1. Counting Proper Colors Given k  N 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.  (K n ;k)=k n.  (K n ;k)=k(k-1)…(k-n+1). KnKn KnKn

2. Proposition 5.3.3 If T is a tree with n vertices, then  (T;k)=k(k-1) n-1. k K-1

3. Proposition 5.3.4 Let x (r) =x(x-1)…(x-r+1). If p r (G) denotes the number of partitions of V(G) into r nonempty independent sets, then  (G;k)=  r=1 r=n(G) p r (G)k (r), which is a polynomial in k of degree n(G).

4. Example 5.3.5 1. Consider G=C 4. 2. p 1 (G)=0. p 4 (G)=1. p 3 (G)=2. p 2 (G)=1. 3.  (C 4 ;k)=1  k(k-1)+2  k(k-1)(k-2)+1  k(k-1)(k-2)(k-3)= k(k-1)(k 2 -3k+3).

5. Theorem 5.3.6 If G is a simple graph and e  E(G), then  (G;k)=  (G-e;k) -  (G  e;k)

6. Example 5.3.7 1. Consider G=C 4. 2. G-e=P 4. 3. G  e=K 3. 4.  (P 4 ;k)=k(k-1) 3 5.  (K 3 ;k)=k(k-1)(k-2). 6.  (C 4 ;k)=  (P 4 ;k)-  (K 3 ;k)=k(k-1)(k 2 -3k+3). C4C4 P4P4 K3K3 e since P 4 is a tree.

7. Example 5.3.9 Find  (K n -e;k). 1.  (K n -e;k)=  (K n ;k)+  (K n-1 ;k). 2.  (K n ;k)=k(k-1)…(k-n+1). 3.  (K n-1 ;k)=k(k-1)…(k-(n-1)+1)=k(k-1)…(k-n+2). 4.  (K n -e;k)= k(k-1)…(k-n+2)(k-n+1+1).

8. Theorem 5.3.10 Let c(G) denote the number of components of a graph G. Given a set S  E(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)=  S  E(G) (-1) |S| k c(G(S)).

9. 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|. k c(G(S))  |S|=1 (-1) |S|. k c(G(S)) =-5k 3. =k 4.

10. 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|. k c(G(S)) =10k 2.  |S|=3 (-1) |S|. k c(G(S)) =-(2k 2 +8k).

11. Example 5.3.11 Every spanning subgraph with 4 edges has 1 components. There is five set of 4 edges.  (G;k)=k 4 -5k 3 +10k 2 -(2k 2 +8k 1 )+5k-k.  |S|=4 (-1) |S|. k c(G(S)) =5k.  |S|=5 (-1) |S|. k c(G(S)) =-k.

12. Example 5.3.11 Let G be s kite (four vertices, five edges). Find  (G;k).

13. Example 5.3.11

17. Proof of Theorem 5.3.8 1. Iterating the chromatic recurrence yields 2 e(G) terms in  (G;k). Each term is k the 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 k c(G(S)). 3. The sign is positive if and only if |S| is even. So,  (G;k)=  S  E(G) (-1) |S| k c(G(S)).