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On the Largest Convexity Number of Co-Finite Sets in the Plane
arXiv:2601.00414v1 Announce Type: cross
Abstract: The convexity number of a set $X \subset \mathbb{R}^2$ is the minimum number of convex subsets required to cover it. We study the following question: what is the largest possible convexity number $f(n)$ of $\mathbb{R}^2 \setminus S$, where $S$ is a set of $n$ points in general position in the plane? We prove that for all $n \geq 4$, $\lfloor\frac{n+5}{2}\rfloor \leq f(n) \leq \frac{7n+44}{11}$. We also show that for every $n \geq 4$, if the points of $S$ are in convex position then the convexity number of $\mathbb{R}^2 \setminus S$ is $\lfloor\frac{n+5}{2}\rfloor$. This solves a problem of Lawrence and Morris [Finite sets as complements of finite unions of convex sets, Disc. Comput. Geom. 42 (2009), 206-218].
Abstract: The convexity number of a set $X \subset \mathbb{R}^2$ is the minimum number of convex subsets required to cover it. We study the following question: what is the largest possible convexity number $f(n)$ of $\mathbb{R}^2 \setminus S$, where $S$ is a set of $n$ points in general position in the plane? We prove that for all $n \geq 4$, $\lfloor\frac{n+5}{2}\rfloor \leq f(n) \leq \frac{7n+44}{11}$. We also show that for every $n \geq 4$, if the points of $S$ are in convex position then the convexity number of $\mathbb{R}^2 \setminus S$ is $\lfloor\frac{n+5}{2}\rfloor$. This solves a problem of Lawrence and Morris [Finite sets as complements of finite unions of convex sets, Disc. Comput. Geom. 42 (2009), 206-218].
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