11.2 Chromatic Polynomials

Let G be a graph on n vertices. Let P⁢(G,q),q∈ℕ∗ 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)⁢∀u∼v}|.

We shall define P⁢(G,x),x∈ℝ to be the polynomial (if it exists) such that P⁢(G,q) is the number of proper q-colourings of G for q∈ℕ∗. 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∈ℕ then g≡P(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 G−e 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 v∼x,w∼y with v′∼x,v′∼y. 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⁢(G−e,q)−P⁢(G/e,q),q∈ℕ.
Proof.

The proof follows by two simple observations that any proper q-colouring of G is a proper q-colouring of G−e in which v and w receive distinct colours and any proper q-colouring of G−e 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⁢(G−e,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∈ℕ:P⁢(G,k)>0}.

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

  1. 3.

    ∑i=1nai=0 or P⁢(G,x)=xn.

  2. 4.

    an−i=(−1)i⁢|an−i|.

  3. 5.

    an−1=−|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⁢(G−e,q)−P⁢(G/e,q) for all q∈ℕ and so P⁢(G,x):=P⁢(G−e,x)−P⁢(G/e,x) is the chromatic polynomial of G. By induction hypothesis, P⁢(G−e,x) is a monic polynomial of degree n and P⁢(G/e,x) is a monic polynomial of degree n−1. 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⁢(G−e,x)=∑i=0n(−1)i⁢|bn−i|⁢xn−i,P⁢(G/e,x)=∑i=0n−1(−1)i⁢|cn−1−i|⁢xn−1−i=∑i=1n(−1)i−1⁢|cn−i|⁢xn−i.

    Then by DC principle we have that

    P⁢(G,x)=|bn|⁢xn+∑i=1n(−1)i⁢[|bn−i|+|cn−i|]⁢xn−i.

    (5) follows easily as an−i=(−1)i⁢[|bn−i|+|cn−i|].

  5. 5.

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

    |E⁢(G)|=|E⁢(G−e)|+1=−bn−1+cn−1=|bn−1|+|cn−1|=−an−1,

    where in the second equality we have used (6) for bn−1.

∎

Lemma 11.5.

If G has k-components G1,…,Gk then

P⁢(G,x)=∏i=1kP⁢(Gi,x)

and further a0=…=ak−1=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∈ℕ∗.

Since d⁢e⁢g⁢(P⁢(Gi,x))≥1, we have that a0=…=ak−1=0 ∎

Theorem 11.6.

P⁢(G,x)=∑X⊂E(−1)|X|⁢xβ0⁢(X) where β0⁢(X)=β0⁢((V,X)).

Proof.

Enough to show that P⁢(G,q)=∑X⊂E(−1)|X|⁢qβ0⁢(X) for q∈ℕ∗. Let c:V→[q] be an improper colouring if c⁢(u)=c⁢(v) for some u∼v. Let I⁢C be the set of all improper colourings. For e=(u,v) define Be:={c∈I⁢C:c⁢(u)=c⁢(v)}. Then we have that

P⁢(G,q) =qn−|I⁢C|=qn−|∪e∈EBe|
=qn−∑∅⊋X⊂E(−1)|X|−1⁢|∩e∈XBe|
=qn+∑∅⊋X⊂E(−1)|X|⁢|∩e∈XBe|.

To complete the proof, enough to prove that for ∅⊋X⊂E, |∩e∈XBe|=qβ0⁢(X). Suppose X={e1,…,ek} where ei=(ui,vi),1≤i≤k. 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

∩e∈XBe ={c:V→[q]:c⁢(ui)=c⁢(vi),1≤i≤k}
={c:V→[q]:c is a constant on each Ci,1≤i≤m}

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 xn−2 in the chromatic polynomial of a n vertex graph is m⁢(m−1)2−T 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⁢(x−1)n−1.