proof of the log n flow cut gap.

main.tex

@@ -132,14 +132,26 @@ Note that $D$ is routable iff the optimum of the LPs is at least 1. Then the the
 The flow-cut gap is $\opt(\ip{})/\opt(\lp{})$ \cite{leighton_multicommodity_1999}.
 
 Suppose that $G$ satisfies the cut condition, that is, $c(\delta(S))$ is at least the demand separated by $\delta(S)$ for all $S\subset V$. This implies $\opt(\ip{})\geq 1$ and in this case the largest integrality gap is $1/\opt(\lp{})$.
-
 For 1 and 2-commodity flow problem the gap is 1 \cite{Ford_Fulkerson_1956,Hu_1963}.
 However, for $k\geq 3$ the gap becomes larger\footnote{\url{https://en.wikipedia.org/wiki/Approximate_max-flow_min-cut_theorem}}.
 It is mentioned in \cite{leighton_multicommodity_1999} that \cite{schrijver_homotopic_1990} proved if the demand graph does not contain either  three disjoint edges or a triangle and a disjoint edge, then the gap is 1.
 
 For the $\Omega(\log n)$ lowerbound consider an uniform \scut{} instance on some 3-regular graph $G$ with unit capacity. In \cite{leighton_multicommodity_1999} they further required that for any $S\subset V$ and small constant $c$, $|\delta(S)|\geq c \min(|S|,|\bar S|)$. Then the value of the sparsest cut is at least $\frac{c}{n-1}$. Observe that for any fixed vertex $v$, there are at most $n/2$ vertices within distance $\log n-3$ of $v$. Thus at least half of the $\binom{n}{2}$ demand pairs are connected with shortest path of length at least $\log n-2$. To sustain a flow $f$ we need at least $\frac{1}{2}\binom{n}{2}(\log n -2)f\leq 3n/2$. Any feasible flow satisfies $f\leq \frac{3n}{\binom{n}{2}(\log n -2)}$ and the gap is therefore $\Omega(\log n)$.
+
 For the upperbound it suffices to show there exists a cut of ratio $O(f\log n)$.
-\cite{leighton_multicommodity_1999} gave an algorithmic proof based on \metric{}. This can also be proven using metric embedding theorem, see \url{https://courses.grainger.illinois.edu/cs598csc/fa2024/Notes/lec-sparsest-cut.pdf}. (I believe the later method is more general and works for \nonuscut{}, while the former method is limited to uniform \scut{}. However, the proof in \cite{leighton_multicommodity_1999} may have connections with the proof of Bourgain's thm? why does the method in \cite{leighton_multicommodity_1999} fail to work on \nonuscut{}?)
+\cite{leighton_multicommodity_1999} gave an algorithmic proof based on \metric{}. This can also be proven using metric embedding results. 
+We can solve \metric{} in polynomial time and get a metric on $V$. Then there is an embedding of $V$ into $\R^d$ with $\ell_1$ metric such that the distortion is $O(\log n)$. 
+Since $\ell_1$ metric is in the cut cone, our metric on $\R^d$ is a conic combination of cut metrics, which implies\footnote{This requires some work. See \url{https://courses.grainger.illinois.edu/cs598csc/fa2024/Notes/lec-sparsest-cut.pdf}} that there is a cut in the conic combination with value at most $O(\log n)\opt(\metric{})$. 
+To find such a cut it suffices to compute a conic combination of cut metrics which is exactly our $\ell_1$ metric in $\R^d$. One way to do this is test $(n-1)d$ cuts by observing the followings,
+\begin{enumerate}
+\item Every coordinate of $\R^d$ corresponds to a line metric;
+\item $\ell_1$ metric in $\R^d$ is the sum of those line metrics;
+\item Every line metric on $n$ points can be represented as some conic combination of $n-1$ cut metrics.
+\end{enumerate}
+
+\begin{remark}
+I believe the later method is more general and works for \nonuscut{}, while the former method is limited to uniform \scut{}. However, the proof in \cite{leighton_multicommodity_1999} may have connections with the proof of Bourgain's thm? why does the method in \cite{leighton_multicommodity_1999} fail to work on \nonuscut{}?
+\end{remark}
 
 \subsection{SDP $O(\sqrt{\log n})$}
 SDP approximation follows from metric embedding results.