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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."

35 of 130 comments (clear)

  1. Really hard to understand for someone by after · · Score: 5, Interesting
    At least for me, I never heard of this ECC-109 method before. The summery clears it up though.
    Project Overview

    This is a distributed computing project to solve the Certicom ECC2-109 challenge.In general it can be described as a whole lot of computers from all over the world, all of them working together in an attempt to have a collision. Each of the computers runs a program that computes DP (Distinguished Points) values. Each time a computer finds a DP, it is uploaded to a main server. The main server then checks to see if anyone has uploaded a matching DP (this is known as a collision). If a collision happens (two matching DP values exist), then the Certicom challenge is solved!


    Yea ... but what practical use does this have? Was this done _just_ because it could have been done?
    1. Re:Really hard to understand for someone by mphase · · Score: 4, Informative

      Price money. About $10,000 I believe.

    2. Re:Really hard to understand for someone by MBAFK · · Score: 2, Interesting

      "Was this done _just_ because it could have been done?"

      Reading around it seems that the idea is to prove that the encryption method is good rather than just theoretically sound. Probably makes it easier to sell stuff based on ECC if you can show how hard it is to crack.

    3. Re:Really hard to understand for someone by tomstdenis · · Score: 3, Insightful

      Like the RC5 challenges [and DES before it] just to say "yes, yes you can do this".

      So when someone says "64-bits of security ought to be enough" you can say "no, no it isn't." ;-)

      Though yeah, if they do more challenges it's just getting futile.

      Tom

      --
      Someday, I'll have a real sig.
    4. Re:Really hard to understand for someone by after · · Score: 2, Insightful

      But does this have any use? Prize money for what? For being lucky enough to get some random number twice? I didnt read teh full description, so I dont really know if there is reale use for this, but from what the Introduction tells me: "We are getting a ton of computers to generate numbers, and if two computers generate the same number, then we win." Hmm, huh? I still dont see what the point of this is? Does this advance some sort of research? Does this support some other principal of theory?

    5. Re:Really hard to understand for someone by cyb97 · · Score: 2, Insightful

      It proves that it is possible with commodity hardware (and a lot of time) to break ciphers that are regarded as pretty strong.

      This ofcourse is nothing to what one can imagine that national agencies have at their disposal. If a gang of internetusers can break a cipher (brute forcing it) using spare cpu-cycles, imagine what a dedicated cluster of highend computers using an algorithm more efficient than bruteforcing it would be.

    6. Re:Really hard to understand for someone by LostCluster · · Score: 2, Informative

      I think it's enough to say the theory is proven... no number of bits of security is ever going to be "enough" to be uncrackable. However, a LARGE number of processor-time units on even the fastest processors available is needed to crack strong encryption.

      I think the only remaining point of the challenges are just to prove that the ability to legally aquire $10,000 and the point of saying it's been done is enough to motivate people to contribute their processor time to such a project. We should have stats prediction formulas that should be able to give pretty reliable statements for how many processor-time units a security key will require to discover by now... the challenges just give us real-world datapoints to verify theories.

    7. Re:Really hard to understand for someone by LostCluster · · Score: 3, Interesting

      I'm not sure the government has more efficient algorithms for most of the commerically-used encryption schemes... I seriously doubt the government has encryption-busters publishing classified theory discoveries that we don't know about.

      However, they most certainly have the resources to create a reduced instruction set processor dedicated to breaking an instance of a popular encryption scheme, and the ability to buy lots of such to build a supercomputer.

    8. Re:Really hard to understand for someone by lightspawn · · Score: 2, Interesting

      I think it's enough to say the theory is proven... no number of bits of security is ever going to be "enough" to be uncrackable

      Pet peeve: You can't prove something is always true by providing anecdotal evidence - but you can disprove it.

      Anyway, to the main point: you live in a universe with a finite number of particles and a finite TTL (physics majors are welcome to argue if they want). Or let's put it this way: The total amount of resources available to humanity is limited. Given that, there are problems that, although possible to solve in theory, are impossible in practice - take, for example, the problem of playing perfect chess by fully computing the game tree. With enough bits encryption becomes unbruteforcable too. And most secrets themselves really have a TTL of only a few years or decades and there's no point in trying to protect them any longer.

      So while a "Digital Fortress"-style theoretically uncrackable encryption may not be possible, practically uncrackable encryption certainly is.

    9. 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.

    10. Re:Really hard to understand for someone by tomstdenis · · Score: 4, Informative

      ECC-109 was not brute forced. They used a cycle finding algorithm to find a collision [blah blah read site].

      The real "contribution" of this attack is that these cycle finding attacks are not just theoretical.

      Against RC5/DES brute force was used and shown to be effective for small key lengths. Against ECC a new attack was used [and a newer attack will be required to really continue]. Similarly for RSA they wanted to show factoring was possible [e.g. QS, GNFS, etc...]

      There is a point where continuing to work is pointless. E.g. RC5-72 is useless. people have learned and use 128-bit symmetric keys now as a "lower boundary". So even if they do break RC5-72 [which is likely to take a shitload of time] we'd already be ahead of the curve.

      In the case of ECC-109 [over GF(p)], 163-bit [over GF(2)]] ECC curves are still "recommended" by NIST document.

      A 163-bit key provides roughly 80-bits of security which is enough to not be insecure against most modest attacks but I doubt many cryptographers would recommend it with a straight face [personally I draw the line at 256-bit curves].

      Point is showing all these theoretical attacks working is important cuz not one attack fits all.

      --
      Someday, I'll have a real sig.
    11. Re:Really hard to understand for someone by bofkentucky · · Score: 2, Insightful

      Last time I checked, the Brits had a implementation of RSA long befor R, S, and A did, it just happened to be classified. Polish mathmeticians broke enigma in what 30, 31? Didn't help them much, but their techniques trained the first generation of computer cryptographers (Turing included). There was no point in having the listening/intercept nets that the US, England, and the former USSR maintained during the cold war had and China and the US have today if all you get to listen to was essentially white noise.

      There are advantadges and disadvantadges to this though, Bin Laden was supposedly tracked to Tora Bora b/c he was using a "failed" brit military scheme, but, Just like with Soviet nuke engineers, there are very good cryptanalysts/cyrtographers for hire out there, and stable, 1st world nations occasionally get outbid for their services.

      --
      09f911029d74e35bd84156c5635688c0
    12. Re:Really hard to understand for someone by dj245 · · Score: 4, Funny
      The NSA is the largest hirer of math majors.

      Gee and I thought the cell phone companies hired the most math majors.

      Seriously, you have to have taken Analytical Calculus I-III, Differential Equations, Geophysical Fluid Dynamics, and be Phd standing in order to understand the rate plans. Think of the guy who must have made them up!

      --
      Even those who arrange and design shrubberies are under considerable economic stress at this period in history.
  2. Question for the more cryptically inclined crowd: by jamonterrell · · Score: 5, Interesting

    I understand finding a collision (two things that when crypted yield the same result) is considered a goldmine in breaking an encryption algorithm.

    How does finding a collision help break the encryption? Does anyone know the technicalities of why this allows you to break an encryption algorithm, to me, who has no clue, this seems just like a coincidence and not very useful, but i'd like to be enlightened.

    --
    I can count to 1023 on my hands. Ask me about #132.
  3. Re:Question for the more cryptically inclined crow by Anonymous Coward · · Score: 2, Informative

    Finding a collision breaks the HASH function associated with the elliptic curve.
    --
    AC

  4. Re:If I recall correctly... by cyb97 · · Score: 3, Informative

    Elliptic Curve Cryptography is a much younger branch of mathematics and cryptography than plain old factor-based ciphers like RSA and friends.

    That's probably why it isn't as well known and well deployed as factor-based encryption. The number of implementations is also much smaller.
    Be aware however that NSA NSA Turns to commercial software for crypto chose ECC as (one of) the way(s) to go for the future not long ago...

  5. What is Elliptic curve cryptography.... by Bytal · · Score: 4, Informative

    Basically it's a cryptographic method that allows the same or nearly the same level of security as a regular public-key encryption scheme(based on factoring large numbers) but makes it computationally cheaper to encrypt the data. So while the bad guys still, theoretically, need nearly the same amounts of processing power and time as regular asymmetric crypto to decrypt the message, the good guys save significantly in encryption. This is of course extremely important for, let's say mobile devices with limited processing power.

    1. Re:What is Elliptic curve cryptography.... by cyb97 · · Score: 4, Informative

      Not only in processing power, but also in keylength. A typical RSA key would be 512-1024bits to be considered secure, an equivalent ECC key would be 140-200 bits. Which leads to smaller circuts and inturn cheaper implementations.

  6. This reminds me of a Bob Newhart sketch by Anonymous Coward · · Score: 3, Funny

    If you put enough computers on the problem, you will eventually find a solution by random chance. (correct me if I'm wrong.)

    The sketch goes something like this:

    If you put enough monkeys at enough typewriters, they will eventually, by random chance, type all the works of Shakespeare. Unfortunately, this would require an infinite number of people going around checking the monkey typing.

    "Bob, Bob, come here, I think we've got something! Yes this is really it! 'To be, or not to be, that is the gzortmanplatz...'"

    1. Re:This reminds me of a Bob Newhart sketch by kpearson · · Score: 2, Interesting

      Believe it or not, there is a distributed computing project for this, too: the Monkey Shakespeare Simulator. So far the best the virtual monkeys have done is the first 14 letters from "Coriolanus."

    2. Re:This reminds me of a Bob Newhart sketch by cynic+pi · · Score: 2, Informative

      "If you put enough computers on the problem, you will eventually find a solution by random chance. (correct me if I'm wrong.)" Yeah, but the number of primes 2^109(109 bit number) is about 2^109 / ln(2^109), and after some ballparking this is 2^109 / 109 (pretend e = 2) which is still a 100+ bit number of primes. Now if you have 1,000,000 computers doing 1,000,000,000 checks per second, this would be a total of 1 quadrillion checks per second. or about 2^40 checks per second, so then this would take 2^60 seconds to check all primes this way. This is equivalent to 2^25 years, a little longer than most of us will live. And the above was only for RSA, and ECC is more secure than that, so enough computers doing this randomly is not a viable option.

  7. This is my basic understanding by jbelcher56 · · Score: 2, Informative

    I'm not an expert, but this is my basic understanding. ECC is a public key encryption algorithm. The challenge was to find the private keys using a list of public keys and a set of system rules. If two computers find keys (DP points) that are the same, then theoretically, the encryption has been cracked, i.e. the private key has been found. It's a brute force method. This shows, however, that it is very hard to crack the encryption and would be generally useless, since by the time you cracked the encryption, whatever you were trying to read is probably not relevant anymore. However, with a fast enough computer or group of computers, it could be useful.

    --
    Don't get off the boat. Absolutely, goddamn right.
  8. Re:If I recall correctly... by Coryoth · · Score: 4, Informative

    The most widely used assymetric system is RSA, which is indeed based on factoring (or calculating the Euler Phi function - it amounts to the same thing).

    Next on the list is Diffie-Hellman, which just a key exchange algorithm (you can't encrypt with it, it simply allows both parties to communiccate in public to agree on a private session key. RSA is slow enough that this is all RSA gets used for mostly anyway though (agreeing on a symmetric session key). Diffie-Hellman is based on the difficulty of the discrete logarothm problem. That is, given a large prime p, and a numbers x, y find a such that a^x mod p = y.

    If you want to do encryption with a Diffie-Hellman liem system, you can, and that system is known as El-Gamal. It works very similarly, and is based on the same problem (Discrete Log Problem).

    Elliptic Curve Cryptography is simply Diffie-Hellman or El-Gamal, except that instead of using Z_p as the group in which you do calculations, you use the group formed by the points of an elliptic curve over a given finite field. Mostly that means that multiplication is much more complicated, and the Discrete Log Problem itself becomes much harder (partly due to multiplication being harder, partly due to other properties of the group that it would be tedious and not very illuminating to explain).

    The advantage of Elliptic Curve systems is that because the DLP is much harder on the group used (elliptic curve group), you can use a much smaller key size and still have strong encryption. Note that it was only a 109bit key that was cracked after years of effort - compare that to the RSA factoring challenge where much larger key sizes have been cracked.

    You have extra benefits in ECC as well - you get to choose the base field, and the curve itself to determine the group, rather than picking a large prime. As the properties of elliptic curve groups can vary dramatically given a change in field or curve this means if you can choose your curve randomly you get even more security (for very few extra bits - elliptic curves are very complicated objects, but simple to describe).

    What all of that means is that, while current systems are based on factoring (RSA), that system require slarger keys, is less secure and - given recent developments by Biham, Bernstein and the like - is looking potentially surprisingly crackable even at some of the larger key sizes. That is to say, Elliptic Curve Cryptography is very much the future of Asymmetric Cryptosystems. Being able to break this key size gives a decent benchmark of the security of current systems (which don't use randomly chosen curves yet - there are still issues with that).

    That is to say - this is very important, but given the complexity and the effort involved, looks like a good sign for the security of Elliptic Curve Cryptography.

    Jedidiah.

  9. 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/

  10. 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.
  11. Re:If I recall correctly... by LostCluster · · Score: 3, Informative

    having non-prime factors makes the cipher much weaker

    Actually, using a non-prime makes your encryption bullet-proof against any check against the domain of all known prime numbers...

    Therefore, the much larger domain of near-primes also needs to be considered. Afterall, you can get a known near-prime just by multiplying two known primes by each other, the only factors will be 1, itself, and the two primes by each other. To check all of those, it'll take quite the load of processor time too.

    And that leads to a guessing situation as to which subset of the possible numbers to check first. If the "blue team" checks all the prime known numbers in a situation where the "red team" has has selected a near-prime... the near-prime will take longer to solve than had any known prime been selected. If the "blue team" checks the near-primes with the same priority of as the primes while the "red team" has selected a prime number, then the possiblity of the near-primes has lead to a time-costing distraction from the real solution.

  12. Re:What is Elliptic curve cryptography... by Coryoth · · Score: 4, Informative

    Basically it's a cryptographic method that allows the same or nearly the same level of security as a regular public-key encryption scheme(based on factoring large numbers) but makes it computationally cheaper to encrypt the data.

    Mostly right. ECC is based on the Discrete Log Problem, not factoring. The Discrete Log Problem is basically: given x, y find g such that g^x = y. That's easy for real numbers - you just take a log. The problem becomes rather more difficult in the case where you are working with integers mod some prime - that is, find an integer g, such that g^x mod p = y. That gives you Diffie-Hellman and El-Gamal. ECC is the same problem, but over the group of points of an elliptic curve over a finite field. You can show that this class of groups effectively maximises the difficulty of the Discrete Log Problem, and that's why the key sizes and computational efficiency is so much better.

    Jedidiah

  13. Re:If I recall correctly... by Coryoth · · Score: 2, Informative

    That and they're only finding collisions. Collisions are next to useless unless you want somebody to accidentally download a file with seemingly random bits instead of something they wanted. (That's just one example, but collisions are not very useful a good 99.99999999% of the time).

    That would be hash collisions you are thinking of, this is rather different, given the nature of the system. I would direct you to this very well written post, which explains the significance of collisions in ECC. It effectively breaks the system.

    Jedidiah.

  14. Waste of cycles by gumpish · · Score: 3, Informative

    It seems that a better use of spare CPU cycles would be distributed.net's Optimal Golomb Ruler search, or Stanford University's Folding @ Home project.

    Make an actual contribution to science - much more satisfying than looking for a very improbable E.T. or senselessly cracking encryption schemes.

    1. Re:Waste of cycles by Drakonian · · Score: 3, Insightful

      Yeah, because only a few trillion dollars in transactions are protected every day by encryption schemes. Nothing much at stake there.

      --
      Random is the New Order.
    2. Re:Waste of cycles by gumpish · · Score: 2, Insightful

      Why spend millions of mips-hours cracking 64-bit encryption when much stronger encryption is available?

      And isn't it trivial to calculate the probability of a solution being found when using a known alogrithm and expending a certain amount of CPU time?

      What is learned?

  15. Some answers, maybe.... by crypto257 · · Score: 5, Interesting

    The purpose of all these challenges is to understand how much computing power is necessary to break encryption or signature schemes. EC109 strength is pretty low, but offer a way-point on the curve. Distinguished points are not really distinguished. They just have an easy search pattern such as a number of trailing zeroes or other constant values. These are searched ad-infinitum and when two matches are found, a little math can get you a private key. The death nell for the DES algorithm was heard when distributed.net, in cooperation with the Electronic Freedom Foundation built a machine that could crack it in 27 days (or so). And the cost made you wonder who might want to build such machines. As a result, we have AES and expanding public key lengths. No-one would really use ECC 109 for current cryptographic systems. The results from this test confirms that. The real question is what is the appropriate key length for a The amount of money (n computers over t time) tells us what sort of advisaries these techniques are useful against. It also

  16. curves... by dj245 · · Score: 3, Funny
    Ooh, Curves! Can I touch?

    I am too dumb to say anything further on the subject.

    --
    Even those who arrange and design shrubberies are under considerable economic stress at this period in history.
  17. Uh... November 2002? by 1000StonedMonkeys · · Score: 3, Interesting

    Okay, so why does the linked webpage indicate that the 109 challenge was Completed in November of 2002?

    1. Re:Uh... November 2002? by Anonymous Coward · · Score: 4, Informative

      That was ECCp-109.

      This is ECC2-109.

      Read the entire challenge list at

      http://www.certicom.com/index.php?action=res,ecc _s olution