Equivalence of regular expressions over a partially commutative alphabet

Abstract

The equivalence problem is considered for regular expressions over a partially commutative alphabet. The alphabet is decomposed into disjoint subsets of noncommutative elements. The special case of the problem when the cardinal number of only one subset is larger than 1 and the cardinal numbers of the other subsets are equal to 1 is proved to be algorithmically solvable.The algorithmic decidability of the equivalence problem for regular expressions over a commutative alphabet is proved by V. N. Red'ko in 1964. It follows from the result obtained by V. A. Tuzov in 1971 that, in the general case, the equivalence problem for regular expressions over a partially commutative alphabet is algorithmically undecidable. A partially commutative alphabet is understood to be the union of disjoint subsets in which any two symbols from the same subset are not permutable and any two symbols from different subsets are permutable. In a definite sense, this alphabet is a generalization of a commutative alphabet that corresponds to the case when the cardinality of each subset is equal to unity. V. A. Tuzov showed that if there are at least two subsets of cardinality greater than one, then the equivalence problem is algorithmically undecidable. Thus, it is only one special case in which it was unknown whether the equivalence problem is decidable, namely, the case when the cardinality of one of subsets in a partially commutative alphabet is more than unity and the cardinality of all other subsets is equal to unity. It is proved in this article that, in this case, the equivalence problem for regular expressions is also algorithmically decidable.Let Y be a finite alphabet, and let Y Y n 1 , , ¼ be a partition of the alphabet Y with respect to Q into disjoint subsets. If Y is a set, then we denote the cardinality of this set by | | Y . Let R 1 and R 2 be regular expressions [1] in the alphabet Y . If, for each w R Î 1 , there is a word w R ¢Î 2 coincident with w to within the permutation of its adjacent symbols belonging to different partition classes of the alphabet Y and, on the contrary, for each word w R ¢Î 2 , there is a word w R Î 1 coincident with ¢ w to within the permutation of its adjacent symbols belonging to partition classes of the alphabet Y , then we call the regular expressions R 1 and R 2 Q-equivalent and denote by R R 1 2 ( ) Q .On the one hand, V. N. Redko showed [2] that if all partition classes of an alphabet are one-element, then the problem of Q-equivalence is algorithmically decidable. On the other hand, as follows from the result of V. A. Tuzov [3], if a partition of an alphabet contains at least two classes with the number of symbols that is not smaller than two, then the problem of Q-equivalence is reduced to the problem of equivalence of nondeterministic multitape automata [4] and, hence, it is algorithmically undecidable.In this paper, an "intermediate case" is considered when all partition classes of the alphabet Y except for one are one-element. This investigation is based on the use of some results of the ...