Author: Robert Hyatt
Date: 15:37:38 10/19/99
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On October 19, 1999 at 17:45:39, Inmann Werner wrote: >On October 19, 1999 at 07:47:56, Robert Hyatt wrote: > > > >>it is simply the 64 bit hash signature... I use the rightmost N bits as the >>probe address, and store the entire 64 bit signature as the 'checksum' as you >>call it... >> >>>My question is: with, at most, 128bits per position (that I'll never would use >>>in fact), isn't it possible to make mistakes? I know the probability is small, >>>but eventually there could be a very strange move coming from a hash table >>>mistake. Is this true? What am I missing? >> >>mistakes are possible. But they are _very_ improbable using 64 bits. The >>way to confirm this is to have a debug mode where you do store the simple 256 >>bit actual board position. When you get a match on the 64 bit value, then >>check the full 256 bit value. If it doesn't match, you have a false match. I >>ran such a test for many hours and got zero. >> >>>Is, anyway, safe to use a 64bit checksum? And if so, is the checksum generated >>>the same way as the hash code, but with different random numbers? >> > >Do we really need a 64 bit checksum? >How much risk a 32 bit checksum would be, you think? > >Werner If you mean "can I compute a 64 bit hash signature, use the right end for the probe address, and only store the left end?" then the answer is maybe. I don't like the collision rate I get when I test with that. If you mean "Can we just use a pure 32 bit hash signature?" the answer is absolutely not...
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