NP-hardness

For a gentler introduction, see Complexity classes P and NP. In computational complexity theory, NP-hard (Non-deterministic Polynomial-time hard) refers to the class containing all problems H, such that for every decision problem L in NP there exists a polynomial-time many-one reduction to H, written $L\leq _{p}H$. If H itself is in NP, then H is called NP-complete. Informally, this class can be described as containing the problems that are at least as hard as any decision problem in NP (note that NP-hard problems are not necessarily decision problems). This intuition is supported by the fact that if we can find an algorithm A that solves one of these problems H in polynomial time, we can construct a polynomial time algorithm for any problem L in NP by first performing the reduction from L to H and then running the algorithm A.

So, formally, a language L is NP-hard if $\forall L^{\prime }\in \mathbf {NP} ,L^{\prime }\leq _{p}L\!$. If it is also the case that L is in NP, then L is called NP-complete.

The notion of NP-hardness plays an important role in the discussion about the relationship between the Complexity classes P and NP. The class NP-hard can be understood as the class of problems that are NP-complete or harder.

A common mistake is to think that the "NP" in "NP-hard" stands for "non-polynomial". Although it is widely suspected that there are no polynomial-time algorithms for these problems, this has never been proven. Furthermore, one should also remember that polynomial complexity problems are contained in the complexity class NP (though are not NP-hard unless P=NP).

Examples

An example of an NP-hard problem is the decision problem SUBSET-SUM which is this: given a set of integers, does any non empty subset of them add up to zero? That is a yes/no question, and happens to be NP-complete.

There are also decision problems that are NP-hard but not NP-complete, for example the halting problem. This is the problem "given a program and its input, will it run forever?" That's a yes/no question, so this is a decision problem. It is easy to prove that the halting problem is NP-hard but not NP-complete. For example the Boolean satisfiability problem can be reduced to the halting problem by transforming it to the description of a Turing machine that tries all truth value assignments and when it finds one that satisfies the formula it halts and otherwise it goes into an infinite loop. It is also easy to see that the halting problem is not in NP since all problems in NP are decidable and the halting problem is not.

Alternative definitions

An alternative definition of NP-hard that is often used replaces polynomial-time many-one reductions with polynomial-time Turing reductions. This notion of NP-hardness can be formulated for function problems and not just decision problems.

In this sense, the problem H is NP-hard if and only if every decision problem L in NP can be solved in polynomial time by an oracle machine with an oracle for H. (Informally we can think of such a machine as an algorithm that can call a subroutine for solving H and solves L in polynomial time if the subroutine call takes only one step to compute.) If we find a polynomial-time algorithm for an NP-hard problem then we have a polynomial-time algorithm for all problems in NP-easy and indeed for all problems in NP.

Whether this definition of NP-hardness is equivalent to the one at the beginning of this article (for the case of decision problems) is still an open problem and is discussed in more detail in the article on NP-completeness.