Counting Proper Colors

• Given kN and a graph G, the value (G;k) is the number of proper colorings f: V(G)[k], where the k colors need not all be used in a coloring f.

(Kn;k)=kn.

(Kn;k)=k(k-1)…(k-n+1).

KnKn

Proposition 5.3.3

If T is a tree with n vertices, then (T;k)=k(k-1)n-1.

k

K-1

K-1

K-1

K-1

K-1

K-1

Proposition 5.3.4

Let x(r)=x(x-1)…(x-r+1).

If pr(G) denotes the number of partitions of V(G) into r nonempty independent sets, then (G;k)=r=1

r=n(G) pr(G)k(r) , which is a polynomial in k of degree n(G).

Example 5.3.5

1. Consider G=C4.

2. p1(G)=0. p4(G)=1. p3(G)=2. p2(G)=1.

3. (C4;k)=1k(k-1)+2 k(k-1)(k-2)+1k(k-1)(k-2)(k-3)= k(k-1)(k2-3k+3).

Theorem 5.3.6

If G is a simple graph and eE(G), then (G;k)=(G-e;k) - (Ge;k)

Example 5.3.7

1. Consider G=C4.

2. G-e=P4.

3. Ge=K3.

4. (P4;k)=k(k-1)3

5. (K3;k)=k(k-1)(k-2).

6. (C4;k)= (P4;k)- (K3;k)=k(k-1)(k2-3k+3).

C4 P4 K3

e

since P4 is a tree.

Example 5.3.9

Find (Kn-e;k).

1. (Kn-e;k)=(Kn;k)+ (Kn-1;k).

2. (Kn;k)=k(k-1)…(k-n+1).

3. (Kn-1;k)=k(k-1)…(k-(n-1)+1)=k(k-1)…(k-n+2).

4. (Kn-e;k)= k(k-1)…(k-n+2)(k-n+1+1).

Theorem 5.3.10

Let c(G) denote the number of components of a graph G.

Given a set SE(G) of edges in G, let G(S) denote the spanning subgraph of G with edge set S.

Then the number (G;k) of proper k-coloring of G is given by (G;k)=SE(G)(-1)|S|kc(G(S)).

Example 5.3.11

• Find the chromatic polynomial for a kite (four vertices, five edges)

• Every spanning subgraph with 0 edge has 4 components. There is one set of 0 edge.

• Every spanning subgraph with 1 edge has 3 components. There is five set of 1 edge.

|S|=0(-1)|S|.kc(G(S))

|S|=1(-1)|S|.kc(G(S)) =-5k3.

=k4.

Example 5.3.11

• Every spanning subgraph with 2 edges has 2 components. There is ten set of 2 edges.

• There are ten sets of 3 edges. For two of these yield spanning subgraphs with one component. The other eight sets of three edges yield spanning subgraphs with one component.

|S|=2(-1)|S|.kc(G(S)) =10k2.

|S|=3(-1)|S|.kc(G(S)) =-(2k2+8k).

Example 5.3.11

• Every spanning subgraph with 4 edges has 1 components. There is five set of 4 edges.

• Every spanning subgraph with 4 edges has 1 components. There is five set of 4 edges.

(G;k)=k4-5k3+10k2-(2k2+8k1)+5k-k.

|S|=4(-1)|S|.kc(G(S)) =5k.

|S|=5(-1)|S|.kc(G(S)) =-k.

Example 5.3.11

Let G be s kite (four vertices, five edges). Find (G;k).

Example 5.3.11

Example 5.3.11

Example 5.3.11

Example 5.3.11

Proof of Theorem 5.3.8

1. Iterating the chromatic recurrence yields 2e(G) terms in (G;k). Each term is kthe number of remaining vertices.

2. Let S be the set of edges that were contracted. The remaining vertices correspond to the component of G(S). It implies Each term is kc(G(S)).

3. The sign is positive if and only if |S| is even. So, (G;k)=SE(G)(-1)|S|kc(G(S)).