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| Format: | Preprint |
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2026
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| Online Access: | https://arxiv.org/abs/2605.01592 |
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| _version_ | 1866917454708473856 |
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| author | Jana, Satyabrata Pal, Debabrata Roy, Bodhayan Roy, Sasanka |
| author_facet | Jana, Satyabrata Pal, Debabrata Roy, Bodhayan Roy, Sasanka |
| contents | We study the Witness Set problem, a natural dual to the classical Art Gallery problem. In the Witness Set problem, we are given a polygon $P$ and an integer $k$ as input, and the objective is to determine whether $P$ has a witness set of size at least $k$. A point set $X$ in $P$ is called a witness set if every point in $P$ is visible from at most one point in $X$.
For simple polygons, we show that Witness Set lies in both $NP$ and $XP$. This stands in sharp contrast to its dual, the Art Gallery problem, which was recently shown to be $\exists \mathbb{R}$-complete by Abrahamsen et al. and is therefore neither in $NP$ nor admits a polynomial-size discretization unless $NP=\exists \mathbb{R}$. In contrast, we prove that Witness Set for simple polygons admits a finite discretization of size $n^{f(k)}$ for some function $f$.
For comparison, even for simple polygons, Efrat and Har-Peled gave an algorithm for Art Gallery running in time $n^{O(k)}$ using tools from real algebraic geometry, and it appears difficult to obtain such algorithms without this machinery. On the other hand, our approach for Witness Set is purely combinatorial and relies on discretization, leading to an $n^{f(k)}$-time algorithm.
Although Amit et al. claimed more than fifteen years ago that Witness Set is $NP$-hard, no proof or reference was provided. We show that the discrete version of the Witness Set problem - where the witness set must be chosen from a given finite point set $Q$ (instead of allowing witnesses to be chosen anywhere in the polygon), referred to as Discrete Witness Set - is $NP$-complete, even when the input is restricted to rectilinear polygons with holes. However, for simple polygons, Discrete Witness Set admits a polynomial-time algorithm by Das et al. Thus, it remains an open question whether the Witness Set problem is $NP$-hard. |
| format | Preprint |
| id |
arxiv_https___arxiv_org_abs_2605_01592 |
| institution | arXiv |
| publishDate | 2026 |
| record_format | arxiv |
| spellingShingle | Witness Set: A Visibility Problem in $NP\cap XP$ Jana, Satyabrata Pal, Debabrata Roy, Bodhayan Roy, Sasanka Computational Geometry We study the Witness Set problem, a natural dual to the classical Art Gallery problem. In the Witness Set problem, we are given a polygon $P$ and an integer $k$ as input, and the objective is to determine whether $P$ has a witness set of size at least $k$. A point set $X$ in $P$ is called a witness set if every point in $P$ is visible from at most one point in $X$. For simple polygons, we show that Witness Set lies in both $NP$ and $XP$. This stands in sharp contrast to its dual, the Art Gallery problem, which was recently shown to be $\exists \mathbb{R}$-complete by Abrahamsen et al. and is therefore neither in $NP$ nor admits a polynomial-size discretization unless $NP=\exists \mathbb{R}$. In contrast, we prove that Witness Set for simple polygons admits a finite discretization of size $n^{f(k)}$ for some function $f$. For comparison, even for simple polygons, Efrat and Har-Peled gave an algorithm for Art Gallery running in time $n^{O(k)}$ using tools from real algebraic geometry, and it appears difficult to obtain such algorithms without this machinery. On the other hand, our approach for Witness Set is purely combinatorial and relies on discretization, leading to an $n^{f(k)}$-time algorithm. Although Amit et al. claimed more than fifteen years ago that Witness Set is $NP$-hard, no proof or reference was provided. We show that the discrete version of the Witness Set problem - where the witness set must be chosen from a given finite point set $Q$ (instead of allowing witnesses to be chosen anywhere in the polygon), referred to as Discrete Witness Set - is $NP$-complete, even when the input is restricted to rectilinear polygons with holes. However, for simple polygons, Discrete Witness Set admits a polynomial-time algorithm by Das et al. Thus, it remains an open question whether the Witness Set problem is $NP$-hard. |
| title | Witness Set: A Visibility Problem in $NP\cap XP$ |
| topic | Computational Geometry |
| url | https://arxiv.org/abs/2605.01592 |