Millennium Prize Problems 

The P versus NP problem is a major unsolved problem in theoretical computer science. Informally, it asks whether every problem whose solution can be quickly verified can also be quickly solved.
Here, quickly means an algorithm that solves the task and runs in polynomial time exists, meaning the task completion time is bounded above by a polynomial function on the size of the input to the algorithm (as opposed to, say, exponential time). The general class of questions that some algorithm can answer in polynomial time is "P" or "class P". For some questions, there is no known way to find an answer quickly, but if provided with an answer, it can be verified quickly. The class of questions where an answer can be verified in polynomial time is NP, standing for "nondeterministic polynomial time".^{[Note 1]}
An answer to the P versus NP question would determine whether problems that can be verified in polynomial time can also be solved in polynomial time. If P ≠ NP, which is widely believed, it would mean that there are problems in NP that are harder to compute than to verify: they could not be solved in polynomial time, but the answer could be verified in polynomial time.
The problem has been called the most important open problem in computer science.^{[1]} Aside from being an important problem in computational theory, a proof either way would have profound implications for mathematics, cryptography, algorithm research, artificial intelligence, game theory, multimedia processing, philosophy, economics and many other fields.^{[2]}
It is one of the seven Millennium Prize Problems selected by the Clay Mathematics Institute, each of which carries a US$1,000,000 prize for the first correct solution.
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16. Complexity: P, NP, NPcompleteness, Reductions

Computation in Complex Systems: P versus NP : P versus NP Problem

P vs. NP and the Computational Complexity Zoo
Transcription
Example
Consider the following yes/no problem: given an incomplete Sudoku grid of size , is there at least one legal solution where every row, column, and square contains the integers 1 through ? It is straightforward to verify "yes" instances of this generalized Sudoku problem given a candidate solution. However, it is not known whether there is a polynomialtime algorithm that can correctly answer "yes" or "no" to all instances of this problem. Therefore, generalized Sudoku is in NP (quickly verifiable), but may or may not be in P (quickly solvable). (It is necessary to consider a generalized version of Sudoku, as any fixed size Sudoku has only a finite number of possible grids. In this case the problem is in P, as the answer can be found by table lookup.)
History
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]}).
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, speculating that cracking a sufficiently complex code would require time exponential in the length of the key.^{[5]} If proved (and Nash was suitably skeptical), this would imply what is now called P ≠ NP, since a proposed key can 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 theoremproving (now known to be coNPcomplete) could be solved in quadratic or linear time,^{[6]} and pointed out one of the most important consequences—that if so, then the discovery of mathematical proofs could be automated.
Context
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 decision problems (defined below) solvable on a deterministic sequential machine in a duration polynomial in the size of the input; the class NP consists of all decision problems whose positive solutions are verifiable in polynomial time given the right information, or equivalently, whose solution can be found in polynomial time on a nondeterministic machine.^{[7]} 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.^{[8]}^{[9]}^{[10]} 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% believed P ≠ NP.^{[10]} These polls do not imply whether P = NP, Gasarch himself stated: "This does not bring us any closer to solving P=?NP or to knowing when it will be solved, but it attempts to be an objective report on the subjective opinion of this era."
NPcompleteness
To attack the P = NP question, the concept of NPcompleteness is very useful. NPcomplete problems are problems that any other NP problem is reducible to in polynomial time and whose solution is still verifiable in polynomial time. That is, any NP problem can be transformed into any NPcomplete problem. Informally, an NPcomplete problem is an NP problem that is at least as "tough" as any other problem in NP.
NPhard problems are those at least as hard as NP problems; i.e., all NP problems can be reduced (in polynomial time) to them. NPhard problems need not be in NP; i.e., they need not have solutions verifiable in polynomial time.
For instance, the Boolean satisfiability problem is NPcomplete by the Cook–Levin theorem, so any instance of any problem in NP can be transformed mechanically into a Boolean satisfiability problem in polynomial time. The Boolean satisfiability problem is one of many NPcomplete problems. If any NPcomplete problem is in P, then it would follow that P = NP. However, many important problems are NPcomplete, and no fast algorithm for any of them is known.
From the definition alone it is unintuitive that NPcomplete problems exist; however, a trivial NPcomplete problem can be formulated as follows: given a Turing machine M guaranteed to halt in polynomial time, does a polynomialsize input that M will accept exist?^{[11]} It is in NP because (given an input) it is simple to check whether M accepts the input by simulating M; it is NPcomplete because the verifier for any particular instance of a problem in NP can be encoded as a polynomialtime 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 NPcomplete was the Boolean satisfiability problem, also known as SAT. As noted above, this is the Cook–Levin theorem; its proof that satisfiability is NPcomplete contains technical details about Turing machines as they relate to the definition of NP. However, after this problem was proved to be NPcomplete, proof by reduction provided a simpler way to show that many other problems are also NPcomplete, 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.^{[12]} This in turn gives a solution to the problem of partitioning tripartite graphs into triangles,^{[13]} which could then be used to find solutions for the special case of SAT known as 3SAT,^{[14]} which then provides a solution for general Boolean satisfiability. So a polynomialtime 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 NPproblem 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".
Harder problems
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 EXPTIMEcomplete if it is in EXPTIME, and every problem in EXPTIME has a polynomialtime manyone reduction to it. A number of problems are known to be EXPTIMEcomplete. 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^{[15]} and similar problems for other board games.^{[16]}
The problem of deciding the truth of a statement in Presburger arithmetic requires even more time. Fischer and Rabin proved in 1974^{[17]} that every algorithm that decides the truth of Presburger statements of length n has a runtime of at least 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 lineartime) P problems.^{[18]} 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 #Pcomplete, 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.
Problems in NP not known to be in P or NPcomplete
In 1975, Richard E. Ladner showed that if P ≠ NP, then there exist problems in NP that are neither in P nor NPcomplete.^{[19]} Such problems are called NPintermediate problems. The graph isomorphism problem, the discrete logarithm problem, and the integer factorization problem are examples of problems believed to be NPintermediate. They are some of the very few NP problems not known to be in P or to be NPcomplete.
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, NPcomplete, or NPintermediate. The answer is not known, but it is believed that the problem is at least not NPcomplete.^{[20]} If graph isomorphism is NPcomplete, 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 NPcomplete. The best algorithm for this problem, due to László Babai, runs in quasipolynomial time.^{[22]}
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 coNP (and even in UP and coUP^{[23]}). If the problem is NPcomplete, the polynomial time hierarchy will collapse to its first level (i.e., NP = coNP). The most efficient known algorithm for integer factorization is the general number field sieve, which takes expected time
to factor an nbit integer. The best known quantum algorithm for this problem, Shor's algorithm, runs in polynomial time, although this does not indicate where the problem lies with respect to nonquantum complexity classes.
Does P mean "easy"?
All of the above discussion has assumed that P means "easy" and "not in P" means "difficult", an assumption known as Cobham's thesis. It is a common assumption in complexity theory; but there are caveats.
First, it can be false in practice. A theoretical polynomial algorithm may have extremely large constant factors or exponents, 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(n^{2}),^{[25]} 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 (using Knuth's uparrow notation), and where h is the number of vertices in H.^{[26]}
On the other hand, even if a problem is shown to be NPcomplete, and even if P ≠ NP, there may still be effective approaches to the problem in practice. There are algorithms for many NPcomplete problems, such as the knapsack problem, the traveling salesman problem, and the Boolean satisfiability problem, that can solve to optimality many realworld instances in reasonable time. The empirical averagecase 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 worstcase time complexity, it runs on par with the best known polynomialtime algorithms.^{[27]}
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.
Reasons to believe P ≠ NP or P = NP
Cook provides a restatement of the problem in The P Versus NP Problem as "Does P = NP?"^{[28]} According to polls,^{[8]}^{[29]} 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 polynomialtime algorithm for any of more than 3,000 important known NPcomplete problems (see List of NPcomplete problems). These algorithms were sought long before the concept of NPcompleteness was even defined (Karp's 21 NPcomplete problems, among the first found, were all wellknown existing problems at the time they were shown to be NPcomplete). Furthermore, the result P = NP would imply many other startling results that are currently believed to be false, such as NP = coNP and P = PH.
It is also intuitively argued that the existence of problems that are hard to solve but whose solutions are easy to verify matches realworld experience.^{[30]}
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.
On the other hand, some researchers believe that it is overconfident to believe P ≠ NP and that researchers should also explore proofs of P = NP. For example, in 2002 these statements were made:^{[8]}
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.
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.
DLIN vs NLIN
When one substitutes "linear time on a multitape Turing machine" for "polynomial time" in the definitions of P and NP, one obtains the classes DLIN and NLIN. It is known^{[31]} that DLIN ≠ NLIN.
Consequences of solution
One of the reasons the problem attracts so much attention is the consequences of the possible answers. Either direction of resolution would advance theory enormously, and perhaps have huge practical consequences as well.
P = NP
A proof that P = NP could have stunning practical consequences if the proof leads to efficient methods for solving some of the important problems in NP. The potential consequences, both positive and negative, arise since various NPcomplete problems are fundamental in many fields.
It is also very possible that a proof would not lead to practical algorithms for NPcomplete problems. The formulation of the problem does not require that the bounding polynomial be small or even specifically known. A nonconstructive proof might show a solution exists without specifying either an algorithm to obtain it or a specific bound. Even if the proof is constructive, showing an explicit bounding polynomial and algorithmic details, if the polynomial is not very loworder the algorithm might not be sufficiently efficient in practice. In this case the initial proof would be mainly of interest to theoreticians, but the knowledge that polynomial time solutions are possible would surely spur research into better (and possibly practical) methods to achieve them.
A solution showing P = NP could upend the field of cryptography, which relies on certain problems being difficult. A constructive and efficient solution^{[Note 2]} to an NPcomplete problem such as 3SAT would break most existing cryptosystems including:
 Existing implementations of publickey cryptography,^{[32]} a foundation for many modern security applications such as secure financial transactions over the Internet.
 Symmetric ciphers such as AES or 3DES,^{[33]} used for the encryption of communications data.
 Cryptographic hashing, which underlies blockchain cryptocurrencies such as Bitcoin, and is used to authenticate software updates. For these applications, finding a preimage that hashes to a given value must be difficult, ideally taking exponential time. If P = NP, then this can take polynomial time, through reduction to SAT.^{[34]}
These would need modification or replacement with informationtheoretically secure solutions that do not assume P ≠ NP.
There are also enormous benefits that would follow from rendering tractable many currently mathematically intractable problems. For instance, many problems in operations research are NPcomplete, such as types of integer programming and the travelling salesman problem. Efficient solutions to these problems would have enormous implications for logistics. Many other important problems, such as some problems in protein structure prediction, are also NPcomplete;^{[35]} making these problems efficiently solvable could considerably advance life sciences and biotechnology.
These changes could be insignificant compared to the revolution that efficiently solving NPcomplete problems would cause in mathematics itself. Gödel, in his early thoughts on computational complexity, noted that a mechanical method that could solve any problem would revolutionize mathematics:^{[36]}^{[37]}
If there really were a machine with φ(n) ∼ k⋅n (or even ∼ k⋅n^{2}), this would have consequences of the greatest importance. Namely, it would obviously mean that in spite of the undecidability of the Entscheidungsproblem, the mental work of a mathematician concerning YesorNo questions could be completely replaced by a machine. After all, one would simply have to choose the natural number n so large that when the machine does not deliver a result, it makes no sense to think more about the problem.
Similarly, Stephen Cook (assuming not only a proof, but a practically efficient algorithm) says:^{[28]}
... 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 guaranteed to find a proof if a "reasonable" size proof exists, would essentially end this struggle.
Donald Knuth has stated that he has come to believe that P = NP, but is reserved about the impact of a possible proof:^{[38]}
[...] if you imagine a number M that's finite but incredibly large—like say the number 10↑↑↑↑3 discussed in my paper on "coping with finiteness"—then there's a humongous number of possible algorithms that do n^{M} bitwise or addition or shift operations on n given bits, and it's really hard to believe that all of those algorithms fail. My main point, however, is that I don't believe that the equality P = NP will turn out to be helpful even if it is proved, because such a proof will almost surely be nonconstructive.
P ≠ NP
A proof of P ≠ NP would lack the practical computational benefits of a proof that P = NP, but would represent a great advance in computational complexity theory and guide future research. It would demonstrate 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.^{[39]}
P ≠ NP still leaves open the averagecase complexity of hard problems in NP. For example, it is possible that SAT requires exponential time in the worst case, but that almost all randomly selected instances of it are efficiently solvable. Russell Impagliazzo has described five hypothetical "worlds" that could result from different possible resolutions to the averagecase complexity question.^{[40]} These range from "Algorithmica", where P = NP and problems like SAT can be solved efficiently in all instances, to "Cryptomania", where P ≠ NP and generating hard instances of problems outside P is easy, with three intermediate possibilities reflecting different possible distributions of difficulty over instances of NPhard problems. The "world" where P ≠ NP but all problems in NP are tractable in the average case is called "Heuristica" in the paper. A Princeton University workshop in 2009 studied the status of the five worlds.^{[41]}
Results about difficulty of proof
Although the P = NP problem itself remains open despite a milliondollar 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 insufficient for answering the question, suggesting 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, all insufficient to prove P ≠ NP:
Classification  Definition 

Relativizing proofs  Imagine a world where every algorithm is allowed to make queries to some fixed subroutine called an oracle (which can answer a fixed set of questions in constant time, such as an oracle that solves any traveling salesman problem in 1 step), and the running time of the oracle is not counted against the running time of the algorithm. Most proofs (especially classical ones) apply uniformly in a world with oracles regardless of what the oracle does. These proofs are called relativizing. In 1975, Baker, Gill, and Solovay showed that P = NP with respect to some oracles, while P ≠ NP for other oracles.^{[42]} As relativizing proofs can only prove statements that are true for all possible oracles, these techniques cannot resolve P = NP. 
Natural proofs  In 1993, Alexander Razborov and Steven Rudich defined a general class of proof techniques for circuit complexity lower bounds, called natural proofs.^{[43]} At the time, all previously known circuit lower bounds were natural, and circuit complexity was considered a very promising approach for resolving P = NP. However, Razborov and Rudich showed that if oneway functions exist, P and NP are indistinguishable to natural proof methods. Although the existence of oneway functions is unproven, most mathematicians believe that they do, and a proof of their existence would be a much stronger statement than P ≠ NP. Thus it is unlikely that natural proofs alone can resolve P = NP. 
Algebrizing proofs  After the Baker–Gill–Solovay result, new nonrelativizing proof techniques were successfully used to prove that IP = PSPACE. However, in 2008, Scott Aaronson and Avi Wigderson showed that the main technical tool used in the IP = PSPACE proof, known as arithmetization, was also insufficient to resolve P = NP.^{[44]} Arithmetization converts the operations of an algorithm to algebraic and basic arithmetic symbols and then uses those to analyze the workings. In the IP = PSPACE proof, they convert the black box and the Boolean circuits to an algebraic problem.^{[44]} As mentioned previously, it has been proven that this method is not viable to solve P = NP and other time complexity problems. 
These barriers are another reason why NPcomplete problems are useful: if a polynomialtime algorithm can be demonstrated for an NPcomplete problem, this would solve the P = NP problem in a way not excluded by the above results.
These barriers lead some computer scientists to suggest the P versus NP problem may be independent of standard axiom systems like ZFC (cannot be proved or disproved within them). An independence result could imply that either P ≠ NP and this is unprovable in (e.g.) ZFC, or that P = NP but it is unprovable in ZFC that any polynomialtime algorithms are correct.^{[45]} However, if the problem is undecidable even with much weaker assumptions extending the Peano axioms for integer arithmetic, then nearly polynomialtime algorithms exist for all NP problems.^{[46]} Therefore, assuming (as most complexity theorists do) some NP problems don't have efficient algorithms, proofs of independence with those techniques are impossible. This also implies proving independence from PA or ZFC with current techniques is no easier than proving all NP problems have efficient algorithms.
Logical characterizations
The P = NP problem can be restated as 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 are expressible in firstorder logic with the addition of a suitable least fixedpoint combinator. Recursive functions can be defined with this and the order relation. 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 secondorder logic—that is, secondorder logic restricted to exclude universal quantification over relations, functions, and subsets. The languages in the polynomial hierarchy, PH, correspond to all of secondorder logic. Thus, the question "is P a proper subset of NP" can be reformulated as "is existential secondorder logic able to describe languages (of finite linearly ordered structures with nontrivial signature) that firstorder logic with least fixed point cannot?".^{[47]} 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 = coNP, which in turn implies that NP = PH).
Polynomialtime algorithms
No known algorithm for a NPcomplete problem runs in polynomial time. However, there are algorithms known for NPcomplete problems that if P = NP, 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 NPcomplete language SUBSETSUM. It runs in polynomial time on inputs that are in SUBSETSUM if and only if P = NP:
// Algorithm that accepts the NPcomplete language SUBSETSUM. // // this is a polynomialtime algorithm if and only if P = NP. // // "Polynomialtime" 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
This is a polynomialtime algorithm accepting an NPcomplete language only if P = NP. "Accepting" means it gives "yes" answers in polynomial time, but is allowed to run forever when the answer is "no" (also known as a semialgorithm).
This algorithm is enormously impractical, even if P = NP. If the shortest program that can solve SUBSETSUM in polynomial time is b bits long, the above algorithm will try at least 2^{b} − 1 other programs first.
Formal definitions
P and NP
A decision problem is a problem that takes as input some string w over an alphabet Σ, and outputs "yes" or "no". If there is an algorithm (say a Turing machine, or a computer program with unbounded memory) that produces the correct answer for any input string of length n in at most cn^{k} steps, where k and c are 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. Formally, P is the set of languages that can be decided by a deterministic polynomialtime Turing machine. Meaning,
where
and a deterministic polynomialtime Turing machine is a deterministic Turing machine M that satisfies two conditions:
 M halts on all inputs w and
 there exists such that , where O refers to the big O notation and
NP can be defined similarly using nondeterministic Turing machines (the traditional way). However, a modern approach uses the concept of certificate and verifier. Formally, NP is the set of languages with a finite alphabet and verifier that runs in polynomial time. The following defines a "verifier":
Let L be a language over a finite alphabet, Σ.
L ∈ NP if, and only if, there exists a binary relation and a positive integer k such that the following two conditions are satisfied:
 For all , such that (x, y) ∈ R and ; and
 the language over is decidable by a deterministic Turing machine in polynomial time.
A Turing machine that decides L_{R} is called a verifier for L and a y such that (x, y) ∈ R is called a certificate of membership of x in L.
Not all verifiers must be polynomialtime. However, for L to be in NP, there must be a verifier that runs in polynomial time.
Example
Let
Whether a value of x is composite is equivalent to of whether x is a member of COMPOSITE. It can be shown that COMPOSITE ∈ NP by verifying that it satisfies the above definition (if we identify natural numbers with their binary representations).
COMPOSITE also happens to be in P, a fact demonstrated by the invention of the AKS primality test.^{[48]}
NPcompleteness
There are many equivalent ways of describing NPcompleteness.
Let L be a language over a finite alphabet Σ.
L is NPcomplete if, and only if, the following two conditions are satisfied:
 L ∈ NP; and
 any L' in NP is polynomialtimereducible to L (written as ), where if, and only if, the following two conditions are satisfied:
 There exists f : Σ* → Σ* such that for all w in Σ* we have: ; and
 there exists a polynomialtime Turing machine that halts with f(w) on its tape on any input w.
Alternatively, if L ∈ NP, and there is another NPcomplete problem that can be polynomialtime reduced to L, then L is NPcomplete. This is a common way of proving some new problem is NPcomplete.
Claimed solutions
While the P versus NP problem is generally considered unsolved,^{[49]} many amateur and some professional researchers have claimed solutions. Gerhard J. Woeginger compiled a list of 116 purported proofs from 1986 to 2016, of which 61 were proofs of P = NP, 49 were proofs of P ≠ NP, and 6 proved other results, e.g. that the problem is undecidable.^{[50]} Some attempts at resolving P versus NP have received brief media attention,^{[51]} though these attempts have been refuted.
Popular culture
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.^{[52]}
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".^{[53]}^{[54]}
In the second episode of season 2 of Elementary, "Solve for X" Sherlock and Watson investigate the murders of mathematicians who were attempting to solve P versus NP.^{[55]}^{[56]}
Similar problems
 R vs. RE problem, where R is analog of class P, and RE is analog class NP. These classes are not equal, because undecidable but verifiable problems do exist, for example, Hilbert's tenth problem which is REcomplete.^{[57]}
 A similar problem exists in the theory of algebraic complexity: VP vs. VNP problem. This problem has not been solved yet.^{[58]}^{[57]}
See also
 Game complexity
 List of unsolved problems in mathematics
 Unique games conjecture
 Unsolved problems in computer science
Notes
 ^ A nondeterministic Turing machine can move to a state that is not determined by the previous state. Such a machine could solve an NP problem in polynomial time by falling into the correct answer state (by luck), then conventionally verifying it. Such machines are not practical for solving realistic problems but can be used as theoretical models.
 ^ Exactly how efficient a solution must be to pose a threat to cryptography depends on the details. A solution of with a reasonable constant term would be disastrous. On the other hand, a solution that is in almost all cases would not pose an immediate practical danger.
References
 ^ Fortnow, Lance (2009). "The status of the P versus NP problem" (PDF). Communications of the ACM. 52 (9): 78–86. CiteSeerX 10.1.1.156.767. doi:10.1145/1562164.1562186. S2CID 5969255. Archived from the original (PDF) on 24 February 2011. Retrieved 26 January 2010.
 ^ Fortnow, Lance (2013). The Golden Ticket: P, NP, and the Search for the Impossible. Princeton, NJ: Princeton University Press. ISBN 9780691156491.
 ^ Cook, Stephen (1971). "The complexity of theorem proving procedures". Proceedings of the Third Annual ACM Symposium on Theory of Computing. pp. 151–158. doi:10.1145/800157.805047. ISBN 9781450374644. S2CID 7573663.
 ^ L. A. Levin (1973). Универсальные задачи перебора [Problems of Information Transmission]. Пробл. передачи информ (in Russian). 9 (3): 115–116.
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 Rachel Crowell (28 May 2021). "The Top Unsolved Questions in Mathematics Remain Mostly Mysterious Just one of the seven Millennium Prize Problems named 21 years ago has been solved". www.scientificamerican.com. Retrieved 21 June 2021.
This problem concerns the issue of whether questions that are easy to verify (a class of queries called NP) also have solutions that are easy to find (a class called P).
 Hosch, William L (11 August 2009). "P versus NP problem mathematics". Encyclopædia Britannica. Retrieved 20 June 2021.
 "P vs NP Problem". www.claymath.org (Cook, Levin). Retrieved 20 June 2021.
Suppose that you are organizing housing accommodations for a group of four hundred university students. Space is limited and only one hundred of the students will receive places in the dormitory. To complicate matters, the Dean has provided you with a list of pairs of incompatible students, and requested that no pair from this list appear in your final choice. This is an example of what computer scientists call an NPproblem...
Further reading
 Cormen, Thomas (2001). Introduction to Algorithms. Cambridge: MIT Press. ISBN 9780262032933.
 Garey, Michael R.; Johnson, David S. (1979). Computers and Intractability: A Guide to the Theory of NPCompleteness. Series of Books in the Mathematical Sciences (1st ed.). New York: W. H. Freeman and Company. ISBN 9780716710455. MR 0519066. OCLC 247570676.
 Goldreich, Oded (2010). P, NP, and NPCompleteness. Cambridge: Cambridge University Press. ISBN 9780521122542. Online drafts
 Immerman, Neil (1987). "Languages that Capture Complexity Classes". SIAM Journal on Computing. 16 (4): 760–778. CiteSeerX 10.1.1.75.3035. doi:10.1137/0216051.
 Papadimitriou, Christos (1994). Computational Complexity. Boston: AddisonWesley. ISBN 9780201530827.
External links
 Fortnow, L.; Gasarch, W. "Computational complexity".
 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. Archived from the original on 24 November 2021 – via YouTube.