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Subject: Re: Computational question for mathematicians, philosophers & computer-geeks

Author: Reinhard Scharnagl

Date: 02:18:27 03/03/05

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On March 03, 2005 at 04:57:19, Dieter Buerssner wrote:

>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.

Hi Dieter,

in each file there will be 15 possible valid pP configurations, not only 10:

01: -p----P-
02: -p---P--
03: -p--P---
04: -p-P----
05: -pP-----
06: --p---P-
07: --p--P--
08: --p-P---
09: --pP----
10: ---p--P-
11: ---p-P--
12: ---pP---
13: ----p-P-
14: ----pP--
15: -----pP-

Reinhard.



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