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@@ -154,7 +154,7 @@ 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*}
+\begin{equation}\label{lp:cutinterdict}
 \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)}\\
@@ -162,7 +162,7 @@ s.t.&       &   \sum_{e\in T} x_e+y_e&\geq 1   &   &\forall T\quad \text{($x+y$
 % &       &   x_e&\geq y_e   &   &\forall e\quad(F\subset C)\\
 &       &   y_e,x_e&\in\{0,1\}      &   &\forall e
 \end{aligned}
-\end{equation*}
+\end{equation}
 
 Note that now this is almost a positive covering LP. Let $L(\lambda)= \min \{ w(C\setminus F)-\lambda(b-c(F)) | \forall \text{cut $C$}\;\forall F\subset C 
 % \land c(F)\leq b
@@ -203,7 +203,22 @@ We are interested in the upperbound $\e$ of $\lambda$ such that the optimal $F$
 Now we focus on $L(\lambda)=\min \{w(C\setminus F)-\lambda(b-c(F)) | \forall \text{cut } C\;\forall F\subset C\}$. We can still assume that $G$ is connected and see that $L(\lambda)$ is pwl concave (1 and 2 still hold). Let $\lambda^*$ be a breakpoint on $L$. Suppose that there are two optimal solutions $(C_1,F_1)$ and $(C_2,F_2)$ at $\lambda^*$. For fixed $C$ ($C_1=C_2$), the same argument for principal partition still works. However, the difficult part is that $C$ might not be the same. So it's unlikely that 3 and 4 hold. For cut interdiction problem, 5 shows connections between normalized mincut and the original interdiction problem. Recall that we observe the denominator in normalized min-cut can be relaxed (that is, we can use $\frac{w(C\setminus F)}{B-c(F)}$ for any $B>b$, instead of restricting to $B=b+1$) and the analysis still works. Now following the previous argument for 5, we assume $\lambda\in [0,\e]$ for small enough positive $\e$. For any $C$, we have $F=C$ since $w(C\setminus F)$ is dominating. For the remaining term $-\lambda(b-c(F))$ we are selecting a cut $F$ with smallest cose with respect to $c$. We can assume that any cut in $G$ has larger cost than $b$ since otherwise the optimum is simply 0. Now we can see that $B$ in the denominator $B-c(F)$ should be the cost of mincut in $G$.
 
 \subsection{integrality gap}
-I guess the 2-approximate min-cut enumeration algorithm implies a 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.
+
+First consider the dual of linear relaxation of \autoref{lp:cutinterdict}.
+
+\begin{equation}\label{lp:dualcutint}
+\begin{aligned}
+\max&       &       \sum_T z_T &- b\lambda          &       &\\
+s.t.&       &       \sum_{T\ni e} z_T &\leq w(e)    &   &\forall e\in E\\
+    &       &       \sum_{T\ni e} z_T &\leq c(e)\lambda &   &\forall e \in E\\
+    &       &                           z_T,\lambda &\geq 0 &   &
+\end{aligned}
+\end{equation}
+We want to prove something like tree packing for \autoref{lp:dualcutint}.
+\begin{conjecture}
+    The optimum of \autoref{lp:dualcutint} is $\min \set{\frac{w(C\setminus F)}{B-c(F)}| \forall \text{cut $C$}, c(F)\leq b}$, where $B$ is the cost of mincut in $G$ and $b$ is the budget.
+\end{conjecture}
 
 \section{Random Stuff}