# Quantified satisfiability is in PSPACE

###### Quantified satisfiability is in PSPACE
Dr.AlexanderTyler,India,Teacher
Published Date:21-07-2017
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9. PSPACE PSPACE complexity class ‣ quantified satisfiability ‣ planning problem ‣ PSPACE-complete ‣ Lecture slides by Kevin Wayne Copyright © 2005 Pearson-Addison Wesley Copyright © 2013 Kevin Wayne http://www.cs.princeton.edu/wayne/kleinberg-tardos Last updated on Sep 8, 2013 6:54 AMGeography game 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. Does Alice have a forced win? Ex. Budapest → Tokyo → Ottawa → Ankara → Amsterdam → Moscow → Washington → Nairobi → … 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. Can first player guarantee to make the last legal move? Remark. Some problems (especially involving 2-player games and AI) defy classification according to NP, EXPTIME, NP, and NP-Complete. 29. PSPACE PSPACE complexity class ‣ quantified satisfiability ‣ planning problem ‣ PSPACE-complete ‣ SECTION 9.1PSPACE P. Decision problems solvable in polynomial time. PSPACE. Decision problems solvable in polynomial space. Observation. P ⊆ PSPACE. poly-time algorithm can consume only polynomial space 4PSPACE n Binary counter. Count from 0 to 2 – 1 in binary. Algorithm. Use n bit odometer. Claim. 3-SAT ∈ PSPACE. Pf. n Enumerate all 2 possible truth assignments using counter. Check each assignment to see if it satisfies all clauses. ▪ Theorem. NP ⊆ PSPACE. Pf. Consider arbitrary problem Y ∈ NP. Since Y ≤ 3-SAT, there exists algorithm that solves Y in poly-time plus P polynomial number of calls to 3-SAT black box. Can implement black box in poly-space. ▪ 59. PSPACE PSPACE complexity class ‣ quantified satisfiability ‣ planning problem ‣ PSPACE-complete ‣ SECTION 9.3Quantiﬁed satisﬁability QSAT. Let Φ(x , …, x ) be a boolean CNF formula. Is the following 1 n propositional formula true? ∃ x ∀ x ∃ x ∀ x … ∀ x ∃ x Φ(x , …, x ) 1 2 3 4 n-1 n 1 n assume n is odd Intuition. Amy picks truth value for x , then Bob for x , then Amy for x , 1 2 3 and so on. Can Amy satisfy Φ no matter what Bob does? € Ex. (x ∨ x ) ∧ (x ∨ x ) ∧ (x ∨ x ∨ x ) 1 2 2 3 1 2 3 Yes. Amy sets x true; Bob sets x ; Amy sets x to be same as x . 1 2 3 2 (x ∨ x ) ∧ (x ∨ x ) ∧ (x ∨ x ∨ x ) Ex. 1 2 2 3 1 2 3 € No. If Amy sets x false; Bob sets x false; Amy loses; 1 2 No. if Amy sets x true; Bob sets x true; Amy loses. 1 2 € 7Quantiﬁed satisﬁability is in PSPACE Theorem. Q-SAT ∈ PSPACE. Pf. Recursively try all possibilities. Only need one bit of information from each subproblem. Amount of space is proportional to depth of function call stack. ∃ return true iff both x = 0 x = 1 1 1 subproblems are true ∀ ∀ return true iff either x = 1 x = 0 2 subproblem is true 2 ∃ ∃ ∃ ∃ x = 0 x = 1 3 3 Φ(0, 0, 0) Φ(0, 0, 1) Φ(0, 1, 0) Φ(0, 1, 1) Φ(1, 0, 0) Φ(1, 0, 1) Φ(1, 1, 0) Φ(1, 1, 1) 89. PSPACE PSPACE complexity class ‣ quantified satisfiability ‣ planning problem ‣ PSPACE-complete ‣ SECTION 9.415-puzzle 8-puzzle, 15-puzzle. Noyes Chapman 1874 Board: 3-by-3 grid of tiles labeled 1–8. Legal move: slide neighboring tile into blank (white) square. Find sequence of legal moves to transform initial configuration into goal configuration. 1 2 3 1 2 3 1 2 3 ? move 6 . . . 4 5 6 4 5 4 5 6 8 7 8 7 6 7 8 goal conﬁguration initial conﬁguration 10Planning problem Conditions. Set C = C , …, C . 1 n Initial configuration. Subset c ⊆ C of conditions initially satisfied. 0 Goal configuration. Subset c ⊆ C of conditions we seek to satisfy. Operators. Set O = O , …, O . 1 k To invoke operator O , must satisfy certain prereq conditions. i After invoking O certain conditions become true, and certain conditions i become false. PLANNING. Is it possible to apply sequence of operators to get from initial configuration to goal configuration? Examples. 15-puzzle. Rubik's cube. Logistical operations to move people, equipment, and materials. 11Planning problem: 8-puzzle Planning example. Can we solve the 8-puzzle? C means tile i is in square j ij Conditions. C , 1 ≤ i, j ≤ 9. 1 2 3 ij 4 5 6 Initial state. c = C , C , …, C , C , C , C . 0 11 22 66 78 87 99 8 7 9 Goal state. c = C , C , …, C , C , C , C . 11 22 66 77 88 99 O i Operators. 1 2 3 Precondition to apply O = C , C , …, C , C , C , C . i 11 22 66 78 87 99 4 5 6 After invoking O , conditions C and C become true. i 79 97 8 9 7 After invoking O , conditions C and C become false. i 78 99 Solution. No solution to 8-puzzle or 15-puzzle 12Diversion: Why is 8-puzzle unsolvable? 8-puzzle invariant. Any legal move preserves the parity of the number of pairs of pieces in reverse order (inversions). 3 1 2 3 1 2 3 1 2 4 5 6 4 5 6 4 6 8 7 8 7 8 5 7 3 inversions 3 inversions 5 inversions 1-3, 2-3, 7-8 1-3, 2-3, 7-8 1-3, 2-3, 7-8, 5-8, 5-6 1 2 3 1 2 3 4 5 6 4 5 6 7 8 8 7 0 inversions 1 inversion: 7-8 13Planning problem: binary counter Planning example. Can we increment an n-bit counter from the all-zeroes state to the all-ones state? C corresponds to bit i = 1 i Conditions. C , …, C . 1 n all 0s Initial state. c = φ. 0 all 1s Goal state. c = C , …, C . 1 n Operators. O , …, O . 1 n i-1 least significant bits are 1 To invoke operator O , must satisfy C , …, C . i 1 i–1 set bit i to 1 After invoking O , condition C becomes true. i i set i-1 least After invoking O , conditions C , …, C become false. i 1 i–1 significant bits to 0 Solution. ⇒ C ⇒ C ⇒ C , C ⇒ C ⇒ C , C ⇒ … 1 2 1 2 3 3 1 n Observation. Any solution requires at least 2 – 1 steps. 14Planning problem is in EXPTIME Configuration graph G. n Include node for each of 2 possible configurations. Include an edge from configuration c' to configuration c'' if one of the operators can convert from c' to c''. PLANNING. Is there a path from c to c in configuration graph? 0 Claim. PLANNING ∈ EXPTIME. Pf. Run BFS to find path from c to c in configuration graph. ▪ 0 n Note. Configuration graph can have 2 nodes, and shortest path can n be of length = 2 – 1. binary counter 15Planning problem is in PSPACE Theorem. PLANNING is in PSPACE. Pf. Suppose there is a path from c to c of length L. 1 2 Path from c to midpoint and from c to midpoint are each ≤ L / 2. 1 2 Enumerate all possible midpoints. Apply recursively. Depth of recursion = log L. ▪ 2 boolean hasPath(c , c , L) 1 2 if (L ≤ 1) return correct answer enumerate using binary counter foreach configuration c' boolean x = hasPath(c , c', L/2) 1 boolean y = hasPath(c , c', L/2) 2 if (x and y) return true return false 169. PSPACE PSPACE complexity class ‣ quantified satisfiability ‣ planning problem ‣ PSPACE-complete ‣ SECTION 9.5PSPACE-complete PSPACE. Decision problems solvable in polynomial space. PSPACE-Complete. Problem Y ∈ PSPACE-complete if (i) Y ∈ PSPACE and (ii) for every problem X ∈ PSPACE, X ≤ Y. P Theorem. Stockmeyer-Meyer 1973 QSAT ∈ PSPACE-complete. Theorem. PSPACE ⊆ EXPTIME. Pf. Previous algorithm solves QSAT in exponential time; and QSAT is PSPACE-complete. ▪ Summary. P ⊆ NP ⊆ PSPACE ⊆ EXPTIME. it is known that P ≠ EXPTIME, but unknown which inclusion is strict; conjectured that all are 18PSPACE-complete problems More PSPACE-complete problems. Competitive facility location. Natural generalizations of games. Othello, Hex, Geography, Rush-Hour, Instant Insanity Shanghai, go-moku, Sokoban Given a memory restricted Turing machine, does it terminate in at most k steps? Do two regular expressions describe different languages? Is it possible to move and rotate complicated object with attachments through an irregularly shaped corridor? Is a deadlock state possible within a system of communicating processors? 19Competitive facility location Input. Graph G = (V, E) with positive edge weights, and target B. Game. Two competing players alternate in selecting nodes. Not allowed to select a node if any of its neighbors has been selected. Competitive facility location. Can second player guarantee at least B units of profit? 10 1 5 15 5 1 5 1 15 10 yes if B = 20; no if B = 25 20
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