1993-10-14 - Re: DES

Header Data

From: “Perry E. Metzger” <pmetzger@lehman.com>
To: cypherpunks@toad.com
Message Hash: 358a0e33cbfdcdc562c02f5d34382e74be3e0eade95aa04bbe155dcde6938200
Message ID: <9310141850.AA11890@snark.lehman.com>
Reply To: <9310141814.AA14779@pawpaw.mitre.org>
UTC Datetime: 1993-10-14 18:51:59 UTC
Raw Date: Thu, 14 Oct 93 11:51:59 PDT

Raw message

From: "Perry E. Metzger" <pmetzger@lehman.com>
Date: Thu, 14 Oct 93 11:51:59 PDT
To: cypherpunks@toad.com
Subject: Re: DES
In-Reply-To: <9310141814.AA14779@pawpaw.mitre.org>
Message-ID: <9310141850.AA11890@snark.lehman.com>
MIME-Version: 1.0
Content-Type: text/plain



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





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