Graph Theory Lecture Notes 6

Chromatic Polynomials

For a given graph G, the number of ways of coloring the vertices with x or fewer colors is denoted by P(G, x) and is called the chromatic polynomial of G (in terms of x).

Examples:

G = chain of length n-1 (so there are n vertices)
P(G, x) = x(x-1)n-1

G = K4
P(G, x) = x(x-1)(x-2)(x-3) = x(4)

G = Star5
P(G, x) = x(x-1)5

G = C4 ( or Z4)
P(G, x) = x(x-1)2 + x(x-1)(x-2)2 = x4 - 4x3 + 6x2 - 3x

P(Kn, x) = x(n) = x(x-1)(x-2) ...(x-n+1)

P(In, x) = xn, where In is the graph on n isolated vertices (no edges, called the empty graph).

Theorem: If G decomposes into k connected components, G1, G2, ..., Gk then
P(G, x) = P( G1, x)P(G2, x) ... P(Gk, 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

P(G, x) = P(G'ß, x) - P(G"ß, x).

Example: Consider C4, 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 K33, so by the theorem:
P(C4, x) = x(x-1)3 - x(x-1)(x-2) = (x4 - 3x3 + 3x2 - x) - (x3 -3x2 + 2x) = x4 - 4x3 + 6x2 - 3x.

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

Properties of the Chromatic Polynomial

Theorem: Pf: Let P(G, x) = apxp + ap-1xp-1 + ... + a1x + a0.

c) Consider P(G,0) = a0 = the number of ways to color G with no colors, but this can't be done, so a0 = 0.

d) Consider P(G, 1) = ap + ap-1 + ... + a1 = 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 In, which has chromatic polynomial xn.

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) = xn-2(x)(x-1) = xn - xn-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,

P(G, x) = P(G'ß, x) - P(G"ß, x).
Now, since G'ß has fewer than k edges, by the induction hypothesis, P(G'ß, x) is monic of degree n. Since G"ß has fewer than k edges and only n-1 vertices, P(G"ß, x) is monic of degree n-1. Upon subtraction of these polynomials, we see that there is no term in P(G"ß, x) that can remove the xn term in P(G'ß, x), so P(G, x) will be monic of degree n.

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 xn - xn-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'ß, x) = xn - an-1xn-1 + an-2xn-2 - + ... and
P(G"ß, x) = xn-1 - bn-2xn-2 + bn-3xn-3 - + ... .
Now, by the fundamental reduction theorem,
P(G, x) = P(G'ß, x) - P(G"ß, x)
= (xn - an-1xn-1 + an-2xn-2 - + ... ) - (xn-1 - bn-2x-2n + bn-3xn-3 - + ...)
= xn - (an-1 + 1)xn-1 + (an-2 + bn-2)xn-2 - + ...
and so, the signs alternate and none of the coefficients are zero.

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

Theorem: |an-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 xn - xn-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'ß, x) = xn - an-1xn-1 + an-2xn-2 - + ... and
P(G"ß, x) = xn-1 - bn-2xn-2 + bn-3xn-3 - + ... ,
where an-1 is the number of edges in G'ß and bn-2 is the number of edges in G"ß.

Now, by the fundamental reduction theorem,

P(G, x) = P(G'ß, x) - P(G"ß, x)
= (xn - an-1xn-1 + an-2xn-2 - + ... ) - (xn-1 - bn-2xn-2 + bn-3xn-3 - + ...)
= xn - (an-1 + 1)xn-1 + (an-2 + bn-2)xn-2 - + ...
and so, since G'ß is obtained from G by removal of just one edge, an-1 = k - 1, so an-1 + 1 = k, and the statement is true.

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, G1, G2, ..., Gk then P(G, x) = P( G1, x)P(G2, x) ... P(Gk, x). Thus, the smallest exponent in the chromatic polynomial of G is the sum of the smallest exponents appearing in each of the P(Gi, 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 x1 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 xn - xn-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'ß and G"ß both have fewer than k edges and are connected, by the induction hypothesis we may assume that their chromatic polynomials have the form:

P(G'ß, x) = xn - an-1xn-1 + an-2xn-2 - + ... +/- a1x and
P(G"ß, x) = xn-1 - bn-2xn-2 + bn-3xn-3 - + ... -/+ b1x.
Now, by the fundamental reduction theorem,
P(G, x) = P(G'ß, x) - P(G"ß, x)
= (xn - an-1xn-1 + an-2xn-2 - + ... +/- a1x) - (xn-1 - bn-2xn-2 + bn-3xn-3 - + ... -/+ b1x)
= xn - (an-1 + 1)xn-1 + (an-2 + bn-2)xn-2 - + ... +/-(a1 + b1)x
and so, the statement is true.

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'ß breaks up into two connected components, but G"ß remains connected and both have fewer than k edges, so, by the induction hypothesis we may assume that their chromatic polynomials have the form:

P(G'ß, x) = xn - an-1xn-1 + an-2xn-2 - + ... -/+ a2x2 and
P(G"ß, x) = xn-1 - bn-2xn-2 + bn-3xn-3 - + ... -/+ b1x.
Now, by the fundamental reduction theorem,
P(G, x) = P(G'ß, x) - P(G"ß, x)
= (xn - an-1xn-1 + an-2xn-2 - + ...-/+ a2x2) - (xn-1 - bn-2xn-2 + bn-3xn-3 - + ... -/+ b1x)
= xn - (an-1 + 1)xn-1 + (an-2 + bn-2)xn-2 - + ... +/- b1x
and so, the statement is true.

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) = x4 - 4x3 + 3x2 is not the chromatic polynomial of any graph.