Author: Dieter Buerssner
Date: 01:57:19 03/03/05
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On March 03, 2005 at 02:39:16, Reinhard Scharnagl wrote: >there might be possibilities to store chess positions in some few less bits if >always THE COMPLETE PIECESET is used for both sides. But I hardly could imagine >that 100 bits should be sufficient even then. Only little more than 100 bits will be needed for positions with all pieces on the board. On each file, there will be 2 pawns. There are only 10 possible pp configurations. Of these we can count 2 twice for a possible ep-target, so we have 12. There are less than 4096 KK-configurations including castling rights (I forgot the actual number). So only 14 pieces are missing. Even if I don't use the fact, that both bishops will have different square color, I get: 2*12^8*4096*46!/(32!*2*2*2*2*2*2) = ~2^110 46!/32! is for the 14 remaining pieces on 46 remaining squares. Six times the factor 2 is for the indistinguishable pieces. Factor 2 at the start for side to move. I doubt, that this can be reduced by many orders of magnitude. 100 bits might be possible. Regards, Dieter
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