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Foundations of Data Science

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For a small number i <strong>of</strong> steps, the probability distribution <strong>of</strong> the size <strong>of</strong> the set <strong>of</strong><br />

discovered vertices at time i is p(k) = e −di (di) k<br />

and has expected value di. Thus, the<br />

k!<br />

expected size <strong>of</strong> the frontier is (d − 1)i. For the frontier to be empty would require that<br />

the size <strong>of</strong> the set <strong>of</strong> discovered vertices be smaller than its expected value by (d − 1)i.<br />

That is, the size <strong>of</strong> the set <strong>of</strong> discovered vertices would need to be di − (d − 1)i = i. The<br />

probability <strong>of</strong> this is<br />

e −di (di) i<br />

= e −di di i i<br />

e i = e −(d−1)i d i −(d−1−ln d)i<br />

= e<br />

i! i i<br />

which drops <strong>of</strong>f exponentially fast with i provided d > 1. Since d − 1 − ln d is some<br />

constant c > 0, the probability is e −ci which for i = ln n is e −c ln n = 1 n c . Thus, with high<br />

probability, the largest small component in the graph is <strong>of</strong> size at most ln n.<br />

Illustration 4.1<br />

at most t − 1 is exactly a i t−1. With this we get:<br />

a t = p 0 +<br />

s∑<br />

p i a i t−1 =<br />

i=1<br />

s∑<br />

p i a i t−1.<br />

i=0<br />

We have a 1 = p 0 < 1. There is a constant α ∈ [p 0 , 1) such that whenever a t−1 ≤ α, the<br />

above recursion implies that a t ≤ α. This would finish the pro<strong>of</strong> since then a 1 ≤ α implies<br />

a 2 ≤ α which implies a 3 ≤ α etc. and so q = lim t→∞ a t ≤ α.<br />

To prove the claim, consider the polynomial<br />

h(x) = x −<br />

s∑<br />

p i x i .<br />

i=0<br />

We see that h(1) = 0 and h ′ (1) = 1 − ∑ s<br />

i=1 ip i ≈ 1 − sd , which is at most a strictly<br />

n<br />

negative constant. By continuity <strong>of</strong> h(·), there exists some x 0 < 1 such that h(x) ≥ 0 for<br />

x ∈ [x 0 , 1]. Take α = Max(x 0 , p 0 ). Now since ∑ s<br />

i=0 p ix i has all nonnegative coefficients,<br />

it is an increasing function <strong>of</strong> x and so if a t−1 is at least α, then, ∑ s<br />

∑ i=0 p ia i t−1 is at least<br />

s<br />

i=0 p iα i ≥ α. Now, if a t−1 ≤ α,<br />

proving the claim.<br />

a t =<br />

s∑<br />

p i a i t−1 ≥<br />

i=0<br />

s∑<br />

p i α i = α − h(α) ≤ α,<br />

i=1<br />

We now prove in Theorem 4.8 that in G(n, d ), d > 1, there is one giant component<br />

n<br />

containing a fraction <strong>of</strong> the n vertices and that the remaining vertices are in components<br />

<strong>of</strong> size less than some constant c 1 times log n. There are no components greater than<br />

c 1 log n other than the giant component.<br />

93

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