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Unformatted text preview: t; 3, the expected delivery time of any decentralized algorithm is
at least βα n(α−2) for some constant βα .
Decentralized search with α = 2 requires time that grows like a polynomial
in log(n ), while any other exponent requires time that grows like a
polynomial in n – exponentially worse.
9 Networks: Lecture 7 Proof Idea for α = 0
The basic idea is depicted in the ﬁgure below: We randomly pick a target
node in the grid and consider the (square) box around t of side length n2/3 . With high probability, the source node lies outside the box.
Because long range contacts are created uniformly at random (since α = 0),
the probability that any one node has a long range contact inside the box is
given by the size of the box divided by n2 , i.e., nn2 = n−2/3 .
Therefore, any decentralized search algorithm will need at least n2/3 steps in
expectation to ﬁnd a node with a long-range contact in the box. Moreover, as long as it doesn’t ﬁnd a long range link leading into the box, it cannot reach the target in less than n2/3 steps (using only local contacts). This shows that the expected time for any decentralized search algorithm must take at least n2/3 steps to reach the target node. 10 Networks: Lecture 7 Proof Idea for 0 ≤ α < 2 We use a similar construction for this range: we consider a (square) box
(2− α )
around the target with side length nγ , where γ ≡ 3 .
Recall that a node v forms a long-range link with node w with probability
proportional to d (v , w )−α . Here, the constant of proportionality is 1/Z
with Z = ∑w d (v , w )−α .
Note that in the 2-dimensional grid, a node has ≈ d neighbors at distance d
from itself. This implies that
n Z= ∑ d .d −α ≈ n2−α = n3γ , d =1 where the last relation follows using an integral approximation.
Hence, the probability that any one node has a long range contact inside the
box satisﬁes ≤ n3γ = n−γ (since there are n2γ nodes inside the box)
This shows that any decentralized search algorithm will need at least
(2− α )
3 nγ = n
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- Fall '09