Re: cryptology, complexity, and quantum cryptology
From: Ralph Hartley (hartley_at_aic.nrl.navy.mil)
Date: 02/15/05
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Date: Tue, 15 Feb 2005 09:22:50 -0500
Mitch Harris wrote:
> Ralph Hartley wrote:
>> A *proof* that a cypher is NP complete to break, can make it
>> vulnerable to known (or is it chosen?) plaintext attacks. The
>> reduction used in the proof lets you solve the NP complete problem if
>> you can get plaintext cyphertext pairs.
>
> That's neat. How does that work?
I can only reconstruct this from vague memories, so excuse me if it is
garbled. I also wish I could remember where I read this, but there is no
chance of that.
Suppose you have a public key cryptosystem where the secret is the solution
to a (public) problem x from the class X. You want to be sure that applying
the secret key is at least as hard as finding that solution, otherwise
knowing that x is hard to solve is irrelevant.
To prove that, you find a reduction from X to applying the secret key.
Assuming x really is a hard, randomly chosen, instance of X, from a purely
computational viewpoint you have proven the system hard to break.
But suppose I can trick you into applying the secret key for me. I can
apply the reduction to x, let you apply the secret key, and you will tell
me the solution to x.
Depending on whether the secret key is being used for signing or
decryption, this is either a chosen plaintext or a chosen cyphertext attack.
You would have been much better off with a system where all attempts over a
long time to apply the secret key without knowing the solution to x had
failed, but no proof by reduction was known or considered likely.
No one knows how to break RSA without doing the factoring problem, but to
prove that breaking RSA is really as hard as factoring, by finding a
reduction, would be very bad news.
There may be ways other than reduction to prove a problem hard, but they
are very unusual.
Ralph Hartley
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