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## Non-autonomous lattice systems with switching effects and delayed recovery.(English)Zbl 1345.34128

The authors study the following non-autonomous lattice dynamical system with a discontinuous nonlinear term, and a special type of recoverable delay \begin{aligned} &\dot u_i=\nu(u_{i-1}-2u_i+u_{i+1})+f_i(t,u_i)\cdot {\mathcal \chi}[L_i-\max_{-\theta\leq s\leq 0}u_i(t+s)], \quad i\in \mathbb Z, \quad t>t_0, \\ &u_i(\tau)=\phi_i(\tau-t_0), \quad \forall \tau\in [t_0-\theta, t_0], \quad i\in \mathbb Z, \quad t_0\in \mathbb R, \end{aligned} where $$\nu =1/R>0$$ is the coupling coefficient, $$R$$ is the intercellular resistance, $${\mathcal \chi}$$ is the Heaviside function, and for each $$i\in \mathbb Z u_i\in \mathbb R$$ represents the membrane potential of the cell at the $$i$$-th active site, $$L_i\in \mathbb R$$ is the threshold triggering, the switch-off at the $$i$$-th site, $$u_i(t+\cdot)\in C([-\theta,0],\mathbb R)$$ is the segment of $$u_i$$ on time interval $$[t-\theta,t]$$ where $$\theta$$ is a positive constant, and $$f_i \in C(\mathbb R\times\mathbb R, \mathbb R)$$ satisfies proper dissipative conditions.
The problem is formulated as an evolution inclusion with delays and the existence of weak and strong solutions is established. The long term behavior is investigated and the existence of a pullback attractor is proved.

### MSC:

 34K31 Lattice functional-differential equations 34K05 General theory of functional-differential equations 34K09 Functional-differential inclusions 34K30 Functional-differential equations in abstract spaces 92C37 Cell biology
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### References:

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# Is this symmetry factor in Peskin wrong?

I am trying to compute the symmetry factor of a Feynman diagram in $\phi^4$ but i do not get the result Peskin Claims. This is the diagram I am considering

$$\left(\frac{1}{4!}\right)^3\phi(x)\phi(y)\int{}d^4z\,\phi\phi\phi\phi\int{}d^4w\,\phi\phi\phi\phi\int{}d^4v\,\phi\phi\phi\phi$$

my attempt is the following: there are 4 ways to join $\phi(x)$ with $\phi(z)$. There are then 3 ways to connect $\phi(y)$ with $\phi(z)$. Then, there are 8 ways to connect $\phi(z)$ with $\phi(w)$ and 4 ways to contract the remainning $\phi(w)$ with $\phi(v)$. Finally the there are 6 ways to contract the $\phi(w)$ and the three $\phi(v)$ in pairs

$$\left(\frac{1}{4!}\right)^3\dot{}4\dot{}3\dot{}8\dot{}4\dot{}6=\frac{1}{6}$$

but the result claimed in Peskin (page 93) is $1/12$. What am I doing wrong?

• It would help if you indicated which points are z, w, and v... I guess w and v are the equivalent points?...
– hft
May 5 '15 at 18:27
• @hft yeah w and v are the equivalent points May 5 '15 at 20:05
• Hmm... actually, now that I'm trying to work it out explicitly, I am ending up with 1/6 as well... are you sure the answer in Peskin is correct?
– hft
May 5 '15 at 22:54
• @hft no I am not May 5 '15 at 23:10
• I got it. It got it. The points z,w,v all need to be considered as the symmetrical point and then there is an extra 1/3! from expanding the exponential. I'll write up an answer.
– hft
May 5 '15 at 23:18

The expansion of $e^x$ is: $$e^x=1+x+x^2/2+x^3/3!+\ldots$$
From expanding the expression: $$\left<\phi_x\phi_y\exp{\left(-\frac{\lambda}{4!}\int dz \phi_z^4\right)}\right>\;,$$ the third order term is: $$\left< \phi_x\phi_y\frac{1}{3!}{\left(\frac{-\lambda}{4!}\right)}^3\int dz \int dw \int dv \phi_z\phi_z\phi_z\phi_z \phi_w\phi_w\phi_w\phi_w \phi_v\phi_v\phi_v\phi_v \right>$$
So the overall symmetry factor is: $$4*3*4*4*2*3*2*3\frac{1}{3!}\frac{1}{4!^3}=\frac{1}{12}$$
Contrary to your previous question Problem understanding the symmetry factor in a feynman diagram the roles of the three vertices are not different so you have from the expansion of the exponential a $1/3!$ but it does not get compensated by the choice of role assignement. Here this choice amounts to deciding who connects directly to $x$ and $y$ and that's it. So you have $\frac{3}{3!}=\frac{1}{2}$. Also the symmetry group has order 12 because you can permute the 3 lines of the embedded sunshine diagram and you can also rotate it with repect to a vertical axis.

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### Wordof the DayTuesday,February 18

Definition of cinquefoil The name of several different species of the genus Potentilla;-- also called five-finger,because of the resemblance of its le

1:2:

### 2 words is found which contain cinquefoil word in database

Words with defination found in database when searching for cinquefoil.

Cinquefoil

n.

The name of several different species of the genus Potentilla;-- also called five-finger,because of the resemblance of its leaves to the fingers of the hand.

Cinquefoil

n.

An ornamental foliation having five points or cups,used in windows,panels,etc.

The wordcinquefoiluses 10 total alphabets with white space

The wordcinquefoiluses 10 total alphabets with white out space

The wordcinquefoiluses 9 unique alphabets:CEFILNOQU

Number of all permutationsnpr forcinquefoil362880

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### 3 same alphabet containing word For cinquefoil

CINCIQCIUCIECIFCIOCIICILCNICQICUICEICFICOICIICLICNQCNUCNECNFCNOCNICNLCQNCUNCENCFNCONCINCLNCQUCQECQFCQOCQICQLCUQCEQCFQCOQCIQCLQCUECUFCUOCUICULCEUCFUCOUCIUCLUCEFCEOCEICELCFECOECIECLECFOCFICFLCOFCIFCLFCOICOLCIOCLOCILCLIICNICQICUICEICFICOICIICLNCIQCIUCIECIFCIOCIICILCINCQNCUNCENCFNCONCINCLQCNUCNECNFCNOCN

### All permutations word for cinquefoil

CEFIILNOQUCEFIILNOUQCEFIILNQOUCEFIILNQUOCEFIILNUOQCEFIILNUQOCEFIILONQUCEFIILONUQCEFIILOQNUCEFIILOQUNCEFIILOUNQCEFIILOUQNCEFIILQNOUCEFIILQNUOCEFIILQONUCEFIILQOUNCEFIILQUNOCEFIILQUONCEFIILUNOQCEFIILUNQOCEFIILUONQCEFIILUOQNCEFIILUQNOCEFIILUQONCEFIINLOQUCEFIINLOUQCEFIINLQOUCEFIINLQUOCEFIINLUOQCEFIINLUQOCEFIINOLQUCEFIINOLUQCEFIINOQLUCEFIINOQULCEFIINOULQCEFIINOUQLCEFIINQLOUCEFIINQLUOCEFIINQOLUCEFIINQOULCEFIINQULOCEFIINQUOLCEFIINULOQCEFIINULQOCEFIINUOLQCEFIINUOQLCEFIINUQLOCEFIINUQOLCEFIIOLNQUCEFIIOLNUQCEFIIOLQNUCEFIIOLQUNCEFIIOLUNQCEFIIOLUQNCEFIIONLQUCEFIIONLUQCEFIIONQLUCEFIIONQULCEFIIONULQCEFIIONUQLCEFIIOQLNUCEFIIOQLUNCEFIIOQNLUCEFIIOQNULCEFIIOQULNCEFIIOQUNLCEFIIOULNQCEFIIOULQNCEFIIOUNLQCEFIIOUNQLCEFIIOUQLNCEFIIOUQNLCEFIIQLNOUCEFIIQLNUOCEFIIQLONUCEFIIQLOUNCEFIIQLUNOCEFIIQLUONCEFIIQNLOUCEFIIQNLUOCEFIIQNOLUCEFIIQNOULCEFIIQNULOCEFIIQNUOLCEFIIQOLNUCEFIIQOLUNCEFIIQONLUCEFIIQONULCEFIIQOULNCEFIIQOUNLCEFIIQULNOCEFIIQULONCEFIIQUNLOCEFIIQUNOLCEFIIQUOLNCEFIIQUONLCEFIIULNOQCEFIIULNQOCEFIIULONQCEFIIULOQN

### All combinations word for cinquefoil

CINQUEFOILCICNCQCUCECFCOCICLINIQIUIEIFIOIIILNQNUNENFNONINLQUQEQFQOQIQLUEUFUOUIULEFEOEIELFOFIFLOIOLILCINCIQCIUCIECIFCIOCIICILCNQCNUCNECNFCNOCNICNLCQUCQECQFCQOCQICQLCUECUFCUOCUICULCEFCEOCEICELCFOCFICFLCOICOLCILINQINUINEINFINOINIINLIQUIQEIQFIQOIQIIQLIUEIUFIUOIUIIULIEFIEOIEIIELIFOIFIIFLIOIIOLIILNQUNQENQFNQONQINQLNUENUFNUONUINULNEFNEONEINELNFONFINFLNOINOLNILQUEQUFQUOQUIQULQEFQEOQEIQELQFOQFIQFLQOIQOLQIL

### All similar letter combinations related to cinquefoil

CINQUEFOIL<KEY>

## Wiktionary Result

Contents

• 1 English
• 1.1 Alternative forms
• 1.2 Etymology
• 1.3 Pronunciation
• 1.4 Noun
• 1.4.1 Synonyms
• 1.4.2 Coordinate terms
• 1.4.3 Related terms
• 1.4.4 Translations
• 1.4.5 Further reading English[ edit] Fan-leaf cinquefoil (Potentilla flabellifolia ) Alternative forms[ edit]
• cinqfoil Etymology[ edit] From Old Frenchcinqfoil,fromcinq ( “ five ” ) +foil ( “ leaf ” ) Pronunciation[ edit]
• ( UK,US ) IPA (key): /ˈsɪŋk.fɔɪl/ Noun[ edit] cinquefoil (pluralcinquefoils )
1. A potentilla ( flower ) .
2. ( heraldry ) A stylized flower or leaf with five lobes.
3. ( topology ) A particular knot of five crossings. Synonyms[ edit]
• ( flower ):fiveleaf
• ( knot ):cinquefoil knot,${\displaystyle 5_{1}},T(5,2){\displaystyle T(5,2)}Coordinate terms[ edit] trefoil quatrefoil sexfoil octofoil Related terms[ edit] cinque multifoil polyfoil Translations[ edit] potentilla —see potentilla heraldry:stylized flower or leaf with five lobesCatalan:quinquefoli French:quintefeuille (fr) German:Fünfpass Italian:cinquefoglie (it) topology:knot Further reading[ edit]cinquefoil knot on Wikipedia. WikipediaWikipedia Result Redirect to:PotentillaThis page is a redirect:To scientific name of a plant:This is a redirect from a vernacular ("common") name to the scientific name of a plant (or group of plants). When appropriate,protection levels are automatically sensed,described and categorized.Previous Wordsof the DayDaintyConfrontDiametrallyChatoymentConducingFerrerDeusedExiguousFileGoWords that start with...ABCDEFGHIJKLMNOPQRSTUVWXYZ (adsbygoogle= window.adsbygoogle||[]).push({});Word by length...Privacy PolicyDisclamerContact Us ©definitionfinder.com 2020$


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MathsGee is Zero-Rated (You do not need data to access) on: Telkom |Dimension Data | Rain | MWEB

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If the difference between the roots of the equation $x^{2}+a x+1=0$ is less than $\sqrt{5}$, then the set possible values of a is

1) $(-3,3)$
2) $(-3, \infty)$
3) $(3, \infty)$
4) $(-\infty,-3)$
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(1)

Explanation

$x^{2}+a x+1=0$
$\alpha+\beta=-a \quad \alpha \beta=1$
$\alpha-\beta \mid=\sqrt{(\alpha+\beta)^{2}-4 \alpha \beta}$
$\Rightarrow|\alpha-\beta|=\sqrt{a^{2}-4}$
$\sqrt{a^{2}-4}<\sqrt{5}$
$\Rightarrow a^{2}-4<5$
$\Rightarrow a^{2}-9<0$
$\Rightarrow a \in(-3,3)$
by Gold Status (10,261 points)

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Research.Publish.Connect.

## Paper

Author: <NAME>

Affiliation: Comenius University, Slovak Republic

ISBN: 978-989-8565-77-8

Keyword(s): Product Logic, DPLL Procedure, Many-valued Logics, Automated Deduction.

Related Ontology Subjects/Areas/Topics: Approximate Reasoning and Fuzzy Inference ; Artificial Intelligence ; Computational Intelligence ; Fuzzy Systems ; Mathematical Foundations: Fuzzy Set Theory and Fuzzy Logic ; Soft Computing

Abstract: In the paper, we investigate the deduction problem of a formula from a finite theory in the propositional Product logic from a perspective of automated deduction.Our approach is based on the translation of a formula to an equivalent satisfiable finite order clausal theory, consisting of order clauses. An order clause is a finite set of order literals of the form $\varepsilon_1\diamond \varepsilon_2$ where $\varepsilon_i$ is either a conjunction of propositional atoms or the propositional constant $\gz$ (false) or $\gu$ (true), and $\diamond$ is a connective either $<$ or $=$. $<$ and $=$ are interpreted by the equality and standard strict linear order on $[0,1]$, respectively. A variant of the DPLL procedure, operating over order clausal theories, is proposed. The DPLL procedure is proved to be refutation sound and complete for finite order clausal theories.

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Paper citation in several formats:
<NAME>. (2013). A DPLL Procedure for the Propositional Product Logic.In Proceedings of the 5th International Joint Conference on Computational Intelligence - Volume 1: FCTA, (IJCCI 2013) ISBN 978-989-8565-77-8, pages 213-224. DOI: 10.5220/0004557402130224

@conference{fcta13,
author={<NAME>},
title={A DPLL Procedure for the Propositional Product Logic},
booktitle={Proceedings of the 5th International Joint Conference on Computational Intelligence - Volume 1: FCTA, (IJCCI 2013)},
year={2013},
pages={213-224},
publisher={SciTePress},
organization={INSTICC},
doi={10.5220/0004557402130224},
isbn={978-989-8565-77-8},
}

TY - CONF

JO - Proceedings of the 5th International Joint Conference on Computational Intelligence - Volume 1: FCTA, (IJCCI 2013)
TI - A DPLL Procedure for the Propositional Product Logic
SN - 978-989-8565-77-8
AU - <NAME>.
PY - 2013
SP - 213
EP - 224
DO - 10.5220/0004557402130224


text:
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 Re: xparse <NAME> <[log in to unmask]> Tue, 1 Sep 2009 17:10:30 +0200 text/plain (33 lines) <NAME> skrev: > <NAME> wrote: >> I was thinking more about single spaces, as in >> >> \moveto 0 0 \curveto 47 0 100 53 100 100 >> >> (the idea being to express a bunch of graphic data compactly while still >> allowing the code to survive reflowing in a text editor), but this is of >> course on the boundary of what can be considered LaTeX2e-ish syntax. > > Personally, I'm not a fan of that input syntax: I prefer something like > the pgf approach. The idea was indeed that these should boil down to \pgfpathqmoveto and \pgfpathqcurveto respectively; I just wanted something more compact at FMi-level -2 (or thereabout). (So everything would be in a special environment, and instead of \moveto the command name might really be \M.) > However, I did a quick test and as I hoped you can do > this with > > \DeclareDocumentCommand \moveto { u{~} u{~} } { ... } > > Not sure how robust this is, but if you really want to do it you can at > least have a go. I'll hopefully do it using \def fairly soon (need to generate the curve data first), but my concern was rather for someone in the far future who might not have \def readily available and thus wanting to go via \DeclareDocumentCommand instead. <NAME>


text:
### pikmike's blog

By pikmike, history, 2 years ago, translation,

1009A - Game Shopping

Tutorial
Solution (Vovuh)

1009B - Minimum Ternary String

Tutorial
Solution (Vovuh)

1009C - Annoying Present

Tutorial
Solution (PikMike)

1009D - Relatively Prime Graph

Tutorial
Solution (PikMike)

1009E - Intercity Travelling

Tutorial
Solution (BledDest)

1009F - Dominant Indices

Tutorial
Solution (BledDest)

1009G - Allowed Letters

Tutorial
Solution (Bleddest)

• +69

 » 2 years ago, # |   +8 How did you got that diff_i formulae in E?
•  » » 2 years ago, # ^ |   +27 Suppose there is a rest site right before the kilometer we are trying to cross. The probability of this is 0.5 (and then we get a1 as the difficulty).Ok, suppose then there is no rest site at position i - 1, but it is at position i - 2. The probability of this is 0.25, and then we get a2 as the difficulty.Then we consider the option when there is a rest site at position i - 3, but no rest sites at position i - 2 and i - 1. The probability of this is 0.125, the difficulty is a3.And so on, until we get to position 0.
•  » » » 2 years ago, # ^ |   +2 Roger that ^_^
 » 2 years ago, # |   0 Is there a reason for bounds on F to be 10^6 instead of something a bit smaller like 2*10^5? The problem ended up being pretty hard to solve in Java even with the intended solution. I understand not wanting more naive solutions to pass though.
•  » » 2 years ago, # ^ |   +8 Probably to disallow the solution with euler tour and sqrt-decomposition.
•  » » 2 years ago, # ^ |   0 why would a linear solution fail? maybe you lost time in reading input just like me.
•  » » » 2 years ago, # ^ | ← Rev. 2 →   -8 Java is about 5x slower than C++, so many times, perfectly fine solutions in C++ fail in java. I've had java solutions fail many times because I was using long instead of int, or an arraylist/customdatastructure instead of an array, or recursion instead of iteration. I've also gotten accepted verdicts by rewriting slow java code in c++.Though you can lose time reading input with scanner, there are many other ways to lose time, ways that C++ or even python uses don't need to care about. I don't think any accepted solutions in java for this problem used recursion, but many did in C++.Case and point: When I made this comment, there were 25 accepted c++ solutions and 6 TLE c++ solutions. Java had 36 TLE solutions (from 6 distinct people) and only 10 accepted solutions (from 6 distinct people).
 » 2 years ago, # | ← Rev. 2 →   +3 As mentioned by redux on Announcement page, problem D can be solved by using recurrence relation such that if (a,b) is a co-prime pair then (2a - b, a), (2a + b, a) and (a + 2b, b) are also co-prime where a > b.Base cases are (2, 1) for even-odd pairs and (3, 1) for odd-odd pairs. The list is also exhaustive as mentioned in this link so there is no overlapping also. The only breaking condition will be a > n or when m co-prime pairs are stored.In this way, we will always jump to only co-prime pairs rather than checking every other pair.So time complexity will also be much less than one given in editorial (Correct me if I am wrong).
•  » » 2 years ago, # ^ |   +8 Lets say you check that m>=n-1 and phi(2)+...+phi(n)>=m. The probability of gcd being 1 is O(1) so we can randomly generate edges and check if gcd is 1. If we use map complexity will be: O(n+mlogm). But one can also prove that without randomizatiom complexity is O(n+m).
•  » » » 2 years ago, # ^ |   0 We don't need to check phi sum as one can just store the pairs and if the no. of pairs are less than m, then it is impossible.
 » 2 years ago, # |   0 Problem G is almost same as the problem Poetic Word from ACM-ICPC Amritapuri Regionals 2017.
•  » » 2 years ago, # ^ |   +20 Is it, really?Yes, the main idea of solution is similar, but checking if we can build the suffix when n = 200 and when n = 105 is, I suppose, "a bit" different.
•  » » » 2 years ago, # ^ |   0 Yeah, the main idea is the same. The difference is in the way to optimize it and using the fact that the alphabet size is small to create less number of nodes by using masks.
 » 2 years ago, # | ← Rev. 2 →   -8 My solution 40344721 for problem E. I think it's easier #include #include using namespace std; typedef long long int ll; const ll mod = 998244353; int main() { ll pow2 = 1, p=0, curp; int n; scanf("%d", &n); vector cost(n); for (int i = n-1; i >= 0; i--) scanf("%lld", &cost[i]); p += cost[0]; for (int i = 1; i < n; i++) { curp = (pow2*(i + 2)) % mod; curp = (curp*cost[i])%mod; p = (p + curp) % mod; pow2 = (pow2 * 2) % mod; } printf("%lld", p); return 0; }
•  » » 2 years ago, # ^ |   -8 I also use this method.But I don't know how to prove that
•  » » » 2 years ago, # ^ | ← Rev. 4 →   -8 Sequence from https://oeis.org/A001792.My proof: (n-segment means segment of length n).a[i] = 1, 3, 8, 20, 48 ... a[0] = 1, because a n-segment can be paved with a n-segment in one way. a[i] is equal the number of ways to pave the free parts of the n-segment for all possible positions of the segment of length >= n-i. (statement 1)System: a[0] = 1 a[i] = a[i-1]+2*2^(i-1) + Σ 2^(k-1)*2^(i-k-1) (from k=1 to i-1) = a[i-1]+(i+3)*2^(i-2), for every i>=1.Clarification: a[i-1] — because we count for all segments of length >=n-i.2*2^(i-1) is equal the number of ways to pave the free parts in case 1 in pic.Σ 2^(k-1)*2^(i-k-1) (from k=1 to i-1) is equal the number of ways to pave the free parts in case of movement (n-i)segment, where 2^(k-1) — left part, 2^(i-k-1) — right part. (case 2 in pic)Proof of "The number of ways to pave the n-segment by k segments is ((n-1)!/(k-1)!)/(n-k)!" you can find here http://kvant.mccme.ru/pdf/1999/05/kv0599levin.pdf after 11-task (it is on russian lang, but i cant find this in english). The number of ways to pave the n-segment by different segment is sum and equals 2^(n-1).Sorry for my bad english, i used google translate.
•  » » » 2 years ago, # ^ |   -8 Sorry, i forgot to attach the pic
•  » » » » 2 years ago, # ^ | ← Rev. 3 →   -8 I got a more intuitive prove!define F(n,k) as number of distance notless than k. So the final result is Σ(k=1~n) F(n,k)*ak.F(n,n) = 1F(n,n-1) = 3F(n,n-2) = 8F(n,n-3) = 20...It's easy to see that F(n,k) = F(n-x,k-x) when k>x.F(1,1) = F(n,n) = 1F(2,1) = F(n,n-1) = 3F(3,1) = F(n,n-2) = 8F(4,1) = F(n,n-3) = 20...So, rewrite the result as Σ(k=1~n) F(k,1) * akF(1,1) =1for F(k,1) , we calculate F(k+1,1): insert a new gap 1' between 0 and 1, it can be a stop , or not stop, and this covers all situations,if the 1' is a stop, it contribute F(k,1) + 2^k , 2^k from 0 to 1', the others remain F(k,1)if the 1' is not a stop, it contribute F(k,1) also, because, F(k,1) means count all gap nums, insert a nonstop don't change the gap nums.Together we get F(k+1,1) = F(k,1)*2 + 2^k.✿✿ヽ(°▽°)ノ✿
 » 2 years ago, # | ← Rev. 3 →   0 What does " "pull" our depth arrays upwards" means? Thanks in advance..
•  » » 2 years ago, # ^ |   0 Assume that u is a child of v and the depth array of u is d[u] = [1, 2, 2]You want to create the depth array for v (d[v] = {1, 1, 2, 2]), but without creating it from scratch (which would take O(n) time).Using maps instead of arrays allows to do this in time, and using vectors with the reversed depth arrays allows doing it in even O(1).
•  » » » 2 years ago, # ^ |   0 In the add method of struct state, if (*a)[i] == (*a)[cur_max] then don't we need the min(i, cur_max) as dominant index according to given conditions? Thanks in advance..
•  » » » » 2 years ago, # ^ |   0 Yes, but he did compared also the index with pairs, so this case is already included in the editorial solution.
•  » » » » » 2 years ago, # ^ | ← Rev. 2 →   0 In the add method we are comparing pairs (val_at_i, i) and (val_at_cur_max, cur_max), if both the values are same then we are updating cur_max if i > cur_max. If the values are same we want to change the value of cur_max only if i < cur_max according to given conditions. Did I miss something?
•  » » » » » » 2 years ago, # ^ |   +3 His depth vector is reversed.
•  » » » » » » » 2 years ago, # ^ |   0 Ohh.. thanks for replying, got my mistake
•  » » » » » » » 2 years ago, # ^ | ← Rev. 3 →   -8 code In this solution of problem F,he is just forming depth array of all the nodes in the form of array of maps ,and using the factor of logn ,pulling up the maps and merging smaller into bigger at all stages ,but why is he not getting MLE ....as if tree is linear,then space would be O(n^2) UPD:GOT IT.....EVERY TIME WE MERGE PARENT AND ITS CHILD ,WE ARE SWAPPING THEIR MAPS(IF SMALLER) AND THEN MERGE,SO SIZE OF PARENT INCREASES AND ITS CHILD DECREASES....
 » 2 years ago, # |   +5 Problem G , author's solution . We can just rebuild the graph with brute force and run dinic on it . This could pass all tests if we do not use vector .
•  » » 2 years ago, # ^ |   +8 I tried your idea, and it worked :) I used vectors though, if you were interested in it, you may have a peek: Submission 40553125.I suppose that if the alphabet size was 7 then this approach would've been to slow (graph would have about 140 edges) while Authors' solution will fit in the limits.Thank you for your comment and best regards :)
•  » » » 2 years ago, # ^ |   0 Wow,you have very nice optimize technique !This is my solutionWhich I got TLE*7 before :)
 » 2 years ago, # |   0 I have a question about the memory complexity of F: since every node's memory is its max "height", when the tree becomes a chain,it will rise to n(n+1)/2, And I can't run the data on my computer,so I wonder if it will get MLE
•  » » 2 years ago, # ^ |   0 If the tree is a chain, I believe you will actually only have 1 (mutable) array. All the nodes will point to that same array. Each time you go up the tree you just copy that array and add 1 to it. This works since you never need to use arrays lower down in the tree again
 » 2 years ago, # |   -7 I am a little confused by the D solution given in the tutorial.In my opinion the O(N^2*log(N)) is not going to fit in time because, as i know the time formula = TimeComplexity(MAXN)/10^9 which is O(N^2*log(N))/10^9 and converting N to MAXN(10^5) is (10^5)^2*log(10^5)/10^9 seconds, 10^10*16/10^9 = 160 seconds. Correct me if i am wrong(i am 100% wrong but i would like an explanation why). As i know from my little experience a O(N^2) solution for a TimeLimit of 2 second a MAXN of 10^5 is way to big.
•  » » 2 years ago, # ^ |   +1 The program in this case terminates much faster because only at n = 600, we already get M >= 100000.
•  » » » 2 years ago, # ^ |   0 thanks, i didn't get that right from editorial
•  » » » » 2 years ago, # ^ |   0 Can we say that the complexity is O(m) since it the loop runs atmost m number of times?
 » 2 years ago, # |   0 For problem F, first example, why is the output 3 2 1 0 wrong? There may be several dominant indices for given node, right?
•  » » 2 years ago, # ^ |   +1 No,dominant index needs to have largest frequency and as closest as possible to the parent node
•  » » » 2 years ago, # ^ |   0 Yeah, thanks, I misread it.
•  » » 2 years ago, # ^ |   +1 The problem is asking for the minimum index of the maximum element in the array. Therefore, there is only 1 dominant index.The array of vertex 1 is: [1, 1, 1, 1, 0, ...] => answer is 0. The array of vertex 2 is: [1, 1, 1, 0, ...] => answer is 0. The array of vertex 3 is: [1, 1, 0, ...] => answer is 0. The array of vertex 4 is: [1, 0, ...] => answer is 0.
 » 2 years ago, # |   0 For problem 1009 D, how is it sufficient to prove that graph has m valid edges in order to make sure that graph is connected? (Like what I can't get is when is the program making sure that graph that it prints is "connected"?)
•  » » 2 years ago, # ^ | ← Rev. 2 →   +6 It isn't sufficient, you need to do both checks.
•  » » » 2 years ago, # ^ |   0 The graph which program prints is connected, right? But how can we say that?
•  » » 2 years ago, # ^ |   +1 In the line, forn(i, n) for (int j = i + 1; j < n; ++j), we are making sure that the graph we built is connected because in the very first loop when i = 0, we are connecting the vertex 1 with all other vertices for j = 1 to n-1, thus making it a connected graph.
•  » » » 2 years ago, # ^ |   0 okay. Thank you.
•  » » 2 years ago, # ^ |   +1 1 is coprime of all numbers, hence in the first iteration itself(if m>=n-1) you are making the graph connected.
 » 2 years ago, # |   -8 In this hacked solution, I used long double for the answer, and added up the intermediate means. The editorial says 10 byte float will pass. So, how many bytes does long double uses? 8? If so, how can we use a 10 byte or larger float in C++? I know I could've done this by first calculating sum using long long int, but just want to know out of curosity, it may help in some future contests. Thanks :)
 » 2 years ago, # |   -8 My submission for E: http://codeforces.com/contest/1009/submission/40471843 Can someone tell me where the overflow is happenning and how to fix it?
•  » » 2 years ago, # ^ | ← Rev. 2 →   0 There's a subtraction in line 26, and if it could be negative, you have to consider it becomes negative. If your procedure is correct, you may fix it by adding mod in line 26.In C++11 or above, modulo (by using '%') for negative integer, it will be negative integer from (1-mod) to 0.(But sadly, I checked your solution with simple fix got TLE. And finally, I cannot remove TLE... My solution is calculating 2^n every time (and this method is slower than yours) but got AC, so I think the cause of TLE is twice multiplication in loop... Sorry for failing to get AC. Someone, please help!)My submission for checking (removed WA, but TLE) : https://codeforces.com/contest/1009/submission/40481944EDIT : I had to say the cause of WA of your original submission is not overflow.
•  » » » 2 years ago, # ^ |   0 ai is always positive. Since diff[i] is sum of ai divided by positive numbers(2^n), it is also always positive. So I don't understand what could be negative and why we should add mod. Also where are taking mod for a negative integer, the minus sign is outside the multiplication?
•  » » » » 2 years ago, # ^ |   0 On modulo, increasing numbers cannot be always increasing numbers. For example, [10, 20, 30] mod 11 is [10, 9, 8]. Similary, (ai * 2^(n-i)) > (ai * 2^(n- i-1)) is true on integer, but not always true on modulo.
•  » » » » » 2 years ago, # ^ |   0 Understood. Then shouldn't we multiply by -1? By adding mod we will get something else right?
•  » » » » » » 2 years ago, # ^ |   0 You can multiply negative numbers, and also subtract, divide(if mod is prime).When the number is negative, add mod and you can get right value.And more, considering the value is big or already positive, you can write like this.ans = (ans % mod + mod) % mod;I often write something like this!
•  » » » » » » » 2 years ago, # ^ |   +5 Understood what you're saying. Thanks for being patient!!!
•  » » 2 years ago, # ^ |   +3 (nested deeply, so I wrote here)I finally got AC by fixing your submission!https://codeforces.com/contest/1009/submission/40486361I added this.ios::sync_with_stdio(false), cin.tie(0);Just with this, the TLE is removed.Note that if there are many inputs, you must write it or use only scanf and printf. It is faster!good luck!
•  » » » 2 years ago, # ^ |   +5 Thanks!
 » 2 years ago, # |   -20 Problem B : I never understood this line.Even till nowString a is lexicographically less than string b (if strings a and b have the same length) if there exists some position i ( 1 ≤ i ≤ | a | , where | s | is the length of the string s ) such that for every j < i holds a [j] = b [j] , and a [i] < b [i] .What is string a ans string b, and indices i and j. What I understood was that string a is the output, and string b is the input string..!!! I cant get the relation between editorial explanation with the above statement given in the question. Help me pls regards ;)
•  » » 2 years ago, # ^ |   +3 String a is lexicographically less than string b if the person with name a is earlier in a phone book than the person with name b.
•  » » » 2 years ago, # ^ |   0 ok.... thanks for your answer... :)so it means in their world, there are only alphabets {0,1,2} 0 < 1 < 2 in the order. i got that. But the conditions : i < j and a[j] = b[j].... and a[i] < b[i].i/p : 100210o/p : 001120what is i and j in this case .... ? Regards :)
•  » » » » 2 years ago, # ^ |   0 i is 0
 » 2 years ago, # |   -8 40500124,(AC),40499958,(MLE)For the problem 1009F's solution, and in the Struct state, why i used "vector a" and got the MLE while i used "vector *a" and got the AC? Thanks in advance...
 » 2 years ago, # | ← Rev. 5 →   0 I would like to share another approach to solve D, though it runs slower and harder to code. (Sorry for the poor English beforehand.)Let rp[i] be a set storing all j <= n, such that gcd(i, j) = 1.When constructing rp[i], we can count total number of pair (i, j) such that j > i and j is in rp[i], and terminate when this number >= m.It is clear that the total number of found numbers won't exceed 2m + n when we stop.The basic case, rp[1], is all integers in [1, n].To construct rp[i], first we pick maximum prime factor p (can be found by sieve) of i, "remove" it from i. Formally, let i = p1k1 * p2k2 * ... * pzkz * pk, pi are primes, then we calculate the value temp = p1k1 * p2k2 * ... * pzkz, it can be found in O(logpn) by dividing i by p while i % p == 0.Since gcd(i, j) = 1 if and only if (i, j) share no common prime factor, we haverp[i] = rp[temp]  -  set of all numbers in rp[temp] which has prime factor p.We can simply iterate through rp[temp] from small to large, when we found a number x, check whether it can be divided by p, and add it to rp[i] if it cannot.By doing so, we may fail (found a number x but doesn't add it to rp[i]) many times, but for at most O(logpn) failures there must be one success (add x to rp[i]) before all those failures, because when we fail for some x = p1k1 * p2k2 * ... * pzkz * pk, we must have add y = p1k1 * p2k2 * ... * pzkz to rp[i] before. Since x <= n, k can only be numbers picked from [1, floor(logpn)], thus each success corresponds to at most O(logpn) failures.Because p must be greater than 2, and success always comes before failures, this algorithm runs in O((n + m) * log2n).Here is my submissionPlease correct me if I am wrong.
•  » » 2 years ago, # ^ | ← Rev. 4 →   0 UPD.I am stupid, simply let temp = i / p and the arguments still hold.UPD2.Also we can store j only for j > i and gcd(i, j) = 1 in rp[i]. The number of operations is smaller than or equal to the previous one.submission
 » 2 years ago, # |   0 Question 1009C: Annoying PresentI am getting a relative error = '0.0000206' in test case 9. I am using a double type variable to store the mean values for each new transformation. I have no clue how to reduce the error for the result. My code is here : https://codeforces.com/contest/1009/submission/40511515.Any help is appreciated.
•  » » 2 years ago, # ^ | ← Rev. 3 →   0 Every single floating point operation causes precision error. Try to replace them by integer operations as many as possible.For example, you can sum up all x by integer operations, and add it to mean at last. In this way we use just a single floating point operation rather than O(m) floating point operations, therefore the precision error can be reduced.The same trick can also be applied to d.
 » 2 years ago, # | ← Rev. 2 →   -19 My solution for Prob.E. Maybe it's better?? #include #include #include #define P 998244353 using namespace std; typedef long long LL; LL n, a[1000005], ans, p = 1; int main() { scanf("%I64d", &n); for(int i = 1; i <= n; i++) scanf("%I64d", a + i); for(int i = n - 1; i > 0; p = (p << 1) % P, i--) ans = (ans + (n + 2 - i) * p % P * a[i] % P) % P; printf("%I64d\n", (ans + a[n]) % P); return 0; }//Rhein_E One Way to prove this:
 » 2 years ago, # | ← Rev. 2 →   -8 For problem F (editorial solution), aren't there a lot of memory leaks? Ex: in the pull method, when we return (a new allocated vector in the heap) and save it in s[x], where s[x] already has allocated a new vector in the beginning of the dfs method. Why not use unique_pointers to prevent memory leaks? Other than that code looks rad. :)


text:
# Monitoring progress of nested function calls

I wrote some code where my output function is defined in terms of various functions, themselves defined in terms of various functions, schematically like $$f(x) = g(h(x))$$, $$h(x) = k(m(x))$$, etc., as you do.

For a given x, the evaluation of each of these functions might take a while, and I would like to keep an eye on the stage my computation is at, e.g. (temporarily) printing "evaluating h", when $$h(k)$$ starts being computed, similarly to how Monitor[] does this for e.g. the iterator in a table.

I am looking for tips on how to do this nicely.

Currently I am defining e.g.

h[x_] := h[x_] = { PrintTemporary["evaluating h"] , (actual expression for the function) }[[2]]

which does the job. The construction with the list seems rather roundabout, but I am not sure how else to make sure the print happens every time the function is called (with a distinct argument). Is there a better, optimal, or even built-in option for this?

• Perhaps Progress Indicator for NestList is what you are looking for Aug 14 '20 at 13:07
• Don't use List brackets, use parentheses to enclose the compound expression: h[x_] := h[x] = (PrintTemporary[StringForm["Evaluating h[]", x]]; calculation ); Aug 14 '20 at 13:26
• Thanks a lot @BobHanlon! That is the appropriate way to implement what I tried to do above. I had a feeling something was off, but could not remember for the life of me how to do it.. Aug 14 '20 at 13:44
• Thanks also @UlrichNeumann. I think the NestList progress indicator would have been very useful if my nesting/functions were indexed, but that is not the structure I am currently dealing with. Aug 14 '20 at 13:47
• I would of course still be interested in solutions that avoid defining my individual functions to signal themselves.. Aug 14 '20 at 13:48


text:
# Proportional samples

• Apr 15th 2009, 10:07 AM
Proportional samples
A company wants to survey 500 of its employees about job satisfaction. The company employs 860 people in British Columbia, 1100 people in Ontario, and 560 people in New Brunswick. How many employees should be sampled in each province so that the number in each provincial sample is proportional to the number of employees in that province?

Draft work:

500 required

BC = 860
ON = 1100
NB = 560
TOTAL: 2520

STEP 1
BC: 860/2520 x 100 = 34.1%
ON: 1100/2520 x 100 = 43.7%
NB: 560/2520 x 100 = 22.2%

STEP 2
BC: 34.1% / 100 = .341 x 860 = 293 (People)
ON: 43.7% / 100 = .437 x 1100 = 480 (People)
NB: 22.2% / 100 = .222 x 560 = 124 (People)

STEP 3
293 + 480 + 124 = way more than 500

HELP!
• Apr 15th 2009, 10:43 PM
Twig
hi

You have to take the percentages times 500.

For ex BC was 34.1%, -> $0.341 \cdot 500 = 170.5$
$0.437 \cdot 500 = 218.5$
$0.222 \cdot 500 = 111$
$170.5 + 218.5 + 111 = 500$
• Apr 16th 2009, 02:35 AM


text:
Get iterative solution of simultaneous differential equations?

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 Dears, I try to implement a program for evaluating the behavior of the stator and rotor currents of a three-phase induction motor , through the Eule´rs method.The rotor is locked and thus s=1. Nonetheless, I have not succeeded in plotting them, simultaneously as well the rms value of the stator currentt. Would someone help me, please ? iL10 = iL11 = iL20 = iL21 = 0.0; iL2ef0 = iL2ef1 = 0.0; Delta = 0.0001; L1 = 5.0/(120.0*Pi); L2 = 5.0/(120.0*Pi); Lm = 20.0/(120*Pi); R1 = 1.5; R2 = 5; s = 1.0; Vmax = 220.0*Sqrt[2]; w = 120.0 Pi; tmax = 0.05; Phi = 0.0; t1 = 0.0; to = 0.0; ListPlot[Table[to = t1; iL10 = iL11; iL20 = iL21; iL2ef0 = iL2ef1; iL11 = iL10 + (Delta *( L2 + Lm)/( L1*L2 + L2*Lm + L1*Lm))*(Vmax * Sin[w *to + Phi* Pi/180.0] - R1*iL10 - (Lm/(L2 + Lm))*R2 *iL20 /s); iL21 = iL20 + (Delta /( L2 + Lm))*( Lm*( L2 + Lm)/( L1 *L2 + L2 *Lm + L1*Lm)*(Vmax *Sin[w *to + Phi* Pi/180.0] - R1*iL10 - (Lm/(L2 + Lm))*R2 *iL20 /s) - R2* iL20 / s ); iL2ef1 = Sqrt[ ( (iL2ef0^2)/(t1/Delta + 1) + (iL21^2) + (iL20^2))/(t1/ Delta + 2) ]; t1 = to + Delta; {t1, iL21}, {t1, 0, tmax, Delta}; {t1, iL11}, {t1, 0, tmax, Delta} ]]
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Posted 1 year ago
 The way you are writing the last few arguments to Table Table[ ... t1 = to + Delta; {t1, iL21}, {t1, 0, tmax, Delta}; {t1, iL11}, {t1, 0, tmax, Delta}] do not match the documentation for Table. Table[expr, {i, imin, imax, di}] Perhaps try this instead z = Table[... t1 = to + Delta; {t1, iL21, iL11}, {t1, 0, tmax, Delta}]; which produces a row of three items t1,iL21,iL11 for each t1 and then ListPlot uses the desired columns. ListPlot[{z[[All, {1, 2}]], z[[All, {1, 3}]]}] Then Sqrt[Mean[z[[All, 2]]^2]] and Sqrt[Mean[z[[All, 3]]^2]] will calculate the square root of the mean of the squares of iL21 and iL11Please check this carefully, perhaps on smaller examples, until you are certain that you trust that this is precisely correct.
Posted 1 year ago
 Dear Bill. Thank you very much and again, for your holy help.I really got confused with the resulting graphics from my proposition. One of them was correct, while the other not. You are right with the need in respecting the structure. But I thought a message of error should appear to my proposition... I suppose your suggestion creates a table of numbers that is somehow stored anywhere and I wonder if this can be a problem if the number of iterations goes too far. Gonna try it. Your proposition for the RMS gives a general result. But what if I want to evaluate the accumulated RMS value at each iteration? This is what I need. thank you so much !!!!!
Posted 1 year ago
 In fact, I would expect to have the graphics like this. Please, note that the blue line is the Rms value of iL2, the green sinusoidal.
Posted 1 year ago
 . il2sq = z[[All, 2]]^2; rmsiL2 = Table[{z[[i, 1]], Sqrt[Mean[Take[il2sq, i]]]}, {i, 1, Length[z]}]; Show[{ListPlot[z[[All, {1, 3}]], PlotStyle -> Green], ListPlot[z[[All, {1, 2}]], PlotStyle -> Orange], ListPlot[rmsiL2, PlotStyle -> Blue]}]
Posted 1 year ago
 Bravissimo ! You are great ! Please, allow me mentioning that Mathematica language is not so intuitive as the programming languages used to be, some decades ago. Of course, this is to facilitate new and bigger achievements that are constantly demanded .
Posted 1 year ago
 Dear Bill, maybe you help me for my last lap ! During each iteration, I need to take the rms value of the current iL2 at that time, for evaluating the power and thus the torque given by the motor. I mean, I cannot wait until the end of all the calculations for taking the general rms value of iL2. I need to do it during each iteration.Please, how can I do that ? Thank you in advance, Scabral iL10 = iL11 = iL20 = iL21 = 0.0; iL2ef0 = 0.0001; Delta = 0.0001; J = 0.5; wr0 = 0.0; tau = 0.03; L1 = 5.0/(120.0*Pi); L2 = 5.0/(120.0*Pi); Lm = 20.0/(120*Pi); m = 0; R1 = 1.5; R2 = 5; s = 1.0; Vmax = 220.0*Sqrt[2]; w = 120.0 Pi; tmax = 0.01; Phi = 0.0; t1 = 0.0; to = 0.0; z = Table[to = t1; iL10 = iL11; iL20 = iL21; wr0 = wr1; iL11 = iL10 + (Delta *( L2 + Lm)/( L1*L2 + L2*Lm + L1*Lm))*(Vmax * Sin[w *to + Phi* Pi/180.0] - R1*iL10 - (Lm/(L2 + Lm))*R2 *iL20 /s); iL21 = iL20 + (Delta /( L2 + Lm))*( Lm*( L2 + Lm)/( L1 *L2 + L2 *Lm + L1*Lm)*(Vmax * Sin[w *to + Phi* Pi/180.0] - R1*iL10 - (Lm/(L2 + Lm))*R2 *iL20 /s) - R2* iL20 / s ); il2sq = z[[All, 2]]^2; rmsiL2 = Table[{z[[i, 1]], Sqrt[Mean[Take[il2sq, i]]]}, {i, 1, Length[z]}]; wr1 = wr0 + Delta*((3*R2*rmsiL2[[1, 1]]^2)/(120.0*Pi*s) - tau*wr0)/J; s = (120*Pi - wr1)/(120*Pi); t1 = to + Delta; {to, iL11, iL21, rmsiL2, wr1}, {to, 0, tmax, Delta}]; how[{ListPlot[z[[All, {1, 3}]], PlotStyle -> Green], ListPlot[z[[All, {1, 2}]], PlotStyle -> Orange], ListPlot[rmsiL2, PlotStyle -> Blue]}, PlotRange -> {{0, tmax}, {-50, 50}}] Show[{ListPlot[z[[All, {1, 3}]], PlotStyle -> Green], ListPlot[z[[All, {1, 4}]], PlotStyle -> Orange]}] Messages of errors : Mean::rectt: Rectangular array expected at position 1 in Mean[{0.,{}}]. Mean::rectt: Rectangular array expected at position 1 in Mean[{0.,{},{}}]. Mean::rectt: Rectangular array expected at position 1 in Mean[{0.,{},{},{}}]. General::stop: Further output of Mean::rectt will be suppressed during this calculation.
 . ... rms = 0; z = Table[ ... rms = Sqrt[(rms^2*(t1/Delta) + iL21^2)/(t1/Delta + 1)]; {t1, iL21, iL11, rms}, {t1, 0, tmax, Delta}]; Show[{ListPlot[z[[All, {1, 3}]], PlotStyle -> Green], ListPlot[z[[All, {1, 2}]], PlotStyle -> Orange], ListPlot[z[[All, {1, 4}]], PlotStyle -> Blue]}] This attempts to do an "online" calculation or RMS, where each subsequent value depends only on the previous value. Please test all this very carefully until you are certain that I have made no mistakes. And study this until you can understand how and why I did what I did. That may help you when you need to make further changes.I am very concerned about the line t1 = to + Delta; that you have inside your Table when you also have {t1, 0, tmax, Delta} controlling the construction of your Table. This is not the way this is usually done. I would remove that t1 = to + Delta; unless you understand something about this that I do not see.You commented about how the Mathematica language is different from other languages that you have seen. The Mathematica language is much more different than you imagine it to be. Even when it looks somewhat like other languages, even when you have learned more about it and think you might understand it, it will still be much more different than you imagine it to be.


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# Task 5 – Distance from Point of Similarity

In the figure below, $\triangle ABC \sim A'B'C'$ and point $P$ is the point of similarity of the two triangles.

1.) What is the ratio of the lengths of the corresponding sides?

2.) What is the ratio of the distance from point P to the corresponding vertices?

3.) What conjecture can you make from your answer in 1 and 2? Verify your conjectures by drawing a different pair of similar triangles.

Explanation

The ratio of the corresponding sides of two similar triangles is the same as the ratio of the distance of their corresponding vertices to the center of similarity. The ratio of the corresponding sides of two similar triangles is the same as the ratio of the distance of their corresponding vertices. In the figure above, ∆ABC ~ ∆A’B’C’ so

and


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# Estimation

• ### Probable Error

The term probable error is seen only in older literature of statistics where it used to donate the product of the product of the standard error and a constant factor 0.6745, thus Where P.E and S.E stand for the probable error and the standard error respectively, by using the above formula the probable error of […]

• ### Standard Error of Statistic

The term standard error has already been introduced very briefly in pervious tutorials while discussing about the sampling distribution of means. In this tutorial we will try to make some more comments on standard error. In the broader sense of the term standard errors of mean, median, standard deviation, coefficient of correlation, regression coefficients etc. […]

• ### Confidence Interval Estimate of Variance

The variance of a population can be estimated using the chi-square variate explained in pervious tutorials. Unlike the and distributions the value of chi-square variate are defined only for positive values. At a level of significance the or are those values of the variate which give an area in the right tail. Also or are […]

• ### Confidence Interval Estimate for Difference Between Means

(1) Large Samples: The difference between two means is of considerable importance in testing the homogeneity of populations. In this tutorial we are concerned about the confidence interval estimate for the difference between two population means. With a non-rigorous logic from central limit theorem we can state that “If we have two populations with means […]

• ### Confidence Interval Estimate of Mean with Small Sample

As long as is known the confidence interval estimate of population mean can be obtained by the method discussed earlier, provided the sample is large. Even if is not known we could replace it by its unbiased estimate , defined by The pervious method of estimation fails to provide good estimations if the sample size […]

• ### Confidence Interval Estimate of Mean

From central limit theorem we know that, is a standard normal variate. From the discussion of introduction to interval estimation we know that P (–1.96 < Z< 1.96) = 0.95 has the least possible range. With this inequality we can construct a 95% confidence interval estimate of the population mean, if we replace Z by […]

• ### Introduction to Interval Estimation

From the discussion on the point estimation we know that is the best possible estimator of the population mean , which is a fixed, usually unknown parameter. It is a matter of common-sense that we have to be extremely lucky to have a sample which has a mean exactly equal to the population mean , […]

• ### Consistent Estimator

An estimator is said to be a consistent estimator of the parameter , if it holds the following conditions: is an unbiased estimator of , i.e. if is biased, it should be unbiased for large values of (in the limit sense) i.e. . Variance of   approaches to zero as becomes very large, i.e., . […]

• ### Efficiency of an Estimator

Among a number of estimators of the same class, the estimator having the least variance is called an efficient estimator. Thus, if we have two estimators and with variances and  respectively, also if then will be an efficient estimator. The ratio of the variances of two estimators denoted by is known as the efficiency of  […]

• ### Unbiasedness of an Estimator

This is probably the most important property that a good estimator should possess. According to this property if the statistic is an estimator of will be an unbiased estimator if the expected value of   equals the true value of the parameter               i.e. Consider the following worked example Example: Show that the sample mean […]

• ### Maximum Likelihood Estimation

The maximum likelihood method is a very rigorous statistical method of estimation. The word likelihood has the same meaning as the word probability. The method of maximum likelihood is not restricted to some specific type of analysis as the least square method; rather its application is universal provided the probability distribution of the population is […]

• ### Point Estimation

In point estimation procedure we make an attempt to compute a numerical value, from sample observations, which could be taken as an approximation to the parameter. The estimators, which are also referred to as statistics (plural of statistic), since based on observations which are random variables are themselves random variables. A number of estimation methods […]


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# [WSS18] Predicting the Scale of Satellite Images Using Neural Network.

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# Introduction

Many interesting projects have been done with satellite images using Neural Networks such as semantic image segmentation for finding roads and homes, and more. In this project, we are taking a similar and broader look on satellite images by using a 50-layer convolution neural network (CNN) to predict the scale of satellite images.

# Getting started

When doing a project with neural networks, the first thing comes to one's mind is almost always the data, because data is one of the most important parts of the neural network. After failing to find the right data set for the project on the internet, I generated the data set (satellite images) in Wolfram Language using GeoImage. Firstly, I decided that I will only focus on one country (the United States) with satellite images that are scaled between 0.2 to 1200 Miles.

geoPositionOfCountry[entities_,numberOfPosition_Integer,folderName_String] :=
Module[
{countries, mesh},
countries = entities;
mesh = DiscretizeGraphics@EntityValue[countries,"Polygon"];
SeedRandom[Hash@{folderName,countries,RandomReal[{1,5}]}];
Reverse[RandomPoint[mesh,numberOfPosition],{2}]
]
geoPositionOfCountry[{Entity["Country","UnitedStates"]},500,"training"]

Random GeoPosition in the United States:

Trying this data set on a pre-trained convolution neural network (CNN) by doing transfer learning in Machine learning, did not give a good result, the mean absolute error was 5.8 which was more than 500%. Finding out this made me question the way my data set was set up. The pre-trained convolution neural network I had was trained on only smaller scaled city images for image segmentation. That could be the reason I am not getting any promising result. I then decided to decrease the scale of the satellite images to between 0.2 to 4 Miles instead of having 0.2 to 1200 Miles and focus on only three cities (Dallas, Chicago, and Houston) in the United States

geoPositionOfCity[{Entity[
"City", {"Dallas", "Texas", "UnitedStates"}],
Entity["City", {"Chicago", "Illinois", "UnitedStates"}],
Entity["City", {"Houston", "Texas",
"UnitedStates"}]}, 700, "training"]

Then, I uninitialized the pre-trained net model and retrain it with 11854 training, 1975 validation and 990 testing images on Amazon EC2 GPU which took 8 hours to complete.

The convolution neural network (CNN) I obtained from a paper (can be found in the summary section) was chopped and added needed layers for the project:

Before we go further, I also want to mention that the scale distribution of satellite images is exponential, meaning that there is more image close to earth than above. For instance, we have more images that are close 0.2 GeoRange than 4 Miles. The reason I chose this way of setting data so CNN can predict smaller scaled images better.

After the training with the above CNN, I got 29% mean absolute error. The error was better than what I had before. However, It was still bad to predict the scale of satellite images. To decrease the error percentage, I added more data, 9814 training, and 1050 validation to train it on top what I had before, and tested it with 420 images. The new error rate was 20% which indicated that with more data and more time, the CNN can learn and predict better.

Some of the prediction from 20% mean absolute error CNN:

encodeID[expr_]:=StringReplace[DeveloperEncodeBase64@BinarySerialize@expr,"/"->"~"];
decodeID[expr_]:=BinaryDeserialize@DeveloperDecodeBase64ToByteArray@StringReplace[expr,"~"->"/"];

getFileNames[folderName_] :=
FileNames["*.png",FileNameJoin[{NotebookDirectory[],folderName}],Infinity];

fromFileNameGetGeoRange[fileName_] :=
decodeID[FileBaseName@fileName]["GeoRange"];

associateFilesToGeoRange[fileNames_] :=
Map[
File[#] -> fromFileNameGetGeoRange[#] &,
getFileNames[fileNames]
]

testing2 = RandomSample@associateFilesToGeoRange["testing"];

sortPerformance[net_,data_] :=
With[
{img = Map[Import,data[[All,1]]], correspondingScale = data[[All,2]]},
SortBy[
Abs[First[#] - Last[#]]&
]
]

dataset2 = sortPerformance[tnet2, testing2];
Normal@dataset2[[10 ;; 12]]
Normal@dataset2[[-2 ;;]]

Now, I was wondering to see how the CNN would behave with bigger scale satellite images. I know that in the first attempt with a bigger scale (0.2 to 1200 Miles) It failed, but that time the images were from all of the United States. So, I generated a new dataset ranged from 1.3 to 10 Miles that were only in city images from Dallas, Chicago, and Houston. This time training took 4 hours with 11494 training and 1050 validation. After testing the CNN with 420 testing images, I got a really low mean absolute error rate of 0.08 which meant that the CNN could predict the scale of satellite images by up to 92% correctly. This was a success.

# Summary

In this project, we discovered that a convolution neural network (CNN) might fail to predict satellite images that have really big scales. However, when scales are decreased and satellite images are focused on cities CNN can predict better.

The convolution neural network (CNN) that was used in this project can be found here: http://community.wolfram.com/groups/-/m/t/1250199


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Public Group

# Stuck on Classes

This topic is 3791 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

## Recommended Posts

So I looked through my college text book, no good examples of classes. They show examples but none in a complete program, not to mention all their examples are pretty poorly explained. So I turned to google as usual when reading this book and I cant seem to find any good info on classes on there either and was wondering if anyone had some nice sites that they use to explain syntax. I was using CProgramming.com and some other sites that looked like they were from universities but everything I come across seems to have different syntax. So I guess I have a two part question: 1. Does anyone know of any sites they prefer that has tutorials on it? 2. Anyone willing to write me a super simple class? :D I believe I understand the concept, its just the syntax I am hung up on. I see some places use constructors and destructors, others dont, and the syntax seems to be different on each page that I go to. EDIT: I forgot to mention its C++. [Edited by - Chrono1081 on February 6, 2008 1:45:09 AM]

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its been a long time since I looked at a entry level C++ book (I'm assuming C++), so I can't think of any suggestions off the top of my head.

But gamedev.net has a large collection of book reviews / ratings.

Books

There are 42 book listed for C++. If your looking for C# or Java you could probably find plenty of those too.

I guess my point is get a book. If the book you looked at isn't clear you just need another book.

But books will get you only so far, looking at other peoples code will teach you more, but only after you get the basics. You'll get the basics from a good book.

Hopefully you can find a book in the link I provided.

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Thank you for your suggestion :) However unfortunatly where I am currently at a new book isnt an option : / I already have a few on the way but they take about a month to get here.

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I'm about to head to school, but I'll try to give you a small example.

A class is a set of instructions for what an object can do. Let's say we want to create a "ball" object. It can bounce and roll (for this examples purposes).

First we write the instructions prototype (often written in a headerfile), which means we're writing the short-version of the class which we will implement later.

For a ball-class it might look something like this:

class Ball{public:/* Public variables and functions are written here, you should always include both constructor and deconstructor here, even if they are empty, to show that you want them to be. */Ball(); // Constructor.~Ball(); // Deconstructor.void Bounce(); // A public bounce function.void Roll(); // A public roll function.};

That is what you might find in a headerfile. That is the short-hand version of a class, the definition. Next, we need to implement the class, which is often done in a .cpp file, which includes the headerfile.

#include "Ball.h"Ball::Ball(){// An empty constructor, but we still write it so other programmers won't misunderstand.}Ball::~Ball(){// An empty deconstructor, but we still write it so other programmers won't misunderstand.}void Ball::Bounce(){/* Here we put the code needed for the ball to bounce. It could be something as simple as "std::cout << "Boing!";". */}void Ball::Roll(){// Same as the ball-function.}

Now we have implemented the Ball-class. But all this code doesn't do anything by itself, we need a main-function in order to run this.

#include "Ball.h"int main(){Ball aTestBall;Ball.Bounce();Ball.Roll();return 0;}

In that main we created a ball object, called both its functions, then exited.

Now I have to run, hope this gives you a small idea of how a simple class can work.

My teacher once said that a class is a "blueprint" for how something should work, and as you said, the headerfile and .cpp file in this example served that purpose.

Have a nice day and good luck with your studies. :)

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You don't want a "tutorial." None of them are any good. I'm sure some have their good points, but they just can't stand up to the books on the subject.

As for books, "Thinking in C++" is one of the better ones and better yet (for you especially), it's available for free from mindview.net. The thing about Thinking in C++ is it's not a real nuts and bolts "this is how things should look in your source code" kind of book. It has long-winded (and very well written and thought out) discussions about object oriented programming and its use in C++. I think it's exactly what you're looking for.

Edit: I didn't mean to imply all C++ books are good either. There's a lot of junk out there..

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Best entry level C++ book I know is "Accelerated C++". Go down to a library or a book store or something and check it out. If you like the way it's written get a copy, it's worth the very reasonable price.

Accelerated C++ teaches C++ starting with a coverage of the higher level concepts and working down to the lower level concepts. Imho this is the way that a programming language should be taught.

A super simple class might look like this:

class SuperSimple {};

in order to add member data, you would declare it inside as in:

class SuperSimple {    int memberData;};

member functions (methods) work in a similar fasion:

class SuperSimple {    void memberFunction( void )    {        return;    }};

It's important that you understand public, protected and private access. You should read a book to understand those, because it takes a lot of time to figure them out but here's the syntax for a public member function and public member data for example:

class SuperSimple {public:    void publicFunction( void )    {        return;    }    int publicData;};

NOTE: It's generally a VERY bad idea to declare any member data public or protected because it's free for anyone to mess around with, potentially breaking any invariants in your code. There are exceptions to this rule of course, but for general use, make sure you declare your member data as private.

Constructors and destructors are kind-of, sort-of optional. You can choose not to define them but it's usually not a good idea to leave them out. Constructors are simply member functions which have the name of the class and no return type. Destructors are simply member functions which have a ~ followed by the name of the class and no return type or parameters. Constructors are automatically called when an instance of a class is created and destructors are automatically called when an instance of a class goes out of scope or is deleted. In general, you use constructors for setting the initial values for member data and allocating memory for member data stored on the free store (heap). Destructors are generally used for deleting member data stored on the free store. You can declare constructors and destructors like so:

class SuperSimple {public:    SuperSimple( void )    {        // Do setup stuff in here    }    ~SuperSimple()    {        // Do cleanup stuff in here    }};

Using a class:

int main( int argc, char **argv ){    // Create an instance of class SuperSimple    SuperSimple myInstance; // The constructor is called here    // Accessing public member data    myInstance.publicData = 0;    // Calling a public member function    myInstance.publicFunction();    return 0;}   // The class goes out of scope when the function it is created in ends// The destructor is called here

When accessing data using a pointer to an object, a different syntax is used. For example:

int main( int argc, char **argv ){    // Create an instance of class SuperSimple    SuperSimple myInstance; // The constructor is called here    SuperSimple *pointerToMyInstance = &myInstance;    // Accessing public member data    pointerToMyInstance->publicData = 0;    // Calling a public member function    pointerToMyInstance->publicFunction();    return 0;}

Of course, the above are simply one way of declaring functions in a class, it's possible to define a function outside of the class definition, or even in another file, if you use a different syntax (like the syntax described in one of the posts above). Once you get into more complex stuff, the syntax becomes more complex so take your time to learn all of these and use what you're comfortable with. Learning C++ is a gradual progression, you can use what you know and learn as you go. :)

[Edited by - Leo_E_49 on February 6, 2008 3:27:22 AM]

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1) First learn to use existing classes. The library stream classes (fstream, stringstream, etc.) and containers (string, vector, etc.) are a good place to start.

2) Lots of class features are "optional", or at least, a default is commonly provided that does what you want. You use the parts that are necessary for the class you're working on.

3) A class represents a data type. Think of 'Foo bar;' exactly like you would think of 'int bar;', except that there is also syntax for accessing members, which can be either data (just like parts of a struct) or functions (which are "bound" to that Foo instance). When you call a member function of an object, it's like telling that particular object of the class to do something.

4) The point of constructors is to initialize the instances. Think of 'Foo bar(x, y, z);' just like 'int bar = 7;'. In fact, in C++ you can write 'int bar(7);', though it's not common. The point of destructors is to clean up the object's mess when the object goes out of scope. You will practically never call it directly. You can think of primitive types as having a destructor that does nothing (because there is nothing about an int that needs to be "cleaned up"). There are very few ways to "make a mess", and there are pre-defined classes (again, lots of them are in the standard library) that handle most of the basic kinds of messes, so you will very rarely have to write a destructor yourself, if you use proper tools.

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# Limit of the ratio of two divergent sequences

Let $$∑a_n$$ be a series of real numbers which converges, but not absolutely.

Let $$p_n = \frac{a_n +|a_n|}{2} \ge 0, \ q_n = \frac{a_n - |a_n|}{2} \ \le 0$$.

Let $$P_n = \sum_{k=1}^{n}{p_k}, \ Q_n = \sum_{k=1}^{n}{q_k}$$.

Show that $${ \lim_{n\to\infty} \frac{Q_n}{P_n} } = -1$$.

I was able to show that both $${Q_n}$$ and $${P_n}$$ diverges. But I couldn't proof the ratio between them (I'm not sure if it's helpful). How can it be proofed?

Note that $$P_n + Q_n = \sum_{k=1}^n a_k$$ so $$1+ \frac{Q_n}{P_n} = \frac{\sum_{k=1}^n a_k}{P_n} \rightarrow \frac{const.}{\infty} = 0$$
• We just subtract 1 from both sides. The limit of a sum is a sum of the limits, in this case $$0 = \lim_{n\rightarrow\infty}(1+\frac{Q_n}{P_n}) = (\lim_{n\rightarrow\infty}1)+(\lim_{n\rightarrow\infty}\frac{Q_n}{P_n}) = 1+ \lim_{n\rightarrow\infty}\frac{Q_n}{P_n}$$ May 8 '19 at 10:32
• Thanks! It might be out of this question's scope, but how do we show that $P_n$ necessarily goes to infinity? Is it enough to say the the series $P_n$ is monotonically increasing and diverges? May 8 '19 at 13:43


text:
G13 Chapter Contents
G13 Chapter Introduction
NAG Library Manual

# NAG Library Routine DocumentG13CFF

Note:  before using this routine, please read the Users' Note for your implementation to check the interpretation of bold italicised terms and other implementation-dependent details.

## 1  Purpose

For a bivariate time series, G13CFF calculates the gain and phase together with lower and upper bounds from the univariate and bivariate spectra.

## 2  Specification

 SUBROUTINE G13CFF ( XG, YG, XYRG, XYIG, NG, STATS, GN, GNLW, GNUP, PH, PHLW, PHUP, IFAIL)
 INTEGER NG, IFAIL REAL (KIND=nag_wp) XG(NG), YG(NG), XYRG(NG), XYIG(NG), STATS(4), GN(NG), GNLW(NG), GNUP(NG), PH(NG), PHLW(NG), PHUP(NG)

## 3  Description

Estimates of the gain $G\left(\omega \right)$ and phase $\varphi \left(\omega \right)$ of the dependency of series $y$ on series $x$ at frequency $\omega$ are given by
 $G^ω= Aω fxxω ϕ^ω=arccos cfω Aω , if qfω≥0 ϕ^ω=2π-arccos cfω Aω , if qfω<0.$
The quantities used in these definitions are obtained as in Section 3 in G13CEF.
Confidence limits are returned for both gain and phase, but should again be taken as very approximate when the coherency $W\left(\omega \right)$, as calculated by G13CEF, is not significant. These are based on the assumption that both $\left(\stackrel{^}{G}\left(\omega \right)/G\left(\omega \right)\right)-1$ and $\stackrel{^}{\varphi }\left(\omega \right)$ are Normal with variance
 $1d 1Wω -1 .$
Although the estimate of $\varphi \left(\omega \right)$ is always given in the range $\left[0,2\pi \right)$, no attempt is made to restrict its confidence limits to this range.

## 4  References

<NAME> (1976) Fourier Analysis of Time Series: An Introduction Wiley
<NAME> and <NAME> (1968) Spectral Analysis and its Applications Holden–Day

## 5  Parameters

1:     XG(NG) – REAL (KIND=nag_wp) arrayInput
On entry: the NG univariate spectral estimates, ${f}_{xx}\left(\omega \right)$, for the $x$ series.
2:     YG(NG) – REAL (KIND=nag_wp) arrayInput
On entry: the NG univariate spectral estimates, ${f}_{yy}\left(\omega \right)$, for the $y$ series.
3:     XYRG(NG) – REAL (KIND=nag_wp) arrayInput
On entry: the real parts, $cf\left(\omega \right)$, of the NG bivariate spectral estimates for the $x$ and $y$ series. The $x$ series leads the $y$ series.
4:     XYIG(NG) – REAL (KIND=nag_wp) arrayInput
On entry: the imaginary parts, $qf\left(\omega \right)$, of the NG bivariate spectral estimates for the $x$ and $y$ series. The $x$ series leads the $y$ series.
Note:  the two univariate and the bivariate spectra must each have been calculated using the same method of smoothing. For rectangular, Bartlett, Tukey or Parzen smoothing windows, the same cut-off point of lag window and the same frequency division of the spectral estimates must be used. For the trapezium frequency smoothing window, the frequency width and the shape of the window and the frequency division of the spectral estimates must be the same. The spectral estimates and statistics must also be unlogged.
5:     NG – INTEGERInput
On entry: the number of spectral estimates in each of the arrays XG, YG, XYRG and XYIG. It is also the number of gain and phase estimates.
Constraint: ${\mathbf{NG}}\ge 1$.
6:     STATS($4$) – REAL (KIND=nag_wp) arrayInput
On entry: the four associated statistics for the univariate spectral estimates for the $x$ and $y$ series. ${\mathbf{STATS}}\left(1\right)$ contains the degrees of freedom, ${\mathbf{STATS}}\left(2\right)$ and ${\mathbf{STATS}}\left(3\right)$ contain the lower and upper bound multiplying factors respectively and ${\mathbf{STATS}}\left(4\right)$ holds the bandwidth.
Constraint: ${\mathbf{STATS}}\left(1\right)\ge 3.0$.
7:     GN(NG) – REAL (KIND=nag_wp) arrayOutput
On exit: the NG gain estimates, $\stackrel{^}{G}\left(\omega \right)$, at each frequency $\omega$.
8:     GNLW(NG) – REAL (KIND=nag_wp) arrayOutput
On exit: the NG lower bounds for the NG gain estimates.
9:     GNUP(NG) – REAL (KIND=nag_wp) arrayOutput
On exit: the NG upper bounds for the NG gain estimates.
10:   PH(NG) – REAL (KIND=nag_wp) arrayOutput
On exit: the NG phase estimates, $\stackrel{^}{\varphi }\left(\omega \right)$, at each frequency $\omega$.
11:   PHLW(NG) – REAL (KIND=nag_wp) arrayOutput
On exit: the NG lower bounds for the NG phase estimates.
12:   PHUP(NG) – REAL (KIND=nag_wp) arrayOutput
On exit: the NG upper bounds for the NG phase estimates.
13:   IFAIL – INTEGERInput/Output
On entry: IFAIL must be set to $0$, $-1\text{ or }1$. If you are unfamiliar with this parameter you should refer to Section 3.3 in the Essential Introduction for details.
For environments where it might be inappropriate to halt program execution when an error is detected, the value $-1\text{ or }1$ is recommended. If the output of error messages is undesirable, then the value $1$ is recommended. Otherwise, if you are not familiar with this parameter, the recommended value is $0$. When the value $-\mathbf{1}\text{ or }\mathbf{1}$ is used it is essential to test the value of IFAIL on exit.
On exit: ${\mathbf{IFAIL}}={\mathbf{0}}$ unless the routine detects an error or a warning has been flagged (see Section 6).

## 6  Error Indicators and Warnings

If on entry ${\mathbf{IFAIL}}={\mathbf{0}}$ or $-{\mathbf{1}}$, explanatory error messages are output on the current error message unit (as defined by X04AAF).
Errors or warnings detected by the routine:
${\mathbf{IFAIL}}=1$
 On entry, ${\mathbf{NG}}<1$, or ${\mathbf{STATS}}\left(1\right)<3.0$.
${\mathbf{IFAIL}}=2$
A bivariate spectral estimate is zero. For this frequency the gain and the phase and their bounds are set to zero.
${\mathbf{IFAIL}}=3$
A univariate spectral estimate is negative. For this frequency the gain and the phase and their bounds are set to zero.
${\mathbf{IFAIL}}=4$
A univariate spectral estimate is zero. For this frequency the gain and the phase and their bounds are set to zero.
${\mathbf{IFAIL}}=5$
A calculated value of the squared coherency exceeds $1.0$. For this frequency the squared coherency is reset to $1.0$ in the formulae for the gain and phase bounds.
If more than one failure of types $2$, $3$, $4$ and $5$ occurs then the failure type which occurred at lowest frequency is returned in IFAIL. However the actions indicated above are also carried out for failures at higher frequencies.

## 7  Accuracy

All computations are very stable and yield good accuracy.

The time taken by G13CFF is approximately proportional to NG.

## 9  Example

This example reads the set of univariate spectrum statistics, the two univariate spectra and the cross spectrum at a frequency division of $\frac{2\pi }{20}$ for a pair of time series. It calls G13CFF to calculate the gain and the phase and their bounds and prints the results.

### 9.1  Program Text

Program Text (g13cffe.f90)

### 9.2  Program Data

Program Data (g13cffe.d)

### 9.3  Program Results

Program Results (g13cffe.r)


text:
### <NAME>

Machine Learning, Artificial Intelligence. On the Move.

# Public-Key Cryptosystems and Digital Signatures

### Introduction

Sometimes I would get some interesting questions from my friends about internet security, such as whether it is possible to leak the bank account password when logging to the online bank via an untrusted network, whether it is possible that the message you received from someone via an untrusted network got modified maliciously, and whether it is possible that someone pretend to be you and send messages in your name. My answer to those questions are as long as your computer and cell phone are uncontaminated they are almost impossible thanks to our modern public-key crytosystems and digital signatures.

There are a couple of introductions to the public-key crystosystems and digital signature available online. However, I think most of them are incomplete or hard to understand. In this blog post, I am going to describe the public-key crystosystems and digital signatures using extremely simple math, so that there would be no ambiguity at all.

### Usages

The modern cryptosystems are public-key encryption systems in which everyone has a public-key for encryption and a private key for decryption. The public-key is seen by everyone, but the private is only accessible by the owner. The public-key encryption systems could also generate digital signatures which could be used to verify whether the message you received is unmodified and truly sent from the sender.

RSA (Rivest–Shamir–Adleman) is a typical algorithm for the public-key cryptosystems used by modern computers to encrypt and decrypt messages. However, we are not going to introduce the RSA algorithm in this blog post. Instead, we would describe the public-key cryptosystems and digital signatures at a high level.

### Essential Features

Each user has their own encryption and decryption functions, $E$ and $D$, using public-key and private key respectively. We use $M$ to represent the message to be encrypted and sent. There are four features that are essential to a public-key cryptosystem.

• Decrypting an encrypted message gives you the original message (Of course!). Specifically,
• Encrypting an decrypted message gives you the original message (Hmm…). Specifically,
• $E$ and $D$ are easy to compute. This means the encryption and decryption process should be fast.

• The publicity of $E$ does not compromise the secrecy of $D$. This means you could hardly find the way to decrypt the encrypted message, even if you know how to encrypt the message.

We would ignore how to satisfy the four features in this blog post.

### Message Encryption and Decryption

Suppose we have two people, Alice and Bob. Both of them are using the same public-key cryptosystems. This means Alice and Bob both have their private keys stored secretly and have their public key published to some authorities. From the authorities, we could find the public keys of Alice and Bob unambiguously. We denote encrypting the message using Alice and Bob’s public keys to be $E_A$ and $E_B$ respectively, and decrypting the message using Alice and Bob’s private keys to be $D_A$ and $D_B$ respectively.

One day, Alice wanted to send a private message $M$ to Bob. Alice found the public key of Bob, encrypted the message $M$ using Bob’s public key. The encrypted message for Bob is denoted as $C$.

Once Bob received the encrypted message $C$, he could decrypt $C$ using his private key.

Even if the network was compromised and someone intercepted $C$, it is still almost impossible to decrypt $C$ because of that $D_B$ is unknown and the feature “the publicity of $E$ does not compromise the secrecy of $D$” for public-key cryptosystems.

The message content is safe because of the public-key encryption systems. However, it does not provide any assurance about the sender. For example, James could send Bob a message $M^\prime$ which specifically says the message is from Alice, encrypt it using $E_B$, and send it to Bob. If there is no author verification procedure and Bob is not careful enough, Bob might actually think the message is sent from Alice. In some other scenarios, James might have intercepted the encrypted message $C$ sent out from Alice to Bob, prevented the message transmission to Bob, replaced the original message to $M^\prime$ which specifically says the message is from Alice, encrypt it using $E_B$, and send it to Bob. Bob might also be convinced that the message content $M^\prime$ was actually the original message Alice have sent. Specifically,

Digital signatures, derived from the public-key cryptosystems, are designed to solve these authentication problems.

### Digital Signatures

In addition to the encrypted message $C$ that Alice sent to Bob, Alice would also have to send her digital signature $S$ to Bob. Namely,

Alice could find $E_B$ using Bob’s public key and $D_A$ using her private key.

Once Bob received both $S$ and $C$, he could decrypt both $S$ and $C$ using his private key and Alice’s public key. Specially,

We found actually the two decrypted messages from $S$ and $C$ are exactly the same. This is expected if the message was sent from Alice and the content of the message was not modified. Let’s further see what will happen if someone pretend to be Alice to send message to Bob, or the content of the message has been modified.

James, again, wanted to send Bob a message $M^\prime$ which specifically says the message is from Alice, or had intercepted an encrypted message $C$ from Alice, blocked it and created a message $M^\prime$ which specifically says the message is from Alice. To make the message readable by Bob, James encrypted $M^\prime$ using $E_B$, send send the encrypted message $C^\prime$ to Bob.

Because Bob does not accept any message without signature, James had to make up a signature. However, because James knew nothing about Alice’s private key, he used an decryption function $D_J$ which is different to Alice’s secrete $D_A$. The signature James generated would be

Once Bob received both $S^\prime$ and $C^\prime$, he could decrypt both $S$ and $C$ using his private key and Alice’s public key as usual. Specially,

In this case, Bob would see the two decrypted messages are not the same. Bob would then realize that there is something unusual happened and he should not trust anything about the message.

### Hacking the Public-Key Cryptosystems

As long as the four features of the public-key cryptosystems hold, cracking it is almost impossible. If some day, when the almighty quantum computer is available, the feature “the publicity of $E$ does not compromise the secrecy of $D$” would be compromised, therefore the modern public-key cryptosystems would no longer be reliable. I may talk about this topic in the future.

There is another way to send fake messages to Bob in name of Alice, without using a quantum computer. If James could somehow crack the account name and password of Alice on the web application, replace the Alice’s public encryption function from $E_A$ to $E_J$, when Bob tried to retrieve Alice’s encryption function, he would get $E_J$ instead of $E_A$. The decryption of signature $S^\prime$ would become $M^\prime$ instead of $M^{\prime\prime}$ then. Concretely,

In this case the two decrypted messages match. Bob would be convinced that the message is from Alice and it has not been modified.

It should be noted that James was replacing Alice’s public encryption function from $E_A$ to $E_J$, instead of replacing his private decryption function from $D_J$ to $D_A$. In principle, $D_A$ would only be kept on Alice local computer and not anywhere else. Even if James has Alice’s account name and password on the web application, he would not get a copy of $D_A$ unless he specifically hacked Alice’s physical computer.

### Final Remarks

This remind us that actually keeping our password safe is the most important.


text:
# Coefficients in Feynman Diagrams

Consider the 4-point correlation function, $$G(x_1,x_2,x_3,x_4)$$, in $$\phi^4$$ theory. Let us consider the term which is $$\propto \lambda^2$$, represented by the following Feynman's diagram:

According to symmetry rules, I should get a $$1/2$$ symmetry factor, due to the middle lines. Hence, I would suppose that

$$G_2(x_1,x_2,x_3,x_4) = -\frac{\lambda}{2}\int\int d^4x_5 d^4x_6\Delta_F(x_1-x_5)\Delta_F(x_2-x_5)\Delta_F(x_4-x_6)\Delta_F(x_3-x_6)\left(\Delta_F(x_5-x_6)\right)^2$$

where $$x_5$$ and $$x_6$$ denote de vertices. However, if I carry out the calculations by hand, I should have something like $$\frac{1}{2\times 24}\times 4 \times 3 \times 2 \times 6 \times 2 = 6 \neq \frac{1}{2}$$. What am I doing wrong here?

The factor is $$\frac{1}{2!} \times \frac{1}{4!} \times \frac{1}{4!} \times 8 \times 3 \times 4 \times 3 \times 2 = \frac{1}{2}$$.
The $$\frac{1}{2!}$$ comes from expanding $$e^x$$ to second order. The two factors of $$\frac{1}{4!}$$ come from the interaction term (which is $$\frac{\lambda}{4!}$$). The remaining factors come from counting contractions so lets look at that. The fields to be contracted are $$\phi_1 \phi_2 \phi_3 \phi_4 \, \, \phi_x\phi_x\phi_x\phi_x \, \, \phi_y\phi_y\phi_y\phi_y$$ First $$\phi_1$$ can contract with 8 different fields. Everything is completely symmetrical, so lets assume it contracts with $$\phi_x$$. $$\phi_2$$ must then contract with $$\phi_x$$ also which happens in 3 different ways. $$\phi_3$$ contracts with $$\phi_y$$ in 4 different ways and $$\phi_4$$ contracts with another $$\phi_y$$ in 3 different ways. Finally, the remaining un-contracted fields are $$\phi_x \phi_x \,\, \phi_y \phi_y$$. The first $$\phi_x$$ can contract with a $$\phi_y$$ in 2 different ways and the final contraction is then fixed. Multiplying all the numbers in this paragraph, we find that the number of contractions is $$8\times3\times4\times3\times2$$.