Plane and Planarity Thresholds for Random Geometric Graphs

Abstract

A random geometric graph, $G(n,r)$, is formed by choosing n points independently and uniformly at random in a unit square; two points are connected by a straight-line edge if they are at Euclidean distance at most r. For a given constant k, we show that $n^{\frac{-k}{2k-2}}$ is a distance threshold function for $G(n,r)$ to have a connected subgraph on k points. Based on that, we show that $n^{-2/3}$ is a distance threshold function for $G(n,r)$ to be plane, and $n^{-5/8}$ is a distance threshold function for $G(n,r)$ to be planar.

Research supported by NSERC.

1 Introduction

Wireless networks are usually modeled as disk graphs in the plane. Given a set P of points in the plane and a positive parameter r, the disk graph is the geometric graph with vertex set P which has a straight-line edge between two points $p,q\in P$ if and only if $|pq|\le r$, where |pq| denotes the Euclidean distance between p and q. If $r=1$, then the disk graph is referred to as unit disk graph. A random geometric graph, denoted by $G(n,r)$, is a geometric graph formed by choosing n points independently and uniformly at random in a unit square; two points are connected by a straight-line edge if and only if they are at Euclidean distance at most r, where $r=r(n)$ is a function of n and $r \rightarrow 0$ as $n\rightarrow \infty$.

We say that two line segments in the plane cross each other if they have a point in common that is interior to both edges. Two line segments are non-crossing if they do not cross. Note that two non-crossing line segments may share an endpoint. A geometric graph is said to be plane if its edges do not cross, and non-plane, otherwise. A graph is planar if and only if it does not contain $K_5$ (the complete graph on 5 vertices) or $K_{3,3}$ (the complete bipartite graph on six vertices partitioned into two parts each of size 3) as a minor. A non-planar graph is a graph which is not planar.

A graph property $\mathcal {P}$ is increasing if a graph G satisfies $\mathcal {P}$, then by adding edges to G, the property $\mathcal {P}$ remains valid in G. Similarly, $\mathcal {P}$ is decreasing if a graph G satisfies $\mathcal {P}$, then by removing edges from G, the property $\mathcal {P}$ remains valid in G. $\mathcal {P}$ is called a monotone property if $\mathcal {P}$ is either increasing or decreasing. Connectivity and “having a clique of size k” are increasing monotone properties, while planarity and “being plane” are decreasing monotone properties in $G(n,r)$, where the value of r increases.

By [13] any monotone property of a random geometric graphs has a threshold function. The thresholds in random geometric graphs are expressed by the distance r. In the sequel, the term w.h.p. (with high probability) is to be interpreted to mean that the probability tends to 1 as $n \rightarrow \infty$. For an increasing property $\mathcal {P}$, the threshold is a function t(n) such that if $r=o(t(n))$ then w.h.p. $\mathcal {P}$ does not hold in $G(n,r)$, and if $r=\omega (t(n))$ then w.h.p. $\mathcal {P}$ holds in $G(n,r)$. Symmetrically, for a decreasing property $\mathcal {P}$, the threshold is a function t(n) such that if $r=o(t(n))$ then w.h.p. $\mathcal {P}$ holds in $G(n,r)$, and if $r=\omega (t(n))$ then w.h.p. $\mathcal {P}$ does not hold in $G(n,r)$. Note that a threshold function may not be unique. It is well known that $\sqrt{\ln n/n}$ is a connectivity threshold for $G(n,r)$; see [14, 19, 20]. In this paper we investigate thresholds in random geometric graphs for having a connected subgraph of constant size, being plane, and being planar.

1.1 Related Work

Random graphs were first defined and formally studied by Gilbert in [10] and Erdös and Rényi [8]. It seems that the concept of a random geometric graph was first formally suggested by Gilbert in [11] and for that reason is also known as Gilbert’s disk model. These classes of graphs are known to have numerous applications as a model for studying communication primitives (broadcasting, routing, etc.) and topology control (connectivity, coverage, etc.) in idealized wireless sensor networks as well as extensive utility in theoretical computer science and many fields of the mathematical sciences.

An instance of Erdös-Rényi graph [8] is obtained by taking n vertices and connecting any two with probability p, independently of all other pairs; the graph derived by this scheme is denoted by $G_{n,p}$. In $G_{n,p}$ the threshold is expressed by the edge existence probability p, while in $G(n,r)$ the threshold is expressed in terms of r. In both random graphs and random geometric graphs, property thresholds are of great interest [4, 7, 9, 13, 18]. Note that edge crossing configurations in $G(n,r)$ have a geometric nature, and as such, have no analogues in the context of the Erdös-Rényi model for random graphs. However, planarity, and having a clique of specific size are of interest in both $G_{n,p}$ and $G(n,r)$.

Bollobás and Thomason [5] showed that any monotone property in random graphs has a threshold function. See also a result of Friedgut and Kalai [9], and a result of Bourgain and Kalai [6]. In the Erdös-Rényi random graph $G_{n,p}$, the connectivity threshold is $p = \log n/n$ and the threshold for having a giant component is $p= 1/n$; see [1]. The planarity threshold for $G_{n,p}$ is $p =1/n$; see [4, 23].

A general reference on random geometric graphs is [22]. There is extensive literature on various aspects of random geometric graphs of which we mention the related work on coverage by [15, 16] and a review on percolation, connectivity, coverage and colouring by [3]. As in random graphs, any monotone property in geometric random graphs has a threshold function [7, 13, 17, 18].

Random geometric graphs have a connectivity threshold of $\sqrt{\ln n/n}$; see [14, 19, 20]. Gupta and Kumar [14] provided a connectivity threshold for points that are uniformly distributed in a disk. By a result of Penrose [21], in $G(n,r)$, any threshold function for having no isolated vertex (a vertex of degree zero) is also a connectivity threshold function. Panchapakesan and Manjunath [19] showed that $\sqrt{\ln n/n}$ is a threshold for being an isolated vertex in $G(n,r)$. This implies that $\sqrt{\ln n/n}$ is a connectivity threshold for $G(n,r)$. For $k\ge 2$, the details on the k-connectivity threshold in random geometric graphs can be found in [21, 22]. Connectivity of random geometric graphs for points on a line is studied by Godehardt and Jaworski [12]. Appel and Russo [2] considered the connectivity under the $L_\infty$-norm.

1.2 Our Results

In this paper we investigate thresholds for some monotone properties in random geometric graphs. In Sect. 2 we show that for a constant k, the distance threshold for having a connected subgraph on k points is $n^{\frac{-k}{2k-2}}$. We show that the same threshold is valid for the existence of a clique of size k. Based on that, we prove the following thresholds for a random geometric graph to be plane or planar. In Sect. 3, we prove that $n^{-2/3}$ is a distance threshold for a random geometric graph to be plane. In Sect. 4, we prove that $n^{-5/8}$ is a distance threshold for a random geometric graph to be planar.

2 The Threshold for Having a Connected Subgraph on k Points

In this section, we look for the distance threshold for “existence of connected subgraphs of constant size”; this is an increasing property. For a given constant k, we show that $n^{\frac{-k}{2k-2}}$ is the threshold function for the existence of a connected subgraph on k points in $G(n,r)$. Specifically, we show that if $r=o(n^{\frac{-k}{2k-2}})$, then w.h.p. $G(n,r)$ has no connected subgraph on k points, and if $r=\omega (n^{\frac{-k}{2k-2}})$, then w.h.p. $G(n,r)$ has a connected subgraph on k points. We also show that the same threshold function holds for the existence of a clique of size k.

Theorem 1

Let $k\ge 2$ be an integer constant. Then, $n^{\frac{-k}{2k-2}}$ is a distance threshold function for $G(n,r)$ to have a connected subgraph on k points.

Proof

Let $P_1, \dots , P_{n\atopwithdelims ()k}$ be an enumeration of all subsets of k points in $G(n,r)$. Let $DG[P_i]$ be the subgraph of $G(n,r)$ that is induced by $P_i$. Let $X_i$ be the random variable such that

$$\begin{aligned} X_i = {\left\{ \begin{array}{ll} 1 &{} \quad \text {if } DG[P_i]\text { is connected,}\\ 0 &{} \quad \text {otherwise.}\\ \end{array}\right. } \end{aligned}$$

Let the random variable X count the number of sets $P_i$ for which $DG[P_i]$ is connected. It is clear that

$$\begin{aligned} X=\sum _{i=1}^{n\atopwithdelims ()k}X_i. \end{aligned}$$

(1)

Observe that $\mathrm {E}[X_i]=\Pr [X_i=1]$. Since the random variables $X_i$ have identical distributions, we have

$$\begin{aligned} \mathrm {E}[X]={n \atopwithdelims ()k}\mathrm {E}[X_1]. \end{aligned}$$

(2)

We obtain an upper bound and a lower bound for $\Pr [X_i=1]$. First, partition the unit square into squares of side equal to r. Let $\{s_1, \dots , s_{1/r^2}\}$ be the resulting set of squares. For a square $s_t$, let $S_t$ be the $kr\times kr$ square which has $s_t$ on its left bottom corner; see Fig. 1(a). $S_t$ contains at most $k^2$ squares each of side length r (each $S_t$ on the boundary of the unit square contains less than $k^2$ squares). Let $A_{i,t}$ be the event that all points in $P_i$ are contained in $S_t$. Observe that if $DG[P_i]$ is connected then $P_i$ lies in $S_t$ for some $t\in \{1,\dots , 1/{r^2}\}$. Therefore,

$$\mathrm{if }\,DG[P_i]\,\mathrm{is\, connected,\, then}\, (A_{i,1} \vee A_{i,2}\vee \dots \vee A_{i,1/{r^2}}),$$

and hence we have

$$\begin{aligned} \Pr [X_i=1]\le \sum _{t=1}^{1/{r^2}} \Pr [A_{i,t}]\le \sum _{t=1}^{1/{r^2}} (k^2r^2)^k=k^{2k} r^{2k-2}. \end{aligned}$$

(3)

Now, partition the unit square into squares with diagonal length equal to r. Each such square has side length equal to $r/\sqrt{2}$. Let $\{s_1, \dots , s_{2/r^2}\}$ be the resulting set of squares. Let $B_{i,t}$ be the event that all points of $P_i$ are in $s_t$. Observe that if all points of $P_i$ are in the same square, then $DG[P_i]$ is a complete graph and hence connected. Therefore,

$$\mathrm{if }\,(B_{i,1} \vee B_{i,2}\vee \dots \vee B_{i,2/{r^2}})\,\mathrm{, then }\, DG[P_i]\,\mathrm{is\, connected},$$

and hence we have

$$\begin{aligned} \Pr [X_i=1]\ge \sum _{t=1}^{2/{r^2}} \Pr [B_{i,t}]=\sum _{t=1}^{2/{r^2}} \left( \frac{r^2}{2}\right) ^k=\frac{1}{2^{k-1}} r^{2k-2}. \end{aligned}$$

(4)

Since $k\ge 2$ is a constant, Inequalities (3) and (4) and Eq. (2) imply that

$$\begin{aligned} \mathrm {E}[X_i]&= \varTheta (r^{2k-2}),\end{aligned}$$

(5)

$$\begin{aligned} \mathrm {E}[X]&=\varTheta (n^kr^{2k-2}). \end{aligned}$$

(6)

If $n\rightarrow \infty$ and $r=o(n^{\frac{-k}{2k-2}})$ we conclude that the following inequalities are valid

$$\begin{aligned} \Pr [X\ge 1]&\le \nonumber \mathrm {E}[X] \, \mathrm{(by Markov's\, Inequality)}\nonumber \\&= \nonumber \varTheta (n^k r^{2k-2})\, \mathrm{(by~(6))} \nonumber \\&= o(1). \end{aligned}$$

(7)

Therefore, w.h.p. $G(n,r)$ has no connected subgraph on k points.

Fig. 1.

(a) The square $S_t$ has $s_t$ on its left bottom corner. (b) The square $S_x$ which is centered at $s_x$.

In the rest of the proof, we assume that $r=\omega (n^{\frac{-k}{2k-2}})$. In order to show that w.h.p. $G(n,r)$ has at least one connected subgraph on k vertices, we show, using the second moment method [1], that $\Pr [X=0]\rightarrow 0$ as $n\rightarrow \infty$. Recall from Chebyshev’s inequality that

$$\begin{aligned} \Pr [X=0] \le \frac{\mathrm {Var}(X)}{\mathrm {E}[X]^2}. \end{aligned}$$

(8)

Therefore, in order to show that $\Pr [X=0] \rightarrow 0$, it suffices to show that

$$\begin{aligned} \frac{\mathrm {Var}(X)}{\mathrm {E}[X]^2}\rightarrow 0. \end{aligned}$$

(9)

In view of Identity (1) we have

$$\begin{aligned} \mathrm {Var}(X) = \sum _{1\le i,j\le {n\atopwithdelims ()k}} \mathrm {Cov}(X_i, X_j), \end{aligned}$$

(10)

where $\mathrm {Cov}(X_i,X_j)=\mathrm {E}[X_iX_j]-\mathrm {E}[X_i]\mathrm {E}[X_j]\le \mathrm {E}[X_iX_j]$. If $|P_i\cap P_j|=0$ then $DG[P_i]$ and $DG[P_j]$ are disjoint. Thus, the random variables $X_i$ and $X_j$ are independent, and hence $\mathrm {Cov}(X_i,X_j)=0$. It is enough to consider the cases when $P_i$ and $P_j$ are not disjoint. Assume $|P_i\cap P_j|=w$, where $w\in \{1,\dots ,k\}$. Thus, in view of Eq. (10), we have

$$\begin{aligned} \mathrm {Var}(X)&=\nonumber \sum _{w=1}^k\sum _{|P_i\cap P_j|=w} \mathrm {Cov}(X_i, X_j)\nonumber \\&\le \sum _{w=1}^k\sum _{|P_i\cap P_j|=w} \mathrm {E}[X_iX_j]. \end{aligned}$$

(11)

The computation of $\mathrm {E}[X_i,X_j]$ involves some geometric considerations which are being discussed in detail below. Since $X_i$ and $X_j$ are 0–1 random variables, $X_iX_j$ is a 0–1 random variable and

$$\begin{aligned} X_iX_j = {\left\{ \begin{array}{ll} 1 &{} \quad \text {if both } DG[P_i]\text { and }DG[P_j]\text { are connected,}\\ 0 &{} \quad \text {otherwise.}\\ \end{array}\right. } \end{aligned}$$

By the definition of the expected value we have

$$\begin{aligned} \mathrm {E}[X_iX_j]&=\nonumber \Pr [X_j=1|X_i=1]\Pr [X_i=1]\nonumber \\&=\Pr [X_j=1|X_i=1]\mathrm {E}[X_i]. \end{aligned}$$

(12)

By (5), $\mathrm {E}[X_i]= \varTheta (r^{2k-2})$. It remains to compute $\Pr [X_j=1|X_i=1]$, i.e., the probability that $DG[P_j]$ is connected given that $DG[P_i]$ is connected. Consider the k-tuples $P_i$ and $P_j$ under the condition that $DG[P_i]$ is connected. Let x be a point in $P_i\cap P_j$. Partition the unit square into squares of side length equal to r. Let $s_x$ be the square containing x. Let $S_x$ be the $(2k-1)r\times (2k-1)r$ square centered at $s_x$. $S_x$ contains at most $(2k-1)^2$ squares each of side length r (if $S_x$ is on the boundary of the unit square then it contains less than $(2k-1)^2$ squares); see Fig. 1(b). The area of $S_x$ is at most $(2kr)^2$, and hence the probability that a specific point of $P_j$ is in $S_t$ is at most $4k^2r^2$. Since $P_i$ and $P_j$ share w points, in order for $DG[P_j]$ to be connected, the remaining $k-w$ points of $P_j$ must lie in $S_x$. Thus, the probability that $DG[P_j]$ is connected given that $DG[P_i]$ is connected is at most $(4k^2r^2)^{k-w}\le c_wr^{2k-2w}$, for some constant $c_w>0$. Thus, $\Pr [X_j=1|X_i=1]\le c_wr^{2k-2w}$. In view of Eq. (12), we have

$$\begin{aligned} \mathrm {E}[X_iX_j]\le c'_w\cdot r^{2k-2w}\cdot r^{2k-2}= c'_wr^{4k-2w-2}, \end{aligned}$$

(13)

for some constant $c'_w>0$.

Since $P_i$ and $P_j$ are k-tuples which share w points, $|P_i\cup P_j|=2k-w$. There are ${n \atopwithdelims (){2k-w}}$ ways to choose $2k-w$ points for $P_i\cup P_j$. Since we choose w points for $P_i\cap P_j$, $k-w$ points for $P_i$ alone, and $k-w$ points for $P_j$ alone, there are ${{2k-w}\atopwithdelims (){w, k-w, k-w}}$ ways to split the $2k-w$ chosen points into $P_i$ and $P_j$. Based on this and Inequality (13), Inequality (11) turns out to

$$\begin{aligned} \mathrm {Var}(X)&\le \nonumber \sum _{w=1}^k\sum _{|P_i\cap P_j|=w} \mathrm {E}[X_iX_j]\\&\le \nonumber \sum _{w=1}^{k} {n \atopwithdelims (){2k-w}}{{2k-w}\atopwithdelims (){w, k-w, k-w}} c'_wr^{4k-2w-2}\\&\le \nonumber \sum _{w=1}^{k} c''_w{n^{2k-w}}r^{4k-2w-2}. \end{aligned}$$

for some constants $c''_w>0$. Consider (9) and note that by (6), $\mathrm {E}[X]^2\ge c''n^{2k}r^{4k-4}$, for some constant $c''>0$. Thus,

$$\begin{aligned} \frac{\mathrm {Var}(X)}{\mathrm {E}[X]^2}&\le \nonumber \sum _{w=1}^{k} \frac{c''_w{n^{2k-w}}r^{4k-2w-2}}{c''n^{2k}r^{4k-4}} = \nonumber \sum _{w=1}^{k} \frac{c''_w}{c''}\cdot \frac{1}{n^{w}r^{2w-2}}\nonumber \\&= \frac{c''_1}{c''}\cdot \frac{1}{n^{1}r^{0}}+\frac{c''_2}{c''}\cdot \frac{1}{n^{2}r^{2}}+\dots +\frac{c''_k}{c''}\cdot \frac{1}{n^{k}r^{2k-2}} \end{aligned}$$

(14)

Since $r=\omega (n^{\frac{-k}{2k-2}})$, all terms in (14) tend to zero. This proves the convergence in (9). Thus, $\Pr [X=0]\rightarrow 0$ as $n\rightarrow \infty$. This implies that if $r=\omega (n^{\frac{-k}{2k-2}})$, then $G(n,r)$ has a connected subgraph on k vertices with high probability.    $\blacksquare$

In the following theorem we show that if $k=O(1)$, then $n^{\frac{-k}{2k-2}}$ is also a threshold for $G(n,r)$ to have a clique of size k; this is an increasing property.

Theorem 2

Let $k\ge 2$ be an integer constant. Then, $n^{\frac{-k}{2k-2}}$ is a distance threshold function for $G(n,r)$ to have a clique of size k.

Proof

By Theorem 1, if $r=o(n^{\frac{-k}{2k-2}})$, then w.h.p. $G(n,r)$ has no connected subgraph on k vertices, and hence it has no clique of size k. This proves the first statement. We prove the second statement by adjusting the proof of Theorem 1, which is based on the second moment method. Assume $r=\omega (n^{\frac{-k}{2k-2}})$. Let $P_1,\dots , P_{n\atopwithdelims ()k}$ be an enumeration of all subsets of k points. Let $X_i$ be equal to 1 if $DG[P_i]$ is a clique, and 0 otherwise. Let $X=\sum X_i$.

Partition the unit square into a set $\{s_1,\dots ,s_{1/{r^2}}\}$ of squares of side length r. Let $S_t$ be the $2r\times 2r$ square which has $s_t$ on its left bottom corner. If $DG[P_i]$ is a clique then $P_i$ lies in $S_t$ for some $t\in \{1,\dots , 1/{r^2}\}$. Therefore,

$$ \Pr [X_i=1]\le 4^{k} r^{2k-2}. $$

Now, partition the unit square into a set $\{s_1,\dots ,s_{2/{r^2}}\}$ of squares with diagonal length r. If all points of $P_i$ fall in the square $s_t$, then $DG[P_i]$ is a clique. Thus,

$$ \Pr [X_i=1]\ge \frac{1}{2^{k-1}} r^{2k-2}. $$

Since $k\ge 2$ is a constant, we have

$$\begin{aligned} \mathrm {E}[X_i]&=\nonumber \varTheta (r^{2k-2}),\\ \mathrm {E}[X]&=\nonumber \varTheta (n^kr^{2k-2}). \end{aligned}$$

In view of Chebyshev’s inequality we need to show that $\frac{\mathrm {Var}(X)}{\mathrm {E}[X]^2}$ tends to 0 as n goes to infinity. We bound $\mathrm {Var}(X)$ from above by Inequality (11). Consider the k-tuples $P_i$ and $P_j$ under the condition that $DG[P_i]$ is a clique. Let $|P_i\cap P_j|=w$, and let x be a point in $P_i\cap P_j$. Partition the unit square into squares of side length r. Let $s_x$ be the square containing x. Let $S_x$ be the $3r\times 3r$ square centered at $s_x$. In order for $DG[P_j]$ to be a clique, the remaining $k-w$ points of $P_j$ must lie in $S_x$. Thus,

$$ \mathrm {E}[X_iX_j]\le c'_wr^{4k-2w-2}, $$

for some constant $c'_w>0$. By a similar argument as in the proof of Theorem 1, we can show that for some constants $c'', c''_w>0$ the followings inequalities are valid:

$$\begin{aligned} \mathrm {Var}(X)&\le \nonumber \sum _{w=1}^{k} c''_w{n^{2k-w}}r^{4k-2w-2},\\ \frac{\mathrm {Var}(X)}{\mathrm {E}[X]^2}&\le \nonumber \sum _{w=1}^{k} \frac{c''_w}{c''}\cdot \frac{1}{n^{w}r^{2w-2}}. \end{aligned}$$

Since $r=\omega (n^{\frac{-k}{2k-2}})$, the last inequality tends to 0 as n goes to infinity. This completes the proof for the second statement.    $\blacksquare$

As a direct consequence of Theorem 2, we have the following corollary.

Corollary 1

$n^{-1}$ is a threshold for $G(n,r)$ to have an edge, and $n^{-\frac{3}{4}}$ is a threshold for $G(n,r)$ to have a triangle.

3 The Threshold for $G(n,r)$ to be Plane

In this section we investigate the threshold for a random geometric graph to be plane; this is a decreasing property. Recall that $G(n,r)$ is plane if no two of its edges cross. As a warm-up exercise we first prove a simple result which is based on the connectivity threshold for random geometric graphs, which is known to be $\sqrt{\ln n/n}$.

Fig. 2.

An square of diameter r which is partitioned into nine sub-squares.

Theorem 4

If $r \ge \sqrt{\frac{c \ln n}{n}}$, with $c \ge 36$, then w.h.p. $G(n,r)$ is not plane.

Proof

In order to prove that w.h.p. $G(n,r)$ is not plane, we show that w.h.p. it has a pair of crossing edges. Partition the unit square into squares each with diagonal length r. Then subdivide each such square into nine sub-squares as depicted in Fig. 2. There are $\frac{18}{r^2}$ sub-squares, each of side length $\frac{r}{3 \sqrt{2}}$. The probability that no point lies in a specific sub-square is $(1 - \frac{r^2}{18})^n$. Thus, the probability that there exists an empty sub-square is at most

$$ \frac{18}{r^2} \left( 1 - \frac{r^2}{18} \right) ^n \le n \left( 1 - \frac{c \ln n}{18 n} \right) ^n \le n^{1-c/18} \le \frac{1}{n}, $$

when $c \ge 36$. Therefore, with probability at least $1-\frac{1}{n}$ all sub-squares contain points. By choosing four points a, b, c, and d as depicted in Fig. 2, it is easy to see that the edges (a, b) and (c, d) cross. Thus, w.h.p. $G(n,r)$ has a pair of crossing edges, and hence w.h.p. it is not plane.    $\blacksquare$

In fact, Theorem 3 ensures that w.h.p. there exists a pair of crossing edges in each of the squares. This implies that there are $\varOmega \left( \frac{n}{\ln n} \right)$ disjoint pair of crossing edges, while for $G(n,r)$ to be not plane we need to show the existence of at least one pair of crossing edges. Thus, the value of r provided by the connectivity threshold seems rather weak. By a different approach, in the rest of this section we show that $n^{-\frac{2}{3}}$ is the correct threshold.

Lemma 1

Let (a, b) and (c, d) be two crossing edges in $G(n,r)$, and let Q be the convex quadrilateral formed by a, b, c, and d. Then, two adjacent sides of Q are edges of $G(n,r)$.

Proof

Refer to Fig. 3. At least one of the angles of Q, say $\angle cad$, is bigger than or equal to $\pi /2$. It follows that in the triangle $\triangle cad$ the side cd is the longest, i.e., $|cd|\ge \max \{|ac|,|ad|\}$. Since $|cd|\le r$, both |ac| and |ad| are at most r. Thus, ac and ad—which are adjacent—are edges of $G(n,r)$.    $\blacksquare$

Fig. 3.

(a) Illustration of Lemma 1. (b) Crossing edges (a, b) and (c, d) form an anchor.

In the proof of Lemma 1, a is connected to b, c, and d. So the distance between a to each of b, c, and d is at most r. Thus, we have the following corollary.

Corollary 2

The endpoints of every two crossing edges in $G(n,r)$ are at distance at most 2r from each other. Moreover, there exists an endpoint which is within distance r from other endpoints.

Based on the proof of Lemma 1, we define an anchor as a set $\{a, b, c, d\}$ of four points in $G(n,r)$ such that three of them form a triangle, say $\triangle cad$, and the fourth vertex, b, is connected to a by an edge which crosses cd; see Fig. 3(b). We call a as the crown of the anchor. The crown is within distance r from the other three points. Note that bc and bd may or may not be edges of $G(n,r)$. In view of Lemma 1, two crossing edges in $G(n,r)$ form an anchor. Conversely, every anchor in $G(n,r)$ introduces a pair of crossing edges.

Observation 1

$G(n,r)$ is plane if and only if it has no anchor.

Theorem 5

$n^{-\frac{2}{3}}$ is a threshold for $G(n,r)$ to be plane.

Proof

In order to show that $G(n,r)$ is plane, by Observation 1, it is enough to show that it has no anchors. Every anchor has four points and it is connected. By Theorem 1, if $r=o(n^{-\frac{2}{3}})$, then w.h.p. $G(n,r)$ has no connected subgraph on 4 points, and hence it has no anchors. This proves the first statement.

We prove the second statement by adjusting the proof of Theorem 1 for $k=4$. Assume $r=\omega (n^{-\frac{2}{3}})$. Let $P_1,\dots , P_{n\atopwithdelims ()4}$ be an enumeration of all subsets of 4 points. Let $X_i$ be equal to 1 if $DG[P_i]$ contains an anchor, and 0 otherwise. Let $X=\sum X_i$. In view of Chebyshev’s inequality we need to show that $\frac{\mathrm {Var}(X)}{\mathrm {E}[X]^2}$ tends to 0 as n goes to infinity.

Partition the unit square into a set $\{s_1,\dots ,s_{2/{r^2}}\}$ of squares with diagonal length r. Then, subdivide each square $s_j$, into nine sub-squares $s_j^1,\dots , s_j^9$ as depicted in Fig. 2. If each of $s_j^1,s_j^3,s_j^7, s_j^9$ or each of $s_j^2,s_j^4,s_j^6, s_j^8$ contains a point of $P_i$, then $DG[P_i]$ is a convex clique of size four and hence it contains an anchor. Thus,

$$ \Pr [X_i=1]\ge \frac{r^{6}}{2^{3}}\cdot \frac{2}{9^4}. $$

This implies that $\mathrm {E}[X_i]=\varOmega (r^{6})$, and hence $\mathrm {E}[X] =\varOmega (n^4r^6)$. Therefore,

$$\mathrm {E}[X]^2\ge c''n^8r^{12},$$

for some constant $c''>0$. By a similar argument as in the proof of Theorem 1 we bound the variance of X from above by

$$\mathrm {Var}(X)\le c''_1 n^{7}r^{12}+c''_2 n^{6}r^{10}+c''_3 n^{5}r^{8}+c''_4 n^{4}r^{6}.$$

Since $r=\omega (n^{-\frac{2}{3}})$, $\frac{\mathrm {Var}(X)}{\mathrm {E}[X]^2}$ tends to 0 as n goes to infinity. That is, w.h.p. $G(n,r)$ has an anchor. By Observation 1, w.h.p. $G(n,r)$ is not plane.    $\blacksquare$

As a direct consequence of the proof of Theorem 4, we have the following:

Corollary 3

With high probability if a random geometric graph is not plane, then it has a clique of size four.

Note that every anchor introduces a crossing and each crossing introduces an anchor. Since, every anchor is a connected graph and has four points, by (6) we have the following corollary.

Corollary 4

The expected number of crossings in $G(n,r)$ is $\varTheta (n^4 r^6)$.

4 The Threshold for $G(n,r)$ to be Planar

In this section we investigate the threshold for the planarity of a random geometric graph; this is a decreasing property. By Kuratowski’s theorem, a finite graph is planar if and only if it does not contain a subgraph that is a subdivision of $K_5$ or of $K_{3,3}$. Note that any plane random geometric graph is planar too; observe that the reverse statement may not be true. Thus, the threshold for planarity seems to be larger than the threshold of being plane. By a similar argument as in the proof of Theorem 3 we can show that if $r\ge \sqrt{{c \ln n}/{n}}$, then w.h.p. each square with diagonal length r contains $K_5$, and hence $G(n,r)$ is not planar.

Theorem 5 $n^{-\frac{5}{8}}$ is a threshold for $G(n,r)$ to be planar.

Proof By Theorem 2, if $r=\omega (n^{-\frac{5}{8}})$, then w.h.p. $G(n,r)$ has a clique of size 5. Thus, w.h.p. $G(n,r)$ contains $K_5$ and hence it is not planar. This proves the second statement of the theorem.

If $r=o(n^{-\frac{5}{8}})$, then by Theorem 1, w.h.p. $G(n,r)$ has no connected subgraph on 5 points, and hence it has no $K_5$. Similarly, if $r=o(n^{-\frac{3}{5}})$, then w.h.p. $G(n,r)$ has no connected subgraph on 6 points, and hence it has no $K_{3,3}$. Since $n^{-\frac{5}{8}} < n^{-\frac{3}{5}}$, it follows that if $r=o(n^{-\frac{5}{8}})$, then w.h.p. $G(n,r)$ has neither $K_5$ nor $K_{3,3}$ as a subgraph.

Note that, in order to prove that $G(n,r)$ is planar, we have to show that it does not contain any subdivision of either $K_5$ or $K_{3,3}$. Any subdivision of either $K_5$ or $K_{3,3}$ contains a connected subgraph on $k\ge 5$ vertices. Since $n^{-5/8} < n^{-k/(2k-2)}$ for all $k \ge 5$, in view of Theorem 1, we conclude that if $r =o(n^{-\frac{5}{8}})$, then w.h.p. $G(n,r)$ has no subdivision of $K_5$ and $K_{3,3}$, and hence $G(n,r)$ is planar. This proves the first statement of the theorem.    $\blacksquare$

As a direct consequence of the proof of Theorem 5, we have the following:

Corollary 5

With high probability if a random geometric graph does not contain a clique of size five, then it is planar.

5 Conclusion and Further Results

We presented thresholds for random geometric graphs to have a connected subgraph of constant size, to be plane, and to be planar. A natural open problem is to extend Theorem 1 for connected subgraphs of k vertices where k is not necessarily a constant, and for connected subgraphs of k vertices which have diameter $\delta$.