The boundary-boundary art-gallery problem asks, given a polygon $P$ representing an art-gallery, for a minimal set of guards that can see the entire boundary of $P$ (the wall of the art gallery), where the guards must be placed on the boundary. That is, for each point on the boundary, there should be a line segment connecting it to one of the guards that is contained in $P$. We show that this art-gallery variant is in NP, even if the polygon can have holes. In order to prove this, we develop a constraint-propagation procedure for continuous constraint satisfaction problems where each constraint involves at most 2 variables. The X-Y variant of the art-gallery problem is the one where the guards must lie in X and need to see all of Y. Each of X and Y can be either the vertices of the polygon, the boundary of the polygon, or the entire polygon, giving 9 different variants. Previously, it was known that X-vertex and vertex-Y variants are all NP-complete and that the point-point, point-boundary, and boundary-point variants are $\exists \mathbb{R}$-complete [Abrahamsen, Adamaszek, and Miltzow, JACM 2021][Stade, SoCG 2025]. However, the boundary-boundary variant was only known to lie somewhere between NP and $\exists \mathbb{R}$. The X-vertex and vertex-Y variants can be straightforwardly reduced to discrete set-cover instances. In contrast, we give example to show that a solution to an instance of the boundary-boundary art-gallery problem sometimes requires placing guards at irrational coordinates, so it unlikely that the problem can be easily discretized.

翻译：边界-边界美术馆警卫问题要求：给定一个表示美术馆的多边形$P$，在边界上布置最少的警卫，使得他们能够看到$P$的整个边界（美术馆的墙壁）。也就是说，对于边界上的每个点，都应存在一条连接该点与某个警卫的线段，且该线段完全包含在$P$内部。我们证明，即使多边形允许存在孔洞，该美术馆问题的变体仍属于NP类。为证明这一点，我们针对连续约束满足问题开发了一种约束传播程序，其中每个约束最多涉及2个变量。美术馆问题的X-Y变体是指警卫必须位于X集合中，且需要看到Y集合的全部区域。X和Y各自可以是多边形的顶点集合、边界集合或整个多边形，从而产生9种不同的变体。此前已知X-顶点和顶点-Y两类变体均为NP完全问题，且点-点、点-边界和边界-点变体是$\exists \mathbb{R}$完全问题[Abrahamsen, Adamaszek, and Miltzow, JACM 2021][Stade, SoCG 2025]。然而，边界-边界变体仅被确认处于NP与$\exists \mathbb{R}$之间的某处。X-顶点和顶点-Y变体可直接归约为离散集合覆盖实例。相比之下，我们通过示例证明：边界-边界美术馆警卫问题实例的解有时需要将警卫布置在无理坐标点上，因此该问题不太可能被简单离散化。