\documentclass{beamer}
\usepackage{nicefrac}
\DeclareMathOperator*{\opt}{OPT}

\title[Edge Conn Interdiction]{Faster FPTAS for Edge Connectivity Interdiction}
\date{\today}
\author{丛宇}
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\usetheme{Simple}
\usepackage{algo}
    
\begin{document}
\begin{frame}[plain]
    % Print the title page as the first slide
    \titlepage
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\begin{frame}[plain]{Plan}
    \tableofcontents
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\section{Connectivity Interdiction}

\begin{frame}{Edge Connectivity \& Cut}
\begin{definition}
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 $G$.
\end{definition}

The edge connectivity of $G$ = $|\text{mincut of $G$}|-1$.
\newline

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.
\newline

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?
\end{frame}


\begin{frame}{Interdiction}
\begin{problem}[edge connectivity interdiction]
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 $G-F$.
\end{problem}

How to solve this problem if\dots
\begin{itemize}
    \item the optimal $F$ is given?
    \item the optimal $C$ is given?
\end{itemize}
\end{frame}


\begin{frame}{Example}
\begin{figure}
\includegraphics[width=0.8\textwidth]{images/knapsack.png}
\end{figure}
\end{frame}

\begin{frame}{Prevous Works}
Zenklusen \citep{zenklusen_connectivity_2014} first studied this problem and showed the following results:
\begin{itemize}
\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;
\item A $\tilde{O}(m^2 n^4)$ algorithm for the unit cost case\footnote{$\tilde{O}$ hides polylog factors}.
\end{itemize}

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 $\tilde{O}(m^2 n^4/\epsilon)$.
\end{frame}


\section{FPTAS}

\begin{frame}{Intermediate Problem}
\begin{problem}[Normalized Mincut]
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.
\end{problem}

Let $\tau$ be the optimum of Normalized Mincut. Consider a truncated weight $w_\tau(e)= \min \{w(e),c(e)\tau\}$.

\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}

\begin{frame}{Algorithm}
\begin{algo}
\underline{\textsc{FPTAS for Connectivity Interdiction}}$(G,w,c,b)$\\
1. estimate Normalized Mincut\\
2. enumerate all 2-approximate mincut with weight $w_\tau$\\
3. \quad for each cut $C$ solve a knapsack to compute $F$\\
return $(C,F)$ with smallest objective value.
\end{algo}

1 takes $O(\log_{1+\epsilon}(poly(n)))$ time;\newline
2 takes $O(n^4)$;\newline
3 takes $O(m^2/\epsilon)$.
\newline

complexity: $\tilde{O}(m^2n^4/\epsilon)$.
\end{frame}


\section{LP Perspective}

\begin{frame}{LP Method}
\citep{vygen_fptas_2024} gives a strong framework but the intuition behind is vague.
\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)}\\
&       &   \sum_{e} y_e c(e) &\leq b   & & &   &\text{(budget for $F$)}\\
% &       &   x_e&\geq y_e   &   &\forall e\quad(F\subset C)\\
&       &   y_e,x_e&\in\{0,1\}      &   &\forall e & &
\end{aligned}
\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}{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}
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\end{document}