...

beamerthemeSimple.sty

@@ -167,8 +167,9 @@
 }
 
 % more theorem env
-\newtheorem{observation}{Observation}
-\newtheorem{question}{Question}
+\newtheorem{conjecture}[theorem]{Conjecture}
+\newtheorem{observation}[theorem]{Observation}
+\newtheorem{question}[theorem]{Question}
 
 
 % ----------------------------------------------------------------------

main.tex

@@ -1,5 +1,6 @@
 \documentclass{beamer}
 \usepackage{nicefrac}
+\DeclareMathOperator*{\opt}{OPT}
 
 \title[Edge Conn Interdiction]{Faster FPTAS for Edge Connectivity Interdiction}
 \date{\today}
@@ -78,7 +79,7 @@ The input is a graph $G=(V,E)$ with edge weights $w:E\to \Z_+$ and edge removal
 
 Let $\tau$ be the optimum of Normalized Mincut. Consider a truncated weight $w_\tau(e)= \min \{w(e),c(e)\tau\}$.
 
-\begin{theorem}
+\begin{theorem}\label{thm:2approx}
 The optimal cut $C^*$ for Connectivity Interdiction is a 2-approximation of global mincut with weights $w_\tau$.
 \end{theorem}
 \end{frame}
@@ -168,7 +169,21 @@ s.t.&       &       \sum_{T\ni e} z_T &\leq w(e)    &   &\forall e\in E\\
 Again we first assume the $\mu$ is fixed. Then for each pair of constraints $\sum z_T \leq w(e)$ and $\sum z_T \leq c(e)\mu$ only one of them works. The real capacity for this fractional tree packing is $\min\{w(e),c(e)\mu\}$, which is exactly the truncated weight $w_\tau$.
 \end{frame}
 
+\begin{frame}{Integrality Gap}
+\begin{conjecture}
+ILP\ref{IP} has an integrality gap of 4.
+\end{conjecture}
 
+Suppose $\mu^*$ is the optimal solution to LP\ref{lp:dualcutint}. Let $\lambda^{fr}$ be the fractional mincut with capacity $w_\mu$.
+
+we have
+\begin{align*}
+4\opt(LP)   &=4\lambda^{fr}-4b\mu^*\\
+            &\geq 2\lambda^{int} - 4b\mu^*\\
+            &\geq w_{\mu^*}(C^*)-b\mu^*,
+\end{align*}
+which implies Theorem \autoref{thm:2approx}.
+\end{frame}
 \begin{frame}{References}
 \bibliographystyle{plainnat}
 \bibliography{ref}