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Let $G_1=\langle V, E_1 \rangle$ be an undirected graph, with at most one odd cycle. Let $G_2=\langle V, E_2 \rangle$ be an undirected tree. Consider $G=\langle V, E_1\cup E_2\rangle$.
Prove:

  • $\chi(G)\leqslant 6$
  • $\chi(G)\leqslant 5$

How does one prove the two claims?

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    Suppose you know $\chi(G_1)\leqslant 2$ and $\chi(G_2)\leqslant 3$. Can you prove $\chi(G)\leqslant 2\cdot 3=6$? – 2017-02-26

1 Answers 1

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For $G_2$, since it is a tree, we have immediately $\chi(G_2) \leq 2$. Suppose the color of each vertex $v \in V$ is $\mathsf{color}_2(v)$ in $G_2$. For $G_1$, we discuss in two cases, namely, the case that $G_1$ contains no odd cycle and the case that $G_1$ contains one odd cycle.

  • If $G_1$ does not contain any odd cycle, then $G_1$ is bipartite. Then $\chi(G_1) \leq 2$. Let $\mathsf{color}_1(v)$ be the color of $v \in V$ in $G_2$. For the union graph $G_1 \cup G_2$, if for each vertex $v \in V$ we assign color $\mathsf{color}_\cup(v) = (\mathsf{color}_1(v), \mathsf{color}_2(v))$ to $v$, then it is guaranteed that for any adjacent $u$, $v$, $\mathsf{color}_\cup(u) \neq \mathsf{color}_\cup(v)$. Therefore, $\chi(G_1 \cup G_2) \leq 2 \times 2 = 4$.

  • If $G_1$ contains one odd cycle, let $e \in E_1$ be an edge in the odd cycle and let $G_3 = G_1 \backslash \{e\}$. That is, $G_3$ is the resulting graph if we remove $e$ from $G_1$. Clearly, $G_3$ does not contain any odd cycle. By our discussion above, $\chi(G_3 \cup G_2) \leq 4$. We have then $$\chi(G_1 \cup G_2) = \chi(G_3 \cup G_2 \cup \{e\}) \leq \chi(G_3 \cup G_2) + 1 = 5$$