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?)
Neil,
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.)
-hpa
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