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]. The set of available colors is [k]={1,…,k}; the k colors need not all be used in a coloring f. Changing the names of colors that are used produces a different coloring.  (K n ;k)=k n and  (K n ;k)=k(k-1)…(k-n+1)

2. Proposition 5.3.3 If T is a tree with n vertices, then  (T;k)=k(k-1) n-1. Proof. 1. Let v be the root of T. We can color v in k ways. 2. If we extend a proper coloring to new vertices as we grow the tree from v, at each step only the color of the parent is forbidden, and we have k-1 choices for the color of the new vertex.

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). Proof. 1. When r colors are used in a proper coloring, the color classes partition V(G) into exactly r independent sets, which can happen in p r (G) ways. 2. When k colors are available, there are exactly k (r) ways to choose colors and assign them to the classes.

4. Proposition 5.3.4 3. All the proper colorings arises in this way, so the formula for  (G;k) is correct. 4. Since k (r) is a polynomial in k and p r (G) is a constant for each r, this formula implies that  (G;k) is a polynomial function of k. 5. There is exactly one partition of G into n(G) independent sets and no partition using more sets, so the leading term in k n(G).

5. Example 5.3.5 1. Consider G=C 4. 2. p 1 (G)=0. p 4 (G)=1. p 2 (G)=1. p 3 (G)=2. 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).

6. Theorem 5.3.6 If G is a simple graph and e  E(G), then  (G;k)=  (G- e;k) -  (G  e;k) Proof. 1. Every proper k-coloring of G is a proper coloring of G-e. 2. A proper k-coloring of G-e is a proper k-coloring of G if and only if it gives distinct colors to the endpoints u, v of e. 3. Hence we can count the proper k-coloring of G by subtracting from  (G;k) the number of proper k- coloring of G-e that give u and v the same color.

7. Theorem 5.3.6 4. Coloring of G-e in which u and v have the same color correspond directly to proper k-coloring of G  e, in which the color of the contracted vertex is the common color of u and v.

8. 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 since P 4 is a tree. 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).

9. 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).

10. Theorem 5.3.8 The chromatic polynomial  (G;k) of a simple graph G has degree n(G), with integer coefficients alternating in sign and beginning 1, -e(G), . Proof. 1. We use induction on e(G). 2. The claims hold when e(G)=0, where  (k n ;k)=k n. 3. For the induction step, let G be an n-vertex graph with e(G)>=1. 4. Each of G-e and G  e has fewer edges than G, and G  e has n-1 vertices.

11. Theorem 5.3.8 5. By induction hypothesis, there are nonnegative integers {a i } and {b i } such that  (G-e;k)=  i=0 i=n (- 1) i a i k n-i and  (G  e;k) )=  i=0 i=n-1 (-1) i b i k n-1-i. 6. By the chromatic recurrence, 7. Hence,  (G;k) is a polynomial with leading coefficient a 0 =1 and next coefficient -e(G), and its coefficients alternate in sign.

12. Theorem 5.3.8 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)).

13. Example 5.3.11 Every spanning subgraph with 0, 1, or 2 edges has 4, 3, or 2 components, respectively. There are ten sets of three edges. For two of these yield spanning subgraphs with one component. The other eight sets of three edges yield spanning subgraphs with one component. All spanning subgraphs with four or five edges have only one component.  (G;k)=k 4 -5k 3 +10k 2 -(2k 2 +8k 1 )+5k-k=  (C 4 ;k)-  (P 3 ;k)

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

15. Example 5.3.11

19. 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)).