From: Roger Bryner <bryner@atlas.chem.utah.edu>
Date: Mon, 4 Jul 94 15:30:59 PDT
To: wcs@anchor.ho.att.com
Subject: Re: Password entropy
In-Reply-To: <9407042147.AA17444@anchor.ho.att.com>
Message-ID: <Pine.3.89.9407041650.A7942-0100000@atlas.chem.utah.edu>
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<thanks for the analysis above>
On Mon, 4 Jul 1994 wcs@anchor.ho.att.com wrote:
> If you still *are* worried about it, however, you can scramble things a bit;
> since MD5 produces 128 bits of output but uses 448 bits of input+padding,
> you can add a different constant to the input at each step.
> If you're using it as a salt, put it at the beginning; if you're
> just doing it for multiple iterations it doesn't matter much.
This is not correct.  You still have the same problem that you don't know 
if the transformation is 1=>1.  You have added a lot of "psudo-random" 
stuff but unless you keep this in your head, it is laying around for your 
oppenent to grab(assuming non-secrecy of the algorithim).

Assuming a random function for MD5, it is simple to calculate the loss of 
entropy by calculating the number of collisions on adverage(intigrate the 
probilility of n collisions) and assumeing indipendence between rounds.  

I might point out that a better "buisy work" function would be to use to 
output of a RNG as a key for multiple idea incryptions, or some such 
scheme as this, as you are guarenteed of not loosing any entropy if you 
can (theoretically) decrypt the result.

The problem with such a "buisy work" function is that it sould be hard to 
simplify, ie xoring with the sequence 1010101010101010101010101... is 
easy to calculate dirrectly, without going through all the steps.  This, 
I would guess, gets into a whole other ball of wax.

Roger, Mad Dog Libertarian, Bryner.