Complexity Classes P and NP

The relationship between the complexity classes P and NP is an unsolved question in theoretical computer science. It is generally agreed to be the most important such unsolved problem, and one of the most important unsolved problems in all of mathematics. The Clay Mathematics Institute has offered a USD 1,000,000 prize for a correct solution.

In essence, the P = NP question asks: if positive solutions to a YES/NO problem can be verified quickly in polynomial time, can the answers also be computed quickly in polynomial time and space? Consider, for instance, the subset-sum problem, an example of a problem which is easy to verify, but whose answer is believed (but not proven) to be difficult to compute. Given a set of integers, does any subset of them sum to 0? For instance, does a subset of the set {-2, -3, 15, 14, 7, -10} add up to 0? The answer is YES, though it may take a little while to find a subset that does - and if the set were larger, it might take a very long time to find a subset that does. On the other hand, if someone claims that the answer is "YES, because {-2, -3, -10, 15} add up to zero", then we can quickly check that with a few additions. Verifying that the subset adds up to zero is much faster than finding the subset in the first place. The information needed to verify a positive answer is also called a certificate. So we conclude that given the right certificates, positive answers to our problem can be verified quickly (in polynomial time) and that's why this problem is in NP.

An answer to the P=NP question would determine whether problems like SUBSET-SUM are really harder to compute than to verify (this would be the case if P does not equal NP), or that they are as easy to compute as to verify (this would be the case if P=NP). The answer would apply to all such problems, not just the specific example of SUBSET-SUM.

The restriction to YES/NO problems doesn't really make a difference; even if we allow more complicated answers, the resulting problem (whether FP = FNP) is equivalent.

Context of the problem

The relation between the complexity classes P and NP is studied in computational complexity theory, the part of the theory of computation dealing with the resources required during computation to solve a given problem. The most common resources are time (how many steps it takes to solve a problem) and space (how much memory it takes to solve a problem).

In such analysis, a model of the computer for which time must be analyzed is required. Typically, such models assume that the computer is deterministic - that, given the computer's present state and any inputs, there is only one possible action that the computer might take - and sequential - it performs actions one after the other. These assumptions reflect the behavior of all practical computers yet devised, even including machines featuring parallel computing.

In this theory, the class P consists of all those decision problems that can be solved on a deterministic sequential machine in an amount of time that is polynomial in the size of the input; the class NP consists of all those decision problems whose positive solutions can be verified in polynomial time given the right information, or equivalently, whose solution can be found in polynomial time on a non-deterministic machine. Arguably, the biggest open question in theoretical computer science concerns the relationship between those two classes:

Is P equal to NP?

In a poll of 100 researchers, 61 believed the answer is no, 9 believed the answer is yes, 22 were unsure, and 8 believed the question may be independent of the currently accepted axioms, and so impossible to prove or disprove.

Formal definitions

More precisely, a decision problem is a problem that takes as input some string and requires as output either YES or NO. If there is an algorithm (say a Turing machine, or a Lisp or Pascal program with unbounded memory) which is able to produce the correct answer for any input string of length n in at most steps, where k and c are some constants independent of the input string, then we say that the problem can be solved in polynomial time and we place it in the class P. Intuitively, we think of the problems in P as those that can be solved reasonably quickly.

Now suppose there is an algorithm A(w,C) which takes two arguments, a string w which is an input string to our decision problem, and a string C which is a "proposed certificate", and such that A produces a YES/NO answer in at most steps (where n is the length of w and neither c nor k depend on w). Suppose furthermore that: w is a YES instance of the decision problem if and only if there exists C such that A(w,C) returns YES. Then we say that the problem can be solved in non-deterministic polynomial time and we place it in the class NP. We think of the algorithm A as a verifier of proposed certificates which runs reasonably fast. (Note that the abbreviation NP stands for "Non-deterministic Polynomial" and not for "Non-Polynomial".)

NP-complete

To attack the P = NP question, the concept of NP-completeness is very useful. Informally, the NP-complete problems are the "toughest" problems in NP in the sense that they are the ones most likely not to be in P. NP-hard problems are those into which any problem in NP can be transformed, in polynomial time. NP-complete problems are those NP-hard problems which are in NP. For instance, the decision problem version of the traveling salesman problem is NP-complete. So any instance of any problem in NP can be transformed mechanically into an instance of the traveling salesman problem, in polynomial time. So, if the traveling salesman problem turned out to be in P, then P = NP. The traveling salesman problem is one of many such NP-complete problems. If any NP-complete problem is in P, then it would follow that P = NP. Unfortunately, many important problems have been shown to be NP-complete and not a single fast algorithm for any of them is known.

It may be surprising to the reader that many such problems exist. The first NP-complete problem discovered was the boolean satisfiability problem, which was proved to be NP-complete by Stephen Cook in 1971 - a result that became known as Cook's theorem. Cook's proof that satisfiability is NP-complete is very complicated. However, once this first problem was proved to be NP-complete, proof by reduction can be used to show that many other problems are in this class.

Still harder problems

Although it is unknown whether P = NP, problems outside of P are known. A number of succinct problems, that is, problems which operate not on normal input but on a computational description of the input, are known to be EXPTIME-complete. Because it can be shown that P ⊂ EXPSPACE, these problems are outside P, and so require more than polynomial time. In fact, by the time hierarchy theorem, they cannot be solved in significantly less than exponential time.

The problem of deciding the truth of a statement in Presburger arithmetic is even harder. Fischer and Rabin proved in 1974 that every algorithm which decides the truth of Presburger statements has a runtime of at least for some constant c. Here, n is the length of the Presburger statement. Hence, the problem is known to need more than exponential run time. Even more difficult are the undecidable problems, such as the halting problem. They cannot be solved in general given any amount of time.

Is P really practical?

All of the above discussion has assumed that P means "easy" and "not in P" means "hard". While this is a common and reasonably accurate assumption in complexity theory, it is not always true in practice, for several reasons:

• It ignores constant factors. A problem that takes time 101000n is in P (it is linear time), but is completely impractical in practice. A problem that takes time 10-100002n is not in P (it is exponential time), but is very practical for values of n up into the thousands.
• It ignores the size of the exponents. A problem with time n1000 is in P, yet impractical. Problems have been proven to exist in P that require arbitrarily large exponents. A problem with time 2n/1000 is not in P, yet is practical for n up into the thousands.
• It only considers worst-case times. There might be a problem that arises in the real world such that most of the time, it can be solved in time n, but on very rare occasions you'll see an instance of the problem that takes time 2n. This problem might have an average time that is polynomial, but the worst case is exponential, so the problem wouldn't be in P.
• It only considers deterministic solutions. Imagine a problem that you can solve quickly if you accept a tiny error probability, but a guaranteed correct answer is much harder to get. The problem would not belong to P even though in practice it can be solved quickly. This is in fact a common approach to attack problems in NP not known to be in P. Even if P=BPP, as many researchers believe, it is often considerably easier to find probabilistic algorithms.
• New computing models such as quantum computers may be able to quickly solve some problems not known to be in P; however, none of the problems they are known to be able to solve are NP-hard. However, the definition of P and NP are in terms of classical computing models like Turing machines. Therefore, even if a quantum computer algorithm were discovered to efficiently solve an NP-hard problem, we would only have a way of physically solving difficult problems quickly, not a proof that the mathematical classes P and NP are equal.

Why do computer scientists think P ≠ NP?

Most computer scientists believe that P≠NP. A key reason for this belief is that after decades of studying these problems, no one has been able to find a polynomial-time algorithm for an NP-hard problem. Moreover, these algorithms were sought long before the concept of NP-completeness was even known (Karp's 21 NP-complete problems, among the first found, were all well-known existing problems). Furthermore, the result P = NP would imply many other startling results that are currently believed to be false, such as NP = co-NP and P = PH.

It is also intuitively argued that the existence of problems that are hard to solve but for which the solutions are easy to verify matches real-world experience.

On the other hand, some researchers believe that we are overconfident in P ≠ NP and should explore proofs of P = NP as well. For example, in 2002 these statements were made:

"The main argument in favour of P≠NP is the total lack of fundamental progress in the area of exhaustive search. This is, in my opinion, a very weak argument. The space of algorithms is very large and we are only at the beginning of its exploration. [. . .] The resolution of Fermat's Last Theorem also shows that very simply [sic] questions may be settled only by very deep theories."

—Moshe Vardi, Rice University

"Being attached to a speculation is not a good guide to research planning. One should always try both directions of every problem. Prejudice has caused famous mathematicians to fail to solve famous problems whose solution was opposite to their expectations, even though they had developed all the methods required."

—Anil Nerode, Cornell University

Consequences of proof

One of the reasons why the problem attracts so much attention are the consequences of the answer.

A proof of P = NP could have stunning practical consequences, if the proof leads to efficient methods for solving some of the important problems in NP. Many NP-complete problems are fundamental in many fields. One negative impact is that many forms of cryptography, notably public-key cryptography, would become trivially easy to "crack" and would therefore be useless. However, there are also enormous positive consequences that would follow from rendering tractable many currently mathematically intractable problems. For instance, many problems in operations research are NP-complete, such as some types of integer programming, and the travelling salesman problem, to name two of the most famous examples. Efficient solutions to these problems would have enormous implications for logistics. But it is by no means the only field that would be profoundly changed. To take one of very many examples, important problems in Protein structure prediction are NP-complete ; if these problems were solvable efficiently it could spur considerable advances in biology.

But such changes may pale into insignificance compared to the revolution an efficient method for solving NP-complete problems would cause in mathematics itself. According to Stephen Cook (PDF),

...it would transform mathematics by allowing a computer to find a formal proof of any theorem which has a proof of a reasonable length, since formal proofs can easily be recognized in polynomial time. Example problems may well include all of the CMI prize problems.

Research mathematicians spend their careers trying to prove theorems, and some proofs have taken decades or even centuries to find after problems have been stated - for instance, Fermat's Last Theorem took over three centuries to prove. A method that is guaranteed to find proofs to theorems, should one exist of a "reasonable" size, would essentially end this struggle and transform mathematics into a search for useful things to prove.

A proof that showed that P ≠ NP, while lacking the practical computational benefits of a proof that P = NP, would represent a massive advance in computational complexity theory and provide guidance for future research. It would allow one to show in a formal way that many common problems cannot be solved efficiently, so that the attention of researchers can be focused on partial solutions or solutions to other problems. Due to widespread belief in P ≠ NP, much of this focusing of research has already taken place.

Results about difficulty of proof

A million-dollar prize and a huge amount of dedicated research with no substantial results are enough to show the problem is difficult. There have also been some formal results demonstrating why the problem might be difficult to solve.

One of the most frequently-cited is a result involving oracles. Imagine you have a magical machine called an oracle that can solve only one problem, such as determining if a given number is prime, but can solve it in constant time. Our new question is now, if we're allowed to use this oracle as much as we want, are there problems we can verify in polynomial time that we can't solve in polynomial time? It turns out that, depending on the problem that the oracle solves, with certain oracles one has P = NP, while for other oracles one has P ≠ NP. The practical consequence of this is that any proof which can be modified to account for the existence of these oracles cannot solve the problem. Unfortunately, most known methods and nearly all classical methods can be modified in such a way (we say they are relativizing).

Furthermore, a 1993 result by Alexander Razborov and Steven Rudich showed that, given a certain credible assumption, proofs that are "natural" in a certain sense cannot solve the P = NP problem (see natural proof). This demonstrated that some of the most seemingly-promising methods of the time were also unlikely to succeed. As more theorems of this kind are proved, a potential proof of the theorem has more and more traps to avoid.

This is actually another reason why NP-complete problems are useful: if a polynomial-time algorithm can be demonstrated for an NP-complete problem, this would solve the P = NP problem in a way which is not excluded by the above results.

Polynomial-time algorithms

No one knows whether polynomial-time algorithms exist for NP-complete languages. But if such algorithms do exist, we already know some of them! For example, the following algorithm correctly accepts an NP-complete language, but no one knows how long it takes in general. This is a polynomial-time algorithm if and only if P = NP.

If P = NP, then this is a polynomial-time algorithm accepting an NP-Complete language. "Accepting" means it gives "YES" answers in polynomial time, but is allowed to run forever when the answer is "NO".

Perhaps we want to "solve" the SUBSET-SUM problem, rather than just "accept" the SUBSET-SUM language. That means we want it to always halt and return a "YES" or "NO" answer. Does any algorithm exist that can provably do this in polynomial time? No one knows. But if such algorithms do exist, then we already know some of them! Just replace the IF statement in the above algorithm with this:

Logical characterizations

The P=NP problem can be restated in terms of the expressibility of certain classes of logical statements. All languages in P can be expressed in first-order logic with the addition of a least fixed point operator (effectively, this allows the definition of recursive functions). Similarly, NP is the set of languages expressible in existential second-order logic — that is, second-order logic restricted to exclude universal quantification over relations, functions, and subsets. The languages in the polynomial hierarchy, PH, correspond to all of second-order logic. Thus, the question "is P a proper subset of NP" can be reformulated as "is existential second-order logic able to describe languages that first-order logic with least fixed point cannot?"