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Connectivity of a Gaussian

Network

P. Balister

Department of Mathematical Sciences,

University of Memphis, Memphis TN 38152, USA

E-mail: pbalistr@msci.memphis.edu

B. Bollob´ as

Department of Mathematical Sciences,

University of Memphis, Memphis TN 38152, USA

E-mail: bollobas@msci.memphis.edu

and

Trinity College,

University of Cambridge, Cambridge CB2 1TQ, UK

A. Sarkar*

Department of Mathematical Sciences,

University of Memphis, Memphis TN 38152, USA

E-mail: asarkar@memphis.edu

* Corresponding author

M. Walters

Peterhouse,

University of Cambridge, Cambridge, CB2 1RD, UK

E-mail: mjw1009@cam.ac.uk

Abstract: Following Etherington, Hoge and Parkes, we consider a network consisting

of (approximately) N transceivers in the plane R2distributed randomly with density

given by a Gaussian distribution about the origin, and assume each transceiver can

communicate with all other transceivers within distance s. We give bounds for the

distance from the origin to the furthest transceiver connected to the origin, and that

of the closest transceiver that is not connected to the origin.

Keywords: Wireless sensor network; Gaussian distribution; Transceiver; Gilbert

model; Continuum percolation.

Biographical notes: B´ ela Bollob´ as currently holds the Jabie Hardin Chair of Ex-

cellence in Graph Theory and Combinatorics at the University of Memphis and is a

Fellow of Trinity College, Cambridge, England. He received PhDs from Eotvos Uni-

versity, Budapest, Hungary and the University of Cambridge, England. He works in

extremal and probabilistic combinatorics, polynomials of graphs, combinatorial prob-

ability and percolation theory. He has written eight books, including Extremal Graph

Theory, Random Graphs and Modern Graph Theory, and over 300 research papers.

He has supervised over forty doctoral students, many of whom hold leading positions

in prominent universities throughout the world.

Paul Balister is a Professor at the University of Memphis, where he has been since

1998. He obtained his PhD from the University of Cambridge, England in 1992 and

subsequently held positions at Harvard and Cambridge. He has worked in algebraic

number theory, combinatorics, graph theory and probability theory.

Amites Sarkar is a Visiting Assistant Professor at the University of Memphis. He ob-

tained his PhD from the University of Cambridge in 1998, and has been at Memphis

since 2003. He has worked in combinatorics and probability theory.

Mark Walters is a Teaching Fellow at Peterhouse College, Cambridge. He obtained his

PhD from the University of Cambridge in 2000, and was a Research Fellow of Trinity

College, Cambridge. He has worked in combinatorics and probability theory.

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1INTRODUCTION

In 1961, E.N. Gilbert defined and studied the following

model of a random geometric graph, known as the disc

model or Gilbert model (4). Let P be a Poisson process

in the plane of intensity one, and join every point of P to

every other point of P within distance r, for some fixed

r > 0. For small r, most points are isolated, that is, not

connected to any other points. However, as r increases, the

points form small connected clusters, which then connect

up to each other (as r increases still further), eventually

forming a (connected) giant component, which contains a

positive fraction of points in any large region.

speaking, Gilbert derived upper and lower bounds for the

smallest value of r such that the probability of the last

eventuality (also known as percolation) is one.

In recent years there has been renewed interest in such

graphs, which are now being used to model sensor net-

works and wireless ad-hoc networks in general. However,

for some applications, it is desirable that the initial dis-

tribution of sensors is non-uniform. One such model was

recently proposed by Etherington, Hoge and Parkes (3), in

which the locations of the sensors are modeled as a Pois-

son process whose intensity is given by a two dimensional

Gaussian distribution. Such a network might arise, for in-

stance, if the sensors were dropped from an aircraft.

Specifically, Etherington, Hoge and Parkes defined the

following random geometric graph G = G(N,σ,s). We

start with a Poisson process in R2with intensity at radius

r given by

ρ(r) =

Loosely

N

2πσ2e−r2/2σ2,

for some constant σ. In addition, place a point at the

origin, so that the expected total number of points is N+1.

Then connect each point to every other point at distance

less than s. We now ask two questions. First, what is the

largest value of r such that every point in Dr(0), the (open)

disc of radius r centered at the origin, is joined to every

other point of Dr(0)? Second, what is the smallest value

of r such that Dr(0) contains all the points of C = C(0),

the component of G containing the origin? (C(0) consists

of all the points of G that can be connected to the origin.)

Formally, we define

r−= sup{r : Dr(0) ∩ V (G) ⊆ V (C)},

and

r+= inf{r : V (C) ⊆ Dr(0)}.

In this paper we derive lower and upper bounds for r−

and r+, for various ranges of values of s. Since r−and r+

are random variables, our bounds will only hold with high

probability (whp), that is, with probability tending to one

as N → ∞. By a uniform scaling of R2, we may without

loss of generality fix σ2= 1/2, so that the density of points

is given by

ρ(r) =N

πe−r2.

Copyright c ? 200x Inderscience Enterprises Ltd.

We imagine N to be very large, and consider various func-

tions s = s(N).

First we give a heuristic argument which yields the

asymptotically correct value of r−for a large range of val-

ues of s. Assuming that the density of points in a small

disc of radius s is approximately constant, the probability

that a point at distance r from the origin is an isolated

vertex in G can be approximated by

e−ρ(r)πs2= e−Ns2e−r2

.

Therefore, provided Rs ≪ 1, the expected number of iso-

lated vertices in DR(0) is approximately

?R

0

2Nre−r2e−Ns2e−r2

dr = s−2?

e−Ns2e−R2

− e−Ns2?

. (1)

When R ≤ r−, there will be no isolated vertices in DR(0).

On the other hand, it seems reasonable to suppose, by

analogy with other similar problems, that the closest point

to the origin not belonging to C is, with probability tending

to one as N → ∞, an isolated vertex in G. Therefore we

may assume that if R > r−, there will be at least one

isolated vertex in DR(0). Consequently, if we choose R so

that the expected number of isolated vertices in DR(0) is

one, we might expect that r− ≈ R. Hence, if N−1/2≪

s ≪ (logN)−1/2,

r2

−≈ log(Ns2/log(1/s2)).(2)

One approach to proving rigorous bounds for r−is to ap-

ply methods and results for the related disc model Gs(A),

described above for the case A = R2. Here, given a region

A ⊂ R2, we consider a Poisson process of intensity one in

A, and join each point to all other points within a radius

s to obtain Gs(A). We have the following result of Gupta

and Kumar (5), which was proved for the square SN of

area N by Penrose (6).

Theorem 1. Let AN be a disc of area N, and let s =

s(N) satisfy πs2= logN + ω(1). Then whp Gs(AN) is

connected.

However, it turns out that applying this result yields

only the weak bound

r2

−? log(Ns2/logN).(3)

A heuristic explanation is as follows. For the result of

Gupta and Kumar, the obstruction to connectivity is the

existence of isolated vertices, and it seems reasonable to

suppose that this is also true for our model. If we choose s

so that the probability that a vertex is isolated is o(N−1),

then the expected number of isolated vertices is o(1), so

whp there are no isolated vertices and (by the above fact)

whp Gs(A) is connected. For our model, even if we choose

r so that the probability that a vertex at distance r from

the origin is isolated is Θ(N−1), as in (3), then the only

vertices in Dr(0) which have probability Θ(N−1) of being

isolated lie very close to the boundary of Dr(0), and there

2

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are far fewer than N such vertices. Consequently, it is

likely that r− is somewhat larger than suggested by (3),

and we will argue directly in the proof of Lemma 5 to show

that indeed (2) is much closer to the truth. Nevertheless,

we will use Theorem 1 in deriving a lower bound for r−

when s is very small.

Next we turn to r+. For the disc model Gs(R2), with an

additional point at the origin, results from (2) (see also (4))

show that there is a critical density γ ≈ 4.512 so that if

πs2< γ then C(0) (defined as above) is finite with proba-

bility one, while if πs2> γ there is a non-zero probability

that C(0) is infinite. This suggests that whp

γ ≈ πs2ρ(r+) = Ns2e−r2

+,

so that whp

r2

+= O(log(Ns2)), (4)

and we will show in Section 2 that in fact r2

for a large range of values of s.

For Rs ≫ 1, the above arguments fail as the density of

points in a disc of radius s at distance R from the origin is

far from constant. Of particular interest is the value of s

for which G becomes connected. We show that this occurs

when

+= Θ(Ns2)

2s

?

logN ≈ loglogN −1

2logloglogN.

The main obstacle to connectivity is the presence of iso-

lated vertices that are among the furthest points from the

origin.These vertices are at distance R ≈

Rs ≫ 1.

√logN, so

2 Precise results

Let rminand rmaxbe the distance of the nearest (respec-

tively furthest) point from the origin.

Theorem 2. Define G = G(N,s), r−and r+as above.

Then the following statements hold with high probability.

1. If 2s√logN ≥ loglogN −1

G is connected.

2. If 2s√logN = loglogN −1

each of the following events has probability bounded

below by some positive constant:

2logloglogN + ω(1) then

2logloglogN +O(1) then

(a) G is connected,

(b) G is connected except for the furthest point from

0 which is isolated,

(c) G is connected except for one isolated point that

is not furthest from 0,

(d) G has more than three components.

3. If 2s√logN ≤ loglogN −1

G is disconnected and r+< rmaxwhp. Moreover, if

2logloglogN − ω(1) then

2s√logN ≤ C loglogN for some C < 1 and Ns2≥

logN,

r2

−=log(Ns2/log(1/s2)) + 2s

−3

r2

+ O(1).

[If 2s√logN = o(1/loglogN), these simplify to

?

logN

2logmax{1,s

?

?

logN} + O(1)

logN loglogN)1/2)

+=log(Ns2) + Θ((s

r2

r2

−= log(Ns2/log(1/s2)) + O(1)

+= log(Ns2) + O(1).]

4. If Ns2= C logN for some constant C > 0 then

(a) if C > 1, r2

(b) if C = 1, r2

−= Θ(1),

−= (1 + o(1))loglogN/logN,

(c) if C < 1, r2

−= NC−1+o(1).

+= log(Ns2) + O(1) = loglogN + O(1)

In all cases r2

as above.

5. If Ns2→ ∞ then r−/s → ∞, while r2

O(1) as above.

+= log(Ns2)+

6. If Ns2= C > 0, then r−/s has a limiting distribu-

tion.

(a) If C > γ, where γ is the critical density for

disc percolation, then with positive probability

r2

+= (1 + o(1))log(C/γ). Conditional on this

not occurring r+/s has a limiting distribution.

(b) If C ≤ γ then r+/s has a limiting distribution.

In both cases there is a positive probability that r+=

0.

7. If Ns2= o(1) then whp the origin is isolated, so that

r−= rminand r+= 0.

Part 1 of the theorem is proved as part of Lemma 5 in

Section 2.1, and part 2 follows from the remarks following

the proof of Lemma 5. Part 3 is contained in Lemmas 5,

7, 8, and 9, except for the assertion relating to rmax, which

follows from the remarks following the proof of Lemma 5.

The remainder of the theorem is proved in Section 3.

3General bounds

First we prove two easy bounds on rmaxgiving an idea

of the scale of this distribution.

Lemma 3. P(r2

logN − ω(1) < r2

Proof. The number of points outside radius R is Poisson

distributed with mean Ne−R2, and thus the probability

that there are no points further than R from the origin is

exp(−Ne−R2). The result follows.

max≤ logN + α) = e−e−α, so whp

max< logN + ω(1).

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Next we prove some bounds on r−. Before we do this, we

first prove a simple lemma concerning the mean number of

points in a disc, which takes into account the variation in

density across the disc.

Lemma 4. Fix a point z of R2at distance r ≥ s from

the origin. Then the expected number Er,sof vertices of G

lying in Ds(z) is given by

Er,s= E|V (G) ∩ Ds(z)|

= Ne−(r−s)2f(r,s)θ(r,s)

where

f(r,s) = min

?

1

2,s2,

1

√4π(r−s),

?

s

4π(r−s)3

?

and c ≤ θ(r,s) ≤ 1 for some c > 0 independent of r and s.

Remark. Numerical calculations show that we can take

c = 0.3055.

Proof. We can calculate the expected number exactly as

N

π

?+s

−s

?√s2−x2

−√s2−x2e−(r+x)2−y2dy dx.

?z

Writing x = −s + ε the above

expression becomes

We can estimate

0 < c1 ≤ θ1(z) ≤ 1.

−ze−y2dy = min{√π,2z}θ1(z) where

N

πe−(r−s)2?2s

0

e−2(r−s)ε−ε2m(ε,s)θ1(

?

2εs − ε2)dε,

where

m(ε,s) = min{√π,2

?

2εs − ε2}.

We can bound the above integral by

?∞

0

2√2εse−2(r−s)εdε =

?

πs

4(r − s)3.

Since this integral is dominated by the contribution when

ε ∼ (r−s)−1, this will give the correct order of magnitude

when r−s ≫ 1,s−1,s. Similarly, we can bound the integral

by?∞

order of magnitude when s ≫ r − s ≫ 1. We can bound

the integral by?∞

order when s ≫ 1 ≫ r − s. Finally we can bound the

integral by

?2s

correct order when 1 ≫ s and s−1≫ r − s. Thus we have

bounded Er,sas required, and shown that the bound is of

the right order except on a compact region in R2

the result follows by continuity (see Figure 1).

0

√π e−2(r−s)εdε =

√π

2(r−s), and this gives the correct

0e−ε2√π dε =π

2and this gives the right

02√2εs − ε2dε = πs2, and this gives the

+, where

3.1Bounds for r−

With the aid of Lemma 4, we can obtain fairly tight

bounds for r−.

0

1

2

012

.8

s

.7

.6

.5

.5

.6

.7

.4

.4

.9

r-s

s2

s

4π(r-s)3

1

4π(r-s)

1

2

Figure 1: The function θ(r,s). Formulae shown are values

of the minimum in the expression in Lemma 4.

Lemma 5. Suppose that s = s(N) = o(1) and s ≥

?(logN)/N. Then whp

r2

−≥log(Ns2/log(1/s2)) + 2s

−3

Moreover, if

?

logN

2logmax{1,3s

?

logN} − 3.

2s

?

logN ≥ loglogN −1

2logloglogN + ω(1)

then G is connected whp.

Proof. For any point x at distance r ≥ s from 0, let

the region A(x) be the intersection of the ball Ds(x) and

Dr(0). Note that all of A(x) is closer to 0 than x is. Let

ER be the expected number of vertices x within R of 0

with no points in the region A(x). If ER→ 0 then whp

r−≥ R since whp we can find a sequence of points joining

any point x ∈ DR(0) to 0.

Now |A(x)| ≥ |Ds(x)|/3, and since the density of points

is higher in A(x) than in Ds(x) \ A(x), E|V (G) ∩ A(x)| ≥

1

3Er,s. By Lemma 4, the probability that A(x) is empty is

at most

exp?−1

3Er,s

?≤ exp

s2,

?

?

−Nµe−(r−s)2?

,

provided µ ≤c

in the minimum in Lemma 4 being redundant when s =

o(1). For r < 3s we can take µ =c

3min

?

s

4π(r−s)3

?

, the other two terms

3s2. Then

E3s≤

?3s

s

?3s

s

2Nre−r2e−Nµe−(r−s)2

dr

≤

≤ 8Ns2e−(c/3−o(1))Ns2.

2Nre−Nµe−4s2

dr

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By assumption Ns2→ ∞, so E3s→ 0. For 3s ≤ r ≤ R set

µ =

logµ + 2rs − s2(note that z is decreasing in r) we have

dz

dr= −2(r − s)Nµe−(r−s)2+ 2s. Thus we have |dz

rNµe−(r−s)2provided r ≥ 3s and Nµe−(r−s)2> 2. Then,

writing

c

3s2min?1,(3Rs)−3/2?. Writing z = Nµe−(r−s)2+

dr| >

z0= Nµe−(R−s)2+ logµ + 2Rs − s2,(5)

we have

ER− E3s≤

?R

3s

?∞

z0

?∞

z0

2Nre−r2e−Nµe−(r−s)2

dr

≤

2µ−1e(r−s)2−r2−Nµe−(r−s)2

dz

=

2e−zdz = 2e−z0,

so it is enough that z0→ ∞, and

Nµe−(R−s)2> 2.(6)

Set (R − s)2= log(Nµ/log(1/s2)) − α, α > 0.

Nµe−(R−s)2= eαlog(1/s2) > 2 for small s, and

Then

z0= (eα− 1)log(1/s2) + log(µ/s2) + 2Rs − s2.

Now log(µ/s2)=−3

log(µ/s2)+2Rs is bounded below. Since α > 0 and s → 0,

we have z0→ ∞ as required.

If Rs > 1, then R2= (1 + o(1))logN, so log(µ/s2) ≥

−3

R2=log(Nµ/log(1/s2)) + 2Rs − s2− 0.1

≥log(Ns2/log(1/s2)) + 2s

−3

and r−≥ R since ER→ 0. If Rs ≤ 1 then the same bound

applies since 2s√logN < 0.1 unless R2= (1+o(1))logN.

For the last part, set R2= logN + α where α is con-

stant. Assume 2s√logN = loglogN −1

β, where β→

2Rs = (1 + o(1))loglogN, so µ = c′(2Rs)1/2R−2=

Θ(√loglogN/logN). Thus z0≥ logµ + 2Rs − s2= β +

O(1) → ∞. Similarly Nµe−(R−s)2= elog µ+2Rs−α+o(1)=

eβ−α+O(1)→ ∞. Hence r2

α. Thus r−> rmax, so G is connected whp.

This establishes part 1 of Theorem 2. For part 2, sup-

pose that

2logmax{1,3Rs} + log(c/3), so

2logmax{1,3s√logN} − 2.5. Thus, taking α = 0.1,

?

logN

2logmax{1,3s

?

logN} − 3,

2logloglogN +

o(loglogN).∞, but β= Then

−≥ logN +α whp for any fixed

2s

?

logN = loglogN −1

2logloglogN + β,

where we will initially suppose that β is constant. The

proof of Lemma 5 shows that there exists an α and C, de-

pending only on β such that ER≤ C when R2= logN+α.

Since the expected number of points at distance greater

than R from the origin is e−α, E∞ ≤ C′for some con-

stant C′≤ C +e−αdepending only on β. By dividing the

plane into, say, 10C′sectors, we see that there is a large

probability that all the points in any one particular sector

can be connected to a point nearer the origin, since in any

one sector the expected number of points which can’t be

connected to a point nearer the origin is at most 0.1. Since

events in different sectors are almost independent, it fol-

lows (by, for example, the Lov´ asz Local Lemma) that G is

connected with probability bounded away from zero. On

the other hand, let us estimate the probability p(α,β) that

a vertex at distance R is isolated, where R2= logN + α

and α is constant. We have

p(α,β) = exp

?

−Ne−(R−s)2?

s

4π(R−s)3θ(R,s)

?

≥ exp?−c0Ne−log N−α+2Rs?s

≥ exp

≥ exp?−c2eβ−α?= c3.

Hence the probability that the furthest, or second furthest

point from the origin is isolated is bounded below by a

constant. Since these points are likely to be far from one

another, they are isolated almost independently. This, to-

gether with the first observation, establishes part 2. Fi-

nally, if β → −∞ with α fixed, we see that p(α,β) → 1

and so r+< rmaxwhp, establishing the first statement of

part 3.

If s is not o(1) then G is connected, so r− = ∞ and

r+ = rmax. If s ≤

Lemma 5 is negative, so trivially true. We shall show later

that in this case r−= o(1).

We call a point x ∈ V (G) isolated if Ds(x)∩V (G) = {x}.

Clearly any isolated point x ?= 0 cannot lie in V (C), so

must be at distance at least r−from 0. To obtain an upper

bound for r−, we follow the proof of Lemma 5 but instead

estimate E′

distance R of 0. We require the following lemma.

R3

?

?

?

−c1e−α+2s√log N

√log log N

(log N)

?(logN)/N, the bound for r2

−in

R, the expected number of isolated points within

Lemma 6. Let E′

lated points in G within distance R of 0. If E′

r−≤ R whp.

Rdenote the expected number of iso-

R→ ∞ then

Proof. First note that any two isolated points must be

at least distance s apart. Hence there is at most one iso-

lated point in any square of side length 2s/3. On the other

hand, the existence of isolated points in two such squares is

independent if the squares are at distance at least 2s, since

the event that a square contains an isolated point depends

only on the process within distance s of that square. Tile

R2with squares Sij = [0,2s/3]2+ (2si/3,2sj/3). Parti-

tion this collection into 16 classes according to the value

of (i mod 4,j mod 4) ∈ Z2

number of isolated points within R of 0 and in one of the

squares in class (k,l). The Xkl are dependent, but still

?

Xk′l′, with EXk′l′ ≥ E′

dependent bernoulli random variables with mean pi, say.

4.Let Xkl, k,l ∈ Z4 be the

k,lEXkl = E′

R. Hence there is at least one Xk,l, say

R/16. Now Xk′l′ is a sum of in-

5

Page 6

Thus

P(Xk′l′ = 0) =

?

i

(1 − pi) ≤

?

i

exp(−pi)

= exp(−EXk′l′) ≤ exp(−E′

R/16),

so if E′

distance R of 0, and r−≤ R whp.

R→ ∞, then whp there is an isolated point within

Lemma 7. Suppose that s ≥?(logN)/N and also that

2s√logN ≤ (1 − ε)loglogN for some ε > 0. Then whp

r2

−≤log(Ns2/log(1/s2)) + 2s

−3

?

logN

2logmax{1,2s

?

logN} + log(2/ε).

Moreover, if

2s

?

logN ≤ loglogN −1

2logloglogN − ω(1)

then G is disconnected whp.

Proof. We estimate E′

bility that x ∈ V (G) is isolated is at least exp(−Er,s) ≥

exp(−Nµe−(r−s)2), where µ = min{s2,?

is distance at least R′+ s from the origin. Hence

Rand use Lemma 6. The proba-

s

4πR′3}, when x

E′

R≥

?R

R′+s

?R−s

R′

2Nre−r2e−Nµe−(r−s)2

dr

=

2N(z + s)e−z2−2sz−s2e−Nµe−z2

dz

≥ e−2Rs

?R−s

R′

µ−1e−Nµe−z2?R−s

= µ−1e−2Rs?

2Nze−z2e−Nµe−z2

dz

= e−2Rs?

z=R′

e−Nµe−(R−s)2

− e−Nµe−R′2?

.

Set (R − s)2= log(Nµ/log(1/s2)) − α, where α < 0 is

constant. Consider the case when s√logN = O(1) first.

Take R′= 0, so µ = s2. Then E′

s−2e−Ns2. By assumption on s, s−2e−Ns2≤ 1/logN → 0,

and Rs = O(1), so E′

sume s√logN is large. Then R2= (1 + o(1))logN. Take

R′= (1 − ε)R. Now Nµe−(R−s)2= eαlog(1/s2) → ∞.

Also eR′2/e(R−s)2= o(1), so logE′

eαlog(1/s2)+o(1) = (1−eα+o(1))loglogN−2Rs+o(1) ≥

(ε − eα+ o(1))loglogN → ∞ for α < logε.

Now, if (R − s)2= log(Nµ/log(1/s2)) − α then

R≥ e−2Rs−log(1/s2)(eα−1)−

R→ ∞ when α < 0.Now as-

R≥ −logµ − 2Rs −

R2≤log(Ns2/log(1/s2)) + 2s

−3

?

logN

2logmax{1,2s

?

logN} − α + o(1),

and r−≤ R.

For the last part, set R2= logN − α. Then logµ +

2Rs → −∞. Thus Nµe−(R−s)2= elog µ+2Rs+α+o(1)→ 0.

Therefore, using the approximation e−θ≈ 1 − θ for small

θ,

E′

R≥ (1 − o(1))µ−1e−2Rs(Nµe−R′2− Nµe−(R−s)2)

= (1 − o(1))(elog N−R′2−2Rs− eα−s2).

If R′2= (1 − ε)R2, then logN − R′2− 2Rs = εlogN +

O(loglogN) so E′

all α > 0, r−< rmaxand G is disconnected whp.

This completes the proof of the estimate for r−in part

3 of Theorem 2.

R→ ∞ as required. Since this holds for

3.2Bounds for r+

Now we turn our attention to r+. In this case we are

interested in the existence of at least one point at distance

R which is joined to the origin.

Lemma 8. Suppose that 2s√logN ≤ loglogN and

r−/s → ∞ whp. Then, whp,

?

+ O(s

?

Proof. First assume s√logN loglogN = O(1). Then

the statement reduces to r2

the disc DR(0) where R is given by

r2

+≥log(Ns2) +1

2

s

?

logN loglogN

logN + 1).

+≥ log(Ns2) + O(1). Consider

R2= log(Ns2) − 3.

The Poisson process restricted to DR(0) stochastically

dominates a Poisson process in DR(0) with constant in-

tensity ρ(R).Cover the disc with a square tessellation

where the squares have side length s/√5. The number

of points inside any of the squares which are wholly in-

side DR(0) dominates a Poisson distribution with mean

λ = ρ(R)s2/5. Substituting for ρ and R we get

λ =N

5πe−R2s2=e3

5π> 1.27.

The probability that any such square contains no points is

at most e−λ.

We compare this process to a site percolation on Z2by

declaring a site to be open if its corresponding square con-

tains at least one point. The site percolation dominates

our process in the sense that percolation in the site model

implies percolation in our model. Since 1 − e−λ> 0.7 >

pc(site) the probability that some square inside Dr−(0) is

in an infinite component tends to one, and thus the prob-

ability that the origin in our process is in an infinite com-

ponent tends to one. Thus, whp, the origin is joined to

some point of the process at distance at least R − s from

the origin. Thus

r2

+≥ (R − s)2≥ R2− 2Rs

≥ log(Ns2) − 3 − 2s

≥ log(Ns2) + O(1)

?

logN

6

Page 7

SRR−1

S

Ai

Figure 2: Regions S and Aiin the proof of Lemma 8. The

discs shown have radius s/2, so that discs corresponding to

adjacent vertices overlap. With high probability, at least

one of the columns of small (approximate) squares contains

a point in every square.

and the result follows.

Now assume s√logN loglogN → ∞. We aim to show

that whp C(0) extends some distance beyond DR(0) in

at least one narrow sector, where R2= log(Ns2) − 3

as before. By Lemma 5, R2− r2

so R − r− = o(1).

about 1/√logN. Consider the approximately square re-

gion formed by intersecting one of these sectors with an

annulus with radii R − 1 and R (see Figure 2). Subdivide

this region S into sectors of angle s/√5logN and annuli

of thickness s/√5. The region is thus subdivided into ap-

proximately square regions of side length at most s/√5.

We aim to show that with a reasonable probability there

is a point in S∩C(0) at distance at least R−s/√5 from 0.

Each small square-like region contains at least one point

with probability at least p0= 1−exp(−s2ρ(R)/5.1) > psite.

Moreover, points in any two adjacent squares are within

distance s, so are connected in G. Comparing this process

with a site percolation with probability p0, we see that

with positive probability there is a square near the centre

of S which is connected to the side r = R of S, and hence

there is a point within r−of 0 joined to a point at least

R − s/√5 of 0 in S.

−= O(loglogN), and

Divide DR(0) into sectors of angle

Consider a sector of angle s/√5logN containing such a

point. For i = 0,1,2,... let ri = R + (i − 1)s/√5. Let

Aibe the intersection of the sector with the annulus with

inner radius ri and outer radius ri+1. Suppose that k is

such that all the regions Aifor 1 ≤ i ≤ k contain at least

one point of the process. Then, for 0 ≤ i < k, points in

adjacent regions Aiand Ai+1are adjacent in G, so we may

conclude that r+is at least R + (k − 1)s/√5.

The density in the region Ai is at least ρ(ri+1). The

condition on s implies that R2= (1 + o(1))logN. Thus,

the area of the region Aiis at least s2/5.1. Hence, setting

λ0= (s2/5.1)ρ(R), the expected number λiof points in Ai

satisfies λi≥ λ0e−αiwhere α = 2rk+1s/√5 (the definition

of R implies that λ0= e3/5.1π + o(1) > 1.25).

The probability pithat Aicontains at least one point is

1−e−λi≥1

Aifor 0 ≤ i ≤ k contain at least one point is at least

ee−αi. Thus the probability that all the regions

k?

i=1

pi≥ exp(−α?k+1

2

?− k).

If

k =

??

0.98(loglogN)/α

?

− 1

then this probability is at least (logN)−0.49−o(1). (Note

that s√logN loglogN → ∞ ensures that αk → ∞.) How-

ever we have at least R = Θ((logN)1/2) disjoint sectors so,

whp, at least one of the sectors satisfies this. Thus whp,

r2

+≥R2+ 2(k − 1)Rs/√5 ≥ log(Ns2)

+1

2

s

?

?

logN loglogN + O(s

?

logN + 1).

Lemma 9. Suppose that s = o(1) and Ns2→ ∞. Then,

whp,

r2

+≤log(Ns2) + 2

+ O(s

?

s

?

logN loglogN

?

logN + 1).

Proof. Let the radius R be defined by πs2ρ(R) = 1/3.

We define a sequence of areas Ai. Let A0be DR(0), and for

i ≥ 1 let Ai= DR+is(0) \ DR+(i−1)s(0). Let V1= V ∩ A1

and for i ≥ 2 let Vibe the vertices in Aijoined to a vertex

in V1wholly inside Dr+is(0).

We want to bound the size of Vi. Let Wibe the points

in Aithat are neighbours of vertices in Vi−1. Now, since

the density in Ai+1is bounded above by ρ(R+is) we have

E(|Wi+1| | |Vi|) ≤ πs2ρ(R + is)|Vi|.

Also, any vertex in Viis either in Wior is a descendant

of a vertex in Wi. Moreover any vertex in Vi\ Wihas an

ancestor in Wiwhich can be reached from this vertex by a

path using no vertex of any Vjfor j < i. Note that these

vertices may lie in Ajfor j < i but cannot lie in A1or A0.

However the expected number of descendants of a vertex

7

Page 8

in Wi which can be reached wholly outside DR(0) is at

most?∞

E(|Vi| | |Wi|) ≤3

Combining these we see that, substituting for R,

j=13−j= 1/2. Hence

2|Wi|.

E(|Vi+1| | |Vi|) ≤3

Also |V1| is at most the number of points outside DR(0),

which is dominated by a Poisson distribution with mean

1

3s2. Hence

2πs2ρ(R + is)|Vi| ≤1

2e−2Ris|Vi|.

E(|Vj|) ≤ 2−(j−1)exp(−2Rs?j

2

?)

1

3s2.

Thus if

j =

??

(Rs)−1log(1/s2)

?

≥ (j − 1)2+ 1 for

+ 1

then E(|Vj|) ≤ 2−je−Rs(using 2?j

j ≥ 2). Now since s = o(1), (Rs)j2→ ∞. Thus either j

or Rs is large and so E(|Vj|) = o(1). Therefore whp r+≤

R + js, so that whp r2

result follows since we may assume log(1/s2) ≤ loglogN

and R ≤√logN.

This completes the proof of the estimate for r+in part

3 of Theorem 2.

Corollary 10. If s = o(loglogN/√logN) and Ns2→

∞, then whp

r2

2

?

+≤ log(Ns2) + 2jRs + O(1). The

max− r2

+≥ (1 + o(1))log(1/s2).

Proof. From Lemma 9, we have whp

r2

+≤logN − log(1/s2) + 2

+ O(s

?

≤logN − log(1/s2) + o(loglogN).

Thus for any ε > 0, whp r2

ω(1). The result now follows from Lemma 3.

?

s

?

logN loglogN

logN + 1)

++ (1− ε)log(1/s2) = logN −

4Small s

The results we have proved so far say very little about

the case Ns2= O(logN). In this section we address this

case.

Lemma 11. Suppose that R is such that πR2ρ(R) →

∞, Ns2= O(logN), and

Ns2e−R2− logN − logR2→ ∞.

Then, whp, G|DR(0)is connected.

Proof. The process restricted to DR(0) stochastically

dominates a Poisson process of mean ρ(R). The expected

number of points of the Poisson process inside the disc is

πR2ρ(R).

(7)

The hypotheses of the theorem imply that R = O(1)

(since if R → ∞ then eventually Ns2e−R2− logN −

logR2< −1

for some constant C. We modify the original process inside

DR(0) by keeping points of the process at distance r ≤ R

from the origin with probability ρ(R)/ρ(r). Then the mod-

ified process has density exactly ρ(R) in DR(0). Denote

by G′the graph obtained from the modified process by

joining two points at distance less than s.

Now suppose that H = G|DR(0)is not connected. Then

H will contain two vertices v1and v2which are not joined

by a path in H. These vertices will be in H′= G′|DR(0)

with probability at least 1/C2, and they will certainly not

be joined by a path in H′. Hence if H is not connected with

probability at least p infinitely often as N → ∞, then H′

is not connected with probability at least p/C2infinitely

often as N → ∞. Thus if H′is connected whp, H is also

connected whp.

Provided that πR2ρ(R) tends to infinity as N tends to

infinity, we can apply Theorem 1 to see that H′is con-

nected, whp, if πs2ρ(R) ≥ log(πR2ρ(R)) + ω(1). Substi-

tuting for ρ(R) and rearranging gives the result.

2logN −logR2→ −∞), so that ρ(0) ≤ Cρ(R)

Corollary 12. Suppose C > 1 and Ns2= C logN.

Then whp

r2

−≥ logC − o(1).

Proof. Fix α > 0 with α < logC. Substitute R2=

logC − α > 0 in (7). Then πR2ρ(R) → ∞ and the left

hand side of (7) becomes

eαlogN − logN − logR2= (eα− 1)logN + O(1)

which tends to infinity.

Lemma 13. If Ns2< (1−ε)logN and Ns2→ ∞ then

whp

r2

−= N−1exp{Ns2(1 + o(1))}.

Proof. First we prove r2

Let R2= N−1exp{Ns2(1 + δ)}.

R = o(1) and the expected number of isolated points in

DR(0) is at least

−≤ N−1exp{Ns2(1 + o(1))}.

Then if 0 < δ < ε,

N(1 − e−R2)exp{−Ns2} ≥1

2NR2exp{−Ns2}

=1

2exp(δNs2) → ∞.

The result follows from Lemma 6.

To prove r2

N−1exp{Ns2(1 − δ)} and apply Lemma 11.

πR2ρ(R) → ∞ and we have

Ns2e−R2− log(NR2) =Ns2(1 − o(1)) − Ns2(1 − δ)

−≥ N−1exp{Ns2(1 + o(1))}, let R2=

Then

+ o(1) → ∞.

The result follows, since if G|DR(0)is connected then r−≥

R.

Lemma 14. If Ns2= logN then whp

r2

−= (1 + o(1))loglogN

logN

.

8

Page 9

Proof. For the lower bound we use Lemma 11. Take

R2= (1 − α)loglogN/logN for α > 0.

πR2ρ(R) = NR2e−R2→ ∞. Now

Ns2e−R2− logN − logR2

≥ (e−R2− 1)logN − log

≥ −R2logN − log(1 − α)

− logloglogN + loglogN

= αloglogN + O(logloglogN) → ∞

so that whp r−≥ R.

For the upper bound, we use the proof of Lemma 7. Take

R2= (1 + α)loglogN/logN for α > 0. We may assume

that (R − s)2≥ (1 +α

that θ(1+α

2) > 1 and assume that N is large enough that

e−(R−s)2≤ 1 − θ(R − s)2. Then with notation as in the

proof of Lemma 7, it is easy to see that there is an absolute

constant C such that

Certainly

?(1 − α)loglogN

logN

?

2)loglogN/logN. Choose θ such

E′

R≥C

s2e−log N·e−(R−s)2

CN

logNe−log N·e

≥

−(1+α

2)log log N

log N

.

Therefore,

logE′

R≥

?

+ logC − loglogN

?

+ logC − loglogN

=?θ(1 +α

1 − e−(1+α

2)log log N

log N

?

logN

≥θ(1 +α

2)log log N

log N

?

logN

2) − 1?loglogN + logC → ∞

and so whp r−≤ R by Lemma 6.

The last three lemmas together establish part 4 of The-

orem 2. Part 7 is immediate, and for part 5, if Ns2→ ∞

then r−/s → ∞ by Lemma 13 and so Lemma 8 applies and

it together with Lemma 9 give r2

6 follows from standard results on branching processes: we

note only that if Ns2= C > γ, where γ is the critical den-

sity for disc percolation, and if R2= logC−logγ, then the

probability that the origin belongs to the giant component

in G|DR(0)lies strictly between 0 and 1, and, conditional

on this event occurring, r+= (1 + o(1))R.

+= log(Ns2)+O(1). Part

5Conclusion

In this paper we have analyzed a model of a random ge-

ometric graph whose vertices are given by a Poisson pro-

cess of Gaussian intensity: two such vertices are connected

in the graph if they lie within distance s of each other.

We have given precise bounds for two parameters which,

loosely speaking, describe the width of the central com-

ponent of such a graph. It is our hope that some of our

methods will find application to other problems in this

area.

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