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\begin{document}
\lecture{23 --- November 21, 2017}{Fall 2017}{Prof.\ Eric Price}{Changyong Hu, Andrew Russell}
\section{Overview}
This lecture continues our discussion on Markov chains; specifically, chains where
\begin{equation*}
P_{uv} = \begin{cases}
\frac{1}{d(u)} & (u,v) \in E\\
0 & \mbox{Otherwise.}
\end{cases}
\end{equation*} for all edges $(u,v)$. That is, we cover random walks on graphs.
\section{Definitions}
Here we recall a few definitions from last time.
\begin{itemize}
\item The hitting time $h_{u,v}$ is the expected number of steps to go from vertex $u$ to a vertex $v$.
\item The commute time $C_{u,v} = h_{u,v} + h_{v,u}$ is the expected number of steps to return to a vertex $u$ after hitting vertex $v$.
\item $C_u(G)$ is the expected time to tour the entire graph $G$ starting at $u$.
\item $C(G) = \max_u C_u(G)$ is the graph's cover time.
\end{itemize}
In this lecture we are interested in proving some bounds about these properties.
\section{Motivating Example: The Lollipop Graph}
\begin{figure}[h]
\caption{The Lollipop Graph on 14 Vertices}
\centering
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minimum size=16pt, inner sep=0pt, draw}]
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\end{figure}
The lollipop graph on $n$ vertices is a clique of $\frac{n}{2}$ vertices connected to a path of $\frac{n}{2}$ vertices. Let $u$ be any vertex in the clique that does not neighbour a vertex in the path, and $v$ be the vertex at the end of the path that does not neighbour the clique. Then $h_{u,v} = \theta(n^3)$ while $h_{v,u} = \theta(n^2)$. This is because it takes $\theta(n)$ time to go from one vertex in the clique to another, and $\theta(n^2)$ time to successfully proceed up the path, but when travelling from $u$ to $v$ the walk will fall back into the clique $\theta(1)$ times as often as it makes it a step along the path to the right, adding an extra factor of $n$ to the hitting time.\\
We now wish to prove this formally.
\section{Electrical Resistance and Commute Time of a Graph}
View graph $G$ as an electrical network with unit resistors as edges. Let $R_{u,v}$ be the effective resistance between vertices $u$ and $v$.
Then the commute time between $u$ and $v$ in a graph is related to $R_{u,v}$ by $$C_{u,v} = 2mR_{u,v}$$ We get the following inequalities assuming this relation.
If $(u,v) \in E$,
\begin{equation*}
R_{u,v} \leq 1 \therefore C_{u,v} \leq 2m
\end{equation*}
In general, $\forall\ u,v \in V$,
\begin{equation*}
R_{u,v} \leq n-1 \therefore C_{u,v} \leq 2m(n-1) < n^3
\end{equation*}
Here is the high level idea behind the proof:
We inject $d(v)$ amperes of current into all vertices $v \in V$. Now fix some vertex $u \in V$ and remove $2m$ current from $u$ leaving net $d(u) - 2m$ current at $u$. Now we get voltages $x_v$ $\forall v \in V$. We will show that $x_v -x_u = h_{v,u}$ $\forall v \neq u \in V$ which will give us a relation between
commute time and resistance.
\subsection{Lollipop Graph}
Let us revisit the lollipop graph with the electrical network view and compute $h_{u,v}$ and $h_{v,u}$ with $u$ and $v$ as before. To compute $h_{u,v}$. Let $u'$ be the vertex common to the clique and the path. Clearly, the path has resistance $\theta(n)$. $\theta(n)$ current is injected in the path and $\theta(n^2)$ current is injected in the clique.
Consider draining current from $v$. The current in the path is $\theta(n^2)$ as $2m -1 = \theta(n^2)$ current is drained from $v$ which enters $v$ through the path implying $x_u' - x_v = \theta (n^3)$ using Ohm's law ($V=IR$). Now consider draining current from $u$ instead. The current in the path is now $\theta(n)$ implying $x_v - x_u' = \theta(n^2)$ by the same argument.
Since the effective resistance between any edge in the clique is less than 1 and $\theta(n^2)$ current is injected, there can be only $\theta(n^2)$ voltage gap between any 2 vertices in the clique. We get $h_{u,v} = x_u - x_v = \theta(n^3)$ in the former case and $h_{v,u} = x_v - x_u = \theta(n^2)$ in the latter.
\subsection{Proof of Relation}
Recall that the Laplacian of $G$ is defined by:
\begin{equation*}
L_{uv} = \begin{cases}
d(u) & v = u\\
-1 & v \neq u
\end{cases}
\end{equation*}
and the degree matrix $D$ is:
\begin{equation*}
D_{uv} = \begin{cases}
d(u) & v = u\\
0 & v \neq u
\end{cases}
\end{equation*}
Consider adding $d(v)$ amps of current to every vertex $v$, and then removing $2m$ amps from a vertex $u$. Let $x$ be the voltage vector for the resulting graph. Then we have
\begin{equation*}
L x = i_u = D - 2m \mathbbm{1}_u
\end{equation*}
\begin{equation}
\forall v \in V, \sum_{(u,v) \in E} x_v - x_u = d(v)
\end{equation}
Define $h_{v,u} = 0$ when $v = u$. We can then write
\begin{align*}
h_{v,u} &= 1 + \sum_{(v,w) \in E} \frac{1}{d(v)} h_{w,u}
\end{align*}
because we take one step in our random walk out of $v$ to another vertex $h_{w,u}$ with probability $\frac{1}{d(v)}$ and then have $h_{w,u}$ expected time to reach $u$. Multiplying through by $d(v)$, we get:
\begin{equation}
d(v) = \sum_{(v,w) \in E} h_{v,u} - h_{w,u}
\end{equation}
Equations 1 and 2 are linear systems with unique solutions and are identical under $x_v - x_u = h_{v,u}$ (up to same additive shift to each entry). $x_v = h_{v,u}$ if $x_u = 0$.
We have shown that for $i_u = D - 2m \mathbbm{1}_u$ with $x = L^{+}i_u$ that $x_v - x_u = h_{v,u}$. For $u'$, we have $x' = L^{+}i_{u'}$. Now, we have,
\begin{equation*}
x - x' = L^{+}(i_u - i_{u'}) = 2m L^{+}(e_{u'} - e_u)
\end{equation*}
where $e_v$ is $1$ at the entry corresponding to $v$ and $0$ elsewhere.
The above is equivalent to $2m$ times voltage obtained if you inject 1 ampere at $u'$ and remove 1 ampere from $u$. Using Kirchoff's law and our previously proven equality that
$x_v - x_u = h_{v,u}$ we get
\begin{align*}
2m R_{u,u'} &= (x-x')_{u'} - (x-x')_{u} \\
&= (x_{u'} - x_u) - (x'_u - x'_{u'}) \\
&= h_{u',u} + h_{u,u'} = C_{u,u'}
\end{align*}
\section{Cover Time of a Graph}
We define $C_u(G)$ as the expected time for a random walk starting at $u$ to visit all vertices in a graph. $C(G)$ is the maximum of $C_u(G)$ over all $u \in V$.
\subsection{Bound for $C(G)$}
We have $\forall u \in V$,
\begin{equation*}
C_u(G) \leq 2m(n-1)
\end{equation*}
Consider the spanning tree $T$ of graph $G$. The cover time is bounded by traversing the edges of the tree in both directions (as we could just do a DFS on the spanning tree), and hitting time gives the expected time of moving along an edge, we get
\begin{align*}
C_u(G) &\leq \sum_{(u,v) \in E(T)} h_{u,v} + h_{v,u}\\
&= \sum_{(u,v) \in E(T)} C_{u,v} \\
& \leq (n-1) \max_u C_{u,v} \\
& \leq 2m(n-1)
\end{align*}
This above inequality is tight for lollipop ($\theta(n^3)$) but not for cliques which has $O(n\log n)$ as we can model it as a coupon collector problem.
\subsection{Using resistance for a better bound}
Let $R_{max} = \max_{u,v \in V} R_{u,v}^{eff}$. Then:
\begin{equation*}
mR_{max} \leq C(G) \lesssim mR_{max}\log n
\end{equation*}
Let $(u,v)$ have $R_{u,v}^{eff} = R_{max}$
\begin{equation*}
C(G) \geq \max(h_{uv}, h_{vu}) \geq \frac{h_{uv} + h_{vu}}{2} = \frac{C_{uv}}{2} = M\cdot R_{uv}^{eff} = M\cdot R_{max}
\end{equation*}
\subsection{Expected time from node $u$ to any node $v$}
\begin{equation*}
h_{uv} \leq C_{uv} = 2M\cdot R_{uv}^{eff} \leq 2M\cdot R_{max}
\end{equation*}
So after $8M\cdot R_{max}$ steps, you will be reached $v$ $w.p. 3/4$
Let's repeat the process $log(n)$ times,
\begin{equation*}
Pr(never\ reach\ v) \leq (\frac{1}{4})^{log(n)} = \frac{1}{n^2}
\end{equation*}
\begin{equation*}
Pr(any\ v not reached) \leq \frac{1}{n^2} \cdot n = \frac{1}{n}
\end{equation*}
\begin{align*}
E[T] &\leq Pr(T \leq B) \cdot B + Pr(T \geq B) \cdot E[T|T\geq B]\\
&\leq 8M\cdot R_{max}log(n) + \frac{1}{n^2} \cdot (2Mn + MR_{max}log(n))\\
& = \Theta(MR_{max}log(n))
\end{align*}
\begin{thebibliography}{42}
\bibitem[MR]{MR}
Rajeev~Motwani, Prabhakar~Raghavan
\newblock Randomized Algorithms.
\newblock {\em Cambridge University Press}, 0-521-47465-5, 1995. %pub - isbn - year
%@book{Motwani:1995:RA:211390,
% author = {Motwani, Rajeev and Raghavan, Prabhakar},
% title = {Randomized Algorithms},
% year = {1995},
% isbn = {0-521-47465-5, 9780521474658},
% publisher = {Cambridge University Press},
% address = {New York, NY, USA},
%}
\end{thebibliography}
\end{document}