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%%%  EDIT THE FOLLOWING TO UPDATE DATE AND LECTURE NUMBER

\lecturehead{CS G399}{Algorithmic Power Tools I}{Fall 2007}{26
October 2007}{Zhifeng Sun}

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\noindent {\bf Lecture Outline:}
\begin{itemize}
 \item Leighton-Rao sparsest cut theorem
\end{itemize}

%%% Itemized outline of the lecture

\section{Leighton-Rao's sparsest cut theorem}
In the previous lecture, we already saw that Max-Flow $f^*$ could be
much smaller than Min-Cut $\eta$ for the Uniform Multicommodity Flow
problem, $f^* \le O\left( \frac{\eta}{\log n}\right)$. In this
lecture, we are going to show this inequality is tight up to constant
factors in Uniform Multicommodity Flow problem. This was proved by
Leighton-Rao in 1990. The result was proved for multicommodity flow
with general demands later by Linial-London-Rabinovich and
Aumann-Rabani.

First, let's recall the LP for this problem. Let $P_i$ be the set of
paths between pair $(s_i,t_i)$, $p_j^i$ be the $j^{th}$ path in $P_i$,
$dem(i)$ be the demands of pair $(s_i,t_i)$, and $f_j^i$ be the amount
of commodity $i$ sent on path $p_j^i$. Then we get the following LP
for Demands Multicommodity Flow problem.
\begin{eqnarray*}
& \max f \\
& \mbox{s.t.} \sum_{p_j^i} f_j^i \ge f\cdot dem(i) & \forall i \\
& \sum_{p_j^i:e\in p_j^i} f_j^i \le c_e & \forall e \\
\end{eqnarray*}
And its dual is
\begin{eqnarray*}
& \min \sum_e c_ed_e \\
& \mbox{s.t.} \sum_{e\in p_j^i} d_e \ge l_i & \forall i,j\\
& \sum_{i} l_i\cdot dem(i) \ge 1 \\
\end{eqnarray*}

In the remainder, we assume uniform demands. That is, we assume we
have a commodity for each pair $(u,v)$ with $demand=1$. For
simplicity, we also assume in this lecture that $c_e=1$.  The
arguments can be easily generalized to arbitrary $c_e$'s.  (We could
also replace the edge with capacity $c_e$ with $c_e$ parallel edges of
capacity $1$.  While this will suffice for proving the
max-flow-min-cut relationship in the above theorem, it would not
establish the polynomial-time computability.)

\begin{Theorem}
For any $n$, there is an $n$-node uniform multicommodity flow problem with Max-Flow $f^*$ and Min-Cut $\eta$ for which $\eta = O\left(f^*\log n\right)$. And a cut of value $O\left(f^*\log n\right)$ can be found in polynomial time. In other words, this is an $O\left(\log n\right)$-approximation for sparsest cut.
\end{Theorem}

To provide some intuition, let us consider the following proof of the
max-flow min-cut theorem for the single commodity flow problem.

\textbf{Single Commodity Flow:} After solving the dual LP, we have
$d_e$ values. Treat these values as the distance assignments of each
edge. From the dual LP constraint, we know that the distance between
the source and the sink is $1$. Now taking the source as center, begin
to grow a ball continously with radius $t$, $t$ ranging from $0$ to
$1$. The set of nodes inside the ball are all nodes that are within
distance $t$ from the source, according to the $d_e$ labels. From
the complimentary slackness, we know if $d_e>0$ then the corresponding
edge is saturated, so the min-cut is some collection of this saturated
edges. 

Let $C(t)$ denote the set of edges that cross the ball of radius $t$.
At time $t$ $(t:0\rightarrow 1)$, let $x(t)$ be the value of cut
$C(t)$.  Then during the period $(t, t+dt)$, the cut $C(t)$
contributes $x(t)\cdot dt$ to $\sum_e c_ed_e$. So $\int_0^1 x(t)dt =
\sum_e c_ed_e$. From an averaging argument, we know $\exists t$ at
which $C(t) \le \sum_e c_ed_e=f^*$.

\textbf{Multicommodity Flow:} We generalize the idea for single
commodity flow to uniform multicommodity flow. After solving the dual
LP, we know $f^*=\sum_e d_e$. The average $d_e$ value $\bar d=f^*/m$
(here $m$ is the number of edges whose $d_e>0$). Now replace every
edge $e$ in the original graph $G$ by a path of $\lceil d_e/\bar
d\rceil$ edges, each with distance $\bar d$. We could have a new
graph, $G^+$. And the total distance values of edges in this graph
won't exceed $2f^*$.  Furthermore, the total number of edges in $G^+$
does not exceed $\sum_{e: d_e > 0} \lceil d_e/ \bar d \rceil$, which
is at most $2m$.

\[\sum_{all\ original\ edges}\lceil \frac{d_e}{\bar d}\bar\rceil \le \sum \frac{d_e}{\bar d}\bar d +\sum \bar d \le f^*+f^*=2f^*\]

Our goal is to find a cut of value $O\left(f^*\log n\right)$. We
define set $S$ as balanced if $\frac{2n}{3}\ge |S|\ge
\frac{n}{3}$. Otherwise, $S$ is unbalanced. The following is a region
growing approach similar to the single commodity flow. Set
\[\alpha = \frac{O\left(n^2\cdot f^*\log n\right)}{m} \]
Pick a source in graph $G^+$, grow the region hop by hop. Each hop we grow the region by $\bar d$ which is the distance weight of edges in $G^+$. After each hop, we compare the ratio $\frac{\#crossing\ edges}{\#internal\ edges}$ with $\alpha$. If the ratio is greater then $\alpha$, we keep on growing. If not, we stop. Pick a new source, which is not covered by any region, using the same approach to grow, until we all the sources are covered by a region. After the region growing step, there are two possible cases.

\textit{case 1: For all the regions $R_i's$, the number of nodes from
graph $G$ is smaller or equal to $\frac{2n}{3}$.}  In this case, we
can group the regions into balanced set in the following way. If there
is a region that is larger than $\frac{n}{3}$, then we can just pick
this region as $S$ and group the other regions as $\bar S$. If all of
the regions are smaller than $\frac{n}{3}$, then we can greedily pack
regions to $S$ until the size of $S$ is larger than
$\frac{n}{3}$. Because each region has the property that
$\frac{\#crossing\ edges}{\#internal\ edges}\le \alpha$, we know if we
group some regions together to a bigger region $R$, it still have this
property, which means
\[\frac{\#crossing\ edges}{\#internal\ edges}\le \alpha =\frac{O\left(n^2\cdot f^*\log n\right)}{m} \]
So,
\[\delta (R)=\#crossing\ edges \le \frac{O\left(n^2\cdot f^*\log n\right)}{m}\cdot (\#internal\ edges) \le O\left(n^2\cdot f^*\log n\right) \]
And $|R|\cdot |\bar R|=O\left(n^2\right)$. We know this cut
\[\eta =\frac{\delta (R)}{|R|\cdot |\bar R|}=O\left(f^*\log n\right) \]
This means we find a cut of value $O\left(f^*\log n\right)$.

\textit{case 2: There is a region $R$ that contains more than $\frac{2n}{3}$ nodes from graph $G$.}
Each step we grow the region $R$, its ratio $\frac{\#corssing\ edges}{\#internal\ edges} \ge \alpha$ until $i^{th}$ step. So the total number of edges included in this region after each step is at least $1$, $(1+\alpha)$, $(1+\alpha)^2$, \dots, $(1+\alpha)^i$. From $(1+\alpha)^i\le m$, we have
\[i\le \frac{log 2m}{\log (1+\alpha)} \approx \frac{\log 2m}{\alpha}=\frac{m\cdot \log 2m}{n^2\cdot f^*\log n} = O\left(\frac{m}{f^*n^2}\right) \]
which mean the number of hops of this region is at most $\frac{m}{f^*n^2}$. Then the radius $r$ of this region 
\[r \le \bar d\cdot \frac{m}{f^*n^2} = \frac{f^*}{m}\cdot \frac{m}{f^*n^2} = O\left(\frac{1}{n^2}\right)\]

We can have $r\le \frac{1}{2n^2}$ by setting appropriate $\alpha$ value. By metric property, we know for any nodes $u$ and $v$,
\[dist(u,v)\le dist(u,R)+dist(v,R)+\frac{1}{2n^2} \]

From the constraint $\sum_{i} l_i\ge 1$ in the dual, we have
\[1\le \sum_{u,v}dist(u,v)\le 2n\sum_u dist(u,R)+\frac{1}{2} \]
So, 
\[\sum_{u\notin R}dist(u,R)\ge \frac{1}{4n}\]

Look at figure 1. Let $n_i$ be the number of nodes that are at least $i$ hops away from the region $R$, and $r_i$ be the ratio of \#edges crossing cut $i$ to the demand crossing cut $i$. Then $\#crossing\ edges\ge r_i\cdot \frac{2n}{3}\cdot n_i$ since the demand crossing cut is $\le \frac{2n}{3}\cdot n_i$. And its contribution to the left hand side of the above inequality is $n_i\cdot \frac{f^*}{m}$. By average argument (contribution per edge), there is at least one cut $i$ for which
\[\frac{n_i\cdot \frac{f^*}{m}}{r_i\cdot \frac{2n}{3}\cdot n_i}\ge \frac{1/4n}{m} \]
So we have
\[r_i\le 6f^* \]

Combining the above two cases, we can conclude that $\eta \le O\left(f^*\log n\right)$.

\begin{figure}[htp]
\center \epsfig{file=./region.eps, width =4in}
%\includegraphics{region.eps}
\caption{One region has more than $2n/3$ nodes}
\end{figure}

%% Section 1

\end{document}
