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/
http://www.zazzle.com/MathematicalOrchid*
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