NODE ab834cf7Re: DES
Joe Thomas <jthomas@pawpaw.mitre.org>Thu, 14 Oct 93 11:16:58 PDT
> > 2) How much longer would it take to break triple DES versus
standard DES
> > using one of the key-breaking machines described?
>
> Using brute force, it would take the cube of the
> time it takes to break single DES.
Hmm... I can't figure out what it would mean to cube time. For
two-key (112 bit) triple DES, it should be 2^56 times longer to
exhaustively search the keyspace, with three keys, 2^112 times
longer. This assumes the keysearch engine is pipelined, so each
trial encryption takes the same amount of time, despite the
additional rounds. Such a machine would cost more, of course, since
it would have a longer pipeline, but wouldn't otherwise be
significantly different in design.
Back-of-the-envelope calculation: a design like the "7-hour
exhaustive keysearch" engine for 2-key triple DES would take 50
trillion years or so to exhaust the keyspace. That's for a cost on
the order of $1 million (it should be buildable for less than three
times the cost of the 56-bit key version).
Seems secure, but as Perry says,
> Whether a more sophisticated techinque is possible is unknown.
Joe
NODE 358a0e33Re: DES
"Perry E. Metzger" <pmetzger@lehman.com>Thu, 14 Oct 93 11:51:59 PDT
Joe Thomas says:
> > > 2) How much longer would it take to break triple DES versus
> > > standard DES using one of the key-breaking machines described?
>
> > Using brute force, it would take the cube of the
> > time it takes to break single DES.
>
> Hmm... I can't figure out what it would mean to cube time. For
> two-key (112 bit) triple DES, it should be 2^56 times longer to
> exhaustively search the keyspace, with three keys, 2^112 times
> longer.
Lets assume we are using three keys, which I what I meant. Lets say 1
is the time do one encryption. (On a parallel machine, just think of
things as being on a uniprocessor going N times faster.) It would take
2^56*N time to break single DES. My claim is that it should take
(2^56)^3 = 2^56*2^56*2^56 = 2^168. Your claim, which is that it would
take 2^56*2^112=2^168, which is the same. The only difference is that
I didn't assume piplineing so there is a constant factor different
floating around somewhere.
Perry