Author: KarinsDad
Date: 12:15:00 10/20/99
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On October 19, 1999 at 21:13:16, Dann Corbit wrote: >Here's a sort of RLL notion... >If the first bit in a sequence is zero, then the square is blank. Else, >interpret the next five bits as follows: >10000 - black king >11000 - white king (white to move) >11111 - white king (black to move) >10001 - black queen >11001 - white queen >10010 - black rook >11010 - white rook >10011 - black rook (may castle) >11011 - white rook (may castle) >10100 - black knight >11100 - white knight >10101 - black bishop >11110 - white bishop >10110 - black pawn >11110 - white pawn >10111 - en passant pawn > So, 64 bits bitboard + 4 bits per piece means that this representation can be used whenever there are: (160 - 1 (bit to indicate algorithm) - 1 bit for side to move - 64) / 4 = 23 or fewer pieces on the board when you want to compress this below 20 bytes. Now this would be a good thing except for one other issue: The minimum number of pieces on the board when handling the maximum number of possible promotions is 28 (i.e. if all possible pawns were promoted and no pieces were captured, there would be 28 pieces on the board). Hence, this algorithm does not help (for the 20 byte attempt) over other algorithms since it can only handle 23 bytes (whereas the smallest maximum promotion cases require a minimum of 28 bytes). In other words, out of the many different multiple promotion cases (the cases that take up so many bits in the first place), this algorithm would not help out on any of them (which if it could, it could possibly enable more compression on one of the other algorithms). Shoot. Another one bites the dust. KarinsDad :)
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