Let be a graph on vertices. Let be the number of ways of colouring (properly) a graph with colours. More formally,
We shall define to be the polynomial (if it exists) such that is the number of proper -colourings of for . is called the chromatic polynomial of . Observe that if is a polynomial such that for all then and hence is unique if it exists. We will now show that this is a monic polynomial of degree and other interesting properties.
Let denote the graph with the edge removed. Let denote the graph with contracted i.e., if then we replace the vertices with and edges with . Observe that 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.
(Delection-contraction principle ) For any edge ,
The proof follows by two simple observations that any proper -colouring of is a proper -colouring of in which and receive distinct colours and any proper -colouring of in which and receive same colours is a proper -colouring of . ∎
If , trivially and so . Thus the chromatic polynomial exists and is monic in this case. Now using Proposition 11.3, we can inductively show that is a polynomial in . 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 is a polynomial with some properties for .
Using uniqueness of chromatic polynomial and DCI, conclude that is the chromatic polynomial of .
Using induction hypothesis and DCI, show that satisfies appropriate property.
For abbreviation, we call this DCI argument in this section.
is a monic polynomial of degree .
.
Now, let us assume that (as trivially). Then we have that
or .
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This claim holds for . Assume that it holds for and . Let be a graph with . By DC principle for all and so is the chromatic polynomial of . By induction hypothesis, is a monic polynomial of degree and is a monic polynomial of degree . So is a monic polynomial of degree .
This follows trivially by the two definitions.
As for (4), observe that no. of proper -colouring of .If , there exists no proper -colouring of and so . Else as required.
We will apply the DCI argument again and prove (4),(5) and (6) together. The three identities can be verified for or has or vertices easily. Let
Then by DC principle we have that
(5) follows easily as .
As for (6), observe by monicity of the chromatic polynomial, induction and DC principle that that
where in the second equality we have used (6) for .
∎
If has -components then
and further .
First equality follows trivially because any combination of -colourings of each component gives a -colouring of the full graph i.e.,
Since , we have that ∎
where .
Enough to show that for . Let be an improper colouring if for some . Let be the set of all improper colourings. For define . Then we have that
To complete the proof, enough to prove that for , . Suppose where . Let be the components in i.e., . It is possible that or . Thus, we have that
The latter is obtained by choosing one colour for each component and so its cardinality if as required. ∎
Find the chromatic polynomial of complete graph , cycle , wheel graph (i.e., the graph obtained by adding a new vertex to and connecting it to all the vertices of ), path graph and the Petersen graph.
Show that the coefficient of in the chromatic polynomial of a vertex graph is where is the number of triangles and is the number of edges.
Show that a graph with vertices is a tree iff .