Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Lecture 12: Polynomial Hierarchy II Arijit Bishnu 06.04.2010
Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Outline 1 Complete Problems for levels of PH 2 Alternating Turing Machines 3 Time versus Alternations
Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Outline 1 Complete Problems for levels of PH 2 Alternating Turing Machines 3 Time versus Alternations
Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Completeness Σ p i -complete We say that a language L is Σ p i -complete if L ∈ Σ p i and for every L ′ ∈ Σ p i , L ′ ≤ P L .
Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Completeness Σ p i -complete We say that a language L is Σ p i -complete if L ∈ Σ p i and for every L ′ ∈ Σ p i , L ′ ≤ P L . Examples
Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Completeness Σ p i -complete We say that a language L is Σ p i -complete if L ∈ Σ p i and for every L ′ ∈ Σ p i , L ′ ≤ P L . Examples SAT is complete for the class NP = Σ p 1 .
Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Completeness Σ p i -complete We say that a language L is Σ p i -complete if L ∈ Σ p i and for every L ′ ∈ Σ p i , L ′ ≤ P L . Examples SAT is complete for the class NP = Σ p 1 . Recall a QBF has the form Q 1 x 1 Q 2 x 2 . . . Q n x n ϕ ( x 1 , x 2 , . . . , x n ) where each Q i ∈ {∃ , ∀} and ϕ is an unquantified boolean formula.
Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Completeness Σ p i -complete We say that a language L is Σ p i -complete if L ∈ Σ p i and for every L ′ ∈ Σ p i , L ′ ≤ P L . Examples SAT is complete for the class NP = Σ p 1 . Recall a QBF has the form Q 1 x 1 Q 2 x 2 . . . Q n x n ϕ ( x 1 , x 2 , . . . , x n ) where each Q i ∈ {∃ , ∀} and ϕ is an unquantified boolean formula. The language TQBF is the set of QBFs that are TRUE, i.e. TQBF = { < φ > | φ is a TRUE fully QBF. }
Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Examples continued Examples
Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Examples continued Examples Define Σ i SAT = ∃ x 1 ∀ x 2 ∃ · · · Q i x i ϕ ( x 1 , x 2 , . . . , x i ) = 1
Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Examples continued Examples Define Σ i SAT = ∃ x 1 ∀ x 2 ∃ · · · Q i x i ϕ ( x 1 , x 2 , . . . , x i ) = 1 For every i , Σ i SAT is a special case of the TQBF problem.
Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Examples continued Examples Define Σ i SAT = ∃ x 1 ∀ x 2 ∃ · · · Q i x i ϕ ( x 1 , x 2 , . . . , x i ) = 1 For every i , Σ i SAT is a special case of the TQBF problem. One can prove that for every i , Σ i SAT is a complete problem for the class Σ p i .
Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Examples continued Examples Define Σ i SAT = ∃ x 1 ∀ x 2 ∃ · · · Q i x i ϕ ( x 1 , x 2 , . . . , x i ) = 1 For every i , Σ i SAT is a special case of the TQBF problem. One can prove that for every i , Σ i SAT is a complete problem for the class Σ p i . Similarly, a problem Π i SAT can be shown to be Π p i -complete.
Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Examples continued Examples Define Σ i SAT = ∃ x 1 ∀ x 2 ∃ · · · Q i x i ϕ ( x 1 , x 2 , . . . , x i ) = 1 For every i , Σ i SAT is a special case of the TQBF problem. One can prove that for every i , Σ i SAT is a complete problem for the class Σ p i . Similarly, a problem Π i SAT can be shown to be Π p i -complete. So, complete problems exist for each class Σ p i .
Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Completeness for PH Definition: Polynomial Hierarchy i Σ p The polynomial hierarchy is the set PH = � i .
Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Completeness for PH Definition: Polynomial Hierarchy i Σ p The polynomial hierarchy is the set PH = � i . PH-complete We say that a language L is PH-complete if L ∈ PH and for every L ′ ∈ PH, L ′ ≤ P L .
Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Completeness for PH Definition: Polynomial Hierarchy i Σ p The polynomial hierarchy is the set PH = � i . PH-complete We say that a language L is PH-complete if L ∈ PH and for every L ′ ∈ PH, L ′ ≤ P L . Does PH have a complete problem? Each class Σ p i Σ p i has complete problems. What about PH = � i ?
Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Completeness for PH Claim If there exists a PH-complete language L , then the polynomial hierarchy collapses to some finite level, i.e. PH = Σ p i .
Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Completeness for PH Claim If there exists a PH-complete language L , then the polynomial hierarchy collapses to some finite level, i.e. PH = Σ p i . Proof idea
Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Completeness for PH Claim If there exists a PH-complete language L , then the polynomial hierarchy collapses to some finite level, i.e. PH = Σ p i . Proof idea Suppose, for some i , Σ p i = Σ p i +1 . Then, ∀ j ≥ i , Σ p j = Π p j = Σ p i . (Proof as an exercise.)
Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Completeness for PH Claim If there exists a PH-complete language L , then the polynomial hierarchy collapses to some finite level, i.e. PH = Σ p i . Proof idea Suppose, for some i , Σ p i = Σ p i +1 . Then, ∀ j ≥ i , Σ p j = Π p j = Σ p i . (Proof as an exercise.) i Σ p i , ∃ i such that L ∈ Σ p Since L ∈ PH = � i .
Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Completeness for PH Claim If there exists a PH-complete language L , then the polynomial hierarchy collapses to some finite level, i.e. PH = Σ p i . Proof idea Suppose, for some i , Σ p i = Σ p i +1 . Then, ∀ j ≥ i , Σ p j = Π p j = Σ p i . (Proof as an exercise.) i Σ p i , ∃ i such that L ∈ Σ p Since L ∈ PH = � i . Since L is PH-complete, ∀ L ′ ∈ Σ p i +1 , L ′ ≤ P L . So, L ′ ∈ Σ p i and Σ p i = Σ p i +1 .
Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Completeness for PH Claim If there exists a PH-complete language L , then the polynomial hierarchy collapses to some finite level, i.e. PH = Σ p i . Proof idea Suppose, for some i , Σ p i = Σ p i +1 . Then, ∀ j ≥ i , Σ p j = Π p j = Σ p i . (Proof as an exercise.) i Σ p i , ∃ i such that L ∈ Σ p Since L ∈ PH = � i . Since L is PH-complete, ∀ L ′ ∈ Σ p i +1 , L ′ ≤ P L . So, L ′ ∈ Σ p i and Σ p i = Σ p i +1 . So, the hierarchy collapses to the level i .
Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Completeness for PH Claim If there exists a PH-complete language L , then the polynomial hierarchy collapses to some finite level, i.e. PH = Σ p i . Proof idea Suppose, for some i , Σ p i = Σ p i +1 . Then, ∀ j ≥ i , Σ p j = Π p j = Σ p i . (Proof as an exercise.) i Σ p i , ∃ i such that L ∈ Σ p Since L ∈ PH = � i . Since L is PH-complete, ∀ L ′ ∈ Σ p i +1 , L ′ ≤ P L . So, L ′ ∈ Σ p i and Σ p i = Σ p i +1 . So, the hierarchy collapses to the level i . So, people believe that complete problems do not exist for the class PH.
Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Space Complexity Classes and Polynomial Hierarchy Relation between PH and PSPACE PH ⊆ PSPACE
Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Space Complexity Classes and Polynomial Hierarchy Relation between PH and PSPACE PH ⊆ PSPACE Proof We can prove just the way we proved NP ⊆ PSPACE by writing down all short (polynomial sized) certificates in the work tape and reusing that space.
Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Space Complexity Classes and Polynomial Hierarchy Claim If PH = PSPACE, the polynomial hierarchy collapses.
Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Space Complexity Classes and Polynomial Hierarchy Claim If PH = PSPACE, the polynomial hierarchy collapses. Proof
Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Space Complexity Classes and Polynomial Hierarchy Claim If PH = PSPACE, the polynomial hierarchy collapses. Proof We know that TQBF is PSPACE-complete.
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