Decision Procedures for Satisfiability and Validity in Propositional Logic

Transcription

1 Decision Procedures for Satisfiability and Validity in Propositional Logic Meghdad Ghari Institute for Research in Fundamental Sciences (IPM) School of Mathematics-Isfahan Branch Logic Group Logic Short Course I, Computational Propositional Logic November 24, 2016 Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

2 Outline 1 Conjunctive Normal Form 2 Disjunctive Normal Form 3 Horn formulas 4 A linear SAT solver 5 Binary Decision Diagrams Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

3 Logical reasoning Is the following argument logically valid? A 1 A 2. A n B Artificial Language = Various Semantics Various Proof systems Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

4 Language Formulas of propositional logic are defined by the following grammar: Formula ::= Atomic Formula TRUE FALSE ( Formula) (Formula Formula) (Formula Formula) (Formula Formula). In Backus Naur form (BNF): φ ::= p ( φ) (φ φ) (φ φ) (φ φ), p Atom Atom denotes a non-empty (finite or infinite) set of atomic propositions (or propositional variables). Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

5 Language Formulas of propositional logic are defined by the following grammar: Formula ::= Atomic Formula TRUE FALSE ( Formula) (Formula Formula) (Formula Formula) (Formula Formula). In Backus Naur form (BNF): φ ::= p ( φ) (φ φ) (φ φ) (φ φ), p Atom Atom denotes a non-empty (finite or infinite) set of atomic propositions (or propositional variables). Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

6 Semantics Valuations: v : Atom {T, F} Valuations can be extended to the set of all formulas as follows: φ φ T F F T T F φ ψ φ ψ T T T T F F F T F F F F φ ψ φ ψ T T T T F T F T T F F F φ ψ φ ψ T T T T F F F T T F F T Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

7 Principle of compositionality The meaning of a complex expression is determined by the meanings of its constituent expressions and the rules used to combine them. v( ) = T v( ) = F v( φ) = 1 v(φ) v(φ ψ) = min(v(φ), v(ψ)) v(φ ψ) = max(v(φ), v(ψ)) v(φ ψ) = min(1, 1 v(φ) + v(ψ)) Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

8 Semantic consequence Definition If for all valuations in which all φ 1, φ 2,..., φ n evaluate to T, ψ evaluates to T as well, we say that φ 1, φ 2,..., φ n = ψ holds and call = the semantic entailment relation. Example p q, p = q Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

9 Syntactic consequence Definition If ψ is provable (in a proof system) from premises φ 1, φ 2,..., φ n, we say that the sequent φ 1, φ 2,..., φ n ψ is valid. Example p q, p q Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

10 Logical consequence Two approaches to logical consequence: Semantic: φ 1, φ 2,..., φ n = ψ. Syntactic: φ 1, φ 2,..., φ n ψ. Theorem (Soundness) A 1 A 2. A n B φ 1, φ 2,..., φ n ψ φ 1, φ 2,..., φ n = ψ Theorem (Completeness) φ 1, φ 2,..., φ n = ψ φ 1, φ 2,..., φ n ψ Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

11 Logic in Computer Science Two approaches: Semantic: M. Huth and M. Ryan, Logic in Computer Science modelling and reasoning about systems, Cambridge University Press, 2004 Syntactic: M. Fitting, First-Order Logic and Automated Theorem Proving, Springer, J. H. Gallier, Logic for Computer Science Foundations of Automatic Theorem Proving, John Wiley, Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

12 Satisfiability, validity, equivalence Definition (Satisfiable) Given a formula φ in propositional logic, we say that φ is satisfiable (has a model) if it has a valuation in which it evaluates to true. Definition (Valid) Given a formula φ in propositional logic, we say that φ is valid if under every valuation it evaluates to true. In that case we write = φ. Definition (Equivalence) Let φ and ψ be formulas of propositional logic. We say that φ and ψ are semantically equivalent iff φ = ψ and φ = ψ hold. In that case we write φ ψ. Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

13 Decision procedures Definition A decision problem is a question in some formal system with a yes-or-no answer, depending on the values of some input parameters. Definition A decision procedure is an algorithm that, given a decision problem, terminates with a correct yes/no answer. Definition A computer program that searches for a model for a propositional formula is called a SAT Solver. Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

14 Some applications of SAT solvers Test for the functional equivalence of two circuits. Identifying defects of integrated circuits. Model checking of hardware and software systems. Planning in artificial intelligence. Joao Marques-Silva, Practical Applications of Boolean Satisfiability, In Workshop on Discrete Event Systems (WODES), IEEE Press, Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

15 Boolean functions Definition A boolean variable x is a variable ranging over the values 0 and 1. A boolean function f of n arguments is a function from {0, 1} n to {0, 1}. Example 0 def def = 1 and 1 = 0; x y def = 1 if x = y = 1; otherwise x y def = 0; x + y def = 0 if x = y = 0; otherwise x + y def = 1. Various representations of boolean functions: Truth tables Propositional formulas. Binary decision diagrams. Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

16 Boolean functions Definition A boolean variable x is a variable ranging over the values 0 and 1. A boolean function f of n arguments is a function from {0, 1} n to {0, 1}. Example 0 def def = 1 and 1 = 0; x y def = 1 if x = y = 1; otherwise x y def = 0; x + y def = 0 if x = y = 0; otherwise x + y def = 1. Various representations of boolean functions: Truth tables Propositional formulas. Binary decision diagrams. Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

17 Truth tables Truth tables are very space-inefficient. Once you have computed a truth table, it is easy to see whether the boolean function represented is satisfiable or valid. Once you have computed a truth table, comparing whether two ordered truth tables represent the same boolean function also seems easy. Checking satisfiability of a function with n atoms requires of the order of 2 n operations if the function is represented as a truth table. We conclude that checking satisfiability, validity and equivalence is highly inefficient with the truth-table representation. Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

18 Propositional formulas Propositional formulas often provide a compact and efficient presentation of boolean functions. Deciding whether two arbitrary propositional formulas f and g denote the same boolean function is suspected to be exponentially expensive. Deciding whether an arbitrary propositional formula is satisfiable is a famous problem in computer science: No efficient algorithms for SAT are known, and it is strongly suspected that there aren t any. P? = NP NP? = co-np Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

19 Theorem Let φ be a formula of propositional logic. Then φ is satisfiable iff φ is not valid. Theorem Given formulas φ 1, φ 2,..., φ n and ψ of propositional logic, φ 1, φ 2,..., φ n = ψ iff = φ 1 (φ 2... (φ n ψ)) Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

20 Theorem Let φ be a formula of propositional logic. Then φ is satisfiable iff φ is not valid. Theorem Given formulas φ 1, φ 2,..., φ n and ψ of propositional logic, φ 1, φ 2,..., φ n = ψ iff = φ 1 (φ 2... (φ n ψ)) Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

21 Conjunctive Normal Form (CNF) L ::= p p, p Atom Literal D ::= L L D Clause C ::= D D C Conjunctive normal form Example ( q p r) ( p r) q. ( q p r q) ( p p). (p r) ( p r). (q p r) ( q p). Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

22 Conjunctive Normal Form (CNF) L ::= p p, p Atom Literal D ::= L L D Clause C ::= D D C Conjunctive normal form Example ( q p r) ( p r) q. ( q p r q) ( p p). (p r) ( p r). (q p r) ( q p). Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

23 Conjunctive Normal Form (CNF) L ::= p p, p Atom Literal D ::= L L D Clause C ::= D D C Conjunctive normal form Example ( q p r) ( p r) q. ( q p r q) ( p p). (p r) ( p r). (q p r) ( q p). Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

24 Conjunctive Normal Form (CNF) L ::= p p, p Atom Literal D ::= L L D Clause C ::= D D C Conjunctive normal form Example ( q p r) ( p r) q. ( q p r q) ( p p). (p r) ( p r). (q p r) ( q p). Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

25 Conjunctive Normal Form (CNF) L ::= p p, p Atom Literal D ::= L L D Clause C ::= D D C Conjunctive normal form Example ( q p r) ( p r) q. ( q p r q) ( p p). (p r) ( p r). (q p r) ( q p). Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

26 CNF and validity Theorem A disjunction of literals L 1 L 2... L m is valid iff there are 1 i, j m such that L i is L j. Example ( q p r q) ( p p) is valid, since both q p r q and p p are valid. Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

27 CNF and validity Theorem A disjunction of literals L 1 L 2... L m is valid iff there are 1 i, j m such that L i is L j. Example ( q p r q) ( p p) is valid, since both q p r q and p p are valid. Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

28 Converting formulas to CNF I Converting using truth tables p q? T T T T F F F T T F F F φ = ( p q) ( p q). Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

29 Converting formulas to CNF II Converting φ to CNF, using algorithm CNF(NNF(IMPL FREE(φ))). Step 1. φ ψ φ ψ Step 2. De Morgan rules: Step 3. Distributivity rules: (φ ψ) φ ψ (φ ψ) φ ψ φ (ψ σ) (φ ψ) (φ σ) φ (ψ σ) (φ ψ) (φ σ) Exercise Compute CNF(NNF(IMPL FREE((p q) p))). Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

30 Algorithm IMPL-FREE function IMPL FREE (φ): /* precondition: φ an arbitrary formula */ /* postcondition: IMPL FREE (φ) computes an implication-free equivalent for φ */ begin function case φ is a literal: return φ φ is φ 1 : return IMPL FREE(φ 1 ) φ is φ 1 φ 2 : return IMPL FREE (φ 1 ) IMPL FREE (φ 2 ) φ is φ 1 φ 2 : return IMPL FREE (φ 1 ) IMPL FREE (φ 2 ) φ is φ 1 φ 2 : return IMPL FREE (φ 1 ) IMPL FREE (φ 2 ) end case end function Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

31 Algorithm NNF function NNF (φ): /* precondition: φ is implication free */ /* postcondition: NNF (φ) computes a NNF for φ */ begin function case φ is a literal: return φ φ is φ 1 : return NNF (φ 1 ) φ is φ 1 φ 2 : return NNF (φ 1 ) NNF (φ 2 ) φ is φ 1 φ 2 : return NNF (φ 1 ) NNF (φ 2 ) φ is (φ 1 φ 2 ): return NNF ( φ 1 ) NNF ( φ 2 ) φ is (φ 1 φ 2 ): return NNF ( φ 1 ) NNF ( φ 2 ) end case end function Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

32 Algorithm DISTR function DISTR (η 1, η 2 ): /* precondition: η 1 and η 2 are in CNF */ /* postcondition: DISTR (η 1, η 2 ) computes a CNF for η 1 η 2 */ begin function case η 1 is η 11 η 12 : return DISTR (η 11, η 2 ) DISTR (η 12, η 2 ) η 2 is η 21 η 22 : return DISTR (η 1, η 21 ) DISTR (η 1, η 22 ) otherwise (= no conjunctions): return η 1 η 2 end case end function Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

33 Algorithm CNF function CNF (φ): /* precondition: φ implication free and in NNF */ /* postcondition: CNF (φ) computes an equivalent CNF for φ */ begin function case φ is a literal: return φ φ is φ 1 φ 2 : return CNF (φ 1 ) CNF (φ 2 ) φ is φ 1 φ 2 : return DISTR (CNF (φ 1 ), CNF (φ 2 )) end case end function Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

34 Davis-Putnam Procedure function DP-sat(clause set S) repeat for each unit clause L in S do /* unit propagation */ delete from S every clause containing L delete L from every clause of S if S is empty return true else if S contains the empty clause return false until no changes occur in S /* splitting */ choose a literal L in S if DP-sat(S {L}) return true else if DP-sat(S { L}) return true else return false. J. J. Lu, E. Rosenthal, Logic-Based Reasoning for Intelligent Systems, in Computer Science Handbook, Allen B. Tucker (Editor), Chapter 61, Chapman and Hall/CRC, Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

35 Davis-Putnam Procedure function DP-sat(clause set S) repeat for each unit clause L in S do /* unit propagation */ delete from S every clause containing L delete L from every clause of S if S is empty return true else if S contains the empty clause return false until no changes occur in S /* splitting */ choose a literal L in S if DP-sat(S {L}) return true else if DP-sat(S { L}) return true else return false. Example: clause set S = {p q, p q, q r, q r} Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

36 Davis-Putnam Procedure function DP-sat(clause set S) repeat for each unit clause L in S do /* unit propagation */ delete from S every clause containing L delete L from every clause of S if S is empty return true else if S contains the empty clause return false until no changes occur in S /* splitting */ choose a literal L in S if DP-sat(S {L}) return true else if DP-sat(S { L}) return true else return false. Splitting: {p q, p q, q r, q r, p} Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

37 Davis-Putnam Procedure function DP-sat(clause set S) repeat for each unit clause L in S do /* unit propagation */ delete from S every clause containing L delete L from every clause of S if S is empty return true else if S contains the empty clause return false until no changes occur in S /* splitting */ choose a literal L in S if DP-sat(S {L}) return true else if DP-sat(S { L}) return true else return false. Propagation: {p q, p q, q r, q r, p} p {q, q r, q r} Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

38 Davis-Putnam Procedure function DP-sat(clause set S) repeat for each unit clause L in S do /* unit propagation */ delete from S every clause containing L delete L from every clause of S if S is empty return true else if S contains the empty clause return false until no changes occur in S /* splitting */ choose a literal L in S if DP-sat(S {L}) return true else if DP-sat(S { L}) return true else return false. Propagation:{q, q r, q r} q { r} Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

39 Davis-Putnam Procedure function DP-sat(clause set S) repeat for each unit clause L in S do /* unit propagation */ delete from S every clause containing L delete L from every clause of S if S is empty return true else if S contains the empty clause return false until no changes occur in S /* splitting */ choose a literal L in S if DP-sat(S {L}) return true else if DP-sat(S { L}) return true else return false. Propagation: { r} r {} Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

40 Davis-Putnam Procedure function DP-sat(clause set S) repeat for each unit clause L in S do /* unit propagation */ delete from S every clause containing L delete L from every clause of S if S is empty return true else if S contains the empty clause return false until no changes occur in S /* splitting */ choose a literal L in S if DP-sat(S {L}) return true else if DP-sat(S { L}) return true else return false. Model: {p, q, r} v(p) = v(q) = T, v(r) = F. Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

41 Disjunctive Normal Form (DNF) L ::= p p, p Atom Literal D ::= L L D C ::= D D C Disjunctive normal form Example ( q p r) ( p r) q. Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

42 Disjunctive Normal Form (DNF) L ::= p p, p Atom Literal D ::= L L D C ::= D D C Disjunctive normal form Example ( q p r) ( p r) q. Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

43 DNF and satisfiability Theorem A conjunction of literals L 1 L 2... L m is satisfiable iff there are no 1 i, j m such that L i is L j. Example ( q p r) ( r r) is satisfiable, since q p r is satisfiable. Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

44 DNF and satisfiability Theorem A conjunction of literals L 1 L 2... L m is satisfiable iff there are no 1 i, j m such that L i is L j. Example ( q p r) ( r r) is satisfiable, since q p r is satisfiable. Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

45 Horn formulas P ::= p, A ::= P P A C ::= A P H ::= C C H p Atom Horn clause Horn formula Example (p q s ) (q r p) ( s). (q ) (p q). (p p) ( s r). ( q p). Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

46 Horn formulas P ::= p, A ::= P P A C ::= A P H ::= C C H p Atom Horn clause Horn formula Example (p q s ) (q r p) ( s). (q ) (p q). (p p) ( s r). ( q p). Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

47 Horn formulas P ::= p, A ::= P P A C ::= A P H ::= C C H p Atom Horn clause Horn formula Example (p q s ) (q r p) ( s). (q ) (p q). (p p) ( s r). ( q p). Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

48 Horn formulas P ::= p, A ::= P P A C ::= A P H ::= C C H p Atom Horn clause Horn formula Example (p q s ) (q r p) ( s). (q ) (p q). (p p) ( s r). ( q p). Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

49 Horn formulas P ::= p, A ::= P P A C ::= A P H ::= C C H p Atom Horn clause Horn formula Example (p q s ) (q r p) ( s). (q ) (p q). (p p) ( s r). ( q p). Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

50 Algorithm HORN function HORN (φ): /* precondition: φ is a Horn formula */ /* postcondition: HORN (φ) decides the satisfiability for φ */ begin function mark all occurrences of in φ; while there is a conjunct P 1 P 2 P ki P of φ such that all P j are marked but P isn t do mark P end while if is marked then return unsatisfiable else return satisfiable end function φ = (p q s ) (q r p) ( s) Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

51 Algorithm HORN function HORN (φ): /* precondition: φ is a Horn formula */ /* postcondition: HORN (φ) decides the satisfiability for φ */ begin function mark all occurrences of in φ; while there is a conjunct P 1 P 2 P ki P of φ such that all P j are marked but P isn t do mark P end while if is marked then return unsatisfiable else return satisfiable end function φ = (p q s ) (q r p) ( s) Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

52 Algorithm HORN function HORN (φ): /* precondition: φ is a Horn formula */ /* postcondition: HORN (φ) decides the satisfiability for φ */ begin function mark all occurrences of in φ; while there is a conjunct P 1 P 2 P ki P of φ such that all P j are marked but P isn t do mark P end while if is marked then return unsatisfiable else return satisfiable end function φ = (p q s ) (q r p) ( s) Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

53 Algorithm HORN function HORN (φ): /* precondition: φ is a Horn formula */ /* postcondition: HORN (φ) decides the satisfiability for φ */ begin function mark all occurrences of in φ; while there is a conjunct P 1 P 2 P ki P of φ such that all P j are marked but P isn t do mark P end while if is marked then return unsatisfiable else return satisfiable end function φ = (p q s ) (q r p) ( s) is satisfiable. Assign T to all marked atoms and F to all unmarked atoms. Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

54 A linear SAT solver φ ::= p φ φ φ, p Atom The parse tree of formula p ( q p): Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

55 Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

56 Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

57 Binary Decision Diagrams (BDD) Binary decision tree for f (x, y) = x + y (NOR): x y f (x, y) Dashed line value 0. Solid line value 1. Binary decision tree = truth table Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

58 Some examples B 0. The BDD of : B 1. The BDD of : B x. The BDD of a variable x: Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

59 C1. Removal of duplicate terminals C1. If a BDD contains more than one terminal 0-node, then we redirect all edges which point to such a 0-node to just one of them. We proceed in the same way with terminal nodes labelled with 1. Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

60 C2. Removal of redundant tests C2. If both outgoing edges of a node n point to the same node m, then we eliminate that node n, sending all its incoming edges to m. Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

61 C3. Removal of duplicate non-terminals C3. If two distinct nodes n and m in the BDD are the roots of structurally identical subbdds, then we eliminate one of them, say m, and redirect all its incoming edges to the other one. Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

62 C3. Removal of duplicate non-terminals C3. If two distinct nodes n and m in the BDD are the roots of structurally identical subbdds, then we eliminate one of them, say m, and redirect all its incoming edges to the other one. Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

63 Ordered Binary Decision Diagrams (OBDD) A BDD which does not have an ordering of variables: Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

64 Definition A BDD is said to be reduced if none of the optimisations C1 C3 can be applied (i.e. no more reductions are possible). An ordered BDD (OBDD) is a BDD which has an ordering for some list of variables. Theorem The reduced OBDD representing a given boolean function f is unique. OBDDs have a canonical form, namely their unique reduced OBDD. Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

65 Definition A BDD is said to be reduced if none of the optimisations C1 C3 can be applied (i.e. no more reductions are possible). An ordered BDD (OBDD) is a BDD which has an ordering for some list of variables. Theorem The reduced OBDD representing a given boolean function f is unique. OBDDs have a canonical form, namely their unique reduced OBDD. Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

66 Definition A BDD is said to be reduced if none of the optimisations C1 C3 can be applied (i.e. no more reductions are possible). An ordered BDD (OBDD) is a BDD which has an ordering for some list of variables. Theorem The reduced OBDD representing a given boolean function f is unique. OBDDs have a canonical form, namely their unique reduced OBDD. Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

67 The importance of canonical form OBDDs allow compact representations of certain classes of boolean functions which only have exponential representations in other systems, such as truth tables and conjunctive normal forms. Example. Consider the even parity function f even (x 1, x 2,..., x n ) which is defined to be 1 if there is an even number of variables x i with value 1; otherwise, it is defined to be 0. Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

68 Absence of redundant variables If the value of the boolean function f (x 1, x 2,..., x n ) does not depend on the value of x i, then any reduced OBDD which represents f does not contain any x i -node. Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

69 Test for semantic equivalence f and g denote the same boolean functions iff the reduced OBDDs have identical structure. Example. The reduced OBDD for both boolean functions f (x, y) = x + y and g(x, y) = x ȳ. Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

70 Test for validity The function f is valid iff its reduced OBDD is B 1. A BDD represents a valid function if no 0-terminal node is reachable from the root along a consistent path in a BDD which represents it. (A consistent path is one which, for every variable, has only dashed lines or only solid lines leaving nodes labelled by that variable.) Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

71 Test for satisfiability The function f is satisfiable iff its reduced OBDD is not B 0. A BDD represents a satisfiable function if a 1-terminal node is reachable from the root along a consistent path in a BDD which represents it. Example. An OBDD for boolean function f (x 1, x 2,..., x 6 ) = (x 1 + x 2 ) (x 3 + x 4 ) (x 5 + x 6 ). Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

72 Test for implication We can test whether f (x 1, x 2,..., x n ) implies g(x 1, x 2,..., x n ) (i.e. whenever f computes 1, then so does g) by computing the reduced OBDD for f ḡ. This is B 0 iff the implication holds. Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57

73 Comparing efficiency of five representations of boolean functions Representation of test for boolean operations boolean functions compact SAT validity + Prop. formulas often hard hard easy easy easy Formulas in DNF sometimes easy hard hard easy hard Formulas in CNF sometimes hard easy easy hard hard Ordered truth tables never hard hard hard hard hard Reduced OBDDs often easy easy medium medium easy Meghdad Ghari (IPM) SAT Solvers Logic Short Course / 57