Given an undirected connected graph $G = (V, E)$, which may contain parallel edges but excludes self-loops, it can be decomposed into $k$ subgraphs $G_1 = (V_1, E_1), \ldots, G_k = (V_k, E_k)$, where each $G_i$ is a complete graph or, under weaker conditions, a connected graph. The graph exhibits the following properties:
- $V = \bigcup_{i=1}^k V_i$ with the condition that for any two distinct subgraphs $G_i$ and $G_j$, the intersection of their vertex sets $V_i \cap V_j$ is not empty for all $i \neq j$ and for each $|V_i| \ge 2$
- the edge sets $E = \bigcup_{i=1}^k E_i$ are such that $E_i \cap E_j = \emptyset$ for all $i \neq j$, that is $\{E_i \}$ is a partition of $E$.
How to prove that it is possible to color the vertices of $G$ using at most three distinct colors in such a way that each $G_i$ contains vertices of at least two different colors.
PS: my attempts
I've illustrated some specific examples with small sizes, where I was able to find a feasible coloring method. However, I haven't discerned a general construction method. As for algorithmic constructions like backtracking, I can't prove that such an algorithm would always find a solution that meets the criteria. For non-constructive approaches, the probabilistic method comes to mind, where each vertex is randomly assigned one of three colors, 1, 2, or 3, with equal probability. Let $X_i$ be the number of different colors in $G_i$. If I can demonstrate that $E[\min(X_1, \ldots, X_k)] > 1$, it would prove the existence of a solution. However, calculating this expected value involves the minimum of random variables, which is challenging to compute.
I'm not an expert in graph theory, but these are the tools I can think of.
Best Answer
The question is stated in a rather convoluted form but it boils down to the following:
Given a finite set $V$ of vertices and given subsets $V_1,\dots,V_k\subseteq V$ with $|V_i|\ge2$ and $V_i\cap V_j\ne\varnothing$ for $i\ne j$, show that the vertices can be colored with three colors so that each set $V_i$ contains vertices of at least two different colors.
Proof. We can assume that $V_1$ is a minimal element of the family $\{V_1,\dots,V_k\}$, i.e., no $V_i$ is a proper subset of $V_1$. Choose a vertex $x\in V_1$. Color $x$ red, color the elements of $V_1\setminus\{x\}$ blue, and color the elements of $V\setminus V_1$ green.
If $x\in V_i$ then $V_i$ contains the red vertex $x$ and at least one other vertex, which can't be red. If $x\notin V_i$ then $V_i$ contains a green vertex (as otherwise $V_i$ would be a proper subset of $V_1$) and also a blue vertex (since $V_i\cap V_1\ne\varnothing$).
Remark. This question is essentially a duplicate of a closed question which I answered in the same way. I also mentioned some generalizations, culminating in the following:
Theorem. Let $H=(V,\mathcal E)$ be a hypergraph with vertex set $V$ and edge set $\mathcal E$ such that every pair of edges has nonempty intersection. Then $H$ is $3$-colorable if and only if $\{U\subseteq V:\exists X\in\mathcal E,\ X\subseteq U\}$ is not an ultrafilter on $V$.