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The Polynomial Hierarchy The Polynomial Hierarchy Xin Li [email protected] Generalization of P, NP, coNP • INDSET={<G, k>: G has an independent set of size k.} ∈ NP. • EXACT INDSET={<G, k>: the largest independent set of G has size k.} • Question: EXACT INDSET ∈ NP? • No easy short certificate: <G, k> ∈ EXACT INDSET iff ∃ an I.S. of size k and ∀ other I.S. has size ≤ k. Generalization of P, NP, coNP • Given a CNF formula Φ, try to find the smallest CNF ψ equivalent to Φ. • MIN-EQ-CNF ={<Φ, k>: ∃ CNF formula ψ of size ≤ k s.t. ψ=Φ.} • Question: MIN-EQ-CNF ∈ NP? • No easy short certificate: <Φ, k> ∈ MIN-EQ-CNF iff ∃ CNF formula ψ of size ≤ k s.t. ∀ x, ψ(x)=Φ(x). Generalization of P, NP, coNP • Seems to need not just one “∃/∀” quantifier, but both. p • Definition: The class Σ2 is the set of all languages L s.t. ∃ a polynomial time TM M and a polynomial function q s.t. ∀ x ∈ {0, 1}* x ∈ L iff ∃ u ∈ {0, 1}q(\x\) ∀ v ∈ {0, 1}q(\x\) M(x, u, v)=1 p • Example: EXACT INDSET, MIN-EQ-CNF ∈ Σ2 . Generalization of P, NP, coNP p • Definition: The class Σ2 is the set of all languages L s.t. ∃ a polynomial time TM M and a polynomial function q s.t. ∀ x ∈ {0, 1}* x ∈ L iff ∃ u ∈ {0, 1}q(\x\) ∀ v ∈ {0, 1}q(\x\) M(x, u, v)=1 p p • NP ⊆ Σ2 (M just ignores v) and coNP ⊆ Σ2 (M just ignores u). p • Similarly, can generalize to Σi (for some constant i) where we have i alternating quantifiers starting with ∃. Generalization of P, NP, coNP p • Similarly, can generalize to Σi (for some constant i) where we have i alternating quantifiers starting with ∃. p p p p • Define Πi = coΣi i.e., all languages whose complement is in Σi . Πi has i alternating quantifiers starting with ∀. p • Example: The class Π2 is the set of all languages L s.t. ∃ a polynomial time TM M and a polynomial function q s.t. ∀ x ∈ {0, 1}* x ∈ L iff ∀ u ∈ {0, 1}q(\x\) ∃ v ∈ {0, 1}q(\x\) M(x, u, v)=1 Generalization of P, NP, coNP p p p p p p p p • Simple relations: Σi ⊆ Σi+1 and Πi ⊆ Σi+1 . Similarly, Σi ⊆ Πi+1 and Πi ⊆ Πi+1 • p p NP= Σ1 and coNP=Π1 . p Πi p • Σi+1 : ∃ u1 ∈ {0, 1}q(\x\) ∀ u2 ∈ {0, 1}q(\x\) … Qi+1 ui+1 M(x, u1, …, ui+1)=1 p p p p • So Σi+1 =NP^Πi (NP using an oracle that decides Πi languages). Similarly, Πi+1 p =coNP^ Σi . The Polynomial Hierarchy p p p p p p p p • Since Σi ⊆ Σi+1 and Πi ⊆ Σi+1 . Σi ⊆ Πi+1 and Πi ⊆ Πi+1 ∞ p ∞ p • Can define PH= Ui=1 Σi = Ui=1 Πi .This is known as the polynomial hierarchy. • We believe P ≠NP and NP ≠ coNP. p p p p p • Generalize to conjecture: ∀ i, Σi ≠ Σi+1 , Σi ≠ Πi or Σi ⊂PH (PH does not collapse). The Polynomial Hierarchy • The smaller i is, the weaker the conjecture is, hence more believable. p p p • If for some i we have Σi = Σi+1 , then this implies that PH= Σi and we say PH collapses to the i’th level. p p p • Theorem: (1) ∀ i, if Σi = Πi , then PH= Σi , the reverse is also true. (2) If P=NP, then PH=P. The Polynomial Hierarchy p p p • Theorem: (1) ∀ i, if Σi = Πi , then PH= Σi , the reverse is also true. (2) If P=NP, then PH=P. p p • Proof of (2): Assume P=NP. Prove by induction that ∀ i, Σi , Πi ⊆P. • Base case: i=1 holds since P=NP=coNP. p p p p • Assume holds for i-1, show Σi ⊆P. Πi follows since Πi = coΣi . The Polynomial Hierarchy p • Let L ∈ Σi , by definition, ∃ a polynomial time TM M and a polynomial function q s.t. ∀ x ∈ {0, 1}* x ∈ L iff ∃ u1 ∈ {0, 1}q(\x\) ∀ u2 ∈ {0, 1}q(\x\) … Qi ui M(x, u1, …, ui)=1 • Define the language L’ as <x, u1 > ∈ L’ iff ∀ u2 ∈ {0, 1}q(\x\) … Qi ui M(x, u1, u2 ,…, ui)=1 p • Thus L’ ∈Πi-1 ⊆P and ∃ a polynomial time TM M’ deciding L’. The Polynomial Hierarchy p • Let L ∈ Σi , by definition, ∃ a polynomial time TM M and a polynomial function q s.t. ∀ x ∈ {0, 1}* x ∈ L iff ∃ u1 ∈ {0, 1}q(\x\) ∀ u2 ∈ {0, 1}q(\x\) … Qi ui M(x, u1, …, ui)=1 • Hence x ∈ L iff ∃ u1 ∈ {0, 1}q(\x\) M’(x, u1)=1 • Thus L ∈NP⊆P. The Polynomial Hierarchy p p • Like NP and coNP, we can define complete problems for Σi and Πi . p p p • Say a language L is Σi complete if L ∈ Σi and ∀ L’ ∈ Σi , L’ ≤p L. • For any i, let Σi SAT={all QBFs of the form ∃ u1 ∀ u2 … Qi ui Φ(u1, …, ui) that are true} where Φ is a Boolean formula. p • Σi SAT is Σi complete. The Polynomial Hierarchy • However, unlikely to have PH-complete languages. p • Theorem: if ∃ a language L that is PH-complete, then ∃ i s.t. PH=Σi . p • Proof: there must ∃ i s.t. L ∈ Σi . If L is PH-complete, then ∀ L’ ∈ PH, L’ ≤p L and thus p p L’ ∈ Σi . Therefore PH=Σi . The Polynomial Hierarchy • Not hard to show that PH⊆PSPACE. • Can use the same algorithm for TQBF to decide any language in PH, since the # of quantifiers is a constant (although can be arbitrarily large). • Corollary: If PH does not collapse, then PH⊂PSPACE. Otherwise TQBF is a PH- complete language. • Question: Does P=NP imply P=PSPACE? why?.
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