Cyclic colorings of plane graphs with independent faces^{†}^{†}thanks: This research was supported by the CzechSlovenian bilateral project MEB 090805 (on the Czech side) and BICZ/0809005 (on the Slovenian side).
Abstract
Let be a plane graph with maximum face size . If all faces of with size four or more are vertex disjoint, then has a cyclic coloring with colors, i.e., a coloring such that all vertices incident with the same face receive distinct colors.
1 Introduction
In 1965, Ringel [23] introduced the notion of planar graphs. These are graphs that can be drawn in the plane such that every edge is crossed by at most one other edge. Ringel [23] proved that planar graphs are colorable and conjectured that they are colorable. Ringel’s conjecture was shown to be true by Borodin [5, 7] in the 1980’s.
Ringel’s problem fits a framework of cyclic colorings, vertex colorings of embedded graphs such that any two vertices incident with the same face receive distinct colors. It is easy to see that every edgemaximal planar graph can be obtained from a plane graph with faces of size three and four by inserting a pair of crossing edges into each face of size four. In the other direction, removing pairs of crossing edges in an edgemaximal planar graph yields a plane graph with faces of size three and four. Hence, Borodin’s result [5, 7] asserts that every plane graph with maximum face size four has a cyclic coloring using at most six colors.
Borodin’s result had been conjectured as one of the cases () in the Cyclic Coloring Conjecture of Ore and Plummer [21]. The conjecture asserts that every plane graph with maximum face size has a cyclic coloring with colors. The statement of the conjecture for is equivalent to the Four Color Theorem, proved in [4, 24]. For , the best known bound of has been obtained by Sanders and Zhao [25] improving earlier bounds of Borodin [6, 8]. A major evidence that the conjecture is true is a recent breakthrough of Amini, Esperet and van den Heuvel [3] which extends an approach of Havet, van den Heuvel, McDiarmid and Reed [11, 12]; Amini et al. [3] showed that the Cyclic Coloring Conjecture is asymptotically true, i.e., for every , there exists such that every plane graph of maximum face size admits a cyclic coloring with at most colors.
The Cyclic Coloring Conjecture stipulated a lot of research, in particular, several restrictions and generalizations of the conjecture has been considered. Plummer and Toft [22] conjectured that the asserted bound can be improved for connected plane graphs to . The conjecture of Plummer and Toft is known [9, 15, 16, 17] to be true for and . In another direction, a possible generalization avoiding the restriction of face sizes, the Facial Coloring Conjecture, was proposed in [18]. This generalization asserts that vertices of every plane graph can be colored with at most colors in such a way that every two vertices joined by a facial walk of length at most receive distinct colors. Partial results towards proving this conjecture, which implies the Cyclic Coloring Conjecture for odd values of , can be found in [13, 14, 18, 19].
In this paper, we consider a different restriction of the Cyclic Coloring Conjecture which is also motivated by colorings of graphs drawn in the plane with restricted structure of crossings, originally introduced by Albertson [1]. Two distinct crossings are independent if the endvertices of every pair of crossing edges are mutually different. In particular, if all crossings are independent, then each edge is crossed by at most one edge, i.e., graphs with mutually independent crossings are plane graphs. Albertson conjectured [1, 2] that every graph that can be drawn in the plane with all its crossings independent is colorable and provided partial results towards the proof of his conjecture (other partial results can be found in [10, 26]). In the cyclic coloring setting, Albertson’s conjecture says that every plane graph with faces of size three and four such that all faces of size four are vertexdisjoint is colorable.
Albertson’s conjecture has been verified by two of the authors in [20]. A natural question is what is the least number of colors needed if the maximum face size is larger than four and the faces of size four or more are still vertex disjoint. The wheels are plane graphs of this type and thus the number of colors needed is at least . We prove that this number also suffices.
2 Overview
Let us first introduce some additional notation. A vertex of degree is a vertex and a face incident with vertices is a face. The graphs we consider throughout the proof have no loops and no faces but they can have parallel edges, in which case the degree of the vertex is considered to be the number of edges incident with it, not the number of its neighbors. Two vertices are cyclic neighbors if they are incident with the same face. The cyclic degree of a vertex is the number of distinct cyclic neighbors of .
A plane graph is minimal if it has no cyclic coloring with at most colors, it has maximum face size at most , all its faces of size four or more are vertexdisjoint, and has the minimal number of vertices subject to the previous constraints. Clearly, a minimal graph is connected and has no separating cycles of length two or three. We will use these facts implicitly throughout the paper.
Our goal is to show that there is no minimal graph with (see Theorem 12). This will combine with the previous results to the following:
Theorem 1.
Every plane graph with maximum face size whose all faces of size four or more are vertexdisjoint has a cyclic coloring with at most colors.
The general structure of the proof is the following. We first identify configurations that cannot appear in a minimal graph; these configurations will be called reducible configurations. Using the knowledge of reducible configurations, we exclude the existence of a minimal graph by assigning each vertex and face charge in such a way that the total amount of charge is negative. The assigned charge is then redistributed using rules preserving its amount. The original amount of total charge will be and we will be able to show that the final amount of charge of all vertices and faces is nonnegative. This will exclude the existence of a minimal graph.
3 Reducible configurations
In this section, we study configurations that cannot appear in a minimal graph . Let us start with a simple observation on the minimum degree of a minimal graph.
Lemma 2.
The minimum degree of every minimal graph , , is at least four.
Proof.
It is straightforward to show that has no vertex. Assume that has a vertex , . If is incident with faces only, then proceed as follows: remove from and consider a cyclic coloring of the resulting graph which exists by the minimality of . This coloring can be extended to since the cyclic degree of is at most . Hence, we assume that is incident with an face, .
Let and be the neighbors of incident with the face and the graph obtained from by removing and adding the edge if the degree of is three. Observe that the maximum face size of does not exceed the maximum face size of and the faces of size four and more are still vertexdisjoint.
Consider a cyclic coloring of which exists by the minimality of . We now construct a cyclic coloring of . The vertices of distinct from preserve their colors. There are at most colors that cannot be assigned to : the colors of the colors incident with the face and the color of the third neighbor of if is a vertex. We conclude that there is a color that can be assigned to and thus the coloring can be completed to a cyclic coloring of . ∎
In the next lemma, we look at vertices of degree four and five in minimal graphs.
Lemma 3.
Let be a minimal graph, . Every vertex, is incident with an face, .
Proof.
Consider a vertex , , contained only in faces. Let be the graph obtained from by removing and triangulating the new face. By the minimality of , the graph has a cyclic coloring. We now extend this coloring to . The vertices distinct from keep their colors. Since the cyclic degree of in is at most , the coloring can be extended to which contradicts our assumption that is minimal. ∎
Next, we show that vertices can be incident with faces and faces, , only.
Lemma 4.
Let be a minimal graph, . No face of contains a vertex.
Proof.
Let be a vertex incident with a face and let be the vertex of the face not adjacent to . By removing from and triangulating the resulting face with edges incident with , we obtain a graph (see Figure 1). By the minimality of , the constructed graph has a cyclic coloring. Since the cyclic degree of is and , there is a color that can be assigned to . This completes the coloring to a cyclic coloring of . ∎
Our next goal is to exclude the cases that a face or a face is incident with too many vertices. This is done in the next two lemmas.
Lemma 5.
Let be a minimal graph, . No face of contains three vertices.
Proof.
Assume that contains a face incident with three vertices , and (in this order on the boundary). Let be the common neighbor of and (see Figure 2). Remove the vertices , and from and triangulate the new face with edges originating from . Note that the obtained graph has no loops since has no separating triangles. By the minimality of , the graph has a cyclic coloring.
We now extend this coloring to . Since , there are at least colors in total which can be used in the coloring. Let be the color of . Color the vertex with . We next color the vertices and . Each of these two vertices has cyclic degree but two of its cyclic neighbors ( and ) have the same color. As , the coloring can be extended to a cyclic coloring. ∎
Lemma 6.
Let be a minimal graph, . No face of contains a vertex of degree five adjacent to a vertex of degree four or five.
Proof.
Let , , , and be the vertices of (in this order) and assume that the vertex is a vertex and is a vertex or a vertex (see Figure 3). Let be the common neighbor of and , the common neighbor of and and the remaining neighbor of . We now modify the graph to another graph . Remove the vertex , identify the vertices and , and and . The resulting graph is and is loopless since has no separating triangles. By the minimality of , has a cyclic coloring.
Before extending the coloring of to , we might have to recolor the vertex (its color can coincide with the color of the vertex or the color of the vertex ). As the cyclic degree of is at most , it has an uncolored neighbor (the vertex ) and two cyclic neighbors with the same color (the vertices and ), it is possible to recolor it. Finally, since the cyclic degree of is and two pairs of its cyclic neighbors (the vertices and , and and ) have the same color, the coloring can be extended to . The existence of a cyclic coloring of contradicts the minimality of . ∎
In the remaining three lemmas, we consider degrees of consecutive vertices on an face, . We first exclude the existence of a face with two consecutive vertices.
Lemma 7.
Let be a minimal graph, . No face , , of contains two consecutive vertices.
Proof.
Let be the vertices incident with listed in the order on its boundary and assume that and are vertices (see Figure 4). Further, let be the common neighbor of and . Form a graph by removing and , adding the edge and triangulating the new face by adding edges originating from as in Figure 4. Since has no separating triangles, is loopless. Consequently, the minimality of implies that has a cyclic coloring.
Let be the color assigned to the vertex . If the color is assigned to none of the vertices , color with . Otherwise color, with any available color (as the cyclic degree of is and has no color, there is a color that can be used). Observe that two cyclic neighbors of now have the color . Since the cyclic degree of is and two of its cyclic neighbors have the same color, the coloring can be completed to a cyclic coloring of . ∎
In the final two lemmas of this section, we exclude that one of three consecutive vertices on an face, , would have degree four and the remaining two would have degree four or five.
Lemma 8.
Let be a minimal graph, . No face , , of contains three consecutive vertices with degrees or (in this order).
Proof.
Let be the vertices incident with listed in the order on its boundary and assume that is a vertex, is a vertex and is a vertex, (see Figures 5 and 6). Let be the graph obtained by removing the vertices , and , adding the edge and triangulating the new face by adding edges originating from (see the figures) where is the common neighbor of and . Again, has no loops as has no separating cycles of length at most three, and the minimality of implies that is cyclically colorable.
We extend a cyclic coloring of to . Let be the color assigned to the vertex . If the color is not assigned to any of the vertices , assign to . Otherwise, color with any available color (as the cyclic degree of is at most and two of its cyclic neighbors are uncolored, there is such an available color). Color now the vertex : the cyclic degree of is but two of its cyclic neighbors have the same color (the color ) and one of its cyclic neighbors (the vertex ) is uncolored. Finally, we color the vertex : since its cyclic degree is and two of its cyclic neighbors have the same color, there is a color that can be assigned to . The existence of a cyclic coloring of contradicts the minimality of . ∎
Lemma 9.
Let be a minimal graph, . No face , , of contains three consecutive vertices with degrees (in this order).
Proof.
The proof follows the lines of the proof of Lemma 8. We assume that has a face with vertices such that and are vertices and is a vertex and we let to be the common neighbor of and (see Figure 7). Remove the vertices , and , add the edge and triangulate the obtained new face with edges incident with . The obtained graph , which is loopless, is cyclically colorable by the minimality of .
The cyclic coloring of can now be extended to . If the color of is not assigned to any of the vertices , we color with . Otherwise, we color with any available color (as the cyclic degree of is and two of its cyclic neighbors are uncolored, there is an available color). We next color the vertex (its cyclic degree is , it has an uncolored cyclic neighbor and has two cyclic neighbors colored with ) and the vertex (its cyclic degree is and has two cyclic neighbors colored with ). ∎
4 Discharging phase
In this section, we present the second part of the proof of our result. At the beginning, every vertex of a minimal graph is assigned charge of units and every face is assigned charge of units. The Euler formula implies that the total amount of charge assigned to all the vertices and faces of the graph is equal to . The initial charge is then redistributed based on the following two rules:
 Rule 1

Every face, , sends units of charge to each incident vertex.
 Rule 2

Every face, , sends unit of charge to each incident vertex.
First, we show that the final charge of every vertex is nonnegative.
Lemma 10.
The final amount of charge of every vertex of a minimal graph , , is nonnegative.
Proof.
Let be the degree of . By Lemma 2, the degree is at least four. If , then the initial amount of charge of is nonnegative and its final amount of charge is also nonnegative as neither receives nor sends out any charge. If , then is incident with an face, , by Lemma 3. If the degree of is four, then receive two units of charge by Rule 1, and if its degree is five, then it receive one unit of charge by Rule 2. In either of the two cases, the final amount of charge of is zero. ∎
We next show that the final charge of every face is nonnegative.
Lemma 11.
The final amount of charge of every face of a minimal graph , , is nonnegative.
Proof.
If is face, its final amount of charge is zero. If is a face, then it is incident with no vertex by Lemma 4 and with at most two vertices by Lemma 5. Hence, sends out at most two units of charge (twice one unit by Rule 2) and the final charge of is nonnegative.
Assume that is a face. By Lemma 6, is is incident with at most two vertices of degree four or five. Consequently, Rules 1 and 2 apply at most twice and sends out at most four units of charge to incident vertices. Since the amount of initial charge of is equal to four units, the final charge of is nonnegative.
The remaining case is that is an face, . Let be vertices on the boundary of . If all vertices of received charge from , then they all would be vertices by Lemmas 7 and 8. Hence, would send out units of charge and its final charge would be nonnegative in this case.
In what follows, we assume that does not send out charge to all incident vertices. Let be maximal consecutive intervals of vertices that receive charge from . Observe that
(1) 
We claim that the total amount of charge sent by to the vertices is at most units for every .
If , then sends at most two units of charge to the only vertex of and the claim holds. If , then sends at most three units of charge to the two vertices of as they both cannot be vertices by Lemma 7. The claim also holds in this case. If , then none of the vertices of is a vertex by Lemmas 7, 8 and 9. We conclude that sends to the vertices of exactly units of charge as they all are vertices.
Since the vertices of receive at most units of charge from , the total amount of charge sent out by is at most which is at most . Since the initial amount of charge of is and , the final amount of charge of is nonnegative. ∎
Theorem 12.
There is no minimal graph, . Every plane graph with maximum face size whose all faces of size four or more are vertexdisjoint has a cyclic coloring with at most colors.
Acknowledgement
The authors would like to thank Riste Škrekovski for discussions on cyclic colorings of plane graphs.
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