Piercing axis-parallel boxes

TL;DR

This paper proves new bounds on piercing axis-parallel boxes in high-dimensional space under specific intersection conditions, showing that a small number of points suffices to intersect all boxes.

Contribution

It introduces novel geometric conditions under which axis-parallel boxes can be pierced by a linear or near-linear number of points, extending previous results.

Findings

Families with no $k+1$ disjoint boxes can be pierced by $O(k)$ points under certain intersection conditions.

In 2D with bounded side ratios, $O(k)$ points suffice for piercing.

Special cases with nested boxes or cubes require $O(k ext{ log log } k)$ or $O(k)$ points respectively.

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 $F\in \F$ 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 $O(k)$.

Equations

$$F_d(\nu (\mathcal{F}))$$

$$\le F_d(\nu ({\mathcal{F}}_1))+F_d(\nu ({\mathcal{F}}_3))+F_{d-1}(\nu ({\mathcal{F}}_2))$$

$$\displaystyle\leq F_{d}(\Big{\lceil}\frac{\nu(\mathcal{F})}{2}\Big{\rceil}-1)+F_{d}(\Big{\lfloor}\frac{\nu(\mathcal{F})}{2}\Big{\rfloor})+F_{d-1}(\nu(\mathcal{F}))$$

$$\le 2\frac{\nu (\mathcal{F})}{2}{\left(1+\log \left(\frac{\nu (\mathcal{F})}{2}\right)\right)}^{d-1}+\nu (\mathcal{F}){\left(1+\log \left(\nu (\mathcal{F})\right)\right)}^{d-2}$$

$$\le \nu (\mathcal{F}){\left(1+\log \left(\nu (\mathcal{F})\right)\right)}^{d-1},$$

$$\tau (\mathcal{B}(M^{\prime }))=\tau (\left\{F\cap H:F\in \mathcal{B}(M^{\prime })\right\})\le d$$

$$\tau (\mathcal{B})\le \sum _{M^{\prime }:M{M}^{\prime }\in E(G_1)}\tau (\mathcal{B}(M^{\prime }))\le d^{+}(M)-1+d,$$

$$\tau ({\mathcal{F}}_i)$$

$$=d|V(G_1)|+|E(G_1)|\le d|\mathcal{M}|+4|\mathcal{M}|=(4+d)\nu (\mathcal{F}),$$

Full text

M. Chudnovsky: Princeton University, Princeton, NJ 08544. E-mail: mchudnov@math.princeton.edu. Supported by NSF grant DMS-1550991 and US Army Research Office Grant W911NF-16-1-0404.

S. Spirkl: Princeton University, Princeton, NJ 08544. E-mail: sspirkl@math.princeton.edu.

S. Zerbib: Department of Mathematics, University of Michigan, Ann Arbor. E-mail: zerbib@umich.edu. Partly supported by the New-England Fund, Technion.

1. Introduction

A matching in a hypergraph $H=(V,E)$ on vertex set $V$ and edge set $E$ is a subset of disjoint edges in $E$, and a cover of $H$ is a subset of $V$ that intersects all edges in $E$. The matching number $\nu(H)$ of $H$ is the maximal size of a matching in $H$, and the covering number $\tau(H)$ of $H$ is the minimal size of a cover. The fractional relaxations of these numbers are denoted as usual by $\nu^{*}(H)$ and $\tau^{*}(H)$. By LP duality we have that $\nu^{*}(H)=\tau^{*}(H)$.

Let $\mathcal{F}$ be a finite family of axis-parallel boxes in $\mathbb{R}^{d}$. We identify $\mathcal{F}$ with the hypergraph consisting with vertex set $\mathbb{R}^{d}$ and edge set $\mathcal{F}$. Thus a matching in $\mathcal{F}$ is a subfamily of pairwise disjoint boxes (also called an independent set in the literature) and a cover in $\mathcal{F}$ is a set of points in $\mathbb{R}^{d}$ intersecting every box in $\mathcal{F}$ (also called a hitting set).

An old result due to Gallai is the following (see e.g. [8]):

Theorem 1.1** (Gallai).**

If $\mathcal{F}$ is a family of intervals in $\mathbb{R}$ (i.e., a family of boxes in $\mathbb{R}$) then $\tau(\mathcal{F})=\nu(\mathcal{F})$.

A rectangle is an axis-parallel box in $\mathbb{R}^{2}$. In 1965, Wegner [12] conjectured that in a hypergraph of axis-parallel rectangles in $\mathbb{R}^{2}$, the ratio $\tau/\nu$ is bounded by 2. Gýarfás and Lehel conjectured in [7] that the same ratio is bounded by a constant. The best known lower bound, $\tau=\lfloor 5\nu/3\rfloor$, is attained by a construction due to Fon-Der-Flaass and Kostochka in [6]. Károlyi [9] proved that in families of axis-parallel boxes in $\mathbb{R}^{d}$ we have $\tau(\mathcal{F})\leq\nu(\mathcal{F})\left(1+\log\left(\nu(\mathcal{F})\right)\right)^{d-1}$, where $\log=\log_{2}$. Here is a short proof of Károlyi’s bound.

Theorem 1.2** (Károlyi [9]).**

If $\mathcal{F}$ is a finite family of axis-parallel boxes in $\mathbb{R}^{d}$, then $\tau(\mathcal{F})\leq\nu(\mathcal{F})\left(1+\log\left(\nu(\mathcal{F})\right)\right)^{d-1}$.

Proof.

We proceed by induction on $d$ and $\nu(\mathcal{F})$. Note that if $\nu(\mathcal{F})\in\{0,1\}$ then the result holds for all $d$. Now let $d,n\in\mathbb{N}$. Let $F_{d^{\prime}}:\mathbb{R}\to\mathbb{R}$ be a function for which $\tau(\mathcal{T})\leq F_{d^{\prime}}(\nu(\mathcal{T}))$ for every family $\mathcal{T}$ of axis-parallel boxes in $\mathbb{R}^{d^{\prime}}$ with $d^{\prime}<d$, or with $d=d^{\prime}$ and $\nu(\mathcal{T})<n$.

Let $\mathcal{F}$ be a family of axis-parallel boxes in $\mathbb{R}^{d}$ with $\nu(\mathcal{F})=n$. For $a\in\mathbb{R}$, let $H_{a}$ be the hyperplane $\left\{x=(x_{1},\dots,x_{d}):x_{1}=a\right\}$. Write $L_{a}=\left\{x=(x_{1},\dots,x_{d}):x_{1}\leq a\right\}$, and let $\mathcal{F}_{a}=\left\{F\in\mathcal{F}:F\subseteq L_{a}\right\}$. Define $a^{*}=\min\left\{a:\nu(F_{a})\geq\lceil\nu/2\rceil\right\}$. The hyperplane $H_{a^{*}}$ gives rise to a partition $\mathcal{F}=\bigcup_{i=1}^{3}\mathcal{F}_{i}$, where $\mathcal{F}_{1}=\left\{F\in\mathcal{F}:F\subseteq L_{a^{*}}\setminus H_{a^{*}}\right\}$, $\mathcal{F}_{2}=\left\{F\in\mathcal{F}:F\cap H_{a^{*}}\neq\emptyset\right\}$, and $\mathcal{F}_{3}=\mathcal{F}\setminus(\mathcal{F}_{1}\cup F_{2})$. It follows from the choice of $a^{*}$ that $\nu(\mathcal{F}_{1})\leq\lceil\nu(\mathcal{F})/2\rceil-1$, $\nu(\mathcal{F}_{2})\leq\nu(\mathcal{F})$, and $\nu(\mathcal{F}_{3})\leq\lfloor\nu(\mathcal{F})/2\rfloor$.

Therefore,

[TABLE]

implying the result. ∎

For $\nu(\mathcal{F})=1$, this implies the following well-known result (see e. g. [6]).

Observation 1.3**.**

Let $\mathcal{F}$ be a family of axis-parallel boxes with $\nu(\mathcal{F})=1$. Then $\tau(\mathcal{F})=1$.

Note that for $\nu(\mathcal{F})=2$, we have that $\mathcal{F}_{1}=\emptyset$, $\nu(\mathcal{F}_{2})=1$ and so $\tau(\mathcal{F})\leq F_{d-1}(2)+1$. Therefore, we have the following, which was also proved in [6].

Observation 1.4** (Fon-der-Flaass and Kostochka [6]).**

Let $\mathcal{F}$ be a family of axis-parallel boxes in $\mathbb{R}^{d}$ with $\nu(\mathcal{F})=2$. Then $\tau(\mathcal{F})\leq d+1$.

The bound from Theorem 1.2 was improved by Akopyan [2] to $\tau(\mathcal{F})\leq(1.5\log_{3}2+o(1))\nu(\mathcal{F})\left(\log_{2}\left(\nu(\mathcal{F})\right)\right)^{d-1}$.

A corner of a box $F$ in $\mathbb{R}^{d}$ is a zero-dimensional face of $F$. We say that two boxes in $\mathbb{R}^{d}$ intersect at a corner if one of them contains a corner of the other.

A family $\mathcal{F}$ of connected subsets of $\mathbb{R}^{2}$ is a family of pseudo-disks, if for every pair of distinct subsets in $\mathcal{F}$, their boundaries intersect in at most two points. In [4], Chan and Har-Peled proved that families of pseudo-disks in $\mathbb{R}^{2}$ satisfy $\tau=O(\nu)$. It is easy to check that if $\mathcal{F}$ is a family of axis-parallel rectangles in $\mathbb{R}^{2}$ in which every two intersecting rectangles intersect at a corner, then $\mathcal{F}$ is a family of pseudo-disks. Thus we have:

Theorem 1.5** (Chan and Har-Peled [4]).**

There exists a constant $c$ such that for every family $\mathcal{F}$ of axis-parallel rectangles in $\mathbb{R}^{2}$ in which every two intersecting rectangles intersect at a corner, we have that $\tau(\mathcal{F})\leq c\nu(\mathcal{F})$.

Here we prove a few different generalizations of this theorem. In Theorem 1.6 we prove the bound $\tau(\mathcal{F})\leq c\nu(\mathcal{F})\log\log(\nu(\mathcal{F}))$ for families $\mathcal{F}$ of axis-parallel boxes in $\mathbb{R}^{d}$ in which every two intersecting boxes intersect at a corner, and in Theorem 1.7 we prove $\tau(\mathcal{F})\leq c\nu(\mathcal{F})$ for families $\mathcal{F}$ of axis-parallel cubes in $\mathbb{R}^{d}$, where in both cases $c$ is a constant depending only on the dimension $d$. We further prove in Theorem 1.8 that in families $\mathcal{F}$ of axis-parallel boxes in $\mathbb{R}^{d}$ satisfying certain assumptions on their pairwise intersections, the bound on the covering number improves to $\tau(\mathcal{F})\leq c\nu(\mathcal{F})$. For $d=2$, these assumptions are equivalent to the assumption that there is a maximum matching $\mathcal{M}$ in $\mathcal{F}$ such that every intersection between a box in $\mathcal{M}$ and a box in $\mathcal{F}\setminus\mathcal{M}$ occurs at a corner. We use this result to prove our Theorem 1.10, asserting that for every $r$, if $\mathcal{F}$ is a family of axis-parallel rectangles in $\mathbb{R}^{2}$ with the property that the ratio between the side lengths of every rectangle in $\mathcal{F}$ is bounded by $r$, then $\tau(\mathcal{F})\leq c\nu(\mathcal{F})$ for some constant $c$ depending only on $r$.

Let us now describe our results in more detail. First, for general dimension $d$ we have the following.

Theorem 1.6**.**

There exists a constant $c$ depending only on $d$, such that for every family $\mathcal{F}$ of axis-parallel boxes in $\mathbb{R}^{d}$ in which every two intersecting boxes intersect at a corner we have $\tau(\mathcal{F})\leq c\nu(\mathcal{F})\log\log(\nu(\mathcal{F}))$.

For the proof, we first prove the bound $\tau^{*}(\mathcal{F})\leq 2^{d}\nu(\mathcal{F})$ on the fractional covering number of $\mathcal{F}$, and then use Theorem 1.11 below for the bound $\tau(\mathcal{F})=O(\tau^{*}(\mathcal{F})\log\log(\tau^{*}(\mathcal{F})))$.

An axis-parallel box is a cube if all its side lengths are equal. Note that if $\mathcal{F}$ consists of axis-parallel cubes in $\mathbb{R}^{d}$, then every intersection in $\mathcal{F}$ occurs at a corner. Moreover, for axis-parallel cubes we have $\tau(\mathcal{F})=O(\tau^{*}(\mathcal{F}))$ by Theorem 1.11, and thus we conclude the following.

Theorem 1.7**.**

If $\mathcal{F}$ be a family of axis-parallel cubes in $\mathbb{R}^{d}$, then $\tau(\mathcal{F})\leq c\nu(\mathcal{F})$ for some constant $c$ depending only on $d$.

To get a constant bound on the ratio $\tau/\nu$ in families of axis-parallel boxes in $\mathbb{R}^{d}$ which are not necessarily cubes, we make a more restrictive assumption on the intersections in $\mathcal{F}$.

Theorem 1.8**.**

Let $\mathcal{F}$ be a family of axis-parallel boxes in $\mathbb{R}^{d}$. Suppose that there exists a maximum matching $\mathcal{M}$ in $\mathcal{F}$ such that for every $F\in\mathcal{F}$ and $M\in\mathcal{M}$, at least one of the following holds:

1. (1)

$F$* contains a corner of $M$;* 2. (2)

$F\cap M=\emptyset$; or 3. (3)

$M$* contains $2^{d-1}$ corners of $F$.*

Then $\tau(\mathcal{F})\leq(2^{d}+(4+d)d)\nu(\mathcal{F})$.

For $d=2$, this theorem implies the following corollary.

Corollary 1.9**.**

Let $\mathcal{F}$ be a family of axis-parallel rectangles in $\mathbb{R}^{2}$. Suppose that there exists a maximum matching $\mathcal{M}$ in $\mathcal{F}$ such that for every $F\in\mathcal{F}$ and $M\in\mathcal{M}$, if $F$ and $M$ intersect then they intersect at a corner. Then $\tau(\mathcal{F})\leq 16\nu(\mathcal{F})$.

Note that Corollary 1.9 is slightly stronger than Theorem 1.5. Here we only need that the intersections with rectangles in some fixed maximum matching $\mathcal{M}$ occur at corners, but we do not restrict the intersections of two rectangles $F,F^{\prime}\notin\mathcal{M}$.

Given a constant $r>0$, we say that a family $\mathcal{F}$ of axis-parallel boxes in $\mathbb{R}^{d}$ has an $r$-bounded aspect ratio if every box $F\in\mathcal{F}$ has $l_{i}(F)/l_{j}(F)\leq r$ for all $i,j\in\left\{1,\dots,d\right\}$, where $l_{i}(F)$ is the length of the orthogonal projection of $F$ onto the $i$th coordinate.

For families of rectangles with bounded aspect ratio we prove the following.

Theorem 1.10**.**

Let $\mathcal{F}$ be a family of axis-parallel rectangles in $\mathbb{R}^{2}$ that has an $r$-bounded aspect ratio. Then $\tau(\mathcal{F})\leq(14+2r^{2})\nu(\mathcal{F})$.

A result similar to Theorem 1.10 was announced in [1], but to the best of our knowledge the proof was not published.

An application of Theorem 1.10 is the existence of weak $\varepsilon$-nets of size $O\left(\frac{1}{\varepsilon}\right)$ for axis-parallel rectangles in $\mathbb{R}^{2}$ with bounded aspect ratio. More precisely, let $P$ be a set of $n$ points in $\mathbb{R}^{d}$ and let $\mathcal{F}$ be a family of sets in $\mathbb{R}^{d}$, each containing at least $\varepsilon n$ points of $P$. A weak $\varepsilon$-net for $\mathcal{F}$ is a cover of $\mathcal{F}$, and a strong $\varepsilon$-net for $\mathcal{F}$ is a cover of $\mathcal{F}$ with points of $P$. The existence of weak $\varepsilon$-nets of size $O\left(\frac{1}{\varepsilon}\right)$ for pseudo-disks in $\mathbb{R}^{2}$ was proved by Pyrga and Ray in [11]. Aronov, Ezra and Sharir in [3] showed the existence of strong $\varepsilon$-nets of size $O\left(\frac{1}{\varepsilon}\log\log\frac{1}{\varepsilon}\right)$ for axis-parallel boxes in $\mathbb{R}^{2}$ and $\mathbb{R}^{3}$, and the existence of weak $\varepsilon$-nets of size $O\left(\frac{1}{\varepsilon}\log\log\frac{1}{\varepsilon}\right)$ for all $d$ was then proved by Ezra in [5]. Ezra also showed that for axis-parallel cubes in $\mathbb{R}^{d}$ there exist an $\varepsilon$-net of size $O\left(\frac{1}{\varepsilon}\right)$. These results imply the following.

Theorem 1.11** (Aronov, Ezra and Sharir [3]; Ezra [5]).**

If $\mathcal{F}$ is a family of axis-parallel boxes in $\mathbb{R}^{d}$ then $\tau(\mathcal{F})\leq c\tau^{*}(\mathcal{F})\log\log(\tau^{*}(\mathcal{F}))$ for some constant $c$ depending only on $d$. If $\mathcal{F}$ consists of cubes, then this bound improves to $\tau(\mathcal{F})\leq c\tau^{*}(\mathcal{F})$.

An example where the smallest strong $\varepsilon$-net for axis-parallel rectangles in $R^{2}$ is of size $\Omega\left(\frac{1}{\varepsilon}\log\log\frac{1}{\varepsilon}\right)$ was constructed by Pach and Tardos in [10]. The question of whether weak $\varepsilon$-nets of size $O(\frac{1}{\varepsilon})$ for axis-parallel rectangles in $\mathbb{R}^{2}$ exist was raised both in [3] and in [10].

Theorem 1.10 implies a positive answer for the family of axis-parallel rectangles in $\mathbb{R}^{2}$ satisfying the $r$-bounded aspect ratio property:

Corollary 1.12**.**

For every fixed constant $r$, there exists a weak $\varepsilon$-net of size $O(\frac{1}{\varepsilon})$ for the family $\mathcal{F}$ of axis-parallel rectangles in $\mathbb{R}^{2}$ with aspect ratio bounded by $r$.

Proof.

Given a set $P$ of $n$ points, there cannot be $\frac{1}{\varepsilon}+1$ pairwise disjoint rectangles in $\mathcal{F}$, each containing at least $\varepsilon n$ points of $P$. Therefore $\nu(\mathcal{F})\leq\frac{1}{\varepsilon}$. Theorem 1.10 implies that there is a cover of $\mathcal{F}$ of size $O(\frac{1}{\varepsilon})$. ∎

This paper is organized as follows. In Section 2 we prove Theorem 1.6. Section 3 contains definitions and tools. Theorem 1.8 is then proved in Section 4 and Theorem 1.10 is proved in Section 5.

2. Proofs of Theorems 1.6 and 1.7

Let $\mathcal{F}$ be a finite family of axis-parallel boxes in $\mathbb{R}^{d}$, such that every intersection in $\mathcal{F}$ occurs at a corner. By performing small perturbations on the boxes, we may assume that no two corners of boxes of $\mathcal{F}$ coincide.

Proposition 2.1**.**

We have $\tau^{*}(\mathcal{F})\leq 2^{d}\nu(\mathcal{F})$.

Proof.

Write $\nu(\mathcal{F})=k$, and let $g:\mathcal{F}\to\mathbb{R}^{+}$ be a rational approximation of a maximal fractional matching for $\mathcal{F}$. By removing boxes $F\in\mathcal{F}$ for which $g(F)=0$ and duplicating boxes if necessary, we may assume that $g(F)=\frac{1}{r}$ for all $F\in\mathcal{F}$, where $r$ is the maximal size of a subset of boxes in $\mathcal{F}$ intersecting in a single point. Letting $n$ be the number of boxes in $\mathcal{F}$ we have $\tau^{*}(\mathcal{F})=\nu^{*}(\mathcal{F})=\frac{n}{r}$, and thus our aim is to show that $\frac{n}{r}\leq 2^{d}k$.

Since $\nu(\mathcal{F})=k$, it follows from Turán’s theorem that there are at least $n(n-k)/(2k)$ unordered intersecting pairs of boxes $\mathcal{F}$. Each such unordered pair contributes at least two pairs of the form $(x,F)$, where $x$ is a corner of a box $F^{\prime}\in\mathcal{F}$, $F$ is box in $\mathcal{F}$ different from $F^{\prime}$, and $x$ pierces $F$. Therefore, since there are altogether $2^{d}n$ corners of boxes in $\mathcal{F}$, there must exist a corner $x$ of a box $F\in\mathcal{F}$ that pierces at least $(n-k)/2^{d}k$ boxes in $\mathcal{F}$, all different from $F$. Together with $F$, $x$ pierces at least $n/2^{d}k$ boxes, implying that $n/2^{d}k\leq r$. Thus $\frac{n}{r}\leq 2^{d}k$, as desired. ∎

Combining this bound with Theorem 1.11, we obtain the proofs of Theorems 1.6 and 1.7.

3. Definitions and tools

Let $R$ be an axis-parallel box in $\mathbb{R}^{d}$ with $R=[x_{1},y_{1}]\times\dots\times[x_{d},y_{d}]$. For $i\in\left\{1,\dots,d\right\}$, let $p_{i}(R)=[x_{i},y_{i}]$ denote the orthogonal projection of $R$ onto the $i$-th coordinate. Two intervals $[a,b],[c,d]\subseteq\mathbb{R}$, are incomparable if $[a,b]\not\subseteq[c,d]$ and $[c,d]\not\subseteq[a,b]$. We say that $[a,b]\prec[c,d]$ if $b<c$. For two axis-parallel boxes $Q$ and $R$ we say that $Q\prec_{i}R$ if $p_{i}(Q)\prec p_{i}(R)$.

Observation 3.1**.**

Let $Q,R$ be disjoint axis-parallel boxes in $\mathbb{R}^{d}$. Then there exists $i\in\left\{1,\dots,d\right\}$ such that $Q\prec_{i}R$ or $R\prec_{i}Q$.

Lemma 3.2**.**

Let $Q,R$ be axis-parallel boxes in $\mathbb{R}^{d}$ such that $Q$ contains a corner of $R$ but $R$ does not contain a corner of $Q$. Then, for all $i\in\left\{1,\dots,d\right\}$, either $p_{i}(R)$ and $p_{i}(Q)$ are incomparable, or $p_{i}(R)\subseteq p_{i}(Q)$, and there exists $i\in\left\{1,\dots,d\right\}$ such that $p_{i}(R)\subsetneq p_{i}(Q)$.

Moreover, if $R\not\subseteq Q$, then there exists $j\in\left\{1,\dots,d\right\}\setminus\left\{i\right\}$ such that $p_{i}(R)$ and $p_{i}(Q)$ are incomparable.

Proof.

Let $x=(x_{1},\dots,x_{d})$ be a corner of $R$ contained in $Q$. By symmetry, we may assume that $x_{i}=\max(p_{i}(R))$ for all $i\in\left\{1,\dots,d\right\}$. Since $x_{i}\in p_{i}(Q)$ for all $i\in\left\{1,\dots,d\right\}$, it follows that $\max(p_{i}(Q))\geq\max(p_{i}(R))$ for all $i\in\left\{1,\dots,d\right\}$. If $\min(p_{i}(Q))\leq\min(p_{i}(R))$, then $p_{i}(R)\subseteq p_{i}(Q)$; otherwise, $p_{i}(Q)$ and $p_{i}(R)$ are incomparable. If $p_{i}(Q)$ and $p_{i}(R)$ are incomparable for all $i\in\left\{1,\dots,d\right\}$, then $y=(y_{1},\dots,y_{d})$ with $y_{i}=\min(p_{i}(Q))$ is a corner of $Q$ and since $\min(p_{i}(Q))>\min(p_{i}(R))$, it follows that $y\in R$, a contradiction. It follows that there exists an $i\in\left\{1,\dots,d\right\}$ such that $p_{i}(R)\subsetneq p_{i}(Q)$.

If $p_{i}(R)\subsetneq p_{i}(Q)$ for all $i\in\left\{1,\dots,d\right\}$, then $R\subseteq Q$; this implies the result. ∎

Observation 3.3**.**

Let $\mathcal{F}$ be a family of axis-parallel boxes in $\mathbb{R}^{d}$. Let $\mathcal{F}^{\prime}$ arise from $\mathcal{F}$ by removing every box in $\mathcal{F}$ that contains another box in $\mathcal{F}$. Then $\nu(\mathcal{F})=\nu(\mathcal{F}^{\prime})$ and $\tau(\mathcal{F})=\tau(\mathcal{F}^{\prime})$.

Proof.

Since $\mathcal{F}^{\prime}\subseteq\mathcal{F}$, it follows that $\nu(\mathcal{F}^{\prime})\leq\nu(\mathcal{F})$ and $\tau(\mathcal{F}^{\prime})\leq\tau(\mathcal{F})$. Let $\mathcal{M}$ be a matching in $\mathcal{F}$ of size $\nu(\mathcal{F})$. Let $\mathcal{M}^{\prime}$ arise from $\mathcal{M}$ by replacing each box $R$ in $\mathcal{M}\setminus\mathcal{F}^{\prime}$ with a box in $\mathcal{F}^{\prime}$ contained in $R$. Then $\mathcal{M}^{\prime}$ is a matching in $\mathcal{F}^{\prime}$, and so $\nu(\mathcal{F}^{\prime})=\nu(\mathcal{F})$. Moreover, let $P$ be a cover of $\mathcal{F}^{\prime}$. Since every box in $\mathcal{F}$ contains a box in $F^{\prime}$ (possibly itself) which, in turn, contains a point in $P$, we deduce that $P$ is a cover of $\mathcal{F}$. It follows that $\tau(\mathcal{F}^{\prime})=\tau(\mathcal{F})$. ∎

A family $\mathcal{F}$ of axis-parallel boxes is clean if no box in $\mathcal{F}$ contains another box in $\mathcal{F}$. By Observation 3.3, we may restrict ourselves to clean families of boxes.

4. Proof of Theorem 1.8

Throughout this section, let $\mathcal{F}$ be a clean family of axis-parallel boxes in $\mathbb{R}^{d}$, and let $\mathcal{M}$ be a matching of maximum size in $\mathcal{F}$. We let $\mathcal{F}(\mathcal{M})$ denote the subfamily of $\mathcal{F}$ consisting of those boxes $R$ in $\mathcal{F}$ for which for every $M\in\mathcal{M}$, either $M$ is disjoint from $R$ or $M$ contains at least $2^{d-1}$ corners of $R$. Our goal is to bound $\tau(\mathcal{F}(\mathcal{M}))$.

Lemma 4.1**.**

Let $R\in\mathcal{F}(\mathcal{M})$. Then $R$ intersects at least one and at most two boxes in $\mathcal{M}$. If $R$ intersects two boxes $M_{1},M_{2}\in\mathcal{M}$, then there exists $j\in\left\{1,\dots,d\right\}$ such that $M_{1}\prec_{j}M_{2}$ or $M_{2}\prec_{j}M_{1}$, and for all $i\in\left\{1,\dots,d\right\}\setminus\left\{j\right\}$, we have that $p_{i}(R)\subseteq p_{i}(M_{1})$ and $p_{i}(R)\subseteq p_{i}(M_{2})$.

Proof.

If $R$ is disjoint from every box in $\mathcal{M}$, then $\mathcal{M}\cup\left\{R\right\}$ is a larger matching, a contradiction. So $R$ intersects at least one box in $\mathcal{M}$. Let $M_{1}$ be in $\mathcal{M}$ such that $R\cap M_{1}\neq\emptyset$. We claim that there exists $j\in\left\{1,\dots,d\right\}$ such that $M_{1}$ contains precisely the set of corners of $R$ with the same $j$th coordinate.

By Lemma 3.2, there exists $j\in\left\{1,\dots,d\right\}$ such that $p_{j}(R)=[a,b]$ and $p_{j}(M_{1})$ are incomparable. By symmetry, we may assume that $a\in p_{j}(M_{1})$, $b\not\in p_{j}(M_{1})$. This proves that $M_{1}$ contains all $2^{d-1}$ corners of $R$ with $a$ as their $j$th coordinate, and our claim follows.

Consequently, $p_{i}(R)\subseteq p_{i}(M_{1})$ for all $i\in\left\{1,\dots,d\right\}\setminus\left\{j\right\}$. Since $R$ has exactly $2^{d}$ corners, and members of $\mathcal{M}$ are disjoint, it follows that there exist at most two boxes in $\mathcal{M}$ that intersect $R$. If $M_{1}$ is the only one such box, then the result follows. Let $M_{2}\in\mathcal{M}\setminus\left\{M_{1}\right\}$ such that $R\cap M_{1}\neq\emptyset$. By our claim, it follows that $M_{2}$ contains $2^{d-1}$ corners of $R$; and since $M_{1}$ is disjoint from $M_{2}$, it follows that $M_{2}$ contains precisely those corners of $R$ with $j$th coordinate equal to $b$. Therefore, $p_{i}(R)\subseteq p_{i}(M_{2})$ for all $i\in\left\{1,\dots,d\right\}\setminus\left\{j\right\}$. We conclude that $p_{i}(M_{2})$ is not disjoint from $p_{i}(M_{1})$ for all $i\in\left\{1,\dots,d\right\}\setminus\left\{j\right\}$, and since $M_{1},M_{2}$ are disjoint, it follows from Observation 3.1 that either $M_{1}\prec_{j}M_{2}$ or $M_{2}\prec_{j}M_{1}$. ∎

For $i\in\left\{1,\dots,d\right\}$, we define a directed graph $G_{i}$ as follows. We let $V(G_{i})=\mathcal{M}$, and for $M_{1},M_{2}\in\mathcal{M}$ we let $M_{1}M_{2}\in E(G_{i})$ if and only if $M_{1}\prec_{i}M_{2}$ and there exists $R\in\mathcal{F}(\mathcal{M})$ such that $R\cap M_{1}\neq\emptyset$ and $R\cap M_{2}\neq\emptyset$. In this case, we say that $R$ witnesses the edge $M_{1}M_{2}$. For $i=\left\{1,\dots,d\right\}$, we say that $R$ is $i$-pendant at $M_{1}\in\mathcal{M}$ if $M_{1}$ is the only box of $\mathcal{M}$ intersecting $R$ and $p_{i}(R)$ and $p_{i}(M_{1})$ are incomparable. Note that by Lemma 4.1, every box $R$ in $\mathcal{F}(\mathcal{M})$ satisfies exactly one of the following: $R$ witnesses an edge in exactly one of the graphs $G_{i}$, $i\in\left\{1,\dots,d\right\}$; or $R$ is $i$-pendant for exactly one $i\in\left\{1,\dots,d\right\}$.

Lemma 4.2**.**

Let $i\in\left\{1,\dots,d\right\}$. Let $Q,R\in\mathcal{F}(\mathcal{M})$ be such that $Q$ witnesses an edge $M_{1}M_{2}$ in $G_{i}$, and $R$ witnesses an edge $M_{3}M_{4}$ in $G_{i}$. If $Q$ and $R$ intersect, then either $M_{1}=M_{4}$, or $M_{2}=M_{3}$, or $M_{1}M_{2}=M_{3}M_{4}$.

Proof.

By symmetry, we may assume that $i=1$. Let $p_{1}(M_{1})=[x_{1},y_{1}]$ and $p_{1}(M_{2})=[x_{2},y_{2}]$. It follows that $p_{1}(Q)\subseteq[x_{1},y_{2}]$. Let $a=(a_{1},a_{2},\dots,a_{d})\in Q\cap R$. It follows that $a_{j}\in p_{j}(Q)\subseteq p_{j}(M_{1})\cap p_{j}(M_{2})$ and $a_{j}\in p_{j}(R)\subseteq p_{j}(M_{3})\cap p_{j}(M_{4})$ for all $j\in\left\{2,\dots,d\right\}$.

If $M_{1}\in\left\{M_{3},M_{4}\right\}$ and $M_{2}\in\left\{M_{3},M_{4}\right\}$, then $M_{1}M_{2}=M_{3}M_{4}$, and the result follows. Therefore, we may assume that this does not happen. If $M_{1}\in\left\{M_{3},M_{4}\right\}$, we reflect every rectangle in $\mathcal{F}$ along the origin. When constructing $G_{1}$ for this family, we have $M_{4}M_{3},M_{2}M_{1}\in E(G_{1})$, and $M_{2}\not\in\left\{M_{3},M_{4}\right\}$. Thus, by symmetry, we may assume that $M_{1}$ is distinct from $M_{3}$ and $M_{4}$.

It follows that $a\not\in M_{1}$, for otherwise $R$ intersects three distinct members of $\mathcal{M}$, contrary to Lemma 4.1. Since $R$ is disjoint from $M_{1}$, it follows that either $M_{1}\prec_{1}R$ or $R\prec_{1}M_{1}$. But $p_{1}(Q)\subseteq[x_{1},y_{2}]$, and since $Q\cap R\neq\emptyset$, it follows that $M_{1}\prec_{1}R$.

Since $M_{3}\neq M_{1}$ and $p_{j}(M_{3})\cap p_{j}(M_{1})\ni a_{j}$ for all $j\in\left\{2,\dots,d\right\}$, it follows that either $M_{1}\prec_{1}M_{3}$ or $M_{1}\prec_{1}M_{3}$. Since $M_{1}\prec_{1}R$ and $R\cap M_{3}\neq\emptyset$, it follows that $M_{1}\prec_{1}M_{3}$.

Suppose that $a\in M_{3}$. Then $Q\cap M_{3}\neq\emptyset$, and since $M_{1}\prec_{1}M_{3}$, we have that $M_{3}=M_{2}$ as desired.

Therefore, we may assume that $a\not\in M_{3}$, and thus $p_{1}(M_{1})\prec p_{1}(M_{3})\prec[a_{1},a_{1}]$. Since $[y_{1},a_{1}]\subseteq p_{1}(Q)$, it follows that $p_{1}(M_{3})\cap p_{1}(Q)\neq\emptyset$. But $p_{j}(M_{3})\cap p_{j}(Q)\ni a_{j}$ for all $j\in\left\{2,\dots,d\right\}$, and hence $Q\cap M_{3}\neq\emptyset$. But then $M_{3}\in\left\{M_{1},M_{2}\right\}$, and thus $M_{3}=M_{2}$. This concludes the proof. ∎

The following is a well-known fact about directed graphs; we include a proof for completeness.

Lemma 4.3**.**

Let $G$ be a directed graph. Then there exists an edge set $E\subseteq E(G)$ with $|E|\geq|E(G)|/4$ such that for every vertex $v\in V(G)$, either $E$ contains no incoming edge at $v$, or $E$ contains no outgoing edge at $v$.

Proof.

For $A,B\subseteq V(G)$, let $E(A,B)$ denote the set of edges of $G$ with head in $A$ and tail in $B$.

Let $X_{0}=Y_{0}=\emptyset$, $V(G)=\left\{v_{1},\dots,v_{n}\right\}$. For $i=1,\dots,n$ we will construct $X_{i},Y_{i}$ such that $X_{i}\cup Y_{i}=\left\{v_{1},\dots,v_{i}\right\}$, $X_{i}\cap Y_{i}=\emptyset$ and $|E(X_{i},Y_{i})|+|E(Y_{i},X_{i})|\geq|E(G|(X_{i}\cup Y_{i}))|/2$, where $G|(X_{i}\cup Y_{i})$ denotes the induced subgraph of $G$ on vertex set $X_{i}\cup Y_{i}$. This holds for $X_{0},Y_{0}$. Suppose that we have constructed $X_{i-1},Y_{i-1}$ for some $i\in\left\{1,\dots,n\right\}$. If $|E(X_{i-1},\left\{v_{i}\right\})|+|E(\left\{v_{i}\right\},X_{i-1})|\geq|E(Y_{i-1},\left\{v_{i}\right\})|+|E(\left\{v_{i}\right\},Y_{i-1})|$, we let $X_{i}=X_{i-1},Y_{i}=Y_{i-1}\cup\left\{v_{i}\right\}$; otherwise, let $X_{i}=X_{i-1}\cup\left\{v_{i}\right\},Y_{i}=Y_{i-1}$. It follows that $X_{i},Y_{i}$ still have the desired properties. Thus, $|E(X_{n},Y_{n})|+|E(Y_{n},X_{n})|\geq|E(G)|/2$. By symmetry, we may assume that $|E(X_{n},Y_{n})|\geq|E(G)|/4$. But then $E(X_{n},Y_{n})$ is the desired set $E$; it contains only incoming edges at vertices in $X_{n}$, and only outgoing edges at vertices in $Y_{n}$. This concludes the proof. ∎

Theorem 4.4**.**

For $i\in\left\{1,\dots,d\right\}$, $|E(G_{i})|\leq 4\nu(\mathcal{F})$.

Proof.

Let $E\subseteq E(G_{i})$ as in Lemma 4.3. For each edge in $E$, we pick one box witnessing this edge; let $\mathcal{F}^{\prime}$ denote the family of these boxes. We claim that $\mathcal{F}^{\prime}$ is a matching. Indeed, suppose not, and let $Q,R\in\mathcal{F}^{\prime}$ be distinct and intersecting. Let $Q$ witness $M_{1}M_{2}$ and $R$ witness $M_{3}M_{4}$. By Lemma 4.2, it follows that either $M_{1}M_{2}=M_{3}M_{4}$ (impossible since we picked exactly one witness per edge) or $M_{1}=M_{4}$ (impossible because $E$ does not contain both an incoming and an outgoing edge at $M_{1}=M_{4}$) or $M_{2}=M_{3}$ (impossible because $E$ does not contain both an incoming and an outgoing edge at $M_{2}=M_{3}$). This is a contradiction, and our claim follows. Now we have $\nu(\mathcal{F})\geq|\mathcal{F}^{\prime}|=|E|\geq|E(G)|/4$, which implies the result. ∎

A matching $\mathcal{M}$ of a clean family $\mathcal{F}$ of boxes is extremal if for every $M\in\mathcal{M}$ and $R\in\mathcal{F}\setminus\mathcal{M}$, either $(\mathcal{M}\setminus\left\{M\right\})\cup\left\{R\right\}$ is not a matching or there exists an $i\in\left\{1,\dots,d\right\}$ such that $\max(p_{i}(R))\geq\max(p_{i}(M))$. Every family $\mathcal{F}$ of axis parallel boxes has an extremal maximum matching. For example, the maximum matching $\mathcal{M}$ minimizing $\sum_{M\in\mathcal{M}}\sum_{i=1}^{d}\max(p_{i}(M))$ is extremal.

Theorem 4.5**.**

For $i\in\left\{1,\dots,d\right\}$, let $\mathcal{F}_{i}$ denote the set of boxes in $\mathcal{F}(\mathcal{M})$ that either are $i$-pendant or witness an edge in $G_{i}$. Then $\tau(\mathcal{F}_{i})\leq(4+d)\nu(\mathcal{F})$. If $\mathcal{M}$ is extremal, then $\tau(\mathcal{F}_{i})\leq(3+d)\nu(\mathcal{F})$.

Proof.

By symmetry, it is enough to prove the theorem for $i=1$. For $M\in\mathcal{M}$, let $\mathcal{F}_{M}$ denote the set of boxes in $\mathcal{F}_{1}$ that either are 1-pendant at $M$, or witness an edge $MM^{\prime}$ of $G_{1}$. It follows that $\bigcup_{M\in\mathcal{M}}\mathcal{F}_{M}=\mathcal{F}_{1}$. For $M\in\mathcal{M}$, let $d^{+}(M)$ denote the out-degree of $M$ in $G_{1}$. We will prove that $\tau(\mathcal{F}_{M})\leq d^{+}(M)+d$ for all $M\in\mathcal{M}$.

Let $M\in\mathcal{M}$, and let $\mathcal{A}$ denote the set of boxes that are $1$-pendant at $M$. Suppose that $\mathcal{A}$ contains two disjoint boxes $M_{1},M_{2}$. Then $(\mathcal{M}\setminus\left\{M\right\})\cup\left\{M_{1},M_{2}\right\}$ is a larger matching than $\mathcal{M}$, a contradiction. So every two boxes in $\mathcal{A}$ pairwise intersect. By Observation 1.3, it follows that $\tau(A)=1$.

Let $\mathcal{B}=\mathcal{F}_{M}\setminus\mathcal{A}$. Suppose that there is an edge $MM^{\prime}\in E(G_{1})$ such that the set $\mathcal{B}(M^{\prime})$ of boxes in $\mathcal{B}$ that witness the edge $MM^{\prime}$ satisfies $\nu(\mathcal{B}(M^{\prime}))\geq 3$. Then $\mathcal{M}$ is not a maximum matching, since removing $M$ and $M^{\prime}$ from $\mathcal{M}$ and adding $\nu(\mathcal{B}(M^{\prime}))$ disjoint rectangles in $\mathcal{B}(M^{\prime})$ yields a larger matching. Moreover, for distinct $M^{\prime},M^{\prime\prime}\in\mathcal{M}$, every box in $\mathcal{B}(M^{\prime})$ is disjoint from every box in $\mathcal{B}(M^{\prime\prime})$ by Lemma 4.2. Thus, if there exist $M^{\prime},M^{\prime\prime}$ such that $\nu(\mathcal{B}(M^{\prime}))=\nu(\mathcal{B}(M^{\prime\prime}))=2$ and $M^{\prime}\neq M^{\prime\prime}$, then removing $M,M^{\prime}$ and $M^{\prime\prime}$ and adding two disjoint rectangles from each of $\mathcal{B}(M^{\prime})$ and $\mathcal{B}(M^{\prime\prime})$ yields a bigger matching, a contradiction.

Let $p_{1}(M)=[a,b]$. Two boxes in $\mathcal{B}(M^{\prime})$ intersect if and only if their intersections with the hyperplane $H=\left\{(x_{1},\dots,x_{d}):x_{1}=b\right\}$ intersect. If $\nu(\mathcal{B}(M^{\prime}))=1$, then $\tau(\mathcal{B}(M^{\prime}))=1$ by Observation 1.3. If $\nu(\mathcal{B}(M^{\prime}))=2$, then $\nu(\left\{F\cap H:F\in\mathcal{B}(M^{\prime})\right\})=2$ and so

[TABLE]

by Observation 1.4.

Therefore,

[TABLE]

and since $\tau(\mathcal{A})\leq 1$, it follows that $\tau(\mathcal{F}_{M})\leq d^{+}(M)+d$ as claimed.

Summing over all rectangles in $\mathcal{M}$, we obtain

[TABLE]

which proves the first part of the theorem.

If $\mathcal{M}$ is extremal, then every $1$-pendant box at $M$ also intersects $H$. Let $M^{\prime}$ be such that $\nu(\mathcal{B}(M^{\prime}))$ is maximum. It follows that $\nu(\mathcal{A}\cup\mathcal{B}(M^{\prime}))\leq 2$ and thus $\tau(\mathcal{A}\cup\mathcal{B}(M^{\prime}))\leq d$, implying $\tau(\mathcal{F}_{M})\leq d^{+}(M)+d-1$. This concludes the proof of the second part of the theorem. ∎

Theorem 4.6**.**

Let $\mathcal{F}^{\prime}\subseteq\mathcal{F}$ be the set of boxes $R\in\mathcal{F}$ such that for each $M\in\mathcal{M}$, either $M\cap R=\emptyset$, or $M$ contains $2^{d-1}$ corners of $R$, or $R$ contains a corner of $M$. Then $\tau(\mathcal{F}^{\prime})\leq(2^{d}+(4+d)d)\nu(\mathcal{F})$. If $\mathcal{M}$ is extremal, then $\tau(\mathcal{F}^{\prime})\leq(2^{d}+(3+d)d)\nu(\mathcal{F})$.

Proof.

We proved in Theorem 4.5 that $\tau(\mathcal{F}_{i})\leq(4+d)\nu(\mathcal{F})$ for $i=1,\dots,d$. Let $\mathcal{F}^{\prime\prime}=\mathcal{F}^{\prime}\setminus\mathcal{F}(\mathcal{M})$. Then $\mathcal{F}^{\prime\prime}$ consists of boxes $R$ such that $R$ contains a corner of some box $M\in\mathcal{M}$. Let $P$ be the set of all corners of boxes in $\mathcal{M}$. It follows that $P$ covers $\mathcal{F}^{\prime\prime}$, and so $\tau(\mathcal{F}^{\prime\prime})\leq 2^{d}\nu(\mathcal{F})$. Since $\mathcal{F}^{\prime}=\mathcal{F}^{\prime\prime}\cup\mathcal{F}_{1}\cup\dots\cup\mathcal{F}_{d}$, it follows that $\tau(\mathcal{F}^{\prime})\leq(2^{d}+(4+d)d)\nu(\mathcal{F})$. If $\mathcal{M}$ is extremal, the same argument yields that $\tau(\mathcal{F}^{\prime})\leq(2^{d}+(3+d)d)\nu(\mathcal{F})$, since $\tau(\mathcal{F}_{i})\leq(3+d)\nu(\mathcal{F})$ for $i=1\dots,d$ by Theorem 4.5. ∎

We are now ready to prove our main theorems.

Proof of Theorem 1.8.

Let $\mathcal{F}$ be a family of axis-parallel boxes in $\mathbb{R}^{d}$, and let $\mathcal{M}$ be a maximum matching in $\mathcal{F}$ such that for every $F\in\mathcal{F}$ and $M\in\mathcal{M}$, either $F\cap M=\emptyset$, or $F$ contains a corner of $M$, or $M$ contains $2^{d-1}$ corners of $F$. It follows that $\mathcal{F}=\mathcal{F}^{\prime}$ in Theorem 4.6, and therefore, $\tau(\mathcal{F})\leq(2^{d}+(4+d)d)\nu(\mathcal{F})$. ∎

5. Proof of Theorem 1.10

Let $\mathcal{M}$ be a maximum matching in $\mathcal{F}$, and let $\mathcal{M}$ be extremal. Observe that each rectangle $R\in\mathcal{F}$ satisfies one of the following:

• •

$R$ contains a corner of some $M\in\mathcal{M}$;

• •

some $M\in\mathcal{M}$ contains two corners of $R$; or

• •

there exists $M\in\mathcal{M}$ such that $M\cap R\neq\emptyset$, and $p_{i}(R)\supseteq p_{i}(M)$ for some $i\in\left\{1,2\right\}$.

By Theorem 4.6, $14\nu(\mathcal{F})$ points suffice to cover every rectangle satisfying at least one of the first two conditions. Now, due to the $r$-bounded aspect ratio, for each $M\in\mathcal{M}$ and for each $i\in\{1,2\}$, at most $r^{2}$ disjoint rectangles $R\in\mathcal{F}$ can satisfy the third condition for $M$ and $i$. Thus the family of projections of the rectangles satisfying the third condition for $M$ and $i$ onto the $(3-i)$th coordinate have a matching number at most $r^{2}$. Since all these rectangles intersect the boundary of $M$ twice, by Theorem 1.1, we need at most $r^{2}$ additional points to cover them. We conclude that $\tau(\mathcal{F})\leq(14+2r^{2})\nu(\mathcal{F})$. ∎

Acknowledgments

We are thankful to Paul Seymour for many helpful discussions.