breakpoints & weight truncation

main.tex

@@ -105,7 +105,7 @@ complexity: $\tilde{O}(m^2n^4/\epsilon)$.
 
 \begin{frame}{LP Method}
 \citep{vygen_fptas_2024} gives a strong framework but the intuition behind is vague.
-\begin{equation}
+\begin{equation}\label{IP}
 \begin{aligned}
 \min&   &                   \sum_{e} x_e w(e)          &   &  & &\\
 s.t.&       &   \sum_{e\in T} x_e+y_e&\geq 1   &   &\forall T & &\text{($x+y$ is a cut)}\\
@@ -116,6 +116,59 @@ s.t.&       &   \sum_{e\in T} x_e+y_e&\geq 1   &   &\forall T & &\text{($x+y$ is
 \end{equation}
 \end{frame}
 
+
+\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
+%  \land c(F)\leq b
+\right\}
+\end{equation*}
+
+\begin{lemma}
+$L(\mu)$ is piecewise linear and concave.
+\end{lemma}
+
+For fixed $\mu$, how to solve $\min\limits_{C,F} \left\{ w(C-F)-\mu(b-c(F))
+\right\}$?
+
+For small enough $\mu\geq 0$, $w(C-F)$ term is dominanting.
+Thus the optimal solution must be $C=F=\text{mincut with weight $c$}$.
+\end{frame}
+
+
+\begin{frame}{Breakpoints on $L(\mu)$}
+We have see that the first line segment is $L(\mu)=(\lambda_c-b)\mu$ where $\lambda_c$ is the value of mincut with capacity $c$. 
+
+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.
+\newline
+
+What about other breakpoints? Currently I can only prove the following :(
+\begin{lemma}
+$\lambda_i=\min \frac{w(C- F)-w(C_{i-1}- F_{i-1})}{c(F_{i-1})-c(F)}$, where the minimum is taken over all cut $C$ and $F\subset C$ such that both the numerator and denominator are positive.
+\end{lemma}
+\end{frame}
+
+
+\begin{frame}{Weight Truncation from LP}
+Consider the dual of linear relaxation of ILP\ref{IP}.
+
+\begin{equation}\label{lp:dualcutint}
+\begin{aligned}
+\max&       &       \sum_T z_T &- b\mu          &       &\\
+s.t.&       &       \sum_{T\ni e} z_T &\leq w(e)    &   &\forall e\in E\\
+    &       &       \sum_{T\ni e} z_T &\leq c(e)\mu &   &\forall e \in E\\
+    &       &                           z_T,\mu &\geq 0 &   &
+\end{aligned}
+\end{equation}
+
+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}{References}
 \bibliographystyle{plainnat}
 \bibliography{ref}