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diff --git a/main.tex b/main.tex
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--- a/main.tex
+++ b/main.tex
@@ -1,5 +1,8 @@
 \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}