189 lines
6.8 KiB
TeX
189 lines
6.8 KiB
TeX
\documentclass{beamer}
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\DeclareMathOperator*{\opt}{OPT}
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\title[Edge Conn Interdiction]{Faster FPTAS for Edge Connectivity Interdiction}
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\date{\today}
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\author{丛宇}
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% \AtBeginSection[]{
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% \frame{\frametitle{Outline}\tableofcontents[currentsection,
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% subsectionstyle=show/show/shaded]}
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% }
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\usetheme{Simple}
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\usepackage{algo}
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\begin{document}
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\begin{frame}[plain]
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% Print the title page as the first slide
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\titlepage
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\end{frame}
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\begin{frame}[plain]{Plan}
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\tableofcontents
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\end{frame}
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\section{Connectivity Interdiction}
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\begin{frame}{Edge Connectivity \& Cut}
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\begin{definition}
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Let $G=(V,E)$ be a undirected connected graph. The minimum cut of $G$ is the minimum set of edges whose removal breaks the connectivity of $G$.
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\end{definition}
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The edge connectivity of $G$ = $|\text{mincut of $G$}|-1$.
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\newline
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Edge connectivity is an important measure of network reliability. The greater the edge connectivity, the more difficult it is to break the network's connectivity.
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\newline
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Now suppose that we want to attack the network. To what extent can we decrease the size of the min-cut by removing a limited set of edges?
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\end{frame}
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\begin{frame}{Interdiction}
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\begin{problem}[edge connectivity interdiction]
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The input is a graph $G=(V,E)$ with edge weights $w:E\to \Z_+$ and edge removal cost $c:E\to \Z_+$ and a budget $b\in \Z_+$. The goal is to find a interdiction set $F\subset E$ with $c(F)\leq b$ that minimizes the mincut in $G-F$.
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\end{problem}
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How to solve this problem if\dots
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\begin{itemize}
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\item the optimal $F$ is given?
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\item the optimal $C$ is given?
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\end{itemize}
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\end{frame}
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\begin{frame}{Example}
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\begin{figure}
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\includegraphics[width=0.8\textwidth]{images/knapsack.png}
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\end{figure}
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\end{frame}
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\begin{frame}{Prevous Works}
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Zenklusen \citep{zenklusen_connectivity_2014} first studied this problem and showed the following results:
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\begin{itemize}
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\item A PTAS\footnote{polynomial time approximation scheme. The running time is polynomial in the input size if $\epsilon$ is fixed.} for edge connectivity interdiction;
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\item A $\tilde{O}(m^2 n^4)$ algorithm for the unit cost case\footnote{$\tilde{O}$ hides polylog factors}.
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\end{itemize}
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Later \citep{vygen_fptas_2024} discovered an FPTAS\footnote{Fully PTAS. The running time is polynomial in both the input size and $1/\epsilon$} with time complexity $\tilde{O}(m^2 n^4/\epsilon)$.
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\end{frame}
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\section{FPTAS}
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\begin{frame}{Intermediate Problem}
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\begin{problem}[Normalized Mincut]
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The input is a graph $G=(V,E)$ with edge weights $w:E\to \Z_+$ and edge removal cost $c:E\to \Z_+$ and a budget $b\in \Z_+$. Find an edge set $F\subset E$ with $c(F)\leq b$ and a cut $C$ such that $\frac{w(C-F)}{b+1-c(F)}$ is minimized.
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\end{problem}
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Let $\tau$ be the optimum of Normalized Mincut. Consider a truncated weight $w_\tau(e)= \min \{w(e),c(e)\tau\}$.
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\begin{theorem}\label{thm:2approx}
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The optimal cut $C^*$ for Connectivity Interdiction is a 2-approximation of global mincut with weights $w_\tau$.
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\end{theorem}
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\end{frame}
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\begin{frame}{Algorithm}
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\begin{algo}
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\underline{\textsc{FPTAS for Connectivity Interdiction}}$(G,w,c,b)$\\
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1. estimate Normalized Mincut $\tau$\\
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2. \quad enumerate all 2-approximate mincut with weight $w_\tau$\\
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3. \quad \quad for each cut $C$ solve a knapsack to compute $F$\\
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4. return $(C,F)$ with smallest objective value.
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\end{algo}
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1 takes $O(\log_{1+\epsilon}(poly(n)))$ time;\newline
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2 takes $O(n^4)$;\newline
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3 takes $O(m^2/\epsilon)$.
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\newline
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complexity: $\tilde{O}(m^2n^4/\epsilon)$.
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\end{frame}
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\section{LP Perspective}
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\begin{frame}{LP Method}
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\citep{vygen_fptas_2024} gives a strong framework but the intuition behind is vague.
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\begin{equation}\label{IP}
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\begin{aligned}
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\min& & \sum_{e} x_e w(e) & & & &\\
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s.t.& & \sum_{e\in T} x_e+y_e&\geq 1 & &\forall T & &\text{($x+y$ is a cut)}\\
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& & \sum_{e} y_e c(e) &\leq b & & & &\text{(budget for $F$)}\\
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% & & x_e&\geq y_e & &\forall e\quad(F\subset C)\\
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& & y_e,x_e&\in\{0,1\} & &\forall e & &
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\end{aligned}
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\end{equation}
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\end{frame}
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\begin{frame}{Normalized Mincut from LP}
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One standard trick for dealing LPs with knapsack constraints is to consider its Lagrangian dual.
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\begin{equation*}
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\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
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% \land c(F)\leq b
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\right\}
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\end{equation*}
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\begin{lemma}
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$L(\mu)$ is piecewise linear and concave.
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\end{lemma}
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For fixed $\mu$, how to solve $\min\limits_{C,F} \left\{ w(C\setminus F)-\mu(b-c(F))
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\right\}$?
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For small enough $\mu\geq 0$, $w(C\setminus F)$ term is dominanting.
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Thus the optimal solution must be $C=F=\text{mincut with capacity $c$}$.
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\end{frame}
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\begin{frame}{Breakpoints on $L(\mu)$}
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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 $c$.
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What is the first breakpoint on $L(\mu)$?
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\newline
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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.
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\newline
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What about other breakpoints?
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\begin{lemma}
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$\mu^*\in [\min\limits_{C,F} \frac{w(C\setminus F)}{\lambda_c - c(F)},\min\limits_{C,F} \frac{w(C\setminus F)}{b - c(F)}]$
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\end{lemma}
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\begin{lemma}
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$\mu_i=\min \frac{w(C\setminus F)-w(C_{i-1}\setminus F_{i-1})}{c(F_{i-1})-c(F)}$, where the minimum is taken over all cut $C$ and $F\subset C$ such that both the numerator and denominator are positive.
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\end{lemma}
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\end{frame}
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\begin{frame}{Weight Truncation from LP}
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Consider the dual of the linear relaxation of IP\ref{IP}.
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\begin{equation}\label{lp:dualcutint}
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\begin{aligned}
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\max& & \sum_T z_T &- b\mu & &\\
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s.t.& & \sum_{T\ni e} z_T &\leq w(e) & &\forall e\in E\\
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& & \sum_{T\ni e} z_T &\leq c(e)\mu & &\forall e \in E\\
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& & z_T,\mu &\geq 0 & &
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\end{aligned}
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\end{equation}
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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 $w_\tau$.
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\end{frame}
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\begin{frame}{Integrality Gap}
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\begin{conjecture}\label{conj:gap2}
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IP\ref{IP} has an integrality gap of 2.
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\end{conjecture}
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Suppose that $\mu^*$ is the optimal solution to LP\ref{lp:dualcutint}. Let $\lambda^{fr}$ and $\lambda^{int}$ be the fractional and integral mincut with capacity $w_{\mu^*}$.
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\newline
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Conjecture \ref{conj:gap2} implies $w_{\mu^*}(C^*)+b\mu^* \geq (\text{value of mincut with $w_{\mu^*}$})$, \newline which is stronger than Theorem \ref{thm:2approx}.
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\end{frame}
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\begin{frame}{References}
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\bibliographystyle{plainnat}
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\bibliography{ref}
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\end{frame}
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\end{document} |