where V
n+1
= Λv
n
∩ Λu
n
, u
n
4 − Ru
n
4 ∪ V
n
∩ Λu
n+1
. The process {T
u,v
k : k ≥ 0} tracks the position of trees after the k
th
step of the algorithm defined above along with the history carried at each stage. Clearly, if
u
k
= v
k
for some k ≥ 1, the trees emanating from
u and v meet while the event that the trees emanating from u and v never meet
corresponds to the event that { u
k
6= v
k
: k ≥ 1}. A formal construction of the above process is achieved in the following manner. We start with an
independent collection of i.i.d. random variables, {W
w
1
z, W
w
2
z : z ∈ Λw, w ∈ Z
4
}, defined on some probability space Ω, F , P, with each of these random variables being uniformly distributed
on [0, 1]. Starting with two open vertices
u and v, set u = maxu, v, v
= minu, v and V = ;.
For ω ∈ Ω, we define k
u
= k
u
ω as k
u
:= min{k : W
u
1
z p for some z ∈ Hu , k}
and N
u
:= { z ∈ Hu
, k
u
: W
u
1
z p}.
We pick R
u ∈ N
u
such that W
u
2
Ru = min{W
u
2
z : z ∈ N
u
}. Define, as earlier,
u
1
= maxRu , v
, v
1
= minRu , v
and V
1
= Λv ∩ Λu
, k
u
∪ V
∩ Λu
1
= Λv ∩ Λu
, k
u
. Further, for
z ∈ V
1
, define W
1 H
z = W
u
1
z.
Having defined { u
k
, v
k
, V
k
, {W
k H
z : z ∈ V
k
} : 1 ≤ k ≤ n}, we define u
n+1
, v
n+1
, V
n+1
, {W
n+1 H
z : z ∈ V
n+1
} as follows: for ω ∈ Ω, we define k
u
n
= k
u
n
ω as k
u
n
:= min{k : W
n H
z p for some z ∈ V
n
∩ Hu
n
, k or W
u
n
1
z p for some z ∈ Hu
n
, k \ V
n
} and
N
u
n
:= { z ∈ Hu
n
, k
u
n
: W
n H
z p if z ∈ V
n
∩ Hu
n
, k or W
u
n
1
z p if z ∈ Hu
n
, k \ V
n
}. We pick
R u
n
∈ N
u
n
such that W
u
n
2
Ru
n
= min{W
u
n
2
z : z ∈ N
u
n
}. Finally, define
u
n+1
= maxRu
n
, v
n
, v
n+1
= minRu
n
, v
n
and V
n+1
= Λv
n
∩ Λu
n
, k
u
n
∪ V
n
∩ Λu
n
. 2178
For z ∈ V
n+1
, define W
n+1 H
z =
W
n H
z
if z ∈ V
n
∩ Λu
n
W
u
n
1
z if Λ
v
n
∩ Λu
n
, k
u
n
. This construction shows that the {
u
k
, v
k
, V
k
, {W
k H
z : z ∈ V
k
} : k ≥ 0} is a Markov chain starting at
u ,
v , ;, ;. A formal proof that this Markov chain describes the joint distribution of the trees
emanating from the vertices u and v can be given in the same manner as in Lemma 2.1.
For z ∈ Z
4
, define
kzk
1
= |z1| + |z2| + |z3|
where zi is the i
th
co-ordinate of z. Fix n ≥ 1, 0
ε 13 and two open vertices u, v and consider
the trees emanating from u and v. Define the event,
A
n, ε
= A
n, ε
u ,
v :=
u
k
6= v
k
for 1 ≤ k ≤ n
4
− 1, V
n
4
= ;, n
21− ε
≤ ku
n
4
− v
n
4
k
1
≤ n
21+ ε
, 0 ≤
u
n
4
4 − v
n
4
4 logn
2
for which we show that the following Lemma holds:
Lemma 3.1. For 0 ε 13 there exist constants C
1
, β 0 and n
≥ 1 such that, for all n ≥ n ,
inf
n
1− ε
≤ku−vk
1
≤n
1+ ε
, 0≤
u4−v4
log n
P A
n, ε
| u ,
v , V
= u, v, ;
≥ 1 − C
1
n
−β
.
First we prove the result using the Lemma 3.1. First, for fixed 0 ε 13, we choose n
from the above Lemma. Now, fix any n ≥ n
and u such that n
1− ε
≤ kuk
1
≤ n
1+ ε
, 0 ≤ u4
log n. With positive probability, the vertices
0 and u are both open. On the event that both the vertices 0 and u
are both open, we consider the trees emanating from u and v = 0. We want to show that P{u
k
6= v
k
for k ≥ 1} 0.
Let τ
= 0 and for i ≥ 1, let τ
i
:= n
4
+ n
2 4
+ · · · + n
2
i−1
4
. For i ≥ 1, define the event
B
i
=
u
k
6= v
k
, for τ
i−1
k ≤ τ
i
, V
τ
i
= ;, n
2
i
1−ε
≤ ku
τ
i
− v
τ
i
k
1
≤ n
2
i
1+ε
, and 0 ≤
u
τ
i
4 − v
τ
i
4 log n
2
i
.
Then, we have P
for all k ≥ 1, u
k
6= v
k
= lim
i→∞
P for 1 ≤ k ≤ τ
i
, u
k
6= v
k
≥ lim sup
i→∞
P for 1 ≤ k ≤ τ
i
, u
k
6= v
k
, and for 1 ≤ l ≤ i, V
τ
l
= ;, n
2
l
1−ε
≤ ku
τ
l
− v
τ
l
k
1
≤ n
2
l
1+ε
and 0 ≤ u
τ
l
4 − v
τ
l
4 log n
2
l
= lim sup
i→∞
P ∩
i l=1
B
l
= lim sup
i→∞ i
Y
l=2
P B
l
| ∩
l−1 j=1
B
j
PB
1
. 32
2179
For l ≥ 2, B
l
is a event which involves the random variables u
k
, v
k
, V
k
for k = τ
l−1
+ 1, . . . , τ
l
. Using the Markov property, we have that the probability P B
l
| ∩
l−1 j=1
B
j
depends only on
u
τ
l−1
, v
τ
l−1
, V
τ
l−1
. Furthermore, on the set ∩
l−1 j=1
B
j
, we note that n
2
l−1
1−ε
≤ ku
τ
l−1
− v
τ
l−1
k
1
≤ n
2
l−1
1+ε
, 0 ≤ u
τ
l−1
4 − v
τ
l−1
4 log n
2
l−1
and V
τ
l−1
= ;. Therefore we have that, for l ≥ 2, P
B
l
| ∩
l−1 j=1
B
j
≥ inf
n
2l−11− ε
≤kz
1
−z
2
k
1
≤n
2l−11+ ε
, 0≤
z
1
4−z
2
4log n
P B
l
| u
τ
l−1
, v
τ
l−1
, V
τ
l−1
= z
1
, z
2
, ; =
inf
n
2l−11− ε
≤kz
1
−z
2
k
1
≤n
2l−11+ ε
, 0≤
z
1
4−z
2
4log n
P A
n
2l−1
, ε
| u
, v
, V
= z
1
, z
2
, ; ≥ 1 − C
1
n
−2
l−1
β
33 and since n
1− ε
≤ kuk
1
≤ n
1+ ε
, 0 ≤ u4
log n. P
B
1
= P A
n, ε
| u ,
v , V
= u, 0, ;
≥ 1 − C
1
n
−β
34 Therefore, from 32, 33 and 34, we have,
P {G is disconnected} ≥ P
u, 0 are both open × lim
Kategori