Are Short Proofs Narrow? QBF Resolution is not so Simple. Meena Mahajan The Institute of Mathematical Sciences, HBNI, Chennai. Dagstuhl 18051 Proof Complexity Meena Mahajan
Credits partially supported by the EU Marie Curie IRSES grant CORCON. joint work with Olaf Beyersdorff Univ of Leeds, UK Leroy Chew Univ of Leeds, UK Anil Shukla formerly at IMSc, Chennai now at IIT Jodhpur In ACM Transactions on Computational Logic 19(1) 2018, 1:1-1:26. (preliminary version in STACS 2016) Dagstuhl 18051 Proof Complexity Meena Mahajan
The Resolution Proof System C : bag of clauses. ˜ a : assignment to the variables. If ˜ a satisfies Then ˜ a satisfies . . . . . . A ∨ x A ∨ x C ′ = C = B ∨ ¬ x B ∨ ¬ x . . . . . . A ∨ B Dagstuhl 18051 Proof Complexity Meena Mahajan
The Resolution Proof System C : bag of clauses. ˜ a : assignment to the variables. If ˜ a satisfies Then ˜ a satisfies . . . . . . A ∨ x A ∨ x C ′ = C = B ∨ ¬ x B ∨ ¬ x . . . . . . A ∨ B C 0 ∈ SAT = ⇒ C 1 ∈ SAT = ⇒ . . . = ⇒ C t − 1 ∈ SAT = ⇒ C t ∈ SAT C 0 �∈ SAT ⇐ . . . ⇐ C i �∈ SAT ⇐ . . . ⇐ C t �∈ SAT ⇐ � ∈ C t Dagstuhl 18051 Proof Complexity Meena Mahajan
Extending Resolution to QBF s QBFs: Quantified Boolean Formulas x · F ( � W.l.o.g., QBF in prenex CNF: Q � x ); F a set of clauses. Resolution is sound: If Q � x · F ( x ) is true, and we add a clause C to F through resolution to get F ′ , then Q � x · F ′ ( x ) is also true. Dagstuhl 18051 Proof Complexity Meena Mahajan
Extending Resolution to QBF s QBFs: Quantified Boolean Formulas x · F ( � W.l.o.g., QBF in prenex CNF: Q � x ); F a set of clauses. Resolution is sound: If Q � x · F ( x ) is true, and we add a clause C to F through resolution to get F ′ , then Q � x · F ′ ( x ) is also true. But Resolution alone is not enough. Consider ∃ x ∀ u ( ¬ u ∨ x ) ( u ∨ ¬ x ) . Resolution can add ( x ∨ ¬ x ) or ( u ∨ ¬ u ). Useless. Universal variable u has to be handled differently. • Two ways to proceed, modelling CDCL-based solvers • expansion-based solvers Dagstuhl 18051 Proof Complexity Meena Mahajan
The Evaluation Game on QBF s QBF Q � x · F ( � x ) Two players, Red and Blue, step through quantifier prefix left-to-right. Red picks values for ∃ variables, Blue for ∀ variables. Assignment constructed: ˜ a . Red wins a run of the game if F (˜ a ) true. Otherwise Blue wins. Dagstuhl 18051 Proof Complexity Meena Mahajan
The Evaluation Game on QBF s QBF Q � x · F ( � x ) Two players, Red and Blue, step through quantifier prefix left-to-right. Red picks values for ∃ variables, Blue for ∀ variables. Assignment constructed: ˜ a . Red wins a run of the game if F (˜ a ) true. Otherwise Blue wins. example: ∃ x ∀ u ( ¬ u ∨ x ) ( u ∨ ¬ x ) . Dagstuhl 18051 Proof Complexity Meena Mahajan
The Evaluation Game on QBF s QBF Q � x · F ( � x ) Two players, Red and Blue, step through quantifier prefix left-to-right. Red picks values for ∃ variables, Blue for ∀ variables. Assignment constructed: ˜ a . Red wins a run of the game if F (˜ a ) true. Otherwise Blue wins. example: ∃ x ∀ u ( ¬ u ∨ x ) ( u ∨ ¬ x ) . Red: x = 1, Blue: u = 1: Red wins Red: x = 1, Blue: u = 0: Blue wins Red: x = 0, Blue: u = 1: Blue wins Blue can always win: set u � = x . Dagstuhl 18051 Proof Complexity Meena Mahajan
The Evaluation Game on QBF s QBF Q � x · F ( � x ) Two players, Red and Blue, step through quantifier prefix left-to-right. Red picks values for ∃ variables, Blue for ∀ variables. Assignment constructed: ˜ a . Red wins a run of the game if F (˜ a ) true. Otherwise Blue wins. example: ∃ x ∀ u ( ¬ u ∨ x ) ( u ∨ ¬ x ) . Red: x = 1, Blue: u = 1: Red wins Red: x = 1, Blue: u = 0: Blue wins Red: x = 0, Blue: u = 1: Blue wins Blue can always win: set u � = x . Q � x · F ( � x ) false if and only if Blue has a winning strategy. Use this to extend Resolution. Dagstuhl 18051 Proof Complexity Meena Mahajan
The ∀ reduction rule Consider this scenario: x · F ( � Q � x ) is true. So Red has a winning strategy. F ( � x ) has a clause C in which the rightmost variable (as per order in Q � x ) is a universal variable u . i.e. C = A ∨ ℓ ; ℓ ∈ { u , ¬ u } ; all variables in A are left of u . Dagstuhl 18051 Proof Complexity Meena Mahajan
The ∀ reduction rule Consider this scenario: x · F ( � Q � x ) is true. So Red has a winning strategy. F ( � x ) has a clause C in which the rightmost variable (as per order in Q � x ) is a universal variable u . i.e. C = A ∨ ℓ ; ℓ ∈ { u , ¬ u } ; all variables in A are left of u . Then by the time Blue has to fix u , Red’s strategy must ensure that sub-clause A is already satisfied. That is, Red has a winning strategy on Q � x · [ F ( � x ) ∧ A ]. So Q � x · [ F ( � x ) ∧ A ] is also true. Equivalently: If Blue has a winning strategy on Q � x · [ F ( � x ) ∧ A ], x · F ( � then Blue has a winning strategy on Q � x ). Dagstuhl 18051 Proof Complexity Meena Mahajan
The proof system Q-Res Q � x · C Grow the bag of clauses C using Resolution with existential pivots: If A ∨ x and B ∨ ¬ x are in the bag, can add A ∨ B (provided not a tautology), ∀ -Reduction for universal variables: If A ∨ ℓ ( u ) in the bag and and all variables in A are left of u , can add A , until the empty clause � is added. Dagstuhl 18051 Proof Complexity Meena Mahajan
The proof system Q-Res (cont’d) Sound: A derivation of � reveals a winning strategy for Blue. [KleineB¨ uningKarpinskiFl¨ ogel 1995] Complete: Use a winning strategy of Blue to decide which clauses to derive. Suffices to eliminate variables in right-to-left order of quantification blocks (Level-ordered Q-Res) Dagstuhl 18051 Proof Complexity Meena Mahajan
A derivation in Q-Res ∃ e ∀ u ∃ c ∃ d [( ¬ e ∨ c ) , ( ¬ u ∨ c ) , ( e ∨ d ) , ( u ∨ d ) , ( ¬ c ∨ ¬ d )] Dagstuhl 18051 Proof Complexity Meena Mahajan
A derivation in Q-Res ∃ e ∀ u ∃ c ∃ d [( ¬ e ∨ c ) , ( ¬ u ∨ c ) , ( e ∨ d ) , ( u ∨ d ) , ( ¬ c ∨ ¬ d )] e 0 1 u u 1 0 c c d d d d c ∨ ¯ c ∨ ¯ (¯ u ∨ c ) ( e ∨ d ) (¯ (¯ e ∨ c ) ( u ∨ d ) (¯ (¯ u ∨ c ) d ) (¯ e ∨ c ) d ) ( e ∨ d ) ( u ∨ d ) Dagstuhl 18051 Proof Complexity Meena Mahajan
A derivation in Q-Res ∃ e ∀ u ∃ c ∃ d [( ¬ e ∨ c ) , ( ¬ u ∨ c ) , ( e ∨ d ) , ( u ∨ d ) , ( ¬ c ∨ ¬ d )] e 0 1 u u 1 0 c c u ∨ c ¯ e ∨ ¯ c e ∨ c ¯ u ∨ ¯ c d d d d c ∨ ¯ c ∨ ¯ (¯ u ∨ c ) ( e ∨ d ) (¯ (¯ e ∨ c ) ( u ∨ d ) (¯ (¯ u ∨ c ) d ) (¯ e ∨ c ) d ) ( e ∨ d ) ( u ∨ d ) Dagstuhl 18051 Proof Complexity Meena Mahajan
A derivation in Q-Res ∃ e ∀ u ∃ c ∃ d [( ¬ e ∨ c ) , ( ¬ u ∨ c ) , ( e ∨ d ) , ( u ∨ d ) , ( ¬ c ∨ ¬ d )] e 0 1 u u 1 0 u ∨ e ¯ e ∨ u ¯ c c u ∨ c ¯ e ∨ ¯ c e ∨ c ¯ u ∨ ¯ c d d d d c ∨ ¯ c ∨ ¯ (¯ u ∨ c ) ( e ∨ d ) (¯ (¯ e ∨ c ) ( u ∨ d ) (¯ (¯ u ∨ c ) d ) (¯ e ∨ c ) d ) ( e ∨ d ) ( u ∨ d ) Dagstuhl 18051 Proof Complexity Meena Mahajan
A derivation in Q-Res ∃ e ∀ u ∃ c ∃ d [( ¬ e ∨ c ) , ( ¬ u ∨ c ) , ( e ∨ d ) , ( u ∨ d ) , ( ¬ c ∨ ¬ d )] e 0 1 u e u e ¯ 1 0 u ∨ e ¯ e ∨ u ¯ c c u ∨ c ¯ e ∨ ¯ c e ∨ c ¯ u ∨ ¯ c d d d d c ∨ ¯ c ∨ ¯ (¯ u ∨ c ) ( e ∨ d ) (¯ (¯ e ∨ c ) ( u ∨ d ) (¯ (¯ u ∨ c ) d ) (¯ e ∨ c ) d ) ( e ∨ d ) ( u ∨ d ) Dagstuhl 18051 Proof Complexity Meena Mahajan
A derivation in Q-Res ∃ e ∀ u ∃ c ∃ d [( ¬ e ∨ c ) , ( ¬ u ∨ c ) , ( e ∨ d ) , ( u ∨ d ) , ( ¬ c ∨ ¬ d )] � e 0 1 u e u e ¯ 1 0 u ∨ e ¯ e ∨ u ¯ c c u ∨ c ¯ e ∨ ¯ c e ∨ c ¯ u ∨ ¯ c d d d d c ∨ ¯ c ∨ ¯ (¯ u ∨ c ) ( e ∨ d ) (¯ (¯ e ∨ c ) ( u ∨ d ) (¯ (¯ u ∨ c ) d ) (¯ e ∨ c ) d ) ( e ∨ d ) ( u ∨ d ) Dagstuhl 18051 Proof Complexity Meena Mahajan
Expansion-Based Systems ∀ uQ � x · F ( u , � x ) is true � x · F (0 , � x )] ∧ [ Q � x · F (1 , � [ Q � x )] is true � x u / 1 · x u / 0 Q � x u / 0 ) ∧ F (1 , � x u / 1 ) � � Q � F (0 , � is true Expand the initial formula judiciously, on the fly. Then use standard resolution. Expansion-based systems: ∀ Exp+Res [Janota,Marques-Silva 2015] , IR [Beyersdorff,Chew,Janota 2014] . Dagstuhl 18051 Proof Complexity Meena Mahajan
Expansion-Based Systems (cont’d) τ, σ : partial assignments to universal variables. l [ τ ] denotes l σ , where σ is the sub-assignment of τ setting variables left of l . Dagstuhl 18051 Proof Complexity Meena Mahajan
Expansion-Based Systems (cont’d) τ, σ : partial assignments to universal variables. l [ τ ] denotes l σ , where σ is the sub-assignment of τ setting variables left of l . ∀ Exp+Res: ∀ Expansion and Resolution For initial clause C , use any of the clauses { l [ τ ] | l ∈ C , l is existential } ∪ { τ ( l ) | l ∈ C , l is universal } . τ : assignment to all universal variables. Use Resolution. Dagstuhl 18051 Proof Complexity Meena Mahajan
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