where is the set of decision problems solvable in polynomial time by a Turing machine augmented by an oracle for some complete problem in class A; the classes and are defined analogously. For example, , and is the class of problems solvable in polynomial time with an oracle for some NP-complete problem.
where is some standard encoding of the pair of binary strings x and w as a single binary string. L represents a set of ordered pairs of strings, where the first string x is a member of , and the second string w is a "short" () witness testifying that x is a member of . In other words, if and only if there exists a short witness w such that . Similarly, define
Let be a class of languages. Extend these operators to work on whole classes of languages by the definition
Again, De Morgan's laws hold: and , where .
The classes NP and co-NP can be defined as , and , where P is the class of all feasibly (polynomial-time) decidable languages. The polynomial hierarchy can be defined recursively as
Note that , and .
This definition reflects the close connection between the polynomial hierarchy and the arithmetical hierarchy, where R and RE play roles analogous to P and NP, respectively. The analytic hierarchy is also defined in a similar way to give a hierarchy of subsets of the real numbers.
An equivalent definition in terms of alternating Turing machines defines (respectively, ) as the set of decision problems solvable in polynomial time on an alternating Turing machine with alternations starting in an existential (respectively, universal) state.
Relations between classes in the polynomial hierarchy
Commutative diagram equivalent to the polynomial time hierarchy. The arrows denote inclusion.
The definitions imply the relations:
Unlike the arithmetic and analytic hierarchies, whose inclusions are known to be proper, it is an open question whether any of these inclusions are proper, though it is widely believed that they all are. If any , or if any , then the hierarchy collapses to level k: for all , . In particular, if P = NP, then the hierarchy collapses completely.
The union of all classes in the polynomial hierarchy is the complexity class PH.
If the polynomial hierarchy has any complete problems, then it has only finitely many distinct levels. Since there are PSPACE-complete problems, we know that if PSPACE = PH, then the polynomial hierarchy must collapse, since a PSPACE-complete problem would be a -complete problem for some k.
Each class in the polynomial hierarchy contains -complete problems (problems complete under polynomial-time many-one reductions). Furthermore, each class in the polynomial hierarchy is closed under -reductions: meaning that for a class in the hierarchy and a language , if , then as well. These two facts together imply that if is a complete problem for , then , and . For instance, . In other words, if a language is defined based on some oracle in , then we can assume that it is defined based on a complete problem for . Complete problems therefore act as "representatives" of the class for which they are complete.
An example of a natural problem in is circuit minimization: given a number k and a circuit A computing a Boolean functionf, determine if there is a circuit with at most k gates that computes the same function f. Let be the set of all boolean circuits. The language
is decidable in polynomial time. The language
is the circuit minimization language. because is decidable in polynomial time and because, given , if and only if there exists a circuit such that for all inputs , .
A complete problem for is satisfiability for quantified Boolean formulas with k alternations of quantifiers (abbreviated QBFk or QSATk). This is the version of the boolean satisfiability problem for . In this problem, we are given a Boolean formula f with variables partitioned into k sets X1, ..., Xk. We have to determine if it is true that
That is, is there an assignment of values to variables in X1 such that, for all assignments of values in X2, there exists an assignment of values to variables in X3, ... f is true?
The variant above is complete for . The variant in which the first quantifier is "for all", the second is "exists", etc., is complete for .