Nonrepetitive edge-colorings of trees
A. K\"undgen, T. Talbot

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 for the Thue edge-chromatic number of trees of maximum degree due to Alon, Grytczuk, Ha{\l}uszczak and Riordan (2002) by providing a simple nonrepetitive edge-coloring with colors.
| 1 | 2 | 3 | 4 | 5 | 6-10 | ||
| 1 | 1 | 2 | 2 | 3 | 3 | 3 | 3 |
| 2 | 2 | 4 | 4 | 5 | 5,7 | ||
| 3 | 3 | 5 | 6,9 | 6,10 | 6,10 | ||
| 4 | 4 | 7 | 7,10 | 7,12 | 7,13 | 7,13 | |
| 5 | 5 | 8 | 9,10 | 9,13 | 9,15 | 9,16 | 9,16 |
| +1 |
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Taxonomy
TopicsAdvanced Graph Theory Research · Graph Labeling and Dimension Problems · semigroups and automata theory
\publicationdetails
1920171182651
Nonrepetitive edge-colorings of trees
André Kündgen\affiliationmark1 Supported by ERC Advanced Grant GRACOL, project no. 320812.
Tonya Talbot\affiliationmark1
California State University San Marcos, San Marcos, CA, USA
(2017-1-17; 2017-6-4)
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 for the Thue edge-chromatic number of trees of maximum degree due to Alon, Grytczuk, Haluszczak and Riordan (2002) by providing a simple nonrepetitive edge-coloring with colors.
keywords:
Thue coloring, Repetition-free coloring, Square-free coloring
1 Introduction
A repetition is a sequence of even length (for example ), 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 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 is nonrepetitive. We will use to denote the Thue chromatic index of a graph , which is the minimum number of colors in a nonrepetitive edge-coloring of . In [1] the notation 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 . Their paper contains many interesting ideas and questions, the most intriguing of which is if is bounded by a constant when is planar. The best result in this direction is due to Dujmović, Frati, Joret, and Wood [7] who show that for planar graphs on vertices is . Conjecture 2 from [1] was settled by Currie [6] who showed that for the -cycle , when . One of the conjectures from [1] that remains open is whether when is a graph of maximum degree . At least 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 is the path on vertices, then . (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 has a nonrepetitive edge-coloring with colors and stated that the same method can be used to obtain a nonrepetitive vertex-coloring with 4 colors. However, while the star 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 there are trees that have no nonrepetitive vertex-coloring from lists of size .
Up to this point the only paper we are aware of that narrows the large gap between the trivial lower bound of colors in a nonrepetitive edge-coloring of a tree of maximum degree and the 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 is a tree of maximum degree , then .
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 -ary tree is a tree with a designated root and the property that every vertex that is not a leaf has exactly children. The -ary tree in which the distance from the root to every leaf is is denoted by . For convenience we will assume that the vertices in are labeled as suggested in Figures 1 and 2 with the root labeled 1, its children labeled , their children and so on. This allows us to write if is to the left or above , 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 of maximum degree it suffices to study -ary trees for , since is a subgraph of for sufficiently large . Of course the Thue sequence shows that for we have , and it is similarly obvious that . It is easy to see that the next smallest tree 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 extends to a unique 4-coloring of whereas the type I coloring extends to exactly 5 non-isomorphic 4-colorings of , one of which we show in Figure 2. It is furthermore shown that none of these 6 colorings can be extended to . In fact 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 through , 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 (compare also Theorem 4 of [4]) and . Their lower bounds follow from counting arguments, whereas the construction for consists of giving the edges at the first level colors and using all the remaining colors below each vertex at level 1. The remaining edges below the edge of color are colored with , in other words cyclically.
To explain the general upper bound of Alon, Grytczuk, Hałuszczak and Riordan [1] we let denote the infinite -ary tree. It is not difficult to see that is the minimum number of colors needed to color for every . They prove that by giving a nonrepetitive edge-coloring of on colors as follows:
Starting with a Thue-sequence insert 4 as every third symbol to obtain a nonrepetitive sequence 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 from the root with different copies of the symbol in position of . For example, the type II coloring in Figure 1 is isomorphic to the first two levels of this coloring of if we replace by 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 is nonrepetitive, and none with a turning point (i.e. a vertex whose two neighbors on the path are its children) since is palindrome-free.
Sudeep and Vishwanathan noted the gap between the bounds , and stated their belief that both can be improved. Even for the gap is large. Whereas obviously is not hard to obtain, the specific question of showing that is already raised in [1] at the end of Section 4.2. Theorem 1 implies that indeed . 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 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 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 be a sequence. The edge-coloring of a -ary tree derived from is obtained as follows: The edges incident with the root receive colors going from left to right in this order. If is any vertex other than the root and if the edge between and its parent has color , then the edges between and its children receive colors again going from left to right in this order.
To color the edges of the infinite -ary tree in this fashion we need to be infinite. To color the edges of it suffices for the length of to be at least (which is rather small considering that there about edges) as each level will use entries of more than the previous level (on the edges incident with the right-most vertex). For example the type I coloring of is the coloring derived from , whereas the coloring of in Figure 2 is derived from . The next definition will enable us to characterize infinite sequences whose derived coloring is nonrepetitive.
Definition 2
Let be a (finite or infinite) sequence. A sequence of indices is called -bad* for if there is an with such that the following four conditions hold:*
- a)
* is a repetition* 2. b)
** 3. c)
* for all with * 4. d)
* if .*
* is called -special if it has no -bad sequence of indices.*
The following proposition says something about the structure of a -special sequence, namely that identical entries must be at least apart.
Proposition 1
A sequence has a -bad sequence of length at most four with if and only if for some .
Proof 3.2**.**
For the back direction observe that if , then the sequence of indices is -bad with . If , then the sequence is -bad with .
For the forward direction, observe that if is -bad (necessarily with ), then we can let and . If is -bad with then we let and and observe that . So we may assume that is -bad with . If , then we let and and obtain as desired. Otherwise are distinct numbers with and we can let .
We are now ready to prove the following.
Theorem 3.3**.**
An infinite sequence is -special if and only if the edge-coloring of derived from is nonrepetitive.
Proof 3.4**.**
* Suppose that a -special sequence creates a repetition on a path in , that is satisfies for . Observe that where , for some . There are two possibilities; is monotone or it has a single turning point.*
**Case 1: Suppose is monotone.
If is monotone then we may assume . Since we know that is the child of so we have that and . The subsequence is a repetition, so that is -bad with , a contradiction.**
Case 2: Suppose has a turning point for some with . By the definition of a turning point and are the children of , and thus . We may also assume without loss of generality that . Observe that is moving towards the root and is moving away from the root. Let . We will show that and that this sequence is -bad for . Since for we know that is the child of and the parent of so we have and . Similarly, since for we know that is the child of and the parent of so and . Finally, since is the parent of and so and since we assumed . The subsequence is a repetition, leading to the contradiction that is -bad.
* We proceed by contrapositive. So suppose has a -bad sequence . We will show that there is a path on vertices with where the color pattern is a repetition in the derived edge-coloring of . The left child of a vertex is the child with the smallest label, and we will denote this child as . Observe that if , then .*
If then we start at the root and successively go to the left child of the current vertex until we find a vertex such that and let . Let be the child of with (this exists since ). We continue in this way until we have found . Now observe that the color pattern of is as desired.
If then we start at the root and successively go to the left child of the current vertex until we find a vertex such that and let . Let be the child of with (this exists since ). Now, for we successively find a child of such that . The existence of is guaranteed by the fact as in the case . For we successively find a child of such that which we can do since . Now observe that the color pattern of is as desired.
Remark 3.5**.**
Observe that the proof of the forward direction also works for the finite case , 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 in Figure 2 is nonrepetitive (see also in Proposition 5.13), but this sequence is not 2-special, because the index-sequence is -bad.
Thus to get a good upper bound on we just need an infinite -special sequence with few symbols. As every consecutive elements must be distinct, the following simple idea turns out to be useful: from a sequence on symbols we can form a sequence on symbols by replacing each symbol in by a block of symbols. In [11] it is shown that if is nonrepetitive and palindrome-free then is -special. This gives a new proof of the result from [1] that . In the next section we will improve on that.
4 Main result
We begin with the simple observation, that if is a sequence then has the property that if are indices with and then implies that either and , or and are in consecutive blocks of and . 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 are distinct.) As a starting point we immediately get the following result.
Corollary 4.6**.**
For all , .
Proof 4.7**.**
It is enough to show that on is -special whenever is an infinite Thue sequence on 3 symbols. Suppose there is a -bad sequence of indices . Since every sequence of consecutive symbols in is distinct we get that by Proposition 1. If , then we can find an index such that and with and . Indeed, if we let , and otherwise we let . In this case and would violate , whereas would violate . Similarly if , then would violate , whereas would violate .
It remains to observe that in the case when the sequence in yields a repetition in by erasing the superscripts and merging identical consecutive terms where necessary.
This bound can be improved to by removing all symbols of the form from for one of the symbols from and showing that the resulting sequence is still -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 -special sequences on symbols.
One difficulty is that removing two symbols from can easily result in the sequence not being -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 that do not contain or .
To give an idea of how such a sequence can be found, observe that it must be built out of blocks of the form and which we denote by , respectively. (In fact, Thue primarily studied two-way infinite sequences, but for our purposes we may simply assume our sequence starts with .) We first build a sufficiently long sequence on the 5 symbols 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 , , , , . Lastly we replace , , and as aforementioned. For example, from we obtain , and then after a second step . This translates to the intermediate sequence
, which gives us the desired sequence
.
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 in the following proof have nothing to do with the 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 in the form of Theorem 4.9 and replace each occurrence of by a block of consecutive symbols , whereas we replace each occurrence of or by shorter blocks and respectively. We claim that the resulting sequence on symbols is -special. So suppose there is a -bad sequence of indices . As before when the sequence in yields a repetition in by erasing the superscripts and merging identical consecutive terms where necessary, as we can not ”jump” over any of the blocks or in . So we may assume that , and since every consecutive elements are distinct Proposition 1 implies that .
Claim:* If there is an index with such that and , then and for and . Consequently, .*
Indeed, and for some . If , then would violate , whereas would violate . Thus . Now would violate , whereas would violate . So we may assume that . If , then this would violate (as the presence of means that the distance is ). Similarly if , then this violates . Hence we must have finishing the proof of the claim.
If , then we can apply the claim with and obtain consequently that , in direct contradiction to condition d) from Definition 2.
So we suppose that . In this case we will let in our claim and we may assume due to the symmetry of in that and . Thus for some with we get and . If , then we may apply the claim again with to obtain that . However, the fact that correspond to symbols means that must have consecutive blocks , yielding a contradiction to the fact that in we had no consecutive symbols .
So we may assume that . Since and and we have that for either all are of the form or there is a smallest index such that for some . In the first case it follows that there must be consecutive blocks (yielding a contradiction) such that and are in the block, are in the first -block and are in the second. In the second case it follows that since there must be blocks with in and in , that must be in the block again, that is . However, since it follows that there must be consecutive blocks in (our final contradiction), such that is in the first block, in the second and are in the block.
5 -special sequences on at most symbols
One possible way to improve on Theorem 1 is to study -special sequences on at most symbols. The sequence for turns out to be a key example in this situation.
Recall that by Proposition 1 the entries in a block of length of a -special sequence must all be distinct. Thus, if we let denote the maximum length of a -special sequence on symbols, then this observation immediately implies that when and up to isomorphism the only sequence achieving this value is . When we can furthermore assume without loss of generality that if is nonrepetitive on symbols, then for (just like .)
If then it follows from Proposition 1 that a sequence achieving must be of the form . It is easy to check is in fact -special, whereas contains the -bad index sequence , which yields the repetition . Thus with being the unique sequence achieving this value. This -bad index sequence also explains why we could not have consecutive blocks or in our construction for Theorem 4.8 . For the remaining range we get
Proposition 5.11**.**
- a)
If , then has a -bad sequence only when and such a sequence must have . 2. b)
If , then .
Proof 5.12**.**
It suffices to prove the first statement, as it immediately implies the second. So suppose and is a -bad sequence of indices for some . If , then is decreasing and so the fact that for all implies that and , yielding the contradiction . So we may assume that .
If , then let . Since and , it follows that . Since and for all we have it follows that there must be some with such that . Since yields a repetition with , but the symbol is unique in we conclude that . It follows that , since otherwise and would contradict as the sets and are disjoint. Now implies that , and since we get , a contradiction.
If , then let . It follows again that , and that there must be some such that and . Thus this time. It follows that , since otherwise and would contradict as the sets and are still disjoint. Now implies that , and since we get , a contradiction unless . In this case also for some .
If we have then and we consider . Since and implies that . Similarly implies that . Since it now follows that this value must be . Hence and thus . This implies the contradiction . Hence and the fact that follows from Proposition 1 and the fact that the distance between identical labels is .
We believe that for in Proposition 5.11 b) equality holds when . An exhaustive search by computer shows that this is the case when with . Moreover turns out to be the unique sequence achieving , whereas for a typical sequence achieving is obtained by permuting the last entries of .
Proposition 5.13**.**
The coloring of derived from is nonrepetitive.
Proof 5.14**.**
If the coloring of derived from contains a repetition of length , then as in the proof of Theorem 3.3 it follows that there must be a -bad sequence of indices. From Proposition 5.11 a) it now follows that . Since a longest path in has 6 edges we must have . However, any repetition of length 6 would have to connect two leaves and turn around at the root, and as such would have , a contradiction.
Combining everything we know so far we get
Corollary 5.15**.**
If , then .
Proof 5.16**.**
If , then the result follows from Proposition 5.13. For we can apply Proposition 5.11 b) with . Since it now follows from Remark 3.5 that the coloring of derived from is nonrepetitive.
The bound in Corollary 5.15 is better than that derived from Theorem 4.8 when and we obtain the following table of values for , 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 in general. Theorem 3.3 implies that the infinite sequence from which we derive the coloring must be -special, and while we were able to provide such a sequence on symbols, it seems unlikely that there are such sequences on symbols. An exhaustive search shows that for the maximum length of a -special sequence on symbols is , which is only 3 more than the length of . The examples achieving this value are all of the strange form where and denotes . In other words they are with the last entries permuted and with 1 and inserted after positions and .
A more promising next step would be to try to improve the lower bounds for for .
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