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@@ -228,7 +228,7 @@ 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. I think normalized min-cut should not require $c(F)\leq b$. Further checks are needed. What we can guarantee is that $c(F)\leq B$.}
+% \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{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.
@@ -257,6 +257,9 @@ s.t.&       &   \sum_{e\in T} x_e+y_e&\geq 1   &   &\forall T\\
 \end{aligned}
 \end{equation}
 Note that we are forcing $b+1-\sum_e c(e)y_e=1$.
+\autoref{lp:normalized} is exactly the fractional optimum of \autoref{lp:conninterdict} since we have shown that to satisfy the complementary slackness condition the budget constraint has to be tight.
+
+How is this related to the optimal $\lambda$ in \autoref{lp:dualcutint}? The integrality gap of \autoref{ip:normalized} is also not clear.
 
 \subsection{integrality gap}
 I guess the 2-approximate min-cut enumeration algorithm implies an integrality gap of 2 for cut interdiction problem.