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\textcolor{red}{\textbf{NOTE:} THESE NOTES HAVE NOT BEEN EDITED OR CHECKED FOR CORRECTNESS}
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\begin{document}
\lecture{22 --- November 16, 2017}{Fall 2017}{Prof.\ Eric Price}{Chenyang Wang}
\section{Overview}
In the last lecture we discussed the basics of network coding.
In this lecture we discuss the application of network coding to solving edge
connectivity on a DAG, as well as introductory material on Markov chains and
random walks.
\section{Network Coding Analysis}
Let $s$ is the source vertex and $t$ the destination vertex. Let $X_v \subset
\mathbb{F}_q^{l'}$ be the vector space $v$ knows about. If $v$ is aware of
$w \in \mathbb{F}_q^{l'}$, then there exists a basis vector $b \in X_v$ such
that $\inprod{b, v} = 0$. Let $n$ be the number of vertices in the graph. Then
after $R$ rounds,
$$
\Pr[t \text{ aware of } w] \geq 1 - (\frac{2e}{q})^{R |C_{s, t}| (1 - \frac{n}{R})}
$$
Then if $R \geq \frac{n}{\epsilon}$, and $q \geq 2^{1 /\epsilon}$, then
$$
\Pr[t \text{ aware of } w] \geq 1 - q^{-R |C_{s, t}| (1 - O(\epsilon))}
$$
If $t$ is aware of all $w \in X_s$, then $X_t = X_s$. Therefore,
$$
\Pr[X_t = X_s] \geq 1 - q^k q^{-R |C_{s, t}| (1 - O(\epsilon))}
$$
We want this probability to be greater than $1 - \frac{1}{q}$, which occurs when
$k < R |C_{s, t}| (1 - O(\epsilon))$. If $l \geq \frac{k}{\epsilon}$, then this
result will be optimal.
\section{Edge Connectivity on a DAG}
Our goal is to find the size of the minimum cut $|C_{s, t}|$ in $O(md^2)$ time,
where $m$ is the number of edges, and $d$ is the max degree. The idea is to
run one round of network coding on the DAG, then measure the throughput at
$t$.
The algorithm is as follows: we let $X_s = \mathbb{F}_q^d$. Note that $d \geq
|C_{s, t}|$. We then scan through the vertices of the DAG in topological sort
order, and at each $(u, v)$, we send $w \in X_v$, which is chosen uniformly
at random. Finally, we compute the $\text{Dim}(X_t)$, which we claim is
equal to $|C_{s, t}|$ w.h.p.
\subsection{Analysis}
To see that this is true, we first note that $\text{Dim}(X_t) \leq |C_{s, t}|$.
Let $S$, $S^c$ be the minimum $s-t$ cut in the graph, such that $s \in S$,
$t \in S^c$. Then the number of unique vectors that are transmitted over the
cut no greater than $|C_{s, t}|$, so the dimensionality of $X_t$ can be at most
$|C_{s, t}|$. With this in mind, it is sufficient to show that $\text{Dim}(X_t)
\geq |C_{s, t}|$ w.h.p.
Next, we analyze $\Pr[t \text{ aware of } w]$ for all $w \in \mathbb{F}_q^d$.
If $t$ is aware of $w$, then that awareness must have been transmitted along
some $s \rightarrow t$ path. The probability of awarness failing to pass along a
given edge is $\frac{1}{q}$; then by the union bound, $\Pr[\text{awareness of
$w$ passed along a given path}] \geq 1 - \frac{n}{q}$. Then the probability
awareness passed on any path is
\begin{align*}
\Pr[t \text{ aware of } w] &\geq 1 - (\frac{n}{q})^{|C_{s, t}|} \\
&= 1 - q^{-|C_{s, t}| ( 1 - \log_q n)}
\end{align*}
Let $Z$ be the number of $w$ that $t$ is unaware of. Then $Z = q^{|X_t^\bot|} =
q^{d - \text{dim}(X_t)}$. From the above probability
\begin{align*}
E[Z] &= q^d \Pr[t \text{ aware of } w] \\
&\geq q^{d - |C_{s,t}|(1 - \log_q n)} \\
Z &\leq q^{d - |C_{s,t}|(1 - \log_q n) + 0.5} \text{ w.h.p}\\
\text{dim}(X_t) &\geq |C_{s,t}|(1 - \log_q n) - 0.5 \\
\end{align*}
The second inquality is an application of Markov's inequality, and the third
statement comes from the definition of $Z$. Let $q \geq n^{2d}$, and
$\text{dim}(X_t)$ will be exact.
\subsection{Runtime}
For each edge $(u, v)$, the algorithm selects a $w \in X_u$, which takes
$O(d^2)$ time, and orthogonalizes it with respect to $X_v$, also in $O(d^2)$
time. The total runtime is then $O(md^2)$. Using matrix operations, we can
get this down to $O(md^{1.38})$.
\section{Markov Chains}
We now introduce the idea of Markov chains and random walks. Let $n$ be the
number of states, and $P$ be the \textbf{transition matrix}, such that
$P_{ij} = \Pr[i \rightarrow j]$ in the next round.
A \textbf{Markov chain} is a sequence $x_0, x_1, x_2, \hdots$ with $x_i \in [n]$
, such that
\begin{align*}
\Pr[x_{t+1} = j | (x_1, x_2, \hdots, x_t)] &= \Pr[x_{t+1} = j | x_t = i] \\
&= P_{ij}
\end{align*}
At any time, $t$, the state of $x_t$ can be expressed by a probability vector
$q^{(t)} \in \R^n$, such that $\sum_i q_i^{(t)} = 1$. Note that $q^{(t+1)} =
q^{(t)} P$.
A \textbf{stationary distribution} is a distrubution vector $\Pi$ such that
$\Pi = \Pi P$. In other words, $\Pi$ is a eigenvector of $P$ with eigenvalue
$1$.
The \textbf{hitting time} $h_{ij}$, is the expected number of steps required
to reach state $j$ from state $i$.
\subsection{Fundamental Theorem of Markov Chains}
A transition matrix $P$ is \textbf{ergodic} if it satisfies the following:
\begin{itemize}
\item $n$ is finite
\item Irriducable: $\exists i \rightarrow j$ path for all $i, j \in [n]$.
\item Aperiodic: $\forall$ states, $\text{gcd}(\text{loops}) = 1$.
\end{itemize}
An ergodic chain has the following characteristics:
\begin{itemize}
\item There exists a unique stationary distribution $\Pi$, $\Pi_i > 0$.
\item All distributions $q$ will eventually converge to $\Pi$.
\item $h_{ii} = \Pi_i^{-1}$
\end{itemize}
\subsection{Random Walks}
Consider a random walk on an undirected, unweighted graph such that
$$
P_{u, v} = \begin{cases}
\frac{1}{d(u)} & \text{if } (u, v) \in E \\
0 & \text{otherwise}
\end{cases}
$$
This random walk is ergodic iff the graph is connected (irriducible), and
not bipartite (has an odd cycle $\rightarrow$ Aperiodic). If so, then $\Pi_v
= \frac{d(v)}{2m}$, and $h_{vv} = \frac{2m}{d(v)}$.
\subsection{More terms}
Define $C_{u, v}$ as the expected commute time from $u$ to $v$ and back to $u$.
In other words, $C_{u, v} = h_{u, v} + h_{v, u}$. Note that $h_{u, v}$ and
$h_{v, u}$ are not necessarily equal; consider a lollipop graph.
Define $C_u(G)$ as the expected cover time starting at $u$; that is, the
expected amount of time to visit all vertices in the graph.
\end{document}