13th International Conference on Membrane Computing - MTA Sztaki

On efficient algorithms for SAT
disjunction of literals. A conjunction of elementary disjunctions is a conjunctive
normal form (CNF, we are using also the terms CNF expression/CNF formula).
(For sake of simplicity we use brackets for all elementary disjunctions.) The SAT
problem is the following: given a propositional formula in conjunctive normal
form, decide whether it is satisfiable (or not). If the formula is unsatisfiable, then
it is equivalent to logical falsity. If the given formulae are in n-ary conjunctive
normal form, then the problem is known as the n-SAT problem.
It is well-known that the SAT and n-SAT problems (for n ≥ 3) are NPcomplete
(see [12, 29]).
A Boolean variable is called positive literal, while its negation is called negative
literal.
Let us fix an alphabet (if we want to speak about a language this is usually
the first step). Let F be a formula in CNF over the alphabet. If there is a satisfying
assignment to the variables such that F evaluates to true, then F is in the
SAT language (and vice-versa, if a word is in the language, then it is a satisfiable
formula in CNF form). We will also say that F is SAT expression/formula.
Similarly we define languages n-SAT, which contain only those formulae of the
SAT language in which each elementary disjunction contains exactly n literals
(n ∈ N). In this case F is also an n-SAT expression (or formula).
Note that there is a dual problem for the SAT problem. In the dual problem
the DNF is used. The problem to solve is to decide whether the given formula is
a tautology, (or not). We will refer to this form as the dual SAT-problem. The
dual of the SAT and n-SAT problems are also NP-complete problems (for n ≥ 3).
Actually, an NP-complete problem is to decide if a formula in CNF is satisfiable.
To decide if a formula in CNF is not satisfiable is co-NP-complete. Similarly, to
decide if a formula in DNF is tautology (logical law) is NP-complete. To decide
whether a formula in DNF is not tautology is co-NP-complete. The relation of NP
and co-NP is usually shown by these examples, the power of non-deterministic
computation is to have at least one computation that gives the result. However
to prove the opposite, i.e., the non-existence of such computation can have a
different complexity from the complexity of the original problem. (The class co-
NP also contains the class P. However, if P=NP, then NP=co-NP.) Here we just
want to mention one interesting moment: The problem to decide if a formula in
DNF is satisfiable or not, is almost trivial: the following linear algorithm solves
it.
1. Let i be 1.
2. Let us consider the ith clause. If there is no such a Boolean variable that
occurs both in positive and negative form (we say that a literal is contradictory
literal in a clause if both of its forms occur in the clause), then the formula is
satisfiable, a satisfying assignment is given by the elements of this clause: the
positive literals have true values, the negative literals have false values, thus the
formula is true independently of the values of the remaining Boolean variables.
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