sxlxc/edge_conn_interdiction
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i guess no c(F)≤b. c(F)≤B is enough.

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Yu Cong 2025-04-19 23:50:02 +08:00
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@ -225,7 +225,7 @@ 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.}
\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{integrality gap}
I guess the 2-approximate min-cut enumeration algorithm implies an integrality gap of 2 for cut interdiction problem.
@ -245,6 +245,10 @@ We want to prove something like tree packing theorem for \autoref{lp:dualcutint}
The optimum of \autoref{lp:dualcutint} is $\min \set{\frac{w(C\setminus F)}{B-c(F)}| \forall \text{cut $C$}, c(F)\leq b}$, where $B$ is the cost of mincut in $G$ and $b$ is the budget.
\end{conjecture}
\begin{conjecture}
\autoref{lp:cutinterdict} has an integrality gap of 2.
\end{conjecture}
\section{Random Stuff}
\subsection{remove box constraints}