gap 4 conj doesn't seem correct... fixed some setminus

main.tex

@@ -121,7 +121,7 @@ s.t.&       &   \sum_{e\in T} x_e+y_e&\geq 1   &   &\forall T & &\text{($x+y$ is
 \begin{frame}{Normalized Mincut from LP}
 One standard trick for dealing LPs with knapsack constraints is to consider its Lagrangian dual.
 \begin{equation*}
-\max_{\mu\geq 0} L(\mu)=  \max_{\mu\geq 0} \min \left\{ w(C-F)-\mu(b-c(F)) | \forall \text{cut $C$}\;\forall F\subset C
+\max_{\mu\geq 0} L(\mu)=  \max_{\mu\geq 0} \min \left\{ w(C\setminus F)-\mu(b-c(F)) | \forall \text{cut $C$}\;\forall F\subset C
 %  \land c(F)\leq b
 \right\}
 \end{equation*}
@@ -130,11 +130,11 @@ One standard trick for dealing LPs with knapsack constraints is to consider its
 $L(\mu)$ is piecewise linear and concave.
 \end{lemma}
 
-For fixed $\mu$, how to solve $\min\limits_{C,F} \left\{ w(C-F)-\mu(b-c(F))
+For fixed $\mu$, how to solve $\min\limits_{C,F} \left\{ w(C\setminus F)-\mu(b-c(F))
 \right\}$?
 
-For small enough $\mu\geq 0$, $w(C-F)$ term is dominanting.
-Thus the optimal solution must be $C=F=\text{mincut with weight $c$}$.
+For small enough $\mu\geq 0$, $w(C\setminus F)$ term is dominanting.
+Thus the optimal solution must be $C=F=\text{mincut with capacity $c$}$.
 \end{frame}
 
 
@@ -144,7 +144,7 @@ We have see that the first line segment is $L(\mu)=(\lambda_c-b)\mu$ where $\lam
 What is the first breakpoint on $L(\mu)$?
 \newline
 
-We have $-\mu(b-\lambda_c)\leq w(C- F)-\mu(b-c(F))$ for any cut $C$ and $F\subsetneq C$. Thus the first breakpoint is $\mu=\min \frac{w(C- F)}{\lambda_c-c(F)}$, which is the value of normalized mincut.
+We have $-\mu(b-\lambda_c)\leq w(C\setminus F)-\mu(b-c(F))$ for any cut $C$ and $F\subsetneq C$. Thus the first breakpoint is $\mu=\min \frac{w(C\setminus F)}{\lambda_c-c(F)}$, which is the value of normalized mincut.
 \newline
 
 What about other breakpoints?
@@ -173,19 +173,19 @@ Again we first assume the $\mu$ is fixed. Then for each pair of constraints $\su
 \end{frame}
 
 \begin{frame}{Integrality Gap}
-\begin{conjecture}
-ILP\ref{IP} has an integrality gap of 4.
-\end{conjecture}
+% \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 $w_\mu$.
+% 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}.
+% 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}