11.2 Chromatic Polynomials

Let G be a graph on n vertices. Let P(G,q),q be the number of ways of colouring (properly) a graph with q colours. More formally,

P(G,q)=|{c:V[q]:c(u)c(v)uv}|.

We shall define P(G,x),x to be the polynomial (if it exists) such that P(G,q) is the number of proper q-colourings of G for q. P(G,x) is called the chromatic polynomial of G. Observe that if g is a polynomial such that g(q)=P(G,q) for all q then gP(G,.) and hence P(G,x) is unique if it exists. We will now show that this is a monic polynomial of degree n and other interesting properties.

Let Ge denote the graph with the edge e removed. Let G/e denote the graph with e contracted i.e., if e=(v,w) then we replace the vertices v,w with v and edges vx,wy with vx,vy. Observe that G/e can be a multigraph. But this will not matter to us as proper colourings of a multigraph and its corresponding simple graph are the same.

Proposition 11.3.

(Delection-contraction principle ) For any edge e=(v,w)G,

P(G,q):=P(Ge,q)P(G/e,q),q.
Proof.

The proof follows by two simple observations that any proper q-colouring of G is a proper q-colouring of Ge in which v and w receive distinct colours and any proper q-colouring of Ge in which v and w receive same colours is a proper q-colouring of G/e. ∎

If E=, trivially P(G,q)=qn and so P(G,x)=xn. Thus the chromatic polynomial exists and is monic in this case. Now using Proposition 11.3, we can inductively show that P(G,x) is a polynomial in q. Here are some very important properties. All the proofs will follow from deletion-contraction principle and induction. Our general proof strategy is as follows.

  • Assume P(G,x) is a polynomial with some properties for E=.

  • Using uniqueness of chromatic polynomial and DCI, conclude that P(G,x):=P(Ge,x)P(G/e,x) is the chromatic polynomial of G.

  • Using induction hypothesis and DCI, show that P(G,x) satisfies appropriate property.

For abbreviation, we call this DCI argument in this section.

Lemma 11.4.
  1. 1.

    P(G,x) is a monic polynomial of degree n.

  2. 2.

    χ(G)=min{k:P(G,k)>0}.

Now, let us assume that P(G,x)=i=1naixi (as a0=0 trivially). Then we have that

  1. 3.

    i=1nai=0 or P(G,x)=xn.

  2. 4.

    ani=(1)i|ani|.

  3. 5.

    an1=|E|.

Proof.
  1. 1.

    This claim holds for E=. Assume that it holds for |E|<m and |V|<n. Let G be a graph with |E|=m,|V|=n. By DC principle P(G,q)=P(Ge,q)P(G/e,q) for all q and so P(G,x):=P(Ge,x)P(G/e,x) is the chromatic polynomial of G. By induction hypothesis, P(Ge,x) is a monic polynomial of degree n and P(G/e,x) is a monic polynomial of degree n1. So P(G,x) is a monic polynomial of degree n.

  2. 2.

    This follows trivially by the two definitions.

  3. 3.

    As for (4), observe that i=1nai=P(G,1)= no. of proper 1-colouring of G.If |E|, there exists no proper 1-colouring of G and so P(G,1)=0. Else P(G,x)=xn as required.

  4. 4.

    We will apply the DCI argument again and prove (4),(5) and (6) together. The three identities can be verified for E= or G has 1 or 2 vertices easily. Let

    P(Ge,x)=i=0n(1)i|bni|xni,P(G/e,x)=i=0n1(1)i|cn1i|xn1i=i=1n(1)i1|cni|xni.

    Then by DC principle we have that

    P(G,x)=|bn|xn+i=1n(1)i[|bni|+|cni|]xni.

    (5) follows easily as ani=(1)i[|bni|+|cni|].

  5. 5.

    As for (6), observe by monicity of the chromatic polynomial, induction and DC principle that that

    |E(G)|=|E(Ge)|+1=bn1+cn1=|bn1|+|cn1|=an1,

    where in the second equality we have used (6) for bn1.

Lemma 11.5.

If G has k-components G1,,Gk then

P(G,x)=i=1kP(Gi,x)

and further a0==ak1=0.

Proof.

First equality follows trivially because any combination of q-colourings of each component gives a q-colouring of the full graph G i.e.,

P(G,q)=i=1kP(Gi,q),q.

Since deg(P(Gi,x))1, we have that a0==ak1=0

Theorem 11.6.

P(G,x)=XE(1)|X|xβ0(X) where β0(X)=β0((V,X)).

Proof.

Enough to show that P(G,q)=XE(1)|X|qβ0(X) for q. Let c:V[q] be an improper colouring if c(u)=c(v) for some uv. Let IC be the set of all improper colourings. For e=(u,v) define Be:={cIC:c(u)=c(v)}. Then we have that

P(G,q) =qn|IC|=qn|eEBe|
=qnXE(1)|X|1|eXBe|
=qn+XE(1)|X||eXBe|.

To complete the proof, enough to prove that for XE, |eXBe|=qβ0(X). Suppose X={e1,,ek} where ei=(ui,vi),1ik. Let C1,,Cm be the components in (V,X) i.e., m=β0(X) . It is possible that ui=uj or vj. Thus, we have that

eXBe ={c:V[q]:c(ui)=c(vi),1ik}
={c:V[q]:c is a constant on each Ci,1im}

The latter is obtained by choosing one colour for each component C1,,Cm and so its cardinality if qm as required. ∎

Exercise(A) 11.7.
  1. 1.

    Find the chromatic polynomial of complete graph Kn, cycle Cn, wheel graph Wn (i.e., the graph obtained by adding a new vertex to Cn and connecting it to all the n vertices of Cn), path graph Pn and the Petersen graph.

  2. 2.

    Show that the coefficient of xn2 in the chromatic polynomial of a n vertex graph is m(m1)2T where T is the number of triangles and m is the number of edges.

  3. 3.

    Show that a graph with n vertices is a tree iff P(G,x)=x(x1)n1.