Author: Dann Corbit
Date: 11:55:07 05/16/01
Go up one level in this thread
On May 16, 2001 at 14:35:18, Uri Blass wrote: >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. One way to get an absoulute upper bound is by a minimum encoding. I believe KarinsDad had an encoding that was 164 bits in length (IIRC). If true, then there cannot possibly be more than 2^164th chess positions, and surely less than that since some fraction won't be legal. But as an absolute upper bound, we can say a correct encoding that enumerates all chess positions in n bits implies with certainty that there cannot be more than 2^n possible states and hence positions to that vector. If (for instance) you could somehow encode a chess position that would work for any board position into 100 bits, that would prove that 2^100 is the largest possible number of board positions.
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