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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