\documentclass[12pt]{article} \usepackage{chao} \title{connectivity interdiction} \author{} \date{} \DeclareMathOperator*{\opt}{OPT} \DeclareMathOperator*{\len}{len} \begin{document} \section{``Cut-free'' Proof} \begin{problem}[b-free knapsack]\label{bfreeknap} Consider a set of elements $E$ and two weights $w:E\to \Z_+$ and $c:E\to Z_+$ and a budget $b\in \Z_+$. Given a feasible set $\mathcal F\subset 2^E$, find $\min_{X \in \mathcal F, F\subset E} w(X\setminus F)$ such that $c(F)\leq b$. \end{problem} Always remember that $\mathcal F$ is usually not explicitly given. \begin{problem}[Normalized knapsack]\label{nknap} Given the same input as \autoref{bfreeknap}, find $\min \limits_{X \in \mathcal F, F\subset E} \frac{w(X\setminus F)}{b+1-c(F)}$ such that $c(F)\leq b$. \end{problem} Denote by $\tau$ the optimum of \autoref{nknap}. Define a new weight $w_\tau:\E\to \R$, \[ w_\tau(e)=\begin{cases} w(e) & \text{if $w(e)< \tau\cdot c(e)$ (light elem)}\\ \tau\cdot c(e) & \text{otherwise (heavy elem)} \end{cases} \] \begin{lemma} Let $(X^N,F^N)$ be the optimal solution to \autoref{nknap}. Every element in $F^N$ is heavy. \end{lemma} proof is exactly the same as lemma 1 in \cite{vygen_fptas_2024}. \begin{lemma}\label{lem:lb} For any $X\in \mathcal F$, $w_\tau(X)\ge \tau(1+b)$. \end{lemma} proof is the same. \begin{lemma} % $w_\tau (X^N)\le \tau(b+1)$. $X^N\in \arg\min\limits_{X\in\mathcal F} w_\tau(X)$. \end{lemma} \begin{proof} \begin{align*} w_\tau (X^N) & \le w(X^N\setminus F^N) + w_\tau(F^N)\\ & = \tau \cdot(b+1-c(F^N)) + \tau\cdot c(F^N)\\ & = \tau(b+1) \end{align*} Thus by \autoref{lem:lb}, $X^N$ gets the minimum. \end{proof} Let $(X^*,F^*)$ be the optimal solution to \autoref{bfreeknap}. \begin{lemma} $X^*$ is either an $\alpha$-approximate solution to $\min_{X\in\mathcal F} w_\tau(X)$ for some $\alpha>1$, or $w(X^*\setminus F^*)\geq \tau(\alpha+(\alpha-1)b)$. \end{lemma} The proof is the same. In fact, corollary 1 and theorem 5 are also the same as those in \cite{vygen_fptas_2024}. Finally we get a knapsack version of Theorem 4: \begin{theorem}[Theorem 4 in \cite{vygen_fptas_2024}] Let $X^{\min}$ be the optimal solution to $\min_{X\in\mathcal F} w_\tau(X)$. The optimal set $X^*$ in \autoref{bfreeknap} is a 2-approximation to $X^{\min}$. \end{theorem} Thus to obtain a FPTAS for \autoref{bfreeknap}, one need to design a FPTAS for \autoref{nknap} and a polynomial time alg for finding all 2-approximations to $\min_{X\in\mathcal F} w_\tau(X)$. \paragraph{FPTAS for \autoref{nknap} in \cite{vygen_fptas_2024}} (The name ``FPTAS'' here is not precise since we do not have a approximation scheme but an enumeration algorithm. But I will use this term anyway.) In their settings, $\mathcal F$ is the collection of all cuts in some graph. Let $\opt^N$ be the optimum of \autoref{nknap}. We can assume that there is no $X\in \mathcal F$ s.t. $c(X)\le b$ since this is polynomially detectable (through min-cut on $c(\cdot)$) and the optimum is 0. Thus we have $\frac{1}{b+1} \le \opt^N \le |E|\cdot \max_e w(e)$. Then we enumerate $\frac{(1+\varepsilon)^i}{b+1}$ where $i\in \set{0,1,\ldots,\floor{\log_{1+ \varepsilon}(|E|w_{\max}(b+1))}}$. There is a feasible $i$ s.t. $(1-\varepsilon) \opt^N\le \frac{(1+\varepsilon)^i}{b+1}\leq \opt^N$ since $\frac{(1+\varepsilon)^i}{b+1}\le \opt^N\le \frac{(1+\varepsilon)^{i+1}}{b+1}$ holds for some $i$. Note that this enumeration scheme also holds for arbitrary $\mathcal F$ if we have a non-zero lowerbound on $\opt^N$. \begin{conjecture} Let $(C,F)$ be the optimal solution to connectivity interdiction. The optimum cut $C$ can be computed in polynomial time. In other words, connectivity interdiction is almost as easy as knapsack. \end{conjecture} \section{Connections} For unit costs, connectivity interdiction with budget $b=k-1$ is the same problem as finding the minimum weighted edge set whose removal breaks $k$-edge connectivity. It turns out that \autoref{nknap} is just a necessary ingredient for MWU. Authors of \cite{vygen_fptas_2024} $\subset$ authors of \cite{chalermsook_approximating_2022}. How to derive normalized min cut for connectivity interdiction? \section{Random Stuff} \subsection{remove box constraints} Given a positive covering LP, \begin{equation*} \begin{aligned} LP1=\min& & \sum_e w(e) x_e& & & \\ s.t.& & \sum_{e\in T} c(e)x_e&\geq k & &\forall T\\ & & c(e)\geq x_e&\geq 0 & &\forall e, \end{aligned} \end{equation*} we want to remove constraints $c(e)\geq x_e$. Consider the following LP, \begin{equation*} \begin{aligned} LP2=\min& & \sum_e w(e) x_e& & & \\ s.t.& & \sum_{e\in T} c(e)x_e&\geq k & &\forall T\\ & & \sum_{e\in T\setminus f} c(e)x_e&\geq k - c(f) & &\forall T \; \forall f\in T\\ & & x_e&\geq 0 & &\forall e, \end{aligned} \end{equation*} These two LPs have the same optimum. One can see that any feasible solution to LP1 is feasible in LP2. Thus $\opt(LP1) \geq \opt(LP2)$. Next we show that any $x_e$ in the optimum solution to LP2 is always in $[0,c(e)]$. Let $x^*$ be the optimum and suppose that $c(f)k$ since we assume $x_f>c(f)$, which means we can decrease $x_f$ without violating any constraint. Thus it contradicts the assumption that $x^*$ is optimal. Then we can add redundant constraints $x_e\leq c(e) \;\forall e$ to LP2 and see that LP1 and LP2 have the same optimum. This applies to \cite{chalermsook_approximating_2022} but cannot get an improvement on their algorithm.(MWU does not care the number of constraints.) So does this trick apply to connectivity interdiction? \[ \min_{\text{cut C}, f\in C}\frac{\sum_{e\in C\setminus\set{f}}w(e)x_e}{k-c(f)} \] \bibliographystyle{plain} \bibliography{ref} \end{document}