...

main.tex

@@ -169,7 +169,7 @@ Note that now this is almost a positive covering LP. Let $L(\lambda)= \min \{ w(
 \}$. 
 Consider the Lagrangian dual,
 \begin{equation*}
-\max_{\lambda\geq 0} L(\lambda)=  \max_{\lambda\geq 0} \min \left\{ w(C\setminus F)-\lambda(b-c(F)), \forall \text{cut $C$}\;\forall F\subset C
+\max_{\lambda\geq 0} L(\lambda)=  \max_{\lambda\geq 0} \min \left\{ w(C\setminus F)-\lambda(b-c(F)) | \forall \text{cut $C$}\;\forall F\subset C
 %  \land c(F)\leq b
 \right\}
 \end{equation*}
@@ -197,7 +197,7 @@ The number of breakpoints on $L(\lambda)$ is at most $n-1$.
 (there is a $\pm1$ difference between principal partition and graph strength... but we dont care those $c\lambda$ terms since the difficult part is minimize $L(\lambda)$ for fixed $\lambda$)
 
 \subsection{principal sequence of partitions for cut interdiction}
-I don't expect similar results hold.
+Now we focus One $L(\lambda)=\min \{w(C\setminus F)-\lambda(b-c(F)) | \forall \text{cut } C\;\forall F\subset C\}$. We can still assume that $G$ is connected and see that $L(\lambda)$ is pwl concave. 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.
 
 \subsection{integrality gap}
 I guess the 2 approximation cut enumeration algorithm implies a integrality gap of 2 for cut interdiction problem.