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Subject: Re: Hash Collisions (Bob Hyatt)

Author: Dan Newman

Date: 15:06:48 02/24/99

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On February 24, 1999 at 17:01:45, Robert Hyatt wrote:

>On February 24, 1999 at 15:09:52, Larry Griffiths wrote:
>
>>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   :-)
>
>I think that was Don's post about 1024.  I have exactly 12 * 64 entries in
>mine...  and yes you can safely leave those table entries 0 since they will
>never be used anyway...

I think I may have been the source of that bit of disinformation.  I use
your piece numbering scheme in my 0x88 program (p=1,n=2,k=3,b=5,r=6,q=7)
and use piece number to index my table of random numbers...

-Dan.

>
>but don't forget you need to handle enpassant and castling privileges as well
>for both sides...



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