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\documentclass[12pt]{article}
\usepackage{chao}
\geometry{a4paper,margin=2cm}
\usepackage[breakable, theorems, skins]{tcolorbox}
\tcbset{enhanced}
\DeclareRobustCommand{\note}[2][blue]{%
@ -174,7 +177,7 @@ Now this LP looks similar to the normalized min-cut problem.
A further reformulation (the new $x$ is $x-y$) gives us the following,
\begin{equation}\label{lp:cutinterdict}
\begin{equation}\label{lp:conninterdict}
\begin{aligned}
\min& & \sum_{e} x_e w(e) & & \\
s.t.& & \sum_{e\in T} x_e+y_e&\geq 1 & &\forall T\quad \text{($x+y$ is a cut)}\\
@ -227,13 +230,13 @@ We have $-\lambda(b-B)\leq w(C\setminus F)-\lambda(b-c(F))$ for any cut $C$ and
\note{It remains to show that the optimal solution at $\e$ guarantees $c(F)\leq b$? or maybe we don't need this for normalized mincut. I think normalized min-cut should not require $c(F)\leq b$. Further checks are needed. What we can guarantee is that $c(F)\leq B$.}
\subsection{a fundamental difference}
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. Need to understand this better...
\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.
\subsection{integrality gap}
I guess the 2-approximate min-cut enumeration algorithm implies an integrality gap of 2 for cut interdiction problem.
First consider the dual of linear relaxation of \autoref{lp:cutinterdict}.
First consider the dual of linear relaxation of \autoref{lp:conninterdict}.
\begin{equation}\label{lp:dualcutint}
\begin{aligned}
@ -252,10 +255,10 @@ I believe the previous conjecture is not likely to be true.
\paragraph{Weight truncation} Assuming we know the optimal $\lambda$ to the LP dual, \autoref{lp:dualcutint} in fact gives the idea of weight truncation. The capacity of each edge $e$ in the ``tree packing'' is $\min\{c(e)\lambda,w(e)\}$.
\paragraph{The optimal $\lambda$} Denote by $\lambda^*$ the optimal $\lambda$ that maximizes $L(\lambda)$. From the previous argument on the first segment of $L(\lambda)$ we know that $\lambda^* \geq \min \frac{w(C\setminus F)}{B-c(F)}$. Now assume $\lambda^* > \min_{c(F)\leq b} \frac{w(C\setminus F)}{b-c(F)}$. We have $\min w(C\setminus F)-\lambda^*(b-c(F))<w(C\setminus F)-\min_{c(F)\leq b} \frac{w(C\setminus F)}{b-c(F)}(b-c(F))=0$ since the optimum must be achieved by $F$ such that $0\leq b-c(F)$(the slope). The negative optimum of $L(\lambda)$ contradicts the fact that $L(0)=0$ and $L$ is concave. Hence, the optimal solution $\lambda^*$ is in the range $[\min\frac{w(C\setminus F)}{B-c(F)},\min_{c(F)\leq B}\frac{w(C\setminus F)}{b-c(F)}]$.
\paragraph{The optimal $\lambda$} Denote by $\lambda^*$ the optimal $\lambda$ that maximizes $L(\lambda)$. From the previous argument on the first segment of $L(\lambda)$ we know that $\lambda^* \geq \min \frac{w(C\setminus F)}{B-c(F)}$. Now assume $\lambda^* > \min_{c(F)\leq b} \frac{w(C\setminus F)}{b-c(F)}$. We have $\min w(C\setminus F)-\lambda^*(b-c(F))<w(C\setminus F)-\min_{c(F)\leq b} \frac{w(C\setminus F)}{b-c(F)}(b-c(F))=0$ since the optimum must be achieved by $F$ such that $0\leq b-c(F)$(the slope). The negative optimum of $L(\lambda)$ contradicts the fact that $L(0)=0$ and $L$ is concave. Hence, the optimal solution $\lambda^*$ is in the range $[\min\frac{w(C\setminus F)}{B-c(F)},\min_{c(F)\leq b}\frac{w(C\setminus F)}{b-c(F)}]$.
\begin{conjecture}
\autoref{lp:cutinterdict} has an integrality gap of 2.
\autoref{lp:conninterdict} has an integrality gap of 2.
\end{conjecture}
\section{Random Stuff}