P = NP problem

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Diagram of complexity classes provided that P ≠ NP. The existence of problems outside both P and NP-complete in this case was established by Ladner.^[1]

The relationship between the complexity classes P and NP is an unsolved question in theoretical computer science. It is considered to be the most important problem in the field – the Clay Mathematics Institute has offered a $1 million US prize for the first correct proof.^[2]

In essence, the question P = NP? asks: if 'yes'-answers to a 'yes'-or-'no'-question can be verified "quickly" (in polynomial time), can the answers themselves also be computed quickly?

Consider, for instance, the subset-sum problem, an example of a problem which is "easy" to verify, but whose answer is believed (but not proven) to be "difficult" to compute. Given a set of integers, does some nonempty subset of them sum to 0? For instance, does a subset of the set {−2, −3, 15, 14, 7, −10} add up to 0? The answer "yes, because {−2, −3, −10, 15} add up to zero", can be quickly verified with a few additions. However, finding such a subset in the first place could take much longer. The information needed to verify a positive answer is also called a certificate. Given the right certificates, "yes" answers to our problem can be verified in polynomial time, so this problem is in NP.

An answer to the P = NP question would determine whether problems like the subset-sum problem are as "easy" to compute as to verify. If it turned out P does not equal NP, it would mean that some NP problems are substantially "harder" to compute than to verify.

The restriction to yes/no problems is unimportant; the resulting problem when more complicated answers are allowed (whether FP = FNP) is equivalent.^[3]

Millennium Prize Problems
P versus NP
The Hodge conjecture
The Poincaré conjecture
The Riemann hypothesis
Yang–Mills existence and mass gap
Navier–Stokes existence and smoothness
The Birch and Swinnerton-Dyer conjecture

1 Context of the problem
2 Formal definitions for P and NP
- 2.1 Example
3 NP-complete
4 Formal definition for NP-completeness
5 Still harder problems
6 Is P really practical?
7 Why do many computer scientists think P ≠ NP?
8 Consequences of proof
9 Results about difficulty of proof
10 Polynomial-time algorithms
11 Logical characterizations
12 See also
13 References
14 Further reading
15 External links

[edit] Context of the problem

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

In such analysis, a model of the computer for which time must be analyzed is required. Typically, such models assume that the computer is deterministic (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). As of 2008, these assumptions are satisfied by all practical computers yet devised, even those featuring parallel computing.^{[citation needed]}

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

Is P equal to NP?

In a 2002 poll of 100 researchers, 61 believed the answer is no, 9 believed the answer is yes, 22 were unsure, and 8 believed the question may be independent of the currently accepted axioms, and so impossible to prove or disprove.^[4]

[edit] Formal definitions for P and NP

Conceptually, a decision problem is a problem that takes as input some string, and outputs "yes" or "no". If there is an algorithm (say a Turing machine, or a computer program with unbounded memory) which is able to produce the correct answer for any input string of length $n$ in at most $c \cdot n^k$ 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 defined as the set of all languages which can be decided by a deterministic polynomial-time Turing machine. That is,

P = ${L : L = L (M) for some deterministic polynomial-time Turing machine M}$

where $L(M) = \{ w\in\Sigma^{*}: M \text{ accepts } w \}$

and a deterministic polynomial-time Turing machine is a deterministic Turing machine $M$ which satisfies the following two conditions:

$M halts on all input w$ ; and
there exists $k \in N$ such that $T_{M}(n)\in\;$ O $(n k)$ ,

where $T_{M}(n) = \max\{ t_{M}(w) : w\in\Sigma^{*}, \left|w\right| = n \}$

and

t M (w) = number of steps M takes to halt on input w .

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, $Σ$ .

$L\in\mathbf{NP}$ if, and only if, there exists a binary relation $R\subset\Sigma^{*}\times\Sigma^{*}$ and a positive integer $k$ such that the following two conditions are satisfied:

For all $x\in\Sigma^{*}$ , $x\in L \Leftrightarrow\exists y\in\Sigma^{*}$ such that $(x,y)\in R\;$ and $\left|y\right|\in\;$ O $(\left|x\right|^{k})$ ; and
the language $L_{R} = \{ x\# y:(x,y)\in R\}$ over $\Sigma\cup\{\#\}$ is decidable by a Turing machine.

A Turing machine that decides $L R$ is called a verifier for $L$ and a $y$ such that $(x,y)\in R$ 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.

[edit] Example

Let $\mathit{COMPOSITE} = \{x\in N:x=pq \;\text{for integers}\; p, q > 1 \}$ and $R = \{(x,y)\in N\times N: 1<y< \sqrt x\; ; \;y\; \text{divides}\; x\}$ .

Clearly, the question of whether a given $x$ is a composite is equivalent to the question of whether $x$ is a member of $C O M P O S I T E$ . It can be shown that $\mathit{COMPOSITE}\in\mathbf{NP}$ by verifying that $C O M P O S I T E$ satifies the above definition.

$C O M P O S I T E$ also happens to be in P ^[5]^[6].

[edit] NP-complete

To attack the P = NP question, the concept of NP-completeness is very useful. Informally, the NP-complete problems are the "toughest" problems in NP in the sense that they are the ones most likely not to be in P. NP-complete problems are those NP-hard problems which are in NP, where NP-hard problems are those to which any problem in NP can be reduced in polynomial time. For instance, the decision problem version of the traveling salesman problem is NP-complete, so any instance of any problem in NP can be transformed mechanically into an instance of the traveling salesman problem, in polynomial time. The traveling salesman problem is one of many such NP-complete problems. If any NP-complete problem is in P, then it would follow that P = NP. Unfortunately, many important problems have been shown to be NP-complete and as of 2008, not a single fast algorithm for any of them is known.

Based on the definition alone, it's not obvious that NP-complete problems exist. A trivial and contrived NP-complete problem can be formulated as: 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?^[7] It is in NP because, given an input, it is simple to check whether or not M accepts the input by simulating M; it is NP-hard 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. This result came to be known as 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 in this class. Thus, a vast class of seemingly unrelated problems are all reducible to one another, and are in a sense the "same problem".

[edit] Formal definition for NP-completeness

Although there are many equivalent ways of describing NP-completeness, in the context of the P vs NP question, it is best to define NP-complete problems in terms of NP problems.^{[citation needed]}

Let $\ L$ be a language over a finite alphabet $\ \Sigma$ .

$\ L$ is NP-complete if, and only if, the following two conditions are satisfied:

$L\in\mathbf{NP}$ ; and
any is polynomial time reducible to (written as ), where if, and only if, the following two conditions are satisfied:
1. There exists $f : \Sigma^{*}\rightarrow\Sigma^{*}$ such that $\forall w\in\Sigma^{*}(w\in L^{'}\Leftrightarrow f(w)\in L)$ ; and
2. there exists a polynomial time Turing machine which halts with $\ f(w)$ on its tape on any input $\ w$ .

[edit] Still harder problems

See also: Complexity class

Although it is unknown whether P = NP, problems outside of P are known. A number of succinct problems, that is, problems which operate not on normal input but on a computational description of the input, are known to be EXPTIME-complete. Because it can be shown that P $\subsetneq$ 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.^{[citation needed]}

The problem of deciding the truth of a statement in Presburger arithmetic requires even more time. Fischer and Rabin proved in 1974 that every algorithm which decides the truth of Presburger statements has a runtime of at least $2^{2^{cn}}$ for some constant c. Here, n is the length of the Presburger statement. Hence, the problem is known to need more than exponential run time. Even more difficult are the undecidable problems, such as the halting problem. They cannot be 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, but either produce the wrong answer or run forever without producing an answer.

[edit] Is P really practical?

Main article: Cobham's thesis

All of the above discussion has assumed that P means "easy" and "not in P" means "hard". While this is a common and reasonably accurate assumption in complexity theory, it is not always true in practice. See Cobham's thesis for an in-depth discussion of this point, but the main arguments are:

It ignores the size of the exponents. In fields where practical problems have millions of variables (such as Operations Research or Electronic Design Automation), even O(n³) algorithms are often impractical.
It ignores constant factors, which can be arbitrarily large and important in practice.
It only considers worst-case times. The simplex algorithm often (or even usually) solves problem instances in time $n$ , but on very rare occasions takes time $2 n$ . Since the worst case time is exponential, the algorithm is not polynomial time, but still very practical.
It only considers deterministic solutions. There might be a problem that can be solved quickly if a tiny error probability is acceptable, but is much harder to solve exactly.
Advances in technology may make exponential-time algorithms efficient for practical ranges of problem sizes.

[edit] Why do many computer scientists think P ≠ NP?

Most computer scientists believe that P≠NP. A key reason for this belief is that after decades of studying these problems, no one has been able to find a polynomial-time algorithm for any NP-hard problem. These algorithms were sought long before the concept of NP-completeness was even known (Karp's 21 NP-complete problems, among the first found, were all well-known existing problems 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.^{[citation needed]}

On the other hand, some researchers believe that we are overconfident in P ≠ NP and should explore proofs of P = NP as well. For example, in 2002 these statements were made:^[4]

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

—Moshe Y. Vardi, Rice University

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

—Anil Nerode, Cornell University

[edit] Consequences of proof

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

A proof of P = NP could have stunning practical consequences, if the proof leads to efficient methods for solving some of the important problems in NP. Various NP-complete problems are fundamental in many fields. 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, to name two of the most famous examples. 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;^[8] if these problems were solvable efficiently it could spur considerable advances in biology.

But such changes may pale in significance compared to the revolution an efficient method for solving NP-complete problems would cause in mathematics itself. According to Stephen Cook,^[9]

...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.

A proof that showed that P ≠ NP, while lacking the practical computational benefits of a proof that P = NP, would also represent a massive advance in computational complexity theory and provide guidance for future research. It would allow one to show in a formal way that many common problems cannot be solved efficiently, so that the attention of researchers can be focused on partial solutions or solutions to other problems. Due to widespread belief in P ≠ NP, much of this focusing of research has already taken place.^[10]

[edit] Results about difficulty of proof

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

One of the most frequently-cited is a result involving oracles. Suppose there is a magical machine called an oracle that can solve only one problem, such as determining if a given number is prime, but can solve it in constant time. Our new question is now, if there is no limit on the number of times that the oracle can be used, are there problems that can be verified in polynomial time that cannot be solved in polynomial time? Depending on the problem that the oracle solves, with certain oracles P = NP, while for other oracles P ≠ NP. The practical consequence of this is that any proof which can be modified to account for the existence of these oracles cannot solve the problem. Unfortunately, most known methods and nearly all classical methods can be modified in such a way (we say they are relativizing).^{[citation needed]}

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

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

[edit] Polynomial-time algorithms

No one knows whether polynomial-time algorithms exist for NP-complete languages. But if such algorithms do exist, some of them are already known. For example, the following algorithm (due to Levin) correctly accepts an NP-complete language, but as of 2008, it is unknown how long it takes in general.

   // 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 P" is the program obtained by
   //         writing the integer P 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 N = 1...infinity
       FOR P = 1...N
           Run program number P for N 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".

Perhaps we want to "solve" the SUBSET-SUM problem, rather than just "accept" the SUBSET-SUM language. That means we want the algorithm to always halt and return a "yes" or "no" answer. As of 2008, it is unknown whether an algorithm exists that does this in polynomial time. But if there is an algorithm that provably does this in polynomial time, then so does the algorithm that is obtained by replacing the IF statement in the above algorithm with this:

           IF the program outputs a complete math proof
               AND each step of the proof is legal
               AND the conclusion is that S does (or does not) have a subset summing to 0
           THEN
               OUTPUT "yes" (or "no") and HALT

[edit] Logical characterizations

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

[edit] See also

[edit] References

^ R. E. Ladner "On the structure of polynomial time reducibility," J.ACM, 22, pp. 151–171, 1975. Corollary 1.1. ACM site.
^ Millennium Prize Problems (2000-05-24). Retrieved on 2008-01-12.
^ Scott Aaronson. Complexity Zoo: FP. "FP = FNP if and only if P = NP". http://qwiki.stanford.edu/wiki/Complexity_Zoo#fp
^ ^a ^b William I. Gasarch (June 2002). "The P=?NP poll.". SIGACT News 33 (2): 34–47. doi:10.1145/1052796.1052804.
^ M. Agrawal, N. Kayal, N. Saxena. Primes is in P.
^ AKS primality test
^ Scott Aaronson. PHYS771 Lecture 6: P, NP, and Friends. Retrieved on 2007-08-27.
^ Berger B, Leighton T (1998). "Protein folding in the hydrophobic-hydrophilic (HP) model is NP-complete". J. Comput. Biol. 5 (1): 27–40. PMID 9541869.
^ Stephen Cook. The P versus NP Problem. Retrieved on 2007-08-27.
^ L. R. Foulds (Oct., 1983). "The Heuristic Problem-Solving Approach". The Journal of the Operational Research Society 34 (10): pp. 927–934.

[edit] Further reading

A. S. Fraenkel and D. Lichtenstein, Computing a perfect strategy for n*n chess requires time exponential in n, Proc. 8th Int. Coll. Automata, Languages, and Programming, Springer LNCS 115 (1981) 278–293 and J. Comb. Th. A 31 (1981) 199–214.
E. Berlekamp and D. Wolfe, Mathematical Go: Chilling Gets the Last Point, A. K. Peters, 1994. D. Wolfe, Go endgames are hard, MSRI Combinatorial Game Theory Research Worksh., 2000.
Neil Immerman. Languages Which Capture Complexity Classes. 15th ACM STOC Symposium, pp.347–354. 1983.
Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein (2001). "Chapter 34: NP-Completeness", Introduction to Algorithms, Second Edition, MIT Press and McGraw-Hill, pp.966–1021. ISBN 0-262-03293-7.
Christos Papadimitriou (1993). "Chapter 14: On P vs. NP", Computational Complexity, 1st edition, Addison Wesley, pp.329–356. ISBN 0-201-53082-1.

[edit] External links

The Clay Mathematics Institute Millennium Prize Problems
The Clay Math Institute Official Problem DescriptionPDF (118 KiB)
Ian Stewart on Minesweeper as NP-complete at The Clay Math Institute
Gerhard J. Woeginger. The P-versus-NP page. A list of links to a number of purported solutions to the problem. Some of these links state that P equals NP, some of them state the opposite. It is probable that all these alleged solutions are incorrect.
Computational Complexity of Games and Puzzles
Scott Aaronson's Complexity Zoo: P, NP
Qeden, a wiki that aims to solve the Millennium Prize Problems
Scott Aaronson's Shtetl Optimized blog: Reasons to believe, a list of justifications for the belief that P ≠ NP

v • d • e Important complexity classes (more)

P • NP • co-NP • NP-C • co-NP-C • NP-hard • UP • #P • #P-C • L • NL • NC • P-C • PSPACE • PSPACE-C • EXPTIME • NEXPTIME • EXPSPACE • 2-EXPTIME • PR • RE • Co-RE • RE-C • Co-RE-C • R • BQP • BPP • RP • ZPP • PCP • IP • PH

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