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:在艺术画廊问题中,我们得到了一个具有合理坐标和整数美元的封闭多边形美元。我们被问及是否有可能找到一套(警卫)美元大小的一套(警卫)美元美元。我们被问及是否有可能找到一套(警卫)美元大小的一套(警卫)美元美元,这样一分一美元就能看到任何一分一美元。我们说两分一美元,如果线段的美元部分包含在1美元之内,美元一美元就可以看到对方。Abrahamsen、Adamaszek和Miltzow都表明,有一个多边形可以由三名警卫加以保护,但如果警卫需要合理的坐标,则需要四名警卫。换句话说,3号尺寸的最佳解决办法可能需要不合理。我们表明,二号的最佳解决办法可能需要不合理。请注意,任何能够用一名警卫加以保护的多元形都有一个合理的坐标。因此,我们的工作缩小了可能发生不合理的警卫时的距离。