1. Proof of Kleinberg’s small-world theorems

2. Kleinberg’s Small-World Model
Consider an (n x n) grid. Each node has links to every node at lattice distance p (short range neighbors) & q long range links. Choose long-range links s.t. the prob. to have a long range contact at lattice distance d is proportional to 1/dr n
p = 1, q = 2
r = 2 n
Recall Kleinberg’s
results.Is there a
justification?

3. Results Theorem 1 There is a constant α0 (depending
on p and q but independent of n), so that when
r = 0, the expected delivery time of any
decentralized algorithm is at least α0.n2/3

4. Proof of theorem 1 U .. s t
Probability that the source s will lie outside U
= (n n 4/3) / n2
= 1 - 2/n2/3 (w.h.p)
n2/3
Probability that a node outside U has a long distance link inside U
= 2. n 4/3 / n2 = 2/n -2/3 . So, roughly in O(n2/3) steps, the query will
enter U, and thereafter, it can take at most n2/3 steps.

5. Results
Theorem 2. There is a decentralized algorithm A and a constant α2 dependent on p and q but independent of n, so that when r = 2 and p = q = 1, the expected delivery time of A is at most α2.(log n)2

6. Proof of theorem 2 Phase j means
Distance from t is between 2j and 2j+1 21 20
target t 22
source s

7. Observation 4. j How many nodes are at a lattice
distance j from a given node?
4. j
How many nodes are at a lattice
distance j or less from a given node?
1 + 4.j + 4. (j-1) + 4. (j-2) + …
= j.(j+1)/2
= j.(j+1) = 1 + 2j2 + 2j

8. Proof
Main idea
We show that in phase j, the expected no of hops before the current message holder discovers a long-range contact within lattice distance 2j of t is O(log n); After this, phase j comes to an end. As there are at most log n phases, a bound proportional to log2n follows.

9. Proof Probability (node u chooses node v as its long-range contact) is
There are 4j nodes
at distance j
But
So, the probability (u chooses v) is

10. Proof Phase j  2j+1 ≤ (distance to v) < 2j
Zone Bj consists
Of all nodes within
Lattice distance
2j from the target
Phase j  2j+1 ≤ (distance to v) < 2j
The maximum value of j is log 2n (⋍ log n)
When will phase j end? What is the prob
that it will end in the next step? v
No of nodes in Zone Bj is u
each within distance 2j+1 + 2j < 2j+2
from a node like u

11. Proof So each has a probability of
Zone Bj consists
Of all nodes within
Lattice distance
2j from the target
So each has a probability of
of being a long-distance contact of u, So, the search
enters Bj with a probability of at least v u
So, the expected number of steps
spent in phase j is 128 ln (6n). Since
There are at most log n phases, the
Expected time to reach v is O(log n)2