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Session 1 Verification of Combinational Circuits Satisfiability Solver Albert-Ludwigs-Universitt Freiburg HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 1 / 129 Literatur Albert-Ludwigs-Universitt

  1. Converting a Propositional Formula into CNF Albert-Ludwigs-Universität Freiburg Proof. Consider the following conversion method: m � F m = ( L j , 1 ∧ L j , 2 ) j = 1 where L j , 1 � = L j , 2 , and in this case only containing positive literals L 1 , 1 , L 1 , 2 , ... , L m , 1 , L m , 2 . The size of such a formula is obviously ( 2 · m − 1 ) . A minimal formula that is equivalents, and in conjunctive normal is shown in F ′ m . � F ′ m = ( L 1 , k 1 ∨ ... ∨ L m , k m ) k 1 ,..., k m ∈{ 1 , 2 } m has a total of 2 m clauses. For the conjunctions of all the clauses, ( 2 m − 1 ) AND F ′ operators are needed. Since every clause has m literals, ( m − 1 ) OR operators are needed for every clause. Therefore, the size of formula F ′ m is: m | = 2 m − 1 + 2 m · ( m − 1 ) = m · 2 m − 1 . | F ′ HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 16 / 129

  2. Converting a Propositional Formula into CNF Albert-Ludwigs-Universität Freiburg Example 1 Given: F 1 = x 1 x 2 ∨ x 3 x 4 m = 2 | F 1 | = ( 2 · m − 1 ) = 3 Conversion of F 1 into an equivalent CNF F ′ , with F 1 ≡ F ′ : F ′ = ( x 1 ∨ x 3 ) ∧ ( x 2 ∨ x 3 ) ∧ ( x 1 ∨ x 4 ) ∧ ( x 2 ∨ x 4 ) | F ′ | = ( m · 2 m − 1 ) = 7 HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 17 / 129

  3. Converting a Propositional Formula into CNF Albert-Ludwigs-Universität Freiburg Example 2 Given F 2 = x 1 x 2 ∨ x 3 x 4 ∨ ... ∨ x 17 x 18 ∨ x 19 x 20 m = 10 | F 2 | = 19 = ( 2 · m − 1 ) For F 2 , the CNF representation F ′′ has a size of: | F ′′ | = ( m · 2 m − 1 ) = ( 10 · 2 10 − 1 ) = ( 10 · 1024 − 1 ) = 10239 HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 18 / 129

  4. Tseitin-Transformation Albert-Ludwigs-Universität Freiburg To avoid the possible exponential size of the CNF representation of a circuit (represented by a function F), the following alternative approach can be applied: Construct a formula F ′ that is satisfiably equivalent to F . Meaning if F is satisfiable, then F ′ is satisfiable. For each gate, intermediate “helper” variables are introduced into the CNF F ′ , which do not appear in F . For each gate a “characteristic function” which is in CNF form will be substituted for every occurrence of that particular gate. The "characteristic function" will evaluate to 1, iff the assignments of the respective gate signal would also cause the output gate to go to 1. To generate the final CNF for the entire circuit, all the gate functions that are used will be ANDed together. ⇒ Tseitin-Transformation HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 19 / 129

  5. Tseitin-Transformation Albert-Ludwigs-Universität Freiburg Gate Function CNF Formula ( ¬ x 3 ∨ x 1 ) ∧ ( ¬ x 3 ∨ x 2 ) ∧ x 1 x 3 ≡ x 1 ∧ x 2 x 3 x 2 ( x 3 ∨¬ x 1 ∨¬ x 2 ) ( x 3 ∨¬ x 1 ) ∧ ( x 3 ∨¬ x 2 ) ∧ x 1 x 3 ≡ x 1 ∨ x 2 x 3 x 2 ( ¬ x 3 ∨ x 1 ∨ x 2 ) ( ¬ x 3 ∨ x 1 ∨ x 2 ) ∧ ( ¬ x 3 ∨¬ x 1 ∨¬ x 2 ) ∧ x 1 x 3 ≡ x 1 ⊕ x 2 x 3 x 2 ( x 3 ∨¬ x 1 ∨ x 2 ) ∧ ( x 3 ∨ x 1 ∨¬ x 2 ) x 2 ≡ ¬ x 1 ( x 2 ∨ x 1 ) ∧ ( ¬ x 2 ∨¬ x 1 ) x 1 x 2 HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 20 / 129

  6. Tseitin-Transformation Albert-Ludwigs-Universität Freiburg x 1 x 5 x 2 x 4 x 6 x 3 F SK = ( x 1 ∧ x 2 ) ∨¬ x 3 F CNF SK = ( ¬ x 5 ∨ x 1 ) ∧ ( ¬ x 5 ∨ x 2 ) ∧ ( x 5 ∨¬ x 1 ∨¬ x 2 ) ∧ ( x 6 ∨ x 3 ) ∧ ( ¬ x 6 ∨¬ x 3 ) ∧ ( x 4 ∨¬ x 5 ) ∧ ( x 4 ∨¬ x 6 ) ∧ ( ¬ x 4 ∨ x 5 ∨ x 6 ) HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 21 / 129

  7. Tseitin-Transformation Albert-Ludwigs-Universität Freiburg x 1 x 5 x 2 x 4 x 6 x 3 F SK = ( x 1 ∧ x 2 ) ∨¬ x 3 F CNF SK = ( ¬ x 5 ∨ x 1 ) ∧ ( ¬ x 5 ∨ x 2 ) ∧ ( x 5 ∨¬ x 1 ∨¬ x 2 ) ∧ ( x 6 ∨ x 3 ) ∧ ( ¬ x 6 ∨¬ x 3 ) ∧ ( x 4 ∨¬ x 5 ) ∧ ( x 4 ∨¬ x 6 ) ∧ ( ¬ x 4 ∨ x 5 ∨ x 6 ) HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 21 / 129

  8. Tseitin-Transformation Albert-Ludwigs-Universität Freiburg x 1 x 5 x 2 x 4 x 6 x 3 F SK = ( x 1 ∧ x 2 ) ∨¬ x 3 F CNF SK = ( ¬ x 5 ∨ x 1 ) ∧ ( ¬ x 5 ∨ x 2 ) ∧ ( x 5 ∨¬ x 1 ∨¬ x 2 ) ∧ ( x 6 ∨ x 3 ) ∧ ( ¬ x 6 ∨¬ x 3 ) ∧ ( x 4 ∨¬ x 5 ) ∧ ( x 4 ∨¬ x 6 ) ∧ ( ¬ x 4 ∨ x 5 ∨ x 6 ) HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 21 / 129

  9. Tseitin-Transformation Albert-Ludwigs-Universität Freiburg x 1 x 5 x 2 x 4 x 6 x 3 F SK = ( x 1 ∧ x 2 ) ∨¬ x 3 F CNF SK = ( ¬ x 5 ∨ x 1 ) ∧ ( ¬ x 5 ∨ x 2 ) ∧ ( x 5 ∨¬ x 1 ∨¬ x 2 ) ∧ ( x 6 ∨ x 3 ) ∧ ( ¬ x 6 ∨¬ x 3 ) ∧ ( x 4 ∨¬ x 5 ) ∧ ( x 4 ∨¬ x 6 ) ∧ ( ¬ x 4 ∨ x 5 ∨ x 6 ) HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 21 / 129

  10. Tseitin-Transformation Albert-Ludwigs-Universität Freiburg As long as the CNF representation of each gate only consist of a fixed number of clauses, the number of clauses required for the entire CNF formula will grow linearly with respect to the number of gates in the circuit (also valid for the size of the formula). HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 22 / 129

  11. Tseitin-Transformation Albert-Ludwigs-Universität Freiburg Equivalence checking using satisfiably equivalent CNF representations. Given: F = x 1 x 2 ∨ x 3 x 4 ∨ ... ∨ x 17 x 18 ∨ x 19 x 20 | F | = 19 = ( 2 · m − 1 ) with m = 10 Conversion of F into an equivalent CNF F ′ with F ≡ F ′ : | F ′ | = ( m · 2 m − 1 ) = ( 10 · 2 10 − 1 ) = ( 10 · 1024 − 1 ) = 10239 Tseitin-Transformation from F into satisfiably equivalent CNF F ′′ : | F ′′ | = + + 18 = 199 100 81 ���� ���� ���� 10 UND − Gatter 9 ODER − Gatter 18 ∧ HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 23 / 129

  12. Verification of Combinational Circuit using SAT Albert-Ludwigs-Universität Freiburg Given the following specification and implementation of a combinational circuit: x 1 x 5 x 2 x 4 x 6 x 3 Specification x 1 x 7 x 8 x 2 x ′ x 9 4 x 3 Implementation Question: Are the specification and implementation functionally equivalent? HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 24 / 129

  13. Verification of Combinational Circuit using SAT Albert-Ludwigs-Universität Freiburg x 1 x 5 x 2 x 4 x 6 x 3 Specification M x 7 x 8 x ′ x 9 4 Implementation F M = ( ¬ x 5 ∨ x 1 ) ∧ ( ¬ x 5 ∨ x 2 ) ∧ ( x 5 ∨¬ x 1 ∨¬ x 2 ) ∧ ( x 6 ∨ x 3 ) ∧ ( ¬ x 6 ∨¬ x 3 ) ∧ ( x 4 ∨¬ x 5 ) ∧ ( x 4 ∨¬ x 6 ) ∧ ( ¬ x 4 ∨ x 5 ∨ x 6 ) ∧ ( ¬ x 7 ∨ x 1 ) ∧ ( ¬ x 7 ∨ x 2 ) ∧ ( x 7 ∨¬ x 1 ∨¬ x 2 ) ∧ ( x 7 ∨ x 8 ) ∧ ( ¬ x 7 ∨¬ x 8 ) ∧ ( ¬ x 9 ∨ x 3 ) ∧ ( ¬ x 9 ∨ x 8 ) ∧ ( x 9 ∨¬ x 3 ∨¬ x 8 ) ∧ ( x 9 ∨ x ′ 4 ) ∧ ( ¬ x 9 ∨¬ x ′ 4 ) ∧ ( ¬ M ∨¬ x 4 ∨¬ x ′ 4 ) ∧ ( ¬ M ∨ x 4 ∨ x ′ 4 ) ∧ ( M ∨¬ x 4 ∨ x ′ 4 ) ∧ ( M ∨ x 4 ∨¬ x ′ 4 ) ∧ ( M ) F M is unsatisfiable ⇒ specification and implementation equivalent! HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 25 / 129

  14. Verification of Combinational Circuit using SAT Albert-Ludwigs-Universität Freiburg x 1 x 5 x 2 x 4 x 6 x 3 Specification M x 7 x 8 x ′ x 9 4 Implementation F M = ( ¬ x 5 ∨ x 1 ) ∧ ( ¬ x 5 ∨ x 2 ) ∧ ( x 5 ∨¬ x 1 ∨¬ x 2 ) ∧ ( x 6 ∨ x 3 ) ∧ ( ¬ x 6 ∨¬ x 3 ) ∧ ( x 4 ∨¬ x 5 ) ∧ ( x 4 ∨¬ x 6 ) ∧ ( ¬ x 4 ∨ x 5 ∨ x 6 ) ∧ ( ¬ x 7 ∨ x 1 ) ∧ ( ¬ x 7 ∨ x 2 ) ∧ ( x 7 ∨¬ x 1 ∨¬ x 2 ) ∧ ( x 7 ∨ x 8 ) ∧ ( ¬ x 7 ∨¬ x 8 ) ∧ ( ¬ x 9 ∨ x 3 ) ∧ ( ¬ x 9 ∨ x 8 ) ∧ ( x 9 ∨¬ x 3 ∨¬ x 8 ) ∧ ( x 9 ∨ x ′ 4 ) ∧ ( ¬ x 9 ∨¬ x ′ 4 ) ∧ ( ¬ M ∨¬ x 4 ∨¬ x ′ 4 ) ∧ ( ¬ M ∨ x 4 ∨ x ′ 4 ) ∧ ( M ∨¬ x 4 ∨ x ′ 4 ) ∧ ( M ∨ x 4 ∨¬ x ′ 4 ) ∧ ( M ) F M is unsatisfiable ⇒ specification and implementation equivalent! HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 25 / 129

  15. Complexity and SAT Problems Albert-Ludwigs-Universität Freiburg A CNF formula belongs to the class of “k-SAT” problems iff each clause in the formula has exactly k literals. S.A. Cook, 1971: 3-SAT Problem is NP-Complete Therefore, in “general”, the SAT problem is NP-Complete as all CNF formulas can be converted into a 3-SAT problem. In special cases, we can solve the problems in linear or polynomial time: 2-SAT (formulas contain only binary clauses). Horn-Formula (every clause in the formula contains a maximum of one positive literal). ... HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 26 / 129

  16. Complexity and SAT Problems Albert-Ludwigs-Universität Freiburg Observations in practice: Modern SAT algorithms are now able to solve many industrially relevant and academically interesting problems in a reasonable amount of time. Commonly, industrial problems with 100,000’s of variables, and millions of clauses can be solved. Applications for SAT algorithms: Combinational Equivalence Checking Automatic Test Pattern Generation Bounded Model Checking, Model Checking AI Planning ... HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 27 / 129

  17. Overview of SAT Algorithms Albert-Ludwigs-Universität Freiburg Complete Algorithms (vs Incomplete) Ability - Can prove the unsatisfiability of a CNF-formula due to a systematic approach the solver uses. DP Algorithm M. Davis, H. Putnam, 1960 Based on resolution DLL Algorithm M. Davis, G. Logemann, D. Loveland, 1962 Based on a depth first search Modern SAT Algorithms Based on the DLL Algorithm, however, they include powerful resolution techniques, efficient data structures, and many more acceleration techniques. zChaff, MiniSat, MiraXT, precosat, lingeling, antom HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 28 / 129

  18. Overview of SAT Algorithms Albert-Ludwigs-Universität Freiburg Incomplete Algorithms Normally based on local searches. Basic concept: Generate an initial variable assignment. Until the formula is satisfied, keep modifying the assignments using some heuristic (i.e. “flip” the value of a specific variable). GSat, WSat (H.A. Kautz, B. Selman, 1992 & 1996) Cannot in general prove a formula is unsatisfiable. ⇒ Will not be considered further in these talks! HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 29 / 129

  19. Notation Albert-Ludwigs-Universität Freiburg A clause C = ( L 1 ∨ ... ∨ L n ) can be regarded as a set of literals: C = ( L 1 ,..., L n ) . The empty clause, represented by � , describes the empty set of literals and is by definition unsatisfiable. HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 30 / 129

  20. Notation Albert-Ludwigs-Universität Freiburg The union of two clauses ( C 1 and C 2 ) results in a new clause ( C 3 ) that contains all the literals of both previous clauses: C 3 = C 1 ∪ C 2 = { L | ( L ∈ C 1 ) ∨ ( L ∈ C 2 ) } Literals that occur in both C 1 and C 2 , only appear once in C 3 . This is a form of simplification. The difference of two clauses is defined as follows: C 1 − C 2 = { L | ( L ∈ C 1 ) ∧ ( L �∈ C 2 ) } HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 31 / 129

  21. Notation Albert-Ludwigs-Universität Freiburg A CNF formula F = C 1 ∧ C 2 ∧ ... ∧ C m can be regarded as a set of clauses: F = { C 1 , C 2 ,..., C m } An empty Formula describes an empty set of clauses, and by definition is satisfiable. The union of two CNF formulas ( F 1 and F 2 ) results in a CNF formula F 3 that contains all the clauses from both previous formulas: F 3 = F 1 ∪ F 2 = { C | ( C ∈ F 1 ) ∨ ( C ∈ F 2 ) } Again, clauses that appear in both F 1 and F 2 will only be represented once in F 3 . HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 32 / 129

  22. Resolution Albert-Ludwigs-Universität Freiburg Definition (Resolution) Given two clauses C 1 and C 2 , and a literal L with the following property: L ∈ C 1 and ¬ L ∈ C 2 , then it is possible to build a clause R : R = ( C 1 −{ L } ) ∪ ( C 2 −{¬ L } ) R is referred to as the resolvent of the clauses C 1 and C 2 on literal L . Using our notation, this is represented by: R = C 1 ⊗ L C 2 HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 33 / 129

  23. Resolution Albert-Ludwigs-Universität Freiburg Example 3 C 1 = ( x 1 , x 2 , x 3 ) , C 2 = ( x 4 , ¬ x 2 ) ⇒ R 1 = C 1 ⊗ x 2 C 2 = ( x 1 , x 3 , x 4 ) C 3 = ( x 4 , x 2 , x 3 ) , C 4 = ( x 4 , ¬ x 2 ) ⇒ R 2 = C 3 ⊗ x 2 C 4 = ( x 3 , x 4 ) C 5 = ( x 4 , x 2 ) , C 6 = ( ¬ x 4 , ¬ x 2 ) ⇒ R 3 = C 5 ⊗ x 2 C 6 = ( x 4 , ¬ x 4 ) ( x 4 , ¬ x 4 ) is for every assignment of x 4 satisfied, and is therefore referred to as a tautological clause. HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 34 / 129

  24. Resolution Albert-Ludwigs-Universität Freiburg Lemma (Resolution Lemma) Given a CNF formula F and the resolvent R of two clauses C 1 and C 2 from F, then it must be the cast that F and F ∪{ R } are equivalent: F ≡ F ∪{ R } . HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 35 / 129

  25. Resolution Albert-Ludwigs-Universität Freiburg Proof. Given a set of assignments A that satisfies the formula F ∪{ R } : A | = F ∪{ R } . Then it must also be the case that A | = F . So assume that the assignments A satisfies the formula F . This means that all the clauses C i ∈ F are also satisfied. Furthermore, assume the resolvent R was constructed as R = ( C 1 −{ L } ) ∪ ( C 2 −{¬ L } ) , with C 1 , C 2 ∈ F , L ∈ C 1 and ¬ L ∈ C 2 . To prove the equivalence of the two, we have to distinguishes between two cases in A | = F . Either A | = L or A | = ¬ L . 1 A | = L . Because A | = C 2 and A �| = ¬ L it follows that A | = ( C 2 −{¬ L } ) . As such, the resolvent R is satisfied by A , and then of course F ∪{ R } is also satisfied. 2 A | = ¬ L . Because A | = C 1 and A �| = L it follows that A | = ( C 1 −{ L } ) . As such, the resolvent R is satisfied by A , and then of course F ∪{ R } is also satisfied. HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 36 / 129

  26. Resolution Albert-Ludwigs-Universität Freiburg Definition Given a formula F in CNF form, we define Res ( F ) as: Res ( F ) = F ∪{ R | R is the Resolvent of two clauses in F } . Furthermore, we define: Res 0 ( F ) = F Res t + 1 ( F ) = Res ( Res t ( F )) for t ≥ 0 = � Res ∗ ( F ) t ≥ 0 Res t ( F ) HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 37 / 129

  27. Resolution Albert-Ludwigs-Universität Freiburg Theorem (Resolutions Theorem) A CNF formula F is unsatisfiable when � ∈ Res ∗ ( F ) . HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 38 / 129

  28. Resolution Albert-Ludwigs-Universität Freiburg Proof. Assume � ∈ Res ∗ ( F ) . In this case it is enough to prove their that resolution is correct, and therefore, F is unsatisfiable. First, the empty clause can only be produced from two clauses of the form C 1 = ( L ) and C 2 = ( ¬ L ) . Since � is contained in Res ∗ ( F ) , it must be the case that for some t ≥ 0: � , C 1 , C 2 ∈ Res t + 1 ( F ) and C 1 , C 2 ∈ Res t ( F ) Obviously there is no assignment to the literals that can solve both C 1 and C 2 , and as such, Res t ( F ) is unsatisfiable. Furthermore, with the help of the Resolutions-Lemmas, you can argue that: F ≡ Res 1 ( F ) ≡ Res 2 ( F ) ≡ ... ≡ Res t ( F ) ≡ Res t + 1 ( F ) ≡ ... Which allows us to reason that the unsatisfiability of Res t ( F ) is equal to the unsatisfiability of F . Now, all that is left is to show that resolution is complete for all possible CNF formulas. Using induction, it can be shown that for any unsatisfiable CNF formula F , we can recursively apply the resolution rule to arrive at the empty clause... HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 39 / 129

  29. Resolution Albert-Ludwigs-Universität Freiburg Using the resolutions lemmas and proofs described earlier, it is now possible to construct a simple complete SAT solver. Given: A CNF formula F Procedure: Calculate F = Res 0 ( F ) for t > 0, and keep increasing t until the empty clause is produced, or there are no clauses left to resolve. Result: In the case that t > 0: � ∈ Res t ( F ) ⇒ F is unsatisfiable. Or, in the case that t > 0: � �∈ Res t ( F ) = Res t + 1 ( F ) ⇒ F is satisfiable. HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 40 / 129

  30. Resolution Albert-Ludwigs-Universität Freiburg Complexity of this naive procedure: Since variables can only appear as positive Literals, negative Literals, or not at all in a clause, the run time of this algorithm for a formula with n Variables is in the worst case O ( 3 n ) . In other words, with n variables there is a maximum of 3 n clauses that can be produced. HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 41 / 129

  31. Resolution Albert-Ludwigs-Universität Freiburg Example 4 Is the following CNF formula F satisfiable? F = ( x 1 , x 2 ) ∧ ( x 1 , ¬ x 3 ) ∧ ( ¬ x 1 , x 3 ) ∧ ( ¬ x 1 , ¬ x 2 ) ∧ ( x 3 , ¬ x 2 ) ∧ ( ¬ x 3 , x 2 ) Using the procedure outlined earlier: Res 0 ( F ) = F Res 1 ( F ) = Res 0 ( F ) ∪{ ( x 2 , x 3 ) , ( x 1 , x 3 ) , ( ¬ x 2 , ¬ x 3 ) , ( x 1 , ¬ x 2 ) , ( ¬ x 1 , x 2 ) , ( ¬ x 1 , ¬ x 3 ) } Res 2 ( F ) = Res 1 ( F ) ∪{ ..., ( x 1 ) ,..., ( ¬ x 1 ) ,... } Res 3 ( F ) = Res 2 ( F ) ∪{ � } ⇒ F is unsatisfiable! HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 42 / 129

  32. Resolution Albert-Ludwigs-Universität Freiburg Example 4 Is the following CNF formula F satisfiable? F = ( x 1 , x 2 ) ∧ ( x 1 , ¬ x 3 ) ∧ ( ¬ x 1 , x 3 ) ∧ ( ¬ x 1 , ¬ x 2 ) ∧ ( x 3 , ¬ x 2 ) ∧ ( ¬ x 3 , x 2 ) Using the procedure outlined earlier: Res 0 ( F ) = F Res 1 ( F ) = Res 0 ( F ) ∪{ ( x 2 , x 3 ) , ( x 1 , x 3 ) , ( ¬ x 2 , ¬ x 3 ) , ( x 1 , ¬ x 2 ) , ( ¬ x 1 , x 2 ) , ( ¬ x 1 , ¬ x 3 ) } Res 2 ( F ) = Res 1 ( F ) ∪{ ..., ( x 1 ) ,..., ( ¬ x 1 ) ,... } Res 3 ( F ) = Res 2 ( F ) ∪{ � } ⇒ F is unsatisfiable! HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 42 / 129

  33. Resolution Albert-Ludwigs-Universität Freiburg Example 5 Is the following CNF formula F satisfiable? F = ( x 1 , x 2 , x 3 ) ∧ ( x 2 , ¬ x 3 , ¬ x 4 ) ∧ ( ¬ x 2 , x 5 ) Using the procedure outlined earlier: Res 0 ( F ) = F Res 1 ( F ) = Res 0 ( F ) ∪{ ( x 1 , x 3 , x 5 ) , ( ¬ x 3 , ¬ x 4 , x 5 ) , ( x 1 , x 2 , ¬ x 4 ) } Res 2 ( F ) = Res 1 ( F ) ∪{ ( x 1 , ¬ x 4 ) , ( x 1 , ¬ x 4 , x 5 ) , ( x 1 , ¬ x 4 , x 2 , x 5 ) } Res 3 ( F ) = Res 2 ( F ) = Res ∗ ( F ) ⇒ F is satisfiable! HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 43 / 129

  34. Resolution Albert-Ludwigs-Universität Freiburg Example 5 Is the following CNF formula F satisfiable? F = ( x 1 , x 2 , x 3 ) ∧ ( x 2 , ¬ x 3 , ¬ x 4 ) ∧ ( ¬ x 2 , x 5 ) Using the procedure outlined earlier: Res 0 ( F ) = F Res 1 ( F ) = Res 0 ( F ) ∪{ ( x 1 , x 3 , x 5 ) , ( ¬ x 3 , ¬ x 4 , x 5 ) , ( x 1 , x 2 , ¬ x 4 ) } Res 2 ( F ) = Res 1 ( F ) ∪{ ( x 1 , ¬ x 4 ) , ( x 1 , ¬ x 4 , x 5 ) , ( x 1 , ¬ x 4 , x 2 , x 5 ) } Res 3 ( F ) = Res 2 ( F ) = Res ∗ ( F ) ⇒ F is satisfiable! HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 43 / 129

  35. DP Algorithm Albert-Ludwigs-Universität Freiburg The SAT algorithm introduced by M. Davis und H. Putnam in 1960 was based on the previous procedures but included the following optimizations: Subsumption Pure Literal Variable Elimination Note: these optimizations improve the run time of the solver, and can decrease the average complexity. However, the worst case complexity remains the same as the naive approach. HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 44 / 129

  36. DP Algorithm Albert-Ludwigs-Universität Freiburg “The superiority of the present procedure over those previously available is indicated in part by the fact that a formula on which Gilmore’s routine for the IBM 704 1 causes the machine to compute for 21 minutes without obtaining a result was worked successfully by hand computation using the present method in 30 minutes.” – M. Davis and H. Putnam 1 The IBM 704 was one of the first commercial scientific computers. It had a maximum memory capacity of 4096 × 36 bit words (excluding magnetic tape storage), and could execute up to 40,000 instructions per HI second. Between 1955 and 1960, IBM sold over 120 of these machines. Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 45 / 129

  37. DP Algorithm Albert-Ludwigs-Universität Freiburg Definition (Subsumption) Assume we have two clauses C 1 and C 2 . C 1 then subsumes C 2 when all the literals in C 1 are also in C 2 : C 1 ⊆ C 2 . Idea: To satisfy a CNF formula F , we must satisfy all the clauses. Therefore, if F is satisfiable, both C 1 and C 2 . Since C 1 ⊆ C 2 , every satisfying assignment for C 1 will automatically solve C 2 . This means that we can delete C 2 from F without changing the satisfiability of the formula. The idea of subsumption is used in all modern SAT solvers. Most only perform subsumption checks during preprocessing, but other do it continually. HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 46 / 129

  38. DP Algorithm Albert-Ludwigs-Universität Freiburg Re-examine Example 5 Is the following CNF formula F satisfiable? F = ( x 1 , x 2 , x 3 ) ∧ ( x 2 , ¬ x 3 , ¬ x 4 ) ∧ ( ¬ x 2 , x 5 ) Using the naive procedure outlined earlier: Res 0 ( F ) = F Res 1 ( F ) = Res 0 ( F ) ∪{ ( x 1 , x 3 , x 5 ) , ( ¬ x 3 , ¬ x 4 , x 5 ) , ( x 1 , x 2 , ¬ x 4 ) } Res 2 ( F ) = Res 1 ( F ) ∪{ ( x 1 , ¬ x 4 ) , ( x 1 , ¬ x 4 , x 5 ) , ( x 1 , ¬ x 4 , x 2 , x 5 ) } � �� � subsummed from ( x 1 , ¬ x 4 ) Res 3 ( F ) = Res 2 ( F ) = Res ∗ ( F ) ⇒ F is satisfiable! HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 47 / 129

  39. DP Algorithm Albert-Ludwigs-Universität Freiburg Definition (Pure Literal) Let F be a CNF formula and L a literal contained in F . We say L is pure literal iff it is only present in its positive or negative form in F . In other words, F contains L or ¬ L , but not both. Idea: Remove all the clauses from F that contain the pure literal L. This can be done because L will only satisfy clauses if assigned correctly. ¬ L on the other hand will only “unsatisfy” clauses. Normally, this step is only used during the preprocessing of a CNF formula for a SAT solver, and is no longer used during the solving process. However, on harder QBF problems, this technique is still used. HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 48 / 129

  40. DP Algorithm Albert-Ludwigs-Universität Freiburg Again, repeating example 5 Is the following CNF formula F satisfiable? F = ( x 1 , x 2 , x 3 ) ∧ ( x 2 , ¬ x 3 , ¬ x 4 ) ∧ ( ¬ x 2 , x 5 ) ⇒ x 1 , ¬ x 4 und x 5 are pure literals. ⇒ Delete clauses containing x 1 , ¬ x 4 or x 5 . ⇒ F = {} ⇒ F is satisfiable! HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 49 / 129

  41. DP Algorithm Albert-Ludwigs-Universität Freiburg Within the DP algorithm resolution is used to completely remove a variable x i from the formula (i.e. delete all positive and negative occurances of x i from a CNF formula F ). ⇒ Variable Elimination Goal: Reduce the number of variables occuring in the CNF formula F , while maintaining a relatively constant number of clauses. HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 50 / 129

  42. DP Algorithm Albert-Ludwigs-Universität Freiburg Definition Let F be a CNF formula, and x i the variable we wish to eliminate (where L = x i and ¬ L = ¬ x i ). Then we need to define P , N and W as follows: Let P be the set of all clauses in F that include L : P = { C | ( L ∈ C ) ∧ ( C ∈ F ) } Let N be the set of all clauses in F that include ¬ L : N = { C | ( ¬ L ∈ C ) ∧ ( C ∈ F ) } Let W be the set of all clauses in F that do not contain L or ¬ L : W = { C | ( L �∈ C ) ∧ ( ¬ L �∈ C ) ∧ ( C ∈ F ) } As such: F = P ∧ N ∧ W . HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 51 / 129

  43. DP Algorithm Albert-Ludwigs-Universität Freiburg Definition Given the clause partitioning mentioned previously, P ⊗ x i N defines the set of clauses that are generated through the pairwise resolution on variable x i from all combinations of clauses from P and N : P ⊗ x i N = { R | ( R = C 1 ⊗ x i C 2 ) ∧ ( C 1 ∈ P ) ∧ ( C 2 ∈ N ) } HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 52 / 129

  44. DP Algorithm Albert-Ludwigs-Universität Freiburg Theorem Let F be a CNF formula, and let x i be a Variable. Assume the possitve occurance (L = x i ) and the negative occurance ¬ L = ¬ x i of the variable x i appear in the formula F. Futhermore, let the clause sets P, N and W be used as defined earlier. Then, it must be the case that F = P ∧ N ∧ W and F ′ = ( P ⊗ x i N ) ∧ W are equally satisfiable. HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 53 / 129

  45. DP Algorithm Albert-Ludwigs-Universität Freiburg Conclusion of the previous sentence: The question of the satisfiability of a formula F can be attributed to the satisfiability of F ′ , where F ′ is the constructed from F through the elimination of the variable x i . As such, if F ′ is unsatisfiable, so is F . Otherwise both are satisfiable. HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 54 / 129

  46. DP Algorithm Albert-Ludwigs-Universität Freiburg Basic procedure for variable elimination: Select a variable x i , then perform resolution between all the pairs of clauses containing x i and ¬ x i . Then replace all the clauses in the sets P and N with the new clauses generated during resolution. If done blindly, the amount of new clauses that are produced normally is much greater than the sum of P and N . As such, the total number of clauses in the formula usually increases. Variable elimination is done in modern SAT solvers during preprocessing. Furthermore, using heuristics, variables are only selected for removal when they results in the total number of clauses remaining the same or decreasing. HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 55 / 129

  47. DP Algorithm Albert-Ludwigs-Universität Freiburg Example 6 Is the following CNF formula F satisfiable? F = ( x 1 , x 2 ) ∧ ( x 1 , ¬ x 3 ) ∧ ( ¬ x 1 , x 3 ) ∧ ( ¬ x 1 , ¬ x 2 ) ∧ ( x 3 , ¬ x 2 ) ∧ ( ¬ x 3 , x 2 ) Elimination of x 1 from formula F : P = { ( x 1 , x 2 ) , ( x 1 , ¬ x 3 ) } = { ( ¬ x 1 , x 3 ) , ( ¬ x 1 , ¬ x 2 ) } N W = { ( x 3 , ¬ x 2 ) , ( ¬ x 3 , x 2 ) } P ⊗ x 1 N = { ( x 2 , x 3 ) , ( x 2 , ¬ x 2 ) , ( ¬ x 3 , x 3 ) , ( ¬ x 3 , ¬ x 2 ) } F ′ = ( P ⊗ x 1 N ) ∧ W = ( x 2 , x 3 ) ∧ ( ¬ x 3 , ¬ x 2 ) ∧ ( x 3 , ¬ x 2 ) ∧ ( ¬ x 3 , x 2 ) Elimination of x 2 from formula F : P ′ = { ( x 2 , x 3 ) , ( ¬ x 3 , x 2 ) } N ′ = { ( ¬ x 3 , ¬ x 2 ) , ( x 3 , ¬ x 2 ) } W ′ = / 0 P ′ ⊗ x 2 N ′ = { ( x 3 , ¬ x 3 ) , ( x 3 ) , ( ¬ x 3 ) , ( ¬ x 3 , x 3 ) } = ( P ′ ⊗ x 2 N ′ ) ∧ W ′ = ( x 3 ) ∧ ( ¬ x 3 ) ⇒ F ′′ and F are unsatisfiable! F ′′ HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 56 / 129

  48. DP Algorithm Albert-Ludwigs-Universität Freiburg Example 6 Is the following CNF formula F satisfiable? F = ( x 1 , x 2 ) ∧ ( x 1 , ¬ x 3 ) ∧ ( ¬ x 1 , x 3 ) ∧ ( ¬ x 1 , ¬ x 2 ) ∧ ( x 3 , ¬ x 2 ) ∧ ( ¬ x 3 , x 2 ) Elimination of x 1 from formula F : P = { ( x 1 , x 2 ) , ( x 1 , ¬ x 3 ) } = { ( ¬ x 1 , x 3 ) , ( ¬ x 1 , ¬ x 2 ) } N W = { ( x 3 , ¬ x 2 ) , ( ¬ x 3 , x 2 ) } P ⊗ x 1 N = { ( x 2 , x 3 ) , ( x 2 , ¬ x 2 ) , ( ¬ x 3 , x 3 ) , ( ¬ x 3 , ¬ x 2 ) } F ′ = ( P ⊗ x 1 N ) ∧ W = ( x 2 , x 3 ) ∧ ( ¬ x 3 , ¬ x 2 ) ∧ ( x 3 , ¬ x 2 ) ∧ ( ¬ x 3 , x 2 ) Elimination of x 2 from formula F : P ′ = { ( x 2 , x 3 ) , ( ¬ x 3 , x 2 ) } N ′ = { ( ¬ x 3 , ¬ x 2 ) , ( x 3 , ¬ x 2 ) } W ′ = / 0 P ′ ⊗ x 2 N ′ = { ( x 3 , ¬ x 3 ) , ( x 3 ) , ( ¬ x 3 ) , ( ¬ x 3 , x 3 ) } = ( P ′ ⊗ x 2 N ′ ) ∧ W ′ = ( x 3 ) ∧ ( ¬ x 3 ) ⇒ F ′′ and F are unsatisfiable! F ′′ HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 56 / 129

  49. DP Algorithm Albert-Ludwigs-Universität Freiburg Davis-Putnam Algorithm bool DP(CNF F ) { if ( F = / 0 ) { return SATISFIABLE ; } // Empty clause set. if ( � ∈ F ) { return UNSATISFIABLE ; } // Empty clause. if ( F contains a unit clause ( L ) ) // Unit clause rule. { // Unit Subsumption. F ′ = F −{ C | ( L ∈ C ) ∧ ( C ∈ F ) ∧ ( C � = ( L )) } ; // Unit Resolution. P = { ( L ) } ; N = { C | ( ¬ L ∈ C ) ∧ ( C ∈ F ′ ) } ; W = F ′ − P − N ; return DP ([ P ⊗ L N ] ∧ W ) ; } if ( F contains a pure literal L ) // Pure literal rule. { // Delete from F every clause containing L . F ′ = F −{ C | ( L ∈ C ) ∧ ( C ∈ F ) } ; return DP( F ′ ); } L = S ELECT L ITERAL ( F ); // Select a literal. P = { C | ( L ∈ C ) ∧ ( C ∈ F ) ; // Variablen elimination. N = { C | ( ¬ L ∈ C ) ∧ ( C ∈ F ) } ; W = F − P − N ; return DP ([ P ⊗ L N ] ∧ W ) ; } HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 57 / 129

  50. DLL Algorithm Albert-Ludwigs-Universität Freiburg Due to the possibly exponential growth in memory requirements, the basic DP Algorithm has only seen little use. However, two years later in 1962, M. Davis, G. Logemann und D. Loveland introduced the DLL-Algorithm, which replaced the physical variable elimination with a depth first search. Idea: If a CNF formula F is satisfiable, a satisfying assignment of the variables in F must included either x i = 1 oder x i = 0 ⇒ Check both paths one after another. In literature, the DLL algorithm is often referred to as the DPLL algorithm. HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 58 / 129

  51. DLL Algorithm Albert-Ludwigs-Universität Freiburg Davis-Logemann-Loveland Algorithm bool DLL(CNF F ) { if ( F = / 0 ) { return SATISFIABLE ; } // Empty clause set. if ( � ∈ F ) { return UNSATISFIABLE ; } // Empty clause. if ( F contains a unit clause ( L ) ) // Unit clause rule. { // Unit Subsumption. F ′ = F −{ C | ( L ∈ C ) ∧ ( C ∈ F ) ∧ ( C � = ( L )) } ; // Unit Resolution. P = { ( L ) } ; N = { C | ( ¬ L ∈ C ) ∧ ( C ∈ F ′ ) } ; W = F ′ − P − N ; return DLL ([ P ⊗ L N ] ∧ W ) ; } if ( F contains a pure literal L ) // Pure literal rule. { // Delete from F every clause containing L . F ′ = F −{ C | ( L ∈ C ) ∧ ( C ∈ F ) } ; return DLL( F ′ ); } L = S ELECT L ITERAL ( F ); // Select a literal. if (DLL( F ∪{ ( L ) } ) == SATISFIABLE ) // Path selection. { return SATISFIABLE ; } else { return DLL( F ∪{ ( ¬ L ) } ); } } HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 59 / 129

  52. DLL Algorithm Albert-Ludwigs-Universität Freiburg ( ¬ 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 ) Path selection HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 60 / 129

  53. DLL Algorithm Albert-Ludwigs-Universität Freiburg x 1 1 ( ¬ 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 ) Select path x 1 = 1 HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 60 / 129

  54. DLL Algorithm Albert-Ludwigs-Universität Freiburg x 1 1 ( ¬ 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 ) Select path x 1 = 1 HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 60 / 129

  55. DLL Algorithm Albert-Ludwigs-Universität Freiburg x 1 1 x 2 1 ( ¬ 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 ) Select path x 2 = 1 HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 60 / 129

  56. DLL Algorithm Albert-Ludwigs-Universität Freiburg x 1 1 x 2 1 ( ¬ 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 ) Select path x 2 = 1 HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 60 / 129

  57. DLL Algorithm Albert-Ludwigs-Universität Freiburg x 1 1 x 2 1 ( ¬ 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 ) Unit clause rule x 3 = 0 and x 3 = 1 HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 60 / 129

  58. DLL Algorithm Albert-Ludwigs-Universität Freiburg x 1 1 x 2 1 ( ¬ 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 ) Try opposite path HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 60 / 129

  59. DLL Algorithm Albert-Ludwigs-Universität Freiburg x 1 1 x 2 0 1 ( ¬ 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 ) Select path x 2 = 0 HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 60 / 129

  60. DLL Algorithm Albert-Ludwigs-Universität Freiburg x 1 1 x 2 0 1 ( ¬ 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 ) Select path x 2 = 0 HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 60 / 129

  61. DLL Algorithm Albert-Ludwigs-Universität Freiburg x 1 1 x 2 0 1 ( ¬ 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 ) Unit clause rule x 3 = 0 and x 3 = 1 HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 60 / 129

  62. DLL Algorithm Albert-Ludwigs-Universität Freiburg x 1 1 x 2 0 1 ( ¬ 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 ) Try opposite path HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 60 / 129

  63. DLL Algorithm Albert-Ludwigs-Universität Freiburg x 1 0 1 x 2 0 1 ( ¬ 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 ) Select path x 1 = 0 HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 60 / 129

  64. DLL Algorithm Albert-Ludwigs-Universität Freiburg x 1 0 1 x 2 0 1 ( ¬ 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 ) Select path x 1 = 0 HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 60 / 129

  65. DLL Algorithm Albert-Ludwigs-Universität Freiburg x 1 0 1 x 2 x 2 0 0 1 ( ¬ 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 ) Pure literal rule x 2 = 0 HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 60 / 129

  66. DLL Algorithm Albert-Ludwigs-Universität Freiburg x 1 0 1 x 2 x 2 0 0 1 ( ¬ 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 ) Pure literal rule x 2 = 0 HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 60 / 129

  67. DLL Algorithm Albert-Ludwigs-Universität Freiburg x 1 0 1 x 2 x 2 0 0 1 ( ¬ 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 ) Formula is satisfiable with x 1 = x 2 = 0 HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 60 / 129

  68. Whats next? Albert-Ludwigs-Universität Freiburg Data structures and algorithmic implementation details have so far been largely omitted. ⇒ Modern SAT algorithms Preprocessing Decision heuristics and strategies Boolean constraint propagation Conflict analysis & non-chronological backtracking Conflict clause deletion Restarts HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 61 / 129

  69. Summary: DLL-Algorithm Albert-Ludwigs-Universität Freiburg It is a recursive procedure where at every recursive step the following are checked: 1 Termination: Empty clause set ⇒ formula is satisfiable Empty clause ⇒ Current (partial) formula unsatisfiable 2 Unit clause / pure literal rule In the case that there is the unit clause ( L ) in the current sub-formula F , we can simplify F to F ′ by deleting all clauses containing L , and by removing all occurrences of ¬ L in the formula. In the case that the sub-formula F contains a pure literal, we can delete the occurrence of these variables from all clauses in F , and in the process, produce F ′ . 3 Search both branches For the current formula F which still contains the literal L , we must call the DLL algorithm recursively on for both F ∪{ ( L ) } and F ∪{ ( ¬ L ) } . HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 62 / 129

  70. From the DLL to modern SAT Algorithms Albert-Ludwigs-Universität Freiburg Approach: DLL Algorithm Recursive procedure From one recursive level r to r + 1 the algorithm modifies the formula (satisfied clauses are deleted, and falsified literals are removed from clauses). When jumping back from recursive level r + 1 to r , the algorithm has to re-insert all deleted clauses and literal occurrences in the formula. HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 63 / 129

  71. From the DLL to modern SAT Algorithms Albert-Ludwigs-Universität Freiburg Approach: Modern SAT Algorithms No longer a recursive procedure Except for special cases, clauses and literals are not physically removed from the CNF formula during the search process. In general, the pure literal rule is no longer used (QBF?) HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 64 / 129

  72. From the DLL to modern SAT Algorithms Albert-Ludwigs-Universität Freiburg Unit Clause DLL Algorithm A clause which contains exactly one literal. HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 65 / 129

  73. From the DLL to modern SAT Algorithms Albert-Ludwigs-Universität Freiburg Unit Clause Modern SAT Algorithms In addition to the previous definition, a clauses that contains only falsified literals and one unassigned literal under the current search space evaluation is also a unit clause. Example: The assignments x 1 = 0 , x 2 = 1 turns ( x 1 , ¬ x 2 , x 3 ) into a unit clause. In the example, by adding the assignment x 3 = 1 to the previous assignments x 1 = 0 , x 2 = 1, the clause ( x 1 , ¬ x 2 , x 3 ) becomes satisfied. This use of the unit clause rule in this case implies x 3 = 1. As such x 3 = 1 is referred to as an Implication. HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 66 / 129

  74. From the DLL to modern SAT Algorithms Albert-Ludwigs-Universität Freiburg Unit Clause Modern SAT Algorithms ... Determining all the implications (i.e. unit propagations) of assigning a variable a value is in modern solvers done by the Boolean Constraint Propagation (BCP) procedure. In its original form, the DLL would recursively call itself after every implication. Example: In F = ( x 1 , ¬ x 2 ) ∧ ( x 1 , x 2 , x 3 ) ∧ ( ¬ x 3 , x 4 ) the assignment x 1 = 0 results in the implications x 2 = 0 , x 3 = 1 , x 4 = 1 HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 67 / 129

  75. From the DLL to modern SAT Algorithms Albert-Ludwigs-Universität Freiburg Unsatisfiable paths / Conflicts DLL Algorithm: Empty clause. Modern SAT Algorithm: A clause where all its literals are falsely assigned. Example: The assignments x 1 = 0 , x 2 = 1 , x 3 = 0 result in the clause ( x 1 , ¬ x 2 , x 3 ) becoming falsified. Since all our formulas are in CNF form, the entire formulas under this assignment is also unsatisfied. HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 68 / 129

  76. From the DLL to modern SAT Algorithms Albert-Ludwigs-Universität Freiburg Unsatisfiable paths / Conflicts DLL Algorithm A conflict is always the result of the previous variable selection, and the resulting unit implications. Backtracking, to a previous recursion level, in which both possible cases of a variable have not been checked, allows the solver to remove the existing conflict. In the case that their exist no unchecked path, and the solver must recursive backtrack to its first call DLL call, the CNF formula is unsatisfiable. HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 69 / 129

  77. From the DLL to modern SAT Algorithms Albert-Ludwigs-Universität Freiburg Unsatisfiable paths / Conflicts Modern SAT Algorithms Current solvers perform a more indepth analysis of every conflict as it is often the case that multiple variable selections play a role in each conflict. Generation (by resolution) and addition of new conflict clauses to the formula allow the solver to learn important information about the problem. These conflict clauses contain a list of literals that are responsible for the current conflict. With the use of the conflict claues, the algorithm can in many cases backtrack past multiple variable selections. The procedure can also produce the the empty or null clause resulting in a final evaluation of UNSATISFIABLE . HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 70 / 129

  78. Modern SAT Algorithms Albert-Ludwigs-Universität Freiburg Basic procedures of a moderns SAT-Solver Preprocessing Main routines: ... Selection of decision variables Boolean constraint propagation / unit propagation Conflict analysis & backtracking Ever now and then during the search: Reduce size of conflict clause set (delete clauses) Restarts If the formula is satisfiable: Output a variable assignment that satisfies all the clauses (i.e. a model) HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 71 / 129

  79. Modern SAT Algorithms Albert-Ludwigs-Universität Freiburg Main procedure of a modern sequential SAT algorithm bool S EQUENTIAL S AT E NGINE (CNF F) { if (P REPROCESS CNF(F) == CONFLICT) // Simplify the CNF formula. { return UNSATISFIABLE ; } // Problem is unsatisfiable. while (true) { if (D ECIDE N EXT B RANCH ()) // Select a free variable and assign it a value. { while (BCP() == CONFLICT) // Boolean constraint propagation. { BLevel = A NALYZE C ONFLICT (); // Conflict analysis. if (BLevel > 0) { B ACKTRACK (BLevel); } // Backtrack to a previous decision. else { return UNSATISFIABLE ; } // Problem unsatisfiable. } } else { return SATISFIABLE ; } // All variables are assigned, problem satisfiable. } } Not explicitly shown: Deletion of conflict clauses, restarts, or outputted model. HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 72 / 129

  80. Modern SAT Algorithms Albert-Ludwigs-Universität Freiburg Main procedure of a modern sequential SAT algorithm bool S EQUENTIAL S AT E NGINE (CNF F) { if (P REPROCESS CNF(F) == CONFLICT) // Simplify the CNF formula. { return UNSATISFIABLE ; } // Problem is unsatisfiable. while (true) { if (D ECIDE N EXT B RANCH ()) // Select a free variable and assign it a value. { while (BCP() == CONFLICT) // Boolean constraint propagation. { BLevel = A NALYZE C ONFLICT (); // Conflict analysis. if (BLevel > 0) { B ACKTRACK (BLevel); } // Backtrack to a previous decision. else { return UNSATISFIABLE ; } // Problem unsatisfiable. } } else { return SATISFIABLE ; } // All variables are assigned, problem satisfiable. } } Not explicitly shown: Deletion of conflict clauses, restarts, or outputted model. HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 73 / 129

  81. Preprocessing Albert-Ludwigs-Universität Freiburg Goal: Prior to actually starting the search, try to simplify the formula as much as possible. Practical observations: In many cases, the size of the input formula directly correlates to the run time of the SAT Algorithm. A reduction of more than 75% in the number of clauses & variables in the input formula can be achieved. Identification and processing of unit clauses that are contained with the original clause set has always been a part of a modern SAT algorithm. The trick is to find a balance between: the simplification that preprocessing is able to achieve;the time required by the preprocessor; and the performance increases gained by the SAT search algorithm. HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 74 / 129

  82. Preprocessing Albert-Ludwigs-Universität Freiburg Unit Propagation Lookahead (UPLA) For a variable x i test x i = 0 and x i = 1. During the test monitor what each assignment leads to: ( x i = 0 → conflict ) ∧ ( x i = 1 → conflict ) ⇒ UNSAT ( x i = 0 → conflict ) ⇒ x i = 1 ( x i = 1 → conflict ) ⇒ x i = 0 ( x i = 0 → x j = 1 ) ∧ ( x i = 1 → x j = 1 ) ⇒ x j = 1 ( x i = 0 → x j = 0 ) ∧ ( x i = 1 → x j = 0 ) ⇒ x j = 0 ( x i = 0 → x j = 0 ) ∧ ( x i = 1 → x j = 1 ) ⇒ x i ≡ x j HI Lewis / Schubert / Becker Verification and SAT Solving Satisfiability Solver 75 / 129

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