I am considering the following problem:
Given a graph $G=(V,E)$ with $|V|=4n$ vertices, can we color the vertices in two colors (red and blue) with the additional constraint that for each vertex subsets: $\{1,2,3,4\}, \{5,6,7,8\}, \ldots, \{4n-3,4n-2,4n-1,4n\}$, exactly two of them are colored red?
- The usual constraint that two vertices connected by an edge must always have different colors AND
- The additional constraint that for all of the following vertex subsets: $\{1,2,3,4\}, \{5,6,7,8\}, \ldots, \{4n-3,4n-2,4n-1,4n\}$, exactly two vertices are colored red (and thus exactly two are blue)?
I think this problem has two flavors. On one hand it's obviously the "bit more constrained version" of 2-coloring, which makes me feel that this problem could be in $\mathsf{P}$ because 2-coloring is. On the other hand, the "exactly two colors" constraint looks like a generic 4-SAT-like relation that in the general case creates an $\mathsf{NP}$-hard constraint satisfaction problem from Schaefer's dichotomy theorem.
Because of this intermediate nature, I feel like maybe this kind of problem is well studied even. Can anybody come up with a $\mathsf{P}$ or $\mathsf{NP}$-hardness proof, or point to some literature concerning this kind of problem? It's quite hard to search for these weird variant problems without knowing their name... I understand that it's a bit similar to problems that were discussed here and there, but I don't see an obvious reduction.
I am actually interested in the more general case where we group the vertices into non-overlapping subsets of size $2k$, where each subsets need to have exactly $k$ reds. The case of $k=1$ is easily seen to be in $\mathsf{P}$, because it is essentially 2-SAT. The problem I wrote above corresponds to the case of $k=2$, which is the smallest case I cannot prove either way.