We study the Witness Set problem, a natural dual to the classical Art Gallery problem. In the Witness Set problem, we are given a polygon $P$ and an integer $k$ as input, and the objective is to determine whether $P$ has a witness set of size at least $k$. A point set $X$ in $P$ is called a witness set if every point in $P$ is visible from at most one point in $X$. For simple polygons, we show that Witness Set lies in both $NP$ and $XP$. This stands in sharp contrast to its dual, the Art Gallery problem, which was recently shown to be $\exists \mathbb{R}$-complete by Abrahamsen et al. and is therefore neither in $NP$ nor admits a polynomial-size discretization unless $NP=\exists \mathbb{R}$. In contrast, we prove that Witness Set for simple polygons admits a finite discretization of size $n^{f(k)}$ for some function $f$. For comparison, even for simple polygons, Efrat and Har-Peled gave an algorithm for Art Gallery running in time $n^{O(k)}$ using tools from real algebraic geometry, and it appears difficult to obtain such algorithms without this machinery. On the other hand, our approach for Witness Set is purely combinatorial and relies on discretization, leading to an $n^{f(k)}$-time algorithm. Although Amit et al. claimed more than fifteen years ago that Witness Set is $NP$-hard, no proof or reference was provided. We show that the discrete version of the Witness Set problem - where the witness set must be chosen from a given finite point set $Q$ (instead of allowing witnesses to be chosen anywhere in the polygon), referred to as Discrete Witness Set - is $NP$-complete, even when the input is restricted to rectilinear polygons with holes. However, for simple polygons, Discrete Witness Set admits a polynomial-time algorithm by Das et al. Thus, it remains an open question whether the Witness Set problem is $NP$-hard.

翻译：我们研究了见证集问题，它是经典美术馆问题的自然对偶。在见证集问题中，我们给定一个多边形 $P$ 和一个整数 $k$ 作为输入，目标是判断 $P$ 是否包含一个大小至少为 $k$ 的见证集。集合 $X\subseteq P$ 被称为见证集，如果 $P$ 中每个点至多能被 $X$ 中的一个点看到。对于简单多边形，我们证明见证集问题属于 $NP$ 和 $XP$。这与其对偶问题——美术馆问题——形成鲜明对比，后者近期被 Abrahamsen 等人证明为 $\exists \mathbb{R}$-完全问题，因此除非 $NP=\exists \mathbb{R}$，它既不属于 $NP$，也不允许多项式大小的离散化。反之，我们证明简单多边形上的见证集问题允许一个大小为 $n^{f(k)}$ 的有限离散化（其中 $f$ 为某个函数）。作为对比，即使对于简单多边形，Efrat 和 Har-Peled 借助实代数几何工具给出了一个运行时间为 $n^{O(k)}$ 的美术馆问题算法，若不借助这些工具很难得到同类算法。另一方面，我们解决见证集问题的方法纯粹基于组合学且依赖于离散化，从而得到一个 $n^{f(k)}$ 时间算法。尽管 Amit 等人在十五年前声称见证集问题是 $NP$-难的，但并未提供证明或参考文献。我们证明见证集问题的离散版本——即见证集必须从给定的有限点集 $Q$ 中选择（而非允许在多边形内任意选择），称为离散见证集问题——是 $NP$-完全的，即使输入限制为带孔直线多边形。然而，对于简单多边形，Das 等人给出了离散见证集问题的多项式时间算法。因此，见证集问题是否为 $NP$-难仍是未解问题。