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TheP versus NP problem is a majorunsolved problem intheoretical computer science. Informally, it asks whether every problem whose solution can be quickly verified can also be quickly solved.
Here, "quickly" means an algorithm exists that solves the task and runs inpolynomial time (as opposed to, say,exponential time), meaning the task completion time isbounded above by apolynomial function on the size of the input to the algorithm. The general class of questions that somealgorithm 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 beverified 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 incomputer science.[1] Aside from being an important problem incomputational 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 sevenMillennium Prize Problems selected by theClay Mathematics Institute, each of which carries a US$1,000,000 prize for the first correct solution.
Consider the following yes/no problem: given an incompleteSudoku 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 polynomial-time 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.)
The precise statement of the P versus NP problem was introduced in 1971 byStephen Cook in his seminal paper "The complexity of theorem proving procedures",[3] and independently byLeonid Levin in 1973.[4]
Although the P versus NP problem was formally defined in 1971, there were previous inklings of the underlying problems involved. In 1955, mathematicianJohn Nash wrote a letter to theNational Security Agency, speculating that the time required to crack a sufficiently complex code would increase exponentially with the length of the key.[5] If proved (and Nash was skeptical), this would imply what is now called P ≠ NP, since a proposed key can be verified in polynomial time. In a 1956 letter written byKurt Gödel toJohn von Neumann, Gödel asked whether theorem-proving (now known to beco-NP-complete) could be solved inquadratic orlinear time,[6] and posited that if so, then the discovery of mathematical proofs could be automated.
The relation between thecomplexity classes P and NP is studied incomputational complexity theory, the part of thetheory 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 isdeterministic (given the computer's present state and any inputs, there is only one possible action that the computer might take) andsequential (it performs actions one after the other).
In this theory, the class P consists of alldecision problems (definedbelow) solvable on a deterministic sequential machine in a durationpolynomial in the size of the input; the classNP consists of all decision problems whose positive solutions are verifiable inpolynomial time given the right information, or equivalently, whose solution can be found in polynomial time on anon-deterministic machine.[7] Clearly, P ⊆ NP. Arguably, the biggest open question intheoretical computer science concerns the relationship between those two classes:
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 2002, 61% believed P ≠ NP, as opposed to 83% in 2012 and 88% in 2019. 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."
To attack the P = NP question, the concept of NP-completeness is very useful. NP-complete 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 NP-complete problem. 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, theBoolean satisfiability problem is NP-complete by theCook–Levin theorem, soany instance ofany problem in NP can be transformed mechanically into a Boolean satisfiability problem in polynomial time. The Boolean satisfiability problem is one of many NP-complete problems. If any NP-complete problem is in P, then it would follow that P = NP. However, many important problems are NP-complete, and no fast algorithm for any of them is known.
From the definition alone it is unintuitive that NP-complete problems exist; however, a trivial NP-complete problem can be formulated as follows: given aTuring machineM guaranteed to halt in polynomial time, does a polynomial-size input thatM will accept exist?[11] It is in NP because (given an input) it is simple to check whetherM accepts the input by simulatingM; it is NP-complete because the verifier for any particular instance of a problem in NP can be encoded as a polynomial-time machineM 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 completeLatin squares in polynomial time.[12] This in turn gives a solution to the problem of partitioningtri-partite graphs into triangles,[13] which could then be used to find solutions for the special case of SAT known as 3-SAT,[14] 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 classEXPTIME is the set of all decision problems that haveexponential running time. In other words, any problem in EXPTIME is solvable by adeterministic Turing machine inO(2p(n)) time, wherep(n) is a polynomial function ofn. A decision problem isEXPTIME-complete if it is in EXPTIME, and every problem in EXPTIME has apolynomial-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 thetime hierarchy theorem, they cannot be solved in significantly less than exponential time. Examples include finding a perfect strategy forchess positions on anN ×N board[15] and similar problems for other board games.[16]
The problem of deciding the truth of a statement inPresburger arithmetic requires even more time.Fischer andRabin proved in 1974[17] that every algorithm that decides the truth of Presburger statements of lengthn has a runtime of at least for some constantc. Hence, the problem is known to need more than exponential run time. Even more difficult are theundecidable problems, such as thehalting 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.[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#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.[19] Such problems are called NP-intermediate problems. Thegraph isomorphism problem, thediscrete logarithm problem, and theinteger 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 finitegraphs areisomorphic. 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, thepolynomial 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 toLászló Babai, runs inquasi-polynomial time.[22]
The integer factorization problem is the computational problem of determining theprime factorization of a given integer. Phrased as a decision problem, it is the problem of deciding whether the input has a factor less thank. No efficient integer factorization algorithm is known, and this fact forms the basis of several modern cryptographic systems, such as theRSA algorithm. The integer factorization problem is in NP and inco-NP (and even inUP and co-UP[23]). If the problem is NP-complete, the polynomial time hierarchy will collapse to its first level (i.e., NP = co-NP). The mostefficient known algorithm for integer factorization is thegeneral number field sieve, which takes expected time
to factor ann-bit integer. The best knownquantum algorithm for this problem,Shor's algorithm, runs in polynomial time, although this does not indicate where the problem lies with respect to non-quantum complexity classes.
All of the above discussion has assumed that P means "easy" and "not in P" means "difficult", an assumption known asCobham'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 ofdeciding whether a graphG containsH as aminor, whereH is fixed, can be solved in a running time ofO(n2),[25] wheren is the number of vertices inG. However, thebig O notation hides a constant that depends superexponentially onH. The constant is greater than (usingKnuth's up-arrow notation), and whereh is the number of vertices inH.[26]
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 the problem in practice. There are algorithms for many NP-complete problems, such as theknapsack problem, thetraveling salesman problem, and theBoolean satisfiability problem, that can solve to optimality many real-world instances in reasonable time. The empiricalaverage-case complexity (time vs. problem size) of such algorithms can be surprisingly low. An example is thesimplex algorithm inlinear programming, which works surprisingly well in practice; despite having exponential worst-casetime complexity, it runs on par with the best known polynomial-time 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 asquantum computation andrandomized algorithms.
Cook provides a restatement of the problem inThe 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 polynomial-time algorithm for any of more than 3,000 important known NP-complete problems (seeList 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 whose solutions are easy to verify matches real-world 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 ofFermat'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.
When one substitutes "linear time on a multitape Turing machine" for "polynomial time" in the definitions of P and NP, one obtains the classesDLIN andNLIN.It is known[31] that DLIN ≠ NLIN.
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.
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 NP-complete problems are fundamental in many fields.
It is also very possible that a proof wouldnot lead to practical algorithms for NP-complete problems. The formulation of the problem does not require that the bounding polynomial be small or even specifically known. Anon-constructive 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 low-order 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 ofcryptography, which relies on certain problems being difficult. A constructive and efficient solution[Note 2] to an NP-complete problem such as3-SAT would break most existing cryptosystems including:
These would need modification or replacement withinformation-theoretically 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 inoperations research are NP-complete, such as types ofinteger programming and thetravelling salesman problem. Efficient solutions to these problems would have enormous implications for logistics. Many other important problems, such as some problems inprotein structure prediction, are also NP-complete;[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 NP-complete 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⋅n2), this would have consequences of the greatest importance. Namely, it would obviously mean that in spite of the undecidability of theEntscheidungsproblem, the mental work of a mathematician concerning Yes-or-No questions could be completely replaced by a machine. After all, one would simply have to choose the natural numbern 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 theCMI 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 numberM 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 donM bitwise or addition or shift operations onn 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.
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 theaverage-case 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 average-case 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 NP-hard problems. The "world" where P ≠ NP but all problems in NP are tractable in the average case is called "Heuristica" in the paper. APrinceton University workshop in 2009 studied the status of the five worlds.[41]
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 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 incomputational 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 anoracle (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 calledrelativizing. In 1975, Baker, Gill, andSolovay 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 andSteven Rudich defined a general class of proof techniques for circuit complexity lower bounds, callednatural 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 ifone-way functions exist, P and NP are indistinguishable to natural proof methods. Although the existence of one-way functions is unproven, most mathematicians believe that they exist, 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 non-relativizing proof techniques were successfully used to prove thatIP = PSPACE. However, in 2008,Scott Aaronson andAvi Wigderson showed that the main technical tool used in the IP = PSPACE proof, known asarithmetization, was also insufficient to resolve P = NP.[44] Arithmetization converts the operations of an algorithm to algebraic and basicarithmetic symbols and then uses those to analyze the workings. In theIP = PSPACE proof, they convert theblack 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 othertime complexity problems. |
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 lead some computer scientists to suggest the P versus NP problem may beindependent of standard axiom systems likeZFC (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 polynomial-time algorithms are correct.[45] However, if the problem is undecidable even with much weaker assumptions extending thePeano axioms for integer arithmetic, then nearly polynomial-time 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.
The P = NP problem can be restated as certain classes of logical statements, as a result of work indescriptive complexity.
Consider all languages of finite structures with a fixedsignature including alinear order relation. Then, all such languages in P are expressible infirst-order logic with the addition of a suitable leastfixed-point 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 existentialsecond-order logic—that is, second-order logic restricted to excludeuniversal quantification over relations, functions, and subsets. The languages in thepolynomial 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?".[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 = co-NP, which in turn implies that NP = PH).
No known algorithm for a NP-complete problem runs in polynomial time. However, there are algorithms known for NP-complete 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 toLevin (without any citation), is such an example below. It correctly accepts the NP-complete languageSUBSET-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
This is a polynomial-time algorithm accepting an NP-complete 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 asemi-algorithm).
This algorithm is enormously impractical, even if P = NP. If the shortest program that can solve SUBSET-SUM in polynomial time isb bits long, the above algorithm will try at least2b − 1 other programs first.
Adecision problem is a problem that takes as input somestringw over an alphabet Σ, and outputs "yes" or "no". If there is analgorithm (say aTuring machine, or acomputer program with unbounded memory) that produces the correct answer for any input string of lengthn in at mostcnk steps, wherek andc are constants independent of the input string, then we say that the problem can be solved inpolynomial time and we place it in the class P. Formally, P is the set of languages that can be decided by a deterministic polynomial-time Turing machine. Meaning,
where
and a deterministic polynomial-time Turing machine is a deterministic Turing machineM that satisfies two conditions:
NP can be defined similarly using nondeterministic Turing machines (the traditional way). However, a modern approach uses the concept ofcertificate andverifier. Formally, NP is the set of languages with a finite alphabet and verifier that runs in polynomial time. The following defines a "verifier":
LetL be a language over a finite alphabet, Σ.
L ∈ NP if, and only if, there exists a binary relation and a positive integerk such that the following two conditions are satisfied:
A Turing machine that decidesLR is called averifier forL and ay such that (x,y) ∈R is called acertificate of membership ofx inL.
Not all verifiers must be polynomial-time. However, forL to be in NP, there must be a verifier that runs in polynomial time.
Let
Whether a value ofx iscomposite is equivalent to of whetherx 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 theAKS primality test.[48]
There are many equivalent ways of describing NP-completeness.
LetL be a language over a finite alphabet Σ.
L is NP-complete if, and only if, the following two conditions are satisfied:
Alternatively, ifL ∈ NP, and there is another NP-complete problem that can be polynomial-time reduced toL, thenL is NP-complete. This is a common way of proving some new problem is NP-complete.
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.
The filmTravelling 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 ofThe 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 ofElementary,"Solve for X" Holmes and Watson investigate the murders of mathematicians who were attempting to solve P versus NP.[55][56]
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).
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 NP-problem...
