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\begin{document}
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\thispagestyle{empty} \vspace*{-0.75in} {\bssten MS\&E 337 \hfill
Lecture \#\arabic{lecture} Notes \\ Information Networks \hfill
Spring 2010
\\ Prof. Amin Saberi \hfill Page 1 of \pageref{totalpag} \\}


\topic{Conductance and Expansion}



{\bf Expansion}  The expansion $\rho(G)$ of a graph $G$ is the
minimum edge cut between two sets divided by the size of the smaller
set, \[\rho(G) = \min_S \frac{C(S,\bar S)}{\min(|S|,|\bar S|)}\]

where $C(S,\bar{S})$ denotes the number of edges between $S$ and $\bar{S}$ in $G$. A similar notion for capturing connectivity is conductance. Define the \emph{sparsity} of a cut $sp(S,\bar
S)$ to be
\[sp(S,\bar S) = \frac{C(S,\bar S)}{\min(vol(S),vol(\bar S))},\]
where $vol(S) = \sum_{i \in S} d_i$.



{\bf Conductance}  The \emph{conductance} of a graph
$\phi(G)$ is
\[\phi(G) = \min_{s \subset V} sp(S,\bar S)\]




It is easy to see that the connected  graph with the smallest conductance is the dumbbell graph below: 
\begin{figure}[h]
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\end{figure}


Obviously the complete graph $K_n$ has the highest conductance.  Which d-regular graph has the highest conductance or expansion? 
This question is much harder to answer.


We will try to do something easier. We will construct  a d-regular graph with constant expansion.  We will do that via the probabilistic method:

\begin{thm} For all $d \geq 3$, there exists a constant $\alpha > 0$
such that with high probability a random $d$-regular graph has
expansion $\rho \geq \alpha$.
\end{thm}

\begin{proof}  
To simplify  the calculations, we will present the proof for
$d$ sufficiently large instead of $d \geq 3$. The proof for $d = 3$ is similar. 

We can generate $d$ regular graphs on $n$ vertices via the “configurational model”: we split each vertex into
$d$ mini-vertices, and find the edges by generating a random perfect matching on the mini-vertices.
Note that a graph produced in this way may have multiple edges and self-loops. 

To prove the theorem we need to show that for each set $S \subset V$ , with $|S| = k$, the
probability that $|C(S, \bar{S})| < \alpha k$ is sufficiently small.
Assume that there exists such an $S$ of size $k$. For a given $k$, there are ${n \choose
k}$ possible choices for $S$. For a given $S$, there are ${dk \choose \alpha k}{dn-dk \choose \alpha k}$
ways to choose the minivertices in $S$ and minivertices
in $\bar{S}$ involved in a cut of size $\alpha k.$

Let us start by calculating the number of $d$ regular graphs on $n$
vertices, i.e., the number of perfect matchings of $K_{dn}$.

\begin{align*}
f(nd) &= {nd \choose 2} {nd - 2 \choose 2} \ldots {2 \choose 2}
\frac{1}{(nd/2)!}\\
 &=  \frac{(nd)!}{2^{nd/2}(nd/2)!}
\end{align*}

We will use the following Stirling approximation for factorials: 

$$n! = \sqrt{2 \pi n}~{\left( \frac{n}{e} \right)}^n \left( 1 + O \left( \frac{1}{n} \right) \right). $$

So 
\begin{align*}
f(nd) = c (nd)^{nd/2}e^{-nd/2} \left( 1 + O \left( \frac{1}{n} \right) \right),
\end{align*}

for some constant $c$. Then the probability of the event that only a certain $\alpha k$ of minivertices match outside of
their proper subset is at most

$$ \frac{f(dk-\alpha k)f(dn - dk - \alpha k)f(2\alpha k)}{f(dn)} $$

Therefore, the probability $p_k$ that there is  a subset $S$, with $|S|=k$ and expansion less than $ \alpha$ is 

\begin{align*}
p_k  \leq \alpha k {n \choose k} {dk \choose \alpha k} {dn-dk \choose \alpha
k}\frac{f(dk-\alpha k)f(dn - dk - \alpha k)f(2\alpha k)}{f(dn)}
\end{align*}

Using the simple but useful inequality, 

$$ \left(\frac{n}{k}\right)^k  \leq {n \choose k} \leq \left(\frac{ne}{k}\right)^k,$$ 

we have

\begin{align*}
p_k 
&\leq c \alpha k \left(\frac{en}{k}\right)^k\left(\frac{edn}{2\alpha k}\right)^{2\alpha k}
\frac{(dk-\alpha k)^{\frac{dk-\alpha k}{2}}(dn - dk - \alpha
k)^{(dn-dk-\alpha k)/2} (2 \alpha k)^{\alpha k}}{(nd)^{nd/2}}\\
&\leq c \alpha k e^k 
\left(\frac{ed}{2\alpha}\right)^{2\alpha k}
\left(\frac{n}{k}\right)^{k + 2\alpha k}
\left(\frac{k}{n}\right)^{(dk-\alpha k)/2}\\
&\leq c\alpha k
\left(e\left(\frac{ed}{2\alpha}\right)^{2\alpha}\right)^k
\left(\frac{k}{n}\right)^{((d-2)k-5\alpha k)/2}\\
&\leq c \alpha k (ce)^k (k/n)^{3k}
\end{align*}
for $d \geq 15$, $\alpha < 1/100$.  Therefore, 

$$ \sum_{k =1}^{n/2} p_k = O(1/n^2).$$
\end{proof}

\topic{Application I: Routing}

Consider an undirected graph $G=(V,E)$. Suppose you would like to route $f$ units of flow between every pair of vertices in $G$. What is the maximum value $f$ that  can be attained if the capacity of every edge is at most $1$? The answer is  the solution to the following ``multicommodity flow'' linear program. Let $P$ be the set of paths
in $G$ and $P_{ij}$ be the set of paths between any $i$ and $j$. 

\begin{align*}
\hbox{maximize}&\quad f&\\
\hbox{subject to}&\quad \sum_{p\in P_{ij}}f_p\ge f &\forall i, j \\
&\quad \sum_{p: e\in p}f_p\le 1 &\forall e\in E\\
&\quad f_p\ge 0&.
\end{align*}

\begin{thm}
Let $f^*$ be the optimal solution of the above LP. Then,
$$\frac{2}{n\log n}\rho(n)\le f^*\le \frac{2}{n}\rho(n).$$
\end{thm}
\begin{proof}
By definition $$f^*\le \min_S\frac{C(S,\bar{S})}{|S||\bar{S}|}\le
\frac{2}{n}\rho(n).$$ Thus, we obtain the upper bound.

To prove the lower bound, we consider the dual of the above LP:
\begin{align*}
\hbox{minimize}&\quad \sum d_e&\\
\hbox{subject to}&\quad \sum_{e\in p}d_e\ge l_{ij} &\forall p\in P_{ij} \\
&\quad \sum_{i,j}l_{ij}\ge1 &\forall i,j\\
&\quad d_e,l_{ij}\ge 0&.
\end{align*}


It is easy to see that the above LP has an optimal solution that satisfies the triangle inequality.  In other words,  the dual program
yields a metric $(V,\bf{d})$ that minimizes \begin{equation} \label{metric} \frac{\sum_{e\in
E}d_e}{\sum_{i,j\in V}d(i,j)},\end{equation}
%By strong duality theorem, we obtain $$f^*=\min_{\bf
%d}\frac{\sum_{e\in E}d_e}{\sum_{i,j\in V}d(i,j)}.$$
where $d(i,j)$ denotes the distance between $i$ and $j$ in $\bf{d}$.  The following theorem shows that it is possible to represent $\bf{d}$ in $l_1$ with small distortion:

\begin{thm}(Bourgain, 1985)
Every metric can be embedded into an $l_1$-metric with distortion
$O(\log n)$.
\end{thm}

The rest of the proof is pretty straightforward. First notice that since $l_1$ is additive across dimensions, we will not lose anything further if we restrict ourselves to $l_1$ metrics in one dimension. Now, consider the vertices embedded on a line and notice that the quantity in equation (\ref{metric}) is optimized when the vertices are clustered together at two points on the line. The expansion of  corresponding cut gives the desired lower bound. 
\end{proof}
\label{totalpag}
\end{document}