Suppose I have a collection of good convex sets and bad convex sets in $\mathbb{R}^d$ (where $d$ can be big). Each convex set is defined by a series of closed ranges in each dimension $d$ - a hyperrectangle if you will. For 2D an example set could be $\{(x, y) : x \in [1, 5]\wedge y \in [-2, 3]\}$. All sets are sparse, that is, for the majority of the dimensions the convex set in that dimension just covers range $[0, 0]$, each set only has a non-zero width in a couple dimensions. This means each hyperrectangle could be efficiently described by listing its vertices despite the high dimensionality.
If the convex hull of the good sets is disjoint from all bad convex sets, there exists a set of hyperplanes (AKA inequalities) that together separate the good from the bad sets. The convex hull immediately gives you such a set of hyperplanes, but it's not minimal - it uses more hyperplanes than necessary.
Can we efficiently compute a minimal set of hyperplanes that separates the good from the bad sets? If one hyperplane suffices we can find it using linear programming, but I don't know how to compute it if more than one hyperplane is needed.
If the answer is no, can we get at least an efficient heuristic/greedy algorithm?
