Size of giant component in
Random Geometric Graphs
-G. Ganesan
Indian Statistical Institute, Delhi.
Random Geometric Graph
• n nodes uniformly distributed in S = [-½, ½]2
• Two nodes u and v connected by an edge if
d(u, v) < rn
• Resulting graph
 random geometric graph (RGG)
Figure 1: Random Geometric Graph
Radius of Connectivity
Giant Component Regime
Sparse, no
Giant Comp.
nr  
2
n
Intermediate,
Dense,
Has giant component fully connected
  nrn2  A log n
nrn2  
nrn2  A logn
rn
Intermediate range
Theorem 1 (Ganesan, 2012)
Proof Sketch of Theorem 1
S
• Divide S into small squares {Si}i
each of size rn  rn


• Choose 4 < ∆ < 5 so that nodes in adjacent
squares are joined by an edge.
• Say that Si is occupied if it contains at least
one node and vacant otherwise.
• Also, divide the unit square S
into ``horizontal” rect.
rn
each of size MK n  1
MK

where
log n
Kn  2
nrn
• Fix the bottom rectangle R
S
n
rn

R
Say that a sequence of occupied squares
L = (S1,…,St) form a occupied left-right crossing
of R if:
(i) Si and Si+1 share an edge for each i.
(ii) S1 intersects left edge of R.
(iii) St intersects right edge of R.
rn
MK n

S
St
S1
1
rn
MK n

1
Probability of left-right crossing
• Let LR(R) denote the event that R has an
occupied left-right crossing…
• Lemma (Ganesan, 2012): We have that
Pr( LR( R ))  1 
1
n
M
for some positive constant δ.
• Thus we have that
Pr(
1
)  1  10
n
if M is sufficiently large.
MK n
rn

1
• What is the advantage of identifying occupied
left-right crossings..
• Ans: We get a path of edges from left to
right…
rn
MK n

1
rn
MK n

1
Thus
Pr(
1
)  1  10
n
• What is the probability that each rectangle
has such a path?
• Ans:
# rect .
 1  10
n
Number of rectangles is O( n )
• The number of rectangles is less than
 5
n
 5
rn rn
c
since by our choice of rn, we have
nr  c
2
n
• Thus…
• The event that each rectangle has an occupied
left-right crossing occurs with prob.
O( n )
1
 1  10  1  9
n
n
• And we then get a network of paths from left
to right…
rn
MK n

1
Thus
Pr(
Pr(
1
)  1  10
n
O( n )
1
)  1  10
 1 9
n
n
• Perform an analogous procedure vertically…
rn
MK n

1
Thus
Pr(
1
)  1  10
n
Pr(
O( n )
1
)  1  10
 1 9
n
n
Pr(
2
)  1 9
n
Why is this useful?
• We have obtained a connected “backbone” of
paths in G…
• We have essentially “trapped” isolated
components of G in boxes of size…
rn
rn
2 MK n  2 MK n


rn
2 MK n

Isolated
component
rn
2 MK n

• So, we know that if backbone occurs…
then all components not attached to
backbone are “boxable” ,i.e.,
rn
rn
can be fitted in a 2MK n  2MK n


box…
• Let X denote the sum of size of all components
of G that are boxable…
nrn2
(1) Pr(X  ne
) e
 nrn2
for some positive constants  , 
• Recall that backbone occurs with prob..
Pr(
2
)  1 9
n
Thus
 X  ne
Pr(
nrn2

2
 nrn2
)  1 9  e
n
Note: Whenever backbone occurs, X denotes total sum of sizes
of components not attached to the backbone…
Therefore
P r( com ponentattached to backbone
nrn2
contains at least n  ne
2
 nrn2
nodes)  1  9  e
n
• We need to prove the estimate (1) regarding
sum of sizes of boxable components…
i.e. to prove that
nrn2
Pr(X  ne
) e
 nrn2
Recall: X sum of size of all components of G
that are boxable
Proof of (1)
• How to compute the size of a component that
can be fitted in a
rn
rn
2 MK n  2 MK n


size box?
• Main idea…count the number of vacant
squares attached to the component…
rn
2 MK n

rn
2 MK n

• Observation from figure…
• “Boxable” components have a circuit of vacant
squares attached to them…
• For any square Si
Pr( Si vacant )  exp(nrn2 )
(proved using standard binomial estimates)
• We therefore cannot have large boxable
components…
• Because, such components have a lot of
vacant squares ``attached” to them…
More precise computation
• Let S0 be the square containing the origin…
• Define C0 to be the maximal connected set of
occupied squares containing S0…
• We say that C0 is the cluster containing S0…
S0
C0
S0
• We count the number of vacant squares Vs
attached to C0
• Suppose C0 contains k squares
and the outermost boundary
∂C0 of C0 contains
L edges…(thick line in fig.)
The following hold:
• (1) The number of distinct vacant squares
attached to ∂C0 is at least Vs > L/8…
• (2) Since ∂C0 contains k squares in its interior,
we must have L  4k
• (3) Also, the ``last edge” of ∂C0 can cut the X
axis at at most L distinct points…
• (4) And ∂C0 has a self-avoiding path of L-1
steps.
• (5) The number of choices of ∂C0 is therefore
at most L.4L-1
S0
Last
edge
• Thus (1)-(5) implies
Pr(#C0  k ) 
4
l
.l. exp(nr l / 8)
l 1
2
n
k /4
 exp( nr
2
n
k)
• Thus….
E # C0  exp(1nr )
2
n
• And if N(C0) denotes the number of nodes is
C0, we also have
EN(C0 )  nr exp(1nr )
2
n
2
n
• Recall C0 is the occupied cluster containing S0…
• Define Ci for each square Si
• Same conclusion holds…
• Use Markov’s inequality to get…

nrn2 
  2 nrn2
Pr  N (Ci )  ne
  Ae
 i

for θ sufficiently small…

N (Ci ) is the sum of size of all
• But
i
boxable components…
• And we are done…
Main summary of steps
• (1) We constructed a backbone of paths with
high prob..…
• (2) We deduced that each component not
attached to the backbone is “boxable”
• (3) We showed that the sum of sizes of all 2
nrn
“boxable” components is less than ne
with high prob…
References
• M. Franceschetti, O. Dousse, D. N. C. Tse and P. Thiran. (2007).
Closing Gap in the Capacity of Wireless Networks via
Percolation Theory. IEEE Trans. Inform. Theory, 53, 1009–
1018.
• G. Ganesan. (2012). Size of the giant component in a
random geometric graph. Accepted for publ. Ann.
Inst. Henri Poincare.
• A. Gopalan, S. Banerjee, A. K. Das and S. Shakottai. (2011).
Random Mobility and the Spread of Infection. Proc. IEEE
Infocomm, pp. 999–1007.
• P. Gupta and P. R. Kumar. (1998). Critical Power for
Asymptotic Connectivity in Wireless Networks. Stoch.
Proc. and Appl., 2203–2214.
• M. Penrose. (2003). Random Geometric Graphs.
Oxford.