impossible via bourgain's construction

algo.sty

@@ -0,0 +1,11 @@
+\def\begin@lg{\begin{minipage}{1in}\begin{tabbing}
+        \quad\=\qquad\=\qquad\=\qquad\=\qquad\=\qquad\=\qquad\=\kill}
+\def\end@lg{\end{tabbing}\end{minipage}}
+
+\newenvironment{algorithm}
+{\begin{tabular}{|l|}\hline\begin@lg}
+{\end@lg\\\hline\end{tabular}}
+
+\newenvironment{algo}
+{\begin{center}\begin{algorithm}}
+{\end{algorithm}\end{center}}

distribution.tex

@@ -1,5 +1,6 @@
 \documentclass[12pt]{article}
 \usepackage{chao}
+\usepackage{algo}
 
 \title{Outlier Embedding Notes}
 
@@ -43,7 +44,9 @@ We first ignore the outlier condition and see if stochastic embeddings break the
 For any metric space $(X,d)$ and for any $p$, there is an embedding of $(X,d)$ into $\ell_p^{O(\log^2 n)}$ with distortion $O(\log n)$.
 \end{theorem}
 
-Bourgain develops a randomized algorithm that finds a desired embedding.\footnote{The expansion bound always holds. The contraction bound holds with probability at least $1/2$. See \url{https://home.ttic.edu/~harry/teaching/pdf/lecture3.pdf}} For the $\ell_2$ case, the embedding has the following bounds:
+Bourgain develops a randomized algorithm that finds a desired embedding.\footnote{The expansion bound always holds. The contraction bound holds with probability at least $1/2$. See \url{https://home.ttic.edu/~harry/teaching/pdf/lecture3.pdf}}
+Can we get better expected distortion by repeating the algorithm and uniformly selecting an embedding?
+For the $\ell_2$ case, the embedding has the following bounds:
 \begin{enumerate}
 \item Expansion. $\|f(x)-f(y)\|_2\leq O(\log n) d(x,y)$
 \item Contraction. $\|f(x)-f(y)\|_2 \geq \frac{d(x,y)}{O(1)}$
@@ -60,18 +63,24 @@ The contraction bound is almost tight. Let $K$ be the dimension of the target sp
 \end{aligned}
 \end{equation*}
 
-One thing we can try is to tighten the second line. 
-Recall that for each dimension $i$ a random subset $S_i\subset X$ is selected and the value of $f_i(x)$ is $\min_{s\in S_i} d(x,s)$.
-We want to show that for any fixed $x,y\in X$ and any dimension $i$ the event that distance $|f_i(x)-f_i(y)|^2$ is much smaller than $d(x,y)^2$ happens with high probability.
+One thing we can try is to tighten the second line.
 
-Now consider a subset $S_j$ by sampling each node in $X$ iid with probability $2^{-j}$. We independently repeat this process $m=576\log n$ times and denote by $S_{ij}$ the sampled set for $i\in [m]$. A~free lemma is the following.
+\begin{algo}
+\underline{Bourgain's construction}:\\
+$m=576\log n$\\
+for $j=1$ to $\log n$:\\
+\quad for $i=1$ to $m$:\\
+\quad \quad choose set $S_{ij}$ by sampling each node in $X$ independently with probability $2^{-j}$\\
+\quad \quad $f_{ij}(x)=\min_{s\in S_{ij}} d(x,s)$\\
+$f(x)=\bigoplus_{j=1}^{\log n} \bigoplus_{i=1}^m f_{ij}(x)$ for all $x\in X$.
+\end{algo}
 
-\begin{lemma}
-For fixed $x,y\in X$ and $j$,
+% Recall that for each dimension $i$ a random subset $S_i\subset X$ is selected and the value of $f_i(x)$ is $\min_{s\in S_i} d(x,s)$.
+We want to show that for any fixed $x,y\in X$ and $j$,
 \[
-\Pr[\text{for at least $18\log n$ values of $i$, $|f_{ij}(x)-f_{ij}(y)|\geq (\rho_j -\rho_{j-1})$}]\geq 1-\frac{1}{n^3},
+\Pr[|f_{ij}(x)-f_{ij}(y)|\leq \frac{d(x,y)}{\polylog n}]\geq ???
 \]
-where $\rho_j$ is the smallest radius for which $|B(x,\rho_j)|\geq 2^j$ and $|B(y,\rho_j)|\geq 2^j$.
-\end{lemma}
+
+One can see that our desired event does not happen with high probability for any pair of $x,y$. Let the original metric space be a line metric with $n$ points. $x,y$ locate on two endpoints of an interval and the rest $n-2$ points locate on the middle of $xy$. Then our metric in the target space $|f_{ij}(x)-f_{ij}(y)|$ is a $\polylog n$ factor smaller than $d(x,y)$ if and only if both $x$ and $y$ are selected in $S_{ij}$, which happens with probability $4^{-j}$. This example shows that Bourgain's construction is tight up to a constant factor for some metric space.
 
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