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\head{CS G399}{Algorithmic Power Tools I}{Fall 2007}{22 November
2007}{Tao Jin}


\newcommand{\p}{\mathcal{P}}
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\noindent {\bf Lecture Outline:}
\begin{itemize}
\item
Review of Sparsest-Cut problem
\item
Frechet's-Embedding
\item
Bourgain's Theorem

\begin{itemize}
\item
Construction of Bourgain's Embedding
\item
Proof of Bourgain's Theorem
\end{itemize}
\end{itemize}

In this lecture we will first review the sparsest cut problem and its relation
with metric embeddings. Then we will show that for any metric space, there
is an isometric Frechet's embedding. Last, we will prove the Bourgain's Theorem with
the construction of Frechet's Embedding, which guarantee us a $O(\log n)$ approximation
factor to the optimal solution of the sparsest-cut problem.

\section{Sparsest-Cut Review}

In the last lecture we've defined the sparsest-cut problem. Recall that:

{\bf Sparsest-Cut Problem:}

Given a graph $G = (V,E)$ and $w: E \rightarrow Z^+$. Determine cut
$(S,\bar{S})$ such that $\frac{E(S,\bar{S})}{|S||\bar{S}|}$ is
minimized where $E(S,\bar{S})$ is number of edges crossing $S$ and
$\bar{S}$.


For any cut $(S,\bar{S})$, we defined the distance between two vertices in the graph as follow,
\[
d_S(i,j)  = \left\{ {\begin{array}{*{20}c}
   1  \\
   0  \\
\end{array}} \right.\begin{array}{*{20}c}
   {(i,j)\ is\ cross\ cut\ edge}  \\
   {otherwise}  \\
\end{array}
\]

We claim that any cut induces a so-called cut metric $d_S$. Also, we have proved that we could relax any metric cut to $l_1$ metric. Then we have,
\begin{eqnarray}
Sparsest Cut &=& \min_S \frac{E(S,\bar{S})}{|S||\bar{S}|}\nonumber\\
             &=& \min_{d_S} \frac {\sum_{i,j \in E} d_S(i,j)}{\sum_{i,j}
             d_S(i,j)} \nonumber\\
             &=& \min_{d_{l_1}} \frac {\sum_{i,j \in E} d_{l_1}(i,j)}{\sum_{i,j}
             d_{l_1}(i,j)}
\end{eqnarray}

Suppose there is a metric $d$ which embeds in $l_1$ with distortion $D$. According to the definition of distortion,
\[
\frac {d(i,j)}{D} \nonumber \le d_{l_1}(i,j) \le d(i,j)
\]
Then we have,
\begin{eqnarray}
d(i,j) \ge d_{l_1}(i,j) \Rightarrow \sum_{i,j \in E} d(i,j) \ge \sum_{i,j \in E} d_{l_1}(i,j) \label{eqn:dist_1}\\
d(i,j) \le D*d(i,j) \Rightarrow \sum_{i,j} d(i,j) \le D*\sum_{i,j} d_{l_1}(i,j) \label{eqn:dist_2}
\end{eqnarray}
So by inequalities \ref{eqn:dist_1} and \ref{eqn:dist_2}, for any metric d
\[
\frac{\sum_{i,j \in E} d(i,j)}{\sum_{i,j} d(i,j)}\nonumber \ge
\frac{1}{D}\nonumber \frac {\sum_{i,j \in E} d_{l_1}(i,j)}{\sum_{i,j} d_{l_1}(i,j)} \nonumber
\]

Since we have proved the equality (1), it is easy to see that for any metric $d$ which embeds in $l_1$ with distortion $D$, the optimal solution to minimize $\frac{\sum_{i,j \in E} d(i,j)}{\sum_{i,j} d(i,j)}\nonumber$ has approximation factor of D to the optimal solution of the sparsest-cut problem.

By Bourgain's Theorem, we can choose $d$ to have dist($d$) = $O(\log n)$, thus we could obtain a $O(\log n)$-approximate solution by solving the linear program below,
\[
 \min \sum_{i,j \in E} d_{l}(i,j)
\]
\[
 \mbox{such that } \sum_{i,j} d_{l}(i,j) =1
\]
\[
 \forall i, j, k d_l(i,j)+d_l(j,k) \ge d_l(i,k)
\]

\section{Frechet's-Embedding}

{\bf Frechet's Theorem:}
Let($V$,$d$) be an arbitrary n-point metric space. Then there is an isometric
embedding $\Phi$:$V \rightarrow l_\infty^n$.

{\bf Proof:}
Let's consider the following Frechet's embedding.
Given a metric $(V, d)$, $|V| = n$. Define
$\Phi: v_i \rightarrow <d(v_i,v_1), d(v_i,v_2), ... , d(v_i,v_n)> $.

By triangle inequality,
\begin{eqnarray}
\|\Phi(v_i)-\Phi(v_j)\|_\infty &=& \max_t |d(v_i,v_t)-d(v_j, v_t)|\nonumber\\
                               &\le& d(v_i, v_j) \nonumber
\end{eqnarray}
so the embedding is nonexpanding. On the other hand,

for all $u \in V$:
\[
\|\Phi(v_i)-\Phi(v_j)\|_\infty \ge |\Phi_j(v_i)-\Phi_j(v_j)| = d(v_i, v_j)
\]
where $\Phi_j(v_i)$ denotes the $j$-$th$ coordinate of $v$ in the new metric space.
By the above two inequalities, it is easy to see that the embedding is isometric.

Generally, it is rare to find isometric embeddings between two spaces of interst,
so we relax and allow the embedding to alter the distances. By Bourgain's Theorem,
any metric embeds in $l_1$ with $O(\log n)$ distortion. In the rest of the lecture, we
will prove Bourgain's Theorem with Frechet's embedding, which is pretty similar as
what we discussed above.

\section{Bourgain's Theorem}

\begin{Theorem}\label{Bourgain}
Let $(V, d)$ be a metric space, and let $n$ denote $|V|$. There exists an embedding $\phi$ from
$(V, d)$ into $l_1^h$, where $h = O(\log n)$, such that the distortion is $O(\log n)$. Moreover, $\phi$
can be computed in polynomial time by a randomized algorithm.
\end{Theorem}

\subsection{Construction of Bourgain's Embedding}

Given a metric $(V, d)$ on $n$ points, we will randomly pick $k$ sets of nodes:
$S_1,...,S_k \subseteq V$ and define the embedding as
\[
\Phi: v_i \rightarrow <d(v_i, S_1), d(v_i, S_2), ..., d(v_i, S_k)>,
\]
where $d(v_i, S) = \min_{s \in S}{d(v_i, s)}$. Let $k = O(\log^2 n)$. For simplicity, we assume that
$n$ is power of two.

{\bf How to pick the subsets:}
\begin{center} pick $\log n$ random sets of size 1 \\
        pick $\log n$ random sets of size 2 \\
        pick $\log n$ random sets of size 4 \\
        ... \\
        pick $\log n$ random sets of size n/2
\end{center}
for each $j = 0,1,2,...,{\log n}-1$, we randomly and independently choose
$\log n$ subsets of $V$, each of cardinality $n/2^j$, i.e. each of the
$\left( {\begin{array}{*{20}c}
   n  \\
   {n/2^j }  \\
\end{array}} \right)$
substes of $V$ of cardinality $n/2^j$ is equally likely to be chosen.

\subsection{Proof of Bourgain's Theorem}

\begin{Proof}
Based on the embedding construction above, we claim that
\begin{eqnarray}
\Omega(\log n)*d(v_i, v_j) \le \|\phi(v_i), \phi(v_j)\|_1 \le k*d(v_i, v_j) \label{claim1}
\end{eqnarray}

\begin{Lemma}\label{lma:tri_ineq}
For any $S \subseteq V, |d(v_i, S)-d(v_j, S)| \le d(v_i, v_j)$ .
\end{Lemma}
\begin{figure}
\center \epsfig{file=tri_ineq.eps, width =150px} \caption{$|d(v_i, S)-d(v_j, S)| \le d(v_i, v_j)$} \label{fig:tri_ineq}
\end{figure}
As shown in Figure \ref{fig:tri_ineq}, for any $S \subseteq V$, assume $d(v_i, S) = d(v_i, v_m), d(v_j, S) = d(v_j, v_n)$, where $n$ and $m$ could be the same,
\begin{eqnarray}
|d(v_i, S)-d(v_j, S)| &=& |d(v_i, v_m)-d(v_j, v_n)| \nonumber\\
                      &\le& |d(v_i, v_m)-d(v_j, v_m)| \label{eqn:tri_1}\\
                      &\le& d(v_i, v_j) \label{eqn:tri_2}
\end{eqnarray}
It is clear to see that the inequality \ref{eqn:tri_1} holds due to the definition of the $d(v, S)$, and the
inequality \ref{eqn:tri_2} holds due to the triangle inequality.

{\bf Upper Bound:}
\begin{eqnarray}
\|\phi(v_i), \phi(v_j)\|_1 &=& \sum_{t=1..\log^2 n}|d(v_i, S_t)-d(v_j, S_t)| \nonumber \\
                           &\le& \log^2 n*d(v_i, v_j) \label{eqn:upr_bound}
\end{eqnarray}
Since we have proved the Lemma \ref{lma:tri_ineq}, it is clear to see that
the inequality \ref{eqn:upr_bound} holds. Then we have proved the upper bound for our claim
of inequality \ref{claim1}.

{\bf Lower Bound:}
\begin{figure}[htp]
\centering
\includegraphics[width=0.4\textwidth,width=100px,clip]{bad_ball.eps} \label{fig:bad_ball} \caption{$d(u,A) = d(v,A)$}
\end{figure}

First we consider a simple example shown in Figure \ref{fig:bad_ball}. Let $A_{ij}$ denotes the $j$th set selected with cardinality $2^{i-1}$. As shown in Figure \ref{fig:eg_1}, if the distance between $u$ and set $A_{ij}$ is the same as the one between $v$ and $A_{ij}$, that means $|d(u, A_{ij}) - d(v, A_{ij})| = 0$, which doesn't contributes to new metric distance $\|\phi(u), \phi(v)\|_1$. In order to fine the lower bound, we are more interested in the case that how the selected sets could contribute to $\|\phi(u), \phi(v)\|_1$.

\begin{Definition}
\label{def:ball}$B(x,r) = \{y \in V | d(x,y) \le r\}$. The ball includes all nodes the distance between which and $x$ is
no greater than $r$.
\end{Definition}
\begin{Definition}
\label{def:radius}for fixed $u$ and $v$, $R_t = \min\{r \colon |B(u, r)| \ge 2^t \& |B(v, r)| \ge 2^t\}$. Further, $R_t$ is defined if and only if $R_t \le d(u,v)/2$, i.e. we don't consider the case that $B(u,r)$ and $B(v,r)$ intersects, but they could touch.
\end{Definition}

\begin{figure}[htp]
\centering
\includegraphics[width=0.4\textwidth,width=200px,clip]{good_ball.eps} \label{fig:good_ball} \caption{good ball example}
\end{figure}
Now we consider the case shown in Figure \ref{fig:good_ball}. The set $A$ of cardinality $2^i$ intersects the ball $B(u, R_{i-1})$ and $B(v, R_i)$, then it is easy to see that $d(u, A) \le R_{i-1}$ and $d(v, A) \ge R_i$. Thus, $|d(u, A) - d(v, A)| \ge R_i-R_{i-1}$.

\begin{Claim}
When we select set $A$ of cardinality $2^i$, for any $i$, the probability that we chose a set that intersects $B(u, R_{i-1})$ and disjoints from $B(v, R_i)$ is constant.
\end{Claim}
\begin{Proof}
By Definition \ref{def:radius}, $|B(u,R_{i-1})| \le 2^i$ and $|B(v,R_i)| \ge 2^i$. So the probability of choosing a set A which intersects $|B(u,R_{i-1})|$ but misses $|B(v,R_i)|$ is
\begin{eqnarray}
Pr[A \cap B(u,R_{i-1}) \neq \emptyset \& A \cap B(v,R_i) = \emptyset]
                            &=&Pr[A \cap B(u,R_{i-1}) \neq \emptyset]*Pr[A \cap B(v,R_i) = \emptyset] \nonumber\\
                            &\ge& (1-(1-\frac{1}{2^i})^{2^{i-1}})*(1-\frac{1}{2^i})^{2^i} \nonumber\\
                            &\ge& (1-(1-\frac{1}{2^i})^{2^i/2})*(1-\frac{1}{2^i})^{2^i} \nonumber\\
                            &\ge& (1-e^{-1/2})*\frac{1}{4}\nonumber
\end{eqnarray}

Therefore, 
\begin{eqnarray}
E[d(u,A)-d(v,A)] &\ge& \sum_{t\ is\ defined} \Omega(1)(R_t-R_{t-1})  \label{eqn:telescope}\nonumber\\
                 &\ge& \Omega(1)(R_m-R_0) \label{eqn:reduction}\\
                 &\ge& \Omega(1)\frac{d(u,v)}{2} \label{eqn:result}\nonumber
\end{eqnarray}
It is easy to see that the inequality \ref{eqn:telescope} is of telescope style, and the 
inequality \ref{eqn:reduction} is the result of reducing \ref{eqn:telescope}.
By the definition of $R_m$, we know that the inequality \ref{eqn:result} holds for certain.

\begin{eqnarray}
E[\|\phi(u), \phi(v)\|_1] &\ge& E[\sum_{i=1..\log n} d(u,A)-d(v,A)] \nonumber\\
                          &\ge& \Omega(\log n)*d(u,v) \nonumber\\
\end{eqnarray}
\end{Proof}

Now we have proved $\Omega(\log n)*d(v_i, v_j) \le \|\phi(v_i), \phi(v_j)\|_1 \le k*d(v_i, v_j)$. Then according to the Linearity of a norm under scalar multiplication, i.e.$ \|\lambda\vec{v}\| = |\lambda|\|\vec{v}\|$,
we have
\[
\frac{d(v_i, v_j)}{\Omega(\log n)} \le \|\phi^\prime(v_i), \phi^\prime(v_j)\|_1 \le d(v_i, v_j),
\]
where $\phi^\prime(v_i) = \phi(v_i)/{log^2 n}$. Thus, this embedding has distortion of $O(log n)$ on $O(log^2 n)$ dimension.

\end{Proof}



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