It is well-known that if a graph has maximum degree $d$, then it is $d+1$ colorable. Say we have $d+1$ graphs $G_1,\ldots, G_{d+1}$ on the same vertex set $V$, and say each $G_i$ has maximum degree at most $d$.

A coloring of $\textbf{G}:=\{G_1,\ldots, G_{d+1}\}$ is just a labelling of the common vertex set of $\textbf{G}$ with $\{1,\ldots, k\}$, for some $k\in\mathbb{N}$. This coloring is proper if for any $i\in\{1,\ldots, k\}$, no edge of $G_i$ can be found between two vertices colored $i$.

My question is whether $\textbf{G}$ admits a proper coloring with just $d+1$ labels.

Another way to formulate this is that we are looking for a partition of the common vertex set $V=V_1\cup \ldots \cup V_{d+1}$, where $V_i$ is an independent set in $G_i$.

The simplest case is when $d=1$. In this case, $\textbf{G}=\{G_1, G_2\}$, and both $G_1$ and $G_2$ are matchings. I claim that $\textbf{G}$ can be properly $2$-colored. Proceed with induction - base case is clear. If there exists a vertex that is isolated in $G_1$, we can remove that vertex, apply induction, and label that vertex $1$. Thus, we can assume every vertex is non-isolated in both $G_1$ and $G_2$, which is only possible if $\textbf{G}$ is an even cycle, its edges alternating between $G_1$ and $G_2$. It follows that $\textbf{G}$ can be properly $2$-colored.

  • $\begingroup$ "no edge of $G_i$": do you instead mean "no edge of any $G_j$"? $\endgroup$
    – RobPratt
    Sep 23, 2020 at 14:22
  • $\begingroup$ Correct me if I am wrong, but is your problem equivalent to the following: for each $G_i$ you want to find an independent set $V_i\subseteq V$ ($V$ being the common vertex set), so that $V_1\cup\cdots\cup V_{d+1}=V$. To see the equivalence: interpret the $V_i$ as vertices that are colored with color $i$. We can assume that the $V_i$ are disjoint: if they are not, remove the doubled vertex from one of the problematic sets. $\endgroup$
    – M. Winter
    Sep 23, 2020 at 15:39
  • $\begingroup$ @RobPratt No, I mean no edge of $G_i$ as written. The vertices of the $i^{th}$ label should be an independent set in $G_i$. $\endgroup$
    – alpmu
    Sep 23, 2020 at 20:27
  • $\begingroup$ @M.Winter That is correct, that's an equivalent formulation. $\endgroup$
    – alpmu
    Sep 23, 2020 at 20:28

1 Answer 1


The concept you introduce is called a cooperative coloring. Check out, e.g., this paper. Theorem 1 (with a reference to another paper) claims a negative answer to your question; but there is other information you may find relevant.


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