More on Bernstein's Number Field Sieve

Russ Nelson writes "Dan Bernstein has a response to Bernstein's NFS analyzed by Lenstra and Shamir, entitled Circuits for integer factorization. He notes that the issue of the cost of factorization is still open, and that it may in fact be inexpensive to factor 1024-bit keys. We don't know, and that's what his research is intended to explore."

Comments

Re:Cool but

[God! Awful]

Could you elaborate more on the "reverse log" problem... If you know the base and the result of [Log x], whats the problem?

The OP got his terminology wrong. It's the discrete log problem that's hard.

Pick a number (e.g. 3.81482), plug it into your calculator and press e^x (result = 45.36874). Now it's easy to get back the original number by using the "ln" key. But imagine instead that you only had the fractional portion of the result (.36874). Now it's next to impossible to figure out what the original number was. The discrete log problem is basically the same this, but using discrete arithmetic instead of real arithmetic.

-a

Re:Why?

[God! Awful]

I didn't really think there was any need for anything better than 128 bit encryption. It would take a lot of factoring that is practically impossible by human standards to figure out the key for a 32 bit encrypted code, and this site [stack.nl] seems to tell me that 128 bit encryption is nearly impossible to break by any standards.

128-bit private key encryption is considered virtually unbreakable. 128-bit public key encryption is not. AES is an example of private key encryption; RSA is an example of public key encryption.

-a

Re:Cool but

[colmore]

discrete logs actually have to do with modular spaces (remainder math, in mod 4, you divide any number by 4 and take the remainder. counting in mod 4 goes like: 0, 1, 2, 3, 0, 1, 2, 3...)

the discrete log problem is specifically, given integers y, g, p, find a (preferably minimal)solution x to the problem

y = g^x mod p, 0 = y p

actually the problem is more general than that, but that's the case that most people talk about and has direct application to cryptanalysis.

it doesn't look too hard, but sit down and try. the algorithms that solve the problem amount to basically highly erudite mathematical guess-and-check. if you can find a P time solution to this, you're a billionaire.

It's also a fun problem because, like Fermat's Last Theorem, Goldbach's Conjecture, and the 4-Color problem, it's easy for an amateur to work on, understand, and make some elementary discoveries and proofs, but the problems have difficulties that test the furthest extent of mathematical knowledge.

here's a fun, related problem:

if you shuffle a 52 card deck perfectly 7 times (divide the deck exactly in half, always have the top half drop the first card, drop exactly one card after another) then you end up with the original order of the deck. Given a deck of n cards, how many shuffles are required for the same effect?