\documentclass[12pt,reqno]{article} \usepackage[usenames]{color} \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsthm} \usepackage{amsfonts} \usepackage{amscd} \usepackage{graphicx} \usepackage[colorlinks=true, linkcolor=webgreen, filecolor=webbrown, citecolor=webgreen]{hyperref} \definecolor{webgreen}{rgb}{0,.5,0} \definecolor{webbrown}{rgb}{.6,0,0} \usepackage{color} \usepackage{fullpage} \usepackage{float} \usepackage{psfig} \usepackage{graphics} \usepackage{latexsym} \usepackage{epsf} \usepackage{breakurl} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.1in} \setlength{\textheight}{8.4in} \def\vtm{{\bf vtm}} \newcommand{\seqnum}[1]{\href{https://oeis.org/#1}{\underline{#1}}} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} \end{center} \theoremstyle{plain} \newtheorem{theorem}{Theorem} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{conjecture}[theorem]{Conjecture} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \begin{center} \vskip 1cm{\LARGE\bf Finite Test Sets for Morphisms That Are \\ Squarefree on Some of Thue's \\ \vskip .1in Squarefree Ternary Words } \vskip 1cm \large James D. Currie\\ Department of Mathematics and Statistics\\ University of Winnipeg\\ Winnipeg, Manitoba R3B 2E9 \\ Canada\\ \href{mailto:j.currie@uwinnipeg.ca}{\tt j.currie@uwinnipeg.ca} \\ \end{center} \vskip .2 in \begin{abstract} Let $S$ be one of $\{aba,cbc\}$ and $\{aba, aca\}$, and let $w$ be an infinite squarefree word over $\Sigma=\{a,b,c\}$ with no factor in $S$. Suppose that $f:\Sigma\rightarrow T^*$ is a non-erasing morphism. We prove that the word $f(w)$ is squarefree if and only if $f$ is squarefree on factors of $w$ of length 8 or less. \end{abstract} \section{Introduction} The papers of Thue on squarefree words \cite{thue06,thue12} are foundational to the area of combinatorics on words. A word $w$ is {\em squarefree} if we cannot write $w=xyyz$, where $y$ is a non-empty word. The longest squarefree words over the 2-letter alphabet $\{a,b\}$ are $aba$ and $bab$, each of length 3, but Thue showed that arbitrarily long squarefree words exist over the 3-letter alphabet $\{a,b,c\}$. Infinite squarefree words over finite alphabets are routinely encountered in combinatorics on words, and are frequently used as building blocks in constructions. (See, for example, \cite{loth97,loth02}.) Let $w$ be an infinite squarefree word over $\Sigma=\{a,b,c\}$. Thue showed that $w$ must contain every squarefree word of length 2 over $\Sigma$ . However, he showed that the same is not true for squarefree words of length 3 over $\Sigma$. For each of $S_1=\{aba,cbc\}, S_2=\{aba, aca\}$, and $S_3=\{aba,bab\}$, Thue constructed an infinite squarefree word over $\Sigma$ with no factor in $S_i$. Constructions giving squarefree words equivalent to Thue's word with no factors in $S_1$ were independently discovered by Braunholtz \cite{braunholtz} and Istrail \cite{istrail}; Berstel \cite{berstel78} shows this equivalence. Their word is called $\vtm$ (for `variation of Thue-Morse') by Blanchet-Sadri et al.~\cite{blanchetsadri}, and has been used as the basis for various constructions \cite{blanchetsadri,currie91,currieetal}. These constructions require showing that $f(\vtm)$ is squarefree for particular morphisms $f$. In this paper, we give a testable characterization of morphisms $f$ such that $f(\vtm)$ is squarefree; we do the same in the case where $\vtm$ is replaced by an infinite squarefree word over $\Sigma$ with no factors in $S_2$. We leave as an open problem whether there is a characterization when we replace $\vtm$ by a word over $\Sigma$ with no factor in $S_3$. \begin{theorem} Let $w$ be an infinite squarefree word over $\Sigma$ such that either $w$ has no factor in $S_1$, or $w$ has no factor in $S_2$. Suppose that $f:\Sigma\rightarrow T^*$ is a non-erasing morphism. The word $f(w)$ is squarefree if and only if $f$ is squarefree on factors of $w$ of length 8 or less. \end{theorem} Our theorem says that to establish squarefreeness of $f(w)$, the morphism $f$ need only be checked for squarefreeness on a finite test set. Crochemore \cite{crochemore} proved a variety of similar theorems; in particular, a morphism $f$ defined on $\Sigma^*$ preserves squarefreeness exactly when $f$ preserves squarefreeness on words of $\Sigma^*$ of length at most 5. Note that while Crochemore's theorem requires testing the squarefreeness of $f(v)$ for every squarefree word $v\in\Sigma^*$ up to a certain length, we only test words $v$ that are factors of $w$. Thus, while $aba$ is squarefree, we do not require $f(aba)$ to be squarefree, for example. Finite test sets for morphisms preserving overlap-freeness have also been well-studied \cite{richomme}. \section{Preliminaries} Let $S=S_1$ or $S=S_2$. For the remainder of this paper, let $w$ be an infinite squarefree word over $\Sigma$ with no factor in $S$. Write $w=a_0a_1a_2a_3\cdots$ with $a_i\in\Sigma.$ For the remainder of this section suppose that $f:\Sigma\rightarrow T^*$ is a non-erasing morphism that is squarefree on factors of $w$ of length 8 or less. \begin{lemma}\label{at most 3}Suppose $f(\xi)$ is a factor of $f(x)$, where $x\in\Sigma$ and $\xi$ is a factor of $w$. Then $|\xi|\le 3$. \end{lemma} \begin{proof} If $x$ is a letter of $\xi$, but $f(\xi)$ is a factor of $f(x)$, we have \begin{eqnarray*} |f(x)|&\le&|f(\xi)|\\ &\le&|f(x)|. \end{eqnarray*} Since $f$ is non-erasing, this forces $x=\xi$, giving $|\xi|=1$. If $x$ is not a letter of $\xi$, then $\xi$ is a squarefree word over a two-letter alphabet, so that $|\xi|\le 3.$ \end{proof} \begin{lemma}\label{no prefix}Suppose that $f(x)$ is a prefix or a suffix of $f(y)$ where $x,y\in\Sigma$. Then $x=y$. \end{lemma} \begin{proof} We give the proof where $f(x)$ is a prefix of $f(y)$. (The other case is similar.) Suppose $x\ne y$. Then $xy$ must be a factor of $w$, and $xy$ is squarefree. However, $f(xy)$ begins with the square $f(x)f(x)$, contradicting the squarefreeness of $f$ on factors of $w$ of length at most 8.\end{proof} \begin{lemma}\label{suffix} There is no solution $(\alpha,\beta,\gamma,\xi,p,s,t,x,y,z)$ to \begin{eqnarray*} &&\alpha\xi xyz\beta\mbox{ is a factor of }w;\\ &&\alpha,\beta,\gamma,x,y,z\in\Sigma, \xi\in\Sigma^*;\\ &&t\mbox{ is a suffix of }f(\gamma);\\ &&s\mbox{ is a suffix of } f(\alpha);\\ &&p\mbox{ is a non-empty prefix of }f(\beta);\\ &&t=sf(\xi xyz)p. \end{eqnarray*} \end{lemma} \begin{proof} Suppose, for the sake of getting a contradiction, that $(\alpha,\beta,\gamma,\xi,p,s,t,x,y,z)$ is a solution. Here $p$ is a prefix of $f(\beta)$, but also a suffix of $t$, that is a suffix of $f(\gamma)$. Thus we see that $f(\gamma\beta)$ contains square $pp$, so that $\gamma\beta$ is not a factor of $w$. This forces $\gamma=\beta$. On the other hand, since $z\beta$ is a factor of $w$, we conclude that $z\ne \gamma$. Again, $f(z)p$ is a prefix of $f(z\beta)$, but also a suffix of $f(\gamma)$, so that $f(\gamma z\beta)$ contains a square. Since $yz\beta$ is a factor of $w$, it follows that $y\ne \gamma$. Similarly, $f(\gamma yz\beta)$ contains a square, but $xyz\beta$ is a factor of $w$, so that that $x\ne \gamma$. Finally, we see that $f(\gamma xyz\beta)$ contains a square. Let $\delta$ be the last letter of $\alpha\xi$. We conclude that $\delta\ne \gamma$. However, now $\delta xyz$ is a squarefree word of length 4 over the two-letter alphabet $\Sigma-\{\gamma\}$. This is impossible. \end{proof} The symmetrical lemma is proved analogously: \begin{lemma}\label{prefix} There is no solution $(\alpha,\beta,\gamma,\xi,p,s,t,x,y,z)$ to \begin{eqnarray*} &&\alpha\xi xyz\beta\mbox{ is a factor of }w;\\ &&\alpha,\beta,\gamma,x,y,z\in\Sigma, \xi\in\Sigma^*;\\ &&t\mbox{ is a suffix of }f(\gamma);\\ &&s\mbox{ is a non-empty suffix of } f(\alpha);\\ &&p\mbox{ is a non-empty of }f(\beta);\\ &&t=sf(xyz\xi)p. \end{eqnarray*} \end{lemma} Suppose that $f(w)$ contains a non-empty square $xx$, with $|x|$ as short as possible. Write $f(w)=uxxv$, such that \begin{align*} u &=A_0A_1\cdots A_i' \\ ux &=A_0A_1\cdots A_j' \\ uxx &=A_0A_1\cdots A_k', \end{align*} where $i\le j\le k$ are non-negative integers, and for each non-negative integer $\ell$, $A_\ell=f(a_\ell)$, and $A_\ell'$ is a prefix of $A_\ell$, but $A_\ell'\ne A_\ell$. This notation is not intended to exclude the possibilities that $i=0$, $i=j$ and/or $j=k$. \begin{remark} Since $f$ is squarefree on factors of $w$ of length at most 8, but $f(a_i\cdots a_k)$ contains the square $xx$, we must have $k-i\ge 8$. \end{remark} \begin{remark} We cannot have $i=j$; otherwise suffix $x$ of $ux$ is a factor of $A_j=A_i$, and $A_{i+1}\cdots A_{k-1}$ is a factor of suffix $x$ of $uxx$. Then, $f(a_{i+1}a_{i+2}\cdots a_{k-1})$ is a factor of $f(a_i)$, forcing $(k-1)-(i+1)+1\le 3$ by Lemma~\ref{at most 3}, so that $k-i\le 4$, a contradiction. Reasoning, in the same way, we show that $i|x|$. Then $|A_j^{\prime\prime}|>|x|-|A_j'|=|A_i^{\prime\prime}A_{i+1}\cdots A_{j-1}|$. It follows that $A_j^{\prime\prime}=A_i^{\prime\prime}A_{i+1}\cdots A_{j-1}A_j^{\prime\prime\prime}$ for some non-empty prefix $A_j^{\prime\prime\prime}$ of $A_j$. Similarly, one shows that $A_j^{\prime}=A_j^{\prime\prime\prime\prime}A_{j+1}\cdots A_{k-1}A_k'$ for some non-empty suffix $A_j^{\prime\prime\prime\prime}$ of $A_j$. Now $|a_{i+1}\cdots a_{j-1}|=(j-1)-(i+1)+1=j-i-1$, and $|a_{j+1}\cdots a_{k-1}|=(k-1)-(j+1)+1=k-j-1$. However, either $j-i-1\ge 3$, or $k-j-1\ge 3$. If $j-i-1\ge 3$, then $A_j^{\prime\prime}=A_i^{\prime\prime}A_{i+1}\cdots A_{j-1}A_j^{\prime\prime\prime}$ for some non-empty prefix $A_j^{\prime\prime\prime}$ of $A_j$ contradicts Lemma~\ref{suffix}, letting $s=A_i^{\prime\prime}$, $\alpha = a_i$, $\xi= a_{i+1}\cdots a_{j-4}$, $xyz=a_{j-3}a_{j-2}a_{j-1}$, $p=A_j^{\prime\prime\prime}$, $\beta=a_j$. In the case where $k-j-1\ge 3$, we get the analogous contradiction using Lemma~\ref{prefix}. \end{proof} \begin{lemma}\label{lineup} We have $j-i=k-j$, $A_i^{\prime\prime}=A_j^{\prime\prime}$, $A_j^{\prime}=A_k^{\prime}$, and $A_{i+\ell}=A_{j+\ell}$, $1\le \ell\le j-i-1$. \end{lemma} \begin{proof} To begin with we show that $A_j^{\prime}=A_k^{\prime}$. Since both words are suffixes of $x$, it suffices to show that $|A_j^{\prime}|=|A_k^{\prime}|$. Suppose not. Suppose that $|A_k^{\prime}|<|A_j^{\prime}|$. By the previous lemma, $|A_j^{\prime}|\le|A_{j+1}\cdots A_{k-2}A_{k-1}A_k'|$. Let $m$ be greatest such that $|A_m\cdots A_{k-1}A_k'|\ge |A_j^{\prime}|$. Thus $j+1\le m\le k-1$, and $$|A_{m+1}\cdots A_{k-1}A_k'|<|A_j^{\prime}|\le |A_mA_{m+1}\cdots A_{k-1}A_k'|.$$ If $A_j^{\prime}=A_mA_{m+1}\cdots A_{k-1}A_k'$, then $A_m$ is a non-empty prefix of $A_j^{\prime}$, forcing $a_m=a_j$ so that $A_m=A_j^{\prime}=A_j$, by Lemma~\ref{no prefix}. This is contrary to our choice of $A_j^{\prime}$. Similarly, suppose $|A_k^{\prime}|<|A_j^{\prime}|$. Suppose that $|A_k^{\prime}|<|A_j^{\prime}|$. Again by the previous lemma, $|A_k^{\prime}|\le|A_{i+1}\cdots A_{j-2}A_{j-1}A_j'|$. Let $m$ be greatest such that $|A_m\cdots A_{j-1}A_j'|\ge |A_k^{\prime}|$. Thus $i+1\le m\le j-1$, and $$|A_{m+1}\cdots A_{j-1}A_j'|<|A_k^{\prime}|\le |A_mA_{m+1}\cdots A_{j-1}A_j'|.$$ If $A_k^{\prime}=A_mA_{m+1}\cdots A_{j-1}A_j'$, then $A_m$ is a non-empty prefix of $A_k^{\prime}$, forcing $a_m=a_k$ so that $A_m=A_k^{\prime}=A_k$, by Lemma~\ref{no prefix}, contrary to our choice of $A_k^{\prime}$. We thus conclude that $A_j'=A_k'$, as desired. Next, we show that $j-1=k-j$ and $A_{i+\ell}=A_{j+\ell}$, $1\le \ell\le j-i-1$. Suppose that $j-i\le k-j$. (The other case is similar.) Suppose now that we have shown that for some $\ell$, $0\le \ell< j-i-1$ that \begin{equation}\label{IHOP}A_{j-\ell}\cdots A_{j-1}A_j'=A_{k-\ell}\cdots A_{k-1}A_k'.\end{equation} This is true when $\ell=0$; i.e., $A_j^{\prime}=A_k^{\prime}$. From (\ref{x=x}), one of $A_{j-\ell-1}A_{j-\ell}\cdots A_{j-1}A_j'$ and $A_{k-\ell-1}A_{k-\ell}\cdots A_{k-1}A_k'$ is a suffix of the other. Together with (\ref{IHOP}), this implies that one of $A_{j-\ell-1}$ and $A_{k-\ell-1}$ is a suffix of the other. By Lemma~\ref{no prefix}, this implies that $a_{j-\ell-1}=a_{k-\ell-1}$, and by combining this with (\ref{IHOP}), \begin{equation}A_{j-\ell-1}\cdots A_{j-1}A_j'=A_{k-\ell-1}\cdots A_{k-1}A_k'.\end{equation} By induction we conclude that $$A_{j-\ell}\cdots A_{j-1}A_j'=A_{k-\ell}\cdots A_{k-1}A_k',0\le\ell\le j-i-1,$$ which implies $A_{j-\ell}=A_{k-\ell}$, $1\le \ell\le j-i-1$. In particular, we note that \begin{equation}\label{var x=x}A_{i+1}\cdots A_{j-1}=A_{k-j+i+1}\cdots A_{k-1}.\end{equation} If we now have $k-j>j-i$, then $k-j+i>j$, and (\ref{x=x}) and (\ref{var x=x}) imply that $A_i^{\prime\prime}=A_j^{\prime\prime}A_{j+1}\cdots A_{k-j+i}$. Then $A_{k-j+i}$ is a suffix of $A_i^{\prime\prime}$ and Lemma~\ref{no prefix} forces $A_{k-j+i}=A_i$. Then (\ref{x=x}) and (\ref{var x=x}) force $A_i=A_i^{\prime}$, contrary to our choice of $A_i^{\prime}$. We conclude that $k-j=j-i$. From (\ref{x=x}) and (\ref{var x=x}) we conclude that $A_i^{\prime\prime}=A_j^{\prime\prime}$, as desired. \end{proof} \begin{corollary}The word $w$ contains a factor $\alpha z \beta z \gamma$, $\alpha,\beta,\gamma\in\Sigma$, $\alpha,\gamma\ne \beta$, $|z|\ge 3$, such that $\alpha\beta\gamma$ is not a factor of $w$. \end{corollary} \begin{proof} By Lemma~\ref{lineup}, $w$ contains a factor $\alpha z \beta z \gamma$, where $\alpha=a_i$, $\beta=a_j$, $\gamma=a_k$, $z=a_{i+1}\cdots a_{j-1}=a_{j+1}\cdots a_{k-1}$. This gives $|z|=j-i-1=k-j-1\ge 3$. Since $w$ is squarefree, we cannot have $a_i=a_j$; otherwise $w$ contains the square $(a_0z)^2$; similarly, $a_j\ne a_k$. To see that $\alpha\beta\gamma$ is not a factor of $w$, we note that $f$ is squarefree on factors of $w$ of length at most 8, but $f(\alpha\beta\gamma)=A_iA_jA_k$ contains the square $(A_i^{\prime\prime} A_j')(A_j^{\prime\prime} A_k')$; this is a square since $A_i^{\prime\prime}=A_j^{\prime\prime}$ and $A_j'=A_k'$. Since $|A_iA_jA_k|=3\le 8$, we conclude that $A_iA_jA_k$ is not a factor of $w$. \end{proof} \section{Results} \begin{lemma}\label{aba,cbc} If $S=\{aba,cbc\}$, the only squarefree words of length 3 that are not factors of $w$ are $aba$ and $cbc$. In addition, $w$ contains no factor of the form $azbza$ or $czbzc$. \end{lemma} \begin{proof} Thue \cite{thue12} showed that a squarefree word over $\Sigma$ not containing $aba$ or $cbc$ as a factor contains every other length 3 squarefree word as a factor. Suppose $w$ contains a factor $azbza$. Since $aba$ is not a factor of $w$, $z\ne \epsilon$. Since $az$ and $bz$ are factors of $w$, and hence squarefree, the first letter of $z$ cannot be $a$ or $b$, and must be $c$. Similarly, the last letter of $z$ must be $c$. But $w$ contains $zbz$, and thus $cbc$. This is a contradiction. Therefore $w$ contains no factor $azbza$. Replacing $a$ by $c$ and vice versa in the preceding argument shows that $w$ contains no factor $czbzc$. \end{proof} Combining the last two lemmas gives this corollary: \begin{corollary} If $S=\{aba,cbc\}$ and $f:\Sigma^*\rightarrow T^*$ is squarefree on factors of $w$ of length at most 8, then $f(w)$ is squarefree. \end{corollary} \begin{lemma}\label{aba,aca} If $S=\{aba,aca\}$, the only squarefree words of length 3 that are not factors of $w$ are $aba$ and $aca$. In addition, $w$ contains no factor of the form $azbza$ or $azcza$, $|z|\ge 3$. \end{lemma} \begin{proof} Thue \cite{thue12} showed that a squarefree word over $\Sigma$ not containing $aba$ or $aca$ as a factor contains every other length 3 squarefree word as a factor. Suppose $w$ contains a factor $azbza$, $|z|\ge 3$. Since $aba$ is not a factor of $w$, $z\ne \epsilon$. Since $az$ and $bz$ are factors of $w$, and hence squarefree, the first letter of $z$ cannot be $a$ or $b$, and must be $c$. Similarly, the last letter of $z$ must be $c$. However, since $az$ is a factor of $w$, but $aca$ is not, the second letter of $z$ cannot be $a$ and must be $b$. Write $z=cbz'c$. Then $azbza=acbz'cbcbz'ca$ contains the square $cbcb$, that is impossible. We conclude that $w$ contains no factor $azbza$, $|z|\ge 3$. Replacing $c$ by $b$ and vice versa in the preceding argument shows that $w$ contains no factor $azcza$, $|z|\ge 3$. \end{proof} \begin{corollary} If $S=\{aba,aca\}$ and $f:\Sigma^*\rightarrow T^*$ is squarefree on factors of $w$ of length at most 8, then $f(w)$ is squarefree. \end{corollary} \begin{lemma} The squarefree word $azbza$ where $z=cabcbacabcacbacabcbac$ has no factors $aba$ or $bab$. It follows that any analogous theorem for $S_3$, with an analogous proof, would require us to replace 8 by a value of at least $|azbza|=45.$ \end{lemma} \begin{proof} This is established by a finite check. \end{proof} \section{Acknowledgments} The author's research was supported by the Natural Sciences and Engineering Research Council of Canada (NSERC), [funding reference number 2017-03901]. The author thanks the careful referee and editor for their comments. \begin{thebibliography}{3} \bibitem{berstel78} J. Berstel, Sur la construction de mots sans carr\'e, {\it S\'em. Th\'eor. Nombres Bordeaux} (1978--1979), 18.01--18.15. \bibitem{berstel95} J. Berstel, {\em Axel Thue's Papers on Repetitions in Words: a Translation}, Publications du Laboratoire de Combinatoire et d'Informatique Math\'ematique {\bf 20} Universit\'e du Qu\'ebec \`a Montr\'eal, 1995. \bibitem{blanchetsadri} F. Blanchet-Sadri, J. Currie, N. Fox, and N. 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Lothaire, {\em Combinatorics on Words}, Cambridge University Press, 1997. \bibitem{loth02}M. Lothaire, {\em Algebraic Combinatorics on Words}, Cambridge University Press, 2002. \bibitem{richomme} G. Richomme and F. Wlazinski, Overlap-free morphisms and finite test-sets, {\em Discrete Appl. Math.} {\bf 143} (2004), 92--109. \bibitem{thue06} Axel Thue, \"{U}ber unendliche Zeichenreihen, {\em Norske Vid. Selsk. Skr. Mat. Nat. Kl.} {\bf 7} (1906), 1--22. \bibitem{thue12} Axel Thue, \"{U}ber die gegenseitige Lage gleicher Teile gewisser Zeichentreihen, {\em Norske Vid. Selske Skr. Mat. Nat. Kl.} {\bf 1} (1912), 1--67. \end{thebibliography} \bigskip \hrule \bigskip \noindent 2010 {\it Mathematics Subject Classification}: Primary 68R15. \noindent \emph{Keywords: } Thue-Morse sequence, squarefree word, nonrepetitive word, factor. \bigskip \hrule \bigskip \noindent (Concerned with sequences \seqnum{A010060} and \seqnum{A036577}.) \bigskip \hrule \bigskip \vspace*{+.1in} \noindent Received October 3 2019; revised versions received November 30 2019; December 7 2019. Published in {\it Journal of Integer Sequences}, December 9 2019. \bigskip \hrule \bigskip \noindent Return to \htmladdnormallink{Journal of Integer Sequences home page}{https://cs.uwaterloo.ca/journals/JIS/}. \vskip .1in \end{document} .