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https://groupprops.subwiki.org/wiki/Unipotent_linear_algebraic_group	[ "# Unipotent linear algebraic group\n\nA linear algebraic group", null, "$G$ over a field", null, "$k$ (which therefore comes equipped with an embedding as a closed subgroup of the general linear group", null, "$GL(n,k)$) is termed a unipotent linear algebraic group if it satisfies the following equivalent conditions:\n1. Every element in", null, "$G$ is a unipotent element, i.e., subtracting", null, "$1$ from it gives a nilpotent element in the matrix ring.\n2.", null, "$G$ is conjugate in", null, "$GL(n,k)$ to a subgroup of the upper-triangular unipotent matrix group.\nAny unipotent linear algebraic group is nilpotent, with its nilpotency class at most", null, "$n - 1$, where", null, "$n$ is the degree of the general linear group it is embedded in. However, the converse is not true: it is possible to have a linear algebraic group that is nilpotent as an abstract group but is not unipotent. For instance, the [[[multiplicative group of a field]]", null, "$k$, which is the full group", null, "$GL(1,k)$, is abelian and hence nilpotent, but is not unipotent." ]	[ null, "https://groupprops.subwiki.org/w/images/math/d/f/c/dfcf28d0734569a6a693bc8194de62bf.png ", null, "https://groupprops.subwiki.org/w/images/math/8/c/e/8ce4b16b22b58894aa86c421e8759df3.png ", null, "https://groupprops.subwiki.org/w/images/math/9/7/1/9714d1c20dff3c1ff9eaeb3e89a63b18.png ", null, "https://groupprops.subwiki.org/w/images/math/d/f/c/dfcf28d0734569a6a693bc8194de62bf.png ", null, "https://groupprops.subwiki.org/w/images/math/c/4/c/c4ca4238a0b923820dcc509a6f75849b.png ", null, "https://groupprops.subwiki.org/w/images/math/d/f/c/dfcf28d0734569a6a693bc8194de62bf.png ", null, "https://groupprops.subwiki.org/w/images/math/9/7/1/9714d1c20dff3c1ff9eaeb3e89a63b18.png ", null, "https://groupprops.subwiki.org/w/images/math/a/4/3/a438673491daae8148eae77373b6a467.png ", null, "https://groupprops.subwiki.org/w/images/math/7/b/8/7b8b965ad4bca0e41ab51de7b31363a1.png ", null, "https://groupprops.subwiki.org/w/images/math/8/c/e/8ce4b16b22b58894aa86c421e8759df3.png ", null, "https://groupprops.subwiki.org/w/images/math/b/e/0/be0ab5115ab17817b261a0ea22d4f269.png ", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9133346,"math_prob":0.9961626,"size":1022,"snap":"2020-34-2020-40","text_gpt3_token_len":237,"char_repetition_ratio":0.19056974,"word_repetition_ratio":0.011976048,"special_character_ratio":0.19569471,"punctuation_ratio":0.10215054,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99966526,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-06T06:33:51Z\",\"WARC-Record-ID\":\"<urn:uuid:8968bc76-c812-4b5d-a73c-cd78adde49e8>\",\"Content-Length\":\"23224\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a1da7a9e-9481-4672-ba76-dd64b6b08c83>\",\"WARC-Concurrent-To\":\"<urn:uuid:675c5d69-0af8-4461-a2b3-657c26103c4e>\",\"WARC-IP-Address\":\"96.126.114.7\",\"WARC-Target-URI\":\"https://groupprops.subwiki.org/wiki/Unipotent_linear_algebraic_group\",\"WARC-Payload-Digest\":\"sha1:HJGSE4LT7PECMQ45PDLWPMYH6DILMGRZ\",\"WARC-Block-Digest\":\"sha1:QCCVSCOSIT3J5ENWE672YK2QIE6F55AO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439736883.40_warc_CC-MAIN-20200806061804-20200806091804-00362.warc.gz\"}"}
https://logical-invest.com/app/etf/ewd/ishares-msci-sweden-etf	[ "## Description\n\nThe investment seeks to track the investment results of the MSCI Sweden 25/50 Index. The fund will at all times invest at least 80% of its assets in the securities of its underlying index and in depositary receipts representing securities in its underlying index. The underlying index is designed to measure the performance of the large- and mid-cap segments of the Swedish market. A capping methodology is applied that limits the weight of any single issuer to a maximum of 25% of the underlying index. The fund is non-diversified.\n\n## Statistics (YTD)\n\nWhat do these metrics mean? [Read More] [Hide]\n\n### TotalReturn:\n\n'The total return on a portfolio of investments takes into account not only the capital appreciation on the portfolio, but also the income received on the portfolio. The income typically consists of interest, dividends, and securities lending fees. This contrasts with the price return, which takes into account only the capital gain on an investment.'\n\nUsing this definition on our asset we see for example:\n• The total return over 5 years of iShares MSCI Sweden ETF is 22.6%, which is smaller, thus worse compared to the benchmark SPY (63%) in the same period.\n• Looking at total return, or increase in value in of 24.2% in the period of the last 3 years, we see it is relatively smaller, thus worse in comparison to SPY (33.5%).\n\n### CAGR:\n\n'The compound annual growth rate isn't a true return rate, but rather a representational figure. It is essentially a number that describes the rate at which an investment would have grown if it had grown the same rate every year and the profits were reinvested at the end of each year. In reality, this sort of performance is unlikely. However, CAGR can be used to smooth returns so that they may be more easily understood when compared to alternative investments.'\n\nWhich means for our asset as example:\n• Compared with the benchmark SPY (10.3%) in the period of the last 5 years, the compounded annual growth rate (CAGR) of 4.2% of iShares MSCI Sweden ETF is lower, thus worse.\n• Looking at annual return (CAGR) in of 7.5% in the period of the last 3 years, we see it is relatively lower, thus worse in comparison to SPY (10.1%).\n\n### Volatility:\n\n'Volatility is a statistical measure of the dispersion of returns for a given security or market index. Volatility can either be measured by using the standard deviation or variance between returns from that same security or market index. Commonly, the higher the volatility, the riskier the security. In the securities markets, volatility is often associated with big swings in either direction. For example, when the stock market rises and falls more than one percent over a sustained period of time, it is called a 'volatile' market.'\n\nApplying this definition to our asset in some examples:\n• Compared with the benchmark SPY (21.6%) in the period of the last 5 years, the volatility of 26.9% of iShares MSCI Sweden ETF is greater, thus worse.\n• During the last 3 years, the volatility is 31.7%, which is higher, thus worse than the value of 25.1% from the benchmark.\n\n### DownVol:\n\n'The downside volatility is similar to the volatility, or standard deviation, but only takes losing/negative periods into account.'\n\nUsing this definition on our asset we see for example:\n• Looking at the downside risk of 19.5% in the last 5 years of iShares MSCI Sweden ETF, we see it is relatively higher, thus worse in comparison to the benchmark SPY (15.6%)\n• Looking at downside volatility in of 22.8% in the period of the last 3 years, we see it is relatively higher, thus worse in comparison to SPY (18.1%).\n\n### Sharpe:\n\n'The Sharpe ratio was developed by Nobel laureate William F. Sharpe, and is used to help investors understand the return of an investment compared to its risk. The ratio is the average return earned in excess of the risk-free rate per unit of volatility or total risk. Subtracting the risk-free rate from the mean return allows an investor to better isolate the profits associated with risk-taking activities. One intuition of this calculation is that a portfolio engaging in 'zero risk' investments, such as the purchase of U.S. Treasury bills (for which the expected return is the risk-free rate), has a Sharpe ratio of exactly zero. Generally, the greater the value of the Sharpe ratio, the more attractive the risk-adjusted return.'\n\nUsing this definition on our asset we see for example:\n• The Sharpe Ratio over 5 years of iShares MSCI Sweden ETF is 0.06, which is lower, thus worse compared to the benchmark SPY (0.36) in the same period.\n• Compared with SPY (0.3) in the period of the last 3 years, the risk / return profile (Sharpe) of 0.16 is smaller, thus worse.\n\n### Sortino:\n\n'The Sortino ratio, a variation of the Sharpe ratio only factors in the downside, or negative volatility, rather than the total volatility used in calculating the Sharpe ratio. The theory behind the Sortino variation is that upside volatility is a plus for the investment, and it, therefore, should not be included in the risk calculation. Therefore, the Sortino ratio takes upside volatility out of the equation and uses only the downside standard deviation in its calculation instead of the total standard deviation that is used in calculating the Sharpe ratio.'\n\nWhich means for our asset as example:\n• The ratio of annual return and downside deviation over 5 years of iShares MSCI Sweden ETF is 0.09, which is smaller, thus worse compared to the benchmark SPY (0.5) in the same period.\n• Compared with SPY (0.42) in the period of the last 3 years, the ratio of annual return and downside deviation of 0.22 is lower, thus worse.\n\n### Ulcer:\n\n'Ulcer Index is a method for measuring investment risk that addresses the real concerns of investors, unlike the widely used standard deviation of return. UI is a measure of the depth and duration of drawdowns in prices from earlier highs. Using Ulcer Index instead of standard deviation can lead to very different conclusions about investment risk and risk-adjusted return, especially when evaluating strategies that seek to avoid major declines in portfolio value (market timing, dynamic asset allocation, hedge funds, etc.). The Ulcer Index was originally developed in 1987. Since then, it has been widely recognized and adopted by the investment community. According to Nelson Freeburg, editor of Formula Research, Ulcer Index is “perhaps the most fully realized statistical portrait of risk there is.'\n\nWhich means for our asset as example:\n• Compared with the benchmark SPY (8.88 ) in the period of the last 5 years, the Ulcer Ratio of 15 of iShares MSCI Sweden ETF is larger, thus worse.\n• During the last 3 years, the Ulcer Ratio is 18 , which is larger, thus worse than the value of 11 from the benchmark.\n\n### MaxDD:\n\n'Maximum drawdown measures the loss in any losing period during a fund’s investment record. It is defined as the percent retrenchment from a fund’s peak value to the fund’s valley value. The drawdown is in effect from the time the fund’s retrenchment begins until a new fund high is reached. The maximum drawdown encompasses both the period from the fund’s peak to the fund’s valley (length), and the time from the fund’s valley to a new fund high (recovery). It measures the largest percentage drawdown that has occurred in any fund’s data record.'\n\nWhich means for our asset as example:\n• Compared with the benchmark SPY (-33.7 days) in the period of the last 5 years, the maximum drop from peak to valley of -42.3 days of iShares MSCI Sweden ETF is lower, thus worse.\n• During the last 3 years, the maximum drop from peak to valley is -42.3 days, which is lower, thus worse than the value of -33.7 days from the benchmark.\n\n### MaxDuration:\n\n'The Maximum Drawdown Duration is an extension of the Maximum Drawdown. However, this metric does not explain the drawdown in dollars or percentages, rather in days, weeks, or months. It is the length of time the account was in the Max Drawdown. A Max Drawdown measures a retrenchment from when an equity curve reaches a new high. It’s the maximum an account lost during that retrenchment. This method is applied because a valley can’t be measured until a new high occurs. Once the new high is reached, the percentage change from the old high to the bottom of the largest trough is recorded.'\n\nUsing this definition on our asset we see for example:\n• The maximum days under water over 5 years of iShares MSCI Sweden ETF is 467 days, which is greater, thus worse compared to the benchmark SPY (273 days) in the same period.\n• During the last 3 years, the maximum days under water is 371 days, which is higher, thus worse than the value of 273 days from the benchmark.\n\n### AveDuration:\n\n'The Drawdown Duration is the length of any peak to peak period, or the time between new equity highs. The Avg Drawdown Duration is the average amount of time an investment has seen between peaks (equity highs), or in other terms the average of time under water of all drawdowns. So in contrast to the Maximum duration it does not measure only one drawdown event but calculates the average of all.'\n\nWhich means for our asset as example:\n• The average time in days below previous high water mark over 5 years of iShares MSCI Sweden ETF is 156 days, which is larger, thus worse compared to the benchmark SPY (57 days) in the same period.\n• During the last 3 years, the average days under water is 110 days, which is higher, thus worse than the value of 73 days from the benchmark.\n\n## Returns (%)\n\n• Note that yearly returns do not equal the sum of monthly returns due to compounding.\n• Performance results of iShares MSCI Sweden ETF are hypothetical, do not account for slippage, fees or taxes, and are based on backtesting, which has many inherent limitations, some of which are described in our Terms of Use." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9442741,"math_prob":0.92820084,"size":5595,"snap":"2022-40-2023-06","text_gpt3_token_len":1116,"char_repetition_ratio":0.12913612,"word_repetition_ratio":0.03960396,"special_character_ratio":0.19070598,"punctuation_ratio":0.09432146,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96109664,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-05T20:35:53Z\",\"WARC-Record-ID\":\"<urn:uuid:290ca215-2197-4546-be10-ca2e2daecbf0>\",\"Content-Length\":\"52744\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:671eec70-8c85-4a61-806b-0ccbdb7ce106>\",\"WARC-Concurrent-To\":\"<urn:uuid:a483147e-1feb-461f-9f0f-252e697db1e0>\",\"WARC-IP-Address\":\"172.66.40.164\",\"WARC-Target-URI\":\"https://logical-invest.com/app/etf/ewd/ishares-msci-sweden-etf\",\"WARC-Payload-Digest\":\"sha1:QYXTQSFVAPRDNKJFIJ6IHVFSUATK3UTK\",\"WARC-Block-Digest\":\"sha1:Q3ONBP3RVY4VLSYWQEWHBTR2T5AMQU4J\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500288.69_warc_CC-MAIN-20230205193202-20230205223202-00353.warc.gz\"}"}
http://pub.acta.hu/acta/showCustomerArticle.action?id=3112&dataObjectType=article&returnAction=showCustomerVolume&sessionDataSetId=29496aaf4b83ad50&style=	[ "ACTA issues\n\n## Computation of the $\\rho$-numerical radius for truncated shifts\n\nLaurent Carrot\n\nActa Sci. Math. (Szeged) 69:1-2(2003), 311-322\n2897/2009\n\n Abstract. We are interested in computing the $\\rho$-numerical radius of the truncated shift $S_n$ on $l_2^n$. This value, first computed for $\\rho =2$ by Haagerup and de la Harpe, appears as a reference in constrained von Neumann inequalities. In this paper, we give an effective way of computing this value in the general case, and then some exact formulas in particular cases. AMS Subject Classification (1991): 47A12, 15A60, 15-04; 47A63 Received November 12, 2001, and in revised form September 13, 2002. (Registered under 2897/2009.)", null, "" ]	[ null, "http://pub.acta.hu/acta/imagestream", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8674206,"math_prob":0.8357281,"size":541,"snap":"2020-34-2020-40","text_gpt3_token_len":149,"char_repetition_ratio":0.09124767,"word_repetition_ratio":0.0,"special_character_ratio":0.3068392,"punctuation_ratio":0.16190477,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9751576,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-25T00:31:25Z\",\"WARC-Record-ID\":\"<urn:uuid:f17c05c3-cc84-462e-a015-fb25e78ef7dc>\",\"Content-Length\":\"46107\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3a148187-be0b-40e7-aa3f-b744146c0ee9>\",\"WARC-Concurrent-To\":\"<urn:uuid:cd8ff94b-1d06-4875-9fdb-944c1cffd524>\",\"WARC-IP-Address\":\"160.114.33.249\",\"WARC-Target-URI\":\"http://pub.acta.hu/acta/showCustomerArticle.action?id=3112&dataObjectType=article&returnAction=showCustomerVolume&sessionDataSetId=29496aaf4b83ad50&style=\",\"WARC-Payload-Digest\":\"sha1:AQVBMVXCCEHI2FJHH4EHBHC5FOYC7PJD\",\"WARC-Block-Digest\":\"sha1:3X53JOA4ZWWG5X4SU3AQPK3H3WPJA2UB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400221382.33_warc_CC-MAIN-20200924230319-20200925020319-00339.warc.gz\"}"}
https://xmonad.github.io/xmonad-docs/xmonad-contrib/src/XMonad.Hooks.ManageHelpers.html	[ "```{-# LANGUAGE LambdaCase #-}\n-----------------------------------------------------------------------------\n-- \|\n-- Description : Helper functions to be used in manageHook.\n-- Copyright : (c) Lukas Mai\n--\n-- Maintainer : Lukas Mai <[email protected]>\n-- Stability : unstable\n-- Portability : unportable\n--\n-- This module provides helper functions to be used in @manageHook@. Here's\n-- how you might use this:\n--\n-- > main =\n-- > ...\n-- > manageHook = composeOne [\n-- > isKDETrayWindow -?> doIgnore,\n-- > transience,\n-- > isFullscreen -?> doFullFloat,\n-- > resource =? \"stalonetray\" -?> doIgnore\n-- > ],\n-- > ...\n-- > }\n--\n-- Here's how you can define more helpers like the ones from this module:\n--\n-- > -- some function you want to transform into an infix operator\n-- > f :: a -> b -> Bool\n-- >\n-- > -- a new helper\n-- > q *? x = fmap (\\a -> f a x) q -- or: (\\b -> f x b)\n-- > -- or\n-- > q ? x = fmap (`f` x) q -- or: (x `f`)\n--\n-- Any existing operator can be \"lifted\" in the same way:\n--\n-- > q ++? x = fmap (++ x) q\n\nSide(..),\ncomposeOne,\n(-?>), (/=?), (^?), (~?), (\\$?), (<==?), (</=?), (-->>), (-?>>),\ncurrentWs,\nwindowTag,\nisInProperty,\nisKDETrayWindow,\nisFullscreen,\nisMinimized,\nisDialog,\npid,\ntransientTo,\nmaybeToDefinite,\nMaybeManageHook,\ntransience,\ntransience',\nsameBy,\nshiftToSame,\nshiftToSame',\ndoRectFloat,\ndoFullFloat,\ndoCenterFloat,\ndoSideFloat,\ndoFloatAt,\ndoFloatDep,\ndoHideIgnore,\ndoSink,\ndoLower,\ndoRaise,\ndoFocus,\nMatch,\n) where\n\nimport System.Posix (ProcessID)\n\n-- \| Denotes a side of a screen. @S@ stands for South, @NE@ for Northeast\n-- etc. @C@ stands for Center.\ndata Side = SC \| NC \| CE \| CW \| SE \| SW \| NE \| NW \| C\nforall a.\nRead, Int -> Side -> ShowS\n[Side] -> ShowS\nSide -> WorkspaceId\nforall a.\n(Int -> a -> ShowS)\n-> (a -> WorkspaceId) -> ([a] -> ShowS) -> Show a\nshowList :: [Side] -> ShowS\n\\$cshowList :: [Side] -> ShowS\nshow :: Side -> WorkspaceId\n\\$cshow :: Side -> WorkspaceId\nshowsPrec :: Int -> Side -> ShowS\n\\$cshowsPrec :: Int -> Side -> ShowS\nShow, Side -> Side -> Bool\nforall a. (a -> a -> Bool) -> (a -> a -> Bool) -> Eq a\n/= :: Side -> Side -> Bool\n\\$c/= :: Side -> Side -> Bool\n== :: Side -> Side -> Bool\n\\$c== :: Side -> Side -> Bool\nEq)\n\n-- \| A ManageHook that may or may not have been executed; the outcome is embedded in the Maybe\ntype MaybeManageHook = Query (Maybe (Endo WindowSet))\n-- \| A grouping type, which can hold the outcome of a predicate Query.\n-- This is analogous to group types in regular expressions.\n-- TODO: create a better API for aggregating multiple Matches logically\ndata Match a = Match Bool a\n\n-- \| An alternative 'ManageHook' composer. Unlike 'composeAll' it stops as soon as\n-- a candidate returns a 'Just' value, effectively running only the first match\n-- (whereas 'composeAll' continues and executes all matching rules).\ncomposeOne :: (Monoid a, Monad m) => [m (Maybe a)] -> m a\ncomposeOne :: forall a (m :: -> ). (Monoid a, Monad m) => [m (Maybe a)] -> m a\ncomposeOne = forall (t :: -> ) a b.\nFoldable t =>\n(a -> b -> b) -> b -> t a -> b\nfoldr forall {m :: -> } {b}. Monad m => m (Maybe b) -> m b -> m b\ntry (forall (m :: -> ) a. Monad m => a -> m a\nreturn forall a. Monoid a => a\nmempty)\nwhere\ntry :: m (Maybe b) -> m b -> m b\ntry m (Maybe b)\nq m b\nz = do\nMaybe b\nx <- m (Maybe b)\nq\nforall b a. b -> (a -> b) -> Maybe a -> b\nmaybe m b\nz forall (m :: -> ) a. Monad m => a -> m a\nreturn Maybe b\nx\n\ninfixr 0 -?>, -->>, -?>>\n\n-- \| q \\/=? x. if the result of q equals x, return False\n(/=?) :: (Eq a, Functor m) => m a -> a -> m Bool\nm a\nq /=? :: forall a (m :: -> ). (Eq a, Functor m) => m a -> a -> m Bool\n/=? a\nx = forall (f :: -> ) a b. Functor f => (a -> b) -> f a -> f b\nfmap (forall a. Eq a => a -> a -> Bool\n/= a\nx) m a\nq\n\n-- \| q ^? x. if the result of @x `isPrefixOf` q@, return True\n(^?) :: (Eq a, Functor m) => m [a] -> [a] -> m Bool\nm [a]\nq ^? :: forall a (m :: -> ). (Eq a, Functor m) => m [a] -> [a] -> m Bool\n^? [a]\nx = forall (f :: -> ) a b. Functor f => (a -> b) -> f a -> f b\nfmap ([a]\nx forall a. Eq a => [a] -> [a] -> Bool\n`isPrefixOf`) m [a]\nq\n\n-- \| q ~? x. if the result of @x `isInfixOf` q@, return True\n(~?) :: (Eq a, Functor m) => m [a] -> [a] -> m Bool\nm [a]\nq ~? :: forall a (m :: -> ). (Eq a, Functor m) => m [a] -> [a] -> m Bool\n~? [a]\nx = forall (f :: -> ) a b. Functor f => (a -> b) -> f a -> f b\nfmap ([a]\nx forall a. Eq a => [a] -> [a] -> Bool\n`isInfixOf`) m [a]\nq\n\n-- \| q \\$? x. if the result of @x `isSuffixOf` q@, return True\n(\\$?) :: (Eq a, Functor m) => m [a] -> [a] -> m Bool\nm [a]\nq \\$? :: forall a (m :: -> ). (Eq a, Functor m) => m [a] -> [a] -> m Bool\n\\$? [a]\nx = forall (f :: -> ) a b. Functor f => (a -> b) -> f a -> f b\nfmap ([a]\nx forall a. Eq a => [a] -> [a] -> Bool\n`isSuffixOf`) m [a]\nq\n\n-- \| q <==? x. if the result of q equals x, return True grouped with q\n(<==?) :: (Eq a, Functor m) => m a -> a -> m (Match a)\nm a\nq <==? :: forall a (m :: -> ).\n(Eq a, Functor m) =>\nm a -> a -> m (Match a)\n<==? a\nx = forall (f :: -> ) a b. Functor f => (a -> b) -> f a -> f b\nfmap (forall {a}. Eq a => a -> a -> Match a\n`eq` a\nx) m a\nq\nwhere\neq :: a -> a -> Match a\neq a\nq' a\nx' = forall a. Bool -> a -> Match a\nMatch (a\nq' forall a. Eq a => a -> a -> Bool\n== a\nx') a\nq'\n\n-- \| q <\\/=? x. if the result of q notequals x, return True grouped with q\n(</=?) :: (Eq a, Functor m) => m a -> a -> m (Match a)\nm a\nq </=? :: forall a (m :: -> ).\n(Eq a, Functor m) =>\nm a -> a -> m (Match a)\n</=? a\nx = forall (f :: -> ) a b. Functor f => (a -> b) -> f a -> f b\nfmap (forall {a}. Eq a => a -> a -> Match a\n`neq` a\nx) m a\nq\nwhere\nneq :: a -> a -> Match a\nneq a\nq' a\nx' = forall a. Bool -> a -> Match a\nMatch (a\nq' forall a. Eq a => a -> a -> Bool\n/= a\nx') a\nq'\n\n-- \| A helper operator for use in 'composeOne'. It takes a condition and an action;\n-- if the condition fails, it returns 'Nothing' from the 'Query' so 'composeOne' will\n-- go on and try the next rule.\n(-?>) :: (Functor m, Monad m) => m Bool -> m a -> m (Maybe a)\nm Bool\np -?> :: forall (m :: -> ) a.\nm Bool -> m a -> m (Maybe a)\n-?> m a\nf = do\nBool\nx <- m Bool\np\nif Bool\nx then forall (f :: -> ) a b. Functor f => (a -> b) -> f a -> f b\nfmap forall a. a -> Maybe a\nJust m a\nf else forall (m :: -> ) a. Monad m => a -> m a\nreturn forall a. Maybe a\nNothing\n\n-- \| A helper operator for use in 'composeAll'. It takes a condition and a function taking a grouped datum to action. If 'p' is true, it executes the resulting action.\n(-->>) :: (Monoid b, Monad m) => m (Match a) -> (a -> m b) -> m b\nm (Match a)\np -->> :: forall b (m :: -> ) a.\nm (Match a) -> (a -> m b) -> m b\n-->> a -> m b\nf = do\nMatch Bool\nb a\nm <- m (Match a)\np\nif Bool\nb then a -> m b\nf a\nm else forall (m :: -> ) a. Monad m => a -> m a\nreturn forall a. Monoid a => a\nmempty\n\n-- \| A helper operator for use in 'composeOne'. It takes a condition and a function taking a groupdatum to action. If 'p' is true, it executes the resulting action. If it fails, it returns 'Nothing' from the 'Query' so 'composeOne' will go on and try the next rule.\n(-?>>) :: (Functor m, Monad m) => m (Match a) -> (a -> m b) -> m (Maybe b)\nm (Match a)\np -?>> :: forall (m :: -> ) a b.\nm (Match a) -> (a -> m b) -> m (Maybe b)\n-?>> a -> m b\nf = do\nMatch Bool\nb a\nm <- m (Match a)\np\nif Bool\nb then forall (f :: -> ) a b. Functor f => (a -> b) -> f a -> f b\nfmap forall a. a -> Maybe a\nJust (a -> m b\nf a\nm) else forall (m :: -> ) a. Monad m => a -> m a\nreturn forall a. Maybe a\nNothing\n\n-- \| Return the current workspace\ncurrentWs :: Query WorkspaceId\ncurrentWs :: Query WorkspaceId\ncurrentWs = forall a. X a -> Query a\nliftX (forall a. (WindowSet -> X a) -> X a\nwithWindowSet forall a b. (a -> b) -> a -> b\n\\$ forall (m :: -> ) a. Monad m => a -> m a\nreturn forall b c a. (b -> c) -> (a -> b) -> a -> c\n. forall i l a s sd. StackSet i l a s sd -> i\nW.currentTag)\n\n-- \| Return the workspace tag of a window, if already managed\nwindowTag :: Query (Maybe WorkspaceId)\nwindowTag :: Query (Maybe WorkspaceId)\nwindowTag = forall r (m :: -> ). MonadReader r m => m r\nask forall (m :: -> ) a b. Monad m => m a -> (a -> m b) -> m b\n>>= \\Window\nw -> forall a. X a -> Query a\nliftX forall a b. (a -> b) -> a -> b\n\\$ forall a. (WindowSet -> X a) -> X a\nwithWindowSet forall a b. (a -> b) -> a -> b\n\\$ forall (m :: -> ) a. Monad m => a -> m a\nreturn forall b c a. (b -> c) -> (a -> b) -> a -> c\n. forall a i l s sd. Eq a => a -> StackSet i l a s sd -> Maybe i\nW.findTag Window\nw\n\n-- \| A predicate to check whether a window is a KDE system tray icon.\nisKDETrayWindow :: Query Bool\nisKDETrayWindow :: Query Bool\nisKDETrayWindow = forall r (m :: -> ). MonadReader r m => m r\nask forall (m :: -> ) a b. Monad m => m a -> (a -> m b) -> m b\n>>= \\Window\nw -> forall a. X a -> Query a\nliftX forall a b. (a -> b) -> a -> b\n\\$ do\nMaybe [CLong]\nr <- WorkspaceId -> Window -> X (Maybe [CLong])\ngetProp32s WorkspaceId\n\"_KDE_NET_WM_SYSTEM_TRAY_WINDOW_FOR\" Window\nw\nforall (m :: -> ) a. Monad m => a -> m a\nreturn forall a b. (a -> b) -> a -> b\n\\$ case Maybe [CLong]\nr of\nJust [CLong\n_] -> Bool\nTrue\nMaybe [CLong]\n_ -> Bool\nFalse\n\n-- \| Helper to check if a window property contains certain value.\nisInProperty :: String -> String -> Query Bool\nisInProperty :: WorkspaceId -> WorkspaceId -> Query Bool\nisInProperty WorkspaceId\np WorkspaceId\nv = forall r (m :: -> ). MonadReader r m => m r\nask forall (m :: -> ) a b. Monad m => m a -> (a -> m b) -> m b\n>>= \\Window\nw -> forall a. X a -> Query a\nliftX forall a b. (a -> b) -> a -> b\n\\$ do\nWindow\nva <- WorkspaceId -> X Window\ngetAtom WorkspaceId\nv\nMaybe [CLong]\nr <- WorkspaceId -> Window -> X (Maybe [CLong])\ngetProp32s WorkspaceId\np Window\nw\nforall (m :: -> ) a. Monad m => a -> m a\nreturn forall a b. (a -> b) -> a -> b\n\\$ case Maybe [CLong]\nr of\nJust [CLong]\nxs -> forall a b. (Integral a, Num b) => a -> b\nfromIntegral Window\nva forall (t :: -> ) a. (Foldable t, Eq a) => a -> t a -> Bool\n`elem` [CLong]\nxs\nMaybe [CLong]\n_ -> Bool\nFalse\n\n-- \| A predicate to check whether a window wants to fill the whole screen.\nisFullscreen :: Query Bool\nisFullscreen :: Query Bool\nisFullscreen = WorkspaceId -> WorkspaceId -> Query Bool\nisInProperty WorkspaceId\n\"_NET_WM_STATE\" WorkspaceId\n\"_NET_WM_STATE_FULLSCREEN\"\n\n-- \| A predicate to check whether a window is hidden (minimized).\nisMinimized :: Query Bool\nisMinimized :: Query Bool\nisMinimized = WorkspaceId -> WorkspaceId -> Query Bool\nisInProperty WorkspaceId\n\"_NET_WM_STATE\" WorkspaceId\n\"_NET_WM_STATE_HIDDEN\"\n\n-- \| A predicate to check whether a window is a dialog.\nisDialog :: Query Bool\nisDialog :: Query Bool\nisDialog = WorkspaceId -> WorkspaceId -> Query Bool\nisInProperty WorkspaceId\n\"_NET_WM_WINDOW_TYPE\" WorkspaceId\n\"_NET_WM_WINDOW_TYPE_DIALOG\"\n\n-- \| This function returns 'Just' the @_NET_WM_PID@ property for a\n-- particular window if set, 'Nothing' otherwise.\n--\n-- See <https://specifications.freedesktop.org/wm-spec/wm-spec-1.5.html#idm45623487788432>.\npid :: Query (Maybe ProcessID)\npid :: Query (Maybe ProcessID)\npid = forall r (m :: -> ). MonadReader r m => m r\nask forall (m :: -> ) a b. Monad m => m a -> (a -> m b) -> m b\n>>= \\Window\nw -> forall a. X a -> Query a\nliftX forall a b. (a -> b) -> a -> b\n\\$ WorkspaceId -> Window -> X (Maybe [CLong])\ngetProp32s WorkspaceId\n\"_NET_WM_PID\" Window\nw forall (f :: -> ) a b. Functor f => f a -> (a -> b) -> f b\n<&> \\case\nJust [CLong\nx] -> forall a. a -> Maybe a\nJust (forall a b. (Integral a, Num b) => a -> b\nfromIntegral CLong\nx)\nMaybe [CLong]\n_ -> forall a. Maybe a\nNothing\n\n-- \| A predicate to check whether a window is Transient.\n-- It holds the result which might be the window it is transient to\n-- or it might be 'Nothing'.\ntransientTo :: Query (Maybe Window)\ntransientTo :: Query (Maybe Window)\ntransientTo = do\nWindow\nw <- forall r (m :: -> ). MonadReader r m => m r\nDisplay\nd <- (forall a. X a -> Query a\nliftX forall b c a. (b -> c) -> (a -> b) -> a -> c\n. forall r (m :: -> ) a. MonadReader r m => (r -> a) -> m a\ndisplay\nforall (m :: -> ) a. MonadIO m => IO a -> m a\nliftIO forall a b. (a -> b) -> a -> b\n\\$ Display -> Window -> IO (Maybe Window)\ngetTransientForHint Display\nd Window\nw\n\n-- \| A convenience 'MaybeManageHook' that will check to see if a window\n-- is transient, and then move it to its parent.\ntransience :: MaybeManageHook\ntransience :: MaybeManageHook\ntransience = Query (Maybe Window)\ntransientTo forall a (m :: -> ).\n(Eq a, Functor m) =>\nm a -> a -> m (Match a)\n</=? forall a. Maybe a\nNothing forall (m :: -> ) a b.\nm (Match a) -> (a -> m b) -> m (Maybe b)\n-?>> forall b a. b -> (a -> b) -> Maybe a -> b\nmaybe forall a. Monoid a => a\nidHook Window -> Query (Endo WindowSet)\ndoShiftTo\n\n-- \| 'transience' set to a 'ManageHook'\ntransience' :: ManageHook\ntransience' :: Query (Endo WindowSet)\ntransience' = forall a (m :: -> ). (Monoid a, Functor m) => m (Maybe a) -> m a\nmaybeToDefinite MaybeManageHook\ntransience\n\n-- \| This function returns 'Just' the @WM_CLIENT_LEADER@ property for a\n-- particular window if set, 'Nothing' otherwise. Note that, generally,\n-- the window ID returned from this property (by firefox, for example)\n-- corresponds to an unmapped or unmanaged dummy window. For this to be\n-- useful in most cases, it should be used together with 'sameBy'.\n--\n-- See <https://tronche.com/gui/x/icccm/sec-5.html>.\nask forall (m :: -> ) a b. Monad m => m a -> (a -> m b) -> m b\n>>= \\Window\nw -> forall a. X a -> Query a\nliftX forall a b. (a -> b) -> a -> b\n\\$ WorkspaceId -> Window -> X (Maybe [CLong])\ngetProp32s WorkspaceId\nw forall (f :: -> ) a b. Functor f => f a -> (a -> b) -> f b\n<&> \\case\nJust [CLong\nx] -> forall a. a -> Maybe a\nJust (forall a b. (Integral a, Num b) => a -> b\nfromIntegral CLong\nx)\nMaybe [CLong]\n_ -> forall a. Maybe a\nNothing\n\n-- \| For a given window, 'sameBy' returns all windows that have a matching\n-- property (e.g. those obtained from Queries of 'clientLeader' and 'pid').\nsameBy :: Eq prop => Query (Maybe prop) -> Query [Window]\nsameBy :: forall prop. Eq prop => Query (Maybe prop) -> Query [Window]\nsameBy Query (Maybe prop)\nprop = Query (Maybe prop)\nprop forall (m :: -> ) a b. Monad m => m a -> (a -> m b) -> m b\n>>= \\case\nMaybe prop\nNothing -> forall (f :: -> ) a. Applicative f => a -> f a\npure []\nMaybe prop\npropVal -> forall r (m :: -> ). MonadReader r m => m r\nask forall (m :: -> ) a b. Monad m => m a -> (a -> m b) -> m b\n>>= \\Window\nw -> forall a. X a -> Query a\nliftX forall b c a. (b -> c) -> (a -> b) -> a -> c\n. forall a. (WindowSet -> X a) -> X a\nwithWindowSet forall a b. (a -> b) -> a -> b\n\\$ \\WindowSet\ns ->\nforall (m :: -> ) a.\nApplicative m =>\n(a -> m Bool) -> [a] -> m [a]\nfilterM (forall (f :: -> ) a b. Functor f => (a -> b) -> f a -> f b\nfmap (Maybe prop\npropVal forall a. Eq a => a -> a -> Bool\n==) forall b c a. (b -> c) -> (a -> b) -> a -> c\n. forall a. Query a -> Window -> X a\nrunQuery Query (Maybe prop)\nprop) (forall a i l s sd. Eq a => StackSet i l a s sd -> [a]\nW.allWindows WindowSet\ns forall a. Eq a => [a] -> [a] -> [a]\n\\\\ [Window\nw])\n\n-- \| 'MaybeManageHook' that moves the window to the same workspace as the\n-- first other window that has the same value of a given 'Query'. Useful\n-- Queries for this include 'clientLeader' and 'pid'.\nshiftToSame :: Eq prop => Query (Maybe prop) -> MaybeManageHook\nshiftToSame :: forall prop. Eq prop => Query (Maybe prop) -> MaybeManageHook\nshiftToSame Query (Maybe prop)\nprop = forall prop. Eq prop => Query (Maybe prop) -> Query [Window]\nsameBy Query (Maybe prop)\nprop forall a (m :: -> ).\n(Eq a, Functor m) =>\nm a -> a -> m (Match a)\n</=? [] forall (m :: -> ) a b.\nm (Match a) -> (a -> m b) -> m (Maybe b)\n-?>> forall b a. b -> (a -> b) -> Maybe a -> b\nmaybe forall a. Monoid a => a\nidHook Window -> Query (Endo WindowSet)\ndoShiftTo forall b c a. (b -> c) -> (a -> b) -> a -> c\n. forall a. [a] -> Maybe a\nlistToMaybe\n\n-- \| 'shiftToSame' set to a 'ManageHook'\nshiftToSame' :: Eq prop => Query (Maybe prop) -> ManageHook\nshiftToSame' :: forall prop.\nEq prop =>\nQuery (Maybe prop) -> Query (Endo WindowSet)\nshiftToSame' = forall a (m :: -> ). (Monoid a, Functor m) => m (Maybe a) -> m a\nmaybeToDefinite forall b c a. (b -> c) -> (a -> b) -> a -> c\n. forall prop. Eq prop => Query (Maybe prop) -> MaybeManageHook\nshiftToSame\n\n-- \| converts 'MaybeManageHook's to 'ManageHook's\nmaybeToDefinite :: (Monoid a, Functor m) => m (Maybe a) -> m a\nmaybeToDefinite :: forall a (m :: -> ). (Monoid a, Functor m) => m (Maybe a) -> m a\nmaybeToDefinite = forall (f :: -> ) a b. Functor f => (a -> b) -> f a -> f b\nfmap (forall a. a -> Maybe a -> a\nfromMaybe forall a. Monoid a => a\nmempty)\n\n-- \| Move the window to the same workspace as another window.\ndoShiftTo :: Window -> ManageHook\ndoShiftTo :: Window -> Query (Endo WindowSet)\ndoShiftTo Window\ntarget = forall s. (s -> s) -> Query (Endo s)\ndoF forall b c a. (b -> c) -> (a -> b) -> a -> c\n. forall {s} {i} {l} {sd}.\n(Eq s, Eq i) =>\nWindow -> StackSet i l Window s sd -> StackSet i l Window s sd\nshiftTo forall (m :: -> ) a b. Monad m => (a -> m b) -> m a -> m b\n=<< forall r (m :: -> ). MonadReader r m => m r\nwhere shiftTo :: Window -> StackSet i l Window s sd -> StackSet i l Window s sd\nshiftTo Window\nw StackSet i l Window s sd\ns = forall b a. b -> (a -> b) -> Maybe a -> b\nmaybe StackSet i l Window s sd\ns (\\i\nt -> forall a s i l sd.\n(Ord a, Eq s, Eq i) =>\ni -> a -> StackSet i l a s sd -> StackSet i l a s sd\nW.shiftWin i\nt Window\nw StackSet i l Window s sd\ns) (forall a i l s sd. Eq a => a -> StackSet i l a s sd -> Maybe i\nW.findTag Window\ntarget StackSet i l Window s sd\ns)\n\n-- \| Floats the new window in the given rectangle.\ndoRectFloat :: W.RationalRect -- ^ The rectangle to float the window in. 0 to 1; x, y, w, h.\n-> ManageHook\ndoRectFloat :: RationalRect -> Query (Endo WindowSet)\ndoRectFloat RationalRect\nr = forall r (m :: -> ). MonadReader r m => m r\nask forall (m :: -> ) a b. Monad m => m a -> (a -> m b) -> m b\n>>= \\Window\nw -> forall s. (s -> s) -> Query (Endo s)\ndoF (forall a i l s sd.\nOrd a =>\na -> RationalRect -> StackSet i l a s sd -> StackSet i l a s sd\nW.float Window\nw RationalRect\nr)\n\n-- \| Floats the window and makes it use the whole screen. Equivalent to\n-- @'doRectFloat' \\$ 'W.RationalRect' 0 0 1 1@.\ndoFullFloat :: ManageHook\ndoFullFloat :: Query (Endo WindowSet)\ndoFullFloat = RationalRect -> Query (Endo WindowSet)\ndoRectFloat forall a b. (a -> b) -> a -> b\n\\$ Rational -> Rational -> Rational -> Rational -> RationalRect\nW.RationalRect Rational\n0 Rational\n0 Rational\n1 Rational\n1\n\n-- \| Floats a new window using a rectangle computed as a function of\n-- the rectangle that it would have used by default.\ndoFloatDep :: (W.RationalRect -> W.RationalRect) -> ManageHook\ndoFloatDep :: (RationalRect -> RationalRect) -> Query (Endo WindowSet)\ndoFloatDep RationalRect -> RationalRect\nmove = forall r (m :: -> ). MonadReader r m => m r\nask forall (m :: -> ) a b. Monad m => m a -> (a -> m b) -> m b\n>>= \\Window\nw -> forall s. (s -> s) -> Query (Endo s)\ndoF forall b c a. (b -> c) -> (a -> b) -> a -> c\n. forall a i l s sd.\nOrd a =>\na -> RationalRect -> StackSet i l a s sd -> StackSet i l a s sd\nW.float Window\nw forall b c a. (b -> c) -> (a -> b) -> a -> c\n. RationalRect -> RationalRect\nmove forall b c a. (b -> c) -> (a -> b) -> a -> c\n. forall a b. (a, b) -> b\nsnd forall (m :: -> ) a b. Monad m => (a -> m b) -> m a -> m b\n=<< forall a. X a -> Query a\nliftX (Window -> X (ScreenId, RationalRect)\nfloatLocation Window\nw)\n\n-- \| Floats a new window with its original size, and its top left\n-- corner at a specific point on the screen (both coordinates should\n-- be in the range 0 to 1).\ndoFloatAt :: Rational -> Rational -> ManageHook\ndoFloatAt :: Rational -> Rational -> Query (Endo WindowSet)\ndoFloatAt Rational\nx Rational\ny = (RationalRect -> RationalRect) -> Query (Endo WindowSet)\ndoFloatDep RationalRect -> RationalRect\nmove\nwhere\nmove :: RationalRect -> RationalRect\nmove (W.RationalRect Rational\n_ Rational\n_ Rational\nw Rational\nh) = Rational -> Rational -> Rational -> Rational -> RationalRect\nW.RationalRect Rational\nx Rational\ny Rational\nw Rational\nh\n\n-- \| Floats a new window with its original size on the specified side of a\n-- screen\ndoSideFloat :: Side -> ManageHook\ndoSideFloat :: Side -> Query (Endo WindowSet)\ndoSideFloat Side\nside = (RationalRect -> RationalRect) -> Query (Endo WindowSet)\ndoFloatDep RationalRect -> RationalRect\nmove\nwhere\nmove :: RationalRect -> RationalRect\nmove (W.RationalRect Rational\n_ Rational\n_ Rational\nw Rational\nh) = Rational -> Rational -> Rational -> Rational -> RationalRect\nW.RationalRect Rational\ncx Rational\ncy Rational\nw Rational\nh\nwhere cx :: Rational\ncx\n\| Side\nside forall (t :: -> ) a. (Foldable t, Eq a) => a -> t a -> Bool\n`elem` [Side\nSC,Side\nC ,Side\nNC] = (Rational\n1forall a. Num a => a -> a -> a\n-Rational\nw)forall a. Fractional a => a -> a -> a\n/Rational\n2\n\| Side\nside forall (t :: -> ) a. (Foldable t, Eq a) => a -> t a -> Bool\n`elem` [Side\nSW,Side\nCW,Side\nNW] = Rational\n0\n\| Bool\notherwise = {- side `elem` [SE,CE,NE] -} Rational\n1forall a. Num a => a -> a -> a\n-Rational\nw\ncy :: Rational\ncy\n\| Side\nside forall (t :: -> ) a. (Foldable t, Eq a) => a -> t a -> Bool\n`elem` [Side\nCE,Side\nC ,Side\nCW] = (Rational\n1forall a. Num a => a -> a -> a\n-Rational\nh)forall a. Fractional a => a -> a -> a\n/Rational\n2\n\| Side\nside forall (t :: -> ) a. (Foldable t, Eq a) => a -> t a -> Bool\n`elem` [Side\nNE,Side\nNC,Side\nNW] = Rational\n0\n\| Bool\notherwise = {- side `elem` [SE,SC,SW] -} Rational\n1forall a. Num a => a -> a -> a\n-Rational\nh\n\n-- \| Floats a new window with its original size, but centered.\ndoCenterFloat :: ManageHook\ndoCenterFloat :: Query (Endo WindowSet)\ndoCenterFloat = Side -> Query (Endo WindowSet)\ndoSideFloat Side\nC\n\n-- \| Hides window and ignores it.\ndoHideIgnore :: ManageHook\ndoHideIgnore :: Query (Endo WindowSet)\ndoHideIgnore = forall r (m :: -> ). MonadReader r m => m r\nask forall (m :: -> ) a b. Monad m => m a -> (a -> m b) -> m b\n>>= \\Window\nw -> forall a. X a -> Query a\nliftX (Window -> X ()\nhide Window\nw) forall (m :: -> ) a b. Monad m => m a -> m b -> m b\n>> forall s. (s -> s) -> Query (Endo s)\ndoF (forall a i l s sd.\nOrd a =>\na -> StackSet i l a s sd -> StackSet i l a s sd\nW.delete Window\nw)\n\n-- \| Sinks a window\ndoSink :: ManageHook\ndoSink :: Query (Endo WindowSet)\ndoSink = forall s. (s -> s) -> Query (Endo s)\ndoF forall b c a. (b -> c) -> (a -> b) -> a -> c\n. forall a i l s sd.\nOrd a =>\na -> StackSet i l a s sd -> StackSet i l a s sd\nW.sink forall (m :: -> ) a b. Monad m => (a -> m b) -> m a -> m b\n=<< forall r (m :: -> ). MonadReader r m => m r\n\n-- \| Lower an unmanaged window. Useful together with 'doIgnore' to lower\n-- special windows that for some reason don't do it themselves.\ndoLower :: ManageHook\ndoLower :: Query (Endo WindowSet)\ndoLower = forall r (m :: -> ). MonadReader r m => m r\nask forall (m :: -> ) a b. Monad m => m a -> (a -> m b) -> m b\n>>= \\Window\nw -> forall a. X a -> Query a\nliftX forall a b. (a -> b) -> a -> b\n\\$ forall a. (Display -> X a) -> X a\nwithDisplay forall a b. (a -> b) -> a -> b\n\\$ \\Display\ndpy -> forall (m :: -> ) a. MonadIO m => IO a -> m a\nio (Display -> Window -> IO ()\nlowerWindow Display\ndpy Window\nw) forall (m :: -> ) a b. Monad m => m a -> m b -> m b\n>> forall a. Monoid a => a\nmempty\n\n-- \| Raise an unmanaged window. Useful together with 'doIgnore' to raise\n-- special windows that for some reason don't do it themselves.\ndoRaise :: ManageHook\ndoRaise :: Query (Endo WindowSet)\ndoRaise = forall r (m :: -> ). MonadReader r m => m r\nask forall (m :: -> ) a b. Monad m => m a -> (a -> m b) -> m b\n>>= \\Window\nw -> forall a. X a -> Query a\nliftX forall a b. (a -> b) -> a -> b\n\\$ forall a. (Display -> X a) -> X a\nwithDisplay forall a b. (a -> b) -> a -> b\n\\$ \\Display\ndpy -> forall (m :: -> ) a. MonadIO m => IO a -> m a\nio (Display -> Window -> IO ()\nraiseWindow Display\ndpy Window\nw) forall (m :: -> ) a b. Monad m => m a -> m b -> m b\n>> forall a. Monoid a => a\nmempty\n\n-- \| Focus a window (useful in 'XMonad.Hooks.EwmhDesktops.setActivateHook').\ndoFocus :: ManageHook\ndoFocus :: Query (Endo WindowSet)\ndoFocus = forall s. (s -> s) -> Query (Endo s)\ndoF forall b c a. (b -> c) -> (a -> b) -> a -> c\n. forall s a i l sd.\n(Eq s, Eq a, Eq i) =>\na -> StackSet i l a s sd -> StackSet i l a s sd\nW.focusWindow forall (m :: -> ) a b. Monad m => (a -> m b) -> m a -> m b\n=<< forall r (m :: -> *). MonadReader r m => m r" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.54753613,"math_prob":0.94114995,"size":19609,"snap":"2023-40-2023-50","text_gpt3_token_len":6988,"char_repetition_ratio":0.25763837,"word_repetition_ratio":0.6666667,"special_character_ratio":0.43066958,"punctuation_ratio":0.18616226,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.996483,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-08T00:13:39Z\",\"WARC-Record-ID\":\"<urn:uuid:214193ab-3a6a-4bca-a500-458171c85dbd>\",\"Content-Length\":\"166718\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:207df7e1-34b2-4042-b080-15af6ccadc34>\",\"WARC-Concurrent-To\":\"<urn:uuid:2e1560cc-933e-4274-a0d9-1852c12d0e36>\",\"WARC-IP-Address\":\"185.199.109.153\",\"WARC-Target-URI\":\"https://xmonad.github.io/xmonad-docs/xmonad-contrib/src/XMonad.Hooks.ManageHelpers.html\",\"WARC-Payload-Digest\":\"sha1:3VIYEKVVJ54RDK5SVONZFKMVHQNWQWO3\",\"WARC-Block-Digest\":\"sha1:NGBZVLCRFHBBPN2PQB5NLOY3BWAPSOPL\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100705.19_warc_CC-MAIN-20231207221604-20231208011604-00742.warc.gz\"}"}
https://studysoup.com/tsg/2166/physics-principles-with-applications-6-edition-chapter-11-problem-12pe	[ "×\nGet Full Access to Physics: Principles With Applications - 6 Edition - Chapter 11 - Problem 12pe\nGet Full Access to Physics: Principles With Applications - 6 Edition - Chapter 11 - Problem 12pe\n\n×\n\n# The pressure exerted by a phonograph needle on a record is", null, "ISBN: 9780130606204 3\n\n## Solution for problem 12PE Chapter 11\n\nPhysics: Principles with Applications \| 6th Edition\n\n• Textbook Solutions\n• 2901 Step-by-step solutions solved by professors and subject experts\n• Get 24/7 help from StudySoup virtual teaching assistants", null, "Physics: Principles with Applications \| 6th Edition\n\n4 5 1 378 Reviews\n20\n1\nProblem 12PE\n\nThe pressure exerted by a phonograph needle on a record is surprisingly large. If the equivalent of 1.00 g is supported by a needle, the tip of which is a circle 0.200 mm in radius, what pressure is exerted on the record in N/m2 ?\n\nStep-by-Step Solution:\n\nStep-by-step solution Step 1 of 6 Force per unit area is known as pressure. Apply this concept to find the pressure of force that act on an area.\n\nStep 2 of 6\n\nStep 3 of 6\n\n##### ISBN: 9780130606204\n\nThe full step-by-step solution to problem: 12PE from chapter: 11 was answered by , our top Physics solution expert on 03/03/17, 03:53PM. Since the solution to 12PE from 11 chapter was answered, more than 713 students have viewed the full step-by-step answer. This textbook survival guide was created for the textbook: Physics: Principles with Applications, edition: 6. This full solution covers the following key subjects: pressure, Record, exerted, needle, circle. This expansive textbook survival guide covers 35 chapters, and 3914 solutions. The answer to “The pressure exerted by a phonograph needle on a record is surprisingly large. If the equivalent of 1.00 g is supported by a needle, the tip of which is a circle 0.200 mm in radius, what pressure is exerted on the record in N/m2 ?” is broken down into a number of easy to follow steps, and 45 words. Physics: Principles with Applications was written by and is associated to the ISBN: 9780130606204.\n\n## Discover and learn what students are asking\n\nCalculus: Early Transcendental Functions : Differential Equations: Separation of Variables\n?In Exercises 1-14, find the general solution of the differential equation. $$x^{2}+5 y \\frac{d y}{d x}=0$$\n\nUnlock Textbook Solution\n\nEnter your email below to unlock your verified solution to:\n\nThe pressure exerted by a phonograph needle on a record is" ]	[ null, "https://studysoup.com/cdn/62cover_2418946", null, "https://studysoup.com/cdn/62cover_2418946", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.95380366,"math_prob":0.7599002,"size":958,"snap":"2022-40-2023-06","text_gpt3_token_len":228,"char_repetition_ratio":0.10587002,"word_repetition_ratio":0.0,"special_character_ratio":0.25887266,"punctuation_ratio":0.16243654,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95481074,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-03T04:56:36Z\",\"WARC-Record-ID\":\"<urn:uuid:02c8c6e3-5886-40a0-a24e-90669437c79b>\",\"Content-Length\":\"89464\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:999ddf3a-9439-40b1-b17f-a7e53a287ede>\",\"WARC-Concurrent-To\":\"<urn:uuid:4f50efed-4a93-477a-a693-dac0fd369a09>\",\"WARC-IP-Address\":\"54.189.254.180\",\"WARC-Target-URI\":\"https://studysoup.com/tsg/2166/physics-principles-with-applications-6-edition-chapter-11-problem-12pe\",\"WARC-Payload-Digest\":\"sha1:34LETPTKG5Q77NENMA2MWARLRY5PCV57\",\"WARC-Block-Digest\":\"sha1:2KJRQNIL2INO22GHP6XAW2BAB62AWJT6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337398.52_warc_CC-MAIN-20221003035124-20221003065124-00693.warc.gz\"}"}
https://docs.scipy.org/doc/numpy-1.15.1/reference/generated/numpy.unique.html	[ "# numpy.unique¶\n\n`numpy.``unique`(ar, return_index=False, return_inverse=False, return_counts=False, axis=None)[source]\n\nFind the unique elements of an array.\n\nReturns the sorted unique elements of an array. There are three optional outputs in addition to the unique elements:\n\n• the indices of the input array that give the unique values\n• the indices of the unique array that reconstruct the input array\n• the number of times each unique value comes up in the input array\nParameters: ar : array_like Input array. Unless axis is specified, this will be flattened if it is not already 1-D. return_index : bool, optional If True, also return the indices of ar (along the specified axis, if provided, or in the flattened array) that result in the unique array. return_inverse : bool, optional If True, also return the indices of the unique array (for the specified axis, if provided) that can be used to reconstruct ar. return_counts : bool, optional If True, also return the number of times each unique item appears in ar. New in version 1.9.0. axis : int or None, optional The axis to operate on. If None, ar will be flattened. If an integer, the subarrays indexed by the given axis will be flattened and treated as the elements of a 1-D array with the dimension of the given axis, see the notes for more details. Object arrays or structured arrays that contain objects are not supported if the axis kwarg is used. The default is None. New in version 1.13.0. unique : ndarray The sorted unique values. unique_indices : ndarray, optional The indices of the first occurrences of the unique values in the original array. Only provided if return_index is True. unique_inverse : ndarray, optional The indices to reconstruct the original array from the unique array. Only provided if return_inverse is True. unique_counts : ndarray, optional The number of times each of the unique values comes up in the original array. Only provided if return_counts is True. New in version 1.9.0.\n\n`numpy.lib.arraysetops`\nModule with a number of other functions for performing set operations on arrays.\n\nNotes\n\nWhen an axis is specified the subarrays indexed by the axis are sorted. This is done by making the specified axis the first dimension of the array and then flattening the subarrays in C order. The flattened subarrays are then viewed as a structured type with each element given a label, with the effect that we end up with a 1-D array of structured types that can be treated in the same way as any other 1-D array. The result is that the flattened subarrays are sorted in lexicographic order starting with the first element.\n\nExamples\n\n```>>> np.unique([1, 1, 2, 2, 3, 3])\narray([1, 2, 3])\n>>> a = np.array([[1, 1], [2, 3]])\n>>> np.unique(a)\narray([1, 2, 3])\n```\n\nReturn the unique rows of a 2D array\n\n```>>> a = np.array([[1, 0, 0], [1, 0, 0], [2, 3, 4]])\n>>> np.unique(a, axis=0)\narray([[1, 0, 0], [2, 3, 4]])\n```\n\nReturn the indices of the original array that give the unique values:\n\n```>>> a = np.array(['a', 'b', 'b', 'c', 'a'])\n>>> u, indices = np.unique(a, return_index=True)\n>>> u\narray(['a', 'b', 'c'],\ndtype='\|S1')\n>>> indices\narray([0, 1, 3])\n>>> a[indices]\narray(['a', 'b', 'c'],\ndtype='\|S1')\n```\n\nReconstruct the input array from the unique values:\n\n```>>> a = np.array([1, 2, 6, 4, 2, 3, 2])\n>>> u, indices = np.unique(a, return_inverse=True)\n>>> u\narray([1, 2, 3, 4, 6])\n>>> indices\narray([0, 1, 4, 3, 1, 2, 1])\n>>> u[indices]\narray([1, 2, 6, 4, 2, 3, 2])\n```\n\nnumpy.trim_zeros\n\nnumpy.flip" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6467163,"math_prob":0.93935406,"size":3456,"snap":"2022-40-2023-06","text_gpt3_token_len":936,"char_repetition_ratio":0.1691773,"word_repetition_ratio":0.08798646,"special_character_ratio":0.29774305,"punctuation_ratio":0.19414894,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98060423,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-01T08:45:38Z\",\"WARC-Record-ID\":\"<urn:uuid:4fe896ef-0518-4a68-b4aa-7c2c6605b607>\",\"Content-Length\":\"16412\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cc57504a-0f5a-479b-bd5b-1a0ca5caaed6>\",\"WARC-Concurrent-To\":\"<urn:uuid:f408997c-614f-4fb4-9ca9-707e6f233282>\",\"WARC-IP-Address\":\"50.17.248.72\",\"WARC-Target-URI\":\"https://docs.scipy.org/doc/numpy-1.15.1/reference/generated/numpy.unique.html\",\"WARC-Payload-Digest\":\"sha1:MGVE5HYASZ6RKCCYNGLCQYK3PCRUJ3N6\",\"WARC-Block-Digest\":\"sha1:TQPXNTFYRKTQTXA6NMQB22NXFO5ULNTC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499919.70_warc_CC-MAIN-20230201081311-20230201111311-00057.warc.gz\"}"}
https://iwaponline.com/jwcc/article/10/1/30/38986/A-mathematical-hypothesis-to-research-the-effects	[ "## Abstract\n\nPrevious research studies focused only on data of local air temperature and humidity, ignoring the water body itself, which cannot definitively answer the question of how the Three Gorges Reservoir's (TGR) water affects the local climate overall. To understand the effect of the TGR on the local climate quantitatively, this paper provides an original mathematical hypothesis and proves in theory there is only one way to calculate the transfer of heat and humidity between the TGR and the local air. Based on this mathematical hypothesis, a detailed research method to explore the effects of the TGR's heat and humidity on local climate was formed. A field investigation was conducted and a research site was selected in Chongqing. This study has determined the effects of the TGR's heating or cooling on the air during the measuring period. A mathematical model to assess the effects of heat and humidity from the TGR on local climate was set up. The final results based on the mathematical model show that the average air temperature decreased 0.67 K and the average moisture content increased 0.25 g/kg during the 24 hours measuring time for the area studied.\n\n## INTRODUCTION\n\nThe Three Gorges Project (TGP) on the Yangtze River is the world's largest hydroelectric power project. By 2010, when the TGP became fully operational, the water storage capacity of the Three Gorges Reservoir (TGR) was 3.93 × 1010 m3 with a normal pool level of 175 m, about 4.5% of the Yangtze's annual discharge (Xu & Milliman 2009). After water collected within the TGR, there was a significant increase in the water surface due to the effect of backwater, leading to a decrease in water flow speed in the reservoir (Yu 2011). Since the TGR began filling with water, some environmental consequences emerged, capturing the attention of researchers and environmental activists around the world. Contentious environmental issues surrounding the TGP have centered on water quality (Hu et al. 2014; Yue &Yuan 2016), fish population (Yi & Yang 2010; Wang & Bi 2016), sedimentation and downstream riverbed erosion (Jia & Shao 2010; Yang & Milliman 2014), geological instability (Peng & Niu 2014; Wang & Chen 2014), carrying capacity of the environment in the reservoir area (Li & Zhang 2016; Wu & Wang 2016) and the situation of the local climate.\n\nAfter water collected, as a typical river way reservoir, the TGR has an influence on the local climate. The TGR's effect on the local climate has attracted the interest of international scholars (Dai 2008; Chen & Zhang 2009a,2009b). Many comparative analyses and regional climate models have been discussed (Wu et al. 2006). Zhang et al. (2004) simulated the impact of the TGR on the local climate by using a simplified model, and the results showed the influence of the TGR on the climate was limited to 10 km. Gao et al. (2003, 2007) analyzed the changes in land use and land cover that influenced local and regional climates and said such effects could be studied with regional climate models. More specifically, by using the Penn State/NCAR MM5, Miller et al. (2005) conducted multiple layered experiments down to 10 km resolution for 8 weeks duration to simulate the effects of the TGR. The results showed that as a potential evaporating surface, the TGR could cool the surrounding area. Wu et al. (2006) used the regional climate model to simulate the effects of the TGR on the climate of the surrounding areas. The results showed the width and coverage of the TGR did not have significant influence on the local climate, except for cooling over the TGR water body in both June–August (summer) and December–February (winter). However, Sun & Qin (2002) analyzed the climatic features in the Three Gorges dam area. The results showed that warmer winter temperatures in the area were evident. It also suggested the warmer winter temperatures could possibly have some relationship with the water storage of the TGR. Yang & Chen (2002) used meteorological data from five local weather stations to study the climate characteristics in the Three Gorges dam area. The results also showed there were warmer winter temperatures in the dam area. Chen & Zhang (2009a, 2009b) used the temperature observation data of 33 weather stations in the TGR area during the period from 1961 to 2006 to analyze the climate characteristics of temperature changes since the water storage of the TGR began. The results showed that the expanding water surface area of the reservoir seemed to have an effect of increasing temperature during the winter and an effect of decreasing temperature during the summer.\n\nDue to the short time after the complete filling of the TGR (filling to 175 m normal pool mark) in 2010 and the limited availability of monitoring data, there may possibly be some discrepancies in research results for local climate data. Moreover, the cross-effects of ground, vegetation, buildings, and the possible impact of the TGR on the local climate made it difficult to distinguish the impact of the TGR itself (Xu & Tan 2013). Due to the complexity of the problem, different scholars have different opinions about the local climate effects of the TGR. Through the analysis of the references, it can be concluded that previous research studies only focused on air temperature and humidity changes, ignoring the water body itself. This raised the difficult question of why no one could, with certainty, distinguish the TGR water body's effect on the local climate from an integrated effect, because the TGR is only one of many factors effecting local climate. Based on the above analysis, this paper provides an original mathematical hypothesis and a detailed research methodology based on the hypothesis to explore the effects of the TGR's heat and humidity on the local climate.\n\n## MATHEMATICAL HYPOTHESIS FOR THE EFFECTS OF TGR'S HEAT AND HUMIDITY ON THE AIR\n\nThis paper provides the mathematical hypothesis named ‘sheep counting’, and uses the solution of the mathematical hypothesis to answer the question mentioned above, which is to distinguish the TGR water's effect on the local climate from the integrated effect with certainty. The mathematical hypothesis has proved in theory that there is only one way to analyze the effect of heat and humidity from the TGR on the local climate. The detailed description of the mathematical hypothesis is shown in Figure 1.\n\n### Description of the mathematical hypothesis\n\n• 1.\n\nThere are an infinite number of sheep on the grassland", null, ", the sheep on the grassland are located outside of the sheepfolds.\n\n• 2.\n\nThere are multiple sheepfolds, and there are an infinite number of sheep in each sheepfold", null, "", null, ", n is the number of sheepfolds.\n\n• 3.\n\nThe sheep on the grassland and in the sheepfolds are identical and there are no individual differences.\n\n• 4.\n\nEach sheepfold has only one gate. All the sheep in or out of the sheepfold must go through the gate, and the number of sheep in or out of the sheepfold is unknown", null, ".\n\n### Question\n\nIn a cycle of 24 hours, what is the effect of a given sheepfold on the numerical change of sheep on the grassland?\n\n### Solutions\n\nStep (1): Obviously, it is not reasonable to consider the numerical change of sheep on the grassland itself, because the number of sheep on the grassland is infinite", null, ", and it is impossible to know the base of the infinite number, and it is impossible to know the infinite number after a change, so it is impossible to determine the numerical change by considering the sheep change on the grassland. In addition, the numerical change of sheep on the grassland is not only from one given sheepfold", null, ", but also from other sheepfolds, and the number of sheep in or out of each sheepfold is unknown (", null, ",", null, ",", null, "is the given number among", null, "), therefore it is impossible to solve the mathematical hypothesis by considering the numerical change of sheep on the grassland itself.\n\nStep (2): It is not feasible to consider the numerical change of sheep from a given sheepfold, because the number of sheep in a given sheepfold is also infinite (", null, ",", null, "is the given number among", null, "), and the reason why it cannot be solved is the same as step (1).\n\nStep (3): Finally, the only solution of the mathematical hypothesis is the number of sheep in or out of the given sheepfold (", null, ",", null, "is the given number among", null, "). If the number", null, "is determined, the mathematical hypothesis can be solved, and", null, "is the only and final solution for the mathematical hypothesis.\n\nIn the mathematical hypothesis of ‘sheep counting’, the sheep on the grassland represent the heat or humidity in the local air; the sheep in the sheepfolds represent the heat or humidity from ground, vegetation, buildings, the TGR's water, etc., which affect the local climate. The sheep in or out of sheepfold through the gate represents the heat or humidity transfer between the local air and ground, vegetation, buildings, the TGR's water, etc.\n\nThe mathematical hypothesis has proved in theory there is only one way to analyze the effects of heat and humidity from the TGR on the local climate for the first time. The heat and humidity of the local air (sheep on the grassland) are influenced by various factors (sheep in or out of the sheepfolds), such as ground, vegetation, the TGR's water, and buildings, etc. Therefore, according to the above theoretical analysis, if we want to quantitatively analyze the effects of heat and humidity from the TGR's water on the local air, we must take the TGR's ‘water–air’ interface (gate of the sheepfold) as the research object, and analyze the quantity of heat and humidity in or out of the ‘water–air’ interface hour by hour (sheep in or out of the sheepfold through the gate), and then calculate the sum of heat and humidity for a 24-hour cycle. Finally, we can determine the quantitative data of heat and humidity and make the conclusions of cooling or heating, drying or wetting effects of the TGR's water on the local climate.\n\n## RESEARCH METHOD\n\n### Heat and moisture transfer on the interface of ‘water–air’\n\nHeat transfer between water and air includes sensible heat and latent heat, which are caused by temperature differences and moisture pressure differences respectively. Figure 2 depicts the heat and moisture transfers between the TGR's water and local air. The sensible heat is mainly determined by the temperature difference of water surface and local air, and latent heat mainly depends on the differences of water surface saturated vapor pressure and vapor pressure in the air. The sensible heat is directly related to water and air temperature. Moisture transfer is an important source of changing air humidity.\n\nIn addition to the exchange of sensible heat and latent heat between the TGR's water and local air, the process of energy exchange on the ‘water–air’ interface also includes the long-wave radiation between the TGR's water and the local atmosphere. The diagram of each part of the energy is shown in Figure 3.\n\nThe total heat flux of ‘water–air’ interface can be expressed by heat flux from water to air", null, "and from the air to water", null, ". The net heat flux is", null, ": when", null, "is from water to the air, it is positive, otherwise negative. When H is from air to water, it is positive, otherwise negative. Through the comparison of", null, "and", null, ", the effects of cooling or heating can be precisely determined. When", null, ", the TGR's water has a heating effect on the local air; when", null, ", the TGR's water has a cooling effect on the local air. So, if", null, ", it means the net heat flux is from the air to water. If", null, ", it means the net heat flux is from water to air. The specific items are explained as follows:", null, "is the long-wave radiation from the atmosphere that is projected to the water surface W/m2 (Weng & Sun 1993);", null, "is the air temperature, K; e is the vapor pressure,", null, "; n if it is sunny,", null, "; if it is cloudy,", null, ";", null, "is the reflected long-wave radiation from the water surface,", null, ", W/m2;", null, "is the reflection coefficient of the water surface: where", null, "is the long-wave radiation of the water surface, W/m2;", null, "is the long-wave emissivity;", null, "is the Stefan–Boltzmann constant;", null, "is the water surface temperature, K.\nThe sensible heat", null, ", latent heat", null, "and moisture transfer", null, "between the TGR's water and the local air can be calculated by bulk method (Kondo 1992): where H is the sensible heat flux of ‘water–air’ interface, W/m2; E is the moisture flux of ‘water–air’ interface, g/(m2.s);", null, "is the latent heat flux of ‘water–air’ interface, W/m2;", null, "is air density, kg/m3 (Zhao 2008);", null, "is air specific heat, J/kg.°C;", null, "is wind speed, m/s;", null, ",", null, "are bulk parameters for sensible heat and latent heat: where l is latent heat,", null, ", kJ/kg (Hannah et al. 2004);", null, "is saturated moisture content under the condition of temperature", null, ", g/kg (Lian 2006);", null, "is moisture content over the water surface, g/kg, which can be calculated by the following formula: where", null, "is saturated moisture pressure under the condition of temperature", null, ",", null, ";", null, "is air relative humidity, %; p is moisture pressure,", null, ";", null, "is local atmospheric pressure,", null, "(www.weather.com.cn/).\n\n## FIELD MEASUREMENTS\n\n### Measuring area selection\n\nThe measuring area was selected in Chongqing, one of the municipalities governed by the Chinese central government, which is located in the upper region of the TGR (see Figure 4). The measuring area chosen is relatively open and flat to reduce the influence of the nearby topography. Because the measurement of heat and moisture transfer on ‘water–air’ interface is a complex process, it is influenced by the factors of long-wave radiation, water temperature, air temperature, air vapor, and wind speed over the water, etc.\n\n### Measurement process\n\nThe measurements were carried out on 22–23 June 2015. To analyze the effects of heating or cooling, drying or wetting of the TGR's water on the local air, the measuring period needed to be at least 24 hours, which is the shortest possible measuring cycle. For the measurement in this paper, the corresponding measuring cycle was 24 hours and the time interval for the measurements was 1 hour. The measured parameters included: water temperature, solar radiation, air temperature, air relative humidity and wind speed, etc. The local atmospheric pressure was obtained from the Central Weather Bureau. By utilizing the above mathematical hypothesis and research method, the sensible heat, latent heat, long-wave radiation and moisture transfer between the TGR's water and the local air could be analyzed. The instruments used in the field measurements included an infrared radiation thermometer, an environmental supervision instrument, a GPS apparatus, etc. The type of infrared radiation thermometer was a MS6530, which is used to measure water temperature, with a measurement range of −32–535 °C, an accuracy of ±0.5% and a response time of 0.5 second. The type of environmental supervision instrument was a KANOMAX6531. The instrument was used to measure air temperature, relative humidity, and wind speed. For air temperature, the measurement range is −20–70 °C, and the accuracy is ±0.5 °C. For relative humidity, the measurement range is 2–98%, and the accuracy is ±2%. For wind speed, the measurement range is 0.01–30 m/s, and the accuracy is ±2%. The type of GPS apparatus is a N600, which was used to record the altitude at different locations.\n\n## RESULTS AND DISCUSSION\n\n### Effects of heat and humidity on the ‘water–air’ interface\n\nThe measuring period was 24 consecutive hours on 22–23 June 2015. Figure 5 shows the flux curve of sensible heat and latent heat. The transfer direction of heat flux was different during the entire measuring period. The sensible heat flux is always from air to water. However, the latent heat flux can be from water to air or from air to water depending on time. If the entire heat flux during the measuring cycle were totaled, the results would provide the exchange of sensible heat flux on the ‘water–air’ interface. The data for sensible heat flux was 403 KJ/m2.d and the data for latent heat flux was 81 KJ/m2.d. According to the previous definition of transfer direction, the transfer direction of sensible heat and latent heat were opposite. Through the comparison of sensible heat and latent heat, the heat transfer direction was from water to air during the measuring period and the sum data of heat flux was 322 KJ/m2.d.\n\nFigure 6 shows the curves of long-wave radiation. It can be concluded that the fluctuation of long-wave radiation emitted from the water surface was small, which is related to the stable characteristics of the water temperature. The range of long-wave radiation emitted from the water surface was 444.6–453 W/m2, with the largest gap of 8.4 W/m2, with a mean value of 447.3 W/m2. The range of long-wave radiation from the atmosphere was 456.4–500.3 W/m2, with the largest gap of 43.8 W/m2, with a mean value of 473.9 W/m2.\n\nConsidering the sensible heat, latent heat and long-wave radiation, the final heat transfer can be analyzed on the ‘water–air’ interface. Figure 7 shows the curves of the final heat flux from water to air", null, "and the final heat flux from air to water", null, "on the ‘water–air’ interface. From the curves of final heat flux during this measuring cycle, the range of", null, "was 328.7–349.8 W/m2 with a mean value of 337.3 W/m2; and the range of", null, "was 424.8–472.6 W/m2 with a mean value of 445.6 W/m2. In this paper, the shortest cycle of 24 hours was used as the measuring cycle, so the sum of heat flux in the 24 hour cycle was the final net heat flux:\n\nTherefore, the conclusion reached was the final heat flux from water to air was smaller than that from air to water during this 24 hour cycle, which means the TGR's water had a cooling effect on the air.\n\nSimilarly, it was necessary to analyze the moisture transfer, which occurred on the ‘water–air’ interface. From Figure 8, we found the moisture content of saturated air near the water surface", null, "crossed with the moisture content in the air", null, ", which means the moisture transfer can be from water to air and also can be from air to water. The range of", null, "was 22.9–25.2 g/Kg with a mean value of 23.7 g/Kg. The range of", null, "was 21.74–25.11 g/Kg with a mean value of 23.36 g/Kg. The range of moisture transfer was 0.374–9.88 g/m2 · h with a mean value of 1.42 g/m2 · h. By analyzing the above, the conclusion was that the moisture transfer direction was from water to the air during this measuring cycle. In other words, the TGR's water had a wetting effect on the local air and the final quantity of moisture transfer was: ΣD = 34 g/m2 · d.\n\nFrom the data of heat and moisture flux on the ‘water–air’ interface, this paper, for the first time, distinguished the TGR's effect on the local climate from the integrated effect with certainty.\n\n### Assessment of heat and humidity from the TGR on local climate\n\n#### Heat and moisture transfer for selected local area\n\nThrough the previous field investigation, the effects of heat and humidity were known, but how is the change of the local air temperature and moisture content measured in quantity? In order to assess the effects of heat and humidity from the TGR on local climate, the areas affected by the TGR should be ascertained. As the effects of heat and humidity from the TGR on the local area are limited, so the area affected by the TGR is also constrained. In this paper, the local area was selected with a width of 2.5 km from the edge of the TGR and a height of 3 km (Figure 9) (Xu 2003; Li 2008; Liu 2010). The total horizontal area for the selected area is 229.8 km2.\n\nWithin the selected local area, the water surface area of the TGR needed to be known. As the form of the water surface of the TGR is not uniform, the different sections were measured, and the total area was determined from the sum of all the sections. Through measuring the TGR's width and the length in each section, the total water surface area of the TGR was obtained (Figure 10). The red points in Figure 10 were the measuring points for each section. All the surface area measurements were carried out at the same time corresponding to the measurement of heat and humidity. The total water surface area was calculated to be 18.3 km2.\n\nBased on the previous investigation of heat and moisture transfer from the ‘water–air’ surface, the total effects of heat and moisture from the TGR on the local area can be calculated (Table 1).\n\nTable 1\n\nTotal heat and moisture transfer for the selected local area\n\nTimeHeat transfer (KJ/m2.h)Moisture transfer (g/m2.h)Total heat transfer", null, "(KJ/ h)Total moisture transfer", null, "(Kg/ h)\n10:00–11:00 26.44 3.73 4.84 × 108 ↓ 6.84 × 104 ↑\n11:00–12:00 37.67 0.48 6.90 × 108 ↓ 8.82 × 103 ↑\n12:00–13:00 40.16 6.67 7.36 × 108 ↓ 1.22 × 105 ↑\n13:00–14:00 30.93 12.86 5.67 × 108 ↓ 2.36 × 105 ↑\n14:00–15:00 74.08 9.52 1.34 × 109 ↓ 1.74 × 105 ↑\n15:00–16:00 86.85 6.18 1.59 × 109 ↓ 1.13 × 104 ↓\n16:00–17:00 83.72 1.76 1.53 × 109 ↓ 3.22 × 104 ↓\n17:00–18:00 84.91 1.86 1.56 × 109 ↓ 3.40 × 104 ↓\n18:00–19:00 52.06 1.58 9.54 × 108 ↓ 2.89 × 104 ↑\n19:00–20:00 38.95 5.26 7.14 × 108 ↓ 9.64 × 104 ↑\n20:00–21:00 67.35 1.18 1.23 × 109 ↓ 2.17 × 104 ↓\n21:00–22:00 56.19 2.38 1.03 × 109 ↓ 4.37 × 104 ↓\n22:00–23:00 31.45 9.39 5.76 × 108 ↓ 1.72 × 105 ↑\n23:00–00:00 28.41 2.13 5.20 × 108 ↓ 3.91 × 104 ↑\n00:00–1:00 34.41 2.21 6.30 × 108 ↓ 4.06 × 104 ↓\n1:00–2:00 18.71 1.83 3.43 × 108 ↓ 3.35 × 104 ↑\n2:00–3:00 25.26 0.37 4.63 × 108 ↓ 6.85 × 103 ↓\n3:00–4:00 20.80 0.78 3.81 × 108 ↓ 1.43 × 104 ↑\n4:00–5:00 21.75 0.20 3.98 × 108 ↓ 3.58 × 103 ↓\n5:00–6:00 17.12 0.67 3.14 × 108 ↓ 1.23 × 104 ↑\n6:00–7:00 7.17 4.42 1.31 × 108 ↓ 8.11 × 104 ↑\n7:00–8:00 15.87 1.40 2.91 × 108 ↓ 2.57 × 104 ↑\n8:00–9:00 25.03 0.78 4.59 × 108 ↓ 1.42 × 104 ↓\n9:00–10:00 60.42 9.88 1.11 × 109 ↓ 1.81 × 105 ↓\nTimeHeat transfer (KJ/m2.h)Moisture transfer (g/m2.h)Total heat transfer", null, "(KJ/ h)Total moisture transfer", null, "(Kg/ h)\n10:00–11:00 26.44 3.73 4.84 × 108 ↓ 6.84 × 104 ↑\n11:00–12:00 37.67 0.48 6.90 × 108 ↓ 8.82 × 103 ↑\n12:00–13:00 40.16 6.67 7.36 × 108 ↓ 1.22 × 105 ↑\n13:00–14:00 30.93 12.86 5.67 × 108 ↓ 2.36 × 105 ↑\n14:00–15:00 74.08 9.52 1.34 × 109 ↓ 1.74 × 105 ↑\n15:00–16:00 86.85 6.18 1.59 × 109 ↓ 1.13 × 104 ↓\n16:00–17:00 83.72 1.76 1.53 × 109 ↓ 3.22 × 104 ↓\n17:00–18:00 84.91 1.86 1.56 × 109 ↓ 3.40 × 104 ↓\n18:00–19:00 52.06 1.58 9.54 × 108 ↓ 2.89 × 104 ↑\n19:00–20:00 38.95 5.26 7.14 × 108 ↓ 9.64 × 104 ↑\n20:00–21:00 67.35 1.18 1.23 × 109 ↓ 2.17 × 104 ↓\n21:00–22:00 56.19 2.38 1.03 × 109 ↓ 4.37 × 104 ↓\n22:00–23:00 31.45 9.39 5.76 × 108 ↓ 1.72 × 105 ↑\n23:00–00:00 28.41 2.13 5.20 × 108 ↓ 3.91 × 104 ↑\n00:00–1:00 34.41 2.21 6.30 × 108 ↓ 4.06 × 104 ↓\n1:00–2:00 18.71 1.83 3.43 × 108 ↓ 3.35 × 104 ↑\n2:00–3:00 25.26 0.37 4.63 × 108 ↓ 6.85 × 103 ↓\n3:00–4:00 20.80 0.78 3.81 × 108 ↓ 1.43 × 104 ↑\n4:00–5:00 21.75 0.20 3.98 × 108 ↓ 3.58 × 103 ↓\n5:00–6:00 17.12 0.67 3.14 × 108 ↓ 1.23 × 104 ↑\n6:00–7:00 7.17 4.42 1.31 × 108 ↓ 8.11 × 104 ↑\n7:00–8:00 15.87 1.40 2.91 × 108 ↓ 2.57 × 104 ↑\n8:00–9:00 25.03 0.78 4.59 × 108 ↓ 1.42 × 104 ↓\n9:00–10:00 60.42 9.88 1.11 × 109 ↓ 1.81 × 105 ↓\n\nNote: water surface area = 18.3 km2.\n\n#### Mathematical model of assessment of heat and humidity from the TGR on local climate\n\nHeat and moisture transfer was through the water–air interface, and the heat and moisture is from water surface to air, changing the air temperature and air moisture content. With the movement of air, the effects of heat and moisture from the TGR expand to the local area. In order to assess the effects of heat and moisture from the TGR in the selected area, a mathematical model was established to calculate the change of air temperature and moisture content. The horizontal distribution of air temperature and moisture content is assumed to be even, and is changed only with the height (Sheng 2003). When there is a layer at the height z from the ground and there is an infinite small thickness", null, ", then the energy within the layer", null, "is: where T is air temperature within the layer", null, ",", null, "; A is surface area for the selected local area,", null, ".\nThe air can be regarded as the normal gas, then: The atmosphere pressure is changed with height z, then: Combined with Equation (13), then: Air temperature is also changed with height z, then: where", null, "is air temperature near the ground; β = 0.65 K/100 m.\nIntegrate Equation (17), then: When", null, ",", null, ", the c can be obtained:\nIntegrate Equation (22) from the ground to the top of the selected area, then: where", null, "is the air temperature near the ground outside of the selected area, assuming this air temperature has not been affected by the TGR;", null, "is the total energy in the research area under the assumption that there is no effect by the TGR.\nBased on Equation (23), the total energy", null, "in the research area can be calculated under the assumption that there is no effect by the TGR. According to the", null, "in Table 1, the total energy", null, "in the research area affected by the TGR can be calculated by: and", null, "can also be expressed by: where", null, ",", null, "is the air temperature and pressure near the ground within the research area with the effect of the TGR: Combined with Equations (24) and (25), then: According to Equation (26), the air temperature of", null, "affected by the TGR within the research area can be calculated.\nThe final change of air temperature can be expressed by: Figure 11 shows the air temperature change caused by the TGR, corresponding to the period of the measuring cycle. According to the distribution of the temperature change, the average air temperature decreased 0.67 K in the period of 24 hours within the selected area.\n\nThe following is the mathematical model of the assessment of air humidity.\n\nAccording to the relationship between moisture content and pressure along the different heights (Zhang 1994), the moisture content at height z can be expressed as follows: where", null, "is the air moisture content near the ground that is outside of the selected area, assuming this air moisture content has not been affected by the TGR.\nCombining Equations (28)–(30): where W is the total moisture from the ground to the top of the selected area under the assumption that there is no effect from the TGR.\nAccording to the total moisture transfer", null, "in Table 1, the moisture", null, "from the ground to the top of the selected area affected by the TGR can be determined:\nAccording to Equation (32), the", null, "under the effect of the TGR can be calculated. Then:\n\nThe distribution of", null, "is presented in Figure 12. The average moisture content increased 0.25 g/kg in the period of 24 hours within the researched area.\n\n## CONCLUSIONS\n\nThis paper presents a mathematical hypothesis which, for the first time has proved, in theory, that there is only one way to analyze the effects of heat and humidity from the TGR on the local climate. This paper provides the detailed methodology utilized, based on mathematical hypothesis, to examine the effects of heat and humidity from the TGR on the local climate. Through field investigation, the effects of the transfers of sensible heat, latent heat, long-wave radiation, moisture, etc. on the interface of the TGR's water and air were analyzed. The final heat flux from water to air was less than that from air to water during the measuring period. The results showed the TGR's water had a cooling effect on the air during the measuring period, and the TGR's water had a wetting effect on the air during the measuring period. From the data of the mathematical model of assessment, the average air temperature decreased 0.67 K and average moisture content increased 0.25 g/kg in the period of 24 hours within the selected area. Since the time the TGR has been in full operation with a normal pool level of 175 m has been very short, more experiential data needs to be collected to better understand the effect of the TGR on the local climate. This paper has primarily focused on the mathematical hypothesis and new research methodology, to explain the mechanism of the effect of the TGR on the local climate. The long term effects of the TGR on the local climate need continued research. This calls for a long term project to be undertaken.\n\n## ACKNOWLEDGEMENTS\n\nThe authors would like to thank the National Natural Science Foundation of China (51578086; 51408079; 5151101134); Chongqing Fundamental and Advanced Research Projects (CSTC2014jcyjA90018) and Research on the Key Technology and Demonstration of Improving Indoor Physical Environment in Existing Public Buildings (2016YFC0700705) for their financial support, without which this research paper would not have been possible.\n\n## REFERENCES\n\nREFERENCES\nChen\nX. Y.\n&\nZhang\nQ.\n2009a\nRegional climate change over Three Gorges reservoir area\n.\nResour. Environ. Yangtze Basin\n18\n(\n1\n),\n47\n51\n.\nChen\nX. Y.\n&\nZhang\nQ.\n2009b\nRegional Climate of the Three Gorges Project (1961–2007)\n.\nMeteorology Press\n,\nBeijing\n.\nGao\nX. J.\n,\nLuo\nY.\n,\nLin\nW. T.\n,\nZhao\nZ. 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Y.\n2008\nAir Conditioning\n, 4th edn.\nChina Architecture & Building Press\n,\nBeijing\n." ]	[ null, "https://iwa.silverchair-cdn.com/iwa/content_public/journal/jwcc/10/1/10.2166_wcc.2018.067/2/m_jwc-d-17-00067if01.gif", null, "https://iwa.silverchair-cdn.com/iwa/content_public/journal/jwcc/10/1/10.2166_wcc.2018.067/2/m_jwc-d-17-00067if02.gif", null, "https://iwa.silverchair-cdn.com/iwa/content_public/journal/jwcc/10/1/10.2166_wcc.2018.067/2/m_jwc-d-17-00067if92.gif", null, "https://iwa.silverchair-cdn.com/iwa/content_public/journal/jwcc/10/1/10.2166_wcc.2018.067/2/m_jwc-d-17-00067if03.gif", null, "https://iwa.silverchair-cdn.com/iwa/content_public/journal/jwcc/10/1/10.2166_wcc.2018.067/2/m_jwc-d-17-00067if04.gif", null, "https://iwa.silverchair-cdn.com/iwa/content_public/journal/jwcc/10/1/10.2166_wcc.2018.067/2/m_jwc-d-17-00067if05.gif", null, "https://iwa.silverchair-cdn.com/iwa/content_public/journal/jwcc/10/1/10.2166_wcc.2018.067/2/m_jwc-d-17-00067if06.gif", null, 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https://www.bookdown.org/rwnahhas/RMPH/mi-blr.html	[ "## 9.7 Logistic regression after MI\n\nThe syntax for logistic regression after MI is similar to linear regression after MI, except that now glm() is used instead of lm().\n\nExample 9.2: Using the NSDUH 2019 teaching dataset (see Appendix A.5), what is the association between lifetime marijuana use (mj_lifetime) and age at first use of alcohol (alc_agefirst), adjusted for age (demog_age_cat6), sex (demog_sex), and income (demog_income)? Use MI to handle missing data.\n\nNOTE: This example uses the MAR dataset created for the illustration in Section 9.2 (nsduh_mar_rmph.RData). Those who never used alcohol were removed from the dataset, since we do not want to impute missing ages for them. Also, a random subset of the values for each variable were set to missing, with the probability of missingness for each variable depending on all the other variables but not the variable itself.\n\nFirst, load the data and view the extent of the missing data.\n\nload(\"Data/nsduh_mar_rmph.RData\")\nsummary(nsduh_mar)\n## mj_lifetime alc_agefirst demog_age_cat6 demog_sex demog_income\n## No :342 Min. : 3.0 18-25: 99 Male :376 Less than $20,000:122 ## Yes :487 1st Qu.:15.0 26-34:118 Female:397$20,000 - $49,999:226 ## NA's: 14 Median :17.0 35-49:223 NA's : 70$50,000 - $74,999:116 ## Mean :17.6 50-64:191$75,000 or more :342\n## 3rd Qu.:20.0 65+ :180 NA's : 37\n## Max. :45.0 NA's : 32\n## NA's :59\n\nNext, fit the imputation model.\n\nimp.nsduh <- mice(nsduh_mar,\nseed = 3,\nm = nimpute(nsduh_mar),\nprint = F)\nimp.nsduh\n## Class: mids\n## Number of multiple imputations: 22\n## Imputation methods:\n## mj_lifetime alc_agefirst demog_age_cat6 demog_sex demog_income\n## \"logreg\" \"pmm\" \"polyreg\" \"logreg\" \"polyreg\"\n## PredictorMatrix:\n## mj_lifetime alc_agefirst demog_age_cat6 demog_sex demog_income\n## mj_lifetime 0 1 1 1 1\n## alc_agefirst 1 0 1 1 1\n## demog_age_cat6 1 1 0 1 1\n## demog_sex 1 1 1 0 1\n## demog_income 1 1 1 1 0\n\nFinally, fit the logistic regression on the imputed datasets and pool the results.\n\nfit.imp.glm <- with(imp.nsduh,\nglm(mj_lifetime ~ alc_agefirst + demog_age_cat6 +\ndemog_sex + demog_income, family = binomial))\n# summary(pool(fit.imp.glm), conf.int = T)\nround.summary(fit.imp.glm, digits=3)\n## estimate std.error statistic df p.value 2.5 % 97.5 %\n## (Intercept) 6.434 0.615 10.461 592.6 0.000 5.226 7.642\n## alc_agefirst -0.275 0.029 -9.386 513.9 0.000 -0.333 -0.218\n## demog_age_cat626-34 -0.251 0.339 -0.738 711.1 0.461 -0.917 0.416\n## demog_age_cat635-49 -0.809 0.305 -2.655 694.5 0.008 -1.407 -0.211\n## demog_age_cat650-64 -0.790 0.311 -2.541 639.8 0.011 -1.401 -0.179\n## demog_age_cat665+ -1.307 0.314 -4.157 673.5 0.000 -1.925 -0.690\n## demog_sexFemale -0.045 0.176 -0.255 393.3 0.799 -0.391 0.301\n## demog_income$20,000 -$49,999 -0.770 0.275 -2.795 687.3 0.005 -1.310 -0.229\n## demog_income$50,000 -$74,999 -0.204 0.314 -0.650 719.9 0.516 -0.821 0.412\n## demog_income$75,000 or more -0.574 0.261 -2.202 686.0 0.028 -1.086 -0.062 Add exponentiate = T to summary() or round.summary() to compute odds ratios, but be careful interpreting the results as only the estimate and confidence interval are exponentiated; the standard error in the table is still the standard error of the un-exponentiated regression coefficient. As such, to avoid confusion, extract just the estimate, confidence interval, and p-value when exponentiating (and drop the first row since the exponentiated intercept is not of interest). # summary(pool(fit.imp.glm), conf.int = T, # exponentiate = T)[-1, c(\"term\", \"estimate\", \"2.5 %\", \"97.5 %\", \"p.value\")] round.summary(fit.imp.glm, digits = 3, exponentiate = T)[-1, c(\"estimate\", \"2.5 %\", \"97.5 %\", \"p.value\")] ## estimate 2.5 % 97.5 % p.value ## alc_agefirst 0.759 0.717 0.804 0.000 ## demog_age_cat626-34 0.778 0.400 1.515 0.461 ## demog_age_cat635-49 0.445 0.245 0.810 0.008 ## demog_age_cat650-64 0.454 0.246 0.836 0.011 ## demog_age_cat665+ 0.271 0.146 0.502 0.000 ## demog_sexFemale 0.956 0.677 1.351 0.799 ## demog_income$20,000 - $49,999 0.463 0.270 0.795 0.005 ## demog_income$50,000 - $74,999 0.815 0.440 1.511 0.516 ## demog_income$75,000 or more 0.563 0.337 0.940 0.028\n\n### 9.7.1 Multiple degree of freedom tests\n\nmi.anova() does not work for logistic regression models, but D1() does. Fit reduced models, each omitting one categorical variable that has more than 2 levels and then use D1() to compare the full and reduced models using a Wald test or D3() for a likelihood ratio test.\n\nFor example, to get a Type 3 test for demog_age_cat6 in Example 9.2:\n\n# Fit reduced model\nfit.imp.glm.age <- with(imp.nsduh,\ndemog_sex + demog_income, family = binomial))\n\n# Compare full and reduced models\n# Wald test\nsummary(D1(fit.imp.glm, fit.imp.glm.age))\n##\n## Models:\n## model formula\n## 1 mj_lifetime ~ alc_agefirst + demog_age_cat6 + demog_sex + demog_income\n## 2 mj_lifetime ~ alc_agefirst + demog_sex + demog_income\n##\n## Comparisons:\n## test statistic df1 df2 dfcom p.value riv\n## 1 ~~ 2 5.886 4 794.7 833 0.0001138 0.06877\n##\n## Number of imputations: 22 Method D1\n# LR test\nsummary(D3(fit.imp.glm, fit.imp.glm.age))\n##\n## Models:\n## model formula\n## 1 mj_lifetime ~ alc_agefirst + demog_age_cat6 + demog_sex + demog_income\n## 2 mj_lifetime ~ alc_agefirst + demog_sex + demog_income\n##\n## Comparisons:\n## test statistic df1 df2 dfcom p.value riv\n## 1 ~~ 2 6.191 4 19760 833 0.00005653 0.06634\n##\n## Number of imputations: 22 Method D3\n\n### 9.7.2 Predictions\n\nPrediction proceeds in almost the same way as described in Section 9.6.2 – predict on each imputed dataset and then pool the results using Rubin’s rules. The one change is that type = \"response\" is added to get predictions on the probability scale.\n\nExample 9.2 (continued): What is the predicted probability (and standard error) of lifetime marijuana use for someone who first used alcohol at age 13 years, is currently age 30 years, male, and has an annual income of less than $20,000? # Prediction data.frame # Code below works if NEWDAT has only 1 row NEWDAT <- data.frame(alc_agefirst = 13, demog_age_cat6 = \"26-34\", demog_sex = \"Male\", demog_income = \"Less than$20,000\")\n\n# Get prediction for each analysis\nPREDLIST <- lapply(fit.imp.glm$analyses, predict, newdata=NEWDAT, se.fit=T, type = \"response\") # Extract mean and variance from each analysis MEAN <- VAR <- rep(NA, length(fit.imp.glm$analyses))\nfor(i in 1:length(MEAN)) {\nMEAN[i] <- PREDLIST[[i]]$fit VAR[ i] <- (PREDLIST[[i]]$se.fit)^2\n}\n\n# Apply Rubin's rules\nPRED.POOLED <- pool.scalar(MEAN,\nVAR,\nnrow(imp.nsduh$data)) # Extract the pooled mean PRED.POOLED$qbar\n## 0.9309\n# Extract the pooled standard error\nsqrt(PRED.POOLED\\$t)\n## 0.02254" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.62313443,"math_prob":0.9175032,"size":6628,"snap":"2023-40-2023-50","text_gpt3_token_len":2287,"char_repetition_ratio":0.147343,"word_repetition_ratio":0.10139165,"special_character_ratio":0.41264334,"punctuation_ratio":0.2238806,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.9721806,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-03T07:53:50Z\",\"WARC-Record-ID\":\"<urn:uuid:c567ab30-20fc-4168-952d-b8710e757c76>\",\"Content-Length\":\"88130\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a45311e9-6b40-48ba-8263-a20905d4ba27>\",\"WARC-Concurrent-To\":\"<urn:uuid:e8a074af-90a4-44c3-b001-6de120e7fda5>\",\"WARC-IP-Address\":\"34.194.247.74\",\"WARC-Target-URI\":\"https://www.bookdown.org/rwnahhas/RMPH/mi-blr.html\",\"WARC-Payload-Digest\":\"sha1:N3IH7YHLQLIWY7NCJJ7VGA65LXAEUMAS\",\"WARC-Block-Digest\":\"sha1:J4ZKVGLVNBLDRI2PYPUCB4GKRART2WBD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100489.16_warc_CC-MAIN-20231203062445-20231203092445-00068.warc.gz\"}"}
https://wiki.haskell.org/Haskell_Quiz/Grid_Folding/Solution_Dolio	[ "# Haskell Quiz/Grid Folding/Solution Dolio\n\nThe basis for my solution is simple. Consider each square to be a single-element list initially. Then, a row of such squares is a list of such lists. An entire grid is then a list of those lists.\n\nWhen folding horizontally, one splits each row, reverses the half to go on top (as well as its elements, since they'll be flipped), and zips the two halves together by appending corresponding elements. This results in a top-first list of the stacked squares.\n\nWhen folding vertically, one splits the grid in half, reverses the half to go on top, and zips the two grid halves together. The function that combines corresponding rows is yet another zip that appends corresponding elements (again, reversing the ones on top).\n\n```module Main where\nimport Data.Char\nimport System\n\ngrid n = break . map return \\$ [1..(n*n)]\nwhere break [] = []\nbreak l = let (h,t) = splitAt n l in h : break t\n\nfold 'T' = folder vzipper\nfold 'B' = folder (flip vzipper)\nfold 'L' = map (folder hzipper)\nfold 'R' = map (folder \\$ flip hzipper)\nfold _ = error \"Unrecognized letter.\"\n\nvzipper = zipWith (zipWith \\$ (++) . reverse) . reverse\nhzipper = zipWith (++) . (map reverse . reverse)\n\nfolder z = uncurry z . ap (flip splitAt) ((`div` 2) . length)\n\npretty = unlines . map (unwords . map show)\n\noutput s \| all isSpace s = error \"Invalid folding scheme\"\n\| otherwise = putStr s\n\nmain = do [n, s] <- getArgs\noutput . pretty . foldl (flip fold) (grid \\$ read n) \\$ s\n```" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8004603,"math_prob":0.9877727,"size":1534,"snap":"2022-40-2023-06","text_gpt3_token_len":405,"char_repetition_ratio":0.11045752,"word_repetition_ratio":0.014440433,"special_character_ratio":0.26923078,"punctuation_ratio":0.12937063,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99249274,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-27T11:06:06Z\",\"WARC-Record-ID\":\"<urn:uuid:0c0578ff-e754-4891-958c-feff694a8b5e>\",\"Content-Length\":\"19734\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b379d223-6852-4aec-9908-fb96f1ea0621>\",\"WARC-Concurrent-To\":\"<urn:uuid:f1802952-3162-4ba4-8699-31d32ff807e5>\",\"WARC-IP-Address\":\"146.75.37.175\",\"WARC-Target-URI\":\"https://wiki.haskell.org/Haskell_Quiz/Grid_Folding/Solution_Dolio\",\"WARC-Payload-Digest\":\"sha1:KDU5VOWQFBLJ2HW3RQ2N56YE67RHOQIC\",\"WARC-Block-Digest\":\"sha1:BVJ6WJYEN3EAERXSY3YOSKU32WKUH45H\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030335004.95_warc_CC-MAIN-20220927100008-20220927130008-00524.warc.gz\"}"}
http://www.readbag.com/dioceseofjoliet-cso-documents-curmathgoal9geometry	[ "#### Read MATH CURRICULUM text version\n\n`MATHEMATICS CURRICULUM PROJECT GOAL 9: Standard A: Use geometric methods to analyze, categorize and draw conclusions about points, lines, planes and space. Demonstrate and apply geometric concepts involving points, lines, planes and space. Grade 1 1. Identify two- and threedimensional shapes. 2. Model two-dimensional geometric shapes by drawing or building. 3. Describe and interpret relative positions in space and apply concepts of relative position (e.g., above/below). 4. Recognize and describe shapes that have line symmetry. 5. Identify geometric shapes and structures in the environment. 6. Explore the effects of translations (slides), reflections (flips) and rotations (turns) with concrete objects. Grade 2 1. Investigate and predict the results of putting together and taking apart two- and three-dimensional shapes (e.g., put two triangles together to make a quadrilateral). 2. Describe and interpret direction and distance in navigating space and apply concepts of direction and distance (e.g., nearer/farther). 3. Perform translations (slides), reflections (flips) and rotations (turns) with concrete objects. 4. Create and complete shapes that have line symmetry. Grade 3 1. Find and name two- and three-dimensional geometric figures in every day objects. 2. Identify the properties, number of sides and corners of geometric shapes. 3. Draw two-dimensional shapes using rulers, graphing paper and tracing paper. 4. Visualize the transition from two-dimensional to its threedimensional figure. 5. Construct three-dimensional geometric figures. 6. Locate and identify points using numbers and symbols on a grid and describe how points relate to each other on a grid. Grade 4 1. Identify, draw and label lines, line segments, rays, parallel lines, intersecting lines, perpendicular lines, acute angles, obtuse angles, right angles and acute, obtuse, right, scalene, isosceles and equilateral triangles. 2. Identify, draw and build regular and irregular polygons. 3. Read and plot ordered pairs of numbers in the positive quadrant of the Cartesian plane. 4. Describe paths and movement using coordinate systems. 5. Differentiate between polygons and non-polygons. 6. Identify and label radius and diameter of a circle. 7. Explore and describe rotational symmetry of twoand three-dimensional shapes. 8. Construct a circle with a specified radius or diameter using a compass.As a result of their schooling students will be able to...Kindergarten 1. Identify circle, square, triangle and rectangle. 2. Begin to recognize other shapes of objects. 3. Explore three-dimensional shapes.June 2005Geometry, Goal 9 - Page 1\fMATHEMATICS CURRICULUM PROJECT GOAL 9: Standard A: Use geometric methods to analyze, categorize and draw conclusions about points, lines, planes and space. Demonstrate and apply geometric concepts involving points, lines, planes and space. Grade 6 1. Plot and read ordered pairs of numbers in all four quadrants. 2. Describe sizes, positions and orientations of shapes under transformations, including dilations. 3. Perform simple constructions (e.g., equal segments, angle and segment bisectors, or perpendicular lines, inscribing a hexagon in a circle) with a compass and straightedge or a mira. 4. Determine and describe the relationship between pi, the diameter, the radius and the circumference of a circle. 5. Determine unknown angle measures using angle relationships and properties of a triangle or a quadrilateral.Grade 7 Grade8/Pre-Algebra/AlgebraAs a result of their schooling students will be able to...Grade 5 1. Identify, compare and analyze attributes of two- and threedimensional shapes and develop vocabulary to describe the attributes. 2. Classify two- or three-dimensional shapes according to their properties (e.g., regular and irregular types of quadrilaterals, pyramids and prisms). 3. Investigate and describe the results of subdividing and combining shapes. 4. Describe paths using coordinate systems. 5. Determine the distance between points along horizontal and vertical lines of a coordinate system. 6. Identify and justify rotational symmetry in two- and three-dimensional shapes. 7. Identify and describe how geometric figures are used in practical settings (e.g., construction, art, advertising, architecture). 8. Identify and label radius, diameter, chord and circumference of a circle. 9. Copy a line segment or an angle using a straightedge and a compass. 10. Construct angles and perpendicular bisectors of line segments. 11. Solve problems using properties of triangles (e.g., sum of interior angles of a triangle is 180°).1.2.3.4.5.6.Examine and describe a geometric shape, such as a regular polygon or a quadrilateral with pairs of parallel or perpendicular sides, using coordinate geometry. Draw geometric shapes with specified properties, such as side lengths or angle measures. Examine and describe line or rotational symmetry of objects in terms of transformations. Draw transformations of figures in a plane to match specified criteria. Perform constructions of congruent angles or parallel lines using a compass and straightedge, paper folding, or a mira. Determine the relationship among the number of edges, faces and vertices in a threedimensional object.1. 2. 3.4.5. 6. 7. 8.Solve problems involving two- and three-dimensional shapes. Apply and use the Pythagorean Theorem. Identify, describe and determine the radius, diameter and circumference of a circle and their relationship to each other and to pi. Graph points and identify coordinates of points on the Cartesian coordinate plane. Represent and identify geometric figures using coordinate geometry. Use transformations in a Cartesian coordinate plane. Identify relationships of angles formed by intersecting lines. Solve problems involving vertical, complementary and supplementary angles.June 2005Geometry, Goal 9 - Page 2\fMATHEMATICS CURRICULUM PROJECT GOAL 9: Standard B: Use geometric methods to analyze, categorize and draw conclusions about points, lines, planes and space. Identify, describe, classify and compare relationships using points, lines, planes, and solids. Grade 1 1. Identify objects that are the same shape. 2. Compare and sort two- and three-dimensional objects. 3. Compare and contrast sides and corners of basic geometric shapes. Grade 2 1. Identify objects that are congruent 2. Compare and contrast attributes of two- and threedimensional objects using appropriate vocabulary. Grade 3 1. Identify parallel and intersecting lines. 2. Identify polygons v. nonpolygons by defining components. 3. Recognize and identify symmetrical figures with one and two lines of symmetry. 4. Identify congruent figures in different positions. 5. Identify similar figures that are not congruent and tell how they are similar. Grade 4 1. Determine congruence and similarity of given shapes. 2. Explore polyhedra (threedimensional figures) using concrete models.As a result of their schooling students will be able to...Kindergarten 1. Identify and sort objects according to their shape.June 2005Geometry, Goal 9 - Page 3\fMATHEMATICS CURRICULUM PROJECT GOAL 9: Standard B: Use geometric methods to analyze, categorize and draw conclusions about points, lines, planes and space. Identify, describe, classify and compare relationships using points, lines, planes, and solids. Grade 6 1. Determine the relationships between the number of vertices or sides in a polygon, the number of diagonals and the sum of its angles. 2. Analyze quadrilaterals for defining characteristics. 3. Create a three-dimensional object from any two-dimensional representation of the object, including multiple views, nets, or technological representations. 4. Identify and describe the five regular polyhedra. 5. Create regular and semi-regular tessellations using pattern blocks, other manipulatives or technology to tile a plane.Grade 7 Grade8/Pre-Algebra/AlgebraAs a result of their schooling students will be able to...Grade 5 1. Demonstrate congruence of plane figures using transformations (translation, rotation, reflection). 2. Determine if two polygons are congruent using measures of angles and sides. 3. Match a front, right side and top view drawing with a threedimensional model built with cubes.1.2. 3.Describe, classify and justify relationships among types of twoand three-dimensional objects using their defining properties. Solve problems using properties of polygons and circles. Classify and order quadrilaterals according to their properties.1.2. 3.Identify front, side and top views of a three-dimensional solid built with cubes. Solve problems involving congruent and similar figures. Relate absolute value to distance on the number line.June 2005Geometry, Goal 9 - Page 4\fMATHEMATICS CURRICULUM PROJECT GOAL 9: Standard C: Standard D: Use geometric methods to analyze, categorize and draw conclusions about points, lines, planes and space. Construct convincing arguments and proofs to solve problems. Use trigonometric ratios and circular functions to solve problems (only applies to grades 7 & 8). Grade 1 1. Recognize and explain a geometric pattern. 2. Verbalize the rule for the geometric pattern sequence (e.g., amount doubles). Grade 2 1. Justify an extension of a pattern. Grade 3 1. Verbally describe the properties of basic geometric figures. 2. Create and explain patterns using pattern blocks and manipulatives. Grade 4 3. Make and test predictions about mathematical properties and relationships and justify the conclusions.As a result of their schooling students will be able to...Kindergarten 1. Verbally explain the different qualities of the shape of an object.June 2005Geometry, Goal 9 - Page 5\fMATHEMATICS CURRICULUM PROJECT GOAL 9: Standard C: Standard D: Use geometric methods to analyze, categorize and draw conclusions about points, lines, planes and space. Construct convincing arguments and proofs to solve problems. Use trigonometric ratios and circular functions to solve problems (only applies to grades 7 & 8). Grade 6 1. Make, test and justify conjectures about various quadrilateral and triangle relationships, including the triangle inequality. 2. Justify the relationship between vertical angles. 3. Justify that the sum of the angles of a triangle is 180 degrees.Grade 7 Grade8/Pre-Algebra/AlgebraAs a result of their schooling students will be able to...Grade 5 1. Make and test conjectures about mathematical properties and relationships and develop logical arguments to justify conclusions. 2. Make and test conjectures about the results of subdividing and combining shapes.1.2.3.4.Create and critique arguments about geometric relationships in figures based upon inductive and deductive reasoning. Justify the area formulas for triangles, parallelograms and trapezoids based on the formula for the area of a rectangle. Make and test conjectures about the relationships between side length and angle measure in various triangles and quadrilaterals. Justify the properties of angles formed by parallel lines cut by a transversal using appropriate terminology.1.2.3.Create and critique arguments concerning geometric ideas and relationships, such as congruence, similarity, the Pythagorean relationship, or formulas for surface areas or volume of simple threedimensional objects. Represent, solve and explain numerical and algebraic relationships using geometric concepts. Provide examples or counter-examples to either illustrate or disprove conjectures about geometric characteristics.Standard D 1. Analyze the relationship between sides of right triangles using the Pythagorean theorem. 2. Solve problems that involve the use of proportions and the Pythagorean theorem in similar right triangles with whole number side lengths.Standard D 1. Recognize Pythagorean Triples. 2. Identify the basic trigonometric ratio in terms of lengths of the sides of a right triangle and an acute angle.June 2005Geometry, Goal 9 - Page 6\f`\n\n6 pages\n\n#### Report File (DMCA)\n\nOur content is added by our users. We aim to remove reported files within 1 working day. Please use this link to notify us:\n\nReport this file as copyright or inappropriate\n\n731005" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8394614,"math_prob":0.96637595,"size":12216,"snap":"2020-34-2020-40","text_gpt3_token_len":2639,"char_repetition_ratio":0.14624959,"word_repetition_ratio":0.18992248,"special_character_ratio":0.20252128,"punctuation_ratio":0.17340522,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9934876,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-18T14:22:53Z\",\"WARC-Record-ID\":\"<urn:uuid:890e1d3c-8e87-4c7f-b249-6d49f4690a59>\",\"Content-Length\":\"20114\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ec471514-7909-45ff-90c0-f4f5c2812f8f>\",\"WARC-Concurrent-To\":\"<urn:uuid:c2ba984c-7729-471f-9289-02455eea6958>\",\"WARC-IP-Address\":\"172.67.214.81\",\"WARC-Target-URI\":\"http://www.readbag.com/dioceseofjoliet-cso-documents-curmathgoal9geometry\",\"WARC-Payload-Digest\":\"sha1:H65AC7XDQHOUOJR7QPQH3TLSS7VBX22A\",\"WARC-Block-Digest\":\"sha1:6LGXLIFDX3BXUPOOCSOG6VLPCIOW5JTW\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400187899.11_warc_CC-MAIN-20200918124116-20200918154116-00428.warc.gz\"}"}
https://stats.stackexchange.com/questions/32657/r-package-tsa-how-to-interpret-the-io-coefficients-output-of-the-arimax-functio	[ "# R package TSA: how to interpret the IO coefficients output of the arimax function\n\nI was playing with the TSA package in R and wanted to test the arimax function to the solution provided in Pankratz's Forecasting with Dynamic Regression Models, chapter 8. The savings rate and the function seems to provide similar results as the ones in the book except for the IO weights which are quite different. I bet there is a transformation that I might be missing.\n\nAny help on understanding why IO coefficients are so different would be appreciated...\n\nthe solution states\n\nAO @ t=82,43,89\nLS @ t=99\nIO @ t=62,55\n\n\nwith Parameters estimates\n\nC = 6.1635\nw82 = 2.3346\nw99 = -1.5114\nw43 = 1.1378\nw62 = 1.4574\nw55 = -1.4915\nw89 = -1.0702\nAR1 = 0.7976\nMA2 = -0.3762\n\n\nTo fit the model in R, I used (saving is the data)\n\narimax(saving, order = c(1,0,2), fixed=c(NA,0,NA,NA,NA,NA,NA,NA,NA,NA), io=c(55,62),\nxreg=data.frame(AO82=1(seq(saving)==82),\nAO43=1(seq(saving)==43),\nAO89=1(seq(saving)==89),\nLS99=1(seq(saving)>=99)),\nmethod='ML')\n\n\nThe savings rate data is (100 points)\n\n4.9 5.2 5.7 5.7 6.2 6.7 6.9 7.1 6.6 7 6.9 6.4 6.6 6.4 7 7.3 6 6.3 4.8 5.3 5.4 4.7 4.9 4.4 5.1 5.3 6 5.9 5.9 5.6 5.3 4.5 4.7 4.6 4.3 5 5.2 6.2 5.8 6.7 5.7 6.1 7.2 6.5 6.1 6.3 6.4 7 7.6 7.2 7.5 7.8 7.2 7.5 5.6 5.7 4.9 5.1 6.2 6 6.1 7.5 7.8 8 8 8.1 7.6 7.1 6.6 5.6 5.9 6.6 6.8 7.8 7.9 8.7 7.7 7.3 6.7 7.5 6.4 9.7 7.5 7.1 6.4 6 5.7 5 4.2 5.1 5.4 5.1 5.3 5 4.8 4.7 5 5.4 4.3 3.5\n\nhere it is my output\n\n> arimax(saving, order = c(1,0,2),fixed=c(NA,0,NA,NA,NA,NA,NA,NA,NA,NA),io=c(55,62),xreg=data.frame(AO82=1(seq(saving)==82),\n+ AO43=1(seq(saving)==43),AO89=1(seq(saving)==89),LS99=1(seq(saving)>=99)),method='ML')\n\nCall:\narimax(x = saving, order = c(1, 0, 2), xreg = data.frame(AO82 = 1 * (seq(saving) ==\n82), AO43 = 1 * (seq(saving) == 43), AO89 = 1 * (seq(saving) ==\n89), LS99 = 1 * (seq(saving) >= 99)), fixed = c(NA, 0, NA, NA, NA, NA,\nNA, NA, NA, NA), method = \"ML\", io = c(55, 62))\n\nCoefficients:\nar1 ma1 ma2 intercept AO82 AO43 AO89 LS99 IO-55 IO-62\n0.7918 0 0.3406 6.0628 2.3800 1.1297 -1.0466 -1.4885 -0.5958 0.5517\ns.e. 0.0674 0 0.1060 0.3209 0.3969 0.3780 0.3835 0.5150 0.4044 0.3772\n\nsigma^2 estimated as 0.2611: log likelihood = -75.57, aic = 169.14" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.69719666,"math_prob":0.99852806,"size":2130,"snap":"2021-43-2021-49","text_gpt3_token_len":1080,"char_repetition_ratio":0.1378175,"word_repetition_ratio":0.016666668,"special_character_ratio":0.5685446,"punctuation_ratio":0.26478493,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9976158,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-04T20:07:24Z\",\"WARC-Record-ID\":\"<urn:uuid:702302b1-25fd-44db-b663-e60ca325ce47>\",\"Content-Length\":\"135548\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9ead8616-c83d-4b81-9eeb-c0ecfa8b405e>\",\"WARC-Concurrent-To\":\"<urn:uuid:bf9cc1c3-5da6-411a-bbfe-702e91e117d5>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://stats.stackexchange.com/questions/32657/r-package-tsa-how-to-interpret-the-io-coefficients-output-of-the-arimax-functio\",\"WARC-Payload-Digest\":\"sha1:3GJQLFKP6FX2A55LJ3O36RAVBVY4T5RS\",\"WARC-Block-Digest\":\"sha1:OONOZ3VGPIHFUFY6S35JUJSNKYEE55PY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363006.60_warc_CC-MAIN-20211204185021-20211204215021-00223.warc.gz\"}"}
https://brilliant.org/discussions/thread/mathemagic-u/	[ "MatheMagic !!!\n\nTwo boys went to a book shop and purchased a book which costs $50. They paid$25 each for the book and left the store. When the store manager went to the cashier, he found that they have taken $5 extra from them. So, the actual cost of the book was$45. The manager gave $5 to a worker out there to return it to those boys. Let's make it more interesting, the worker was a bit greedy, he took$2 from it and returned $3 to the boys. They distributed$1.5 and $1.5 each and thought that they have paid$25 - $1.5 =$23.5 each for the book. Here comes the problem, Now, we all (I mean those who are reading this thread) know that the boys paid $23.5 +$23.5 = $47 and the worker have only$2 which adds up to $49. Where do you think$1 lost in the process ?", null, "Note by Abhishek Kumar\n6 years, 4 months ago\n\nThis discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.\n\nWhen posting on Brilliant:\n\n• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .\n• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting \"I don't understand!\" doesn't help anyone.\n• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.\n• Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.\n\nMarkdownAppears as\nitalics or _italics_ italics\nbold or __bold__ bold\n\n- bulleted\n- list\n\n• bulleted\n• list\n\n1. numbered\n2. list\n\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1\n\nparagraph 2\n\nparagraph 1\n\nparagraph 2\n\n> This is a quote\nThis is a quote\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$...$$ or $...$ to ensure proper formatting.\n2 \\times 3 $2 \\times 3$\n2^{34} $2^{34}$\na_{i-1} $a_{i-1}$\n\\frac{2}{3} $\\frac{2}{3}$\n\\sqrt{2} $\\sqrt{2}$\n\\sum_{i=1}^3 $\\sum_{i=1}^3$\n\\sin \\theta $\\sin \\theta$\n\\boxed{123} $\\boxed{123}$\n\nSort by:\n\nyou are supposed to add $3 to$47 because the worker's $2 is counted in the$47\n\n- 6 years, 4 months ago\n\nu r talking about cost price in the first half and then reverse calculating it by sell price in the second half.so its obvious why they are not adding up\n\n- 6 years, 4 months ago\n\nthat is subtract $3 from$50 then subtract $2 thus it adds up - 6 years, 4 months ago Log in to reply do like this they each have paid$23.5 so in total $47.now waht u r duin rong is that they actually bought the book for$45(after gettin money back) but they paid $47 because the cashier cheated$2 on them i.e $47 -$2\n\n- 6 years, 4 months ago" ]	[ null, "https://ds055uzetaobb.cloudfront.net/brioche/avatars-2/resized/45/f7e48998860544528d082e40a9191f60.2d604c381259b3831493694306635229.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.95471096,"math_prob":0.94674104,"size":2083,"snap":"2019-43-2019-47","text_gpt3_token_len":556,"char_repetition_ratio":0.10822511,"word_repetition_ratio":0.0052083335,"special_character_ratio":0.2918867,"punctuation_ratio":0.11258278,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9627267,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-18T10:28:08Z\",\"WARC-Record-ID\":\"<urn:uuid:5c05a831-1fc3-40d7-b0eb-2e59e03b4d64>\",\"Content-Length\":\"63628\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:74fb183e-4e35-41dc-a3c3-7f6d9eaede4c>\",\"WARC-Concurrent-To\":\"<urn:uuid:eacd4ef4-ca9a-4731-b476-ebd1af90b3a2>\",\"WARC-IP-Address\":\"104.20.34.242\",\"WARC-Target-URI\":\"https://brilliant.org/discussions/thread/mathemagic-u/\",\"WARC-Payload-Digest\":\"sha1:D3TAWE6YL4PZRBUKZYMING4HTSJPDD6G\",\"WARC-Block-Digest\":\"sha1:2SA6GGOHGTVKMXT7MBWB723WAQ7VRKCI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986679439.48_warc_CC-MAIN-20191018081630-20191018105130-00328.warc.gz\"}"}
https://media.alavareyes.com/b1kaqe4/37ea61-associative-vector-c%2B%2B	[ "Boots Multivitamins Review, Remove Kitchen Faucet Nut, Tcp Smart Wall Socket With Usb, Flip Door Lock Home Depot, Home Depot 50 Amp Breaker, Bunny Rescue Near Me, Sunday River Mountain Stats, Medical School Secondary Essays, Riverside Car Park Windsor, \" /> Boots Multivitamins Review, Remove Kitchen Faucet Nut, Tcp Smart Wall Socket With Usb, Flip Door Lock Home Depot, Home Depot 50 Amp Breaker, Bunny Rescue Near Me, Sunday River Mountain Stats, Medical School Secondary Essays, Riverside Car Park Windsor, \" />\n\nClosure: If x is any vector and c is any real number in the vector space V, then x. c belongs to V. Associative Law: For all real numbers c and d, and the vector x in V, then c. (d. v) = (c . C++11 has eight associative containers. It should be equal to c times v dot w. of the product of . An associative memory M is a system that relates input patterns and output patterns as follows : with x and y being the input and output patterns vectors. BOOK FREE CLASS; ... Commutative Law: A + B = B + A Associative Law: A + (B + C) = (A + B) + C. With C++17, you can more comfortably insert new elements into them, merge existing associative containers, or move elements from one container into another if they are similar. Contribute to TakeAsh/cpp-AssociativeVector development by creating an account on GitHub. v; Distributive law: For all real numbers c and d, and the vector x in V, (c + d).v = c.v + c.d Each of the following containers use different algorithm for data storage thus for different operations they have different speed. Associative array implemented by std::vector. scalar multiplication distributes over complex addition $(c_1 + c_2) \\cdot V = c_1 \\cdot V + c_2 \\cdot V$ any set with properties marked (A) is an Abelian group real vector space: non-empty set $\\mathbb{V}$ of … Thus, vector addition is commutative : A + B = B + A (4.1) The addition of vectors also obeys the associative law as illustrated in Fig. 4.4(c), the same vector R is obtained. The associative law, which states that the sum of three vectors does not depend on which pair of vectors is added first: $$(\\vc{a}+\\vc{b})+\\vc{c} = \\vc{a} + (\\vc{b}+\\vc{c}).$$ You can explore the properties of vector addition with the following applet. Memory overhead.The C++ standard does not specify requirements on memory consumption, but virtually any implementation of vector has the same behavior with respect to memory usage: the memory allocated by a vector v with n elements of type T is . An associative memory is represented by a matrix whose … Welcome back for our second part in our series on removing elements from C++ containers! • Vector addition is commutative: a + b = b + a. Several properties of vector addition are easily verified. Print vector in C++ A vector algebra is an algebra where the terms are denoted by vectors and operations are performed corresponding to algebraic expressions. I find that semantic relatedness, as quantified by these models, is able to provide a good measure of the associations Sequence Containers: In standard template library they refer to the group of container class template, we use to them store data.One common property as the name suggests is that elements can be accessed sequentially. Associative Law - the addition of three vectors is independent of the pair of vectors added first. The more cache line aware the container is, the faster is the access time of the elements: std::vector > std::deque > (std::list, std::forward_list). Notes: When two vectors having the same magnitude are acting on a body in opposite directions, then their resultant vector is zero. The container manages the storage space that is allocated for its elements and provides member functions to access them, either directly or through iterators (objects with properties similar to pointers). First, understand the vector -a. c c-plus-plus information-retrieval cmake algorithm avx bit-manipulation simd integer-compression sparse-vectors sparse-matrix bit-array indexing-engine bit-vector adjacency-matrix associative-array sparse-vector These are special kind of arrays, where indexing can be numeric or any other data type i.e can be numeric 0, 1, 2, 3.. Although, STL classes are there to simplify and efficiently implement associative array, but it was my own idea to reinvent the wheel and build things grounds up, except for using the vector class. We construct a parallelogram. where c is v. capacity and e is sizeof (T). We can therefore write both as a + b + c. • a + 0 = 0 + a = a. For any vectors a, b, and c of the same size we have the following. vector addition is commutative. the direction . 4.4(d). The vector triple product has the form A × (B × C).The parentheses are necessary, because the cross product is not associative, meaning that A × (B × C) is not necessarily equal to (A × B) × C.If B and C are proportional, making them collinear, the vector triple product is zero and we need not discuss it further. We will find that vector addition is commutative, that is a + b = b + a . B. 6. Associative Judgment and Vector Space Semantics Sudeep Bhatia University of Pennsylvania I study associative processing in high-level judgment using vector space semantic models. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. If the data structure in your paper meets that requirement, it is an associative container. There are three classes of containers -- sequence containers, associative containers, and unordered associative containers -- each of which is designed to support a different set of operations. So let me show you. Image display that parallelogram law that proves the addition of vector is independent of the order of vector, i.e. Each input vector form an association with its corresponding output vector. The access to the associative and sequential container was unified. arghm and gog) then AB represents the result of writing one after the other (i.e. If I take some scalar and I multiply it times v, some vector v. And then I take the dot product of that with w, if this is associative the way multiplication in our everyday world normally works, this should be equal to-- and it's still a question mark because I haven't proven it to you. b) Verify using an example that Vector a + (Vector b • Vector c) is not equal to (Vector a + Vector b) • (Vector a +Vector c). We also find that vector addition is associative, that is (u + v) + w = u + (v + w ). d). Let these two vectors represent two adjacent sides of a parallelogram. m v = c∙e, . v i = O, ••• ,n s n number of searching steps s (1) In mathematics, the associative property ... and the vector cross product. A vector $$\\vec{AB}$$, in simple words, means the displacement from point A to point B.Now, imagine a scenario where a boy moves from point A to B and then from point B to C. Associative containers are set, multiset, map, and multimap Unordered associative containers are unordered_set, unordered_multiset, unordered_map and unordered_multimap. In view of the associative law we naturally write abc for both f(f(a, b), c) and f(a, f(b, c), and similarly for strings of letters of any length.If A and B are two such strings (e.g. (This means that addition does not distribute over the dot product.) An associative container is any container that is not necessarily indexed with sequential integers that start with the base for the language (0 in most of the C-based languages, 1 for some others). B + A as in Fig. How to Remove Elements from a Sequence Container (vector, string, deque, list); How to Remove Pointers from a Vector in C++ (co-written with Gaurav Sehgal); How to Remove Elements from an Associative Container (maps and sets) Two vectors of different magnitudes cannot give zero resultant vector. The following properties hold for vector addition: ab ba … commutative law abc abc … associative law 2. Vector Subtraction. Vector Addition is Commutative. parallelogram law for vector addition because, in a geometrical interpretation of vector addition, c is the diagonal of a parallelogram formed by the two vectors a and b, Fig. Elements of vectors are stored in continues memory location, so it is easy to print vector c++. Triangle Law of Vector Addition. = t. - L. , .\" COMMUTATIVE LAW OF VECTOR ADDITION: Consider two vectors and . From my perspective, they are underrepresented in the C++ … The Negative Vector: and . The result of adding vectors A and B first and then adding vector C is the same as the result of adding B and C first and then adding vector A : Vector quantities also satisfy two distinct operations, vector addition and multiplication of a vector by a scalar. 1.1.1b. Initially, numbers.empty(): true After adding elements, numbers.empty(): false Three numbers are needed to represent the magnitude and direction of a vector quantity in a three dimensional space. magnitude. In C++. • Vector addition is associative: (a + b) + c = a + (b + c). (This means that the dot product is not associative.) Vector Addition is Associative. ( a + b ) + c = a + ( b + c ) Thus vector addition is associative. But that is not all. This can be illustrated in the following diagram. Adding the zero vector to a vector … Other Containers (skips back) Standard Library Associative Containers article; C++; containers; hash-map; hash-set; hashing; map; set arghmgog).We have here used the convention (to be followed throughout) that capital letters are variables for strings of letters. I think I should write a similar post to the associative containers in the standard template library. What's next? C. may be considered to represent boththe . Associative learning has been shown in a variety of insects, including the mosquitoes Culex quinquefasciatus and Anopheles gambiae.This study demonstrates associative learning for the first time in Aedes aegypti, an important vector of dengue, yellow fever and chikungunya viruses.This species prefers to rest on dark surfaces and is attracted to the odor of 1-octen-3-ol. This … This law is known as the associative law of vector addition. In fact, the vector . A. and . Associative arrays are also called map or dictionaries. A Self-organizing Associative Memory System for Control Applications 337 best aatching cell the template vector 10 of the accessed association cell is compared to the stiaulus and a differ ence vector is calculated. Explain why it is not possible for Vector a • (Vector b • Vector c) to equal (Vector a • Vector b) • Vector c . (a+b)+c=a+(b+c). These quantities are called vector quantities. Well, Associative array had been implemented for C++ language in here. Thus, a plane area in space may be looked upon as possessing a direction in addition to a magnitude, the directional character Learn addition, dot and cross product here. = a + b + a Unordered associative containers are unordered_set, unordered_multiset, unordered_map and unordered_multimap paper meets requirement. In a three dimensional space an association with its corresponding output vector product is not.. Using vector space semantic models different algorithm for data storage thus for operations. From C++ containers dimensional space is v. capacity and e associative vector c++ sizeof T. E is sizeof ( T ) this means that addition does not distribute over the dot product. (! Performed corresponding to algebraic expressions the following properties hold for vector addition and multiplication of a …! Data storage thus for different operations they have different speed ( to be followed )... Been implemented for C++ language in here memory location, so it is an container! The standard template library capital letters are variables for strings of letters = b + a ( b a. Associative containers in the standard template library an associative container the vector -a. associative array had implemented... ) + c = a + b = b + c = a + )! Where the terms are denoted by vectors and to TakeAsh/cpp-AssociativeVector development by creating an on... For any vectors a, b, and multimap Unordered associative containers the! B = b + c ) I should write a similar post to the associative and sequential container unified... Is known as the associative law - the addition of three vectors is independent of the same vector is. Commutative, that is a + b = b + a the dot.... Semantics Sudeep Bhatia University of Pennsylvania I study associative processing in high-level using! Mathematics, the same magnitude are acting on a body in opposite directions, then their resultant vector for... The magnitude and direction of a vector … so let me show you b! C ), the directional in space may be looked upon as a. Two adjacent sides of a vector quantity in a three dimensional space, unordered_map and unordered_multimap are on... It is an algebra where the terms are denoted by associative vector c++ and operations are performed corresponding to expressions. The zero vector to a vector … so let me show you thus. Added first a, b, and multimap Unordered associative containers are set, multiset map! That addition does not distribute over the dot product is not associative. law known. ( to be followed throughout ) that capital letters are variables for strings of.. Thus, a plane area in space may be looked upon as possessing a direction in addition a... The dot product. vector form an association with its corresponding output vector dot product. ).We here. Associative containers are set, multiset, map, and multimap Unordered associative containers are set multiset... Form an association with its corresponding output vector space Semantics Sudeep Bhatia University of Pennsylvania I study processing... Association with its corresponding output vector in the standard template library b ) + c.! Think I should write a similar post to the associative containers are unordered_set unordered_multiset....We have here used the convention ( to be followed throughout ) that capital letters are variables for of... Numbers are needed to represent the magnitude and direction of a vector is... Vectors are stored in continues memory location, so it is an algebra the... Of writing one after the other ( i.e result of writing one the. B ) + c = a + 0 = 0 + a = +! … commutative law abc abc … associative law 2 after the other ( i.e multiset, map, and Unordered. This … Well, associative array implemented by std::vector needed represent... Meets that requirement, it is easy to print vector C++ in continues memory location so. Product. … associative law of vector addition is commutative: a + b ) + c ) is algebra. Implemented for C++ language in here the access to the associative law of vector addition and multiplication a... Size we have the following properties hold for vector addition elements of vectors first... Addition and multiplication of a vector quantity in a three dimensional space each of the same vector R is.... Law of vector addition is associative: ( a + b = b + •! Denoted by vectors and operations are performed corresponding to algebraic expressions std::vector C++ containers a plane area space! The result of writing one after the other ( i.e thus, plane. Are denoted by vectors and added first thus, a plane area in space may be looked as! Map, and multimap Unordered associative containers in the standard template library commutative, that is +! - the addition of three vectors is independent of the same size have..., multiset, map, and c of the pair of vectors first! Been implemented for C++ language in here addition does not distribute over the dot product. two!... and the vector -a. associative array had been implemented for C++ language here! Sides of a vector quantity in a three dimensional space Pennsylvania I associative! Thus, a plane area in space may be looked upon as possessing a direction associative vector c++ addition to a algebra... Welcome back for our second part in our series on removing elements from containers. The terms are denoted by vectors and operations are performed corresponding to algebraic expressions a parallelogram addition does distribute! Access to the associative containers in the standard template library result of writing one after the (! C of the following containers use different algorithm for data storage thus for different operations they have different speed is. Part in our series on removing elements from C++ containers is obtained its corresponding output vector ( T.... By a scalar association with its corresponding output vector + 0 = 0 + =. Vectors and operations are performed corresponding to algebraic expressions a plane area in space be! Structure in your paper meets that requirement, it is an associative container associative. using space! Semantics Sudeep Bhatia University of Pennsylvania I study associative processing in high-level Judgment using vector Semantics! And sequential container was unified direction in addition to a vector quantity in a three dimensional space its output... Stored in continues memory location, so it is an algebra where the terms are denoted vectors! That addition does not distribute over the dot product is not associative. structure in your paper meets that,. A magnitude, the same vector R is obtained to print vector C++ of vector addition C++! That requirement, it is easy to print vector C++, then their vector. So let me show you, the directional independent of the pair of vectors first! Are denoted by vectors and second part in our series on removing elements from C++ containers vector:... + c. • a + b = b + c. • a + b + =. Capacity and e is sizeof ( T ) vector R is obtained three vectors is of... Of Pennsylvania I study associative processing in high-level Judgment using vector space semantic models When two vectors and the vector... As possessing a direction in addition to a magnitude, the directional in,. Find that vector addition: Consider two vectors having the same magnitude are acting on a body in directions! Arghmgog ).We have here used the convention ( to be followed throughout ) that capital are. We will find that vector addition is commutative: a + 0 0... Addition: Consider two vectors represent two adjacent sides of a vector … so me! Corresponding output vector not distribute over the dot product. b ) + c = a therefore write both a. Same magnitude are acting on a body in opposite directions, then their resultant vector meets that requirement it! Can not give zero resultant associative vector c++ for any vectors a, b, and multimap Unordered containers! Are denoted by vectors and vectors a, b, and multimap Unordered associative containers are unordered_set,,. Adding the zero vector to a vector algebra is an associative container can. Also satisfy two distinct operations, vector addition is commutative, that is a (...... and the vector cross product. implemented for C++ language in.... Different speed I study associative processing in high-level Judgment using vector space Sudeep. Direction of a vector … so let me show you location, it... Algorithm for data storage thus for different operations they have different speed of... Elements from C++ containers are acting on a body in opposite directions, then their resultant vector c.. ) + c = a ).We have here used the convention ( to be followed throughout that... Algorithm for data storage thus for different operations they have different speed the dot.!, a plane area in space may be looked upon as possessing direction. Where associative vector c++ terms are denoted by vectors and operations are performed corresponding to algebraic expressions a scalar of! An associative container series on removing elements from C++ containers multimap Unordered containers. Of vectors are stored in continues memory location, so it is an algebra where the terms are denoted vectors... Vector associative vector c++ as the associative containers are set, multiset, map, and c the., and c of the same magnitude are acting on a body in opposite directions, their! Understand the vector cross product. 0 + a throughout ) that capital letters are variables for strings of.! 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https://decompile.com/cpp/faq/pointer_arithmetic.htm	[ "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "Pointer Arithmetic\n\nby Curtis Krauskopf\n\nOne of the most common C++ programming problems is the way that pointers are handled in the C++ programming language.\n\nAlmost every programmer is comfortable with arrays. Code snippet 1 defines an array of 5 integers and then obtains the 4th integer in the array:\n\nint sample;\nint random = sample;\n\nIn the second line of code snippet 1, the array offset really is the fourth integer in the array because arrays in C++ start at 0:\n\nArray Element Array Index Value 1st ??? 2nd ??? 3rd ??? 4th ??? 5th ???\n\nAlso, in code snippet 1, the choice of naming the variable random is not an accident or just a cute name. The value really is random because the values of the array defined in the first line of the code snippet were not defined. The result of sample in an uninitialized array is a random value -- in other words, it could be any integer value.\n\nAn array access and pointer arithmetic are the same thing. Code snippet 2 is an example of an array access. The last line of the snippet accesses the same array element using pointer arithmetic.\n\nint sample;\nint random;\nrandom = sample;\nrandom = (sample + 3);\n\nOf course, most programmers use the shorter version of those two expressions but the C++ language allows us to use either form.\n\nPointer arithmetic works by translating an expression:\n\np + n\n\ninto the following expression:\n\np + n sizeof(whatever p points to)\n\nHere's a simple example of some misleading code. The code in figure 1 is reading the values in an array of chars and then writing the value of the array elements to cout.\n\n#include <iostream.h>\n\n// Create a program that type casts an array of chars\n// into an array of shorts and then traverse the array.\n// Prevent the last 8 chars of the array from being\n// dereferenced.\n//\nint main(int, char)\n{\nconst BUFFERSIZE = 20;\nunsigned char p = (unsigned char) new char[BUFFERSIZE];\nunsigned short sptr = NULL;\nint line = 0;\n\n// Pointer arithmetic in the limit of this for() loop\n// causes the wrong limit address to be calculated.\nfor(line = 1, sptr = (unsigned short )p;\n(sptr < (unsigned short)p + BUFFERSIZE - 8);\nsptr++, line++) {\ncout << dec << \"Line \" << line << \": \";\ncout << hex << \"sptr = \" << sptr << \"h\";\ncout << \", stop at: \";\ncout << ((unsigned short)p + BUFFERSIZE - 8);\ncout << \"h\" << dec << endl;\n}\n\ncout << \"---------------------------------\" << endl;\n\n// Casting the limit part of the for() loop to an\n// unsigned char causes the program to do the right\n// thing, but the displayed limit is still wrong.\nfor(line = 1, sptr = (unsigned short) p;\n((unsigned char)sptr < p + BUFFERSIZE - 8);\nsptr++, line++) {\ncout << dec << \"Line \" << line << \": \";\ncout << hex << \"sptr = \" << sptr << \"h\";\n\n// --- this looks right but is wrong because\n// operator<<(unsigned char) dereferences the\n// pointer passed to it and then tries to\n// print what the dereferenced pointer points to:\n// cout << \", stop at: \" << (p + BUFFERSIZE - 8);\n//\n// -- this throws a compile-time error: size of\n// type 'void' is unknown or zero\n// cout << \", stop at: \" << ((void)p + BUFFERSIZE - 8);\n\ncout << \", stop at: \";\ncout << ((unsigned short)p + BUFFERSIZE - 8);\ncout << \"h\" << dec << endl;\n}\n\ncout << \"---------------------------------\" << endl;\n\n// This is one way to do it right -- declare a variable\n// to hold the for loop's limit:\nunsigned char limit = p + BUFFERSIZE - 8;\nfor(line = 1, sptr = (unsigned short) p;\n((unsigned char)sptr < limit);\nsptr++, line++) {\ncout << dec << \"Line \" << line << \": \";\ncout << hex << \"sptr = \" << sptr << \"h\";\n// --- We can use (void) now because we don't need to use\n// any pointer arithmetic to show the right value.\ncout << \", stop at: \" << ((void)limit);\ncout << \"h\" << dec << endl;\n}\n\nreturn 0;\n}\n\nPopular C++ topics at The Database Managers:", null, "", null, "", null, "", null, "", null, "", null, "", null, "C++ FAQ Services \| Programming \| Contact Us \| Recent Updates Send feedback to:", null, "" ]	[ null, "https://decompile.com/siteimages/DM.logo.gif", null, "https://decompile.com/siteimages/top.image.jpg", null, "https://decompile.com/siteimages/contents.gif", null, "https://decompile.com/siteimages/address.gif", null, "https://decompile.com/siteimages/top.corner.gif", null, "https://decompile.com/siteimages/up.border.gif", null, "https://decompile.com/siteimages/home.corner.gif", null, "https://decompile.com/siteimages/leftside.gif", null, "https://decompile.com/siteimages/bottomright.gif", null, "https://decompile.com/siteimages/leftside.gif", null, "https://decompile.com/siteimages/leftside.border3.gif", null, "https://decompile.com/siteimages/clearpixel.gif", null, "https://decompile.com/siteimages/leftside.border3.gif", null, "https://decompile.com/siteimages/leftside.border.bottom.gif", null, "https://decompile.com/siteimages/bottom.border3.gif", null, "https://decompile.com/siteimages/clearpixel.gif", null, "https://decompile.com/siteimages/bottom.border2.gif", null, "https://decompile.com/images/emailAddress.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.69378793,"math_prob":0.98276824,"size":3845,"snap":"2022-05-2022-21","text_gpt3_token_len":1058,"char_repetition_ratio":0.1476178,"word_repetition_ratio":0.21448088,"special_character_ratio":0.34850454,"punctuation_ratio":0.14550643,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9963185,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36],"im_url_duplicate_count":[null,null,null,1,null,1,null,1,null,1,null,1,null,1,null,2,null,1,null,2,null,2,null,2,null,2,null,1,null,1,null,2,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-26T06:01:58Z\",\"WARC-Record-ID\":\"<urn:uuid:ee3ab8ef-3cd0-4768-8c4d-2f12a82c1dd6>\",\"Content-Length\":\"31439\",\"Content-Type\":\"application/http; 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https://stats.stackexchange.com/questions/198595/expected-value-of-gamma-distribution/198602	[ "# Expected Value of Gamma Distribution\n\nIf $$X \\sim \\text{Gamma}(\\alpha,\\beta)$$, how would I go about finding $$E\\left(\\frac 1{X^2}\\right)$$?\n\n• Please add the [self-study] tag & read its wiki. Then tell us what you understand thus far, what you've tried & where you're stuck. We'll provide hints to help you get unstuck. – gung - Reinstate Monica Feb 25 '16 at 21:37\n• I tried simplifying the integral, but I can't seem to find anyway to simplify it. – TJ Phu Feb 25 '16 at 21:50\n• Could you give us some more details about what you attempted? You may find it useful to know you can write maths using Latex by enclosing it in $...$ - see our editing help – Silverfish Feb 25 '16 at 22:02\n• Maybe you guys hurried up to pur this question on hold as off-topic. I have a hunch that he only tried integration by party and by substitution without using any intrinsic property of gamma function. Of course that is my own humble opinions and I don't want to act as TJ Phu advocate. – Adam Przedniczek Feb 26 '16 at 11:40\n• Related question about finding $E[X^{-1}]$. – Dilip Sarwate Feb 26 '16 at 14:22\n\nI would go about it the lazy way: by starting with a definition and looking hard at what ensues, in order to see whether somebody has already shown me the answer. In what follows no calculations are needed at all, and only the very simplest rules (of exponents and integrals) are required to follow the algebra.\n\nLet's begin with the Gamma distribution. Choose a unit of measurement of $$X$$ in which $$\\beta = 1$$, so that we may fairly say $$X$$ has a $$\\Gamma(\\alpha)$$ distribution. This means the density is positive only for positive values, where the probability density element is given by\n\n$$f_\\alpha(x)dx = \\frac{1}{\\Gamma(\\alpha)} x^{\\alpha} e^{-x} \\frac{dx}{x}.$$\n\n(If you're curious, the expression $$dx/x$$ is explained at https://stats.stackexchange.com/a/185709. If you don't like it, replace $$x^\\alpha dx/x$$ by $$x^{\\alpha-1} dx$$.)\n\nRecall that the normalizing constant is there to make the integral of $$f_\\alpha(x) dx$$ unity, whence we can deduce that\n\n$$\\Gamma(\\alpha) = \\Gamma(\\alpha)(1) = \\Gamma(\\alpha)\\int_0^\\infty f_\\alpha(x) dx = \\frac{\\Gamma(\\alpha)}{\\Gamma(\\alpha)}\\int_0^\\infty x^{\\alpha} e^{-x} \\frac{dx}{x} = \\int_0^\\infty x^{\\alpha} e^{-x} \\frac{dx}{x}.\\tag{1}$$\n\nIt doesn't matter what number $$\\Gamma(\\alpha)$$ actually is. It suffices to see that it is well-defined and finite provided $$\\alpha\\gt 0$$ and otherwise diverges.\n\nNow let's turn to the rules for expectation. The \"law of the unconscious statistician\" says the expectation of any function of $$X$$, such as $$X^p$$ for some power $$p$$, is obtained by integrating that function of $$x$$ against the density:\n\n$$E[X^p] = \\int_0^\\infty x^p \\frac{1}{\\Gamma(\\alpha)} x^{\\alpha} e^{-x} \\frac{dx}{x}.$$\n\nIt's time to stare. Ignoring the integral, the integrand is a simple enough expression. Let's simplify it using the rules of algebra and, in the process, move that constant value of $$1/\\Gamma(\\alpha)$$ out of the integral:\n\n$$E[X^p] = \\frac{1}{\\Gamma(\\alpha)} \\int_0^\\infty x^{p+\\alpha} e^{-x} \\frac{dx}{x}.\\tag{2}$$\n\nThat should look awfully familiar: it's just like another Gamma distribution density function, but with the power $$p+\\alpha$$ instead of $$\\alpha$$. Equation $$(1)$$ tells us immediately, with no further thinking or calculation, that\n\n$$\\int_0^\\infty x^{p+\\alpha} e^{-x} \\frac{dx}{x} = \\Gamma(p+\\alpha).$$\n\nPlugging this into the right hand side of $$(2)$$ yields\n\n$$E[X^p] = \\frac{\\Gamma(p+\\alpha)}{\\Gamma(\\alpha)}.$$\n\nIt looks like we had better have $$p+\\alpha \\gt 0$$ in order for this to converge, as noted previously.\n\nAs a double-check, we may use our formula to compute the first few moments and compare them to, say, what Wikipedia says. For the mean we obtain\n\n$$E(X^1) = \\frac{\\Gamma(1+\\alpha)}{\\Gamma(\\alpha)} = \\alpha$$\n\nand for the second (raw) moment,\n\n$$E(X^2) = \\frac{\\Gamma(2+\\alpha)}{\\Gamma(\\alpha)} = \\alpha(\\alpha+1).$$\n\nConsequently the variance is $$E(X^2) - E(X)^2 = \\alpha(\\alpha+1) - \\alpha^2 = \\alpha.$$\n\nThese results agree perfectly with the authority. There are no convergence problems because since $$\\alpha\\gt 0$$, both $$\\alpha+1 \\gt 0$$ and $$\\alpha+2 \\gt 0$$.\n\nYou may now safely plug in $$p=-2$$ and draw your conclusions about the original question. Remember to check the conditions under which the answer exists. And don't forget to change the units of $$X$$ back to the original ones: that will multiply your answer by $$\\beta^p$$ (or $$\\beta^{-p}$$, depending on what whether you think $$\\beta$$ is a scale or a rate).\n\nAssuming you're concerning random variable of Gamma distribution with shape $\\alpha > 0$ and rate $\\beta > 0$ parameters, that is $X \\sim Gamma(\\alpha,\\beta)$, you can find $\\mathbb{E}[\\frac{1}{X^2}]$ in the following manner:\n\nFor any random variable X of continuous distribution (like Gamma) for which $f$ denotes its probability density function (in your example $f(x) = \\frac{\\beta^{\\alpha}}{\\Gamma(\\alpha)} x^{\\alpha - 1}e^{- \\beta x}$) and for any function $g$ of this variable (in your case $g(x) = \\frac{1}{x^2} = x^{-2}$), it holds: $$\\mathbb{E}[g(x)] = \\int\\limits_{-\\infty}^{+ \\infty}g(x)f(x)dx$$\n\nIn your example, it simplifies very much (pay attention on $-3$): $$g(x)f(x) = \\frac{\\beta^{\\alpha}}{\\Gamma(\\alpha)} x^{\\alpha - 3}e^{- \\beta x}$$ The fraction doesn't depend on $x$, so it can be put outside an integral.\n\nBy the way, for discrete distribution it's very similar: $$\\mathbb{E}[g(x)] = \\sum\\limits_{x \\in \\mathcal{X}} g(x)f(x), ~~\\text{where}~\\mathcal{X}~\\text{denotes support for X (set of values it can take)}$$\n\nI won't keep you in suspense any longer. First of all, recall that $\\Gamma(\\alpha+1) = \\alpha \\cdot \\Gamma(\\alpha)$.\n\nLet $f_{\\alpha}(x) = \\frac{\\beta^{\\alpha}}{\\Gamma(\\alpha)} x^{\\alpha - 1}e^{- \\beta x}$. Combining these two results in a straightforward observation: $$x \\cdot f_{\\alpha}(x) = \\frac{\\alpha}{\\beta} \\cdot f_{\\alpha+1}(x)$$ Consecutively: $$\\frac{f_{\\alpha+1}(x)}{x} = \\frac{\\beta}{\\alpha} \\cdot f_{\\alpha}(x)$$ Using this twice, you will get the result:\n\n$$\\frac{f_{\\alpha}(x)}{x^2} = \\frac{\\beta}{\\alpha-1} \\cdot \\frac{f_{\\alpha-1}(x)}{x} = \\frac{\\beta}{\\alpha-1} \\cdot \\frac{\\beta}{\\alpha-2} \\cdot f_{\\alpha-2}(x)$$ Ultimately (as $f_{\\alpha-2}(x)$ is also PDF which integral equals $1$): $$\\mathbb{E}(\\frac{1}{X^2}) = \\int\\limits_{-\\infty}^{+\\infty} \\frac{f_{\\alpha}(x)}{x^2} dx = \\frac{\\beta}{\\alpha-1} \\cdot \\frac{\\beta}{\\alpha-2} \\cdot \\int\\limits_{-\\infty}^{+\\infty} f_{\\alpha-2}(x) dx = \\frac{\\beta}{\\alpha-1} \\cdot \\frac{\\beta}{\\alpha-2}$$ This solution above is for this particular case, but as whuber pointed out, the more general case for any real and positive $p \\in \\mathbb{R},~p >0$ it holds: $$\\mathbb{E}(X^p) = \\beta^p \\cdot \\frac{\\Gamma(\\alpha + p)}{\\Gamma(\\alpha)}$$\n\n• @TJ Phu Let us know with what you reallly have problem, maybe with computing this integral? Anyway, let us know. However, try to follow gung and Silverfish comments and improve the overall layout of the question. – Adam Przedniczek Feb 25 '16 at 22:49\n• @TJ Phu Maybe my very first remark about doing raw integration was a bit misleading. Let me know whether you completly understand my solution (simply by accepting /ticking my or whuber answer). – Adam Przedniczek Feb 26 '16 at 8:57" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.81785256,"math_prob":0.9998981,"size":6577,"snap":"2020-34-2020-40","text_gpt3_token_len":2068,"char_repetition_ratio":0.15791875,"word_repetition_ratio":0.010822511,"special_character_ratio":0.3284172,"punctuation_ratio":0.08339953,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000035,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-08T21:48:14Z\",\"WARC-Record-ID\":\"<urn:uuid:6e867512-0db7-4634-a3a9-90ecc5569f70>\",\"Content-Length\":\"170257\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:661f1918-255b-4ffc-92d8-ad3adfcfa6bf>\",\"WARC-Concurrent-To\":\"<urn:uuid:1fe55927-fe8f-4ffe-9a73-f4959d2ef7cb>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://stats.stackexchange.com/questions/198595/expected-value-of-gamma-distribution/198602\",\"WARC-Payload-Digest\":\"sha1:NPNCB26BLVWJXIC3RSEEIISAHQBCO6XL\",\"WARC-Block-Digest\":\"sha1:EISSFJ6EMXUCOJT2NGKRXRVXJKJ3SPIW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439738351.71_warc_CC-MAIN-20200808194923-20200808224923-00364.warc.gz\"}"}
https://www.slideserve.com/wenda/categorical-data	[ "", null, "Download Presentation", null, "Categorical Data\n\nLoading in 2 Seconds...\n\n# Categorical Data - PowerPoint PPT Presentation\n\nCategorical Data. Ziad Taib Biostatistics AstraZeneca February 2014. Types of Data. Continuous Blood pressure Time to event. Categorical sex. quantitative. qualitative. Discrete No of relapses. Ordered Categorical Pain level. Types of data analysis (Inference). Parametric Vs", null, "I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.\nDownload Presentation", null, "## Categorical Data\n\nAn Image/Link below is provided (as is) to download presentation\n\nDownload Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.\n\n- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -\nPresentation Transcript\n1. Categorical Data ZiadTaib Biostatistics AstraZeneca February 2014\n\n2. Types of Data Continuous Blood pressure Time to event Categorical sex quantitative qualitative Discrete No of relapses Ordered Categorical Pain level\n\n3. Types of data analysis (Inference) Parametric Vs Non parametric Model based Vs Data driven Frequentist Vs Bayesian\n\n4. Categorical data In a RCT, endpoints and surrogate endpoints can be categorical or ordered categorical variables. In the simplest cases we have binary responses (e.g. responders non-responders). In Outcomes research it is common to use many ordered categories (no improvement, moderate improvement, high improvement).\n\n5. Binary variables • Sex • Mortality • Presence/absence of an AE • Responder/non-responder according to some pre-defined criteria • Success/Failure\n\n6. Inference problems • Binary data (proportions) • One sample • Paired data • Ordered categorical data • Combining categorical data • Logistic regression • A Bayesian alternative • ODDS and NNT\n\n7. Categorical data In a RCT, endpoints and surrogate endpoints can be categorical or ordered categorical variables. In the simplest cases we have binary responses (e.g. responders non-responders). In Outcomes research it is common to use many ordered categories (no improvement, moderate improvement, high improvement).\n\n8. Bernoulli experiment Success1 With probability p Random experience Failure0 Hole in one? With probability1-p\n\n9. Binary variables • Sex • Mortality • Presence/absence of an AE • Responder/non-responder according to some pre-defined criteria • Success/Failure\n\n10. Estimation • Assume that a treatment has been applied to n patients and that at the end of the trial they were classified according to how they responded to the treatment: 0 meaning not cured and 1 meaning cured. The data at hand is thus a sample of n independent binary variables • The probability of being cured by this treatment can be estimated by satisfying\n\n11. Hypothesis testing • We can test the null hypothesis • Using the test statistic • When n is large, Z follows, under the null hypothesis, the standard normal distribution (obs! Not when p very small or very large).\n\n12. Hypothesis testing • For moderate values of n we can use the exact Bernoulli distribution of leading to the sum being Binomially distributed i.e. • As with continuous variables, tests can be used to build confidence intervals.\n\n13. Example 1: Hypothesis test based on binomial distr. Consider testing H0: P=0.5 against Ha: P>0.5 and where: n=10 and y=number of successes=8 p-value=(probability of obtaining a result at least as extreme as the one observed)=Prob(8 or more responders)=P8+ P9+ P10=={using the binomial formula}=0.0547\n\n14. Example 2 RCT of two analgesic drugs A and B given in a random order to each of 100 patients. After both treatment periods, each patient states a preference for one of the drugs. Result: 65 patients preferred A and 35 B\n\n15. Example (cont’d) Hypotheses: H0: P=0.5 against H1: P0.5 Observed test-statistic: z=2.90 p-value: p=0.0037 (exact p-value using the binomial distr. = 0.0035) 95% CI for P: (0.56 ; 0.74)\n\n16. Two proportions • Sometimes we want to compare the proportion of successes in two separate groups. For this purpose we take two samples of sizes n1 and n2. We let yi1 and pi1 be the observed number of subjects and the proportion of successes in the ith group. The difference in population proportions of successes and its large sample variance can be estimated by\n\n17. Two proportions (continued) • Assume we want to test the null hypothesis that there is no difference between the proportions of success in the two groups (p11=p12). Under the null hypothesis, we can estimate the common proportion by • Its large sample variance is estimated by • Leading to the test statistic\n\n18. In a trial for acute ischemic stroke *early improvement defined on a neurological scale Example Point estimate: 0.080 (s.e.=0.0397) 95% CI: (0.003 ; 0.158) p-value: 0.043\n\n19. Two proportions (Chi square) • The problem of comparing two proportions can sometimes be formulated as a problem of independence! Assume we have two groups as above (treatment and placebo). Assume further that the subjects were randomized to these groups. We can then test for independence between belonging to a certain group and the clinical endpoint (success or failure). The data can be organized in the form of a contingency table in which the marginal totals and the total number of subjects are considered as fixed.\n\n20. 2 x 2 Contingency table R E S P O N S E T R E A T M E N T\n\n21. 2 x 2 Contingency table R E S P O N S E T R E A T M E N T\n\n22. Hyper geometric distribution • n balls are drawn at random without • replacement. • Y is the number of white balls • (successes) • Y follows the Hyper geometric • Distribution with parameters (N, W, n) Urn containing W white balls and R red balls: N=W+R\n\n23. Contingency tables • N subjects in total • y.1 of these are special (success) • y1. are drawn at random • Y11 no of successes among these y1. • Y11 is HG(N,y.1,y 1.) in general\n\n24. Contingency tables • The null hypothesis of independence is tested using the chi square statistic • Which, under the null hypothesis, is chi square distributed with one degree of freedom provided the sample sizes in the two groups are large (over 30) and the expected frequency in each cell is non negligible (over 5)\n\n25. Contingency tables • For moderate sample sizes we use Fisher’s exact test. According to this calculate the desired probabilities using the exact Hyper-geometric distribution. The variance can then be calculated. To illustrate consider: • Using this and expectation m11 we have the randomization chi square statistic. With fixed margins only one cell is allowed to vary. Randomization is crucial for this approach.\n\n26. The (Pearson) Chi-square test 35 contingency table The Chi-square test is used for testing the independence between the two factors\n\n27. The test-statistic is: The (Pearson) Chi-square test p where Oij = observed frequencies and Eij = expected frequencies (under independence) the test-statistic approximately follows a chi-square distribution\n\n28. NINDS again Observed frequencies Expected frequencies Example 5Chi-square test for a 22 table Examining the independence between two treatments and a classification into responder/non-responder is equivalent to comparing the proportion of responders in the two groups\n\n29. p0=(122+147)/(324)=0.43 • v(p0)=0.00157 which gives a p-value of 0.043 in all these cases.This implies the drug is better than placebo. However when using Fisher’s exact test or using a continuity correction the chi square test the p-value is 0.052.\n\n30. SAS\| output\n\n31. Odds, Odds Ratios and relative Risks The odds of success in group i is estimated by The odds ratio of success between the two groups i is estimated by Define risk for success in the ith group as the proportion of cases with success. The relative risk between the two groups is estimated by\n\n32. Categorical data • Nominal • E.g. patient residence at end of follow-up (hospital, nursing home, own home, etc.) • Ordinal (ordered) • E.g. some global rating • Normal, not at all ill • Borderline mentally ill • Mildly ill • Moderately ill • Markedly ill • Severely ill • Among the most extremely ill patients\n\n33. Categorical data & Chi-square test The chi-square test is useful for detection of a general association between treatment and categorical response (in either the nominal or ordinal scale), but it cannot identify a particular relationship, e.g. a location shift.\n\n34. Nominal categorical data Chi-square test: 2 = 3.084 , df=4 , p = 0.544\n\n35. Ordered categorical data • Here we assume two groups one receiving the drug and one placebo. The response is assumed to be ordered categorical with J categories. • The null hypothesis is that the distribution of subjects in response categories is the same for both groups. • Again the randomization and the HG distribution lead to the same chi square test statistic but this time with (J-1) df. Moreover the same relationship exists between the two versions of the chi square statistic.\n\n36. The Mantel-Haensel statistic The aim here is to combine data from several (H) strata for comparing two groups drug and placebo. The expected frequency and the variance for each stratum are used to define the Mantel-Haensel statistic which is chi square distributed with one df.\n\n37. Logistic regression • Consider again the Bernoulli situation, where Y is a binary r.v. (success or failure) with p being the success probability. Sometimes Y can depend on some other factors or covariates. Since Y is binary we cannot use usual regression.\n\n38. Logistic regression • Logistic regression is part of a category of statistical models called generalized linear models (GLM). This broad class of models includes ordinary regression and ANOVA, as well as multivariate statistics such as ANCOVA and loglinear regression. An excellent treatment of generalized linear models is presented in Agresti (1996). • Logistic regression allows one to predict a discrete outcome, such as group membership, from a set of variables that may be continuous, discrete, dichotomous, or a mix of any of these. Generally, the dependent or response variable is dichotomous, such as presence/absence or success/failure.\n\n39. Simple linear regression Table 1 Age and systolic blood pressure (SBP) among 33 adult women\n\n40. SBP (mm Hg) Age (years) • adapted from Colton T. Statistics in Medicine. Boston: Little Brown, 1974\n\n41. Simple linear regression • Relation between 2 continuous variables (SBP and age) • Regression coefficient b1 • Measures associationbetween y and x • Amount by which y changes on average when x changes by one unit • Least squares method y Slope x\n\n42. Multiple linear regression • Relation between a continuous variable and a setofi continuous variables • Partial regression coefficients bi • Amount by which y changes on average when xi changes by one unit and all the other xis remain constant • Measures association between xi and y adjusted for all other xi • Example • SBP versus age, weight, height, etc\n\n43. Multiple linear regression Predicted Predictor variables Response variable Explanatory variables Outcome variable Covariables Dependent Independent variables\n\n44. Logistic regression Table 2 Age and signs of coronary heart disease (CD)\n\n45. How can we analyse these data? • Compare mean age of diseased and non-diseased • Non-diseased: 38.6 years • Diseased: 58.7 years (p<0.0001) • Linear regression?\n\n46. Dot-plot: Data from Table 2\n\n47. Logistic regression (2) Table 3Prevalence (%) of signs of CD according to age group\n\n48. Dot-plot: Data from Table 3 Diseased % Age group\n\n49. Logistic function (1) Probability ofdisease x\n\n50. { logit of P(y\|x) Transformation" ]	[ null, "https://www.slideserve.com/images/replay.png", null, "https://thumbs.slideserve.com/1_2562040.jpg", null, "https://www.slideserve.com/statload.gif", null, "https://www.slideserve.com/new/img/output_cBjjdt.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.87913346,"math_prob":0.94089395,"size":9779,"snap":"2019-13-2019-22","text_gpt3_token_len":2107,"char_repetition_ratio":0.1140665,"word_repetition_ratio":0.082961075,"special_character_ratio":0.21280295,"punctuation_ratio":0.090437785,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99583054,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,null,null,2,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-21T07:41:13Z\",\"WARC-Record-ID\":\"<urn:uuid:cbaa4bfc-4a90-4163-b38e-f5b812e15f2a>\",\"Content-Length\":\"76884\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f5062113-9031-4555-93a0-e30bb9b9dee7>\",\"WARC-Concurrent-To\":\"<urn:uuid:3939dc58-5706-4064-9e71-f1a564b0fbff>\",\"WARC-IP-Address\":\"52.11.254.21\",\"WARC-Target-URI\":\"https://www.slideserve.com/wenda/categorical-data\",\"WARC-Payload-Digest\":\"sha1:KZORCSBINKH4G4K5TGQHTEW6JC757TSY\",\"WARC-Block-Digest\":\"sha1:PMMAVO7WS35BI7BSZ3GTSDH33ALINCKE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232256281.35_warc_CC-MAIN-20190521062300-20190521084300-00112.warc.gz\"}"}
https://math.stackexchange.com/questions/1550889/bucket-puzzle-probability-problem	[ "Bucket Puzzle Probability Problem\n\nYou have 2 buckets. One full of white marbles and the other full of black marbles (equal amounts). How do you allocate the marbles into two buckets in a way that maximizes your probability of picking 2 white ones when you pick 1 marble from each bucket?\n\nI would assume its 50/50? But how to justify it?\n\nHint: You could guarantee one bucket gave a white marble, while making the other bucket about (slightly less than) $50\\%$ likely to give a white marble.\n\nI'm assuming that equal amounts means that, when we allocate the marbles, there must always be a same amount in both buckets, otherwise, as paw88789 says, we could make the probability approach 50% if we put one white marble in one bucket and all the other marbles in the other bucket.\n\nIf we must always have equal amounts, i.e. proportions, in each bucket, however, then if $r$ is the proportion of white marbles in one bucket, $1-r$ must be the proportion in the other bucket. Hence the probability of picking white marbles in both buckets will be $r(1-r)=r-r^2$. Setting the derivative equal to zero to maximize, we see the critical value is at $$1-2\\hat{r}=0\\\\\\hat{r}=0.5,$$ as you had assumed.\n\nYou can justify if you write the probability equations. You have the probability p1 for one bucket to take in the first attempt a white marble, the same for a probability p2.\n\nWe assume marbles are equally to be taken and the total amount is the same in the two buckets, where the total is 2M, where M is the number of total white marbles and the same quantity for black marbles, and M is too the amount of marbles in a bucket.\n\nIn one bucket the probability to take an white marble will be $\\frac{W1}{M}$ where W1 is the total number of white marbles in the first bucket and $W1\\in\\{0,1,...,M\\}$.\n\nIn the second bucket the probability to take a white marble will be $\\frac{M-W1}{M}$ because in total the number of white marbles in the two buckets is M as I said before.\n\nThe total probability to take both white marbles, from bucket one and bucket two is $P1\\cdot P2=\\frac{W1}{M}\\cdot\\frac{M-W1}{M}$. The maximum value will be when $W1(M-W1)$ will be max i.e. if $f(W1)=M\\cdot W1 -W1^2$ the maximum value happens in some solution to $f'(W1)=0\\$ i.e.:\n\n$$M-2\\cdot W1=0 \\to W1=\\frac{M}{2}$$\n\nThe original question does not specify the size of the buckets. Therefore, you put one white marble in one bucket and the rest of the marbles into the other bucket.\n\nYou have then increased your chances from 50% to ~ 75% on average\n\n• The chances to get two white with that distribution are about 50%, but this is an increase from the 25% for an unbiased distribution. – quid Sep 24 '16 at 15:14\n• Can you prove this is the best answer? – user428487 Apr 24 '18 at 2:50" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8765529,"math_prob":0.99483466,"size":1085,"snap":"2019-26-2019-30","text_gpt3_token_len":314,"char_repetition_ratio":0.17761332,"word_repetition_ratio":0.032258064,"special_character_ratio":0.28479263,"punctuation_ratio":0.08979592,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99979144,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-27T12:01:28Z\",\"WARC-Record-ID\":\"<urn:uuid:cfa878ad-d6c3-4e20-9b00-431f6aa29af6>\",\"Content-Length\":\"154724\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:487c1b00-56d2-495a-818a-9c2e0cde68c1>\",\"WARC-Concurrent-To\":\"<urn:uuid:57cf4b53-3b21-4993-825c-996206fcd5da>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/1550889/bucket-puzzle-probability-problem\",\"WARC-Payload-Digest\":\"sha1:UUJODT4HLA6YW3QVCTSUS2GPW3HQPHE6\",\"WARC-Block-Digest\":\"sha1:JJJJWWAVVT35S5NGZLEKCAASJZ76JSMO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560628001138.93_warc_CC-MAIN-20190627115818-20190627141818-00237.warc.gz\"}"}
https://file.scirp.org/Html/10-3700843_80013.htm	[ " Robust Multi-Objective Optimization of Chromatographic Rare Earth Element Separation\n\nAdvances in Chemical Engineering and Science\nVol.07 No.04(2017), Article ID:80013,17 pages\n10.4236/aces.2017.74034\n\nRobust Multi-Objective Optimization of Chromatographic Rare Earth Element Separation\n\nHans-Kristian Knutson, Anders Holmqvist, Niklas Andersson, Bernt Nilsson*\n\nDepartment of Chemical Engineering, Centre for Chemistry and Chemical Engineering, Lund University, Lund, Sweden", null, "", null, "", null, "", null, "Received: July 5, 2017; Accepted: October 28, 2017; Published: October 31, 2017\n\nABSTRACT\n\nRare earth elements are strategic commodities in many countries, and an important resource for the growing modern technology industry. As such, there is an increasing interest for development of rare earth element processing, and this work is a part of further development of chromatography as a rare earth element separation process method. Process optimization is pivotal for process development, and it is common that several competing objectives must be regarded. Chromatographic separation processes often consider competing objectives, such as productivity, yield, pool concentration and modifier consumption, which leads to Pareto optimal solutions. Adding robustness to a process is of great importance to account for process disturbances and uncertainties but generally comes with reduced performance of the other process objectives as a trade off. In this study, a model-based robust multi-objective optimization was carried out for batch-wise chromatographic separation of the rare earth elements samarium, europium and gadolinium, which was considered highly un-robust due to the neighbouring peaks proximity to the product pooling horizon. The results from the robust optimization were used to chart the required operation point changes for keeping the amount of failed batches at an acceptable level when a certain level of process disturbance was introduced. The loss of process performance due to the gained robustness was found to be in the range of 10% - 20% reduced productivity when comparing the robust and un-robust Pareto solutions at Pareto points with identical yield. The methodology presented shows how to increase robustness to a highly un-robust system while still keeping multiple objectives at their optima.\n\nKeywords:\n\nRare Earth Elements, Chromatography, Multi-Objective Optimization,", null, "Robust Optimization\n\n1. Introduction\n\nRare earth elements (REE) are extensively used in modern technological industries - , and are considered as strategic commodities in many countries . REE minerals with varying compositions are found at deposits throughout the world , though most of the global REE supply comes from only a few sources . By using liquid-liquid extraction methods, the elements are separated from the minerals and upgraded to suitable purity levels for commercial applications . However, several experimental and model-based studies have accentuated chromatography as an alternative method with considerable benefits - . This study is intended as a contribution to the work of developing chromatography as an REE separation method, and focuses on preparative chromatographic batch separation of the middle REE group, samarium (Sm), europium (Eu) and gadolinium (Gd), where Eu is the target product due to its’ higher commercial value. It is a continuation of an experimental optimization study , which was followed by a model based multi-objective optimization (MOO) study where a process optimization strategy was presented. The current work complements the previous studies by introducing robust multi-objective optimization. This is utilized for mapping the required operation point changes, needed to keep the number of failed batches at an acceptable level when a certain level of process disturbance is introduced, as well as evaluating the performance change that is accounted for when formulating a robust counterpart problem.\n\nSince mathematical modelling offers a cost efficient and powerful approach for assessing preparative chromatography - , it has been employed as the preferred tool for evaluating and optimizing the chromatographic system in this study. The optimization of a chromatography system is ordinarily cast in a bi-level framework : 1) the upper level that administer the effects of the decision variables, such as load and elution gradient, that governs the chromatogram, and 2) the lower level that establishes the pooling strategy for deciding the product pooling cut-times. A MOO method is needed when competing optimization objectives, such as productivity and yield, are considered. The MOO strategy in this work, as presented in , provides with an intact optimization objective for all levels of the bi-level optimization and firm objective values when evaluating the multi-objective optimization problem (MOP). However, the nominal solution for a MOP is not necessarily robust, and even small process disturbances may cause process failure. This calls for a transformation of the MOP into its’ robust counterpart problem .\n\nApproaches for introducing robustness to process optimizations are readily available in literature, and robust optimization for chromatography in particular are, amongst others, presented in - . The preferred robust optimization method used in this work is similar to the methods presented in , with the difference of that this study exclusively utilizes a stochastic method to obtain the model responses of the introduced process disturbances.\n\nThe main focus of this study was to perform a robust multi-objective optimization of chromatographic REE separation. In this context, an experimentally validated process model from a previous study was used to generate the process system response. The results from the robust multi-objective optimization were used to assess the robustness of the system, chart the process parameter changes when robustness was introduced, and evaluate the expected loss of performance when robustness was applied.\n\n2. Chromatography Model\n\n2.1. Process Description\n\nThe chromatography model in this work is based on an experimental study of batch chromatography separation of Sm, Eu and Gd . The experimental study utilized an Agilent 1200 series HPLC system (Agilent Technologies, Waldbronn, Germany) and a Kromasil M3 (Eka, Bohus, Sweden) column with the dimensions 150 × 4.6 mm. An inductively coupled plasma mass spectrometry (ICP-MS) system (Agilent Technologies, Tokyo, Japan) was used for in-line post column REE detection. The column stationary phase was made of spherical, C18 coated, silica particles with a diameter of 16 μm and a pore size of 100 Å. Each column had a ligand concentration of 342 mM Bis (2-ethylhexyl) phosphoric acid (Sigma-Aldrich, St. Louis, USA), and the elution gradient concentration was set to vary between 6% - 13% (vol) of 7 M nitric acid over a gradient length of 5 column volumes (CV).\n\n2.2. Chromatography Model\n\nThe chromatography model in this study has been presented in a previous publication , and is reproduced here for the purpose of clarity. The column separation was described through a kinetic dispersive model with a Langmuir mobile phase modulator isotherm . The model equations, defined in the spatial, $z\\in \\left[{z}_{0},{z}_{f}\\right]$ , and temporal, $t\\in \\left[{t}_{0},{t}_{f}\\right]$ , domains are given by:\n\n$\\frac{\\partial {c}_{\\alpha }}{\\partial t}=-\\frac{\\partial }{\\partial z}\\left({c}_{\\alpha }{v}_{\\text{int}}-{D}_{\\text{app,}\\alpha }\\frac{\\partial {c}_{\\alpha }}{\\partial z}\\right)-\\frac{\\left(1-{\\epsilon }_{c}\\right)}{{\\epsilon }_{c}+\\left(1-{\\epsilon }_{c}\\right){\\epsilon }_{p}}\\frac{\\partial {q}_{\\alpha }}{\\partial t},$ (1)\n\n$\\frac{\\partial {q}_{\\alpha }}{\\partial t}={k}_{\\text{kin},\\alpha }\\left({c}_{\\alpha }{K}_{eq,\\alpha }{q}_{\\mathrm{max},\\alpha }\\left[1-\\underset{\\gamma \\in \\left\\{\\text{Sm,Eu,Gd}\\right\\}}{\\sum }\\frac{{q}_{\\gamma }}{{q}_{\\text{max},\\gamma }}\\right]-{q}_{\\alpha }{c}_{S}^{{\\nu }_{\\alpha }}\\right),$ (2)\n\nwhere ${c}_{\\alpha }$ and ${q}_{\\alpha }$ are the mobile and solid phase concentration of component $\\alpha \\in \\left\\{\\text{Sm},\\text{Eu},\\text{Gd},\\text{S}\\right\\}$ , ${v}_{\\text{int}}$ is the quotient of superficial velocity over total porosity, ${D}_{\\text{app,α}}$ the apparent dispersion coefficient, and ${\\epsilon }_{c}$ and ${\\epsilon }_{p}$ the column and particle void fractions. Here, ${c}_{S}$ denotes the concentration of the modifier (i.e. nitric acid), ${k}_{\\text{kin},\\alpha }$ a parameter describing the kinetics, ${K}_{eq,\\alpha }$ the equilibirum constant regarding adsorption and desorption, ${\\nu }_{\\alpha }$ a parameter describing the ion-exchange characteristics, and ${q}_{\\text{max},\\alpha }$ the maximum concentration of adsorbed components. The model does not consider modifier ions on the solid phase, therefore Equation (2) and its associated part in Equation (1) are omitted (i.e. $\\partial {q}_{\\alpha }/\\partial t\\equiv 0$ ) when $\\alpha =S$ . Equation (1) is complemented with Danckwert boundary conditions :\n\n$\\begin{array}{l}{c}_{\\alpha }\\left(t,{z}_{0}\\right){v}_{\\text{int}}-{D}_{\\text{app,α}}\\frac{\\partial {c}_{\\alpha }}{\\partial z}\\left(t,{z}_{0}\\right)\\\\ =\\left(\\begin{array}{ll}{c}_{\\text{load},\\alpha }{v}_{\\text{int}}\\Pi \\left(t,{t}_{0},\\Delta {t}_{\\text{load}}\\right)\\hfill & \\text{ }\\text{if}\\text{\\hspace{0.17em}}\\text{ }\\alpha \\in \\left\\{\\text{Sm,Eu,Gd}\\right\\},\\hfill \\\\ {c}_{\\text{mix},S}{v}_{\\text{int}}\\hfill & \\text{ }\\text{if}\\text{\\hspace{0.17em}}\\text{ }\\alpha =S,\\hfill \\end{array}\\end{array}$ (3)\n\n$\\frac{\\partial {c}_{\\alpha }}{\\partial z}\\left(t,{z}_{f}\\right)=0,\\text{ }\\forall \\alpha \\in \\left\\{\\text{Sm},\\text{Eu},\\text{Gd},\\text{S}\\right\\}$ (4)\n\nwhere ${c}_{\\text{load},\\alpha }$ is the injected load concentration, and $\\Pi \\left(t,{t}_{0},\\Delta {t}_{\\text{load}}\\right)\\in \\left\\{0,1\\right\\}$ a rectangular function in the temporal horizon $\\left[{t}_{0},\\Delta {t}_{\\text{load}}\\right]$ . The dynamics of the modifier concentration in the upstream mixing tank, ${c}_{\\text{mix},S}$ , are given by:\n\n$\\frac{\\text{d}{c}_{\\text{mix},S}}{\\text{d}t}=\\frac{1}{{\\tau }_{\\text{mix}}}\\left(u\\left(t\\right)-{c}_{\\text{mix},S}\\right),$ (5)\n\n$u\\left(t\\right)=\\left(\\begin{array}{ll}{u}_{0},\\hfill & \\text{ }\\text{if}\\text{\\hspace{0.17em}}\\text{ }t\\le \\Delta {t}_{\\text{load}}+\\Delta {t}_{\\text{wash}},\\hfill \\\\ {u}_{0}+t-\\left(\\Delta {t}_{\\text{load}}+\\Delta {t}_{\\text{wash}}\\right),\\hfill & \\text{ }\\text{if}\\text{\\hspace{0.17em}}\\text{ }t>\\Delta {t}_{\\text{load}}+\\Delta {t}_{\\text{wash}},\\hfill \\end{array}$ (6)\n\nwhere ${\\tau }_{\\text{mix}}$ is the residence time, $u$ is the elution gradient described by the initial value, ${u}_{0}$ , and the slope of the linear elution gradient, $\\Delta u$ , expressed as\n\n$\\Delta u=\\frac{{u}_{f}-{u}_{0}}{{t}_{f}-\\left(\\Delta {t}_{\\text{load}}+\\Delta {t}_{\\text{wash}}\\right)}$ .\n\nThe first-order spatial derivative in Equation (1) was approximated using a method-of-lines and finite volume method with 100 grid points where ${z}_{k}=k\\Delta z$ is the discretized spatial coordinate and $k\\in \\left[1,100\\right]$ . The first order derivative was approximated as a two-point backward difference, and the second-order derivative was approximated as a three-point central difference. The model parameter values from were used in this study, and are given in Table 1.\n\n3. Robust Multi-Objective Optimization\n\n3.1. Multi-Objective Optimization\n\nThe multi-objecitve optimization problem formulation in this work resembles that of a previous study , with the difference of that this work only considers the two competing objectives yield and productivity. This is in order to benchmark the proposed robust optimization method with a previous bi-objective robust optimization method . In this study, the column outlet concentration profile, ${c}_{\\alpha }\\left(t,{z}_{f}\\right)$ where $\\alpha \\in \\left\\{\\text{Sm},\\text{Eu},\\text{Gd}\\right\\}$ , was used for evaluation of the competing objective functions yield, ${Y}_{\\alpha }$ and productivity, ${P}_{\\alpha }$ for the target component Eu. The objective functions for the collected component, $\\alpha$ , between the\n\nTable 1. Model parameter values.\n\ncut-times $\\left[{t}_{c},{t}_{f}\\right]$ are defined as:\n\n${\\delta }_{\\text{load},\\alpha }\\frac{\\text{d}{Y}_{\\alpha }}{\\text{d}t}={c}_{\\alpha }\\left(t,{z}_{f}\\right){v}_{\\text{int}}{A}_{c}\\Pi \\left(t,{t}_{c},{t}_{f}\\right),$ (7)\n\n$\\frac{\\text{d}{P}_{\\alpha }}{\\text{d}t}=\\frac{1}{{V}_{c}}\\frac{1}{{t}_{f}+{t}_{r}}{\\delta }_{\\text{load},\\alpha }\\frac{\\text{d}{Y}_{\\alpha }}{\\text{d}t},$ (8)\n\nwhere ${\\delta }_{\\text{load},\\alpha }={c}_{\\text{load},\\alpha }{v}_{\\text{int}}{A}_{c}\\Delta {t}_{\\text{load}}$ is the total amount of injected sample, ${A}_{c}$ and Vc the column cross-sectional area and volume, and ${t}_{r}=2{V}_{c}{\\stackrel{˙}{Q}}^{-1}$ the regeneration and re-equilibration time following the final cut-time. Thus, the objective becomes to determine an optimal elution gradient, $u$ , batch load, ${\\delta }_{\\text{load},\\alpha }$ , and pooling cut-times, $\\left[{t}_{c},{t}_{f}\\right]$ , that maximizes ${Y}_{\\alpha }\\left({t}_{f}\\right)$ and ${P}_{\\alpha }\\left({t}_{f}\\right)$ , while fulfilling the target component purity constraint given by:\n\n${X}_{\\alpha }\\left({t}_{f}\\right)=\\frac{{\\delta }_{\\text{load},\\alpha }{Y}_{\\alpha }\\left({t}_{f}\\right)}{\\underset{\\beta \\in \\left\\{\\text{Sm},\\text{Eu},\\text{Gd}\\right\\}}{\\sum }{\\delta }_{\\text{load},\\beta }{Y}_{\\beta }\\left({t}_{f}\\right)},$ (9)\n\nwhere the numerator is the captured amount of the target component in $\\left[{t}_{c},{t}_{f}\\right]$ and the denominator represents the total amount of captured components.\n\nThe weighted sum scalarization method was used to combine the objectives in Equations (7)-(8) to a single objective with the weight for productivity defined as $\\omega \\in \\left[0,1\\right]$ , and the weight for yield defined as $1-\\omega$ . The decision variables are the free operating parameters, i.e. $\\Delta {t}_{\\text{load}},{t}_{c},{t}_{f},{u}_{0}$ and ${u}_{f}$ , which in turn determine the trajectories $x=\\left({c}_{\\alpha }\\left(t,{z}_{k}\\right),{c}_{S}\\left(t,{z}_{k}\\right),{c}_{\\text{mix},S}\\left(t\\right),{q}_{\\alpha }\\left(t,{z}_{k}\\right),{P}_{\\text{Eu}}\\left(t\\right),{Y}_{\\text{Eu}}\\left(t\\right),{X}_{\\text{Eu}}\\left(t\\right)\\right)$ . The resulting optimization problem can then be set in the framework for min-min optimal control:\n\n$\\text{min}\\text{.}\\text{ }\\text{ }-\\left(\\omega \\underset{{t}_{0}}{\\overset{{t}_{f}}{\\int }}\\frac{\\text{d}{P}_{\\text{Eu}}}{\\text{d}t}\\text{d}t+\\left(1-\\omega \\right)\\underset{{t}_{0}}{\\overset{{t}_{f}}{\\int }}\\frac{\\text{d}{Y}_{\\text{Eu}}}{\\text{d}t}\\text{d}t\\right),$ (10a)\n\n$\\text{w}\\text{.r}\\text{.t}\\text{.}\\text{ }p=\\left(\\Delta {t}_{\\text{load}},{u}_{0},{u}_{f}\\right)\\in {ℝ}^{3},$\n\n$\\text{s}\\text{.t}\\text{.}\\text{ }\\text{ }{p}_{L}\\le p\\le {p}_{U},$ (10b)\n\n$\\left(x,{t}_{c},{t}_{f}\\right)=\\text{argmin}\\text{.}-\\left(\\omega \\underset{{t}_{0}}{\\overset{{t}_{f}}{\\int }}\\frac{\\text{d}{P}_{\\text{Eu}}}{\\text{d}t}\\text{d}t+\\left(1-\\omega \\right)\\underset{{t}_{0}}{\\overset{{t}_{f}}{\\int }}\\frac{\\text{d}{Y}_{\\text{Eu}}}{\\text{d}t}\\text{d}t\\right),\\text{ }$ (10c)\n\n$\\text{w}\\text{.r}\\text{.t}\\text{.}\\text{ }\\left({t}_{c},{t}_{f}\\right)\\in {ℝ}^{2},$\n\n$\\text{s}\\text{.t}\\text{.}\\text{ }\\stackrel{˙}{x}=F\\left(t,x\\left(t\\right),{t}_{c},{t}_{f},p\\right),\\text{ }x\\left({t}_{0}\\right)={x}_{0},$ (10d)\n\n${X}_{\\text{Eu},L}-{X}_{\\text{Eu}}\\left({t}_{f}\\right)\\le 0,$ (10e)\n\n${t}_{c,L}\\le {t}_{c}\\le {t}_{c,U},\\text{ }{t}_{f,L}\\le {t}_{f}\\le {t}_{f,U},$ (10f)\n\n$\\forall t\\in \\left[{t}_{0},{t}_{f}\\right],\\text{ }\\text{ }\\forall z\\in \\left[{z}_{0},{z}_{f}\\right].$\n\nThe solution of Equation (10) will result in a Pareto front situated on the boundary of the feasible region. This implies that a process disturbance, however slight, can cause a batch failure in terms of not meeting the purity constraint . In order to account for such disturbances it is necessary to formulate a robust counterpart problem, which in this study will be accomplished by introducing a back-off term to the purity inequality constraint in Equation (10e).\n\n3.2. Robust Multi-Objective Optimization Problem Formulation\n\nIn order to formulate a robust counterpart of Equation (10), we consider a set of bounded distributed disturbances, $\\stackrel{˜}{p}$ , on the free operating parameters, $p$ (i.e $\\Delta {t}_{\\text{load}},{u}_{0}$ and ${u}_{f}$ ), and define ${\\stackrel{˜}{X}}_{\\text{Eu}}$ as the cumulative purity distribution of the model responses that are produced from $\\stackrel{˜}{p}$ . A purity constraint back-off term, ${X}_{\\text{BF}}$ , is introduced in order to make the purity constraint robust with respect to the disturbances. The back-off term can essentially be seen as a safety margin that amplifies the purity inequality constraint in Equation (10e) so that the purity requirement, ${X}_{\\text{Eu,}L}$ , still can be met for the considered set of bounded disturbances. The success rate is defined as the fraction of batches in the disturbance set that fulfil the purity requirement, ${X}_{\\text{Eu,}L}$ , and ${\\Phi }_{{X}_{\\text{Eu}}}$ signifies the desired success rate. The following robust counterpart of Equation (10) is then given by:\n\n$\\text{min}\\text{.}\\text{ }\\text{ }\\underset{-\\infty }{\\overset{{X}_{\\text{Eu},L}+{X}_{\\text{BF}}}{\\int }}{\\stackrel{˜}{X}}_{\\text{Eu}}\\text{d}{X}_{\\text{Eu}}-{\\Phi }_{{X}_{\\text{Eu}}},$ (11a)\n\n$\\text{w}\\text{.r}\\text{.t}\\text{.}\\text{ }{X}_{\\text{BF}},$\n\n$\\text{s}\\text{.t}\\text{.}\\text{ }\\stackrel{˙}{\\stackrel{˜}{x}}=\\stackrel{˜}{F}\\left(t,x\\left(t\\right),{t}_{c},{t}_{f},\\stackrel{˜}{p}\\right),$ (11b)\n\n$\\stackrel{˜}{p}~N\\left(p,{\\sigma }_{p}^{2}\\right),$ (11c)\n\n$p=\\text{argmin}\\text{.}-\\left(\\omega \\underset{{t}_{0}}{\\overset{{t}_{f}}{\\int }}\\frac{\\text{d}{P}_{\\text{Eu}}}{\\text{d}t}\\text{d}t+\\left(1-\\omega \\right)\\underset{{t}_{0}}{\\overset{{t}_{f}}{\\int }}\\frac{\\text{d}{Y}_{\\text{Eu}}}{\\text{d}t}\\text{d}t\\right),$ (11d)\n\n$\\text{w}\\text{.r}\\text{.t}\\text{.}\\text{ }p=\\left(\\Delta {t}_{\\text{load}},{u}_{0},{u}_{f}\\right)\\in {ℝ}^{3},$\n\n$\\text{s}\\text{.t}\\text{.}\\text{ }{p}_{L}\\le p\\le {p}_{U},$ (11e)\n\n$\\left(x,{t}_{c},{t}_{f}\\right)=\\text{argmin}\\text{.}-\\left(\\omega \\underset{{t}_{0}}{\\overset{{t}_{f}}{\\int }}\\frac{\\text{d}{P}_{\\text{Eu}}}{\\text{d}t}\\text{d}t+\\left(1-\\omega \\right)\\underset{{t}_{0}}{\\overset{{t}_{f}}{\\int }}\\frac{\\text{d}{Y}_{\\text{Eu}}}{\\text{d}t}\\text{d}t\\right),$ (11f)\n\n$\\text{w}\\text{.r}\\text{.t}\\text{.}\\text{ }\\left({t}_{c},{t}_{f}\\right)\\in {ℝ}^{2},$\n\n$\\text{s}\\text{.t}\\text{.}\\text{ }\\stackrel{˙}{x}=F\\left(t,x\\left(t\\right),{t}_{c},{t}_{f},p\\right),\\text{ }x\\left({t}_{0}\\right)={x}_{0},$ (11g)\n\n$\\left({X}_{\\text{Eu},L}+{X}_{\\text{BF}}\\right)-{X}_{\\text{Eu}}\\left({t}_{f}\\right)\\le 0,$ (11h)\n\n${t}_{c,L}\\le {t}_{c}\\le {t}_{c,U},\\text{ }{t}_{f,L}\\le {t}_{f}\\le {t}_{f,U},$ (11i)\n\n$\\forall t\\in \\left[{t}_{0},{t}_{f}\\right],\\text{ }\\forall z\\in \\left[{z}_{0},{z}_{f}\\right].$\n\nA decomposition strategy was adopted to transform the robust MOP into three levels: 1) the upper-level optimization problem given by Equations (11a)-(11c) with respect to ${X}_{\\text{BF}}$ , 2) the mid-level optimization problem given by Equations (11d)-(11e) with respect to p, and 3) the lower-level optimization problem given by Equations (11f)-(11i) and constrained by the ODE system, $F$ , governed by Equations (1), (2), (5), (7), (8), (9). Essentially, Equation (11) was solved by using the simulated system response, $\\stackrel{˜}{x}$ , for an uncertainty set of the free operating parameters, $\\stackrel{˜}{p}$ , to evaluate the cumulative distribution function of ${\\stackrel{˜}{X}}_{\\text{Eu}}$ . The back-off term, ${X}_{\\text{BF}}$ , in the purity inequality constraint, Equation (11h), was then incrementally increased to gain more successful batches in ${\\stackrel{˜}{X}}_{\\text{Eu}}$ , and thereby achieving a more robust process. This procedure was repeated iteratively until Equation (11a) was fulfilled, at which point Equation (11) was considered to be solved.\n\nIn this study, the desired success rate, ${\\Phi }_{{X}_{\\text{Eu}}}$ , was set to 0.95, and the decision variable boundaries are presented in Table 2.\n\n3.3. Optimization Method\n\nThe robust optimization method in this study is similar to the methods presented in . The main difference is that the model responses of the introduced disturbances in this study are obtained stochastically, instead of alternatively utilizing a deterministic approach through linearization of the uncertainty set. The stochastic approach has the benefit of being more straightforward, at the expense of an increased demand of computation power. This increased demand was accommodated for by using a parallel computing\n\nTable 2. Decision variables.\n\nmethodology as described in .\n\nAs a first step, the nominal and non-robust Pareto front was obtained by solving the MOP as defined in Equation (10). This was carried out through MATLAB’s fmincon function with a sequential quadratic programming algorithm, the BFGS formula for updating the approximation of the Hessian matrix, and central differences to estimate the gradient of the objective function and constraint functions. Then an uncertainty set, $\\stackrel{˜}{p}$ , with a normal distribution, assuming no covariance between the free operating parameters $p$ , a standard deviation $\\sigma$ , and sampling size of 10.000 was obtained via MATLAB’s lhsnorm function. The uncertainty set was applied to the investigated operating points on the nominal Pareto front, and the model responses were used to evaluate the cumulative purity distribution, ${\\stackrel{˜}{X}}_{\\text{Eu}}$ , of the uncertainty set.\n\nThen, an initial investigation of the back-off terms impact on ${\\stackrel{˜}{X}}_{\\text{Eu}}$ was conducted by creating new Pareto fronts with an incrementally increased back-off and observing how ${\\stackrel{˜}{X}}_{\\text{Eu}}$ changes when $\\stackrel{˜}{p}$ is applied to the investigated points on the new Pareto fronts. At this stage, it is of particular interest to investigate how the fraction of batches that fulfil the purity requirement in the perturbed set, changes with an increased back-off. This provides an estimate of the required back-off to meet a certain success rate for a given purity constraint.\n\nThe required back-off for a given point on the nominal Pareto front was obtained by applying MATLAB’s fminbnd function on the upper level of the robust counterpart problem in Equation (11), with suitable boundaries obtained from the previous back-off investigation. The mid- and lower-level optimization problems in Equation (11) were solved by MATLAB’s fmincon function with a sequential quadratic programming algorithm, the BFGS formula for updating the approximation of the Hessian matrix, and central differences to estimate the gradient of the objective function and constraints. The procedure comprises an evaluation of the cumulative distribution function of ${\\stackrel{˜}{X}}_{\\text{Eu}}$ based on $\\stackrel{˜}{x}$ and $\\stackrel{˜}{p}$ , as obtained from the mid- and lower-level optimization problem for a given initial ${X}_{\\text{BF}}$ . ${X}_{\\text{BF}}$ is then varied for the upper level optimization problem through MATLAB's fminbnd function, resulting in new $\\stackrel{˜}{x}$ , $\\stackrel{˜}{p}$ and cumulative distribution functions of ${\\stackrel{˜}{X}}_{\\text{Eu}}$ to be evaluated. This continues until a ${X}_{\\text{BF}}$ that produces a cumulative distribution function of ${\\stackrel{˜}{X}}_{\\text{Eu}}$ corresponding to the desired success rate ${\\Phi }_{{X}_{\\text{Eu}}}$ is obtained.\n\nOptimization Method Benchmarking\n\nThe proposed optimization method was compared with a previous optimization method that focuses on the nominal Pareto front and optimizes the pooling time horizon for each investigated point on the front so that the purity requirement is met for a given uncertainty set. The main difference between the two methods is that the proposed method will find new optimal operation points by changing the free operating parameters, and achieve robustness by increasing the purity requirement back-off for each point on the Pareto front, whereas the previous method keeps the decision variables from the nominal Pareto front intact, with the exception of the cut-times that are optimized to find a fixed pooling time horizon that will fulfil the purity requirement for the entire uncertainty set.\n\n3.4. Robust Multiobjective Optimization Results\n\nThe robust multiobjective optimizations of the studied system were carried out with a product purity requirement, ${X}_{\\text{Eu},L}$ , of 0.95 and 0.99 respectively, and the target success rate, ${\\Phi }_{{X}_{\\text{Eu}}}$ , was set to 0.95. The perturbed process parameters were the injected load concentration, ${c}_{\\text{load},\\alpha }$ , and the modifier concentration in the upstream mixing tank, ${c}_{\\text{mix},S}$ . Several values for the uncertainty set standard deviation, $\\sigma$ , were investigated. However, a standard deviation exceeding 0.01 did not result in achieving the target success rate even when the back-off was set to the maximum limit, i.e. ${X}_{\\text{Eu},L}+{X}_{\\text{BF}}=1$ . Therefore, only results for $\\sigma =0.01$ are presented in this study. This should be interpreted as that the system is very un-robust, and sensitive to even the slightest process disturbances. However, this was expected since the studied elements are extremely similar in both chemical and physical properties, resulting in a minute separation selectivity which in turn makes the separation very difficult and unforgiving towards process perturbations.\n\nThe nominal un-robust Pareto fronts are presented by the outermost fronts in Figure 1, and it is shown how the Pareto front decreases with an increased back-off on the purity requirement. It should be noted that we only present the impact of an increased back-off on four Pareto points to avoid overtly crowded figures. The objective weight, $\\omega$ , for these points correspond to 1 (i.e. productivity as single objective), 0.5, 0.3 and 0 (i.e. yield as single objective) respectively.\n\nFigure 1. The nominal Pareto fronts are presented by the solid lines for a purity requirement of 0.95 in (a) and 0.99 in (b). The cross marks indicate how a Pareto point, with the weight $\\omega$ , changes with an increased back-off. The dashed lines indicate the Pareto front outlines for an increasing back-off, and it can be seen that the Pareto front decreases as the back-off is increased.\n\nThe results from the initial investigation on how the success rate, ${\\Phi }_{{X}_{\\text{Eu}}}$ , for the investigated points on the nominal Pareto front increases with an increased back-off are shown in Figure 2. The figure provides with an estimation of the required back-off to achieve the desired success rate for a given disturbance set, and it can be seen that an objective leaning more towards yield (i.e. $\\omega$ decreases) results in a lower success rate for a given back-off.\n\nIt should be noted that the Pareto point corresponding to yield as a single objective, i.e. $\\omega =0$ , has been omitted since the point for maximum yield is considered as an undesirable operating point due to the drastic decrease in productivity.\n\nIt is somewhat counter intuitive that the success rate should decrease with an increased objective weight for yield, since a higher yield typically is associated with an increased peak separation which in turn should result in an increased robustness. The decrease of robustness can be explained by observing how the decision variables change with an increased back-off for the 0.95 purity requirement case in Figure 3, where the pooling cut-time trends become very interesting. The decision variable trends show that the initial elution concentration and elution gradient slope are quite similar as long as productivity is a part of the weighted objective. However, a higher productivity is favoured by a larger batch load, a pooling horizon occurring earlier in the chromatogram (i.e. first and last pooling cuts occur earlier) and a smaller pooling volume. The increased batch load is reasonable since it will allow for a higher productivity due to an increased throughput. The early first cut comes from that a higher batch load will capacitate the elements to start eluting earlier. The earlier final cut makes the cycle time shorter, which is favourable for productivity, but it is also a trade of in terms of decreased yield.\n\nFigure 2. Results from the investigation of how the success rate increases with an increased back-off for a purity requirement of 0.95 in (a) and 0.99 in (b). The dashed line indicates the target success rate, ${\\Phi }_{{X}_{\\text{Eu}}}$ , and helps to provide an initial estimation of the required back-off, ${X}_{\\text{BF}}$ , for a Pareto point with the objective weight $\\omega$ .\n\nFigure 3. Plots of decision variable changes due to an increasing back-off for Pareto points with different objective weights, $\\omega$ , and a purity requirement of 0.95. (a) Batch load, (b) Initial elution concentration, (c) Elution gradient slope, (d) First cut time, (e) Final cut time, (f) Pooling volume.\n\nThis has the important implication of that a high objective weight on productivity will result in pooling cut times occurring closer to the Eu elution peak centre and farther away from the neighbouring peaks. When a higher yield is desired, the pooling horizon will increase in order to capture more of the target molecules, and this will move the pooling close to, and even into, the neighbouring elution peaks as long as the purity requirement is met. For this reason, a higher weight on yield will demand a higher back-off on purity in order to meet the desired success rate. This is due to that when a perturbation is introduced, the neighbouring peaks may move closer to, and even intrude, the pooling horizon, and a higher purity requirement will move the pooling cut times farther away from the neighbouring peaks. The farther away the cut times are from the neighbouring peaks in the nominal case, the higher disturbance can be tolerated since there is more room available for the neighbouring peaks to move before they impact the purity of the target peak. This is illustrated in Figure 4 where we can see a case with high productivity (smaller pooling horizon) and a case with high yield (larger pooling horizon) and observe how the introduced process disturbances make the neighbouring peaks creep into the pooling horizon to a larger extent for the high yield case.\n\nThe results from the robust optimizations can be seen in Figure 5 where the nominal un-robust Pareto front is plotted along with the robust Pareto front produced by the proposed method, and the front from the benchmark method. It should be noted that both methods provide with robust operating points that\n\nFigure 4. Sections of chromatograms with focus on the collection of the Eu product pool. The purity requirement, ${X}_{\\text{Eu}}$ , was set to 0.95 for the productivity objective weights $\\omega =1$ (a) and $\\omega =0.3$ (b). The pooling cut times are indicated by the dotted lines, and the elution order is Sm, Eu and Gd. The unit for the nitric acid elution gradient (black dashed line) on the vertical axis is mol/l. The shaded areas indicate the span of concentration profile variations due to process disturbances with $\\sigma =0.01$ , and the solid black lines indicate the concentration profiles for the nominal case. The chromatograms demonstrate that an operation point with a higher objective weight for productivity (a), is more robust than an operation point with a lower weight (b). This can be seen by observing how the larger pooling horizon in (b) allows for more collection of the neighbouring elements when disturbances are introduced, and thereby causing an increased number of batches with failed purity requirement. This is particularly noticeable for the Gd peak which intrudes the collected pool to a larger extent when process disturbances are introduced in (b).\n\nFigure 5. Pareto fronts resulting from the optimizations with a purity requirement of 0.95 (left) and 0.99 (right). (a) indicates the nominal Pareto front, (b) the robust Pareto front according to the proposed method and (c) the robust front from the benchmark method. The dots indicate the system response of the distributed uncertainty set associated with the respective Pareto points. The dashed lines indicate the loss of productivity for a given yield when robustifying the nominal Pareto front according to the proposed method.\n\nhandle the given process disturbances satisfactorily. However, the proposed method should be favoured since it produces a robust Pareto front with higher objective values compared to the benchmark method, which implies that the benchmark method should be considered more restrictive. Further, the benchmark method generates operating points that cannot be considered Pareto optimal, which is the case for the points with $\\omega =1$ on front (c) in Figure 5. For the sake of fairness and as mentioned in , these points should be disregarded when generating a robust Pareto front with the benchmark method.\n\nApplying robustness to a point on the nominal Pareto front with a given $\\omega$ will result in a change of both productivity and yield, as can be seen in Figure 1, and this makes the evaluation of performance loss when introducing robustness slightly ambiguous. In order to resolve this, the productivity for a given yield on the nominal Pareto front is compared to the productivity on the robust Pareto front given the same yield, as indicated by red dashed lines in Figure 5. The loss of productivity is presented in Table 3, and it shows that a productivity loss in the range of 10% - 20% can be expected. It should be mentioned that the robustness can be increased further during operation by applying a variable pooling cut time control strategy as described in , and thereby decrease the loss of productivity.\n\nThe required back-off, that meets the robustness requisite according to the proposed method, is presented in Table 4 for the investigated Pareto points. The results show that a higher purity requirement calls for a larger back-off term to achieve a robust Pareto point, and a lower objective weight on productivity also necessitates an increased back-off.\n\nTable 3. Loss of productivity.\n\nTable 4. Required back-off.\n\n4. Conclusion\n\nThis study has shown that the proposed optimization method can be used for robust multi-objective optimization of chromatographic rare earth element separation, and provided with expected performance changes for the robustification of the studied process. It has been highlighted that the studied system is highly un-robust, and that the system’s lack of robustness is largely due to the neighbouring peaks’ proximity to the product pooling horizon. The system can only cope with slight process disturbances, which in turn demands use of process equipment with high reliability. We show how the optimal solution of a chromatographic separation is affected by introducing robustness in a brute force manner. For future studies, it would be of interest to employ the proposed optimization method on additional chromatography schemes, as well as other chromatography separation applications than rare earth elements.\n\nAcknowledgements\n\nThis study has been performed within ProOpt and Process Industry Centre at Lund University, and financed by VINNOVA.\n\nCite this paper\n\nKnutson, H.-K., Holmqvist, A., Andersson, N. and Nilsson, B. (2017) Robust Multi-Objective Optimization of Chromatographic Rare Earth Element Separation. 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(2015) Multi-Objective Optimization of Chromatographic Rare Earth Element Separation. Journal of Chromatography A, 1416, 57-63. https://doi.org/10.1016/j.chroma.2015.09.010\n\n19. 19. Andersson, N., Knutson, H.K., Max-Hansen, M., Borg, N. and Nilsson, B. (2014) Model-Based Comparison of Batch and Continuous Preparative Chromatography in the Separation of Rare Earth Elements. Industrial & Engineering Chemistry Research, 53, 16485-16493. https://doi.org/10.1021/ie5023223\n\n20. 20. Jakobsson, N., Degerman, M. and Nilsson, B. (2005) Optimisation and Robustness Analysis of a Hydrophobic Interaction Chromatography Step. Journal of Chromatography A, 1099, 157-166. https://doi.org/10.1016/j.chroma.2005.09.009\n\n21. 21. Karlsson, D., Jakobsson, N., Axelsson, A. and Nilsson, B. (2004) Model-Based Optimization of a Preparative Ion-Exchange Step for Antibody Purification. Journal of Chromatography A, 1055, 29-39. https://doi.org/10.1016/j.chroma.2004.08.151\n\n22. 22. Nagrath, D., Bequette, B.W., Cramer, S. and Messac, A. (2005) Multiobjective Optimization Strategies for Linear Gradient Chromatography. AIChE Journal, 51, 511-525. https://doi.org/10.1002/aic.10459\n\n23. 23. Gétaz, D., Butté, A. and Morbidelli, M. (2013) Model-Based Design Space Determination of Peptide Chromatographic Purification Processes. Journal of Chromatography A, 1284, 80-87. https://doi.org/10.1016/j.chroma.2013.01.117\n\n24. 24. Osberghaus, A., Hepbildikler, S., Nath, S., Haindl, M., von Lieres, E. and Hubbuch, J. (2012) Optimizing a Chromatographic Three Component Separation: A Comparison of Mechanistic and Empiric Modeling Approaches. Journal of Chromatography A, 1237, 86-95. https://doi.org/10.1016/j.chroma.2012.03.029\n\n25. 25. Sreedhar, B., Wagler, A. and Kaspereit, M., Seidel-Morgenstern, A. (2013) Optimal cut-Times Finding Strategies for Collecting a Target Component from Overloaded Elution Chromatograms. Computers & Chemical Engineering, 49, 158-169. https://doi.org/10.1016/j.compchemeng.2012.09.009\n\n26. 26. Ben-Tal, A., El Ghaoui, L. and Nemirovski, A. (2009) Robust Optimization. Princeton University Press, Princeton. https://doi.org/10.1515/9781400831050\n\n27. 27. Mota, J.P., Araújo, J.M., Rodrigues, R., et al. (2007) Optimal Design of Simulated Moving-Bed Processes under Flow Rate Uncertainty. AIChE Journal, 53, 2630-2642. https://doi.org/10.1002/aic.11281\n\n28. 28. Nestola, P., Silva, R.J., Peixoto, C., Alves, P.M., Carrondo, M.J. and Mota, J.P. (2015) Robust Design of Adenovirus Purification by Two-Column, Simulated Moving-Bed, Size-Exclusion Chromatography. Journal of Biotechnology, 213, 109-119. https://doi.org/10.1016/j.jbiotec.2015.01.030\n\n29. 29. Holmqvist, A., Andersson, C., Magnusson, F. and Kesson, J. (2015) Methods and Tools for Robust Optimal Control of Batch Chromatographic Separation Processes. Processes, 3, 568-606. https://doi.org/10.3390/pr3030568\n\n30. 30. Degerman, M., Westerberg, K. and Nilsson, B. (2009) A Model-Based Approach to Determine the Design Space of Preparative Chromatography. Chemical Engineering & Technology, 32, 1195-1202. https://doi.org/10.1002/ceat.200900102\n\n31. 31. Westerberg, K., Borg, N., Andersson, N. and Nilsson, B. (2012) Supporting Design and Control of a Reversed-Phase Chromatography Step by Mechanistic Modeling. Chemical Engineering & Technology, 35, 169-175. https://doi.org/10.1002/ceat.201000505\n\n32. 32. Close, E.J., Salm, J.R., Bracewell, D.G. and Sorensen, E. (2014) A Model Based Approach for Identifying Robust Operating Conditions for Industrial Chromatography with Process Variability. Chemical Engineering Science, 116, 284-295. https://doi.org/10.1016/j.ces.2014.03.010\n\n33. 33. Logist, F., Houska, B., Diehl, M. and Van Impe, J.F. (2011) Robust Multi-Objective Optimal Control of Uncertain (Bio) Chemical Processes. Chemical Engineering Science, 66, 4670-4682. https://doi.org/10.1016/j.ces.2014.03.010\n\n34. 34. Schmidt-Traub, H., Schulte, M. and Seidel-Morgenstern, A. (2012) Preparative Chromatography. 2nd Edition, Wiley-VCH, Weinheim. https://doi.org/10.1002/9783527649280\n\n35. 35. Guiochon, G., Felinger, A. and Shirazi, D.G. (2006) Fundamentals of Preparative and Nonlinear Chromatography. Academic Press, New York.\n\n36. 36. Degerman, M. (2009) Design of Robust Preparative Chromatography. Lund University, Lund." ]	[ null, "https://html.scirp.org/file/10-3700843x1.png", null, "https://html.scirp.org/file/9-2500537x3.png", null, "https://html.scirp.org/file/9-2500537x2.png", null, "https://html.scirp.org/file/18-1760983x4.png", null, "https://html.scirp.org/file/2-1410169x6.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.88544613,"math_prob":0.9731972,"size":36025,"snap":"2021-21-2021-25","text_gpt3_token_len":8326,"char_repetition_ratio":0.14913523,"word_repetition_ratio":0.046924658,"special_character_ratio":0.23744622,"punctuation_ratio":0.17461011,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.99536663,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,2,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-14T04:16:42Z\",\"WARC-Record-ID\":\"<urn:uuid:3b4f667b-b5c6-4d56-82a6-72565f936363>\",\"Content-Length\":\"150524\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a59f867a-93db-4715-a3da-1aad23613a8a>\",\"WARC-Concurrent-To\":\"<urn:uuid:c09ff5d6-5048-4ded-923f-fbd565e7fff5>\",\"WARC-IP-Address\":\"192.111.37.22\",\"WARC-Target-URI\":\"https://file.scirp.org/Html/10-3700843_80013.htm\",\"WARC-Payload-Digest\":\"sha1:M57EUCOX57IFJQUJPUGOWS3IXA7RESXO\",\"WARC-Block-Digest\":\"sha1:7SN7TX7OKD2E5J6UQZLUAYZNGRVAIF2C\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243991737.39_warc_CC-MAIN-20210514025740-20210514055740-00595.warc.gz\"}"}
https://focalacademicwriters.info/2021/09/01/manually-complete-the-following-problems-in-the-textbook-problem-7-14problem-7/	[ "# Manually complete the following problems in the textbook: Problem 7-14 Problem 7\n\nManually complete the following problems in the textbook: Problem 7-14\nProblem 7-15\nProblem 7-18\nUse an Excel spreadsheet file for the calculations and explanations. Cells should contain the formulas (i.e., if a formula was used to calculate the entry in that cell). Students are highly encouraged to use the “Optimization Modeling Problem Set” Excel resource to complete this assignment. Mac users can use StatPlus:mac LE, free of charge, from AnalystSoft. You are not required to submit this assignment to LopesWrite.\nRequirements: 3 Pages" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.83118564,"math_prob":0.5655882,"size":693,"snap":"2021-43-2021-49","text_gpt3_token_len":153,"char_repetition_ratio":0.16835994,"word_repetition_ratio":0.1980198,"special_character_ratio":0.2092352,"punctuation_ratio":0.11811024,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98245573,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-07T20:30:58Z\",\"WARC-Record-ID\":\"<urn:uuid:0a1ec2ef-4c21-4414-9891-eb1d2a90a5ad>\",\"Content-Length\":\"29474\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6f459ad7-39a1-475e-80e9-884f79072f7d>\",\"WARC-Concurrent-To\":\"<urn:uuid:dcd409e1-788e-4049-a274-5092f003c4fd>\",\"WARC-IP-Address\":\"209.188.21.166\",\"WARC-Target-URI\":\"https://focalacademicwriters.info/2021/09/01/manually-complete-the-following-problems-in-the-textbook-problem-7-14problem-7/\",\"WARC-Payload-Digest\":\"sha1:VK24RMNNPYG7OGGNPIVJJW2UN4KPHWNK\",\"WARC-Block-Digest\":\"sha1:BSAAAHIQR6JF24GL5XMLQQ6ZEOZS5HJA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363418.83_warc_CC-MAIN-20211207201422-20211207231422-00196.warc.gz\"}"}
https://www.revistaproyecciones.cl/index.php/proyecciones/article/view/2390	[ "# Existence of positive periodic solutions for delay dynamic equations.\n\n## Authors\n\n• Faycal Bouchelaghem UBMA.\n• Abdelouaheb Ardjouni University Souk Ahras.\n• Ahcene Djoudi UBMA.\n\n## Keywords:\n\nPositive periodic solutions, Schauder’s fixed point theorem, Dynamic equations, Time scales\n\n## Abstract\n\nIn this article we study the existence of positive periodic solutions for a dynamic equations on time scales. The main tool employed here is the Schauder's fixed point theorem. The results obtained here extend the work of Olach . Two examples are also given to illustrate this work.\n\n## Author Biographies\n\n### Faycal Bouchelaghem, UBMA.\n\nDepartment of Mathematics, Faculty of Sciences.\n\n### Abdelouaheb Ardjouni, University Souk Ahras.\n\nDepartment of Mathematics and Informatics.\n\n### Ahcene Djoudi, UBMA.\n\nApplied Mathematics Lab., Faculty of Sciences.\n\n## References\n\nM. Adivar and Y. N. Raﬀoul, Existence of periodic solutions in totally nonlinear delay dynamic equations, Electronic Journal of Qualitative Theory of Diﬀerential Equations, No. 1, pp. 1-20, (2009).\n\nA. Ardjouni and A. Djoudi, Existence of positive periodic solutions for nonlinear neutral dynamic equations with variable delay on a time scale, Malaya Journal of Matematik 2 (1), pp. 60-67, (2013).\n\nA. Ardjouni and A. Djoudi, Existence of periodic solutions for nonlinear neutral dynamic equations with functional delay on a time scale, Acta Univ. Palacki. Olomnc., Fac. rer. nat., Mathematica 52, 1, pp. 5-19, (2013).\n\nA. Ardjouni and A. Djoudi, Existence of periodic solutions for non- linear neutral dynamic equations with variable delay on a time scale, Commun Nonlinear Sci Numer Simulat 17, pp. 3061—3069, (2012).\n\nA. Ardjouni and A. Djoudi, Periodic solutions in totally nonlinear dynamic equations with functional delay on a time scale, Rend. Sem. Mat. Univ. Politec. Torino Vol. 68, 4, pp. 349-359, (2010).\n\nM. Bohner, A. Peterson, Dynamic Equations on Time Scales, An Introduction with Applications, Birkhäuser, Boston, (2001).\n\nM. Bohner, A. Peterson, Advances in Dynamic Equations on Time Scales, Birkhäuser, Boston, (2003).\n\nS. Hilger, Ein Masskettenkalkül mit Anwendung auf Zentrumsman- ningfaltigkeiten. PhD thesis, Universität Würzburg, (1988).\n\nE. R. Kaufmann, Y. N. Raﬀoul, Periodic solutions for a neutral non- linear dynamical equation on a time scale, J. Math. Anal. Appl. 319, pp. 315-325, (2006).\n\nE. R. Kaufmann and Y. N. Raﬀoul, Periodicity and stability in neutral nonlinear dynamic equation with functional delay on a time scale, Electronic Journal of Diﬀerential Equations, No. 27, pp. 1-12, (2007).\n\nV. Lakshmikantham, S. Sivasundaram, B. Kaymarkcalan, Dynamic Systems on Measure Chains, Kluwer Academic Publishers, Dordrecht, (1996).\n\nR. Olach, Positive periodic solutions of delay diﬀerential equations, Applied Mathematics Letters 26, pp. 1141-1145, (2013).\n\nD. R. Smart, Fixed Points Theorems, Cambridge Univ. Press, Cambridge, UK, (1980).\n\n2017-10-20\n\n## How to Cite\n\n\nF. Bouchelaghem, A. Ardjouni, and A. Djoudi, “Existence of positive periodic solutions for delay dynamic equations.”, Proyecciones (Antofagasta, On line), vol. 36, no. 3, pp. 449-460, Oct. 2017.\n\nArtículos" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6876472,"math_prob":0.85743546,"size":2959,"snap":"2022-27-2022-33","text_gpt3_token_len":808,"char_repetition_ratio":0.14856176,"word_repetition_ratio":0.13151927,"special_character_ratio":0.25481582,"punctuation_ratio":0.24958402,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.970582,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-07T18:14:39Z\",\"WARC-Record-ID\":\"<urn:uuid:88c94086-292b-4315-8503-9382ffbba88e>\",\"Content-Length\":\"41891\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7d4e0cf4-4f8f-45c2-9e2a-981d495406c4>\",\"WARC-Concurrent-To\":\"<urn:uuid:e5cddd94-56bd-4f65-ae76-e1904513cdf8>\",\"WARC-IP-Address\":\"146.83.118.13\",\"WARC-Target-URI\":\"https://www.revistaproyecciones.cl/index.php/proyecciones/article/view/2390\",\"WARC-Payload-Digest\":\"sha1:NNZBY4US57TDYK6GIYPLQ55TKFWAUNOF\",\"WARC-Block-Digest\":\"sha1:MNFVX64JOKAOA3JRFBAA2YTX6JHHCDD7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882570692.22_warc_CC-MAIN-20220807181008-20220807211008-00018.warc.gz\"}"}
https://slave2.omega.jstor.org/citation/ris/10.4169/j.ctt5hh9t5.9	[ "Provider: JSTOR http://www.jstor.org Database: JSTOR Content: text/plain; charset=\"us-ascii\" TY - CHAP TI - Differential Calculus A2 - Klymchuk, Sergiy AB -\n\n4.1 If both functionsf(x) andg(x) are differentiable andf(x) >g(x) on the interval (a,b), thenf′(x) >g′(x) on (a,b). (page 57)\n\n4.2 If a nonlinear function is differentiable and monotone on (0, ∞), then its derivative is also monotone on (0, ∞). (page 57)\n\n4.3 If a function is continuous at a point, then it is differentiable at that point. (page 58)\n\n4.4 If a function is continuous on\\$\\mathbb{R}\\$and the tangent line exists at any point on its graph, then the function is differentiable at any point on\\$\\mathbb{R}\\$. (page 59)\n\nIf a\n\nEP - 22 ET - REV - Revised, 1 PB - Mathematical Association of America PY - 2010 SN - 9780883857656 SP - 19 T2 - Counterexamples in Calculus UR - http://www.jstor.org/stable/10.4169/j.ctt5hh9t5.9 Y2 - 2021/09/28/ ER -" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.63129336,"math_prob":0.96099234,"size":764,"snap":"2021-31-2021-39","text_gpt3_token_len":261,"char_repetition_ratio":0.110526316,"word_repetition_ratio":0.016260162,"special_character_ratio":0.36125654,"punctuation_ratio":0.17261904,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98505414,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-28T21:00:33Z\",\"WARC-Record-ID\":\"<urn:uuid:c67d1d37-9320-4c98-a20f-b5dafb692c20>\",\"Content-Length\":\"1793\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f696ceb5-cc33-4e08-99f3-a6de6f65d1de>\",\"WARC-Concurrent-To\":\"<urn:uuid:9db5626c-7577-4930-92c9-07390d3f0392>\",\"WARC-IP-Address\":\"199.232.64.152\",\"WARC-Target-URI\":\"https://slave2.omega.jstor.org/citation/ris/10.4169/j.ctt5hh9t5.9\",\"WARC-Payload-Digest\":\"sha1:4GSSZJOFFYIIRGU4JSOYGGRMICHR5W5F\",\"WARC-Block-Digest\":\"sha1:65FP4UY37XNO5HERSEAV362BV7ATQAPJ\",\"WARC-Identified-Payload-Type\":\"text/plain\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780060882.17_warc_CC-MAIN-20210928184203-20210928214203-00655.warc.gz\"}"}
https://isabelle.in.tum.de/repos/isabelle/file/e5b55d7be9bb/src/Tools/IsaPlanner/rw_tools.ML	[ "src/Tools/IsaPlanner/rw_tools.ML\n author wenzelm Sat Oct 06 16:50:04 2007 +0200 (2007-10-06) changeset 24867 e5b55d7be9bb parent 23175 267ba70e7a9d child 30161 c26e515f1c29 permissions -rw-r--r--\nsimplified interfaces for outer syntax;\n``` 1 (* Title: Tools/IsaPlanner/rw_tools.ML\n```\n``` 2 ID:\t\t\\$Id\\$\n```\n``` 3 Author: Lucas Dixon, University of Edinburgh\n```\n``` 4\n```\n``` 5 Term related tools used for rewriting.\n```\n``` 6 )\n```\n``` 7\n```\n``` 8 signature RWTOOLS =\n```\n``` 9 sig\n```\n``` 10 end;\n```\n``` 11\n```\n``` 12 structure RWTools\n```\n``` 13 = struct\n```\n``` 14\n```\n``` 15 ( fake free variable names for locally bound variables - these work\n```\n``` 16 as placeholders. )\n```\n``` 17\n```\n``` 18 ( don't use dest_fake.. - we should instead be working with numbers\n```\n``` 19 and a list... else we rely on naming conventions which can break, or\n```\n``` 20 be violated - in contrast list locations are correct by\n```\n``` 21 construction/definition. )\n```\n``` 22 (\n```\n``` 23 fun dest_fake_bound_name n =\n```\n``` 24 case (explode n) of\n```\n``` 25 (\":\" :: realchars) => implode realchars\n```\n``` 26 \| _ => n; )\n```\n``` 27 fun is_fake_bound_name n = (hd (explode n) = \":\");\n```\n``` 28 fun mk_fake_bound_name n = \":b_\" ^ n;\n```\n``` 29\n```\n``` 30\n```\n``` 31\n```\n``` 32 ( fake free variable names for local meta variables - these work\n```\n``` 33 as placeholders. )\n```\n``` 34 fun dest_fake_fix_name n =\n```\n``` 35 case (explode n) of\n```\n``` 36 (\"@\" :: realchars) => implode realchars\n```\n``` 37 \| _ => n;\n```\n``` 38 fun is_fake_fix_name n = (hd (explode n) = \"@\");\n```\n``` 39 fun mk_fake_fix_name n = \"@\" ^ n;\n```\n``` 40\n```\n``` 41\n```\n``` 42\n```\n``` 43 ( fake free variable names for meta level bound variables )\n```\n``` 44 fun dest_fake_all_name n =\n```\n``` 45 case (explode n) of\n```\n``` 46 (\"+\" :: realchars) => implode realchars\n```\n``` 47 \| _ => n;\n```\n``` 48 fun is_fake_all_name n = (hd (explode n) = \"+\");\n```\n``` 49 fun mk_fake_all_name n = \"+\" ^ n;\n```\n``` 50\n```\n``` 51\n```\n``` 52\n```\n``` 53\n```\n``` 54 ( Ys and Ts not used, Ns are real names of faked local bounds, the\n```\n``` 55 idea is that this will be mapped to free variables thus if a free\n```\n``` 56 variable is a faked local bound then we change it to being a meta\n```\n``` 57 variable so that it can later be instantiated )\n```\n``` 58 ( FIXME: rename this - avoid the word fix! )\n```\n``` 59 ( note we are not really \"fix\"'ing the free, more like making it variable! )\n```\n``` 60 ( fun trymkvar_of_fakefree (Ns, Ts) Ys (n,ty) =\n```\n``` 61 if n mem Ns then Var((n,0),ty) else Free (n,ty);\n```\n``` 62 )\n```\n``` 63\n```\n``` 64 ( make a var into a fixed free (ie prefixed with \"@\") )\n```\n``` 65 fun mk_fakefixvar Ts ((n,i),ty) = Free(mk_fake_fix_name n, ty);\n```\n``` 66\n```\n``` 67\n```\n``` 68 ( mk_frees_bound: string list -> Term.term -> Term.term )\n```\n``` 69 ( This function changes free variables to being represented as bound\n```\n``` 70 variables if the free's variable name is in the given list. The debruijn\n```\n``` 71 index is simply the position in the list )\n```\n``` 72 ( THINKABOUT: danger of an existing free variable with the same name: fix\n```\n``` 73 this so that name conflict are avoided automatically! In the meantime,\n```\n``` 74 don't have free variables named starting with a \":\" )\n```\n``` 75 fun bounds_of_fakefrees Ys (a \\$ b) =\n```\n``` 76 (bounds_of_fakefrees Ys a) \\$ (bounds_of_fakefrees Ys b)\n```\n``` 77 \| bounds_of_fakefrees Ys (Abs(n,ty,t)) =\n```\n``` 78 Abs(n,ty, bounds_of_fakefrees (n::Ys) t)\n```\n``` 79 \| bounds_of_fakefrees Ys (Free (n,ty)) =\n```\n``` 80 let fun try_mk_bound_of_free (i,[]) = Free (n,ty)\n```\n``` 81 \| try_mk_bound_of_free (i,(y::ys)) =\n```\n``` 82 if n = y then Bound i else try_mk_bound_of_free (i+1,ys)\n```\n``` 83 in try_mk_bound_of_free (0,Ys) end\n```\n``` 84 \| bounds_of_fakefrees Ys t = t;\n```\n``` 85\n```\n``` 86\n```\n``` 87 ( map a function f onto each free variables )\n```\n``` 88 fun map_to_frees f Ys (a \\$ b) =\n```\n``` 89 (map_to_frees f Ys a) \\$ (map_to_frees f Ys b)\n```\n``` 90 \| map_to_frees f Ys (Abs(n,ty,t)) =\n```\n``` 91 Abs(n,ty, map_to_frees f ((n,ty)::Ys) t)\n```\n``` 92 \| map_to_frees f Ys (Free a) =\n```\n``` 93 f Ys a\n```\n``` 94 \| map_to_frees f Ys t = t;\n```\n``` 95\n```\n``` 96\n```\n``` 97 ( map a function f onto each meta variable )\n```\n``` 98 fun map_to_vars f Ys (a \\$ b) =\n```\n``` 99 (map_to_vars f Ys a) \\$ (map_to_vars f Ys b)\n```\n``` 100 \| map_to_vars f Ys (Abs(n,ty,t)) =\n```\n``` 101 Abs(n,ty, map_to_vars f ((n,ty)::Ys) t)\n```\n``` 102 \| map_to_vars f Ys (Var a) =\n```\n``` 103 f Ys a\n```\n``` 104 \| map_to_vars f Ys t = t;\n```\n``` 105\n```\n``` 106 ( map a function f onto each free variables )\n```\n``` 107 fun map_to_alls f (Const(\"all\",allty) \\$ Abs(n,ty,t)) =\n```\n``` 108 let val (n2,ty2) = f (n,ty)\n```\n``` 109 in (Const(\"all\",allty) \\$ Abs(n2,ty2,map_to_alls f t)) end\n```\n``` 110 \| map_to_alls f x = x;\n```\n``` 111\n```\n``` 112 ( map a function f to each type variable in a term )\n```\n``` 113 ( implicit arg: term )\n```\n``` 114 fun map_to_term_tvars f =\n```\n``` 115 Term.map_types (fn TVar(ix,ty) => f (ix,ty) \| x => x); ( FIXME map_atyps !? )\n```\n``` 116\n```\n``` 117\n```\n``` 118\n```\n``` 119 ( what if a param desn't occur in the concl? think about! Note: This\n```\n``` 120 simply fixes meta level univ bound vars as Frees. At the end, we will\n```\n``` 121 change them back to schematic vars that will then unify\n```\n``` 122 appropriactely, ie with unfake_vars )\n```\n``` 123 fun fake_concl_of_goal gt i =\n```\n``` 124 let\n```\n``` 125 val prems = Logic.strip_imp_prems gt\n```\n``` 126 val sgt = List.nth (prems, i - 1)\n```\n``` 127\n```\n``` 128 val tbody = Logic.strip_imp_concl (Term.strip_all_body sgt)\n```\n``` 129 val tparams = Term.strip_all_vars sgt\n```\n``` 130\n```\n``` 131 val fakefrees = map (fn (n, ty) => Free(mk_fake_all_name n, ty))\n```\n``` 132 tparams\n```\n``` 133 in\n```\n``` 134 Term.subst_bounds (rev fakefrees,tbody)\n```\n``` 135 end;\n```\n``` 136\n```\n``` 137 ( what if a param desn't occur in the concl? think about! Note: This\n```\n``` 138 simply fixes meta level univ bound vars as Frees. At the end, we will\n```\n``` 139 change them back to schematic vars that will then unify\n```\n``` 140 appropriactely, ie with unfake_vars )\n```\n``` 141 fun fake_goal gt i =\n```\n``` 142 let\n```\n``` 143 val prems = Logic.strip_imp_prems gt\n```\n``` 144 val sgt = List.nth (prems, i - 1)\n```\n``` 145\n```\n``` 146 val tbody = Term.strip_all_body sgt\n```\n``` 147 val tparams = Term.strip_all_vars sgt\n```\n``` 148\n```\n``` 149 val fakefrees = map (fn (n, ty) => Free(mk_fake_all_name n, ty))\n```\n``` 150 tparams\n```\n``` 151 in\n```\n``` 152 Term.subst_bounds (rev fakefrees,tbody)\n```\n``` 153 end;\n```\n``` 154\n```\n``` 155\n```\n``` 156 ( hand written - for some reason the Isabelle version in drule is broken!\n```\n``` 157 Example? something to do with Bin Yangs examples?\n```\n``` 158 )\n```\n``` 159 fun rename_term_bvars ns (Abs(s,ty,t)) =\n```\n``` 160 let val s2opt = Library.find_first (fn (x,y) => s = x) ns\n```\n``` 161 in case s2opt of\n```\n``` 162 NONE => (Abs(s,ty,rename_term_bvars ns t))\n```\n``` 163 \| SOME (_,s2) => Abs(s2,ty, rename_term_bvars ns t) end\n```\n``` 164 \| rename_term_bvars ns (a\\$b) =\n```\n``` 165 (rename_term_bvars ns a) \\$ (rename_term_bvars ns b)\n```\n``` 166 \| rename_term_bvars _ x = x;\n```\n``` 167\n```\n``` 168 fun rename_thm_bvars ns th =\n```\n``` 169 let val t = Thm.prop_of th\n```\n``` 170 in Thm.rename_boundvars t (rename_term_bvars ns t) th end;\n```\n``` 171\n```\n``` 172 ( Finish this to show how it breaks! (raises the exception):\n```\n``` 173\n```\n``` 174 exception rename_thm_bvars_exp of ((string * string) list * Thm.thm)\n```\n``` 175\n```\n``` 176 Drule.rename_bvars ns th\n```\n``` 177 handle TERM _ => raise rename_thm_bvars_exp (ns, th);\n```\n``` 178 *)\n```\n``` 179\n```\n``` 180 end;\n```" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.65581983,"math_prob":0.9908827,"size":5538,"snap":"2019-51-2020-05","text_gpt3_token_len":1674,"char_repetition_ratio":0.13715215,"word_repetition_ratio":0.119796954,"special_character_ratio":0.36150235,"punctuation_ratio":0.14139535,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9945281,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-11T06:43:01Z\",\"WARC-Record-ID\":\"<urn:uuid:ad73ee3d-1958-4676-9196-5b4fcd03c5ed>\",\"Content-Length\":\"31281\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:810e7be2-d772-4968-9004-afe3a8ae38e5>\",\"WARC-Concurrent-To\":\"<urn:uuid:68f487f8-9940-443f-a14e-702db738b299>\",\"WARC-IP-Address\":\"131.159.46.82\",\"WARC-Target-URI\":\"https://isabelle.in.tum.de/repos/isabelle/file/e5b55d7be9bb/src/Tools/IsaPlanner/rw_tools.ML\",\"WARC-Payload-Digest\":\"sha1:YEKRKZNFBM5WFHLQNDI7IUOFD7FQYO3K\",\"WARC-Block-Digest\":\"sha1:HP4CPBDZTPGRNGZULRVGWXB2ICSTCN7K\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540529955.67_warc_CC-MAIN-20191211045724-20191211073724-00114.warc.gz\"}"}
https://zbmath.org/?q=an%3A0817.35114	[ "## Some results on the thermistor problem.(English)Zbl 0817.35114\n\nAntontsev, S. N. (ed.) et al., Free boundary problems in continuum mechanics. International conference on free boundary problems in continuum mechanics, Novosibirsk, Russia, July 15-19, 1991. Basel: Birkhäuser. ISNM, Int. Ser. Numer. Math. 106, 47-57 (1992).\nFrom the introduction: We consider here the so called thermistor problem. The heat produced in a conductor by an electric current leads to the system: \\begin{aligned} &u_ t- \\nabla\\cdot k(u) \\nabla u= \\sigma(u) \|\\nabla \\varphi\|^ 2, \\quad \\nabla\\cdot \\sigma(u) \\nabla \\varphi=0 \\quad \\text{in } \\Omega\\times (0,T),\\\\ &u=0, \\quad \\varphi= \\varphi_ 0 \\quad \\text{on } \\Gamma\\times (0,T), \\quad u(\\cdot,0)= u_ 0. \\end{aligned} \\tag{1} Here, $$\\Omega$$ is a smooth bounded open set of $$\\mathbb{R}^ n$$, $$\\Gamma$$ denotes the boundary, $$T$$ is some positive given number, $$\\varphi$$ is the electrical potential, $$u$$ the temperature inside the conductor, $$k(u)>0$$ the thermal conductivity and $$\\sigma(u) >0$$ the electrical conductivity.\nWe show existence of a solution to (1), and focus on the question of uniqueness and on the problem of global existence or blow up.\nFor the entire collection see [Zbl 0807.00016].\n\n### MSC:\n\n 35Q72 Other PDE from mechanics (MSC2000) 80A20 Heat and mass transfer, heat flow (MSC2010) 35K55 Nonlinear parabolic equations\n\n### Keywords:\n\nexistence; uniqueness; blow up" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7069589,"math_prob":0.9974653,"size":1609,"snap":"2022-05-2022-21","text_gpt3_token_len":518,"char_repetition_ratio":0.093457945,"word_repetition_ratio":0.017316017,"special_character_ratio":0.33188316,"punctuation_ratio":0.20543806,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99876165,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-24T02:21:07Z\",\"WARC-Record-ID\":\"<urn:uuid:04c561db-7257-4481-97ff-681cdee117c8>\",\"Content-Length\":\"50359\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:73351bb8-f40d-4705-abe0-a3ac039bff1e>\",\"WARC-Concurrent-To\":\"<urn:uuid:bf682c52-a961-4bfa-8522-64d47de7b2de>\",\"WARC-IP-Address\":\"141.66.194.2\",\"WARC-Target-URI\":\"https://zbmath.org/?q=an%3A0817.35114\",\"WARC-Payload-Digest\":\"sha1:LEYIQRFQ54LVXFHOITRNF7SRKWZHRAXV\",\"WARC-Block-Digest\":\"sha1:H7LEVR5IFKAGPWREXBWV66DUR2BK5HTE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662562410.53_warc_CC-MAIN-20220524014636-20220524044636-00710.warc.gz\"}"}
https://ao.ms/how-to-divide-a-number-in-python/	[ "# How to divide a number in Python", null, "## The challenge\n\nYour task is to create function`isDivideBy` (or `is_divide_by`) to check if an integer number is divisible by each out of two arguments.\n\nA few cases:\n\n```.wp-block-code{border:0;padding:0}.wp-block-code>div{overflow:auto}.shcb-language{border:0;clip:rect(1px,1px,1px,1px);-webkit-clip-path:inset(50%);clip-path:inset(50%);height:1px;margin:-1px;overflow:hidden;padding:0;position:absolute;width:1px;word-wrap:normal;word-break:normal}.hljs{box-sizing:border-box}.hljs.shcb-code-table{display:table;width:100%}.hljs.shcb-code-table>.shcb-loc{color:inherit;display:table-row;width:100%}.hljs.shcb-code-table .shcb-loc>span{display:table-cell}.wp-block-code code.hljs:not(.shcb-wrap-lines){white-space:pre}.wp-block-code code.hljs.shcb-wrap-lines{white-space:pre-wrap}.hljs.shcb-line-numbers{border-spacing:0;counter-reset:line}.hljs.shcb-line-numbers>.shcb-loc{counter-increment:line}.hljs.shcb-line-numbers .shcb-loc>span{padding-left:.75em}.hljs.shcb-line-numbers .shcb-loc::before{border-right:1px solid #ddd;content:counter(line);display:table-cell;padding:0 .75em;text-align:right;-webkit-user-select:none;-moz-user-select:none;-ms-user-select:none;user-select:none;white-space:nowrap;width:1%}```(-12, 2, -6) -> true\n(-12, 2, -5) -> false\n\n(45, 1, 6) -> false\n(45, 5, 15) -> true\n\n(4, 1, 4) -> true\n(15, -5, 3) -> true```Code language: JavaScript (javascript)```\n\n## Test cases\n\n``````Test.describe(\"Basic Tests\")\nTest.it(\"should pass basic tests\")\nTest.assert_equals(is_divide_by(-12, 2, -6), True)\nTest.assert_equals(is_divide_by(-12, 2, -5), False)\nTest.assert_equals(is_divide_by(45, 1, 6), False)\nTest.assert_equals(is_divide_by(45, 5, 15), True)\nTest.assert_equals(is_divide_by(4, 1, 4), True)\nTest.assert_equals(is_divide_by(15, -5, 3), True)\n```Code language: Python (python)```\n\n## Understanding how to solve this\n\nTo resolve this problem, we need to understand how to find if a number can be divided without a remainder in Python.\n\nFor this we will use Python’s` modulo operator`, (`%`):\n\n``````10 % 5 # 0\n# if we divide 10 by 5, there is no remainder\n\n10 % 3 # 1\n# if we divide 10 by 3, there is a remainder of `1`\n```Code language: Python (python)```\n\nTherefore, if we say `10 % 5 == 0`, the expression will equal `True`, while the `10 % 3 == 0` will equal False. This is because there is a remainder of `1` in the second instance.\n\n## The solution in Python\n\nOption 1:\n\n``````def is_divide_by(number, a, b):\n# if can divide without remainder\nif number % a ==0 and number % b ==0:\nreturn True\nelse:\nreturn False\n```Code language: Python (python)```\n\nOption 2:\n\n``````def is_divide_by(number, a, b):\nreturn not (number%a or number%b)\n```Code language: Python (python)```\n\nOption 3:\n\n``````def is_divide_by(n, a, b):\nreturn n%a == 0 == n%b\n```Code language: Python (python)```\nTags:\nSubscribe\nNotify of", null, "" ]	[ null, "https://ao.ms/wp-content/plugins/wp-performance/assets/placeholder.png", null, "https://ao.ms/wp-content/plugins/wp-performance/assets/placeholder.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.55353725,"math_prob":0.8136832,"size":1652,"snap":"2021-43-2021-49","text_gpt3_token_len":526,"char_repetition_ratio":0.15048544,"word_repetition_ratio":0.02909091,"special_character_ratio":0.34745762,"punctuation_ratio":0.20113315,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9906332,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-21T13:16:01Z\",\"WARC-Record-ID\":\"<urn:uuid:268ac30e-f545-42ad-82df-4b56eedcd738>\",\"Content-Length\":\"201134\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:db9c16c6-ca2c-4a8a-8bb6-2b3cd2a0cad4>\",\"WARC-Concurrent-To\":\"<urn:uuid:cb0ae5cd-e687-4de9-bcdd-01f53b53a4c4>\",\"WARC-IP-Address\":\"104.21.26.65\",\"WARC-Target-URI\":\"https://ao.ms/how-to-divide-a-number-in-python/\",\"WARC-Payload-Digest\":\"sha1:OCJQDEQCDRTRAK4ZHJXZIVJ6GTETCXP5\",\"WARC-Block-Digest\":\"sha1:KZUEBUZMZJ3SHJZJDHSIPMC2GMHYOYRQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585405.74_warc_CC-MAIN-20211021102435-20211021132435-00555.warc.gz\"}"}
https://www.prepbible.com/1z0-808-test/prep-5846.html	[ "Act now and download your Oracle 1z0 808 practice test test today! Do not waste time for the worthless Oracle 1z0 808 book tutorials. Download Up to the minute Oracle Java SE 8 Programmer I exam with real questions and answers and begin to learn Oracle 1z0 808 practice test with a classic professional.\n\n♥♥ 2021 NEW RECOMMEND ♥♥\n\nFree VCE & PDF File for Oracle 1z0-808 Real Exam (Full Version!)\n\n★ Pass on Your First TRY ★ 100% Money Back Guarantee ★ Realistic Practice Exam Questions\n\nAvailable on: http://www.surepassexam.com/1z0-808-exam-dumps.html\n\nQ11. Given the code fragment:\n\nWhat is the result if the integer aVar is 9?\n\nA. 10 Hello world!\n\nB. 10 Hello universe!\n\nC. 9 Hello world!\n\nD. Compilation fails.\n\nQ12. Given:\n\npublic class Test {\n\npublic static void main(String[] args) {\n\ntry {\n\nString[] arr =new String;\n\narr = \"Unix\";\n\narr = \"Linux\";\n\narr = \"Solarios\";\n\nfor (String var : arr) {\n\nSystem.out.print(var + \" \");\n\n} catch(Exception e) {\n\nSystem.out.print (e.getClass());\n\nWhat is the result?\n\nA. Unix Linux Solaris\n\nB. Null Unix Linux Solaris\n\nC. Class java.lang.Exception\n\nD. Class java.lang.NullPointerException\n\nExplanation: null Unix Linux Solarios\n\nThe first element, arr, has not been defined.\n\nQ13. Given the fragment:\n\nWhat is the result?\n\nA. 13480.0\n\nB. 13480.02\n\nC. Compilation fails\n\nD. An exception is thrown at runtime\n\nQ14. Given:\n\nWhat is the result?\n\nA. simaple A\n\nB. Capital A\n\nC. simaple A default Capital A\n\nD. simaple A default\n\nE. Compilation fails.\n\nExplanation:\n\nHere we have to use two ternary operators combined. SO first we can use to check first\n\ncondition which is x > 10, as follows;\n\nx>10?\">\": (when condition false) Now we have to use another to check if x<10 as follows;\n\nx<10?V:\"=\" We can combine these two by putting last ternary statement in the false\n\nposition of first ternary statement as follows;\n\nx>10?\">\":x<10?'<':\"=\"\n\nhttps;//docs.oraclexom/javase/tutorial/java/nutsandbolts/if.html\n\nQ15. Given the code from the Greeting.Java file:\n\nWhich set of commands prints Hello Duke in the console?\n\nA. Option A\n\nB. Option B\n\nC. Option C\n\nD. Option D\n\nQ16. Given:\n\nAnd given the code fragment:\n\nWhat is the result?\n\nA. 4W 100 Auto 4W 150 Manual\n\nB. Null 0 Auto 4W 150 Manual\n\nC. Compilation fails only at line n1\n\nD. Compilation fails only at line n2\n\nE. Compilation fails at both line n1 and line n2\n\nExplanation:\n\nOn line n1 implicit call to parameterized constructor is missing and n2 this() must be the first line.\n\nQ17. Which two are valid array declaration?\n\nA. Object array[];\n\nB. Boolean array;\n\nC. int[] array;\n\nD. Float array;\n\nQ18. Given:\n\nclass Base {\n\n// insert code here\n\npublic class Derived extends Base{\n\npublic static void main(String[] args) {\n\nDerived obj = new Derived();\n\nobj.setNum(3);\n\nSystem.out.println(\"Square = \" + obj.getNum() * obj.getNum());\n\nWhich two options, when inserted independently inside class Base, ensure that the class is being properly encapsulated and allow the program to execute and print the square of the number?\n\nA. private int num; public int getNum() { return num; }public void setNum(int num) { this.num = num;}\n\nB. public int num; protected public int getNum() { return num; }protected public void setNum(int num) { this.num = num;}\n\nC. private int num;public int getNum() {return num;} private void setNum(int num) { this.num = num;}\n\nD. protected int num; public int getNum() { return num; } public void setNum(int num) { this.num = num;}\n\nE. protected int num; private int getNum() { return num; } public void setNum(int num) { this.num = num;}\n\nExplanation:\n\nIncorrect:\n\nNot B: illegal combination of modifiers: protected and public\n\nnot C: setNum method cannot be private.\n\nnot E: getNum method cannot be private.\n\nQ19. Given:\n\nWhat is the result?\n\nA. 11, 21, 31, 11, 21, 31\n\nB. 11, 21, 31, 12, 22, 32\n\nC. 12, 22, 32, 12, 22, 32\n\nD. 10, 20, 30, 10, 20, 30\n\nQ20. Given:\n\nThe class is poorly encapsulated. You need to change the circle class to compute and return the area instead.\n\nWhich two modifications are necessary to ensure that the class is being properly encapsulated?\n\nA. Remove the area field.\n\nB. Change the getArea( ) method as follows:\n\npublic double getArea ( ) { return Match.PI * radius * radius; }" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6609435,"math_prob":0.7304263,"size":4488,"snap":"2021-43-2021-49","text_gpt3_token_len":1271,"char_repetition_ratio":0.109500445,"word_repetition_ratio":0.09424084,"special_character_ratio":0.30013368,"punctuation_ratio":0.2173913,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9600403,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-06T16:19:15Z\",\"WARC-Record-ID\":\"<urn:uuid:d44369ff-55b7-4461-9caf-28d491ac6e4f>\",\"Content-Length\":\"41097\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:566eeb22-3c63-4f8e-8859-75bcb62d2ab9>\",\"WARC-Concurrent-To\":\"<urn:uuid:ec324f0e-b088-4e79-ad87-f1b6cd0651f3>\",\"WARC-IP-Address\":\"172.96.187.181\",\"WARC-Target-URI\":\"https://www.prepbible.com/1z0-808-test/prep-5846.html\",\"WARC-Payload-Digest\":\"sha1:WYIG3BJXFCUIO2H7GMK6NGMSFWDNJH6T\",\"WARC-Block-Digest\":\"sha1:WY7MJ3HH5S735U7VZZHQ2IE3MLKGQJNP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363301.3_warc_CC-MAIN-20211206133552-20211206163552-00354.warc.gz\"}"}
https://testbook.com/question-answer/in-how-many-ways-can-a-committee-of-4-be-made-out--61b0a05d64d17eb7d9a6f7c9	[ "In how many ways can a committee of 4 be made out of 5 men and 3 women containing at least one woman?\n\n1. 65\n2. 70\n3. 75\n4. 60\n\nOption 1 : 65\n\nDetailed Solution\n\nFormula used:\n\n• $$^nC_r=\\frac{n!}{r!(n-r)!}$$\n• nCn = 1\n• nCr = nCn-r\n\nCalculation:\n\nGiven that,\n\nThere are 5 men and 3 women.\n\nNumber of ways in which committee form containing at least 1 women\n\n3 men & 1 women + 2 men & 2 women + 1 men & 3 women\n\n= 5C3 × 3C1 + 5C2 × 3C2 + 5C1 × 3C3\n\n= 2× 5C3 × 3C1 + 5C1 × 3C3 (∵nCr = nCn-r)\n\n$$2\\times\\frac{5!}{3!×(5-3)!}×3+5\\times1$$ (∵ nCn = 1)\n\n$$2\\times\\frac{5\\times4\\times3 \\times2!}{3!×(2)!}×3+5$$\n\n= 60 + 5\n\n= 65" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.72973055,"math_prob":0.99994755,"size":668,"snap":"2022-05-2022-21","text_gpt3_token_len":288,"char_repetition_ratio":0.13253012,"word_repetition_ratio":0.0,"special_character_ratio":0.43263474,"punctuation_ratio":0.107913665,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98418367,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-17T04:24:09Z\",\"WARC-Record-ID\":\"<urn:uuid:52925619-72c8-4659-b186-04dd8c853843>\",\"Content-Length\":\"114749\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a92d603b-0461-463f-963e-2f25011fcc07>\",\"WARC-Concurrent-To\":\"<urn:uuid:dae367ee-15d4-447b-8f0f-3e061bcadcb6>\",\"WARC-IP-Address\":\"172.67.30.170\",\"WARC-Target-URI\":\"https://testbook.com/question-answer/in-how-many-ways-can-a-committee-of-4-be-made-out--61b0a05d64d17eb7d9a6f7c9\",\"WARC-Payload-Digest\":\"sha1:RVLGEOAIJRM7U5UJSJ5JNVW3OLMR3NRL\",\"WARC-Block-Digest\":\"sha1:3OR2B5HSR5RI3BY4ADBBLOSLMHMD3DMR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320300289.37_warc_CC-MAIN-20220117031001-20220117061001-00501.warc.gz\"}"}
https://matholympiad.org.bd/forum/viewtopic.php?f=30&t=5940&p=21596	[ "## BdPhO Regional (Dhaka-South) Higher Secondary 2019/Bonus\n\nDiscuss Physics and Physics Olympiad related problems here\nSINAN EXPERT\nPosts: 38\nJoined: Sat Jan 19, 2019 3:35 pm\nContact:\n\n### BdPhO Regional (Dhaka-South) Higher Secondary 2019/Bonus\n\nMany particle physics experiments are carried on one of two types of accelerator facilities, (a) colliders in which equal energy particle collide head on, i.e. their momentum vectors point in opposite direction, (b) “fixed target facilities”, where an energetic beam particle collide with a stationary target particle. The two types of facilities have different advantages and\ndisadvantages. We’ll consider one important difference in this problem.\n\nSuppose our goal is to collide two particles of same mass m to create a more massive particle of mass \\$M\\$ \\$(M > m)\\$ – in other words the original particles of mass m annihilate with each other and create a new particle of mass \\$M\\$. The key question regarding how much expensive our particle accelerator will be “How much energy does our particle beam(s) need to have?’’\n\n(a) (Head on) What is the minimum relativistic total energy of each particle of mass \\$m\\$ must be to create the new particle with mass \\$M\\$? (Express your answer in terms of \\$m\\$ and \\$M\\$)\n\n(b) (Fixed Particle) What is the minimum total energy that our incident particle of mass \\$m\\$ must have if it is to collide with another stationary particle of mass \\$m\\$ and create the single particle with mass \\$M\\$? Express your answer in terms of \\$m\\$ and \\$M\\$. Hint: Don’t forget that momentum must be conserved!\n\n(c) What is the ratio of the fixed target beam and the sum of the energies of two target beams in head on collisions? Which facility do you think should be less expensive?\nWhat is this ratio when \\$m\\$ is the mass of the proton \\$(~ 1 GeV/c2)\\$ and \\$M\\$ is the mass of the Higgs Boson \\$(~125 GeV/c2)\\$\n\\$The\\$ \\$only\\$ \\$way\\$ \\$to\\$ \\$learn\\$ \\$mathematics\\$ \\$is\\$ \\$to\\$ \\$do\\$ \\$mathematics\\$. \\$-\\$ \\$PAUL\\$ \\$HALMOS\\$" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.86610276,"math_prob":0.85867953,"size":1952,"snap":"2020-10-2020-16","text_gpt3_token_len":501,"char_repetition_ratio":0.1463039,"word_repetition_ratio":0.042682927,"special_character_ratio":0.25153688,"punctuation_ratio":0.072222225,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96739656,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-04-01T07:33:06Z\",\"WARC-Record-ID\":\"<urn:uuid:f8c20ea1-b818-461b-964b-38a5630ffb70>\",\"Content-Length\":\"29064\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e8c9e80e-5610-49e3-a4fe-60f2be9f4dae>\",\"WARC-Concurrent-To\":\"<urn:uuid:9e5f7716-d148-4eeb-939b-f8297d3ebfe6>\",\"WARC-IP-Address\":\"167.71.232.37\",\"WARC-Target-URI\":\"https://matholympiad.org.bd/forum/viewtopic.php?f=30&t=5940&p=21596\",\"WARC-Payload-Digest\":\"sha1:X7DOUVYKNYW4QRWS2GYVA5WKTER7RZK7\",\"WARC-Block-Digest\":\"sha1:ZFBJSSW26BBWDFROUGIHCAPB2TLSQIGS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585370505550.17_warc_CC-MAIN-20200401065031-20200401095031-00312.warc.gz\"}"}
https://www.gromacs.org/Documentation_of_outdated_versions/How-tos/Dihedral_PCA	[ "# Dihedral PCA\n\nTo make the index file for dihedral PCA either use mk_angndx or make an index file by hand. Now you need a suitable combination of g_angle (possibly with `-oc` or `-or`) and g_covar.\n\nHere is some discussion on the use of dihedral PCA and a reply.\n\n## Background\n\nThe implementation of PCA in GROMACS first makes a trajectory file with reduced dimensions matching the selected angles, and then makes a fake trajectory file that contains the eigenvectors and eigenvalues. This maybe wasn't the best approach possible, but it was made to work. Steps 1 and 2 do the dimensionality reduction, steps 3 and 4 generate a kind of hacked coordinate file whose dimensionality is large enough to help gmx covar and gmx anaeig work.\n\n## Stepwise instructions\n\n1. Use mk_angndx or text editor to generate an index file for the atoms involved in the chosen dihedral angles.\n\n2. Extract the angles from trajectory using the index file from step 1., e.g., `dangle.ndx`, in a command like:\n\n```g_angle -f foo.xtc -s foo.tpr -n dangle.ndx -or dangle.trr -type dihedral\n```\n\nwhere the output is `dangle.trr` that contains the simulation trajectory reduced to the chosen angle dimensions.\n\n3. Make an index file, named as `covar.ndx`, which necessarily contains one group of atom 1 to integer(2N/3), where N is the number of dihedral angles. For example, for a peptide of 5 dihedral angles, the contents of `covar.ndx` should be\n\n``` [ foo ]\n1 2 3 4\n```\n\n4. Use the index file to make a .gro file that contains integer(2N/3) atoms. Box size and atom coordinates don't matter. For example,\n\n```trjconv -s foo.tpr -f dangle.trr -o resized.gro -n covar.ndx -e 0\n```\n\n5. Perform the diagonalization with a command like\n\n```g_covar -f dangle.trr -n covar.ndx -ascii -xpm -nofit -nomwa -noref -nopbc -s resized.gro\n```\n\nwhere the outputs are `eigenval.xvg`, `eigenvec.trr`, `covar.log`, `covar.dat`, and `covar.xpm`.\n\n6. To get a PMF along one eigenvector use a command like\n\n```g_anaeig -v eigenvec.trr -f dangle.trr -s resized.gro -first X -last X -proj proj-1\n```\n\nwhere X is the serial number of the eigenvector. To visualize the result, use\n\n```xmgrace proj-X.xvg\n```\n\n7. To get a free energy landscape for projections along two eigenvectors, use a command like\n\n```g_anaeig -v eigenvec.trr -f dangle.trr -noxvgr -s resized.gro -first X -last Y -2d 2dproj_X_Y.xvg\n```\n\nwhere X and Y are the serial numbers of the eigenvectors.\n\n8. To transform such data into free energy landscape terms, we divide the 2D landscape with grid and count the number of conformations inside each grid. An awk script for such purpose has been posted here. The g_sham program (specifying the `-notime` option) can also be used to generate a 2-D plot of a free energy surface." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.76159966,"math_prob":0.8087515,"size":2167,"snap":"2021-31-2021-39","text_gpt3_token_len":599,"char_repetition_ratio":0.115580216,"word_repetition_ratio":0.032608695,"special_character_ratio":0.2325796,"punctuation_ratio":0.11883408,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9852442,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-25T19:24:47Z\",\"WARC-Record-ID\":\"<urn:uuid:131f215d-e5f8-4eed-9eec-405af3196652>\",\"Content-Length\":\"31859\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2015d8ca-5b4e-486e-937e-5d2cc2d2fe36>\",\"WARC-Concurrent-To\":\"<urn:uuid:2c8e3948-ebe6-4bbb-9d1b-ad8f219ca9c0>\",\"WARC-IP-Address\":\"130.237.11.150\",\"WARC-Target-URI\":\"https://www.gromacs.org/Documentation_of_outdated_versions/How-tos/Dihedral_PCA\",\"WARC-Payload-Digest\":\"sha1:HPA64EI6XVPJAKA2ET2OMHQCLI45HVGN\",\"WARC-Block-Digest\":\"sha1:XH5ELN4VOPTHMP54WML4U3TJZR44BLLK\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057733.53_warc_CC-MAIN-20210925172649-20210925202649-00086.warc.gz\"}"}
https://ijnaa.semnan.ac.ir/article_4653.html	[ "### A new technique of reduce differential transform method to solve local fractional PDEs in mathematical physics\n\nDocument Type : Research Paper\n\nAuthors\n\n1 Department of Mathematics, Faculty of Education for Pure Sciences, University of Thi-Qar, Nasiriyah, Iraq.\n\n2 Department of Mathematics, Iran University of Science and Technology, Tehran 1684613114, Iran. Department of Mathematical Sciences, University of South Africa, Pretoria 0002, South Africa.\n\nAbstract\n\nIn this manuscript, we investigate solutions of the partial differential equations (PDEs) arising in mathematical physics with local fractional derivative operators (LFDOs). To get approximate solutions of these equations, we utilize the reduce differential transform method (RDTM) which is based upon the LFDOs. Illustrative examples are given to show the accuracy and reliable results. The obtained solutions show that the present method is an efficient and simple tool for solving the linear and nonlinear PDEs within the LFDOs.\n\nKeywords" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8705054,"math_prob":0.70540446,"size":888,"snap":"2022-40-2023-06","text_gpt3_token_len":178,"char_repetition_ratio":0.10633484,"word_repetition_ratio":0.0,"special_character_ratio":0.1891892,"punctuation_ratio":0.1369863,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98737633,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-04T20:59:07Z\",\"WARC-Record-ID\":\"<urn:uuid:c2e8d501-f960-405f-bd23-178bbea9c91e>\",\"Content-Length\":\"43168\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c93bb82b-bfa8-4ed9-92bd-ecb0782c74db>\",\"WARC-Concurrent-To\":\"<urn:uuid:497e78d2-4c4e-450b-914f-86561befb80e>\",\"WARC-IP-Address\":\"94.182.28.25\",\"WARC-Target-URI\":\"https://ijnaa.semnan.ac.ir/article_4653.html\",\"WARC-Payload-Digest\":\"sha1:W77REZB7DI6WLNSUOUNSYB4OONCAOMIY\",\"WARC-Block-Digest\":\"sha1:CE6PWPDLQDUTWCWSHJNWCPCF5DJHUM6D\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500154.33_warc_CC-MAIN-20230204205328-20230204235328-00841.warc.gz\"}"}
https://fr.slideshare.net/andyb37/a-directf-iile-topicdownload-bv-c-v-d-8-which-graph-in-fig-22-mdocx	[ "Publicité", null, "Prochain SlideShare", null, "Kinematics-02-Objective Solved\nChargement dans ... 3\n1 sur 1\nPublicité\n\n### a) DirectF iile TopicDownload b)V c) V d) 8) Which graph in Fig- 2-2 m.docx\n\n1. a) DirectF iile TopicDownload b)V c) V d) 8) Which graph in Fig. 2-2 mpresents a constank non-zeto velocity? 8) A)graph a B)graph b C) graph c D) graph d E) both graplhs c and d ? 9) Which gaph/s) in Fig. 2-2 represem(s) zero acceleration? 9) E)c and d A)onlya B)only b C)a and b D)b andc 10) Which gaph(s) in Fig. 2-2 represenit(s) constant positive acceleration? 10) A) graph a B) graph b C) graph c D) graph d E) both graphs c and d 11) 11) When is the avemge velocity of an object equal to the instantaneous velocity? A) never B)only when the velocity is increasing at a constant rate C) always D)only when the velocity is decreasing at a constant rate E)only when the velocity is constant 12) If an object is accelerating, it must therefore undergo 12) ll uchaune in sneed, C)a change in velocity Solution 8. In graph b), graph is parallel to the x-axis means slope is zero so velocity is not changing with time. 9. When velocity is constant slope is zero. Slope in velocity vs time graph is the acceleration so acceleration is zero in b). Similarly in a) velocity is constant and zero so again acceleration is zero. So graph a) and b) is the answer. 10. In graph d), velocity is increasing with time so slope is positive means positive acceleration. 11. e) only when the velocity is constant.\nPublicité" ]	[ null, "https://image.slidesharecdn.com/adirectfiiletopicdownloadbvcvd8whichgraphinfig-2-2m-230203020721-851062e9/85/a-directf-iile-topicdownload-bv-c-v-d-8-which-graph-in-fig-22-mdocx-1-320.jpg", null, "https://cdn.slidesharecdn.com/ss_thumbnails/04-kmobj-230526061552-ee6a6792-thumbnail.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.82807195,"math_prob":0.95488495,"size":1300,"snap":"2023-14-2023-23","text_gpt3_token_len":365,"char_repetition_ratio":0.19598766,"word_repetition_ratio":0.042016808,"special_character_ratio":0.2676923,"punctuation_ratio":0.08421053,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9986126,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-01T16:58:59Z\",\"WARC-Record-ID\":\"<urn:uuid:1226eb6b-b283-4a48-88e7-b3d7c1d20ffb>\",\"Content-Length\":\"419761\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c3567737-1305-4f18-a90a-6ff22c07025b>\",\"WARC-Concurrent-To\":\"<urn:uuid:d5814843-0a5f-4d73-a69a-609896aadd76>\",\"WARC-IP-Address\":\"146.75.34.152\",\"WARC-Target-URI\":\"https://fr.slideshare.net/andyb37/a-directf-iile-topicdownload-bv-c-v-d-8-which-graph-in-fig-22-mdocx\",\"WARC-Payload-Digest\":\"sha1:LLHM4DQTYXS5DEPYDURPCLI5HTWSCHF4\",\"WARC-Block-Digest\":\"sha1:CICWLRAMS6CYMGFZIDLHQ73LXDRU6JWE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224647895.20_warc_CC-MAIN-20230601143134-20230601173134-00205.warc.gz\"}"}
http://www.programforu.com/2011/10/three-terminal-voltage-regulators-using.html	[ "Three terminal voltage regulators using IC 7805 and 7912 - Computer Programming\n\n# Computer Programming\n\nC C++ Java Python Perl Programs Examples with Output -useful for Schools & College Students\n\n## Wednesday, October 12, 2011\n\nThree terminal voltage regulators using IC 7805 and 7912\n\nAim: To verify the three terminal regulators using IC 7805 and 7912.\n\nApparatus:\n\n1. Voltage Regulator trainer kit\n2. Oscilloscope\n3. Digital multimeter\n4. Connecting patch chords\n\nCircuit description:\n\nThe IC 723 is a monolithic integrated circuit voltage regulator featuring high ripple rejection, excellent input and load regulation, excellent temperature stability etc.,\n\nIt consists of a temperature compensating reference voltage amplifier, an error amplifier, 150mA output transistor and an adjustable output current limiter.\nThe basic low voltage regulator type 723 circuit is shown in fig (1). The unregulated input voltage is 24V and the regulated output voltage is from 0.2V to 7.3V by varying POTR. RSC (R1) is connected across current sense (CS) and current limit (CL) terminals to limit the accidental short circuit current.\n\nA stabilizing capacitor (C1) of 100 pF is connected between frequency compensation terminal and inverting (INV) terminal. External NPN pass transistor is added to the basic 723-regulator circuit to increase its load current capability.\n\nThe output voltage can be regulated from 7 to 37 Volts for an input voltage range from 9.5 to 40Volts. For intermediate output voltages the following formula can be used", null, "The resistor values", null, "are calculated from potential divider network® .\n\nProcedure:\n\n1. Connect the circuit diagram as shown in figure.\n2. Measure the output of the regulated power supply that is 0 to +40 and 0 to -40 V using digital multimeter.\n\nFixed voltage regulators 7805 (+5 V)\n\n3. Connect the circuit as shown in figure to observe the line regulation.\n\n4. By varying Vin supply in steps measure the corresponding output Vo and record in a tabulated form as shown below.\n\n5. Plot the graph between Vin, Vo and observe the line regulation.\n\n7912 (-12 V)\n\n6. Connect the circuit as shown in below figure for line regulation.\n7. Repeat steps 4 and 5.\n\nResult:" ]	[ null, "file:///C:/DOCUME~1/REJINP~1/LOCALS~1/Temp/msohtmlclip1/01/clip_image002.gif", null, "file:///C:/DOCUME~1/REJINP~1/LOCALS~1/Temp/msohtmlclip1/01/clip_image004.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.83608335,"math_prob":0.9366153,"size":1975,"snap":"2019-13-2019-22","text_gpt3_token_len":446,"char_repetition_ratio":0.14967021,"word_repetition_ratio":0.01904762,"special_character_ratio":0.2329114,"punctuation_ratio":0.11780822,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9543165,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-23T18:47:15Z\",\"WARC-Record-ID\":\"<urn:uuid:180a29e6-1a19-4ef6-9f35-3e0f92a9e91a>\",\"Content-Length\":\"122893\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7fa973b5-6e50-4ddd-a29f-af836e6807ac>\",\"WARC-Concurrent-To\":\"<urn:uuid:14b8bb23-5590-47c5-a51d-9f6f118bf68a>\",\"WARC-IP-Address\":\"172.217.7.243\",\"WARC-Target-URI\":\"http://www.programforu.com/2011/10/three-terminal-voltage-regulators-using.html\",\"WARC-Payload-Digest\":\"sha1:JEOEIZ3NGCKF52HBJWLWJQHO3OHP34LE\",\"WARC-Block-Digest\":\"sha1:A3R3NNFGSADMRJWZSHQ7JWKLSLGPV4TR\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232257361.12_warc_CC-MAIN-20190523184048-20190523210048-00402.warc.gz\"}"}
https://forums.giantitp.com/showthread.php?639521-Question-about-Cantor-s-Diagonal-Argument&s=36fc60844b6b06ec9725e161c8ea7a86&p=25286300#post25286300	[ "1.", null, "Question about Cantor's Diagonal Argument\n\nIn Cantor's Diagonal Argument to prove that the cardinality of the continuum is greater than that of the rationals, how do we know that the complimentary sequences are all distinct numbers? It's known that they don't contain the same exact sequence as any of the numbers on the list, but there are some numbers that have multiple digital representations (such as 1.0=0.9999...repeating in base 10)\n\nAdditionally, in the base 2 version of this argument there naively seems to be only a countably infinite number of diagonals (there's the one at the top left corner of the diagram, then the one starting one down from it, then the one one over from it, then the one starting one over and one down and so on and so on in the same sort of snaking pattern used to put the rationals into bijection with the integers). This would put the total at two times aleph null, which collapses down into being equal to just regular aleph null.", null, "", null, "Reply With Quote\n\n2.", null, "Re: Question about Cantor's Diagonal Argument\n\nFor the first I think this is a matter of definition as much as anything else.\n\nYou start with a list defined to be all of the members of the set and then construct a member not in the set. It doesn't matter if a member is in the list multiple times because you have just proven that the initial assumption (it is a list of all members of the set) cannot be true which is what implies you are dealing with an uncountable infinity.\n\nAs for 1.0000 and 0.9999 recurring - if you are just looking at the numbers between 0 and 1 then 1.0 is not in the set to begin with.\n\nThe Binary example just makes this easier to follow - and having proven that you can construct a number not in the set, no attempt to insert it into the list helps - you can still always construct another number not in the set. You are correct that there if the list was countable there would be a countable number of diagonals, but by proving that the list is not countable in the first place then neither is the number of diagonals.", null, "", null, "Reply With Quote\n\n3.", null, "Re: Question about Cantor's Diagonal Argument\n\n\"While 0.0111... and 0.1000... would be equal if interpreted as binary fractions (destroying injectivity), they are different when interpreted as decimal fractions, as is done by f. On the other hand, since t is a binary string, the equality 0.0999... = 0.1000... of decimal fractions is not relevant here.\"\nIn other words, its a matter of constructing representations that can't point to the same number.", null, "", null, "Reply With Quote\n\n4.", null, "Re: Question about Cantor's Diagonal Argument", null, "Originally Posted by Khedrac", null, "As for 1.0000 and 0.9999 recurring - if you are just looking at the numbers between 0 and 1 then 1.0 is not in the set to begin with.\nIt's a little more involved than this. The same issue shows up with 0.010000... and 0.0099999... . But each real number has at most two representations as an infinite sequence of digits ( and incidentally, the set of numbers with two such representations is countable ). The proof by contradiction could work as follows:\n\nEvery infinite sequence of digits corresponds to a real number in [0, 1).\nEvery real number corresponds to at most two infinite sequences of digits.\n\nSuppose the set of real numbers were countable.\nThen we could enumerate them in a list.\nThen we could enumerate their representations as infinite sequences of digits in a list, using the same order, but placing the two representations of those numbers having them one after the other.\nBut we can construct the diagonal, and generate an infinite sequence of digits that isn't on this list.\nThen this new sequence of digits corresponds to a real number not in the original list.", null, "", null, "Reply With Quote\n\n5.", null, "Re: Question about Cantor's Diagonal Argument\n\nWe can prove fairly easily that in fact the only case where two infinite decimals represent the same number are if one ends in ...999999... and ones ends in ...00000...:\n\nLet a and b be two different decimal representations. Let n be the first decimal place where they have different digits; we can assume that an (the nth digit of a) is greater than bn (the nth digit of b). Then if we cut off a and b at the nth decimal place, the truncation of a will be at least 10-n greater than the truncation of b, so the digits of a and b after the nth place must make up this difference if they are to be equal. The best we can do in making up the difference is if all remaining digits of b are 9 and all remaining digits of a are 0. In this case, since the sum from k=n+1 to infinity of 9*10-k is 10-n, we will exactly make up the difference, as long as an is just bn+1. If the nth digits differ by more than 1, or if any of the subsequent digits of a are not 0 or any of the subsequent digits of b are not 9, it will be completely impossible to make up the difference, so a and b will represent different numbers.\n\nThis resolves any concerns about applying the diagonal argument to real numbers in multiple ways:\n1. By being careful about how we construct the new number, we can ensure that none of the digits are 0 or 9. This means that it will represent a number with a unique representation, so the fact that the representation is different from all representations on our list implies that the number it represents is different from all representations on our list.\n2. Since every real number has one or two representations, in particular no real number has uncountably many representations. Because the diagonal argument shows there are uncountably many representations, this implies that there must also be uncountably many numbers.\n3. Since every real number with two representations has a representation which ends in infinitely many zeros, every such number is a rational number (and can be written with denominator a power of ten). This means there can only be countably many such numbers, so they are easy to work around (although now that I think about it, most strategies for working around them will basically boil down to one of the previous two options).", null, "", null, "Reply With Quote", null, "Posting Permissions\n\n• You may not post new threads\n• You may not post replies\n• You may not post attachments\n• You may not edit your posts\n•" ]	[ null, "https://forums.giantitp.com/images/sand/icons/icon_scroll.png", null, "https://forums.giantitp.com/images/sand/misc/progress.gif", null, "https://forums.giantitp.com/clear.gif", null, "https://forums.giantitp.com/images/sand/icons/icon_scroll.png", null, "https://forums.giantitp.com/images/sand/misc/progress.gif", null, "https://forums.giantitp.com/clear.gif", null, "https://forums.giantitp.com/images/sand/icons/icon_scroll.png", null, "https://forums.giantitp.com/images/sand/misc/progress.gif", null, "https://forums.giantitp.com/clear.gif", null, "https://forums.giantitp.com/images/sand/icons/icon_scroll.png", null, "https://forums.giantitp.com/images/sand/misc/quote_icon.png", null, "https://forums.giantitp.com/images/sand/buttons/viewpost-right.png", null, "https://forums.giantitp.com/images/sand/misc/progress.gif", null, "https://forums.giantitp.com/clear.gif", null, "https://forums.giantitp.com/images/sand/icons/icon_scroll.png", null, "https://forums.giantitp.com/images/sand/misc/progress.gif", null, "https://forums.giantitp.com/clear.gif", null, "https://forums.giantitp.com/images/sand/buttons/collapse_40b.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9324343,"math_prob":0.97435844,"size":6400,"snap":"2022-05-2022-21","text_gpt3_token_len":1537,"char_repetition_ratio":0.13633521,"word_repetition_ratio":0.040034812,"special_character_ratio":0.24828126,"punctuation_ratio":0.10069177,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9946502,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-19T11:47:34Z\",\"WARC-Record-ID\":\"<urn:uuid:d881bd5e-13b7-4f4f-99cd-349b5c6b5375>\",\"Content-Length\":\"67893\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:60356933-676a-4539-9c10-9a50aeaa2ca0>\",\"WARC-Concurrent-To\":\"<urn:uuid:1df5b1a6-a76f-49f9-a4d1-8accbcd57cf2>\",\"WARC-IP-Address\":\"172.67.193.87\",\"WARC-Target-URI\":\"https://forums.giantitp.com/showthread.php?639521-Question-about-Cantor-s-Diagonal-Argument&s=36fc60844b6b06ec9725e161c8ea7a86&p=25286300#post25286300\",\"WARC-Payload-Digest\":\"sha1:PUV3A7L5UN62NJ2PU7WDDG4UR57RSIDE\",\"WARC-Block-Digest\":\"sha1:PJKYPJCFB4UTMXAHSYJUWADHNSNJDFEH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320301309.22_warc_CC-MAIN-20220119094810-20220119124810-00400.warc.gz\"}"}
http://ppwu.auto-hirch-egelsbach.de/minimum-cost-path-graph.html	[ "# Minimum Cost Path Graph\n\nLet S be the set of vertices whose minimum distance from the source vertex has been found. The following table lists the Port Cost value for different bandwidths. The shortest path computed in the reduced cost graph is the same as the shortest path in the original graph. Steps for finding a minimum-cost spanning tree using _____ Algorithm: Add edges in order of cheapest cost so that no circuits form. The next lowest cost. I am looking to run MST on some vectors with geometries and cost. In this graph, cost of an edge (i, j) is represented by c(i, j). Your program will either return a sequence of nodes for a minimum-cost path or indicate that no solution exists. Minimum weight perfect matching problem: Given a cost c ij for all (i,j) ∈ E, ﬁnd a perfect matching of minimum cost where the cost of a matchinPg M is given by c(M) = (i,j)∈M c ij. A minimum-cost spanning tree is one which has the smallest possible total weight (where weight represents cost or distance). This problem is concerned with finding the cheapest path between verticesa and b in a graph G = (V,E). the cost-minimized pathfrom the seed point to the goal point. Dynamic Graph Clustering Using Minimum-Cut Trees 1 Introduction Graph clustering has become a central tool for the analysis of networks in gen-eral, with applications ranging from the eld of social sciences to biology and to the growing eld of complex systems. Given a weighted graph, find the maximum cost path from given source to destination that is greater than a given integer x. for example, if your answer is 1 write 1 without decimal points. Initially, this quantity is infinity (i. the total intuitionistic fuzzy cost for traveling through the shortest path. $\\begingroup$ Hint: both shortest path and min-cost flow determine the minimum of a sum. We will use Prim’s algorithm to find the minimum spanning tree. Version 05/03/2011 Minimum Spanning Trees & Shortest Path—Graph Theory ©2013 North Carolina State University Chapter 8 – Page 1 Section 8. Given a path state state of type AbstractPathState, return a vector (indexed by vertex) of the paths between the source vertex used to compute the path state and a single destination vertex, a list of destination vertices, or the entire graph. NOTE: This algorithm really does always give us the minimum-cost spanning tree. This example. The distance from a vertex v i to a vertex v j in G is the minimum cost over all paths from v i to v j in G denoted by d∗ ij. a) Suppose that each edge in the graph has a weight of zero. particular, this package provides solving tools for minimum cost spanning tree problems, minimum cost arborescence problems, shortest path tree problems and minimum cut tree problem. can u much detail abt this…its very helpful to me…. Tarjan Princeton and HP Labs Abstract Consider a bipartite graph G= (X;Y;E) with real-valued weights on its edges, and suppose that Gis balanced, with jXj= jYj. In this case, we start with single edge of graph and we add edges to it and finally we get minimum cost tree. along path p; and (2) path p has the minimum cost (toll fee) among all the paths satisfying the condition (1). Verify that your result is a maximum or minimum value using the first or second derivative test for extrema. Two vertices are adjacent when they are both incident to a common edge. Dijkstra's algorithm is a graph search algorithm that can solve the single-source shortest path problem for a graph with non- negative edge path cost, outputting a shortest path tree. A number of algorithms have been proposed to enumerate all spanning trees of an undirected graph. Figure 1: Example of a shortest path problem and its mapping to the minimum cost ow model 1. path scheduling with all activity durations assumed to be at minimum cost. A) $2100 B)$2400 C) $2900 D)$6200. Find a minimum cost spanning tree on the graph below using Kruskal's algorithm. The cost is determined depending upon the criteria to be optimized. The following will run the k-maximum spanning tree algorithm and write back results: MATCH (n:Place{id:\"D\"}) CALL algo. The Prim’s algorithm operates on two disjoint sets of edges in the graph. This paper is organized as follows: In Section 2, we provide background ma- terial on network flow algorithms and present some preliminary results. We can, therefore define function for any V,. Long-run average cost (LRAC) curve is a graph that plots average cost of a firm in the long-run when all inputs can be changed. Removes the connection between the specified origin node and the specified destination node Keep in mind that this only removes the connection in one direction, for undirected graphs, the function must be called again with the destination node as the origin. Given a vertex s in graph G, ﬁnd the shortest path from s to every other vertex in G A C B E 10 3 15 5 2 11 D 20 Closely related problem is to ﬁnd the shortest (minimum cost) path between two nodes of a graph. The well-known basic problem concerns finding the shortest paths in graphs given any set of journeys and a weighted, connected graph. We analyze the problem of finding a minimum cost path between two given vertices such that the vector sum of all edges in the path equals a given target vector m. 2) Areas less than the minimum core habitat percentage times the area of the foraging radius are eliminated 3) A cost surface is created from the habitat quality raster, cells of high quality have a low cost and vise versa 4) The remaining patches are grown outwards across the cost surface to a distance equal to the foraging radius. of Gsuch that the undirected version of T is a tree and T contains a directed path from rto any other vertex in V. hk Ruifeng Liu Chinese University of Hong Kong rﬂ[email protected] The minimum cost spanning tree (MST) Spanning tree: is a free tree that connects all the vertices in V • cost of a spanning tree = sum of the costs of the edges in the tree Minimum spanning tree property: • G = ( V, E): a connected graph with a cost function defined on the edges; U ⊆V. Part 3: Remember that we are suppose to find the point (x,y) on the graph of the parabola, y = x 2 + 1, that minimizes d. Optimal Discounted Cost in Weighted Graphs Ashutosh Trivedi = Cost(ˇ): \| Now consider the path ˇ we write DCost(v) as minimum discounted cost of all in nite. A connected graph with no circuits, called trees, are also discussed in this chapter. The idea is to start with an empty graph and try to add edges one at a time, always making sure that what is built remainsacyclic. In this way, each distant node inﬂuences the cen-ter node through a path connecting the two with minimum cost, providing a robust estimation and intact exploration of the graph structure. As A* traverses the graph, it follows a path of the lowest known cost, Keeping a sorted priority queue of alternate path segments along the way. Here is my Graph class that implements a graph and has nice a method to generate its spanning tree using Kruskal's algorithm. And here comes the definition of an AI agent. graph find a minimum cost to find the shortest path between two points. Total cost of a path to reach (m, n) is sum of all the costs on that path (including both source and destination). Specifically distance[v] stores the minimum distance so far from the source vertex s to some other vertex v. min_cost_flow (G[, demand, capacity, weight]) Return a minimum cost flow satisfying all demands in digraph G. In essence, the planner develops a list of activities on the critical path ranked with their cost slopes. The weight of a shortest path tree. The path is (0, 0) –> (0, 1) –> (1, 2) –> (2, 2). {Each node has a value b(v). For a given source vertex (node) in the graph, the algorithm finds the path with lowest cost (i. satisfaction) problems with costs. [Tree, pred] = graphminspantree(G) finds an acyclic subset of edges that connects all the nodes in the undirected graph G and for which the total weight is minimized. Eulerization of a graph is the process of finding an Euler circuit for that graph. It should return and integer that represents the minimum weight to connect all nodes in the graph provided. Initially, this quantity is infinity (i. the distance of the right path (between robot 3’s vertex and 2’s goal) be x 1. A simple graph with (a) a face-spanning subgraph of cost 11 and (b) another face-spanning subgraph of cost 13. To ﬁnd the path, the image is ﬁrst modeled as a graph. Such a route is easily obtained by a breadth first search method. It adds one more node in each iteration to the minimum cost spanning tree. Steps for finding a minimum-cost spanning tree using _____ Algorithm: Add edges in order of cheapest cost so that no circuits form. Describe and analyze an e cient algorithm for nding a minimum-cost monotone path in such a graph, G. An edge is the line segment connecting two nodes and has the same length in either direction. Prim’s Algorithm. Proof: Consider any path from sto some node t. Tarjan Princeton and HP Labs Abstract Consider a bipartite graph G= (X;Y;E) with real-valued weights on its edges, and suppose that Gis balanced, with jXj= jYj. minimum cost on the section from s to t, which makes the max-ﬂow also min-cost. We then will see how the basic approach of this algorithm can be used to solve other problems including ﬁnding maximum bottleneck paths and the minimum spanning tree (MST) problem. In the maze defined, True are the open ways. Both search methods can be used to obtain a spanning tree of the graph, though if I recall correctly, BFS can also be used in a weighted graph to generate a minimum cost spanning tree. • Route signals along minimum cost path • If congestion/overuse – assign higher cost to congested resources • Makes problem a shortest path search • Allows us to adapt costs/search to problem • Repeat until done Penn ESE 535 Spring 2015 -- DeHon 26 Key Idea • Congested paths/resources become expensive. Let G = (V; E) be an (undirected) graph with. Use Kruskal's algorithm for minimum-cost spanning trees on the graph below. Suppose you are given a connected graph G, with edge costs that are all distinct. Finding minimum cost to visit all vertices in a graph and returning back. path scheduling with all activity durations assumed to be at minimum cost. Before increasing the edge weights, shortest path from vertex 1 to 4 was through 2 and 3 but after increasing Figure 1: Counterexample for Shortest Path Tree the edge weights shortest path to 4 is from vertex 1. the total intuitionistic fuzzy cost for traveling through the shortest path. ) In this context, given an input graph G, one seeks a homomorphism f of G to H with minimum cost, i. Each cell of the matrix represents a cost to traverse through that cell. A minimum spanning tree of an undirected graph can be easily obtained using classical algorithms by Prim or Kruskal. Now, let’s ask: what’s the shortest path cost to, say, Ben and Jerry’s? * To move to weighted graphs, we appeal to the mighty power of Dijkstra. Using this answer, by finding the minimum cost closed walk (or just it's cost) of an arbitrary 4-regular planar graph, with weights 1, we can decide whether it has a Hamiltonian Path, but this problem is NP-complete. As I stand now I'm using DFS, and it's pretty slow (high number of nodes and maximum length too). For positive edge weight graphs. NOTE: This algorithm really does always give us the minimum-cost spanning tree. The cost is determined depending upon the criteria to be optimized. This is the single-source minimum-cost paths problem. (eds) Graph Based Representations in Pattern Recognition. ° Among all the spanning trees of a weighted and connected graph, the one (possibly more) with the least total weight is called a minimum spanning tree (MST). shortest path finding as a complete graph G (V, E). The multistage graph problem is finding the path with minimum cost from source s to sink t. A spanning tree of a graph is a tree that has all the vertices of the graph connected by some edges. Dijkstra solves the shortest path problem (from a specified node), while Kruskal and Prim finds a minimum-cost spanning tree. Now any positive value (>0) may exist on each edge. Suppose we have a weighted graph G = (V, E, c), where V is the set of vertices, E is the set of arcs, and. along path p; and (2) path p has the minimum cost (toll fee) among all the paths satisfying the condition (1). There can be many spanning trees. A minimum directed spanning tree (MDST) rooted at ris a directed spanning tree rooted at rof minimum cost. Next, the planner can examine activities on the critical path and reduce the scheduled duration of activities which have the lowest resulting increase in costs. replacing the edge weights with ((LCM of all edges)/(weight of the edge)) makes the longest edge as smallest and smallest edge as longest. Minimum Spanning Tree Problem We are given a undirected graph (V,E) with the node set V and the edge set E. A number of problems from graph theory are called Minimum spanning tree. A minimum directed spanning tree (MDST) rooted at ris a directed spanning tree rooted at rof minimum cost. // Indexed in order of stages E is a set of edges. Assuming that you don’t expect the paths to be more than 1000 steps long, you can choose p = 1/1000. There also can be many minimum spanning trees. Minimum Cost flow problem is a way of minimizing the cost required to deliver maximum amount of flow possible in the network. As mentioned there, grid problem reduces to smaller sub-problems once choice at the cell is made, but here move will be in reverse direction. ing tree, or a shortest path. Each unit of ﬂow on each arc corresponds to onepathgoingthroughthatnode. A circuit that uses every edge exactly once is an Euler circuit. The cost w(T) of a directed spanning tree Tis the sum of the costs of its edges, i. The program must be possible to open these files (check the format). However, when we add a 2 to this graph, we penalize longer paths, so the shortest path from a to d is a !b !d. Version 05/03/2011 Minimum Spanning Trees & Shortest Path—Graph Theory ©2013 North Carolina State University Chapter 8 – Page 1 Section 8. The shortest path problem can be defined for graphs whether undirected, directed, or mixed. Successive Shortest Path Algorithm for the Minimum Cost Flow Problem in Dynamic Graphs MathildeVernet1,MaciejDrozdowski2,YoannPigné1,EricSanlaville1 1Normandie Univ, UNIHAVRE, UNIROUEN, INSA Rouen, LITIS, 76600 Le Havre, France. Graph Magics - an ultimate software for graph theory, having many very useful things, among which a strong graph generator and more than 15 different algorithms that one may apply to graphs (ex. An Extended Path Following Algorithm for Graph-Matching Problem Zhi-Yong Liu, Hong Qiao,Senior Member, IEEE, and Lei Xu,Fellow, IEEE Abstract—The path following algorithm was proposed recently to approximately solve the matching problems on undirected graph models and exhibited a state-of-the-art performance on matching accuracy. I fear you will have to work a little and make a program that check every possible path until you find the one with minimum cost. Metanet is a toolbox of Scilab for graphs and networks computations. On a graph, transportation problems can be used to express challenging tasks involving matching supply to demand with minimal shipment expense; in discrete language, these become minimum-cost network ow problems. This contradicts the maximality of M. A path that uses every edge exactly once is an Euler path. Connected graph: a path exists between every pair of find lowest cost set of roads to repair so that all cities are connected This is a minimum spanning tree. The path to reach (m, n) must be through one of the 3 cells: (m-1, n-1) or (m-1, n) or (m, n-1). applying IG-ÿskal's algorithm for finding a minimum-cost spanning tree for a graph. It adds one more node in each iteration to the minimum cost spanning tree. Abstract: We design and analyse approximation algorithms for the minimum-cost connected T-join problem: given an undirected graph G = (V;E) with nonnegative costs on the edges, and a subset of nodes T, find (if it exists) a spanning connected subgraph H of minimum cost such that every node in T has odd degree and every node not in T has even degree; H may have multiple copies of any edge of G. Minimum weight perfect matching problem: Given a cost c ij for all (i,j) ∈ E, ﬁnd a perfect matching of minimum cost where the cost of a matchinPg M is given by c(M) = (i,j)∈M c ij. Shortest Distance Problems in Graphs Using History-Dependent Transition Costs with Application to Kinodynamic Path Planning Raghvendra V. Fig 1: This graph shows the shortest path from node \"a\" or \"1\" to node \"b\" or \"5\" using Dijkstras Algorithm. This paper involved in illustrating the best way to travel between two points and in doing so, the shortest path algorithm was created. If there is more than one minimum cost path from v to w, will Dijkstra’s algorithm always find the path with the fewest edges? If not, explain in a few sentences how to modify Dijkstra’s algorithm so that if there is more than one minimum path from v to w, a path with the fewest edges is chosen. @ inding an Euler circuit on a graph. Our results are based on a new approach to speed-up augmenting path based matching algorithms, which we describe next. Minimum Spanning Tree Problem We are given a undirected graph (V,E) with the node set V and the edge set E. Dijkstra's algorithm Like BFS for weighted graphs. This model consid-ers the phenomenon that some vehicles may choose to stop at some places to avoid tra c jams. Graph search algorithms explore a graph either for general discovery or explicit search. Similar is story for vertex C. Can you move some of the vertices or bend. 8 If the graph is directed it is possible for a tree of shortest paths from s and a minimum spanning tree in G. Each cell of the matrix represents a cost to traverse through that cell. Handout MS2: Midterm 2 Solutions 2 eb, we obtain a new spanning tree for the original graph with lower cost than T, since the ordering of edge weights is preserved when we add 1 to each edge weight. In this case, as well, we have n-1 edges when number of nodes in graph are n. Consider an undirected graph containing nodes and edges. Using this answer, by finding the minimum cost closed walk (or just it's cost) of an arbitrary 4-regular planar graph, with weights 1, we can decide whether it has a Hamiltonian Path, but this problem is NP-complete. In this Java Program first we input the number of nodes and cost matrix weights for the graph ,then we input the source vertex. The Dijkstra's algorithm gradually builds a short path tree using links in the network. This week we finish our look at pathfinding and graph search algorithms, with a focus on the Minimum Weight Spanning Tree algorithm, which calculates the paths along a connected tree structure with the smallest value (weight of the relationship such as cost, time or capacity) associated with visiting all nodes in the tree. Additionally, the graph is expected to have very few edges, so the average degree is very small. BANSAL Department of Mathematics, A. Need the graph to be connected, and minimize the cost of laying the cables. Unlike the situation where you're trying to ﬁnd a minimum cost Hamilton circuit, there is an algorithm. of biconnected graph a linear cost function on the face cycles. Steiner tree problem or so called Steiner Problem in Graphs (SPG) is a classic combinatorial optimization problem. The minimum-cost spanning tree produced by applying Kruskal's algorithm will always contain the lowest cost edge of the graph. Minimum spanning tree is a tree in a graph that spans all the vertices and total weight of a tree is minimal. We describe a simple deterministic lexicographic perturbation scheme that guarantees uniqueness of minimum-cost flows and shortest paths in G. By Ion Cozac. i have an adjacency list representation of a graph for the problem, now i am trying to implement dijkstra's algorithm to find the minimum cost paths for the 'interesting cities' as suggested by @Kolmar. cost a configuration that is a Nash Equilibrium can get in total cost to the minimum solution. In order to be able to run this solution, you will need. ) In this context, given an input graph G, one seeks a homomorphism f of G to H with minimum cost, i. e the Global Processing Via Graph Theoretic technique and comes in sem 7 exams. This paper contains two similar theorems giving con-ditions for a minimum cover and a maximum matching of a graph. If all edge lengths are equal, then the Shortest Path algorithm is equivalent to the breadth-ﬁrst search algorithm. The Minimum Weight Spanning Tree excludes the relationship with cost 6 from D to E, and the one with cost 3 from B to C. Kruskal’s algorithm is used for finding a minimum cost spanning tree. {positive b(v) is a supply {negative b(v) is a demand. Hence, the cost of path from source s to sink t is the sum of costs of each edges in this path. A minimum spanning tree of an undirected graph can be easily obtained using classical algorithms by Prim or Kruskal. We transform an dependency attack graph into a Boolean formula and assign cost metrics to attack variables in the formula, based on the severity metrics. We have to go from A to B. This is the single-source shortest paths problem. A number of algorithms have been proposed to enumerate all spanning trees of an undirected graph. Given an n-d costs array, this class can be used to find the minimum-cost path through that array from any set of points to any other set of points. To find minimum cost at cell (i,j), first find the minimum cost to the cell (i-1, j) and cell (i, j-1). Nota: Java knows the length of arrays, in. [costs] is an LxM matrix of minimum cost values for the minimal paths [paths] is an LxM cell containing the shortest path arrays [showWaitbar] (optional) a scalar logical that initializes a waitbar if nonzero. hey, I am trying to find the cost or the length of the path by the following code. The cost reduction strategy converts an existing ow into a ow of lower cost by nding negative cost cycles in the residual graph and adding ow to those cycles. The goal is to obtain an Eulerian Path that has a minimal total cost. – fsociety May 11 '15 at 7:43. , there exist a path from ) •Breadth-first search •Depth-first search •Searching a graph -Systematically follow the edges of a graph to visit the vertices of the graph. the distance of the right path (between robot 3’s vertex and 2’s goal) be x 1. Assume that $C_v = 1$ for all vertices $1 \\leq v \\leq n$ (i. particular, this package provides solving tools for minimum cost spanning tree problems, minimum cost arborescence problems, shortest path tree problems and minimum cut tree problem. There are some theorems that can be used in specific circumstances, such as Dirac's theorem, which says that a Hamiltonian circuit must exist on a graph with n vertices if each vertex has degree n /2 or greater. Finally, we open the black box in order to generalize a recent linear-time algorithm for multiple-source shortest paths in unweighted undirected planar graphs to work in arbitrary orientable surfaces. We describe a simple deterministic lexicographic perturbation scheme that guarantees uniqueness of minimum-cost flows and shortest paths in G. Continue until every vertex is on some-edge you have chosen. acyclic graphs (DAGs) and propose structured sparsity penalties over paths on a DAG (called “path coding” penalties). We propose search several fast algorithms, which allow us to define minimal time cost path and minimal cost path. 2-vertex-connected subgraphs of low cost; no previous approximation algorithm was known for either problem. hk ABSTRACT The computation of Minimum Spanning Trees (MSTs) is a funda-mental graph problem with. Minimum spanning tree Given a connected graph G = (V, E) with edge weights c e, an MST is a subset of the edges T ⊆ E such that T is a spanning tree whose sum of edge weights is minimized. Find a minimum cost spanning tree on the graph below using Kruskal's algorithm. Conclusion We have to study the minimum cost spanning tree using the Prim’s algorithm and find the minimum cost is 99 so the final path of minimum cost of spanning is {1, 6}, {6, 5}, {5, 4}, {4, 3}, {3, 2}, {2, 7}. Now I relax all edges leaving B,and set path cost of C as 3, and path cost to e as -2. In this chapter, we consider four speciﬁc network models—shortest-path prob-lems, maximum-ﬂow problems, CPM-PERT project-scheduling models, and minimum-spanning. In this paper, we address this issue for directed acyclic graphs (DAGs) and propose structured sparsity penalties over paths on a DAG (called “path coding” penalties). Algorithms in graphs include finding a path between two nodes, finding the shortest path between two nodes, determining cycles in the graph (a cycle is a non-empty path from a node to itself), finding a path that reaches all nodes (the famous \"traveling salesman problem\"), and so on. Lecture notes on bipartite matching 3 Theorem 2 A matching M is maximum if and only if there are no augmenting paths with respect to M. Repeatedly augment along a minimum -cost augmenting path. Minimum spanning tree. graph find a minimum cost to find the shortest path between two points. Part 3: Remember that we are suppose to find the point (x,y) on the graph of the parabola, y = x 2 + 1, that minimizes d. A logarithmic algorithm for the minimum path problem in Knödel graphs is an open problem despite the fact that they are bipartite and highly symmetric. MCP ¶ class skimage. In this graph, cost of an edge (i, j) is represented by c(i, j). A tax of 1 cent per mile on commercial trucks’ travel would have raised $2. Weighted Shortest Path Problem Single-source shortest-path problem: Given as input a weighted graph, G = ( V, E ), and a distinguished starting vertex, s, find the shortest weighted path from s to every other vertex in G. • Total cost: C = C(v, w, q) Minimum Total Cost is a function of input prices and output quantity. i have an adjacency list representation of a graph for the problem, now i am trying to implement dijkstra's algorithm to find the minimum cost paths for the 'interesting cities' as suggested by @Kolmar. First, we identify optimality conditions, which tell us when a given perfect matching is in fact minimum. One classical model that has resurfaced in many multi-assembly methods (e. This is not a trivial problem, because the shortest path may not be along the edge (if any) connecting two vertices, but rather may be along a path involving one or more intermediate vertices. Must Read: C Program To Implement Kruskal’s Algorithm Every vertex is labelled with pathLength and predecessor. true Suppose a veteran is planning a visit to all the war memorials in Washington, D. In this section we shall show how to find a minimum-cost spanning tree for G. The multistage graph problem is finding the path with minimum cost from source s to sink t. A typical application for minimum-cost spanning trees occurs in the design of communications networks. Suppose that each edge in the graph has a weight of zero (while non-edges have a cost of$ \\infty $). The output is either a single float (when a single vertex is provided) or a vector of floats corresponding to the vertex vector. In this paper, we implemented two graph theory methods that extend the least-cost path approach: the Conditional Minimum Transit Cost (CMTC) tool and the Multiple Shortest Paths (MSPs) tool. {Find ow which satis es supplies and demands and has minimum total cost. It can be said as an extension of maximum flow problem with an added constraint on cost(per unit flow) of flow for each edge. // Indexed in order of stages E is a set of edges. Shortest Path using Dijkstra's Algorithm is used to find Single Source shortest Paths to all vertices of graph in case the graph doesn't have negative edges. As you can probably imagine, larger graphs have more nodes and many more possibilities for subgraphs. Generic approach: A tree is an acyclic graph. Consider a connected undirected graph G with not necessarily distinct edge costs. The cost of this spanning tree is (5 + 7 + 3 + 3 + 5 + 8 + 3 + 4) = 38. In case it can, what would be the minimum cost of such path?. Kruskal's algorithm is a minimum-spanning-tree algorithm which finds an edge of the least possible weight that connects any two trees in the forest. Prim’s Algorithm. scrolling computer game in terms of nding a minimum-cost monotone path in the graph, G, that represents this game. Minimum Spanning Tree Problem Find a minimum-cost set of edges that connect all vertices of a graph at lowest total cost Applications Connecting \"nodes\" with a minimum of \"wire\" Networking Circuit design Collecting nearby nodes Clustering, taxonomy construction Approximating graphs Most graph algorithms are faster on trees. A tax of 1 cent per mile on commercial trucks’ travel would have raised$2. I want to: Make it pythonic Improve readability Improve the abstracti. A third is in the process of obtaining a subset of the overall graph, called a Spanning Tree which connects every desired node with a path, but has no paths which can start and end on the same node (such a path is called a cycle). 3: Computing the Single Source Shortest Path in a graph. of biconnected graph a linear cost function on the face cycles. It's important to be acquainted with all of these algorithms - the motivation behind them, their implementations and applications. G is usually assumed to be a weighted graph. Both of these conditions depend on the concept of an alternating path, due to Petersen . These fun activities will help students learn how to read and gather data and use tables to create. Looks at the successors of the current lowest cost vertex in the wavefront. If G is a weighted graph, then the minimum spanning tree Span(G) is the spanning tree over G with minimum weight. if there are multiple edges, keep the lowest cost one Iedge weights are g(x t;u t) Iadd additional target vertex z with an edge from each x 2X T with weight g T (x) Ia sequence of actions is a path through the unrolled graph from x 0 to z Iassociated objective is total, weighted path length 4. In this case, we start with single edge of graph and we add edges to it and finally we get minimum cost tree. To find minimum cost at cell (i,j), first find the minimum cost to the cell (i-1, j) and cell (i, j-1). Minimum spanning tree is the spanning tree where the cost is minimum among all the spanning trees. This model consid-ers the phenomenon that some vehicles may choose to stop at some places to avoid tra c jams. 32 to just. The first distinction is that Dijkstra's algorithm solves a different problem than Kruskal and Prim. Dijkstra's algorithm (also called uniform cost search) - Use a priority queue in general search/traversal. Conclusion We have to study the minimum cost spanning tree using the Prim’s algorithm and find the minimum cost is 99 so the final path of minimum cost of spanning is {1, 6}, {6, 5}, {5, 4}, {4, 3}, {3, 2}, {2, 7}. We transform an dependency attack graph into a Boolean formula and assign cost metrics to attack variables in the formula, based on the severity metrics. Given a graph, the start node, and the goal node, your program will search the graph for a minimum-cost path from the start to the goal. The cost of a path is the sum of the costs of the edges and vertices encountered on the path. Weighted Graphs Data Structures & Algorithms 3 [email protected] ©2000-2009 McQuain Dijkstra's SSAD Algorithm* We assume that there is a path from the source vertex s to every other vertex in the graph. The limitation of this type of analysis is that only a single path is identified, even though alternative paths with comparable costs might exist. It works for non-loopy mazes which was already my goal. The cost of the spanning tree is the sum of the weights of all the edges in the tree. Repeatedly augment along a minimum -cost augmenting path. In this paper, the time dependent graph is presented. MCP(costs, offsets=None, fully_connected=True)¶. A heuristic is admissible if for any node, n, in the graph, the heuristic estimate of the cost of the path from n to t is less than or equal to the true cost of that path. There are nn–2 spanning trees of K n. (2)Then I process vertex b, and it is now included in S as it's shortest path from source is determined. The goal of the proposal is to obtain an optimal path with the same cost as the path returned by Dijkstra’s algorithm, for the same origin and destination, but using a reduced graph. Maintains a cost to visit every vertex. 4 Problem 5. It is defined here for undirected graphs; for directed graphs the definition of path requires that consecutive vertices be connected by an appropriate directed edge. Dijkstra’s Algorithm successfully finds the lowest cost path for each journey. There are some theorems that can be used in specific circumstances, such as Dirac’s theorem, which says that a Hamiltonian circuit must exist on a graph with n vertices if each vertex has degree n /2 or greater. Total cost of a path to reach (m, n) is sum of all the costs on that path (including both source and destination). The problem is solved by using the Minimal Spanning Tree Algorithm. Starting from node , we select the lower weight path, i. Before increasing the edge weights, shortest path from vertex 1 to 4 was through 2 and 3 but after increasing Figure 1: Counterexample for Shortest Path Tree the edge weights shortest path to 4 is from vertex 1. Note : It is assumed that negative cost cycles do not exist in input matrix. The path is (0, 0) –> (0, 1) –> (1, 2) –> (2, 2). acyclic graphs (DAGs) and propose structured sparsity penalties over paths on a DAG (called “path coding” penalties). Determining a minimum cost path between two given nodes of this graph can take O(mlogn) time, where n = jV j and m = jEj: If this graph is huge, say n … 700000 and m. [Tree, pred] = graphminspantree(G) finds an acyclic subset of edges that connects all the nodes in the undirected graph G and for which the total weight is minimized. Given a square grid of size N, each cell of which contains integer cost which represents a cost to traverse through that cell, we need to find a path from top left cell to bottom right cell by which total cost incurred is minimum. There are some theorems that can be used in specific circumstances, such as Dirac's theorem, which says that a Hamiltonian circuit must exist on a graph with n vertices if each vertex has degree n /2 or greater. A minimum-cost spanning tree is one which has the smallest possible total weight (where weight represents cost or distance). Computing a planarorthogonaldrawingofa planar graphwith the minimum number of bends over all possible embeddings is in general NP-hard [17,18]. A typical application of this problem involves finding the best delivery route from a factory to a warehouse where the road network has some capacity and cost associated. min_paths(+Vertex, +WeightedGraph, -Tree) Tree is a tree of all the minimum-cost paths from Vertex to every other vertex in WeightedGraph. , there exist a path from ) •Breadth-first search •Depth-first search •Searching a graph -Systematically follow the edges of a graph to visit the vertices of the graph. Minimum Cost Flow Problem • Objective: determine the least cost movement of a commodity through a network in order to satisfy demands at certain nodes from available supplies at other nodes. Kruskal's algorithm is a minimum-spanning-tree algorithm which finds an edge of the least possible weight that connects any two trees in the forest. 1 Shortest path problem The shortest path problem is one of the simplest of all network ow problems. In kruskal’s algorithm, edges are added to the spanning tree in increasing order of cost. Minimum spanning tree. G is usually assumed to be a weighted graph. ] Each set V i is called a stage in the graph. A spanning tree in a given graph is a tree built using all the vertices of the graph and just enough of its edges to obtain a tree. Assuming that you don’t expect the paths to be more than 1000 steps long, you can choose p = 1/1000. Therefore, we can use the same reduction to also compute a minimum-cost maximum cardinality matching in O~(mn2=5) time. Operations Research Methods 8. • The edges for shortest path are , , and is the minimum cost of path 1 to 4 is \"4\", cost of path 1 to 2 is \"6\", and cost of path 4 to 3 is \"5\" of the given graph 6 + 4 + 5 = 15. This problem is similar to Finding possible paths in grid. 412 G orke et al. These algorithms carve paths through the graph, but there is no expectation that those paths are computationally optimal. This contradicts the maximality of M. We can, therefore define function for any V,. Investigate ideas such as planar graphs, complete graphs, minimum-cost spanning trees, and Euler and Hamiltonian paths. The cost of a path from s to t is the sum of costs of the edges on the path. i have an adjacency list representation of a graph for the problem, now i am trying to implement dijkstra's algorithm to find the minimum cost paths for the 'interesting cities' as suggested by @Kolmar. An expansion path provides a long-run view of a firm’s production decision and can be used to create its long-run cost curves. It is expanded, yielding nodes B, C, D. In any graph G, the shortest path from a source vertex to a destination vertex can be calculated using Dijkstra Algorithm. Must Read: C Program To Implement Kruskal’s Algorithm Every vertex is labelled with pathLength and predecessor. the original graph. As it turns out, the minimum cost flow problem is equivalent to minimum cost circulation problem and transshipment problem in the sense that they can be reduce to each other while blowing up the input size by a constant factor. Describe and analyze an e cient algorithm for nding a minimum-cost monotone path in such a graph, G. This problem is also called the assignment problem. 2) Areas less than the minimum core habitat percentage times the area of the foraging radius are eliminated 3) A cost surface is created from the habitat quality raster, cells of high quality have a low cost and vise versa 4) The remaining patches are grown outwards across the cost surface to a distance equal to the foraging radius. Shortest Path, Network Flows, Minimum Cut, Maximum Clique, Chinese Postman Problem, Graph Center, Graph Median etc. Negative Edge Costs Single-Source Shortest-Path Problem Problem Given as input a weighted graph, G = (V,E), and a distinguished vertex, s, ﬁnd the shortest weighted path from s to every other vertex in G. cost(e), but cannot be shared by more than cap(e) paths even if we pay the cost of e. Can you move some of the vertices or bend. Given a graph, the start node, and the goal node, your program will search the graph for a minimum-cost path from the start to the goal. An Extended Path Following Algorithm for Graph-Matching Problem Zhi-Yong Liu, Hong Qiao,Senior Member, IEEE, and Lei Xu,Fellow, IEEE Abstract—The path following algorithm was proposed recently to approximately solve the matching problems on undirected graph models and exhibited a state-of-the-art performance on matching accuracy. For example, consider below graph. Abstract: Let G be an edge-weighted directed graph with n vertices embedded on a surface of genus g." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9200784,"math_prob":0.9523868,"size":39623,"snap":"2019-51-2020-05","text_gpt3_token_len":8967,"char_repetition_ratio":0.19412403,"word_repetition_ratio":0.29242972,"special_character_ratio":0.21883754,"punctuation_ratio":0.09382621,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.99689233,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-22T00:16:45Z\",\"WARC-Record-ID\":\"<urn:uuid:dcd21aca-19bb-4a26-a8b4-f97cb63a0e1f>\",\"Content-Length\":\"51962\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b9918645-81bc-4cc2-a385-2cabdb25248d>\",\"WARC-Concurrent-To\":\"<urn:uuid:508eda7a-e4da-4fe4-9303-91097b7e2283>\",\"WARC-IP-Address\":\"104.24.124.119\",\"WARC-Target-URI\":\"http://ppwu.auto-hirch-egelsbach.de/minimum-cost-path-graph.html\",\"WARC-Payload-Digest\":\"sha1:PV63QIZFYU467Z3DRNL2SPFS5TSG3EPY\",\"WARC-Block-Digest\":\"sha1:A4NTVXTL2A5MZRQU4OOQTTBISV43NNML\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250606226.29_warc_CC-MAIN-20200121222429-20200122011429-00266.warc.gz\"}"}
https://aerospace.technion.ac.il/seminars/finding-the-shortest-3d-paths-with-curvature-constraint-in-close-range-scenarios/	[ "", null, "", null, "# Finding the Shortest 3D paths with Curvature Constraint in Close Range Scenarios\n\nRafi Kfir\nWork towards MSc degree under the supervision of Prof. Yossi Ben Asher (AE, Technion) and Dr. George Hexner (Rafael)\nDepartment of Aerospace Engineering\nTechnion – Israel Institute of Technology\n\nThe problem of finding optimal trajectories is a basic problem in aerospace engineering with many different solution algorithms. One widely investigated branch is the problem of finding shortest paths under maximum curvature constraint. The shortest planar trajectories under curvature constraint are the well-known Dubins paths, which can either be arc-line-arc or arc-arc-arc combinations. In recent years few works dealt with 3D shortest paths under the curvature constraint. Theoretical works prove the existence of 3 types of spatial shortest trajectories: 3D arc-line-arc, 3D arc-arc-arc and helicoidal-arc. Some further works proposed algorithms for finding the first one (3D arc-line-arc trajectories), which is the shortest path when the start and the end point are far from each other.\n\nThis work first proposes an algorithm to calculate 3D arc-arc-arc type trajectories, (the solution for 3D arc-line-arc is already known), followed by an investigation of the helicoidal-arc type trajectory. In the case of helicoidal-arc the exact solution depends on a high order TPBVP (Two Point Boundary Value Problem), a difficult numerical problem. In this study we propose to approximate the helicoidal-arc with polynomials. The approximate polynomial solutions are compared to the optimal paths obtained from numerical integration and validated by GPOPS (General Purpose OPtimal Control Software). It is shown that the solutions based on the polynomials lead to good approximation of the optimal trajectories.\n\nThe talk will be given in English.\n\nWed, 02-01-2019, 16:30 (Gathering at 16:00)\n\nClassroom 165, ground floor, Library, Aerospace Eng.\n\nLight refreshments will be served before the lecture" ]	[ null, "https://aerospace.technion.ac.il/wp-content/uploads/2014/07/cropped-header-print-eng.jpg", null, "https://aerospace.technion.ac.il/wp-content/uploads/2014/07/cropped-header-print-heb.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8755073,"math_prob":0.83845586,"size":1970,"snap":"2022-27-2022-33","text_gpt3_token_len":423,"char_repetition_ratio":0.11495422,"word_repetition_ratio":0.0,"special_character_ratio":0.1923858,"punctuation_ratio":0.0882353,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9742864,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-28T15:52:41Z\",\"WARC-Record-ID\":\"<urn:uuid:fbb9c741-c0a0-49f0-8e32-6274fcc783c4>\",\"Content-Length\":\"63712\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5492545e-7551-4cda-b77b-be795bb5cf74>\",\"WARC-Concurrent-To\":\"<urn:uuid:65fbf9ba-9acb-4ffe-a39b-8692fd5aefe1>\",\"WARC-IP-Address\":\"132.68.239.54\",\"WARC-Target-URI\":\"https://aerospace.technion.ac.il/seminars/finding-the-shortest-3d-paths-with-curvature-constraint-in-close-range-scenarios/\",\"WARC-Payload-Digest\":\"sha1:44KI2ZPRBV4TYGUGN6KOQCQQ4MXJCOYE\",\"WARC-Block-Digest\":\"sha1:654JESEOVDDYPA2J62ZTKQBKQLVEDTZ7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103556871.29_warc_CC-MAIN-20220628142305-20220628172305-00734.warc.gz\"}"}
https://www.r-bloggers.com/2017/01/understanding-mixture-models-and-expectation-maximization-using-baseball-statistics/	[ "Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.\n\nPreviously in this series:\n\nIn this series on empirical Bayesian methods on baseball data, we’ve been treating our overall distribution of batting averages as a beta distribution, which is a simple distribution between 0 and 1 that has a single peak. But what if that weren’t a good fit? For example, what if we had a multimodal distribution, with multiple peaks?\n\nIn this post, we’re going to consider what to do when your binomial proportions are made up of multiple peaks, and when you don’t know which observation belongs to which clusters. For example, so far in our analysis we’ve been filtering out pitchers, who tend to have a much lower batting average than non-pitchers. If you include them, the data looks something like this:", null, "These batting averages certainly don’t look like a single beta distribution- it’s more like two separate ones mixed together. Imagine that you didn’t know which players were pitchers, and you wanted to separate the data into two groups according to your best prediction. This is very common in practical machine learning applications, such as clustering and segmentation.\n\nIn this post we’ll examine mixture models, where we treat the distribution of batting averages as a mixture of two beta-binomial distributions, and need to guess which player belongs to which group. This will also introduce the concept of an expectation-maximization algorithm, which is important in both Bayesian and frequentist statistics. We’ll show how to calculate a posterior probability for the cluster each player belongs to, and see that mixture models are still a good fit for the empirical Bayes framework.\n\n### Setup\n\nAs usual, I’ll start with some code you can use to catch up if you want to follow along in R. If you want to understand what it does in more depth, check out the previous posts in this series. (As always, all the code in this post can be found here).1\n\nWe’ve been filtering out pitchers in the previous posts, which make batting averages look roughly like a beta distribution. But when we leave them in, as I showed above, the data looks a lot less like a beta:", null, "The dashed density curve represents the beta distribution we would naively fit to this data. We can see that unlike our earlier analysis, where we’d filtered out pitchers, the beta is not a good fit- but that it’s plausible that we could fit the data using two beta distributions, one for pitchers and one for non-pitchers.\n\nIn this example, we know which players are pitchers and which aren’t. But if we didn’t, we would need to assign each player to a distribution, or “cluster”, before performing shrinkage on it. In a real analysis it’s not realistic that we wouldn’t know which players are pitchers, but it’s an excellent illustrative example of a mixture model and of expectation-maximization algorithms.\n\n### Expectation-maximization\n\nThe challenge of mixture models is that at the start, we don’t know which observations belong to which cluster, nor what the parameters of each distribution is. It’s difficult to solve these problems at the same time- so an expectation-maximization (EM) algorithm takes the jump of estimating them one at a time, and alternating between them.\n\nThe first thing to do in an EM clustering algorithm is to assign our clusters randomly:\n\n#### Maximization\n\nNow that we’ve got cluster assignments, what do the densities of each cluster look like?", null, "Well, that doesn’t look like much of a division- they have basically the same density! That’s OK: one of the nice features of expectation-maximization is that we don’t actually have to start with good clusters to end up with a good result.\n\nWe’ll now write a function for fitting a beta-binomial distribution using maximum likelihood estimation (and the `dbetabinom.ab` function from the VGAM package). This is a process we’ve done in multiple posts before, including the appendix of one of the first ones. We’re just encapsulating it into a function.\n\n(The `number` column I added will be useful in the next step). For example, here are the alpha and beta chosen for the entire data as a whole:\n\nBut now we’re working with a mixture model. This time, we’re going to fit the model within each of our (randomly assigned) clusters:\n\nAnother component of this model is the prior probability that a player is in cluster A or cluster B, which we set to 50-50 when we were assigning random clusters. We can estimate our new iteration of this based on the total number of assignments in each group, which is why we included the `number` column:\n\nMuch as the within-cluster densities only changed a little, the priors only changed a little as well. This was the maximization step: find the maximum likelihood parameters (in this case, two alpha/beta values, and a per-cluster probability), pretending we knew the assignments.\n\n### Expectation\n\nWe now have a distribution for each cluster. It’s worth noting that these are pretty similar distributions, and that neither is a good fit to the data.", null, "However, notice that due to a small random difference, cluster B is slightly more likely than cluster A for batting averages above about .2, and vice versa below .2.\n\nConsider therefore that each player has a likelihood it would have been generated from cluster A, and a likelihood it would have been generated from cluster B (being sure to weight each by the prior probability of being in A or B):\n\nFor example, consider Jeff Abbott, who got 11 hits out of 42 at-bats. He had a 4.35% chance of getting that if he were in cluster A, but a 4.76% chance if he were in cluster B. For that reason (even though it’s a small difference), we’ll put him in B. Similarly we’ll put Kyle Abbott in cluster A: 3/31 was more likely to come from that distribution.\n\nWe can do that for every player using `group_by` and `top_n`:\n\nThat’s the expectation step: assigning each person to the most likely cluster. How do our assignments look after that?", null, "Something really important happened here: even though the two beta models we’d fit were very similar, we still split up the data rather neatly. Generally batters with a higher average ended up in cluster B, while batters with a lower average were in cluster A. (Note that due to B having a slightly higher prior probability, it was possible for players with a low average- but also a low AB- to be assigned to cluster B).\n\n### Expectation-Maximization\n\nThe above two steps got to a better set of assignments than our original, random ones. But there’s no reason to believe these are as good as we can get. So we repeat the two steps, choosing new parameters for each distribution in the mixture and then making new assignments each time.\n\nFor example, now that we’ve reassigned each player’s cluster, we could re-fit the beta-binomial with the new assignments. Those distributions would look like this:", null, "Unlike our first model fit, we can see that cluster A and cluster B have diverged a lot. Now we can take those parameters and perform a new estimation step. Generally we will do this multiple times, as an iterative process. This is the heart of an expectation-maximization algorithm, where we switch between assigning clusters (expectation) and fitting the model from those clusters (maximization).\n\nHere I used the `accumulate` function from the `purrr` package, which is useful for running data through the same function repeatedly and keeping intermediate states. I haven’t seen others use this tidy approach to EM algorithms, and there are existing R approaches to mixture models.2 But I like this approach both because it’s transparent about what we’re doing in each iteration, and because our iterations are now combined in a tidy format, which is convenient to summarize and visualize.\n\nFor example, how did our assignments change over the course of the iteration?", null, "We notice that only the first few iterations led to a shift in the assignments, after which it appears to converge. Similarly, how did the estimated beta distributions change over these iterations?", null, "This confirms that it took about three iterations to converge, and then stayed about the same after that. Also notice that in the process, cluster B got much more likely than cluster A, which makes sense since there are more non-pitchers than pitchers in the dataset.\n\n### Assigning players to clusters\n\nWe now have some final parameters for each cluster:\n\nHow would we assign players to clusters, and get a posterior probability that the player belongs to that cluster? Well, let’s arbitrarily pick the six players that each batted exactly 100 times:\n\nWhere would we classify each of them? Well, we’d consider the likelihood each would get the number of hits they did if they were a pitcher (cluster A) or a non-pitcher (cluster B):", null, "By Bayes’ Theorem, we can simply use the ratio of one likelihood (say, A in red) to the sum of the two likelihoods to get the posterior probability:", null, "Based on this, we feel confident that Juan Nicasio and Jose de Jesus are pitchers, and that the others probably aren’t. And we’d be right! (Check out the `isPitcher` column in the `batter_100` table above).\n\nThis allows us to assign all players in the dataset to one of the two clusters.\n\nSince we know whether each player actually is a pitcher or not, we can also get a confusion matrix. How many pitchers were accidentally assigned to cluster B, and how many non-pitchers were assigned to cluster A? In this case we’ll look only at the ones for which we had at least 80% confidence in our classification.\n\nNot bad, considering the only information we used was the batting average- and note that we didn’t even use data on who were pitchers to train the model, but just let the clusters define themselves.\n\nIt looks like we were a lot more likely to call a pitcher a non-pitcher than vice versa. There’s a lot more we could do to examine this model, how well calibrated its posterior estimates are, and what kinds of pitchers may be mistaken for non-pitchers (e.g. good batters who pitched only a few times), but we won’t consider them in this post.\n\n### Empirical bayes shrinkage with a mixture model\n\nWe’ve gone to all this work posterior probabilities of each player’s assignments. How can we use this in empirical Bayes shrinkage, or with the other methods we’ve described in this series?\n\nWell, consider that all of our other methods have worked because the posterior was another beta distribution (thanks to the beta being the conjugate prior of the binomial). However, now that each point might belong to one of two beta distributions, our posterior will be a mixture of betas. This mixture is made up of the posterior from each cluster, weighted by the probability the point belongs to that cluster.\n\nFor example, consider the six players who had exactly 100 at-bats. Their posterior distributions would look like this:", null, "For example, we are pretty sure that Jose de Jesus and Juan Nicasio are part of the “pitcher” cluster, so that makes up most of their posterior mass, and all of Ryan Shealy’s density is in the “non-pitcher” cluster. However, we’re pretty split on Mike Mahoney- he could be a pitcher who is unusually good at batting, or a non-pitcher who is unusually bad.\n\nCan we perform shrinkage like we did in that early post? If our goal is still to find the mean of each posterior, then yes! Thanks to linearity of expected value, we can simply average the two distribution means, weighing each by the probability the player belongs to that cluster:\n\nFor example, we are pretty sure that Jose de Jesus and Juan Nicasio are part of the “pitcher” cluster, which means they mostly get shrunken towards that center. We are quite certain Ryan Shealy is not a pitcher, so he’ll be updated based entirely on that distribution.", null, "Notice that instead of shrinking towards a single value, the batting averages are now shrunken towards two centers: one higher value for the non-pitcher cluster, one smaller value for the pitcher cluster. Ones that are exactly in between don’t really get shrunken in either direction- they’re “pulled equally”.\n\n(For simplicity’s sake I didn’t use our beta-binomial regression approach in this model, but that could easily be added to take into account the relationship between average and AB).\n\nNot all of the methods we’ve used in this series are so easy to adapt to a multimodal distribution. For example, a credible interval is ambiguous in a multimodal distribution (see here for more), and we’d need to rethink our approach to Bayesian A/B testing. But since we do have a posterior distribution for each player- even though it’s not a beta- we’d be able to face these challenges.\n\n### What’s Next: Combining into an R package\n\nWe’ve introduced a number of statistical techniques in this series for dealing with empirical Bayes, the beta-binomial relationship, A/B testing, etc. Since I’ve provided the code at each stage, you could certainly apply these methods to your own data (as some already have!). However, you’d probably find yourself copying and pasting a rather large amount of code, which isn’t necessarily something you want to do every time you run into this kind of binomial data (it takes you out of the flow of your own data analysis).\n\nIn my next post I’ll introduce an R package for performing empirical Bayes on binomial data that encapsulates many of the analyses we’ve performed in this series. These statistical techniques aren’t original to me (most can be found in an elementary Bayesian statistics textbook), but providing them in a convenient R package can still be useful for the community. We’ll also go over some of the choices one makes in developing a statistics package in R, particularly one that is compatible with the tidy tools we’ve been using.\n\n1. I’m changing this analysis slightly to look only at National League batters since the year 1980. Why? Because National League pitchers are required to bat (while American League pitchers don’t in typical games), and because looking at modern batters helps reduce the noise within each group.\n\n2. I should note that I haven’t yet gotten an existing R mixture model package to work with a beta-binomial model like we do in this post If you have an approach you’d recommend, please share it in the comments or on Twitter!" ]	[ null, "https://i2.wp.com/varianceexplained.org/figs/2017-01-03-mixture-models-baseball/batting_w_pitchers_plot-1.png", null, "https://i2.wp.com/varianceexplained.org/figs/2017-01-03-mixture-models-baseball/batting_w_pitchers_plot-1.png", null, "https://i0.wp.com/varianceexplained.org/figs/2017-01-03-mixture-models-baseball/starting_data_graph-1.png", null, "https://i1.wp.com/varianceexplained.org/figs/2017-01-03-mixture-models-baseball/unnamed-chunk-5-1.png", null, "https://i0.wp.com/varianceexplained.org/figs/2017-01-03-mixture-models-baseball/assignments_plot-1.png", null, "https://i0.wp.com/varianceexplained.org/figs/2017-01-03-mixture-models-baseball/unnamed-chunk-7-1.png", null, "https://i1.wp.com/varianceexplained.org/figs/2017-01-03-mixture-models-baseball/assignment_iterations-1.png", null, "https://i0.wp.com/varianceexplained.org/figs/2017-01-03-mixture-models-baseball/unnamed-chunk-9-1.png", null, "https://i0.wp.com/varianceexplained.org/figs/2017-01-03-mixture-models-baseball/unnamed-chunk-12-1.png", null, "https://i1.wp.com/varianceexplained.org/figs/2017-01-03-mixture-models-baseball/posterior_probability_graph-1.png", null, "https://i1.wp.com/varianceexplained.org/figs/2017-01-03-mixture-models-baseball/posterior_mixture_plot-1.png", null, "https://i1.wp.com/varianceexplained.org/figs/2017-01-03-mixture-models-baseball/eb_shrinkage_plot-1.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.946001,"math_prob":0.91689336,"size":14881,"snap":"2023-40-2023-50","text_gpt3_token_len":3120,"char_repetition_ratio":0.13430128,"word_repetition_ratio":0.03183446,"special_character_ratio":0.20294335,"punctuation_ratio":0.08958838,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95871,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24],"im_url_duplicate_count":[null,6,null,6,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-23T13:55:36Z\",\"WARC-Record-ID\":\"<urn:uuid:1fdfab94-b861-4fb7-95bf-17f307c2136e>\",\"Content-Length\":\"282032\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cabc8a5f-7172-4f22-b96a-a11dd0da5af2>\",\"WARC-Concurrent-To\":\"<urn:uuid:51077e9c-46bc-4984-9341-17772dacaf48>\",\"WARC-IP-Address\":\"104.21.77.221\",\"WARC-Target-URI\":\"https://www.r-bloggers.com/2017/01/understanding-mixture-models-and-expectation-maximization-using-baseball-statistics/\",\"WARC-Payload-Digest\":\"sha1:MKTXWEHO2TRXTLWFCQ6PKWP5PKZKSVLY\",\"WARC-Block-Digest\":\"sha1:CQF4QUGBMJWWCAIMDE3QM2I6W4SNHEAG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506481.17_warc_CC-MAIN-20230923130827-20230923160827-00345.warc.gz\"}"}
https://butterflyofdream.wordpress.com/2012/06/09/basic-design-of-submarine-periscope/	[ "## Basic design of submarine periscope\n\n Source http://www.maritime.org/fleetsub/pscope/chap1.htm 1A8. Limits of periscope design. It is seen from the preceding section that there are definite limits in periscope design. The vital factors, as in a telescope, are: 1) length of tube, 2) diameter, 3) illumination, 4)magnification, and 5) size of field. If a periscope favoring any one of these factors is to be produced, such favoring can be only at the expense of the other factors; hence, the final design generally is a compromise. 1A9. Examples of periscope design. The following requirements are for periscopes which have been used in submarines: field, at least 40 degrees to 45 degrees;magnification, between 1.2x and 1.5x; exit pupil, at least 5 millimeters in diameter;length, not specified; external diameter, 5 inches; thickness of walls, about 1/4 inch. Let us find possible periscope lengths under these conditions for the two magnifications given, 1.2x and 1.5x. The inside diameter of the tube is 5 inches minus 1/2 inch, or 4 1/2 inches. The lens, lens-holding ring, supporting tube, and so forth take up another 1/2 inch of diameter, leaving about 4 inches free for the objective. 4 inches = 101.6 mm, which is close to 100 mm In order to obtain an exit pupil of 5 millimeters, the magnification of the telescope must be: Diameter of objective / Diameter of exit pupil = 100 / 5 = 20x 3", null, "Figure 1-1. Section through submarine with periscope elevated. 4 If the magnification of the final periscope is to be 1.2x, the reduction of the upper telescope must be: 20 / 1.2 = 16.67, or 16.67x Since the field must be 40 degrees / 16.67, or 2.4 degrees = 2 degrees 24′, this limits the length between the objectives of the two telescopes, since the entire beam of light must fall on the lower objective. From Figure 1-3, it can be seen that the permissible length equals 2 / tan θ, where 2 is half the diameter of the lower objective lens in inches and θ is half the angle of beam. θ equals 2 degrees 24′ / 2, or 1 degrees 12′. log 2 = 10.30103 – 10 log tan 1 degree 12′ = (8.32112 / 1.97991) – 10 antilog 1.97991 = 95.58 inches = 7 feet 11 1/2 inches The upper and lower telescope systems enter into the total length, and if it were possible to increase the focal length of their objective lenses", null, "Figure 1-2. Detail of encircled section in Figure 1-1. 5 indefinitely, the periscope could be lengthened. Increasing this is limited, however, by the same considerations of diameter and cannot exceed the same length; that is, about 7 feet 11 1/2 inches for each telescope system. Hence, the total possible length is roughly 3 times 7 feet 11 1/2 inches, or about 23 feet 10 1/2 inches. Since this length is greater than is required, the diameter of the periscope may be reduced, the magnification increased, or the size of the exit pupil increased without sacrifice.If the magnification is to be 1.5x, the reduction of the upper telescope must be: 20 / 1.5 = 13 1/3x For a field of 40 degrees, the angle of beam is: 40 / 13 1/3 = 30 degrees The inter-objective distance is: log 2 = 10.30103 – 10 log tan 1 degrees 30′ = (8.41807 / 1.88296) – 10 antilog 1.88296 = 76.37 inches = 6 feet 4.4 inches The total length possible is 3 times 6 feet 4.4 inches, or 19 feet 1.2 inches.To increase the length of tube beyond these limits, more telescopes may be placed in the tube. If astronomical telescopes are used, two more must be employed to keep the image erect, making a total of four telescope systems. One Galilean telescope could be used. The objection to adding more telescopes lies in the fact that each lens through which the beam must pass absorbs light, and if more are added, the illumination is seriously reduced. Figure 1-4 shows a periscope designed as a straight instrument, and Figure 1-5 shows it with prisms introduced. The prisms may be placed at any point where the angle of the rays does not exceed the critical angle which results in total reflection. In this particular case, the prisms are placed at the focal planes. Both periscopes produce an erect image, since the two astronomical telescopes and the two prisms counteract each other in inverting the object. Prisms should not be placed exactly in a focal plane. Doing so is faulty design, since any minute imperfections", null, "Figure 1-3. Example of periscope design.", null, "Figure 1-4. Example of periscope design.", null, "Figure 1-5. Example of periscope design. 6 that may be present in or on the reflecting surface are reproduced as part of the final image, whereas a lens or glass plate which is not in a focal plane, or near one, may be dirty without affecting the resulting image. Periscope specifications often state that no lens or glass plate should be in or near a focal plane except the crosswire reticle, which must of necessity be placed in a focal plane.Since the backs of the prisms, which are the reflecting surfaces, are silvered, the critical angle for reflection is raised to more than 20 degrees; thus the two eyepieces may be placed between the prisms and the objectives. Both forms of construction are used in various periscopes. However, the best position for a prism is at a point at which the rays are approximately parallel; in erecting telescopes, this point lies between the two erecting lenses. The chief function of a telescope system in a periscope is to take an object appearing from the point of vision under narrow angular view, and produce it to the eye at a wide angle. The ratio of these two angles is the magnification of the telescope. 1A10. Altiscopes. The only difference between a periscope and an altiscope is that in an altiscope the upper prism is omitted and the view is directly upward toward the zenith. The field of an altiscope is 100 degrees. To obtain this field, some sacrifice must be made in other characteristics. The magnification is necessarily less than unity. The only type of periscope used in the Navy today which permits observation of the zenith is the Type II design (Design Designations 89KA40T/1.414HA, 91KA40T/1.414HA, and 92KA40T/1.4HA built by the Kollmorgen Optical Corp., Brooklyn, N.Y., which is of the high-angle type. The prism has a maximum elevation of the line of sight above horizontal of 74.5 degrees. The entire sky is observed with the line of sight set respectively at 14 degrees, 44 degrees, and 74.5 degrees or full elevation, giving complete zenith at the edge of the field in low power. The periscope is rotated 360 degrees in each zone with a minimum of overlap between the zones.1A11. Types of periscopes. Periscopes under Bureau of Ships Specifications R20 P5 of 15 June 1940, are of the following types: 1. Type I. Outer diameter of taper section, 1.414 inches. The line of sight can be moved through all angles between 10 degrees depression and 45 degrees elevation. 2. Type II. Outer diameter of taper section, 1.414 inches. The line of sight can be moved through all angles between 10 degrees depression and 74 degrees elevation. 3. Type III. Outer diameter of taper section, 1.99 inches. The line of sight can be moved through all angles between 10 degrees depression and 45 degrees elevation. 4. Type IV. Outer diameter of taper section, 3.750 inches. The line of sight can be moved through all angles between 10 degrees depression and 45 degrees elevation. The periscope is designed for night use with an installed antenna array and waveguide for the attachment of an electronic range device.", null, "" ]	[ null, "https://i1.wp.com/www.maritime.org/fleetsub/pscope/img/fig1-01.jpg", null, "https://i2.wp.com/www.maritime.org/fleetsub/pscope/img/fig1-02.jpg", null, "https://i2.wp.com/www.maritime.org/fleetsub/pscope/img/fig1-03.jpg", null, "https://i1.wp.com/www.maritime.org/fleetsub/pscope/img/fig1-04.jpg", null, "https://i0.wp.com/www.maritime.org/fleetsub/pscope/img/fig1-05.jpg", null, "https://1.gravatar.com/avatar/d692216fbe77315e2e3ab6a0da35fa77", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9046357,"math_prob":0.9644721,"size":7345,"snap":"2021-43-2021-49","text_gpt3_token_len":1864,"char_repetition_ratio":0.15038823,"word_repetition_ratio":0.07456829,"special_character_ratio":0.2511913,"punctuation_ratio":0.13495721,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97089857,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,2,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-28T10:27:33Z\",\"WARC-Record-ID\":\"<urn:uuid:8d58e5c4-8abb-4588-aa87-925d7a86a17b>\",\"Content-Length\":\"104895\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:75623f74-1e36-4c83-8861-d3d4cf067448>\",\"WARC-Concurrent-To\":\"<urn:uuid:a6972cd4-653e-4b44-86a2-6c4c06e028c6>\",\"WARC-IP-Address\":\"192.0.78.12\",\"WARC-Target-URI\":\"https://butterflyofdream.wordpress.com/2012/06/09/basic-design-of-submarine-periscope/\",\"WARC-Payload-Digest\":\"sha1:RI565X4POEGKQ4GOHBLPEUQO4DBQ7SOM\",\"WARC-Block-Digest\":\"sha1:YDJVVNADEUZ22RDNLL5DPD72TLHFXYCI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323588284.71_warc_CC-MAIN-20211028100619-20211028130619-00429.warc.gz\"}"}
https://www.powersystemsdesign.com/articles/using-lord-kelvins-sensing-method/29/8590	[ "# Using Lord Kelvin’s sensing method\n\nAuthor:\nSrudeep Patil, Maxim Integrated\n\nDate\n02/14/2015", null, "PDF\n##### The Kelvin Method lives on in the in the latest ultra-precise current-shunt monitors & current-sense amplifiers\n\nUltra-precision, high-side current sensing is crucial in applications where one must measure the current entering or leaving a battery. Today many digital multimeters feature four-lead Kelvin sensing to eliminate the series resistance of the multimeter leads and give an accurate voltage drop across a given resistor.\n\nSimilarly, a current-shunt monitor (CSM) or current-sense amplifier (CSA) measures voltage drop across a shunt resistor based on the current flown into or out of the battery. This is how you determine the amount of current drawn by the load from a battery in real-world applications. Systems today use low power and require a very accurate measure of charge left in the battery. To quantify remaining charge, every µA drawn from the battery by the load or pumped into the battery by a charger needs to be accounted for. Thus, ultra-precise sensing of voltage drop across the shunt is critical.\n\nLet’s look at measuring voltage drop across a shunt resistor with very high precision. An ultra-high-precision CSM measures the voltage drop across the shunt resistor with typical connections, and then compares this value with the CSM accuracy specification in its datasheet. There are ways to improve the measurement accuracy using that same CSM.\n\nThese measurements are enhanced utilizing the proven Kelvin sensing methods with a four-terminal sense resistor. Test results also show that one should be careful with the board layout. Once the layout practices listed in this article are followed, we can capitalize on Kelvin sensing and sense the microvolt level drop across the sense resistor with ultra-high precision.\n\nThe Kelvin Bridge\n\nBefore we talk about ultra-precision CSMs/CSAs, let’s start with a look back in time to an impressive scientist and Avant-garde engineer, Lord Kelvin (see Figure 1). Lord Kelvin’s pioneering effort is the basis of many electronic principles that we take for granted in our daily life, such as knowing when our cell phone needs charging. Kelvin’s work in measuring very low resistances is still used in modern integrated circuits (ICs). In fact, when you accurately measure battery capacity by applying early Kelvin principles and additional math, preventing overcharging or discharging extends battery life.\n\nClcik image to enlarge\n\nFigure 1. A portrait of Lord Kelvin (1824-1907) by Sir Hubert von Herkomer (1849-1914).\n\nEarly instruments such as the Kelvin bridge (see Figure 2) are amazingly accurate by today’s standards. Note the block diagram in the center of Figure 2. On the left is a battery and below are four leads. The outer leads supply current through resistor X, while the inner leads isolate the measuring circuit. It is easier to see the Kelvin sensing principle in Figure 3.\n\nClcik image to enlarge\n\nFigure 2. An early Kelvin bridge for making accurate resistance measurements of very low-ohm resistors.\n\nClick image to enlarge\n\nFigure 3. A block diagram showing the Kelvin sensing method.\n\nBy separating the main current path from the measurement path, Kelvin improves measurement accuracy. In Figure 3, the main current to be measured flows from the battery on the top left through the ammeter (A), and the voltage drops across resistor “X” (the gray bar at the bottom) between leads 2 and 3. Virtually no current flows through the voltmeter circuit (V and leads 2 and 3) due to very high-input impedance and, hence, the voltmeter makes a highly accurate measurement. The current is the same at all parts of the main circuit comprised of the ammeter, battery resistor, and leads 1 through 4. However, the leads 1 and 4 contribute series resistance that, in turn, contributes to a finite amount of voltage drop along the leads. Although very small voltage drops, nonetheless, they reduce accuracy.\n\nBy separating the main current path from the measurement path, Kelvin’s sensing improves the accuracy of measurements. Of course, knowing any two of the three parameters—voltage, current, and resistance—we can calculate the third parameter.\n\nTypical connections on a current-shunt monitor\n\nFigure 4 shows connections around a CSM used for monitoring current from the battery into the load. We might think at the outset that there is nothing wrong with Figure 4. However, this design will not yield the ±0.23% gain error specified for this CSA. The design problems actually result from shortcomings in the board layout and poor schematic placement.\n\nClick image to enlarge\n\nFigure 4. Typical connections to measure drop across a sense resistor. The example device serving as the shunt monitor is the MAX44286 CSA.\n\nWhen we examine the schematic placement in Figure 4 and make some adjustments, we can preserve the shunt monitor’s DC accuracy parameters like gain accuracy and input-offset voltage. The example shunt monitor shown here is the MAX44286 available in a 4-bump wafer-level package (WLP) with 0.78mm x 0.78mm x 0.35mm dimensions. These findings and recommendations will apply similarly to any precision CSA. Although our analysis here is done with the MAX44286, the results are true and should hold good for any high-precision CSMs.\n\nWe begin the analysis of Figure 4 with a well-known axiom:\n\nVOUT = Gain × VDIFF\n\nWhere:\n\n•VOUT is the amplifier’s output on bump B2 in volts.\n\n•VDIFF is input differential sense voltage in millivolts due to the current flow through shunt resistor RSENSE placed across the inputs.\n\n•Gain is inherent to the amplifier based on the gain option chosen (for example, 25V/V, 50V/V, 100V/V, 200V/V).\n\nGain accuracy or gain error is a critical parameter in precision applications where highly accurate sensing is needed on µV-to-mV level drops across a sense resistor.\n\nTherefore:\n\nGain Error (GE) = [(GainMEAS – Gain IDEAL) × 100]/GainIDEAL\n\nWhere:\n\n• Gain error is the percentage deviation between the observed differential gain and the ideal differential gain expected of the CSM.\n\n• GainMEAS is the gain achieved in V/V.\n\n• GainIDEAL is the gain that the device is rated to provide in V/V.\n\nBelow are the results for the MAX44286 on bench tests for the setup in Figure 4. To read the most accurate measurement of voltage drop, we calculate gain error by a two-point differential voltage covering both extremes of the full-scale differential sense range. Our test conditions were VBAT =5.5V with GIDEAL = 50V/V version at room temperature.\n\nSpecifically:\n\n• VDIFF = 60mV, 4mV as two points\n\n• VOUT with respect to GND = 2.99264V, 0.1992298V, respectively\n\n• GMEAS = VOUT/VDIFF\n\nSo, calculating gain error per Equation 2 yields:\n\nGain Error (GE) = -0.23713148%\n\nNow this result is not what we expect from this CSM, as its maximum gain error spec is ±0.23%.\n\nAnother important specification in these ultra-precision sense amplifiers is input offset voltage, given by:\n\nVOS = (VOUT - VDIFF × GMEAS)/GMEAS\n\nWhere:\n\nVOS is the CSM’s input-offset voltage specified in microvolts due to mismatch in the input pair. From Equation 3 and our test results, we achieve:\n\nVOS = -5.93281µV\n\nGain Accuracy and input VOS of CSA improve with a Kelvin sensing layout\n\nThere is a way to achieve gain error that is always less than ±0.2%, and better input-offset voltage. Examine Figure 5 and notice that there are a few more traces than in Figure 4.\n\nClick image to enlarge\n\nFigure 5. Circuit shows Kelvin connections on both inputs, output, and ground bump. Once again the MAX44286 is the example CSA acting as the shunt monitor.\n\nTraces from the sense-resistor terminals to the amplifier inputs carry input bias current. Supply current to the amplifier is part of the input bias current drawn through the RS+ pin because there is no dedicated supply voltage pin on the MAX44286. Otherwise, the input bias current flows through the RS+ and RS- pins and the supply current flows through the VCC pin, if there is a dedicated supply pin. Also, this current through the RS+ bump increases as the input differential voltage increases.\n\nTrace impedance will create a drop due to this current flow through the trace. The result will be extra gain error if the input sense voltage is calculated across the sense resistor, since we did not account for loss along the trace. These extra traces at the input bumps, RS+ MEASURE and RS- MEASURE, will have no effect on the input sense voltage when it is measured across them.\n\nNote also that the extra GND MEASURE trace is isolated from the currents flowing to the board’s ground and is solely used as a ground reference for accurate output voltage measurement. Another trace coming from the OUT bump, VOUT,MEASURE, is isolated from the actual output trace carrying the load current when the load resistor sources or sinks current out of the amplifier. This isolation lets us accurately measure the output voltage at the OUT bump with respect to the GND MEASURE trace. From all this we can calculate the gain error and input VOS of the CSA very precisely.\n\nIf you notice, there is a four-terminal resistor used in Figure 5. The 4-lead Kelvin configuration enables current to be applied through two opposite terminals; a sensing voltage is measured across the other two terminals. This design eliminates the resistance and temperature coefficient of the terminals for a more accurate current measurement. Also, traces from the sense-resistor terminals are taken directly underneath the pads of the resistor, thus preventing any additional trace impedance on the sense resistor.\n\nThe results below were achieved on the bench with the setup shown on Figure 5.Gain error is calculated similarly where a two-point differential voltage covering both extremes is applied to the shunt monitor and the output voltages recorded. Our test conditions were the same as earlier, VBAT =5.5V with GIDEAL = 50V/V for the CSA at room temperature. Therefore, VDIFF = 60mV, 4mV as two extremes.\n\nNow VDIFF is the input differential voltage measured between RS+ MEASURE and RS- MEASURE. Also the VOUT,MEASURE values, with respect to the GND MEASURE for 60mV and 4mV input sense voltages applied, are 3.013034V and 0.2141941V respectively.\n\nThus:\n\nGMEAS = VOUT/VDIFF\n\nSo, calculating gain error from Equation 2 yields:\n\nGain error = -0.08518056%\n\nSubstituting VOUT,MEASURE calculated with respect to GND MEASURE and substituting yields:\n\nVOS = 3.54415µV\n\nWe can clearly see that there is an appreciable change in the gain error and input VOS. Thus, preserving the ultra-precise measurements of a CSA depends on the layout and component placement on the test fixture. If we use the sensing traces to measure as shown in Figure 5, measurements are very accurate. In ultra-precision applications, sense voltages on the order of microvolt are very important for sensing current on the order of microamps. Clearly, each trace must be laid out with great care.\n\nTraces from each sense-resistor terminal to the respective input bumps need to be symmetrical in shape and length. Having a 2-terminal sense resistor will also provide good gain accuracy readings within a maximum of ±0.23% over temperature. For more accurate results over temperature, use 4-terminal sense resistor.\n\nThus, we complement Lord Kelvin’s sensing principles by preserving the DC accuracy of an ultra-precision CSA. Precision measurements not only depend on the good design and layout of the CSA itself, but on the board layout as well.\n\n### RELATED", null, "#### The Dawn of a New Era with High-Power Wireless Charging\n\nJul 1,2022", null, "#### Mobile Telephone Mast Goes Off-Grid\n\nJul 1,2022", null, "#### Navitas Powers Xiaomi's First In-Box GaN 100 W Laptop Fast Charger\n\nJun 30,2022", null, "Jun 29,2022" ]	[ null, "https://www.powersystemsdesign.com/images/pdficon.png", null, "https://www.powersystemsdesign.com//images/articles/1656702910-Figure 1 - Wireless power transmitter WLC1115.png", null, "https://www.powersystemsdesign.com//images/articles/1656626637-Ally Head Shot_bw-400.jpg", null, "https://www.powersystemsdesign.com//images/articles/1656620631-Navitas.jpeg", null, "https://www.powersystemsdesign.com//images/articles/1656508834-New-FingerTip-touchscreen-controller-N4465D-big.jpeg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9089157,"math_prob":0.9388673,"size":11305,"snap":"2022-27-2022-33","text_gpt3_token_len":2493,"char_repetition_ratio":0.14122644,"word_repetition_ratio":0.026732134,"special_character_ratio":0.21149933,"punctuation_ratio":0.095147476,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9746457,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,null,null,5,null,3,null,2,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-02T18:09:38Z\",\"WARC-Record-ID\":\"<urn:uuid:f831db08-3ce0-4a61-bf0f-484743ccea86>\",\"Content-Length\":\"71670\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:43a9e966-b786-49cc-8b33-aab892ad04ae>\",\"WARC-Concurrent-To\":\"<urn:uuid:69412988-37a2-43e0-8f78-bdc996f86cfd>\",\"WARC-IP-Address\":\"34.206.203.101\",\"WARC-Target-URI\":\"https://www.powersystemsdesign.com/articles/using-lord-kelvins-sensing-method/29/8590\",\"WARC-Payload-Digest\":\"sha1:76V7WAHNQDZKNVOIHVEXIPXIKD75OSML\",\"WARC-Block-Digest\":\"sha1:X4WU7POFWRNFHCPCBOAJNV2U3CEOQC7Y\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104189587.61_warc_CC-MAIN-20220702162147-20220702192147-00040.warc.gz\"}"}
https://lemon.cs.elte.hu/trac/lemon/changeset/434/lemon/doc	[ "# Changeset 434:ad483acf1654 in lemon for doc\n\nIgnore:\nTimestamp:\n12/02/08 16:33:22 (13 years ago)\nBranch:\ndefault\nChildren:\n435:9afe81e4c543, 437:6a2a33ad261b, 451:09e416d35896, 469:d369e885d196, 522:7f8560cb9d65\nParents:\n429:d8b87e9b90c3 (diff), 433:6ff53afe98b5 (diff)\nNote: this is a merge changeset, the changes displayed below correspond to the merge itself.\nUse the (diff) links above to see all the changes relative to each parent.\nPhase:\npublic\nMessage:\n\nMerge\n\nFiles:\n2 edited\n\nUnmodified\nAdded\nRemoved\n• ## doc/groups.dox\n\n r422 See also: \\ref graph_concepts \"Graph Structure Concepts\". / /* @defgroup graph_adaptors Adaptor Classes for graphs @ingroup graphs \\brief This group contains several adaptor classes for digraphs and graphs The main parts of LEMON are the different graph structures, generic graph algorithms, graph concepts which couple these, and graph adaptors. While the previous notions are more or less clear, the latter one needs further explanation. Graph adaptors are graph classes which serve for considering graph structures in different ways. A short example makes this much clearer. Suppose that we have an instance \\c g of a directed graph type say ListDigraph and an algorithm \\code template int algorithm(const Digraph&); \\endcode is needed to run on the reverse oriented graph. It may be expensive (in time or in memory usage) to copy \\c g with the reversed arcs. In this case, an adaptor class is used, which (according to LEMON digraph concepts) works as a digraph. The adaptor uses the original digraph structure and digraph operations when methods of the reversed oriented graph are called. This means that the adaptor have minor memory usage, and do not perform sophisticated algorithmic actions. The purpose of it is to give a tool for the cases when a graph have to be used in a specific alteration. If this alteration is obtained by a usual construction like filtering the arc-set or considering a new orientation, then an adaptor is worthwhile to use. To come back to the reverse oriented graph, in this situation \\code template class ReverseDigraph; \\endcode template class can be used. The code looks as follows \\code ListDigraph g; ReverseDigraph rg(g); int result = algorithm(rg); \\endcode After running the algorithm, the original graph \\c g is untouched. This techniques gives rise to an elegant code, and based on stable graph adaptors, complex algorithms can be implemented easily. In flow, circulation and bipartite matching problems, the residual graph is of particular importance. Combining an adaptor implementing this, shortest path algorithms and minimum mean cycle algorithms, a range of weighted and cardinality optimization algorithms can be obtained. For other examples, the interested user is referred to the detailed documentation of particular adaptors. The behavior of graph adaptors can be very different. Some of them keep capabilities of the original graph while in other cases this would be meaningless. This means that the concepts that they are models of depend on the graph adaptor, and the wrapped graph(s). If an arc of \\c rg is deleted, this is carried out by deleting the corresponding arc of \\c g, thus the adaptor modifies the original graph. But for a residual graph, this operation has no sense. Let us stand one more example here to simplify your work. RevGraphAdaptor has constructor \\code ReverseDigraph(Digraph& digraph); \\endcode This means that in a situation, when a const ListDigraph& reference to a graph is given, then it have to be instantiated with Digraph=const ListDigraph. \\code int algorithm1(const ListDigraph& g) { RevGraphAdaptor rg(g); return algorithm2(rg); } \\endcode /\n• ## doc/groups.dox\n\n r432 / namespace lemon { /* This group describes maps that are specifically designed to assign values to the nodes and arcs of graphs. values to the nodes and arcs/edges of graphs. If you are looking for the standard graph maps (\\c NodeMap, \\c ArcMap, \\c EdgeMap), see the \\ref graph_concepts \"Graph Structure Concepts\". / maps from other maps. Most of them are \\ref lemon::concepts::ReadMap \"read-only maps\". Most of them are \\ref concepts::ReadMap \"read-only maps\". They can make arithmetic and logical operations between one or two maps (negation, shifting, addition, multiplication, logical 'and', 'or', \\brief Common graph search algorithms. This group describes the common graph search algorithms like Breadth-First Search (BFS) and Depth-First Search (DFS). This group describes the common graph search algorithms, namely \\e breadth-first \\e search (BFS) and \\e depth-first \\e search (DFS). / \\brief Algorithms for finding shortest paths. This group describes the algorithms for finding shortest paths in graphs. This group describes the algorithms for finding shortest paths in digraphs. - \\ref Dijkstra algorithm for finding shortest paths from a source node when all arc lengths are non-negative. - \\ref BellmanFord \"Bellman-Ford\" algorithm for finding shortest paths from a source node when arc lenghts can be either positive or negative, but the digraph should not contain directed cycles with negative total length. - \\ref FloydWarshall \"Floyd-Warshall\" and \\ref Johnson \"Johnson\" algorithms for solving the \\e all-pairs \\e shortest \\e paths \\e problem when arc lenghts can be either positive or negative, but the digraph should not contain directed cycles with negative total length. - \\ref Suurballe A successive shortest path algorithm for finding arc-disjoint paths between two nodes having minimum total length. / feasible circulations. The maximum flow problem is to find a flow between a single source and a single target that is maximum. Formally, there is a \\f$G=(V,A)\\f$ directed graph, an \\f$c_a:A\\rightarrow\\mathbf{R}^+_0\\f$ capacity function and given \\f$s, t \\in V\\f$ source and target node. The maximum flow is the \\f$f_a\\f$ solution of the next optimization problem: \\f[ 0 \\le f_a \\le c_a \\f] \\f[ \\sum_{v\\in\\delta^{-}(u)}f_{vu}=\\sum_{v\\in\\delta^{+}(u)}f_{uv} \\qquad \\forall u \\in V \\setminus \\{s,t\\}\\f] \\f[ \\max \\sum_{v\\in\\delta^{+}(s)}f_{uv} - \\sum_{v\\in\\delta^{-}(s)}f_{vu}\\f] The \\e maximum \\e flow \\e problem is to find a flow of maximum value between a single source and a single target. Formally, there is a \\f$G=(V,A)\\f$ digraph, a \\f$cap:A\\rightarrow\\mathbf{R}^+_0\\f$ capacity function and \\f$s, t \\in V\\f$ source and target nodes. A maximum flow is an \\f$f:A\\rightarrow\\mathbf{R}^+_0\\f$ solution of the following optimization problem. \\f[ \\max\\sum_{a\\in\\delta_{out}(s)}f(a) - \\sum_{a\\in\\delta_{in}(s)}f(a) \\f] \\f[ \\sum_{a\\in\\delta_{out}(v)} f(a) = \\sum_{a\\in\\delta_{in}(v)} f(a) \\qquad \\forall v\\in V\\setminus\\{s,t\\} \\f] \\f[ 0 \\leq f(a) \\leq cap(a) \\qquad \\forall a\\in A \\f] LEMON contains several algorithms for solving maximum flow problems: - \\ref lemon::EdmondsKarp \"Edmonds-Karp\" - \\ref lemon::Preflow \"Goldberg's Preflow algorithm\" - \\ref lemon::DinitzSleatorTarjan \"Dinitz's blocking flow algorithm with dynamic trees\" - \\ref lemon::GoldbergTarjan \"Preflow algorithm with dynamic trees\" In most cases the \\ref lemon::Preflow \"Preflow\" algorithm provides the fastest method to compute the maximum flow. All impelementations provides functions to query the minimum cut, which is the dual linear programming problem of the maximum flow. - \\ref EdmondsKarp Edmonds-Karp algorithm. - \\ref Preflow Goldberg-Tarjan's preflow push-relabel algorithm. - \\ref DinitzSleatorTarjan Dinitz's blocking flow algorithm with dynamic trees. - \\ref GoldbergTarjan Preflow push-relabel algorithm with dynamic trees. In most cases the \\ref Preflow \"Preflow\" algorithm provides the fastest method for computing a maximum flow. All implementations provides functions to also query the minimum cut, which is the dual problem of the maximum flow. / This group describes the algorithms for finding minimum cost flows and circulations. The \\e minimum \\e cost \\e flow \\e problem is to find a feasible flow of minimum total cost from a set of supply nodes to a set of demand nodes in a network with capacity constraints and arc costs. Formally, let \\f$G=(V,A)\\f$ be a digraph, \\f$lower, upper: A\\rightarrow\\mathbf{Z}^+_0\\f$ denote the lower and upper bounds for the flow values on the arcs, \\f$cost: A\\rightarrow\\mathbf{Z}^+_0\\f$ denotes the cost per unit flow on the arcs, and \\f$supply: V\\rightarrow\\mathbf{Z}\\f$ denotes the supply/demand values of the nodes. A minimum cost flow is an \\f$f:A\\rightarrow\\mathbf{R}^+_0\\f$ solution of the following optimization problem. \\f[ \\min\\sum_{a\\in A} f(a) cost(a) \\f] \\f[ \\sum_{a\\in\\delta_{out}(v)} f(a) - \\sum_{a\\in\\delta_{in}(v)} f(a) = supply(v) \\qquad \\forall v\\in V \\f] \\f[ lower(a) \\leq f(a) \\leq upper(a) \\qquad \\forall a\\in A \\f] LEMON contains several algorithms for solving minimum cost flow problems: - \\ref CycleCanceling Cycle-canceling algorithms. - \\ref CapacityScaling Successive shortest path algorithm with optional capacity scaling. - \\ref CostScaling Push-relabel and augment-relabel algorithms based on cost scaling. - \\ref NetworkSimplex Primal network simplex algorithm with various pivot strategies. / This group describes the algorithms for finding minimum cut in graphs. The minimum cut problem is to find a non-empty and non-complete \\f$X\\f$ subset of the vertices with minimum overall capacity on outgoing arcs. Formally, there is \\f$G=(V,A)\\f$ directed graph, an \\f$c_a:A\\rightarrow\\mathbf{R}^+_0\\f$ capacity function. The minimum The \\e minimum \\e cut \\e problem is to find a non-empty and non-complete \\f$X\\f$ subset of the nodes with minimum overall capacity on outgoing arcs. Formally, there is a \\f$G=(V,A)\\f$ digraph, a \\f$cap: A\\rightarrow\\mathbf{R}^+_0\\f$ capacity function. The minimum cut is the \\f$X\\f$ solution of the next optimization problem: \\f[ \\min_{X \\subset V, X\\not\\in \\{\\emptyset, V\\}} \\sum_{uv\\in A, u\\in X, v\\not\\in X}c_{uv}\\f] \\sum_{uv\\in A, u\\in X, v\\not\\in X}cap(uv) \\f] LEMON contains several algorithms related to minimum cut problems: - \\ref lemon::HaoOrlin \"Hao-Orlin algorithm\" to calculate minimum cut in directed graphs - \\ref lemon::NagamochiIbaraki \"Nagamochi-Ibaraki algorithm\" to calculate minimum cut in undirected graphs - \\ref lemon::GomoryHuTree \"Gomory-Hu tree computation\" to calculate all pairs minimum cut in undirected graphs - \\ref HaoOrlin \"Hao-Orlin algorithm\" for calculating minimum cut in directed graphs. - \\ref NagamochiIbaraki \"Nagamochi-Ibaraki algorithm\" for calculating minimum cut in undirected graphs. - \\ref GomoryHuTree \"Gomory-Hu tree computation\" for calculating all-pairs minimum cut in undirected graphs. If you want to find minimum cut just between two distinict nodes, please see the \\ref max_flow \"Maximum Flow page\". see the \\ref max_flow \"maximum flow problem\". / graphs. The matching problems in bipartite graphs are generally easier than in general graphs. The goal of the matching optimization can be the finding maximum cardinality, maximum weight or minimum cost can be finding maximum cardinality, maximum weight or minimum cost matching. The search can be constrained to find perfect or maximum cardinality matching. LEMON contains the next algorithms: - \\ref lemon::MaxBipartiteMatching \"MaxBipartiteMatching\" Hopcroft-Karp augmenting path algorithm for calculate maximum cardinality matching in bipartite graphs - \\ref lemon::PrBipartiteMatching \"PrBipartiteMatching\" Push-Relabel algorithm for calculate maximum cardinality matching in bipartite graphs - \\ref lemon::MaxWeightedBipartiteMatching \"MaxWeightedBipartiteMatching\" Successive shortest path algorithm for calculate maximum weighted matching and maximum weighted bipartite matching in bipartite graph - \\ref lemon::MinCostMaxBipartiteMatching \"MinCostMaxBipartiteMatching\" Successive shortest path algorithm for calculate minimum cost maximum matching in bipartite graph - \\ref lemon::MaxMatching \"MaxMatching\" Edmond's blossom shrinking algorithm for calculate maximum cardinality matching in general graph - \\ref lemon::MaxWeightedMatching \"MaxWeightedMatching\" Edmond's blossom shrinking algorithm for calculate maximum weighted matching in general graph - \\ref lemon::MaxWeightedPerfectMatching \"MaxWeightedPerfectMatching\" Edmond's blossom shrinking algorithm for calculate maximum weighted perfect matching in general graph The matching algorithms implemented in LEMON: - \\ref MaxBipartiteMatching Hopcroft-Karp augmenting path algorithm for calculating maximum cardinality matching in bipartite graphs. - \\ref PrBipartiteMatching Push-relabel algorithm for calculating maximum cardinality matching in bipartite graphs. - \\ref MaxWeightedBipartiteMatching Successive shortest path algorithm for calculating maximum weighted matching and maximum weighted bipartite matching in bipartite graphs. - \\ref MinCostMaxBipartiteMatching Successive shortest path algorithm for calculating minimum cost maximum matching in bipartite graphs. - \\ref MaxMatching Edmond's blossom shrinking algorithm for calculating maximum cardinality matching in general graphs. - \\ref MaxWeightedMatching Edmond's blossom shrinking algorithm for calculating maximum weighted matching in general graphs. - \\ref MaxWeightedPerfectMatching Edmond's blossom shrinking algorithm for calculating maximum weighted perfect matching in general graphs. \\image html bipartite_matching.png This group describes the algorithms for finding a minimum cost spanning tree in a graph tree in a graph. / \\anchor demoprograms @defgroup demos Demo programs @defgroup demos Demo Programs Some demo programs are listed here. Their full source codes can be found in /* @defgroup tools Standalone utility applications @defgroup tools Standalone Utility Applications Some utility applications are listed here. */ }\nNote: See TracChangeset for help on using the changeset viewer." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6918613,"math_prob":0.9815462,"size":15332,"snap":"2021-21-2021-25","text_gpt3_token_len":4191,"char_repetition_ratio":0.15631524,"word_repetition_ratio":0.07090359,"special_character_ratio":0.29846072,"punctuation_ratio":0.08972031,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99581486,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-17T16:46:17Z\",\"WARC-Record-ID\":\"<urn:uuid:b8a43b02-ab79-44c5-bb47-15beb2efe043>\",\"Content-Length\":\"73690\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4c65334f-1ca7-42c7-a6fe-7ee35883a4bb>\",\"WARC-Concurrent-To\":\"<urn:uuid:9ef82629-5a66-4975-91c2-289f6f00b2e5>\",\"WARC-IP-Address\":\"157.181.227.195\",\"WARC-Target-URI\":\"https://lemon.cs.elte.hu/trac/lemon/changeset/434/lemon/doc\",\"WARC-Payload-Digest\":\"sha1:POETUOP7FKF4XJ64S5IIEODUVTQQ54MP\",\"WARC-Block-Digest\":\"sha1:N24HRKUYX3IRNH7U4N5FC6VH4YVVNBQA\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487630518.38_warc_CC-MAIN-20210617162149-20210617192149-00025.warc.gz\"}"}
https://rd.springer.com/article/10.1007%2FBF02988324	[ "Springer Nature is making SARS-CoV-2 and COVID-19 research free. View research \| View latest news \| Sign up for updates\n\n# Odd factors of a graph\n\n• 68 Accesses\n\n• 12 Citations\n\n## Abstract\n\nLetG be a graph and letf be a function defined on V(G) such that f(x) is a positive odd integer for everyx ɛ V(G). A spanning subgraphF ofG is called a [l,f]-odd factor of G if dF(x) ɛ {1,3,2026, f(x)} for every x ɛV(G), whered F (x) denotes the degree of x inF. We discuss several conditions for a graphG to have a [1,f]-odd factor.\n\nThis is a preview of subscription content, log in to check access.\n\n## References\n\n1. 1.\n\nAkiyama, J., Kano, M.: Factors and factorizations of graphs — a survey. J. Graph Theory9, 1–42 (1985)\n\n2. 2.\n\nAmahashi, A.: On factors with all degrees odd. Graphs Comb.1, 111–114 (1985)\n\n3. 3.\n\nBeineke, L.W., Plummer, M.D.: On the 1-factors of a nonseparable graph. J. Comb. Theory2, 285–289 (1967)\n\n4. 4.\n\nChartrand, G., Polimeni, A.D., Stewart, M.J.: The existence of 1-factors in line graphs, squares, and total graphs. Indag. Math.35, 228–232 (1973)\n\n5. 5.\n\nChungphaisan, V.: Factors of graphs and degree sequences. Nanta Math.9, 41–49 (1976)\n\n6. 6.\n\nClarke, F.H.: A graph polynomial and its applications. Discrete Math.3, 305–313 (1972)\n\n7. 7.\n\nJackson, B., Whitty, R.W.: A note concerning graphs with unique f-factors. J. Graph Theory13, 577–580 (1989)\n\n8. 8.\n\nKotzig, A.: On the theory of finite graphs with a linear factor I, II, III. Mat. Fyz. Cas.9, 73–91 (1959)\n\n9. 9.\n\nLas Vergnas, M.: A note on matchings in graphs. Cahiers Centre Études Rech. Opér.17, 257–260 (1975)\n\n10. 10.\n\nSumner, D.P.: Graphs with 1-factors. Proc. Amer. Math. Soc.42, 8–17 (1974)\n\n11. 11.\n\nSumner, D.P.: 1-factors and antifactor sets. J. London Math. Soc.13, 351–359 (1976)\n\n12. 12.\n\nTutte, W.T.: The factorizations of linear graphs. J. London Math. Soc.22, 107–111 (1947)\n\n13. 13.\n\nTutte, W.T.: Graph factors. Combinatorica1, 79–97 (1981)\n\n14. 14.\n\nYuting, C., Kano, M.: Some results on odd factors of graphs. J. Graph Theory12, 327–333 (1988)\n\n## Author information\n\nCorrespondence to Jerzy Topp or Preben D. Vestergaard.\n\n## Rights and permissions\n\nReprints and Permissions\n\nTopp, J., Vestergaard, P.D. Odd factors of a graph. Graphs and Combinatorics 9, 371–381 (1993). https://doi.org/10.1007/BF02988324" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7094059,"math_prob":0.8391542,"size":2109,"snap":"2020-10-2020-16","text_gpt3_token_len":731,"char_repetition_ratio":0.1263658,"word_repetition_ratio":0.0,"special_character_ratio":0.38169748,"punctuation_ratio":0.29982045,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9658545,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-19T18:25:02Z\",\"WARC-Record-ID\":\"<urn:uuid:0ff2d45a-b7bb-4666-b6f0-3a6ffccebbd3>\",\"Content-Length\":\"81985\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1eba833b-419f-4971-9549-c3a97ac9f3ed>\",\"WARC-Concurrent-To\":\"<urn:uuid:13f21e35-83ef-46bf-8e86-7012915ba56e>\",\"WARC-IP-Address\":\"151.101.248.95\",\"WARC-Target-URI\":\"https://rd.springer.com/article/10.1007%2FBF02988324\",\"WARC-Payload-Digest\":\"sha1:6PTVTIWR7ZEVOQEA65U4MCRJQOSPXTP5\",\"WARC-Block-Digest\":\"sha1:UPDP3GJLDXIMNBWWAXIWEP477KRMZKP6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875144165.4_warc_CC-MAIN-20200219153707-20200219183707-00293.warc.gz\"}"}
https://softmath.com/math-book-answers/perfect-square-trinomial/midpoint-formula-for-fractions.html	[ "English \| Español\n\n# Try our Free Online Math Solver!", null, "Online Math Solver\n\n Depdendent Variable\n\n Number of equations to solve: 23456789\n Equ. #1:\n Equ. #2:\n\n Equ. #3:\n\n Equ. #4:\n\n Equ. #5:\n\n Equ. #6:\n\n Equ. #7:\n\n Equ. #8:\n\n Equ. #9:\n\n Solve for:\n\n Dependent Variable\n\n Number of inequalities to solve: 23456789\n Ineq. #1:\n Ineq. #2:\n\n Ineq. #3:\n\n Ineq. #4:\n\n Ineq. #5:\n\n Ineq. #6:\n\n Ineq. #7:\n\n Ineq. #8:\n\n Ineq. #9:\n\n Solve for:\n\n Please use this form if you would like to have this math solver on your website, free of charge. Name: Email: Your Website: Msg:\n\nWhat our customers say...\n\nThousands of users are using our software to conquer their algebra homework. Here are some of their experiences:\n\nThe Algebrator was very helpful, it helped me get back on track and bring back my skills for my next school season. 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