Imagine making a MPG with a K4 subgraph where every vertex of the K4 has a degree of four or higher. Start by thinking about the K4 shown below. In order for vertex D to have a degree of four or higher, we must add new vertices inside at least one of the faces.
Note that K4 has four faces: ABD, ACD, BCD, and the outside face ABC corresponding to the infinite area outside. If we add new vertices in at least two of these faces (well, for ABC it would be “out” rather than “in”), we can make a MPG where all of the vertices of the K4 subgraph have a degree of at least four. An example is shown below.
Find A, B, C, and D in the triangle above. This was our original K4, which is now a subgraph of a MPG with 10 vertices. We added vertices E, F, and G in face ABD, and added vertices H, I, and J “in” the outside face ABC (when a face corresponds to the infinite area outside,
the area of the face is “out” of the face rather than “in” it). We then added enough edges to triangulate the graph, turning it into a MPG. For V = 10 vertices, we need a total of E = 3V – 6 = 3(10) – 6 = 24 edges (arranged so that every face has three edges). The way we chose to add the needed edges, note that all 10 vertices of the MPG have a degree of at least four, including all four vertices of the original K4 subgraph.
Every K4 subgraph has a separating triangle. What does this mean? A separating triangle, which we will abbreviate ST, is a triangle consisting of three vertices and three edges, where the triangle isn’t one of the faces. A ST has at least two faces inside of it and at least two faces outside of it. (Note that one of the faces outside of a ST can be the infinite area outside of the MPG.)
In the previous example, ABD is a ST because ABD isn’t a face and because there are at least two faces inside and at least two faces outside of ABD. We call it a ST because vertices A, B, and D divide the MPG into two smaller MPG’s, as seen in the diagram below. Observe that triangle ABD appears in both MPG’s below, and now ABD is a face (whereas it wasn’t in the original graph). Any MPG that has a ST can similarly be split into two separate MPG’s. (There is actually a second ST in the MPG below and on the previous page: triangle ABC.)
If each of the smaller MPG’s is individually four-colorable, it follows that the original MPG is also four-colorable. Why? The only thing that the two smaller MPG’s have in common is the ST, which has three vertices and three edges. When we color the two smaller MPG’s, in each case we may begin by coloring the vertices of the ST. For example, we can choose A to be red, B to be blue, and D to be yellow for each MPG. Since A, B, and D lie on one face of each of the smaller MPG’s, we know that these three vertices must have different colors. After coloring each of the smaller MPG’s, when we put them together to form the original MPG, the same coloring will work for the original MPG because A, B, and D have the same colors in each of the smaller MPG’s.
This means that we don’t need to worry about any graphs that have ST’s. If we can prove the four-color theorem for all MPG’s that don’t have ST’s, it will follow that the four-color theorem also applies to MPG’s that do have ST’s (since a ST can be used to divide a MPG into two smaller MPG’s).
Let’s look at another example of a ST. Can you find a ST in the MPG below?
You can find the ST in the K4 subgraph. The K4 subgraph is AFGL. (Why is AFGL a K4 subgraph? Each of these three vertices connects to all three of the others.) The ST is FGL. Why? FGL isn’t a face. There are at least two faces inside FGL and at least two faces outside of FGL (in this case, there are many more than two faces both inside and outside of the ST; the requirement is for there to be at least two).
At this point, we’ve shown that it isn’t necessary to prove the four-color theorem directly for any MPG’s that contain K4 subgraphs for the following reasons:
• If a K4 subgraph has a vertex with degree three, the vertex with degree three may be removed from the MPG (as discussed in Chapter 11). This subgraph will no longer be a K4. Since we don’t need to prove the four-color theorem directly for any MPG that has a vertex with degree three, we
don’t need to consider any MPG’s with K4’s with a vertex with degree three. If we prove the four-color theorem for all MPG’s without vertices with degree three, it will follow that all MPG’s with vertices with degree three are also four-colorable.
• Even if a K4 subgraph doesn’t have any vertices with degree three, the K4 subgraph will have a ST. The ST may be used to divide the MPG into two smaller MPG’s. If we prove the four-color theorem for all MPG’s without ST’s, it will follow that all MPG’s with ST’s are also four-colorable.
Combining these ideas together, if we prove the four-color theorem for all MPG’s without K4 subgraphs, it will follow that all MPG’s with K4 subgraphs are also four-colorable.
Technically, all K4 subgraphs have ST’s, even if a K4 subgraph has a vertex with degree three. However, if there is a vertex with degree three, we may simply remove that vertex from the MPG (Chapter 11). Imagine the K4 above as a subgraph in a MPG. Vertex D has degree three. If you make the ST ABC to divide the MPG into two graphs, all you are effectively doing anyway is removing vertex D from the MPG.
CHAPTER 12 EXERCISES
1. Identify all of the ST’s in each MPG below.
2. Identify a ST in the MPG below. Draw two smaller graphs to show how this ST may be used to divide the MPG into two separate graphs.
3. The MPG below has been divided into two graphs using its ST. Color the two smaller graphs below and show that this same coloring works for the original graph.
4. A MPG has a ST involving vertices E, K, and P. This ST is used to separate the MPG into two smaller graphs. One graph is four-colored with E green, K blue, and P yellow. The other MPG is four-colored with E red, K green, and P blue. Explain precisely how one or both of these graphs can be recolored such that the original MPG is properly four-colored.
5. What is the greatest number of ST’s possible that a MPG with 10 vertices can have? Draw one MPG with this number of ST’s with 1 red vertex, 1 blue vertex, 4 green vertices, and 4 yellow vertices. Draw another MPG with this number of ST’s with 3 green vertices, 3 yellow vertices, 2 red vertices, and 2 blue vertices.
Challenge problem 1: Are all ST’s (of MPG’s) parts of K4 subgraphs?
Either provide an example with a ST (in a MPG) that isn’t part of a K4 subgraph or explain why all ST’s are contained in K4 subgraphs.
Challenge problem 2: Can we make separating quadrilaterals, pentagons, or any other polygons besides a ST and use them to separate a MPG into two smaller graphs? If so, will the same principle apply that if each of the smaller graphs is four-colorable the original MPG will also be four-colorable? Either provide an example and explain what allows you to do this or explain why this is only possible for a triangle.
Note: The answer key doesn’t include answers to the challenge problems.
These problems are intended to encourage you to think about the ideas.