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fix wrong math

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Yu Cong 2025-05-20 18:52:48 +08:00
parent cecdc0660e
commit 3ae08afcb0
2 changed files with 6 additions and 3 deletions
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@ -171,14 +171,17 @@ s.t.& & \sum_{ij\in V\times V}\|v_i-v_j\|^2&=1 & &\\
\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 integral $x$ with $O(\sqrt{\log n}) \opt(SDP)$ upperbound on the objective.
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
\[
\frac{c(\delta(S))}{|S||V-S|}\leq \frac{\sum_{ij\in E} c_{ij}\|v_i-v_j\|^2}{\sum_{i\in S,j\in T} \|v_i-v_j\|^2}\leq ?
\frac{c(\delta(S))}{|S||V-S|}
\leq n|S| \frac{\sum_{ij\in E} c_{ij}\|v_i-v_j\|^2}{\sum_{i\in S,j\in T} \|v_i-v_j\|^2}
\leq |S| \frac{\sum_{ij\in E} c_{ij}\|v_i-v_j\|^2}{n} O(\sqrt{\log n})
\leq O(\sqrt{\log n}) \opt(SDP).
\]
This is the framework of the proof in \cite{arora_expander_2004}.
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 can move
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