Foundations of Data Science

3k/2
k/2
u
k
t
Ω(k 2 ) vertices at
distance k/2 from t
Figure 4.19: Small worlds.
distance √ n of the destination, for at least half of the starting points the path length will
be at least √ n. Thus, the expected time is at least 1 2√ n and hence not in polylog time.
For the general r < 2 case, we show that a local algorithm cannot find paths of length
O(n (2−r)/4 ). Let δ = (2 − r)/4 and suppose the algorithm finds a path with at most n δ
edges. There must be a long-distance edge on the path which terminates within distance
n δ of t; otherwise, the path would end in n δ grid edges and would be too long. There are
O(n 2δ ) vertices within distance n δ of t and the probability that the long distance edge from
one vertex of the path ends at one of these vertices is at most n 2δ ( 1
n 2−r )
= n (r−2)/2 . To
see this, recall that the lower bound on the normalizing constant is θ(n 2−r ) and hence an
upper bound on the probability of a long distance edge hitting v is θ ( )
1
n independent
2−r
of where v is. Thus, the probability that the long distance edge from one of the n δ vertices
on the path hits any one of the n 2δ vertices within distance n δ of t is n 2δ 1 = n r−2
n 2−r
2 .
The probability that this happens for any one of the n δ vertices on the path is at most
n r−2
2 n δ = n r−2
2 n 2−r
4 = n (r−2)/4 = o(1) as claimed.
Short paths exist for r < 2
Finally we show for r < 2 that there are O(ln n) length paths between s and t. The
proof is similar to the proof of Theorem 4.17 showing O(ln n) diameter for G(n, p) when
p is Ω(ln n/n), so we do not give all the details here. We give the proof only for the case
when r = 0.
For a particular vertex v, let S i denote the set of vertices at distance i from v. Using
only local edges, if i is O( √ ln n), then |S i | is Ω(ln n). For later i, we argue a constant
factor growth in the size of S i as in Theorem 4.17. As long as |S 1 |+|S 2 |+· · ·+|S i | ≤ n 2 /2,
for each of the n 2 /2 or more vertices outside, the probability that the vertex is not in
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