What are the differences between NP, NP-Complete and NP-Hard?
I am aware of many resources all over the web. I'd like to read your explanations, and the reason is they might be different from what's out there, or there is something that I'm not aware of.
I assume that you are looking for intuitive definitions, since the technical definitions require quite some time to understand. First of all, let's remember a preliminary needed concept to understand those definitions.
Now, let us define those complexity classes.
P is a complexity class that represents the set of all decision problems that can be solved in polynomial time.
That is, given an instance of the problem, the answer yes or no can be decided in polynomial time.
Example
Given a connected graph G, can its vertices be coloured using two colours so that no edge is monochromatic?
Algorithm: start with an arbitrary vertex, color it red and all of its neighbours blue and continue. Stop when you run out of vertices or you are forced to make an edge have both of its endpoints be the same color.
NP is a complexity class that represents the set of all decision problems for which the instances where the answer is "yes" have proofs that can be verified in polynomial time.
This means that if someone gives us an instance of the problem and a certificate (sometimes called a witness) to the answer being yes, we can check that it is correct in polynomial time.
Example
Integer factorisation is in NP. This is the problem that given integers n and m, is there an integer f with 1 < f < m, such that f divides n (f is a small factor of n)?
This is a decision problem because the answers are yes or no. If someone hands us an instance of the problem (so they hand us integers n and m) and an integer f with 1 < f < m, and claim that f is a factor of n (the certificate), we can check the answer in polynomial time by performing the division n / f.
NP-Complete is a complexity class which represents the set of all problems X in NP for which it is possible to reduce any other NP problem Y to X in polynomial time.
Intuitively this means that we can solve Y quickly if we know how to solve X quickly. Precisely, Y is reducible to X, if there is a polynomial time algorithm f to transform instances y of Y to instances x = f(y) of X in polynomial time, with the property that the answer to y is yes, if and only if the answer to f(y) is yes.
Example
3-SAT. This is the problem wherein we are given a conjunction (ANDs) of 3-clause disjunctions (ORs), statements of the form
(x_v11 OR x_v21 OR x_v31) AND
(x_v12 OR x_v22 OR x_v32) AND
... AND
(x_v1n OR x_v2n OR x_v3n)
where each x_vij is a boolean variable or the negation of a variable from a finite predefined list (x_1, x_2, ... x_n).
It can be shown that every NP problem can be reduced to 3-SAT. The proof of this is technical and requires use of the technical definition of NP (based on non-deterministic Turing machines). This is known as Cook's theorem.
What makes NP-complete problems important is that if a deterministic polynomial time algorithm can be found to solve one of them, every NP problem is solvable in polynomial time (one problem to rule them all).
Intuitively, these are the problems that are at least as hard as the NP-complete problems. Note that NP-hard problems do not have to be in NP, and they do not have to be decision problems.
The precise definition here is that a problem X is NP-hard, if there is an NP-complete problem Y, such that Y is reducible to X in polynomial time.
But since any NP-complete problem can be reduced to any other NP-complete problem in polynomial time, all NP-complete problems can be reduced to any NP-hard problem in polynomial time. Then, if there is a solution to one NP-hard problem in polynomial time, there is a solution to all NP problems in polynomial time.
Example
The halting problem is an NP-hard problem. This is the problem that given a program P and input I, will it halt? This is a decision problem but it is not in NP. It is clear that any NP-complete problem can be reduced to this one. As another example, any NP-complete problem is NP-hard.
My favorite NP-complete problem is the Minesweeper problem.
This one is the most famous problem in computer science, and one of the most important outstanding questions in the mathematical sciences. In fact, the Clay Institute is offering one million dollars for a solution to the problem (Stephen Cook's writeup on the Clay website is quite good).
It's clear that P is a subset of NP. The open question is whether or not NP problems have deterministic polynomial time solutions. It is largely believed that they do not. Here is an outstanding recent article on the latest (and the importance) of the P = NP problem: The Status of the P versus NP problem.
The best book on the subject is Computers and Intractability by Garey and Johnson.
I've been looking around and seeing many long explanations. Here is a small chart that may be useful to summarise:
Notice how difficulty increases top to bottom: any NP can be reduced to NP-Complete, and any NP-Complete can be reduced to NP-Hard, all in P (polynomial) time.
If you can solve a more difficult class of problem in P time, that will mean you found how to solve all easier problems in P time (for example, proving P = NP, if you figure out how to solve any NP-Complete problem in P time).
____________________________________________________________ | Problem Type | Verifiable in P time | Solvable in P time | Increasing Difficulty ___________________________________________________________| | | P | Yes | Yes | | | NP | Yes | Yes or No * | | | NP-Complete | Yes | Unknown | | | NP-Hard | Yes or No ** | Unknown *** | | ____________________________________________________________ V
Notes on Yes or No entries:
I also found this diagram quite useful in seeing how all these types correspond to each other (pay more attention to the left half of the diagram).
P (Polynomial Time): As name itself suggests, these are the problems which can be solved in polynomial time.
NP (Non-deterministic-polynomial Time): These are the decision problems which can be verified in polynomial time. That means, if I claim that there is a polynomial time solution for a particular problem, you ask me to prove it. Then, I will give you a proof which you can easily verify in polynomial time. These kind of problems are called NP problems. Note that, here we are not talking about whether there is a polynomial time solution for this problem or not. But we are talking about verifying the solution to a given problem in polynomial time.
NP-Hard: These are at least as hard as the hardest problems in NP. If we can solve these problems in polynomial time, we can solve any NP problem that can possibly exist. Note that these problems are not necessarily NP problems. That means, we may/may-not verify the solution to these problems in polynomial time.
NP-Complete: These are the problems which are both NP and NP-Hard. That means, if we can solve these problems, we can solve any other NP problem and the solutions to these problems can be verified in polynomial time.
This is a very informal answer to the question asked.
Can 3233 be written as the product of two other numbers bigger than 1? Is there any way to walk a path around all of the Seven Bridges of Königsberg without taking any bridge twice? These are examples of questions that share a common trait. It may not be obvious how to efficiently determine the answer, but if the answer is 'yes', then there's a short and quick to check proof. In the first case a non-trivial factorization of 51; in the second, a route for walking the bridges (fitting the constraints).
A decision problem is a collection of questions with yes or no answers that vary only in one parameter. Say the problem COMPOSITE={"Is n composite": n is an integer} or EULERPATH={"Does the graph G have an Euler path?": G is a finite graph}.
Now, some decision problems lend themselves to efficient, if not obvious algorithms. Euler discovered an efficient algorithm for problems like the "Seven Bridges of Königsberg" over 250 years ago.
On the other hand, for many decision problems, it's not obvious how to get the answer -- but if you know some additional piece of information, it's obvious how to go about proving you've got the answer right. COMPOSITE is like this: Trial division is the obvious algorithm, and it's slow: to factor a 10 digit number, you have to try something like 100,000 possible divisors. But if, for example, somebody told you that 61 is a divisor of 3233, simple long division is a efficient way to see that they're correct.
The complexity class NP is the class of decision problems where the 'yes' answers have short to state, quick to check proofs. Like COMPOSITE. One important point is that this definition doesn't say anything about how hard the problem is. If you have a correct, efficient way to solve a decision problem, just writing down the steps in the solution is proof enough.
Algorithms research continues, and new clever algorithms are created all the time. A problem you might not know how to solve efficiently today may turn out to have an efficient (if not obvious) solution tomorrow. In fact, it took researchers until 2002 to find an efficient solution to COMPOSITE! With all these advances, one really has to wonder: Is this bit about having short proofs just an illusion? Maybe every decision problem that lends itself to efficient proofs has an efficient solution? Nobody knows.
Perhaps the biggest contribution to this field came with the discovery a peculiar class of NP problems. By playing around with circuit models for computation, Stephen Cook found a decision problem of the NP variety that was provably as hard or harder than every other NP problem. An efficient solution for the boolean satisfiability problem could be used to create an efficient solution to any other problem in NP. Soon after, Richard Karp showed that a number of other decision problems could serve the same purpose. These problems, in a sense the "hardest" problems in NP, became known as NP-complete problems.
Of course, NP is only a class of decision problems. Many problems aren't naturally stated in this manner: "find the factors of N", "find the shortest path in the graph G that visits every vertex", "give a set of variable assignments that makes the following boolean expression true". Though one may informally talk about some such problems being "in NP", technically that doesn't make much sense -- they're not decision problems. Some of these problems might even have the same sort of power as an NP-complete problem: an efficient solution to these (non-decision) problems would lead directly to an efficient solution to any NP problem. A problem like this is called NP-hard.
In addition to the other great answers, here is the typical schema people use to show the difference between NP, NP-Complete, and NP-Hard:
The easiest way to explain P v. NP and such without getting into technicalities is to compare "word problems" with "multiple choice problems".
When you are trying to solve a "word problem" you have to find the solution from scratch. When you are trying to solve a "multiple choice problems" you have a choice: either solve it as you would a "word problem", or try to plug in each of the answers given to you, and pick the candidate answer that fits.
It often happens that a "multiple choice problem" is much easier than the corresponding "word problem": substituting the candidate answers and checking whether they fit may require significantly less effort than finding the right answer from scratch.
Now, if we would agree the effort that takes polynomial time "easy" then the class P would consist of "easy word problems", and the class NP would consist of "easy multiple choice problems".
The essence of P v. NP is the question: "Are there any easy multiple choice problems that are not easy as word problems"? That is, are there problems for which it's easy to verify the validity of a given answer but finding that answer from scratch is difficult?
Now that we understand intuitively what NP is, we have to challenge our intuition. It turns out that there are "multiple choice problems" that, in some sense, are hardest of them all: if one would find a solution to one of those "hardest of them all" problems one would be able to find a solution to ALL NP problems! When Cook discovered this 40 years ago it came as a complete surprise. These "hardest of them all" problems are known as NP-hard. If you find a "word problem solution" to one of them you would automatically find a "word problem solution" to each and every "easy multiple choice problem"!
Finally, NP-complete problems are those that are simultaneously NP and NP-hard. Following our analogy, they are simultaneously "easy as multiple choice problems" and "the hardest of them all as word problems".
I think we can answer it much more succinctly. I answered a related question, and copying my answer from there
But first, an NP-hard problem is a problem for which we cannot prove that a polynomial time solution exists. NP-hardness of some "problem-P" is usually proven by converting an already proven NP-hard problem to the "problem-P" in polynomial time.
To answer the rest of question, you first need to understand which NP-hard problems are also NP-complete. If an NP-hard problem belongs to set NP, then it is NP-complete. To belong to set NP, a problem needs to be
(i) a decision problem,
(ii) the number of solutions to the problem should be finite and each solution should be of polynomial length, and
(iii) given a polynomial length solution, we should be able to say whether the answer to the problem is yes/noNow, it is easy to see that there could be many NP-hard problems that do not belong to set NP and are harder to solve. As an intuitive example, the optimization-version of traveling salesman where we need to find an actual schedule is harder than the decision-version of traveling salesman where we just need to determine whether a schedule with length <= k exists or not.
NP-complete problems are those problems that are both NP-Hard and in the complexity class NP. Therefore, to show that any given problem is NP-complete, you need to show that the problem is both in NP and that it is NP-hard.
Problems that are in the NP complexity class can be solved non-deterministically in polynomial time and a possible solution (i.e., a certificate) for a problem in NP can be verified for correctness in polynomial time.
An example of a non-deterministic solution to the k-clique problem would be something like:
1) randomly select k nodes from a graph
2) verify that these k nodes form a clique.
The above strategy is polynomial in the size of the input graph and therefore the k-clique problem is in NP.
Note that all problems deterministically solvable in polynomial time are also in NP.
Showing that a problem is NP-hard typically involves a reduction from some other NP-hard problem to your problem using a polynomial time mapping: http://en.wikipedia.org/wiki/Reduction_(complexity)
There are really nice answers for this particular question, so there is no point to write my own explanation. So I will try to contribute with an excellent resource about different classes of computational complexity.
For someone who thinks that computational complexity is only about P and NP, here is the most exhaustive resource about different computational complexity problems. Apart from problems asked by OP, it listed approximately 500 different classes of computational problems with nice descriptions and also the list of fundamental research papers which describe the class.
As I understand it, an np-hard problem is not "harder" than an np-complete problem. In fact, by definition, every np-complete problem is:
-- Intro. to Algorithms (3ed) by Cormen, Leiserson, Rivest, and Stein, pg 1069
Condition 1. and 2. translated to English:
