gap 4

main.tex

@@ -105,7 +105,7 @@ where the second inequality uses \autoref{lem:conditionalLB}.
 One can see that if $\alpha>2$, $\frac{w(C^N\setminus F^N)}{w(C^*\setminus F^*)}\leq \frac{B}{\alpha B-b} <1$ which implies $(C^*,F^*)$ is not optimal. Thus for $\alpha >2$, $X^*$ must be a $2$-approximate solution to $\min_{X\in\mathcal F} w_\tau(X)$.
 
 Finally we get a knapsack version of Theorem 4:
-\begin{theorem}[Theorem 4 in \cite{vygen_fptas_2024}]
+\begin{theorem}[Theorem 4 in \cite{vygen_fptas_2024}]\label{thm:main}
     Let $X^{\min}$ be the optimal solution to $\min_{X\in\mathcal F} w_\tau(X)$.
     The optimal set $X^*$ in \autoref{bfreeknap} is a 
     2-approximation to $X^{\min}$.
@@ -254,7 +254,7 @@ We want to prove something like tree packing theorem for \autoref{lp:dualcutint}
 
 I believe the previous conjecture is not likely to be true.
 
-\paragraph{Weight truncation} Assuming we know the optimal $\lambda$ to the LP dual, \autoref{lp:dualcutint} in fact gives the idea of weight truncation. The capacity of each edge $e$ in the ``tree packing'' is $\min\{c(e)\lambda,w(e)\}$. Therefore, the optimum of \autoref{lp:dualcutint} is $\Lambda_{w_\tau}-b\lambda$, where $\Lambda_{w_\tau}$ is the fractional mincut on $G$ with weights $w_\tau$.
+\paragraph{Weight truncation} Assuming we know the optimal $\lambda$ to the LP dual, \autoref{lp:dualcutint} in fact gives the idea of weight truncation. The capacity of each edge $e$ in the ``tree packing'' is $\min\{c(e)\lambda,w(e)\}$. Therefore, the optimum of \autoref{lp:dualcutint} is $\Lambda_{w_\tau}^{fr}-b\lambda$, where $\Lambda_{w_\tau}^{fr}$ is the fractional mincut on $G$ with weights $w_\tau$.
 
 \paragraph{The optimal $\lambda$} Denote by $\lambda^*$ the optimal $\lambda$ that maximizes $L(\lambda)$. From the previous argument on the first segment of $L(\lambda)$ we know that $\lambda^* \geq \min \frac{w(C\setminus F)}{B-c(F)}$. Now assume $\lambda^* > \min_{c(F)\leq b} \frac{w(C\setminus F)}{b-c(F)}$. We have $\min w(C\setminus F)-\lambda^*(b-c(F))<w(C\setminus F)-\min_{c(F)\leq b} \frac{w(C\setminus F)}{b-c(F)}(b-c(F))=0$ since the optimum must be achieved by $F$ such that $0\leq b-c(F)$(the slope). The negative optimum of $L(\lambda)$ contradicts the fact that $L(0)=0$ and $L$ is concave. Hence, the optimal solution $\lambda^*$ is in the range $[\min\frac{w(C\setminus F)}{B-c(F)},\min_{c(F)\leq b}\frac{w(C\setminus F)}{b-c(F)}]$.
 
@@ -264,10 +264,18 @@ $\lambda_i=\min \frac{w(C\setminus F)-w(C_{i-1}\setminus F_{i-1})}{c(F_{i-1})-c(
 \end{lemma}
 The proof is using the argument for showing $\lambda_1=\min \frac{w(C\setminus F)}{B-c(F)}$ and induction. $\lambda_i$ looks similar to normalized mincut but is related to the slope and vertical intercept of a previous segment.
 
-\begin{conjecture}
-    \autoref{lp:conninterdict} has an integrality gap of 2.
+\begin{conjecture}\label{conj:intgap}
+    \autoref{lp:conninterdict} has an integrality gap of 4.
 \end{conjecture}
 
+\paragraph{The plan} Assume the optimal $\lambda$ in \autoref{lp:dualcutint} is indeed $\frac{w(C\setminus F)}{b+k-c(F)}$, \autoref{conj:intgap} implies a stronger version of \autoref{thm:main}. Let $\lambda^*$ be the optimal solution to interdiction LP\ref{lp:dualcutint}. Let $w_{\lambda^*}$ be the truncated weight. Denote the fractional and integral solution to the mincut problem(tree packing) on weight $w_{\lambda^*}$ by $\Lambda_{w_{\lambda^*}}^{fr}$ and $\Lambda_{w_{\lambda^*}}^{int}$(subscripts will be ignored). We have $2\Lambda^{fr}\geq \Lambda^{int}$ by the integrality gap of graph cut. Let $(C^*,F^*)$ be the optimal solution to \autoref{lp:conninterdict} and let $\opt(IP)$ be the optimum. Recall that we have $w_{\lambda^*}(C^*)\leq b\lambda^*+\opt(IP)$. \autoref{conj:intgap} implies that
+\begin{align*}
+4\opt(LP)   &=4\Lambda^{fr}-4b\lambda^*\\
+            &\geq 2\Lambda^{int} - 4b\lambda^*\\
+            &\geq w_{\lambda^*}(C^*)-b\lambda^*.
+\end{align*}
+The last inequality shows \autoref{thm:main}.
+
 \section{Random Stuff}
 
 \subsection{remove box constraints}