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- From: bs@gauss.mitre.org (Robert D. Silverman)
- Newsgroups: sci.crypt
- Subject: Re: RSA-129 contest
- Message-ID: <1992Aug20.134949.4740@linus.mitre.org>
- Date: 20 Aug 92 13:49:49 GMT
- References: <9208200145.AA29470@ucbvax.Berkeley.EDU>
- Sender: news@linus.mitre.org (News Service)
- Organization: Research Computer Facility, MITRE Corporation, Bedford, MA
- Lines: 54
- Nntp-Posting-Host: gauss.mitre.org
-
- In article <9208200145.AA29470@ucbvax.Berkeley.EDU> jbs@WATSON.IBM.COM writes:
- >
- > D. J. Berstein wrote:
- >:To factor RSA-129 with the general number field sieve (GNFS) requires a
- >:fifth-degree polynomial
- >:
- >: f(x) = c_5 x^5 + c_4 x^4 + c_3 x^3 + c_2 x^2 + c_1 x + c_0,
- >:
- >:with integer coefficients c_i, and a positive integer m such that f(m)
- >:is a small positive integer multiple of RSA-129. Given any random choice
- >:of m you can easily expand RSA-129 in base m and find the coefficients
- >:c_i. However, for GNFS to finish in a reasonable time, the coefficients
- >:must be significantly smaller than random.
- >
- > How small do the coefficients have to be? I have been led to
- >believe that the speed of the GNFS overall does not depend much on the
- >efficiency of the search for a good polynomial. If this is the case it
- >would seem to be unwise to devote a lot of effort to finding a good poly-
- >nomial. What is the point of this contest?
- > James B. Shearer
-
- This is NOT the case. The performance is VERY sensitive to the size
- of the coefficients and it is worthwhile spending a significant amount
- of time trying to find small coefficients.
-
- For a 5th degree polynomial, 'random' coefficients will be about 21 or 22
- digits. One needs to factor two sides of a congruence, one in Z, the other
- in Q(alpha). The norms on the integer side will be about 10^32.
- [M ~ 10^(130/5) and the average b will be about 10^6]. On the algebraic
- side they will be about (10^6)^5 * poly coef or about 10^52.
- Of course many will be smaller, [for small b]
-
- I estimate that to be competitive with the Quadratic sieve they need
- to find coefficients of 'about' 14 digits. I am not sure it is
- possible.
-
- Their estimates may, of course, differ depending on their assumptions.
-
- I have already done about 1% of the work to factor the RSA-129 with
- the two-large-prime variation of the Quadratic Sieve. I was going
- to do it completely, but a deal with Sun to use their internal net
- [3000 workstations!] fell through due to management difficulties.
- I estimate it was going to take between 5000 and 6000 MIPS-YEARS
- to do the computation.
-
-
- Now if I could just scrape together 3000 Sparcs for 2-3 months, I
- could finish the factorization.
-
- --
- Bob Silverman
- These are my opinions and not MITRE's.
- Mitre Corporation, Bedford, MA 01730
- "You can lead a horse's ass to knowledge, but you can't make him think"
-