Author: Uri Blass
Date: 11:35:18 05/16/01
Go up one level in this thread
On May 16, 2001 at 14:07:18, Robert Hyatt wrote: >On May 16, 2001 at 13:05:13, J. Wesley Cleveland wrote: > >>On May 15, 2001 at 22:11:15, Robert Hyatt wrote: >> >>>On May 15, 2001 at 12:18:43, J. Wesley Cleveland wrote: >>> >>[snip] >> >>>>>First, how do you conclude 10^25? assuming alpha/beta and sqrt(N)? >>>> >>>>It is a classic alpha-beta search with a transposition table large enough to >>>>hold *all* positions found in the search. I'm guessing at the number of >>>>positions, but I feel that the same logic should hold, as only positions with >>>>one side playing perfectly would be seen. >>> >>>I don't follow. We know that within the 50 move rule, the longest game that >>>can be played is something over 10,000 plies. IE 50 moves, then a pawn push >>>or capture, then 50 more, etc. Eventually you run out of pieces and it is a >>>draw. But 38^10000 and 10^25 seem to have little in common. The alpha/beta >>>algorithm is going to search about 38^50000 nodes to search that tree to max >>>depth of 10,000. >> >>Look at it another way. The only positions that are visited by an alpha/beta >>search (with perfect move ordering) are those where one side plays perfectly. >>The question is what fraction of the total number of positions that is. >> > >The precise formula is: > > N = W^floor(D/2) + W^ceil(D/2) for all D. floor means round down in integer >math, ceil means round up. For the cases where D is even: > > N = 2 * W^(D/2) which is 2 * sqrt(minimax). > >If you assume that the total number of positions is roughly 2^168, then you >get 2 * sqrt(2^168) or 2 * 2^84. Which is fairly close to the number of atoms >in the universe. Note that 168 is not cast in stone either. It might be a >few bits more or less, but it is probably close. It is proved to be clearly less than 2^168. I believe my counting program proved that it is less than 2^160 but I have not the numbers near me. I guess it is between 2^140 and 2^150. I had an idea how to get a good estimate for it by a program but nobody tried to calculate an estimate for it. My idea is simply to generate a lot of random pseudo legal positions(1000000 is enough) and try to find the number of legal positions. In order to generate pseudo legal position you need to the following steps: 1)Calculate the number of pseudo legal positions for every possible legal material structure 2)generate a random pseudo legal position. 3)check if the pseudo legal position can be achieved by a chess game. If you find that the number of pseudo legal positions is 10^47 and you also find that 173 out of 1000000 pseudo legal positions are legal then 10^47*173\1000000 is a very good estimate for the number of the legal positions. (If you have enough pseudo legal positions that are legal you can be sure with 95% confidence that the mistake in the estimate is less than 10%) It seems that nobody is really interested in the number of legal positions so we are not going to know a good estimate. Uri
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