Bounding and computing obstacle numbers of graphs

Martin Balko, Steven Chaplick, Robert Ganian, Siddharth Gupta, Michael (M.) Hoffmann, Pavel Valtr, Alexander Wolff

Research output: Working paper / PreprintPreprint

Abstract

An obstacle representation of a graph $G$ consists of a set of pairwise disjoint simply-connected closed regions and a one-to-one mapping of the vertices of $G$ to points such that two vertices are adjacent in $G$ if and only if the line segment connecting the two corresponding points does not intersect any obstacle. The obstacle number of a graph is the smallest number of obstacles in an obstacle representation of the graph in the plane such that all obstacles are simple polygons. It is known that the obstacle number of each $n$-vertex graph is $O(n \log n)$ [Balko, Cibulka, and Valtr, 2018] and that there are $n$-vertex graphs whose obstacle number is $\Omega(n/(\log\log n)^2)$ [Dujmovi\'c and Morin, 2015]. We improve this lower bound to $\Omega(n/\log\log n)$ for simple polygons and to $\Omega(n)$ for convex polygons. To obtain these stronger bounds, we improve known estimates on the number of $n$-vertex graphs with bounded obstacle number, solving a conjecture by Dujmovi\'c and Morin. We also show that if the drawing of some $n$-vertex graph is given as part of the input, then for some drawings $\Omega(n^2)$ obstacles are required to turn them into an obstacle representation of the graph. Our bounds are asymptotically tight in several instances. We complement these combinatorial bounds by two complexity results. First, we show that computing the obstacle number of a graph $G$ is fixed-parameter tractable in the vertex cover number of $G$. Second, we show that, given a graph $G$ and a simple polygon $P$, it is NP-hard to decide whether $G$ admits an obstacle representation using $P$ as the only obstacle.
Original languageEnglish
DOIs
Publication statusPublished - 30 Jun 2022

Keywords

  • cs.CG
  • math.CO

Fingerprint

Dive into the research topics of 'Bounding and computing obstacle numbers of graphs'. Together they form a unique fingerprint.

Cite this