An Expressive Model for Instance Decomposition Based Parallel SAT Solvers
- 1. An Expressive Model for Instance Decomposition Based Parallel SAT Solvers Tobias Philipp Knowledge Representation and Reasoning Group Technische Universität Dresden
- 2. Outline (1) The Instance Decomposition Approach (2) Incorrect UNSAT Answers (3) A Formal Model (4) Unsatisfiability Proofs 1
- 3. The Instance Decomposition Approach F 2
- 4. The Instance Decomposition Approach F 2
- 5. The Instance Decomposition Approach F F0 x F1 x 2
- 6. The Instance Decomposition Approach F F0 x F1 x 2
- 7. The Instance Decomposition Approach F F0 x F1 F10 y F11 y x 2
- 8. The Instance Decomposition Approach Solver1 Solver2 Solver3 Solver4 Solver5 F F0 F1 F10 F11 F F0 x F1 F10 y F11 y x 2
- 9. The Instance Decomposition Approach Solver1 Solver2 Solver3 Solver4 Solver5 F F0 F1 F10 F11 SAT F F0 x F1 F10 y F11 y x 2
- 10. The Instance Decomposition Approach Solver1 Solver2 Solver3 Solver4 Solver5 F F0 F1 F10 F11 UNSAT UNSAT F F0 x F1 F10 y F11 y x 2
- 11. The Instance Decomposition Approach Solver1 Solver2 Solver3 Solver4 Solver5 F F0 F1 F10 F11 UNSAT UNSAT UNSATF F0 x F1 F10 y F11 y x 2
- 12. The Instance Decomposition Approach Solver1 Solver2 Solver3 Solver4 Solver5 F F0 F1 F10 F11 UNSAT UNSAT UNSATF F0 x F1 F10 y F11 y x Clause Sharing Formula Simplifications 2
- 14. UNSAT may be Incorrect: Sharing Partition Constraints F0 is satisfiable. p(F0) = (F ∧ x), (F ∧ x) F0 ∧ x ∧ x is unsatisfiable. F0 F0 ∧ x F0 ∧ x F0 ∧ x F0 ∧ x F0 ∧ x F0 ∧ x ∧ x F0 ∧ x F0 ∧ x UNSAT 3
- 15. Blocked Clauses C is blocked in F iff there is L ∈ C st all resolvents of C with resolution candidates in F upon L are tautologies F = (x ∨ y) ∧ (x ∨ y) ∧ (x ∨ z) ∧ (y ∨ z) C = (x ∨ z) is blocked in F by z D = (y ∨ z) is blocked in F by z 4
- 16. UNSAT can be Incorrect: Applying Clause Addition Techniques in Two Solvers F = (x ∨ y) ∧ (x ∨ y) ∧ (x ∨ z) ∧ (y ∨ z) C = (x ∨ z) is blocked in F by z D = (y ∨ z) is blocked in F by z F is satisfiable: I = {x, y, z} J = {x, y, z} I |= C, J |= D F ∧ C ∧ D is unsatisfiable F F F ∧ C F F ∧ C F ∧ D F ∧ C ∧ D F ∧ D UNSAT 5
- 17. UNSAT can be Incorrect: Applying Clause Elimination and Addition Techniques in One Solver F = (x ∨ y) ∧ (x ∨ y) ∧ (x ∨ z) ∧ (y ∨ z) C = (x ∨ z) is blocked in F by z D = (y ∨ z) is blocked in F by z F is satisfiable: I = {x, y, z} J = {x, y, z} F ∧ C is satisfiable: J = {x, y, z} F ∧ C ∧ D is unsatisfiable, since J |= D F ∧ C F ∧ C F F ∧ C F ∧ D F ∧ C F ∧ D ∧ C F ∧ C UNSAT 6
- 18. Our Formalism With Clause Elimination and Label-Based Clause Sharing
- 19. Label Based Clause Sharing can Handle Partition Constraints Label function C : N × Clauses → 2Labels F : N → 2Labels Label based clause sharing Solveri Solverj C C(j, C) ⊆ F(i) Partition constraints C(C, i) = {unsafe}, if C is a partition constraint C(C, i) = ∅, otherwise F(i) = ∅ 7
- 20. Consistent Label Functions Can Handle Clause Elimination Techniques is consistent for F0, . . . , Fn iff Fi ≡sat Fi ∧ (j,C)⊆ (i), C∈Fj, 0≤j≤n C Proposition: Position Based Tagging and Boolean Tagging are consistent label functions. 8
- 21. Instance Decomposition Based Solvers can be Described As State Transition Systems Partition function pn(F0) = (F1, F2, . . . , Fn) F0 ≡ F1 ∨ F2 ∨ . . . ∨ Fn. Cooperation tree E E ⊆ {0, . . . , n} × {0, . . . , n} F0 is satisfiable, if Fi is satisfiable, Fi is unsatisfiable, if Fj is unsatisfiable for all children j of i. Label functions: 9
- 22. The States in the Instance Decomposition Model States F0 F1 . . . Fn E Initial state initpn, ,E(F0) = (F0, . . . , Fn, , E) pn(F0) = (F1, . . . , Fn) s.t. F0 ≡ F1 ∨ . . . ∨ Fn. is a consistent label function for F0, . . . , Fn. E is a cooperation tree for F0, . . . , Fn, where F0 is the root. Terminal states SAT, UNSAT 10
- 23. SAT Termination Rule F0 F1 . . . Fi . . . Fn E SAT some Fi is satisfiable 11
- 24. UNSAT Termination Rule F0 F1 . . . Fi . . . Fn E UNSAT ∅ ∈ F0 12
- 25. LOCUNSAT Rule F0 F1 . . . Fi . . . Fn E F0 F1 . . . Fi ∧ ∅ . . . Fn E all children of Fi in E contain ∅, (i, ∅) := (i) 13
- 26. Labeled Resolution Rule F0 F1 . . . Fi . . . Fn E F0 F1 . . . Fi ∧ D . . . Fn E D = C ⊗ C , C, C ∈ Fi, (i, D) := (i, C) ∪ (i, C ) 14
- 27. Clause Deletion Rule F0 F1 . . . Fi . . . Fn E F0 F1 . . . Fi . . . Fn E Fi ⊂ Fi and Fi ≡sat Fi 15
- 28. Label Based Clause Sharing Rule F0 F1 . . . Fi . . . Fn E F0 F1 . . . Fi ∧ C . . . Fn E C ∈ Fj and (j, C) ⊆ (i), (i, C) := (j, C) 16
- 29. Properties of the Instance Decomposition Model Clause Elimination Techniques: Blocked Clause Elimination Variable Elimination Instances: a restricted form of PCASSO Key Invariant: If initpn,E, (F0) ∗ ; (F0, . . . , Fn, , E): F0 ≡sat F0, is consistent for F0, . . . , Fn, E is a cooperation tree for F0, . . . , Fn. Theorem: The Instance Decomposition Model is sound. We can use clause elimination techniques in PCASSO. 17
- 30. Unsatisfiability Proof F SAT Solver SAT, I UNSAT, P Checker Checker Execution traces correspond to unsatisfiability results Record changes in the formulas 18
- 31. Conclusion Contributions A sound formalism for instance decomposition based parallel SAT solvers consistent label functions A proof format Future Work How can we use clause elimination and addition methods in all solver incarnations? How can we construct clausal proofs? 19
- 32. An Expressive Model for Instance Decomposition Based Parallel SAT Solvers Tobias Philipp Knowledge Representation and Reasoning Group Technische Universität Dresden Thank you for your attention.
- 34. Position-Based Tagging F F0 x F1 F10 y F11 y x Position-based label function F( ) = { } F(0) = { , 0} C(0, {x}) = {0} F(1) = { , 1} C(w, {x}) = {1} for w ∈ {1, 10, 11} F(10) = { , 1, 10} C(10, {y}) = {10} F(11) = { , 1, 11} C(11, {y}) = {11} 20
- 35. F = ˙{{x, y}˙} F0 = ˙{{x, y}, {x}˙} F1 = ˙{{x, y}, {x}˙} F10 = ˙{{x, y}, {x}, {y}˙} F11 = ˙{{x, y}, {x}, {y}˙} F F0 x F1 F10 y F11 y x 21
- 36. Proof Format Position-based tagging Labeled and extended resolution derivation of Cn in F (Ci | 1 ≤ i ≤ n) such Ci ∈ F, Ci is a labeled resolvent from two previous clauses Cj and Ck where j < i and k < n, or Ci is the empty clause with label u and for every a ∈ Σ there is the empty clause labeled with ua. Proof Checking Check correct partition constraints and labels in F Check derivation Contains empty clause with no labels attached 22
- 38. Partition Functions Partition function pf (F) = (F1, . . . , Fn) F ≡ F1 ∨ . . . ∨ Fn F = ˙{{x, y}˙} F0 = ˙{{x, y}, {x}˙} F1 = ˙{{x, y}, {x}˙} F10 = ˙{{x, y}, {x}, {y}˙} F11 = ˙{{x, y}, {x}, {y}˙} F F0 x F1 F10 y F11 y x 24
- 39. Clause Addition Techniques can make Formulas Unsatisfiable F0 ≡ (F0 ∧ x) ∨ (F0 ∧ x) Assume F0 ≡sat F0 ∧ x. F0 ∧ x ∧ x is unsatisfiable. Result: There is no consistent label function that allows to share x. F F F F F ∧ F F UNSAT 25