P Np

The existence of problems outside both P and NP-complete in this case was established by Ladner.

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.
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 three additions. However, finding such a subset in the first place could take more time.

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 question when more complicated answers are allowed (whether FP = FNP) is equivalent.
The Clay Mathematics Institute has offered a $1 million US prize for the first correct proof.


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 2009, these assumptions are satisfied by all practical computers yet devised.
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. 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.
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 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
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 such that O(nk),


where
and tM(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.
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 and O; and
the language over is decidable by a Turing machine in polynomial time.
A Turing machine that decides LR is called a verifier for L and a y such that is called a certificate of membership of x in L.
In general, a verifier does not have to be polynomial-time. It can be shown that by verifying that COMPOSITE satisfies the above definition.
COMPOSITE also happens to be in P.
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. If any NP-complete problem is in P, then it would follow that P = NP.Vardi, Rice University

Being attached to a speculation is not a good guide to research planning. 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
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.

(It is also possible that a proof would not lead directly to efficient methods, perhaps if the proof is non-constructive, or the size of the bounding polynomial is too big to be efficient in practice.) 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. 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.

In fact, 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:
Relativizing proofs: Imagine a world where every algorithm is allowed to make queries to some fixed subroutine called an oracle, 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.

At the time, all previously known circuit lower bounds were natural, and circuit complexity was considered a very promising approach for resolving P = NP. However, Razborov and Rudich showed that in order to prove P ≠ NP using a natural proof, one necessarily must also prove an even stronger statement, which is believed to be false.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.
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 which is not excluded by the above results.
These barriers have also led some computer scientists to suggest that the P versus NP problem may be independent of standard axiom systems like ZFC (cannot be proved or disproved within them). The interpretation of an independence result could be that either no polynomial time algorithm exist for any NP-complete problem, but such a proof cannot be constructed in (say) ZFC, or that polynomial time algorithms for NP-complete problems may exist, but it's impossible to prove (in ZFC) that such algorithms are correct. However, if the problem cannot be decided even with much weaker assumptions extending the Peano axioms (PA) for integer arithmetic, then there would necessarily exist nearly-polynomial-time algorithms for every problem in NP. Therefore, if one believes (as most complexity theorists do) that problems in NP do not have efficient algorithms, it would follow that such notions of independence cannot be possible.

It correctly accepts the NP-complete language SUBSET-SUM, and runs in polynomial time 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 P" is the program obtained by
writing the integer P in binary, then
considering that string of bits to be a
program. "Accepting" means it gives "yes" answers in polynomial time, but is allowed to run forever when the answer is "no".
Note that this is enormously impractical, even if P = NP.

That means we want the algorithm to always halt and return a "yes" or "no" answer. All languages (of finite structures with a fixed signature including a linear order relation) in P can be expressed in first-order logic with the addition of a suitable least fixed point operator (effectively, this, in combination with the order, allows the definition of recursive functions); indeed, (as long as the signature contains at least one predicate or function in addition to the distinguished order relation ), this precisely characterizes P.

The languages in the polynomial hierarchy, PH, correspond to all of second-order logic. http:www.claymath.org/millennium/P_vs_NP/Official_Problem_Description.pdf.

http:www.cse.iitk.ac.in/users/manindra/algebra/primality_v6.pdf. A list of links to a number of purported solutions to the problem.