P versus NP

Revision as of 13:40, 28 October 2007 by Inscrutableroot (talk | contribs) (Overview)

$P$ versus $NP$ is one of the greatest computability and complexity problems of modern mathematics, and one of the Millennium Problems. $P$ stands for polynomial, and it represents the category of problems for which there is an algorithm such that a computer could process it. $NP$ stands for non-deterministic polynomial, and it represents problems such that a computer could not solve with a straightforward general algorithm.


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 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 - 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. These assumptions reflect the behaviour of all practical computers yet devised, even including machines featuring parallel processing.

In this theory, the class $P$ consists of all those decision problems 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.


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 <dollar/>1,000,000 prize for a correct solution, as it has listed it as one of its Millenium Problems.


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 $NP$ and therefore they are the least likely to be in P. More precisely, any problem in $NP$, through some efficient (takes at most a polynomial-bounded number of steps) transformation, can be expressed as a problem in $NP$-complete. Therefore if one finds an efficient (again, polynomial-bounded) solution to any $NP$-complete problem, then every problem in $NP$ can be solved efficiently and therefore must be in $P$, hence proving $P = NP$. (See $NP$-complete for the exact definition.) Most theoretical computer scientists currently believe that the relationship among the classes $P$, $NP$, and $NPC$ is as shown in the picture, with the P and NPC classes disjoint.

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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