Tim Cook wrote:
> 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).
> 
> :(

Indeed. For although the number of possible MD5 hashes may be utterly 
vast, the number of possible 4GB files is so MIND-PARALYSINGLY GIGANTIC 
that you can't shoe-horn all the members of one set into the other.

-- 
http://blog.orphi.me.uk/
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