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

this is my thread in another post:

http://stackoverflow.com/questions/5724167/problem-to-determine-the-chromatic-polynomial-of-a-graph

For the Descomposition Theorem of Chromatic Polynomials. if G=(V,E), is a connected graph and e belong E

P (G, λ) = P (Ge, λ) -P(Ge', λ)

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) ^ 3 - [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?

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2 Answers 2

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Your graph is a 5-cycle. I wouldn't use that theorem, I'd just do it directly.

1st vertex: $\lambda$ options.

2nd vertex: $\lambda-1$ options.

3rd vertex: two cases. If it's the same color as 1st vertex, then $\lambda-1$ options for 4th vertex, $\lambda-2$ options for 5th vertex. So this case contributes $\lambda(\lambda-1)^2(\lambda-2)$. Second case, if it differs from the 1st vertex ($\lambda-2$ options), then two subcases: if 4th vertex is same color as 1st, then $\lambda-1$ options for 5th, making $\lambda(\lambda-1)(\lambda-2)(\lambda-1)$. If 4th differs from 1st ($\lambda-2$ options), then $\lambda-2$ options for 5th, making $\lambda(\lambda-1)(\lambda-2)^3$.

Now add 'em all up.

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I did not know that you can not apply the theorem for graphs 5-cycles. I do not understand why you can not. –  franvergara66 Apr 20 '11 at 1:56
    
You can; all I'm saying is that I wouldn't, because I'd find it easier to just reason my way through this problem. –  Gerry Myerson Apr 20 '11 at 2:16

I think what you need in your answer is $\lambda (\lambda- 1)^4$ for the polynomial of Ge. The 5-cycle with one edge removed is a path of five vertices. $\lambda$ choices for the first, and $(\lambda - 1)$ choices for the other 4.

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