# Is right this chromatic polynomial for this Bridge Graph?

I have the following graph:

And I need to find its chromatic polynomial.

Based in my notes, I have reached the following result:

$$Pg(x) = \frac{((x-1)^4 + (x-1))^2}{x(x-1)}$$.

It is because I have applied this formula:

$$Pg(x) = \frac{Pg1(x) Pg2(x)}{Pkr(x)}$$.

Taking the Bridge Graph and the previous formula, I have replaced with:

$$Pg1(x) = Pg2(x) = Pc4(x)= (x-1)^4+(x-1)$$

Because both sub-graphs are cycles.

On the other hand, the Pkr(x) in this case is equals to Pk2(x). So, the chromatic polynomial for a complete graph with n = 2, could be expressed as :

$$x(x-1)$$

So, my main doubt is if this reasoning is right. Because I'm not sure about my resolution and I think that Ia have missing something in the final response.

Thanks a lot!

The formula $$P_G(x) = \frac{P_{G_1}(x) \cdot P_{G_2}(x)}{P_{K_r}(x)}$$ that you are using is appropriate when $$G$$ is the clique sum of the graphs $$G_1$$ and $$G_2$$ along a copy of $$K_r$$.

The graph you give,

is not a clique sum of two cycles. Instead, it is a clique sum (along a $$K_2$$) of the graph induced by vertices $$\{a,b,c,d,e\}$$ and the graph induced by vertices $$\{d,e,f,g,h\}$$.

Both of these are not cycles, but cycles with an extra leaf vertex. Here is how we deal with that extra vertex. If the cycle has chromatic number $$(x-1)^4+(x-1)$$, then there are always $$x-1$$ ways to color the leaf vertex with a color different from its only neighbor, so in this case we have $$P_{G_1}(x) = P_{G_2}(x) = (x-1)^5 + (x-1)^2$$.

Applying the formula, we get $$P_G(x) = \frac{[(x-1)^5 + (x-1)^2]^2}{x(x-1)} = x^8-9 x^7+36 x^6-82 x^5+114 x^4-96 x^3+45 x^2-9 x.$$ (A quick sanity check which this result passes and yours does not is that the degree of $$P_G(x)$$ must be the number of vertices in $$G$$.)

Using standard techniques:

$$x^8-9 x^7+36 x^6-82 x^5+114 x^4-96 x^3+45 x^2-9 x$$