P versus NP problem

Consider Sudoku, a game where the player is given a partially filled-in grid of numbers and attempts to complete the grid following certain rules. Given an incomplete Sudoku grid, of any size, is there at least one legal solution? Any proposed solution is easily verified, and the time to check a solution grows slowly (polynomially) as the grid gets bigger. However, all known algorithms for finding solutions take, for difficult examples, time that grows exponentially as the grid gets bigger. So, Sudoku is in NP (quickly checkable) but does not seem to be in P (quickly solvable). Thousands of other problems seem similar, in that they are fast to check but slow to solve. Researchers have shown that many of the problems in NP have the extra property that a fast solution to any one of them could be used to build a quick solution to any other problem in NP, a property called NP-completeness. Decades of searching have not yielded a fast solution to any of these problems, so most scientists suspect that none of these problems can be solved quickly. This, however, has never been proven.

The travelling salesman problem (also called the travelling salesperson problem or TSP) asks the following question: "Given a list of cities and the distances between each pair of cities, what is the shortest possible route that visits each city and returns to the origin city?" It is an NP-hard problem in combinatorial optimization, important in theoretical computer science and operations research. Some algorithms solve TSP for small number of cities (under 10 or up to 15) but they start to approximate when the number of cities become high.

The precise statement of the P versus NP problem was introduced in 1971 by Stephen Cook in his seminal paper "The complexity of theorem proving procedures"[3] (and independently by Leonid Levin in 1973[4]) and is considered by many to be the most important open problem in computer science.[5]

Although the P versus NP problem was formally defined in 1971, there were previous inklings of the problems involved, the difficulty of proof, and the potential consequences. In 1955, mathematician John Nash wrote a letter to the NSA, where he speculated that cracking a sufficiently complex code would require time exponential in the length of the key.[6] If proved (and Nash was suitably skeptical) this would imply what is now called P ≠ NP, since a proposed key can easily be verified in polynomial time. Another mention of the underlying problem occurred in a 1956 letter written by Kurt Gödel to John von Neumann. Gödel asked whether theorem-proving (now known to be co-NP-complete) could be solved in quadratic or linear time,[7] and pointed out one of the most important consequences—that if so, then the discovery of mathematical proofs could be automated.

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 (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).

In this theory, the class P consists of all those decision problems (defined below) 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.[8] Clearly, P ⊆ NP. Arguably the biggest open question in theoretical computer science concerns the relationship between those two classes:

Is P equal to NP?

Since 2002, William Gasarch has conducted three polls of researchers concerning this and related questions.[9][10][11] Confidence that P ≠ NP has been increasing — in 2019, 88% believed P ≠ NP, as opposed to 83% in 2012 and 61% in 2002. When restricted to experts, the 2019 answers became 99% believe P ≠ NP.[11]

Euler diagram for P, NP, NP-complete, and NP-hard set of problems (excluding the empty language and its complement, which belong to P but are not NP-complete)

To attack the P = NP question, the concept of NP-completeness is very useful. NP-complete problems are a set of problems to each of which any other NP-problem can be reduced in polynomial time and whose solution may still be verified in polynomial time. That is, any NP problem can be transformed into any of the NP-complete problems. Informally, an NP-complete problem is an NP problem that is at least as "tough" as any other problem in NP.

NP-hard problems are those at least as hard as NP problems, i.e., all NP problems can be reduced (in polynomial time) to them. NP-hard problems need not be in NP, i.e., they need not have solutions verifiable in polynomial time.

For instance, the Boolean satisfiability problem is NP-complete by the Cook–Levin theorem, so any instance of any problem in NP can be transformed mechanically into an instance of the Boolean satisfiability problem in polynomial time. The Boolean satisfiability problem is one of many such NP-complete problems. If any NP-complete problem is in P, then it would follow that P = NP. However, many important problems have been shown to be NP-complete, and no fast algorithm for any of them is known.

Based on the definition alone it is not obvious that NP-complete problems exist; however, a trivial and contrived NP-complete problem can be formulated as follows: given a description of a Turing machine M guaranteed to halt in polynomial time, does there exist a polynomial-size input that M will accept?[12] It is in NP because (given an input) it is simple to check whether M accepts the input by simulating M; it is NP-complete because the verifier for any particular instance of a problem in NP can be encoded as a polynomial-time machine M that takes the solution to be verified as input. Then the question of whether the instance is a yes or no instance is determined by whether a valid input exists.

The first natural problem proven to be NP-complete was the Boolean satisfiability problem, also known as SAT. As noted above, this is the Cook–Levin theorem; its proof that satisfiability is NP-complete contains technical details about Turing machines as they relate to the definition of NP. However, after this problem was proved to be NP-complete, proof by reduction provided a simpler way to show that many other problems are also NP-complete, including the game Sudoku discussed earlier. In this case, the proof shows that a solution of Sudoku in polynomial time could also be used to complete Latin squares in polynomial time.[13] This in turn gives a solution to the problem of partitioning tri-partite graphs into triangles,[14] which could then be used to find solutions for the special case of SAT known as 3-SAT,[15] which then provides a solution for general Boolean satisfiability. So a polynomial time solution to Sudoku leads, by a series of mechanical transformations, to a polynomial time solution of satisfiability, which in turn can be used to solve any other NP-problem in polynomial time. Using transformations like this, a vast class of seemingly unrelated problems are all reducible to one another, and are in a sense "the same problem".

Although it is unknown whether P = NP, problems outside of P are known. Just as the class P is defined in terms of polynomial running time, the class EXPTIME is the set of all decision problems that have exponential running time. In other words, any problem in EXPTIME is solvable by a deterministic Turing machine in O(2^p(n)) time, where p(n) is a polynomial function of n. A decision problem is EXPTIME-complete if it is in EXPTIME, and every problem in EXPTIME has a polynomial-time many-one reduction to it. A number of problems are known to be EXPTIME-complete. Because it can be shown that P ≠ EXPTIME, 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. Examples include finding a perfect strategy for chess positions on an N × N board[16] and similar problems for other board games.[17]

The problem of deciding the truth of a statement in Presburger arithmetic requires even more time. Fischer and Rabin proved in 1974[18] that every algorithm that decides the truth of Presburger statements of length n has a runtime of at least ${\displaystyle 2^{2^{cn}}}$ for some constant c. 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 completely solved by any algorithm, in the sense that for any particular algorithm there is at least one input for which that algorithm will not produce the right answer; it will either produce the wrong answer, finish without giving a conclusive answer, or otherwise run forever without producing any answer at all.

It is also possible to consider questions other than decision problems. One such class, consisting of counting problems, is called #P: whereas an NP problem asks "Are there any solutions?", the corresponding #P problem asks "How many solutions are there?" Clearly, a #P problem must be at least as hard as the corresponding NP problem, since a count of solutions immediately tells if at least one solution exists, if the count is greater than zero. Surprisingly, some #P problems that are believed to be difficult correspond to easy (for example linear-time) P problems.[19] For these problems, it is very easy to tell whether solutions exist, but thought to be very hard to tell how many. Many of these problems are #P-complete, and hence among the hardest problems in #P, since a polynomial time solution to any of them would allow a polynomial time solution to all other #P problems.

In 1975, Richard E. Ladner showed that if P ≠ NP then there exist problems in NP that are neither in P nor NP-complete.[1] Such problems are called NP-intermediate problems. The graph isomorphism problem, the discrete logarithm problem and the integer factorization problem are examples of problems believed to be NP-intermediate. They are some of the very few NP problems not known to be in P or to be NP-complete.

The graph isomorphism problem is the computational problem of determining whether two finite graphs are isomorphic. An important unsolved problem in complexity theory is whether the graph isomorphism problem is in P, NP-complete, or NP-intermediate. The answer is not known, but it is believed that the problem is at least not NP-complete.[20] If graph isomorphism is NP-complete, the polynomial time hierarchy collapses to its second level.[21] Since it is widely believed that the polynomial hierarchy does not collapse to any finite level, it is believed that graph isomorphism is not NP-complete. The best algorithm for this problem, due to László Babai and Eugene Luks, has run time 2^{O(√n log n)} for graphs with n vertices.

The integer factorization problem is the computational problem of determining the prime factorization of a given integer. Phrased as a decision problem, it is the problem of deciding whether the input has a factor less than k. No efficient integer factorization algorithm is known, and this fact forms the basis of several modern cryptographic systems, such as the RSA algorithm. The integer factorization problem is in NP and in co-NP (and even in UP and co-UP[22]). If the problem is NP-complete, the polynomial time hierarchy will collapse to its first level (i.e., NP = co-NP). The best known algorithm for integer factorization is the general number field sieve, which takes expected time

${\displaystyle O\left(\exp \left(\left({\tfrac {64n}{9}}\log(2)\right)^{\frac {1}{3}}\left(\log(n\log(2))\right)^{\frac {2}{3}}\right)\right)}$

to factor an n-bit integer. However, the best known quantum algorithm for this problem, Shor's algorithm, does run in polynomial time, although this does not indicate where the problem lies with respect to non-quantum complexity classes.

The graph shows time (average of 100 instances in ms using a 933 MHz Pentium III) vs.problem size for knapsack problems for a state-of-the-art specialized algorithm. Quadratic fit suggests that empirical algorithmic complexity for instances with 50–10,000 variables is O((log(n))²).[23]

All of the above discussion has assumed that P means "easy" and "not in P" means "hard", an assumption known as Cobham's thesis. It is a common and reasonably accurate assumption in complexity theory; however, it has some caveats.

First, it is not always true in practice. A theoretical polynomial algorithm may have extremely large constant factors or exponents thus rendering it impractical. For example, the problem of deciding whether a graph G contains H as a minor, where H is fixed, can be solved in a running time of O(n²),[24] where n is the number of vertices in G. However, the big O notation hides a constant that depends superexponentially on H. The constant is greater than ${\displaystyle 2\uparrow \uparrow (2\uparrow \uparrow (2\uparrow \uparrow (h/2)))}$ (using Knuth's up-arrow notation), and where h is the number of vertices in H.[25]

On the other hand, even if a problem is shown to be NP-complete, and even if P ≠ NP, there may still be effective approaches to tackling the problem in practice. There are algorithms for many NP-complete problems, such as the knapsack problem, the traveling salesman problem and the Boolean satisfiability problem, that can solve to optimality many real-world instances in reasonable time. The empirical average-case complexity (time vs. problem size) of such algorithms can be surprisingly low. An example is the simplex algorithm in linear programming, which works surprisingly well in practice; despite having exponential worst-case time complexity it runs on par with the best known polynomial-time algorithms.[26]

Finally, there are types of computations which do not conform to the Turing machine model on which P and NP are defined, such as quantum computation and randomized algorithms.

According to polls,[9][27] 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 any of more than 3000 important known NP-complete problems (see List of NP-complete problems). These algorithms were sought long before the concept of NP-completeness was even defined (Karp's 21 NP-complete problems, among the first found, were all well-known existing problems at the time they were shown to be NP-complete). 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.[28]

If P = NP, then the world would be a profoundly different place than we usually assume it to be. There would be no special value in "creative leaps," no fundamental gap between solving a problem and recognizing the solution once it's found.

— Scott Aaronson, UT Austin

On the other hand, some researchers believe that there is overconfidence in believing P ≠ NP and that researchers should explore proofs of P = NP as well. For example, in 2002 these statements were made:[9]

The main argument in favor 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 simple questions may be settled only by very deep theories.

— Moshe Y. 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

One of the reasons the problem attracts so much attention is the consequences of the answer. Either direction of resolution would advance theory enormously, and perhaps have huge practical consequences as well.

Although the P = NP problem itself remains open despite a million-dollar prize and a huge amount of dedicated research, efforts to solve the problem have led to several new techniques. In particular, some of the most fruitful research related to the P = NP problem has been in showing that existing proof techniques are not powerful enough to answer the question, thus suggesting that novel technical approaches are required.

As additional evidence for the difficulty of the problem, essentially all known proof techniques in computational complexity theory fall into one of the following classifications, each of which is known to be insufficient to prove that P ≠ NP:

These barriers are 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 not excluded by the above results.

These barriers have also led some computer scientists to suggest that the P versus NP problem may be independent of standard axiom systems like ZFC (cannot be proved or disproved within them). The interpretation of an independence result could be that either no polynomial-time algorithm exists for any NP-complete problem, and such a proof cannot be constructed in (e.g.) ZFC, or that polynomial-time algorithms for NP-complete problems may exist, but it is impossible to prove in ZFC that such algorithms are correct.[43] However, if it can be shown, using techniques of the sort that are currently known to be applicable, that the problem cannot be decided even with much weaker assumptions extending the Peano axioms (PA) for integer arithmetic, then there would necessarily exist nearly-polynomial-time algorithms for every problem in NP.[44] Therefore, if one believes (as most complexity theorists do) that not all problems in NP have efficient algorithms, it would follow that proofs of independence using those techniques cannot be possible. Additionally, this result implies that proving independence from PA or ZFC using currently known techniques is no easier than proving the existence of efficient algorithms for all problems in NP.

While the P versus NP problem is generally considered unsolved,[45] many amateur and some professional researchers have claimed solutions. Gerhard J. Woeginger maintains a list that, as of 2018, contains 62 purported proofs of P = NP, 50 proofs of P ≠ NP, 2 proofs the problem is unprovable, and one proof that it is undecidable.[46] Some attempts at resolving P versus NP have received brief media attention,[47] though these attempts have since been refuted.

The P = NP problem can be restated in terms of expressible certain classes of logical statements, as a result of work in descriptive complexity.

Consider all languages of finite structures with a fixed signature including a linear order relation. Then, all such languages in P can be expressed in first-order logic with the addition of a suitable least fixed-point combinator. Effectively, this, in combination with the order, allows the definition of recursive functions. As long as the signature contains at least one predicate or function in addition to the distinguished order relation, so that the amount of space taken to store such finite structures is actually polynomial in the number of elements in the structure, this precisely characterizes P.

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 (of finite linearly ordered structures with nontrivial signature) that first-order logic with least fixed point cannot?".[48] The word "existential" can even be dropped from the previous characterization, since P = NP if and only if P = PH (as the former would establish that NP = co-NP, which in turn implies that NP = PH).

No algorithm for any NP-complete problem is known to run in polynomial time. However, there are algorithms known for NP-complete problems with the property that if P = NP, then the algorithm runs in polynomial time on accepting instances (although with enormous constants, making the algorithm impractical). However, these algorithms do not qualify as polynomial time because their running time on rejecting instances are not polynomial. The following algorithm, due to Levin (without any citation), is such an example below. It correctly accepts the NP-complete language SUBSET-SUM. It runs in polynomial time on inputs that are in SUBSET-SUM if and only if P = NP:

// Algorithm that accepts the NP-complete language SUBSET-SUM.
//
// this is a polynomial-time algorithm if and only if P = NP.
//
// "Polynomial-time" means it returns "yes" in polynomial time when
// the answer should be "yes", and runs forever when it is "no".
//
// Input: S = a finite set of integers
// Output: "yes" if any subset of S adds up to 0.
// Runs forever with no output otherwise.
// Note: "Program number M" is the program obtained by
// writing the integer M in binary, then
// considering that string of bits to be a
// program. Every possible program can be
// generated this way, though most do nothing
// because of syntax errors.
FOR K = 1...∞
  FOR M = 1...K
    Run program number M for K steps with input S
    IF the program outputs a list of distinct integers
      AND the integers are all in S
      AND the integers sum to 0
    THEN
      OUTPUT "yes" and HALT

If, and only 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" (also known as a semi-algorithm).

This algorithm is enormously impractical, even if P = NP. If the shortest program that can solve SUBSET-SUM in polynomial time is b bits long, the above algorithm will try at least 2^b − 1 other programs first.

The film Travelling Salesman, by director Timothy Lanzone, is the story of four mathematicians hired by the US government to solve the P versus NP problem.[50]

In the sixth episode of The Simpsons' seventh season "Treehouse of Horror VI", the equation P=NP is seen shortly after Homer accidentally stumbles into the "third dimension".[51][52]

In the second episode of season 2 of Elementary, "Solve for X" revolves around Sherlock and Watson investigating the murders of mathematicians who were attempting to solve P versus NP.[53][54]

• Game complexity
• List of unsolved problems in mathematics
• Unique games conjecture
• Unsolved problems in computer science

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• Aviad Rubinstein's Hardness of Approximation Between P and NP, winner of the ACM's 2017 Doctoral Dissertation Award.
• "P vs. NP and the Computational Complexity Zoo". 26 August 2014 – via YouTube.