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

Abelian-Square-Rich Words

An abelian square is the concatenation of two words that are anagrams of one another. A word of length $n$ can contain at most $\Theta(n^2)$ distinct factors, and there exist words of length $n$ containing $\Theta(n^2)$ distinct abelian-square factors, that is, distinct factors that are abelian squares. This motivates us to study infinite words such that the number of distinct abelian-square factors of length $n$ grows quadratically with $n$. More precisely, we say that an infinite word $w$ is {\it abelian-square-rich} if, for every $n$, every factor of $w$ of length $n$ contains, on average, a number of distinct abelian-square factors that is quadratic in $n$; and {\it uniformly abelian-square-rich} if every factor of $w$ contains a number of distinct abelian-square factors that is proportional to the square of its length. Of course, if a word is uniformly abelian-square-rich, then it is abelian-square-rich, but we show that the converse is not true in general. We prove that the Thue-Morse word is uniformly abelian-square-rich and that the function counting the number of distinct abelian-square factors of length $2n$ of the Thue-Morse word is $2$-regular. As for Sturmian words, we prove that a Sturmian word $s_{\alpha}$ of angle $\alpha$ is uniformly abelian-square-rich if and only if the irrational $\alpha$ has bounded partial quotients, that is, if and only if $s_{\alpha}$ has bounded exponent.Comment: To appear in Theoretical Computer Science. Corrected a flaw in the proof of Proposition