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Why is NP-hard unequal to NP-complete?
My informal understanding of definitions being used:
NP - all problems that can be verified in polynomial time
NP-complete - all problems that are NP and NP-hard
NP-hard - at least as hard as the hardest problem in NP
Decision Problem - A problem that asks a question with regards to an input and outputs a bool value
Confusion:
The problem with unknown solution of P vs NP arises from the fact that we cannot prove or disprove all problems in NP can be solved in polynomial time. It feels like a similar question arises from NP-complete vs NP-hard. How do we know all problems in NP-hard cannot be verified in polynomial time and thus result in NP-hard=NP-complete?
Here is my line of reasoning
From online research the distinction seems that this has something to do with decision problems (a concept I'm entirely new to but seem simple enough). I think this means that problems in NP have complementary decision problems that ask if an input is the solution to the problem. Let's say the problem is to find an optimal solution. I believe the complementary decision problem to be "is the given input the optimal solution?"and I believe that if this decision problem is verifiable in polynomial time then the problem is NP-complete (or in NP). So this means that NP-hard problems that aren't NP-complete problems are those that either have no decision problem (which I believe is never true since any brute force solution can answer this) or a problem is NP-hard and not NP-complete if it has a decision problem that's not verifiable in polynomial time. If it is the latter then it feels like we have the same problem from P vs NP. That is, how do we confirm all decision problems in NP-hard do not have polynomial time solutions?
Sorry if the above phrasing is weird. I will try and clarify any confusion in my question.
notes
I am interested in both an intuitive explanation and a formal explanation (a proof if it's a complicated answer). The formal explanation can certainly be a link to an academic paper. I don't want anyone to invest a significant amount of time into an overly complicated proof that may be beyond the scope of my understanding (I've found complexity theory to become very quickly... complex).
If it helps for the sake of explanation I have done work on the traveling salesman problem and I am currently working on a paper for the nurse scheduling problem (I believe these are NP-hard problems).
NP-Hard includes all problems whose solutions can be used to derive solutions to problems in NP with polynomial overhead.
This includes lots of problems that aren't in NP. For instance, the halting problem - an undecidable problem - is NP-Hard, because any problem in NP can be reduced to it in polynomial time:
Reduce any problem in NP to an instance of the NP-Complete problem 3-SAT
Construct in polynomial time a TM which checks all assignments and halts iff a satisfying assignment is found.
Use a solution to the halting problem to tell whether the TM halts.
If it halts, accept; otherwise, reject.
I get that if you can do a polynomial time reduction from "every" problem then it proves that the problem is at least as hard as every problem in NP. Except, how do we know that we've discovered every problem in NP? Can't there exist problems that we may not have discovered or proven exist in NP but CANNOT be reduced to any np-complete problem? Or is this still an open question?
As others have correctly stated, the existence of the problem that is NP, but is not NP-complete would imply that P != NP, so finding one would bring you a million dollar and eternal glory. One famous problem that is believed to belong in this class is integer factorization. However, your original question was
Can't there exist problems that we may not have discovered or proven
exist in NP but CANNOT be reduced to any np-complete problem?
The answer is no. By definition of NP-completeness, one of two
necessary conditions for a problem A to be NP-complete is that every NP problem needs to be reducible in polynomial time to A. If you want to find out how to prove that every single NP problem can be reducible in polynomial time to some NP-complete problem, have a look at the proof of Cook-Levin theorem that states that 3-SAT problem is NP-complete. It was the first proven NP-complete problem and many other NP-complete problems are later proven to be NP-complete by finding the appropriate reduction from 3-SAT to these problems.
NP consists of all problems that could (theoretically) be solved by being able to make lucky guesses, guessing the solution and checking in polynomial time that the solution is correct. For example, the travelling salesman problem "can I visit the capitols of all 50 states of the USA with a trip of less than 9,825 miles" can be solved by guessing a trip and checking that it is not too long.
And one problem in NP is basically simulating a programmable computer circuit with various inputs and checking whether a certain output can be achieved. And that programmable computer circuit is powerful enough to solve all problems in NP.
So yes, we know all about all problems in NP.
(Then of course an NP complete problem can by definition be used to solve any problem in NP. If there is a problem that it cannot solve, that problem is not in NP).
Except, how do we know that we've discovered every problem in NP?
We don't. The set of all problems in the universe is not only infinite, but uncountable.
Can't there exist problems that we may not have discovered or proven
exist in NP but CANNOT be reduced to any np-complete problem?
We don't know that. We suspect that this is the case, but this hasn't been proven yet. If we were to find a NP problem that is not in NP-Complete, it would be proof that P =/= NP.
It is one of the great unsolved problems in CS. Many brilliant minds have been taking a go at it, but this nut has been one tough one to crack.
Am a bit confused about the relationship between undecidable problems and NP hard problems. Whether NP hard problems are a subset of undecidable problems, or are they just the same and equal, or is it that they are not comparable?
For me, I have been arguing with my friends that undecidable problems are a superset to the NP hard problems. There would exist some problems that are not in NP hard but are undecidable. But i am finding this argument to be weak and am confused a bit. Are there NP-complete problems that are undecidable.? is there any problem in NP hard which is decidable.??
Some discussion would be of great help! Thanks!
Undecidable = unsolvable for some inputs. No matter how much (finite) time you give your algorithm, it will always be wrong on some input.
NP-hard ~= super-polynomial running time (assuming P != NP). That's hand-wavy, but basically NP-hard means it is at least as hard as the hardest problem in NP.
There are certainly problems that are NP-hard which are not undecidable (= are decidable). Any NP-complete problem would be one of them, say SAT.
Are there undecidable problems which are not NP-hard? I don't think so, but it isn't easy to rule it out - I don't see an obvious argument that there must be a reduction from SAT to all possible undecidable problems. There could be some weird undecidable problems which aren't very useful. But the standard undecidable problems (the halting problem, say) are NP-hard.
An NP-hard is a problem that is at least as hard as any NP-complete problem.
Therefore an undecidable problem can be NP-hard. A problem is NP-hard if an oracle for it would make solving NP-complete problems easy (i.e. solvable in polynomial time). We can imagine an undecidable problem such that, given an oracle for it, NP-complete problems would be easy to solve. For example, obviously every oracle that solves the halting problem can also solve an NP-complete problem, so every Turing-complete problem is also NP-hard in the sense that a (fast) oracle for it would make solving NP-complete problems a breeze.
Therefore Turing-complete undecidable problems are a subset of NP-hard problems.
Undecidable problem e.g. Turing Halting Problem is NP-Hard only.
<---------NP Hard------>
|------------|-------------||-------------|------------|--------> Computational Difficulty
|<----P--->|
|<----------NP---------->|
|<-----------Exponential----------->|
|<---------------R (Finite Time)---------------->|
In this diagram, that small pipe shows overlapping of NP and NP-Hard and which shows NP-Completeness, i.e. set of those problems which are NP as well as NP-Hard.
Undecidable problems are NP Hard problems which do not have solution and which are not in NP.
Would it be an polynomial time algorithm to a specific NP-complete problem, or just abstract reasonings that demonstrate solutions to NP-complete problems exist?
It seems that the a specific algoithm is much more helpful. With it, all we'll have to do to polynomially solve an NP problem is to convert it into the specific NP-complete problem for which the proof has a solution, and we are done.
P = NP: "The 3SAT problem is a classic NP complete problem. In this proof, we demonstrate an algorithm to solve it that has an asymptotic bound of (n^99 log log n). First we ..."
P != NP: "Assume there was a polynomial algorithm for the 3SAT problem. This would imply that .... which by ..... implies we can do .... and then ... and then ... which is impossible. This was all predicated on a polynomial time algorithm for 3SAT. Thus P != NP."
UPDATE: Perhaps something like this paper (for P != NP).
UPDATE 2: Here's a video of Michael Sipser sketching out a proof for P != NP
Call me pessimistic, but it will be like this:
...
∴, P ≠ NP
QED
There are some meta-results about what a P=NP or P≠NP proof can not look like. The details are quite technical, but it is known that the proof cannot be
relativizing, which kind of means that the proof must make use of the exact definition of Turing machine used, because with some modifications ("oracles", like very powerful CISC instructions added to the instruction set) P=NP, and with some other modifications, P≠NP. See also this blog post for a nice explanation of relativization.
natural, a property of several classic circuit complexity proofs,
or algebrizing, a generalization of relativizing.
It could take the form of demonstrating that assuming P ≠ NP leads to a contradiction.
It might not be connected to P and NP in a straightforward way... Many theorems now are based on P!=NP, so proving one assumed fact to be untrue would make a big difference. Even proving something like constant ratio approximation for TS should be enough IIRC. I think, existence of NPI (GI) and other sets is also based on P!=NP, so making any of them equal to P or NP might change the situation completely.
IMHO everything happens now on a very abstract level. If someone proves anything about P=/!=NP, it doesn't have to mention any of those sets or even a specific problem.
Probably it would be in the form of a reduction from an NP problem to a P problem. See the Wikipedia page on reductions.
OR
Like this proof proposed by Vinay Deolalikar.
The most straightforward way is to prove that there is a polynomial time solution to the problems in the class NP-complete. These are problems that are in NP and are reducable to one of the known np problem. That means you could give a faster algorithm to prove the original problem posed by Stephen Cook or many others which have also been shown to be NP-Complete. See Richard Karp's seminal paper and this book for more interesting problems. It has been shown that if you solve one of these problems the entire complexity class collapses. edit: I have to add that i was talking to my friend who is studying quantum computation. Although I had no clue what it means, he said that a certain proof/experiment? in the quantum world could make the entire complexity class, i mean the whole thing, moot. If anyone here knows more about this, please reply.
There have also been numerous attempts to the problem without giving a formal algorithm. You could try to count the set. Theres the Robert/Seymore proof. People have also tried to solve it using the tried and tested diagonlization proof(also used to show that there are problems that you can never solve). Razborov also showed that if there are certain one-way functions then any proof cannot give a resolution. That means that new techniques will be required in order to solve this question.
Its been 38 years since the original paper has been published and there still is no sign of a proof. Not only that but lot of problems that mathematicians had been posing before the notion of complexity classes came in has been shown to be NP. Therefor many mathematicians and computer scientists believe that some of the problems are so fundamental that a new kind of maths may be needed to solve the problem. You have to keep in mind that the best minds human race has to offer have tackled this problem without any success. I think it should be at least decades before somebody cracks the puzzle. But even if there is a polynomial time solution the constants or the exponent could be so large that it would be useless in our problems.
There is an excellent survey available which should answer most of your questions: http://www.scottaaronson.com/papers/pnp.pdf.
Certainly a descriptive proof is the most useful, but there are other categories of proof: it is possible, for example, to provide 'existence proofs' that demonstrate that it is possible to find an answer without finding (or, sometimes, even suggesting how to find) that answer.
Set N equal to the multiplicative identity. Then NP = P. QED. ;-)
It would likely look almost precisely like one of these
Good question; it could take either form. Obviously, the specific algorithm would be more helpful, yes, but there's no determining that that would be the way that a theoretical P=NP proof would occur. Given that the nature of NP-complete problems and how common they are, it would seem that more effort has been put into solving those problems than has been put into solving the theoretical reasoning side of the equation, but that's just supposition.
Any nonconstructive proof that P=NP really is not. It would imply that the following explicit 3-SAT algorithm runs in polynomial time:
Enumerate all programs. On round i, run all programs numbered
less than i for one step. If
a program terminates with a
satisfying input to the formula, return true. If a program
terminates with a formal proof that
no such input exists, return
false.
If P=NP, then there exists a program which runs in O(poly(N)) and outputs a satisfying input to the formula, if such a formula exists.
If P=coNP, there exists a program which runs in O(poly(N)) and outputs a formal proof that no formula exists, if no formula exists.
If P=NP, then since P is closed under complement NP=coNP. So, there exists a program which runs in O(poly(N)) and does both. That program is the k'th program in the enumeration. k is O(1)! Since it runs in O(poly(N)) our brute force simulation only requires
k*O(poly(N))+O(poly(N))^2
rounds once it reaches the program in question. As such, the brute force simulation runs in polynomial time!
(Note that k is exponential in the size of the program; this approach is not really feasible, but it suggests that it would be hard to do a nonconstructive proof that P=NP, even if it were the case.)
An interesting read that is somewhat related to this
To some extent, the form such a proof needs to have depends on your philosophical point of view (= the axioms you deem to be true) - e.g., as a contructivist you would demand the construction of an actual algorithm that requires polynomial time to solve an NP-complete problem. This could be done by using reduction, but not with an indirect proof. Anyhow, it really seems to be very unlikely :)
The proof would deduce a contradiction from to the assumption that at least one element (problem) of NP isn't also an element of P.
Assuming some background in mathematics, how would you give a general overview of computational complexity theory to the naive?
I am looking for an explanation of the P = NP question. What is P? What is NP? What is a NP-Hard?
Sometimes Wikipedia is written as if the reader already understands all concepts involved.
Hoooo, doctoral comp flashback. Okay, here goes.
We start with the idea of a decision problem, a problem for which an algorithm can always answer "yes" or "no." We also need the idea of two models of computer (Turing machine, really): deterministic and non-deterministic. A deterministic computer is the regular computer we always thinking of; a non-deterministic computer is one that is just like we're used to except that is has unlimited parallelism, so that any time you come to a branch, you spawn a new "process" and examine both sides. Like Yogi Berra said, when you come to a fork in the road, you should take it.
A decision problem is in P if there is a known polynomial-time algorithm to get that answer. A decision problem is in NP if there is a known polynomial-time algorithm for a non-deterministic machine to get the answer.
Problems known to be in P are trivially in NP --- the nondeterministic machine just never troubles itself to fork another process, and acts just like a deterministic one. There are problems that are known to be neither in P nor NP; a simple example is to enumerate all the bit vectors of length n. No matter what, that takes 2n steps.
(Strictly, a decision problem is in NP if a nodeterministic machine can arrive at an answer in poly-time, and a deterministic machine can verify that the solution is correct in poly time.)
But there are some problems which are known to be in NP for which no poly-time deterministic algorithm is known; in other words, we know they're in NP, but don't know if they're in P. The traditional example is the decision-problem version of the Traveling Salesman Problem (decision-TSP): given the cities and distances, is there a route that covers all the cities, returning to the starting point, in less than x distance? It's easy in a nondeterministic machine, because every time the nondeterministic traveling salesman comes to a fork in the road, he takes it: his clones head on to the next city they haven't visited, and at the end they compare notes and see if any of the clones took less than x distance.
(Then, the exponentially many clones get to fight it out for which ones must be killed.)
It's not known whether decision-TSP is in P: there's no known poly-time solution, but there's no proof such a solution doesn't exist.
Now, one more concept: given decision problems P and Q, if an algorithm can transform a solution for P into a solution for Q in polynomial time, it's said that Q is poly-time reducible (or just reducible) to P.
A problem is NP-complete if you can prove that (1) it's in NP, and (2) show that it's poly-time reducible to a problem already known to be NP-complete. (The hard part of that was provie the first example of an NP-complete problem: that was done by Steve Cook in Cook's Theorem.)
So really, what it says is that if anyone ever finds a poly-time solution to one NP-complete problem, they've automatically got one for all the NP-complete problems; that will also mean that P=NP.
A problem is NP-hard if and only if it's "at least as" hard as an NP-complete problem. The more conventional Traveling Salesman Problem of finding the shortest route is NP-hard, not strictly NP-complete.
Michael Sipser's Introduction to the Theory of Computation is a great book, and is very readable. Another great resource is Scott Aaronson's Great Ideas in Theoretical Computer Science course.
The formalism that is used is to look at decision problems (problems with a Yes/No answer, e.g. "does this graph have a Hamiltonian cycle") as "languages" -- sets of strings -- inputs for which the answer is Yes. There is a formal notion of what a "computer" is (Turing machine), and a problem is in P if there is a polynomial time algorithm for deciding that problem (given an input string, say Yes or No) on a Turing machine.
A problem is in NP if it is checkable in polynomial time, i.e. if, for inputs where the answer is Yes, there is a (polynomial-size) certificate given which you can check that the answer is Yes in polynomial time. [E.g. given a Hamiltonian cycle as certificate, you can obviously check that it is one.]
It doesn't say anything about how to find that certificate. Obviously, you can try "all possible certificates" but that can take exponential time; it is not clear whether you will always have to take more than polynomial time to decide Yes or No; this is the P vs NP question.
A problem is NP-hard if being able to solve that problem means being able to solve all problems in NP.
Also see this question:
What is an NP-complete in computer science?
But really, all these are probably only vague to you; it is worth taking the time to read e.g. Sipser's book. It is a beautiful theory.
This is a comment on Charlie's post.
A problem is NP-complete if you can prove that (1) it's in NP, and
(2) show that it's poly-time reducible to a problem already known to
be NP-complete.
There is a subtle error with the second condition. Actually, what you need to prove is that a known NP-complete problem (say Y) is polynomial-time reducible to this problem (let's call it problem X).
The reasoning behind this manner of proof is that if you could reduce an NP-Complete problem to this problem and somehow manage to solve this problem in poly-time, then you've also succeeded in finding a poly-time solution to the NP-complete problem, which would be a remarkable (if not impossible) thing, since then you'll have succeeded to resolve the long-standing P = NP problem.
Another way to look at this proof is consider it as using the the contra-positive proof technique, which essentially states that if Y --> X, then ~X --> ~Y. In other words, not being able to solve Y in polynomial time isn't possible means not being to solve X in poly-time either. On the other hand, if you could solve X in poly-time, then you could solve Y in poly-time as well. Further, you could solve all problems that reduce to Y in poly-time as well by transitivity.
I hope my explanation above is clear enough. A good source is Chapter 8 of Algorithm Design by Kleinberg and Tardos or Chapter 34 of Cormen et al.
Unfortunately, the best two books I am aware of (Garey and Johnson and Hopcroft and Ullman) both start at the level of graduate proof-oriented mathematics. This is almost certainly necessary, as the whole issue is very easy to misunderstand or mischaracterize. Jeff nearly got his ears chewed off when he attempted to approach the matter in too folksy/jokey a tone.
Perhaps the best way is to simply do a lot of hands-on work with big-O notation using lots of examples and exercises. See also this answer. Note, however, that this is not quite the same thing: individual algorithms can be described by asymptotes, but saying that a problem is of a certain complexity is a statement about every possible algorithm for it. This is why the proofs are so complicated!
I remember "Computational Complexity" from Papadimitriou (I hope I spelled the name right) as a good book
very much simplified: A problem is NP-hard if the only way to solve it is by enumerating all possible answers and checking each one.
Here are a few links on the subject:
Clay Mathematics statement of P vp NP problem
P vs NP Page
P, NP, and Mathematics
In you are familiar with the idea of set cardinality, that is the number of elements in a set, then one could view the question like P representing the cardinality of Integer numbers while NP is a mystery: Is it the same or is it larger like the cardinality of all Real numbers?
My simplified answer would be: "Computational complexity is the analysis of how much harder a problem becomes when you add more elements."
In that sentence, the word "harder" is deliberately vague because it could refer either to processing time or to memory usage.
In computer science it is not enough to be able to solve a problem. It has to be solvable in a reasonable amount of time. So while in pure mathematics you come up with an equation, in CS you have to refine that equation so you can solve a problem in reasonable time.
That is the simplest way I can think to put it, that may be too simple for your purposes.
Depending on how long you have, maybe it would be best to start at DFA, NDFA, and then show that they are equivalent. Then they understand ND vs. D, and will understand regular expressions a lot better as a nice side effect.