\documentclass[a4paper]{article}
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\usepackage{algo}

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\title{A Note on Interdiction of Linear Minimization Problems}
\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}
    The strict $2$-approximation analysis generalizes to non-negative set function objectives.
    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}

\begin{figure}  
\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 $w_{\lambda^*}(S)<2L(\lambda^*)$\\
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}
\caption{Template for solving interdiction version of linear minimization problem.}
\label{fig:alg}
\end{figure}

The theorem gives the a general template shown in \autoref{fig:alg}.
$\lambda^*$ can be found using parametric search techniques.

\begin{lemma}[\cite{salowe_parametric}]\label{lem:para}
Let $S(n)$ be the complexity of computing $L(\lambda)=\min_{H\in\mathcal F} w_\lambda(H)$ for fixed $\lambda$ (where $n$ is the size of the input), then one can compute $\lambda^*$ using parametric search in $O(S(n)^2)$ time.

If there is a parallel algorithm that solves $L(\lambda)$ for fixed $\lambda$ using $P(n)$ processors in time $T(n)$, then one can compute $\lambda^*$ using parametric search in $O( S(n)T(n)\log P(n)+T(n)P(n) )$ 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}.
This algorithmic template shows the interdiction version of a minimization problem can be reduced to polynomially many knapsack problem if all strict $2$-approximations of a reweighted original problem can be enumerated in polynomial time.

\subsection{Application on Connectivity Interdiction}

Let $G=(V,E)$ be an undirected multigraph, and let $\mathcal F$ be the family of
all nontrivial cuts $\delta(U)$, where $\emptyset\neq U\subsetneq V$.  The
interdiction problem is
\[
    \min_{\substack{C\in\mathcal F,\;R\subseteq C\\c(R)\leq b}}
        w(C\setminus R).
\]
If there is a cut $C$ with $c(C)\leq b$, then the optimum is $0$: remove all
edges in $C$.  This case is detected by one min-cut computation under capacities
$c$, so below assume every cut has $c$-cost larger than $b$.

For a fixed $\lambda$, put capacity $w_\lambda(e)=\min\{w(e),\lambda c(e)\}$
on every edge.  Then
$
    L(\lambda)
    =
    \min_{C\in\mathcal F} w_\lambda(C),
$
so evaluating $L(\lambda)$ is just a global min-cut computation in
$(G,w_\lambda)$.  Using the deterministic almost-linear time min-cut algorithm
of Li \cite{Li_2021}, this takes $\tilde O(m)$ time.  Therefore,
\autoref{lem:para} gives a $\tilde O(m^2)$-time algorithm for computing $\lambda^*$.
If randomization is allowed, we can find $\lambda^*$ in near linear time.
Min-cut can be computed with high probability in $O(m\log^2 n)$ work and $O(\log^3 n)$ depth \cite{Anderson_Blelloch_2021}. Brent's law \cite{Gustafson_2011} shows that the running time is $T(n,P)=O(\frac{m\log^2 n}{P}+\log^3 n)$ with $P$ processors. Setting $P=O(m)$ gives a randomized $\tilde O(m)$-time algorithm for $\lambda^*$.

Let $L^*=L(\lambda^*)$.  By \autoref{thm:two-approx-witness}, the optimal
interdiction cut $C^*$ satisfies
$
    w_{\lambda^*}(C^*)<2L^*
$.
Thus $C^*$ is one of the strict $2$-approximate min-cuts in
$(G,w_{\lambda^*})$.  Karger \cite{Karger2000} showed that the number of
$\alpha$-approximate min-cuts is $O(n^{\floor{2\alpha}})$.  Hence the number of
cuts with value strictly smaller than $2L^*$ is $O(n^3)$, and they can be
enumerated in $O(n^3)$ time\footnote{We note that only succinct representation of cuts (with respect to Karger's tree packing) are needed and the cut edges can be recovered while solving the knapsack. However, it is not known if one can deterministically enumerate all $\alpha$-approximate min-cuts in time $O(n^{\floor{2\alpha}})$.}\cite{Karger2000}.
To sum up, our framework gives a FPRAS for connectivity interdiction with running time $\tilde O(m+n^3(m+\frac{1}{\e^2}))$.

Compared with \cite{huang_fptas_2024}, our analyais is based on LP methods and is more intuitive. An algorithmic improvement is that we use parametric search thus avoiding the enumeration of $\lambda^*$ (the optimum of a normalized min-cut problem in \cite{huang_fptas_2024}).

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