[Q] 128-bit symmetric encryption.

Daniel Carrera dcarrera@math.umd.edu
Sat May 31 21:23:02 2003

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On Fri, May 30, 2003 at 09:33:25PM -1000, Denis McCauley wrote:

> Strictly speaking, no. In reality a brute force attack would rarely, if
> ever, have to try 2^128 combinations to break the encryption because
> it's a question of probability. An attacker has a 50-50 chance of
> breaking it in half that number. But the probability of breaking the
> encryption in a lifetime is so remote that it's unlikely a pure brute
> force attack like that would be tried.

Yes, I'm aware of that.  But what I said is that a 128-bit key takes 2**64=
times longer to crack by brute force than a a 64-bit key.  The 50% factor=
gets accounted for in the time that it takes to brute force a 64-bit key.

But that's getting a little off toppic.  What I'm trying to find out is if=
there is any point at all in having a key length longer than 128 bits.

In other words, the key length is not an issue.  Any fault in the=20
encryption will be elsewhere (e.g. algorithm choice, implementation,=20
physical security, etc).

Daniel Carrera         | OpenPGP fingerprint:
Graduate TA, Math Dept | 6643 8C8B 3522 66CB D16C D779 2FDD 7DAC 9AF7 7A88
UMD  (301) 405-5137    | http://www.math.umd.edu/~dcarrera/pgp.html

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