Fast Approximation Algorithms for Piercing Boxes by Points

TL;DR

The duality between the piercing set and independent set (for boxes) to speed up the multiplicative weight-update (MWU) method is exploited and a simpler and slightly more efficient algorithm for constructing a weak $\eps$-net than in [Ezr10] is presented.

Abstract

$\newcommand{\popt}{\mathcal{p}} \newcommand{\Re}{\mathbb{R}}\newcommand{\N}{\mathcal{N}} \newcommand{\BX}{\mathcal{B}} \newcommand{\bb}{\mathsf{b}} \newcommand{\eps}{\varepsilon} \newcommand{\polylog}{\mathrm{polylog}} $ Let $\mathcal{B}=\{\mathsf{b}_1, \ldots ,\mathsf{b}_n\}$ be a set of $n$ axis-aligned boxes in $\Re^d$ where $d\geq2$ is a constant. The \emph{piercing problem} is to compute a smallest set of points $\N \subset \Re^d$ that hits every box in $\mathcal{B}$, i.e., $\N\cap \mathsf{b}_i\neq \emptyset$, for $i=1,\ldots, n$. Let $\popt=\popt(\mathcal{B})$, the \emph{piercing number} be the minimum size of a piercing set of $\mathcal{B}$. We present a randomized $O(d^2\log\log \popt)$-approximation algorithm with expected running time $O(n^{d/2}\polylog n)$. Next, we present a faster $O(n^{\log d+1})$-time algorithm but with a slightly inferior approximation factor of $O(2^{4d}\log\log\popt)$. The running time of both algorithms can be improved to near-linear using a sampling-based technique, if $\popt = O(n^{1/d})$. For the dynamic version of the problem in the plane, we obtain a randomized $O(\log\log\popt)$-approximation algorithm with $O(n^{1/2}\polylog n )$ amortized expected update time for insertion or deletion of boxes. For squares in $\Re^2$, the update time can be improved to $O(n^{1/3}\polylog n )$.

Figures (5)

• Figure 2.1: The two crates charged to $p_j$, and their projection to the $xy$-axis.
• Figure 3.1: We describe a data-structure for maintaining (implicitly) the vertices of an arrangement of boxes, under operations weight , double , halve , insert , delete , see \ref{['def:arr:ds1']}.
• Figure 3.2: LPs for the stabbing and independence problems. By duality, we have $\mathcalb{p}^{*}(\mathcal{B},\mathcal{S})=\mathcalb{i}^*(\mathcal{B},\mathcal{S})$.
• Figure 7.1: $\Delta=I_x \times I_y$ denotes a cell in $\Pi$. $\mathcal{S}_{\varphi}=\{R_1,R_2,R_3,R_4\}$ intersects $\Delta$. Let $R_i=R_i^x\times R_i^y$ for $i\in\{1,2,3,4\}$. $\mathcal{X}_{\Delta}=\{R_2^x, R_3^x, R_4^x\}$, $\mathcal{Y}_{\Delta}=\{R_1^y, R_5^y\}$. Observe that $\mathcalb{w}^{}_{\!\mathcal{S}_{\varphi}}\mleft({\{p_1\}}\mright) = \mathcalb{w}^{}_{\!\mathcal{X}_{\Delta}}\mleft({\{p_1\}_{\downarrow x}}\mright) \cdot \mathcalb{w}^{}_{\!\mathcal{Y}_{\Delta}}\mleft({\{p_1\}_{\downarrow y}}\mright)=16$ and $\mathcalb{w}^{}_{\!\mathcal{S}_{\varphi}}\mleft({\{p_2\}}\mright) = \mathcalb{w}^{}_{\!\mathcal{X}_{\Delta}}\mleft({\{p_2\}_{\downarrow x}}\mright) \cdot \mathcalb{w}^{}_{\!\mathcal{Y}_{\Delta}}\mleft({\{p_2\}_{\downarrow y}}\mright)=32$.
• Figure 7.2: The figure represents the partition $\Pi$ and the tree $T$ built on $\Pi$. $\sigma$ is a strip of $\Pi$, $\Delta$ is a cell of $\sigma$. $T_{\sigma}$ is the tree built on the cells of $\sigma$. $R$ is a rectangle in $\mathcal{B}$. The highlighted nodes represent $C(\mathsf{b})$. ( Note that the axes are flipped.)

Theorems & Definitions (45)

• Lemma 2.1
• Remark 2.2
• Definition 2.3
• Definition 2.4
• Lemma 2.5
• Lemma 2.6
• Lemma 2.8
• Remark 2.9
• Lemma 2.10
• Lemma 2.11