Author: Dann Corbit
Date: 18:52:44 10/19/99
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On October 19, 1999 at 21:44:24, Dann Corbit wrote:
>We interpret sequences of bits as follows:
>0 == go to next square
>1111 = king, white
>1011 = king black
>1101 = white bishop
>1001 = black bishop
>11101 = rook, white
>10100 = queen, black
>11001 = pawn white
>10000 = knight, black
>11100 = queen, white
>10101 = rook, black
>11000 = knight, black
>10001 = pawn, black
>
>We arrive at these encodings as follows:
>if square occupied, then first bit 1
>{
>if color white, then next bit 1, else 0
>if man > bishop then next bit 1, else 0
>if greater than Q/N[1] next bit 1, else 0
>if less than Q/N[2] next bit 1, else 0
>}
>else
>0
>
>[1] if we know it is greater than a biship second question is for queen, else
>knight
>[2] if we know it is greater than a biship second question is for queen, else
>knight
>
>We can improve the mapping in a general sense by giving pawns the value of a
>king and queens the value of a pawn (remember, this is just for mapping, not for
>play). Hence, the common pieces (white and black queens and pawns) will take
>only 4 bits.
This would give 174 bits for the initial position, since we would have:
7*5+4 (7 five bit pieces + queen=4bits)
8*4 (8 four bit pawns)
8 (blank row)
8 (blank row)
8 (blank row)
8 (blank row)
8*4 (8 four bit pawns)
7*5+4 (7 five bit pieces + queen=4bits)
= 39+32+32+32+39=174 bits
Since you have to remove (at least) pawns from the board to promote a pawn (and
a minimum of 4 pawns or pieces to promote them all) I think that may be the
maximal encoding for this scheme.
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