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lp for normalized mincut cont'd

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Yu Cong 2025-04-28 17:08:04 +08:00
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@ -228,7 +228,7 @@ 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. I think normalized min-cut should not require $c(F)\leq b$. Further checks are needed. What we can guarantee is that $c(F)\leq B$.}
% \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{differences}
Consider $L(\lambda)$ for cut problem. One can see that the optimal $\lambda$ is clearly 0 since $L(\lambda)$ is pwl concave and the slope is negative at $\lambda=0$. What we are really interested in is the first segment on $L$. At the left end, $L(0)$ is exactly the weight of minimum cut. (the complementary slackness condition is satisfied.) At the right end, as we have shown in the previous paragraph, $\lambda$ equals to the value of the strength (which is the optimum of the linear relaxation of the cut IP). However, for cut interdiction problems $L(0)$ is not the optimum.
@ -257,6 +257,9 @@ s.t.& & \sum_{e\in T} x_e+y_e&\geq 1 & &\forall T\\
\end{aligned}
\end{equation}
Note that we are forcing $b+1-\sum_e c(e)y_e=1$.
\autoref{lp:normalized} is exactly the fractional optimum of \autoref{lp:conninterdict} since we have shown that to satisfy the complementary slackness condition the budget constraint has to be tight.
How is this related to the optimal $\lambda$ in \autoref{lp:dualcutint}? The integrality gap of \autoref{ip:normalized} is also not clear.
\subsection{integrality gap}
I guess the 2-approximate min-cut enumeration algorithm implies an integrality gap of 2 for cut interdiction problem.