- G = chain of length n-1 (so there are n vertices)
- P(G, x) = x(x-1)
^{n-1} - G = K
_{4} - P(G, x) = x(x-1)(x-2)(x-3) = x
^{(4)} - G = Star
_{5} - P(G, x) = x(x-1)
^{5} - G = C
_{4}( or Z_{4}) - P(G, x) = x(x-1)
^{2}+ x(x-1)(x-2)^{2}= x^{4}- 4x^{3}+ 6x^{2}- 3x

P(**I _{n}**, x) = x

**Theorem**: If G decomposes into k connected components, G_{1}, G_{2}, ..., G_{k} then

P(G, x) = P( G_{1}, x)P(G_{2}, x) ... P(G_{k}, x).

**Def**: *Reduced graph, Condensed graph*

**Theorem**: (*Fundamental Reduction Theorem*) Let ß be an edge of G, G'_{ß} the reduced
graph at ß, and G"_{ß} the condensed graph at ß, then

**Example**: Consider C_{4}, and let ß be any edge. The reduced graph at ß is a chain of length 3
and the condensed graph at ß is a K_{3}3, so by the theorem:

P(C_{4}, x) = x(x-1)^{3} - x(x-1)(x-2) = (x^{4} - 3x^{3} + 3x^{2} - x) - (x^{3} -3x^{2} + 2x) = x^{4} - 4x^{3} + 6x^{2} - 3x.

*Note*: You can use the reduction theorem to either add or remove edges.

- a) degree of P(G,x) = n = |V| (the number of vertices of G).
- b) coefficient of x
^{n}is 1. (a monic polynomial) - c) the constant term of P(G,x) is 0.
- d) either P(G,x) = x
^{n}or the sum of the coefficients is 0.

c) Consider P(G,0) = a_{0} = the number of ways to color G with no colors, but this can't be
done, so a_{0} = 0.

d) Consider P(G, 1) = a_{p} + a_{p-1} + ... + a_{1} = the number of ways to color G with 1 color. If G
contains an edge, then it can not be colored with 1 color so this sum must be 0. If G contains
no edge, then it is the empty graph I_{n}, which has chromatic polynomial x^{n}.

*Review Induction Proofs.*

a) and b). Suppose that G has n vertices. We will proceed by induction on the number of edges.

First, consider the base case, where G has only one edge. Then the n-2 isolated vertices can
be colored in any way, and the two vertices on the edge can be colored in x(x-1) ways, so,
P(G,x) = x^{n-2}(x)(x-1) = x^{n} - x^{n-1}. So, the degree of P(G, x) in this case is n and the polynomial is monic.

We now impose the induction hypothesis and assume that the statement is true for all
graphs with fewer than k edges (strong form of induction). Assume that G has k edges and let
ß be one of them. By the fundamental reduction theorem,

By induction, the statement is true for all graphs with n vertices and any number of edges.

**Theorem:** (Whitney, 1932): The powers of the chromatic polynomial are consecutive and the
coefficients alternate in sign.

**Proof**: We will again proceed by induction on the number of edges of G.
As in the proof of the above theorem, the chromatic polynomial of a graph with n vertices
and one edge is x^{n} - x^{n-1}, so our statement is true for such a graph.

Now, assume true for all graphs on n vertices with fewer than k edges. Let G be a graph
with k edges and consider the fundamental reduction theorem. Since G'_{ß} and G"_{ß} both have
fewer than k edges, by the induction hypothesis we may assume that their chromatic
polynomials have the form:

P(G"

= (x

= x

By induction, the statement is true for all graphs with n vertices and any number of edges.

**Theorem**: |a_{n-1}| = |E| (the number of edges of G).

**Proof**: We will again proceed by induction on the number of edges of G.

As in the proofs of the above theorems, the chromatic polynomial of a graph with n vertices
and one edge is x^{n} - x^{n}-1, so our statement is true for such a graph, |-1| = 1.

Now, assume true for all graphs on n vertices with fewer than k edges. Let G be a graph
with k edges and consider the fundamental reduction theorem. Since G'_{ß} and G"_{ß} both have
fewer than k edges, by the induction hypothesis we may assume that their chromatic
polynomials have the form:

P(G"

Now, by the fundamental reduction theorem,

= (x

= x

By induction, the statement is true for all graphs with n vertices and any number of edges.

**Theorem**: The smallest exponent in the chromatic polynomial is the number of connected
components of G.

**Proof**: If G has k connected components, G_{1}, G_{2}, ..., G_{k} then
P(G, x) = P( G_{1}, x)P(G_{2}, x) ... P(G_{k}, x). Thus, the smallest exponent in the chromatic
polynomial of G is the sum of the smallest exponents appearing in each of the P(G_{i}, x)'s. As
these can be no smaller than 1, we have that the smallest exponent of P(G, x) >= k.

To show that this exponent must be k, we need to show that the chromatic polynomial of
any connected graph has an x^{1} term. We proceed by induction on the number of edges of G.

As in the proofs of the above theorems, the chromatic polynomial of a graph with n
vertices and one edge is x^{n} - x^{n-1}. If the graph is connected, then n = 2 and our statement is
true.

Now, assume true for all connected graphs on n vertices with fewer than k edges. Let G be a connected graph with k edges and consider the fundamental reduction theorem. We have two cases to consider:

*Case I: The removal of ß leaves G' _{ß} connected.*

Since G'

P(G"

= (x

= x

By induction, the statement is true for all graphs with n vertices and any number of edges.

*Case II: The removal of ß disconnects G' _{ß} .*

G'

P(G"

= (x

= x

By induction, the statement is true for all graphs with n vertices and any number of edges.

These properties do *not* characterize the chromatic polynomials amongst all the polynomials
with integer coefficients. For example,

P(x) = x^{4} - 4x^{3} + 3x^{2} is not the chromatic polynomial of any graph.