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.
I have to check out whether my logic is on the right path.
NP-HARD: these are the hardest problems which may/may not be in NP class. If you have an efficient algorithm for these problems you have one for every problem in class NP.
NP COMPLETE: these are the hardest problems in class NP and also if you solve one of these you could solve any problem in class NP. So, NP COMPLETE problem is an NP-HARD problem.
COOK'S THEOREM: If SAT(NP-HARD) has a polynomial time algorithm then so does every problem in class NP.
Now, suppose we have to prove that CDP(clique decision problem) is NP COMPLETE.
->Step 1: Prove that CDP is in the class NP.
It is in class NP because the prover can generate a proof for yes inputs which would enable the verifier to check that it is a CDP (has a clique of size k).
->Step 2: Prove that CDP is NP HARD.
For that, we can convert the SAT to CDP by constructing a graph from clauses and supplying k.
We supply(G,k) to the clique subroutine which would verify is there a clique of size k or not. If it can figure this out in polynomial time then SAT has a polynomial time algorithm as CDP had a polynomial time algorithm and we converted SAT to CDP. So, now we proved that if there is a polynomial time algorithm for CDP then there is for the SAT. Now if we can find a polynomial time algorithm for CDP then it would imply that there is a polynomial time algorithm for SAT. This would imply that there is a polynomial time algorithm for every problem in NP by COOK'S THEOREM.
So we proved that CDP is NP COMPLETE. Once we have added CDP to NP COMPLETE class and now we come up with a new problem which we have again to prove that it is NP COMPLETE we can prove that problem to be in NP and then we could prove that if there is an efficient algorithm for given problem then that implies that there is an efficient algorithm for SAT/CDP(as we have added this to NP COMPLETE). Then as said above we can convert this problem to CDP/SAT and then prove that if there is an efficient algorithm for our problem then there is one for CDP/SAT and then by COOK'S THEOREM again we have that if there is a solution to NP-HARD problem (in this case CDP/SAT) then there is one for every problem in NP. So we again proved our problem as NP-HARD and as now it also belongs to NP as said above it is NP COMPLETE.
So we can add as many problems to the NP COMPLETE class as long as we can convert some problem which is already in NP-HARD class(in this case SAT/CDP) into our problem and we should find an efficient algorithm to our problem which would indirectly find an efficient algorithm to the NP-HARD problem and by COOK's theorem we can say that as some NP-HARD problem has an efficient algorithm we have an efficient algorithm to solve all problems in NP.
You're on the right path, but there your logic is a little incomplete.
The general structure of your proof is correct: First prove a problem is in NP, then prove the problem is NP-Hard. Those two bits of information together prove that a problem is NP-Complete.
Your proof for proving a problem is in NP is incomplete. Here are the key components to proving a problem is in NP:
Reword the problem as a decision problem that can be answered with a yes or a no.
Describe what a "certificate" would be. NOTE: a certificate is an output that can be checked to verify the answer to the decision problem. For CDP it could be a list of vertices and edges that make up the clique of size k.
Prove that this certificate can be verified in polynomial time.
Your proof for proving NP-Hard is incomplete. Here are the key components to proving a problem is NP-Hard:
Transform the input of the known NP-Hard problem into the input for the problem you are trying to prove.
Prove that this transformation can be done in polynomial time.
Transform the output of the problem your trying to prove into the output of the known NP-Hard problem.
Demonstrate how this can be done in polynomial time.
Prove that if you get an answer for the problem you are trying to prove, then you have an answer for the known problem.
Prove that if you get an answer for the known problem you have an answer for the problem you are trying to prove.
Only by meeting those 6 criterion can you say you have completely proven that a problem is NP-Hard.
Besides the specifics on that your logic is sound. Be careful when saying "efficient" if you really mean "can be solved in polynomial time".
I'm learning how to prove something is NP. In Thomas Cormen's intro to algorithm book, he states something is NP if given a solution to some problem, you can verify it is correct in polynomial time.
Say the problem is 2x + 9 = 55, and let's pretend I don't know how long it takes the find the correct x value, but an algorithm to solve the problem returned the solution 23. Then to show it is NP, do I only need to plug 23 in back in the equation, and see if that took a polynomial time and gave me 55? Thanks.
Yes; if you can check the correctness of every correct and every incorrect answer for every instance of this problem in polynomial time, then the problem is in NP.
Adding information to #Mehrdad answer:
Note that NP stands for Nondeterministic Polynomial time - it means that by definition - a problem is in NP if and only if it can be solved polynomially by a Non-deterministic Turing Machine.
It is equivalent to saying that in the standard computation model (RAM machine/ deterministic turing machine) - you can verify an answer in polynomial time (like #Mehrdad answered). The proof for the equivalence is described in the wikipedia page for equivalence of definitions
Bonus:
The question of "is everything that is verifiable (polynomially) on turing machine is also solveable polynomially" and the questions "is everything that is solveable on non-deterministic turing machine polynomially also solveable on deterministic turing machine polynomially" are also equivalent.
The answer is yet unknown and the problem is known as the P vs NP problem, which is the most important open question on computer science.
While the problems above cover the last, and perhaps, most identifiable step of NP-ness, there are 3 basic steps to showing that a problem is in NP.
Decision Problem: Can you turn your problem into a decision problem? In the case of the equation problem,the decision problem would be "is there an x such that 2x+9=55?" ?
Certificate: Can you identify an answer to your decision question? Again, in the case of the equation problem, an answer might be x = 23
Verification: Can you verify a certificate in polynomial time. By verifying in polynomial time, you know that the problem is not NP-Hard. Some verification steps might be: is x a number? is X equal to one half of 55-9?
That is how you know that your problem is completely in NP.
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.