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Robert J. Hansen wrote:
> Alphax wrote:
>
>>- Some hash algorithms are "broken"; it
is possible to generate the
>>inputs for a given input (of a given size), reducing
the number of
>>"possible" inputs by some amount - I'm
guessing by the size of the hash.
>>So even though there are still however many milllion
possible files with
>>that filesize, how many /also/ have that particular
hash value for that
>>particular algorithm?
>
>
> Depends a lot on how they're broken. Broken
algorithms have properties
> very far from the Platonic ideals which make the math
work out nicely.
>
Ok, I wondered if that was the case.
>
>>- What if we have two hash values from different
algorithms? Three?
>>Four? N? Does this make it any easier?
>
>
> Assuming you have N different perfect M-bit hash
algorithms and they're
> all perfectly independent of each other, you've
recovered (N*M) bits.
> Again, if you assume they're not perfectly
independent, your mileage
> will vary--however, it's far more likely you'll
recover fewer than (N*M)
> bits than you'll recover more. (And by 'far' more
likely, I mean 'I
> think Claude Shannon would like to have a talk with you
if this ever
> happens'.)
>
Well, assuming we have two 512-bit hashes which are
perfectly unique,
we've just recovered a 256kB file...
But I see what you're getting at. As the subject line says,
this was a
thought experiment. Like most others, it's been disproved
fairly
quickly. However, this is why we experiment - so that we can
learn.
- --
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