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 p SAT is complete for the class NP = Σ1. Recall a QBF has the form Q1x1Q2x2 ... Qnxn ϕ(x1, x2,..., xn) where each Qi ∈ {∃, ∀} 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.}

Examples

Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Completeness

p Σi -complete p p We say that a language L is Σi -complete if L ∈ Σi and for every 0 p 0 L ∈ Σi , L ≤P L. p SAT is complete for the class NP = Σ1. Recall a QBF has the form Q1x1Q2x2 ... Qnxn ϕ(x1, x2,..., xn) where each Qi ∈ {∃, ∀} 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 Completeness

p Σi -complete p p We say that a language L is Σi -complete if L ∈ Σi and for every 0 p 0 L ∈ Σi , L ≤P L.

Examples Recall a QBF has the form Q1x1Q2x2 ... Qnxn ϕ(x1, x2,..., xn) where each Qi ∈ {∃, ∀} 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 Completeness

p Σi -complete p p We say that a language L is Σi -complete if L ∈ Σi and for every 0 p 0 L ∈ Σi , L ≤P L.

Examples p SAT is complete for the class NP = Σ1. 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 Completeness

p Σi -complete p p We say that a language L is Σi -complete if L ∈ Σi and for every 0 p 0 L ∈ Σi , L ≤P L.

Examples p SAT is complete for the class NP = Σ1. Recall a QBF has the form Q1x1Q2x2 ... Qnxn ϕ(x1, x2,..., xn) where each Qi ∈ {∃, ∀} and ϕ is an unquantified boolean formula. Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Completeness

p Σi -complete p p We say that a language L is Σi -complete if L ∈ Σi and for every 0 p 0 L ∈ Σi , L ≤P L.

Examples p SAT is complete for the class NP = Σ1. Recall a QBF has the form Q1x1Q2x2 ... Qnxn ϕ(x1, x2,..., xn) where each Qi ∈ {∃, ∀} 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.} Define

Σi SAT = ∃x1∀x2∃ · · · Qi xi ϕ(x1, x2,..., xi ) = 1

For every i,Σi SAT is a special case of the TQBF problem.

One can prove that for every i,Σi SAT is a complete problem p for the class Σi . p Similarly, a problem Πi SAT can be shown to be Πi -complete. p So, complete problems exist for each class Σi .

Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Examples continued

Examples For every i,Σi SAT is a special case of the TQBF problem.

One can prove that for every i,Σi SAT is a complete problem p for the class Σi . p Similarly, a problem Πi SAT can be shown to be Πi -complete. p So, complete problems exist for each class Σi .

Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Examples continued

Examples Define

Σi SAT = ∃x1∀x2∃ · · · Qi xi ϕ(x1, x2,..., xi ) = 1 One can prove that for every i,Σi SAT is a complete problem p for the class Σi . p Similarly, a problem Πi SAT can be shown to be Πi -complete. p So, complete problems exist for each class Σi .

Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Examples continued

Examples Define

Σi SAT = ∃x1∀x2∃ · · · Qi xi ϕ(x1, x2,..., xi ) = 1

For every i,Σi SAT is a special case of the TQBF problem. p Similarly, a problem Πi SAT can be shown to be Πi -complete. p So, complete problems exist for each class Σi .

Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Examples continued

Examples Define

Σi SAT = ∃x1∀x2∃ · · · Qi xi ϕ(x1, x2,..., xi ) = 1

For every i,Σi SAT is a special case of the TQBF problem.

One can prove that for every i,Σi SAT is a complete problem p for the class Σi . p So, complete problems exist for each class Σi .

Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Examples continued

Examples Define

Σi SAT = ∃x1∀x2∃ · · · Qi xi ϕ(x1, x2,..., xi ) = 1

For every i,Σi SAT is a special case of the TQBF problem.

One can prove that for every i,Σi SAT is a complete problem p for the class Σi . p Similarly, a problem Πi SAT can be shown to be Πi -complete. Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Examples continued

Examples Define

Σi SAT = ∃x1∀x2∃ · · · Qi xi ϕ(x1, x2,..., xi ) = 1

For every i,Σi SAT is a special case of the TQBF problem.

One can prove that for every i,Σi SAT is a complete problem p for the class Σi . p Similarly, a problem Πi SAT can be shown to be Πi -complete. p So, complete problems exist for each class Σi . PH-complete We say that a language L is PH-complete if L ∈ PH and for every 0 0 L ∈ PH, L ≤P L.

Does PH have a complete problem? p S p Each class Σi has complete problems. What about PH = i Σi ?

Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Completeness for PH

Definition: Polynomial Hierarchy S p The polynomial hierarchy is the set PH = i Σi . Does PH have a complete problem? p S p Each class Σi has complete problems. What about PH = i Σi ?

Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Completeness for PH

Definition: Polynomial Hierarchy S p The polynomial hierarchy is the set PH = i Σi .

PH-complete We say that a language L is PH-complete if L ∈ PH and for every 0 0 L ∈ PH, L ≤P L. Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Completeness for PH

Definition: Polynomial Hierarchy S p The polynomial hierarchy is the set PH = i Σi .

PH-complete We say that a language L is PH-complete if L ∈ PH and for every 0 0 L ∈ PH, L ≤P L.

Does PH have a complete problem? p S p Each class Σi has complete problems. What about PH = i Σi ? p p Suppose, for some i,Σi = Σi+1. Then, ∀j ≥ i, p p p Σj = Πj = Σi . (Proof as an exercise.) S p p Since L ∈ PH = i Σi , ∃i such that L ∈ Σi . 0 p 0 0 p Since L is PH-complete, ∀L ∈ Σi+1, L ≤P L. So, L ∈ Σi p p and Σi = Σi+1. So, the hierarchy collapses to the level i. So, people believe that complete problems do not exist for the class PH.

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 p hierarchy collapses to some finite level, i.e. PH = Σi . p p Suppose, for some i,Σi = Σi+1. Then, ∀j ≥ i, p p p Σj = Πj = Σi . (Proof as an exercise.) S p p Since L ∈ PH = i Σi , ∃i such that L ∈ Σi . 0 p 0 0 p Since L is PH-complete, ∀L ∈ Σi+1, L ≤P L. So, L ∈ Σi p p and Σi = Σi+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 Completeness for PH

Claim If there exists a PH-complete language L, then the polynomial p hierarchy collapses to some finite level, i.e. PH = Σi .

Proof idea S p p Since L ∈ PH = i Σi , ∃i such that L ∈ Σi . 0 p 0 0 p Since L is PH-complete, ∀L ∈ Σi+1, L ≤P L. So, L ∈ Σi p p and Σi = Σi+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 Completeness for PH

Claim If there exists a PH-complete language L, then the polynomial p hierarchy collapses to some finite level, i.e. PH = Σi .

Proof idea p p Suppose, for some i,Σi = Σi+1. Then, ∀j ≥ i, p p p Σj = Πj = Σi . (Proof as an exercise.) 0 p 0 0 p Since L is PH-complete, ∀L ∈ Σi+1, L ≤P L. So, L ∈ Σi p p and Σi = Σi+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 Completeness for PH

Claim If there exists a PH-complete language L, then the polynomial p hierarchy collapses to some finite level, i.e. PH = Σi .

Proof idea p p Suppose, for some i,Σi = Σi+1. Then, ∀j ≥ i, p p p Σj = Πj = Σi . (Proof as an exercise.) S p p Since L ∈ PH = i Σi , ∃i such that L ∈ Σi . 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 Completeness for PH

Claim If there exists a PH-complete language L, then the polynomial p hierarchy collapses to some finite level, i.e. PH = Σi .

Proof idea p p Suppose, for some i,Σi = Σi+1. Then, ∀j ≥ i, p p p Σj = Πj = Σi . (Proof as an exercise.) S p p Since L ∈ PH = i Σi , ∃i such that L ∈ Σi . 0 p 0 0 p Since L is PH-complete, ∀L ∈ Σi+1, L ≤P L. So, L ∈ Σi p p and Σi = Σi+1. 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 Completeness for PH

Claim If there exists a PH-complete language L, then the polynomial p hierarchy collapses to some finite level, i.e. PH = Σi .

Proof idea p p Suppose, for some i,Σi = Σi+1. Then, ∀j ≥ i, p p p Σj = Πj = Σi . (Proof as an exercise.) S p p Since L ∈ PH = i Σi , ∃i such that L ∈ Σi . 0 p 0 0 p Since L is PH-complete, ∀L ∈ Σi+1, L ≤P L. So, L ∈ Σi p p and Σi = Σi+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 p hierarchy collapses to some finite level, i.e. PH = Σi .

Proof idea p p Suppose, for some i,Σi = Σi+1. Then, ∀j ≥ i, p p p Σj = Πj = Σi . (Proof as an exercise.) S p p Since L ∈ PH = i Σi , ∃i such that L ∈ Σi . 0 p 0 0 p Since L is PH-complete, ∀L ∈ Σi+1, L ≤P L. So, L ∈ Σi p p and Σi = Σi+1. So, the hierarchy collapses to the level i. So, people believe that complete problems do not exist for the class PH. 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

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. We know that TQBF is PSPACE-complete. Now, if PH = PSPACE, then TQBF is also a complete problem for the class PH. We have already shown that the existence of a complete problem for the class PH implies that the polynomial hierarchy collapses. So, the widely held belief is that PH 6= PSPACE.

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. We know that TQBF is PSPACE-complete. Now, if PH = PSPACE, then TQBF is also a complete problem for the class PH. We have already shown that the existence of a complete problem for the class PH implies that the polynomial hierarchy collapses. So, the widely held belief is that PH 6= PSPACE.

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 Now, if PH = PSPACE, then TQBF is also a complete problem for the class PH. We have already shown that the existence of a complete problem for the class PH implies that the polynomial hierarchy collapses. So, the widely held belief is that PH 6= PSPACE.

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. We have already shown that the existence of a complete problem for the class PH implies that the polynomial hierarchy collapses. So, the widely held belief is that PH 6= PSPACE.

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. Now, if PH = PSPACE, then TQBF is also a complete problem for the class PH. So, the widely held belief is that PH 6= PSPACE.

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. Now, if PH = PSPACE, then TQBF is also a complete problem for the class PH. We have already shown that the existence of a complete problem for the class PH implies that 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 We know that TQBF is PSPACE-complete. Now, if PH = PSPACE, then TQBF is also a complete problem for the class PH. We have already shown that the existence of a complete problem for the class PH implies that the polynomial hierarchy collapses. So, the widely held belief is that PH 6= PSPACE. 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 Alternating Turing Machines and Classes

Definition: Alternating Time For every T : N → N, we say that an ATM M runs in T (n)-time if for every input x ∈ {0, 1}∗ and for every possible sequence of transition function choices, M halts after at most T (|x|) steps. The configuration Cacc is labeled ACCEPT. If a configuration C is in a state that is labeled ∃ and there is an edge from C to C 0 labeled ACCEPT, we label C ACCEPT. If a configuration C is in a state labeled ∀ and both the configurations C 0, C 00 reachable from it in one step are labeled ACCEPT, then we label C ACCEPT. We say that M accepts x if this repeated application of labeling rules (till they cannot be applied any more) labels Cs , the start configuration ACCEPT.

Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Alternating Turing Machines and Classes

Definition: ATIME(T(n)) A language L is in ATIME(T(n)) if there is a constant c and a c · T (n)-time ATM M s.t. for every x ∈ {0, 1}∗, M accepts x iff x ∈ L. The definition of accept comes via labeling vertices on the configuration graph GM,x as follows: If a configuration C is in a state that is labeled ∃ and there is an edge from C to C 0 labeled ACCEPT, we label C ACCEPT. If a configuration C is in a state labeled ∀ and both the configurations C 0, C 00 reachable from it in one step are labeled ACCEPT, then we label C ACCEPT. We say that M accepts x if this repeated application of labeling rules (till they cannot be applied any more) labels Cs , the start configuration ACCEPT.

Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Alternating Turing Machines and Classes

Definition: ATIME(T(n)) A language L is in ATIME(T(n)) if there is a constant c and a c · T (n)-time ATM M s.t. for every x ∈ {0, 1}∗, M accepts x iff x ∈ L. The definition of accept comes via labeling vertices on the configuration graph GM,x as follows:

The configuration Cacc is labeled ACCEPT. If a configuration C is in a state labeled ∀ and both the configurations C 0, C 00 reachable from it in one step are labeled ACCEPT, then we label C ACCEPT. We say that M accepts x if this repeated application of labeling rules (till they cannot be applied any more) labels Cs , the start configuration ACCEPT.

Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Alternating Turing Machines and Classes

Definition: ATIME(T(n)) A language L is in ATIME(T(n)) if there is a constant c and a c · T (n)-time ATM M s.t. for every x ∈ {0, 1}∗, M accepts x iff x ∈ L. The definition of accept comes via labeling vertices on the configuration graph GM,x as follows:

The configuration Cacc is labeled ACCEPT. If a configuration C is in a state that is labeled ∃ and there is an edge from C to C 0 labeled ACCEPT, we label C ACCEPT. We say that M accepts x if this repeated application of labeling rules (till they cannot be applied any more) labels Cs , the start configuration ACCEPT.

Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Alternating Turing Machines and Classes

Definition: ATIME(T(n)) A language L is in ATIME(T(n)) if there is a constant c and a c · T (n)-time ATM M s.t. for every x ∈ {0, 1}∗, M accepts x iff x ∈ L. The definition of accept comes via labeling vertices on the configuration graph GM,x as follows:

The configuration Cacc is labeled ACCEPT. If a configuration C is in a state that is labeled ∃ and there is an edge from C to C 0 labeled ACCEPT, we label C ACCEPT. If a configuration C is in a state labeled ∀ and both the configurations C 0, C 00 reachable from it in one step are labeled ACCEPT, then we label C ACCEPT. Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Alternating Turing Machines and Classes

Definition: ATIME(T(n)) A language L is in ATIME(T(n)) if there is a constant c and a c · T (n)-time ATM M s.t. for every x ∈ {0, 1}∗, M accepts x iff x ∈ L. The definition of accept comes via labeling vertices on the configuration graph GM,x as follows:

The configuration Cacc is labeled ACCEPT. If a configuration C is in a state that is labeled ∃ and there is an edge from C to C 0 labeled ACCEPT, we label C ACCEPT. If a configuration C is in a state labeled ∀ and both the configurations C 0, C 00 reachable from it in one step are labeled ACCEPT, then we label C ACCEPT. We say that M accepts x if this repeated application of labeling rules (till they cannot be applied any more) labels Cs , the start configuration ACCEPT. Classes related to such ATMs S c S c What is c Σi TIME(n ) and c Πi TIME(n )?

Claim p S c p S c For every i ∈ N,Σi = c Σi TIME(n ) and Πi = c Πi TIME(n ).

Proof Left as an exercise.

Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations ATMs Restricted to a Fixed Number of Alternations

Definition

For every i ∈ N, we define Σi TIME(T (n)) (Πi TIME(T (n))) to be the set of languages accepted by a T (n)-time ATM M whose initial state is labeled ∃ (∀) and for every input and on every directed path from the starting configuration in the configuration graph, M can alternate at most i − 1 times. Claim p S c p S c For every i ∈ N,Σi = c Σi TIME(n ) and Πi = c Πi TIME(n ).

Proof Left as an exercise.

Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations ATMs Restricted to a Fixed Number of Alternations

Definition

For every i ∈ N, we define Σi TIME(T (n)) (Πi TIME(T (n))) to be the set of languages accepted by a T (n)-time ATM M whose initial state is labeled ∃ (∀) and for every input and on every directed path from the starting configuration in the configuration graph, M can alternate at most i − 1 times.

Classes related to such ATMs S c S c What is c Σi TIME(n ) and c Πi TIME(n )? Proof Left as an exercise.

Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations ATMs Restricted to a Fixed Number of Alternations

Definition

For every i ∈ N, we define Σi TIME(T (n)) (Πi TIME(T (n))) to be the set of languages accepted by a T (n)-time ATM M whose initial state is labeled ∃ (∀) and for every input and on every directed path from the starting configuration in the configuration graph, M can alternate at most i − 1 times.

Classes related to such ATMs S c S c What is c Σi TIME(n ) and c Πi TIME(n )?

Claim p S c p S c For every i ∈ N,Σi = c Σi TIME(n ) and Πi = c Πi TIME(n ). Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations ATMs Restricted to a Fixed Number of Alternations

Definition

For every i ∈ N, we define Σi TIME(T (n)) (Πi TIME(T (n))) to be the set of languages accepted by a T (n)-time ATM M whose initial state is labeled ∃ (∀) and for every input and on every directed path from the starting configuration in the configuration graph, M can alternate at most i − 1 times.

Classes related to such ATMs S c S c What is c Σi TIME(n ) and c Πi TIME(n )?

Claim p S c p S c For every i ∈ N,Σi = c Σi TIME(n ) and Πi = c Πi TIME(n ).

Proof Left as an exercise. TQBF is a PSPACE-complete problem. TQBF ∈ AP as we can guess using ∃ and ∀ states of the ATM. So, AP ⊆ PSPACE. To show PSPACE ⊆ AP, use a recursive procedure similar . to the one we used for showing TQBF ∈ PSPACE.

Theorem AP = PSPACE

Proof Idea

Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Unlimited Number of Alternations

The Class AP S c AP = c ATIME(n ). TQBF is a PSPACE-complete problem. TQBF ∈ AP as we can guess using ∃ and ∀ states of the ATM. So, AP ⊆ PSPACE. To show PSPACE ⊆ AP, use a recursive procedure similar . to the one we used for showing TQBF ∈ PSPACE.

Proof Idea

Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Unlimited Number of Alternations

The Class AP S c AP = c ATIME(n ).

Theorem AP = PSPACE TQBF is a PSPACE-complete problem. TQBF ∈ AP as we can guess using ∃ and ∀ states of the ATM. So, AP ⊆ PSPACE. To show PSPACE ⊆ AP, use a recursive procedure similar . to the one we used for showing TQBF ∈ PSPACE.

Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Unlimited Number of Alternations

The Class AP S c AP = c ATIME(n ).

Theorem AP = PSPACE

Proof Idea TQBF ∈ AP as we can guess using ∃ and ∀ states of the ATM. So, AP ⊆ PSPACE. To show PSPACE ⊆ AP, use a recursive procedure similar . to the one we used for showing TQBF ∈ PSPACE.

Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Unlimited Number of Alternations

The Class AP S c AP = c ATIME(n ).

Theorem AP = PSPACE

Proof Idea TQBF is a PSPACE-complete problem. So, AP ⊆ PSPACE. To show PSPACE ⊆ AP, use a recursive procedure similar . to the one we used for showing TQBF ∈ PSPACE.

Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Unlimited Number of Alternations

The Class AP S c AP = c ATIME(n ).

Theorem AP = PSPACE

Proof Idea TQBF is a PSPACE-complete problem. TQBF ∈ AP as we can guess using ∃ and ∀ states of the ATM. To show PSPACE ⊆ AP, use a recursive procedure similar . to the one we used for showing TQBF ∈ PSPACE.

Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Unlimited Number of Alternations

The Class AP S c AP = c ATIME(n ).

Theorem AP = PSPACE

Proof Idea TQBF is a PSPACE-complete problem. TQBF ∈ AP as we can guess using ∃ and ∀ states of the ATM. So, AP ⊆ PSPACE. Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Unlimited Number of Alternations

The Class AP S c AP = c ATIME(n ).

Theorem AP = PSPACE

Proof Idea TQBF is a PSPACE-complete problem. TQBF ∈ AP as we can guess using ∃ and ∀ states of the ATM. So, AP ⊆ PSPACE. To show PSPACE ⊆ AP, use a recursive procedure similar . to the one we used for showing TQBF ∈ PSPACE. Exercise Consider AL to be the languages accepted by ATMs that run using logarithmic space. Show that AL = P.

Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Exercises

Exercise Consider APSPACE to be the languages accepted by ATMs that run using polynomial space. Show that APSPACE = EXP. Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Exercises

Exercise Consider APSPACE to be the languages accepted by ATMs that run using polynomial space. Show that APSPACE = EXP.

Exercise Consider AL to be the languages accepted by ATMs that run using logarithmic space. Show that AL = P. 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 What is that we can do?

The Class TISP For every two functions S, T : N → N, the class TISP(T (n), S(n)) is the set of languages decided by a TM M that on every input x ∈ {0, 1}∗ takes at most O(T (|x|)) steps and uses at most O(S(|x|)) cells of its read/write tape.

SAT and TISP One can show that SAT ∈/ TISP(nc , nd ) for every constants c, d such that c(c + d) < 2.

Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Time-Space TradeOff

As we do not know much about the P vs NP question, we even cannot rule out a deterministic poly-time algo for SAT. The Class TISP For every two functions S, T : N → N, the class TISP(T (n), S(n)) is the set of languages decided by a TM M that on every input x ∈ {0, 1}∗ takes at most O(T (|x|)) steps and uses at most O(S(|x|)) cells of its read/write tape.

SAT and TISP One can show that SAT ∈/ TISP(nc , nd ) for every constants c, d such that c(c + d) < 2.

Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Time-Space TradeOff

As we do not know much about the P vs NP question, we even cannot rule out a deterministic poly-time algo for SAT. What is that we can do? SAT and TISP One can show that SAT ∈/ TISP(nc , nd ) for every constants c, d such that c(c + d) < 2.

Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Time-Space TradeOff

As we do not know much about the P vs NP question, we even cannot rule out a deterministic poly-time algo for SAT. What is that we can do?

The Class TISP For every two functions S, T : N → N, the class TISP(T (n), S(n)) is the set of languages decided by a TM M that on every input x ∈ {0, 1}∗ takes at most O(T (|x|)) steps and uses at most O(S(|x|)) cells of its read/write tape. Complete Problems for levels of PH Alternating Turing Machines Time versus Alternations Time-Space TradeOff

As we do not know much about the P vs NP question, we even cannot rule out a deterministic poly-time algo for SAT. What is that we can do?

The Class TISP For every two functions S, T : N → N, the class TISP(T (n), S(n)) is the set of languages decided by a TM M that on every input x ∈ {0, 1}∗ takes at most O(T (|x|)) steps and uses at most O(S(|x|)) cells of its read/write tape.

SAT and TISP One can show that SAT ∈/ TISP(nc , nd ) for every constants c, d such that c(c + d) < 2.