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Yu Cong 2025-04-18 16:29:28 +08:00
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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 $\lambda$) (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} \section{Random Stuff}
@ -220,7 +225,7 @@ s.t.& & \sum_{e\in T} c(e)x_e&\geq k & &\forall T\\
\end{aligned} \end{aligned}
\end{equation*} \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 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 $[0,c(e)]$. $x_e$ in the optimum solution to LP2 is always in $[0,c(e)]$.
Let $x^*$ be the optimum and Let $x^*$ be the optimum and