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Subject: Re: Question about Bit storage

Author: Dann Corbit

Date: 15:54:07 01/28/02

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On January 28, 2002 at 18:44:15, Vincent Diepeveen wrote:

>On January 28, 2002 at 18:17:11, Dann Corbit wrote:
>
>that only shows how to store KRK as far as i see Dann,
>not a random position with nearly all pieces on the board.
>
>Not a single example of a full board position is inside the document.
>
>please encode next position for me, ignore castling rights doing it:
>
>nr3qrk/2QRp1Np/2p1Pp1n/2Pp3P/pp1P1K1P/3B1P2/PP1BNbp1/R7 w - - 0 1

I assume that you can read his simple encoding system.  Now, take that method
and compose the position for it.  Then consider that that position (together
with its eval, ce, pv, etc) are identical to these, if you have read "Through
the Looking Glass":

7r/1PBnb1pp/2p1b3/p1k1p1PP/p3Pp2/N1Pp1P2/Pn1Prq2/KRQ3RN b - -
krq3rn/pN1pRQ2/n1pP1p2/P3pP2/P1K1P1pp/2P1B3/1pbNB1PP/7R w - -
nr3qrk/2QRp1Np/2p1Pp1n/2Pp3P/pp1P1K1P/3B1P2/PP1BNbp1/R7 w - -
r7/pp1bnBP1/3b1p2/PP1p1k1p/2pP3p/2P1pP1N/2qrP1nP/NR3QRK b - -

Since we encode 4 positions and need to store only 1 (the one that is lexically
smallest on top) we divide the number of bits needed by 4.  It is a trick so
simple that I am surprised anyone would not grasp the notion instantly.



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