Sharpness in the k-nearest neighbours random geometric graph model

0.21 MB·English

by  Victor Falgas-Ravry

k
Sharpness in the -nearest neighbours random
geometric graph model
Victor Falgas–Ravry Mark Walters
∗† ∗‡
1
1
January 18, 2011
0
2

n
a
J
Abstract
6
1
Let S denote the random geometric graph obtained by placing
n,k
] points in a square box of area n according to a Poisson process of
R
intensity 1 and joining each point to its k nearest neighbours. In [1]
P
Balister, Bollob´as, Sarkar and Walters conjectured that for every 0 <
.
h ε<1 and all n sufficiently large there exists C =C(ε) such that if
t
a
m P(S connected) ε
n,k
≥
[
then

1 P(S connected)>1 ε.
v n,k+C −
3 In this paper we prove this conjecture.
8
AsacorollaryweprovethatthereisaconstantC′ suchthatwhen-
0
3 ever k(n) is a sequence of integers with
.
1
0 P(Sn,k(n) connected) 1 as n ,
→ →∞
1
1 then for any integer sequence s(n) with s(n)=o(logn),
:
v
i P(Sn,k(n)+⌊C′sloglogn⌋ s-connected) 1 as n .
X → →∞
r This proves another conjecture of Balister, Bollob´as, Sarkar and Wal-
a
ters [3].
∗SchoolofMathematicalSciences,QueenMary,UniversityofLondon,LondonE14NS,
England
†[email protected]
‡[email protected]
1
2
Introduction
Let S be the square [0,√n] [0,√n] R2 and let k be an integer. Place
n
× ⊂
points in S according to a Poisson process of intensity 1 and put an undi-
n
rected edge between each point and its k nearest neighbours. Let S be
n,k
the resulting random geometric graph.
Several authors (see below) have considered the following question: for
which k is S connected? Of course, it is always possible for S to fail to
n,k n,k
beconnected, nomatter how large k is; the bestwe can hopefor is that S
n,k
is connected ‘asymptotically’. Formally, given a function k: N N and a
→
property of geometric graphs, we say that S has a property with
n,k(n)
Q Q
high probability (abbreviated to whp) if
lim P(S has property ) = 1.
n,k(n)
n Q
→∞
We remark that what this is saying is that the probability a random point
set gives rise to a graph with property tends to one.
Q
Elementary arguments indicate that there exist constants c and c such
l u
that for every c < c , S is whp not connected while for every c> c
l n, clogn u
⌈ ⌉
S is whp connected. Using a result of Penrose [6], Xue and Ku-
n, clogn
⌊ ⌋
4π/3+√3/2
mar [10] showed that c 5.1774. A bound of c 2log
u ≤ u ≤ π+3√3/4 ≈
3.8597canalsobereadoutofearlierworkbyGonza´les-Barrios(cid:16)andQuiro(cid:17)z[5].
These results were significantly improved by Balister, Bollob´as, Sarkar
and Walters in [1, 2] where they established the existence of a critical con-
stant c : 0.3043 < c < 1/log7 0.5139 such that for any c < c S
n, clog
∗ ∗ ≈ ∗ ⌈ ⌉
is whp not connected and for any c > c S is whp connected. They
n, clogn
∗ ⌊ ⌋
also made the following conjecture about the sharpness of the transition.
Conjecture (Conjecture3of[1]). For any 0 < ε< 1, there existsan integer
constant C(ε) such that for all n sufficiently large, if
P(S is connected) ε
n,k
≥
then
P(S is connected)> 1 ε.
n,k+C(ε)
−
The main result of this paper is the following theorem which proves the
conjecture for an explicit function C(ε).
Theorem 1. There exist absolute constants C > 0 and γ > 0 such that for
every 0 < ε < 1 and all n > ε γ, if
−
P(S is connected) ε
n,k
≥
3
then
P(S is connected) > 1 ε.
n,k+ Clog(1/ε)
⌊ ⌋ −
In [3] Balister, Bollob´as, Sarker and Walters proved a weaker variant
of their conjecture which they used to show that if k = k(n) is such that
S is connected whp then for any s = o(logn) the graphs S where
n,k(n) n,k′(n)
k (n) = k(n)+ 6 (s 1)logn are whp s-connected in a technical senseof
′
⌊ − ⌋
‘on average’. As an immediate corollary to Theorem 1, we may remove the
p
somewhatcomplicated hypothesisthattheyneededinthestatementoftheir
result: Theorem 10 of [3] (admittedly with a weaker constant). Moreover,
in the final section we strengthen this substantially proving the following
theorem.
Theorem 2. Whenever k(n) is an integer sequence such that S is
n,k(n)
whp connected and s(n) is an integer sequence with s(n) = o(logn), then
S is whp s-connected.
n,k(n)+ 2Csloglogn
⌊ ⌋
This proves the main conjecture in [3].
Before we describe the structure of our paper, we briefly contrast the
k nearest neighbours model with another classical random geometric graph
model introduced by Gilbert [4]. As before, let S be the square [0,√n]
n
×
[0,√n] R2. Let r be a real number. Again, place points in S according
n
⊂
to a Poisson process of intensity 1 but this time put an undirected edge
between any pair of points which lie at a distance of at most r from one
another. We denote by G the resulting random geometric graph model.
n,r
G is often known as the Gilbert disc model. Penrose [6, 7, 8] proved very
n,r
precise results on the connectivity of G . In particular he showed that
n,r
isolated vertices are the main obstacle to connectivity in the sense that whp
inf r 0: G is connected =inf r 0 : G has no isolated vertices .
n,r n,r
{ ≥ } { ≥ }
Thesituationisquitedifferentforthek nearestneighboursmodel,which
has no isolated vertices nor any immediately apparent analogous family of
geometric obstructions to connectivity — indeed the the value of the crit-
ical constant c is not known (although it may well be the lower bound of
∗
0.3043... proved in [1]).
One motivation for the study of S (and the Gilbert disc model) comes
n,k
from the theory of ad-hoc wireless networks. We imagine that we have
various radio transmitters (nodes) that wish to communicate using multiple
hops. The transmitters could have fixed range which naturally corresponds
to the Gilbert disc model, or they could adjust their power so that each
node has some fixed number of neighbours which is exactly the k-nearest
4
neighbour model. In this context Theorem 2 is a result about the fault
toleranceofsuchanetwork: itsaysthatwecanhaveafaulttolerantnetwork
for very little additional cost over the minimum needed for communication.
Outline of Paper
In the first section, we adapt techniques first introduced in [2] to relate the
global property of connectivity to certain families of local events: these will
be events determined by the Poisson process inside a square of area of order
logn.
In the second section we prove a geometric lemma which is crucial to
our argument, establishing that ‘small’ connected components in S have
n,k
a region of ‘high point density’.
In the third section we show that removing points from such a dense
region results in a much more likely configuration which still gives rise to a
small connected component in the k -nearest neighbour graph for some k a
′ ′
little smaller than k. In other words the graph S is much more likely to
n,k′
be disconnected than S which is exactly Theorem 1.
n,k
In the final section we prove Theorem 2.
1 Local obstacles to connectivity
Following[2],weshallrelatetheglobalconnectivityofS tocertainfamilies
n,k
of local events. Let M be an integer constant which we shall specify later
on. Let U be the square
n
M√logn M√logn M√logn M√logn
U = − , − , R2.
n
2 2 × 2 2 ⊂
(cid:20) (cid:21) (cid:20) (cid:21)
We shall refer to the subsquare 1U as the central subsquare of U . Place
2 n n
points in U according to a Poisson process of intensity 1, and put an undi-
n
rected edge between any point and the k points nearest to it to obtain the
random geometric graph U .
n,k
We define A to be the event that U has a connected component
k n,k
wholly contained inside the central subsquare 1U . First, note that our A
2 n k
event is slightly different from the family of events defined in [2]: there the
size of the box corresponding to U varied with k rather than logn. One of
n
the advantages of our definition of U is that the A -events are nested: if
n k
k k , then A A . We shall cover most of S with copies of U and
′ k′ k n n
≤ ⊆
show (approximately) that S is disconnected if and only if the event A
n,k k
occurs in one of these copies.
5
For this argument to work we need to ensure that whp S contains
n,k
no ‘long’ edges (relative to M√logn) and only one connected component of
‘large’ diameter. The following result is exactly what we want.
Lemma 3 (Lemma 1 of [2]). For any fixed α ,α with 0 < α < α and
1 2 1 2
any β > 0, there exists c = c(α ,α ,β) > 0, depending only on α ,α and
1 2 1 2
β, such that for any k with α logn k α logn, the probability that S
1 2 n,k
≤ ≤
contains two components each of diameter at least c√logn or any edge of
length at least c√logn is O(n β).
−
Remark: In this paper we use the O notation in a slightly non-standard
way. Most of our results depend on n and k where k = k(n) is a function
of n. When we say f(n,k) = O(n) we mean ‘uniformly in k’: that is there
is a constant B such that f(n,k) Bn for all n and k (satisfying our other
≤
constraints).
Let M = max( 16c(0.3,0.6,2) ,30). In our argument we shall also need
⌈ ⌉
the following lemma, which is an easy modification of Corollary 6 of [2].
Lemma 4. For any n and any integer k with 0.3logn < k < 0.6logn, the
probability that U contains an edge of length at least M√logn is O(n 6).
n,k 8 −
Proof. This is very similar to the proof of Corollary 6 of [2], but we have to
make allowances for the slight difference in our definition of the event A .
k
Let k < 0.6logn. Suppose some vertex x U has its kth nearest
n
∈
neighbour lying at a distance of at least M√logn. Then there must be fewer
8
thank < 0.6lognpointswithinaquarter-discaboutxofarea πM2logn. (We
256
need to consider quarter-discs since x may beclose to a corner of U .) Since
n
we picked M 30, we have πM2logn > 10logn. Let X Poisson(10logn).
≥ 256 ∼
Then,
(10logn)s
P(X < 0.6logn)= e 10logn
−
s!
s<0.6logn
X
0.6logn
10logn
< (0.6logn) e 10logn
−
0.6logn/e
(cid:18) (cid:19)
< 0.6(logn)e(0.6log(50e/3) 10)logn
−
< e 7logn for n sufficiently large.
−
Thus the probability that any vertex x U has its kth nearest neighbour
n
∈
lying at distance at least M√logn away is at most
8
E number of vertices in U P(X < 0.6logn)< M2(logn)e 7logn
n −
{ } ×
= O n 6 ,
−
as required. (cid:0) (cid:1)
6
We also need to define what we meant by ‘most’ of S . Let
n
2
√n
T = M logn, 1 M logn .
n M√logn −
The nice feature of Th ips that it(cid:16)ijs not verky clo(cid:17)se topany ofithe boundary of
n
S . The following lemma is a minor restatement of Theorem 1 of [9].
n
Lemma 5. There is a positive constant 0< c < 2 such that if k > 0.3logn
1
then the probability that S contains any component of diameter O(√logn)
n,k
not wholly contained in T is O(n c1).
n −
We now define two covers of T by copies of U . The independent cover
n n
of T is obtained by covering T with copies of U with disjoint interiors.
1 n n n
C
The dominating cover of T is obtained from by replacing each square
2 n 1
C C
V by the sixteen translates V +(iM√logn,jM√logn), i,j 0,1,2,3 .
∈ C1 4 4 ∈ { }
By construction, we have and the copies of 1U corresponding to
C1 ⊆ C2 4 n
elements of cover the whole of T . Also < 16 n .
C2 n |C2| M2logn
We shall write ‘A occurs in ’ as a convenient shorthand for ‘there is
k i
C
a copy V of U in for which the event corresponding to A occurs’. We
n i k
C
shall also write V for the k-nearest neighbour graph on V, and 1V for the
k 2
centre subsquare of V.
Lemmas 3 and 4 allow us to relate, up to some small error, the global
connectivity to thelocal events A . Before wemake this relationship precise
k
we need a technical lemma.
Lemma 6. Suppose S contains noedge of length greater than M√logn and
n,k 16
that V is a copy of U such that V contains no edge of length greater
2 n k
∈ C
than M√logn. Then S has a connected component contained inside 1V
8 n,k 2
whenever the event corresponding to A occurs in V.
k
Proof. LetΓ denotethesubgraphof V consisting of alledges withatleast
V k
one end in 1V, and let Γ be the subgraph of S consisting of all edges
2 S n,k
with at least one end in 1V. We aim to show that Γ = Γ . Obviously this
2 V S
will imply the lemma.
Trivially, S [V]is a subsetof V . What extra edges can there bein V ?
n,k k k
We are assumingthat S contains no edges of length greater than M√logn.
n,k 16
Thus only the vertices within distance M√logn of the boundary of V may
16
be joined in S to points in S V. So every edge in V S [V] (i.e., all
n,k n k n,k
\ \
extra edges) must meet one of these vertices.
Now V contains no edges of length greater than M√logn , so that all the
k 8
vertices meeting an edge of V S [V] must lie a distance at most
k n,k
\
M√logn M√logn M√logn
+ <
8 16 4
7
from the boundary of V. Since the vertices inside the central subsquare 1V
2
all lie at distance at least M√logn fromthe boundaryof V, they donot meet
4
any extra edges, and we have Γ = Γ as claimed.
V S
Theorem 7. For all n N and all integers k with 0.3logn < k < 0.6logn,
∈
and c as given by Lemma 5,
1
P(S not connected)= P(A occurs in )+O(n c1).
n,k k 2 −
C
Proof. Supposethat A occurs in . Then there is a copy V of U in for
k 2 n 2
C C
which A occurs; in other words, V has a connected component X wholly
k k
contained inside the central subsquare 1V. By Lemma 3 and our choice
2
of M, the probability that S contains an edge of length at least M√logn
n,k 16
is O(n 2). Let us assume this does not happen. Then there are no edges
−
between 1V and S V in S . It follows that X is a connected component
2 n\ n,k
in S as well as in V , so that S is disconnected. Thus
n,k k n,k
P(S not connected) P(A occurs in )+O(n 2).
n,k k 2 −
≥ C
Conversely, suppose S is not connected. It must contain at least two
n,k
connected components. By Lemma 3 and our choice of M, the probability
that T contains any edge of length at least M√logn or two components of
n,k 16
diameter at least M√logn is at most O(n 2). By Lemma 5 the probability
16 −
that thereis asmall componentnot contained entirely within T is O(n c1).
n −
AlsobyLemma4,theprobabilitythatU hasanyedgelongerthan M√logn
n,k 8
isO(n 6). TheprobabilitythatV hasanedgelongerthan M√logn forsome
− k 8
copy V of U in is therefore at most O(n 6) = O(n 5). Thus the
n 2 2 − −
C |C |
probability of any of the above occuring in S is at most O(n c1).
n,k −
From now on let us assume none of the above occur. Then at least one
of the connected components of S is contained in T and has diameter
n,k n
less than M√logn. Let X be such a component and x be a vertex of X. By
16
our definition of there is a copy V of U such that x 1V. For any point
C2 n ∈ 4
x / 1V, we have d(x,x ) > M√logn. By our assumption on the diameter
′ ∈ 2 ′ 8
of X, we have that x / X and hence X 1V. So X is contained entirely
′ ∈ ⊆ 2
inside the central subsquare 1V. Now V and S satisfy the hypotheses of
2 k n,k
Lemma 6, hence the event corresponding to A occurs in V, and
k
P(S not connected) P(A occurs in )+O(n c1).
n,k k 2 −
≤ C
The theorem follows.
8
Roughly speaking P(A occurs in C ) is of order n P(A ) so, from a
k 2 logn k
heuristicperspective, Theorem 7 tells usthat as we increase k the transition
of S from whpnotconnected to whpconnected happensat thesame time
n,k
as the transition from P(A ) logn to P(A ) logn. The following is a
k ≫ n k ≪ n
precise statement of this relationship.
Corollary 8. There exists a constant c >0 such that for all ε: 0 < ε 1,
2 ≤ 2
all integers n > ε c2 and all integers k : 0.3logn < k < 0.6logn, if
−
P(S connected) ε
n,k
≥
holds then
1 M2logn
P(A ) elog .
k
≤ ε n
(cid:18) (cid:19)
Conversely, if
ε M2logn
P(A ) ,
k ≤ e4 n
then
P(S connected) > 1 ε.
n,k
−
Remark: There is nothing special about the constants e and e4: we picked
these values for later convenience, but all we needed was e > 2 and e4 > 16.
Proof. Suppose P(S is connected) ε. The copies of U contained in
n,k n 1
≥ C
have disjoint interiors, hence the event corresponding to A occurs in each
k
of them independently. Therefore
P(Ak occurs in 1) = 1 (1 P(Ak))|C1|.
C − −
Now,
P(A occurs in ) P(A occurs in ) since
k 1 k 2 1 2
C ≤ C C ⊂ C
= P(S not connected)+O(n c1) by Theorem 7
n,k −
1 ε+O(n c1).
−
≤ −
Hence,
(1 P(Ak))|C1| ε+O(n−c1).
− ≥
Hence, provided we chose c large enough, we see that, for all n> ε c2, the
2 −
right hand side is at least ε. Taking logarithms on both sides and using the
2
inequality log(1 x) x for 0 x 1 yields
− ≤− ≤ ≤
P(A ) log(ε/2)
1 k
−|C | ≥
9
so
1 1 1 1
P(A ) log = log +log2 .
k
≤ ε/2 ε
|C1|(cid:18) (cid:19) |C1|(cid:18) (cid:19)
Now contains n (1+O( logn)) copies of U , 0 < ε 1 and e > 2.
C1 M2logn n n ≤ 2
Hence, provided that we chooseqour constant c2 sufficiently large, for all
n > ε c2 we have
−
eM2logn 1
P(A ) log .
k
≤ n ε
For the converse supposethat P(A ) εM2logn. By Theorem 7 we have
k ≤ e4n
P(S not connected)= P(A occurs in )+O(n c1)
n,k k 2 −
C
P(A )+O(n c1)
2 k −
≤ |C |
M2logn
ε +O(n c1)
≤ |C2| e4n −
16
ε +O(n c1) since < 16n .
≤ e4 − |C2| M2logn
Since 0 < ε 1 and 16 < 1, we have (again providing we chose c
≤ 2 e4 2
sufficiently large) for all n > ε c2,
−
P(S not connected) < ε.
n,k
2 Small components have high point density
Having madeprecisetherelationship between P(A )andP(S connected),
k n,k
weturnourattention toA . Ouraiminthissectionistoshowthatprovided
k
k > 0.3logn, small connected components in U witnessing A must have
n,k k
a region with ‘high point density’.
Let N be an integer constant whose value we shall specify later. We
consider a perfect tiling of U by square tiles of area logn. (Such a perfect
n N2
tiling exists as U has area M2logn and M, N are integers.) The expected
n
number of points of the Poisson point process on U in each tile is logn.
n N2
Fix 0 < η 1. Given a tile Q, we say that the event A occurs if A
≤ 2 k,Q k
occurs and the tile Q receives more than (1+η)logn points. Similarly, we
N2
say that the event A occurs if A occurs and the tile Q receives more
′k,Q k
than (1+ η)logn points.
2 N2
Lemma 9. Suppose k [0.3logn,0.6logn]. Then
∈
P(A A ) = O(n 1.1).
k k,Q −
\
Q
[
10
The main idea of the proof of this geometric lemma is the following:
suppose X is a connected component of U wholly contained inside 1U ,
n,k 2 n
and suppose x is a vertex of X which lies ‘on the boundary’ of X. Write r
for the distance between x and its k-th nearest neighbour.
If U contains no tile with high density (i.e. no tile receiving more
n,k
than (1+η) times the expected number of points), then then intersection
of the ball of radius r centred at x with the ‘convex hull’ of X must have
large area (about k o(k)). In particular looking outwards from X at x
1+η −
there must be quite a few empty tiles. Doing the above in several different
directions one gets that X is surrounded by a wide ‘sea’ of empty tiles of
area at least 1.1logn. Since the number of tiles M2N2 is a constant, the
probability that such a collection of empty tiles exists is O(n 1.1), yielding
−
the desired result.
Before we start, we need the following technical result.
Lemma 10. Let γ :[0,1] U be a closed continuously differentiable curve
n
→
in U . Let l(Γ) be the length of the curve Γ = γ([0,1]), and let D be the
n
number of tiles it meets. Then
9l(Γ)
D .
≤ √logn/N
Proof. We define a graph G on the set of tiles of U by setting an edge
n
between tiles Q and Q if they meet in at least one point. (G is just the
′
usual square integer lattice on 1,2,...MN 2 with diagonal edges added.)
{ }
Every tile has at most 8 neighbours in this graph. Let S be the set of tiles
met by Γ. Greedily pick a maximal subset S S which is independent
′
⊆
in G: pick the tile Q with γ(0) Q , then pick the first nonadjacent tile
1 1
∈
Q which γ(t) next meets and so on. We have D = S 9S . Now Γ
2 ′
| | ≤ | |
is continuous and cycles through the tiles of S before coming back to Q .
′ 1
Since the minimum distance between points lying in nonadjacent tiles is at
least one tile length (i.e., √logn), it follows that the length of Γ satisfies
N
√logn
l(Γ) S .
′
≥ | | N
Substituting D 9S and rearranging terms, we get the desired inequality
′
≤ | |
9l(Γ)
D .
≤ √logn/N
Proof of Lemma 9. Let k be an integer with 0.3logn < k < 0.6logn. By
Lemma 4 the probability of U containing any edge of length at least
n,k