l2 grid. how to avoid the dimension?

distribution.tex

@@ -103,4 +103,19 @@ We want to show that for any fixed $x,y\in X$ and $j$,
 
 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.
 
+\section{Grid}
+
+Recall that we need an algorithm that outputs an embedding which extends a $(k,c)$-outlier embedding into $\ell_2$ and we want the extended embedding to have a good (expected) expansion bound.
+
+\begin{conjecture}\label{conj:expansion}
+Let $(X,d)$ be a metric space such that $|X|=n$ and $\alpha: X\setminus K \to \R^d$ be a $(k,c)$-outlier embedding of $(X,d)$ into $\ell_2^{d}$, where $K\subset X$ is the outlier set. Then there exist an embedding $\beta: X\to \R^d$ such that $\beta$ completes $\alpha$ and has expansion bound
+\[
+\max_{x,y\in X} \frac{\norm{\beta(x)-\beta(y)}_2}{d(x,y)}\leq O(c\sqrt{\log k}).
+\]
+\end{conjecture}
+
+In their bi-criteria approximation the dimension $d$ is not important and therefore is considered as a fixed parameter. \autoref{conj:expansion} provides more tools than theorem 2.6, i.e. we know the coordinates of non-outlier points in the embedding $\beta$ and we can use coordinates in $\R^d$ instead of simply mapping non-outlier points to outliers.
+
+A common and powerful method is to use grid. We divide $\R^d$ into identical hypercubes of some sidelength $s$ and working with grid cells instead of points. However, this method often involves the dimension $d$, which is not desirable...
+
 \end{document}