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@@ -225,7 +225,7 @@ Now we focus on $L(\lambda)=\min \{w(C\setminus F)-\lambda(b-c(F)) | \forall \te
 Let $B$ be the minimum cost of cuts in $G$. 
 We have $-\lambda(b-B)\leq w(C\setminus F)-\lambda(b-c(F))$ for any cut $C$ and $F\subsetneq C$. Thus the upperbound is $\e=\min \frac{w(C\setminus F)}{B-c(F)}$.
 
-\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.}
+\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{integrality gap}
 I guess the 2-approximate min-cut enumeration algorithm implies an integrality gap of 2 for cut interdiction problem.
@@ -245,6 +245,10 @@ We want to prove something like tree packing theorem for \autoref{lp:dualcutint}
     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}
 
+\begin{conjecture}
+    \autoref{lp:cutinterdict} has an integrality gap of 2.
+\end{conjecture}
+
 \section{Random Stuff}
 
 \subsection{remove box constraints}