Wesley R. Elsberry
Posts: 4945 Joined: May 2002

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^(v2).A
where v is the number of trials (or generations), H and F are matrices, A is a vector and . is the 1norm. 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 5e02=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
