problem to determine the chromatic polynomial of a graph

for a homework graph theory, I'm asked to determine the chromatic polynomial of the following graph

For the Descomposition Theorem of Chromatic Polynomials. if G=(V,E), is a connected开发者_运维知识库 graph and e belong E

P (G, λ) = P (Ge, λ) -P(Ge', λ)

where Ge denotes de subgraph obtained by deleting de edge e from G (Ge= G-e) and Ge' is the subgraph obtained by identifying the vertices {a,b} = e

When calculating chromatic Polynomials, i shall place brackets about a graph to indicate its chromatic polynomial. removes an edge any of the original graph to calculate the chromatic polynomial by the method of decomposition.

 P (G, λ) = P (Ge, λ)-P (Ge', λ) = λ (λ-1)^4 - [λ(λ-1)*(λ^2 - 3λ + 3)]

But the response from the answer key and the teacher is:

P (G, λ) = λ (λ-1)(λ-2)(λ^2-2λ-2)

I have operated on the polynomial but I can not reach the solution that I ask .. what am I doing wrong?

math.stackexchange.com told me as a way to solve my problem. Here's the solution:

https://math.stackexchange.com/questions/33946/problem-to-determine-the-chromatic-polynomial-of-a-graph

Your answer is correct, and so is the teacher's --they are equal. [By the way, nice picture and explanation.]

An odd-cycle can have no 2-coloring, hence the 5-cycle can have no 2-coloring, so its chromatic polynomial, f(x), must have x * [x - 1] * [x - 2]

as a divisor. If you combine your expression for f(x) and divide out the

x * [x - 1]

then you'll find that what remains is divisible by [x - 2], and the quotient is what your teacher wrote. -Jonathan King

In the book I am following (Graph Theory with Applications - Deo Prentice Hall) it is done differently. Instead of excluding the edge they connect two non-adjacent vertices.

Using this technique I am getting

P (G, λ) = 2λ(λ-1)^2(λ-2) + 2λ(λ-1)(λ-2)(λ-3) + λ(λ-1)(λ-2)(λ-3)(λ-4) which is also not equal to either of your results.