Foundations of Data Science

Proof: Let |S i | = k. For each vertex u not in S 1 ∪ S 2 ∪ . . . ∪ S i , the probability that
u is not in S i+1 is (1 − p) k and these events are independent. So, |S i+1 | is the sum of
n − (|S 1 | + |S 2 | + · · · + |S i |) independent Bernoulli random variables, each with probability
of
1 − (1 − p) k −ck ln n/n
≥ 1 − e
of being one. Note that n − (|S 1 | + |S 2 | + · · · + |S i |) ≥ 999n/1000. So,
Subtracting 200k from each side
E(|S i+1 |) ≥ 999n
ln n
(1 − e−ck n ).
1000
E(|S i+1 |) − 200k ≥ n 2
(
)
ln n
−ck k
1 − e n − 400 .
n
Let α = k n and f(α) = 1 − e−cα ln n − 400α. By differentiation f ′′ (α) ≤ 0, so f is concave
and the minimum value of f over the interval [0, 1/1000] is attained at one of the end
points. It is easy to check that both f(0) and f(1/1000) are greater than or equal to
zero for sufficiently large n. Thus, f is nonnegative throughout the interval proving that
E(|S i+1 |) ≥ 200|S i |. The lemma follows from Chernoff bounds.
Theorem 4.17 For p ≥ c ln n/n, where c is a sufficiently large constant, almost surely,
G(n, p) has diameter O(ln n).
Proof: By Corollary 4.2, almost surely, the degree of every vertex is Ω(np) = Ω(ln n),
which is at least 20 ln n for c sufficiently large. Assume this holds. So, for a fixed vertex
v, S 1 as defined in Lemma 4.16 satisfies |S 1 | ≥ 20 ln n.
Let i 0 be the least i such that |S 1 |+|S 2 |+· · ·+|S i | > n/1000. From Lemma 4.16 and the
union bound, the probability that for some i, 1 ≤ i ≤ i 0 −1, |S i+1 | < 2(|S 1 |+|S 2 |+· · ·+|S i |)
is at most ∑ n/1000
k=20 ln n e−10k ≤ 1/n 4 . So, with probability at least 1 − (1/n 4 ), each S i+1 is
at least double the sum of the previous S j ’s, which implies that in O(ln n) steps, i 0 + 1
is reached.
Consider any other vertex w. We wish to find a short O(ln n) length path between
v and w. By the same argument as above, the number of vertices at distance O(ln n)
from w is at least n/1000. To complete the argument, either these two sets intersect in
which case we have found a path from v to w of length O(ln n) or they do not intersect.
In the latter case, with high probability there is some edge between them. For a pair of
disjoint sets of size at least n/1000, the probability that none of the possible n 2 /10 6 or
more edges between them is present is at most (1−p) n2 /10 6 = e −Ω(n ln n) . There are at most
2 2n pairs of such sets and so the probability that there is some such pair with no edges
is e −Ω(n ln n)+O(n) → 0. Note that there is no conditioning problem since we are arguing
this for every pair of such sets. Think of whether such an argument made for just the n
subsets of vertices, which are vertices at distance at most O(ln n) from a specific vertex,
would work.
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