Re: Whirlpool 512-bit collisions?
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Re: Whirlpool 512-bit collisions?

From: Alan <a__l__a__n@hotmail.com>
Date: Wed Jun 15 2005 - 19:40:23 CEST



Gregory G Rose wrote:


> Assuming that the hash function acts as a perfect


> random function, about 1/e ~= 0.37 of the


> possible outputs won't appear.



Are you speaking of W or of complete Whirlpool?  That seems like a


reasonable simplifying assumption (ie perfect random function) for W but not


for Whirlpool.  Assuming that is what you are saying..



Let P(M) = C be a perfect random permutation of M.


Then, for


   H = P(M) XOR M


Given 2^512 possible M, you are saying there would be (2^512)/e different


outputs H.  Correct?



How do you arrive at 1/e?



Alan


Received on Thu Sep 29 21:44:02 2005