This question is inspired by a recent related question.
Given a bipartition $V=B \cup C$ of the vertices of a graph, call an edge $B$-monochromatic (or $C$-monochromatic) if both ends are in $B$ (or in $C$) and monochromatic if either holds. Discuss the minimum number of monochromatic edges from a bipartition.
A graph is bipartite (by definition) if there is a bipartition with no monochromatic edges. An early theorem of graph theory is that this happens exactly when there are no odd cycles.
If there is a set of $k$ edge disjoint odd cycles then there must be at least $k$ monochromatic edges in any bipartition. Let $j$ be the total number of odd cycles. Also, let $\ell$ be the minimum size for a set of edges $F$ so that every odd cycle includes at least one edge from $F.$
The linked question is
Q1: If we insist on no B-monochromatic edges can we get a bipartition with no more than $j$ C-monochromatic edges?
With this formulation it might be more natural to ask
Q2: Can we get a bipartition with no more than $j$ monochromatic edges?
I suspect that $Q1$ (and hence $Q2$) may be true simply because either the structure with respect to odd cycles is quite simple or else there are a large number of odd cycles. So the right question might be something else, perhaps replacing $j$ with $\ell.$
Before continuing I'll note that this is related to the max-cut problem, minimizing the number of monochromatic edges is the same as maximizing the number of edges separating a vertex subset $B$ from its complement $C.$
In the graph below we can handle all the odd cycles by making the edges $bc$ and $b'c'$ $C$-monochromatic but then the same is true for edge $cc'.$