13th International Conference on Membrane Computing - MTA Sztaki

A new approach for solving SAT by P systems with active membranes
The SAT problem. Let X = {x 1 , x 2 , x 3 , . . .} be a recursively enumerable set
of propositional variables (variables, to be short), and, for every n ∈ N, where N
denotes the set of natural numbers, let X n := {x 1 , . . . , x n }. An interpretation of
the variables in X n (or just an interpretation if X n is clear from the context) is
a function I : X n → {true, false}.
The variables and their negations are called literals. A clause C is a disjunction
of finitely many pairwise different literals satisfying the condition that there
is no x ∈ X such that both x and ¯x occur in C, where ¯x denotes the negation
of x. The set of all clauses over the variables in X n is denoted by C n . A formula
in conjunctive normal form (CNF) is a conjunction of finitely many clauses. We
will often treat formulas in CNF as finite sets of clauses, where the clauses are
finite sets of literals. A clause C ∈ C n is called a complete clause if, for every
x ∈ X n , x ∈ C or ¯x ∈ C.
Let ϕ be a formula in CNF over the variables in X n (n ∈ N) and let I
be an interpretation for ϕ. We say that I satisfies ϕ, denoted by I |= ϕ, if ϕ
evaluates to true under the truth assignment defined by I. Note that I |= ϕ if
and only if, for every C ∈ ϕ, I |= C. We say that ϕ is satisfiable, if there is
an interpretation I such that I |= ϕ. The SAT problem (boolean satisfiability
problem of propositional formulas in CNF) can be defined as follows. Given
a formula ϕ in CNF, decide whether or not there is an interpretation I such
that I |= ϕ. Let ϕ 1 and ϕ 2 be two formulas in CNF over the variables in X n
(n ∈ N). We say that ϕ 1 and ϕ 2 are equivalent, denoted by ϕ 1 ∼ ϕ 2 , if, for every
interpretation I, I |= ϕ 1 if and only if I |= ϕ 2 .
Let ϕ be a formula in CNF. The set of variables occurring in ϕ, denoted by
var(ϕ), is defined by var(ϕ) := {x ∈ X | ∃C ∈ ϕ : x ∈ C or ¯x ∈ C}. For a
clause C ∈ C n and a set Y ⊆ X n (n ∈ N) such that var(C) ∩ Y = ∅, let C Y be
the following set of clauses. Assume that Y = {x i1 , . . . , x ik } (k ≤ n, 1 ≤ i 1 <
. . . < i k ≤ n). Then let C Y := {C ∪ {l 1 , . . . , l k } | 1 ≤ j ≤ k : l j ∈ {x ij , ¯x ij }}.
For a formula ϕ = {C 1 , . . . , C m } in CNF over the variables in X n (m, n ∈ N),
let ϕ ′ := ⋃ C∈ϕ C Y , where Y := X n − var(C).
The correctness of the P systems that we are going to construct to solve SAT
is based on the following statement which can be proved by standard arguments
of propositional logic.
Proposition 1. For a formula ϕ = {C 1 , . . . , C m } in CNF over the variables in
X n (m, n ∈ N), ϕ ′ contains every full clause in C n if and only if ϕ is unsatisfiable.
Proof. Let ϕ := {C 1 , . . . , C m } be a formula in CNF over the variables in X n
(m, n ∈ N). We prove the above statement in two steps. First, we show that
ϕ ∼ ϕ ′ , then we show that ϕ ′ is unsatisfiable if and only if it contains every full
clause in C n .
To see that ϕ ∼ ϕ ′ we show that, for every interpretation I, I |= ϕ if
and only if I |= ϕ ′ . Let I be an interpretation and assume first that I |= ϕ.
Let C ∈ ϕ. Then I |= C and, for every C ′ ∈ C Y , where Y = X n − var(C),
var(C) ⊆ var(C ′ ). This clearly implies that, for every C ′ ∈ C Y , I |= C ′ . It
follows then that I |= ϕ ′ as well.
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