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)?
> [...] 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
[ reply ]