Lecture 10

Notes below are meant to supplement the scribed notes.

Commute Time by Effective Resistance (Section 10.2.1)

When we consider the effective resistance between two nodes \(u\) and \(v\) in a graph \(G\), we can think of setting each edge in \(G\) as a resistor with resistance 1 ohm and then connecting a battery with voltage 1 volt between nodes \(u\) and \(v\). Let the resulting current (sum of currents on edges incident to \(u\)) that flows from \(u\) to \(v\) be \(I_{uv}\). Then the effective resistance \(R_{uv}\) is simply given by Ohm's law: \(R_{uv} = 1/I_{uv}\).

Interpretation: The effective resistance \(R_{uv}\) can be interpreted as a measure of connectivity between nodes \(u\) and \(v\). A smaller \(R_{uv}\) indicates more parallel paths (more connectivity) between \(u\) and \(v\), while a larger \(R_{uv}\) indicates fewer paths (less connectivity).

The effective resistance of an edge \((u,v)\) is defined as the effective resistance between nodes \(u\) and \(v\). The effective resistance of each edge is the probability that the edge is chosen in a uniform spanning tree distribution. (1)

1. The probability is given by \(Pr(e\in T) = \tau(G\cdot e)/ \tau(G)\), where \(\tau(G)\) is the number of spanning trees in graph \(G\), and \(G\cdot e\) is the graph obtained by contracting edge \(e\) in \(G\). The ratio can be derived using Kirchoff's matrix-tree theorem.

Cover Time for Complete Graphs

We can improve the bound in the lecture notes for the cover time of complete graphs. This is equivalent to the coupon collector's problem (Section 6.5), where the expected time to collect all \(n\) coupons is \(\Theta(n\log n)\).