Foundations of Data Science

Keep in mind that the leading terms here for large k are the last two and, in fact, at k = n,
they cancel each other so that our argument does not prove the fallacious statement for
c ≥ 1 that there is no connected component of size n, since there is. Let
f(k) = ln n + k + k ln ln n − 2 ln k + k ln c − ck ln n + ck 2 ln n
n .
Differentiating with respect to k,
f ′ (k) = 1 + ln ln n − 2 k
+ ln c − c ln n +
2ck ln n
n
and
f ′′ (k) = 2 2c ln n
+ > 0.
k2 n
Thus, the function f(k) attains its maximum over the range [2, n/2] at one of the extreme
points 2 or n/2. At k = 2, f(2) ≈ (1 − 2c) ln n and at k = n/2, f(n/2) ≈ −c n ln n. So
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f(k) is maximum at k = 2. For k = 2, E(x k ) = e f(k) is approximately e (1−2c) ln n = n 1−2c
and is geometrically falling as k increases from 2. At some point E(x k ) starts to increase
but never gets above n − c 4 n . Thus, the expected sum of the number of components of size
k, for 2 ≤ k ≤ n/2 is
⎛ ⎞
n/2
∑
E ⎝ x k
⎠ = O(n 1−2c ).
k=2
This expected number goes to zero for c > 1/2 and the first-moment method implies that,
almost surely, there are no components of size between 2 and n/2. This completes the
proof of Theorem 4.15.
4.5.3 Threshold for O(ln n) Diameter
We now show that within a constant factor of the threshold for graph connectivity,
not only is the graph connected, but its diameter is O(ln n). That is, if p is Ω(ln n/n), the
diameter of G(n, p) is O(ln n).
Consider a particular vertex v. Let S i be the set of vertices at distance i from v. We
argue that as i grows, |S 1 | + |S 2 | + · · · + |S i | grows by a constant factor up to a size of
n/1000. This implies that in O(ln n) steps, at least n/1000 vertices are connected to v.
Then, there is a simple argument at the end of the proof of Theorem 4.17 that a pair of
n/1000 sized subsets, connected to two different vertices v and w, have an edge between
them.
Lemma 4.16 Consider G(n, p) for sufficiently large n with p = c ln n/n for any c > 0.
Let S i be the set of vertices at distance i from some fixed vertex v. If |S 1 |+|S 2 |+· · ·+|S i | ≤
n/1000, then
Prob ( |S i+1 | < 2(|S 1 | + |S 2 | + · · · + |S i |) ) ≤ e −10|S i| .
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