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lp for normalized mincut

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Yu Cong 2025-04-28 15:42:20 +08:00
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@ -233,6 +233,31 @@ We have $-\lambda(b-B)\leq w(C\setminus F)-\lambda(b-c(F))$ for any cut $C$ and
\subsection{differences} \subsection{differences}
Consider $L(\lambda)$ for cut problem. One can see that the optimal $\lambda$ is clearly 0 since $L(\lambda)$ is pwl concave and the slope is negative at $\lambda=0$. What we are really interested in is the first segment on $L$. At the left end, $L(0)$ is exactly the weight of minimum cut. (the complementary slackness condition is satisfied.) At the right end, as we have shown in the previous paragraph, $\lambda$ equals to the value of the strength (which is the optimum of the linear relaxation of the cut IP). However, for cut interdiction problems $L(0)$ is not the optimum. Consider $L(\lambda)$ for cut problem. One can see that the optimal $\lambda$ is clearly 0 since $L(\lambda)$ is pwl concave and the slope is negative at $\lambda=0$. What we are really interested in is the first segment on $L$. At the left end, $L(0)$ is exactly the weight of minimum cut. (the complementary slackness condition is satisfied.) At the right end, as we have shown in the previous paragraph, $\lambda$ equals to the value of the strength (which is the optimum of the linear relaxation of the cut IP). However, for cut interdiction problems $L(0)$ is not the optimum.
\subsection{normalized mincut}
I read this trick in sparsest cut notes\footnote{\url{https://courses.grainger.illinois.edu/cs598csc/fa2024/Notes/lec-sparsest-cut.pdf}}. First we write a IP for normalized mincut.
\begin{equation}
\begin{aligned}\label{ip:normalized}
\min& & \frac{\sum_e w(e)x_e}{b+1-\sum_e c(e)y_e}& & &\\
s.t.& & \sum_{e\in T} x_e+y_e&\geq 1 & &\forall T\\
& & \sum_{e} y_e c(e) &\leq b & &\\
% & & x_e&\geq y_e & &\forall e\quad(F\subset C)\\
& & y_e,x_e&\in\{0,1\} & &\forall e
\end{aligned}
\end{equation}
A lp relaxation would be the following,
\begin{equation}
\begin{aligned}\label{lp:normalized}
\min& & \sum_e w(e)x_e& & &\\
s.t.& & \sum_{e\in T} x_e+y_e&\geq 1 & &\forall T\\
& & \sum_{e} y_e c(e) &= b & &\\
% & & x_e&\geq y_e & &\forall e\quad(F\subset C)\\
& & y_e,x_e&\geq 0 & &\forall e
\end{aligned}
\end{equation}
Note that we are forcing $b+1-\sum_e c(e)y_e=1$.
\subsection{integrality gap} \subsection{integrality gap}
I guess the 2-approximate min-cut enumeration algorithm implies an integrality gap of 2 for cut interdiction problem. I guess the 2-approximate min-cut enumeration algorithm implies an integrality gap of 2 for cut interdiction problem.