Obstacles, Slopes, and Tic-Tac-Toe: An excursion in discrete geometry and combinatorial game theory
| Autor: | Mukkamala, V S Padmini |
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| Rok vydání: | 2011 |
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| Druh dokumentu: | Working Paper |
| Popis: | 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. Comment: This is Padmini Mukkamala's PhD thesis |
| Databáze: | arXiv |
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