Foundations of Data Science

4.5.2 Full Connectivity
As p increases from p = 0, small components form. At p = 1/n a giant component
emerges and swallows up smaller components, starting with the larger components and
ending up swallowing isolated vertices forming a single connected component at p = ln n,
n
at which point the graph becomes connected. We begin our development with a technical
lemma.
Lemma 4.14 The expected number of connected components of size k in G(n, p) is at
most ( n
k)
k k−2 p k−1 (1 − p) kn−k2 .
Proof: The probability that k vertices form a connected component consists of the product
of two probabilities. The first is the probability that the k vertices are connected,
and the second is the probability that there are no edges out of the component to the
remainder of the graph. The first probability is at most the sum over all spanning trees
of the k vertices, that the edges of the spanning tree are present. The ”at most” in the
lemma statement is because G (n, p) may contain more than one spanning tree on these
nodes and, in this case, the union bound is higher than the actual probability. There are
k k−2 spanning trees on k nodes. See Section 12.9.6 in the appendix. The probability of
all the k − 1 edges of one spanning tree being present is p k−1 and the probability that
there are no edges connecting the k vertices to the remainder of the graph is (1 − p) k(n−k) .
Thus, the probability of one particular set of k vertices forming a connected component
is at most k k−2 p k−1 (1 − p) kn−k2 . Thus, the expected number of connected components of
size k is at most ( n
k)
k k−2 p k−1 (1 − p) kn−k2 .
, the giant component has absorbed all small compo-
We now prove that for p = 1 2
nents except for isolated vertices.
ln n
n
Theorem 4.15 For p = c ln n with c > 1/2, almost surely there are only isolated vertices
n
and a giant component. For c > 1, almost surely the graph is connected.
Proof: We prove that almost surely for c > 1/2, there is no connected component with
k vertices for any k, 2 ≤ k ≤ n/2. This proves the first statement of the theorem since, if
there were two or more components that are not isolated vertices, both of them could not
be of size greater than n/2. The second statement that for c > 1 the graph is connected
then follows from Theorem 4.6 which states that isolated vertices disappear at c = 1.
We now show that for p = c ln n , the expected number of components of size k,
n
2 ≤ k ≤ n/2, is less than n 1−2c and thus for c > 1/2 there are no components, except
for isolated vertices and the giant component. Let x k be the number of connected components
of size k. Substitute p = c ln n into ( n
( n
k)
k k−2 p k−1 (1 − p) kn−k2 and simplify using
n
)
k ≤ (en/k) k , 1 − p ≤ e −p , k − 1 < k, and x = e ln x to get
(
E(x k ) ≤ exp ln n + k + k ln ln n − 2 ln k + k ln c − ck ln n + ck 2 ln n )
.
n
104