In a complete graph, every vertex connects to each of the other vertices. The notation KV represents a complete graph with V vertices. For example, the complete graph shown below, which has V = 8 vertices, is represented by K8.
A complete graph can be used to represent the following classic handshaking problem. If V people are in a room and each person shakes hands once with each of the V – 1 other people, how many handshakes will there be all together? The answer is given by the handshaking lemma: V(V – 1)/2 equals the number of handshakes. Why? Multiply V by (V – 1) and then divide by 2 to correct for double counting. To better understand the formula associated with the handshaking lemma, let’s consider the K6 graph shown below.
Let’s list the handshakes:
• A’s handshakes include AB, AC, AD, AE, and AF. That’s five.
• B’s handshakes include BA, BC, BD, BE, and BF. That’s also five, but note that BA is the same as AB.
• C’s handshakes include CA, CB, CD, CE, and CF. That’s also five, but CA and CB are the same as AC and BC.
• D’s handshakes include DA, DB, DC, DE, and DF. That’s another five, but DA, DB, and DC are the same as AD, BD, and CD.
• E’s handshakes include EA, EB, EC, ED, and EF. That’s five more, but EA, EB, EC, and ED are the same as AE, BE, CE, and DE.
• F’s handshakes include FA, FB, FC, FD, and FE. That’s five again, but all of these are repeats. They are the same as AF, BF, CF, DF, and EF.
The previous graph, K6, has V = 6 vertices. According to the handshaking lemma, there are V(V – 1)/2 = 6(5)/2 = 15 handshakes. Look at our list. For A thru F, there are 5 handshakes each, giving us 6(5) = 30 handshakes, but 15 of these are repeated so that there really are only 30/2 = 15 handshakes.
That’s what we meant about dividing by 2 to correct for double counting. For example, BC and CB are the same handshake because they involve the same two people. A complete graph is really asking: given V vertices, how many ways are there to make pairs of them. The answer is V(V – 1)/2.
Complete graphs with 1 to 10 vertices are illustrated on the following page.
A couple of noteworthy complete graphs concerning the four-color theorem include:
• K4 is the largest complete graph that is a MPG. K4 is also the largest
complete graph that can be colored such that the coloring satisfies the four-color theorem. As we will explore in later chapters, MPG’s with K4
subgraphs tend to have more restrictive coloring (meaning that there tend to be fewer ways to color the graphs compared to graphs that lack K4’s), vertices with degree three tend to take part in K4 subgraphs, and even K4 subgraphs without any vertices with degree three have separating triangles.
K4 is sometimes referred to as the tetrahedral graph.
• K5 is the smallest complete graph that isn’t a MPG. K5 is also the smallest complete graph that isn’t four-colorable. As we will explore in Chapter 6, K5 plays an important role in determining whether or not a graph is a PG or if it is nonplanar. K5 is sometimes referred to as the pentatope graph.
The number of edges on a complete graph with V vertices is given by the formula from the handshaking lemma: E = V(V – 1)/2. For example, you can verify that the K6 graph shown above with V = 6 vertices has E = 6(6 – 1)/2 = 6(5)/2 = 30/2 = 15 edges.
A complete bipartite graph has two sets of vertices where every vertex of one set connects to every vertex of the other set. The notation KX,Y denotes a complete bipartite graph where set P has X vertices and set Q has Y vertices.
A complete bipartite graph has E = XY edges because each of the X vertices in set P connects to each of the Y vertices in set Q. For example, the complete bipartite graph K3,4 shown on the following page has X = 3 vertices in set P, Y = 4 vertices in set Q, and E = XY = 3(4) = 12 edges.
A bipartite graph is also called a bigraph. A bipartite graph (without the word “complete”) is defined to have two sets of vertices where no two vertices of a single set connect to one another. A complete bipartite graph has every vertex of one set connected to every vertex of the other set and vice-versa.
The K3,3 graph shown above is noteworthy as it concerns PG’s. In Chapter 6, we will see that K3,3 is nonplanar and that K3,3 and K5 play an important role in determining whether or not a graph is a PG or if it is nonplanar. The K3,3 graph is also referred to as the utility graph, based on the following puzzle [Ref. 5]. Can you connect three cottages (A, B, and C) to three utilities (D, E, and F) without crossing the lines? Since the utility graph isn’t planar (as we will show in Chapter 6), the answer is no. Recall that we are defining PG to include any graph that can be drawn in the plane without crossings (even if the graph happens to be drawn in a form that has avoidable crossings).
Observe that complete bipartite graphs are two-colorable. Since the vertices of set P don’t connect to one another, every vertex in set P can be one color (such as red). Similarly, since the vertices of set Q don’t connect to one another, every vertex in set Q can be another color (such as blue). For example, in the K3,4 graph shown above, A, B, and C can each be red and D, E, F, and G can each be blue.
CHAPTER 5 EXERCISES
1. A total of 12 people attend a conference. If every person at the conference shakes hands once with every other person, how many handshakes will occur in total?
2. A research project is started with 9 female mathematicians and 7 male mathematicians. If each female shakes hands once with all of the males, each male shakes hands once with all of the females, no female shakes hands with another female, and no male shakes hands with another male, how many handshakes will occur in total?
3. Draw K4 as it was drawn in this chapter. Now redraw K4 to show that it is a PG. Is K4 a MPG? How can you tell?
4. Draw K5. Now redraw the graph with one of its edges removed. Is this new graph (with one edge removed) planar? Is it a MPG? How can you tell?
5. Draw K6. How many edges must be removed from K6 in order to for the new graph (with edges removed) to be a MPG? Draw the new graph (with edges removed), showing that it can be a MPG. Draw a second graph with the same number of edges removed, which is also a MPG, but where the vertices have different degrees from the first MPG. Now draw a third graph with the same number of edges removed, but which isn’t a MPG.
6. How many edges must be removed from a complete graph with 9 vertices in order to make a MPG with 9 vertices? Will the resulting graph necessarily be a MPG?
7. Show that if you remove (V – 3)(V – 4)/2 edges from a complete graph with V vertices that the resulting graph will have the right number of edges to be a MPG with V vertices. Will the resulting graph necessarily be a MPG?
8. Draw K5. How many edges must be removed from K5 in order to for the new graph (with edges removed) to be K2,3? Draw K2,3 by removing this
number of edges from K5.
9. Draw K6. How many edges must be removed from K6 in order to for the new graph (with edges removed) to be K3,3? Draw K3,3 by removing this number of edges from K6. Can you find more than one way to remove edges from K6 to draw a graph that is isomorphic to K3,3?
10. If V is an even number, show that V(V – 2)/4 edges need to be removed from KV in order for the new graph (with edges removed) to be KN,N with N = V/2. Will the resulting graph necessarily be KN,N?