RSA-576 Factored
An anonymous reader writes "I thought Slashdot would have picked this up
several days ago, but apparently not. Although
you still won't see any mention of it on the
RSA challenge site, Mathworld is carrying the news that a team at the German Bundesamt fur Sicherheit in der Informationstechnik submitted a factorization of
RSA-576 on December 3. RSA-576 is the smallest challenge number that RSA Security offers a cash prize for, to the tune of $10,000"
Ontday oyay inkthay osay?
I think that composite numbers everywhere will sleep just a little bit less securely tonight, knowing that the Bundesamt fur Sicherheit in der Informationstechnik is out there, somewhere, waiting for them.
Yup.
Irritable, left-wing and possibly humorous bumper stickers and t-shirts
Wow, I havn't really read in to it, but is that very big? I mean, they were talking about not too long ago that 128bit encryption is "almost impossiable" to break. If this is 576bit encryption, and they've broken it, doesn't this mean that 1024bit is looking slightly weak? Whats the 'difficulty' of breaking this key on a relative scale?
Chris
Look! I did it too!
1 2 3 4 6 8 9 12 16 18 24 32 36 48 64 72 96 144 192 288 576
Does anyone know the relative computational difficulty of cracking RC5-72 vs. trying to factor one of the RSA numbers? Given the higher monetary payoff, I'm wondering if I wouldn't be better off implementing and running a prime factor sieve, rather than running the RC5 client (which only runs on my W2k workstation, because the distributed folks never rewrote the older cores that run on my pre-OSX Macs.)
Mersenne primes are a number of the form 2^n - 1, where n is some prime number. Mersenne primes are one of the easiest to find, and there is a quick (relatively) algorithm for checking whether it is prime. Not all Mersenne numbers are prime.
Don't waste your vote! Vote for whoever you want, unless you live in a swing state it won't matter anyways
Not sure if this is a troll, but I may as well offer a simple explanation.
The RSA public-key cryptosystem takes advantage of the theory that factoring composite numbers is a computationally difficult problem. I'm not going to get into specifics, but the depth of the problem is in that the composite number acts more or less as a public key, and encoded within that composite number (as one of the factors) is the private key.
Being able to factor an RSA number is big news because it says that an RSA encoded message with a number of that size (576) can be defeated. Whether or not this is economical to defeat (i.e. time and resources put into the factoring effort) is really the key to this exercise, but one can now assume that a properly funded entity (most likely government) has the ability to defeat RSA-576.
Hope this helps.
I think I speak for 99% of the population when I say...
"Oh."
GL
They probably just looked in the back of the book.
Now there needs to be cash prize for discovering the discoverer when the challenge has been met. I just checked that site today.
Well i_am_syco, articles are there for reading. They can even increase your knowledge, and one day you may even learn how to spell psycho properly.
"She's a West Texas girl, just like me" - G.W Bush Iraqis
I think I speak for 99% of the population when I say... "Oh."
...
I think I speak for the other 1% when I say
"Um."
-kgj
-kgj
It wasn't a) anti-Microsoft or b) anti-SCO, so no one on /. cares.
Well, the computational complexity of the General Number Field Sieve is:
O(exp(c*log(n)^(1/3)*log(log(n))^(2/3)))
where the value of c is reflected by the specific flavor of the NFS you're using, but in each case c>1
I don't know the complexity of RC5, but I can imagine it's not exponential like the NFS.
Factoring composite is probably somewhat easier in the sense that it doesn't require more hardware/brute force computational power, however, its probably harder in the sense that you need to be very famillar with the very latest in factoring algorithms and essentially be skilled and willing enough to implement the very leading edge known algorithms.
In other words, factoring RC5-72 requires Moore's law, plus a massively growing audience of people willing to participate. But even with the most optimistic projections, RC5-72 will not fall for decades. Factoring something like RSA-574 requires that you be up on the latest in computational mathematics as it relates to number theory. If you think you can bone up on the latest in factoring techniques in less than 10 years, then you probably have a better shot at the RSA prizes.
How could they *factor* ME without *my* own knowledge?! Somebody call the doctor... -RSA-576
In order to win the prize, you must submit your result to RSA, they don't actively seek out winners. That's why RSA's page hasn't been updated.
They can submit their answer here.
If anything, it deserves a few more insightfuls. He has politely asked a number of important questions about RSA encryption. Obviously Chris doesn't understand how this "breakthrough" matters in the big scheme of internet security, and he damn well deserves a polite response or two to help teach him about the impact.
Asshole mod: I hope you're satisfied.
Chris: I hope some of the comments from the other users have been useful to you.
Your first factor is composite, slick.
/. revolution, instead of spelling nazis, we now have composite number nazis.
This is a
Oh my lord. Get up, walk away from your computer, go into the bathroom, and drown yourself in the toilet. Seriously, you'd be doing humanity a favor.
I'm not suprised that someone has done it. Even the RSA site suggested 576 would fall soon. What I do find interesting is that it took 4 days for word to get out, and that the factorization was done in Germany. More interesting would be knowing what algorithm was used - is it new, or just further refinement of GNFS or MPQS with faster hardware?
They're busy multiplying the two 87-digit factors by hand, just to be sure.
Crap, there go my plans to factor it myself.
Ron Paul 2012
...sorry...
......if you get lucky!
.1% of keyspace you search.
There is nothing that says you won't find the correct key in the 1st
Veramocor
They're all composites, einstein. The first, third and fourth are even numbers. The second is a multiple of 5. Joke is on you.
one day you may even learn how to spell psycho properly.
:P
Not a big Suicidal Tendencies fan are you?
- Nasty
- Offtopic
what the hell is that supposed to mean ? your carrying encrypted data in your suitcase which happened to use that same prime number ?
When 128-bit cyphers are described as "secure", they're almost certainly talking about symmetriccyphers - that is, the key you use to encrypt the message is the same as the key you use to decrypt the message. There are no known ways to break currently acceptable symmetric cyphers (such as 3-DES and AES) faster than brute force - that is, trying each key one at a time. If you have a 128-bit key, this will on average take (2^128 / 2 = 2^127 ~= 10^38) tries before you get the key. This will take billions of years to do, even using a massively parallel computer.
The other sort of encryption, the sort we are talking about here, is public-key encryption, where you use two different keys to scramable and descramble the message. The advantage of this method is you can create a key pair, and give one key to everyone who wants to send you a message (the public key), and while they can send you message securely, it is very difficult for them to figure out your private key (and thhus read messages other people have sent you).
The bad news with public-key encryption is that the algorithms are considerably weaker than with secret-key cyphers. You can mount considerably quicker attacks than just brute-forcing the keyspace. Therefore, you need longer keys for equivalent levels of security. With RSA, the most common method, figuring out your private key from your public key is done by trying to figure out the factor of a very, very large number that is the product of two very large prime numbers. This is still very difficult to do, but it is a simpler problem than brute-forcing an entire keyspace. These Germans have just demonstrated the ability to factor a larger such number than anyone else has done before.
Whilst this is interesting, from what (little) I understand of cryptography it's still a very long way from here to cracking 1024 bit RSA keys. In any case, as the hardware makes it easier for the attackers, it makes it practical to go with longer encryption keys, so faster hardware is neither a help nor hindrance to attackers. The one proviso is, of course, the security of data encrypted by older cyphers.
Any sufficiently advanced technology is indistinguishable from a rigged demo
--Andy Finkel (J. Klass?)
-- Funny
-- On topic
I don't know the answer to your question, but I must show off some knowledge, so here's something I do know! But the answer to you question probably has nothing to do with what I just said.
Oh hell, that's very insightful! There's numbers after all!
My key is 1024 too... but it's ElGamal. Is the ElGamal algorithm inherently more or less secure than RSA?
Karma: It's all a bunch of tree-huggin' hippy crap!
RU-486 the Next Generation?
--
"Outlook not so good." That magic 8-ball knows everything! I'll ask about Exchange Server next.
Yeah, I noticed that after posting, but it's not trivial to generate a 173-digit prime, so I just faked it with rand(). On second thought, I should have at least made sure the last digit was 1, 3, 7, or 9. But what's done is done.
No, not many are. Which band member are you related to?
If a woodchuck could, would it be too lazy to?
The sad thing is that just like Axl Rose, they probably thought they were spelling it correctly.
"She's a West Texas girl, just like me" - G.W Bush Iraqis
Fellowship 9/11
attracting only comments from old troll accounts?
No one knows anything about how you go about factoring huge composite numbers, or can read German, or even knows the difference between breaking RSA-576 and breaking RC5-72.
So all that's left are people trying to find clever ways of linking to the prime number shitting goatse, and a surprising dearth of posts by abandoned troll accounts.
Care to explain?
Fuck Beta. Fuck Dice
Granted, there are some dnet trojans out there, but I think Norton goes a little overboard on this one. I don't see how dnet is less legit than SETI or Folding. Why not call those trojans, too?
"To confine our attention to terrestrial matters would be to limit the human spirit." -Stephen Hawking
And I quoth from the article:
3980750 8642406493 7397125500 5503864911 9906436234 2526708406 3851895759 4638895726 1768583317
x
4727721 4610743530 2536223071 9730482246 3291469530 2097116459 8521711305 2071125636 3590397527
which can easily be multiplied to verify that they do indeed give the original number.
Does anyone have a calculator that can "easily" multiply these two numbers... Holy Cow!
I don't see anything about how long it took them! I do know that RSA-155 took 7 months of computation to break...
not being the math wiz that most /.ers are I was wondering what this meant for me...I found the below statement on RSA's FAQs and it answered my question that I'm sure many here like me have..
***************
What does it mean when a Challenge Number is factored?
Users of the RSA public-key cryptosystem may wonder what the factoring of a challenge number implies about the security of their keys. Should they immediately replace their keys with larger ones? Should they stop using RSA altogether?
Clearly, the factoring of a challenge-number of specific length does not mean that the RSA cryptosystem is "broken." It does not even mean, necessarily, that keys of the same length as the factored challenge number must be discarded. It simply gives us an idea of the amount of work required to factor a modulus of a given size. This can be translated into an estimate of the cost of breaking a particular RSA key pair.
Suppose, for example, that in the year 2010 a factorization of RSA-768 is announced that requires 6 months of effort on 100,000 workstations. In this hypothetical situation, would all 768-bit RSA keys need to be replaced? The answer is no. If the data being protected needs security for significantly less than six months, and its value is considerably less than the cost of running 100,000 workstations for that period, then 768-bit keys may continue to be used.
Applications that require longer-term security or have data with a high financial value should migrate to longer keys before the factoring of the corresponding challenge number is announced. In either case, the results of the Factoring Challenge provide real data to help the cryptosystem user choose the appropriate key size.
RSA Laboratories' Frequently Asked Questions About Today's Cryptography provides more information on choosing RSA key lengths for various applications. RSA Laboratories Bulletin #13 discusses key length requirements for various cryptosystems.
***********************
And honsetly I think for most people the idea of someone devoting a cluster of computers just so they can read some documents you may have on your hard drive kindof egotistical for the end user...but hey we all know that the NSA breaks every key they can right?...even ones from people just trying to protect their data from average joe hackers...
Norton AV's flagging of dnetc is fully in error. The distributed.net staff is aware of the situation and is working with Norton to resove the issue.
Suicidal Tendencies spells it Cyco, not Syco. As in "Born to be Cyco." So the guy's still a bad speller.
What will be the face of the next from of Crypto? Only one-time pads? That sounds way painful.
-- Multics
not the number factoring methods. I was speculating that most of the posters had no way of intelligently commenting on the story, as significant as it once. Thanks for the links, though.
Fuck Beta. Fuck Dice
The authors of Microsoft-product-exploiting worms might have more computing resources available than any government by now.
Alrighty, let's have Hilary Rosen and Jack Valenti sue them for violating the DMCA.
Wh47 d1d j00 541, 31337 15n't t3h r0xor5 ne m0r3???
Dunno if this is authorized, but: Appropriate Link
LOAD "SIG",8,1
Python can do this, and any arithmetic with arbitrarily large numbers (up to the limit of available memory).
I don't know the complexity of RC5, but I can imagine it's not exponential like the NFS.
The complexity of RC5 is O(n). Encryption time is constant but key setup time is linear, so the whole process is linear.
However, that's not relevant. What you need to compare is the complexity of a brute-force search of an n-bit keyspace, which is O(2^n). Definitely exponential.
Note to ACs: I usually delete AC replies without reading them. If you want to talk to me, log in.
MD5 has been breakable for ages. There are tons of tools out there (john the ripper and such). It's still an industry standard (forum encryption for passwords, hell, even the /etc/shadow). It's just a matter of whether or not it's economically reasonable to break these passwords...(i.e., most people cant reverse a password like 'a2430234234mfxmiff' since they lack the power to brute force it).
This looks like a prime job for distributed computing. Code a program that does this distributed, offer people xxx dollars per work calculation with a $1000 bonus if their calculation is correct. Deduct a few thousand for yourself, and voila. Crack the hardest code.
Here is the latest cipher with perfect secrecy, with an added assumption on the memory limits of the adversary.
The real threat is Shamir's TWIRL and TWINKLE optical factorization-assistance gadgets. They aren't necessarily a threat to 1024-bit keys, but they're a definite threat to 768-bit keys, and it's time to start thinking seriously about threat models before embedding 1024-bit limitations in hardware. That's probably big enough that only national governments are likely to be funding, rather than bank robbers and other Mafiosi.
Bill Stewart
New Fast-Compression-only CPR http://preview.tinyurl.com/dy575ks
Just a clarification...I know you probably know what you meant, but a casual reader may not...this is not a 576 digit number, but a 576 bit number - the value is on the order of 2^576 - a f*'in big number to factor.
No, Thursday's out. How about never - is never good for you?
Yeah, but3 980716356337941 73827007633564229888\7 43045317388011303396716 19969232120573403187\5 02570599 906436234252 67084063851895759463\7 214610743530253622307197304822463291469530209 71164598521711305207\
RSA-576=
1881988129206079638386972394616504
59715234665485319060606504
95506569962213051687593076
=
3980750864240649373971255005503864911
88957261768583317
x
4727
11256363590397527
What if one day someone discovers an miraculous way to easily factorizes a large number in no time. Would it be outlawed by the DMCA?
Factoring RSA-2048 might considerably contribute to global warming due to computer activity. Got anybody an idea how big the temperature rise will be assuming the efficiency given by the RSA-576 challenge?
I still remember homework about a driver being stopped because he ignored a red traffic light. He claimed the light was green because of Doppler shift caused by high speed of the car. The amount of fuel required to reach this speed was by far higher than the total consumption by mankind.
It's still a 576-digit number, binary digits, to be precise, not decimal digits ;-)
And: 2^576 is somewhat smaller than 10^576 would have been.
Google accepts the sum but only gives 9 sig figs in the answer. Oh well.
wot no sig
I'm certain that NSA is following the news from RSA competition pretty closely. I wonder how long RSA keys they can crack?...
The Ruby will help you.
This is an OO interpreter which can multiply big integers without any additional modules.
RSA's security is based on there being only two primes that make up the magic numbers. The RSA contests are about the difficult problem of factoring primes however that isn't the only hard problem that RSA encryption depends on. Another problem is the Euclidian algorithm which I feel is the main weakness. The problem is that there isn't a 1:1 ratio of public keys to private keys and there may be a very large number of private keys that will decrypt a given public key.
Haven't you guys heard of those new quantum code-majiggers? Apparently it's impossible to crack because when you look at a photon it is different (I dunno where they come in). Saw it in the paper.
So far they seem give you the dates of the completions, but not the time / cpu hours taken.
Even the distributed.net doesn't seem to have seti-like stats on cpu hours used.
This seems to take more than a single cpu and workaday time.
Without those numbers on actual cpu time needed, this seems like flagpole-sitting.
The public perception would be that anyone with a workstation and some time on their hands might do this if the methods are 'just math' and made public.
"Win treats sysadmins better than users. Mac treats users better than sysadmins. Linux treats everyone like sysadmins."
...I'm going back to school to study cryptography. It's just too damned cool (and I used to be a rocket scientist, a cool vocation itself).
Is it just my observation, or are there way too many stupid people in the world?
> Those people could be concerned at this point that
> their old messages could be cracked.
Who would want to spend zillions of hours of computer time to read some geek's old messages?
"Great news, today I have finally managed to install the latest 0.99.1 kernel and boy is it great! I'm so glad I picked SLS instead of slackware, whose installer sucks big time. With my beloved SLS all I had to do was swap four floppies in and out and everything works beautifully! No crashes yet. I never realized how much of a pain DOS was! I just finished getting my printer to work (sure was tough, see attached comments on my 7551 hack) and am bored stiff. Do you know if there are any games for Linux? Moria just doesn't cut it after King's Quest."
Umm no. By choosing a public exponent that is relatively prime to (p-1)*(q-1), (as is currently done), the private exponent will be unique, making the private key unique.
4 * 13 = ?
you just fucking regurgitate a webpage to seem smart. you know shit. you know nothing about math or computer science and you have to come to /. to flex that muscle in your skull.
what. a. fucking. loser.
hey fag. that isnt very funny is it. no it isnt.
i think what they guy meant is this:
if you publish an algorithm that is strong so that other people can use it to produce a fairly strong encryption, no matter what you publish, the algorithm will be crackable.
so you stupid fag, what did you use to come up with that line of text?
you are such a mouthy little fag loser with no life that gets no pussy. shut the fuck up.
just to let you know, Dr Weird rules, as does about 50% of the episodes in the first and second season, with revenge of the mooninites (the BELT) being the best. the third season has had a tinge of gayness, i would say that only 15-20% are very good, with the rest being sawdust.
but ATHF rules. GENTLEMAN. BEHOLD.
which only runs on my W2k workstation, because the distributed folks never rewrote the older cores that run on my pre-OSX Macs.
Have you tried mailing d.net about porting the clients to older Macs? They should be able to find someone willing to do it, although I admit the latest clients on their download page are very old.
I know for a fact that RC5-72 cores for PPC were made, because I wrote some of them :-)
Man, I had no idea that was so funny...
... hmmm ....
It is funny. Or at least, I'd rather laugh than cry. (note: I'm trying to be funny here).
But now I have at least one clue: both factors were of equal length, and were both half the length of the full number
The length of a product is equal to the sum of the lengths of it's factors (1/3 of the time it's less one). That part shouldn't be a surprise. Of the 45 different single digit multiplications (I'm excluding the use of 0), 30 produce a double digit product. 1*anything results in a single digit product, 2*2,3,4 also and 3*3. The sum_of_lengths idea for multiplication applies to all bases, not just decimal.
When I read their faq, I got the impression that the factors will be approximately of equal length (when read in binary). They also want to make it 'impossible' to crack it brute-force using a prime number list. If one factor is much smaller, then it'll be less effort using the bottom-up approach we all used in 3rd grade (x/2, x/3, x/5, x/7, etc).
I put a lot of time into the project so I won't just give it all away (see *NOTE* below). But I'll provide a couple ideas:
1. The two factors are prime. Being that both are odd values, their average will always be a whole number.
2. The average is always greater-than or equal-to the square-root of the product.
Now look up Fermat's method. It's the basis for most/all factoring algos. Here's a link. I don't necessarily like the way they describe it (there are easier ways), but it's a hint - just like you asked.
I regularly get new ideas on how to improve my approach, but each one shaves (on average) half the time. Half of ~10^74 years is still ~10^74 years. I am only using one computer, so that's one hinderence...
Here's another hint: don't use bottom-up as your main approach, it's a bigger waste of time than 10^74 years since you're guaranteed that the lower value is not small. Recall that RSA implements the use of two LARGE factors. 'Large' is definitely subjective but is relative to the given value.
I looked at your journal. The hints above help provide ways to skip large sections. 'Large' is again subjective.
Since I mentioned the journal, sorry - I'm not a fan of Descent. I just couldn't get used to the controls and navigating upside-down/sideways and so forth.
I'm sorry if this disappoints you, but I'm sure there's some quick and dirty way to find the answer without brute force. My opinion is that you just have to REALLY think different (like rejecting division or something just as radical), since no one's done better with the tools available. A historical analogy would be something like the introduction of polar coordinates or logarithmic scales - departures from conventional techniques to simplify a particular difficulty.
NOTE: I need to feel that it's actually worth something even though it's probably not.
This is not my sig.
Too bad then that there's no karma bonuses on ++funny
Nothing to see here; Move along.