1995-02-09 - Re: Lucky primes–third time’s the charm?

Header Data

From: Hal <hfinney@shell.portal.com>
To: cypherpunks@toad.com
Message Hash: 56e8b048efea191c21c88f7d7318a675526b8e0611e53038cfcd6758b7b4c918
Message ID: <199502091641.IAA12454@jobe.shell.portal.com>
Reply To: <Pine.3.89.9502090512.B21224-0100000@lia.bga.com>
UTC Datetime: 1995-02-09 16:41:47 UTC
Raw Date: Thu, 9 Feb 95 08:41:47 PST

Raw message

From: Hal <hfinney@shell.portal.com>
Date: Thu, 9 Feb 95 08:41:47 PST
To: cypherpunks@toad.com
Subject: Re: Lucky primes--third time's the charm?
In-Reply-To: <Pine.3.89.9502090512.B21224-0100000@lia.bga.com>
Message-ID: <199502091641.IAA12454@jobe.shell.portal.com>
MIME-Version: 1.0
Content-Type: text/plain


Nathan Zook <nzook@bga.com> writes:

>-----BEGIN PGP SIGNED MESSAGE-----
> 
>The algorithm I posted the second time works, (nice improvement, eh?) but
>is likely to take several thousand years to complete.  And when it does, we
>can expect weak primes.  The enhancement I propose should fix that.
> 

I did not realize before that p was an output of your algorithm, rather
than an input.  That explains better what you were trying to do.  You are
in effect trying to search for a prime such that e's multiplicative
inverse has a lot of 0's.

This looks like it will work pretty well, with the caveat as we discussed
before that going too far with this could make searching for the primes
easier.  But the only obvious attack would be to try to reproduce your
prime-finding algorithm to find a p which divides the modulus n, and that
is basically a sqrt(n) algorithm, which is far from the worst-case attack
we face.  The search space can be reduced by a considerable factor before
it becomes competitive with modern algorithms.

I guess another point is that if i is 2 or 4 then p itself will likely be
0-rich and conceivably there could be some attacks against a modulus
known to be the product of two 0-rich primes even when the primes are not
weak in the normal sense.  (p = (ed-1)/i+1, d is 0-rich, and e has only
2 bits on so ed is likely also to be somewhat 0-rich; dividing by i is
just a shift right if i is 2 or 4, and adding 1 won't make much
difference.)

Restricing i to other numbers would still give p a simple arithmetical
relation to a 0-rich number (i=3 --> p*3 is 0-rich).  Maybe you could
choose a d such that d itself was 0-rich while ed happens not to be
0-rich; this might feel safer since p would have less of an
arithmetical relation to a 0-rich number.

(Admittedly, I don't know of any factoring attacks directly applicable to
0-rich factors but there is at least a superficial similarity to weak
primes and that suggests caution.)

Hal





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