Author: Dan Andersson
Date: 11:01:20 03/29/04
Go up one level in this thread
> >in any case, you can now assume that you find the best move 1st in the fraction >F (~0.9), and in the remaining case at random, this leads to > >nodes(depth) ~ (N*(1*F+(N/2*(1-F)))^D/2. This is a flat distribution for the rest of the moves. A reasonable argument can be made for some other kind of distribution. F.ex.: The next move has the success rate of F. Thus the mean number of moves tried will be close to one for any reasonably large F. MvH Dan Andersson > >plugging in some numbers like N=20 (reduce from 40, trying to account for >nullmove?!), D=10, F=0.90 and 0.91 i get > >nodes(10, F=0.90) = 79 million >nodes(10, F=0.91) = 62 million > >so for 1% better move ordering i get 22% less nodes?! seems rather excessive. >let me try again using sqrt(N) instead of N/2 - that still gets a 12% reduction. >
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