1993-10-18 - Re: crypto technique

Header Data

From: Karl Lui Barrus <klbarrus@owlnet.rice.edu>
To: cypherpunks@toad.com
Message Hash: a4507afec9111d87a9277d44a196385aaeba99fbf8927d4283aaa1f9bb254f75
Message ID: <9310181535.AA27722@great-gray.owlnet.rice.edu>
Reply To: <kgkO9UG00awF4Pr0lK@andrew.cmu.edu>
UTC Datetime: 1993-10-18 15:37:15 UTC
Raw Date: Mon, 18 Oct 93 08:37:15 PDT

Raw message

From: Karl Lui Barrus <klbarrus@owlnet.rice.edu>
Date: Mon, 18 Oct 93 08:37:15 PDT
To: cypherpunks@toad.com
Subject: Re: crypto technique
In-Reply-To: <kgkO9UG00awF4Pr0lK@andrew.cmu.edu>
Message-ID: <9310181535.AA27722@great-gray.owlnet.rice.edu>
MIME-Version: 1.0
Content-Type: text/plain


Matthew J Ghio wrote:

>In summary, I see no method which would yeild the original input without
>knowing the values added to the nested polynomials (the private key),
>and there is no way to determine the private key if the modulus is
>applied to the resulting function.

I've been thinking about this (well, not too much since I'm in the
midst of midterms week :-).

I beleive the equation leaks information.  When you expand the
equation symbolically, it is easy to solve for the constants by
matching the coefficients of the highest powers and working backwards.
If the constants can be negative as well as positive, the signs of
some of the terms will reflect this.

We know the magnitude of the constants must be less than P, which is
public.  But can they be negative - will the decoding process still
work?  Or, will you obtain the correct decoding for the correct choice
and an incorrect decoding for the incorrect choice?  If it turns out
that either choice will decode a number to the same value, or if the
decoding won't work with negative numbers, then this method is too
easy to invert.

If the constants can't be negative, or if they can be but it doesn't
make a difference in the decoding, then taking the modulus doesn't
obscure anything at all.

For example, suppose c = 127 and d = -225.  Then

y1 = 7903. + 63.75 x + 63.875 x  + 0.25 x  + 0.125 x

which becomes (after mod 256)

y1 = 223. + 63.75 x + 63.875 x  + 0.25 x  + 0.125 x

However, if c = 127, d = 31, then

y2 = 8159. + 63.75 x + 63.875 x  + 0.25 x  + 0.125 x

which becomes (after mod 256)

y2 = 223. + 63.75 x + 63.875 x  + 0.25 x  + 0.125 x

So y1 = y2.  Here, d = -225 and d = 31 yeild the same equation (after
mod operation).  

Now I need to try the decoding process to see if d = -225 or d = 31
yeild the same or different answers.

-- 
Karl L. Barrus: klbarrus@owlnet.rice.edu         
keyID: 5AD633 hash: D1 59 9D 48 72 E9 19 D5  3D F3 93 7E 81 B5 CC 32 

"One man's mnemonic is another man's cryptography" 
  - my compilers prof discussing file naming in public directories




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