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@@ -257,6 +257,12 @@ I believe the previous conjecture is not likely to be true.
 
 \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 $[\min\frac{w(C\setminus F)}{B-c(F)},\min_{c(F)\leq b}\frac{w(C\setminus F)}{b-c(F)}]$.
 
+It would be nice if we can prove that any breakpoint is of the form $\min \frac{w(C\setminus F)}{b'-c(F)}$ for some $b'\in [b,B]$. However, this seems incorrect. Let $\{(C_0,F_0),\dots,(C_h,F_h)\}$ be the sequence of solutions for each segment on $L(\lambda)$ and let $\lambda_1< \dots <\lambda_{h}$ be the sequence of breakpoints. ($\lambda_i$ is the intersection of the corresponding segments of $(C_{i-1},F_{i-1})$ and $(C_i,F_i)$.)
+\begin{lemma}
+$\lambda_i=\min \frac{w(C\setminus F)-w(C_{i-1}\setminus F_{i-1})}{c(F_{i-1})-c(F)}$, where the minimum is taken over all cut $C$ and $F\subset C$ such that both the numerator and denominator are positive.
+\end{lemma}
+The proof is using the argument for showing $\lambda_1=\min \frac{w(C\setminus F)}{B-c(F)}$ and induction. $\lambda_i$ looks similar to normalized mincut but is related to the slope and vertical intercept of a previous segment.
+
 \begin{conjecture}
     \autoref{lp:conninterdict} has an integrality gap of 2.
 \end{conjecture}