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notes.tex

@@ -0,0 +1,228 @@
+\documentclass[a4paper]{article}
+\usepackage{chao}
+\usepackage{algo}
+
+\geometry{margin=2cm}
+
+\title{A Note on Minimization Interdiction}
+\author{Yu Cong \and Kangyi Tian}
+\date{\today}
+
+\DeclareMathOperator*{\opt}{OPT}
+
+\begin{document}
+\maketitle
+
+\begin{abstract}
+Motivated by the FPTAS for connectivity interdiction of Huang \etal{}
+\cite{huang_fptas_2024}, we isolate the part of the argument that does not use
+cuts.  The setting is a minimization problem over a feasible-set family
+$\mathcal F$ with a linear objective $w(S)=\sum_{e\in S}w(e)$.  After dualizing
+the interdiction budget, deletion can be absorbed into truncated weights
+$w_\lambda(e)=\min\{w(e),\lambda c(e)\}$.  At an optimal Lagrange multiplier
+$\lambda^*$, the unknown optimal interdiction witness is a strict
+$2$-approximate minimizer of the reweighted problem.  Thus an exact algorithm
+can be obtained whenever one can optimize $w_{\lambda^*}$ over $\mathcal F$,
+enumerate all its $2$-approximate minimizers, and solve the remaining knapsack problem.
+\end{abstract}
+
+\section{Model}
+
+Let $E$ be a finite ground set and let $\mathcal F\subseteq 2^E$ be the family
+of feasible sets of an underlying linear minimization problem
+\[
+    \min_{S\in\mathcal F} w(S),
+\]
+where $w(S)=\sum_{e\in S}w(e)$ for weights $w:E\to \Z_+$.  We are also given
+an interdiction cost $c:E\to \Z_+$ and a budget $b\in \Z_+$.
+
+The linear minimization interdiction problem considered here is
+\begin{equation}\label{eq:interdiction}
+    \opt
+    =
+    \min_{\substack{S\in\mathcal F,\; R\subseteq S\\ c(R)\leq b}}
+        w(S\setminus R),
+\end{equation}
+where $c(R)=\sum_{e\in R}c(e)$.  This is equivalent to first deleting an
+arbitrary set $R$ with $c(R)\leq b$ and then minimizing $w(S\setminus R)$ over
+$S\in\mathcal F$, because only $R\cap S$ affects the value of a chosen feasible
+set $S$.
+
+Connectivity interdiction is the special case where $\mathcal F$ is the family
+of cuts of a graph.  In that case \eqref{eq:interdiction} is exactly the
+$b$-free min-cut formulation
+\[
+    \min_{\substack{\text{cut } C,\;R\subseteq C\\c(R)\leq b}}
+        w(C\setminus R).
+\]
+
+The argument below uses both minimization and linearity.  Minimization gives the
+near-minimizer certificate in \autoref{thm:two-approx-witness}; linearity gives
+the truncation formula $w_\lambda(e)=\min\{w(e),\lambda c(e)\}$.
+
+\section{Lagrangian relaxation}
+
+It is helpful to first look at \eqref{eq:interdiction} as an integer program,
+even though we do not write variables for the feasible-set constraints:
+\[
+\begin{aligned}
+\opt=\min&\quad w(S\setminus R)\\
+\text{s.t.}&\quad S\in\mathcal F,\quad R\subseteq S,\\
+&\quad c(R)\leq b.
+\end{aligned}
+\tag{IP}
+\]
+The set $S$ is the object chosen by the original minimization problem, and
+$R$ is the part of $S$ removed by the interdiction budget.  For a fixed $S$,
+choosing $R$ is a knapsack-like deletion problem; the global difficulty is that
+we do not know which feasible set $S$ will be optimal after deletion.
+
+The Lagrangian relaxation prices the single budget constraint.  For a multiplier
+$\lambda\geq 0$, move the constraint $c(R)\leq b$ into the objective:
+\[
+    \Phi(\lambda)
+    =
+    \min_{\substack{S\in\mathcal F\\R\subseteq S}}
+        \bigl(w(S\setminus R)+\lambda(c(R)-b)\bigr).
+\]
+For every fixed $\lambda$, this is a lower bound on the interdiction optimum:
+budget-feasible pairs get a non-positive correction term
+$\lambda(c(R)-b)$.  The Lagrangian dual is $\Lambda=\max_{\lambda\geq 0}\Phi(\lambda)$ and we denote a maximizer by $\lambda^*$. We assume $\Lambda>0$ since otherwise every feasible set $S$ would satisfy $c(S)\leq b$, so the interdiction problem would reduce to the original optimization problem.
+
+Now remove the constant term $-\lambda b$ from $\Phi(\lambda)$ and focus on the
+inner minimization
+\[
+    L(\lambda)
+    =
+    \min_{\substack{S\in\mathcal F\\R\subseteq S}}
+        \bigl(w(S\setminus R)+\lambda c(R)\bigr),
+\]
+so that $\Phi(\lambda)=L(\lambda)-\lambda b$.  For a fixed $S$ and $\lambda$,
+each element $e\in S$ has only two relevant choices: If $e$ is included in $R$, its cost will be $\lambda c(e)$; Otherwise the cost is $w(e)$.
+Thus the best truncated weight of whether including $e$ in $R$ is $w_\lambda(e)=\min\{w(e),\lambda c(e)\}$.
+Because both $w$ and $c$ are linear,
+\[
+\begin{aligned}
+    \min_{R\subseteq S}\bigl(w(S\setminus R)+\lambda c(R)\bigr)
+    &=\sum_{e\in S}\min\{w(e),\lambda c(e)\}\\
+    &=w_\lambda(S).
+\end{aligned}
+\]
+Therefore we have $L(\lambda)=\min_{S\in\mathcal F} w_\lambda(S)$, which is the original linear optimization problem under the truncated weight.
+
+The function $\Phi$ is the lower envelope of finitely many lines in $\lambda$,
+so it is piecewise-linear and concave.
+
+\begin{lemma}\label{lem:lagrangian-lower-bound}
+For every $\lambda\geq 0$, $\Phi(\lambda)\leq \opt$.  Hence
+$\Lambda\leq \opt$.
+\end{lemma}
+
+\begin{proof}
+Let $(S,R)$ be feasible for \eqref{eq:interdiction}.  Since $c(R)\leq b$, $w(S\setminus R)+\lambda(c(R)-b)\leq w(S\setminus R)$.
+Minimizing the left-hand side over all pairs $(S,R)$, and then minimizing the
+right-hand side only over budget-feasible pairs, gives
+$\Phi(\lambda)\leq \opt$.
+\end{proof}
+
+\begin{lemma}\label{lem:feasible-active}
+Assume $\lambda^*$ is a finite maximizer of $\Phi$.  Then there is a pair
+$(S^{LD},R^{LD})$ attaining $L(\lambda^*)$ such that $c(R^{LD})\leq b$.
+Consequently,
+\[
+    L(\lambda^*)\geq \opt.
+\]
+\end{lemma}
+
+\begin{proof}
+If every pair attaining $L(\lambda^*)$ had $c(R)>b$, then every active line in
+the lower envelope defining $\Phi$ would have positive slope at $\lambda^*$.
+For sufficiently small $\delta>0$, the value of the lower envelope would then
+increase from $\lambda^*$ to $\lambda^*+\delta$, contradicting the optimality of
+$\lambda^*$.
+
+Thus some active pair $(S^{LD},R^{LD})$ has $c(R^{LD})\leq b$.  This pair is
+feasible for \eqref{eq:interdiction}, so
+$w(S^{LD}\setminus R^{LD})\geq \opt$.  Therefore
+\[
+    L(\lambda^*)
+    =
+    w(S^{LD}\setminus R^{LD})+\lambda^*c(R^{LD})
+    \geq \opt.
+\]
+\end{proof}
+
+\section{The main observation}
+
+\begin{theorem}\label{thm:two-approx-witness}
+Let $(S^*,R^*)$ be an optimal solution to \eqref{eq:interdiction}.  If
+$\Lambda>0$, then
+\[
+    L(\lambda^*)
+    \leq
+    w_{\lambda^*}(S^*)
+    \leq
+    L(\lambda^*)+b\lambda^*
+    <
+    2L(\lambda^*).
+\]
+In particular, $S^*$ is a strict $2$-approximate minimizer of
+$\min_{S\in\mathcal F} w_{\lambda^*}(S)$.
+\end{theorem}
+
+\begin{proof}
+The lower bound follows immediately from the definition of $L(\lambda^*)$:
+\[
+    L(\lambda^*)=\min_{S\in\mathcal F}w_{\lambda^*}(S)
+    \leq w_{\lambda^*}(S^*).
+\]
+For the upper bound, use the particular deletion set $R^*$ inside the definition
+of $w_{\lambda^*}(S^*)$:
+\[
+\begin{aligned}
+    w_{\lambda^*}(S^*)
+    &\leq w(S^*\setminus R^*)+\lambda^*c(R^*) \\
+    &= \opt+\lambda^*c(R^*) \\
+    &\leq \opt+\lambda^*b \\
+    &\leq L(\lambda^*)+\lambda^*b,
+\end{aligned}
+\]
+where the last inequality is \autoref{lem:feasible-active}.  Finally,
+$\Lambda=L(\lambda^*)-\lambda^*b>0$, so
+$L(\lambda^*)+\lambda^*b<2L(\lambda^*)$.
+\end{proof}
+
+\begin{remark}
+    For connectivity interdiction, $S^*$ is the optimal interdiction cut, so it is
+    among the strict $2$-approximate min-cuts in the graph with capacities
+    $w_{\lambda^*}$.
+\end{remark}
+
+\section{Algorithmic template}
+
+The theorem gives the following general template.
+
+\begin{algo}
+\underbar{\textsc{Linear-Minimization-Interdiction}}$(E,\mathcal F,w,c,b)$:\\
+compute a maximizer $\lambda^*$ of $\Phi(\lambda)=L(\lambda)-\lambda b$\\
+compute the truncated weight $w_{\lambda^*}$\\
+enumerate every $S\in\mathcal F$ with $w_{\lambda^*}(S)<2L^*$\\
+for each enumerated $S$:\\
+\;\; compute $g_b(S)=\min\{w(S\setminus R):R\subseteq S,\ c(R)\leq b\}$\\
+return the pair $(S,R)$ with minimum value
+\end{algo}
+
+$\lambda^*$ can be found using parametric search techniques.
+
+\begin{lemma}[\cite{salowe_parametric}]\label{lem:para}
+Let $T(n)$ be the complexity of computing $L(\lambda)=\min_{S\in\mathcal F} w_\lambda(S)$ for fixed $\lambda$ (where $n$ is the size of the input), then one can compute $\lambda^*$ using parametric search in $O(T(n)^2)$ time.
+\end{lemma}
+
+Computing $g_b(S)$ is essentially solving a knapsack problem on groundset $S$ and takes $\tilde O(m+\frac{1}{\e^2})$ time for an $(1+\e)$-approximation \cite{10.1145/3618260.3649730}.
+
+\paragraph{Application on Connecticity Interdiction} $L(\lambda)$ can be obtained via min-cut in deterministic $\tilde O(m)$ time \cite{Li_2021}. Combining with \autoref{lem:para} gives a $\tilde O(m^2)$-time algorithm for $\lambda^*$. Karger \cite{Karger2000} showed the number of $\alpha$-approximate min-cut is $O(n^{\floor{2\alpha}})$. Thus the number of enumerated $S$ is $O(n^3)$ and we can enumerate them in ...
+
+\bibliographystyle{plain}
+\bibliography{ref}
+
+\end{document}

ref.bib

@@ -102,3 +102,18 @@
 	pages         = {683–695},
 	numpages      = {13}
 }
+
+@article{Karger2000,
+	title = {Minimum cuts in near-linear time},
+	volume = {47},
+	issn = {00045411},
+	url = {http://portal.acm.org/citation.cfm?doid=331605.331608},
+	doi = {10.1145/331605.331608},
+	number = {1},
+	journal = {Journal of the ACM},
+	author = {Karger, David R},
+	month = jan,
+	year = {2000},
+	pages = {46--76},
+}
+@inproceedings{Li_2021, address={Virtual Italy}, title={Deterministic mincut in almost-linear time}, ISBN={9781450380539}, url={https://dl.acm.org/doi/10.1145/3406325.3451114}, DOI={10.1145/3406325.3451114}, booktitle={Proceedings of the 53rd Annual ACM SIGACT Symposium on Theory of Computing}, publisher={ACM}, author={Li, Jason}, year={2021}, month=june, pages={384–395}, language={en} }