The Spanning Ratio of the Directed $Θ_6$-Graph is 5

Abstract

Given a finite set $P\subset\mathbb{R}^2$, the directed Theta-6 graph, denoted $\vecΘ_6(P)$, is a well-studied geometric graph due to its close relationship with the Delaunay triangulation. The $\vecΘ_6(P)$-graph is defined as follows: the plane around each point $u\in P$ is partitioned into $6$ equiangular cones with apex $u$, and in each cone, $u$ is joined to the point whose projection on the bisector of the cone is closest. Equivalently, the $\vecΘ_6(P)$-graph contains an edge from $u$ to $v$ exactly when the interior of $\nabla_u^v$ is disjoint from $P$, where $\nabla_u^v$ is the unique equilateral triangle containing $u$ on a corner, $v$ on the opposite side, and whose sides are parallel to the cone boundaries. It was previously shown that the spanning ratio of the $\vecΘ_6(P)$-graph is between $4$ and $7$ in the worst case (Akitaya, Biniaz, and Bose \emph{Comput. Geom.}, 105-106:101881, 2022). We close this gap by showing a tight spanning ratio of 5. This is the first tight bound proven for the spanning ratio of any $\vecΘ_k(P)$-graph. Our lower bound models a long path by mapping it to a converging series. Our upper bound proof uses techniques novel to the area of spanners. We use linear programming to prove that among several candidate paths, there exists a path satisfying our bound.

Figures (18)

• Figure 1: Left: The dotted circle represents the unit disk in the standard Euclidean norm ($\|\cdot\|_2$). The shaded inner hexagon is the unit disk in the $\|\cdot\|_{ }$-norm. The outer tipped hexagon is the unit disk in the $\|\cdot\|_{ }$-norm. Right: The six cones about $p$ are labeled in counterclockwise order from $C_p^0$ below $p$. The dotted green bisector is $B_p^0$ and the red ray is $R_p^0$. The empty shaded triangle in $C_p^1$ is $\nabla_p^q$. Each shaded region is empty.
• Figure 2: Graph construction achieving a spanning ratio lower bound of $5-\epsilon$. The shortest path from $s$ to $t$ is the greedy path $s,a,b,c,p^0,q^0,p^1,q^1,...,,q^{k-1},p^k,t$. Top Left: The colour coding for edges based on direction.
• Figure 3: In Claim \ref{['claim:lb']}, we define $r^1$ and $r^2$ in order to show that $r$ approaches $(\frac{1}{2},\frac{\sqrt{3}}{2})$. The red lines have slope $\pm (\sqrt{3}-\delta)$, whereas the black lines have slope $\pm\sqrt{3}$. In this example, the line through $r^1 p^0$ has a larger vertical distance to $r$ than the line through $p^k r^1$. This distance is at most $\delta$. Since $r^1$ is to the right of $r$ in this example, $r^1$ must lie on the blue segment, which is inside the green disk centered at $r$ with radius at most $\delta$ in the $\|\cdot\|_1$-norm. If instead $r^1$ were to the left of $r$, then the diagram would be similar.
• Figure 4: Lemma \ref{['lem:hexthex']} with $v\in C_u^0$. The dotted vertical ray is the bisector $B_u^0$. The horizontal green dashed segment represents all points $p\in C_u^0$ such that $\|up\|_{ }=\|uv\|_{ }$. The red dashed segments represent all points $p\in C_u^0$ such that $\|up\|_{ }=\|uv\|_{ }$.
• Figure 5: For the proof of Lemma \ref{['lem:WithinR']}, we assume $i=0$ and show the case when $v\in C_u^4$ on the left and $v\in C_u^3$ on the right.

Theorems & Definitions (26)

• Lemma 1
• proof
• Claim 1
• proof : Proof
• Theorem 1
• Lemma 2
• proof
• Lemma 3
• proof
• Definition 1: Empty Region $R$