fix wrong math

main.tex

@@ -152,6 +152,36 @@ I believe the later method is more general and works for \nonuscut{}, while the
 \subsection{SDP $O(\sqrt{\log n})$ - \uscut{}}
 This $O(\sqrt{\log n})$ approximation via SDP is developed in \cite{arora_expander_2004}. This is also described in \cite[section 15.4]{Williamson_Shmoys_2011}.
 
+\begin{equation*}
+\begin{aligned}
+\min&   &   \frac{\sum_{ij\in E}c_{ij}(x_i-x_j)^2}{\sum_{ij\in V\times V}(x_i-x_j)^2}&  &   &\\
+s.t.&   &   (x_i-x_j)^2 + (x_j-x_k)^2&\geq (x_i-x_k)^2  &   &\forall i,j,k\in V\\
+    &   &                   x_i&\in \{+1,-1\}   &   &\forall i \in V
+\end{aligned}
+\end{equation*}
+
+This SDP models \uscut{} since every assignment of $x$ corresponds to a cut and the objective is the sparsity of the cut (up to a constant factor, but we don't care since we cannot achieve a constant factor approximation anyway). Now we consider a relaxation which is similar to \lp{}.
+
+\begin{equation*}
+\begin{aligned}
+\min&   &   \sum_{ij\in E}c_{ij}\|v_i-v_j\|^2&  &   &\\
+s.t.&   &                   \sum_{ij\in V\times V}\|v_i-v_j\|^2&=1  &   &\\
+    &   &   \|v_i-v_j\|^2 + \|v_j-v_k\|^2&\geq \|v_i-v_k\|^2  &   &\forall i,j,k\in V\\
+    &   &                   v_i&\in \R^n   &   &\forall i \in V
+\end{aligned}
+\end{equation*}
+
+To get a $O(\sqrt{\log n})$ (randomized) approximation algorithm we need to first solve the SDP and then round the solution to get a cut $\delta(S)$ with $c(\delta(S))=|S| \opt(SDP) O(n\sqrt{\log n})$. If we can find two sets $S,T\subset V$ both of size $\Omega(n)$ that are well-separated, in the sense that for any $s\in S$ and $t\in T$, $\|v_s-v_t\|^2=\Omega(1/\sqrt{\log n})$, then we have
+
+\[
+\frac{c(\delta(S))}{|S||V-S|}
+\leq n|S| \frac{\sum_{ij\in E} c_{ij}\|v_i-v_j\|^2}{\sum_{i\in S,j\in T} \|v_i-v_j\|^2}
+\leq |S| \frac{\sum_{ij\in E} c_{ij}\|v_i-v_j\|^2}{n} O(\sqrt{\log n}) 
+\leq O(\sqrt{\log n}) \opt(SDP).
+\]
+
+This is the framework of the proof in \cite{arora_expander_2004}.
+
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 \bibliography{ref}
 \end{document}