123729-20201129153729-00095.warc.gz\"}"}
|
https://kr.mathworks.com/help/simulink/slref/sldiscmdl.html
|
[
"Documentation\n\n# sldiscmdl\n\nDiscretize model that contains continuous blocks\n\n## Syntax\n\n```sldiscmdl('model_name',sample_time) sldiscmdl('model_name',sample_time,method) sldiscmdl('model_name',sample_time,options) sldiscmdl('model_name',sample_time,method,freq) sldiscmdl('model_name',sample_time,method,options) sldiscmdl('model_name',sample_time,method,freq,options) [old_blks,new_blks] = sldiscmdl('model_name',sample_time,method,freq,options) ```\n\n## Description\n\n`sldiscmdl('model_name',sample_time)` discretizes the model named `'model_name'` using the specified `sample_time`. The model does not need to be open, and the units for `sample_time` are simulation seconds.\n\n`sldiscmdl('model_name',sample_time,method)` discretizes the model using `sample_time` and the transform method specified by `method`.\n\n`sldiscmdl('model_name',sample_time,options)` discretizes the model using `sample_time` and criteria specified by the `options` cell array. This array consists of four elements: {`target`, `replace_with`, `put_into`, `prompt`}.\n\n`sldiscmdl('model_name',sample_time,method,freq)` discretizes the model using `sample_time`, `method`, and the critical frequency specified by `freq`. The units for `freq` are Hz. When you specify `freq`, `method` must be `'prewarp'`.\n\n`sldiscmdl('model_name',sample_time,method,options)` discretizes the model using `sample_time`, `method`, and `options`.\n\n`sldiscmdl('model_name',sample_time,method,freq,options)` discretizes the model using `sample_time`, `method`, `freq`, and `options`. When you specify `freq`, `method` must be `'prewarp'`.\n\n```[old_blks,new_blks] = sldiscmdl('model_name',sample_time,method,freq,options)``` discretizes the model using `sample_time`, `method`, `freq`, and `options`. When you specify `freq`, `method` must be `'prewarp'`. The function also returns two cell arrays that contain full path names of the original, continuous blocks and the new, discretized blocks.\n\n## Input Arguments\n\n`model_name`\n\nName of the model to discretize.\n\n`sample_time`\n\nSample-time specification for the model:\n\n Scalar value Sample time with zero offset, such as `1` Two-element vector Sample time with nonzero offset, such as ```[1 0.1]```\n\n`method`\n\nMethod of converting blocks from continuous to discrete mode:\n\n `'zoh'` (default) Zero-order hold on the inputs `'foh'` First-order hold on the inputs `'tustin'` Bilinear (Tustin) approximation `'prewarp'` Tustin approximation with frequency prewarping `'matched'` Matched pole-zero methodFor single-input, single-output (SISO) systems only\n\n`freq`\n\nCritical frequency in Hz. This input applies only when the `method` input is `'prewarp'`.\n\n`options`\n\nCell array {`target`, `replace_with`, `put_into`, `prompt`}, where each element can take the following values:\n\n `target` `'all'` (default) Discretize all continuous blocks `'selected'` Discretize only selected blocks in the model `'full_blk_path'` Discretize specified block `replace_with` `'parammask'` (default) Create discrete blocks whose parameters derive from the corresponding continuous blocks `'hardcoded'` Create discrete blocks with hard-coded parameters placed directly into each block dialog box `put_into` `'copy'` (default) Create discretization in a copy of the original model `'configurable'` Create discretization candidate in a configurable subsystem `'current'` Apply discretization to the current model `'untitled'` Create discretization in a new untitled window `prompt` `'on'` (default) Show discretization information at the command prompt `'off'` Do not show discretization information at the command prompt\n\n## Examples\n\nDiscretize all continuous blocks in the `slexAircraftExample` model using a 1-second sample time:\n\n```sldiscmdl('slexAircraftExample',1); ```\n\nDiscretize the `Aircraft Dynamics Model` subsystem in the `slexAircraftExample` model using a 1-second sample time, a 0.1-second offset, and a first-order hold transform method:\n\n```sldiscmdl('slexAircraftExample',[1 0.1],'foh',... {'slexAircraftExample/Aircraft Dynamics Model',... 'parammask','copy','on'}); ```\n\nDiscretize the `Aircraft Dynamics Model` subsystem in the `slexAircraftExample` model and retrieve the full path name of the second discretized block:\n\n```[old_blks,new_blks] = sldiscmdl('slexAircraftExample',[1 0.1],... 'foh',{'slexAircraftExample/Aircraft Dynamics Model','parammask',... 'copy','on'}); % Get full path name of the second discretized block new_blks{2} ```"
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.5885287,"math_prob":0.8756224,"size":2878,"snap":"2019-51-2020-05","text_gpt3_token_len":826,"char_repetition_ratio":0.21329159,"word_repetition_ratio":0.28032345,"special_character_ratio":0.23558027,"punctuation_ratio":0.22908367,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.968075,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-22T04:50:37Z\",\"WARC-Record-ID\":\"<urn:uuid:19f183d8-e62c-408f-9d33-d93c6d4e696b>\",\"Content-Length\":\"75964\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4ec5110a-f61d-4e57-9ad4-f39b9932703d>\",\"WARC-Concurrent-To\":\"<urn:uuid:2aec8244-b028-454a-8c45-8cfd32d7dbbf>\",\"WARC-IP-Address\":\"23.50.228.199\",\"WARC-Target-URI\":\"https://kr.mathworks.com/help/simulink/slref/sldiscmdl.html\",\"WARC-Payload-Digest\":\"sha1:QBKONZDG6PVWYG2DJIJJL4THDOYFY4CA\",\"WARC-Block-Digest\":\"sha1:Y7O5BWFONLBMTA5XKJMHXFJ7TK6GXIAF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250606696.26_warc_CC-MAIN-20200122042145-20200122071145-00553.warc.gz\"}"}
|
http://haskell.1045720.n5.nabble.com/Type-Operator-td5769872.html
|
[
"# Type Operator",
null,
"Classic",
null,
"List",
null,
"Threaded",
null,
"3 messages",
null,
"Open this post in threaded view\n|\n\n## Type Operator\n\n I have a quick question. Recall that: class Monad m where (>>=) :: m a -> (a -> m b) -> m b ... and suppose I have a data type Sample: data Sample a b = ... how could I define Sample to be an instance of Monad such that: (>>=) :: Sample a c -> (a -> Sample b c) -> Sample b c ? I would like to use a (\\a -> ...)-like operator, but for types. So, something like this: instance Monad (\\a -> Sample a c) where (>>=) :: Sample a c -> (a -> Sample b c) -> Sample b c a >>= f = ... but that obviously doesn't work. Alternatively I would like to use a type declaration and partially apply it: type SampleFlip b a = Sample a b instance Monad (SampleFlip c) where (>>=) :: SampleFlip c a -> (a -> SampleFlip c b) -> SampleFlip c b which translates to: (>>=) :: Sample a c -> (a -> Sample b c) -> Sample b c But this doesn't work either, and ghc extensions don't add this functionality. Can I do this in Haskell?"
] |
[
null,
"http://haskell.1045720.n5.nabble.com/images/view-classic.gif",
null,
"http://haskell.1045720.n5.nabble.com/images/view-list.gif",
null,
"http://haskell.1045720.n5.nabble.com/images/view-threaded.gif",
null,
"http://haskell.1045720.n5.nabble.com/images/pin.png",
null,
"http://haskell.1045720.n5.nabble.com/images/gear.png",
null
] |
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|
https://www.cheenta.com/tag/geometric-series/
|
[
"Categories\n\n## Geometric Progression and Integers | PRMO 2017 | Question 5\n\nTry this beautiful problem from the Pre-RMO, 2017 based on Geometric Progression and integers.\n\n## Geometric Progression and Integers – PRMO 2017\n\nLet u,v,w be real numbers in geometric progression such that u>v>w. Suppose $u^{40}=v^{n}=w^{60}$, find value of n.\n\n• is 107\n• is 48\n• is 840\n• cannot be determined from the given information\n\n### Key Concepts\n\nGeometric series\n\nIntegers\n\nAlgebra\n\nPRMO, 2017, Question 5\n\nHigher Algebra by Hall and Knight\n\n## Try with Hints\n\nFirst hint\n\nLet u=a, v=ar, w=$ar^{2}$\n\nthen $a^{40}$=$(ar)^{n}$=$(ar^{2})^{60}$\n\nSecond Hint\n\n$\\Rightarrow a^{20}=r^{-120}$\n\n$\\Rightarrow a=r^{-6}$\n\nFinal Step\n\nand $r^{-240}=r^{-5n}$\n\n$\\Rightarrow 5n=240$\n\n$\\Rightarrow n=48$."
] |
[
null
] |
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|
https://lists.gnu.org/archive/html/paparazzi-devel/2015-11/msg00005.html
|
[
"paparazzi-devel\n[Top][All Lists]\n\n## Re: [Paparazzi-devel] Use of shifting operators in Paparazzi\n\n From: Felix Ruess Subject: Re: [Paparazzi-devel] Use of shifting operators in Paparazzi Date: Mon, 2 Nov 2015 18:52:18 +0100\n\nHi Srinath,\n\ngv_adapt_Xmeas is the ratio of vertical acceleration over thrust command.\nIf these would be floating point variables you would simply write\nBut since they are fixed point variables, the rounding is done by adding/subtracting 0.5 * thrust_applied before dividing it...\n\nCheers, Felix\n\nOn Sat, Oct 31, 2015 at 8:33 PM, Sreenath Dama wrote:\nHi All,\n\nI have been looking into the code of guidance_v_adapt.c and came across the following lines.\n\nif (g_m_zdd > 0) {\ngv_adapt_Xmeas = (g_m_zdd + (thrust_applied >> 1)) / thrust_applied;\n} else {\ngv_adapt_Xmeas = (g_m_zdd - (thrust_applied >> 1)) / thrust_applied;\n}\n\nI understand that g_m_zdd represents the acceleration measured by the (IMU-9.81). It would be very helpful if someone could explain from where does the formula of gv_adapt_Xmeas come. Also g_m_zdd is left shifted by 24 while thrust is shifted by 0 (on scale of 0 to 9600) Hence, why are we adding variables with different shifts?\n\nThanks,\nSrinath\n\nOn Tue, Oct 27, 2015 at 2:24 PM, Alexandre Bustico wrote:\nLe 26/10/2015 18:41, Flavio Justino a écrit :\nThanks for replying Felix ;) One last question: what about Double instead of float? What is the main difference (not theoretically as a programming type but for the case of paparazzi objectives). Do you think that it will have any interference in the results relatively to float? Thanks man =)\n\nIn term of performance, float (single precision floating point) is done by hardware on stm32f4 class of microcontroler,\nbut double (double precision floating point) is not accelerated and done by a library which come with gcc.\n\nFuture microcontrolers, based on arm cortex m7 core, will have accelerated double precision in hardware.\n\n--\nAlexandre\n\n_______________________________________________\nPaparazzi-devel mailing list"
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.84607637,"math_prob":0.75816673,"size":1903,"snap":"2022-27-2022-33","text_gpt3_token_len":503,"char_repetition_ratio":0.13586098,"word_repetition_ratio":0.015444015,"special_character_ratio":0.29164478,"punctuation_ratio":0.1337386,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95758545,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-05T19:49:49Z\",\"WARC-Record-ID\":\"<urn:uuid:5bb37559-a9ef-4d4d-befb-03d4fd35f95d>\",\"Content-Length\":\"9585\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:67826346-8f0e-4b96-a962-d73bbd55dcc7>\",\"WARC-Concurrent-To\":\"<urn:uuid:e9885547-d9bf-4a5f-96d0-6a172c6d59b4>\",\"WARC-IP-Address\":\"209.51.188.17\",\"WARC-Target-URI\":\"https://lists.gnu.org/archive/html/paparazzi-devel/2015-11/msg00005.html\",\"WARC-Payload-Digest\":\"sha1:D5C2PUSW4JRFLIN42UCOW665WFDZG7DI\",\"WARC-Block-Digest\":\"sha1:RHY47KPHDWWIDBHNNEYVJTBWRKCPWFKC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104597905.85_warc_CC-MAIN-20220705174927-20220705204927-00438.warc.gz\"}"}
|
http://www.fsjusfer.com/Product/145443.html?view=1
|
[
"• 急钮类\n• 水泡钮\n• 波子钮\n• 平面钮\n• 牛仔纽扣\n• 锑底钮\n• 铁线钮\n• 胶底钮\n• 沙发纽扣\n• 动物纽扣\n• 包芯钮\n• 异形纽扣\n• 金圈纽扣\n• 两眼鸡眼纽扣\n\n• 10=6.50MM\n• 14=9.00MM\n• 16=10.00MM\n• 18=11.00MM\n• 20=12.50MM\n• 22=14.00MM\n• 24=15.00MM\n• 26=16.50MM\n• 28=18.00MM\n• 30=19.00MM\n• 32=20.50MM\n• 34=21.50MM\n• 36=23.00MM\n• 40=25.50MM\n• 44=28.00MM\n• 50=32.00MM\n• 54=34.00MM\n• 60=38.00MM\n• 70=44.00MM\n\nQ Q:2277328964\n\n话:0757-86603523\n\n真:0757-86605253\n\n箱:[email protected]\n\n址:广东省佛山市南海区丹灶镇金沙上安永洪五金厂",
null,
"",
null,
"业务 庞小姐",
null,
"电话 86-757-86603523",
null,
"业务 陆小姐",
null,
"电话 86-757-86612272",
null,
"# —— 快速找到适合您需要的产品 ——",
null,
"",
null,
"0757-8661-2272\n\n•",
null,
"•",
null,
"•",
null,
""
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null
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|
https://best-excel-tutorial.com/58-excel-functions/126-rand-function
|
[
"# How to Use Rand Function in Excel\n\nIn this lesson, you can learn how to use RAND function in Excel. The RAND is a function that would return with an equally distributed random number that is either greater than or equal to 0 and less than 1. The new random real number will be returned every time that it is being calculated.\n\nRAND function draws a value between 0 and 1. The Syntax is RAND(), and there are no arguments for it.\n\n## Rand function syntax\n\n=RAND()\n\nSince the operation is done from scratch each time per sheet, it is best to immediately convert it into value. Otherwise it draws a new value each time you change the data in the file. Often the result of this function is multiplied by or divided to obtain an order of magnitude of numbers that we are interested in\n\n• =RAND()*1000\n• =0.5+RAND()/10\n• =RAND()*(100-50)+50\n• =RAND()*(A1-A2)+A2\n• =INT(RAND()*(A1-A2)+A2)\n\nThe function is sometimes used most often to create a variety of data, which look realistic. Most of the data in Best Excel Tutorial was created using this function.\n\n## Examples of Rand function\n\n### Example 1: Standard RAND Formula\n\nWe would like to know the rand, which is why we are using the RAND function.",
null,
"=RAND()\n\n### Example 2: Numbers and RAND\n\nThis is for finding out the even numbers with the rand function.",
null,
"=22+44+RAND()\n\n### Example 3: Evenly 100\n\nThis is for finding out the even numbers for 100, with the rand function.",
null,
"=RAND() *150\n\n### Example 4: RAND Function with Flexibilities\n\nThis is to try and find out the rand, since we have already layout different data.",
null,
"=RAND() +500*0.25+300\n\n### Example 5: SUMPRODUCT and RAND\n\nWe'd have different numbers that have been layout, and would like to know what it would mean for finding answers.",
null,
"=SUMPRODUCT(A2:J2)+RAND()\n\n### Example 6: SUM and RAND\n\nThe data is layout, and we would like to acknowledge the randomly generated number. This is to find out the even numbers.",
null,
"=RAND() *SUM(A2:J2)\n\n### Example 7: Double SUM and RAND\n\nThe data has been spread, and we would like to find out what the even numbers will be, and subtract 15 percent from the total number, in order to get the evenly generated number.",
null,
"=SUM(A2:K2)+RAND()-SUM(A2:K2)*0.15\n\n### Example 8: Double SUM and Double RAND\n\nThe circumstances is we are examining how it would look for the even numbers when we are trying to find the answers that correspond with the data.",
null,
"=SUM(A2:K2)+RAND()-SUM(B2:J2)+RAND()\n\n### Example 9: AVERAGE and RAND\n\nWe are using the previous data to find the answer, with the average at its center. This is why we are using both AVERAGE and RAND formulas.",
null,
"=AVERAGE(A2:K2)*RAND()\n\n### Example 10: Rand and Average\n\nThis time we'd have no data, but we have to find out the random and average functions working together to find the answers.",
null,
"=AVERAGE(-(RAND())))"
] |
[
null,
"data:image/svg+xml;base64,PHN2ZyB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciIHdpZHRoPSI1ODYiIGhlaWdodD0iNzQiPjwvc3ZnPg==",
null,
"data:image/svg+xml;base64,PHN2ZyB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciIHdpZHRoPSI2MDQiIGhlaWdodD0iNzYiPjwvc3ZnPg==",
null,
"data:image/svg+xml;base64,PHN2ZyB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciIHdpZHRoPSI1OTgiIGhlaWdodD0iNzMiPjwvc3ZnPg==",
null,
"data:image/svg+xml;base64,PHN2ZyB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciIHdpZHRoPSI2MTMiIGhlaWdodD0iNzkiPjwvc3ZnPg==",
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null,
"data:image/svg+xml;base64,PHN2ZyB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciIHdpZHRoPSI2MTIiIGhlaWdodD0iMTE0Ij48L3N2Zz4=",
null,
"data:image/svg+xml;base64,PHN2ZyB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciIHdpZHRoPSI1ODYiIGhlaWdodD0iMTA5Ij48L3N2Zz4=",
null,
"data:image/svg+xml;base64,PHN2ZyB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciIHdpZHRoPSI2MDYiIGhlaWdodD0iMTE1Ij48L3N2Zz4=",
null,
"data:image/svg+xml;base64,PHN2ZyB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciIHdpZHRoPSI2MTQiIGhlaWdodD0iMTEyIj48L3N2Zz4=",
null,
"data:image/svg+xml;base64,PHN2ZyB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciIHdpZHRoPSI2MDAiIGhlaWdodD0iMTA3Ij48L3N2Zz4=",
null
] |
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|
https://robotics.stackexchange.com/questions/9962/linearize-a-non-linear-system/10140
|
[
"# Linearize a non linear system\n\nHow do I linearize the following system using taylor series expansion:\n$\\dot x = v cos\\theta \\\\ \\dot y = v sin\\theta \\\\ \\dot \\theta = u$\nHere, $\\theta$ is the heading direction of my robot, measured counter clockwise with respect to $x$ axis.\n$v$ is the linear velocity of the robot,\n$u$ is the angular velocity of the robot.\n\n• for each equation, take the derivative with respect to each state variable which eventually will yield a matrix. May 29 '16 at 18:00\n• The method you are taking is the Jacobian method right? May 29 '16 at 19:05\n• That's right... May 29 '16 at 20:39\n• Should that be written as answer?\n– Ian\nMay 31 '16 at 18:47\n\nThe short answer to this question is that linearization won't work, and here's why:\n\nDifferential drive robots can be modeled with unicycle dynamics of the form: $$\\dot{z}=\\left[\\begin{matrix}\\dot{x}\\\\ \\dot{y} \\\\ \\dot{\\theta} \\end{matrix}\\right] = \\left[\\begin{matrix}cos(\\theta)&0\\\\sin(\\theta)&0\\\\0&1\\end{matrix}\\right] \\left[\\begin{matrix}v\\\\\\omega\\end{matrix}\\right],$$ where $x$ and $y$ are Cartesian coordinates of the robot, and $\\theta \\in (-\\pi,\\pi]$ is the angle between the heading and the $x$-axis. The input vector $u=\\left[v, \\omega \\right]^T$ consists of linear and angular velocity inputs.\n\nWe can re-write this in affine form, such that: $$\\dot{z}=f(z)+g(z)u.$$ Notice that $f(z)=0$, which is equivalent to saying that there is no drift in the system. For clarity, let $\\mathbf{g}(z,u)=g(z)u.$ The Jacobian, which is essentially a matrix of truncated Taylor series expansions, is used to linearize nonlinear systems about an equilibrium point, but the eigenvalues of the Jacobian matrix, evaluated at equilibrium, must be nonzero in order for the linearization to hold (there are some important theorems that come into play here, which are covered in linear algebra and linear systems textbooks). The Jacobian for state matrix A of the possible linearized unicycle system is given by: $$A:=D\\mathbf{g}(z,u)=\\frac{\\partial \\mathbf{g}(z,u)}{\\partial z}=\\left[\\begin{matrix}\\frac{\\partial g_1(z,u)}{\\partial x}&\\frac{\\partial g_1(z,u)}{\\partial y}&\\frac{\\partial g_1(z,u)}{\\partial \\theta}\\\\\\frac{\\partial g_2(z,u)}{\\partial x}&\\frac{\\partial g_2(z,u)}{\\partial y}&\\frac{\\partial g_2(z,u)}{\\partial \\theta}\\\\\\frac{\\partial g_3(z,u)}{\\partial x}&\\frac{\\partial g_3(z,u)}{\\partial y}&\\frac{\\partial g_3(z,u)}{\\partial \\theta}\\end{matrix}\\right]=\\left[\\begin{matrix}0&0&-v\\text{sin}(\\theta)\\\\0&0&v\\text{cos}(\\theta)\\\\0&0&0\\end{matrix}\\right],$$ where $g_n$ is the $n^{th}$ function of $\\mathbf{g}(z,u)$. Each element in the Jacobian is evaluated as follows, using $n=1$ as an example: $$\\frac{\\partial g_1(z,u)}{\\partial x}=\\frac{\\partial (v cos{\\theta})}{\\partial x}=0.$$\n\nBy inspection, we can see that the eigenvalues of the Jacobian are zero for all $z$, never mind at any equilibrium point, so this system cannot be linearized. Despite not being able to linearize the system, non-linear control of unicycle robots is straightforward and well studied.\n\n• How do you expand from a 3x2 state matrix to a 3x3 state matrix? If your states were originally $[v,w]^T$, a 2x1 state vector, what are they now? Jun 21 '16 at 13:21\n• Good question! Correct me if I'm wrong, but this is how I understand the Jacobian. There are 3 state variables ($x$, $y$, and $\\theta$), and there are 3 functions working on the state variables ($\\mathbf{g}(z,u)=g(z)u$). The Jacobian maps from $\\mathbf{g}(z,u)\\in\\mathbb{R}^3$ to the update of $\\dot{z}\\in\\mathbb{R}^3$, and is in $\\mathbb{R}^{3\\times 3}$. Jun 21 '16 at 16:30\n• If the state variables are $[x,y,\\theta]^T$, then what is the state matrix? All zeros? Perhaps this is the point you're trying to make. But then I still don't see how you get to your $Dg(z)$ matrix above - what are the $g_N$ terms in $\\partial g_N$? Is $g(z)$ not a 3x2 matrix? How are you indexing $g(z)$? I'm not trying to nit-pick, I just think you've glossed over a couple steps (maybe?) and I'm genuinely curious how you've gotten your answer. Jun 21 '16 at 18:14\n• The discussion is good, and hopefully other readers will benefit! The goal is to take the nonlinear system $\\dot{z}=\\mathbf{g}(z,u)$ and linearize it to $\\dot{z}=Ax+Bu.$ To do this we must compute the Jacobian matrices $A:=\\frac{\\partial \\mathbf{g}(z,u)}{\\partial z}\\in\\mathbb{R}^{3\\times 3}$ and $B:=\\frac{\\partial \\mathbf{g}(z,u)}{\\partial u}\\in\\mathbb{R}^{3\\times 2}$. After computing A, it was easy to determine that linearization would fail about any equilibrium point (none exist!), so I didn't bother computing B. In short, we are both right--you correctly highlight that I left a step out! Jun 21 '16 at 21:14\n• I found a great paper talking about the problem of linearization of unicycle robot models: dis.uniroma1.it/~labrob/pub/papers/Ramsete01.pdf -- it goes into good detail. In particular, the author highlights the problem in section 2. Jul 1 '16 at 13:16"
] |
[
null
] |
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|
https://talks.cam.ac.uk/talk/index/10346
|
[
"",
null,
"# Multiply robust estimation of statistical interaction parameters\n\nA primary focus of an increasing number of scientific studies is to determine whether two exposures interact in the effect that they produce on an outcome of interest. Interaction is commonly assessed by fitting regression models in which the linear predictor includes the product between those exposures. When the main interest lies in the interaction, this approach is not entirely satisfactory because it is prone to (possibly severe) bias when the main exposure effects or the association between outcome and extraneous factors are misspecified. In this talk, I will therefore consider conditional mean models with identity or log link which postulate the statistical interaction in terms of a finite-dimensional parameter, but which are otherwise unspecified. I will show that estimation of the interaction parameter is often not feasible in this model because it would require nonparametric estimation of auxiliary conditional expectations given high-dimensional variables. I will thus consider `multiply robust estimation’, assuming at least one of several working submodels holds. The proposed approach is novel in that it makes use of information on the joint distribution of the exposures conditional on the extraneous factors in making inferences about the interaction parameter of interest. As such, it essentially encompasses a `propensity score’ approach to the estimation of interaction parameters. In the special case of a randomized trial or a family-based genetic study in which the joint exposure distribution is known by design or by Mendelian inheritance, the procedure leads to asymptotically distribution-free tests of the null hypothesis of no interaction on an additive scale. I will illustrate the methods via simulation and the analysis of a randomized follow-up study. This is based on joint work with Tyler VanderWeele (University of Chicago) and James Robins (Harvard University).\n\nThis talk is part of the MRC Biostatistics Unit Seminars series."
] |
[
null,
"https://talks.cam.ac.uk/image/show/10829/image.png",
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.88379383,"math_prob":0.8669689,"size":3225,"snap":"2022-05-2022-21","text_gpt3_token_len":594,"char_repetition_ratio":0.10990376,"word_repetition_ratio":0.012793177,"special_character_ratio":0.16713178,"punctuation_ratio":0.057086613,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95968944,"pos_list":[0,1,2],"im_url_duplicate_count":[null,5,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-19T00:00:04Z\",\"WARC-Record-ID\":\"<urn:uuid:0584e0ba-ae88-4ea8-b3b8-39912517222e>\",\"Content-Length\":\"18924\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e2f1ecdd-8970-4780-b1a2-eaff4854e912>\",\"WARC-Concurrent-To\":\"<urn:uuid:1104e7f9-0476-4e68-b693-a5ad111dc9e6>\",\"WARC-IP-Address\":\"131.111.150.181\",\"WARC-Target-URI\":\"https://talks.cam.ac.uk/talk/index/10346\",\"WARC-Payload-Digest\":\"sha1:SFUOYR2PEQU7YXBVEMTTHC2RJLCFUGIT\",\"WARC-Block-Digest\":\"sha1:J2M4IZSFQPYATNVQ5VBDYAGUVXV7FLSI\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662522556.18_warc_CC-MAIN-20220518215138-20220519005138-00793.warc.gz\"}"}
|
https://quant.stackexchange.com/questions/tagged/pairs-trading?sort=newest&page=2
|
[
"Pairs trading is a market-neutral trading strategy enabling traders to profit from virtually any market conditions: uptrend, downtrend, or sideways movement. This strategy is categorized as a statistical arbitrage and convergence trading strategy.\n\n84 questions\nFilter by\nSorted by\nTagged with\n584 views\n\n### Need help on cointegration\n\nI tried to test stock pairs for pairs trading. There are two questions I am not sure. I am not using ADF to test the log difference between two stocks. But I also see people using Johansen test. What'...\n244 views\n\n### How to projectP&L or drawdowns on pair trading , trading and portfolios? [closed]\n\nThis is for planning and risk management. I am stuck on the following thoughts - Back-test the trading strategy for a period similar to the one you expect and then project. Do the above using ...\n234 views\n\n### 2 stocks, no shorting vs shorting. (concrete questions, mean-variance)\n\nI'd appreciate help with the following questions. Suppose there are two stocks $A$ and $B$ with expected returns $E_A, E_B >0$ and volatilities $v_A, v_B >0$, respectively . Also, suppose ...\n236 views\n\n### Replicating the short part of a long-short trade using inverse ETFs\n\nI devised a pair trading strategy going long XXX and short B*YYY. B is the quantity of shares of YYY I need to short. The problem is I can’t go short on YYY, but there is an inverse ETF for YYY ...\n1k views\n\n### Ornstein versus AR(1) for modeling stationary data\n\nI've come across several posts regarding parameter estimation for O-U models given some stationary data (say, some sort of mean reverting spread), but I can't seem to find an answer as to why modeling ...\n359 views\n\nI'm constructing money-neutral spread by this formula: Spread = log(P1) - log(P2), where P1 and P2 is prices of two instruments But sometimes spread can get into ...\n2k views\n\n### Position Sizing For Ratio Pairs Trade\n\nOk, let's say I'm trading a spread of two stocks, X & Y, The spread is calculated as a ratio (Spread = X / Y). I use rolling stats to calculate the mean, standard deviation and hence the z-score ...\n391 views\n\nI have calculated a hedge ratio that generates a mean reverting spread (stationary, without trends) 60-70% of the time. But the remaining 30% of the time, it seems like there is a trend in the spread. ...\n371 views\n\n### pairs trading or long short strategy given volatility of the stocks\n\nSuppose I have 2 stocks and the only thing I know about them is their volatility and that they cointegrate. Let's say vol of stock A is 25% and B is 20%. Will I be able to find a hedge ratio only ...\n2k views\n\n### ADF test in R yielding perfect cointegration. How is this possible?\n\nI am using the famous conintegrated pairs tutorial to just different stocks for cointegration. The adf.test yeilds perfect cointegration, which I feel must be incorrect. Here is why: When I run adf....\n349 views\n\n### Stat Arb Equity Pair Position Trigger\n\nI am new to pairs trading and am in the process of constructing the code for backtesting a basic pair trading strategy. While I understand the basic idea behind the pair trading strategy, I am having ...\n694 views\n\nI have constructed a mean reverting spread using two indexes. I know they have to be mean reverting, but when plotted side by side they are mean reverting for a little bit and then deviate and head ...\n798 views\n\n### main arbitrage & statistical arbitrage concepts\n\nCan we please sumarise here some of the basic concepts, tools used in arbitrage and statistical arbitrage in real life? ARB: benefit from price difference on same asset ARB: difference between stock ...\n1k views\n\n### Pairs trade CDS contracts using cointegration\n\nRecently I have looked at some sovereign CDS spreads (of the Nordic countries to be precise) and have tested for cointegration in the levels (i.e. untransformed) and logs of the spreads. Tests ...\n4k views\n\n### What is the proper way to calculate returns for Pair Trading?\n\nEdit I am assuming that I don't need to use margin account to short here: What is a standard way to calculate return for pairs trading strategy? For example, I bought 100 dollars worth of a loser (L)...\n1k views\n\n### Pairs trading: Question on non-negative profits, size of the positions and trading signals\n\nI'm trying to backtest Pairs Trading but have become a bit confused on the different methods of selecting pairs, how to look for trading signals and what size of the positions to take in the assets. ...\n2k views\n\n### What different methods of pairs selection exists? (For Pairs trading)\n\n(I'm quite new to quant finance so I'm not sure if this is an eligible question.) I've decided I want to backtest pairs trading on the Nordic stockmarket. So I would guess there exists different ...\n2k views\n\n### Multiple (linear) regression\n\nI am looking for some inputs on a pair trading strategy that I am trying to improve with some semi-fundamental input. The basic idea is to use multiple linear regression to estimate the price of a ...\n892 views\n\n### Statistical significance of a pair trading strategy\n\nHow can I test the significance of a pair trading strategy, i.e. that the H0 is \"The strategy has no predicting power\". I was considering to use the technique in Evidence Based Technical Analysis ...\n793 views\n\n### Performance Stats of Pairs Trades\n\nThis is something I've been thinking about for a while but I can't reach a clear conclusion. When we calculate, for example, the profit factor for a pairs trading strategy, do we treat each pairs ...\n1k views\n\n### Cointegration trading: Ignoring pairs that aren't economically related\n\nCointegration trading question What's the state of the art when it comes to choosing proper subsets of stocks/assets where cointegrating relationships aren't ignored as (likely to be) spurious? For ...\n1k views\n\n### Calculating the right portfolio(position size for each leg) in a Long/Short Strategy\n\nFor a Long/Short Strategy, I have two stocks with different volatilities. How can I calculate the right position size for each leg? *The pair trading is not coming from co-integration but more as a ...\n1k views\n\n### Searching for pairs-trading in sub O(n^2 t) time\n\nLet there be $n$ stock symbols. Let each stock symbol have exactly $t$ ticks (with all ticks miraculously aligned.) We are now searching for potential pairs for pair trading. A brute-force solution ...\n1k views\n\nSuppose the trade is between Index Options of two Indices X and Y which are quite similar (but not exactly). So for the equivalent strikes, one can quote option on Index X and cover in Index Y. But ...\n2k views\n\n### How should I compute the Sharpe Ratio for mid-frequency pair trading strategy?\n\nI have a pair trading strategy with positions that last 3-5 days and trades 2-3 times a month. By design, all the trades are profitable until the cointegration is broken. Should I calculate the ...\n1k views\n\n### How to calculate the weight of the stocks using the linear regression?\n\nI do a simple example with the follow three series(stocks prices): ...\n6k views\n\n### How to build a mean reverting basket?\n\nI have been playing with mean reverting pairs, but seems that most of the low hanging fruit (ie pairs) have been squeezed already. I would like to start with mean reverting baskets (>2 securities) in ...\n2k views\n\n### How to build an execution trading system with CQG API?\n\nI am currently using CQG for spread trading and have a spread trading strategy in CQG chart. I am trying to automate my spread trading strategy in CQG, but CQG told me to look at CQG API samples to ...\n2k views\n\n### How do different methods and techniques used in pairs trading compare?\n\nI was going through the paper of Avellaneda (2008) on stat arb and I found it interesting that he uses asset returns vs. their respective ETFs to compute the s-score. I am wondering if anyone has ...\n325 views\n\n### How would I calculate a stop on a pair trade? [closed]\n\nI have a trading strategy that I use on single tickers. I'd like to start using it with pairs as well. However, I'm somewhat math challenged and not sure how to best calculate the stops of the ...\n3k views\n\n### How does Kalman filtering of beta in pairs trading model work in R?\n\nCould anyone show how this could be done in R? The dlm package seems to be a good start, but I can't really find any good examples to learn from. Currently I have ...\n678 views\n\n### Should cointegration be tested using close or adjusted close prices?\n\nWhen doing cointegration tests should I use the adjusted close price or just close price for the time series? The dividend of each stock is on different dates and can cause jumps in the data."
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https://im.kendallhunt.com/HS/students/1/2/15/index.html
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[
"# Lesson 15\n\nSolving Systems by Elimination (Part 2)\n\n• Let’s think about why adding and subtracting equations works for solving systems of linear equations.\n\n### 15.1: Is It Still True?\n\nHere is an equation: $$50 + 1 = 51$$.\n\n1. Perform each of the following operations and answer these questions: What does each resulting equation look like? Is it still a true equation?\n\n1. Add 12 to each side of the equation.\n2. Add $$10 + 2$$ to the left side of the equation and 12 to the right side.\n3. Add the equation $$4 + 3 = 7$$ to the equation $$50 + 1 = 51$$.\n2. Write a new equation that, when added to $$50 +1 = 51$$, gives a sum that is also a true equation.\n3. Write a new equation that, when added to $$50 +1 = 51$$, gives a sum that is a false equation.\n\n### 15.2: Classroom Supplies\n\nA teacher purchased 20 calculators and 10 measuring tapes for her class and paid \\$495. Later, she realized that she didn’t order enough supplies. She placed another order of 8 of the same calculators and 1 more of the same measuring tape and paid \\$178.50.\n\nThis system represents the constraints in this situation:\n\n\\begin{cases} \\begin {align}20c + 10m &= 495\\\\ 8c + \\hspace{4.5mm} m &= 178.50 \\end{align}\\end{cases}\n\n1. Discuss with a partner:\n1. In this situation, what do the solutions to the first equation mean?\n2. What do the solutions to the second equation mean?\n3. For each equation, how many possible solutions are there? Explain how you know.\n4. In this situation, what does the solution to the system mean?\n2. Find the solution to the system. Explain or show your reasoning.\n3. To be reimbursed for the cost of the supplies, the teacher recorded: “Items purchased: 28 calculators and 11 measuring tapes. Amount: \\\\$673.50.”\n\n1. Write an equation to represent the relationship between the numbers of calculators and measuring tapes, the prices of those supplies, and the total amount spent.\n2. How is this equation related to the first two equations?\n3. In this situation, what do the solutions of this equation mean?\n4. How many possible solutions does this equation have? How many solutions make sense in this situation? Explain your reasoning.\n\n### 15.3: A Bunch of Systems\n\nSolve each system of equations without graphing and show your reasoning. Then, check your solutions.\n\nA\\begin {cases} \\begin {align}2x + 3y &= \\hspace {2mm}7\\\\ \\text-2x +4y &= 14 \\end {align} \\end {cases}\n\nB\\begin {cases} \\begin {align}2x + 3y &= \\hspace {2mm}7\\\\ 3x -3y &= 3 \\end {align} \\end {cases}\n\nC\\begin {cases} \\begin {align}2x + 3y &= 5\\\\ 2x +4y &= 9 \\end {align} \\end {cases}\n\nD\\begin {cases} \\begin {align}2x + 3y &=16\\\\ 6x -5y &= 20 \\end {align} \\end {cases}\n\nThis system has three equations: $$\\begin{cases}3 x + 2y - z = 7 \\\\ \\text{-} 3x + y +2z =\\text- 14 \\\\ 3x+y-z=10 \\end{cases}$$\n\n1. Add the first two equations to get a new equation.\n2. Add the second two equations to get a new equation.\n3. Solve the system of your two new equations.\n4. What is the solution to the original system of equations?\n\n### Summary\n\nWhen solving a system with two equations, why is it acceptable to add the two equations, or to subtract one equation from the other?\n\nRemember that an equation is a statement that says two things are equal. For example, the equation $$a = b$$ says a number $$a$$ has the same value as another number $$b$$. The equation $$10 + 2 = 12$$ says that $$10+2$$ has the same value as 12.\n\nIf $$a = b$$ and $$10 + 2 = 12$$ are true statements, then adding $$10+2$$ to $$a$$ and adding $$12$$ to $$b$$ means adding the same amount to each side of $$a=b$$. The result, $$a + 10 + 2 = b + 12$$, is also a true statement.\n\nAs long as we add an equal amount to each side of a true equation, the two sides of the resulting equation will remain equal.\n\nWe can reason the same way about adding variable equations in a system like this:\n\n\\begin {cases} \\begin {align} e + f = 17\\\\ \\text-2e + f =\\text-1 \\end{align}\\end {cases}\n\nIn each equation, if $$(e,f)$$ is a solution, the expression on the left of the equal sign and the number on the right are equal. Because $$\\text-2e+f$$ is equal to -1:\n\n• Adding $$\\text-2e + f$$ to $$e+f$$ and adding -1 to 17 means adding an equal amount to each side of $$e+f=17$$. The two sides of the new equation, $$\\text-e + 2f = 16$$, stay equal.\n\nThe $$e$$- and $$f$$-values that make the original equations true also make this equation true.\n\n\\begin {align} e +\\hspace{2mm} f &= 17\\\\ \\text-2e +\\hspace{2mm} f &=\\hspace{0.8mm}\\text-1 \\quad+\\\\ \\overline {\\hspace{2mm}\\text-e + 2f }& \\overline{ \\hspace{1mm}=16} \\end {align}\n\n• Subtracting $$\\text-2e + f$$ from $$e+f$$ and subtracting -1 from 17 means subtracting an equal amount from each side of $$e+f=17$$. The two sides of the new equation, $$3e = 18$$, stay equal.\n\nThe $$f$$-variable is eliminated, but the $$e$$-value that makes both the original equations true also makes this equation true.\n\n\\begin {align} e +\\hspace{2mm} f &= 17\\\\ \\text-2e +\\hspace{2mm} f &=\\hspace{0.8mm}\\text-1 \\quad-\\\\ \\overline{\\hspace{0.8mm}3e \\hspace{9.5mm}} &\\overline{\\hspace{1mm}=18} \\end {align}\n\nFrom $$3e = 18$$, we know that $$e=6$$. Because 6 is also the $$e$$-value that makes the original equations true, we can substitute it into one of the equations and find the $$f$$-value.\n\nThe solution to the system is $$e=6, f=11$$, or the point $$(6,11)$$ on the graphs representing the system. If we substitute 6 and 11 for $$e$$ and $$f$$ in any of the equations, we will find true equations. (Try it!)\n\n### Glossary Entries\n\n• elimination\n\nA method of solving a system of two equations in two variables where you add or subtract a multiple of one equation to another in order to get an equation with only one of the variables (thus eliminating the other variable).\n\n• solution to a system of equations\n\nA coordinate pair that makes both equations in the system true.\n\nOn the graph shown of the equations in a system, the solution is the point where the graphs intersect."
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https://www.cnblogs.com/abatei/archive/2008/06/06/1215114.html
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[
"# C#与数据结构--图的遍历\n\n## 8.2 图的存储结构\n\n### 8.2.1 邻接矩阵表示法",
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"8.10所示的是无向图的邻接矩阵表示法,可以观察到,矩阵延对角线对称,即A(ij)= A(ji)。无向图邻接矩阵的第i行或第i列非零元素的个数其实就是第i个顶点的度。这表示无向图邻接矩阵存在一定的数据冗余。\n\n8.11所示的是有向图邻接矩阵表示法,矩阵并不延对角线对称,A(ij)=1表示顶点Vi邻接到顶点VjA(ji)=1则表示顶点Vi邻接自顶点Vj。两者并不象无向图邻接矩阵那样表示相同的意思。有向图邻接矩阵的第i行非零元素的个数其实就是第i个顶点的出度,而第i列非零元素的个数是第i个顶点的入度,即第i个顶点的度是第i行和第i列非零元素个数之和。\n\n### 8.2.2 邻接表表示法",
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"",
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"",
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"【注意】:观察图8.14可以发现,当删除存储表头结点的数组中的某一元素,有可能使部分表头结点索引号的改变,从而导致大面积修改表结点的情况发生。可以在表结点中直接存放指向表头结点的指针以解决这个问题(在链表中存放类实例即是存放指针,但必须要保证表头结点是类而不是结构体)。在实际创建邻接表时,甚至可以使用链表代替数组存放表头结点或使用顺序表存代替链表存放表结点。对所学的数据结构知识应当根据实际情况及所使用语言的特点灵活应用,切不可生搬硬套。\n\nusing System;\nusing System.Collections.Generic;\n{\nList\n<Vertex<T>> items; //图的顶点集合\npublic AdjacencyList() : this(10) { } //构造方法\n{\nitems\n= new List<Vertex<T>>(capacity);\n}\n\n{ //不允许插入重复值\nif (Contains(item))\n{\n\nthrow new ArgumentException(\"插入了重复顶点!\");\n}\nnew Vertex<T>(item));\n}\n\npublic void AddEdge(T from, T to) //添加无向边\n{\nVertex\n<T> fromVer = Find(from); //找到起始顶点\nif (fromVer == null)\n{\n\nthrow new ArgumentException(\"头顶点并不存在!\");\n}\nVertex\n<T> toVer = Find(to); //找到结束顶点\nif (toVer == null)\n{\n\nthrow new ArgumentException(\"尾顶点并不存在!\");\n}\n\n//无向边的两个顶点都需记录边信息\n}\n\npublic bool Contains(T item) //查找图中是否包含某项\n{\n\nforeach (Vertex<T> v in items)\n{\n\nif (v.data.Equals(item))\n{\n\nreturn true;\n}\n}\n\nreturn false;\n}\n\nprivate Vertex<T> Find(T item) //查找指定项并返回\n{\n\nforeach (Vertex<T> v in items)\n{\n\nif (v.data.Equals(item))\n{\n\nreturn v;\n}\n}\n\nreturn null;\n}\n\n//添加有向边\nprivate void AddDirectedEdge(Vertex<T> fromVer, Vertex<T> toVer)\n{\n\nif (fromVer.firstEdge == null//无邻接点时\n{\nfromVer.firstEdge\n= new Node(toVer);\n}\n\nelse\n{\nNode tmp, node\n= fromVer.firstEdge;\n\ndo\n{\n//检查是否添加了重复边\n{\n\nthrow new ArgumentException(\"添加了重复的边!\");\n}\ntmp\n= node;\nnode\n= node.next;\n}\nwhile (node != null);\ntmp.next\n= new Node(toVer); //添加到链表未尾\n}\n}\n\npublic override string ToString() //仅用于测试\n{ //打印每个节点和它的邻接点\nstring s = string.Empty;\n\nforeach (Vertex<T> v in items)\n{\ns\n+= v.data.ToString() + \":\";\n\nif (v.firstEdge != null)\n{\nNode tmp\n= v.firstEdge;\n\nwhile (tmp != null)\n{\ns\ntmp\n= tmp.next;\n}\n}\ns\n+= \"\\r\\n\";\n}\n\nreturn s;\n}\n\n//嵌套类,表示链表中的表结点\npublic class Node\n{\n\npublic Node next; //下一个邻接点指针域\npublic Node(Vertex<T> value)\n{\n= value;\n}\n}\n\n//嵌套类,表示存放于数组中的表头结点\npublic class Vertex<TValue>\n{\n\npublic TValue data; //数据\npublic Node firstEdge; //邻接点链表头指针\npublic Boolean visited; //访问标志,遍历时使用\npublic Vertex(TValue value) //构造方法\n{\ndata\n= value;\n}\n}\n}\n\nl Vertex类中包含了一个visited成员,它的作用是在图遍历时标识当前节点是否被访问过,这一点在稍后会讲到。\n\n【例8-1 Demo8-1.cs】图的邻接表存储结构测试\n\nusing System;\nclass Demo8_1\n{\n\nstatic void Main(string[] args)\n{\n\n//添加顶点\n'B');\n'C');\n'D');\n\n//添加边\n'A''C');\n'A''D');\n'B''D');\nConsole.WriteLine(a.ToString());\n}\n}\n\nABCD\n\nCA\n\nDAB\n\n## 8.3 图的遍历\n\n### 8.3.1 深度优先搜索遍历",
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"【例8-2 DFSTraverse.cs】深度优先搜索遍历\n\n35 public void DFSTraverse() //深度优先遍历\n36 {\n37 InitVisited(); //将visited标志全部置为false\n38 DFS(items); //从第一个顶点开始遍历\n39 }\n40 private void DFS(Vertex<T> v) //使用递归进行深度优先遍历\n41 {\n42 v.visited = true//将访问标志设为true\n43 Console.Write(v.data + \" \"); //访问\n44 Node node = v.firstEdge;\n45 while (node != null//访问此顶点的所有邻接点\n46 { //如果邻接点未被访问,则递归访问它的边\n48 {\n50 }\n51 node = node.next; //访问下一个邻接点\n52 }\n53 }\n\n98 private void InitVisited() //初始化visited标志\n99 {\n100 foreach (Vertex<T> v in items)\n101 {\n102 v.visited = false//全部置为false\n103 }\n104 }\n\n【例8-2 Demo8-2.cs】深度优先搜索遍历测试\n\nusing System;\nclass Demo8_2\n{\n\nstatic void Main(string[] args)\n{\n\"V1\");\n\"V2\");\n\"V3\");\n\"V4\");\n\"V5\");\n\"V6\");\n\"V7\");\n\"V8\");\n\"V1\"\"V2\");\n\"V1\"\"V3\");\n\"V2\"\"V4\");\n\"V2\"\"V5\");\n\"V3\"\"V6\");\n\"V3\"\"V7\");\n\"V4\"\"V8\");\n\"V5\"\"V8\");\n\"V6\"\"V8\");\n\"V7\"\"V8\");\na.DFSTraverse();\n}\n}\n\nV1 V2 V4 V8 V5 V6 V3 V7\n\n### 8.3.2 广度优先搜索遍历",
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"【例8-3 BFSTraverse.cs】广度优先搜索遍历\n\n54 public void BFSTraverse() //广度优先遍历\n55 {\n56 InitVisited(); //将visited标志全部置为false\n57 BFS(items); //从第一个顶点开始遍历\n58 }\n59 private void BFS(Vertex<T> v) //使用队列进行广度优先遍历\n60 { //创建一个队列\n61 Queue<Vertex<T>> queue = new Queue<Vertex<T>>();\n62 Console.Write(v.data + \" \"); //访问\n63 v.visited = true//设置访问标志\n64 queue.Enqueue(v); //进队\n65 while (queue.Count > 0//只要队不为空就循环\n66 {\n67 Vertex<T> w = queue.Dequeue();\n68 Node node = w.firstEdge;\n69 while (node != null//访问此顶点的所有邻接点\n70 { //如果邻接点未被访问,则递归访问它的边\n72 {\n73 Console.Write(node.adjvex.data + \" \"); //访问\n76 }\n77 node = node.next; //访问下一个邻接点\n78 }\n79 }\n80 }\n\n【例8-3 Demo8-3.cs】广度优先搜索遍历测试\n\nusing System;\nclass Demo8_3\n{\n\nstatic void Main(string[] args)\n{\n\"V1\");\n\"V2\");\n\"V3\");\n\"V4\");\n\"V5\");\n\"V6\");\n\"V7\");\n\"V8\");\n\"V1\"\"V2\");\n\"V1\"\"V3\");\n\"V2\"\"V4\");\n\"V2\"\"V5\");\n\"V3\"\"V6\");\n\"V3\"\"V7\");\n\"V4\"\"V8\");\n\"V5\"\"V8\");\n\"V6\"\"V8\");\n\"V7\"\"V8\");\na.BFSTraverse();\n//广度优先搜索遍历\n}\n}\n\nV1 V2 V3 V4 V5 V6 V7 V8\n\n### 8.3.3 非连通图的遍历",
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"public void DFSTraverse() //深度优先遍历\n{\nInitVisited();\n//将visited标志全部置为false\nforeach (Vertex<T> v in items)\n{\n\nif (!v.visited) //如果未被访问\n{\nDFS(v);\n//深度优先遍历\n}\n}\n}\n\npublic void BFSTraverse() //广度优先遍历\n{\nInitVisited();\n//将visited标志全部置为false\nforeach (Vertex<T> v in items)\n{\n\nif (!v.visited) //如果未被访问\n{\nBFS(v);\n//广度优先遍历\n}\n}\n}\n\nposted @ 2008-06-06 13:40 abatei 阅读(27503) 评论(20编辑 收藏 举报"
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"https://www.cnblogs.com/images/cnblogs_com/abatei/Image00014.jpg",
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"https://www.cnblogs.com/images/cnblogs_com/abatei/Image00016.jpg",
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"https://www.cnblogs.com/images/cnblogs_com/abatei/Image00017.jpg",
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"https://www.cnblogs.com/images/cnblogs_com/abatei/Image00018.jpg",
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"https://www.cnblogs.com/images/cnblogs_com/abatei/Image00019.jpg",
null,
"https://www.cnblogs.com/images/cnblogs_com/abatei/Image00020.jpg",
null
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|
https://smahesh.com/jax/
|
[
"tech,\n\nAn introduction to JAX",
null,
"Mahesh Apr 04, 2021 · 13 mins read\n\nWhat is JAX? JAX is NumPy but more with various functionalities designed to make machine learning research faster. It introduces a functional programming paradigm and has other valuable features for high-performance machine learning training.\n\nBut before we get into all of those details, I want to give an unjaxy introduction to set the stage for those new to ML. We will write a tiny neural network and see where we run into (design) issues, and learn a thing or two about JAX by addressing those issues.\n\nYou may want to skip to the last portion of the following section if you are familiar with the basics.\n\nAn unjaxy introduction\n\nFor now, we can treat JAX just like NumPy.\n\nLet’s take this opportunity to construct a tiny neural network that outputs a probability value.\n\nMatmul\n\nThe most basic building block is matrix multiplication. Remember, JAX is like NumPy. So you can do a lot of the stuff you do in NumPy similarly in JAX.\n\nimport jax.numpy as jnp\n\ninput = jnp.array([[2., 3.], [4., 5.], [6., 7]])\nkernel = jnp.array([[9., 1., 0., 0.], [1., 0., 1., 0]])\n\nIn fact, you can use\n\nimport jax.numpy as np\n\nand use JAX like NumPy. It should work for the vast majority of the cases, but let’s stick with importing as jnp for clarity.\n\nNow, you can multiply these two matrices as follows:\n\njnp.matmul(input, kernel)\n\nor\n\njnp.dot(input, kernel)\n\nIn fact, you can also do the following as well:\n\ninput @ kernel\n\nHere is the output:\n\nDeviceArray([[20., 2., 2., 0.],\n[40., 4., 4., 0.],\n[60., 6., 6., 0.]], dtype=float32)\n\nCuriously, the output is now an instance of DeviceArray instead of a list or NumPy/JAX array. We will learn more about DeviceArray later.\n\nThe forward pass\n\nLet’s specify a list of layers and then run the inputs through the list of layers. Each layer will have a ReLU non-linearity, and the final output will be a sigmoid.\n\ndef predict(input, layers):\nfor layer_index, kernel in enumerate(layers):\ninput = input @ kernel\nif layer_index != len(layers) - 1:\ninput = jax.nn.relu(input)\nelse:\ninput = jax.nn.sigmoid(input)\nreturn input\n\nkernel1 = jnp.array([[9., 1., 0., 0.], [1., 0., 1., 0.]])\nkernel2 = jnp.array([[1.], [-10.], [0.1], [2.]])\n\nlayers = [kernel1, kernel2]\ninput = jnp.array([[2., 3.], [4., 5.], [6., 7.]])\n\npredict(input, layers)\n\nOutput:\n\nDeviceArray([[0.785835 ],\n[0.81757444],\n[0.8455348 ]], dtype=float32)\n\nLoss\n\nWe will use the binary cross-entropy loss:\n\neps = 1e-8\ndef loss_fn(input, target, layers):\noutput = predict(input, layers)\nreturn -jnp.mean(\ntarget * jnp.log(output + eps) + (1 - target) * jnp.log(1 - output + eps))\n\nIn practice, we may use a numerically stable version of the above loss, which will take as input pre-sigmoid values and then calculate the loss.\n\nBackward pass\n\nOk, so now we have to calculate the gradient of the loss with respect to all the variables, and then update the layer weights and biases based on the gradient.\n\nSo how do we calculate the gradient?\n\nIn TensorFlow, the way this is done is with a tf.GradientTape.\n\noutput = predict(input, ..)\nloss = loss_fn(output, label)\n\nand so on.\n\nIn JAX, things are a bit different. All you have to do is to use jax.grad() to calculate the gradient. And, while in TF we calculated the gradient on the actual loss computed, in JAX, we will call jax.grad on the loss function itself (jax.grad(loss_fn)).\n\nBut if you call jax.grad(loss_fn)(input, target, layers), you will see the following output:\n\nDeviceArray([[-0.0397228 , 0.04369506],\n[-0.05030257, 0.05533284],\n[-0.0596227 , 0.06558497]], dtype=float32)\n\nBut what we want is to take the derivative with respect to the weights (i.e. layers).\n\nBy default, jax.grad will take derivative with respect to the 0th argument, which in this case is the input. That’s why what you see above has the same shape as the input array. To fix this, we need to specify a argnums to the argument.\n\nOutput:\n\n[DeviceArray([[ 0.6383921 , -6.3839216 , 0.06383922, 0. ],\n[ 0.78804016, -7.880402 , 0.07880402, 0. ]], dtype=float32),\nDeviceArray([[6.5335693 ],\n[0.6383921 ],\n[0.78804016],\n[0. ]], dtype=float32)]\n\nWe can now see that the shape of the gradients matches the shape of the layers.\n\nUpdate step\n\nPutting all of this together, our update step can look like this:\n\ndef update_step(input, target, layers):\nfor index in range(len(layers)):\nreturn layers\n\nThis is basically using SGD with learning rate of 0.1.\n\nIf you run\n\nupdate_step(input, target, layers)\n\nyou will get\n\n[DeviceArray([[ 8.9361610e+00, 1.6383922e+00, -6.3839220e-03,\n0.0000000e+00],\n[ 9.2119598e-01, 7.8804022e-01, 9.9211961e-01,\n0.0000000e+00]], dtype=float32), DeviceArray([[ 0.34664303],\n[-10.063839 ],\n[ 0.02119599],\n[ 2. ]], dtype=float32)]\n\nWe have successfully taken one training step of our tiny neural network!\n\nReflecting back\n\nLet’s stop here with this toy example and reflect on a few things:\n\n1. From a quick glance at the predict method, it is hard to interpret what the network architecture is. If you want to mix other kind of layers, like convolutional layers, the code will be completely unreadable. If you have used Keras, you will know that there are better ways of expressing models, and the above is not one of them. So we need a better way of expressing the models. Like Keras, there are many other high-level libraries to make the code readable. We will look at them later.\n\n2. The jax.grad example above demonstrates a key feature of JAX — the functional programming paradigm. You can also use composition to make nested functions, e.g. jax.grad(jax.grad(loss_fn)) will create a function that takes the second derivative of loss_fn.\n\n3. This brings to the other philosophy that JAX embodies, that the code should read like mathematical equation. In TensorFlow, we saw that gradient was called on the output of the function. d(loss_fn)/dw makes a lot of sense.\n\nAn actual introduction to JAX\n\nWe can now start to describe JAX in more detail, and you should be able to see how JAX addresses the problems we have seen.\n\nAs we said before, JAX is basically like NumPy, but more. In a nutshell,\n\nJAX is NumPy on the CPU, GPU, and TPU, with great automatic differentiation for high-performance machine learning research.\n\nLet’s dive in:\n\n1. JAX stands for JAX is Autograd and XLA.\n\n2. As the name indicates, a central feature of JAX is Autograd, which is used for automatic differentiation. What is cool is that, with JAX, you can differentiate native Python and NumPy code. You can compute higher-order derivatives by calling the jax.grad function repeatedly.\n\n3. JAX uses XLA to compile and run your code. This means that, it can be run on GPU and TPU as well. In fact, since XLA was originally built for TPUs, you will get very good performance on TPUs when using JAX. Also, even on CPUs, JAX can be much faster than NumPy due to its reliance on XLA. This is because XLA can do all sorts of optimizations. For example, if you express expensive computation over a large array but then slice out only a portion of the array to return, XLA will notice this and run the expensive computation only for the required portion. Another byproduct of using XLA underneath is that the same code can run on CPU/GPU/TPU. As someone who has had a fair amount of experience getting stuff to work on TPUs with TensorFlow, there is quite a bit of work involved in transforming code that was written for CPUs to TPUs.\n\n4. Just in time compilation (JIT). If you profile the matmul of JAX and compare against the profile of matmul of numpy, you will see a speedup, thanks to XLA. But the above code is still underutilizing the benefit of XLA. Every line is compiled by XLA above, but you can compile blocks of code using XLA as well, for further speedup. This is done using jax.jit. You can call it as a function or use it as a decorator.\n\n> %timeit -n 100 update_step(input, target, layers)\n100 loops, best of 5: 10.3 ms per loop\n> update_step_jit = jax.jit(update_step)\n> %timeit -n 100 update_step_jit(input, target, layers)\nThe slowest run took 256.52 times longer than the fastest. This could mean that an intermediate result is being cached\n100 loops, best of 5: 4.39 µs per loop\n1. JAX embraces functional programming paradigm. In particular, when we have functions without side effects, i.e. functions that don’t modify some global state, XLA can do a good job of optimizing these functions.\n\nIntermediate JAX\n\nDeviceArray\n\nWhen you encounter an array stored as DeviceArray, it means that the array is stored on the device (e.g. inside TPU). By default, this value is not returned unless you have requested to print. So this can avoid expensive back and forth.\n\nPRNGKey\n\nFor best performance, it is preferred that functions are without side-effects and don’t store states or access global variables. This means that the usual way of specifying the random number generator’s seed wouldn’t work (we typically just call a function and set the seed to be used by many other functions). So we have to pass in the seed explicitly in JAX.\n\njax.random.PRNGKey(0)\n\njax.vmap\n\njax.vmap is like map by Python but adds vectorization. You can express your computation for a single example, and then use vmap to run the computation for multiple examples at a time (aka adding batch dimension). This can help with performance as well.\n\nFLAX: A high-level library for neural network\n\nAs we saw above, writing matmuls by hand to define neural networks gets old very quickly. There are in fact several high-level libraries to improve the ergonomics, and the two most popular ones are FLAX and Haiku. We will look at FLAX here.\n\nHere is an example of how we can define the same neutral network we had before.\n\nclass Net(nn.Module):\n\[email protected]\ndef __call__(self, x):\nx = nn.Dense(4)(x)\nx = nn.relu(x)\nx = nn.Dense(1)(x)\nreturn nn.sigmoid(x)\n\nAs you can see, this is a lot more readable than what we had. While this looks like an object, which has state, what actually happens is that there are other functions in nn.Module that will convert this class into pure functions.\n\nAnd since we don’t want to store any state in the function, we need to explicitly get the variables of our neural network by calling init.\n\nvars = Net().init(jax.random.PRNGKey(0), input)\n\nYou can see here that we have to pass the random key as we noted before. If you inspect the vars, they will look like this:\n\nFrozenDict({\nparams: {\nDense_0: {\nkernel: DeviceArray([[ 1.1381536 , -1.0838526 , 0.37998098, 0.15393464],\n[ 0.17555283, -0.3848625 , 0.52419275, -1.4104135 ]], dtype=float32),\nbias: DeviceArray([0., 0., 0., 0.], dtype=float32),\n},\nDense_1: {\nkernel: DeviceArray([[ 0.22024584],\n[ 0.5676514 ],\n[ 0.4185372 ],\n[-0.3969197 ]], dtype=float32),\nbias: DeviceArray([0.], dtype=float32),\n},\n},\n})\n\nIt also has the bias terms, which we didn’t include before. One quick thing to note is that the shape of kernel is (2, 4), and the shape of the bias is (4,). If the input is of shape (B, 2) [B was set to 3 in our example], then input times kernel should be of shape (B, 4). So how can we add a vector of shape (4,) to that? The answer is broadcasting. This array will get added to each of the B rows, which is exactly what we want.\n\nThe other thing to note is that the input’s batch size doesn’t matter.\n\nSo, the following, where the input shape is (1, 2), will produce the same result as well:\n\nvars = Net().init(jax.random.PRNGKey(0), jnp.array([[1., 2.]]))\nvars\nFrozenDict({\nparams: {\nDense_0: {\nkernel: DeviceArray([[ 1.1381536 , -1.0838526 , 0.37998098, 0.15393464],\n[ 0.17555283, -0.3848625 , 0.52419275, -1.4104135 ]], dtype=float32),\nbias: DeviceArray([0., 0., 0., 0.], dtype=float32),\n},\nDense_1: {\nkernel: DeviceArray([[ 0.22024584],\n[ 0.5676514 ],\n[ 0.4185372 ],\n[-0.3969197 ]], dtype=float32),\nbias: DeviceArray([0.], dtype=float32),\n},\n},\n})\n\nAnd in fact, if we use an input of shape (2,), we will still see the same output\n\nvars = Net().init(jax.random.PRNGKey(0), jnp.array([1., 2.]))\nvars\nFrozenDict({\nparams: {\nDense_0: {\nkernel: DeviceArray([[ 1.1381536 , -1.0838526 , 0.37998098, 0.15393464],\n[ 0.17555283, -0.3848625 , 0.52419275, -1.4104135 ]], dtype=float32),\nbias: DeviceArray([0., 0., 0., 0.], dtype=float32),\n},\nDense_1: {\nkernel: DeviceArray([[ 0.22024584],\n[ 0.5676514 ],\n[ 0.4185372 ],\n[-0.3969197 ]], dtype=float32),\nbias: DeviceArray([0.], dtype=float32),\n},\n},\n})\n\nTo get the prediction, we can just call:\n\n> Net().apply(vars, input)\n\nDeviceArray([[0.58010364],\n[0.4232422 ],\n[0.24422583],\n[0.60353917],\n[0.505488 ],\n[0.572509 ],\n[0.67478746],\n[0.72071725],\n[0.53566575],\n[0.44790095]], dtype=float32)\n\nYou may be wondering why we are initializing the object of Net again here. This is merely for convenience: since this object is stateless, it is fine to create a new object here.\n\nIn fact, this is a great opportunity to show that jax.vmap can be used as a way to add batch dimension.\n\n> vmap(lambda x: Net().apply(vars, x))(input)\n\nDeviceArray([[0.58010364],\n[0.4232422 ],\n[0.24422583],\n[0.60353917],\n[0.505488 ],\n[0.572509 ],\n[0.67478746],\n[0.72071725],\n[0.53566575],\n[0.44790095]], dtype=float32)\n\nConclusion\n\nSo, there you go. You have learnt the basics of JAX. There is a lot more that we didn’t cover, like using optax for defining optimizers and loss functions, or how to create realistic training loops when using FLAX. Let’s reserve that for another day, but hope this can help you get started!"
] |
[
null,
"https://smahesh.com/assets/images/Mahesh-avatar.jpg",
null
] |
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https://math.stackexchange.com/questions/2591632/p-nilpotent-finite-group-g-is-a-p-normal-group/2591717
|
[
"# $p$-nilpotent finite group $G$ is a $p$-normal group?\n\nIs it true that a $p$-nilpotent finite group $G$ is necessarily a $p$-normal group?\n\nThe definitions are here: p-nilpotent and p-normal (I like the 2nd definition)\n\nLet $p$ be a prime and $P,Q$ be $p$-Sylow subgroups of $G$ such that $Z(Q)\\subseteq P$.\n\nWe have to show $Z(Q)=Z(P)$. Writing $Q=P^{g}$ we can write this as $Z(P)^g=Z(P)$.\n\nThe group $G$ is $p$-nilpotent, so it has a $p$-normal complement by definition. Let $N$ be a $p$-normal complement for $G$, i.e., a normal subgroup $N$ of $G$ such that $G=NP$ and $N\\cap P=1$.\n\nSo again, we need to show $Z(P)^g=Z(P)$. We could write $g=np$ and this becomes $Z(P)^n=Z(P)$, for the center $Z(P)$ is a normal subgroup of $P$. But I'm not sure what to do now.\n\nAny help? Thank you.\n\nSuppose $Q = P^{g}$ is another Sylow $p$-subgroup such that $Z(P) \\le Q$. We show that $Z(P) = Z(Q)$.\nNote first that we may take $g \\in N$.\nLet $z \\in Z(P)$, and $y \\in Q$, so that $y = x^{g}$ for some $x \\in P$.\nThen \\begin{equation} [z, y] = [z, x^{g}] = [z, x [x, g]]= [z, [x, g]] [z, x]^{[x, g]} = [z, [x, g]]. \\end{equation} Now $[z, y] \\in Q$, but also $[z, [x, g]] \\in N$, as $g \\in N\\trianglelefteq G$, so $[z, y] = 1$ as $Q \\cap N = 1$\nThis shows that $Z(P) \\le Z(Q)$, equality follow from the fact that they have the same order."
] |
[
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|
https://gamedev.stackexchange.com/questions/105660/pygame-scrolling-camera-bug
|
[
"# Pygame Scrolling Camera bug\n\nI asked a question the other day about how to implement a scrolling camera. I got a great answer which helped me achieve that... but in doing so I have lost basically all collision detection. Strangely, There are still small, invisible \"platforms\" that can be collided with... Above the first platform in view upon spawning (the second platform down), there is a very small area where I can get collision detection... happens near coordinates (60, 51). Why is collision only happening in that tiny area, and not on any of the platforms throughout the entire level?\n\nThanks for any help, code at github: https://github.com/tear727/Netse/blob/master/game2.py\n\nimport random\nimport pygame\nfrom pygame import *\n\ndisplay_width = 800\ndisplay_height = 640\ncamera = [0,0]\nsize = (display_width, display_height)\nscreen = pygame.display.set_mode(size)\nblack = (0,0,0)\nwhite = (255,255,255)\nbackground = Surface((32, 32))\nbackground.convert()\nbackground.fill(Color('#783131'))\nlevel = [\n\"PPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP\",\n\"P P\",\n\"P P\",\n\"P P\",\n\"P PPPPPPPPPPP P\",\n\"P P\",\n\"P P\",\n\"P P\",\n\"P PPPPPPPP P\",\n\"P P\",\n\"P PPPPPPP P\",\n\"P PPPPPP P\",\n\"P P\",\n\"P PPPPPPP P\",\n\"P P\",\n\"P PPPPPP P\",\n\"P P\",\n\"P PPPPPPPPPPP P\",\n\"P P\",\n\"P PPPPPPPPPPP P\",\n\"P P\",\n\"P P\",\n\"P P\",\n\"P P\",\n\"PPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP\",\n]\n\ndef main():\npygame.init()\nclock = pygame.time.Clock()\nx = 0\ny = 0\nleft_viewbox = display_width/2 - display_width/8\nright_viewbox = display_width/2 + display_width/10\nup_viewbox = display_height/2 - display_height/12\ndown_viewbox = display_height/2 - display_height/12\n\nplatformlst = []\nplayers = []\n\nplayer = Player(camera, camera)\nmap = Map(x, y)\nplatforms = Platform(camera, camera)\nplayers.append(player)\n\npygame.display.set_caption(\"My game\")\n\ndef buildMap(xpos, ypos):\nfor i in level:\nfor j in i:\nif j == \"P\":\nplatformlst.append(platforms)\nscreen.blit(platforms.image, (xpos + camera, ypos + camera))\nxpos += 64\nypos += 64\nxpos = 0\n\ndef follow(shift_x, shift_y):\ncamera += shift_x\ncamera += shift_y\nfor i in players:\ni.rect.x += shift_x\ni.rect.y += shift_y\n\ndef viewbox():\nif player.x <= left_viewbox:\nview_difference = left_viewbox - player.x\nplayer.x = left_viewbox\nfollow(view_difference, 0)\n\nif player.x >= right_viewbox:\nview_difference = right_viewbox - player.x\nplayer.x = right_viewbox\nfollow(view_difference, 0)\n\nif player.y <= up_viewbox:\nview_difference = up_viewbox - player.y\nplayer.y = up_viewbox\nfollow(0, view_difference)\n\nif player.y >= down_viewbox:\nview_difference = down_viewbox - player.y\nplayer.y = down_viewbox\nfollow(0, view_difference)\n\ndef collide():\nif (platforms.rect.x < player.rect.x + player.rect.w and\nplatforms.rect.x + platforms.rect.w > player.rect.x and\nplatforms.rect.y < player.rect.y + player.rect.h and\nplatforms.rect.h + platforms.rect.y > player.rect.y):\nprint \"collided\"\nplayer.grounded = True\n\ngame_exit = False\n\nwhile not game_exit:\nclock.tick(60)\n\nfor event in pygame.event.get():\nif event.type == pygame.QUIT:\ngame_exit = True\n\nif event.type == pygame.KEYDOWN:\nprint \"pressed key\"\n\nif event.key == pygame.K_LEFT:\nplayer.left()\n\nif event.key == pygame.K_RIGHT:\nprint \"move right\"\nplayer.right()\nif event.key == pygame.K_UP:\nprint \"move up\"\nplayer.up()\n\nif event.type == pygame.KEYUP:\nif event.key == pygame.K_LEFT or event.key == pygame.K_RIGHT:\nplayer.stopx()\n\nif event.key == pygame.K_UP or pygame.K_DOWN:\nplayer.stopy()\n\nif event.type == pygame.MOUSEBUTTONDOWN:\nprint \"pressed mouse button\"\nplayer.boost()\n\nif event.type == pygame.MOUSEBUTTONUP:\nprint \"mouse button up\"\nplayer.stop()\n\nprint player.rect.x\nprint player.rect.y\nmap.buildBackground()\nbuildMap(x, y)\nplayer.draw()\nplayer.move()\ncollide()\nviewbox()\npygame.display.update()\n\nclass Map:\ndef __init__(self, x, y):\nself.x = x\nself.y = y\n\ndef buildBackground(self):\nfor i in range(32):\nfor j in range(32):\nscreen.blit(background, (i*32, j*32))\n\nclass Player:\ndef __init__(self, x, y):\nself.grounded = True\nself.x = x\nself.y = y\nself.velx = 0\nself.vely = 0\nself.image = player_image\nself.rect = player_image.get_rect()\n\ndef draw(self):\nscreen.blit(self.image, (self.x, self.y))\n\ndef move(self):\nself.x += self.velx\nself.y += self.vely\nif not self.grounded:\nself.y += 7\nself.grounded = False\n\ndef right(self):\nself.velx = 20\n\ndef left(self):\nself.velx = -20\n\ndef up(self):\nself.vely = -20\n\ndef stopx(self):\nself.velx = 0\n\ndef stopy(self):\nself.vely = 0\n\ndef boost(self):\nself.velx += 20\nself.vely += 20\n\ndef stop(self):\nself.velx = 0\nself.vely = 0\n\nclass Platform:\ndef __init__(self, x, y):\nself.image = Surface((64, 64))\nself.image.convert()\nself.image.fill(Color('#000000'))\nself.rect = Rect(x, y, 64, 64)\n\ndef draw(self):\nscreen.blit(self.image, (x, y))\n\nif __name__ == \"__main__\":\nmain()\n\n\nThe trick is I think you misunderstood my original answer to your previous question. The camera's position should only be taken into account when drawing (using blit). You're offsetting all drawn objects in the screen by the camera's position.\n\nSo two specific problems with the code you have shown:\n\n1. The blit that does use the camera position should subtract it rather than adding it\n2. Every blit needs to subtract the camera's position\n• thanks again. However, I have tried this with subtraction and addition, along with changing all the x,y variables of the objects to their own x, y coordinates, the local x, y coordinates, and the camera's x, y. Still nothing. Addition yields the best results because the camera works perfectly, just nothing else. – thevengefulco Aug 15 '15 at 7:56\n• I can assure you, subtraction is correct. It has nothing to do with what \"looks\" better because likely other aspects of your game are confusing the issue. The reason for subtraction is when your camera moves to the right, everything else moves to the left. If you ADD the camera's position the opposite happens. You can also think of it like, when the camera is at some position, any object at that same position should then be drawn to the screen at (0, 0). That can only be accomplished with subtraction. Your main issue again is that all draws or blits must have the camera position subtracted. – Alex Sherman Aug 16 '15 at 15:53\n\nThis may, or may not, be the correct answer for you,\n\nBUT\n\nIn your code, when you draw the platforms images to the screen, you add the camera's position to their x/y positions. This would not be necessary because you would then have problems with colliding, because your adding the cameras scroll to it so you wouldn't see them in their colliding positions.\n\nIf you wanted to keep that, then I would suggest adding the camera x/y to the collision section.\n\nHope this helps.\n\n• Thanks for the help. That makes a lot of sense. Yet, when the values are not added, the camera no longer scrolls and the player is just stuck inside the viewbox. Maybe viewbox or some x/y coordinates are wrong altogether – thevengefulco Aug 15 '15 at 2:10\n• This maybe also another thing, but it's just a suggestion, instead of drawing the camera and updating it based on the player position when it moves, you might want to draw the camera from the players x/y position (of course subtracting it based on half the screens width/height)? – PlatyPi Aug 15 '15 at 2:23"
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.6336311,"math_prob":0.87695956,"size":4786,"snap":"2020-10-2020-16","text_gpt3_token_len":1374,"char_repetition_ratio":0.20242576,"word_repetition_ratio":0.07401575,"special_character_ratio":0.31487674,"punctuation_ratio":0.26516634,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96114594,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-04-02T14:09:57Z\",\"WARC-Record-ID\":\"<urn:uuid:60013deb-a3a4-45e5-8903-4d111c6e454f>\",\"Content-Length\":\"158617\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8ae06208-fae7-4029-9a1b-980040f0f3c3>\",\"WARC-Concurrent-To\":\"<urn:uuid:e819670c-8a46-46bd-b57a-4a775af48894>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://gamedev.stackexchange.com/questions/105660/pygame-scrolling-camera-bug\",\"WARC-Payload-Digest\":\"sha1:GVOPRPX54MAWV4ARJ5NQV2ZOXN6HSKRW\",\"WARC-Block-Digest\":\"sha1:ECNSRATRTYFPNYMHKVTXPUWY6KIC4RA4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585370506959.34_warc_CC-MAIN-20200402111815-20200402141815-00223.warc.gz\"}"}
|
https://www.vegasledscreens.com/faq/121-led-screen-power-consumption.html
|
[
"## Vegas LED Screens - LED Displays & LED Signs\n\nSales Support: +1 (702) 967 0700\n\nf73hr8q9xi\n\n# LED Screen Power Consumption Explained",
null,
"",
null,
"",
null,
"One of the most asked questions is how to calculate the maximal power consumption of a LED screen. Well first of all, you should know the input current and the voltage of the LEDs on the LED screen. Theoretically, in a labratory environment the greatest input current of the LEDs on a LED screen is 20mA, and the voltage for the LEDs for the LED screens is 5V. But in reality, the current of the LEDs of a LED screen can't reach 20mA. So the power consumption of a LED is 20mA x 5V = 0.1W. So it is very simple how to calculate the power consumption of one LED on a LED display.\n\n## Pixel Configuration of LED Displays\n\nAnd secondly, what you should know is the pixel configuration of the LED display and the resolution per square meter of the LED display. When you get this data you will know how many LEDs are available on the LED display. For example, let's take a LED display with a pixel pitch of P16mm, the pixel configuration is 2R1G1B, which means that each pixel consists of 2 red LED's, 1 green LED and 1 blue LED. So this means that every pixel has 4 LED's. And the resolution per square meter of pitch 16mm LED display is 3906 pixels, which means there are totally 3906 pixels in one square meter. According to the previous data the calculation will be as folows: 4 x 3906 = 15,624 total LED's per square meter for this LED display.\n\n## Maximal Power Consumption for LED displays\n\nThere are two types of driving methods for LED displays, which is the constant static driving method and the scan driving method. First let's explain how to calculate the maximal power consumption and average power consumption of a pitch 16mm LED display using constant static current driving.\n\n### Constant Static Current Driving Method\n\nThe maximal power consumption of a pitch 16mm LED display is 15624 x 0.1W = 1.56Kw pero square meter. The average power consumption of a pitch 16mm LED display is 1.56Kw / 2 = 780W. This calculation way is for static constant current driving LED displays.\n\n### Scan Driving Method\n\nFor scan driving LED displays, you must notice that it has a 1/4 driving, 1/8 driving or 1/2 driving. Taking a pitch 6mm indoor LED display as an example, the usual driving method is 1/8, so if you calculate according to above way, the maximal power consumption of a pitch 6mm indoor LED display will be extremely high and gives us more than 8Kw per square meter. But in reality it is not correct, the reason is that you need to make the total calculation result divided by 8 because the driving method is 1 over 8 (1/8), and like that you get a maximal power consumption of 1.1Kw per square meter."
] |
[
null,
"https://www.vegasledscreens.com/images/M_images/pdf_button.png",
null,
"https://www.vegasledscreens.com/images/M_images/printButton.png",
null,
"https://www.vegasledscreens.com/images/M_images/emailButton.png",
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.9002034,"math_prob":0.9935601,"size":3003,"snap":"2021-04-2021-17","text_gpt3_token_len":735,"char_repetition_ratio":0.16838947,"word_repetition_ratio":0.072088726,"special_character_ratio":0.24875125,"punctuation_ratio":0.087027915,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95483345,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,3,null,4,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-16T18:26:08Z\",\"WARC-Record-ID\":\"<urn:uuid:d9fae640-b1c7-43f2-a3b3-edc7435ae9bb>\",\"Content-Length\":\"27240\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:398620d6-cd89-4f23-bd67-c9714eaf5d09>\",\"WARC-Concurrent-To\":\"<urn:uuid:44d403fd-655f-4b74-b42f-14455a2a88c4>\",\"WARC-IP-Address\":\"192.169.237.50\",\"WARC-Target-URI\":\"https://www.vegasledscreens.com/faq/121-led-screen-power-consumption.html\",\"WARC-Payload-Digest\":\"sha1:YEUE477YZJMNJH6IZM5HPGK27UX6GUGL\",\"WARC-Block-Digest\":\"sha1:W6FJOZVDFISRWGGRQOIVJPZPLPOH3H4D\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038088245.37_warc_CC-MAIN-20210416161217-20210416191217-00018.warc.gz\"}"}
|
https://www.thebearsenal.com/python-programs/write-a-list-comprehension-in-python-to-get-a-list-of-even-numbers-when-a-range-is-given
|
[
"# Write a list comprehension in python to get a list of even numbers when a range is given\n\n## Topic: Write a list comprehension in python to get a list of even numbers when a range is given\n\nSolution\n\n```N = 20\nnumber_list = [ x for x in range(N) if x % 2 == 0]\nprint(f'List of Even Numbers:', number_list)\n```"
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.79441726,"math_prob":0.99303293,"size":321,"snap":"2022-27-2022-33","text_gpt3_token_len":81,"char_repetition_ratio":0.16088328,"word_repetition_ratio":0.46666667,"special_character_ratio":0.2647975,"punctuation_ratio":0.046153847,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98867047,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-08T22:18:15Z\",\"WARC-Record-ID\":\"<urn:uuid:8f49e95f-bb1b-41da-b536-b921cd7bd95a>\",\"Content-Length\":\"4401\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:295b4ef8-7a19-4d0c-b1ae-9fd0a47b2357>\",\"WARC-Concurrent-To\":\"<urn:uuid:e59e2ba5-39bf-4b43-9aef-e733e63c49dd>\",\"WARC-IP-Address\":\"52.73.153.209\",\"WARC-Target-URI\":\"https://www.thebearsenal.com/python-programs/write-a-list-comprehension-in-python-to-get-a-list-of-even-numbers-when-a-range-is-given\",\"WARC-Payload-Digest\":\"sha1:26KWEOQZMYB3I3L5DTVEMB3U2YSSTF5F\",\"WARC-Block-Digest\":\"sha1:7DKLMQWEHTYQTMGTC3WQ3WXDSQGT7G4D\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882570879.1_warc_CC-MAIN-20220808213349-20220809003349-00234.warc.gz\"}"}
|
https://tradingqna.com/t/option-profit-loss/10591
|
[
"# Option profit/loss\n\nif i YES BANK strick price 1400CE BUY @ 50 LOT SIZE 700 (MIS / NORMAL )\n\n`````` so THE PREMIUM 700*50 =35000\n``````\n\nIF spot price 1390 after 1 hour rice to 1395 and Primium price 50 to 60 then i SALE the YES BANK Contract\n\n1. What would happen ?\nI lose all premium because SPOT PRICE(1395) is less then STRICK PRICE (1400CE) ?\ni mean - I only get profit when SPOT PRICE MORE THEN STRICK PRICE WHEN STOP PRICE CROSS BRACKEVEN POINT 1400+50=1450 ? THEN I GET BOOK PROFIT WHEN SPOT PRICE MORE THEN 1450 LIKE 1451 ,1452 ,1453 .\n\nHello,\n\nThere are two values which add to determine an Option’s value, namely Time value and Intrinsic value.\n\nOptions value = Intrinsic value + Time value.\n\nThe Time value of the option keeps reducing as the option moves towards expiry and at expiry, the Time value becomes zero. Only the Intrinsic value of the option will remain at expiry if the option is In-The-Money.\n\nLets take 2 cases for Yesbank 1400CE at expiry -\n\n1. Yesbank Spot closes at 1395\n\nYour option is Out-Of-The-Money(OTM) and it will expire worthless. It’s time value as well as Intrinsic value will become zero.\n\n1. Yesbank Spot closes at 1405\n\nYour option is In-The-Money(ITM) and since the time value of the option is at 0, your Intrinsic value will be 5 which is equal to the Option value.\n\nNow lets consider your case -\n\nYour option has both intrinsic value as well as time value.\n\nIf Yesbank spot goes from 1390 to 1395 and the Premium of 1400CE moves from 50 to 60, then you will make a Rs.10 profit. If you multiply this with your quantity, then your Total Profit =700 * 10 = Rs.7000.\n\nYou make money in options on the changes in the Option Premium values. If you buy an option and the option premium value moves up from you buying price, then you are in a profit."
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.80552274,"math_prob":0.9490983,"size":1748,"snap":"2023-14-2023-23","text_gpt3_token_len":464,"char_repetition_ratio":0.15137614,"word_repetition_ratio":0.0,"special_character_ratio":0.28318077,"punctuation_ratio":0.0726257,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95484793,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-29T19:04:10Z\",\"WARC-Record-ID\":\"<urn:uuid:05512045-96a0-4d15-8d7c-5dceb7fe2f2f>\",\"Content-Length\":\"20434\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d1494cda-8b93-4bef-b3a6-3e3cf7fc79de>\",\"WARC-Concurrent-To\":\"<urn:uuid:7021cb1c-84fb-4299-87a2-54cfbbefb7b6>\",\"WARC-IP-Address\":\"104.21.75.39\",\"WARC-Target-URI\":\"https://tradingqna.com/t/option-profit-loss/10591\",\"WARC-Payload-Digest\":\"sha1:CNDO3QSMXEIIKK6NBKCE4AFI4WZS6QTC\",\"WARC-Block-Digest\":\"sha1:NAV7K3K2P2SCP5M2UYX2LAJUTSZZ5EL5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296949025.18_warc_CC-MAIN-20230329182643-20230329212643-00695.warc.gz\"}"}
|
https://forums.developer.nvidia.com/t/data-streaming-in-2d-array-lbm/62895
|
[
"",
null,
"# Data streaming in 2D array, LBM\n\nHi, I have a problem with an unexplained problem, results in the array on the edges are broken (prints 0). Im not sure whats wrong. Variable ‘a’ is a array filled 0-1 from the host. In the program are a lot of time steps and the kernel is executed many times. Somebody can explain what is maybe wrong in kernel function?\n\nMy array\n\n``````|--------|\n| i | n | *2\n|--------|\n| n | *2 + 1\n``````\n``````__global__ void GPU_CalculateTemp(float *a, float *b) {\n\nint i = blockIdx.x * blockDim.x + threadIdx.x;\n\n//Patterns\n\n//Main loop\nif ((i > 0) && (i < N )) {\n//Collision\nfloat fl = a[i * 2];\nfloat fr = a[i * 2 + 1];\n\nfloat rho = fl + fr;\nfloat feq = 0.5 * rho;\n\n//Streaming on the left and right from i\nb[(i - 1) * 2] = omega * feq + (1 - omega) * fl;\nb[(i + 1) * 2 + 1] = omega * feq + (1 - omega) * fr;\n}\n\n}\n``````\n\nKernel\n\n``````void compute(int rank, float **device_a, float **device_b) {\nint threadsperblock = N; // static size #define N 100\nint blockspergrid = 1;\nsize_t size = N*sizeof(float);\n\ncudaMemcpy(*device_a, *device_b, size, cudaMemcpyDeviceToDevice); //after this device_a is copied to host array and print\n}\n``````\n\nMy left side array, second and third positions are 0:\n\n``````0.05\t 0.00\t 0.10\t 0.00\t 0.25\t 0.11\t 0.23\t 0.14\t 0.37\n``````\n\nsize_t size seems undersized, given you’re accessing up to element\n\nb[(i + 1) * 2 + 1] for i ranging from 1 through N-1\n\nso the last element written to is b[2*N-1]\n\nAre you sure the arrays are allocated (and copied) with size 2Nsizeof(float) ?\n\nsize_t size = N*sizeof(float); seems to suggest otherwise.\n\nChristian\n\nMy dear Christian, this was very helpful."
] |
[
null,
"https://aws1.discourse-cdn.com/nvidia/original/2X/3/3f301944ed0d2d0d779b3eaa251520d35458d467.png",
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.64142585,"math_prob":0.98989195,"size":1250,"snap":"2020-34-2020-40","text_gpt3_token_len":406,"char_repetition_ratio":0.11476725,"word_repetition_ratio":0.042918455,"special_character_ratio":0.3816,"punctuation_ratio":0.1640625,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.971675,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-18T17:52:10Z\",\"WARC-Record-ID\":\"<urn:uuid:65cd6fcf-ae51-44b4-bae1-f793c11f98ec>\",\"Content-Length\":\"22678\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7cd41aec-207f-4fff-95e3-f3b84f3c5747>\",\"WARC-Concurrent-To\":\"<urn:uuid:b632f2cc-f6a4-4dc1-8119-b172d2c33d2b>\",\"WARC-IP-Address\":\"65.19.128.98\",\"WARC-Target-URI\":\"https://forums.developer.nvidia.com/t/data-streaming-in-2d-array-lbm/62895\",\"WARC-Payload-Digest\":\"sha1:QLZPGFTWJ4IVV6BBK52MTKJCQ66I3UIB\",\"WARC-Block-Digest\":\"sha1:V2A72SHLBJS3ASD67SFOA3JNTB7KB6A5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400188049.8_warc_CC-MAIN-20200918155203-20200918185203-00759.warc.gz\"}"}
|
https://math.stackexchange.com/questions/3460809/integrating-int-frac-3x24-dx
|
[
"# Integrating $\\int\\frac{-3}{x^2+4}\\ dx$\n\nI was given the problem: $$\\int\\frac{-3}{x^2+4}\\ dx$$\n\nI am unsure how to integrate it. It seems to me that it requires a u-substitution because of the $$x^2$$ in the denominator, but I cannot figure out what to substitute. I cannot substitute in $$x^2$$ because then I am stuck with an x in that derivative.\nI do know that I can pull the -3 out of the integral and get: $$-3\\int\\frac{1}{x^2+4}$$ but this does not help me - I'm still stuck. I also know that arctan is $$\\int\\frac{1}{x^2+1}\\ dx$$ which seems similar to this, but it is not the same thing.\n\n• Divide by $4$ and define $y = x/2$ then solve as an arctan Dec 3, 2019 at 3:22\n• Let $x=2 \\tan(\\theta)$. Dec 3, 2019 at 3:24\n\nSubstitute $$t = \\dfrac x2\\implies x = 2t,\\mathrm dx = 2\\mathrm dt$$.\n\nTherefore,\n\n$$\\int\\dfrac {-3}{x^2 + 4}\\,\\mathrm dx = -\\int\\dfrac 6{4t^2 + 4}\\,\\mathrm dt = -\\int\\frac6{4(t^2 + 1)}\\,\\mathrm dt = -\\frac 32\\int\\frac1{t^2 + 1}\\,\\mathrm dt$$\n\n$$\\displaystyle\\int\\dfrac 1{t^2 + 1}\\,\\mathrm dt$$ results in $$\\arctan t + C$$. Reverse substitution to get $$\\int\\dfrac {-3}{x^2 + 4}\\,\\mathrm dx = -\\dfrac32\\arctan\\left(\\dfrac x2\\right)+C.$$\n\n• I can't follow how that substitution works. Can you break it down really simply - between $\\int\\dfrac {-3}{x^2 + 4}\\,\\mathrm dx$ and $-\\int\\dfrac 6{4t^2 + 4}\\,\\mathrm dt$\n– Burt\nDec 3, 2019 at 3:35\n• Substitute $x=2t$ Dec 3, 2019 at 3:40\n• @J.W.Tanner what do you mean?\n– Burt\nDec 3, 2019 at 3:47\n• If $x=2t$ then $\\displaystyle\\int\\dfrac{-3}{x^2+4}dx=\\int\\dfrac{-3}{(2t)^2+4}2dt$; I thought that's what you were asking for Dec 3, 2019 at 3:56\n• @Burt Which part are you stuck at?\n– an4s\nDec 3, 2019 at 5:00\n\nYup, $$\\tan^{-1}$$ is a great start.\n\nSo once you have $$\\displaystyle -3\\int \\frac{1}{x^2+4}\\,dx$$, you can turn this into $$\\displaystyle -\\frac{3}{4}\\int\\frac{1}{(\\frac{x}{2})^2+1}\\,dx=-\\frac{3}{2}\\int\\frac{\\frac{1}{2}}{(\\frac{x}{2})^2+1}\\,dx$$.\n\nYou see the perfect u-substitution yet?\n\n$$\\int{\\frac{-3dx}{x^2+4}}=-3\\int{\\frac{dx}{x^2+4}}=-3\\int{\\frac{dx}{4\\left(\\frac{x^2}{4}+1\\right)}}$$\n\n-3/2arctanx/2 is a primitive function of -3/(x^2+4). It is easy to check this because (-3/2arctanx/2)’=-3/(x^2+4)"
] |
[
null
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https://americanfootballdatabase.fandom.com/wiki/Geographic_coordinate_system?oldid=22978
|
[
"## FANDOM\n\n57,099 Pages\n\nFundamentals Geodesy Geodesy · GeodynamicsGeomatics · Cartography Datum · Distance · GeoidFigure of the EarthGeodetic systemGeog. coord. systemHor. pos. representationMap projectionReference ellipsoidSatellite geodesySpatial reference system GNSS · GPS · ... ED50 · ETRS89 · NAD83NAVD88 · SAD69 · SRIDUTM · WGS84 · ... History of geodesyNAVD29 · ... Template:Tnavbar\n\nA geographic coordinate system is a coordinate system that enables every location on the Earth to be specified by a set of numbers. The coordinates are often chosen such that one of the numbers represent vertical position, and two or three of the numbers represent horizontal position. A common choice of coordinates is latitude, longitude and elevation.\n\n## Geographic latitude and longitude",
null,
"Edit\n\nThe geographic latitude (abbreviation: Lat., φ, or phi) of a point on the Earth's surface is the angle between the equatorial plane and a line that passes through that point and is normal to the surface of a reference ellipsoid which approximates the shape of the Earth.[n 1] This line passes a few kilometers away from the center of the Earth except at the poles and the equator where it passes through Earth's center.[n 2] Lines joining points of the same latitude trace circles on the surface of the Earth called parallels, as they are parallel to the equator and to each other. The north pole is 90° N; the south pole is 90° S. The 0° parallel of latitude is designated the equator, the fundamental plane of all geographic coordinate systems. The equator divides the globe into Northern and Southern Hemispheres.\n\nThe Longitude (abbreviation: Long., λ, or lambda) of a point on the Earth's surface is the angle east or west from a reference meridian to another meridian that passes through that point. All meridians are halves of great ellipses (often improperly called great circles), which converge at the north and south poles.\n\nA line passing near the Royal Observatory, Greenwich (near London in the UK) has been chosen as the international zero-longitude reference line, the Prime Meridian. Places to the east are in the eastern hemisphere, and places to the west are in the western hemisphere. The antipodal meridian of Greenwich is both 180°W and 180°E. The zero/zero point is located in the Gulf of Guinea about 625 km south of Tema, Ghana.\n\nIn 1884, the United States hosted the International Meridian Conference and twenty-five nations attended. Twenty-two of them agreed to adopt the location of Greenwich as the zero-reference line. The Dominican Republic voted against the adoption of that motion, while France and Brazil abstained. To date, there exist organizations around the world which continue to use historical prime meridians which existed before the acceptance of Greenwich became common-place.[n 3]\n\nThe combination of these two components specifies the position of any location on the planet, but does not consider altitude nor depth.\n\nThis latitude/longitude \"webbing\" is known as the conjugate graticule.\n\nIn defining an ellipse, the short (vertical) diameter is known as the conjugate diameter, and the long (horizontal) diameter—perpendicular, or \"transverse\", to the conjugate—is the transverse diameter. With a sphere or ellipsoid, the conjugate diameter is known as the polar axis and the transverse as the equatorial axis. The graticule perspective is based on this designation: As the longitudinal rings — geographically defined, all great circles — converge at the poles, it is the poles that the conjugate graticule is defined. If the polar vertex is \"pulled down\" 90°, so that the vertex is on the equator, or transverse diameter, then it becomes the transverse graticule, upon which all spherical trigonometry is ultimately based (if the longitudinal vertex is between the poles and equator, then it is considered an oblique graticule).\n\n1. The surface of the Earth is closer to an ellipsoid than to a sphere, as its equatorial diameter is larger than its north-south diameter.\n2. The greatest distance between an ellipsoid normal and the center of the Earth is 21.9 km at a latitude of 45°, using Earth radius#Radius at a given geodetic latitude and Latitude#Comparison of selected types: (6367.5 km)×tan(11.67')=21.9 km.\n3. The French Institut Géographique National (IGN) maps still use longitude from a meridian passing through Paris, along with longitude from Greenwich.\n\n## UTM and UPS systems",
null,
"Edit\n\nThe Universal Transverse Mercator (UTM) and Universal Polar Stereographic (UPS) coordinate systems both use a metric-based cartesian grid laid out on a conformally projected surface to locate positions on the surface of the Earth. The UTM system is not a single map projection but a series of map projections, one for each of sixty 6-degree bands of longitude. The UPS system is used for the polar regions, which are not covered by the UTM system.\n\n## Stereographic coordinate system",
null,
"Edit\n\nDuring medieval times, the stereographic coordinate system was used for navigation purposes.[citation needed] The stereographic coordinate system was superseded by the latitude-longitude system.\n\nAlthough no longer used in navigations, the stereographic coordinate system is still used in modern times to describe crystallographic orientations in the fields of crystallography, mineralogy and materials science.[citation needed]\n\n## Geodetic height",
null,
"Edit\n\nTo completely specify a location of a topographical feature on, in, or above the Earth, one has to also specify the vertical distance from the centre of the Earth, or from the surface of the Earth. Because of the ambiguity of \"surface\" and \"vertical\", it is more commonly expressed relative to a precisely defined vertical datum which holds fixed some known point. Each country has defined its own datum. For example, in the United Kingdom the reference point is Newlyn, while in Canada, Mexico and the United States, the point is near Rimouski, Quebec, Canada. The distance to Earth's centre can be used both for very deep positions and for positions in space.\n\n## Cartesian coordinates",
null,
"Edit\n\nEvery point that is expressed in ellipsoidal coordinates can be expressed as an x y z (Cartesian) coordinate. Cartesian coordinates simplify many mathematical calculations. The origin is usually the center of mass of the earth, a point close to the Earth's center of figure.\n\nWith the origin at the center of the ellipsoid, the conventional setup is the expected right-hand:\n\nZ-axis along the axis of the ellipsoid, positive northward\nX- and Y-axis in the plane of the equator, X-axis positive toward 0 degrees longitude and Y-axis positive toward 90 degrees east longitude\n\nAn example is the NGS data for a brass disk near Donner Summit, in California. Given the dimensions of the ellipsoid, the conversion from lat/lon/height-above-ellipsoid coordinates to X-Y-Z is straightforward—calculate the X-Y-Z for the given lat-lon on the surface of the ellipsoid and add the X-Y-Z vector that is perpendicular to the ellipsoid there and has length equal to the point's height above the ellipsoid. The reverse conversion is harder: given X-Y-Z we can immediately get longitude, but no closed formula for latitude and height exists. However, using Bowring's formula in 1976 Survey Review the first iteration gives latitude correct within $10^{-11}$ degree as long as the point is within 10000 meters above or 5000 meters below the ellipsoid.\n\n## Shape of the Earth",
null,
"Edit\n\nThe Earth is not a sphere, but an irregular shape approximating a biaxial ellipsoid. It is nearly spherical, but has an equatorial bulge making the radius at the equator about 0.3% larger than the radius measured through the poles. The shorter axis approximately coincides with axis of rotation. Map-makers choose the true ellipsoid that best fits their need for the area they are mapping. They then choose the most appropriate mapping of the spherical coordinate system onto that ellipsoid. In the United Kingdom there are three common latitude, longitude, height systems in use. The system used by GPS, WGS84, differs at Greenwich from the one used on published maps OSGB36 by approximately 112m. The military system ED50, used by NATO, differs by about 120m to 180m.\n\nThough early navigators thought of the sea as a flat surface that could be used as a vertical datum, this is far from reality. The Earth has a series of layers of equal potential energy within its gravitational field. Height is a measurement at right angles to this surface, roughly toward the centre of the Earth, but local variations make the equipotential layers irregular (though roughly ellipsoidal). The choice of which layer to use for defining height is arbitrary. The reference height we have chosen is the one closest to the average height of the world's oceans. This is called the geoid.\n\nThe Earth is not static as points move relative to each other due to continental plate motion, subsidence, and diurnal movement caused by the Moon and the tides. The daily movement can be as much as a metre. Continental movement can be up to 10 cm a year, or 10 m in a century. A weather system high-pressure area can cause a sinking of 5 mm. Scandinavia is rising by 1 cm a year as a result of the melting of the ice sheets of the last ice age, but neighbouring Scotland is rising by only 0.2 cm. These changes are insignificant if a local datum is used, but are significant if the global GPS datum is used.\n\n## Expressing latitude and longitude as linear units",
null,
"Edit\n\nOn the GRS80 or WGS84 spheroid at sea level at the equator, one latitudinal second measures 30.715 metres, one latitudinal minute is 1843 metres and one latitudinal degree is 110.6 kilometres. The circles of longitude, meridians, meet at the geographical poles, with the west-east width of a second naturally decreasing as latitude increases. On the equator at sea level, one longitudinal second measures 30.92 metres, a longitudinal minute is 1855 metres and a longitudinal degree is 111.3 kilometres. At 30° a longitudinal second is 26.76 metres, at Greenwich (51° 28' 38\" N) 19.22 metres, and at 60° it is 15.42 metres.\n\nOn the WGS84 spheroid, the length in meters of a degree of latitude at latitude φ (that is, the distance along a north-south line from latitude (φ - 0.5) degrees to (φ + 0.5) degrees) is about\n\n111132.954 - 559.822(cos 2φ) + 1.175(cos 4φ)\n\n(Those coefficients can be improved, but as they stand the distance they give is correct within a centimeter.)\n\nThe width of one longitudinal degree at latitude $\\scriptstyle{\\phi}\\,\\!$ can be calculated by this formula (to get the width per minute and second, divide by 60 and 3600, respectively):\n\n$\\frac{\\pi}{180}\\cos \\phi (M_r) \\!$\n\nwhere Earth's average meridional radius $\\scriptstyle{M_r}\\,\\!$ approximately equals 6,367,449 m. Due to the average radius value used, this formula is of course not precise. A better approximation of a longitudinal degree at latitude $\\scriptstyle{\\phi}\\,\\!$ is\n\n$\\frac{\\pi}{180}a \\cos \\beta \\,\\!$\n\nwhere Earth's equatorial radius $a$ equals 6,378,137 m and $\\scriptstyle{\\tan \\beta = \\frac{b}{a}\\tan\\phi}\\,\\!$; for the GRS80 and WGS84 spheroids, b/a calculates to be 0.99664719. ($\\scriptstyle{\\beta}\\,\\!$ is known as the parametric or reduced latitude).\n\nAside from rounding, this is the exact distance along a parallel of latitude; getting the distance along the shortest route will be more work, but those two distances are always within 0.6 meter of each other if the two points are one degree of longitude apart.\n\nLongitudinal length equivalents at selected latitudes\nLatitude Town Degree Minute Second ±0.0001°\n60°Saint Petersburg 55.65 km 0.927 km 15.42 m 5.56 m\n51° 28' 38\" NGreenwich 69.29 km 1.155 km 19.24 m 6.93 m\n45°Bordeaux 78.7 km 1.31 km 21.86 m 7.87 m\n30°New Orleans 96.39 km 1.61 km 26.77 m 9.63 m\nQuito 111.3 km 1.855 km 30.92 m 11.13 m\n\n## Datums often encountered",
null,
"Edit\n\nLatitude and longitude values can be based on different geodetic systems or datums, the most common being WGS 84, a global datum used by all GPS equipment.[n 1] Other datums are significant because they were chosen by a national cartographical organisation as the best method for representing their region, and these are the datums used on printed maps. The latitude and longitude on a map may not be the same as on a GPS receiver. Coordinates from the mapping system can sometimes be roughly changed into another datum using a simple translation. For example, to convert from ETRF89 (GPS) to the Irish Grid add 49 metres to the east, and subtract 23.4 metres from the north. More generally one datum is changed into any other datum using a process called Helmert transformations. This involves converting the spherical coordinates into Cartesian coordinates and applying a seven parameter transformation (translation, three-dimensional rotation), and converting back.\n\nIn popular GIS software, data projected in latitude/longitude is often represented as a 'Geographic Coordinate System'. For example, data in latitude/longitude if the datum is the North American Datum of 1983 is denoted by 'GCS North American 1983'.\n\n## Geostationary coordinates",
null,
"Edit\n\nGeostationary satellites (e.g., television satellites) are over the equator at a specific point on Earth, so their position related to Earth is expressed in longitude degrees only. Their latitude is always zero, that is, over the equator.\n\n##",
null,
"Edit\n\nCommunity content is available under CC-BY-SA unless otherwise noted."
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|
https://sass-lang.com/documentation/values/numbers
|
[
"# Numbers\n\nNumbers in Sass have two components: the number itself, and its units. For example, in `16px` the number is `16` and the unit is `px`. Numbers can have no units, and they can have complex units. See Units below for more details.\n\n### SCSS Syntax\n\n``````@debug 100; // 100\n@debug 0.8; // 0.8\n@debug 16px; // 16px\n@debug 5px * 2px; // 10px*px (read \"square pixels\")\n``````\n\n### Sass Syntax\n\n``````@debug 100 // 100\n@debug 0.8 // 0.8\n@debug 16px // 16px\n@debug 5px * 2px // 10px*px (read \"square pixels\")\n``````\n\nSass numbers support the same formats as CSS numbers, including scientific notation, which is written with an `e` between the number and its power of 10. Because support for scientific notation in browsers has historically been spotty, Sass always compiles it to fully expanded numbers.\n\n### SCSS Syntax\n\n``````@debug 5.2e3; // 5200\n@debug 6e-2; // 0.06\n``````\n\n### Sass Syntax\n\n``````@debug 5.2e3 // 5200\n@debug 6e-2 // 0.06\n``````\n\nSass doesn’t distinguish between whole numbers and decimals, so for example `math.div(5, 2)` returns `2.5` rather than `2`. This is the same behavior as JavaScript, but different than many other programming languages.\n\nSass has powerful support for manipulating units based on how real-world unit calculations work. When two numbers are multiplied, their units are multiplied as well. When one number is divided by another, the result takes its numerator units from the first number and its denominator units from the second. A number can have any number of units in the numerator and/or denominator.\n\n### SCSS Syntax\n\n``````@debug 4px * 6px; // 24px*px (read \"square pixels\")\n@debug math.div(5px, 2s); // 2.5px/s (read \"pixels per second\")\n\n// 3.125px*deg/s*em (read \"pixel-degrees per second-em\")\n@debug 5px * math.div(math.div(30deg, 2s), 24em);\n\n\\$degrees-per-second: math.div(20deg, 1s);\n@debug \\$degrees-per-second; // 20deg/s\n@debug math.div(1, \\$degrees-per-second); // 0.05s/deg\n``````\n\n### Sass Syntax\n\n``````@debug 4px * 6px // 24px*px (read \"square pixels\")\n@debug math.div(5px, 2s) // 2.5px/s (read \"pixels per second\")\n\n// 3.125px*deg/s*em (read \"pixel-degrees per second-em\")\n@debug 5px * math.div(math.div(30deg, 2s), 24em)\n\n\\$degrees-per-second: math.div(20deg, 1s)\n@debug \\$degrees-per-second // 20deg/s\n@debug math.div(1, \\$degrees-per-second) // 0.05s/deg\n``````\n\nBecause CSS doesn’t support complex units like square pixels, using a number with complex units as a property value will produce an error. This is a feature in disguise, though; if you aren’t ending up with the right unit, it usually means that something’s wrong with your calculations! And remember, you can always use the `@debug` rule to check out the units of any variable or expression.\n\nSass will automatically convert between compatible units, although which unit it will choose for the result depends on which implementation of Sass you’re using.If you try to combine incompatible units, like `1in + 1em`, Sass will throw an error.\n\n### SCSS Syntax\n\n``````// CSS defines one inch as 96 pixels.\n@debug 1in + 6px; // 102px or 1.0625in\n\n@debug 1in + 1s;\n// ^^^^^^^^\n// Error: Incompatible units s and in.\n``````\n\n### Sass Syntax\n\n``````// CSS defines one inch as 96 pixels.\n@debug 1in + 6px // 102px or 1.0625in\n\n@debug 1in + 1s\n// ^^^^^^^^\n// Error: Incompatible units s and in.\n``````\n\nAs in real-world unit calculations, if the numerator contains units that are compatible with units in the denominator (like `math.div(96px, 1in)`), they’ll cancel out. This makes it easy to define a ratio that you can use for converting between units. In the example below, we set the desired speed to one second per 50 pixels, and then multiply that by the number of pixels the transition covers to get the time it should take.\n\n### SCSS Syntax\n\n``````\\$transition-speed: math.div(1s, 50px);\n\n@mixin move(\\$left-start, \\$left-stop) {\nposition: absolute;\nleft: \\$left-start;\ntransition: left (\\$left-stop - \\$left-start) * \\$transition-speed;\n\n&:hover {\nleft: \\$left-stop;\n}\n}\n\n.slider {\n@include move(10px, 120px);\n}\n``````\n\n### Sass Syntax\n\n``````\\$transition-speed: math.div(1s, 50px)\n\n@mixin move(\\$left-start, \\$left-stop)\nposition: absolute\nleft: \\$left-start\ntransition: left (\\$left-stop - \\$left-start) * \\$transition-speed\n\n&:hover\nleft: \\$left-stop\n\n.slider\n@include move(10px, 120px)\n\n``````\n\n### CSS Output\n\n``````.slider {\nposition: absolute;\nleft: 10px;\ntransition: left 2.2s;\n}\n.slider:hover {\nleft: 120px;\n}\n\n``````\n\nIf your arithmetic gives you the wrong unit, you probably need to check your math. You may be leaving off units for a quantity that should have them! Staying unit-clean allows Sass to give you helpful errors when something isn’t right.\n\nYou should especially avoid using interpolation like `#{\\$number}px`. This doesn’t actually create a number! It creates an unquoted string that looks like a number, but won’t work with any number operations or functions. Try to make your math unit-clean so that `\\$number` already has the unit `px`, or write `\\$number * 1px`.\n\nPercentages in Sass work just like every other unit. They are not interchangeable with decimals, because in CSS decimals and percentages mean different things. For example, `50%` is a number with `%` as its unit, and Sass considers it different than the number `0.5`.\n\nYou can convert between decimals and percentages using unit arithmetic. `math.div(\\$percentage, 100%)` will return the corresponding decimal, and `\\$decimal * 100%` will return the corresponding percentage. You can also use the `math.percentage()` function as a more explicit way of writing `\\$decimal * 100%`.\n\nCompatibility (10 Digit Default):\nDart Sass\nLibSass\nRuby Sass\nsince 3.5.0\n\nLibSass and older versions of Ruby Sass default to 5 digits of numeric precision, but can be configured to use a different number. It’s recommended that users configure them for 10 digits for greater accuracy and forwards-compatibility.\n\nSass numbers are represented internally as 64-bit floating point values. They support up to 10 digits of precision after the decimal point when serialized to CSS and for the purposes of equality. This means a few different things:\n\n• Only the first ten digits of a number after the decimal point will be included in the generated CSS.\n\n• Operations like `==` and `>=` will consider two numbers equivalent if they’re the same up to the tenth digit after the decimal point.\n\n• If a number is less than `0.0000000001` away from an integer, it’s considered to be an integer for the purposes of functions like `list.nth()` that require integer arguments.\n\n### SCSS Syntax\n\n``````@debug 0.012345678912345; // 0.0123456789\n@debug 0.01234567891 == 0.01234567899; // true\n@debug 1.00000000009; // 1\n@debug 0.99999999991; // 1\n``````\n\n### Sass Syntax\n\n``````@debug 0.012345678912345 // 0.0123456789\n@debug 0.01234567891 == 0.01234567899 // true\n@debug 1.00000000009 // 1\n@debug 0.99999999991 // 1\n``````\n\n### 💡 Fun fact:\n\nNumbers are rounded to 10 digits of precision lazily when they’re used in a place where precision is relevant. This means that math functions will work with the full number value internally to avoid accumulating extra rounding errors."
] |
[
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https://discuss.codecademy.com/t/python-unable-to-add-column/544179
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[
"",
null,
"# Python - Unable to add column\n\nI’m using dataset from U.S Medical Insurance and working on Jupyter Notebook. I couldn’t add in a column that averages out 2 columns (please view the last row of code here.)\n\nCan someone point me out where did i do wrongly?",
null,
"Thank you\n\n@raysonng, I see what you were trying to do here, but you ran into the exact reason why using indexing in Pandas (square bracket access) is generally preferred over attribute (dot) access.\n\nYou are getting this error because you are trying to divide by `df_totalcharges.count`.\nWhat you think you are doing is dividing by the `df_totalcharges` column named `count`.\nHowever, Pandas interprets this as you trying to divide by the un-invoked pandas.DataFrame.count method. Hence the error saying you can’t divide a float by a method.\n\nSo, to get what you want, you should use this line:\n\n``````df_totalcharges['avg_charges'] = df_totalcharges['charges'] / df_totalcharges['count']\n``````\n\nMoving forward, remember that dot access in Pandas is a shortcut — but shortcuts aren’t always the best choice for every situation.\n\n4 Likes\n\nWorks like magic. Thank you!\n\nI’m still new to Python and been struggling with when to use brackets, square brackets and dot access.\n\nIf I may ask a follow-up question, i tried to plot a bar chart:\n\nI figured Pandas is not recognizing my “region” column, and i tried to rename the column but its not working. I could use the “region” from another dataframe, but would like to understand what mistake I made that result in this error.\n\nPandas doesn’t recognize a `region` column in `df_totalcharges`, because no such column exists.\n\nWhen you created `df_totalcharges` using this code…\n\n``````totalcharges = data.groupby('region').charges.sum()\ndf_totalcharges = pd.DataFrame(totalcharges)\n``````\n\n… you made the `region` column your index. So now you don’t have a `region` column, but your index is the values of your previous `region` column.\n\nSo, there are two ways you can go about making your bar chart: you can use the DataFrame’s index as your x values, or you can go back and use `groupby()` in a way that doesn’t make your index the region values. Documentation on how to do that is here:\n\n1 Like\n\nHi @el_cocodrilo, I managed to solve the task by adding as_index=False when I use groupby.\n\nThank you for your enlightenment! Appreciate it!\n\n1 Like"
] |
[
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"https://aws1.discourse-cdn.com/codecademy/original/5X/e/0/8/c/e08cf790a5972ac52a036a0fb64e4e139baec615.png",
null,
"https://sjc5.discourse-cdn.com/codecademy/images/emoji/twitter/woozy_face.png",
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