1 research outputs found
Most Complex Deterministic Union-Free Regular Languages
A regular language is union-free if it can be represented by a regular
expression without the union operation. A union-free language is deterministic
if it can be accepted by a deterministic one-cycle-free-path finite automaton;
this is an automaton which has one final state and exactly one cycle-free path
from any state to the final state. Jir\'askov\'a and Masopust proved that the
state complexities of the basic operations reversal, star, product, and boolean
operations in deterministic union-free languages are exactly the same as those
in the class of all regular languages. To prove that the bounds are met they
used five types of automata, involving eight types of transformations of the
set of states of the automata. We show that for each there exists one
ternary witness of state complexity that meets the bound for reversal and
product. Moreover, the restrictions of this witness to binary alphabets meet
the bounds for star and boolean operations. We also show that the tight upper
bounds on the state complexity of binary operations that take arguments over
different alphabets are the same as those for arbitrary regular languages.
Furthermore, we prove that the maximal syntactic semigroup of a union-free
language has elements, as in the case of regular languages, and that the
maximal state complexities of atoms of union-free languages are the same as
those for regular languages. Finally, we prove that there exists a most complex
union-free language that meets the bounds for all these complexity measures.
Altogether this proves that the complexity measures above cannot distinguish
union-free languages from regular languages.Comment: 12 pages, 3 Figures. This version corrects an error in the proof of
Theorem 1 (7c). arXiv admin note: text overlap with arXiv:1701.0394