In the art gallery problem, we are given a closed polygon $P$, with rational coordinates and an integer $k$. We are asked whether it is possible to find a set (of guards) $G$ of size $k$ such that any point $p\in P$ is seen by a point in $G$. We say two points $p$, $q$ see each other if the line segment $pq$ is contained inside $P$. It was shown by Abrahamsen, Adamaszek, and Miltzow that there is a polygon that can be guarded with three guards, but requires four guards if the guards are required to have rational coordinates. In other words, an optimal solution of size three might need to be irrational. We show that an optimal solution of size two might need to be irrational. Note that it is well-known that any polygon that can be guarded with one guard has an optimal guard placement with rational coordinates. Hence, our work closes the gap on when irrational guards are possible to occur.

翻译：在美术馆问题中，给定一个具有有理数坐标的闭合多边形$P$和一个整数$k$，询问是否存在一个大小为$k$的警卫集合$G$，使得多边形$P$内的任意一点$p$都能被$G$中的某一点看到。这里，两点$p$和$q$相互可见的条件是线段$pq$完全包含在$P$内部。Abrahamsen、Adamaszek和Miltzow已证明，存在一个多边形可以用三个警卫守卫，但如果要求警卫坐标必须为有理数，则需要四个警卫。换言之，一个大小为三的最优解可能需要无理数坐标的警卫。本文证明，一个大小为二的最优解也可能需要无理数坐标的警卫。值得注意的是，众所周知，任何可以用一个警卫守卫的多边形都存在具有有理数坐标的最优警卫布置。因此，我们的工作填补了关于无理数坐标警卫可能出现的条件的研究空白。