Notes on Complexity Theory Last updated: September, 2011 Lecture 8 Jonathan Katz 1 The Polynomial Hierarchy We have seen the classes NP and coNP, which are de¯ned as follows: L 2 N P if there is a (deterministic) Turing machine M running in time polynomial in its ¯rst input, such that x 2 L , 9w M(x; w) = 1: L 2 coNP if there is a (deterministic) Turing machine M running in time polynomial in its ¯rst input, such that x 2 L , 8w M(x; w) = 1: It is natural to generalize the above; doing so allows us to capture problems that are \more di±cult" than NP[ coNP. As an example, consider the language IND-SET: IND-SET def= f(G; k): G has an independent set of size ¸ kg : We know that IND-SET 2 N P; a certi¯cate is just an independent set of vertices of size at least k (that the vertices are independent can easily be veri¯ed). How about the following language? MAX-IND-SET def= f(G; k) : the largest independent set in G has size exactly kg : This language does not appear to be in NP: we can certify that some graph G has an independent set of size k, but how do we certify that this is the largest independent set in G? The language does not appear to be in coNP, either: although we could prove that (G; k) 62 MAX-IND-SET if G happened to have an independent set of size larger than k, there is no easy way to prove that (G; k) 62 MAX-IND-SET in case its largest independent set has size smaller than k. As another example, consider the problem of CNF-formula minimization. A CNF formula Á n on n variables naturally de¯nes a function fÁ : f0; 1g ! f0; 1g, where fÁ(x) = 1 i® the given assignment x satis¯es Á. Can we tell when a given formula is minimal? Consider the language def MIN-CNF = fÁ : no formula with fewer clauses than Á computes fÁg: This language does not appear to be in NP or coNP, either. (Even if I give you a smaller formula 0 Á that computes fÁ, there is no obvious way for you to verify that fact e±ciently.) The above examples motivate the following de¯nition: De¯nition 1 Let i be a positive integer. L 2 §i if there is a (deterministic) Turing machine M running in time polynomial in its ¯rst input, such that x 2 L , 9w 8w ¢ ¢ ¢ Q w M(x; w ; : : : ; w ) = 1: | 1 2{z i }i 1 i i times 8-1 where Qi = 8 if i is even, and Qi = 9 if i is odd. L 2 ¦i if there is a (deterministic) Turing machine M running in time polynomial in its ¯rst input, such that x 2 L , 8w 9w ¢ ¢ ¢ Q w M(x; w ; : : : ; w ) = 1: | 1 2{z i }i 1 i i times where Qi = 8 if i is odd, and Qi = 9 if i is even. As in the case of NP, we may assume without loss of generality that the wi each have length polynomial in x. Note also the similarity to TQBF. The crucial di®erence is that TQBF allows an unbounded number of alternating quanti¯ers, whereas §i; ¦i each allow (at most) i quanti¯ers. Since TQBF is PSPACE-complete, this implies that §i; ¦i 2 PSPACE for all i. (One could also prove this directly, via a PSPACE algorithm just like the one used to solve TQBF.) Returning to our examples, note that MAX-IND-SET 2 §2 since (G; k) 2 MAX-IND-SET i® there exists a set of vertices S such that for all sets of vertices S0 the following (e±ciently veri¯able) predicate is true: jSj = k and S is an independent set of G; moreover, either jS0j · k or S0 is not an independent set of G. (It is easier to express the above in English as: \there exists a set S of k vertices, such that for all sets S0 containing more than k vertices, S is an independent set and S0 is not an independent set". But one has to be careful to check that anytime the quanti¯cation is limited [e.g., by quantifying over all sets of size greater than k, rather than all sets], the limitation is e±cient to verify.) We also 0 have MAX-IND-SET 2 ¦2 since we can swap the order of quanti¯ers in this case (since S does not depend on S, above). Turning to the second example, note MIN-CNF 2 ¦2 since Á 2 MIN-CNF i® for all formulas Á0 that are smaller than Á there exists an input x for which Á(x) 6= Á(x0). Here, however, we cannot just swap the order of quanti¯ers (do you see why?), and so we do not know, or believe, that MIN-CNF 2 §2. We make some relatively straightforward observations, leaving their proofs as exercises. ² §1 = NP and ¦1 = coNP. ² For all i, we have co§i = ¦i (and co¦i = §i). ² For all i, we have §i; ¦i ⊆ ¦i+1 and §i; ¦i ⊆ §i+1. The polynomial hierarchy PH consists of all those languages of the form de¯ned above; that is, def S S PH = i §i = i ¦i. Since §i ⊆ PSPACE for all i, we have PH ⊆ PSPACE. Each §i (resp., ¦i) is called a level in the hierarchy. As in the case of NP and coNP, we believe Conjecture 1 §i 6= ¦i for all i. If the above conjecture is not true, then the polynomial hierarchy collapses in the following sense: Theorem 2 If §i = ¦i for some i, then PH = §i. 8-2 Proof We show that for all j > i, we have §j = §j¡1. Let L 2 §j. So there is a polynomial-time Turing machine M such that x 2 L , 9wj8wj¡1 ¢ ¢ ¢ Qw1 M(x; wj; : : : ; w1) = 1; where we have numbered the variables in descending order for clarity in what follows. Assume j ¡i is even. (A symmetric argument works if j ¡ i is odd.) De¯ne language L0 as 0 (x; wj; : : : ; wi+1) 2 L , 9wi8wi¡1 ¢ ¢ ¢ Qw1 M(x; wj; : : : ; w1) = 1; 0 0 0 and note that L 2 §i. By the assumption of the theorem, L 2 ¦i and so there is some machine M running in time polynomial in jxj + jwjj + ¢ ¢ ¢ + jwi+1j (and hence polynomial in jxj) such that 0 0 0 0 0 0 0 0 (x; wj; : : : ; wi+1) 2 L , 8wi9wi¡1 ¢ ¢ ¢ Q w1 M (x; wj; : : : ; wi+1; wi; : : : ; w1) = 1: 0 0 (Note that the fwi; : : : ; w1g need not have any relation with the fwi; : : : ; w1g.) But then 0 x 2 L , 9wj8wj¡1 ¢ ¢ ¢ 8wi+1 (x; wj; : : : ; wi+1) 2 L £ 0 0 0 0 0 0 ¤ , 9wj8wj¡1 ¢ ¢ ¢ 8wi+1 8wi9wi¡1 ¢ ¢ ¢ Q w1 M (x; wj; : : : ; w1) = 1 0 0 0 0 0 0 , 9wj8wj¡1 ¢ ¢ ¢ 8wi+1wi9wi¡1 ¢ ¢ ¢ Q w1 M (x; wj; : : : ; w1) = 1; and there are only j ¡ 1 quanti¯ers in the ¯nal expression. Thus, L 2 §j¡1. A similar argument gives: Theorem 3 If P = NP then PH = P. Each §i has the complete problem §i-SAT, where this language contains all true expressions of the form 9w18w2 ¢ ¢ ¢ Qwi Á(w1; : : : ; wi) = 1 for Á a boolean formula in CNF form. However, if PH has a complete problem, the polynomial hierarchy collapses: If L were PH-complete then L 2 §i 0 for some i; but then every language L 2 §i+1 ⊆ PH would be reducible to L, implying §i+1 = §i. Since PSPACE has complete languages, this indicates that PH is strictly contained in PSPACE. 1.1 Alternating Turing Machines The polynomial hierarchy can also be de¯ned using alternating Turing machines. An alternating Turing machine (ATM) is a non-deterministic Turing machine (so it has two transition functions, one of which is non-deterministically chosen in each step of its computation) in which every state is labeled with either 9 or 8. To determine whether a given ATM M accepts an input x, we look at the con¯guration graph of M(x) and recursively mark the con¯gurations of M(x) as follows: ² The accepting con¯guration is marked \accept". ² A con¯guration whose state is labeled 9 that has an edge to a con¯guration marked \accept" is itself marked \accept". ² A con¯guration whose state is labeled 8, both of whose edges are to con¯gurations marked \accept," is itself marked \accept". 8-3 M(x) accepts i® the initial con¯guration of M(x) is marked \accept". We can de¯ne classes atime(t(n)) and aspace(s(n)) in the natural way. We can also de¯ne §itime(t(n)) (resp., ¦itime(t(n))) to be the class of languages accepted by an ATM running in time O(t(n)), whose initial state is labeled 9 (resp., 8), and that makes at most i ¡ 1 alternations between states labeled 9 and states labeled 8. Note that L 2 §2time(t(n)) i® x 2 L , 9w18w2 M(x; w1; w2); where M is a deterministic Turing machine running in time t(jxj). (The situation for space-bounded ATMs is tricker, since the length of the certi¯cates themselves must be accounted for.) Given the S c S c de¯nitions, it should not be surprising that §i = c §itime(n ) and ¦i = c ¦itime(n ).