1993-10-20 - Re: crypto technique

Header Data

From: Karl Lui Barrus <klbarrus@owlnet.rice.edu>
To: cypherpunks@toad.com
Message Hash: 178b6f8958d267dce8fd747555226861da7039ef0c1d58f6c67789ab6b2cadc0
Message ID: <9310202130.AA09601@flammulated.owlnet.rice.edu>
Reply To: N/A
UTC Datetime: 1993-10-20 21:32:48 UTC
Raw Date: Wed, 20 Oct 93 14:32:48 PDT

Raw message

From: Karl Lui Barrus <klbarrus@owlnet.rice.edu>
Date: Wed, 20 Oct 93 14:32:48 PDT
To: cypherpunks@toad.com
Subject: Re: crypto technique
Message-ID: <9310202130.AA09601@flammulated.owlnet.rice.edu>
MIME-Version: 1.0
Content-Type: text/plain


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One question before I work on this more at a later date.

Can you still decode properly if you take the modulus of each
coefficient?  I ask since if you can't, the modulus makes is harder to
find the constants, but as a side effect it also destroys the message.

Say I use

f = 29/2 x + 40
g = 135/2 f^2 + 135/2 f + 75

I get g = 110775 + 317155/4 x + 113535/8 x^2

This is easy to solve, so Matt takes the modulus of each coefficient
(some power of 2, I pick 32 here to keep it simple).

results in g' = 23 + 99/4 x + 127/8 x^2

Say I encode my message x = 5

g(5) = 6895725/8 mod 32 = 109/8

g'(5) = 4349/8 mod 32 = 253/8

Notice that these two results aren't equal at all!  

Am I misunderstanding the encoding (and decoding) process?  Aren't
these two results supposed to be equal?  I'm not getting the expected
result when each coefficient is reduced mod 32.


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-- 
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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