Parameterized and exact algorithms for finding a read-once resolution refutation in 2CNF formulas

Abstract

In this paper, we discuss algorithms for the problem of finding read-once resolution refutations of unsatisfiable 2CNF formulas within the resolution refutation system. Broadly, a read-once resolution refutation is one in which each constraint (input or derived) is used at most once. Read-once resolution refutations have been widely studied in the literature for a number of constraint system-refutation system pairs. For instance, read-once resolution has been analyzed for boolean formulas in conjunctive normal form (CNF) and read-once cutting planes have been analyzed for polyhedral systems. By definition, read-once refutations are compact, and hence valuable in applications that place a premium on visualization. The satisfiability problem (SAT) is concerned with finding a satisfying assignment for a boolean formula in CNF. While SAT is NP-complete in general, there exist some interesting subclasses of CNF formulas, for which it is decidable in polynomial time. One such subclass is the class of 2CNF formulas, i.e., CNF formulas in which each clause has at most two literals. The existence of efficient algorithms for satisfiability checking in 2CNF formulas (2SAT), makes this class useful from the perspective of modeling selected logic programs. The work in this paper is concerned with the read-once refutability problem (under resolution) in this subclass. Although 2SAT is decidable in polynomial time, the problem of finding a read-once resolution refutation of an unsatisfiable 2CNF formula is NP-complete. We design non-trivial, parameterized and exact exponential algorithms for this problem. Additionally, we study the computational complexity of finding a shortest read-once resolution refutation of a 2CNF formula.

Appendices

Appendix A: Incompleteness of read-once resolution refutation

Example 1

Let Φ be the following 2CNF formula:

$$ \begin{array}{@{}rcl@{}} (x_{1} \vee x_{2}) & (x_{3} \vee x_{4}) & (\neg x_{1} \vee \neg x_{3})\\ (\neg x_{1} \vee \neg x_{4}) & (\neg x_{2} \vee \neg x_{3}) & (\neg x_{2} \vee \neg x_{4}) \end{array} $$

We now show that Φ does not have a read-once resolution refutation.

Assume for the sake of contradiction, that Φ has a read-once resolution refutation R. Observe that Φ has four variables x₁, x₂, x₃, and x₄. Thus, the last resolution step of any read-once resolution refutation of Φ must be one of (1) $(x_{1}) \wedge (\neg x_{1}) {\mid }\!\frac {1}{RES} \sqcup$, (2) $(x_{2}) \wedge (\neg x_{2}) {\mid }\!\frac {1}{RES} \sqcup$, (3) $(x_{3}) \wedge (\neg x_{3}) \mid \frac {1}{RES} \sqcup$, or, (4) $(x_{4}) \wedge (\neg x_{4}) {\mid }\!\frac {1}{RES} \sqcup$.

If the last resolution step of R is $(x_{1}) \wedge (\neg x_{1}) {\mid }\!\frac {1}{RES} \sqcup$, then R must derive both (x₁) and (¬x₁). Let us consider each derivation separately.

The only clause in Φ with the literal x₁ is (x₁ ∨ x₂). Thus, this clause must be used in the derivation of (x₁). Since the literal x₂ is not present in the clause (x₁), the derivation of (x₁) must use a clause with the literal ¬x₂. The only clauses in Φ with the literal ¬x₂ are (¬x₂ ∨¬x₃) and (¬x₂ ∨¬x₄). Thus, using the same argument as before, the derivation of (x₁) must use a clause with either the literal x₃ or the literal x₄. The only clause in Φ with either of these literals is (x₃ ∨ x₄). Thus, this clause must be used in the derivation of (x₁).

The only clauses in Φ with the literal ¬x₁ are (¬x₁ ∨¬x₃) and (¬x₁ ∨¬x₄). Thus, using the same argument as before, the derivation of (¬x₁) must use a clause with either the literal x₃ or the literal x₄. The only clause in Φ with either of these literals is (x₃ ∨ x₄). Thus, this clause must be used in the derivation of (¬x₁). Consequently, the clause (x₃ ∨ x₄) must be used both in the derivation of (x₁) and in the derivation of (¬x₁). This violates the assumption that R is a read-once resolution refutation.

If the last resolution step of R is $(x_{2}) \wedge (\neg x_{2}) {\mid }\!\frac {1}{RES} \sqcup$, then R must derive both (x₂) and (¬x₂). Let us consider each derivation separately.

The only clause in Φ with the literal x₂ is (x₁ ∨ x₂). Thus, this clause must be used in the derivation of (x₂). Since the literal x₁ is not present in the clause (x₂), the derivation of (x₂) must use a clause with the literal ¬x₁. The only clauses in Φ with the literal ¬x₁ are (¬x₁ ∨¬x₃) and (¬x₁ ∨¬x₄). Thus, using the same argument as before, the derivation of (x₂) must use a clause with either the literal x₃ or the literal x₄. The only clause in Φ with either of these literals is (x₃ ∨ x₄). Thus, this clause must be used in the derivation of (x₂).

The only clauses in Φ with the literal ¬x₂ are (¬x₂ ∨¬x₃) and (¬x₂ ∨¬x₄). Thus, using the same argument as before, the derivation of (¬x₂) must use a clause with either the literal x₃ or the literal x₄. The only clause in Φ with either of these literals is (x₃ ∨ x₄). Thus, this clause must be used in the derivation of (¬x₂). Consequently, the clause (x₃ ∨ x₄) must be used both in the derivation of (x₂) and in the derivation of (¬x₂). This violates the assumption that R is a read-once resolution refutation.

If the last resolution step of R is $(x_{3}) \wedge (\neg x_{3}) {\mid }\!\frac {1}{RES} \sqcup$, then R must derive both (x₃) and (¬x₃). Let us consider each derivation separately.

The only clause in Φ with the literal x₃ is (x₃ ∨ x₄). Thus, this clause must be used in the derivation of (x₃). Since the literal x₄ is not present in the clause (x₃), the derivation of (x₃) must use a clause with the literal ¬x₄. The only clauses in Φ with the literal ¬x₄ are (¬x₁ ∨¬x₄) and (¬x₂ ∨¬x₄). Thus, using the same argument as before, the derivation of (x₃) must use a clause with either the literal x₁ or the literal x₂. The only clause in Φ with either of these literals is (x₁ ∨ x₂). Thus, this clause must be used in the derivation of (x₃).

The only clauses in Φ with the literal ¬x₃ are (¬x₁ ∨¬x₃) and (¬x₂ ∨¬x₃). Thus, using the same argument as before, the derivation of (¬x₃) must use a clause with either the literal x₁ or the literal x₂. The only clause in Φ with either of these literals is (x₁ ∨ x₂). Thus, this clause must be used in the derivation of (¬x₃). Consequently, the clause (x₁ ∨ x₂) must be used both in the derivation of (x₃) and in the derivation of (¬x₃). This violates the assumption that R is a read-once resolution refutation.

If the last resolution step of R is $(x_{4}) \wedge (\neg x_{4}) {\mid }\!\frac {1}{RES} \sqcup$, then R must derive both (x₄) and (¬x₄). Let us consider each derivation separately.

The only clause in Φ with the literal x₄ is (x₃ ∨ x₄). Thus, this clause must be used in the derivation of (x₄). Since the literal x₃ is not present in the clause (x₄), the derivation of (x₄) must use a clause with the literal ¬x₃. The only clauses in Φ with the literal ¬x₃ are (¬x₁ ∨¬x₃) and (¬x₂ ∨¬x₃). Thus, using the same argument as before, the derivation of (x₄) must use a clause with either the literal x₁ or the literal x₂. The only clause in Φ with either of these literals is (x₁ ∨ x₂). Thus, this clause must be used in the derivation of (x₄).

The only clauses in Φ with the literal ¬x₄ are (¬x₁ ∨¬x₄) and (¬x₂ ∨¬x₄). Thus, using the same argument as before, the derivation of (¬x₄) must use a clause with either the literal x₁ or the literal x₂. The only clause in Φ with either of these literals is (x₁ ∨ x₂). Thus, this clause must be used in the derivation of (¬x₄). Consequently, the clause (x₁ ∨ x₂) must be used both in the derivation of (x₄) and in the derivation of (¬x₄). This violates the assumption that R is a read-once resolution refutation.

This means that R must either use the clause (x₃ ∨ x₄) twice or the clause (x₁ ∨ x₂) twice. Thus, Φ does not have a read- once resolution refutation.

Appendix : B: Approximation complexity classes

We begin by defining the complexity class NPO [32].

Definition 1

The complexity class NPO is the set of optimization problems such that:

• 1. The set of instances can be recognized in polynomial time.

• 2. Solutions are polynomially sized and can be verified in polynomial time.

• 3. The objective function can be computed in polynomial time.

We next define the complexity class NPO PB [23].

Definition 2

NPO PB is the set of NPO problems for which the value of the objective function is polynomial in the size of the input.

Finally, we introduce the notion of PTAS reductions [32].

Definition 3

A PTAS reduction from problem A to problem B, is a trio of functions f, g, and α computable in polynomial time, such that:

• 1. f maps instances of problem A to instances of problem B.

• 2. g takes an instance x of problem A, an approximate solution to the corresponding problem f(x) in B, and an error parameter ε and produces an approximate solution to x.

• 3. α maps error parameters for solutions to instances of problem A to error parameters for solutions to problem B.

• 4. If the solution y to f(x) (an instance of problem B) is at most (1 + α(ε)) times worse than the optimal solution, then the corresponding solution g(x,y,ε) to x (an instance of problem A) is at most (1 + ε) times worse than the optimal solution.

Definition 4

A problem P is NPO PB-hard under PTAS reductions, if every problem in NPO PB can be reduced to P by a PTAS reduction.

Unless otherwise stated, we assume that NPO PB-hardness is specified with respect to PTAS reductions.

The set of problems which are in the class NPO PB and are NPO PB-hard are called NPO PB-complete. Additionally, for every NPO PB-complete problem P there exists an ε > 0 such that P cannot be approximated to within a factor of O(n^ε) unless P = NP [5]. Thus, if any NPO PB-complete problem can be approximated to within a polylogarithmic factor, then P = NP.

An example of an NPO PB-complete problem is Bounded Minimum 0-1 Programming problem. This problem is formulated as follows:

Given an integer program A ⋅x ≥b, x ∈{0, 1}ⁿ, find the minimum value of 1 ⋅x. This specific form of Minimum 0-1 Programming is known to be NPO PB-complete [23].