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

On the Parikh-de-Bruijn grid

We introduce the Parikh-de-Bruijn grid, a graph whose vertices are fixed-order Parikh vectors, and whose edges are given by a simple shift operation. This graph gives structural insight into the nature of sets of Parikh vectors as well as that of the Parikh set of a given string. We show its utility by proving some results on Parikh-de-Bruijn strings, the abelian analog of de-Bruijn sequences.Comment: 18 pages, 3 figures, 1 tabl

Abelian repetitions in partial words

AbstractWe study abelian repetitions in partial words, or sequences that may contain some unknown positions or holes. First, we look at the avoidance of abelian pth powers in infinite partial words, where p>2, extending recent results regarding the case where p=2. We investigate, for a given p, the smallest alphabet size needed to construct an infinite partial word with finitely or infinitely many holes that avoids abelian pth powers. We construct in particular an infinite binary partial word with infinitely many holes that avoids 6th powers. Then we show, in a number of cases, that the number of abelian p-free partial words of length n with h holes over a given alphabet grows exponentially as n increases. Finally, we prove that we cannot avoid abelian pth powers under arbitrary insertion of holes in an infinite word

Avoiding abelian powers cyclically

We study a new notion of cyclic avoidance of abelian powers. A finite word $w$ avoids abelian $N$-powers cyclically if for each abelian $N$-power of period $m$ occurring in the infinite word $w^\omega$, we have $m \geq |w|$. Let $\mathcal{A}(k)$ be the least integer $N$ such that for all $n$ there exists a word of length $n$ over a $k$-letter alphabet that avoids abelian $N$-powers cyclically. Let $\mathcal{A}_\infty(k)$ be the least integer $N$ such that there exist arbitrarily long words over a $k$-letter alphabet that avoid abelian $N$-powers cyclically.We prove that $5 \leq \mathcal{A}(2) \leq 8$, $3 \leq \mathcal{A}(3) \leq 4$, $2 \leq \mathcal{A}(4) \leq 3$, and $\mathcal{A}(k) = 2$ for $k \geq 5$. Moreover, we show that $\mathcal{A}_\infty(2) = 4$, $\mathcal{A}_\infty(3) = 3$, and $\mathcal{A}_\infty(4) = 2$.</p