233 lines
11 KiB
TeX
233 lines
11 KiB
TeX
\documentclass[12pt]{article}
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\usepackage{chao}
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\title{connectivity interdiction}
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\author{}
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\date{}
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\DeclareMathOperator*{\opt}{OPT}
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\DeclareMathOperator*{\len}{len}
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\begin{document}
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\section{``Cut-free'' Proof}
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\begin{problem}[b-free knapsack]\label{bfreeknap}
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Consider a set of elements $E$ and two weights $w:E\to \Z_+$ and
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$c:E\to Z_+$ and a budget $b\in \Z_+$. Given a feasible set $\mathcal
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F\subset 2^E$, find $\min_{X
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\in \mathcal F, F\subset E} w(X\setminus F)$ such that $c(F)\leq b$.
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\end{problem}
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Always remember that $\mathcal F$ is usually not explicitly given.
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\begin{problem}[Normalized knapsack]\label{nknap}
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Given the same input as \autoref{bfreeknap}, find $\min \limits_{X
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\in \mathcal F, F\subset E} \frac{w(X\setminus F)}{B-c(F)}$ such that
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$c(F)\leq b$.
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\end{problem}
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In \cite{vygen_fptas_2024} the normalized min-cut problem use $B=b+1$. Here we use any integer $B>b$ and see how their method works.
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Denote by $\tau$ the optimum of \autoref{nknap}. Define a new weight
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$w_\tau:\E\to \R$,
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\[
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w_\tau(e)=\begin{cases}
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w(e) & \text{if $w(e)< \tau\cdot c(e)$ (light elem)}\\
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\tau\cdot c(e) & \text{otherwise (heavy elem)}
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\end{cases}
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\]
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\begin{lemma}
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Let $(X^N,F^N)$ be the optimal solution to \autoref{nknap}.
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Every element in $F^N$ is heavy.
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\end{lemma}
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The proof is exactly the same as \cite[Lemma 1]{vygen_fptas_2024}.
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The following two lemmas show (a general version of) that the optimal cut $C^N$ to normalized min-cut is exactly the minimum cut under weights $w_\tau$.
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\begin{lemma}\label{lem:lb}
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For any $X\in \mathcal F$, $w_\tau(X)\ge \tau B$.
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\end{lemma}
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\begin{lemma}
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% $w_\tau (X^N)\le \tau(b+1)$.
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$X^N\in \arg\min\limits_{X\in\mathcal F} w_\tau(X)$.
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\end{lemma}
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\begin{proof}
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\begin{align*}
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w_\tau (X^N) & \le w(X^N\setminus F^N) + w_\tau(F^N)\\
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& = \tau \cdot(B-c(F^N)) + \tau\cdot c(F^N)\\
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& = \tau B
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\end{align*}
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Thus by \autoref{lem:lb}, $X^N$ gets the minimum.
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\end{proof}
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Now we show the counter part of \cite[Theorem 5]{vygen_fptas_2024}, which states the optimal solution to \autoref{bfreeknap} is a $\alpha$-approximate solution to $\min_{F\in \mathcal{F}} w_\tau(F)$.
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\begin{lemma}[Lemma 4 in \cite{vygen_fptas_2024}]\label{lem:conditionalLB}
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Let $(X^*,F^*)$ be the optimal solution to \autoref{bfreeknap}.
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$X^*$ is either an $\alpha$-approximate solution to $\min_{X\in\mathcal F}
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w_\tau(X)$ for some $\alpha>1$, or $w(X^*\setminus F^*)\geq
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\tau(\alpha B-b)$.
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\end{lemma}
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% In fact, corollary 1 and theorem 5 are also the same as those in
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% \cite{vygen_fptas_2024}.
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Then following the argument of Corollary 1 in \cite{vygen_fptas_2024}, assume that $X^*$ is not an $\alpha$-approximate solution to $\min_{X\in\mathcal F}
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w_\tau(X)$ for some $\alpha>1$. We have
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\[
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\frac{w(C^N\setminus F^N)}{w(C^*\setminus F^*)}\leq \frac{\tau(B-c(F^N))}{\tau(\alpha B-b)}\leq \frac{B}{\alpha B-b},
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\]
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where the second inequality uses \autoref{lem:conditionalLB}.
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One can see that if $\alpha>2$, $\frac{w(C^N\setminus F^N)}{w(C^*\setminus F^*)}\leq \frac{B}{\alpha B-b} <1$ which implies $(C^*,F^*)$ is not optimal. Thus for $\alpha >2$, $X^*$ must be a $2$-approximate solution to $\min_{X\in\mathcal F} w_\tau(X)$.
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Finally we get a knapsack version of Theorem 4:
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\begin{theorem}[Theorem 4 in \cite{vygen_fptas_2024}]
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Let $X^{\min}$ be the optimal solution to $\min_{X\in\mathcal F} w_\tau(X)$.
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The optimal set $X^*$ in \autoref{bfreeknap} is a
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2-approximation to $X^{\min}$.
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\end{theorem}
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Thus to obtain a FPTAS for \autoref{bfreeknap}, one need to design a FPTAS for
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\autoref{nknap} and a polynomial time alg for finding all 2-approximations to
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$\min_{X\in\mathcal F} w_\tau(X)$.
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\paragraph{FPTAS for \autoref{nknap} in \cite{vygen_fptas_2024}} (The name
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``FPTAS'' here is not precise since we do not have a approximation scheme but
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an enumeration algorithm. But I will use this term anyway.) In their settings,
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$\mathcal F$ is the collection of all cuts in some graph.
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Let $\opt^N$ be the optimum of \autoref{nknap}. We can assume that there is no
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$X\in \mathcal F$ s.t. $c(X)\le b$ since this is polynomially detectable
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(through min-cut on $c(\cdot)$) and the optimum is 0. Thus we have
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$\frac{1}{b+1} \le \opt^N \le |E|\cdot \max_e w(e)$. Then we enumerate
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$\frac{(1+\varepsilon)^i}{b+1}$ where $i\in \set{0,1,\ldots,\floor{\log_{1+
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\varepsilon}(|E|w_{\max}(b+1))}}$. There is a feasible $i$ s.t. $(1-\varepsilon)
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\opt^N\le \frac{(1+\varepsilon)^i}{b+1}\leq \opt^N$ since
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$\frac{(1+\varepsilon)^i}{b+1}\le \opt^N\le \frac{(1+\varepsilon)^{i+1}}{b+1}$
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holds for some $i$.
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Note that this enumeration scheme also holds for arbitrary $\mathcal F$ if we
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have a non-zero lowerbound on $\opt^N$.
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\begin{conjecture}
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Let $(C,F)$ be the optimal solution to connectivity interdiction. The optimum
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cut $C$ can be computed in polynomial time. In other words, connectivity
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interdiction is almost as easy as knapsack.
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\end{conjecture}
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\section{Connections}
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For unit costs, connectivity interdiction with budget $b=k-1$ is the same
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problem as finding the minimum weighted edge set whose removal breaks $k$-edge
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connectivity.
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It turns out that \autoref{nknap} is just a necessary ingredient for MWU.
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Authors of \cite{vygen_fptas_2024} $\subset$ authors of
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\cite{chalermsook_approximating_2022}.
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How to derive normalized min cut for connectivity interdiction?
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\begin{equation*}
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\begin{aligned}
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\max& & z& & & \\
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s.t.& & \sum_{e} y_e c(e) &\leq B & &\text{(budget for $F$)}\\
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& & \sum_{e\in T} x_e&\geq 1 & &\forall T\quad \text{($x$ forms a cut)}\\
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& & \sum_{e} \min(0,x_e-y_e) w(e)&\geq z & &\\
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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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we can assume that $y_e\leq x_e$.
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\begin{equation*}
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\begin{aligned}
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\min& & \sum_{e} (x_e&-y_e) w(e) & & \\
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% s.t.& & \sum_{e} (x_e-y_e) w(e)&\geq z & &\\
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s.t.& & \sum_{e\in T} x_e&\geq 1 & &\forall T\quad \text{($x$ forms 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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Now this LP looks similar to the normalized min-cut problem.
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A further reformulation (the new $x$ is $x-y$) gives us the following,
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\begin{equation*}
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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\quad \text{($x$ forms 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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Note that now this is almost a positive covering LP. Let $L(\lambda)= \min \{ w(C\setminus F)-\lambda(b-c(F)) | \forall \text{cut $C$}\;\forall F\subset C
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% \land c(F)\leq b
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\}$.
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Consider the Lagrangian dual,
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\begin{equation*}
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\max_{\lambda\geq 0} L(\lambda)= \max_{\lambda\geq 0} \min \left\{ w(C\setminus F)-\lambda(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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% At this point, it becomes clear how the normalized min-cut is implicated in \cite{vygen_fptas_2024}. The optimum of normalized min-cut is exactly the value of $\lambda$ when $L(\lambda)$ is 0.
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We have shown that the budget $B$ in normalized min-cut does not really matter as long as $B>b$. Note that $L(\lambda)$ and the normalized min-cut look similar to the principal sequence of partitions of a graph and the graph strength problem.
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\subsection{graph strength}
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Assume that the graph $G$ is connected (otherwise add dummy edges).
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Given a graph $G=(V,E)$ and a cost function $c:V\to \Z_+$, the strength $\sigma(G)$ is defined as $\sigma(G)=\min_{\Pi}\frac{c(\delta(\Pi))}{|\Pi|-1}$, where $\Pi$ is any partition of $V$, $|\Pi|$ is the number of parts in the partition and $\delta(\Pi)$ is the set of edges between parts. Note that an alternative formulation of strength (using graphic matroid rank) is $\sigma(G)=\min_{F\subset E} \frac{|E-F|}{r(E)-r(F)}$, which in general is the fractional optimum of matroid base packing.
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The principal sequence of partitions of $G$ is a piecewise linear concave curve $L(\lambda)=\min_\Pi c(\delta(\Pi))-\lambda |\Pi|$. Cunningham used principal partition to computed graph strength\cite{cunningham_optimal_1985}. There is a list of good properties mentioned in \cite[Section 6]{chekuri_lp_2020}.
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\begin{itemize}
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\item $L(\lambda)$ is piecewise linear concave since it is the lower envelope of some line arrangement.
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\item Consider two adjacent breakpoints on $L$...
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\end{itemize}
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(there is a $\pm1$ difference between principal partition and graph strength... but we dont care those $c\lambda$ terms since the difficult part is minimize $L(\lambda)$ for fixed $\lambda$)
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\section{Random Stuff}
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\subsection{remove box constraints}
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Given a positive covering LP,
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\begin{equation*}
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\begin{aligned}
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LP1=\min& & \sum_e w(e) x_e& & & \\
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s.t.& & \sum_{e\in T} c(e)x_e&\geq k & &\forall T\\
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& & c(e)\geq x_e&\geq 0 & &\forall e,
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\end{aligned}
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\end{equation*}
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we want to remove constraints $c(e)\geq x_e$. Consider the following LP,
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\begin{equation*}
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\begin{aligned}
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LP2=\min& & \sum_e w(e) x_e& & & \\
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s.t.& & \sum_{e\in T} c(e)x_e&\geq k & &\forall T\\
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& & \sum_{e\in T\setminus f} c(e)x_e&\geq k - c(f) & &\forall T \;
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\forall f\in T\\
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& & x_e&\geq 0 & &\forall e,
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\end{aligned}
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\end{equation*}
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These two LPs have the same optimum. One can see that any feasible solution to
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LP1 is feasible in LP2. Thus $\opt(LP1) \geq \opt(LP2)$. Next we show that any
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$x_e$ in the optimum solution to LP2 is always in $[0,c(e)]$.
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Let $x^*$ be the optimum and
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suppose that $c(f)<x_f\in x^*$. Consider all constraints $\sum_{e\in T\setminus f} c(e)
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x_e\geq k-c(f)$ on $T\ni f$. For any such constraint, we have $\sum_{e\in T} c(e)
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x_e>k$ 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.
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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?
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\[
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\min_{\text{cut C}, f\in C}\frac{\sum_{e\in C\setminus\set{f}}w(e)x_e}{k-c(f)}
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\]
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\bibliographystyle{plain}
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\bibliography{ref}
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\end{document}
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