Given a set of points $P$ and axis-aligned rectangles $\mathcal{R}$ in the plane, a point $p \in P$ is called \emph{exposed} if it lies outside all rectangles in $\mathcal{R}$. In the \emph{max-exposure problem}, given an integer parameter $k$, we want to delete $k$ rectangles from $\mathcal{R}$ so as to maximize the number of exposed points. We show that the problem is NP-hard and assuming plausible complexity conjectures is also hard to approximate even when rectangles in $\mathcal{R}$ are translates of two fixed rectangles. However, if $\mathcal{R}$ only consists of translates of a single rectangle, we present a polynomial-time approximation scheme. For range space defined by general rectangles, we present a simple $O(k)$ bicriteria approximation algorithm; that is by deleting $O(k^2)$ rectangles, we can expose at least $\Omega(1/k)$ of the optimal number of points.

中文翻译：

---

最大暴露问题

给定一组点$ P $和平面中与轴对齐的矩形$ \ mathcal {R} $，如果P $中的点$ p \位于$ \ mathcal {中的所有矩形之外，则称为\ emph {暴露} R} $。在\ emph {max-exposure问题}中，给定整数参数$ k $，我们希望从$ \ mathcal {R} $中删除$ k $个矩形，以最大化暴露点的数量。我们证明了问题是NP难的，并且即使$ \ mathcal {R} $中的矩形是两个固定矩形的平移，假设合理的复杂性猜想也很难近似。但是，如果$ \ mathcal {R} $仅包含单个矩形的平移，则我们提出多项式时间近似方案。对于由常规矩形定义的范围空间，我们提出了一种简单的$ O（k）$双标准近似算法；即删除$ O（k ^ 2）$个矩形，