Foundations of Data Science

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Figure 5.7: Illustration that adding edges to a graph can either increase or decrease hitting
time.
Hitting time is not symmetric even in the case of undirected graphs. In the graph of
Figure 5.7, the expected time, h xy , of a random walk from x to y, where x is the vertex of
attachment and y is the other end vertex of the chain, is Θ(n 3 ). However, h yx is Θ(n 2 ).
Commute time
The commute time, commute(x, y), is the expected time of a random walk starting at
x reaching y and then returning to x. So commute(x, y) = h xy + h yx . Think of going
from home to office and returning home. We now relate the commute time to an electrical
quantity, the effective resistance. The effective resistance between two vertices x and y in
an electrical network is the voltage difference between x and y when one unit of current
is inserted at vertex x and withdrawn from vertex y.
Theorem 5.9 Given an undirected graph, consider the electrical network where each edge
of the graph is replaced by a one ohm resistor. Given vertices x and y, the commute time,
commute(x, y), equals 2mr xy where r xy is the effective resistance from x to y and m is the
number of edges in the graph.
Proof: Insert at each vertex i a current equal to the degree d i of vertex i. The total
current inserted is 2m where m is the number of edges. Extract from a specific vertex j
all of this 2m current. Let v ij be the voltage difference from i to j. The current into i
divides into the d i resistors at vertex i. The current in each resistor is proportional to the
voltage across it. Let k be a vertex adjacent to i. Then the current through the resistor
between i and k is v ij − v kj , the voltage drop across the resister. The sum of the currents
out of i through the resisters must equal d i , the current injected into i.
d i = ∑ k adj
to i
(v ij − v kj ) = d i v ij − ∑ k adj
to i
v kj .
Solving for v ij
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