[rc5] key security question

David McNett (dmcnett at hfdirect.com) dmcnett at hfdirect.com
Thu Jun 5 17:57:56 EDT 1997


> > >Think keyspace: for each first key you have to search all the 2^56 second
> > >keys. This makes 2^112 combinations, IOW a 112-bit key. This presumes that
> > Not true, the work is 2^56 for each, for 2^57 total.
>
>Hmm, well - if I just look at what a brute-force approach has to do, I
>think 2^57 is wrong. For single encryption, we'll have to check 2^56 keys.
>For double encryption this should mean having to check 2^56*2^56 keys i.e.
>2^112 keys - yeah, it looks like it would be 2^112 keys...

The question is, can the brute force decryption effort "know" when it has successfully
cracked the outer layer.  If so, then you're talking about simply doubling the effort
involved in the decryption effort (i.e. the equivalent of adding one bit to the keysize).

If you presume that the decryption effort must try every possible combination of key
for both inner and outer layer, then yes it would be the equivalent to a keysize equal to
the combined keyspaces of each layer.

The question is, aside from the obvious security through obscurity, what is the real
benefit to the added complexity?

-Dave
 nugget at slacker.com
-------------- next part --------------
A non-text attachment was scrubbed...
Name: not available
Type: application/ms-tnef
Size: 1935 bytes
Desc: not available
Url : http://lists.distributed.net/pipermail/rc5/attachments/19970605/e6e79eaf/attachment-0001.bin


More information about the rc5 mailing list