As discussed in Chapter 13, as far as the four-color theorem is concerned, we only need to consider MPG’s that have HC’s. Any MPG that has a HC may be drawn with all of its vertices arranged in the shape of a closed convex polygon, as shown below.
The diagrams above show how a MPG with a HC can be drawn as a closed convex polygon.
• First identify one HC. See the solid lines at the top right:
AHGEDKLJCIBFA.
• Arrange the vertices and edges of the HC in order in the shape of a closed convex polygon. See the diagram at the bottom left.
• Add in the remaining edges. One-half of the remaining edges will fit inside of the polygon and one-half will fit outside of the polygon. Choose
wisely and there won’t be any crossings (since a MPG is a planar graph).
See the diagram at the bottom right.
This allows us to draw all MPG’s with HC’s in a consistent form.
The MPG on the previous page has V = 12 vertices and E = 3V – 6 = 3(12) – 6 = 30 edges.
• V = 12 vertices are arranged in the shape of a convex polygon.
• V = 12 edges connect the V = 12 vertices to form the closed convex polygon.
• V – 3 = 12 – 3 = 9 edges lie inside of the polygon.
• V – 3 = 12 – 3 = 9 edges lie outside of the polygon.
• There are V + 2(V – 3) = 12 + 2(12 – 3) = 12 + 2(9) = 12 + 18 = 30 edges in total, which agrees with the formula from Chapter 4 for a MPG: E = 3V – 6 = 3(12) – 6 = 30.
Every MPG with a HC that is drawn in the shape of a closed convex polygon will have V – 3 edges inside of the polygon and V – 3 edges outside of the polygon. Why? This follows from Euler’s formula.
• Recall from Chapter 4 that a MPG with V vertices has E = 3V – 6 edges.
• The HC connects the V vertices with V edges. Subtract V edges from 3V – 6 to see that there are 2V – 6 edges remaining.
• Divide this by 2 to get V – 3. There are V – 3 edges inside of the polygon and V – 3 edges outside of the polygon.
• As with any MPG, it is possible to draw all of the edges without crossing.
The HC divides the MPG in half in the following sense. There are V – 2 diagram. In this sense, the polygon serves as a “ring of invertibility.”
The two MPG’s shown above are isomorphic: both contain edges AB, AC, AD, AE, AF, BC, BD, BF, CD, DE, DF, and EF. (Recall from Chapter 1 that isomorphic graphs are structurally equivalent in terms of edge-sharing.) As far as which vertices share edges is concerned, the two graphs shown previously may effectively be treated as the same MPG. The only distinction is that “inside” and “outside” of the HC have been inverted.
It is well-known that any PG can be drawn on the surface of a sphere instead of in the plane [Ref. 14]. Imagine drawing a MPG with a HC on the surface of a sphere. We may choose to place the HC along the equator of the sphere.
Then the two sets of V – 3 edges that we interpreted as “inside” and “outside”
edges in the plane would simply correspond to the two hemispheres of a sphere. Inverting “inside” and “outside” in the plane is simply like turning the sphere upside down.
Technically, the HC around the equator would look like its mirror image in the bottom view compared to the top view. To make the comparison of the two views simpler, we mirrored the bottom view image in the diagram above so that the HC around the equator would be consistent with the top view.
CHAPTER 14 EXERCISES
1. Redraw the MPG below in the form of a closed convex polygon.
2. Redraw the MPG below with its inside and outside edges inverted.
3. If a MPG with 24 vertices which has a HC is drawn in the form of a closed convex polygon, how many inside edges and outside edges will it have? How many edges does it have in total? How many inside faces and outside faces will it have? How many faces does it have in total?
Challenge problem 1: If a MPG with a HC is drawn in the form of a closed convex polygon and we remove the V – 3 outside edges, prove that the remaining graph is three-colorable. Similarly, if we remove the V – 3 inside edges (instead of removing the outside edges), prove that the remaining graph is three-colorable. Can you use these proofs to prove that when both sets of V – 3 edges (inside and outside) are included that the complete MPG is four-colorable? Either use this to prove the four-color theorem or explain what makes this task very difficult or impossible to carry out.
Challenge problem 2: Show that the faces (not the vertices) of every MPG with a HC are trivially four-colorable since the set of faces on either side of the HC is two-colorable. Show that we can actually go a step further: The faces of many MPG’s with HC’s are actually three-colorable. Draw a variety of examples to illustrate this. Are the faces of every MPG with a HC three-colorable? Either prove this, find a counterexample, or explain why it would be very difficult to determine. What are the prospects of using either of these results (four-coloring or three-coloring of the faces of MPG’s with HC’s) to prove the four-color theorem (which relates to vertices, not faces)? Now consider that the four-color theorem applies to all maps, where faces (not vertices) are colored. Can these results (regarding the coloring of the faces of MPG’s with HC’s) be used to prove the four-color theorem for maps?
Explain.
Note: The answer key doesn’t include answers to the challenge problems.
These problems are intended to encourage you to think about the ideas.