Author: Dann Corbit
Date: 20:23:32 10/19/99
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On October 19, 1999 at 22:42:26, Dann Corbit wrote: >men values for collation: >b=0, n=1, p=2, r=3, k=4, q=5 >Questions used: >bit0 = occupied? If zero, skip to next sequence, else: >bit1 = color == white >bit2 = greater than pawn? >bit3 = greater than k/n[1] >bit4 = less than k/n[2] > >[1] If greater than pawn, then bit3 question is about king, else knight >[1] If greater than pawn, then bit4 question is about king, else knight > >encodings are as follows: >0 == blank square >1111 == white queen >1011 == black queen >1101 == white pawn >1001 == black pawn >11101 == white rook >10100 == black king >11001 == white bishop >10000 == black knight >11100 == white king >10101 == black rook >11000 == white knight >10001 == black bishop > >Opening board is: >101011000010001101110100100011000010101 >10011001100110011001100110011001 >00000000 >00000000 >00000000 >00000000 >11011101110111011101110111011101 >111011100011001111111100110011100011101 >It would also require castling, ep, and stm which is 4+2+1 7 more bits = 181 >bits >:-( >But I have not thought about squeezing the first and last rows >:-) > >Perhaps some of you clever fellows can crunch it better than I have. We have 174 bits for the pieces, and since we don't have a pawn location question for the first and last rows, we can use that question to squeeze out the ep square, and then add 4 bits for castle and one for stm = 179 bits
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