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Consequences of a Solution to NP Complete Problems?
m00nshyn3 asks: "If a person were to find a O(n) solution to an NP complete
problem, it would obviously be a great advance in computer science,
but what are the consequences of such a discovery? Would our most
popular implementations
of cryptography be useless overnight? It seems like there is a lot
of immediate damage that could occur if such a solution were found.
So if (when) the time comes, what is the responsible way for the
solution to be made public?" If you had such an algorithm in hand,
what could you do with it? It would be interesting to see how many
problems we could map into the NP Complete model.
Salesmen will certainly be happy when such an algorithm is found. For other classes of NP-complete problems, check this book.
No we wouldn't see the whole crypto world come smashing down around us, but a large portion of it. All of the Public Key crypto would be reduced from an assumed set of hard problems to a set of simple ones. No more digital signatures, or communications without pre shared secrets. Although any system using just symmetric ciphers would be immune from this reduction in work effort.
Even if a timly NP complete solution is not found, PK based on factoring or discrete logarithms might get broken by other discoveries. For that reason I'm always watching the emergence of new PK systems such as FAPKC and some of the lattice based codes.
Crypto algorithms are strong for other reasons. Factoring primes is not a difficult problem... the fastest known solution is already O(n), the reason it's time consuming is that you can choose arbitrarily large values of n, in which case, you need to do better than O(n) in order to effectively break it. One time pads are another thing altogether. There is no algorithm of any order that can break a proper OTP (note that an improper OTP, i.e. one who's pad is not truly random or who pad is reused could be cracked). Other algorithms are based on other principles, but in any case, most are based on mathematically difficult or impossible problems, not computationally difficult ones.
I like to play children's songs in minor keys.
"We're all sons of bitches now." --J. Robert Oppenheimer
Considering that the Travelling Salesman Problem is NP complete and affects almost any problem that involves delivering something to several destinations in an optimal fashion. A solution to NP problems ould have widespread ramifications in improving many aspects of businesses that involve deliveries (including the airline business now that I think about it).
Even if someone manages to prove that P=NP, it doesn't mean that a reasonably efficient solution can be found. All it means is that an NP-complete problem can be solved in polynomial time. That polynomial can still be huge, say N^1000. Except for really large N, current exponential-time algorithms could be superior to polynomial-time ones.
So, the short answer is that proving P=NP probably won't ruin your encryption. On the other hand, if someone did prove it, there will probably be a mad scramble to invent some new encryption schemes, just in case.
-Chris
Uh...
If only it was easy to find any decimal of PI with a simple formula, [blah blah]
Well, it almost is. Not decimal, precisely, but any arbitrary hex digit...
sig fault
Regardless of what was said above, this doesn't destroy public key cryptography at all. The two biggest mathematical assumptions used in PK crypto are that:
:)
a) Factoring large numbers is hard
b) Solving discrete log problems is hard
Mind you, these are *not* NP-Complete problems (at the moment). They are believe to be in NP, but that's another story completly. Finding a polynomial algo for an NP-complete problem does not give you an algo for factoring and/or solving DLP problems.
Now on the other hand, if you had a quantum computer, you could factor in quadratic time, and solve DLPs in cubic time. Now *that* would be somewhat bad.
I've actually got quite a few pizza this way. If this NP solution helps to speed up pizza delivery, I don't I would like that, at all.
Several router manufacturers are designing their new equipment with support for MPLS. MPLS, among other things, offers better support for traffic engineering. Traffic engineering means deciding which router hops to take to get from border point A to border point B (it also means you can select an alternate route depending on network congestion or backbone router failures).
Finding the most efficient path between the two endpoints is an NP complete problem.
I think this problem also exists for BGP/IGP networks, but I'm not experienced with them. I do know that the promise of the internet being secure and redundant and safe from one node being bombed is a bunch of nonsense. Although the technology to route around outages exists, the chore of houskeeping for it has turned out to be rather intractable so it is minimally implemented at best. Usually if a large backbone provider suffers a hardware failure and can't swap in a backup, they call in one of their router guru's to look at their topology map and configure their other routers to route data around the bad router at that time. That's the automated process.
All the providers out there would *REALLY* like a tool that could take their network topology map and output a reasonable set of LSPs (MPLS tags that indicate the route the packet is to take) for them in a reasonable amount of time. However, since this problem is NP complete, such a tool would have to compute every possible path and then choose those at the top of the list.
Education is a better safeguard of liberty than a standing army.
Edward Everett (1794 - 1865)
Firstly, an affirmative answer to the NP=P? conjecture only means that there is a solution to every NP-complete problem in P. That is, there exists a solution for every NP-complete problem that is O(n^d), where d is a constant integer. If d > 3, the solution would be practically infeasible anyway. Furthermore, even with an O(n) problem, this only means that the computational complexity approaches C*n in the limit of large n, where C is some constant. If C has to be arbitrarily large, or there exists a large constant additive factor in any potential solution, again the solution is infeasible.
Furthermore, the security of public key cryptography does not rely on NP!=P. It is not known whether the discrete-log and integer factoring problems are in NP (I think ... correct me if that's wrong). In fact, some CS researchers believe public key cryptography to be insecure, since some brilliant person could come up with a feasible factoring algorithm tomorrow, without requiring that NP=P.
Toronto-area transit rider? Rate your ride.
Not all encryption schemes are NP-Complete! Most are actually just NP-Hard because you can't tell whether or not you've found the correct decryption. So, decryption schemes will not be solved even if you can convert into NP complete problems into NP problems. It will be a lot easier though.
That polynomial can still be huge, say N^1000. Except for really large N, current exponential-time algorithms could be superior to polynomial-time ones.
This lacks some of the insight into NP and NPC problems. We only care about the large cases for the most part. On the small scale (small being relative to the problem), exponential solutions are always easy to solve.
There are some amazing implications of this anyway. For instance, we can solve chess (find the best possible game), and all other decidable search problems.
Keep in mind that our computer improvements allow us to make polynomial time reductions in the amount of time that the problem takes.
Mod me down and I will become more powerful than you can possibly imagine!
The thing we would get if someone were to find a polynomial-time algorithm for any NP-complete problem is an immediate, poly-time algorithm for every NP-complete problem. This is because the definition of NP-complete is that there is a (known) poly-time algorithm to turn any one NP-complete algorithm into any other, so just by composing these two you get them all. (I'll attach a glossary at the bottom -- most people on this list probably aren't mathematicians :)
But OK, what does this mean realistically? The good news is that there are several very useful NP-complete problems; probably the best known (as someone has already mentioned) is the travelling salesman, and being able to do fast TS problems could mean incredible reductions in cost for shipping of goods and things like that. All sorts of problems in computer network architecture are also NP-complete; think about trying to design an internet which is both fault-tolerant and maximizing bandwidth.
The bad news: There are two things. First of all: This does not mean encryption of any sort is broken! The heart of public-key crypto is that factorization takes exponential time (or more specifically, the discrete logarithm, which is at the heart of fast factorization, takes exp time) and so if you could do poly-time factorization, you could break various algorithms like RSA. But factorization is only conjectured to be NP-complete; there is no proof, and in particular the explicit algorithm which would be needed to use a poly-time algorithm for some other NP-complete problem for factorization isn't known. This doesn't mean it can't be done; it just means that there's one other significant step between finding such an algorithm and breaking crypto.
The second problem is that even a poly-time algorithm isn't necessarily useful if the coefficients are large. What poly-time really means is that, in the limit of very large inputs, computation time doesn't go completely out of control; the fact that (to the best of current knowledge) factorization isn't poly is what makes adding one digit to key size enough to increase the difficulty of decryption by a factor of two. (i.e., the work increases as an exponential of the input) So this is important when you're trying to create "sufficiently large" inputs to jam up an algorithm. But for real-world problems that people are trying to solve apart from crypto, an O(N^1000) algorithm might technically be "better" than an O(e^N) algorithm but practically still be way out of reach.
In fact, most of the interesting NP-complete problems such as travelling salesman are routinely worked on by methods which give approximate answers in fairly short time; this turns out to be more than good enough for a remarkable range of uses, which means that the advance of getting poly time wouldn't be as earth-shaking for most real-world applications.
I've discovered a wonderful linear way to calculate optimal routing that I'd like to share, but unfortunately my keyboagu;[f=s af\sdfgsv asdfw352.,.f354asf
Prove that it's impossible to solve NP-Complete problems in linear time, before someone figures out how to do it.
dinner: it's what's for beer
false.
Definitions:
P is a class of problems that can be decided by a deterministic turing machine in polynomial time.
NP s a class of problems that can be decided by a non-deterministic turing machine in polynomial time. It means (literally) non-deterministic polynomial.
A problem is NP-Complete if:
1. It is in NP.
2. Any other problem in NP can be reduced (read: "converted") to it in polynomial time.
So, if a polynomial-time solution to an NP-complete problem is ever found, any other problem in NP will automatically have a polynomial-time solution. That includes most of the known algorighms.
Oh, and just to preempt stupid replies saying "it depends on which NP-complete problem is solved" or "perhaps the problem is misclassified": read the definition again. Any problem in NP can be reduced to an NP-complete problem in polynomial time; including another NP-complete problem.
Finally, a description of what P ?= NP question is:
P is a subset of NP (obviously). The question is whether it is a proper subset (i.e. P is strictly smaller than NP) or P and NP are actually coincident (i.e. P = NP). The answer so far is "probably not". There is a large number of NP-complete problems and so far no one has been able to come up with an efficient (i.e. poly-time) algorithm to solve any of them. However, P = NP implies that such algorithm exists (and vice versa).
(Stupid slashdot filter deletes less then signs)
___
If you think big enough, you'll never have to do it.
It doesn't surprise me any more when Slashdot displays large amounts of ignorance about non-computer topics, but I expected better for something like this. I've been skimming through the +2 and higher comments, and not a single poster has defined NP-complete correctly (this may have changed by the time you read this, but it was true when there were >50 comments at 2 or above).
Here's the correct definition of NP-complete:
To be in NP-complete, a problem must be in NP--that is, it must be a concrete decision problem, and have a polynomial-time verification algorithm (i.e., if somebody hands you a solution to the problem, you can verify that it's correct in polynomial time). Furthermore, there must be a polynomial-time mapping from every problem in NP (not just NP-complete) to your problem.
My source for this is Introduction to Algorithms, by Cormen, Leiserson, and Rivest (yes, that's THE Rivest of cryptography fame), which is part of the MIT Electrical Engineering and Computer Science Series.
Now, some people have been getting confused and saying that being in NP-complete only requires there to be a polynomial-time mapping from every problem in NP. This is an understandable mistake, since the way one usually proves a problem is in NP-complete is by finding a polynomial-time mapping from one NP-complete problem to the problem of interest (which must already have been shown to be in NP). However, since you already know that there are polynomial-time mappings from every problem in NP to the NP-complete problem, there is thus also a polynomial-time mapping from every problem in NP to your problem. The difference here is that efficiently solving an NP-complete problem means you can efficiently solve all NP problems, not just the other NP-complete ones.
The second big error--the one that boggles my mind--is that the poster seems to have confused O(n), which is linear time, with polynomial time. True, O(n) implies polynomial running time, but polynomial-time does not necessarily imply O(n)--you could have O(n^2), O(n^3)...
The third mistake is the implications of any polynomial-time solution to an NP-complete problem--even if it is O(n^1000). A few people here are claiming a polynomial-time solution would be irrelevant if the order of the polynomial was large (giving O(n^1000) as an example of large). For moderately large inputs (and we're really not talking that large), O(n^1000) is better than O(e^n). Taking the example of O(n^1000), n only has to be greater than 10,000 for O(n^1000) to beat O(e^n). Exponentials grow fast, people.
Finally, getting to implications, any polynomial-time solution to any NP-complete problem would instantly destroy public-key cryptography. Even if everybody immediately switched cryptosystems, the implications would be staggering, because someone could have archived all sorts of encrypted transactions, the contents of which are still sensitive. My understanding is that prime factorization is in NP, but not in NP-complete (not that this matters too much, as I explained earlier). Not sure about the math other forms of PK crypto are based on, but I suspect it's also NP.
On the plus side, protein folding simulation is suspected to be in NP (this may have been proven--not sure). If you could do protein folding simulations efficiently, it would make finding a cure to most diseases a hell of a lot easier (and we'd finally be able to figure out what the hell the human genome means).
The simplest problem to understand in NP space is optimization. EVERY optimization problem is NP-complete. If you want to find the shortest route to visit every city in the US, this is an optimization problem.
What a problem being in NP-complete space means is that if you can solve one problem in the space, you can solve all problems in that space (by transforming the other problems to look like the solved problem and then solving that).
So if you prove P=NP, then you can solve optimization problems in polynomial time O(n^t). Pick a problem where you could say "I wonder what the fastest/shortest/lightest way of doing this is?" and you could solve it in O(n^t).
4 caveats/common misconceptions:
O(n^t) could be very VERY large. n^237203 still takes a very long time.
As far as I know, encryption algorithms are NOT KNOWN to be NP-complete. So it's UNKNOWN what proving P=NP means to the encryption world.
People who say P=NP is not provable aren't conveying the situation. The situation is that it is UNKNOWN if P=NP or not. They don't know the answer.
There are problems that do not fall into NP-complete class. There are NP problems that are not complete. There are (provably) problems that cannot be solved. So P=NP doesn't really act as a cycle multiplier (you don't just get more pure computation), it just gives you another trick in your bag.
So my answer for what happens if P=NP is proven: Everything in the world gets cheaper.
passetspike!
It's interesting how most of the answers here fail to look at the actual question. The question wasn't so much, "what will break;" the question was, "what should one do."
Although it's interesting and even essential to review what parts of computer science or our economy would be toppled by such a discovery, doing so doesn't answer the question.
Let me rephrase the question for any who missed it: "Suppose you discover that P=NP. What is the right thing to do?" Do I cover it up, or do I release? What if I proved that P=NP, but I don't know of an algorithm to actually convert any known problem? Or, what if I did know the algorithm and the proof, and I believed that the algorithm couldn't be reconstructed from the proof -- should I release the proof?
This is a powerful question!
My feeling is that the truth should be known, but experience shows that information without knowledge, and knowledge without understanding, can be deadly. (I'm afraid you can't affect whether people will get understanding without wisdom, so we'll stop the natural progression before we reach that point.) A little survey of the literature (please see the other posts here; they've got GREAT info) shows that the practical benefits of this discovery would be IMMENSE. The drawbacks seem huge, too; but if a particular algorithm has become easier to destroy, so also do new algorithms open up. Look around -- identify as many existing things which are harmed by your discovery, and try to provide a discussion of how to recover from the harm.
Even if you're not altruistic you want to do this; the person who is most essential in the new world isn't the person who discovered the info, since once the secret's out it's not under his control; it's the person on whom people are depending. Be that person -- but don't be selfish about it. Too selfish ruins the game too; there's always someone with enough power to take your position away from you.
So that's my knee-jerk answer, with a bit of reasoning: research the discovery, find who it hurts, and prepare to help them. Then make it public.
There's another question which is implied by the first one: to whom do I release info about my discovery? I can't answer that. Does anyone have an answer? I can say that you'll HAVE to release some information before you're ready to release it all; for example, you might want to found a corporation, and you'll almost certainly need library assistance. What can I say about that? Pick people you can trust, and don't trust them with too much.
-Billy
I solved it! I solved it!
I have the proof right here! I'm going to post it right now, right here on Slashdot!
Oh wait, there's a knock at my door. Hang in there folks, I'll be right back...
Muffled Yell
Thud
Long Silence
intellectual property law is philosophically incoherent. it is your moral duty to ignore it or sabotage it