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@@ -196,6 +196,11 @@ 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.
+
+\subsection{integrality gap}
+I guess the 2 approximation cut enumeration algorithm implies a integrality gap of 2 for cut interdiction problem.
 
 \section{Random Stuff}
 
@@ -220,7 +225,7 @@ s.t.&   &   \sum_{e\in T} c(e)x_e&\geq k    &   &\forall T\\
 \end{aligned}
 \end{equation*}
 
-These two LPs have the same optimum. One can see that any feasible solution to 
+These two LPs have the same optimum. Any feasible solution to 
 LP1 is feasible in LP2. Thus $\opt(LP1) \geq \opt(LP2)$. Next we show that any 
 $x_e$ in  the optimum solution to LP2 is always in $[0,c(e)]$. 
 Let $x^*$ be the optimum and