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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."

11 of 151 comments (clear)

  1. Fascinating Discussion... by gweihir · · Score: 5, Interesting

    ... this is. I especially like the mixture of theoretical, practical and yet unknowen aspects of the whole problem.

    My impression is that so far DJB has done a good job of being honest and clear. Although "the press" is sadly lacking in experts these days and often will not even notice they have not understood the problem. I have to admit that I did not quite follow
    Lenstra-Shamir-Tomlinson-Tromer, but I think DJB's original proposal is still the best source on what is going on. No real surprises so far for practical purposes, but I will follow this closely.

    Incidentally I don't fear for my 4096/1024 bit ElGamal/DSA gpg key in the near future. I am confident that installing a keyboard sniffer without me noticing is far easier than breaking that key.

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  2. Re:Security ... by Russ+Nelson · · Score: 3, Interesting

    No. Security through obscurity means that your task is made easier on every decryption by knowing something which is intended to be a secret. Needing to factor a number every time isn't obscurity. It's just plain hard.
    -russ

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  3. Re:Cool but by God!+Awful · · Score: 4, Informative


    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

  4. Re:Why? by God!+Awful · · Score: 5, Informative


    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

  5. Re:Cool but by Uller-RM · · Score: 3, Interesting

    What happens if you don't know the base? ^_^

    Shor's algorithm requires a bit of number theory to prove its correctness, but the first part is the important one. You need 2n+1 qubits to factor a n-bit number - two n-bit registers and a keeper bit. (Note, I'm running from memory here, not my notes, so I may have a step wrong.)

    You initialize the first with the Hadamard transform, creating a superposition of all possible n-bit numbers. You then raise that superposition to the power of your number to factor, modulo 2^n, and store the result in the second register, which will itself be a superposition (and entangled with the first register). You then measure the second register - and as long as it doesn't measure to be zero, its collapse will trigger a partial collapse in the first register, resulting in a set of bases which are congruent modulo the second register's collapsed result. You then perform a discrete Fourier transform on the first register, and the rest is all logic and repetition.

  6. Re:Headhunters visiting the site? by larry+bagina · · Score: 3, Funny
    Quantum Computers, Advances in Number Theory; looks like this decade will become interesting

    I'll say! Arthur Andersen has advanced number theory further than anyone had imagined it could go!

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  7. Re:Security ... by dillon_rinker · · Score: 3, Informative

    No. The phrase "security through obscurity" generally means that the security algorithm is kept a secret. The lock on your house is a good example - how the lock works is open, public information. The particular key that you use, however, is not. To unlock your door, someone must either learn what key you use or exploit weaknesses in the lock. A standard locksmith text, some standard locksmith tools, and a little time is all that's required to pick the standard lock. How the lock works is not obscured; what key you use is.

    Since everyone knows how the lock works, everyone can see what the weaknesses are and everyone can try to correct them (this has been done over the last couple of centuries). If you used a non-standard lock with a mode of operation that only you knew, that would be security through obscurity. No one can learn to pick it by reading standard locksmith texts. The weaknesses would have to be learned through trial and error by someone on your front porch. Any lockpicker would have to learn not only what key you used but how to use it in the lock.

    Public key encryption generally uses well-known algoorithms - this is what makes it not obscure. The keys are secret, just as with obscure systems. However, everyone can see whether or not the system is a strong one.

    Go read Cryptonomicon for a fictional introduction to the perils of obscured cryptosystems, as well as the benefits of open ones. Then read Applied Cryptography for the real (ie technical and mathematical and correct) deal. Or just take my word for it like a happy little slashdotizen. :)

    BTW, modern locks are the result of a couple hundred years of improvements and refinements; they are currently about as good as they need to be, given weaknesses of the door itself, the convenience of regularly using the mechanism, the cost of the lock and keys, etc. Cryptosystems are nowhere near that level yet. The internet is still a country town, in the sense that everyone leaves their house unlocked and their keys in the ignition of their car. Once everyone's been hacked a couple of times (and I do mean everyone), adoption of crypto will become widespread, and we'll begin to see sufficient standardization to make crypto comparable to real-life locks. Government officials don't whine about locks the way they do about crypto for good reason - they aren't much more than courtesy and inconvenience devices. Give me a good sledgehammer and I'll be in your door in five minutes. Give me a power saw and I'll make my own door. Betcha the standardized crypto will be similar.

  8. Re:Security ... by Waffle+Iron · · Score: 3, Funny
    Security through obscurity is when you run your relaying smtp server on port 27 instead of port 25.

    That's why I choose real security. My relaying smtp server runs on a prime number port, protected by factoring. In fact, I can go ahead and tell you that the port is one of the prime factors of 899 without reducing my security at all.

    (The example of 27 is a particularly lame choice, since it's 3x3x3. That's not even nearly prime.)

  9. Time to update that crypto by hazyshadeofwinter · · Score: 3, Funny

    Yeah, I once thought it was secure, but it looks like now I might have to replace my rot13 encryption with rot26, or even rot52...

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  10. Cost model by blair1q · · Score: 4, Funny

    Computation time multiplied by the cost of the computer?

    His department comptroller must love him. "No, you can't have a new plastic spoon, because it costs 11 cents and you will be using it for 0.8 years and that's...2.8 million dollar-seconds...we'll buy you a new $40 silver spoon every day and let you use it to stir your coffee for three seconds per...that's only 35K dollar-seconds..." It's pathological.

    Okay, if you fully depreciate the computer to the moment you start the computation, or better yet, market-price it, then watch the price as the computation continues along (could drop 10-20% in a few weeks for a given top-end PC type machine), then you're calculating the average replacement cost of the machine over the life of the computation.

    It still seems a little verschimmelt. The quasi-rent on such a machine is really the depreciation over the term of the computation.

    Need to think more on what cost means to someone who's trying to steal all your base. They probably stole the computer, anyway.

    --Blair

  11. Re:Cool but by colmore · · Score: 4, Informative

    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?

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