Obstacles, Slopes, and Tic-Tac-Toe: An excursion in discrete geometry and combinatorial game theory

TL;DR

This paper explores geometric representations of graphs, establishing bounds on slopes and obstacles needed for certain graph drawings, and analyzes strategies in a combinatorial game on integer lattices.

Contribution

It proves that all cubic graphs can be drawn with only four slopes, characterizes when four slopes suffice for complete graphs, and provides bounds on obstacle numbers and game strategies.

Findings

Cubic graphs can be drawn with four slopes , , 0, .

Graphs can have obstacle numbers as large as n/log n.

Breaker can force a draw with a pairing strategy using at most 2n+o(n) marks.

Abstract

A drawing of a graph is said to be a {\em straight-line drawing} if the vertices of $G$ are represented by distinct points in the plane and every edge is represented by a straight-line segment connecting the corresponding pair of vertices and not passing through any other vertex of $G$. The minimum number of slopes in a straight-line drawing of $G$ is called the slope number of $G$. We show that every cubic graph can be drawn in the plane with straight-line edges using only the four basic slopes $\{0,\pi/4,\pi/2,-\pi/4\}$. We also prove that four slopes have this property if and only if we can draw $K_4$ with them. Given a graph $G$, an {\em obstacle representation} of $G$ is a set of points in the plane representing the vertices of $G$, together with a set of obstacles (connected polygons) such that two vertices of $G$ are joined by an edge if and only if the corresponding points can be connected by a segment which avoids all obstacles. The {\em obstacle number} of $G$ is the minimum number of obstacles in an obstacle representation of $G$. We show that there are graphs on $n$ vertices with obstacle number $\Omega({n}/{\log n})$. We show that there is an $m=2n+o(n)$, such that, in the Maker-Breaker game played on $\Z^d$ where Maker needs to put at least $m$ of his marks consecutively in one of $n$ given winning directions, Breaker can force a draw using a pairing strategy. This improves the result of Kruczek and Sundberg who showed that such a pairing strategy exits if $m\ge 3n$. A simple argument shows that $m$ has to be at least $2n+1$ if Breaker is only allowed to use a pairing strategy, thus the main term of our bound is optimal.