# Construction Of A Rich Word Containing Given Two Factors

## TL;DR

The paper addresses the open problem of determining whether two given rich words can be factors of a larger rich word, providing an explicit bound on the length of such a word for decision purposes.

## Contribution

It establishes a constructive bound on the length of a rich word containing two given rich factors, enabling a finite check for their coexistence within a larger rich word.

## Key findings

- Provides a bound on the length of a rich word containing two given rich factors.
- Shows it is sufficient to check all rich words up to a certain length to decide factor inclusion.
- Addresses an open problem in the combinatorics of rich words.

## Abstract

A finite word $w$ with $\vert w\vert=n$ contains at most $n+1$ distinct palindromic factors. If the bound $n+1$ is attained, the word $w$ is called \emph{rich}. Let $\Factor(w)$ be the set of factors of the word $w$. It is known that there are pairs of rich words that cannot be factors of a common rich word. However it is an open question how to decide for a given pair of rich words $u,v$ if there is a rich word $w$ such that $\{u,v\}\subseteq \Factor(w)$. We present a response to this open question:\\ If $w_1, w_2,w$ are rich words, $m=\max{\{\vert w_1\vert,\vert w_2\vert\}}$, and $\{w_1,w_2\}\subseteq \Factor(w)$ then there exists also a rich word $\bar w$ such that $\{w_1,w_2\}\subseteq \Factor(\bar w)$ and $\vert \bar w\vert\leq m2^{k(m)+2}$, where $k(m)=(q+1)m^2(4q^{10}m)^{\log_2{m}}$ and $q$ is the size of the alphabet. Hence it is enough to check all rich words of length equal or lower to $m2^{k(m)+2}$ in order to decide if there is a rich word containing factors $w_1,w_2$.