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，我们给定一个具有有理坐标的闭合多边形$P$和一个整数$k$。我们需要判断是否存在一个大小为$k$的守卫集合$G$，使得多边形$P$内的任意一点$p$都能被$G$中的某个点看到。如果两点$p$、$q$之间的线段$pq$完全位于$P$内部，则称这两点相互可见。Abrahamsen、Adamaszek和Miltzow的研究表明，存在某个多边形可以用三个守卫守护，但如果要求守卫具有有理坐标，则需要四个守卫。换言之，规模为三的最优解可能需要无理数坐标的守卫。我们证明规模为二的最优解也可能需要无理数坐标的守卫。值得注意的是，众所周知，任何可由单个守卫守护的多边形都存在具有有理坐标的最优守卫布置。因此，我们的工作填补了关于无理数守卫可能出现的场景的认知空白。