On Wed, 11 April 2007 16:23:21 -0700, H. Peter Anvin wrote:
> David Lang wrote:
> >On Thu, 12 Apr 2007, Neil Brown wrote:
> >
> >>For the second.
> >> You say that you " would need at least 96 bits in order to make that
> >> guarantee; 64 bits of hash, plus a 32-bit count value in the hash
> >> collision chain". I think 96 is a bit greedy. Surely 48 bits of
> >> hash and 16 bits of collision-chain-position would plenty. You would
> >> need 65537 entries before a collision was even possible, and
> >> billions before it was at all likely. (How big does a set of 48bit
> >> numbers have to get before the probability that "No subset of 65536
> >> numbers are all the same" drops below 0.95?)
> >
> > you can get a hash collision with two entries.
>
> Yes, but the probability is 2^-n for an n-bit hash, assuming it's
> uniformly distributed.
>
> The probability approaches 1/2 as the number of entries hashes
> approaches 2^(n/2) (birthday number.)
I believe you are both barking up the wrong tree. Neil proposed a 16bit
collision chain. With that, it takes 65537 entries before a collision
chain overflow is possible.
Calling a collision chain overflow "collision" is inviting confusion, of
course. :)
Jörn
--
The competent programmer is fully aware of the strictly limited size of
his own skull; therefore he approaches the programming task in full
humility, and among other things he avoids clever tricks like the plague.
-- Edsger W. Dijkstra
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