Coloration of minor free graphs
Abstract
Hadwiger’s conjecture says that every minor free graph is colorable. This problem has been proved for but remains open for . minor free graphs have been proved to be colorable (Albar & Gonçalves, 2013). We prove here that minor free graphs are colorable, where is the graph obtained from by removing one edge.
1 Introduction
A minor of a graph is a graph obtained from by a succession of edge deletions, edge contractions and vertex deletions. All graphs we consider are simple, i.e. without loops or multiple edges.
Hadwiger’s conjecture says that every chromatic graph (i.e. ) contains has a minor. This conjecture has been proved for , where the case is equivalent to the Four Color Theorem by Wagner’s structure theorem of minor free graphs, and the case has been proved by Robertson, Seymour and Thomas [rst1]. The conjecture remains open for .
In [ag1], the author and D. Gonçalves proved that minor graphs are colorable.
In [kt1], Kawarabayashi and Toft proved that any and minor free graph is colorable by using the fact that a minor free graph contains at most edges. In particular, this implies that it contains some vertices of degree . In their proof they show that most of these vertices in a chromatic critical graph (i.e. such that every strict minor of this graph is colorable) are contained in a subgraph and use these subgraphs and the connectivity of a chromatic critical graph to find a minor.
We use here similar techniques to prove the following theorem.
Theorem 1
Every minor free graph is colorable.
2 Proof of Theorem 1
Let be a minimal counter example to Theorem 1, i.e. a minimal minor free chromatic critical graph,
First we will prove that a lot of vertices of degree are contained in subgraphs and then we will apply some techniques introduced in [kt1] to conclude.
We will use the following theorem of Jakobsen to prove that minor free graphs are degenerate.
Theorem 2 (Jakobsen, 1983, [jakobsen2])
Every graph with at least vertices and at least edges has a minor or is a cockade.
We also need the following theorem of Mader.
Theorem 3 (Mader, 1968, [mader1])
Any chromatic critical graph that is not isomorphic to is connected for .
Hence is connected, and thus is not a cockade. Thus we can deduce the following corollary of these two theorems.
Corollary 4
has less than edges.
We also need the following folklore lemma (see [ag1] for a proof).
Lemma 5 (Folklore)
In a chromatic critical graph , has minimum degree at least and for any vertex of degree (resp. ), then the graph induced by has no stable of size (resp. ).
In particular, this lemma implies that has minimum degree at least because if contains a vertex of degree then has no stable set of size and thus contains a minor, a contradiction. We will use vertices of degree and their neighborhoods to find a minor. The following lemma ensures the existence of such vertices.
Lemma 6
has at least vertices of degree .
Proof. By Corollary 4, has less than edges. Suppose that has at most vertices of degree . By Lemma 5, has no vertices of degree strictly less than , so we have that :
a contradiction.
Lemma 7
Let be a vertex of degree , then either contains as a subgraph or contains the graph , i.e. the circulant graph on vertices with jumps (see Figure 1), as a subgraph.
Before proving Lemma 7, let us introduce some material. The following lemma can be immediatly deduced from the fourcolor theorem.
Lemma 8
Let , and be three vertices of , then is connected and nonplanar.
Proof. The first part of the lemma is obvious by the connectivity of . Suppose now that there exists such that is planar. By the Four Color Theorem, if is planar then it is colorable, thus is colorable, a contradiction.
We need the following definition and theorem introduced by Robertson, Seymour and Thomas in [rst1].
Definition 9
Let be a graph and be a triangle. is said triangular with respect to if one of the following holds.

For some , has maximum valency at most , and either is a circuit or it has no circuit.

All vertices of have valency at most , there is at most one 3valent vertex , and has no circuit.

All vertices of have valency at most , there is a triangle with , every 3valent vertex of is in , and every circuit of except these two triangles meets both and .
Theorem 10 (Robertson, Seymour & Thomas, 1993, [rst1])
Let be a triangle in a connected nonplanar graph . Let be an induced subgraph of such that and is not triangular with respect to . Then has a minor in such that .
Let us now prove Lemma 7.
Proof. Let be a vertex of degree of and suppose that the graph induced by is free.
Claim 11
is connected.
Proof. Let be a minimal separation of . Since there is no stable of size in by Lemma 5, for each pair of vertices of and any vertex in , contain at least one edge. This edge cannot be or because is a separation of . So this must be the edge . We deduce that both and are complete graphs.
If is a separation of order , then either or . By the previous remark, contains a subgraph, a contradiction.
Let suppose that is a separation of order , then in this case . Let . Since the graph induced by is free and since and are triangles, there is one vertex such that is not an edge. In the same way, there is a vertex such that is not an edge. Since is a separation of , then is a stable set of size , a contradiction.
Let now suppose that is a separation of order . By the previous remark, and . Since and , we can assume without loss of generality that and . Let and let . Suppose that there is a vertex , and a vertex , , such that is not an edge, then since has no stable set of size , is adjacent to all the vertices of the triangle but then contains a subgraph, a contradiction. Thus we can assume that and are adjacent to all the vertices of .
Now since is free, is a stable set because if say are adjacent for then would be a subgraph, a contradiction. But then is a stable set of size , a contradiction.
Claim 12
is planar.
Proof. Assume that is nonplanar. Since is connected by Claim 11, then contains a minor by Wagner’s theorem [wagner1]. Since is not isomorphic to as it contain at least vertices, then we can find . By the connectivity of , there is vertex disjoint paths between and . Let denote them by . We can always assume that these paths are minimal in length and thus that these paths intersect in at most one vertex. If there is vertexdisjoint paths between and , then there exists vertexdisjoint paths between and every vertex of . Since contains a minor, then , together with , and the paths between and , contains a minor, a contradiction.
So now, let be the only vertex of which is not contained in any of the paths between and . By Ramsey’s theorem, since has vertices and no stable set of size , then it contains a triangle. Denote by , and its vertices. Since is connected, is not triangular with respect to , and since it is connected and nonplanar, then by Theorem 10, there exists and such that is a minor. Since does not contain any subgraphs, then , so both sets and intersect at least one of the paths , . Thus is a minor, a contradiction.
Claim 13
does not contain any vertex of degree or greater in .
Proof. Suppose that contains a vertex of degree greater or equal than , then the graph induced by in contains no stable set of size , but then by Ramsey’s theorem it contains a triangle. Thus contains a subgraph, a contradiction.
Claim 14
The neighborhood of any vertex of is a path, a cycle or a cycle.
Proof. Let be a vertex of degree in , denote by , , and its neighbors. Suppose that its neighborhood is not a path nor a cycle. Since the neighborhood of is trianglefree and does not contain a stable set of size , it must be two disjoint edges, say and . Denote by , and , the three vertices in . is a triangle because otherwise there is stable set of size with .
Every vertex in sees exactly two vertices in because, either there would be a vertex of degree at most in , contradicting the connectivity of , or if one these vertices is adjacent to the three vertices , and then would contain a subgraph, another contradiction. Then as there are edges between and , there exists one vertex of degree in say . By symmetry, we can assume that is adjacent to , and . Now since every vertex in is adjacent to two vertices in , , and are adjacent to either or . But then is a minor, contradicting Claim 12.
If is a vertex of degree , then since does not contain any stable set of size and any triangle, then can only be isomorphic to the cycle of length .
Since is planar, it has at most edges by Mader’s theorem, so it contains at least one vertex of degree . Let be such a vertex. Denote by , , and its neighbors and , and its nonneighbors. Then can induce either a path or a cycle.
Suppose that is a path. Now the neighborhood of cannot induces a cycle or a cycle because this would contradict that has degree . So has degree and its neighborhood is a path. By symmetry we can assume that ’s neighbors are the path . Moreover is a triangle since otherwise would contain a stable set of size with . Now is adjacent to at least other vertex in because it would be of degree otherwise, contradicting the connectivity of . Planarity forces to be adjacent to , but then contains .
Now suppose that is the cycle . Suppose that has degree and assume that ’s neighborhood is a path, say . Now is also a triangle because otherwise there is a stable set of size with . As and have degree at least in and as then and are both adjacent to at least one vertex in . Moreover and cannot be both adjacent to the same vertex because otherwise there would be a subgraph with . So either is adjacent to and the is adjacent to either is adjacent to and the is adjacent to . In both cases, after removing the edge , the graph is isomorphic to . Note that the same argument applies when ’s neighborhood is a cycle by also removing the edge at the end.
Suppose now that has degree so we can assume that its neighborhood is the cycle . Then is adjacent to because otherwise is a stable set of size . Since has degree at least in it is also adjacent to at least one vertex in the set . But if is adjacent to then after removing the edge , the graph is isomorphic to , and if is adjacent to then after removing the edge , the graph is isomorphic to .
Lemma 15
Let and be two degree vertices of such that and contain the graph as a subgraph, then and are not adjacents.
Proof. Let suppose that and are adjacent. Since every vertex of has degree at least and by hypothesis, denote by , , and the four neighbors of in the subgraph of . These four vertices induce a path in this subgraph, say . Let denote by , and the vertices of in a way that is the only vertex adjacent to both and and is the one adjacent to and . Now consider . is connected and nonplanar by Lemma 8. Let , then is not triangular with respect to because has degree in . Thus by Theorem 10, there exists and such that is a minor in and such that . But then is a minor in , a contradiction.
The following lemma is the key to prove that a lot of degree vertices are contained in a .
Lemma 16
Let and be two vertices of degree such that and contain the graph as a subgraph and , then contains a minor.
Proof. By Lemma 15, we can assume that and are not adjacent. Denote by the vertices of as shown in Figure 1. Since is connected, there is at least internally disjoint paths between and that induce disjoint paths between and . Note that theses paths can be of length if the two neighborhoods intersect. By contracting the nonzero length paths, we obtain a graph with . From now on, we consider only this new graph . By construction of , still contain a subgraph.
By symmetry of , we can assume that is the only neighbor of which is not a neighbor of . In particular, we have that for all . But then is a minor (only and are not adjacent) of and thus a minor of , a contradiction.
Claim 17
Let and be two vertices of degree such that and contain the graph as a subgraph, then .
Proof. Suppose that there exists two vertices and of degree such that . Then we can create a minor in by using the same argument as in the proof of Lemma 16, a contradiction.
Claim 18
At most one vertex of degree have a neighborhood containing the graphs as a subgraph.
Proof. Suppose that there exists two vertices of degree such that their neighborhood contains the graph as a subgraph. By Claim 17, these two vertices have a different neighborhood. By Lemma 16, this imply that there is a minor in , a contradiction.
Lemma 19
There is at least different in .
Proof. By Lemma 6, there is at least vertices of degree and by Lemma 18, there is at most one vertices of degree containing the graph as a subgraph of their neighborhood. By Lemma 7, this imply that there is at least vertices of degree that contains a in their neighbourhood. As every subgraph can contain at most vertices of degree , this finally imply that there is at least different subgraph in .
The following lemma is the last key to the proof. It uses techniques introduced by Kawarabayashi and Toft [kt1].
Lemma 20
There is different copies of , and such that .
Proof. Assume by contradiction that no three copies of , denoted , and , are such that .
The next claim follows easily from the connectivity of .
Claim 21
does not contain a subgraph.
Proof. Suppose that contains a subgraph. Since is not isomorphic to , there exists a vertex that is not contained in this subgraph. Since is connected, by Menger’s theorem there are vertexdisjoint paths between and the vertices of the subgraph. This induces a minor, a contradiction.
Claim 22
Two different intersects on at most vertices.
Proof. Let and be two copies of of and suppose that they intersect on vertices, then contains a as a subgraph, contradicting Claim 21. If they intersect on vertices, then denote by the set of vertices in and by the set of vertices of . By Lemma 8, is connected and nonplanar so by (2.6) of [rst1] there is a minor rooted in and a minor in , a contradiction.
Claim 23
No two are disjoints.
Proof. Assume that and are two disjoint copies of . For any copy of , since two copies of cannot intersect on vertices and since , and . By Claim 22, and . Let and .
Now is connected so by Menger’s theorem there are vertex disjoint paths , and between and but then is a minor, a contradiction.
Claim 24
No two intersect on exactly one vertex.
Proof. Assume that . Let be a copy of different from and . By Claim 23, intersects both and .
Suppose that . Since , . Let . is connected and nonplanar by Lemma 8. Let . Denote . is not triangular with respect to , hence there exists such that is a minor in and such that . Moreover we can assume without loss of generality that is adjacent to . Thus is a minor in (only and may not be adjacent), a contradiction.
Suppose now that . Since and , by Claim 22, we can assume that . Let us denote .
If , let . Now is connected. So by Menger’s theorem, there are vertex disjoint paths , and , between and . Hence is a minor, a contradiction.
If , let . is connected, so by Menger’s theorem, there are vertex disjoint paths and between and . But is a minor, a contradiction.
Claim 25
No two intersect on exactly two vertices.
Proof. Assume that . Let be a different from and . By Claims 22, 23 and 24, intersects each and on two vertices.
Suppose that and let and . Then is a subgraph, a contradiction with Claim 21.
Suppose that and let , , . Now is connected and nonplanar. Let and let . is not triangular with respect to , so there exists and such that is a minor in . Without loss of generality, we can assume that but then is a minor in , a contradiction.
Finally, suppose that , is connected and nonplanar. Let and . is not triangular with respect to so there exists and such that is a minor in , but then is a minor in , a contradiction.
We conclude the proof of Theorem 1 by using the following theorem due to Kawarabayashi and Toft [kt1].
Theorem 26 (Kawarabayashi & Toft, 2005, [kt1])
Let be a connected graph with at least vertices. Suppose that contains three , say , and , such that , then contains a minor.
Applying this theorem to the three given by Lemma 20 gives us a contradiction.
3 Conclusion
We have seen that minor free graphs are colorable. The techniques used here are not sufficient to prove that minor free graphs are colorable because we then have to deal with "sparse" neighborhoods of degree and vertices. However, since connected minor free graphs are degenerated [song1], we wonder whether similar techniques can be extended to prove that minor free graphs are colorable. Currently the best bound for minor free graphs is given by the fact that minor free graphs are colorable [ag1].
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