Advanced Artificial Intelligence, DT4048

Transcription

1 Advanced Artificial Intelligence, DT4048 Part 5 Boolean Satisfiability Federico Pecora federico.pecora@oru.se Center for Applied Autonomous Sensor Systems (AASS) Örebro University, Sweden

2 What is Boolean Satisfiability? George Boole ( ) introduces a way to reduce propositional logic to algerbaic manipulation Propositional logic had been studies since Aristotle provides a foundation for formalizing deductive reasoning today, used to automate such reasoning in AI 2015 F. Pecora / Örebro University aass.oru.se 2 / 49

3 What is Boolean Satisfiability? George Boole ( ) introduces a way to reduce propositional logic to algerbaic manipulation Propositional logic had been studies since Aristotle provides a foundation for formalizing deductive reasoning today, used to automate such reasoning in AI Example [T. Walsh lecture slides, 2002]: you are instructed to organizing the embassy ball the ambassador wants to invite Peru, or exclude Qatar the vice-ambassador wants Qatar, Romania, or both recent diplomatic events make it impossible to invite both Romania and Peru Who do you invite? 2015 F. Pecora / Örebro University aass.oru.se 2 / 49

4 What is Boolean Satisfiability? the ambassador wants to invite Peru, or exclude Qatar P Q the vice-ambassador wants Qatar, Romania, or both Q R impossible to invite both Romania and Peru (R P) Variables: meaningful parts of the problem (P, Q, R) Domains: possible values of variables (T, F) Constraints: conditions restricting possible mutual assignments φ = (P Q) (Q R) ( (R P)) = (P Q) (Q R) ( R P) 2015 F. Pecora / Örebro University aass.oru.se 3 / 49

5 What is Boolean Satisfiability? the ambassador wants to invite Peru, or exclude Qatar P Q the vice-ambassador wants Qatar, Romania, or both Q R impossible to invite both Romania and Peru (R P) Variables: meaningful parts of the problem (P, Q, R) Domains: possible values of variables (T, F) Constraints: conditions restricting possible mutual assignments φ = (P Q) (Q R) ( (R P)) = (P Q) (Q R) ( R P) Constraint Satisfaction Problem: find an assignment of values to variables that satisfies all constraints {P = T, Q = T, R = F} 2015 F. Pecora / Örebro University aass.oru.se 3 / 49

6 Boolean Satisfiability (SAT) Instance: a set U of variables and a collection C of clauses over the variables U variables: all boolean, i.e., domains are all {, } clauses: are disjunction or conjunction of literals literals: a variable or its negation conjunctive normal form (CNF): formula is conjunction of clauses, clauses are disjunctions of literals, e.g., (P Q R) (R S) disjunctive normal form (DNF): formula is disjunction of clauses, clauses are conjunction of literals, e.g., (P Q R) (R S) Question: is there a model for C? model: truth assignment to variables that satisfies all clauses (if CNF) or at least one clause (if DNF) 2015 F. Pecora / Örebro University aass.oru.se 4 / 49

7 Boolean Satisfiability (SAT) Recall a few important opertions with booleans (P Q) P Q (De Morgan s law) (P Q) P Q (De Morgan s law) P Q P Q (implication) P Q (P Q) (Q P) (logical equivalence) P Q Q P (commutative law) P Q Q P (commutative law) P (Q R) (P Q) R (associative law) P (Q R) (P Q) R (associative law) P (Q R) (P Q) (P R) (distributive law) P (Q R) (P Q) (P R) (distributive law) P P, P P P (identity law) 2015 F. Pecora / Örebro University aass.oru.se 5 / 49

8 Clausal Normal Form DNF and CNF, as normal forms, are useful for automated problem solving In both CNF and DNF, negation ( ) can only be used as part of a literal (i.e., it can only precede a propositional variable) Every propositional formula can be converted to an equivalent formula in CNF or DNF k-cnf: each clause has k lilterals k-sat: the problem of finding a model for a k-cnf formula 2015 F. Pecora / Örebro University aass.oru.se 6 / 49

9 k-sat and Complexity problem complexity NP-Hard NP-Complete 3-SAT NP? = P 2-SAT 2-SAT P 2-SAT can be solved in polynomial time 3-SAT NP sol(3-sat) can be verified in polynomial time 3-SAT NP-Complete [Cook, 1971, Garey and Johnson, 1979] all problems in NP can be reduced to 3-SAT in polynomial time 2015 F. Pecora / Örebro University aass.oru.se 7 / 49

10 k-sat and Complexity problem complexity NP-Hard NP-Complete 3-SAT NP? = P 2-SAT 2-SAT P 2-SAT can be solved in polynomial time 3-SAT NP sol(3-sat) can be verified in polynomial time 3-SAT NP-Complete [Cook, 1971, Garey and Johnson, 1979] all problems in NP can be reduced to 3-SAT in polynomial time If a polinomial time algorithm for 3-SAT is found, this algorithm will solve all problems in NP (and more... )! 2015 F. Pecora / Örebro University aass.oru.se 7 / 49

11 3-SAT and Complexity of CSP Recall that π NP-Complete iff sol(π) can be verified in polynomial time all problems in NP can be reduced to π Now we know that 3-SAT is NP-Complete what does this say about CSP? 2015 F. Pecora / Örebro University aass.oru.se 8 / 49

12 3-SAT and Complexity of CSP Recall that π NP-Complete iff sol(π) can be verified in polynomial time all problems in NP can be reduced to π Now we know that 3-SAT is NP-Complete what does this say about CSP? all problems in NP can be reduced to 3-SAT 3-SAT can be reduced to CSP (because 3-SAT is a CSP) a solution to a CSP can be verified in polynomial time (just check all constraints) 2015 F. Pecora / Örebro University aass.oru.se 8 / 49

13 3-SAT and Complexity of CSP Recall that π NP-Complete iff sol(π) can be verified in polynomial time all problems in NP can be reduced to π Now we know that 3-SAT is NP-Complete what does this say about CSP? all problems in NP can be reduced to 3-SAT 3-SAT can be reduced to CSP (because 3-SAT is a CSP) a solution to a CSP can be verified in polynomial time (just check all constraints) Hence, the CSP is NP-Complete 2015 F. Pecora / Örebro University aass.oru.se 8 / 49

14 CSP as SAT Every CSP can be translated to a SAT problem instance every CSP variable (domain size d) is represented by d SAT variables (n d variables) each variable requires a small cluster ( of clauses to ensure (d ) ) that it takes exactly one value (n + 1 clauses) every CSP constraint is represented by a series of mutual exclusions between the SAT variables that participate in the constraint (e (d 2) clauses) Can often lead to very large instances despite size, even these are often efficiently solvable by SAT solvers F. Pecora / Örebro University aass.oru.se 9 / 49

15 CSP as SAT: Example A Domains: A, B, C, D {red, green, blue} C D B Constraints: A B, A C, C B, D B, A D 2015 F. Pecora / Örebro University aass.oru.se 10 / 49

16 CSP as SAT: Example C A D B One boolean var. per CSP domain value: x r A, xg A, xb A, xr B, xg B, xb B, xr C, xg C, xb C, xr D, xg D, xb D Domains: A, B, C, D {red, green, blue} Constraints: A B, A C, C B, D B, A D (12 variables) 2015 F. Pecora / Örebro University aass.oru.se 10 / 49

17 CSP as SAT: Example C A D B Domains: A, B, C, D {red, green, blue} Constraints: A B, A C, C B, D B, A D One boolean var. per CSP domain value: xa r, xg A, xb A, xr B, xg B, xb B, xr C, xg C, xb C, xr D, xg D, xb D (12 variables) Single value clauses: (xa r xg A xb A ) ( xr A xg A ) ( xr A xb A ) ( xg A xb A )... ( 4 = 16 clauses) 2015 F. Pecora / Örebro University aass.oru.se 10 / 49

18 CSP as SAT: Example C A D B Domains: A, B, C, D {red, green, blue} Constraints: A B, A C, C B, D B, A D One boolean var. per CSP domain value: xa r, xg A, xb A, xr B, xg B, xb B, xr C, xg C, xb C, xr D, xg D, xb D (12 variables) Single value clauses: (xa r xg A xb A ) ( xr A xg A ) ( xr A xb A ) ( xg A xb A )... ( 4 = 16 clauses) Binary constraint clauses: ( x r A xr B ) ( xg A xg B ) ( xb A xb B )... ( 5 = 15 clauses) 2015 F. Pecora / Örebro University aass.oru.se 10 / 49

19 Why Reduce to SAT? Many interesting problems can be reduced to SAT planning, scheduling, circuit design, model verification,... polinomial-time reductions model leads to solution of original problem Efficient algorithms can often decide large, interesting problems What if specific solving techniques exists for the original problem? 2015 F. Pecora / Örebro University aass.oru.se 11 / 49

20 Why Reduce to SAT? Many interesting problems can be reduced to SAT planning, scheduling, circuit design, model verification,... polinomial-time reductions model leads to solution of original problem Efficient algorithms can often decide large, interesting problems What if specific solving techniques exists for the original problem? SAT is simple 2015 F. Pecora / Örebro University aass.oru.se 11 / 49

21 Why Reduce to SAT? Many interesting problems can be reduced to SAT planning, scheduling, circuit design, model verification,... polinomial-time reductions model leads to solution of original problem Efficient algorithms can often decide large, interesting problems What if specific solving techniques exists for the original problem? SAT is simple SAT solvers employ very sophisticated search techniques 2015 F. Pecora / Örebro University aass.oru.se 11 / 49

22 Why Reduce to SAT? Many interesting problems can be reduced to SAT planning, scheduling, circuit design, model verification,... polinomial-time reductions model leads to solution of original problem Efficient algorithms can often decide large, interesting problems What if specific solving techniques exists for the original problem? SAT is simple SAT solvers employ very sophisticated search techniques SAT solvers are the workhorses of AI 2015 F. Pecora / Örebro University aass.oru.se 11 / 49

23 Reduction ot SAT: Another Example Correspondence between digital logic circuits and propositional logic Digital design circuit logic gate input wire internal wire voltage level Propositional logic formula propositional connective literal sub-expression boolean value 2015 F. Pecora / Örebro University aass.oru.se 12 / 49

24 Reduction ot SAT: Another Example b a c Correspondence between digital logic circuits and propositional logic a b c φ 1 = b a c φ 2 = (a b) (b c) Are the the two circuits equivalent? Is φ = φ 1 φ 2 SAT? 2015 F. Pecora / Örebro University aass.oru.se 12 / 49

25 Transforming to CNF Unfortunately naïvely applying distribution can cause the size of the formula to grow exponentially original formula: (x 1 y 1 ) (x 2 y 2 )... (x n y n ) naïve transformation gives 2 n clauses: (x 1... x n 1 x n ) (x 1... x n 1 y n )... (y 1... y n 1 y n ) However, we are not usually interested in equivalent CNF formulations We only care about satisfiability and possibly a simple way to extract the model of the original formula 2015 F. Pecora / Örebro University aass.oru.se 13 / 49

26 Transforming to CNF SAT-preserving trasformation to CNF: φ φ φ is SAT φ is SAT (equisatisfiable) φ is only linearly bigger than φ transformation introduces linear number of new variables model of φ is projection of model of φ on original variables 2015 F. Pecora / Örebro University aass.oru.se 14 / 49

27 Transforming to CNF SAT-preserving trasformation to CNF: φ φ φ is SAT φ is SAT (equisatisfiable) φ is only linearly bigger than φ transformation introduces linear number of new variables model of φ is projection of model of φ on original variables Example: (p (q r)) s introduce new variables for sub-formulas: 2015 F. Pecora / Örebro University aass.oru.se 14 / 49

28 Transforming to CNF SAT-preserving trasformation to CNF: φ φ φ is SAT φ is SAT (equisatisfiable) φ is only linearly bigger than φ transformation introduces linear number of new variables model of φ is projection of model of φ on original variables Example: (p (q r)) s introduce new variables for sub-formulas: (p 1 q r) 2015 F. Pecora / Örebro University aass.oru.se 14 / 49

29 Transforming to CNF SAT-preserving trasformation to CNF: φ φ φ is SAT φ is SAT (equisatisfiable) φ is only linearly bigger than φ transformation introduces linear number of new variables model of φ is projection of model of φ on original variables Example: (p (q r)) s introduce new variables for sub-formulas: (p 1 q r) (p 2 p p 1 ) 2015 F. Pecora / Örebro University aass.oru.se 14 / 49

30 Transforming to CNF SAT-preserving trasformation to CNF: φ φ φ is SAT φ is SAT (equisatisfiable) φ is only linearly bigger than φ transformation introduces linear number of new variables model of φ is projection of model of φ on original variables Example: (p (q r)) s introduce new variables for sub-formulas: (p 1 q r) (p 2 p p 1 ) (p 3 p 2 s) 2015 F. Pecora / Örebro University aass.oru.se 14 / 49

31 Transforming to CNF SAT-preserving trasformation to CNF: φ φ φ is SAT φ is SAT (equisatisfiable) φ is only linearly bigger than φ transformation introduces linear number of new variables model of φ is projection of model of φ on original variables Example: (p (q r)) s introduce new variables for sub-formulas: (p 1 q r) (p 2 p p 1 ) (p 3 p 2 s) p F. Pecora / Örebro University aass.oru.se 14 / 49

32 Transforming to CNF SAT-preserving trasformation to CNF: φ φ φ is SAT φ is SAT (equisatisfiable) φ is only linearly bigger than φ transformation introduces linear number of new variables model of φ is projection of model of φ on original variables Example: (p (q r)) s introduce new variables for sub-formulas: (p 1 q r) (p 2 p p 1 ) (p 3 p 2 s) p 3 then transform to CNF: ( p 1 q) ( p 1 r) (p 1 q r) ( p 2 p p 1 ) (p 2 p) (p 2 p 1 ) ( p 3 p 2 ) ( p 3 s) (p 3 p 2 s) p F. Pecora / Örebro University aass.oru.se 14 / 49

33 Davis-Putnam-Logeman-Loveland Procedure Fundamental SAT solving algorithm developed through two seminal papers [Davis and Putnam, 1960], [Davis et al., 1962] latter fixes memory explosion problem in former method known today as DPLL DPLL is essentially a backtracking search algorithm As such, DPLL benefits from propagation propagation in general CSP: enforce some level of consistency while making valid assignments propagation in SAT: transform the set of clauses while preserving satisfiability 2015 F. Pecora / Örebro University aass.oru.se 15 / 49

34 Davis-Putnam-Logeman-Loveland Procedure Propagation in DPLL is based on two transformations on the set of clauses: unit propagation (UP): if a unit clause p appears, remove p from other clauses and remove all clauses including p pure-literal elimination (PLE): if p occurs only negated, or only unnegated, delete all clauses involving p 2015 F. Pecora / Örebro University aass.oru.se 16 / 49

35 Davis-Putnam-Logeman-Loveland Procedure Propagation in DPLL is based on two transformations on the set of clauses: unit propagation (UP): if a unit clause p appears, remove p from other clauses and remove all clauses including p pure-literal elimination (PLE): if p occurs only negated, or only unnegated, delete all clauses involving p UP example: (x 5 ) is a unit clause, i.e., x 5 must be (x 1 x 2 x 3 ) (x 1 x 2 ) ( x 2 x 3 ) (x 1 x 2 x 5 ) ( x 3 x 4 ) (x 4 x 5 ) (x 5 ) 2015 F. Pecora / Örebro University aass.oru.se 16 / 49

36 Davis-Putnam-Logeman-Loveland Procedure Propagation in DPLL is based on two transformations on the set of clauses: unit propagation (UP): if a unit clause p appears, remove p from other clauses and remove all clauses including p pure-literal elimination (PLE): if p occurs only negated, or only unnegated, delete all clauses involving p UP example: (x 5 ) is a unit clause, i.e., x 5 must be (x 1 x 2 x 3 ) (x 1 x 2 ) ( x 2 x 3 ) (x 1 x 2 x 5 ) ( x 3 x 4 ) (x 4 x 5 ) (x 5 ) 2015 F. Pecora / Örebro University aass.oru.se 16 / 49

37 Davis-Putnam-Logeman-Loveland Procedure Propagation in DPLL is based on two transformations on the set of clauses: unit propagation (UP): if a unit clause p appears, remove p from other clauses and remove all clauses including p pure-literal elimination (PLE): if p occurs only negated, or only unnegated, delete all clauses involving p UP example: (x 5 ) is a unit clause, i.e., x 5 must be (x 1 x 2 x 3 ) (x 1 x 2 ) ( x 2 x 3 ) (x 1 x 2 x 5 ) ( x 3 x 4 ) (x 4 x 5 ) (x 5 ) PLE example: x 1 is never negated, i.e., x 1 can be 2015 F. Pecora / Örebro University aass.oru.se 16 / 49

38 Davis-Putnam-Logeman-Loveland Procedure Propagation in DPLL is based on two transformations on the set of clauses: unit propagation (UP): if a unit clause p appears, remove p from other clauses and remove all clauses including p pure-literal elimination (PLE): if p occurs only negated, or only unnegated, delete all clauses involving p UP example: (x 5 ) is a unit clause, i.e., x 5 must be (x 1 x 2 x 3 ) (x 1 x 2 ) ( x 2 x 3 ) (x 1 x 2 x 5 ) ( x 3 x 4 ) (x 4 x 5 ) (x 5 ) PLE example: x 1 is never negated, i.e., x 1 can be (x 1 x 2 x 3 ) (x 1 x 2 ) ( x 2 x 3 ) (x 1 x 2 x 5 ) ( x 3 x 4 ) (x 4 x 5 ) (x 5 ) 2015 F. Pecora / Örebro University aass.oru.se 16 / 49

39 Davis-Putnam-Logeman-Loveland Procedure Unit propagation and pure-literal elimination is a Boolean Constraint Propagation (BCP) procedure Algorithm 1: BCP(φ) : formula or φ unit-propagate(φ) φ eliminate-pure-lit(φ) if empty clause φ then return return φ 2015 F. Pecora / Örebro University aass.oru.se 17 / 49

40 Davis-Putnam-Logeman-Loveland Procedure DPLL: recursive depth-first enumeration of possible models Algorithm 2: DPLL(φ) : model for φ or UNSAT static assignment (current assignment of variables to {, }) φ BCP(φ) if φ = then return UNSAT assignment state(φ) if assignment is complete then return assignment variable choose-unassigned-variable(φ,assignment) return DPLL(φ variable) DPLL(φ variable) 2015 F. Pecora / Örebro University aass.oru.se 18 / 49

41 Davis-Putnam-Logeman-Loveland Procedure: Example x 1 x 2 x 3 x 1 x 4 x 1 x 2 x 2 x 5 x 1 x 3 x 3 x 6 x 2 x 3 x 1 x 7 x 4 x 5 x 6 x 2 x 8 x 4 x 5 x 3 x 9 x 4 x 6 x 1 x 10 x 5 x 6 x 2 x 11 x 7 x 8 x 9 x 3 x 12 x 7 x 8 x 4 x 7 x 7 x 9 x 5 x 8 x 8 x 9 x 6 x 9 x 10 x 11 x 12 x 4 x 10 x 10 x 11 x 5 x 11 x 10 x 12 x 6 x 12 x 11 x 12 x 7 x 10 x 8 x 11 x 9 x 12 branch over x 1 =... x F. Pecora / Örebro University aass.oru.se 19 / 49

42 Davis-Putnam-Logeman-Loveland Procedure: Example x 1 x 2 x 3 1 x 3 x 3 x 6 x 2 x 3 1 x 7 x 4 x 5 x 6 x 2 x 8 x 4 x 5 x 3 x 9 x 4 x 6 1 x 10 x 5 x 6 x 2 x 11 x 7 x 8 x 9 x 3 x 12 x 7 x 8 x 4 x 7 x 7 x 9 x 5 x 8 x 8 x 9 x 6 x 9 x 10 x 11 x 12 x 4 x 10 x 10 x 11 x 5 x 11 x 10 x 12 x 6 x 12 x 11 x 12 x 7 x 10 x 8 x 11 x 9 x 12 x 2 is a unit clause x 2 = x F. Pecora / Örebro University aass.oru.se 19 / 49

43 Davis-Putnam-Logeman-Loveland Procedure: Example x 1 x 2 x 3 1 x 3 x 3 x 6 x 2 x 3 1 x 7 x 4 x 5 x 6 x 2 x 8 x 4 x 5 x 3 x 9 x 4 x 6 1 x 10 x 5 x 6 x 2 x 11 x 7 x 8 x 9 x 3 x 12 x 7 x 8 x 4 x 7 x 7 x 9 x 5 x 8 x 8 x 9 x 6 x 9 x 10 x 11 x 12 x 4 x 10 x 10 x 11 x 5 x 11 x 10 x 12 x 6 x 12 x 11 x 12 x 7 x 10 x 8 x 11 x 9 x 12 x 3 is a unit clause x 3 = x F. Pecora / Örebro University aass.oru.se 19 / 49

44 Davis-Putnam-Logeman-Loveland Procedure: Example x 1 x 2 x 3 1 x 3 x 3 x 6 x 2 x 3 1 x 7 x 4 x 5 x 6 x 2 x 8 x 4 x 5 x 3 x 9 x 4 x 6 1 x 10 x 5 x 6 x 2 x 11 x 7 x 8 x 9 x 3 x 12 x 7 x 8 x 4 x 7 x 7 x 9 x 5 x 8 x 8 x 9 x 6 x 9 x 10 x 11 x 12 x 4 x 10 x 10 x 11 x 5 x 11 x 10 x 12 x 6 x 12 x 11 x 12 x 7 x 10 x 8 x 11 x 9 x 12 x 4 is a unit clause x 4 = x F. Pecora / Örebro University aass.oru.se 19 / 49

45 Davis-Putnam-Logeman-Loveland Procedure: Example x 1 x 2 x 3 1 x 3 x 3 x 6 x 2 x 3 1 x 7 x 4 x 5 x 6 x 2 x 8 x 4 x 5 x 3 x 9 x 4 x 6 1 x 10 x 5 x 6 x 2 x 11 x 7 x 8 x 9 x 3 x 12 x 7 x 8 x 4 x 7 x 7 x 9 x 5 x 8 x 8 x 9 x 6 x 9 x 10 x 11 x 12 x 4 x 10 x 10 x 11 x 5 x 11 x 10 x 12 x 6 x 12 x 11 x 12 x 7 x 10 x 8 x 11 x 9 x 12 x 7 is a unit clause x 7 = x F. Pecora / Örebro University aass.oru.se 19 / 49

46 Davis-Putnam-Logeman-Loveland Procedure: Example x 1 x 2 x 3 1 x 3 x 3 x 6 x 2 x 3 1 x 7 x 4 x 5 x 6 x 2 x 8 x 4 x 5 x 3 x 9 x 4 x 6 1 x 10 x 5 x 6 x 2 x 11 x 7 x 8 x 9 x 3 x 12 x 7 x 8 x 4 x 7 x 7 x 9 x 5 x 8 x 8 x 9 x 6 x 9 x 10 x 11 x 12 x 4 x 10 x 10 x 11 x 5 x 11 x 10 x 12 x 6 x 12 x 11 x 12 x 7 x 10 x 8 x 11 x 9 x 12 x 10 is a unit clause x 10 = x F. Pecora / Örebro University aass.oru.se 19 / 49

47 Davis-Putnam-Logeman-Loveland Procedure: Example x 1 x 2 x 3 1 x 3 x 3 x 6 x 2 x 3 1 x 7 x 4 x 5 x 6 x 2 x 8 x 4 x 5 x 3 x 9 x 4 x 6 1 x 10 x 5 x 6 x 2 x 11 x 7 x 8 x 9 x 3 x 12 x 7 x 8 x 4 x 7 x 7 x 9 x 5 x 8 x 8 x 9 x 6 x 9 x 10 x 11 x 12 x 4 x 10 x 10 x 11 x 5 x 11 x 10 x 12 x 6 x 12 x 11 x 12 x 7 x 10 x 8 x 11 x 9 x 12 BCP does nothing branch over x 5 =... x F. Pecora / Örebro University aass.oru.se 19 / 49

48 Davis-Putnam-Logeman-Loveland Procedure: Example x 1 x 2 x 3 1 x 3 x 3 x 6 x 2 x 3 1 x 7 x 4 x 5 x 6 x 2 x 8 x 4 x 5 x 3 x 9 x 4 x 6 1 x 10 x 5 x 6 x 2 x 11 x 7 x 8 x 9 x 3 x 12 x 7 x 8 x 4 x 7 x 7 x 9 x 5 x 8 x 8 x 9 x 6 x 9 x 10 x 11 x 12 x 4 x 10 x 10 x 11 x 5 x 11 x 10 x 12 x 6 x 12 x 11 x 12 x 7 x 10 x 8 x 11 x 9 x 12 x 5 x F. Pecora / Örebro University aass.oru.se 19 / 49

49 Davis-Putnam-Logeman-Loveland Procedure: Example x 1 x 2 x 3 x 1 x 4 x 1 x 2 x 2 x 5 x 1 x 3 x 3 x 6 x 2 x 3 x 1 x 7 x 4 x 5 x 6 x 2 x 8 x 4 x 5 x 3 x 9 x 4 x 6 x 1 x 10 x 5 x 6 x 2 x 11 x 7 x 8 x 9 x 3 x 12 x 7 x 8 x 4 x 7 x 7 x 9 x 5 x 8 x 8 x 9 x 6 x 9 x 10 x 11 x 12 x 4 x 10 x 10 x 11 x 5 x 11 x 10 x 12 x 6 x 12 x 11 x 12 x 7 x 10 x 8 x 11 x 9 x 12 x 6 is a unit clause x 6 = x 5 x F. Pecora / Örebro University aass.oru.se 19 / 49

50 Davis-Putnam-Logeman-Loveland Procedure: Example x 1 x 2 x 3 x 1 x 4 x 1 x 2 x 2 x 5 x 1 x 3 x 3 x 6 x 2 x 3 x 1 x 7 x 4 x 5 x 6 x 2 x 8 x 4 x 5 x 3 x 9 x 4 x 6 x 1 x 10 x 5 x 6 x 2 x 11 x 7 x 8 x 9 x 3 x 12 x 7 x 8 x 4 x 7 x 7 x 9 x 5 x 8 x 8 x 9 x 6 x 9 x 10 x 11 x 12 x 4 x 10 x 10 x 11 x 5 x 11 x 10 x 12 x 6 x 12 x 11 x 12 x 7 x 10 x 8 x 11 x 9 x 12 x 8 is a unit clause x 8 = x 5 x F. Pecora / Örebro University aass.oru.se 19 / 49

51 Davis-Putnam-Logeman-Loveland Procedure: Example x 1 x 2 x 3 x 1 x 4 x 1 x 2 x 2 x 5 x 1 x 3 x 3 x 6 x 2 x 3 x 1 x 7 x 4 x 5 x 6 x 2 x 8 x 4 x 5 x 3 x 9 x 4 x 6 x 1 x 10 x 5 x 6 x 2 x 11 x 7 x 8 x 9 x 3 x 12 x 7 x 8 x 4 x 7 x 7 x 9 x 5 x 8 x 8 x 9 x 6 x 9 x 10 x 11 x 12 x 4 x 10 x 10 x 11 x 5 x 11 x 10 x 12 x 6 x 12 x 11 x 12 x 7 x 10 x 8 x 11 x 9 x 12 x 11 is a unit clause x 11 = x 5 x F. Pecora / Örebro University aass.oru.se 19 / 49

52 Davis-Putnam-Logeman-Loveland Procedure: Example x 1 x 2 x 3 x 1 x 4 x 1 x 2 x 2 x 5 x 1 x 3 x 3 x 6 x 2 x 3 x 1 x 7 x 4 x 5 x 6 x 2 x 8 x 4 x 5 x 3 x 9 x 4 x 6 x 1 x 10 x 5 x 6 x 2 x 11 x 7 x 8 x 9 x 3 x 12 x 7 x 8 x 4 x 7 x 7 x 9 x 5 x 8 x 8 x 9 x 6 x 9 x 10 x 11 x 12 x 4 x 10 x 10 x 11 x 5 x 11 x 10 x 12 x 6 x 12 x 11 x 12 x 7 x 10 x 8 x 11 x 9 x 12 x 9 is a unit clause x 9 = x 5 x F. Pecora / Örebro University aass.oru.se 19 / 49

53 Davis-Putnam-Logeman-Loveland Procedure: Example x 1 x 2 x 3 x 1 x 4 x 1 x 2 x 2 x 5 x 1 x 3 x 3 x 6 x 2 x 3 x 1 x 7 x 4 x 5 x 6 x 2 x 8 x 4 x 5 x 3 x 9 x 4 x 6 x 1 x 10 x 5 x 6 x 2 x 11 x 7 x 8 x 9 x 3 x 12 x 7 x 8 x 4 x 7 x 7 x 9 x 5 x 8 x 8 x 9 x 6 x 9 x 10 x 11 x 12 x 4 x 10 x 10 x 11 x 5 x 11 x 10 x 12 x 6 x 12 x 11 x 12 x 7 x 10 x 8 x 11 x 9 x 12 x 12 is a unit clause x 12 = x 5 x F. Pecora / Örebro University aass.oru.se 19 / 49

54 Davis-Putnam-Logeman-Loveland Procedure: Example x 1 x 2 x 3 x 1 x 4 x 1 x 2 x 2 x 5 x 1 x 3 x 3 x 6 x 2 x 3 x 1 x 7 x 4 x 5 x 6 x 2 x 8 x 4 x 5 x 3 x 9 x 4 x 6 x 1 x 10 x 5 x 6 x 2 x 11 x 7 x 8 x 9 x 3 x 12 x 7 x 8 x 4 x 7 x 7 x 9 x 5 x 8 x 8 x 9 x 6 x 9 x 10 x 11 x 12 x 4 x 10 x 10 x 11 x 5 x 11 x 10 x 12 x 6 x 12 x 11 x 12 x 7 x 10 x 8 x 11 x 9 x 12 empty clause! backtrack on x 5... x 5 x F. Pecora / Örebro University aass.oru.se 19 / 49

55 Davis-Putnam-Logeman-Loveland Procedure: Example x 1 x 2 x 3 1 x 3 x 3 x 6 x 2 x 3 1 x 7 x 4 x 5 x 6 x 2 x 8 x 4 x 5 x 3 x 9 x 4 x 6 1 x 10 x 5 x 6 x 2 x 11 x 7 x 8 x 9 x 3 x 12 x 7 x 8 x 4 x 7 x 7 x 9 x 5 x 8 x 8 x 9 x 6 x 9 x 10 x 11 x 12 x 4 x 10 x 10 x 11 x 5 x 11 x 10 x 12 x 6 x 12 x 11 x 12 x 7 x 10 x 8 x 11 x 9 x 12 x 5 x F. Pecora / Örebro University aass.oru.se 19 / 49

56 Conflict-Driven Backjumping and Clause Learning BCP detects conflicts when a clause becomes empty Chronological backtracking: go back to most recent branching and flip value of variable Notice that we can make a more informed choice on where to backtrack most recent decision is not necessarily what has caused the conflict can deduce the culprit when BCP finds a conflict Clause learning: learn clauses that will avoid repeating mistakes Conflict-driven backjumping: backtrack to the culprit 2015 F. Pecora / Örebro University aass.oru.se 20 / 49

57 Conflict-Driven Backjumping and Clause Learning x 1 x 4 x 1 x 3 x 8 x 1 x 8 x 12 x 2 x 11 x 7 x 3 x 9 x 7 x 8 x 9 x 7 x 8 x 10 x 7 x 10 x F. Pecora / Örebro University aass.oru.se 21 / 49

58 Conflict-Driven Backjumping and Clause Learning x 1 x 4 x 1 x 3 x 8 x 1 x 8 x 12 x 2 x 11 x 7 x 3 x 9 x 7 x 8 x 9 x 7 x 8 x 10 x 7 x 10 x 12 x 1 x 1 = x 1 = 2015 F. Pecora / Örebro University aass.oru.se 21 / 49

59 Conflict-Driven Backjumping and Clause Learning x 1 x 4 x 1 x 3 x 8 x 1 x 8 x 12 x 2 x 11 x 7 x 3 x 9 x 7 x 8 x 9 x 7 x 8 x 10 x 7 x 10 x 12 x 1 x 1 = x 4 = x 4 = x 1 = 2015 F. Pecora / Örebro University aass.oru.se 21 / 49

60 Conflict-Driven Backjumping and Clause Learning x 1 x 4 x 1 x 3 x 8 x 1 x 8 x 12 x 2 x 11 x 7 x 3 x 9 x 7 x 8 x 9 x 7 x 8 x 10 x 7 x 10 x 12 x 1 x 1 = x 4 = x 3 x 3 = x 4 = x 1 = x 3 = 2015 F. Pecora / Örebro University aass.oru.se 21 / 49

61 Conflict-Driven Backjumping and Clause Learning x 1 x 4 x 1 x 3 x 8 x 1 x 8 x 12 x 2 x 11 x 7 x 3 x 9 x 7 x 8 x 9 x 7 x 8 x 10 x 7 x 10 x 12 x 1 x 1 = x 4 = x 3 x 3 = x 8 = x 4 = x 1 = x 3 = x 8 = 2015 F. Pecora / Örebro University aass.oru.se 21 / 49

62 Conflict-Driven Backjumping and Clause Learning x 1 x 4 x 1 x 3 x 8 x 1 x 8 x 12 x 2 x 11 x 7 x 3 x 9 x 7 x 8 x 9 x 1 x 1 = x 4 = x 3 x 3 = x 8 = x 12 = x 7 x 8 x 10 x 7 x 10 x 12 x 4 = x 1 = x 3 = x 8 = x 12 = 2015 F. Pecora / Örebro University aass.oru.se 21 / 49

63 Conflict-Driven Backjumping and Clause Learning x 1 x 4 x 1 x 3 x 8 x 1 x 8 x 12 x 2 x 11 x 7 x 3 x 9 x 7 x 8 x 9 x 1 x 1 = x 4 = x 3 x 3 = x 8 = x 12 = x 7 x 8 x 10 x 7 x 10 x 12 x 4 = x 2 x 2 = x 1 = x 3 = x 2 = x 8 = x 12 = 2015 F. Pecora / Örebro University aass.oru.se 21 / 49

64 Conflict-Driven Backjumping and Clause Learning x 1 x 4 x 1 x 3 x 8 x 1 x 8 x 12 x 2 x 11 x 7 x 3 x 9 x 7 x 8 x 9 x 1 x 1 = x 4 = x 3 x 3 = x 8 = x 12 = x 7 x 8 x 10 x 7 x 10 x 12 x 4 = x 2 x 2 = x 11 = x 1 = x 3 = x 2 = x 11 = x 8 = x 12 = 2015 F. Pecora / Örebro University aass.oru.se 21 / 49

65 Conflict-Driven Backjumping and Clause Learning x 1 x 4 x 1 x 3 x 8 x 1 x 8 x 12 x 2 x 11 x 7 x 3 x 9 x 7 x 8 x 9 x 7 x 8 x 10 x 7 x 10 x 12 x 1 x 1 = x 4 = x 3 x 3 = x 8 = x 12 = x 4 = x 2 x 2 = x 11 = x 1 = x 2 = x 3 = x 7 = x 7 x 7 = x 11 = x 8 = x 12 = 2015 F. Pecora / Örebro University aass.oru.se 21 / 49

66 Conflict-Driven Backjumping and Clause Learning x 1 x 4 x 1 x 3 x 8 x 1 x 8 x 12 x 2 x 11 x 7 x 3 x 9 x 7 x 8 x 9 x 7 x 8 x 10 x 7 x 10 x 12 x 1 x 1 = x 4 = x 3 x 3 = x 8 = x 12 = x 4 = x 9 = x 2 x 2 = x 11 = x 1 = x 2 = x 3 = x 7 = x 9 = x 7 x 7 = x 9 =? x 11 = x 8 = x 12 = 2015 F. Pecora / Örebro University aass.oru.se 21 / 49

67 Conflict-Driven Backjumping and Clause Learning x 1 x 4 x 1 x 3 x 8 x 1 x 8 x 12 x 2 x 11 x 7 x 3 x 9 x 7 x 8 x 9 x 7 x 8 x 10 x 7 x 10 x 12 x 1 x 2 x 1 = x 4 = x 3 x 3 = x 8 = x 12 = x 2 = x 11 = x 4 = x 9 = x 1 = x 3 = x 7 = x 7 x 7 = x 9 =? x 2 = x 9 = x 11 = x 12 = x 8 = (x 3 x 7 x 8 ) implies conflict 2015 F. Pecora / Örebro University aass.oru.se 21 / 49

68 Conflict-Driven Backjumping and Clause Learning x 1 x 4 x 1 x 3 x 8 x 1 x 8 x 12 x 2 x 11 x 7 x 3 x 9 x 7 x 8 x 9 x 7 x 8 x 10 x 7 x 10 x 12 x 1 x 1 = x 4 = x 3 x 3 = x 8 = x 12 = x 3 x 7 x 8 x 2 x 2 = x 11 = x 4 = x 9 = x 1 = x 2 = x 3 = x 7 = x 9 = x 7 x 7 = x 9 =? x 11 = x 12 = x 8 = learn clause ( x 3 x 7 x 8 ) 2015 F. Pecora / Örebro University aass.oru.se 21 / 49

69 Conflict-Driven Backjumping and Clause Learning x 1 x 4 x 1 x 3 x 8 x 1 x 8 x 12 x 2 x 11 x 7 x 3 x 9 x 7 x 8 x 9 x 7 x 8 x 10 x 7 x 10 x 12 x 1 x 1 = x 4 = x 3 x 3 = x 8 = x 12 = x 3 x 7 x 8 x 2 x 4 = x 1 = x 3 = x 7 x 12 = x 8 = backjump to x F. Pecora / Örebro University aass.oru.se 21 / 49

70 Conflict-Driven Backjumping and Clause Learning Conflict-driven backjumping can significantly reduce the search space Conflict clauses never cease to be useful Clauses are useful in generating new conflict clauses BCP picks up more conflicts, therefore learns more All modern systematic SAT solvers employ these strategies real-world applications thousands, even millions of variables research direction changes towards more efficient implementations 2015 F. Pecora / Örebro University aass.oru.se 22 / 49

71 Performance in SAT Solvers As all systematic CSP strategies, DPLL alternates branching and propagation So performance depends on propagation and ordering heuristics BCP iteratively analyses all clauses at each clause, either the clause is eliminated (SAT), or a literal is eliminated, or a conflict is identified (UNSAT) BCP may have to cycle over clauses several times In general, all DPLL-based algorithms spend most of their time doing BCP (up to 90%!) Modern SAT solvers implement efficient BCP procedures and sophisticated variable ordering heuristics 2015 F. Pecora / Örebro University aass.oru.se 23 / 49

72 BCP and Variable Ordering in CHAFF CHAFF is an efficient implementation of DPLL developed at Princeton University [Moskewicz et al., 2001] Efficiency gains made from fast UP (watched literals) intelligent backtracking conflict resolution (clause learning) effective branching rule (VSIDS) Key point in CHAFF s BCP: avoid revisiting clauses until they can lead to conflict Key point in CHAFF s branching: branch over variables that have recently lead to conflicts 2015 F. Pecora / Örebro University aass.oru.se 24 / 49

73 BCP in CHAFF: Watched Literals A clause (x 1... x n ) in implied iff n 1 literals are During BCP, we wish to quickly find newly implied clauses We could keep a counter for each clause representing the number of -literals, and visit clauses with counter = n 1 However, there is no need to visit a clause when 1, 2,... n 2 literals are We would like to only visit a clause when its counter goes from n 2 to n 1 CHAFF approximates this with watched literals 2015 F. Pecora / Örebro University aass.oru.se 25 / 49

74 BCP in CHAFF: Watched Literals Every clause watches two literals if neither are, clause cannot be implied if one is, clause may be implied We only visit a clause when one of its two watched literals is assigned to When a watched literal is set to the clause is searched for a new unassigned literal if none exists, the other watched literal is forced to otherwise the new unassigned literal becomes watched 2015 F. Pecora / Örebro University aass.oru.se 26 / 49

75 BCP in CHAFF: Watched Literals Watched literals facilitate finding new implications excludes clauses that couldn t possibly lead to a new implication Watched literals require very little overhead during UP skip clauses where watched literals are untouched do nothing when a watched literal is set to simple update of new watched literals Watched literals require no overhead when backtracking the literals that are watched can stay the same, because they were unassigned anyway un-assignment can be done in constant time 2015 F. Pecora / Örebro University aass.oru.se 27 / 49

76 Variable Ordering in DPLL Which unassigned variable should DPLL branch on? dynamic variable ordering i.e., order of expansion changes during search An effective heuristic is not necessarily the one which leads to fewest branches different assignments yield different # BCP operations shorter sequence of assignments may lead to more BCP operations than a longer one Must somehow minimize consequences for BCP as well as keep a low overhead for bookeeping 2015 F. Pecora / Örebro University aass.oru.se 28 / 49

77 Variable Ordering in DPLL: DLIS A simple and intuitive strategy: DLIS (Dynamic Largest Individual Sum) choose the assignment that satisfies the most unsatisfied clauses However, considerable work is required to maintain the statistics necessary for this heuristic maintain counters for each clause must update rankings when counter(s) transition from 0 to 1 need to reverse process when backtracking Total effort is much more than BCP! 2015 F. Pecora / Örebro University aass.oru.se 29 / 49

78 Variable Ordering in CHAFF: VSIDS 1 Each variable in each polarity has a counter, initialized to 0 2 When a conflict clause is added, counter associated with each literal in the clause is incremented 3 The (unassigned) variable and polarity with the highest counter is chosen for branching ties are broken randomly 4 Periodically, all the counters are divided by a constant 2015 F. Pecora / Örebro University aass.oru.se 30 / 49

79 Variable Ordering in CHAFF: VSIDS VSIDS effectively leads DPLL to attempt to satisfy recent conflict clauses difficult problems generate many conflicts conflict clauses dominate the problem in terms of literal count step 4 (division by constant) favors information from literals that often provoke conflicts direct coupling of combinatorial core to search process VSIDS has very low overhead does not depend on state of variable assignments statistics updated only when new conflict is detected 2015 F. Pecora / Örebro University aass.oru.se 31 / 49

80 DPLL-Based Methods So Far First generation (1960s) DP, DPLL 10s of variables Second generation (1980s/1990s) POSIT, Tableau, [... ] 10s 100s of variables Third generation (mid-1990s) SATO, satz, grasp, [... ] 1k 10k of variables Fourth generation (2000s) Chaff, Berkmin, [... ] 10k 1m of variables 2015 F. Pecora / Örebro University aass.oru.se 32 / 49

81 DPLL-Based Methods So Far Is SAT solver technology keeping up with Moore s law? number of transistors doubles every 18 months number of variables reported in random 3-SAT experiments doubles every 3 or 4 years Getting better performance, but primarily due to computing power Do we need a paradigm shift? 2015 F. Pecora / Örebro University aass.oru.se 33 / 49

82 Local Search for SAT DPLL-based methods are complete Systematic methods are inevitably slower than local search aka repair-based methods However, local search based methods cannot prove unsatisfiability often useful, e.g., in verification, planning, F. Pecora / Örebro University aass.oru.se 34 / 49

83 Local Search for SAT: GSAT GSAT: hill-climbing for SAT Algorithm 3: GSAT(φ) : model for φ or failure for i=1 to maxrestarts do assignment random-assignment(φ) for j=1 to maxclimbs do if evaluate(assignment,φ) then return assignment return fail assignment best-successor(assignment,φ) 2015 F. Pecora / Örebro University aass.oru.se 35 / 49

84 Local Search for SAT: GSAT The best successor is the one that makes the most clauses true Algorithm 4: best-successor(assignment,φ) : an assignment maxsatclauses 0 best successors compute-successors(assignment) forall the successor successors do if (numsat get-sat-clauses(φ,successor)) > maxsatclauses then maxsatclauses numsat best successor return best 2015 F. Pecora / Örebro University aass.oru.se 36 / 49

85 SAT Problem Structure Structure of SAT instance provides complexity classification 2-SAT is polinomial k-sat with k > 2 is NP-Complete 2015 F. Pecora / Örebro University aass.oru.se 37 / 49

86 SAT Problem Structure Structure of SAT instance provides complexity classification 2-SAT is polinomial k-sat with k > 2 is NP-Complete Horn Clauses: a clause with at most one positive literal P Q... T U What does a Horn clause really say? 2015 F. Pecora / Örebro University aass.oru.se 37 / 49

87 SAT Problem Structure Structure of SAT instance provides complexity classification 2-SAT is polinomial k-sat with k > 2 is NP-Complete Horn Clauses: a clause with at most one positive literal P Q... T U What does a Horn clause really say? (P Q... T) U to prove U, prove P and Q and... T 2015 F. Pecora / Örebro University aass.oru.se 37 / 49

88 SAT Problem Structure Structure of SAT instance provides complexity classification 2-SAT is polinomial k-sat with k > 2 is NP-Complete Horn Clauses: a clause with at most one positive literal P Q... T U What does a Horn clause really say? (P Q... T) U to prove U, prove P and Q and... T Theorem: If φ consists only of Horn clauses, then φ can be proved to be SAT or UNSAT through unit propagation only HORNSAT is P-complete (i.e., one of the hardest or most expressive problems which is known to be computable in polynomial time) 2015 F. Pecora / Örebro University aass.oru.se 37 / 49

89 HORNSAT Solvable by Unit Propagation: Proof Assume φ has only Horn clauses, and UP has completed UP has removed all clauses that were satisfied, and removed corresponding UNSAT literals from remaining clauses Only clauses of two types remain (all Horn) P 1 P 2... P n Q P 1 P 2... P n Each remaining clause has at least two variables (otherwise UP would have eliminated it, or found a conflict) We can satisfy all remaining clauses by assigning remaining variables to because each remaining clause has at least one negated variable (QED) 2015 F. Pecora / Örebro University aass.oru.se 38 / 49

90 Horn Theories and P? = NP 3-SAT is NP-complete Can we transform a 3-SAT formula into an equisatisfiable formula containing only Horn clauses? 2015 F. Pecora / Örebro University aass.oru.se 39 / 49

91 Horn Theories and P? = NP 3-SAT is NP-complete Can we transform a 3-SAT formula into an equisatisfiable formula containing only Horn clauses? Yes 2015 F. Pecora / Örebro University aass.oru.se 39 / 49

92 Horn Theories and P? = NP 3-SAT is NP-complete Can we transform a 3-SAT formula into an equisatisfiable formula containing only Horn clauses? Yes Does this mean that P = NP? 2015 F. Pecora / Örebro University aass.oru.se 39 / 49

93 Horn Theories and P? = NP 3-SAT is NP-complete Can we transform a 3-SAT formula into an equisatisfiable formula containing only Horn clauses? Yes Does this mean that P = NP? No! Why? 2015 F. Pecora / Örebro University aass.oru.se 39 / 49

94 Horn Theories and P? = NP 3-SAT is NP-complete Can we transform a 3-SAT formula into an equisatisfiable formula containing only Horn clauses? Yes Does this mean that P = NP? No! Why? Transformation to HORNSAT takes exponential time Resulting formula is exponentially longer that the original 2015 F. Pecora / Örebro University aass.oru.se 39 / 49

95 (TANSTAAFL) There Ain t No Such Thing As A Free Lunch 2015 F. Pecora / Örebro University aass.oru.se 40 / 49

96 SAT Problem Structure Structure of SAT instance provides complexity classification 2-SAT is polinomial k-sat with k > 2 is NP-Complete HORNSAT is P-Complete More subtle structural charateristics also predict cost of search Random k-sat instances uniformly sampled from space of all possible k-clauses n variables, l clauses meaningful parameter: ratio l/n of clauses to variables This is true also for general CSPs 2015 F. Pecora / Örebro University aass.oru.se 41 / 49

97 SAT Problem Structure: Phase Transition computational cost SAT UNSAT Hard instances are around l/n = 4.3 Complexity peak occurs around solubility transition l/n < 4.3: problems underconstrained and SAT l/n > 4.3: problems overconstrained and UNSAT l/n 4.3: difficult to prove SAT or UNSAT ratio of clauses to variables l/n 2015 F. Pecora / Örebro University aass.oru.se 42 / 49

98 SAT Problem Structure: Phase Transition median computational cost ratio of clauses to variables l/n Inrease problem size by increasing n Complexity peak invariant of problem size Complexity peak invariant of solving procedure systematic procedures (e.g., DPLL) local search procedures (e.g., GSAT) 2015 F. Pecora / Örebro University aass.oru.se 43 / 49

99 SAT Problem Structure: Phase Transition Why is the phase transition interesting? tells us that all algorithms have similar performance on random instances l/n predicts difficult random problems (and SAT/UNSAT) Phase transition behavior dissapears when problems are more structured e.g., in real-world problems, l/n doesn t tell us anything! The easiest type of test for your new algorithm is a random benchmark however, it is important to test your algorithm on real instances 2015 F. Pecora / Örebro University aass.oru.se 44 / 49

100 Random vs. Real-World Problems The real world is not random Real world problems have other structural features (in addition to random components) real graphs tend to be sparse (random graphs could be both) real graphs tend to have short path lengths (like random graphs) real graphs tend to be clustered (unlike sparse random graphs) Small world graphs: typical of social networks sparse, clustered, short path lengths six degrees of separation (shown to apply to actors DB, neural net of a worm,... ) 2015 F. Pecora / Örebro University aass.oru.se 45 / 49

101 Random vs. Real-World Problems Real-world problems have strong differences with random problems A compelling example [T. Walsh, 2000]: 1994 exam timetable at Edinburgh University 59 nodes, 594 edges (relatively sparse) but contains 10-clique! clique totally dominates search cost Less than chance of a 10-clique in a random graph with same size and density 2015 F. Pecora / Örebro University aass.oru.se 46 / 49

102 Summary Boolean Satisfiability (SAT) is a particular case of constraint reasoning As such, techniques seen for CSPs apply backtracking search, propagation, variable/value ordering SAT is convenient due to its simple formulation SAT solvers have become the workhorses of AI Today, complete algorithms in SAT are as performant as local search algorithms (millions of variables) are we approaching the limit of what we can do? Phase transition behavior for random k-sat predicts difficulty and solubility behavior dissapears in real world problems 2015 F. Pecora / Örebro University aass.oru.se 47 / 49

103 Part 5 Boolean Satisfiability Thank you! 2015 F. Pecora / Örebro University aass.oru.se 48 / 49

104 References Cook, S. A. (1971). The complexity of theorem-proving procedures. In Proceedings of the Third Annual ACM Symposium on Theory of Computing, STOC 71, pages , New York, NY, USA. ACM. Davis, M., Logemann, G., and Loveland, D. (1962). A machine program for theorem-proving. Commun. ACM, 5(7): Davis, M. and Putnam, H. (1960). A computing procedure for quantification theory. J. ACM, 7(3): Garey, M. R. and Johnson, D. S. (1979). Computers and Intractability: A Guide to the Theory of NP-Completeness. W. H. Freeman & Co., New York, NY, USA. Moskewicz, M., Madigan, C., Zhao, Y., Zhang, L., and Malik, S. (2001). Chaff: Engineering an Efficient SAT Solver. In Proceedings of the 38th Design Automation Conference (DAC 01) F. Pecora / Örebro University aass.oru.se 49 / 49

105 Acknowledgements Portions of this course are inspired, compiled or taken from scientific presentations and teaching material by the following authors: Fahiem Bacchus, Amedeo Cesta, Gary Cottrell, Rina Dechter, Simone Fratini, Hector Geffner, Geoff Gordon, Russell Greiner, John Harrison, Joachim Hertzberg, David Kriegman, Sharad Malik, Dana Nau, Peter Norvig, Madhusudan Parthasarathy, Rhys Price Jones, Stuart Russel, Tuomas Sandholm, Stephen F. Smith, Padhraic Smyth, Paolo Traverso, Toby Walsh F. Pecora / Örebro University aass.oru.se 49 / 49