1 Introduction
###### Abstract

MSC: 05C15 (05C55, 05D40)

Keywords: nonrepetitive colouring, total Thue number

Total Thue colourings of graphs

Jens Schreyer111Institute of Mathematics, Faculty of Mathematics and Natural Sciences, Ilmenau University of Technology, Ilmenau, Germany,  [jens.schreyer@tu-ilmenau.de] and Erika Škrabul’áková222Institute of Control and Informatization of Production Processes, Faculty of Mining, Ecology, Process Control and Geotechnology, Technical University of Košice, Košice, Slovakia,  [erika.skrabulakova@tuke.sk]333This work was supported by the Slovak Research and Development Agency under the contract No. APVV-0482-11, by the grant VEGA 1/0130/12 and DAAD.

MSC: 05C15 (05C55, 05D40)

Keywords: nonrepetitive colouring, total Thue number

## 1 Introduction

A finite sequence of symbols is called a repetition if for all . A sequence is called repetitive if it contains a subsequence of consecutive terms that is a repetition. Otherwise is called nonrepetitive. Nonrepetitive sequences were first studied by Axel Thue in the beginning of the last century as a part of the investigation of word structures. In his famous paper from 1906 [16] he showed the existence of arbitrarily long nonrepetitive sequences consisting only of three different symbols. Later these sequences found widespread applications not only in mathematics, but also in informatics, data security management and elsewhere. Their first appearance in graph theory was in 1987 (see Currie [6]) but the investigation of nonrepetitive graph colourings began with the seminal paper of Alon et. al. from 2002 [1].
Let be a colouring of the vertices of a graph . We say that is a nonrepetitive vertex-colouring of if for any simple path on vertices , in the associated sequence of colours is not a repetition. The minimum number of colours in a nonrepetitive vertex-colouring of a graph is the Thue chromatic number . Analogously nonrepetitive edge-colourings and the Thue chromatic index are defined. The original paper of Alon et. al. [1] introduced both variants of colouring however it foccused on the edge variant. Here it was proved that for arbitrary graph it holds , where is the maximum degree of the considered graph. Unhappily the Thue chromatic index of the graph is here called the Thue number of and denoted . This was the cause of a lot of misunderstanding and misprints in several subsequent papers that gave more attention to the vertex-colouring variant for which the upper bound of the same form: , is known. The constant was improved several times and the best known bound today is by Dujmović et. al. [9] who could show that for large graphs even tends to 1. The bound is almost best possible because it is known that there are infinitely many graphs with Thue chromatic number at least . There are some classes of graphs, where the Thue chromatic number is known exactly. For paths it was already shown by Thue himself [16], and Currie [7] showed that for every cycle of length , , , , , and for other lengths of cycles on at least vertices . In [11] various questions concerning nonrepetitive colourings of graphs have been formulated and consequently, also a lot of their variations appeared in the scientific literature (see e.g. Barát and Czap [3], Czerviński and Grytczuk [8], Grytczuk et. al. [12], [13] or Schreyer and Škrabul’áková [15]).
The purpose of this paper is a first look at nonrepetitive total colourings of a graph. A (proper) total colouring of a graph is a colouring of its vertices and edges, where no two adjacent vertices or edges have the same colour and moreover, no edge has the same colour as an incident vertex. We want to apply the concept of nonrepetitive colourings to total graph colourings and do it in two different ways. If a colouring of the vertices and edges of a graph has the property that the colour sequence of consecutive vertex- and edge-colours of every path in is nonrepetitive, we call a weak total Thue colouring. If moreover, both the induced vertex- and edge-colourings are nonrepetitive as well, we call a (strong) total Thue colouring of . The minimum number of colours appearing in such a colouring is called weak total Thue chromatic number for the first case and total Thue chromatic number for the latter case. Note that while every total Thue colouring is a proper total colouring, this does not need to be the case for weak total Thue colourings, because two adjacent vertices or edges may have the same colour.
In this paper we show that the total Thue chromatic number is lesser than , where is the maximum degree of the graph. The bound is extended for the list version of the problem. For the weak total Thue chromatic number we show , that for planar graphs with faces gives . We also give some upper and lower bounds for these parameters considering special classes of graphs.

## 2 Basic observations and preliminary lemmas

We will use the following Lemma, that was proved in [14]. For a sequence of symbols with for some set , for all , the block is denoted by here.

###### Lemma 1.

([14]) Let be a nonrepetitive sequence with for all  . Let ; , be nonrepetitive sequences with for all   and . If , then is a nonrepetitive sequence.

A sequence of length consisting of different symbols is called a rainbow sequence. A rainbow sequence is trivially nonrepetitive and if each sequence ; , from Lemma 1 consists of only one element, then it is also trivially nonrepetitive:

###### Corollary 1.

Let be a rainbow sequence with for all  . For let . Then is a nonrepetitive sequence.

Moreover, it is easy to see that every nonrepetitive vertex-colouring of with colours together with one additional colour (not used for the colouring of the vertices of ) used to colour all the edges of gives a weak total Thue colouring of according to Corollary 1. A similar argument holds for nonrepetitive edge-colourings of a graph . Hence, and . Therefore, we have:

###### Observation 1.

.

Moreover, the upper bound in Observation 1 is tight. To see this it is enough to consider an arbitrary star and its nonrepetitive colouring. Obviously , , for (as there exists no nonrepetitive sequence of length over a symbol alphabet) and .
As every total Thue colouring is also a weak total Thue colouring and in a total Thue colouring both, the edge-colouring and the vertex-colouring of the graph have to be nonrepetitive, we have:

###### Observation 2.

, and .

On the other hand, if we colour all vertices of the graph nonrepetitively with colours and we use another colours to colour all edges of nonrepetitively, by Lemma 1 we obtain a total Thue colouring of . Hence, the following is true:

###### Observation 3.

.

For we have , but , which gives the next observation.

###### Observation 4.

The difference between and can be arbitrarily large.

Weak total Thue colourings are closely related to nonrepetitive vertex-colourings of subdivided graphs as there is an easy 1-1 correspondence between weak total Thue colourings of a graph and nonrepetitive vertex-colourings of the graph which is obtained from by subdividing every edge. Hence, we have:

###### Observation 5.

If is the graph obtained from by subdividing every edge then:

Together with Currie’s result on nonrepetitive colourings of cycles this immediately implies the following:

###### Corollary 2.

For the cycle on vertices it holds if or . Otherwise .

###### Observation 6.

There exists a graph where .

From the above considerations it already follows that . Figure  1 shows, that 4 colours are also enough for a total Thue colouring of the cycle .

## 3 Some general results

###### Theorem 1.

Let be a graph; , . Then .

###### Proof.

Consider a spanning tree of . If every edge of is subdivided by one vertex the tree remains a tree and therefore has a Thue chromatic number equal to 4 (see [5]). Hence (by Observation 5), there is a weak total Thue colouring of using 4 colours. If the remaining edges of are coloured by different colours, all paths obviously remain nonrepetitive. Hence, . ∎

###### Corollary 3.

Let be a plane graph; . Then .

###### Proof.

By the Euler formula for every plane graph with , , it holds . Then from Theorem 1 it follows that . ∎

###### Theorem 2.

Let be an outerplanar graph on vertices. Then for : ,  for :  and for : .

###### Proof.

Let be an outerplanar graph on vertices. Obviously , and by Currie’s theorem [7] . One can obtain all these graphs by subdividing each edge of an outerplanar graph on , or vertices respectively, by one vertex. According to Observation 5 this gives for .

Every outerplanar graph on vertices is a subgraph of a diamond graph - depicted on Figure 2 together with its weak total Thue colouring using colours. Hence . On the other hand every spanning tree of the diamond graph contains as a subgraph a path on edges. Subdividing each edge of it by one vertex one can obtain a path ; , and from Observation 5 it follows that the weak total Thue number of an outerplanar graph on vertices is at least . Hence .
The general result follows from Lemma 1, Observation  1 and the fact, that every outerplanar graph admits a nonrepetitive vertex-colouring with 12 colours (Barát and Varjú [4]). ∎

###### Theorem 3.

Let be a graph containing bridges the removal of which separates into -edge-connected components . Then

###### Proof.

The colouring algorithm:
1. Denote the tree obtained from by contracting the components ,…, into single vertices.
2. Colour the edges of (i.e. ) nonrepetitively. According to the Theorem proved in [1] at most colours are needed.
3. To obtain the weak total Thue colouring of find a weak total Thue colouring of each component (with colours different from the colours used to colour the edges of ); As there is no repetitive path in each block , by Lemma 1 the colouring obtained by the algorithm described above gives a weak total Thue colouring of with the claimed number of colours. ∎

From Thue’s theorem and Observation 5 it is obvious that the weak total Thue chromatic number of paths on at least 3 vertices is 3. From Lemma 1 it can be seen, that a total Thue colouring of every path with 6 colours can be constructed by combining a nonrepetitive vertex-colouring on 3 colours and a nonrepetitive edge-colouring on another 3 colours. The following Theorem improves this bound.

###### Theorem 4.

For every path on at least 4 vertices it holds .

###### Proof.

To see the lower bound, assume there is a total Thue colouring of using only three colours 1,2,3. Consider the colour sequence of the first 3 vertices and edges. Because such a colouring is also a proper total colouring, every colour in the sequence must differ from the two preceding colours. Then up to renaming of the colours the sequence has to be 123123 which is repetitive, a contradiction.
For the upper bound we construct a colouring using 5 colours. Let be a path of length . W.l.o.g. we can suppose that is divisible by , as every other path is subgraph of such a path.

The colouring algorithm:
1. For all divisible by 4 colour the vertex with colour 4.
2. Let be a nonrepetitive sequence on . Then for each colour the vertex with colour 5. That means between any two vertices of colour 4 there is a vertex of colour 5 and the sequence of distances between the colour 5 vertices to the preceding colour 4 vertices is nonrepetitive.
3. Whenever there are two uncoloured vertices between a vertex of colour 4 and 5, colour the edge connecting them with colour 5.
4. Colour all uncoloured edges using a nonrepetitive sequence on .
5. For every vertex that is adjacent to a vertex of colour 4 and 5 use a colour from different from the colours of the neighbouring edges.
6. For two adjacent uncoloured vertices in between two vertices of colour 4 consider the edge-colours that appear between the colour 4 vertices. If one colour from is missing, colour the middle vertex with this colour and the other one with a colour from that is different from this one and the colour of the neighboring edge of colour 1, 2 or 3. If all colours of edges appear, the sequence of vertex- and edge-colours between the two vertices of colour 4 has to be , where are different edge-colours from and and are the vertex-colours to be chosen. Choose and .

From Lemma 1 it immediately follows, that there is no repetitive sequence of edge-colours.
Assume there is a repetitive sequence of vertex-colours and it contains at least one vertex of colour 4. Then it contains an even number of vertices of colour 4 and exactly as many vertices of colour 5. If the first vertex of colour 4 or 5 has colour 4, then the sequence of distances from the vertices of colour 5 to the preceding vertex of colour 4 is repetitive, a contradiction. In case the first vertex of colour 4 or 5 has colour 5, then the sequence of distances of the vertices of colour 5 to the next vertex of colour 4 must be repetitive. This is a contradiction because if the sequence is nonrepetitive, the sequence is nonrepetitive, too. Hence, no repetitive sequence of vertex-colours can contain a vertex of colour 4. That means, a repetition of vertex-colours can contain only two elements and adjacent vertices are coloured differently by construction.
Now assume that there is a colour sequence of consecutive vertex- and edge-colours. If it contains a colour 4, then this is a vertex-colour that can only be repeated by another vertex-colour. If this is the case, vertex-colours are repeated by vertex-colours and edge-colours by edge-colours. Hence, the subsequences of vertex- and edge-colours must be repetitive themselves. This is not possible as every sequence of consecutive edge-colours is nonrepetitive. Therefore, a repetition can contain at most 3 vertex-colours. All vertices are coloured differently from their edge neighbours, and the repetition cannot consist of two vertex- and two edge-colours because otherwise two adjacent edges would have the same colour. The only remaining possibility is a sequence of three consecutive edge- and three consecutive vertex-colours, non of which is colour 4. Now it is easy to see, that these repetitions are excluded by construction steps 5 and 6. Consequently, no repetition of any kind occurs and the constructed colouring is a total Thue colouring. ∎

In general we conjecture the following:

###### Conjecture 1.

There is an integer such that for every path on at least vertices

An immediate consequence for cycles is the following:

###### Corollary 4.

For every cycle on at least 4 vertices it holds

This can be achieved by choosing one edge of a unique colour and colour the remaining path as before. But in many cases, at least if the number of vertices is large enough and divisible by 4, the colouring strategy from the previous theorem can be applied directly to generate a colouring with 5 colours.

The following theorem gives the exact values of the total Thue numbers of stars.

###### Theorem 5.

Let be a star on vertices. Then .

###### Proof.

All the edges of are adjacent to each other, and therefore, they have to be coloured with different colours in every strong total Thue colouring of . They are incident with the central vertex of the star as well, therefore, has to be coloured with a new colour under . Hence, . In order to obtain a strong total Thue colouring of the star colour the uncoloured vertices , as follows: let and for let .
All the vertices of are coloured with different colours and all the paths on vertices and edges of are coloured with a colour sequence of the form or , therefore, is a strong total Thue colouring using colours. ∎

## 4 Bounds depending on the maximum degree

###### Theorem 6.

Let be a graph with maximum degree . Then .

###### Proof.

Let be a graph of maximum degree . Dujmović et. al. [9] proved that . By considering the line graph of (which has a maximum degree of less than this implies for the edge version of the problem . Together with Observation 3 this implies

Note, that the upper bound on the Thue chromatic number used in the proof can be improved for larger . The actual bound given in [9] is , which with the same arguments as above implies .

Our last result is an extension of the above result to list colourings. The graph is nonrepetitively total -choosable if for every list assignment with minimum list size at least there exists a total Thue colouring with colours from the associated lists. The total Thue choice number of a graph is the minimum number , such that is nonrepetitively total -choosable. (One can similarly define also the weak total Thue choice number of a graph). A bound on this parameter cannot be proved by considering vertex- and edge-colourings separately because it cannot be guaranteed, that the used colour sets of both colourings will be distinct.
We will use a probabilistic approach to prove our result. In probability theory, if a large number of events are all independent of one another and each has probability less than , then there is a positive probability that none of the events will occur. The Lovász Local Lemma (see Erdős and Lovász [10]) allows one to relax the independence condition slightly: As long as the events are ”mostly” independent from one another and aren’t individually too likely, then there is a positive probability that none of them occurs. There are several different versions of the lemma - see Alon and Spencer [2]. We will use the asymmetric one formulated below:

###### Lemma 2.

Let be a finite set of events in the probability space . For let denote a subset of such that is independent from the collection of events . If there exists an assignment of reals to the events such that then the probability of avoiding all events in is positive, in particular

###### Theorem 7.

For every graph with maximum degree at most ; , the total Thue choice number is at most .

###### Proof.

Let be a graph with maximum degree at most , where every vertex and edge is endowed with a list of at least colours. To fulfil the conditions of Lovász Local Lemma we suppose that the colour of each vertex and edge is chosen randomly, independently and equiprobably out of its list. We consider the following types of bad events that may happen when this procedure is applied:

• For every path on vertices let denote the event that the colour sequence of the first vertices is the same as the colour sequence of the second vertices. For the probability of the event we have . We assign the number to the event , where .

• For every path on edges let denote the event that the colour sequence of the first edges is the same as the colour sequence of the second edges. For the probability of the event we have .We assign the number to the event , where .

• For every path on vertices together with the internal edges and one edge incident with the final vertex let denote the event that the colour sequence of the first elements (vertices and edges) of is the same as the colour sequence of the second half. For the probability of the event we have .We assign the number to the event , where .

For an arbitrary event let denote the set of paths on vertices sharing at least one vertex with and the set of paths on vertices and edges sharing at least one vertex with . It it easy to see, that and , as . We will show that

 Pr(APt)≤xPt⋅∞∏s=1∏Ps∈As∖{Pt}(1−xPs)∏Rs∈Cs(1−zRt) (1)

Consider the right hand side of inequality (1):

 RHS1 = xPt⋅∞∏s=1∏Ps∈As∖{Pt}(1−xPs)∏Rs∈Cs(1−zRt) = 11+at⋅1+atat⋅∞∏s=1∏Ps∈As(1−11+as)∏Rs∈Cs(1−11+cs) ≥ 1at∞∏s=1(as1+as)2tsΔ2s(cs1+cs)23tsΔ2s ≥ 1at∞∏s=1(e−1as)2tsΔ2s(e−1cs)23tsΔ2s,

since for all positive it holds that . Moreover,

 1at∞∏s=1(e−1as)2tsΔ2s(e−1cs)23tsΔ2s = 1at⋅⎛⎝e−2∞∑s=1s⋅(Δ2a)s−23∞∑s=1s⋅(Δ2c)s⎞⎠t= =1(7.5Δ2)t⋅⎛⎝e−2∞∑s=1s⋅(17.5)s−23∞∑s=1s⋅(110)s⎞⎠t = 1(7.5Δ2)t⋅(e−449010266.75)t,

as . Hence, , what proves inequality (1).

For an arbitrary event let denote the set of paths on edges sharing at least one edge with and the set of paths on vertices and edges sharing at least one edge with . It is easy to see that and . Similarly as in the previous case we will show that

 Pr(BQt)≤yQt⋅∞∏s=1∏Qs∈Bs∖{Qt}(1−yQs)∏Rs∈Cs(1−zRt). (2)

Consider the right hand side of inequality (2):

 RHS2 = yQt⋅∞∏s=1∏Qs∈Bs∖{Qt}(1−yQs)∏Rs∈Cs(1−zRt) = 11+bt⋅1+btbt⋅∞∏s=1∏Qs∈Bs(1−11+bs)∏Rs∈Cs(1−11+cs) ≥ 1bt∞∏s=1(bs1+bs)4tsΔ2s(cs1+cs)43tsΔ2s ≥ 1bt∞∏s=1(e−1bs)4tsΔ2s(e−1cs)43tsΔ2s=1bt⋅⎛⎝e−4∞∑s=1s⋅(Δ2b)s−43∞∑s=1s⋅(Δ2c)s⎞⎠t = 1(7.5Δ2)t⋅⎛⎝e−4∞∑s=1s⋅(17.5)s−43∞∑s=1s⋅(110)s⎞⎠t=1(7.5Δ2)t⋅(e−898010266.75)t

Hence, , what proves inequality (2).

For an arbitrary event let denote the set of paths on vertices sharing at least one vertex with , denote the set of paths on edges sharing at least one edge with and the set of paths on vertices and edges sharing at least one vertex with . It is easy to see that , and . We will show that

 Pr(CRt)≤zRt⋅∞∏s=1∏Ps∈As(1−xPs)∏Qs∈Bs(1−yQs)∏Rs∈Cs∖{Rt}(1−zRt) (3)

Consider the right hand side of inequality (3):

 RHS3 = zRt⋅∞∏s=1∏Ps∈As(1−xPs)∏Qs∈Bs(1−yQs)∏Rs∈Cs∖{Rt}(1−zRt) = 11+ct⋅1+ctct⋅∞∏s=1∏Ps∈As(1−11+as)∏Qs∈Bs(1−11+bs)∏Rs∈Cs(1−11+cs) ≥ 1ct∞∏s=1(as1+as)tsΔ2s(bs1+bs)2tsΔ2s(cs1+cs)13tsΔ2s ≥ 1ct∞∏s=1(e−1as)tsΔ2s(e−1bs)2tsΔ2s(e−1cs)13tsΔ2s = 1ct⋅⎛⎝e−∞∑s=1s⋅(Δ2a)s−2∞∑s=1s⋅(Δ2b)s−13∞∑s=1s⋅(Δ2c)s⎞⎠t = 1(10Δ2)t⋅⎛⎝e−3∞∑s=1s⋅(17.5)s−13∞∑s=1s⋅(110)s⎞⎠t=1(10Δ2)t⋅(e−898010266.75)t

Hence, , what proves inequality (3).

Since the inequalities (1) (2) and (3) are valid, by the Local Lemma with positive probability none of the bad events happens. Hence, there is a total Thue colouring of the graph from lists of size .

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