Author: Dann Corbit
Date: 18:55:27 10/19/99
Go up one level in this thread
On October 19, 1999 at 21:52:44, Dann Corbit wrote:
>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.
You can also eliminate one question for the first and eighth ranks, compressing
further(possibly) since you don't need to find out about pawns.
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