1 Overview

Satisfiability,the firstNP-completeproblem, ¹ isaclassicproblemin constraintsatisfaction. Inthis lecture, we describe complete and incomplete algorithms designed to solve satisfiability. Given a formula of propositional logic, complete methods, such as Davis–Putnam, are guaranteed to find a satisfying assignment, if one exists; on the other hand, incomplete methods such as GSAT and WalkSAT, which are based on local search techniques, need not return a satisfying assignment even if one exists, but perform surprisingly well in practice.

2 Problem Definition

Let A = {x1 ,...,xn} be an alphabet of propositional variables. Elements of this alphabet are called positive literals. For each positive literal x ∈A, there exists the corresponding negative literal ¬x. A set ofliterals,bothpositive and negative,forms a clause, and a formula consists of a set of clauses. By convention, a clause is interpreted as the disjunction of its literals, whereas a formula is interpreted as the conjunction of its clauses. Formulas expressed in this way are said to be in conjunctive normal form.

A truth assignment is a function v : A→{t, f}. Atruth assignment overpropositional variables can be extended to a truth assignment over literals as follows: if v(x)= t, then v(¬x)= f; otherwise, if v(x)= f, then v(¬x)= t. Given a truth assignment, a literal is said to be satisfied iff it is assigned the value t. A conjunction is satisfied iff all its conjuncts are satisfied. A disjunction is satisfied iff at least one of its disjuncts are satisfied. Given a CNF formula φ of propositional logic,the satisfiabilityproblem(SAT) canbe stated asfollows:

¹ An NP-complete problem is a decision problem for which a solution can be verified in polynomial time, but there is no known efficient means of generating such a solution in the first place.

“does there exist a truth assignment under which φ is satisfied?”

If such an assignment exists, then φ is satisfiable. Otherwise, φ is unsatisfiable.

Remark: An arbitrary formula ofpropositionallogic constructed using the connectives ¬, ∧, ∨, → , ←, ↔ and an alphabet of propositional variables can be systematically converted into a CNF formula.

Example: The negation of the formula

(A → (B → C))→ ((A → B)→ (A → C))

can be converted into CNF as follows:

¬((A → (B → C))→ ((A → B)→ (A → C))) iff ¬(¬(¬A ∨ (¬B ∨ C))∨ (¬(¬A ∨ B)∨ (¬A ∨ C))) iff (¬A ∨ (¬B ∨ C))∧ ((¬A ∨ B)∧¬(¬A ∨ C)) iff (¬A ∨¬B ∨ C)∧ (¬A ∨ B)∧ A ∧¬C

The logical laws used in this derivation are as follows: (A → B)↔ (¬A ∨ B)(line 2); the law of double negation: ¬¬A ↔ A (line 3);and, DeMorgan’s law: ¬(A ∨ B)↔¬A ∧¬B (lines 3and 4).

3.1 Davis–Putnam

Given aformulaφ , theDavis-Putnam(DP)procedure(seeTable1)tackles the satisfiabilityproblem by attempting toapply theresolution ruletoeach variablethat appearsin φ in turn. DP simplifies φ down to φ ^′ , until either φ ^′ contains the empty clause, which is not satisfiable, implying that φ itself is not satisfiable; or φ ^′ can be simplified no further but does not contain the empty clause, implying that φ itself is satisfiable.

DP(φ)

Inputs CNF formula φ

Output satisfiable or unsatisfiable

Initialize N = number of propositional variables in φ

1. for i =1 to N

(a)

let x be the ith variable in φ

(b)

initialize the set of resolvents R = ∅

(c)

for all C1 ,C2 ∈ φ s.t. x ∈ C1 and ¬x ∈ C2

i. ψ = resolution(C1 ,C2 ): i.e., ψ = C1 \{x}∪ C2 \ {¬x}

ii. eliminate any redundancies that appear in ψ

iii. if ψ = {}, return unsatisfiable

iv. if ψ is not tautological, R = R ∪{ψ}

(d) Cx = {C ∈ φ | x ∈ C or ¬x ∈ C}

(e) φ = φ \ Cx ∪ R

2. return satisfiable

Table 1: Davis–Putnam.

In addition to the application of resolution, after resolving clauses of the form {x, y} and {¬x, y} on variable x, DP simplifies the resolvent {y, y} to {y}; and after resolving clauses of the form {x, ¬y} and {¬x, y} on variable x, DP omits the resolvent {¬y, y} since it is tautological.

A tautology is a formula whichis true under all truth assignments. Aformulais a tautology iff its negation is unsatisfiable. We can use Davis-Putnam to show that φ ≡ (A → (B → C))→ ((A → B)→ (A → C))is a tautology by determining that the negation of φ is unsatisfiable.

The negation of φ, in conjuctive normal form, is: {{¬A, ¬B, C}, {¬A, B}, {A}, {¬C}}. First select variable A. Resolving all clauses that contain A and ¬A yields: {{¬B, C}, {B}, {¬C}}. Next select variable B. Resolving all clauses that contain B and ¬B yields: {{C}, {¬C}}. Lastly, select variable C. Resolving all clauses that contain C and ¬C yields the empty clause. Therefore, the negation of φ is unsatisfiable: i.e., φ is a tautology.

3.2 Davis–Putnam–Logemann–Loveland

Unlike the originalDavis-Putnam algorithm, whichcannot construct a satisfying assignment,DPLL does return a satisfying assignment, if one exists. (In fact, withthe appropriate extensions, DPLL is capable of returning all satisfying assignments.) DPLL incorporates the following three ideas (whichspecialize forward checking in CSPs to satisfiability):

1. eliminate singleton clauses: Given singleton clause {l}, remove all clauses that contain the literal l, including the singleton clause itself, and remove all occurrences of the literal ¬l in other clauses. Moreover, if l = x, then let v(x)= t; else if l = ¬x, then let v(x)= f. (This step, which implements unit resolution, is called unit propagation.)

DPLL(φ, v)

1. if {} ∈ φ, fail
2. if φ = ∅, return v
1. if singleton clause {l}∈ φ, or, if pure literal l appears in φ
1. if l is positive, let v(l)= t if l is negative, let v(l)= f
2. return (DPLL (Reduce (φ, l),v))
3. choose unassigned variable x ∈ C for some C ∈ φ
4. return (DPLL (φ ∪ {{x}},v)∨ DPLL (φ ∪ {{¬x}},v))

Table 2: DPLL. If DPLL fails, then the formula φ is unsatisfiable.

1. eliminate pure literals:A pure literal is a literal that appears only in positive form or only in negative form; eliminate all clauses containing any pure literals. Moreover, if l = x, then let v(x)= t; else if l = ¬x, then let v(x)= f. (This stepis called purification. Clauses containing pure literals are easily satisfied, and thus can be eliminated.)
2. branch recursively: Choose some variable x ∈ C for some C ∈ φ. Now φ is satisfiable iff φ ∪ {{x}} is satisfiable or φ ∪ {{¬x}} is satisfiable. Checking that φ ∪ {{x}} is satisfiable amounts to setting v(x)= t, while checking that φ ∪ {{¬x}} is satisfiable amounts to setting v(x)= f. Note that this step in the algorithm is nondeterministic.

DPLL can be augmented with heuristics for selecting the next variable on which to branch, such as: select the variable that occurs most often; select the variable that induces the most unit propagations. Branching decisions can have tremendous impact on the size of the search space.

Reduce(φ, l) Inputs CNF formula φ, literal l Output reduced CNF formula φ ^′ Initialize = ∅

φ ′ for all C ∈ φ

1. if ¬l ∈ C, then φ ^′ = φ ^′ ∪{C \ {¬l}}

2. /* else ifl ∈ C, then φ ^′ = φ ^′ */ return φ ^′

Table 3: DPLL: Reduce Subroutine.

As an example, consider the formula φ = {{x, ¬y, ¬z}, {¬x, y, ¬z}, {¬x, ¬y, z}}, which contains no singleton clauses nor pure literals. To branch on the variable x, DPLL recurs on the formula φ ∪ {{x}}, which contains the singleton clause {x}. DPLL eliminates {x}, {x, ¬y, ¬z}, and any occurrences of theliteral ¬x. Next, DPLL checks the satisfiability of φ ^′ = {{y, ¬z}, {¬y, z}}, which contains no singleton clauses or pure literals. To branch on the variable y, DPLL recurs on the formula φ ^′ ∪ {{y}}, which contains the singleton clause {y}. DPLL eliminates {y}, {y, ¬z}, and any occurrences of the literal ¬y. Next, DPLL checks the satisfiability of φ ^′′ = {{z}}. Since φ ^′′ is easily satisfiable, DPLL returns the satisfying assignment v(x)= v(y)= v(z)= t.

Exercise: Trace DPLL on the formula φ = {{¬A, ¬B, C}, {¬A, B}, {A}, {¬C}}.