On Thu, 2005-02-17 at 16:34 -0500, Scovetta, Michael V wrote: > [...] And due to recent discoveries, we can > push those down to 2**50 and 2**55 respectively. Breaking a composition > would still take on the order of 2**55 (the harder of the two) If the two algorithms are different, finding a collision in one of them does not deliver a working collision in the other, no? Don't you have to find a "common" collision between the two? Wouldn't that require an effort of 2**(50+55)? Regards, Frank
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