Complexity of Suffix-Free Regular Languages

Abstract

We study various complexity properties of suffix-free regular languages. The quotient complexity of a regular language L is the number of left quotients of L; this is the same as the state complexity of L, which is the number of states in a minimal deterministic finite automaton (DFA) accepting L. A regular language L ′ is a dialect of a regular language L if it differs only slightly from L (for example, the roles of the letters of L ′ are a permutation of the roles of the letters of L). The quotient complexity of an operation on regular languages is the maximal quotient complexity of the result of the operation expressed as a function of the quotient complexities of the operands. A sequence (L k , L k+1 , . . . ) of regular languages in some class C, where n is the quotient complexity of Ln, is called a stream. A stream is most complex in class C if its languages Ln meet the complexity upper bounds for all basic measures, namely, they meet the quotient complexity upper bounds for star and reversal; they have largest syntactic semigroups; they have the maximal numbers of atoms, each of which has maximal quotient complexity; and (possibly together with their dialects L ′ m ) they meet the quotient complexity upper bounds for boolean operations and product (concatenation). It is known that there exist such most complex streams in the class of regular languages, in the class of prefix-free languages, and also in the classes of right, left, and two-sided ideals. In contrast to this, we prove that there does not exist a most complex stream in the class of suffix-free regular languages. However, we do exhibit one ternary suffix-free stream that meets the bound for product and whose restrictions to binary alphabets meet the bounds for star and boolean operations. We also exhibit a quinary stream that meets the bounds for boolean operations, reversal, size of syntactic semigroup, and atom complexities. Moreover, we solve an open problem about the bound for the product of two languages of quotient complexities m and n in the binary case by showing that it can be met for infinitely many m and n. Two transition semigroups play an important role for suffix-free languages: semigroup T 5 (n) is a suffix-free semigroup that has maximal cardinality for 2 n 5, while semigroup T 6 (n) has maximal cardinality for n = 2, 3, and n 6. We prove that all witnesses meeting the bounds for the star and the second witness in a product must have transition semigroups in T 5 (n). On the other hand, witnesses meeting the bounds for reversal, size of syntactic semigroup, and the complexity of atoms must have semigroups in T 6 (n).

Complexity of bifix-free regular languages

“…Proof. We follow similarly to the proof from [9] for the class of suffix-free languages. If n − 1 ∈ S then A S would be empty, because the quotient of n − 1 is the empty language, and so will not form an atom.…”

“…Then we showed a most complex stream that meets all the upper bounds of all three complexity measures. Suffix-free [9] ≤ 3 and 5…”

“…It may be surprising that such a single witness exists for most of the natural subclasses of regular languages: the class of all regular languages [3], right-, left-, and two-sided ideals [4], and prefixconvex languages [7]. However, there does not exist a single witness for the class of suffix-free languages [9], where two different witnesses must be used.…”

Complexity of Left-Ideal, Suffix-Closed and Suffix-Free Regular Languages

“…By the restricted case, the states of Q ′ m × Q n are reachable and distinguishable using words in {a, b, d, e} * . Let R ∅ ′ = {(∅ ′ , q) | q ∈ Q n } and The complexity of suffix-free languages was studied in detail in [11,15,16,21,22,26]. For completeness we present a short summary of some of those results.…”

“…It is known that there is a most complex stream of left ideals that meets all the bounds in both the restricted [8,13] and unrestricted [13] cases, but a most complex suffix-free stream does not exist [15].…”