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Consider a semi-regular bipartite graph $C$ consisting of two parts $A$ (having each vertex of degree $\Delta$) and $B$ (having each vertex of degree $2$). Let its chromatic polynomial be $C(x)$. Now, consider connecting each of the vertices in the independent sets $A$ and $B$ such that the resulting subgraph $A'$ is $\Delta$ regular and the subgraph $B'$ is $\beta$ regular. Also consider that the maximum of chromatic numbers of $A'$ and $B'$ is $\chi$ and the chromatic polynomials of $A'$ and $B'$ are $A(x)$ and $B(x)$ respectively. Then, what is the chromatic polynomial of the new graph $A'\cup B'\cup C$ where $\cup$ denotes union of vertex and edge sets of the subgraphs.

I think it is of the form $r(x)A(x)B(x)$ where $r(x)$ has some relation with $C(x)$. Specifically, I think the chromatic number of the new graph would be at most $2$ more than $\chi$.

Note that the case when the bipartite graph is just a bridge (edge) reduces $r(x)$ to $\frac{x-1}{x}$ . This may be verified by the deletion-contraction principle. But, what if we have more than one connecting edge between $A$ and $B$. Any hints? Thanks beforehand.

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  • $\begingroup$ Please clarify your terminology. You are calling $A$ and $B$ "components" (implying there are no edges between them) but also "independent sets" (implying there have no edges inside them). Are $A$ and $B$ the biparts of the graph? If so, then they are independent sets, so have chromatic number $1$. What do you then mean by "consider that the maximum of chromatic numbers of $A$ and $B$ is $\chi$". Also, if I understand your construction, then the blown-up graph is bipartite, so have chromatic number $2$. $\endgroup$
    – verret
    Commented Aug 8, 2019 at 2:20
  • $\begingroup$ @verret sorry, clarified the word. But, a downvote is unreasonable, I think? $\endgroup$
    – vidyarthi
    Commented Aug 8, 2019 at 5:58
  • $\begingroup$ Verrret's complaint is still valid. What does "the chromatic polynomial of $A$" mean? I don't understand your "blowing up" operation either. Much more precision is required. $\endgroup$ Commented Aug 8, 2019 at 7:57
  • $\begingroup$ @BrendanMcKay again edited. Is it clear now? $\endgroup$
    – vidyarthi
    Commented Aug 8, 2019 at 9:02
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    $\begingroup$ Complete bipartite graphs are semiregular. $C=K_{2,\Delta}$ satisfies your criteria. Take several copies of it so that you have more freedom to add edges. $\endgroup$ Commented Aug 9, 2019 at 18:40

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