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Subject: Re: Hash Collisions (Bob Hyatt)

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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