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Subject: Re: hash collisions

Author: martin fierz

Date: 03:33:05 02/01/01

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On February 01, 2001 at 05:57:58, Tony Werten wrote:

>On January 31, 2001 at 08:13:52, martin fierz wrote:
>
>>hi,
>>
>>i recently corrected some code in my connect 4 program and now it is able to
>>solve connect 4 in less than a day on a fast PC with a large hashtable (of
>>course, connect 4 has been solved long ago). i tried again with a smaller
>>hashtable and
>>got some strange results, for very long searches (billions of nodes) i don't get
>>the same value for
>>the root position. for not-so-deep searches i get the same values for both
>>versions. i am wondering, [if this is not just a bug :-)] could this be some
>>hashcollision-problem? can anybody give me a probability for a hash collision
>>occuring when using B-byte key & lock, with a hashtable with S entries after
>>searching N nodes? (i am using 4 byte ints for the key and the lock)
>>also, i think i remember somebody mentioning here that one can choose the random
>>numbers for the XORs in a clever way making hashcollision probabilities
>>smaller - can somebody tell me how?
>
>Ideal would be if every number would change half of the number of bits.
>
>Currently I'm working on connect 4,5,6 and 7, using hash scheme of 64bits. I
>have found no problems. The advantage of using zobrist is you can handle
>mirroring very easy. You just have to keep 16 hash values and store the lowest.
>It sounds slow, but getting a hashhit saves such a lot of nodes that it's worth
>it.
you are probably right. i used symmetries to solve solitaire a long time ago
with exactly that scheme - i thought in connect4 they would be less important
since there is only 1 symmetry operation (flip the board left/right), but i
never tested it. why do you use 16 values?


>Using this hashscheme and some intelligent recognisers (recognizing
>win-in-5-moves is a big winner), solving 4 in a row shouldn't cost you more than
>2M nodes (with boards >= 6 by 5, wich is the minimum size for a first player
>win).
i'm using the standard 7x6 board. i can't really imagine solving that with
2Mnodes - what makes you think you can do it with so few nodes?

cheers
  martin



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