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Made clear that the usual coloring constraint is there too.
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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?

  1. The usual constraint that two vertices connected by an edge must always have different colors AND
  2. 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.

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?

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.

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

  1. The usual constraint that two vertices connected by an edge must always have different colors AND
  2. 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.

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Complexity of 2-coloring with extra constraints

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?

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.