1.6 Colorings
1.6.3 The Four Color Problem
chromatic number is 6 (see Exercise 2).
The upper and lower bounds given in Theorem 1.45 concernα(G), the inde- pendence number ofG, defined back in Section 1.4.3. The proofs are left as an exercise (see Exercise 6).
Theorem 1.45. For any graphGof ordern, n
α(G) ≤χ(G)≤n+ 1−α(G).
Exercises
1. Recall thatavgdeg(G)denotes the average degree of vertices inG. Prove or give a counterexample to the following statement:
χ(G)≤1 + avgdeg(G).
2. IfGis the graph in Figure 1.92, prove thatχ(G) = 6andω(G) = 5.
3. Determine a necessary and sufficient condition for a graph to have a 2- colorable line graph.
4. Recall thatτ(G)denotes the number of vertices in a detour path (a longest path) ofG, prove thatχ(G)≤τ(G).
5. Prove that the only graphGof ordernfor whichχ(G) =nisKn. 6. Prove that for any graphGof ordern,
n
α(G) ≤χ(G)≤n+ 1−α(G).
7. IfGis bipartite, prove thatω(G) =χ(G).
8. LetGbe a graph of ordern. Prove that (a) n≤χ(G)χ(G);
(b) 2√
n≤χ(G) +χ(G).
The seemingly simple Four Color Problem was introduced in 1852 by Francis Guthrie, a student of Augustus DeMorgan. The first written reference to the prob- lem is a letter from DeMorgan to Sir William Rowan Hamilton. Despite Hamil- ton’s indifference15, DeMorgan continued to talk about the problem with other mathematicians. In the years that followed, many of the world’s top mathematical minds attempted either to prove or disprove the conjecture, and in 1879 Alfred Kempe announced that he had found a proof. In 1890, however, P. J. Heawood discovered an error in Kempe’s proof. Kempe’s work did have some positive fea- tures, though, for Heawood made use of Kempe’s ideas to prove that five colors always suffice. In this section, we translate the Four Color Problem into a graph theory problem, and we prove the Five Color Theorem.
Any map can be represented by a planar graph in the following way: Repre- sent each country on the map by a vertex, and connect two vertices with an edge whenever the corresponding countries share a nontrivial border (more than just a point). Some examples are shown in Figure 1.93.
FIGURE 1.93. Graph representations of maps.
The regions on the map correspond to vertices on the graph, so a graph col- oring yields a map coloring with no bordering regions colored the same. This natural representation allows us to see that a map is 4-colorable if and only if its associated graph is 4-colorable.
The Four Color Conjecture is equivalent to the following statement. A thorough discussion of this equivalence can be found in [52].
15Perhaps he was too busy perfecting plans for a cool new game that he would release a few years later. See Section 1.4.3.
Theorem 1.46 (Four Color Theorem). Every planar graph is4-colorable.
When Heawood pointed out the error in Kempe’s proof, researchers flocked back to the drawing board. People worked on the Four Color Problem for years and years trying numerous strategies. Finally, in 1976, Kenneth Appel and Wolf- gang Haken, with the help of John Koch, announced that they had found a proof [12]. To complete their proof, they verified thousands of cases with computers, using over 1000 hours of computer time. As you might imagine, people were skeptical of this at first. Was this really a proof? How could an argument with so many cases be verified?
While the Appel–Haken proofisaccepted as being valid, mathematicians still search for alternative proofs. Robertson, Sanders, Seymour, and Thomas [239]
have probably come the closest to finding a short and clever proof, but theirs still requires a number of computer calculations.
In a 1998 article [267], Robin Thomas said the following.
For the purposes of this survey, let me telescope the difficulties with the A&H proof into two points: (1) part of the proof uses a computer and cannot be verified by hand, and (2) even the part that is suppos- edly hand-checkable has not, as far as I know, been independently verified in its entirety. . . . Neil Robertson, Daniel P. Sanders, Paul Seymour, and I tried to verify the Appel–Haken proof, but soon gave up and decided that it would be more profitable to work out our own proof. . . . We were not able to eliminate reason (1), but we managed to make progress toward (2).
As mentioned earlier, Heawood [156] provided a proof of the Five Color Theorem in the late 1890s, and we present his proof here. Some of the ideas in his proof came from Kempe’s attempt [174] to solve the Four Color Problem.
Theorem 1.47 (Five Color Theorem). Every planar graph is5-colorable.
Proof. We induct on the order ofG. LetGbe a planar graph of ordern. Ifn≤5, then the result is clear. So suppose thatn ≥ 6and that the result is true for all planar graphs of ordern−1. From Theorem 1.35, we know thatGcontains a vertex, sayv, havingdeg(v)≤5.
Consider the graphGobtained by removing fromGthe vertexvand all edges incident withv. Since the order ofGisn−1(and sinceGis of course planar), we can apply the induction hypothesis and conclude thatGis 5-colorable. Now, we can assume thatG has been colored using the five colors, named 1, 2, 3, 4, and 5. Consider now the neighbors ofvinG. As noted earlier,vhas at most five neighbors inG, and all of these neighbors are vertices in (the already colored)G. If inGfewer than five colors were used to color these neighbors, then we can properly colorGby using the coloring forGon all vertices other thanv, and by coloringvwith one of the colors that is not used on the neighbors ofv. In doing this, we have produced a 5-coloring forG.
So, assume that inGexactly five of the colors were used to color the neighbors ofv. This implies that there are exactly five neighbors, call themw1,w2,w3,w4, w5, and assume without loss of generality that eachwiis colored with colori(see Figure 1.94).
5
1
2
3 4
v w1
w5
w4 w3
w2
FIGURE 1.94.
We wish to rearrange the colors ofG so that we make a color available for v. Consider all of the vertices ofG that have been colored with color 1 or with color 3.
Case 1. Suppose that inGthere does not exist a path fromw1tow3where all of the colors on the path are 1 or 3. Define a subgraphHofGto be the union of all paths that start atw1and that are colored with either 1 or 3. Note thatw3is not a vertex ofHand that none of the neighbors ofw3are inH(see Figure 1.95).
3
1
1 1
1 3
3 w1
FIGURE 1.95.
Now, interchange the colors inH. That is, change all of the 1’s into 3’s and all of the 3’s into 1’s. The resulting coloring of the vertices ofG is a proper coloring, because no problems could have possibly arisen in this interchange. We now see thatw1is colored 3, and thus color 1 is available to use forv. Thus,Gis 5-colorable.
Case 2. Suppose that inGthere does exist a path fromw1tow3where all of the colors on the path are 1 or 3. Call this pathP. Note now thatP along withv forms a cycle that encloses eitherw2orw4(Figure 1.96).
or
1 1
1
3 3
3 3
1
1 1
3 3 3
3 3 1 1
1 v
w1 w5
w4 w3
w2
v w1 w5
w4 w3
w2
FIGURE 1.96. Two possibilities.
So there does not exist a path fromw2 tow4 where all of the colors on the path are 2 or 4. Thus, the reasoning in Case 1 applies! We conclude thatGis 5-colorable.
Exercises
1. Determine the chromatic number of the graph of the map of the United States.
2. Determine the chromatic number of the graph of the map of the countries of South America.
3. Determine the chromatic number of the graph of the map of the countries of Africa.
4. Determine the chromatic number of the graph of the map of the countries of Australia. Hint: This graph will be quite small!
5. Where does the proof of the Five Color Theorem go wrong for four colors?