Author: KarinsDad
Date: 12:15:31 05/19/99
Go up one level in this thread
On May 19, 1999 at 11:32:51, Heiner Marxen wrote: >On May 18, 1999 at 08:02:09, Peter Klausler wrote: > >>On May 17, 1999 at 21:15:59, Dann Corbit wrote: >> >>>You would also need to store e.p. square, if any, I think. Ten bits more >>>[5ep+4castle+1stm]gives 174 bits. >> >>Two bits are all you need for the en passant file, actually. >>There can be at most four pawns of the opponent on his >>fourth rank next to a pawn of the player (on his fifth). > >Wrong. There can be 5 such pawns. E.g. 8/8/8/8/PpPPpPPp/8/8/8 b - >Even when we know already that there is an e.p. possible, we still >have to distinguish up to at least 5 cases. No, you only need 1 ep bit (based on the structure that this original statement was made). Because, if ep exists, then you can add 4 more bits to the structure to represent the square involved (3 bits) and the direction of capture (1 bit, a file to b file, or b file to a file type of thing) and you can remove 6 bits from the original structure for the pawn that can be taken by ep and the pawn that can take (if you have 2 pawns that can take, you do not care since the other one still shows up in the original structure). Hence, an en-passant case takes up 2 bits less (+4 -6) than a non-en-passant case. So, only 1 ep bit is needed. KarinsDad :)
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