P versus NP problem
If the solution to a problem is easy to check for correctness, must the problem be easy to solve?
Millennium Prize Problems |
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The P versus NP problem is a major unsolved problem in computer science. It asks whether every problem whose solution can be quickly verified can also be solved quickly.
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.
The informal term quickly, used above, means the existence of an
An answer to the P = NP question would determine whether problems that can be verified in polynomial time can also be solved in polynomial time. If it turned out that 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.
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]
Example
Consider
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]) 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.
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
In this theory, the class P consists of all those
- 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]
NP-completeness
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.
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".
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
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 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
Problems in NP not known to be in P or NP-complete
In 1975,
The graph isomorphism problem is the computational problem of determining whether two finite
The
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.
Does P mean "easy"?
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),[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 up-arrow notation), and where h is the number of vertices in H.[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 tackling the problem in practice. There are algorithms for many NP-complete problems, such as the
Finally, there are types of computations which do not conform to the Turing machine model on which P and NP are defined, such as
Reasons to believe P ≠ NP or P = NP
According to polls,[9][28] 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.[29]
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.
—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
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.
Consequences of solution
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.
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. It is also possible that a proof would not lead directly to efficient methods, perhaps if the proof is
Cryptography, for example, relies on certain problems being difficult. A constructive and efficient solution[Note 2] to an NP-complete problem such as 3-SAT would break most existing cryptosystems including:
- Existing implementations of public-key cryptography,[30] a foundation for many modern security applications such as secure financial transactions over the Internet.
- 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, the problem of finding a pre-image that hashes to a given value must be difficult in order to be useful, and ideally should require exponential time. However, if P = NP, then finding a pre-image M can be done in polynomial time, through reduction to SAT.[32]
These would need to be modified or replaced by information-theoretically secure solutions not inherently based on P-NP inequivalence.
On the other hand, there are enormous positive consequences that would follow from rendering tractable many currently mathematically intractable problems. For instance, many problems in operations research are NP-complete, such as some types of integer programming and the travelling salesman problem. 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 NP-complete;[33] if these problems were efficiently solvable it could spur considerable advances in life sciences and biotechnology.
But such changes may pale in significance compared to the revolution an efficient method for 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:[34][35]
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 the Entscheidungsproblem, 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 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 says[36]
... it would transform mathematics by allowing a computer to find a formal proof of any theorem which has a proof of a reasonable length, since formal proofs can easily be recognized in polynomial time. Example problems may well include all of the
CMI prize problems.
Research mathematicians spend their careers trying to prove theorems, and some proofs have taken decades or even centuries to find after problems have been stated—for instance, Fermat's Last Theorem took over three centuries to prove. A method that is guaranteed to find proofs to theorems, should one exist of a "reasonable" size, would essentially end this struggle.
Donald Knuth has stated that he has come to believe that P = NP, but is reserved about the impact of a possible proof:[37]
[...] 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 that showed that P ≠ NP would lack the practical computational benefits of a proof that P = NP, but would nevertheless represent a very significant advance in computational complexity theory and provide guidance for future research. It would allow one to show in a formal way that many common problems cannot be solved efficiently, so that the attention of researchers can be focused on partial solutions or solutions to other problems. Due to widespread belief in P ≠ NP, much of this focusing of research has already taken place.[38]
Also P ≠ NP still leaves open the average-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.[39] 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. A Princeton University workshop in 2009 studied the status of the five worlds.[40]
Results about difficulty of proof
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:
Classification | Definition |
---|---|
Relativizing proofs
|
Imagine a world where every algorithm is allowed to make queries to some fixed subroutine called an oracle (a black box which can answer a fixed set of questions in constant time, such as a black box that solves any given 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.[41] Since relativizing proofs can only prove statements that are uniformly true with respect to all possible oracles, this showed that relativizing techniques cannot resolve P = NP. |
Natural proofs | In 1993, one-way functions exist, then no natural proof method can distinguish between P and NP. Although one-way functions have never been formally proven to exist, 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 non-relativizing 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.[43] |
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
Claimed solutions
While the P versus NP problem is generally considered unsolved,[46] 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.[47] Some attempts at resolving P versus NP have received brief media attention,[48] though these attempts have since been refuted.
Logical characterizations
The P = NP problem can be restated in terms of expressible certain classes of logical statements, as a result of work in
Consider all languages of finite structures with a fixed
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?".[49] 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).
Polynomial-time algorithms
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 2b − 1 other programs first.
Formal definitions
P and NP
Conceptually speaking, 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 can produce the correct answer for any input string of length n in at most cnk 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 defined as the set of all languages that can be decided by a deterministic polynomial-time Turing machine. That is,
where
and a deterministic polynomial-time Turing machine is a deterministic Turing machine M that satisfies the following 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 to define NP is to use the concept of certificate and verifier. Formally, NP is defined as the set of languages over a finite alphabet that have a verifier that runs in polynomial time, where the notion of "verifier" is defined as follows.
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 LR is called a verifier for L and a y such that (x, y) ∈ R is called a certificate of membership of x in L.
In general, a verifier does not have to be polynomial-time. However, for L to be in NP, there must be a verifier that runs in polynomial time.
Example
Let
Clearly, the question of whether a given x is a composite is equivalent to the question 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.[50]
NP-completeness
There are many equivalent ways of describing NP-completeness.
Let L be a language over a finite alphabet Σ.
L is NP-complete if, and only if, the following two conditions are satisfied:
- L ∈ NP; and
- any L′ in NP is polynomial-time-reducible 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 polynomial-time Turing machine that halts with f(w) on its tape on any input w.
Alternatively, if L ∈ NP, and there is another NP-complete problem that can be polynomial-time reduced to L, then L is NP-complete. This is a common way of proving some new problem is NP-complete.
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.[51]
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".[52][53]
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.[54][55]
The game Minecraft has "NP is not in P!" among it's random title screen messages.
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
- ^ a b R. E. Ladner "On the structure of polynomial time reducibility," Journal of the ACM 22, pp. 151–171, 1975. Corollary 1.1. ACM site.
- 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.
{{cite book}}
: External link in
(help); Unknown parameter|chapterurl=
|chapterurl=
ignored (|chapter-url=
suggested) (help) - ^ L. A. Levin (1973). "Универсальные задачи перебора" (in Russian). 9 (3) (Problems of Information Transmission ed.): 115–116.
{{cite journal}}
: Cite journal requires|journal=
(help) - doi:10.1145/1562164.1562186. Archived from the original(PDF) on 24 February 2011. Retrieved 26 January 2010.
- ^ NSA (2012). "Letters from John Nash" (PDF).
- ^ Hartmanis, Juris. "Gödel, von Neumann, and the P = NP problem" (PDF). Bulletin of the European Association for Theoretical Computer Science. 38: 101–107.
- ^ Sipser, Michael: Introduction to the Theory of Computation, Second Edition, International Edition, page 270. Thomson Course Technology, 2006. Definition 7.19 and Theorem 7.20.
- ^ . Retrieved 26 September 2018.
- ^ William I. Gasarch. "The Second P=?NP poll" (PDF). SIGACT News. 74.
- ^ a b "Guest Column: The Third P =? NP Poll1" (PDF).
- ^ Scott Aaronson. "PHYS771 Lecture 6: P, NP, and Friends". Retrieved 27 August 2007.
- ^ "NP-completeness of Sudoku".
- ^ Colbourn, Charles J. (1984). "The complexity of completing partial Latin squares". Discrete Applied Mathematics. 8 (1): 25–30.
- ^ I. Holyer (1981). "The NP-completeness of some edge-partition problems". SIAM J. Comput. 10: 713–717.
- .
- ^ David Eppstein. "Computational Complexity of Games and Puzzles".
- ^ Fischer, Michael J.; Rabin, Michael O. (1974). "Super-Exponential Complexity of Presburger Arithmetic". Proceedings of the SIAM-AMS Symposium in Applied Mathematics. 7: 27–41. Archived from the original on 15 September 2006. Retrieved 15 October 2017.
- doi:10.1137/0208032.
- .
- )
- .
- ^ Lance Fortnow. Computational Complexity Blog: Complexity Class of the Week: Factoring. 13 September 2002.
- ^ Pisinger, D. 2003. "Where are the hard knapsack problems?" Technical Report 2003/08, Department of Computer Science, University of Copenhagen, Copenhagen, Denmark
- doi:10.1016/j.jctb.2011.07.004.)
{{cite journal}}
: CS1 maint: multiple names: authors list (link - .
- .
- ^ Rosenberger, Jack (May 2012). "P vs. NP poll results". Communications of the ACM. 55 (5): 10.
- ^ Scott Aaronson. "Reasons to believe"., point 9.
- ISBN 978-3-540-63890-2.) for a reduction of factoring to SAT. A 512 bit factoring problem (8400 MIPS-years when factored) translates to a SAT problem of 63,652 variables and 406,860 clauses.
{{cite book}}
:|journal=
ignored (help)CS1 maint: multiple names: authors list (link - doi:10.1023/A:1006326723002.) in which an instance of DES is encoded as a SAT problem with 10336 variables and 61935 clauses. A 3DES problem instance would be about 3 times this size.
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- ^ History of this letter and its translation from Michael Sipser. "The History and Status of the P versus NP question" (PDF).
- ^ David S. Johnson. "A Brief History of NP-Completeness, 1954–2012" (PDF). From pages 359–376 of Optimization Stories, M. Grötschel (editor), a special issue of ¨ Documenta Mathematica, published in August 2012 and distributed to attendees at the 21st International Symposium on Mathematical Programming in Berlin.
- ^ Cook, Stephen (April 2000). "The P versus NP Problem" (PDF). Clay Mathematics Institute. Retrieved 18 October 2006.
{{cite journal}}
: Cite journal requires|journal=
(help) - InformIT. Retrieved 20 July 2014.
- JSTOR 2580891.
- ^ R. Impagliazzo, "A personal view of average-case complexity," sct, pp.134, 10th Annual Structure in Complexity Theory Conference (SCT'95), 1995
- ^ "Tentative program for the workshop on "Complexity and Cryptography: Status of Impagliazzo's Worlds"". Archived from the original on 15 November 2013.
- doi:10.1137/0204037.
- .
- doi:10.1145/1374376.1374481.)
{{cite conference}}
: Unknown parameter|lastauthoramp=
ignored (|name-list-style=
suggested) (help - ^ Aaronson, Scott. "Is P Versus NP Formally Independent?" (PDF)..
- ^ Ben-David, Shai; Halevi, Shai (1992). "On the independence of P versus NP". Technical Report. 714. Technion.
{{cite journal}}
: Cite journal requires|journal=
(help). - ^ John Markoff (8 October 2009). "Prizes Aside, the P-NP Puzzler Has Consequences". The New York Times.
- ^ Gerhard J. Woeginger. "The P-versus-NP page". Retrieved 24 June 2018.
- ^ Markoff, John (16 August 2010). "Step 1: Post Elusive Proof. Step 2: Watch Fireworks". The New York Times. Retrieved 20 September 2010.
- ^ Elvira Mayordomo. "P versus NP" Archived 16 February 2012 at the Wayback Machine Monografías de la Real Academia de Ciencias de Zaragoza 26: 57–68 (2004).
- JSTOR 3597229.
- ^ Geere, Duncan (26 April 2012). "'Travelling Salesman' movie considers the repercussions if P equals NP". Wired UK. Retrieved 26 April 2012.
- ^ Hardesty, Larry. "Explained: P vs. NP".
- ^ Shadia, Ajam. "What is the P vs. NP problem? Why is it important?".
- ^ Gasarch, William (7 October 2013). "P vs NP is Elementary? No— P vs NP is ON Elementary". blog.computationalcomplexity.org. Retrieved 6 July 2018.
- ^ Kirkpatrick, Noel (4 October 2013). "Elementary Solve for X Review: Sines of Murder". TV.com. Retrieved 6 July 2018.
Further reading
- Cormen, Thomas (2001). ISBN 978-0-262-03293-3.
- Garey, Michael; Johnson, David (1979). Computers and Intractability: A Guide to the Theory of NP-Completeness. San Francisco: ISBN 978-0-7167-1045-5.
- Goldreich, Oded (2010). P, NP, and NP-Completeness. Cambridge: Cambridge University Press.
- Immerman, N. (1987). "Languages which capture complexity classes". SIAM Journal on Computing. 16 (4): 760–778. doi:10.1137/0216051.
- Papadimitriou, Christos (1994). Computational Complexity. Boston: Addison-Wesley. ISBN 978-0-201-53082-7.
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 – via YouTube.