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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