Invisible wrote:
> Now, here's a thought. MD5 produces a 128-bit hash value. So what are
> the changes of two random datums having the same hash?
>
> 2^128 = 3.40 * 10^38
And yet, it's not possible to just keep a record of MD5 hashes and
backward-compute the file that generated it (or even a reasonably small
list and a note as to which one should be used) (and save an awful lot
of storage space).
:(
--
Tim Cook
http://empyrean.freesitespace.net
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