A Short Proof of the Toughness of Delaunay Triangulations

TL;DR

This paper provides a concise, self-contained proof that Delaunay triangulations of planar point sets are 1-tough, implying they have perfect matchings and confirming a conjecture about blocking points.

Contribution

It offers a new, shorter proof of the toughness of Delaunay triangulations and verifies a related conjecture on blocking points.

Findings

Delaunay triangulations are 1-tough for planar point sets.

They have guaranteed perfect matchings.

At least n points are needed to block any n-vertex Delaunay triangulation.

Abstract

We present a self-contained short proof of the seminal result of Dillencourt (SoCG 1987 and DCG 1990) that Delaunay triangulations, of planar point sets in general position, are 1-tough. An important implication of this result is that Delaunay triangulations have perfect matchings. Another implication of our result is a proof of the conjecture of Aichholzer et al. (2010) that at least $n$ points are required to block any $n$-vertex Delaunay triangulation

Equations

$$\angle prq+\angle psq<180.$$

$$d=180\cdot g+360\cdot b.$$

$$d<180\cdot e=180\cdot (|\mathcal{S}|+b+g-1).$$

$$180\cdot g+360\cdot b<180\cdot (|\mathcal{S}|+b+g-1),$$

Full text

1 Introduction

Let $P$ be a set of points in the plane that is in general position, i.e., no three points on a line and no four points on a circle. The Delaunay triangulation of $P$ is an embedded planar graph with vertex set $P$ that has a straight-line edge between two points $p,q\in P$ if and only if there exists a closed disk that has only $p$ and $q$ on its boundary and does not contain any other point of $P$. A graph is $1$-tough if for any $k$, the removal of $k$ vertices splits the graph into at most $k$ connected components. In 1987, Dillencourt proved the toughness of Delaunay triangulations.

Theorem 1** (Dillencourt [4]).**

Let $T$ be the Delaunay triangulation of a set of points in the plane in general position, and let $S\subseteq V(T)$. Then $T\setminus S$ has at most $|S|$ components.

Dillencourt’s proof of Theorem 1 is nontrivial and employs a large set of combinatorial and structural properties of (Delaunay) triangulations. Using the same proof idea, he showed that if $T$ is a Delaunay triangulation of an arbitrary point set in the plane (not necessarily in general position) then $T\setminus S$ has at most $|S|+1$ components. Combining this with Tutte’s classical theorem that characterizes graphs with perfect matchings [5], implies the following well-known result.

Theorem 2** (Dillencourt [4]).**

Every Delaunay triangulation has a perfect matching.

In this note we present a self-contained short proof of Theorem 1. To that end, we first present an upper bound on the maximum size of an independent set of $T$. To facilitate comparisons we use the same definitions and notations as in [4]. The number of elements of a set $S$ is denoted by $|S|$. For a graph $G$, the vertex set of $G$ is denoted by $V(G)$, and $|G|=|V(G)|$.

Every interior face of $T$ is a triangle, and the boundary of $T$ is a convex polygon; see Figure 1(a). An edge is called a boundary edge if it is on the boundary of $T$, and is called an interior edge otherwise. For any interior edge $(p,q)\in T$ between two faces $pqr$ and $pqs$ it holds that

[TABLE]

2 A combinatorial and a structural property

A direct implication of Theorem 1 gives the upper bound $\lfloor(|T|+1)/2\rfloor$ on the size of any independent set of $T$; see e.g. [1]. We present a different self-contained proof for a slightly better bound.

Theorem 3**.**

Let $T$ be the Delaunay triangulation of a set of points in the plane in general position, and let $I$ be an independent set of $T$. Then $|I|\leqslant\lfloor|T|/2\rfloor$, and this bound is tight.

Proof.

This upper bound is tight as any maximum independent set in the $n$-vertex Delaunay triangulation of Figure 2(b) has exactly $\lfloor n/2\rfloor$ vertices (regardless of the parity of $n$).

Now we prove the upper bound. Set $S:=V(T)\setminus I$, and let $u$ be a vertex of $S$ that is on the boundary of $T$ (observe that such a vertex exists). Let $v,w\notin V(T)$ be two points in the plane such that (i) $T$ lies in the triangle $(u,v,w)$ and (ii) neither of $v$ and $w$ lies in the disks that introduce edges of $T$; see Figure 1(b). Let $\mathcal{T}$ be the Delaunay triangulation of $V(T)\cup\{v,w\}$. Our choice of $v$ and $w$ ensures that any edge of $T$ is also an edge of $\mathcal{T}$, and thus $T\subset\mathcal{T}$. Set $\mathcal{S}:=S\cup\{v,w\}$. In the rest of the proof we show that $|I|\leqslant|\mathcal{S}|-2$. This implies that $|I|\leqslant|S|$ (because $|\mathcal{S}|=|S|+2$) which in turn implies that $|I|\leqslant\lfloor|T|/2\rfloor$ (because $|T|=|S|+|I|$, and $|I|$ and $|T|$ are integers).

To show that $|I|\leqslant|\mathcal{S}|-2$ we use a counting argument similar to that of [4, Lemma 3.8]. Let $\mathcal{T}[\mathcal{S}]$ be the subgraph of $\mathcal{T}$ that is induced by $\mathcal{S}$. In other words, $\mathcal{T}[\mathcal{S}]$ is the resulting graph after removing vertices of $I$ and their incident edges from $\mathcal{T}$. Since $\mathcal{T}$ is a triangulation and $I$ does not contain boundary vertices of $\mathcal{T}$, the removal of every vertex of $I$ creates a hole (a new face which is the union of original faces) whose boundary is a simple polygon. All edges of this polygon belong to $\mathcal{T}[\mathcal{S}]$ because $I$ is an independent set. Therefore, $\mathcal{T}[\mathcal{S}]$ is a connected plane graph, the boundaries of its interior faces are simple polygons, and the boundary of its outer face is the triangle $(u,v,w)$; see Figure 1(b). Each interior face of $\mathcal{T}[\mathcal{S}]$ contains either no point of $I$ or exactly one point of $I$. Interior faces that do not contain any point of $I$ are called good faces, and interior faces that contain a point of $I$ are called bad faces. Each good face is a triangle. Let $g$ and $b$ denote the number of good and bad faces, respectively. Thus the number of interior faces is $g+b$.

Since $|I|=b$, it suffices to show that $b\leqslant|\mathcal{S}|-2$. To do so, we assign to each edge $(p,q)\in\mathcal{T}[\mathcal{S}]$ certain distinguished angles. If $(p,q)$ is an interior edge then we distinguish the two angles of $\mathcal{T}$ that are opposite to $(p,q)$, and if $(p,q)$ is a boundary edge then we distinguish the unique angle of $\mathcal{T}$ that is opposite to $(p,q)$, as in Figure 1(b). Let $d$ be the total measure of all distinguished angles. We compute $d$ in two different ways: once with respect to the number of faces of $\mathcal{T}[\mathcal{S}]$ and once with respect to the number of edges of $\mathcal{T}[\mathcal{S}]$. Each good face contains three distinguished angles, their sum is $180^{\circ}$. The sum of the distinguished angles in each bad face is $360^{\circ}$ because these angles are anchored at the removed vertex in the face. Therefore

[TABLE]

Now we compute $d$ with respect to the number of edges of $\mathcal{T}[\mathcal{S}]$ which we denote by $e$. By Euler’s formula, we have $e=|\mathcal{S}|+b+g-1$. By Inequality (1), the sum of (at most two) distinguished angles assigned to each edge is less than $180^{\circ}$. Therefore

[TABLE]

Combining (2) and (3), we have

[TABLE]

which simplifies to $b<|\mathcal{S}|-1$. Since $b$ and $|\mathcal{S}|$ are integers, $b\leqslant|\mathcal{S}|-2$. ∎

Our proof of Theorem 1 employs Theorem 3 and the following structural property of Delaunay triangulations presented by the author [3]. For the sake of completeness we repeat its proof.

Theorem 4**.**

Let $T$ be the Delaunay triangulation of a set of points in the plane in general position. Let $p$ and $q$ be two vertices of $T$ and let $D$ be any closed disk that has on its boundary only vertices $p$ and $q$. Then there exists a path, between $p$ and $q$ in $T$, that lies in $D$.

Proof.

The proof is by induction on the number of vertices in $D$. If there is no vertex of $V(T)\setminus\{p,q\}$ in the interior of $D$, then $(p,q)$ is an edge of $T$, and so is a desired path. Assume that there exists a vertex $r\in V(T)\setminus\{p,q\}$ in the interior of $D$. Let $c$ be the center of $D$. Consider the ray $\overrightarrow{pc}$ emanating from $p$ and passing through $c$. Fix $D$ at $p$ and then shrink it along $\overrightarrow{pc}$ until $r$ lies on its boundary; see Figure 1(c). Denote the resulting disk $D_{pr}$, and notice that it lies fully in $D$. Compute the disk $D_{qr}$ in a similar fashion by shrinking $D$ along $\overrightarrow{qc}$. The disk $D_{pr}$ does not contain $q$ and the disk $D_{qr}$ does not contain $p$. By induction hypothesis there exists a path, between $p$ and $r$ in $T$, that lies in $D_{pr}$, and similarly there exists a path, between $q$ and $r$ in $T$, that lies in $D_{qr}$. The union of these two paths contains a path, between $p$ and $q$ in $T$, that lies in $D$. ∎

3 Proof of Theorem 1

Recall $T$ and $S$. Pick an arbitrary representative vertex from each component of $T\setminus S$, and let $C$ be the set of these vertices. The number of components is $|C|$. Consider the Delaunay triangulation $T^{\prime}$ of $S\cup C$. Observe that $C$ is an independent set of $T$. We prove by contradiction that $C$ is also an independent set of $T^{\prime}$. Assume that there exists an edge $(c_{1},c_{2})\in T^{\prime}$ such that $c_{1},c_{2}\in C$. Since $T^{\prime}$ is a Delaunay triangulation, by definition there exists a closed disk $D$ that has only $c_{1}$ and $c_{2}$ on its boundary and does not contain any other point of $S\cup C$. Now consider $T$ and $D$. By Theorem 4 there exists a path between $c_{1}$ and $c_{2}$ in $T$, that lies in $D$. Since $D$ does not contain any point of $S$, all edges of this path belong to $T\setminus S$. This contradicts the fact that $c_{1}$ and $c_{2}$ belong to different components of $T\setminus S$. Therefore $C$ is an independent set of $T^{\prime}$. By Theorem 3, we have $|C|\leqslant|T^{\prime}|/2$. This and the fact that $|T^{\prime}|=|S|+|C|$ imply that $|C|\leqslant|S|$.

4 Blocking Delaunay triangulations

In this section, we use Theorem 3 and prove the conjecture of Aichholzer et al. [1] that at least $n$ points are required to block any $n$-vertex Delaunay triangulation. Let $P$ be a set of points in the plane and let $T$ be the Delaunay triangulation of $P$. A point set $B$ blocks or stabs $T$ if in the Delaunay triangulation of $P\cup B$ there is no edge between two points of $P$. In other words, every disk that introduces an edge in $T$ contains a point of $B$. Throughout this section we assume that $P\cup B$ is in general position.

In 2010, Aronov et al. [2] showed that $2n$ points are sufficient to block any $n$-vertex Delaunay triangulation, and if the vertices are in convex position then $4n/3$ points suffice. These bounds have been improved by Aichholzer et al. [1] (2010) to $3n/2$ and $5n/4$, respectively.

For the lower bound, Aronov et al. [2] showed the existence of $n$-vertex Delaunay triangulations that require $n$ points to be blocked, for example see Figure 2(a) in which every disk (representing a Delaunay edge) requires a unique point to be blocked as the disks are interior disjoint. Aichholzer et al. [1] proved that at least $n-1$ points are necessary to block any $n$-vertex Delaunay triangulations, and stated the following conjecture.

Conjecture 1**.**

For any point set $P$ in the plane in convex position, $|P|$ points are necessary and sufficient to block the Delaunay triangulation of $P$.

An implication of Theorem 3 proves the necessity of $|P|$ blocking points in Conjecture 1 (even if $P$ is in general position); the sufficiency remains open.

Theorem 5**.**

Let $P\cup B$ be any set of points in the plane in general position such that $B$ blocks the Delaunay triangulation of $P$. Then $|B|\geqslant|P|$, and this bound is tight.

Proof.

Consider the Delaunay triangulation $T$ of $P\cup B$. Since $B$ blocks the Delaunay triangulation of $P$, the removal of $B$ from $T$ leaves exactly $|P|$ components each consisting of a single point of $P$. Thus $P$ is an independent set of $T$. By Theorem 3, we have $|P|\leqslant\lfloor|T|/2\rfloor\leqslant|T|/2$ which implies that $|B|\geqslant|P|$ (because $|T|=|P|+|B|$).

To verify the tightness of this bound, consider a set of $n$ points in convex position where $n-1$ points are at distances approximately 1 from one point, say $p$, so that no four points lie on a circle. In the Delaunay triangulation of this point set, $p$ is connected to all other points, as depicted in Figure 2(b). This Delaunay triangulation can be blocked by $n$ points that are placed outside the convex hull: two points are placed very close to $p$ and $n-2$ points are placed very close to the $n-2$ convex hull edges that are not incident to $p$. A similar placement has also been used in [1] and [2]. ∎