Author: Janosch Zwerensky
Date: 06:53:47 09/23/01
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>I use the estimate 1/2^64+2/2^64+3/2^64+... 2^32/2^64~=1/2 P.S.: A relatively good efficiently computable lower bound for the probability of getting two equal hash-keys from N randomly generated positions should be (for N that are small compared to 2^64) something like 1-(1-1/2^64)^(N*(N-1)/2) according to my calculations. Regards, Janosch.
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