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Complexity classes P and NPComputational complexity theory is 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 does it take to solve a problem) and space (how much memory does it take to solve a problem).
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. The biggest open question in theoretical computer science concerns the relationship between those two classes:
If P = NP, P would encompass the NP and NP-Complete areas.
An important role in this discussion is played by the set of NP-complete problems (or NPC) which can be loosely described as those problems in NP that are the least likely to be in P. (See NP-complete for the exact definition.) Theoretical computer scientists believe this is how P, NP, and NPC currently exists amongst each other. The intersection of P and NPC equals the empty set.
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 two large numbers X and Y, we might ask whether Y is a multiple of any integers between 1 and X, exclusive. For example, we might ask whether 69799 is a multiple of any integers between 1 and 250. The answer is YES, though it would take a fair amount of work to find it manually. On the other hand, if someone claims that the answer is YES because 223 is a divisor of 69799, then we can quickly check that with a single division. Verifying that a number is a divisor is much easier than finding the divisor 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. It is not known whether the problem is in P. The special case where X=Y was first shown to be in P in 2002 (see references for "PRIMES in P" below).
A decision problem is a problem that takes as input some string and requires as output either YES or NO. If there is an algorithm (say a Turing machine, or a LISP or Pascal program with unbounded memory) which is able to produce the correct answer for any input string of length n in at most nk steps, where k is some constant 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. Intuitively, we think of the problems in P as those that can be solved reasonably fast.
Now suppose there is an algorithm A(w,C) which takes two arguments, a string w which is an input string to our decision problem, and a string C which is a "proposed certificate", and such that A produces a YES/NO answer in at most nk steps (where n is the length of w and k doesn't depend on w). Suppose furthermore that
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. This means that if a single NP-complete problem could be shown to be in P, then it would follow that P = NP. Unfortunately, many important problems have been shown to be NP-complete and not a single fast algorithm for any of them is known.
The P = NP question has also been addressed using oracles.
Although we don't know whether P=NP, we do know of problems outside both P and NP. The problem of finding the best move in Chess or Go (on an n by n board) is EXPTIME-complete. This means it requires exponential time, and so is outside P and NP. The problem of deciding the truth of a statement in Presburger arithmetic is even harder. 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 solved in general given any amount of time.
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, for several reasons:
// 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's "NO". // // Input: S = a finite set of integers // Output: "YES" if any subset of S adds up to 0. // Otherwise, it runs forever with no output. // Note: "Program number P" is the program you get 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 0If 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 it to always halt and return a "YES" or "NO" answer. Does any algorithm exist that can provably do this in polynomial time? No one knows. But if such algorithms do exist, then we already know some of them! Just replace 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 "NO" (or "NO" if that was proved) and HALT
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