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P versus NP

Revision as of 14:02, 25 October 2017 by Scrabbler94 (talk | contribs) (→‎Arguments: more clearly define NP-completeness, give examples of NP/NP-complete problems)

$P$ versus $NP$ is one of the greatest computability and complexity problems of modern mathematics, and one of the Millennium Problems. $P$ the class of decision problems (those whose answer is either "yes" or "no," as opposed to other classes such as counting problems) that can be solved by a deterministic algorithm in polynomial time. $NP$ is the class of decision problems that can be solved by a non-deterministic algorithm in polynomial time. The $P$ versus $NP$ question asks whether these two classes are the same, or whether there are problems in $NP$ that are not in $P$.

Since all modern computers (with the exception of a few quantum computers) are deterministic, non-deterministic algorithms are of theoretical, rather than practical, interest. However, the class $NP$ can also be defined without reference to nondeterminism: This article is a stub. Help us out by expanding it.

Overview

The relation between the complexity classes $P$ and $NP$ is one of the most important open problems in theoretical computer science and mathematics. The most common measurements are time (how many steps it takes to solve a problem as a function of input, usually expressed with big-O notation) 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 - that, 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, such as a deterministic Turing machine. These assumptions reflect the behaviour of all practical computers yet devised, even including machines featuring parallel processing.

A decision problem is a problem that admits a yes or no answer (as opposed to an optimization problem, such as "What is the length of the longest path from $s$ to $t$?"). More formally, a decision problem may be thought of as a language $L$ for which we wish to decide if a given word $w$ belongs to the language.

We say that an algorithm $A$ decides a language $L$ if, for all inputs $w$, $A$ either accepts or rejects $w$.

The class $P$ consists of all those decision problems (languages) that can be decided using a deterministic Turing machine in an amount of time that is polynomial in the size of the input. More formally, $P = \bigcup_{k \ge 0} \text{TIME}(O(n^k))$ where $\text{TIME}(f(n))$ is the set of languages decidable by an $O(f(n))$-time deterministic Turing machine.

The class $NP$ (for non-deterministic polynomial time) consists of all those decision problems that are decidable using a non-deterministic Turing machine. It is equivalent to the set of decision problems for which whose yes instances are efficiently verifiable in polynomial time using a certificate. Examples of problems in $P$ and $NP$ are given below.

Importance

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.

The Clay Mathematics Institute has offered a USD $1,000,000 prize for a correct solution, as it has listed it as one of its Millenium Problems.

Arguments

It is easy to show that $P \subseteq NP$, as if we are given any $L \in P$, a polynomial-time verifier for $L$, given input $w$ and a certificate $c$, can simply ignore the certificate and decide if $w \in L$.

An important role in this discussion is played by the set of $NP$-complete problems (or $NPC$) which can be loosely described as the hardest problems in $NP$. More precisely, a language $L$ is NP-complete if both are true:

  • $L \in NP$
  • Any language in NP has a polynomial-time reduction to $L$ (NP-hardness).

The main idea behind a polynomial-time reduction is this: If we knew how to decide $L$ in polynomial time, then any problem in $NP$ can be converted into an instance of $L$ in polynomial time, and then we can use the algorithm that decides $L$ as a subroutine.

Examples of problems in P, NP, NP-complete problems

The following problems are examples of problems in $P$ (i.e. ones we can answer in polynomial time as a function of input):

  • Given a list of $n$ integers, is it sorted in non-decreasing order?
  • Given a weighted, undirected graph $G = (V,E)$ and two vertices $s, t \in E$, does there exist a path from $s$ to $t$ of weight at most $c$?
  • Given two positive integers $m$ and $n$ and a positive integer $d$, is it true that $d = \gcd(m,n)$?

A classic example of a problem that is $NP$-complete but not known to be in $P$ is the subset sum problem: Given a list $S$ of integers and a number $t$, all encoded in some base $b > 1$, is there some subset of numbers in $S$ whose sum is $t$? For example, is there a subset of $\{-4,2,3,10,-8,7\}$ whose sum is 14? The answer is yes, and it can be checked in polynomial time that the answer is yes (by giving the certificate $\{2,3,10,-8,7\}$, but this is a difficult problem to solve in general, and it is not known if subset sum is in $P$.

In essence, the $P = NP$ question asks: if positive solutions to a $YES/NO$ problem can be verified quickly, can the answers also be computed quickly? Here is an example to get a feeling for the question. Given a set of integers, does any subset of them sum to 0? For instance, does a subset of the set $\{-2, -3, 8, 15, -10\}$ add up to $0$? The answer is $YES$, though it may take a little while to find a subset that does - and if the set was larger, it might take a very long time to find a subset that does. On the other hand, if someone claims that the answer is $YES$, because $\{-2, -3, -10, 15\}$ add up to zero, then we can quickly check that with a few additions. Verifying that the subset adds up to zero is much faster than finding the subset in the first place. The information needed to verify a positive answer is also called a certificate. So we conclude that given the right certificates, positive answers to our problem can be verified quickly (i.e. in polynomial time) and that's why this problem is in $NP$.

The restriction to $YES/NO$ problems doesn't really make a difference; even if we allow more complicated answers, the resulting problem (whether $FP = FNP$) is equivalent.

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