Covering Points by Disjoint Boxes with Outliers

For a set of n points in the plane, we consider the axis--aligned (p,k)-Box Covering problem: Find p axis-aligned, pairwise-disjoint boxes that together contain n-k points. In this paper, we consider the boxes to be either squares or rectangles, and we want to minimize the area of the largest box. For general p we show that the problem is NP-hard for both squares and rectangles. For a small, fixed number p, we give algorithms that find the solution in the following running times: For squares we have O(n+k log k) time for p=1, and O(n log n+k^p log^p k time for p = 2,3. For rectangles we get O(n + k^3) for p = 1 and O(n log n+k^{2+p} log^{p-1} k) time for p = 2,3. In all cases, our algorithms use O(n) space.Comment: updated version: - changed problem from 'cover exactly n-k points' to 'cover at least n-k points' to avoid having non-feasible solutions. Results are unchanged. - added Proof to Lemma 11, clarified some sections - corrected typos and small errors - updated affiliations of two author

Piercing axis-parallel boxes

Let \F be a finite family of axis-parallel boxes in $\R^d$ such that \F contains no $k+1$ pairwise disjoint boxes. We prove that if \F contains a subfamily \M of $k$ pairwise disjoint boxes with the property that for every F\in \F and M\in \M with $F \cap M \neq \emptyset$, either $F$ contains a corner of $M$ or $M$ contains $2^{d-1}$ corners of $F$, then \F can be pierced by $O(k)$ points. One consequence of this result is that if $d=2$ and the ratio between any of the side lengths of any box is bounded by a constant, then \F can be pierced by $O(k)$ points. We further show that if for each two intersecting boxes in \F a corner of one is contained in the other, then \F can be pierced by at most $O(k\log\log(k))$ points, and in the special case where \F contains only cubes this bound improves to $O(k)$