## Abstract

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 ∈ F and M ∈ M with F ∩ M ≠ ∅, 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).

Original language | English (US) |
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Article number | #P1.70 |

Journal | Electronic Journal of Combinatorics |

Volume | 25 |

Issue number | 1 |

DOIs | |

State | Published - Mar 29 2018 |

## All Science Journal Classification (ASJC) codes

- Theoretical Computer Science
- Geometry and Topology
- Discrete Mathematics and Combinatorics
- Computational Theory and Mathematics
- Applied Mathematics