LOS PUNTOS CLAVE DE LA
NÚMERO CÓDIGO PESO NOMBRE
This section contains proofs of all the theorems. First we will prove the existence of a giant
component in supercritical regime.
4.3.1
Proof of Theorem 4.0.1
Proof.
Fixα
0=ε/10,c=ε
3, now for large enoughnthe conclusion of the Lemma 4.2.2 holds.
Now we run the DFS algorithm with a sequence of i.id Bernouli(ρ) random variables. We
consider the situation after
εnmany vertex queries (a vertex will be included inU
or not type
queries) of the algorithm. Assume at some time point
t∈[ε
3n, εn], the set
U
becomes empty.
Then we must have|S ∪ W|=tand
|S|=P
ti=1
Y
i. Then by 2 and 3 in Lemma 4.2.3whp
(1 +
34ε)t
n
6p|S|62ε61/3
for small enough
ε. Now since at that point
U
is empty,
NG(S)
⊂ W. The function
g(x) :=
x−
2xn2, is non-decreasing when
x6n, thus it is non-decreasing atx=np|S|6n/3< n. Now
since we have (1 +
34ε)t6np|S|, hence by Lemma 4.2.2 we havewhp,
|W|>(1−α
0)
np|S| −np
2|S|
22
= (1−α
0)
np|S| −(np|S|)
22n
>(1−
ε
10)(1 +
3ε
4
)t
1−
1
2n(1 +
3ε
4)t
>(1−
ε
10)(1 +
3ε
4
)t
1−
1
6(1 +
3ε
4
)ε
> t,
(4.3.1)
for small enough
ε, contradicting our theorem assumption. Thus
whp
all the vertices that
are being explored in the time frame [ε
3n, εn] belong to the same epoch and hence the same
component. Again using parts 2 and 3 of the Lemma 4.2.3 we get the size of this component is
bounded below by
εp.
4.3.2
Proof of uniqueness under hereditary degree assumption
First we will prove uniqueness of the giant component under an additional assumption that
we will call hereditary degree assumption. It is interesting to note that if in addition to
A1
and
A2we suppose that the following hereditary property (HD) holds for the graph sequence
G
n= (V
n, E
n), then the giant component will be unique when
p=o(1),np
2→ ∞,a
n=o(np),
b
n=np
2. In particular we will not requireA3
andb
n=o(np
3).
Assumption HD.
For each
β >
0 and
n
>
N(β), every large subgraph
U
⊂
V
n, say for
|U|>0.9nsatisfy,
max
v∈U
d(v, U)<(1 +β)p|U|.
Proposition 4.3.1.
In addition to the conditions in Theorem 4.0.1 assume thatGsatisfyHD.
Then there is an unique giant component with size greater than or equal to
εpwhpand all other
components are of size less than
O((lnn)
2).
Proof of Proposition 4.3.1.
Let
C(ρ) be a component with size at least
εp.
Recall that
O
Gn(C(ρ)) :={v∈V\C(ρ) :v
is
nota neighbor of
C(ρ)}. Lemma 4.2.11 gives
|O
G(C(ρ))|6
n(1−ε+
ε22+εl
n). At this end, note that all vertices of a component that is not connected to
C(ρ) must belong toO
G(C(ρ)). Also the largest connected component of
Per
Gn,ρ(O
G(C(ρ)))
is no more than the largest connected component in a setPer
Gn,ρ(P) whereP
is any set with
sizen(1−ε+
ε22+εl
n) containing
O
Gn(C(ρ)). Choose
εsmall enough and
nlarge enough so
that
n(1−ε+
ε22+εl
n)
>0.9n. Now by
HD
max
v∈Pd(v,P)
<(1 +ε
5)p|P|
when
n
is large.
We get thatP
is a graph onn(1−ε+
ε22+εl
n) vertices with maximum degree (1 +ε
5)p|P|and
each vertex is retained with probability
1+npε. It is easy to check for
ε >0 small enough and
n
large enough,
1−ε
2/4
np(1 +ε
5)(1−ε+
ε2 2+εl
n)
>
1 +ε
np
.
Thus the subgraph induced by
G
non
P
is a graph on
n(1−ε+
ε2
2
+εl
n) vertices with
maximum degree (1 +ε
5)p|P|and each of the vertices is retained with probabilityρthat is less
than1−
ε24
/p(1 +ε
5)|P|, hence we appeal directly to Theorem 1 in (Krivelevich, 2016) and
get that the largest connected component in
Per
G,ρ(P) is less than
O
(lnn)
2.
4.3.3
Proving uniqueness under A1, A2, A3
Notice that in the proof of Proposition 4.3.1, we only needed max
v∈Ud(v, U)<(1 +β)p|U|
for a particular set
U, namely for
U
=
O
G(C(ρ)). From Lemma 4.2.6 we have that if
G
n=
(V
n, E
n) satisfiesA1,A2and
A3then for any fixed large setU
⊂V
n, there are not too many
vertices inG
nthat do not satisfyHD.
Proof of Theorem 4.0.3.
The proof is similar to Proposition 4.3.1, except that now we do not
have
max
v∈P
d(v,P)<(1 +ε
5