Nonrepetitive edge-colorings of trees

TL;DR

This paper improves the upper bound on the number of colors needed for nonrepetitive edge-colorings of trees, reducing it from 4Δ-4 to 3Δ-2, thus advancing understanding of graph coloring constraints.

Contribution

The authors present a simpler nonrepetitive edge-coloring method for trees that improves the upper bound on the Thue edge-chromatic number.

Findings

Established a new upper bound of 3Δ-2 colors for trees.

Simplified the coloring method compared to previous approaches.

Enhanced the theoretical understanding of nonrepetitive graph colorings.

Abstract

A repetition is a sequence of symbols in which the first half is the same as the second half. An edge-coloring of a graph is repetition-free or nonrepetitive if there is no path with a color pattern that is a repetition. The minimum number of colors so that a graph has a nonrepetitive edge-coloring is called its Thue edge-chromatic number. We improve on the best known general upper bound of $4\Delta-4$ for the Thue edge-chromatic number of trees of maximum degree $\Delta$ due to Alon, Grytczuk, Ha{\l}uszczak and Riordan (2002) by providing a simple nonrepetitive edge-coloring with $3\Delta-2$ colors.

Tables

$k\backslash h$	1	2	3	4	5	6-10	$h\ge 11$
1	1	2	2	3	3	3	3
2	2	4	4	5	$5^{\ast }$	$5^{\ast }$	5,7
3	3	5	$6^{\ast }$	$6^{\ast }$	6,9	6,10	6,10
4	4	7	$7^{\ast }$	7,10	7,12	7,13	7,13
5	5	8	9,10	9,13	9,15	9,16	9,16
$⋮$	$⋮$	$⋮$	$⋮$	$⋮$	$⋮$	$⋮$	$⋮$
$k$	$k$	$\lfloor 1.5k\rfloor$+1	$1.61k,2k$	$1.61k,\lceil 2.5k\rceil$	$1.61k,3k$	$1.61k,3k+1$	$1.61k,3k+1$

Full text

We improve on the best known general upper bound of $4\Delta-4$ for the Thue edge-chromatic number of trees of maximum degree $\Delta$ due to Alon, Grytczuk, Haluszczak and Riordan (2002) by providing a simple nonrepetitive edge-coloring with $3\Delta-2$ colors.

1 Introduction

A repetition is a sequence of even length (for example $abacabac$), such that the first half of the sequence is identical to the second half. In 1906 Thue [13] proved that there are infinite sequences of 3 symbols that do not contain a repetition consisting of consecutive elements in the sequence. Such sequences are called Thue sequences. Thue studied these sequences as words that do not contain any square words $ww$ and the interested reader can consult Berstel [2, 3] for some background and a translation of Thue’s work using more current terminology. Thue sequences have been studied and generalized in many views (see the survey of Grytczuk [9]), but in this paper we focus on the natural generalization of the Thue problem to Graph Theory.

In 2002 Alon, Grytczuk, Hałuszczak and Riordan [1] proposed calling a coloring of the edges of a graph nonrepetitive if the sequence of colors on any open path in $G$ is nonrepetitive. We will use $\pi^{\prime}(G)$ to denote the Thue chromatic index of a graph $G$, which is the minimum number of colors in a nonrepetitive edge-coloring of $G$. In [1] the notation $\pi(G)$ was used for the Thue chromatic index, but by common practice we will instead use this notation for the Thue chromatic number, which is the minimum number of colors in a nonrepetitive coloring of the vertices of $G$. Their paper contains many interesting ideas and questions, the most intriguing of which is if $\pi(G)$ is bounded by a constant when $G$ is planar. The best result in this direction is due to Dujmović, Frati, Joret, and Wood [7] who show that for planar graphs on $n$ vertices $\pi(G)$ is $O(\log n)$. Conjecture 2 from [1] was settled by Currie [6] who showed that for the $n$-cycle $C_{n}$, $\pi(C_{n})=3$ when $n\geq 18$. One of the conjectures from [1] that remains open is whether $\pi^{\prime}(G)=O(\Delta)$ when $G$ is a graph of maximum degree $\Delta$. At least $\Delta$ colors are always needed, since nonrepetitive edge-colorings must give adjacent edges different colors.

In this paper we study the seemingly easy question of nonrepetitive edge-colorings of trees. Thue’s sequence shows that if $P_{n}$ is the path on $n$ vertices, then $\pi^{\prime}(P_{n})=\pi(P_{n-1})\leq 3$. (Keszegh, Patkós, and Zhu [10] extend this to more general path-like graphs.) Using Thue sequences Alon, Grytczuk, Hałuszczak and Riordan [1] proved that every tree of maximum degree $\Delta\geq 2$ has a nonrepetitive edge-coloring with $4(\Delta-1)$ colors and stated that the same method can be used to obtain a nonrepetitive vertex-coloring with 4 colors. However, while the star $K_{1,t}$ is the only tree whose vertices can be colored nonrepetitively with fewer than 3 colors, it is still unknown which trees need 3 colors, and which need 4 (see Brešar, Grytczuk, Klavžar, Niwczyk, Peterin [5].) Interestingly Fiorenzi, Ochem, Ossona de Mendez, and Zhu [8] showed that for every integer $k$ there are trees that have no nonrepetitive vertex-coloring from lists of size $k$.

Up to this point the only paper we are aware of that narrows the large gap between the trivial lower bound of $\Delta$ colors in a nonrepetitive edge-coloring of a tree of maximum degree $\Delta$ and the $4\Delta-4$ upper bound from [1] is by Sudeep and Vishwanathan [12]. We will describe their results in the next section. The main result of this paper is to give the first nontrivial improvement of the upper bound from [1].

Theorem 1

If $G$ is a tree of maximum degree $\Delta$, then $\pi^{\prime}(G)\leq 3\Delta-2$.

We will give a proof of this theorem in Section 4 using a coloring method we describe in Section 3 . We discuss some possible ways for further improvements in Section 5.

2 Trees of small height

A $k$-ary tree is a tree with a designated root and the property that every vertex that is not a leaf has exactly $k$ children. The $k$-ary tree in which the distance from the root to every leaf is $h$ is denoted by $T_{k,h}$. For convenience we will assume that the vertices in $T_{k,h}$ are labeled as suggested in Figures 1 and 2 with the root labeled 1, its children labeled $2,\dots,k+1$, their children $k+2,\dots k^{2}+k+1$ and so on. This allows us to write $u<v$ if $u$ is to the left or above $v$, and also gives the vertices at each level (distance from the root) a natural left to right order.

To obtain bounds on the Thue chromatic index of general trees $G$ of maximum degree $\Delta\geq 2$ it suffices to study $k$-ary trees for $k=\Delta-1$, since $G$ is a subgraph of $T_{k,h}$ for sufficiently large $h$. Of course the Thue sequence shows that for $h>4$ we have $\pi^{\prime}(T_{1,h})=\pi^{\prime}(P_{h})=3$, and it is similarly obvious that $\pi^{\prime}(T_{k,1})=\pi^{\prime}(K_{1,k})=k$. It is easy to see that the next smallest tree $T_{2,2}$ already requires 4 colors, and Figure 1 shows the only two such 4-colorings up to isomorphism.

The Masters thesis of the second author [11] contains a proof of the fact that the type II coloring of $T_{2,2}$ extends to a unique 4-coloring of $T_{2,3}$ whereas the type I coloring extends to exactly 5 non-isomorphic 4-colorings of $T_{2,3}$, one of which we show in Figure 2. It is furthermore shown that none of these 6 colorings can be extended to $T_{2,4}$. In fact $\pi^{\prime}(T_{2,4})=5$ as we can easily extend the coloring from Figure 2 by using color 5 on one of the two new edges at every vertex from $8$ through $15$, and (for example) using colors 1,1,3,4,2,3,2,3 on the other edges in this order.

On a more general level, Sudeep and Vishwanathan [12] proved that $\pi^{\prime}(T_{k,2})=\lfloor\frac{3}{2}k\rfloor+1$ (compare also Theorem 4 of [4]) and $\pi^{\prime}(T_{k,3})>\frac{\sqrt{5}+1}{2}k>1.618k$. Their lower bounds follow from counting arguments, whereas the construction for $h=2$ consists of giving the edges at the first level colors $0,1,\dots,k-1$ and using all the $\lfloor k/2\rfloor+1$ remaining colors below each vertex at level 1. The remaining $m=\lceil k/2\rceil-1$ edges below the edge of color $i$ are colored with $i+1{~{}\rm mod~{}}k,i+2{~{}\rm mod~{}}k,\dots,i+m{~{}\rm mod~{}}k$, in other words cyclically.

To explain the general upper bound of Alon, Grytczuk, Hałuszczak and Riordan [1] we let $T_{k}$ denote the infinite $k$-ary tree. It is not difficult to see that $\pi^{\prime}(T_{k})$ is the minimum number of colors needed to color $T_{k,h}$ for every $h\geq 1$. They prove that $\pi^{\prime}(T_{k})\leq 4k$ by giving a nonrepetitive edge-coloring of $T_{k}$ on $4k$ colors as follows:

Starting with a Thue-sequence $123231\dots$ insert 4 as every third symbol to obtain a nonrepetitive sequence $S=124324314\dots$ that also does not contain a palindrome, that is a sequence of length at least 2 that reads forwards the same as backwards, such as 121. Now color the edges with a common parent at distance $h-1$ from the root with $k$ different copies $s^{(1)},\dots,s^{(k)}$ of the symbol $s$ in position $h$ of $S$. For example, the type II coloring in Figure 1 is isomorphic to the first two levels of this coloring of $T_{2}$ if we replace $1^{(1)},1^{(2)},2^{(1)},2^{(2)}$ by $1,2,3,4$ respectively. It is now easy to verify that this coloring has no repetitively colored paths that are monotone (i.e. have all vertices at different levels) since $S$ is nonrepetitive, and none with a turning point (i.e. a vertex whose two neighbors on the path are its children) since $S$ is palindrome-free.

Sudeep and Vishwanathan noted the gap between the bounds $1.618k<\pi^{\prime}(T_{k})\leq 4k$, and stated their belief that both can be improved. Even for $k=2$ the gap $3.2<\pi^{\prime}(T_{2})\leq 8$ is large. Whereas obviously $\pi^{\prime}(T_{2})\geq\pi^{\prime}(T_{2,4})=5$ is not hard to obtain, the specific question of showing that $\pi^{\prime}(T_{2})<8$ is already raised in [1] at the end of Section 4.2. Theorem 1 implies that indeed $\pi^{\prime}(T_{2})\leq 7$. On the other hand, improving on the lower bound of 5 (if that is possible) would require different ideas from those in [12] because [11] presents a nonrepetitive 5-coloring of $T_{2,10}$ as Example 3.2.6.

3 Derived colorings

In this section, which can also be found in [11], we present a way to color the edges of $T_{k}$ that is different from that used by Alon, Grytczuk, Hałuszczak and Riordan [1]. While their idea is in some sense the natural generalization of the type II coloring in the sense that the coloring precedes by level, our coloring generalizes the type I coloring by moving diagonally. The fact that the type I colorings could be extended in 5 nonisomorphic ways, whereas the extension of the type II coloring was unique encourages this notion.

Definition 1

Let $S=s_{1},s_{2},\dots$ be a sequence. The edge-coloring of a $k$-ary tree $T$ derived from $S$ is obtained as follows: The edges incident with the root receive colors $s_{1},s_{2},\dots,s_{k}$ going from left to right in this order. If $v$ is any vertex other than the root and if the edge between $v$ and its parent has color $s_{i}$, then the edges between $v$ and its children receive colors $s_{i+1},s_{i+2},\dots,s_{i+k}$ again going from left to right in this order.

To color the edges of the infinite $k$-ary tree $T_{k}$ in this fashion we need $S$ to be infinite. To color the edges of $T_{k,h}$ it suffices for the length of $S$ to be at least $kh$ (which is rather small considering that there about $k^{h}$ edges) as each level will use $k$ entries of $S$ more than the previous level (on the edges incident with the right-most vertex). For example the type I coloring of $T_{2,2}$ is the coloring derived from $S=1,2,3,4$, whereas the coloring of $T_{2,3}$ in Figure 2 is derived from $S=1,2,3,4,1,2$. The next definition will enable us to characterize infinite sequences whose derived coloring is nonrepetitive.

Definition 2

Let $S=s_{1},s_{2},\dots$ be a (finite or infinite) sequence. A sequence of indices $i_{1},i_{2},\dots,i_{2r}$ is called $k$-bad* for $S$ if there is an $m$ with $1<m\leq 2r$ such that the following four conditions hold:*

1. a)

$s_{i_{1}},s_{i_{2}},\dots,s_{i_{2r}}$* is a repetition* 2. b)

$i_{1}>i_{2}>\dots>i_{m}<i_{m+1}<i_{m+2}<\dots<i_{2r}$** 3. c)

$|i_{j}-i_{j+1}|\leq k$* for all $j$ with $1\leq j<2r$* 4. d)

$i_{m+1}<i_{m}+k$* if $m<2r$.*

$S$* is called $k$-special if it has no $k$-bad sequence of indices.*

The following proposition says something about the structure of a $k$-special sequence, namely that identical entries must be at least $2k$ apart.

Proposition 1

A sequence $S$ has a $k$-bad sequence of length at most four with $m\leq 3$ if and only if $s_{i}=s_{j}$ for some $i<j<i+2k$.

Proof 3.2**.**

For the back direction observe that if $j\leq i+k$, then the sequence of indices $j,i$ is $k$-bad with $m=2$. If $i+k\leq j<i+2k$, then the sequence $i+k-1,i,i+k-1,j$ is $k$-bad with $m=2$.

For the forward direction, observe that if $i_{1},i_{2}$ is $k$-bad (necessarily with $m=2$), then we can let $j=i_{1}$ and $i=i_{2}$. If $i_{1},i_{2},i_{3},i_{4}$ is $k$-bad with $m=2$ then we let $i=i_{2}$ and $j=i_{4}$ and observe that $i<i_{3}<j\leq i_{3}+k\leq i+2k-1$. So we may assume that $i_{1},i_{2},i_{3},i_{4}$ is $k$-bad with $m=3$. If $i_{2}=i_{4}$, then we let $i=i_{3}$ and $j=i_{1}$ and obtain $i<i_{2}<j\leq i_{4}+k-1=i_{2}+k-1\leq i+2k-1$ as desired. Otherwise $i_{2},i_{4}$ are distinct numbers $x$ with $i_{3}<x\leq i_{3}+k$ and we can let $\{i,j\}=\{i_{2},i_{4}\}$.

We are now ready to prove the following.

Theorem 3.3**.**

An infinite sequence $S$ is $k$-special if and only if the edge-coloring of $T_{k}$ derived from $S$ is nonrepetitive.

Proof 3.4**.**

$(\Rightarrow)$* Suppose that a $k$-special sequence $S$ creates a repetition on a path $P=v_{0},v_{1},\dots,v_{2r}$ in $T_{k}$, that is $R=c(v_{0}v_{1}),c(v_{1}v_{2}),\dots,c(v_{2r-1}v_{2r})$ satisfies $c(v_{i}v_{i+1})=c(v_{i+r}v_{i+r+1})$ for $0\leq i\leq r-1$. Observe that $c(v_{j}v_{j+1})=s_{i_{j+1}}$ where $0\leq j\leq 2r-1$, for some $s_{i_{j+1}}\in S$. There are two possibilities; $v_{0},v_{1},\dots,v_{2r}$ is monotone or it has a single turning point.*

**Case 1: Suppose $v_{0},v_{1},\dots,v_{2r}$ is monotone.

If $v_{0},v_{1},v_{2}\dots,v_{2r}$ is monotone then we may assume $v_{0}>v_{1}>v_{2}>\dots>v_{2r}$. Since $v_{j}>v_{j+1}$ we know that $v_{j}$ is the child of $v_{j+1}$ so we have that $i_{j}>i_{j+1}$ and $|i_{j}-i_{j+1}|\leq k$. The subsequence $s_{i_{1}},s_{i_{2}},\dots,s_{i_{2r}}$ is a repetition, so that $i_{1},\dots,i_{2r}$ is $k$-bad with $m=2r$, a contradiction.**

Case 2: Suppose $v_{0},v_{1},\dots,v_{2r}$ has a turning point $v_{m}$ for some $m$ with $0<m<2r$. By the definition of a turning point $v_{m-1}$ and $v_{m+1}$ are the children of $v_{m}$, and thus $v_{0}>v_{1}>\dots>v_{m-1}>v_{m}<v_{m+1}<\dots<v_{2r}$. We may also assume without loss of generality that $v_{m-1}<v_{m+1}$. Observe that $v_{0},v_{1},\dots,v_{m}$ is moving towards the root and $v_{m},v_{m+1},\dots,v_{2r}$ is moving away from the root. Let $c(v_{j}v_{j+1})=s_{i_{j+1}}$. We will show that $i_{1}>i_{2}>\dots>i_{m-1}>i_{m}<i_{m+1}<\dots<i_{2r}$ and that this sequence is $k$-bad for $S$. Since $v_{j-1}>v_{j}>v_{j+1}$ for $1\leq j<m$ we know that $v_{j}$ is the child of $v_{j+1}$ and the parent of $v_{j-1}$ so we have $i_{j}>i_{j+1}$ and $|i_{j}-i_{j+1}|\leq k$. Similarly, since $v_{j-1}<v_{j}<v_{j+1}$ for $m<j<2r$ we know that $v_{j}$ is the child of $v_{j-1}$ and the parent of $v_{j+1}$ so $i_{j}<i_{j+1}$ and $|i_{j}-i_{j+1}|\leq k$. Finally, since $v_{m}$ is the parent of $v_{m-1}$ and $v_{m+1}$ so $|i_{m}-i_{m+1}|<k$ and $i_{m}<i_{m+1}$ since we assumed $v_{m-1}<v_{m+1}$. The subsequence $s_{i_{1}},s_{i_{2}},\dots,s_{i_{2r}}$ is a repetition, leading to the contradiction that $i_{1},\dots,i_{2r}$ is $k$-bad.

$(\Leftarrow)$* We proceed by contrapositive. So suppose $S$ has a $k$-bad sequence ${i_{1}},{i_{2}},\dots,{i_{2r}}$. We will show that there is a path on vertices $v_{0},v_{1},v_{2},\dots,v_{2r}$ with $c(v_{j}v_{j+1})=s_{i_{j+1}}$ where the color pattern $c(v_{0}v_{1}),c(v_{1}v_{2})\dots,c(v_{2r-1}v_{2r})$ is a repetition in the derived edge-coloring of $T_{k}$. The left child of a vertex $v$ is the child with the smallest label, and we will denote this child as $v^{\prime}$. Observe that if $c(vp(v))=s_{\alpha}$, then $c(vv^{\prime})=s_{\alpha+1}$.*

If $m=2r$ then we start at the root and successively go to the left child of the current vertex until we find a vertex $v_{2r}$ such that $c(v_{2r}v_{2r}^{\prime})=s_{i_{2r}}$ and let $v_{2r-1}=v_{2r}^{\prime}$. Let $v_{2r-2}$ be the child of $v_{2r-1}$ with $c(v_{2r-1}v_{2r-2})=s_{i_{2r-1}}$ (this exists since $|i_{j}-i_{j+1}|\leq k$). We continue in this way until we have found $v_{0}$. Now observe that the color pattern of $v_{0},v_{1},\dots,v_{2r}$ is $s_{i_{1}},s_{i_{2}},\dots,s_{i_{2r}}$ as desired.

If $m<2r$ then we start at the root and successively go to the left child of the current vertex until we find a vertex $v_{m}$ such that $c(v_{m}v^{\prime}_{m})=s_{i_{m}}$ and let $v_{m-1}=v_{m}^{\prime}$. Let $v_{m+1}$ be the child of $v_{m}$ with $c(v_{m}v_{m+1})=s_{i_{m+1}}$ (this exists since $i_{m}<i_{m+1}<i_{m}+k$). Now, for $0\leq p\leq(m-1)$ we successively find a child $v_{p-1}$ of $v_{p}$ such that $c(v_{p}v_{p-1})=s_{i_{p}}$. The existence of $v_{p-1}$ is guaranteed by the fact $|i_{p}-i_{p-q}|\leq k$ as in the case $m=2r$. For $m+1\leq q\leq 2r$ we successively find a child $v_{q+1}$ of $v_{q}$ such that $c(v_{q}v_{q+1})=s_{i_{q-1}}$ which we can do since $|i_{q}-i_{q+1}|\leq k$. Now observe that the color pattern of $v_{0},v_{1},\dots,v_{2r}$ is $s_{i_{1}},s_{i_{2}},\dots,s_{i_{2r}}$ as desired.

Remark 3.5**.**

Observe that the proof of the forward direction also works for the finite case $T_{k,h}$, a fact we will use in Section 5. However, the back direction need not hold in this case: We already mentioned that the coloring derived from $S=1,2,3,4,1,2$ in Figure 2 is nonrepetitive (see also $k=2$ in Proposition 5.13), but this sequence $S$ is not 2-special, because the index-sequence $3,1,2,3,5,6$ is $2$-bad.

Thus to get a good upper bound on $\pi^{\prime}(T_{k})$ we just need an infinite $k$-special sequence with few symbols. As every $2k$ consecutive elements must be distinct, the following simple idea turns out to be useful: from a sequence $S$ on $q$ symbols we can form a sequence $S^{(w)}$ on $qw$ symbols by replacing each symbol $t$ in $S$ by a block $T=t^{(0)},t^{(1)},\dots t^{(w-1)}$ of $w$ symbols. In [11] it is shown that if $S$ is nonrepetitive and palindrome-free then $S^{(k)}$ is $k$-special. This gives a new proof of the result from [1] that $\pi^{\prime}(T_{k})\leq 4k$. In the next section we will improve on that.

4 Main result

We begin with the simple observation, that if $S$ is a sequence then $S^{(k+1)}=S^{+}$ has the property that if $i,j$ are indices with $s^{+}_{i}=x^{(u)}$ and $s^{+}_{j}=y^{(v)}$ then $i<j\leq i+k$ implies that either $x=y$ and $u<v$, or $s^{+}_{i}$ and $s^{+}_{j}$ are in consecutive blocks $XY$ of $S^{+}$ and $u>v$. In other words we can tell whether we are moving left or right through the sequence just by looking at the superscripts (as long as consecutive symbols in $S$ are distinct.) As a starting point we immediately get the following result.

Corollary 4.6**.**

For all $k\geq 1$, $\pi^{\prime}(T_{k})\leq 3k+3$.

Proof 4.7**.**

It is enough to show that $S^{+}$ on $3(k+1)$ is $k$-special whenever $S$ is an infinite Thue sequence on 3 symbols. Suppose there is a $k$-bad sequence of indices $i_{1},\dots,i_{2r}$. Since every sequence of $2(k+1)$ consecutive symbols in $S^{+}$ is distinct we get that $r>1$ by Proposition 1. If $m<2r$, then we can find an index $j$ such that $i_{j}>i_{j+1}$ and $i_{r+j}<i_{r+j+1}$ with $s_{i_{j}}=s_{i_{r+j}}=x^{(u)}$ and $s_{i_{j+1}}=s_{i_{r+j+1}}=y^{(v)}$. Indeed, if $2<m\leq r$ we let $j=1$, and otherwise we let $j=m-r$. In this case $x=y$ and $u\leq v$ would violate $i_{j}>i_{j+1}\geq i_{j}-k$, whereas $u\geq v$ would violate $i_{r+j}<i_{r+j+1}\leq i_{r+j}+k$. Similarly if $x\neq y$, then $u\geq v$ would violate $i_{j}>i_{j+1}\geq i_{j}-k$, whereas $u\leq v$ would violate $i_{r+j}<i_{r+j+1}\leq i_{r+j}+k$.

It remains to observe that in the case when $m=2r$ the sequence $s_{i_{1}},s_{i_{2}},\dots,s_{i_{2r}}$ in $S^{+}$ yields a repetition in $S$ by erasing the superscripts and merging identical consecutive terms where necessary.

This bound can be improved to $3k+2$ by removing all symbols of the form $a^{(0)}$ from $S^{+}$ for one of the symbols $a$ from $S$ and showing that the resulting sequence is still $k$-special. However, we can do a bit better. In fact, Theorem 1 follows directly from our main result in this section.

Theorem 4.8**.**

There are arbitrarily long $k$-special sequences on $3k+1$ symbols.

One difficulty is that removing two symbols from $S^{+}$ can easily result in the sequence not being $k$-special anymore. To make the proof work we need to start with a Thue sequence with additional properties. The following result was proved by Thue [14] and reformulated by Berstel [2, 3] using modern conventions.

Theorem 4.9**.**

There are arbitrarily long nonrepetitive sequences with symbols $a,b,c$ that do not contain $aba$ or $bab$.

To give an idea of how such a sequence can be found, observe that it must be built out of blocks of the form $ca,cb,cab,$ and $cba$ which we denote by $x,y,z,u$, respectively. (In fact, Thue primarily studied two-way infinite sequences, but for our purposes we may simply assume our sequence starts with $c$.) We first build a sufficiently long sequence on the 5 symbols $A,B,C,D,E$ by starting with the sequence ”B” and then in each step simultaneously replacing each letter as follows:

[TABLE]

In the resulting sequence we then let $A=zuyxu$, $B=zu$, $C=zuy$, $D=zxu$, $E=zxy$. Lastly we replace $x$, $y$, $z$ and $u$ as aforementioned. For example, from $B$ we obtain $BDC$, and then after a second step $BDCBEACBDAE$. This translates to the intermediate sequence

$zuzxuzuyzuzxyzuyxuzuyzuzxuzuyxuzxy$, which gives us the desired sequence

$cabcbacabcacbacabcbacbcabcbacabcacbcabcbacbcacbacabcbacbcabcbacabcacbacabcbacbcacbacabcacb$.

It is worth pointing out that Thue’s work goes deeper in that he essentially characterizes all two-way infinite sequences that meet the conditions from Theorem 4.9 as well as several other related sequences. We also want to mention that the $A,B,C$ in the following proof have nothing to do with the $A,B,C$ in the previous paragraph, but we wanted to maintain the notation used in [2, 3].

Proof 4.10** (of Theorem 4.8).**

Start with an infinite sequence $S$ in the form of Theorem 4.9 and replace each occurrence of $c$ by a block $C$ of $k+1$ consecutive symbols $c^{(0)},c^{(1)},\dots,c^{(k)}$, whereas we replace each occurrence of $a$ or $b$ by shorter blocks $A=a^{(1)},\dots,a^{(k)}$ and $B=b^{(1)},\dots,b^{(k)}$ respectively. We claim that the resulting sequence $S^{\prime}$ on $3k+1$ symbols is $k$-special. So suppose there is a $k$-bad sequence of indices $i_{1},\dots,i_{2r}$. As before when $m=2r$ the sequence $s_{i_{1}},s_{i_{2}},\dots,s_{i_{2r}}$ in $S^{\prime}$ yields a repetition in $S$ by erasing the superscripts and merging identical consecutive terms where necessary, as we can not ”jump” over any of the blocks $A,B$ or $C$ in $S^{\prime}$. So we may assume that $1<m<2r$, and since every $2k$ consecutive elements are distinct Proposition 1 implies that $r>2$.

Claim:* If there is an index $j$ with $0<j<r$ such that $i_{j}>i_{j+1}$ and $i_{r+j}<i_{r+j+1}$, then $s_{i_{j}}=s_{i_{r+j}}=x^{(u)}$ and $s_{i_{j+1}}=s_{i_{r+j+1}}=y^{(u)}$ for $1\leq u\leq k$ and $\{x,y\}=\{a,b\}$. Consequently, $i_{j}-i_{j+1}=k=i_{r+j+1}-i_{r+j}$.*

Indeed, $s_{i_{j}}=s_{i_{r+j}}=x^{(u)}$ and $s_{i_{j+1}}=s_{i_{r+j+1}}=y^{(v)}$ for some $u,v,x,y$. If $x=y$, then $u\leq v$ would violate $i_{j}>i_{j+1}\geq i_{j}-k$, whereas $u\geq v$ would violate $i_{r+j}<i_{r+j+1}\leq i_{r+j}+k$. Thus $x\neq y$. Now $u>v$ would violate $i_{j}>i_{j+1}\geq i_{j}-k$, whereas $u<v$ would violate $i_{r+j}<i_{r+j+1}+k$. So we may assume that $u=v$. If $x=c$, then this would violate $i_{j}>i_{j+1}\geq i_{j}+k$ (as the presence of $c^{(0)}$ means that the distance is $k+1$). Similarly if $y=c$, then this violates $i_{r+j}<i_{r+j+1}\leq i_{r+j}+k$. Hence we must have $\{x,y\}=\{a,b\}$ finishing the proof of the claim.

If $r<m<2r$, then we can apply the claim with $j=m-r$ and obtain consequently that $i_{m+1}-i_{m}=k$, in direct contradiction to condition d) from Definition 2.

So we suppose that $2\leq m\leq r$. In this case we will let $j=m-1$ in our claim and we may assume due to the symmetry of $S$ in $a,b$ that $x=a$ and $y=b$. Thus for some $u$ with $1\leq u\leq k$ we get $s_{i_{m-1}}=a^{(u)}=s_{i_{m+r-1}}$ and $s_{i_{m}}=b^{(u)}=s_{i_{m+r}}$. If $m>2$, then we may apply the claim again with $j=m-2$ to obtain that $s_{i_{m-2}}=b^{(u)}=s_{i_{m+r-2}}$. However, the fact that $i_{m-2}>i_{m-1}>i_{m}$ correspond to symbols $b^{(u)},a^{(u)},b^{(u)}$ means that $S^{\prime}$ must have consecutive blocks $BAB$, yielding a contradiction to the fact that in $S$ we had no consecutive symbols $bab$.

So we may assume that $m=2$. Since $r>2$ and $s_{i_{2}}=b^{(u)}$ and $i_{2}<\dots<i_{r}$ we have that for $3\leq j\leq r$ either all $s_{i_{j}}$ are of the form $b^{(u_{j})}$ or there is a smallest index $j$ such that $s_{i_{j}}=x^{(u_{j})}$ for some $x\neq b$. In the first case it follows that there must be consecutive blocks $BAB$ (yielding a contradiction) such that $i_{1}$ and $i_{r+1}$ are in the $A$ block, $i_{2},\dots i_{r}$ are in the first $B$-block and $i_{r+2},\dots,i_{2r}$ are in the second. In the second case it follows that since there must be blocks $BA$ with $i_{1}$ in $A$ and $i_{2}$ in $B$, that $i_{j}$ must be in the $A$ block again, that is $s_{i_{j}}=a^{(u_{j})}$. However, since $i_{r+1}<\dots<i_{r+j}$ it follows that there must be consecutive blocks $ABA$ in $S^{\prime}$ (our final contradiction), such that $i_{r+1}$ is in the first $A$ block, $i_{r+j}$ in the second and $i_{r+2},\dots,i_{r+j-1}$ are in the $B$ block.

5 $k$-special sequences on at most $3k$ symbols

One possible way to improve on Theorem 1 is to study $k$-special sequences on at most $3k$ symbols. The sequence $S_{n,c}=1,2,\dots,n,1,2,\dots c$ for $n>c\geq 0$ turns out to be a key example in this situation.

Recall that by Proposition 1 the entries in a block of length $2k$ of a $k$-special sequence must all be distinct. Thus, if we let $f_{k}(n)$ denote the maximum length of a $k$-special sequence $S$ on $n$ symbols, then this observation immediately implies that $f_{k}(n)=n$ when $n<2k$ and up to isomorphism the only sequence achieving this value is $S_{n,0}$. When $n\geq 2k$ we can furthermore assume without loss of generality that if $S$ is nonrepetitive on $n$ symbols, then $S_{i}=i$ for $1\leq i\leq 2k$ (just like $S_{n,c}$.)

If $n=2k$ then it follows from Proposition 1 that a sequence achieving $f_{k}(2k)$ must be of the form $S_{2k,c}$. It is easy to check $S_{2k,1}$ is in fact $k$-special, whereas $S_{2k,2}$ contains the $k$-bad index sequence $k+1,1,2,k+1,2k+1,2k+2$, which yields the repetition $k+1,1,2,k+1,1,2$. Thus $f_{k}(2k)=2k+1$ with $S_{2k,1}$ being the unique sequence achieving this value. This $k$-bad index sequence also explains why we could not have consecutive blocks $ABA$ or $BAB$ in our construction for Theorem 4.8 . For the remaining range we get

Proposition 5.11**.**

1. a)

If $n\geq 2k$, then $S_{n,n-k}$ has a $k$-bad sequence only when $n=2k$ and such a sequence must have $2=m<r$. 2. b)

If $n\geq 2k+1$, then $f_{k}(n)\geq 2n-k$.

Proof 5.12**.**

It suffices to prove the first statement, as it immediately implies the second. So suppose $n\geq 2k$ and $I=i_{1},\dots,i_{2r}$ is a $k$-bad sequence of indices for some $m$. If $m=2r$, then $I$ is decreasing and so the fact that $s_{i_{j}}=s_{i_{j+r}}$ for all $1\leq j\leq r$ implies that $i_{1}>\dots>i_{r}\geq n+1$ and $n-k\geq i_{r+1}>\dots>i_{2r}$, yielding the contradiction $i_{r}-i_{r+1}>k$. So we may assume that $m<2r$.

If $m>r$, then let $m^{\prime}=m-r$. Since $s_{i_{m}}=s_{i_{m^{\prime}}}$ and $i_{m^{\prime}}>i_{m}$, it follows that $i_{m}=i_{m^{\prime}}-n\in\{1,\dots,n-k\}$. Since $i_{m^{\prime}}\geq n,$ $i_{m}\leq n-k$ and for all $j$ we have $|i_{j}-i_{j+1}|\leq k$ it follows that there must be some $j$ with $m^{\prime}<j<m$ such that $i_{j}\in\{n-k+1,\dots,n\}$. Since $I$ yields a repetition with $i_{1}>\dots>i_{m}$, but the symbol $s_{i_{j}}=i_{j}$ is unique in $S_{n,n-k}$ we conclude that $i_{j}=i_{j+r}$. It follows that $j=m^{\prime}+1$, since otherwise $i_{m^{\prime}}>i_{j-1}>i_{j}$ and $i_{m}<i_{j+r-1}<i_{j+r}$ would contradict $s_{i_{j-1}}=s_{i_{j+r-1}}$ as the sets $\{s_{i_{j}+1},s_{i_{j}+2},\dots,s_{i_{m^{\prime}}-1}\}$ and $\{s_{i_{m}+1},s_{i_{m}+2},\dots s_{i_{j}-1}\}$ are disjoint. Now $j=m^{\prime}+1$ implies that $i_{m^{\prime}}-k=i_{j-1}-k\leq i_{j}=i_{j+r}=i_{m+1}\leq i_{m}+k-1$, and since $i_{m^{\prime}}=i_{m}+n$ we get $n\leq 2k-1$, a contradiction.

If $m\leq r$, then let $m^{\prime}=m+r$. It follows again that $i_{m^{\prime}}=i_{m}+n$, and that there must be some $j$ such that $i_{j}=i_{j+r}\in\{n-k+1,\dots,n\}$ and $j<m<j+r$. Thus $m^{\prime}>j+r$ this time. It follows that $j=m-1$, since otherwise $i_{j}>i_{j+1}>i_{m}$ and $i_{j+r}<i_{j+r+1}<i_{m^{\prime}}$ would contradict $s_{i_{j+1}}=s_{i_{j+r+1}}$ as the sets $\{s_{i_{m}+1},s_{i_{m}+2},\dots s_{i_{j}-1}\}$ and $\{s_{i_{j}+1},s_{i_{j}+2},\dots,s_{i_{m^{\prime}}-1}\}$ are still disjoint. Now $j=m-1$ implies that $i_{m}+k=i_{j+1}+k\geq i_{j}=i_{j+r}=i_{m^{\prime}-1}\geq i_{m^{\prime}}-k$, and since $i_{m^{\prime}}=i_{m}+n$ we get $n\leq 2k$, a contradiction unless $n=2k$. In this case also $i_{m}+k=i_{j}=i_{j+r}=i_{m^{\prime}}-k=x$ for some $k+1\leq x\leq n=2k$.

If we have $m>2$ then $j-1=m-2\geq 1$ and we consider $i_{j-1}$. Since $i_{j+r-1}<i_{j+r}$ and $k+1=n-k+1\leq s_{i_{j}}\leq n=2k$ implies that $s_{i_{j+r-1}}\in\{x-k,x-k+1,\dots,x-1\}$. Similarly $i_{j-1}>i_{j}$ implies that $s_{i_{j-1}}\in\{x+1,x+2,\dots,n\}\cup\{1,2,\dots,k-(n-x)=x-k\}$. Since $s_{i_{j+r-1}}=s_{i_{j-1}}$ it now follows that this value must be $x-k=i_{m}$. Hence $i_{j+r-1}=i_{m}$ and thus $m=j+r-1=(m-1)+r-1$. This implies the contradiction $2=r\geq m>2$. Hence $m=2$ and the fact that $r>2$ follows from Proposition 1 and the fact that the distance between identical labels is $2k$.

We believe that for in Proposition 5.11 b) equality holds when $2k<n<3k$. An exhaustive search by computer shows that this is the case when $2k<n<3k$ with $n\leq 16$. Moreover $S_{2k+1,k+1}$ turns out to be the unique sequence achieving $f_{k}(2k+1)=3k+2$, whereas for $2k+2\leq n<3k$ a typical sequence achieving $f_{k}(n)$ is obtained by permuting the last $n-k$ entries of $S_{n,n-k}$.

Proposition 5.13**.**

The coloring of $T_{k,3}$ derived from $S_{2k,k}$ is nonrepetitive.

Proof 5.14**.**

If the coloring of $T_{k,3}$ derived from $S_{2k,k}$ contains a repetition of length $2r$, then as in the proof of Theorem 3.3 it follows that there must be a $k$-bad sequence of $2r$ indices. From Proposition 5.11 a) it now follows that $r>m=2$. Since a longest path in $T_{k,3}$ has 6 edges we must have $r=3$. However, any repetition of length 6 would have to connect two leaves and turn around at the root, and as such would have $m=3$, a contradiction.

Combining everything we know so far we get

Corollary 5.15**.**

If $h\geq 3$, then $\pi^{\prime}(T_{k,h})\leq\lceil\frac{h+1}{2}k\rceil$.

Proof 5.16**.**

If $h=3$, then the result follows from Proposition 5.13. For $h>3$ we can apply Proposition 5.11 b) with $n=\lceil\frac{h+1}{2}k\rceil$. Since $2n-k\geq hk$ it now follows from Remark 3.5 that the coloring of $T_{k,h}$ derived from $S_{n,n-k}$ is nonrepetitive.

The bound in Corollary 5.15 is better than that derived from Theorem 4.8 when $h\leq 5$ and we obtain the following table of values for $\pi^{\prime}(T_{h,k})$, where the presence of two values denotes a lower and an upper bound. The values marked by an asterisk were confirmed by computer search. The programs used are based on those found in [11] and the Python code is available at http://public.csusm.edu/akundgen/Python/Nonrepetitive.py

It is worth noting that even though it may be possible to use derived colorings to improve individual columns of this table by a more careful argument (as we did in Proposition 5.13), this seems unlikely to work for $\pi^{\prime}(T_{k})$ in general. Theorem 3.3 implies that the infinite sequence from which we derive the coloring must be $k$-special, and while we were able to provide such a sequence on $3k+1$ symbols, it seems unlikely that there are such sequences on $3k$ symbols. An exhaustive search shows that for $k\leq 5$ the maximum length of a $k$-special sequence on $n=3k$ symbols is $5k+3$, which is only 3 more than the length of $S_{n,n-k}$. The $k!$ examples achieving this value are all of the strange form $[1,2k],1,[2k+1,3k],x_{1},[k+2,2k],1,x_{2},x_{3},\dots,x_{k},x_{1},2k+1$ where $\{x_{1},\dots,x_{k}\}=\{2,\dots,k+1\}$ and $[a,b]$ denotes $a,a+1,a+2\dots,b$. In other words they are $S_{3k,2k+1}$ with the last $2k+1$ entries permuted and with 1 and $x_{1}$ inserted after positions $2k$ and $3k$.

A more promising next step would be to try to improve the lower bounds for $\pi^{\prime}(T_{k,h})$ for $h=3,4,5$.