NP-membership for the boundary-boundary art-gallery problem

TL;DR

The paper investigates the boundary-boundary art-gallery problem, proving NP-membership by reducing to a continuous constraint satisfaction problem called $\text{M}-2\text{SAT}$, where constraints are piecewise fractional-linear. It develops a robust continuous inference system, including function-composition rules and symmetry considerations, and shows refutation completeness. A key technical advance is showing that any satisfiable instance admits a solution with coordinates of the form $p+q\sqrt{r}$ with polynomially bounded bit-size, enabling polynomial-time verification and a quasi-polynomial-time algorithm for $\text{M}-2\text{SAT}$. The authors also provide a concrete irrational-coordinate example and introduce bounded-depth refutations via infinite compositions and path compression, offering a new toolkit for 2-variable continuous CSPs and their applications to geometry-inspired decision problems.

Abstract

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. We show that this art-gallery variant is in NP. 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.

Figures (12)

• Figure 1: Some examples of constraints of the form $x\le f(y)$ for different $f$.
• Figure 2: Left: the portion of the boundary seen by a single guard is a union of at most $n$ connected pieces, since any two different pieces are separated by a vertex of the polygon. There are at most $n$ guards, so in a guarding configuration, each edge is covered by a union of at most $n^2$ intervals that are each visible to a single guard. Right: we guess that a guard at position $\mathbf{a}x+\mathbf{b}$ sees the interval from $\mathbf{c}y+\mathbf{d}$ to $\mathbf{c}z+\mathbf{d}$. In order for this to be true, the vertex $\mathbf{v}$ of $P$ must be on the left side of the $\mathbf{a}x+\mathbf{b}$ to $\mathbf{c}y+\mathbf{d}$, creating a fractional-linear constraint.
• Figure 3: The case where the third inference rule is needed. There appear to be points that satisfy both constraints, but none where $x=-(-x)$. The other inference rules don't know about the relationship between $x$ and $-x$ as literals, so aren't powerful enough to refute this case.
• Figure 4: Illustration of $f^{\infty}$
• Figure 5: Left: an art gallery whose boundary can be guarded with three guards on the boundary, but only if irrational coordinates are allowed. Right: guarding this example with $3$ guards. There are $4$ nooks that represent constraints between pairs of guards.

Theorems & Definitions (38)

• Lemma 1
• proof
• Theorem 2
• Lemma 3
• proof
• Lemma 4
• proof
• proof
• Lemma 5
• proof