Author: Larry Griffiths
Date: 12:09:52 02/24/99
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On February 23, 1999 at 16:40:50, Robert Hyatt wrote: >On February 23, 1999 at 14:33:03, Larry Griffiths wrote: > >>Dan, >> >>I read your post on hash collisions. I have tried some hamming distance >>code and I see little difference when compared to generating random >>numbers. I have been thinking of writing code that does what you said >>where 64 bits must have 32 bits (or 50%) turned on in each number. I also see >>where there are many references to the 12 pieces times 64 squares = 768 >>hash codes. Since the a Bishop can only control 32 possible squares and >>there are 4 bishops, then 128 of the 768 are unused. Also, Pawns only >>use 48 squares for each side, so 32 squares are unused by pawns. >>768-128-48 leaves 592 hashcodes for the piece square table. >>Food for thought? Errors in my thinking? >> >>Larry. > > >A couple. First there are 12 _types_ of pieces. And since each side has >two bishops on opposite colors, all 64 random numbers are needed for the >bishops. > >you can get away with leaving 16 out of the pawn random numbers since none >exist on the 1st/8th rank. There are not 4 sets of 64 numbers for the 4 >bishops. There are two sets of 64, one for black bishops, one for white >bishops. Because the bishops are not 'unique' and are interchangable. In >fact, after promoting to a B, that B is identical to the B that existed early >in the game on the same color square. > >So you _could_ reduce this to 752 numbers... but then you need a few more >for things like castling status, enpassant status, so you take those back >again. Bob, I can get away with leaving 16 out of the pawn random numbers and I have two pawn tables so 16 * 2 = 32. So I could reduce this to 768-32 or 736? I also saw a post that your table is 1024 entries. What are the other 4 tables (256 entries) used for? Thanks for your reply. Larry :-)
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