Author: Uri Blass
Date: 07:22:00 09/23/01
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On September 23, 2001 at 09:43:59, Janosch Zwerensky wrote: > >>I use the estimate 1/2^64+2/2^64+3/2^64+... 2^32/2^64~=1/2 > >I think 1/2^64*(sum(2^i,i=0..32)=1/2^64*(2^33-1)=2^(-31)-2^(-64) is closer to >the truth. I used 1/2^64*(sum i,i=1,2,3,4,5,6,7...2^32-1) and not sum(2^i) and I get 1/2^64*[(2^32)*((2^32)-1)/2] that is almost 1/2 >That said, I don't really see that the left part of your equation is in any way >related to the problem discussed here either. I will explain. 1/2^64 is the probability for hash colission in the second position. 2/2^64 is the probability for hash collision in the third position 3/2^64 is the probability of hash collision in the 4th position. (2^32-1)/2^64 is the probability of hash collision in the last position. The sum of these probability is exactly the expected number of the hash collisions. This formuala is not exactly correct because I assume that there are no hash collisions in calculating the probabilities. If there are hash collisions then it mean that the sum is smaller but the number of hash collisions are so small that it is not important because one hash collision can reduce this sum by only less than 1/(2^32). Regards, Uri > >Regards, >Janosch.
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