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**Automated Theorem Proving**

Lecture 2 Propositional Satisfiability

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**Decision procedures Boolean programs Arithmetic programs**

Propositional satisfiability Arithmetic programs Propositional satisfiability modulo theory of linear arithmetic Memory programs Propositional satisfiability modulo theory of linear arithmetic + arrays

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**Case I: Boolean programs**

Boolean-valued variables and boolean operations Formula := b | | b SymBoolConst

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**SAT First NP-complete problem (Cook 1972)**

Davis-Putnam algorithm (1960) resolution-based may use exponential memory Davis-Logemann-Loveland algorithm (1962) search-based basis for all successful modern solvers Conflict-driven learning and non-chronological backtracking (1996) resolution strikes back! Amazing progress GRASP, SATO, Chaff, ZChaff, BerkMin, …

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**Conjunctive Normal Form**

CNF Formula ::= c1 c2 … cm c Clause ::= l1 l2 … ln l Literal ::= b | b b SymBoolConst Unit clause ( l ) a clause containing a single literal Empty clause ( ) a clause containing no literal equivalent to false

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Conversion into CNF In general, converting into an equivalent CNF formula may result in an exponential blow-up We are only interested in satisfiability of Convert into an equi-satisfiable CNF formula EQCNF() is satisfiable iff EQCNF() is satisfiable size of EQCNF() is polynomial in size of

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**Conversion into CNF Convert formula into normal form NF()**

NF() is polynomial in Convert = NF() into equisatisfiable CNF formula EQCNF() EQCNF() is polynomial in

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**Normal Form Normal form: NF() Negated normal form: NNF() **

NF(b) = b NNF(b) = b NF() = NNF() NNF() = NF() NF(1 2) = NF(1) NF(1) NNF(1 2) = NNF(1) NNF(2)

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**Equi-satisfiable CNF Let be a formula in normal form.**

For each subformula of : - create a fresh symbol v in SymBoolConst Identify vb with b and vb with b Cl(b) = Cl(b) = true Cl() = Cl() Cl() (v v v) (v v) (v v) Cl() = Cl() Cl() (v v v) (v v) (v v) EQCNF() = v Cl()

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**Resolution (c1 b) (c2 b) (c1 c2) c1, c2 independent of b**

clauses (c1 b) (c2 b) (c1 c2) resolvent resolvent(b, c1 b, c2 b) = c1 c2 = b. (c1 b) (c2 b)

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** (c1 b) (c2 b) (c1 c2)**

Theorem (c1 b) (c2 b) iff (c1 b) (c2 b) (c1 c2) Adding the resolvent to the set of clauses does not affect the satisfiability of the clause set.

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**Unit resolution One of the clauses being resolved is a unit clause**

( b ) (c2 b) ( c2 ) ( b ) (c2 b) ( c2 ) Derivation of the empty clause (denoted by ) ( b ) ( b )

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**Davis-Putnam algorithm (I)**

Given clause set C: Rule 1: If a clause (c l l) C, replace it with (c l) Rule 2: If a clause (c b b) C, remove it from C Rule 3a: If b does not occur in any clause in C, remove every clause containing b from C Rule 3b: If b does not occur in any clause in C, remove every clause containing b from C

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**Davis-Putnam algorithm (II)**

Saturate C w.r.t Rules 1, 2, 3a, and 3b while (C is nonempty) { Pick a variable b appearing in some clause in C C’ = { resolvent(b,c1,c2) | c1,c2 C } Saturate C’ w.r.t. Rules 1, 2, 3a, and 3b if ( C’) return unsatisfiable C = C’ } return satisfiable

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**Satisfiable example (a b c) (b c f) (b c) Rule 3a**

(c c f) Resolve on b Rule 2 Clause set is empty

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**Unsatisfiable example**

(a b) (a b) (a c) (a c) Pick b ( a ) (a c) (a c) Pick a ( c ) ( c ) Pick c

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**Correctness Saturate C w.r.t Rules 1, 2, 3a, and 3b**

while (C is nonempty) { Pick a variable b appearing in some clause in C C’ = { resolvent(b,c1,c2) | c1,c2 C } Saturate C’ w.r.t. Rules 1, 2, 3a, and 3b if ( C’) return unsatisfiable C = C’ } return satisfiable Two observations: - Each of the rules 1, 2, 3a, and 3b preserve satisfiability - C’ = b. C

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**Memory explosion Saturate C w.r.t Rules 1, 2, 3a, and 3b**

while (C is nonempty) { Pick a variable b appearing in some clause in C C’ = { resolvent(b,c1,c2) | c1,c2 C } Saturate C’ w.r.t. Rules 1, 2, 3a, and 3b if ( C’) return unsatisfiable C = C’ } return satisfiable Let n be the number of clauses in the input clause set Number of clauses after i-th iteration of loop: O(n^(2^i))

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**Davis-Logemann-Loveland algorithm**

Slides of sat_course1.pdf Download from:

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**Davis-Logemann-Loveland algorithm**

Eliminates exponential memory requirement Might still need exponential time

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**Conflict-driven learning and non-chronological backtracking**

Slides 2-20 of sat_course2.pdf Download from:

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