Foundations of Data Science

Size of
frontier
0 ln n nθ
n
Number of
search steps
Figure 4.9: The solid curve is the expected size of the frontier. The two dashed curves
indicate the high-probability range of possible values for the actual size of the frontier.
For d ≠ 1, d − 1 − ln d > 0. 14 Thus the probability e −(d−1−ln d)i drops off exponentially
with i. For i > c ln n and sufficiently large c, the probability that the breadth first search
starting from a particular vertex terminates with a component of size i is o(1/n) as long
as the Poisson approximation is valid. In the range of this approximation, the probability
that a breadth first search started from any vertex terminates with i > c ln n vertices is
o(1). Intuitively, if the component has not stopped growing within Ω(ln n) steps, it is
likely to continue to grow until it becomes much larger and the expected value of the size
of the frontier again becomes small. While the expected value of the frontier is large, the
probability that the actual size will differ from the expected size sufficiently for the actual
size of the frontier to be zero is vanishingly small.
For i near nθ the absolute value of the expected size of the frontier increases linearly
with |i − nθ|. Thus for the actual size of the frontier to be zero, z i must deviate from its
expected value by an amount proportional to |i − nθ|. For values of i near nθ, the binomial
distribution can be approximated by a Gaussian distribution. The Gaussian falls off
exponentially fast with the square of the distance from its mean. The distribution falls
off proportional to e − k2
σ 2 where σ 2 is the variance and is proportional to n. Thus to have a
non vanishing probability, k must be at most √ n. This implies that the giant component
is in the range [nθ − √ n, nθ + √ n]. Thus a component is either small or in the range
[nθ − √ n, nθ + √ n].
In Theorem 4.8, we prove that there is one giant component of size Ω(n) along with
a number of components of size O(ln n). We first prove a technical lemma stating that
the probability of a vertex being in a small component is strictly less than one and hence
there is a giant component. We refer to a connected component of size O(log n) as a small
component.
14 Let f(d) = d − 1 − ln d. Then ∂f
∂d = 1 − 1 d
at d = 1 and is positive for all other d > 1.
∂f
∂f
and
∂d
< 0 for d < 1 and
∂d
> 0 for d > 1. Now f(d) = 0
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