New Bounds for Facial Nonrepetitive Colouring^{†}^{†}thanks: This research is partially funded by NSERC and the Ontario Ministry of Research and Innovation.
Abstract
We prove that the facial nonrepetitive chromatic number of any outerplanar graph is at most 11 and of any planar graph is at most 22.
NEW BOUNDS FOR FACIAL NONREPETITIVE COLOURING^{†}^{†}thanks: This research is partially funded by NSERC and the Ontario Ministry of Research and Innovation.
Prosenjit Bose,^{†}^{†}thanks: School of Computer Science, Carleton University Vida Dujmović,^{†}^{†}thanks: Department of Computer Science and Electrical Engineering, University of Ottawa Pat Morin,^{2}^{2}footnotemark: 2 Lucas RiouxMaldague^{†}^{†}thanks: Google
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1 Introduction
A sequence , , is a repetition if for each . For example, 1212 is a repetition, while 1213 is not. A block of a sequence is any subsequence of consecutive terms in . A sequence is nonrepetitive if for every nonempty block of , is not a repetition. Otherwise is repetitive. For example, 1312124 is repetitive as it contains the block 1212 which is a repetition, while 123213 is nonrepetitive as it contains no such block. No sequence of length greater than three using only two symbols can be nonrepetitive. A result by Axel Thue in 1906 states that nonrepetitive sequences of infinite length can be created using three symbols [25]. Thue’s work is considered to have initiated the study of the combinatorics of words [1].
A graph colouring variation on this theme was proposed by Alon et al. [2]. A nonrepetitive (vertex) colouring of a graph is an assignment of colours to the vertices of such that, for every path in , the sequence of colours of vertices in is not a repetition. The nonrepetitive chromatic number of , denoted , is the minimum number of colours required to nonrepetitively colour . In this setting, Thue’s result states that, for all , the path on vertices has . Since its introduction, nonrepetitive graph colouring has received much attention [3, 4, 5, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 19, 20, 21, 22, 23, 24].
A wellknown conjecture, due to Alon et al. [2], is that there exists a constant such that, for every planar graph , . The current best upper bound for vertex planar graphs is [9]. No planar graph with nonrepetitive chromatic number greater than 11 is known (see Appendix A in [9]).
More is known about the facial version of the problem for embedded planar graphs. Harant and Jendrol [17] asked if every plane graph can be coloured with a constant number of colours such that every facial path^{1}^{1}1A facial path is a path that is a contiguous subsequence of a facial walk; see Section 2 for a more rigorous definition. is nonrepetitively coloured. Barát and Czap [3] answered this question in the affirmative by showing that colours are sufficient. We reduce this bound to 22 by proving a bound of 11 for facial nonrepetitive colouring of outerplane graphs.
1.1 Related Work
Nonrepetitive Colouring.
It is known that some families of graphs have bounded nonrepetitive chromatic number. In their original work, Alon et al. [2] showed that if has maximum degree and that there are are graphs of maximum degree with nonrepetitive chromatic number . The constants in the upper bound have been steadily improved [10, 14, 15, 17].
Barát and Varjú [4] and Kündgen and Pelsmajer [20] independently showed that if is outerplanar and, more generally, if has treewidth at most . (Barát and Varjú proved the latter bound with while Kündgen and Pelsmajer proved it with .) The bound of for trees is tight if (trees), but it is not known if it is tight for other values of . Even the upper bound of 12 for outerplanar graphs may not be tight, as no outerplanar graph with nonrepetititive chromatic number greater than 7 is known [4].
Facial Nonrepetitive Colouring.
Facial nonrepetitive colouring was first considered by Havet et al [18], who studied the edgecolouring variant of the problem. In this setting, they were able to show that the edges of any plane graph can be 8coloured so that every facial trail^{2}^{2}2A facial trail is a contiguous subsequence of the edges traversed during the boundary walk of a face. is coloured nonrepetitively. For the listcolouring version of this problem, Przybyło [22] showed that lists with at least 12 colours are sufficient to colour the edges of any plane graph so that every facial trail is coloured nonrepetitively.
For the vertexcolouring version we study, Harant and Jendrol [17] proved that if is a plane graph of maximum degree and that if is a Hamiltonian plane graph. They also conjectured that when is any plane graph. As mentioned above, this latter conjecture was confirmed by Barát and Czap [3], who showed that, for any plane graph , . The results of Barát and Czap [3] also extend to graphs embedded in surfaces. They show that a graph embedded on a surface of genus can be facially nonrepetitively coloured. The best lower bounds for facial nonrepetitive chromatic numbers are 5 for plane graphs and 4 for outerplane graphs [3].
2 Preliminary Results and Definitions
We assume the reader is familiar with standard graph theory terminology as used by, e.g., Bondy and Murty [6]. All graphs we consider are undirected, but not necessarily simple; they may contain loops and parallel edges. For a graph, , we use the notations and to denote ’s vertex and edge sets, respectively. For , denotes the subgraph of induced by the vertices in and .
A graph is connected if it contains more than vertices and has no vertex cut of size less than . A connected component of a graph is a maximal subset of vertices of that induces a connected subgraph. A bridge in a graph is an edge whose removal increases the number of connected components. A graph is bridgeless if it has no bridges.
A plane graph is a fixed embedding of a graph in the plane such that its edges intersect only at their common endpoints. An outerplane graph is a plane graph such that all the vertices of are incident on the outer face of . A chord in an outerplane graph is an edge that is not incident to the outer face. A cactus graph is an outerplane graph with no chords. An ear in an simple outerplane graph is an inner face that is incident to exactly one chord. An ear is triangular if it has exactly three vertices. The weak dual of a outerplane graph is a forest whose vertices are the inner faces of and that contains the edge if the face and the face have a chord in common. Note that the ears of are leaves in the weak dual and that, if is biconnected, then its weak dual is a tree.
A walk in a graph is a sequence of vertices such that, for every the edge is in . The walk is closed if is also in . A walk is a path if all its vertices are distinct. A facial walk in a plane graph is a closed walk such that, for every , the edges and occur consecutively in the counterclockwise cyclic ordering of the edges incident to in the embedding of . A facial path is a contiguous subsequence of a facial walk that is also a path in . A facial path is an outerfacial path if it appears in a facial walk of the outer face of and it is an innerfacial path if it appears in a facial walk of some inner face of .
Before proceeding with our results, we introduce a helper lemma due to Havet et al. [18] and two theorems that will be used throughout the paper. The helper lemma provides a way to interlace nonrepetitive sequences.
Lemma 1 (Havet et al. [18]).
Let be a nonrepetitive sequence over an alphabet in which each has size at least 1. For each , let be a (possibly empty) nonrepetitive sequence over an alphabet with . Then is a nonrepetitive sequence.
We will require two results about the nonrepetitive chromatic number of trees and cycles:
Theorem 1 (Alon et al. [2]).
For every tree, , .
Theorem 2 (Currie [8]).
For every , the cycle on vertices has
3 Outerplane Graphs
We begin with a simple lemma that allows us to focus, when convenient, on simple outerplane graphs.
Lemma 2.
Let be a simple outerplane graph and let be an outerplane graph obtained by adding parallel edge and/or loops to . Then, any facially nonrepetitive colouring of is also a facially nonrepetitive colouring of , so .
Proof.
We argue that any facial path (described as a sequence of vertices) in is also a facial path in . Therefore, by facially nonrepetitively colouring , we obtain a facial nonrepetitive colouring of .
First, note that no facial path uses a loop, so the addition of loops does not introduce new facial paths in . When a (nonloop) edge is added parallel to an existing edge of , the union of the embeddings of and form a Jordan curve that does not contain any vertices of (since we require to be outerplane). This implies that any facial path in that uses the new edge exists in as a facial path that uses the edge . ∎
Next, we introduce a definition that is crucial to the rest of the paper. Let be an outerplane graph. A blocking set of is a set of vertices such that for each 2connected component of , is a tree and for each inner face , . See Figure 1 for an example of a blocking set.
The definition of a blocking set is subtle and implies some properties that we will use throughout.
Observation 1.
For any blocking set of , does not include both endpoints of any chord of .
Observation 2.
For any blocking set of and any inner face of , the vertices of occur consecutively on the boundary of . In other words, is a nonempty path.
Observations 1 and 2 are true because, otherwise, would be disconnected for the 2connected component containing or , respectively.
Lemma 3.
For every biconnected outerplane graph, , and any vertex , there exists a blocking set of such that and, for each inner face of , .
Proof.
The proof is by induction on the number of inner faces. If has only one inner face, we take and we are done. Otherwise, select some ear, of whose chord is and such that . Such an ear always exists because has at least two ears. Let . The graph has one less inner face than so, by induction, it has a blocking set that satisfies the conditions of the lemma. There are two cases to consider:

If one of or is in then we take to obtain a blocking set that satisifes the conditions of the lemma.

Otherwise, let be any vertex in and take to obtain a blocking set that satisifes the conditions of the lemma. ∎
Lemma 3 allows us to prescribe that a particular vertex be included in the blocking set, but it will also be convenient to exclude a particular vertex by using Lemma 3 to force the inclusion of ’s neighbour on the outer face (which is also on some inner face with ).
Corollary 1.
For every biconnected outerplane graph, , and any vertex , there exists a blocking set of such that and, for each inner face of , .
At this point we pause to sketch how Lemma 3 can already be used to give an upperbound of 8 on the facial nonrepetitive chromatic number of biconnected outerplane graphs. For a biconnected outerplane graph, , we take a blocking set of using Lemma 3. By Theorem 1, we can nonrepetitively 4colour the tree using the colours , so what remains is to assign colours to the vertices in . To do this, we use Theorem 2 to nonrepetitively 4colour the cycle, , that contains the vertices of in the order they appear on the outer face of using the colours . We claim that the resulting 8colouring of is facially nonrepetitive. No facial path on an inner face is coloured repetitively since each such facial path is also either present in the tree or it contains exactly one vertex of . No facial path on the outer face is coloured repetitively since it is obtained by interleaving a nonrepetitive sequence of colours in with nonrepetitive sequences taken from ; by Lemma 1, a sequence obtained in this way is nonrepetitive.
In Appendix A, we show that the preceding argument can be improved to give a bound of 7 on the facial nonrepetitive chromatic number of biconnected outerplane graphs. This is just a matter of adding vertices to the blocking set so that the cycle does not have length in , so that it can be nonrepetitively 3coloured.
Finally, we remind the reader that, although Lemma 3 and Corollary 1 provide blocking sets that include only one vertex on each inner face, not all blocking sets have this property. It is helpful to keep this in mind in the next section.
3.1 The Blocking Graph
The blocking graph of for a blocking set is the graph, denoted by , whose vertex set is and whose edges are defined as follows: Begin with the (closed) facial walk on the outer face of . Remove every vertex not in from to obtain a cyclic sequence of vertices in . For each consecutive pair of vertices in we add an edge to . This naturally defines the embedding of . See Figure 2 for an example. Note that is a plane graph that is not necessarily simple; it may contain parallel edges (cycles of length two) and selfloops (cycles of length one).
The fact that a blocking set does not contain both endpoints of any chord of (Observation 1) implies the following observation:
Observation 3.
For every outerplane graph and any blocking set of , the blocking graph is a bridgeless cactus graph.
Observation 4.
For every outerplane graph , any blocking set of , and any facial path on the outer face of , the subsequence of containing only the vertices of is a (outer) facial path in .
Observation 5.
For every outerplane graph , any blocking set of and any inner face of , is a nonempty path that is a facial path (on some inner face) in .
In the previous section, we sketched a proof of an upper bound of 8 on the facial chromatic number of biconnected outerplane graphs. This proof works by nonrepetitively 4colouring the tree, , obtained after removing the blocking set and then nonrepetitively 4colouring a cycle, , of vertices in the blocking set. This cycle, , is actually the blocking graph, . The following lemma shows that this strategy generalizes to the situation where we can find a facial nonrepetitive colouring of with few colours.
Lemma 4.
Let be an outerplane graph and be a blocking set of . If there exists a facial nonrepetitive colouring of (the outer face of) , then there exists a facial nonrepetitive colouring of .
Proof.
By Theorem 1 we can colour nonrepetitively using colours and, by assumption, we can facially nonrepetitively colour with colours . These two colourings define a colouring of that we now show is facially nonrepetitive.
Let be a facial path in . If is a facial path of or a path in then there is nothing to prove.
Otherwise consider first the case that is a path on an inner face of . There are two cases to consider:
Finally, consider the case where is a facial path on the outer face of . In this case, the colour sequence obtained from is of the form where each is obtained from a (possibly empty) path in and is obtained from a (outer) facial path in (by Observation 4). Again, Lemma 1 implies that the resulting colour sequence is nonrepetitive. ∎
3.2 Colouring Even Cactus Graphs
We now show how to colour the blocking graph—a cactus graph—of an outerplane graph. By Lemma 4, if we can find a facial nonrepetitive colouring of any cactus graph, we can get a facial nonrepetitive colouring of any outerplane graph.
Recall that the best known upper bound for the facial Thue chromatic number of outerplane graphs is 12, which is the bound for the Thue chromatic number [4, 20]. Thus, to improve this bound, we need to find a facial nonrepetitive 7colouring of the blocking graph. We have been unable to do this unless all cycles of the blocking graph are even. We will eventually address this limitation in Section 3.3 by proving the existence of a blocking set such that has no odd cycles.
For any graph, , a levelling of is a function such that for each , . The level pattern of a path is the sequence .
Lemma 5 (Kündgen and Pelsmajer [20]).
Let be a graph and be a levelling of . Let be a nonrepetitive palindromefree sequence on an alphabet with and be a colouring of defined as . If a path with in is repetitively coloured under , then and have the same level pattern.
Lemma 6.
For every cactus graph, , with no odd cycles (and therefore, for every blocking graph with no odd cycles), .
Proof.
By Lemma 2, we may assume that is simple. We may also assume that is connected as this does not affect its nonrepetitive chromatic number. Also, assume that is neither a cycle nor a tree since for both these classes of graphs. If there exists a vertex of such that , then let the root of be . Otherwise, let be any vertex of of degree at least . Let be a levelling of where is the distance in from to . Let be a graph that contains all vertices such that

is on a cycle of ,

and

.
In other words, contains the vertices of degree 2 that are on the deepest level of a cycle (see Figure 3). Notice that since every cycle of is even, there is at most one vertex of in each cycle of .
If , there must exist at least one face of such that exactly one vertex of has degree greater than two. From our choice of , it follows that is the minimum over all vertices in . Since , has three consecutive degree2 vertices , , and , such that . If , we add to . If , we add both and to . Notice that now, either or .
We now define the edge set of . For each , we add the edge to if there is a facial path on the outer face of with endpoints and that does not contain any other vertices in . Note that is either a cycle or a forest of paths. It can only be a cycle if has no vertices of degree 1, in which case, . In this case, our choice of ensures that the length of this cycle is not in . This implies that can be nonrepetitively coloured using the colour set , either by using the result of Thue [25] or Currie (Theorem 2).
To colour the remaining vertices of , let and be a palindromefree nonrepetitive sequence on . (A nonrepetitive palindromefree sequence can be constructed from any ternary nonrepetitive sequence by adding a fourth symbol between blocks of size 2 [7].) Then, each vertex is assigned the colour .
We will now show that the resulting 7colouring of is a facial nonrepetitive colouring. Suppose that this is not the case. Thus, there exists a path such that the colour sequence corresponding to vertices of is a repetition. Let us first suppose that is on the outer face of . We will need the following claim:
Claim 1.
Let be a path on the outer face of such that . The level sequence corresponding to vertices of must be strictly decreasing, strictly increasing, or strictly decreasing then strictly increasing.
Proof of Claim 1.
Suppose that this is not the case. Then cannot contain two consecutive elements of the form as this can only correspond to an odd cycle of , but all cycles of are even. Thus, must contain a block of the form . Since is on the outer face, we must have that the vertex corresponding to is the highest numbered vertex on some cycle and that . But in this case, must be in , which is a contradiction. ∎
By Lemma 5, and have the same level pattern. However, if this is incompatible with Claim 1. Thus, must contain vertices of . Let be the sequence of vertices of in the same order as in . Notice that is a path in . Therefore, the colour sequence corresponding to is nonrepetitive. Now, observe that the colour sequence formed by is of the form where is a nonrepetitive sequence of colours from and each is a nonrepetitive sequence of colours from . Therefore, by Lemma 1, is coloured nonrepetitively.
Thus, must be a facial path on some inner face of . If then, by Lemma 5, and have the same level pattern. No path on an even cycle has such a pattern using the levelling . Therefore, , so contains 1, 2, or 3 vertices of . If is a repetition it must contain exactly 2 vertices of , thus is a facial path on (since every other inner face contributes at most one vertex to ). Reusing the notation above, cannot contain since has a unique colour in . Therefore, must contain and and, in fact, these are the endpoints of . The colour sequence of must therefore be of the form where and is a nonempty sequence over the alphabet . Such a sequence is not a repetition. ∎
3.3 Making an Even Blocking Graph
Lemma 7.
For every biconnected outerplane graph, , and any vertex :

has a blocking set such that is an even cycle and ; and

has a blocking set such that is an even cycle and .
Proof.
We first obtain a blocking set that contains or does not contain , as appropriate, by applying Lemma 3 or Corollary 1. Recall that contains exactly one vertex on each inner face of . It is simple to verify that is a cycle; if it is an even cycle, then we are done, so we may assume that is an odd cycle.
If has only one inner face, then contains one vertex, , on this face and is an odd cycle (of length 1). In this case, we can select the neighbour, , of such that and let . It is easy to verify that is a blocking set, and either includes or excludes , as appropriate, and that is a cycle of length 2.
Thus, we may assume that has at least two inner faces and we consider several cases:

If contains an ear, , with four or more vertices such that either or is one of the endpoints of the chord of . There is exactly one vertex on the face . Let be a neighbour of on such that is not on the chord of (so has degree 2). Such a exists since has at least four vertices. Set . Now is even, so is an even cycle. Furthermore, is a tree and is a leaf in this tree, so is also a tree. Finally, by choice, contains if and only if contains , so satisifies the conditions of the lemma.

Next, consider the case where contains a triangular ear, , such that one of the endpoints of the chord of is in and is not the degree 2 vertex, , of . By the same argument as above, satisfies the conditions of the lemma.

Refer to Figure 4. For an edge , let and be the two (possibly empty) sets of vertices in the (at most) two connected components of . Let and . If neither of the two previous cases applies, then there exists an edge of such that and the weak dual of is a star whose central vertex is the face, incident on .
We now argue why such an edge exists. Recall that the weak dual, , of is a tree whose vertices are the faces of . Select some face, , of that has on its boundary and root at . This tree has a height, , and some vertex of depth (recall that has at least two inner faces). The face will be the face described above. We now show how to choose the edge .
If (because ), then we take to be an edge of , one of whose endpoints is . Such an edge exists since is on . (Note that, in this case, may be a chord or may be on the outer face.)
If , then we take to be the chord of that separates it from . (In this case, may still be one of or .) In either case, the edge and the face satisfy the condition described above. In particular, the dual of is a star because had height and by our choice of .
Now that we have established the existence of and , we will now show that we can select another vertex, , from so that is a blocking set. This is sufficient since is even so is an even cycle.

By choice, has at least 2 faces and each of them, other than , is a triangular ear incident to and whose degree2 vertex is in (otherwise, one of those faces would have been handled by Case 1 or 2).
Let be the unique vertex of on . The vertex has two neighbours on . We claim that one of these is not in and we take this vertex to be . This claim is valid because otherwise, is a triangle, , with . This case is not possible because by at least one of or is incident on both and a triangular ear of . Both and the degree 2 vertex of are in . This contradicts the fact that includes at most one vertex from each face of , including .
Let . We claim that is a blocking set of . First, note that contains at most two vertices from each face, , of , so . We now show that is a leaf in the tree , so that is also a tree.
First, observe that is not a chord of since, by , the face incident to other than would have two of its vertices in . Thus, in addition to , has at most two neighbours in . One of these, , is on and so . Finally, may have one additional neighbour, which is a degree2 vertex of a triangular ear incident on . In this case, by , this degree2 vertex is in . Thus, is the only edge incident to in the tree so is a leaf in this tree. ∎

Remark 1.
The proof of Lemma 7 can be modified to prove something stronger than just requiring the inclusion or exclusion of . We can specify an edge on the outer face of and obtain a blocking set such that is an even cycle, and . The only difference in the proof is ensuring that is not included in . The resulting proof has the same three cases. Case 1 applies as long as is not on the boundary of the ear . Case 2 applies as long as is not on the boundary of the ear . Otherwise, in Case 3, the edge is in the subgraph , so there is no chance of including in . This stronger version of Lemma 7 is used in Appendix A.
Lemma 8.
Every simple bridgeless outerplane graph has a blocking set such that all cycles in are even.
Proof.
The proof is by induction on the number of 2connected components of . If has no 2connected components, then we take to be the empty blocking set. If has only one 2connected component, then we apply Lemma 7. Otherwise, select a 2connected component, , that shares exactly one vertex, , with the rest of . Let and apply induction on to obtain a blocking set, , of such that has only even cycles. There are two cases to consider:

If contains , then we apply the first part of Lemma 7 to obtain a blocking set of such that is an even cycle and . We take , which clearly forms a blocking set of . The blocking graph is simply the union of the two blocking graphs and , which have only the vertex in common. Thus, every cycle in is also a cycle in one of these two graphs, so it has even length.

If does not contain , then we apply the second part of Lemma 7 to obtain a blocking set of such that is an even cycle and . We take .
Refer to Figure 5. Starting at some appropriate vertex in in the facial walk on the outer face of , there is a last vertex, , encountered before the walk encounters the first vertex and there is a last vertex encountered before the walk returns to the next vertex . The edge is in and the edge is in .
Figure 5: Case 2 in the proof of Lemma 8. Since every blocking graph is a bridgeless cactus graph (Observation 3), each of these edges is part of one even cycle in its respective graph. In these two cycles are merged by removing the edges and and adding the edges and . The resulting cycle is even. Every other cycle in is also a cycle in one of or so it has even length. ∎
Lemma 9.
Every simple outerplane graph has a blocking set such that all cycles in are even.
Proof.
The proof is by induction on the number of bridges of . If has no bridges, then we apply Lemma 8. Otherwise, select some bridge of and contract it to obtain a graph in which corresponds to a single vertex . By induction, we obtain a blocking set of such that has only even cycles (or is empty). There are two cases to consider:

If , then we take . This introduces exactly one new cycle in that is not present in and this cycle has length 2.

If , then we take , so . ∎
Finally, have all the tools to prove our main result on outerplane graphs:
Theorem 3.
For every outerplane graph, , .
4 Plane Graphs
In this section, we show how to reuse the ideas from Barát and Czap [3] to facially nonrepetitively 22colour every plane graph. Some modifications are needed because Barát and Czap use a nonrepetitive 12colouring of outerplanar graphs whereas our Theorem 3 provides an facial nonrepetitive 11colouring of outerplane graphs.
Theorem 4.
Let . Then, for every plane graph , .
Proof.
For any plane graph, , the peeling layering of is a partition of into sets as follows. Let be the vertices on the outer face of and let , , be the vertices on the outer face of .
We augment to obtain a plane graph in the following way. For each inner face of , let be the facial walk of . The walk contains only vertices from and for some . Remove from all vertices in to obtain a cyclic sequence of vertices from . For any two consecutive vertices in , we add the edge to and embed it inside the face . This construction has the following implications: (a) The resulting graph is still plane (though not necessarily simple) and, for all , . From this point onward, we use the notation . (b) The cyclic sequence defined above is a facial walk in and, since it contains only vertices in , it is also a facial walk in .
For each , is outerplane. To colour we use Theorem 3 to facially nonreptitively colour with if is even or if is odd. This defines a colouring of that we now prove is facially nonrepetitive.
Let be a facial path, , in . The graphs and have the same outer face so, if is on the outer face, then the colour sequence of is not a repetition since our colouring is facially nonrepetitive for .
Therefore, is an inner face and all vertices on are in for some . Write as , where each consists of vertices from and each consists of vertices from . Notice that, for each , contains an edge joining the last vertex in to the first vertex in . Indeed, is a facial path in (It is a contiguous subsequence of the sequence defined above.) Next, observe that each is a facial path on the outer face of . Therefore, the colour sequence determined by is of the form (with corresponding to and corresponding to ). The sequence is nonrepetitive and each sequence is nonrepetitive. Therefore, by Lemma 1, the colour sequence determined by is nonrepetitive. ∎
5 Concluding Remarks
We note that the proofs in this paper lead to straightforward lineartime algorithms. After the appropriate decomposition steps, there are essentially two subproblems: (1) finding an appropriate blocking set in a biconnected outerplane graph (Lemma 7) and (2) colouring cactus graphs with no odd cycles (Lemma 6). The proof of Lemma 6 is easily made into a lineartime algorithm. The proof of Lemma 7 can be implemented by a recursive earcutting algorithm that implements Lemma 3 followed by a traversal of the dual tree in order to find the face used in the proof of Lemma 7.
It seems unlikely that our upper bound of 11 for the facial nonrepetitive chromatic number of outerplane graphs (and hence the bound of 22 for plane graphs) is tight. (Recall that the best known lower bounds are 4 and 5, respectively [3].) Thus, an obvious direction for future work is to improve these bounds. Our proof of Lemma 4 uses a nonrepetitive 4colouring of trees, but a facial nonrepetitive colouring would also be sufficient. This leads naturally to the following problem:
Open Problem 1.
Is for every tree, ?
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Appendix A Biconnected Outerplane Graphs
Lemma 10.
If is a biconnected outerplane graph, then has a blocking set such that .
Proof.
If is a cycle, take to be two consecutive vertices on and we are done. Otherwise, select an ear, of and let be ’s chord. Then apply the stronger version of Lemma 9 discussed in Remark 1 to the graph to obtain a blocking set of even size that contains and not .
Note that, since and is on , has no cycles. Furthermore, since , is connected, so is a blocking set of . Since is even, , so if , then we are done with . Otherwise, let be ’s degree2 neighbour on and take . ∎
Corollary 2.
If is an outerplane graph with at most one 2connected component, then .
Proof.
If is a tree, then , so we may assume contains exactly one 2connected component, . Apply Lemma 10 to to obtain a blocking set with and observe that is also a blocking set of . The blocking graph is a cycle, , that has a nonrepetitive 3colouring. The blocking graph consists of and possibly some self loops so, by Lemma 2, has a facial nonrepetitive 3colouring. Therefore, by Lemma 4, . ∎