Finite test sets for morphisms which are square-free on some of Thue's square-free ternary words

TL;DR

This paper characterizes when a morphism preserves square-freeness on infinite words over a ternary alphabet, showing it suffices to check factors up to length 8 for certain sets of factors.

Contribution

It introduces finite test sets for morphisms to determine square-freeness on specific classes of infinite words, simplifying verification processes.

Findings

Square-freeness of morphisms can be tested on finite factors of length 8 or less.

The result applies to words avoiding certain factor sets $S$.

Provides a practical criterion for morphism analysis in combinatorics on words.

Abstract

Let $S$ be one of $\{aba,bcb\}$ and $\{aba, aca\}$, and let $w$ be an infinite square-free word over $\Sigma=\{a,b,c\}$ with no factor in $S$. Suppose that $f:\Sigma\rightarrow T^*$ is a non-erasing morphism. Word $f(w)$ is square-free if and only if $f$ is square-free on factors of $w$ of length 8 or less.

Equations

$$w=a_0{a}_1{a}_2{a}_3\cdots ,a_i\in \Sigma .$$

$$|f(x)|\le |f(\chi )|\le |f(x)|.$$

$$u=A_0{A}_1\cdots A_i^{\prime }$$

$$ux=A_0{A}_1\cdots A_j^{\prime }$$

$$uxx=A_0{A}_1\cdots A_k^{\prime }$$

$$x=A_i^{\prime \prime }{A}_{i+1}\cdots A_{j-1}{A}_j^{\prime }=A_j^{\prime \prime }{A}_{j+1}\cdots A_{k-1}{A}_k^{\prime }.$$

$$|A_{m+1}\cdots A_{k-1}{A}_k^{\prime }|<|A_j^{\prime }|\le |A_m{A}_{m+1}\cdots A_{k-1}{A}_k^{\prime }|.$$

$$A_{j-\ell }\cdots A_{j-1}{A}_j^{\prime }=A_{k-\ell }\cdots A_{k-1}{A}_k^{\prime }.$$

$$A_{j-\ell -1}\cdots A_{j-1}{A}_j^{\prime }=A_{k-\ell -1}\cdots A_{k-1}{A}_k^{\prime }.$$

$$A_{j-\ell }\cdots A_{j-1}{A}_j^{\prime }=A_{k-\ell }\cdots A_{k-1}{A}_k^{\prime },0\le \ell \le j-i-1,$$

$$A_{i+1}\cdots A_{j-1}=A_{k-j+i+1}\cdots A_{k-1}.$$

Full text

The papers of Axel Thue on square-free words [9, 10] are foundational to the area of combinatorics on words. A word $w$ is square-free if we cannot write $w=xyyz$, where $y$ is a non-empty word. The longest square-free words over the 2-letter alphabet $\{a,b\}$ are $aba$ and $bab$, each of length 3, but Thue showed that arbitrarily long square-free words exist over a 3-letter alphabet. Infinite square-free words over finite alphabets are routinely encountered in combinatorics on words, and are frequently used as building blocks in constructions. (See [6, 7] for further background and definitions.)

Let $w$ be an infinite square-free word over $\Sigma=\{a,b,c\}$. Thue showed that $w$ must contain every length 2 square-free word over $\Sigma$ as a factor. However, Thue showed that the same is not true for length 3 square-free words over $\Sigma$. For each of $S_{1}=\{aba,bcb\},S_{2}=\{aba,aca\}$, and $S_{3}=\{aba,bab\}$, he constructed an infinite square-free word over $\Sigma$ with no factor in $S_{i}$. The infinite square-free word over $\Sigma$ with no factor in $S_{1}$ has been called vtm (for ‘variation of Thue-Morse’), and has been used in constructions in several papers [2, 4, 5]. Various constructions involve the necessity of showing $f$(vtm) to be square-free for particular morphisms $f$. In this paper, we give a simple testable characterization of morphisms $f$ such that $f($vtm) is square-free; we do the same in the case where vtm is replaced by an infinite square-free word over $\Sigma$ with no factors in $S_{2}$.

From now on, fix $S$ to be one of $S_{1}$ and $S_{2}$, and let $w$ be an infinite square-free word over $\Sigma$ with no factor in $S$. We establish the following:

Our theorem says that to establish square-freeness of $f(w)$, one need only check $f$ for square-freeness on a finite test set. A variety of similar theorems were proved by Crochemore [3]; in particular, a morphism $g$ defined on $\Sigma$ preserves square-freeness exactly when it preserves square-freeness on words of $\Sigma^{*}$ of length at most 5. Note that the theorem of the present paper tests a weaker property; while $aba$ is square-free, we do not require $f(aba)$ to be square-free, for example. Finite test sets for morphisms preserving overlap-freeness have also been well-studied [8].

The theorem suggests a couple of open problems:

Write

[TABLE]

Suppose that $f:\Sigma\rightarrow T^{*}$ is a non-erasing morphism which is square-free on factors of $w$ of length 7 or less.

[TABLE]

Since $f$ is non-erasing, this forces $x=\chi$, giving $|\chi|=1$.

If $x$ is not a letter of $\chi$, then $\chi$ is a square-free word over a two-letter alphabet, so that $|\chi|\leq 3.$ ∎

Since $p$ is a non-empty prefix of $f(\beta)$, but also a suffix of $f(t)$, we see that $f(t\beta)$ contains square $pp$, so that $t\beta$ is not a factor of $w$. This forces $t=\beta$. On the other hand, since $z\beta$ is a factor of $w$, we conclude that $z\neq t$. Again, $f(z)p$ is a prefix of $f(z\beta)$, but also a suffix of $f(t)$, so that $f(tz\beta)$ contains a square. Since $yz\beta$ is a factor of $w$, it follows that $y\neq t$. Similarly, $f(tyz\beta)$ contains a square, but $xyz\beta$ is a factor of $w$, so that that $x\neq t$. Finally, we see that $f(txyz\beta)$ contains a square. Let $\delta$ be the last letter of $\alpha\chi$. We conclude that $\delta\neq t$. However, now $\delta xyz$ is a squarefree word of length 4 over the two-letter alphabet $\Sigma-\{t\}$. This is impossible. ∎

The symmetrical lemma is proved analogously:

The proof of Theorem 2 will use several lemmas. Suppose that $f(w)$ contains a non-empty square $xx$, with $|x|$ as short as possible. Write $f(w)=uxxv$, such that

[TABLE]

[TABLE]

[TABLE]

where $i\leq j\leq k$ are non-negative integers, and for each non-negative integer $\ell$, $A_{\ell}=f(a_{\ell})$, and $A_{\ell}^{\prime}$ is a prefix of $A_{\ell}$, but $A_{\ell}^{\prime}\neq A_{\ell}$. This notation is not intended to exclude the possibilities that $i=0$, $i=j$ and/or $j=k$.

For $\ell\in\{i,j,k\}$, let $A_{\ell}^{\prime\prime}$ be the suffix of $A_{\ell}$ such that $A_{\ell}=A_{\ell}^{\prime}A_{\ell}^{\prime\prime}$. By our choice of $A_{\ell}^{\prime}$, $A_{\ell}^{\prime\prime}\neq\epsilon.$ Then

[TABLE]

In the case where $k-j-1\geq 3$, we get the analogous contradiction using Lemma 4. ∎

By the previous lemma, $|A_{j}^{\prime}|\leq|A_{j+1}\cdots A_{k-2}A_{k-1}A_{k}^{\prime}|$. Let $m$ be greatest such that $|A_{j}^{\prime}|\leq|A_{m}\cdots A_{k-1}A_{k}^{\prime}|.$ Thus $j+1\leq m\leq k-1$, and

[TABLE]

It follows that $A_{j}^{\prime}=A_{m}^{\prime\prime}A_{m+1}\cdots A_{k-1}A_{k}^{\prime}$, for some non-empty suffix $A_{m}^{\prime\prime}$ of $A_{m}$. Therefore, $f(a_{m}a_{j})$ contains the square factor $A_{m}^{\prime\prime}A_{m}^{\prime\prime}$, forcing $a_{m}=a_{j}$. If $A_{m}^{\prime\prime}=A_{m}$, then $A_{m}$ is a prefix of $A_{j}$, and Lemma 2 forces $A_{m}=A_{j}^{\prime}=A_{j}$. This is contrary to our choice of $A_{j}^{\prime}$. We may therefore write $A_{m}=A_{m}^{\prime}A_{m}^{\prime\prime}$ where $A_{m}^{\prime}$ is non-empty.

Now $A_{m}^{\prime}$ is a suffix of $A_{i}^{\prime\prime}A_{i+1}\cdots A_{j-1}$. Let $n$ be greatest such that $A_{m}^{\prime}$ is a suffix of $A_{n}A_{n+1}\cdots A_{j-1}$; thus $i\leq n\leq j-1$. It follows that a non-empty prefix $A_{m}^{\prime\prime\prime}$ of $A_{m}^{\prime}$ is a suffix of $A_{n}$; this implies $f(a_{n}a_{m})$ contains a square, whence $a_{n}=a_{m}$. Let $q\leq j-1$ be greatest such that $a_{q}=a_{m}$; thus $i\leq n\leq q\leq j-1$. Each of $A_{m}^{\prime}$ and $A_{q}\cdots A_{j-1}$ is a suffix of $A_{n}A_{n+1}\cdots A_{j-1}$. Since $|A_{m}^{\prime}|<|A_{m}|=|A_{q}|$, we conclude that $A_{m}^{\prime}$ is a suffix of $A_{q}\cdots A_{j-1}$. By the choice of $q$, $a_{q+1}\cdots a_{j-1}$ does not have letter $a_{m}$ as a factor; it is thus a square-free word over a 2-letter alphabet, whence $|a_{q+1}\cdots a_{j-1}|\leq 3$. However, $A_{m}^{\prime}$ is a suffix of $A_{q}\cdots A_{j-1}$, so that the square $A_{m}^{\prime}A_{m}^{\prime}$ is a factor of $A_{q}\cdots A_{j-1}A_{m}=A_{q}\cdots A_{j-1}A_{j}.$ This is impossible, since $a_{q}\cdots a_{j-1}a_{j}$ is a factor of $w$ of length at most 4.

The assumption $|A_{k}^{\prime}|<|A_{j}^{\prime}|$ leads to a contradiction. The assumption $|A_{j}^{\prime}|<|A_{k}^{\prime}|$ leads to a similar contradiction, and we conclude that $A_{j}^{\prime}=A_{k}^{\prime}$, as desired.

Next, we show that $j-i=k-j$ and $A_{i+\ell}=A_{j+\ell}$, $1\leq\ell\leq j-i-1$. Suppose that $j-i\leq k-j$. (The other case is similar.) Suppose now that we have shown that for some $\ell$, $0\leq\ell<j-i-1$ that

[TABLE]

This is true when $\ell=0$; i.e., $A_{j}^{\prime}=A_{k}^{\prime}$.

From (1), one of $A_{j-\ell-1}A_{j-\ell}\cdots A_{j-1}A_{j}^{\prime}$ and $A_{k-\ell-1}A_{k-\ell}\cdots A_{k-1}A_{k}^{\prime}$ is a suffix of the other. Together with (2), this implies that one of $A_{j-\ell-1}$ and $A_{k-\ell-1}$ is a suffix of the other. By Lemma 2, this implies that $a_{j-\ell-1}=a_{k-\ell-1}$, and by combining this with (2),

[TABLE]

By induction we conclude that

[TABLE]

which implies $A_{j-\ell}=A_{k-\ell}$, $1\leq\ell\leq j-i-1$. In particular, we note that

[TABLE]

If we now have $k-j>j-i$, then $k-j+i>j$, and (1) and (4) 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 2 forces $A_{k-j+i}=A_{i}$. Then (1) and (4) force $A_{i}=A_{i}^{\prime}$, contrary to our choice of $A_{i}^{\prime}$. We conclude that $k-j=j-i$. From (1) and (4) we conclude that $A_{i}^{\prime\prime}=A_{j}^{\prime\prime}$, as desired. ∎

Since $w$ is square-free, we cannot have $a_{i}=a_{j}$; otherwise $w$ contains the square $(a_{i}z)^{2}$; similarly, $a_{j}\neq a_{k}$. To see that $\alpha\beta\gamma$ is not a factor of $w$, we note that $f$ is square-free on factors of $w$ of length at most 7, but $f(\alpha\beta\gamma)=A_{i}A_{j}A_{k}$ contains the square $(A_{i}^{\prime\prime}A_{j}^{\prime})(A_{j}^{\prime\prime}A_{k}^{\prime})$; this is a square since $A_{i}^{\prime\prime}=A_{j}^{\prime\prime}$ and $A_{j}^{\prime}=A_{k}^{\prime}$. Since $|A_{i}A_{j}A_{k}|=3\leq 7$, we conclude that $a_{i}a_{j}a_{k}$ is not a factor of $w$. ∎

Suppose $w$ contains a factor $azbza$. Since $aba$ is not a factor of $w$, $z\neq\epsilon$. Since $az$ and $bz$ are factors of $w$, and hence square-free, 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 $czbzc$. ∎

Combining Theorem 2 and Lemma refaba,cbc gives this corollary:

Suppose $w$ contains a factor $azbza$, $|z|\geq 3$. Since $az$ and $bz$ are factors of $w$, and hence square-free, 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 $z=cbz^{\prime}c$. Then $azbza=acbz^{\prime}cbcbz^{\prime}ca$ contains the square $cbcb$, which is impossible. We conclude that $w$ contains no factor $azbza$, $|z|\geq 3$.

Replacing $c$ by $b$ and vice versa in the preceding argument shows that $w$ contains no factor $azcza$, $|z|\geq 3$. ∎

Combining Corollaries 1 and 2 establishes our main result.

Theorem 1: Suppose that $f:\Sigma\rightarrow T^{*}$ is a non-erasing morphism. Word $f(w)$ is square-free if and only if $f$ is square-free on factors of $w$ of length 7 or less.