LP method

main.tex

@@ -287,6 +287,64 @@ For LP, we assign $x=0$ and $y_e=\frac{1}{n-2}$ for every edge except $e_1$. The
 
 What is the gap if we only relax $\lambda$ in the Lagrangian dual?
 
+\section{LP method}
+Consider the following IP for connectivity interdiction:
+
+\begin{equation*}
+\begin{aligned}
+\min&   &                   \sum_{e} x_e w(e)          &   & \\
+s.t.&       &   \sum_{e\in T} x_e+y_e&\geq 1   &   &\forall T\quad \text{($x+y$ is a cut)}\\
+&       &   \sum_{e} y_e c(e) &\leq b   &   &\text{(budget for $F$)}\\
+% &       &   x_e&\geq y_e   &   &\forall e\quad(F\subset C)\\
+&       &   y_e,x_e&\in\{0,1\}      &   &\forall e
+\end{aligned}
+\end{equation*}
+ 
+We write its Lagrangian dual.
+\[
+\max_{\lambda\geq 0} \min_{\text{cut $C$ and $F\subseteq C$}} w(C-F)-\lambda(b-c(F))
+\]
+
+For fixed $\lambda$, we define $L(\lambda)=\min\limits_{\text{cut $C$ and $F\subseteq C$}} w(C)-w(F)+\lambda c(F)$. Note that the Lagrangian dual is $\max_\lambda L(\lambda) - \lambda b$. 
+
+If the optimal $C^*$ of $L(\lambda)$ is known, any edge in $C^*$ with $w(e)\geq c(e)\lambda$ will be added to $F^*$. So now the hard part is to find $C^*$. We reweight edges.
+
+\[
+w_\lambda(e)=\begin{cases}
+    w(e) & \text{if $w(e)< \lambda\cdot c(e)$ (light elem)}\\
+    \lambda\cdot c(e) & \text{otherwise (heavy elem)}
+\end{cases}
+\]
+
+Then clearly edges in $F^*$ are heavy and edges in $C^*-F^*$ are light.
+
+Now consider the min cut under capacity $w_\lambda$. For any cut $C$ in $G$, its capacity $w_\lambda(C)$ can be write as the sum of heavy elements ($\lambda c(F)$) and light elements ($w(C-F)$) which is at least $L(\lambda)$. 
+On the other hand, $C^*$ is a feasible cut and the capacity of $C^*$ is exactly $L(\lambda)$. Thus $C^*$ is a mincut in the reweighted graph and we can easily find it.
+
+\subsection{Approximation}
+
+How is $C^*$ related to the optimal solution to IP?
+
+% The argument is almost identical to lyft optimization
+
+One can see that the Lagrangian dual (denoted by LD) is at most the optimum of IP. So we have $\opt(LD) \leq \opt(IP)$.
+
+We can assume WLOG that the optimal $\lambda^*$ in LD is the intersection of two lines with positive and negative slopes. Then there exists an optimal solution $(\lambda^*, C^{LD},F^{LD})$ to LD such that $c(F^{LD})\leq b$. Then we have 
+\[
+\opt(LD)+\lambda^* b -\lambda c(F^{LD}) = w(C^{LD}-F^{LD}) \geq \opt(IP),
+\]
+since $\opt(IP)$ is the smallest $b$-free min cut.
+Thus we can approximate the optimum of IP within an additive error if $\lambda^*$ is known.
+
+We can also recover IPCO alg's 2-approximation.
+\begin{itemize}
+\item $L(\lambda^*)=\opt(LD)+\lambda^* b\geq \opt(IP)$
+\item $L(\lambda^*)-\lambda^* b=LD \leq IP$ and 
+\item $L(\lambda^*)\geq \lambda^* b$ (since $LD \geq 0$)
+\end{itemize}
+
+Then $\opt(IP)\in [L(\lambda^*),2L(\lambda^*)]$.
+
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