13th International Conference on Membrane Computing - MTA Sztaki

B. Nagy
Proposition 1. Every CNF formula is of the following regular form:
[(a + ¬a)(∨(a + ¬a)) ∗ ] (∧ [(a + ¬a)(∨(a + ¬a)) ∗ ]) ∗ .
Every n-ary CNF formula (for the n-SAT languages) is of the form:
[
(a + ¬a)(∨(a + ¬a))
n−1 ] ( ∧ [ (a + ¬a)(∨(a + ¬a)) n−1]) ∗
.
Using finite set of variables: {a 1 , a 2 , ..., a k }, we should substitute each occurrence
of the symbol a above (in each expression) with the regular expression (a 1 + a 2 +
... + a k ).
3.2 Deterministic Finite Automaton for the SAT Languages
In this section, we construct the following automata: an automaton which accepts
exactly the SAT-language and automata accepting the n-SAT languages (for any
fixed n).
Let C be the set of subsets of powerset 2 k , where k is the number of the
variables in the language. We will interpret the elements of C as the sets of the
values of the variables when the given logical expression is false. We use this
part in this construction to know when the longest prefix of the formula which
is syntactically correct CNF expression is not satisfied.
Let Y be the set of the possible states of a DFA A = (Y, T, M A , y 0 , {y f })
which accepts the syntactically correct CNF expressions.
Let D be the set of k + 1 dimensional vectors over {0, 1, 2}. This vector will
count which variables are in the new clause. 0 on the i-th place of a vector d ∈ D
means that the i-th variable is not (yet) in the clause currently being read. 1 and
2 on the i-th place mean the occurrence of the i-th variable without negation and
with negation, respectively. The value 1 on the (k + 1)-th element denotes that
there is a variable in the actual clause with both types of occurrences (positive
and negative).
The states of the automaton are given by the Cartesian product of the sets
C, Y and D, where Y refers to the CNF syntax; and the sets C and D hold the
semantical content, i.e., for which values of the variables the formula is false.
Let the initial state σ 0 = ({}, y 0 , 0), where {} is a value from C, y 0 is the
initial state of A, and 0 is the k + 1 dimensional nullvector containing only 0’s.
For the input alphabet T of the automaton, we use the same alphabet as at
the CNF expressions: {a 1 , a 2 , ..., a k , [, ] , ¬, ∧, ∨}.
Let the transition function be the following: ((c, y, d), t) → (c ′ , y ′ , d ′ )
– if t ∈ {∧, ∨, ¬, [}, then only the syntactical part will change: c ′ = c, d ′ = d
and y ′ is the corresponding state of A, i.e. y ′ = M A (y, t).
– if t is a variable, then c ′ = c, y ′ = M A (y, t) and we have the following cases
for calculating the value of d ′ :
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