Clusters of repetition roots: single chains (Algebraic system, Logic, Language and Related Areas in Computer Sciences II)

This work proposes a new approach towards solving an over 20 years old conjecture regarding the maximum number of distinct squares that a word can contain. To this end we look at clusters of repetition roots, that is, the set of positions where the root u of a repetition u^[l] occurs. We lay the foundation of this theory by proving basic properties of these clusters and establishing upper bounds on the number of distinct squares when their roots form a chain with respect to the prefix order

Sweep Complexity Revisited

We study the sweep complexity of DFA in one-way jumping mode answering several questions posed earlier. This measure is the number of times in the worst case that such machines have to return to the beginning of their input after having skipped some of the symbols. The class of languages accepted by these machines strictly includes the regular class and constant sweep complexity allows exactly the acceptance of regular languages. However, we show that there exist machines with higher than constant complexity still only accepting regular languages and that in general the sweep complexity of an automaton does not distinguish between accepting regular and non-regular languages. We establish separation results for asymptotic classes defined by this complexity measure and give a surprising exponential/logarithmic relation between factors of certain inputs which can be verified by such machines.Comment: 12 pages, 8 figure

Counting distinct squares in partial words

A well known result of Fraenkel and Simpson states that the number of distinct squares in a word of length n is bounded by 2n since at each position there are at most two distinct squares whose last occurrence start. In this paper, we investigate the problem of counting distinct squares in partial words, or sequences over a finite alphabet that may have some "do not know" symbols or "holes" (a (full) word is just a partial word without holes). A square in a partial word over a given alphabet has the form uu' where u is compatible with u, and consequently, such square is compatible with a number of full words over the alphabet that are squares. We consider the number of distinct full squares compatible with factors in a partial word with h holes of length n over a k-letter alphabet, and show that this number increases polynomially with respect to k in contrast with full words, and give bounds in a number of cases. For partial words with one hole, it turns out that there may be more than two squares that have their last occurrence starting at the same position. We prove that if such is the case, then the hole is in the shortest square. We also construct a partial word with one hole over a k-letter alphabet that has more than k squares whose last occurrence start at position zero

Efficient Computation of Descriptive Patterns

A pattern $\alpha$ is a word consisting of constants and variables and the pattern language $L(\alpha)$ (over an alphabet $\Sigma$) is the set of all words that can be obtained from $\alpha$ by uniformly replacing the variables with words over $\Sigma$. We investigate the problem of computing a pattern $\alpha$ that is descriptive of a given finite set $S \subseteq \Sigma^*$ of words, i.\,e., $S \subseteq L(\alpha)$ and there is no other pattern $\beta$ with $S \subseteq L(\beta) \subset L(\alpha)$. A pattern $\alpha$ that is descriptive of a set $S$ represents the structural commonalities of the words in $S$ and, thus, can serve as a classifier with respect to this structure. Furthermore, (polynomial time) computability of descriptive patterns is sufficient for (polynomial time) inductive inference of pattern languages. We investigate the complexity of computing descriptive patterns and, for subclasses of patterns, we present efficient algorithms for computing them

Avoiding abelian squares in partial words

AbstractErdős raised the question whether there exist infinite abelian square-free words over a given alphabet, that is, words in which no two adjacent subwords are permutations of each other. It can easily be checked that no such word exists over a three-letter alphabet. However, infinite abelian square-free words have been constructed over alphabets of sizes as small as four. In this paper, we investigate the problem of avoiding abelian squares in partial words, or sequences that may contain some holes. In particular, we give lower and upper bounds for the number of letters needed to construct infinite abelian square-free partial words with finitely or infinitely many holes. Several of our constructions are based on iterating morphisms. In the case of one hole, we prove that the minimal alphabet size is four, while in the case of more than one hole, we prove that it is five. We also investigate the number of partial words of length n with a fixed number of holes over a five-letter alphabet that avoid abelian squares and show that this number grows exponentially with n

Counting Maximal-Exponent Factors in Words

This article shows tight upper and lower bounds on the number of occurrences of maximal-exponent factors occurring in a word

5-Abelian cubes are avoidable on binary alphabets

A k-abelian cube is a word uvw, where the factors u, v, and w are either pairwise equal, or have the same multiplicities for every one of their factors of length at most k. Previously it has been shown that k-abelian cubes are avoidable over a binary alphabet for k ≥ 8. Here it is proved that this holds for k ≥ 5

A note on the number of squares in a partial word with one hole

A well known result of Fraenkel and Simpson states that the number of distinct squares in a word of length n is bounded by 2n since at each position there are at most two distinct squares whose last occurrence starts. In this paper, we investigate squares in partial words with one hole, or sequences over a finite alphabet that have a “do not know” symbol or “hole”. A square in a partial word over a given alphabet has the form uv where u is compatible with v, and consequently, such square is compatible with a number of words over the alphabet that are squares. Recently, it was shown that for partial words with one hole, there may be more than two squares that have their last occurrence starting at the same position. Here, we prove that if such is the case, then the length of the shortest square is at most half the length of the third shortest square. As a result, we show that the number of distinct squares compatible with factors of a partial word with one hole of length n is bounded by $\frac{7n}{2}$