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Probable Solution Found for ECC2-109 Challenge

kpearson writes "The eCompute ECC2-109 distributed computing project discovered a probable solution to Certicom's ECC2-109 challenge today. The challenge was to defeat a 109-bit Elliptic Curve Cryptosystem (ECC). Since the eCompute ECC2-109 project began on November 8, 2002, 1,981 volunteers have run the project's software and found almost 40.5 million distinguished points. From those points the project found two which matched and caused a collision, enabling the project to find a solution to the ECC. The solution was submitted to Certicom this morning for verification."

3 of 130 comments (clear)

  1. Re:Question for the more cryptically inclined crow by acidblood · · Score: 5, Informative

    The `collision' mentioned here is related to the particular algorithm being used to break ECC, which is called Pollard rho for discrete logarithms.

    Let's work with the integers modulo a prime p -- the algorithm works just the same with elliptic curves. Say you were told that a^b == c mod p (where == means `is congruent to'). You were also given a, c and p, and you need to figure out b. This is the so-called discrete logarithm problem.

    Pollard's rho algorithm solves this problem the following way. Suppose you somehow figure out that a^x c^y == a^w c^z mod p, and of course x != w, y != z (which is the trivial solution). That's the kind of collision they found. Now this yields a solution because, as c == a^b, then a^x c^y == a^x (a^b)^y == a^x a^(by) == a^(x+by), and similarly a^w c^z == a^(w+bz). Thus a^(x+by) == a^(w+bz), so one is left with the very easy task of solving the equation x+by == w+bz modulo the group order, which is p-1 here since we are working with integers modulo p (this is Fermat's Little Theorem). For elliptic curves, it's not so easy (i.e., it may take a couple of hours, maybe days, on a single CPU for a curve of cryptographic interest) to figure out the group order but it's still possible.

    And how is that collision found anyway? That's a bit complicated, but I guess it can be found on the Handbook. It has to do with the theory of random functions.

    --

    Join the NFSNET. Our prime goal is making little numbers out of big ones. http://www.nfsnet.org/

  2. Re:Question for the more cryptically inclined crow by rueba · · Score: 5, Informative

    This write-up from the website linked in the article is fairly informative:

    http://www.ecompute.org/ecc2/ecc2-109.pdf

    The question of why a collision makes solving the problem easy is answered at the top of page 8.

    --
    The only reason all cover-ups appear to fail is that you never hear about the ones that succeed.
  3. Re:Really hard to understand for someone by bluGill · · Score: 5, Informative

    The NSA is the largest hirer of math majors. The NSA is mostly interested in encryption. A custom computer is not nearly as good as an algorithm breakthrough, though the computer is obviously creatable (or not creatable depending on the requirements). We know the NSA is doing algorithm research, but because they are not talking we don't know what or how much.

    Back in the '70s IBM developed DES, and it was proposed that it become the national standard for encryption. In the process the NSA got involved and told IBM to make some changes, IBM did, but didn't understand why. About 10 years latter differential cryptanalysis was developed, and it turns out that the changes the NSA requested made DES much more secure than the original design, just because of resistance to differential cryptanalysis.

    Many researchers believe that with recent private interest in cryptography the gap has been closed. However we do not know that. In any case, the NSA know everything universities know, but they do not contribute back.