Author: Dieter Buerssner
Date: 09:55:49 11/19/01
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>Need to know what the minimum number of bits that are needed to store the worse >case scenario for a pv (ie Qa6xb7# or fxg1=Q+). As you can see both these >strings are 7 characters long and represent the longest possible situation. If >someone is aware of how to represent these 2 conditions with the fewest number >of bits I would appreciate it. And, to answer in another way. If you just want to store the string rather dense, and without any knowledge of the board position, the following should work. The SAN consists of the following: Piece charcter: nothing (=P), N, B, R, Q, K: 6 states. File of the moving piece: nothing or a-h: 9 states. Rank of the moving piece: nothing or 1-7: 9 states. Capture: nothing or x: 2 states. destination square: 64 states. promotion square: nothing, N,B,R,Q: 5 states. Check: nothing, check, checkmate: 3 states. So we get 6*9*9*2*64*5*3 as an upper bound. This is 93312, so it will need 17 bits at most. Again, I think, it can be made even more dense, because this has much redundancy (only pawns can promote, but pawns will never need file and rank, etc.) Regards, Dieter
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