Foundations of Data Science

For at least half of the pairs of vertices, the grid distance, measured by grid edges
between the vertices, is greater than or equal to n/4. Any path between them must have
at least 1 r+2
n/n 2r = 1n r−2
2r edges since there are no edges longer than n r+2
2r and so there is
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no polylog length path.
An algorithm for the r = 2 case
For r = 2, the local algorithm that selects the edge that ends closest to the destination
t finds a path of expected length O(ln n) 3 . Suppose the algorithm is at a vertex u which
is a at distance k from t. Then within an expected O(ln n) 2 steps, the algorithm reaches
a point at distance at most k/2. The reason is that there are Ω(k 2 ) vertices at distance at
most k/2 from t. Each of these vertices is at distance at most k + k/2 = O(k) from u. See
Figure 4.19. Recall that the normalizing constant c r is upper bounded by O(ln n), and
hence, the constant of proportionality is lower bounded by some constant times 1/ ln n.
Thus, the probability that the long-distance edge from u goes to one of these vertices is
at least
Ω(k 2 k −r / ln n) = Ω(1/ ln n).
Consider Ω(ln n) 2 steps of the path from u. The long-distance edges from the points
visited at these steps are chosen independently and each has probability Ω(1/ ln n) of
reaching within k/2 of t. The probability that none of them does is
(
1 − Ω(1/ ln n)
) c(ln n) 2
= c1 e − ln n = c 1
n
for a suitable choice of constants. Thus, the distance to t is halved every O(ln n) 2 steps
and the algorithm reaches t in an expected O(ln n) 3 steps.
A local algorithm cannot find short paths for the r < 2 case
For r < 2 no local polylog time algorithm exists for finding a short path. To illustrate
the proof, we first give the proof for the special case r = 0, and then give the proof for
r < 2.
When r = 0, all vertices are equally likely to be the end point of a long distance edge.
Thus, the probability of a long distance edge hitting one of the n vertices that are within
distance √ n of the destination is 1/n. Along a path of length √ n, the probability that
the path does not encounter such an edge is (1 − 1/n) √n . Now,
(
lim 1 − 1 ) √ n
= lim
(1 − 1 ) n √n
1
n→∞ n n→∞ n
= lim e − √ 1
n = 1.
n→∞
Since with probability 1/2 the starting point is at distance at least n/4 from the destination
and in √ n steps, the path will not encounter a long distance edge ending within
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