There exist infinite cube-free words over any sequence of binary alphabets

TL;DR

The paper resolves the cube-avoidance problem for infinite words over sequences of binary alphabets by proving that for any such sequence there exists an infinite cube-free word respecting it. It delivers two independent, technically intricate approaches: a growth-based method using a regular-language approximation with $p=12$ yielding a $1.35^n$ lower bound on the number of length-$n$ cube-free words, and a computable construction offering a polynomial-time method to extend prefixes to full cube-free words for computable alphabet sequences. A crucial computer-assisted verification provides the eigenvector-like weights that drive both proofs, and the results are strengthened by an algorithmic perspective showing how to compute the next letter efficiently. Together, these contributions advance both the combinatorial and algorithmic understanding of cube-free words and their applicability to nonrepetitive colorings and list-coloring problems.

Abstract

We prove that for any sequence of binary alphabets $\mathcal{A}_1,\mathcal{A}_2,\dots$, there exists a cube-free word $c_1c_2\dots$ so that $c_1\in\mathcal{A}_1,c_2\in\mathcal{A}_2,\dots$. In particular, for every $n$, there are at least $1.35^n$ cube-free words in $\mathcal{A}_1\times\mathcal{A}_2\times\dots\times \mathcal{A}_n$. We also prove that if the list of alphabets is computable then one of these words is computable and its $n$th letter can be computed in time polynomial in $n$.

Theorems & Definitions (23)

• Conjecture 1: rosenfeldThueListCzerwinskiThueChoiceNumberGagolThueChoiceNumberGrytczukSurveyThueChoiceNumberGrytczukThueChoiceNumberWoodThueChoiceNumberZhaoThueChoiceNumberGrytczuk2011JanFiorenzi2011OctSkrabulakova2015Aug
• Conjecture 2
• Theorem 1
• Theorem 2
• Theorem 3
• Lemma 4
• Lemma 5
• Theorem 6: Fine, Wilf
• Lemma 7
• Remark 1