Foundations of Data Science

5.4.1 Using Normalized Conductance to Prove Convergence
We now apply Theorem 5.5 to some examples to illustrate how the normalized conductance
bounds the rate of convergence. Our first examples will be simple graphs. The
graphs do not have rapid converge, but their simplicity helps illustrate how to bound the
normalized conductance and hence the rate of convergence.
A 1-dimensional lattice
Consider a random walk on an undirected graph consisting of an n-vertex path with
self-loops at the both ends. With the self loops, the stationary probability is a uniform 1 n
over all vertices. The set with minimum normalized conductance is the set with probability
π ≤ 1 and the maximum number of vertices with the minimum number of edges leaving
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it. This set consists of the first n/2 vertices, for which total conductance of edges from S
to ¯S is (with m = n/2) π m p m,m+1 = Ω( 1 ) and π(S) = 1. (π n 2 m is the stationary probability
of vertex numbered m.) Thus
Φ(S) = 2π m p m,m+1 = Ω(1/n).
By Theorem 5.5, for ε a constant such as 1/100, after O(n 2 log n) steps, ||a t −π|| 1 ≤ 1/100.
This graph does not have rapid convergence. The hitting time and the cover time are
O(n 2 ). In many interesting cases, the mixing time may be much smaller than the cover
time. We will see such an example later.
A 2-dimensional lattice
Consider the n × n lattice in the plane where from each point there is a transition to
each of the coordinate neighbors with probability 1 / 4 . At the boundary there are self-loops
with probability 1-(number of neighbors)/4. It is easy to see that the chain is connected.
Since p ij = p ji , the function f i = 1/n 2 satisfies f i p ij = f j p ji and by Lemma 5.3 is the
stationary probability. Consider any subset S consisting of at most half the states. Index
states by their x and y coordinates. For at least half the states in S, either row x or
column y intersects ¯S (Exercise 5.4). So at least Ω(|S|/n) points in S are adjacent to
points in ¯S. Each such point contributes π i p ij = Ω(1/n 2 ) to flow(S, ¯S). So
∑ ∑
π i p ij ≥ c|S|/n 3 .
i∈S
j∈ ¯S
Thus, Φ ≥ Ω(1/n). By Theorem 5.5, after O(n 2 ln n/ε 2 ) steps, |a t − π| 1 ≤ 1/100.
A lattice in d-dimensions
Next consider the n × n × · · · × n lattice in d-dimensions with a self-loop at each
boundary point with probability 1 − (number of neighbors)/2d. The self loops make all
π i equal to n −d . View the lattice as an undirected graph and consider the random walk
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