\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 $\min \limits_{X
    \in \mathcal F, F\subset E} \frac{w(X\setminus F)}{B-c(F)}$ such that 
    $c(F)\leq b$.
\end{problem}
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.

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}
The proof is exactly the same as \cite[Lemma 1]{vygen_fptas_2024}.

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 $w_\tau$.

\begin{lemma}\label{lem:lb}
    For any $X\in \mathcal F$, $w_\tau(X)\ge \tau B$.
\end{lemma}


\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-c(F^N)) + \tau\cdot c(F^N)\\
                    & = \tau B
    \end{align*}
    Thus by \autoref{lem:lb}, $X^N$ gets the minimum.
\end{proof}

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 $\min_{F\in \mathcal{F}} w_\tau(F)$. 

\begin{lemma}[Lemma 4 in \cite{vygen_fptas_2024}]\label{lem:conditionalLB}
    Let $(X^*,F^*)$ be the optimal solution to \autoref{bfreeknap}.
    $X^*$ is either an $\alpha$-approximate solution to $\min_{X\in\mathcal F} 
    w_\tau(X)$ for some $\alpha>1$, or $w(X^*\setminus F^*)\geq 
    \tau(\alpha B-b)$.
\end{lemma}

% In fact, corollary 1 and theorem 5 are also the same as those in 
% \cite{vygen_fptas_2024}. 
Then following the argument of Corollary 1 in \cite{vygen_fptas_2024}, assume that $X^*$ is not an $\alpha$-approximate solution to $\min_{X\in\mathcal F} 
    w_\tau(X)$ for some $\alpha>1$. We have
\[
\frac{w(C^N\setminus F^N)}{w(C^*\setminus F^*)}\leq \frac{\tau(B-c(F^N))}{\tau(\alpha B-b)}\leq \frac{B}{\alpha B-b},
\]
where the second inequality uses \autoref{lem:conditionalLB}.
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 $\min_{X\in\mathcal F} w_\tau(X)$.

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?


\begin{equation*}
\begin{aligned}
\max&   &                   z&          &   & \\
s.t.&   &   \sum_{e} y_e c(e) &\leq B   &   &\text{(budget for $F$)}\\
&       &   \sum_{e\in T} x_e&\geq 1   &   &\forall T\quad \text{($x$ forms a cut)}\\
&       &   \sum_{e} \min(0,x_e-y_e) w(e)&\geq z   &   &\\
&       &   y_e,x_e&\in\{0,1\}      &   &\forall e
\end{aligned}
\end{equation*}

we can assume that $y_e\leq x_e$.

\begin{equation*}
\begin{aligned}
\min&   &                   \sum_{e} (x_e&-y_e) w(e)          &   & \\
% s.t.&   &   \sum_{e} (x_e-y_e) w(e)&\geq z   &   &\\
s.t.&       &   \sum_{e\in T} x_e&\geq 1   &   &\forall T\quad \text{($x$ forms 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*}

Now this LP looks similar to the normalized min-cut problem.

A further reformulation (the new $x$ is $x-y$) gives us the following,

\begin{equation*}
\begin{aligned}
\min&   &                   \sum_{e} x_e w(e)          &   & \\
s.t.&       &   \sum_{e\in T} x_e+y_e&\geq 1   &   &\forall T\quad \text{($x$ forms 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*}

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 
% \land c(F)\leq b
\}$. 
Consider the Lagrangian dual,
\begin{equation*}
\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
%  \land c(F)\leq b
\right\}
\end{equation*}


% 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.
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.

\subsection{graph strength}
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.

The principal sequence of partitions of $G$ is a piecewise linear concave curve $L(\lambda)=\min_\Pi c(\delta(\Pi))-\lambda |\Pi|$.

(there is a $\pm1$ difference...)


\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 $c(f)<x_f\in x^*$. Consider all constraints $\sum_{e\in T\setminus f} c(e)
x_e\geq k-c(f)$ on $T\ni f$. For any such constraint, we have $\sum_{e\in T} c(e)
x_e>k$ 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.

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?

\[
\min_{\text{cut C}, f\in C}\frac{\sum_{e\in C\setminus\set{f}}w(e)x_e}{k-c(f)}
\]
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