In this chapter, we will construct and consider ultimate four-colorable graphs, which we will abbreviate U4CG, defined as follows:
• K vertices are blue (B). Each blue vertex connects to every vertex that isn’t blue.
• L vertices are green (G). Each green vertex connects to every vertex that isn’t green.
• M vertices are red (R). Each red vertex connects to every vertex that isn’t red.
• N vertices are yellow (Y). Each yellow vertex connects to every vertex that isn’t yellow.
• The total number of vertices is V = K + L + M + N.
An U4CG is a complete multipartite graph with four sets of vertices where every vertex of one set connects to every vertex of the other sets, but where no vertices of the same set are connected. (We saw an example of a complete multipartite graph with two sets of vertices in Chapter 5, called a complete bipartite graph. The complete multipartite graphs of this chapter are similar, except that they have four sets of vertices rather than two.)
A general U4CG is shown below, where B1, B2, B3, thru BK represent the K regions colored blue; G1, G2, G3, thru GL represent the L regions colored green; R1, R2, R3, thru RM represent the M regions colored red; and Y1, Y2, Y3, thru YN represent the N regions colored yellow.
The … symbol in the above graph means “and so on.”
Let’s count how many edges an U4CG has. Recall that K, L, M, and N represent the numbers of blue, green, red, and yellow vertices, respectively.
• K blue vertices connect with L + M + N vertices that aren’t blue: K(L + M + N).
• L green vertices connect with K + M + N vertices that aren’t green: L(K + M + N).
• M red vertices connect with K + L + N vertices that aren’t red: M(K + L + N).
• N yellow vertices connect with K + L + M vertices that aren’t yellow:
N(K + L + M).
Add these up and divide by two (to correct for double-counting, like we did in Chapters 4-5).
Apply the distributive property of algebra. For example, K(L + M + N) = KL + KM + KN.
We applied the commutative property of multiplication, that order doesn’t matter when multiplying numbers. For example, LK = KL. Now we will combine like terms. For example, KL + KL = 2KL. Note that each term appears twice on the right-hand side.
An alternative formula for the number of edges of an U4CG appears in Exercise 11 of Chapter 17. For example, an U4CG with K = 3 blue vertices, L = 4 green vertices, M = 5 red vertices, and N = 6 yellow vertices has E = for a MPG from Chapter 4). Most U4CG’s are nonplanar because they have too many edges to be MPG’s (but there are special cases where U4CG’s have few enough edges to be PG’s, like the case where K = L = M = N = 1).
The significance of the U4CG is that any four-colorable graph (planar or not) can be drawn by removing edges from an U4CG. The same principle that motivated triangulation (Chapter 3) applies here: Since every U4CG is inherently four-colorable, any graph that is obtained by removing edges from an U4CG is also four-colorable. If we could prove that every possible MPG can be obtained by removing edges from an U4CG, we could use that proof as a lemma to easily prove the four-color theorem.
Let’s consider a specific example of an U4CG with 2 blue vertices, 2 green vertices, 3 red vertices, and 1 yellow vertex. This U4CG is shown below.
• Each blue connects to all 6 non-blue regions.
• Each green connects to all 6 non-green regions.
• Each red connects to all 5 non-red regions.
• The only yellow connects to all 7 non-yellow regions.
• K = 2, L = 2, M = 3, and N = 1.
• There are V = K + L + M + N = 2 + 2 + 3 + 1 = 8 vertices.
• There are E = KL + KM + KN + LM + LN + MN = 2(2) + 2(3) + 2(1) + 2(3) + 2(1) + 3(1) = 4 + 6 + 2 + 6 + 2 + 3 = 23 edges.
• A MPG with V = 8 vertices would have E = 3V – 6 = 3(8) – 6 = 24 – 6 = 18 edges. Thus, the U4CG shown above is nonplanar. (See Chapter 6.) Like all U4CG’s, the graph shown above is four-colorable by construction since it doesn’t connect any edges to two vertices that have the same color.
Any graph that can be drawn by removing edges from this (or any other) U4CG is also four-colorable.
Note that the previous graph isn’t the only type of U4CG that can be drawn for a graph with 8 vertices. An U4CG with V = 8 vertices could have:
• 2 blues, 2 greens, 2 reds, and 2 yellows.
• 3 of one color, 3 of another color, and 1 each of the remaining two colors, such as 3 blues, 3 greens, 1 red, and 1 yellow.
• 3 of one color, 1 of another color, and 2 each of the remaining two colors, such as 3 blues, 1 green, 2 reds, and 2 yellows.
• 4 of one color, 2 of another color, and 1 each of the remaining two colors, such as 4 blues, 2 greens, 1 red, and 1 yellow.
• 5 of one color and 1 each of the remaining three colors, such as 5 blues, 1 green, 1 red, and 1 yellow.
Note that K, L, M, and N must all be positive in order for a graph to be U4CG. If one value is zero, it would be a three-colored graph instead of a four-colored graph. Observe that any graph that is three-colorable is also four-colorable (provided that the graph has at least four vertices). For example, suppose that a graph is colored using red, green, and blue only (and that no two vertices of the same color share an edge). Choose any color that has at least two vertices of that color and change exactly one of its vertices to yellow. The graph will now be four-colored (and the yellow certainly won’t share an edge with any other yellows).
Let’s look at two extreme types of U4CG’s. The first extreme is the case of equitable coloring where V is evenly divisible by 4. In this case, there are equal numbers of vertices for each color: K = L = M = N = V/4. All 6 terms in the formula for the number of edges will be the same:
For example, for V = 12 vertices, equitable coloring would have 3 blues, 3 greens, 3 reds, and 3 yellows. In this case, the number of edges is E = 3(122)/8 = 3(144)/8 = 54. Each vertex connects to all 9 of the other vertices which aren’t the same color as itself. We actually saw this in an example in Chapter 15, where we added 24 edges to the icosahedral graph.
Another important extreme is the case of lopsided coloring. We don’t need to concern ourselves with U4CG’s where half of the vertices are a single color if our interest in U4CG’s relates to the four-color theorem. Any MPG (with V >
2) that can be formed by removing edges from an U4CG won’t have half of its vertices a single color when it is colored. (You may explore the reason for this in one of the exercises at the end of the chapter, if you wish.)
With this in mind, for an even number of vertices, the most lopsided coloring of an U4CG would have V/2 – 1 vertices one color, V/2 – 1 vertices a second color, and the remaining two vertices would be the remaining two colors. To determine the number of edges in this case, set K = L = V/2 – 1 and M = N = 1.
Anything times one equals itself.
Apply the foil method: (W + X)(Y + Z) = WY + WZ + XY + XZ.
Two pairs of V/2’s cancel. The remaining pair equals V (since a half plus a half makes one whole). There are two +1’s and four –1’s, which add up to –2.
For example, for V = 12, there could be 5 blue vertices, 5 green vertices, 1 red vertex, and 1 yellow vertex. (Note that 12/2 – 1 = 6 – 1 = 5.) In this example, E = 122/4 + 12 – 2 = 144/4 + 10 = 36 + 10 = 46 edges. (Compare this to the 54 edges for an U4CG with equitable coloring.)
It is interesting to compare these two types of extreme U4CG’s to two types of extreme MPG’s.
• One type of extreme MPG has a fairly equitable distribution of the degrees of its vertices. One example is the icosahedral MPG where all 12 vertices have degree 5. We saw in Chapter 15 that the icosahedral MPG can be four-colored with an equitable distribution of colors (3 blues, 3 greens, 3 reds, and 3 yellows).
• Another type of extreme MPG has a lopsided distribution of the degrees of its vertices. One example has two vertices with a degree of V – 1, two vertices with degree 3, and all of the remaining vertices have degree 4.
Such a MPG has the maximum number of ST’s (Chapter 12). This MPG has the same coloring as the U4CG with lopsided coloring that we
considered: V/2 – 1 vertices of one color, V/2 – 1 vertices of another color, and 1 vertex each for the two remaining colors.
These two extreme types of MPG’s can be drawn by removing edges from the corresponding types of extreme U4CG’s. If we could show that all MPG’s can be drawn by removing edges from U4CG’s, we could use this to prove the four-color theorem.
In Chapter 17, we will discuss the idea of removing edges from one type of graph to make another type of graph. In Chapter 27, we’ll explore the significance of U4CG’s more fully (but in that chapter we will refer to them as “complete tetrapartite graphs”).
Notation: Since K7 represents a complete graph with 7 vertices and K5,8 represents a complete bipartite graph with one set of 5 vertices and one set of 8 vertices (Chapter 5), we could use the notation KK,L,M,N to represent an U4CG (which is a complete multipartite graph with four sets of vertices) with K blue vertices, L green vertices, M red vertices, and N yellow vertices.
Grammar note: If “an U4CG” and “a MPG” seem backwards to you (that is, you think they should be “a U4CG” and “an MPG”), see the last paragraph in Chapter 6.
CHAPTER 16 EXERCISES
1. For each U4CG described below, draw the graph and also determine how many vertices and edges the graph has.
• 4 blue vertices, 3 green vertices, 2 red vertices, and 1 yellow vertex
• 3 blue vertices, 3 green vertices, 2 red vertices, and 2 yellow vertices 2. Which structurally different types of U4CG’s can be formed with 12 vertices? (If a graph can be made from another by swapping colors, it isn’t structurally different. For example, a graph with 3 blues, 2 greens, 1 red, and 1 yellow has the same structure as a graph with 3 reds, 2 yellows, 1 blue, and 1 green.) Determine the number of edges for each case.
3. Are any U4CG’s planar? If so, which ones? Are any U4CG’s also MPG’s?
If so, which ones. Are there any U4CG’s with few enough edges to be PG’s, but which aren’t planar? If so, which ones?
4. Explain why any MPG that can be formed by removing edges from an U4CG won’t have half of its vertices a single color when it is colored such that no two vertices that share an edge have the same color (of course, V > 2 for a MPG).
5. Show that any U4CG that meets the condition from Problem 4 has a HC, and thus may be redrawn as a closed convex polygon similar to the graphs of Chapter 14 (except that they will generally not be MPG’s).
6. Identify a HC in the U4CG shown below and use it to redraw the graph as a closed convex polygon similar to the graphs of Chapter 14 (except that it isn’t a MPG).
7. An U4CG has K = L = M = (V – 1)/3 and N = 1 where (V – 1) is evenly
9. A MPG has 24 vertices. None of its vertices has a degree less than 5. What is the maximum possible degree than any single vertex can have in this MPG? What would the degrees of the other vertices be if one vertex has this maximum possible degree?
Now consider an U4CG with 24 vertices with equal numbers of blue, green, red, and yellow vertices. What is the degree of each vertex in this U4CG?
Compare this U4CG to the MPG that we asked about earlier in this exercise.
10. A MPG has 100 vertices. None of its vertices has a degree less than 5.
What is the maximum possible degree than any single vertex can have in this MPG? What would the degrees of the other vertices be if one vertex has this maximum possible degree?
Now consider an U4CG with 100 vertices with equal numbers of blue, green, red, and yellow vertices. What is the degree of each vertex in this U4CG?
Compare this U4CG to the MPG that we asked about earlier in this exercise.
Challenge problem 1: Let’s revisit Challenge Problem 1 from Chapter 6 and see if what we now know about U4CG’s has any impact on that approach.
• Any U4CG with at least 6 vertices contains a K3,3 subgraph.
• K3,3 is two-colorable.
• Every U4CG is four-colorable.
• K5 isn’t four-colorable.
• U4CG’s don’t contain any K5 subgraphs.
• MPG’s don’t contain any subgraphs that are subdivisions of K5 or K3,3. Show that U4CG’s don’t contain any K5 subgraphs. Can an U4CG contain a subgraph that is a subdivision of K5? If so, give an example; if not, explain why.
Can you prove that every MPG can be made by removing edges from a suitable U4CG? If so, you could use this prove the four-color theorem.
Alternatively, can you prove that every graph that doesn’t contain a subgraph that is a subdivision of K5 is four-colorable, and if so, can you use this to prove the four-color theorem?
Perhaps you can prove the following alternative: Any graph that doesn’t contain a subgraph that is a subdivision of K5 is a subgraph of a complete multipartite graph with no more than four sets of vertices (the case with four sets of vertices we’re referring to as an U4CG), and every complete multipartite graph with no more than four sets of vertices is four-colorable.
Since no MPG contains a subgraph that is a subdivision of K5, it would then follow that all MPG’s are four-colorable. If you can’t prove this, either explain the flaw in the reasoning or explain why it would be very challenging to prove this.
Note: The answer key doesn’t include answers to the challenge problems.
These problems are intended to encourage you to think about the ideas.
However, you may wish to consider how this problem relates to Chapter 27.