Author: J. Wesley Cleveland
Date: 12:10:33 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. 2^84 = 1.9*10^25 If these were atoms, it would be 32 moles, or about 1kg of silicon.
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