Computational Complexity IV: PSPACE

Seminar on Theoretical Computer Science and Discrete Mathematics Aristotle University of Thessaloniki

Context 1 Section 1: PSPACE 2 3 4

Time Complexity Time complexity of DTM M: - Increasing function t: N Ð N such that for every input string x, Sx S = n, M(x) ends in B t(n) steps.

Time Complexity Time complexity of DTM M: - Increasing function t: N Ð N such that for every input string x, Sx S = n, M(x) ends in B t(n) steps. DTIME[t(n)] P: P is solved in O(t(n)) time

Complexity Class P Complexity Class P Complexity Class P is the set of decision problems that can be solved by a DTM in a polynomial time of steps. (there is a polynomial time algorithm)

Complexity Class P Complexity Class P Complexity Class P is the set of decision problems that can be solved by a DTM in a polynomial time of steps. (there is a polynomial time algorithm) Polynomial time: P kc0 DTIME n k

Time Complexity Space complexity of (multitape) DTM M: - Increasing function s: N Ð N such that for every input string x, Sx S = n, M(x) ends in B s(n) space.

Time Complexity Space complexity of (multitape) DTM M: - Increasing function s: N Ð N such that for every input string x, Sx S = n, M(x) ends in B s(n) space. DSPACE[s(n)] P: P is solved in O(s(n)) space

Complexity Class PSPACE Complexity Class PSPACE Complexity Class PSPACE is the set of decision problems that can be solved by a (multitape) DTM in a polynomial number of SPACEs on the tape.

Complexity Class PSPACE Complexity Class PSPACE Complexity Class PSPACE is the set of decision problems that can be solved by a (multitape) DTM in a polynomial number of SPACEs on the tape. Polynomial time: PSPACE kc0 DSPACE n k

PSPACE 3-SAT: Given a set of clauses C 1,...,C k, each of length 3, over a set of variables X = x 1,...,x n, does there exist a satisfying truth assignment?

PSPACE 3-SAT: Given a set of clauses C 1,...,C k, each of length 3, over a set of variables X = x 1,...,x n, does there exist a satisfying truth assignment? 3-SAT > N P

PSPACE 3-SAT: Given a set of clauses C 1,...,C k, each of length 3, over a set of variables X = x 1,...,x n, does there exist a satisfying truth assignment? 3-SAT > N P 3-SAT > PSPACE

PSPACE 3-SAT: Given a set of clauses C 1,...,C k, each of length 3, over a set of variables X = x 1,...,x n, does there exist a satisfying truth assignment? 3-SAT > N P 3-SAT > PSPACE Proof (Idea): - Enumerate all 2 n possible truth assignments using counter. - Check each assignment to see if it satisfies all clauses.

Quantified satisfiability 3-SAT: Let ϕ(x 1,x 2,...,x n ) be a boolean 3-CNF formula. Is the following propositional formula true? x 1 x 2 x 3... x n ϕ(x 1,x 2,...,x n )

Quantified satisfiability 3-SAT: Let ϕ(x 1,x 2,...,x n ) be a boolean 3-CNF formula. Is the following propositional formula true? x 1 x 2 x 3... x n ϕ(x 1,x 2,...,x n ) QSAT(Quantified 3-SAT): Let ϕ(x 1,x 2,...,x n ) be a boolean 3-CNF formula. (assume n is odd) Is the following propositional formula true? x 1 x 2 x 3... x n ϕ(x 1,x 2,...,x n )

Quantified satisfiability QSAT(Quantified 3-SAT): Let ϕ(x 1,x 2,...,x n ) be a boolean 3-CNF formula. (assume n is odd) Is the following propositional formula true? x 1 x 2 x 3... x n ϕ(x 1,x 2,...,x n ) Intuition: Alice picks truth value for x 1, then Bob for x 2, then Alice for x 3 and so on. Can Alice satisfy ϕ no matter what Bob does?

Quantified satisfiability Example: ϕ(x 1,x 2,x 3 ) = (x 1 - x 2 - x 3 ), (x 1 - x 2 - x 3 ), ( x 1 - x 2 - x 3 ), ( x 1 - x 2 - x 3 ) x 1 x 2 x 3 ϕ(x 1,x 2,x 3 )

Quantified satisfiability Example: ϕ(x 1,x 2,x 3 ) = (x 1 - x 2 - x 3 ), (x 1 - x 2 - x 3 ), ( x 1 - x 2 - x 3 ), ( x 1 - x 2 - x 3 ) x 1 x 2 x 3 ϕ(x 1,x 2,x 3 ) YES (x 1 = 1)

Quantified satisfiability Example: ϕ(x 1,x 2,x 3 ) = (x 1 - x 2 - x 3 ), (x 1 - x 2 - x 3 ), ( x 1 - x 2 - x 3 ), ( x 1 - x 2 - x 3 ) x 1 x 2 x 3 ϕ(x 1,x 2,x 3 ) YES (x 1 = 1) Theorem QSAT > PSPACE

QSAT > PSPACE Idea: Divide and Conquer algorithm and something more! Proof:

QSAT > PSPACE Idea: Divide and Conquer algorithm and something more! Proof: Recursively try all possibilities. - Only need one bit of information from each subproblem. - Amount of space is proportional to depth of function call stack.

PSPACE - complete Complexity Class PSPACE - complete X > PSPACE - complete if: (i) X > PSPACE and (ii) Y > PSPACE, Y BP X.

Section 1: PSPACE PSPACE - complete Theorem Stockmeyer - Meyer (1973) QSAT > PSPACE - complete

Theorem Savitch Theorem Savitch (1970) If a NTM can solve a problem using f(n) space, an ordinary DTM can solve the same problem in the square of that space bound. For any function f > Ω(log(n)), NSPACE(f(n)) b DSPACE(ˆf ˆn 2 ).

PSPACE Theorem PSPACE = NPSPACE

PSPACE Theorem PSPACE = NPSPACE P b N P b PSPACE b EXPTIME

PSPACE Theorem PSPACE = NPSPACE P b N P b PSPACE b EXPTIME _ \_( )")_/ _ Open Problems: P c??? N P c??? PSPACE c??? EXPTIME

Generalized geography Geography: a children s game

Generalized geography Geography: Alice names capital city c of country she is in. Bob names a capital city c that starts with the letter on which c ends. Alice and Bob repeat this game until one player is unable to continue.

Generalized geography Geography on graphs: Given a directed graph G = (V, E) and a start node s, two players alternate turns by following, if possible, an edge out of the current node to an unvisited node.

Generalized geography Players: A and B. A plays first.

Generalized geography Players: A and B. A plays first. GG = @G, sa S A has a winning strategy for the generalized geography game played on graph G starting at node s

Generalized geography Players: A and B. A plays first. GG = @G, sa S A has a winning strategy for the generalized geography game played on graph G starting at node s GG > PSPACE

Generalized geography Players: A and B. A plays first. GG = @G, sa S A has a winning strategy for the generalized geography game played on graph G starting at node s GG > PSPACE Proof (Idea):

Generalized geography Players: A and B. A plays first. GG = @G, sa S A has a winning strategy for the generalized geography game played on graph G starting at node s GG > PSPACE Proof (Idea): - polynomial space recursive algorithm win(g,s): determines if player A wins the game starting from s.

Generalized geography - If (s,v) in G, then return NO.

Generalized geography - If (s,v) in G, then return NO. - Else (s,v) edge in G, call win recursively in (G - s, v) and if some of them answers NO, then return YES else NO.

Generalized geography - If (s,v) in G, then return NO. - Else (s,v) edge in G, call win recursively in (G - s, v) and if some of them answers NO, then return YES else NO. Depth recursive: # rounds = n = # nodes of G

Generalized geography GG > PSPACE - complete Proof (Idea):

Generalized geography GG > PSPACE - complete Proof (Idea): - QSAT BP GG

Generalized geography GG > PSPACE - complete Proof (Idea): - QSAT BP GG Input: Φ = x 1 x 2 x 3... x k ϕ(x 1,x 2,...,x k ), where ϕ = C 1,C 2,...,C m

Generalized geography GG > PSPACE - complete Proof (Idea): - QSAT BP GG Input: Φ = x 1 x 2 x 3... x k ϕ(x 1,x 2,...,x k ), where -Player A ( ) -Player B ( ) ϕ = C 1,C 2,...,C m

Generalized geography -We construct graph G with initial node s such that A wins the game starting from s iff Φ is true.

Generalized geography -We construct graph G with initial node s such that A wins the game starting from s iff Φ is true. -The graph has:

Generalized geography -We construct graph G with initial node s such that A wins the game starting from s iff Φ is true. -The graph has: a piece that corresponds to the quantified variables (players choose values)

Generalized geography Each diamond structure corresponds to a quantified variable. Players take turns deciding paths at each branching node. If exists an unsatisfying clause, then B will choose an unsatisfying clause to win.

Generalized geography Example: Φ = x 1 x 2 x 3 ϕ(x 1,x 2,x 3 ) ϕ(x 1,x 2,x 3 ) = (x 1 - x 2 - x 1 ), (x 2 - x 2 - x 3 ), ( x 1 - x 2 - x 3 )

Other PSPACE - complete problems Í Tic-tac-toe (or noughts and crosses or Xs and Os)

Other PSPACE - complete problems Í 3D tic-tac-toe (or Qubic)