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CS-E4530 Computational Complexity Theory The polynomial time hierarchy Lecture 11: The Polynomial Time Hierarchy and Alternating Turing Alternating Turing machines Machines (C. Papadimitriou: Computational Complexity, Chapters 16 & 17) Aalto University School of Science Department of Computer Science

Spring 2020

CS-E4530 Computational Complexity Theory / Lecture 11 Department of Computer Science 2/16

11.1 The Polynomial Time Hierarchy The polynomial time hierarchy—cont’d

Properties: Definition (11.1) p p Σ0 P ∆1 = P = P = P The polynomial time hierarchy comprises the sequence of classes: p p Σ = NPΣ0 = NPP = NP p p p 1 p p ∆0 = Σ0 = Π0 = P Σ0 Π1 = coNP = coNP p p Σk p p k 0 : ∆k+1 = P ∆ = Σ1 = NP ≥ p p 2 P P Σk p p Σk+1 = NP Σ1 NP p Σ2 = NP = NP p Σ p Π = k p Σ NP k+1 coNP Π = coNP 1 = coNP p 2 p p Cumulative polynomial time hierarchy: PH = k 0 Σk ≥ p Σk p Σk+1 p ∆k p ∆k+1 p ∆k+2 S ⊆ Πk ⊆ ⊆ Πk+1 ⊆

CS-E4530 Computational Complexity Theory / Lecture 11 CS-E4530 Computational Complexity Theory / Lecture 11 Department of Computer Science Department of Computer Science 3/16 4/16 Certificates Certificates—cont’d Theorem (11.1) k+1 p A relation R (Σ ) is said to be polynomially balanced if whenever Let L be a language and k 1. Then L Σ iff there is a polynomially ⊆ ∗ ≥ ∈ k (x,y ,...,y ) R, it holds that y ,..., y x t for some t. balanced relation R such that the language x;y (x,y) R is in 1 k ∈ | 1| | k| ≤ | | p { | ∈ } Πk 1 and − L = x there isa y such that (x,y) R . Corollary (11.3) { | ∈ } p Let L be a language and k 1. Then L Σ iff there is a polynomially ≥ ∈ k Corollary (11.2) balanced, polynomial-time decidable (k + 1)-ary relation R such that p Let L be a language and k 1. Then L Πk iff there is a polynomially L = x y y y Qy such that (x,y ,...,y ) R ≥ ∈ p { | ∃ 1∀ 2∃ 3 ··· k 1 k ∈ } balanced relation R such that the language x;y (x,y) R is in Σk 1 { | ∈ } − and where Q is if k is even and if k is odd. t ∀ ∃ L = x for all y with y x ,(x,y) R . { | | | ≤ | | ∈ }

CS-E4530 Computational Complexity Theory / Lecture 11 CS-E4530 Computational Complexity Theory / Lecture 11 Department of Computer Science Department of Computer Science 5/16 6/16

Complete problems Is PH a proper hierarchy?

QSATk (quantified satisfiability with k alternating quantifiers): Proposition (11.4) Given a Boolean expression φ with its Boolean variables p p p p p p If Σ = Π for some k 1, then ∆ = Σ = Π = Σ for all j > k. partitioned into k sets X1,...,Xk, is it true that k k ≥ j j j k there is a partial truth assignment for the variables X1 such that In this case the polynomial time hierarchy is said to collapse to the kth for all partial truth assignments for X2 level. there is a partial truth assignment for X3 ... φ is satisfied by the overall truth assignment? If P = NP, or if NP = coNP, then the polynomial time hierarchy collapses to the first level. (If P = NP, then actually already to the QSATk: Is the following quantified Boolean expression true zeroth level.) X1 X2 X3 QXk φ P = NP iff P = PH. ∃ ∀ ∃ ··· where Q is if k is even and if k is odd? Notice that it can be the case that P = NP and NP = coNP but ∀ ∃ 6 6 the polynomial time hierarchy collapses to the second level. Example (This is not expected to happen, though.) p q (p q) QSAT but p q (p q) QSAT ∃{ }∀{ } ∨ ∈ 2 ∃{ }∀{ } ∧ 6∈ 2

CS-E4530 Computational Complexity Theory / Lecture 11 CS-E4530 Computational Complexity Theory / Lecture 11 Department of Computer Science Department of Computer Science 7/16 8/16 Complete problems

Theorem (11.5) p For all k 1, QSAT is Σ -complete. ≥ k k Complete problems—cont’d

Example Theorem (11.6) CIRCUIT MINIMISATION: Given a Boolean circuit C and an integer K, If there is a PH-complete problem, then the polynomial time hierarchy is there a circuit C0 with at most K gates that computes the same collapses to some finite level. function as C? p CIRCUIT MINIMISATION is Σ -complete (Buchfuhrer & Umans 2008). p 2 Proof. Assume L is PH-complete. Then L Σk for some k. But then p ∈ any L0 Σk+1 reduces to L (because L is PH-complete), and so p ∈ p p Example Σ Σ (because Σ is closed under reductions). 2 k+1 ⊆ k k INTEGER EXPRESSION EQUIVALENCE: Do two integer expressions E and E , built out of binary numbers by + and , describe the same 1 2 ∪ set? (E.g. (001 011) + (010) = (011 101).) ∪ ∪ p INTEGER EXPRESSION EQUIVALENCE is Π2-complete (Stockmeyer 1973).

CS-E4530 Computational Complexity Theory / Lecture 11 CS-E4530 Computational Complexity Theory / Lecture 11 Department of Computer Science Department of Computer Science 9/16 10/16

Complete problems—cont’d 11.2 Alternating Turing Machines PH PSPACE ⊆ p I L PH iff L Σ for some k iff ∈ ∈ k L = x y y Qy s.t. (x,y ,...,y ) R for some k. { | ∃ 1∀ 2 ··· k 1 k ∈ } It is open whether PH = PSPACE. I If PH = PSPACE, then the polynomial time hierarchy collapses to Alternation is an important generalisation of nondeterminism. some finite level. (Follows from Thm 11.6, because there are In a nondeterministic computation each configuration is an PSPACE-complete problems.) implicit OR of its successor configurations: i.e. a configuration If PH does not collapse, problems are strictly harder at an upper “leads to acceptance” iff at least one of its successors does. p level when compared to the lower level: if L is a Σk+1-complete p The idea is to allow both OR and AND configurations in a tree of language and L Σ , then PH collapses to level k. ∈ k configurations generated by an NTM N computing on input x. Example p Consider some Σ2-complete problem. It cannot be solved with a polynomial overhead on top of a procedure for a problem in NP (unless PH collapses to level 1).

CS-E4530 Computational Complexity Theory / Lecture 11 CS-E4530 Computational Complexity Theory / Lecture 11 Department of Computer Science Department of Computer Science 11/16 12/16 Alternation-based complexity classes Definition (11.2) An alternating Turing machine N is a nondeterministic Turing machine Definition (11.3) where the set of states K is partitioned into two sets K = KAND KOR. An alternating Turing machine N decides a language L iff N accepts ∪ all strings x L and rejects all strings x L. ∈ 6∈ Given the tree of configurations of N on input x, the eventually accepting configurations of N are defined recursively: It is straightforward to define ATIME(f (n)) and ASPACE(f (n)); k 1. Any leaf configuration with state “yes” is eventually accepting. and using them, e.g. AP = ATIME(n ), AL = ASPACE(logn) etc. 2. A configuration with state in KAND is eventually accepting iff all its successors are. Roughly speaking, alternating time classes correspond to deterministic space and alternating space classes correspond to 3. A configuration with state in K is eventually accepting iff at OR deterministic time but one exponential higher. least one of its successors is. N accepts x iff its initial configuration is eventually accepting. Theorem (11.7) AL = P, AP = PSPACE, APSPACE = EXP,...

CS-E4530 Computational Complexity Theory / Lecture 11 CS-E4530 Computational Complexity Theory / Lecture 11 Department of Computer Science Department of Computer Science 13/16 14/16

Learning Objectives Alternation and the polynomial time hierarchy

Denote by Σ P (resp. Π P), k 1, the family of languages decided by k k ≥ polynomially time-bounded alternating Turing machines whose every computation satisfies the following conditions:

The initial state belongs to KOR (resp. KAND). The concept of the polynomial time hierarchy. The computation alternates from a state in KOR to a state in Alternating Turing machines and alternation-based complexity KAND or vice versa at most k 1 times. classes. − By definition, set also Σ0P = Π0P = P.

Theorem (11.8) p p For every k 0, Σ P = Σ and Π P = Π . ≥ k k k k

CS-E4530 Computational Complexity Theory / Lecture 11 CS-E4530 Computational Complexity Theory / Lecture 11 Department of Computer Science Department of Computer Science 15/16 16/16