wrong conj

main.tex

@@ -268,13 +268,23 @@ The proof is using the argument for showing $\lambda_1=\min \frac{w(C\setminus F
     \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}.
+However, \autoref{conj:intgap} is wrong. The integrality gap is unbounded. Consider a cycle $C_n$ of $n$ vertices with two special edges $e_1,e_2$. Let $L$ be a large number.
+\[
+w(e)=\begin{cases}
+    1 & e=e_1\\
+    L & e=e_2\\
+    2 & \text{else}
+\end{cases},\quad
+c(e)=\begin{cases}
+    L & e=e_1\\
+    1 & \text{else}
+\end{cases}, \quad b=2-\epsilon
+\]
+
+For IP, it is clear that $F=\{e_2\}, C\setminus F=\{e_1\}$ and the optimum is 1\newline
+For LP, we assign $x=0$ and $y_e=\frac{1}{n-2}$ for every edge except $e_1$. The optimum is 0.
+
+What is the gap if we only relax $\lambda$ in the Lagrangian dual?
 
 \section{Random Stuff}