Chromatic Polynomials

1.6 Colorings

1.6.4 Chromatic Polynomials

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?

red

white green

blue red

white green

v

1

v

2 blue

v

4

v

3

v

1

v

2

v

4

v

3

FIGURE 1.97. Two different colorings.

If we restrict ourselves to four colors, how many different colorings are there ofK₄? Since there are four choices forv₁, then three forv₂, etc., we see that there are4·3·2·1different colorings ofK4using four colors. If six colors were available, there would be6·5·4·3different colorings. If only two were available, there would be no proper colorings ofK4.

In general, definecG(k)to be the number of different colorings of a graphG using at mostkcolors. So we havecK₄(4) = 24,cK₄(6) = 360, andcK₄(2) = 0.

In fact, ifkandnare positive integers wherek≥n, then cK_n(k) =k(k−1)(k−2)· · ·(k−n+ 1).

Further, ifk < n, then cK_n(k) = 0. We also note thatcE_n(k) = kⁿ for all positive integerskandn.

A simple but important property ofcG(k)is thatGisk-colorable if and only if cG(k)>0. Equivalently,cG(k)>0if and only ifχ(G)≤k.

Finding values ofcG(k)is relatively easy for some well-known graphs. Com- puting this function in general, though, can be hard. Birkhoff and Lewis [27]

developed a way to reduce this hard problem to an easier one. Before we see their method, we need a definition.

LetGbe a graph and letebe an edge ofG. Recall thatG−edenotes the graph whereeis removed fromG. Define the graphG/eto be the graph obtained from Gby removinge, identifying the end vertices ofe, and leaving only one copy of any resulting multiple edges.

As an example, a graphGand the graphsG−bcandG/bcare shown in Fig- ure 1.98.

Theorem 1.48. LetGbe a graph andebe any edge ofG. Then cG(k) =cG−e(k)−cG/e(k).

Proof. Suppose that the end vertices ofeare uandv, and consider the graph G−e.

How manyk-colorings are there ofG−ewhereuandvare assigned the same color? IfCis a such a coloring ofG−e, thenCcan be thought of as a coloring of G/e, sinceuandvare colored the same. Similarly, any coloring ofG/ecan also be thought of as a coloring ofG−ewhereuandvare colored the same. Thus, the answer to this question iscG/e(k).

G

a

b c

w z

x y

G

-

^bc a

b c z

w

x y

G / bc a

b & c

x y

w z

FIGURE 1.98. Examples of the operations.

Now, how manyk-colorings are there ofG−ewhereuandv are assigned different colors? IfCis a such a coloring ofG−e, thenCcan be considered as a coloring ofG, sinceuandvare colored differently. Similarly, any coloring ofG can also be thought of as a coloring ofG−ewhereuandvare colored differently.

Thus, the answer to this second question isc_G(k).

Thus, the total number ofk-colorings ofG−eis c_G−e(k) =c_G/e(k) +c_G(k), and the result follows.

For example, suppose we want to findc_P4(k). That is, how many ways are there to color the vertices ofP₄withkcolors available? We label the edges ofP₄as shown in Figure 1.99.

P

4

e

1

e

2

e

3 FIGURE 1.99. The labeled edges ofP4. The theorem implies that

cP4(k) =cP4−e1(k)−cP₄/e₁(k).

For convenience, let us denoteP4−e1andP4/e1byG11andG12, respectively (see Figure 1.100).

e

₂

e

₃

e

₂

e

₃

G

11

G

12

FIGURE 1.100. The first application.

Applying the theorem again, we obtain

cP4(k) =cG11−e2(k)−cG₁₁/e₂(k)−cG12−e2(k) +cG₁₂/e₂(k).

Denote the graphsG11−e2,G11/e2,G12−e2, andG12/e2byG21,G22,G23, andG24, respectively (see Figure 1.101).

e

3

e

3

e

3

e

3

G

21

G

22

G

24

G

23

FIGURE 1.101. The second application.

Applying the theorem once more yields

c_P4(k) =c_G21−e3(k)−c_G21/e3(k)−c_G22−e3(k) +c_G22/e3(k)

−cG₂₃−e₃(k) +c_G23/e3(k) +cG₂₄−e₃(k)−c_G24/e3(k).

That is,

c_P4(k) =c_E4(k)−c_E3(k)−c_E3(k)+c_E2(k)−c_E3(k)+c_E2(k)+c_E2(k)−c_E1(k).

Thus,

cP₄(k) =k⁴−k³−k³+k²−k³+k²+k²−k

=k⁴−3k³+ 3k²−k.

We should check a couple of examples. How many colorings ofP4 are there with one color?

cP4(1) = 1⁴−3(1)³+ 3(1)²−1 = 0.

This, of course, makes sense. And how many colorings are there with two colors?

cP₄(2) = 2⁴−3(2)³+ 3(2)²−2 = 2.

Figure 1.102 shows these two colorings. Score one for Birkhoff!

red blue red blue blue red blue red

FIGURE 1.102. Two 2-colorings ofP4.

As you can see, chromatic polynomials provide a way to count colorings, and the Birkhoff–Lewis theorem allows you to reduce a problem to a slightly simpler one. We should note that it is not always necessary to work all the way down to empty graphs, as we did in the previous example. Once a graphGis obtained for which the value ofcG(k)is known, there is no need to reduce that one further.

We now present some properties ofcG(k).

Theorem 1.49. LetGbe a graph of ordern. Then 1. cG(k)is a polynomial inkof degreen, 2. the leading coefficient ofcG(k)is1, 3. the constant term ofcG(k)is0,

4. the coefficients ofcG(k)alternate in sign, and

5. the absolute value of the coefficient of thekⁿ⁻¹term is the number of edges inG.

We leave the proof of this theorem as an exercise (Exercise 3). One proof strat- egy is to induct on the number of edges inGand use the Birkhoff–Lewis reduction theorem (Theorem 1.48).

Before leaving this section, we should note that Birkhoff considered chromatic polynomials of planar graphs, and he hoped to find one of them that had 4 as a root. If he had found one, then the corresponding planar graph would not be 4- colorable, and hence would be a counterexample to the Four Color Conjecture.

Although he was unsuccessful in proving the Four Color Theorem, he still de- serves credit for producing a very nice counting technique.

Exercises

1. Find chromatic polynomials for each of the following graphs. For each one, determine how many 5-colorings exist.

(a) K_1,3 (b) K1,5

(c) C4

(d) C5

(e) K4−e (f) K₅−e

2. Show thatk⁴−4k³+ 3k²is not a chromatic polynomial for any graph.

3. Prove Theorem 1.49.

4. Determine the chromatic polynomial for a tree of ordern.