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Wesley R. Elsberry Posts: 4991 Joined: May 2002 (Permalink) Posted: Mar. 06 2015,12:32    Is there such a thing as an open-source Javascript webapp and GUI framework with RAD visual designer (doing for Javascript what Delphi does for Object Pascal or Visual Studio does for C#)? My sole foray into Javascript so far has been a simple DOM thing to implement Weasel and display results in an HTML form. Now there looks to be a plethora of libraries, but little of the info oriented along the lines of interest in my question. -------------- "You can't teach an old dogma new tricks." - Dorothy Parker

k.e.. Posts: 5432 Joined: May 2007 (Permalink) Posted: Mar. 06 2015,21:45    Quote (Wesley R. Elsberry @ Mar. 06 2015,20:32) Is there such a thing as an open-source Javascript webapp and GUI framework with RAD visual designer (doing for Javascript what Delphi does for Object Pascal or Visual Studio does for C#)? My sole foray into Javascript so far has been a simple DOM thing to implement Weasel and display results in an HTML form. Now there looks to be a plethora of libraries, but little of the info oriented along the lines of interest in my question. I've played with Netbeans and Eclipse but you might be interested in Processing (processing.org) check this out http://www.openprocessing.org/sketch....255 -------------- "I get a strong breeze from my monitor every time k.e. puts on his clown DaveTard suit" dogdidit "ID is deader than Lenny Flanks granmaws dildo batteries" Erasmus "I'm busy studying scientist level science papers" Galloping Gary Gaulin

Wesley R. Elsberry Posts: 4991 Joined: May 2002 (Permalink) Posted: April 24 2016,23:07    Avida-ED is pretty close to going to a web application. Currently in alpha test, the new version presents the familiar user interface in your browser. This is made possible by applying the Emscripten compiler to turn the Avida core into a Javascript library. It should be generally available sometime in mid-June. -------------- "You can't teach an old dogma new tricks." - Dorothy Parker

Wesley R. Elsberry Posts: 4991 Joined: May 2002 (Permalink) Posted: June 19 2016,00:17    Avida-ED Web 3.0 is out for beta release now. Rob Pennock gave a formal announcement at the Evolution 2016 meeting. Avida-ED Web access page Mirror -------------- "You can't teach an old dogma new tricks." - Dorothy Parker

midwifetoad Posts: 4003 Joined: Mar. 2008 (Permalink) Posted: June 19 2016,05:49    Doesn't seem to work on iPad Chrome. -------------- Any version of ID consistent with all the evidence is indistinguishable from evolution.

Wesley R. Elsberry Posts: 4991 Joined: May 2002 (Permalink) Posted: June 19 2016,07:58    Tablet compatibility wasn't the goal for this phase, but will be addressed later. The Avida-ED team is planning to get a ChromeBook and tablets for testing. I know it works on Chrome on Android 5.1, which is what my phone is. That's not a great use case, but being able to show it at all there is pretty cool for a demo. -------------- "You can't teach an old dogma new tricks." - Dorothy Parker

Wesley R. Elsberry Posts: 4991 Joined: May 2002 (Permalink) Posted: June 19 2016,08:04    Diane's MacBook Pro that she does the UI development on had issues. When we were at the Apple Store in Tampa for repairs to it, we brought up Avida-ED on an iPad and an iPod there. Probably in Safari, which is not a good fit now because Safari refuses to allow saving blob data to local storage. Fixing that has been in the Safari developers issue list for a long time now. -------------- "You can't teach an old dogma new tricks." - Dorothy Parker

midwifetoad Posts: 4003 Joined: Mar. 2008 (Permalink) Posted: Sep. 20 2016,19:26    Anyone have any info about the whereabouts of Zach, or where one could get a copy of Gregor's Bookkeeper? -------------- Any version of ID consistent with all the evidence is indistinguishable from evolution.

midwifetoad Posts: 4003 Joined: Mar. 2008 (Permalink) Posted: Sep. 20 2016,20:59    Found it on archive.com -------------- Any version of ID consistent with all the evidence is indistinguishable from evolution.

WebHopper Posts: 7 Joined: Feb. 2017 (Permalink) Posted: Feb. 10 2017,11:32    Hello, I am wondering if there is an analytical solution to the Weasel algorithm. I think of a probability distribution of the number of trials necessary to achieve a target sequence, the mean and variance implying the following parameters: length of the alphabet length of the target sequence mutation rate population size Have you seen a site where this is done? On evoinfo.org Dembski proposes a solution for what he calls "partitioned search". But I am looking for a solution of what he names "proximity reward search", which apparently is the Weasel algorithm. To be clear, I am not interested in a numerical  but analytical solution of the mean and variance as a functions of the above-mentioned parameters.

Wesley R. Elsberry Posts: 4991 Joined: May 2002 (Permalink) Posted: Feb. 10 2017,18:30    Quote (WebHopper @ Feb. 10 2017,11:32) Hello, I am wondering if there is an analytical solution to the Weasel algorithm. I think of a probability distribution of the number of trials necessary to achieve a target sequence, the mean and variance implying the following parameters: length of the alphabet length of the target sequence mutation rate population size Have you seen a site where this is done? On evoinfo.org Dembski proposes a solution for what he calls "partitioned search". But I am looking for a solution of what he names "proximity reward search", which apparently is the Weasel algorithm. To be clear, I am not interested in a numerical  but analytical solution of the mean and variance as a functions of the above-mentioned parameters. Further up the thread, I pointed out the difference between partitioned search and "weasel". http://www.antievolution.org/cgi-bin....y140986       Quote "Locking" or "latching" is the same as removing the term that allows for correct bases to mutate to incorrect ones. What remains is an expectation that the number of correct bases can only monotonically increase.   If you have the analytical form you like for partitioned search, then modify to add in the additional element I note for "weasel" and any other adjustments. I left off that project before fully working up the population component for probabilities. I've had some issues with hosted images going stale. I need to look up some of my graphs in this thread and restore them. http://austringer.net/wp....-....-part-1 http://austringer.net/wp....-....-part-2 And somewhere, sometime, I know I did a numerical scan of parameter space to show the likely range of parameters for Dawkins' original runs given his reported generation times for results. I'm not finding where I shared that, though. -------------- "You can't teach an old dogma new tricks." - Dorothy Parker

DiEb Posts: 312 Joined: May 2008 (Permalink) Posted: Feb. 11 2017,03:02    Quote (WebHopper @ Feb. 10 2017,17:32) Hello, I am wondering if there is an analytical solution to the Weasel algorithm. I think of a probability distribution of the number of trials necessary to achieve a target sequence, the mean and variance implying the following parameters: length of the alphabet length of the target sequence mutation rate population size Have you seen a site where this is done? On evoinfo.org Dembski proposes a solution for what he calls "partitioned search". But I am looking for a solution of what he names "proximity reward search", which apparently is the Weasel algorithm. To be clear, I am not interested in a numerical  but analytical solution of the mean and variance as a functions of the above-mentioned parameters. Try  "The probability of formation of a nucleotide sequence by random mutations" by Ulrich Utiger...

WebHopper Posts: 7 Joined: Feb. 2017 (Permalink) Posted: Feb. 11 2017,05:27    Wesley, if I understand you well the expected correct bases after mutation is E = C + (u * (L - C) / K) - (u * C * (K - 1) / K) where C = number of correct bases K = alphabet length L = number of base pairs u = mutation rate I guess with "expected" E you refer to the mean. So if C=L we have E = L (K + u - K u)/K This is a linear law with respect to L if the other parameters are held constant. However, this does not match a Monte Carlo simulation, which shows that E is exponential with respect to L. So there must be some error in your calculation or I don't understand you well. DiEb: Thanks for the link, I will look into this. Seems to be complicated however...

DiEb Posts: 312 Joined: May 2008 (Permalink) Posted: Feb. 11 2017,06:26    Quote (WebHopper @ Feb. 11 2017,11:27) DiEb: Thanks for the link, I will look into this. Seems to be complicated however... Well, there are three different approaches: 1) Simulation 2) Numerical Modelling 3) Analytical Modelling Simulation is the easiest way, but doesn't provide much insights. Unless there aren't some clever simplifications/estimation/approximations, the analytical way gets just to complicated for me... I've preferred to model the weasel as a Markov-Chain, given the mutation rate, the alphabet, the size of the population and the length of the target phrase, \mu and \sigma can be calculated in a straightforward was - and one gets some nice pictures: Which Parameters should Dembski have used?

WebHopper Posts: 7 Joined: Feb. 2017 (Permalink) Posted: Feb. 11 2017,09:51    The next configuration of a Markov chain only depends on the actual configuration and on neighboring sites, right? But the probability of a nucleotide sequence to achieve a target depends on all sites, that is, all sites must be correct. For example the third site of the sequence AAT is not correct if the target is AAA even though the first and second site are correct. So whether the target is achieved or not depends on all three sites. But maybe I don't understand exactly what is a Markov chain. Can you explain? Can you also explain what is on the abscissa and ordinate in your graphic? And what mean the lines? Some constant parameters?

Wesley R. Elsberry Posts: 4991 Joined: May 2002 (Permalink) Posted: Feb. 11 2017,16:30    Quote (WebHopper @ Feb. 11 2017,05:27) Wesley, if I understand you well the expected correct bases after mutation is E = C + (u * (L - C) / K) - (u * C * (K - 1) / K) where C = number of correct bases K = alphabet length L = number of base pairs u = mutation rate I guess with "expected" E you refer to the mean. So if C=L we have E = L (K + u - K u)/K This is a linear law with respect to L if the other parameters are held constant. However, this does not match a Monte Carlo simulation, which shows that E is exponential with respect to L. So there must be some error in your calculation or I don't understand you well. DiEb: Thanks for the link, I will look into this. Seems to be complicated however... Edit: I'll look this over again. I derived it with Monte Carlo ground-truthing, so I'm not sure where we are diverging in expectation. Edited by Wesley R. Elsberry on Feb. 11 2017,16:35 -------------- "You can't teach an old dogma new tricks." - Dorothy Parker

Wesley R. Elsberry Posts: 4991 Joined: May 2002 (Permalink) Posted: Feb. 12 2017,14:28    Quote (Wesley R. Elsberry @ Feb. 11 2017,16:30)   Quote (WebHopper @ Feb. 11 2017,05:27) Wesley, if I understand you well the expected correct bases after mutation is E = C + (u * (L - C) / K) - (u * C * (K - 1) / K) where C = number of correct bases K = alphabet length L = number of base pairs u = mutation rate I guess with "expected" E you refer to the mean. So if C=L we have E = L (K + u - K u)/K This is a linear law with respect to L if the other parameters are held constant. However, this does not match a Monte Carlo simulation, which shows that E is exponential with respect to L. So there must be some error in your calculation or I don't understand you well. DiEb: Thanks for the link, I will look into this. Seems to be complicated however... Edit: I'll look this over again. I derived it with Monte Carlo ground-truthing, so I'm not sure where we are diverging in expectation. I think the issue is that the expectation I derived in the equation is the expected number of correct bases in one possibly-mutated offspring. I have rechecked the equation with a new Monte Carlo analysis, and it checks out. Code Sample u 0.0357142857143 # correct, MC result, equation result, partitioned search result 0 0.0368 0.0370 0.0370 1 1.0028 1.0013 1.0357 2 1.9652 1.9656 2.0344 3 2.9311 2.9299 3.0331 4 3.8931 3.8942 4.0317 5 4.8590 4.8585 5.0304 6 5.8222 5.8228 6.0291 7 6.7859 6.7870 7.0278 8 7.7504 7.7513 8.0265 9 8.7139 8.7156 9.0251 10 9.6783 9.6799 10.0238 11 10.6419 10.6442 11.0225 12 11.6067 11.6085 12.0212 13 12.5706 12.5728 13.0198 14 13.5344 13.5370 14.0185 15 14.5018 14.5013 15.0172 16 15.4639 15.4656 16.0159 17 16.4264 16.4299 17.0146 18 17.3926 17.3942 18.0132 19 18.3574 18.3585 19.0119 20 19.3230 19.3228 20.0106 21 20.2867 20.2870 21.0093 22 21.2499 21.2513 22.0079 23 22.2203 22.2156 23.0066 24 23.1779 23.1799 24.0053 25 24.1475 24.1442 25.0040 26 25.1056 25.1085 26.0026 27 26.0730 26.0728 27.0013 28 27.0409 27.0370 28.0000 Note that by the point we are mutating a string with 2 correct bases, we *expect* fewer than 2 correct bases afterward. Partitioned search mutation expectation is always greater than or equal to the starting correct number of bases. When we start talking about the expected number of generations to increment the best organism in the population to having another correct base, yes, that ends up in an exponentially increasing series with increasing number of correct bases. But the mutation expectation per offspring is a simpler calculation than that. Edited by Wesley R. Elsberry on Feb. 12 2017,15:19 -------------- "You can't teach an old dogma new tricks." - Dorothy Parker

WebHopper Posts: 7 Joined: Feb. 2017 (Permalink) Posted: Feb. 12 2017,16:50    Sorry Wesley, but I don't understand a clue what you are talking about... I started to read Utiger's paper, he explains it quite well. I mean what we need is an equation for the mean number of generations necessary to achieve a target. For instance, for Dawkins' weasel sentence this number is around 60 or so for a population size of 100 and a mutation rate of 0.05 as explained on Wiki. Utiger found a distribution like that of throwing dices: P(v) = q^v-1 p^v where v is the number of generations and p = 1-q is the probability that the dice got the correct number. When several nucleotides and a population size greater than one is involved, p and q become matrices with the same dimension as the length of the sequence. The mean is calculated in the same manner than for dices. This way, Utiger found that the mean is a logarithmic law with respect to the sequence length if the population size is greater than one, otherwise it is exponential. He checks this with Monte Carlo simulations and both the analytical and numerical results perfectly fit.

DiEb Posts: 312 Joined: May 2008 (Permalink) Posted: Feb. 12 2017,17:52    Quote (WebHopper @ Feb. 11 2017,15:51) Can you also explain what is on the abscissa and ordinate in your graphic? And what mean the lines? Some constant parameters? The graph answers the question "What is the expected number of queries for Dawkins's original weasel?" That is, the size of the alphabet (27) and the length of the target (28) are fixed. On the x-axis, there is the mutation rate \mu, on the y-axis, the number of queries, i.e., the size of the population times the number of expected generations. The coloured lines show the relation between \mu and number of queries for certain certain sizes of population. The dot displays the minimal expected number of queries for a certain population size - the black text enlists the relevant information: 1) size of population (e.g., 2 for the red line) 2) most efficient rate of mutation (e.g., 0.000 049 for the read line) 3) number of expected queries of this combination of size and rate (e.g., 21,213) The black line connects the minimal points and allows for extrapolation - though this is quite difficult in this log-log diagram. When I calculated the values nearly eight years ago, I concluded that the most efficient size of population would be 9 with a mutation rate of 0.00901: this would result in 1576 queries on average (or some 175 generations). Edited by DiEb on Feb. 12 2017,23:54

WebHopper Posts: 7 Joined: Feb. 2017 (Permalink) Posted: Feb. 13 2017,03:53    Quote (DiEb @ Feb. 12 2017,17:52) When I calculated the values nearly eight years ago, I concluded that the most efficient size of population would be 9 with a mutation rate of 0.00901: this would result in 1576 queries on average (or some 175 generations). If I understand you well, a pop. size of 9 and mut. rate of 0.00901 yields a minimal average number of queries?   There is an optimal mut. rate. But there is no optimal pop. size yielding a minimal number of trials. So the higher the pop. size, the lower the number of trials. Maybe it would be more useful if you only had the expected number of trials on the y-axis. Why take the number of queries?

DiEb Posts: 312 Joined: May 2008 (Permalink) Posted: Feb. 13 2017,05:40    Quote (WebHopper @ Feb. 13 2017,09:53) Quote (DiEb @ Feb. 12 2017,17:52) When I calculated the values nearly eight years ago, I concluded that the most efficient size of population would be 9 with a mutation rate of 0.00901: this would result in 1576 queries on average (or some 175 generations). If I understand you well, a pop. size of 9 and mut. rate of 0.00901 yields a minimal average number of queries?   There is an optimal mut. rate. But there is no optimal pop. size yielding a minimal number of trials. So the higher the pop. size, the lower the number of trials. Maybe it would be more useful if you only had the expected number of trials on the y-axis. Why take the number of queries? Frankly, I'm not sure what you mean by "trial"... The general idea is that one wishes to reduce the costs of a simulation: you can define the idea of costs in various ways - maybe there is a restriction to the size of a generation and / or the number of generations - but it is standard to define the number of calls to the oracle / evaluations of the fitness function as the cost of a program. That's why I displayed the number of queries, i.e., the number of individuals created for which the fitness function has to be evaluated. The optimal mutation rate depends on the size of the population - there is no overall optimum for all sizes!

WebHopper Posts: 7 Joined: Feb. 2017 (Permalink) Posted: Feb. 14 2017,06:45    Quote (DiEb @ Feb. 13 2017,05:40) Frankly, I'm not sure what you mean by "trial"... The number of trials is the number of tries. When you roll dices for instance, a trial is when you throw the dice once. Or for the Weasel algorithm, it's the number of generations. I calculated the probability distribution for the Weasel algorithm with a mutation rate of 0.05 and a population size of 100 according to Utiger's paper: P(v)=|H.F^(v-2).A| where v is the number of trials (or generations), H and F are matrices, A is a vector and |.| is the 1-norm. This yields As you can see, the numerical blue points and the analytical red curve perfectly fit. The mean calculated analytically is 79.19 and the standard deviation is 24.64. So the number of queries is 100*79.19 = 7919 according to your indications. On your graphic above however, the intersection point of the vertical line at 5e-02=0.05 with the green curve passing through the point 100... is about 2e+05=2*10^5. This is the number of queries if I understand you well. So there is disagreement with both results...

DiEb Posts: 312 Joined: May 2008 (Permalink) Posted: Feb. 14 2017,12:16    Quote (WebHopper @ Feb. 14 2017,12:45) Quote (DiEb @ Feb. 13 2017,05:40) Frankly, I'm not sure what you mean by "trial"... The number of trials is the number of tries. When you roll dices for instance, a trial is when you throw the dice once. Or for the Weasel algorithm, it's the number of generations. I calculated the probability distribution for the Weasel algorithm with a mutation rate of 0.05 and a population size of 100 according to Utiger's paper: P(v)=|H.F^(v-2).A| where v is the number of trials (or generations), H and F are matrices, A is a vector and |.| is the 1-norm. This yields As you can see, the numerical blue points and the analytical red curve perfectly fit. The mean calculated analytically is 79.19 and the standard deviation is 24.64. So the number of queries is 100*79.19 = 7919 according to your indications. On your graphic above however, the intersection point of the vertical line at 5e-02=0.05 with the green curve passing through the point 100... is about 2e+05=2*10^5. This is the number of queries if I understand you well. So there is disagreement with both results... Great - I will try to find out where my error laid...

WebHopper Posts: 7 Joined: Feb. 2017 (Permalink) Posted: Feb. 14 2017,12:55    Quote (WebHopper @ Feb. 12 2017,16:50) Utiger found a distribution like that of throwing dices: P(v) = q^v-1 p^v where v is the number of generations and p = 1-q is the probability that the dice got the correct number. Sorry for this error. The probability distribution for throwing a dice is of course P(v) = q^(v-1) p with p=1/6. You should definitely read Utiger's article even though his conclusion is that natural selection does not work and you guys believe in the exact opposite... But he makes it with honesty and clarity. Furthermore, he does not draw any conclusions in favor of intelligent design or any other creation beliefs. He just says that there is disagreement between the empirical data of 13 million years ago when the split between the Pan and Homo genera occurred and the data furnished by the Weasel-algorithm, that is, billions of years ago... The paper can also be found on ResearchGate.

k.e.. Posts: 5432 Joined: May 2007 (Permalink) Posted: Feb. 14 2017,21:16    If you really want to get a grip on the power and limitations of statistics I suggest Nate Silver's "The Signal and The Noise". My favorite quote: "Economists have predicted nine of the last six recessions". Berlinkski famously conflated  the mathematical meaning of the word limit with the limit of actions available in the physical world. Personal incredulity unfortunately is a limit even for Mathemeticians. If his work was truly useful he could see if the strategy worked in Poker. I predict the results would see him lose. I have yet to see any mathematician change the outcome of a dice rolled in the past. -------------- "I get a strong breeze from my monitor every time k.e. puts on his clown DaveTard suit" dogdidit "ID is deader than Lenny Flanks granmaws dildo batteries" Erasmus "I'm busy studying scientist level science papers" Galloping Gary Gaulin

Wesley R. Elsberry Posts: 4991 Joined: May 2002 (Permalink) Posted: Feb. 14 2017,21:42    I haven't gone through all of Utiger's math, but I can speak some to the biology interactions. Utiger attributes all change to natural selection. His paper is devoid of references to drift, and his only mention of Kimura is an offhand reference to mathematics. Thus, you can readily dismiss any conclusions he makes about plausibility of genetic change; he is not dealing with the full model of how genetics changes. Referencing Meyer as an authority on genetics is always a bad sign. -------------- "You can't teach an old dogma new tricks." - Dorothy Parker

Wesley R. Elsberry Posts: 4991 Joined: May 2002 (Permalink) Posted: Feb. 15 2017,12:06    Quote (DiEb @ Feb. 14 2017,12:16)   Quote (WebHopper @ Feb. 14 2017,12:45)   Quote (DiEb @ Feb. 13 2017,05:40) Frankly, I'm not sure what you mean by "trial"... The number of trials is the number of tries. When you roll dices for instance, a trial is when you throw the dice once. Or for the Weasel algorithm, it's the number of generations. I calculated the probability distribution for the Weasel algorithm with a mutation rate of 0.05 and a population size of 100 according to Utiger's paper: P(v)=|H.F^(v-2).A| where v is the number of trials (or generations), H and F are matrices, A is a vector and |.| is the 1-norm. This yields As you can see, the numerical blue points and the analytical red curve perfectly fit. The mean calculated analytically is 79.19 and the standard deviation is 24.64. So the number of queries is 100*79.19 = 7919 according to your indications. On your graphic above however, the intersection point of the vertical line at 5e-02=0.05 with the green curve passing through the point 100... is about 2e+05=2*10^5. This is the number of queries if I understand you well. So there is disagreement with both results... Great - I will try to find out where my error laid... Dieb, I don't think it is your error. The plot WebHopper has is for population size 100, and should have been run for population size 9 to speak to your numbers. -------------- "You can't teach an old dogma new tricks." - Dorothy Parker

DiEb Posts: 312 Joined: May 2008 (Permalink) Posted: Feb. 15 2017,17:26    Quote (Wesley R. Elsberry @ Feb. 15 2017,18:06) Quote (DiEb @ Feb. 14 2017,12:16)     Quote (WebHopper @ Feb. 14 2017,12:45)     Quote (DiEb @ Feb. 13 2017,05:40) Frankly, I'm not sure what you mean by "trial"... The number of trials is the number of tries. When you roll dices for instance, a trial is when you throw the dice once. Or for the Weasel algorithm, it's the number of generations. I calculated the probability distribution for the Weasel algorithm with a mutation rate of 0.05 and a population size of 100 according to Utiger's paper: P(v)=|H.F^(v-2).A| where v is the number of trials (or generations), H and F are matrices, A is a vector and |.| is the 1-norm. This yields As you can see, the numerical blue points and the analytical red curve perfectly fit. The mean calculated analytically is 79.19 and the standard deviation is 24.64. So the number of queries is 100*79.19 = 7919 according to your indications. On your graphic above however, the intersection point of the vertical line at 5e-02=0.05 with the green curve passing through the point 100... is about 2e+05=2*10^5. This is the number of queries if I understand you well. So there is disagreement with both results... Great - I will try to find out where my error laid... Dieb, I don't think it is your error. The plot WebHopper has is for population size 100, and should have been run for population size 9 to speak to your numbers. Nevertheless, I'll look into it over the next week. Heck, I liked the picture, but something seems to be off. I run a few simulations: WebHopper seems to be correct! Now, I have to go over eight year old code - from when I was young and pretty ;-)

DiEb Posts: 312 Joined: May 2008 (Permalink) Posted: Feb. 16 2017,10:04    Quote (DiEb @ Feb. 15 2017,23:26)   Quote (Wesley R. Elsberry @ Feb. 15 2017,18:06)     Quote (DiEb @ Feb. 14 2017,12:16)         Quote (WebHopper @ Feb. 14 2017,12:45)         Quote (DiEb @ Feb. 13 2017,05:40) Frankly, I'm not sure what you mean by "trial"... The number of trials is the number of tries. When you roll dices for instance, a trial is when you throw the dice once. Or for the Weasel algorithm, it's the number of generations. I calculated the probability distribution for the Weasel algorithm with a mutation rate of 0.05 and a population size of 100 according to Utiger's paper: P(v)=|H.F^(v-2).A| where v is the number of trials (or generations), H and F are matrices, A is a vector and |.| is the 1-norm. This yields As you can see, the numerical blue points and the analytical red curve perfectly fit. The mean calculated analytically is 79.19 and the standard deviation is 24.64. So the number of queries is 100*79.19 = 7919 according to your indications. On your graphic above however, the intersection point of the vertical line at 5e-02=0.05 with the green curve passing through the point 100... is about 2e+05=2*10^5. This is the number of queries if I understand you well. So there is disagreement with both results... Great - I will try to find out where my error laid... Dieb, I don't think it is your error. The plot WebHopper has is for population size 100, and should have been run for population size 9 to speak to your numbers. Nevertheless, I'll look into it over the next week. Heck, I liked the picture, but something seems to be off. I run a few simulations: WebHopper seems to be correct! Now, I have to go over eight year old code - from when I was young and pretty ;-) ARRRRGH! I don't know how I have managed it, but I uploaded two wrong pictures! Those belong to a blogpost from Oct 3, 2009 on Dembski's "Random Search" which had a target length of 100, and an alphabet of size 2 only! The day before I wrote about Dawkin's weasel - here are the correct pictures for Dawkins: The number of queries: The number of generations: i) for a population size of 100 and a mutation rate of 0.044, I get an expected number of generations of 78.47 ii) I still think that the number of individuals created best reflects the cost of the algorithm - a least mathematically: the onerous task is to create/mutate a child, using random numbers - in this sense, 10 generations of 10 children are equally expensive as one generation of 100 children. Edited by DiEb on Feb. 16 2017,16:09

Wesley R. Elsberry Posts: 4991 Joined: May 2002 (Permalink) Posted: Feb. 16 2017,11:32    Dieb: Quote ii) I still think that the number of individuals created best reflects the cost of the algorithm - a least mathematically: the onerous task is to create/mutate a child, using random numbers - in this sense, 10 generations of 10 children are equally expensive as one generation of 100 children. Yes, that is absolutely standard. -------------- "You can't teach an old dogma new tricks." - Dorothy Parker

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