refine normalized knapsack problem formulation and update related lemmas

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Yu Cong 2025-04-12 15:47:32 +08:00
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@ -21,9 +21,10 @@ Always remember that $\mathcal F$ is usually not explicitly given.
\begin{problem}[Normalized knapsack]\label{nknap} \begin{problem}[Normalized knapsack]\label{nknap}
Given the same input as \autoref{bfreeknap}, find $\min \limits_{X 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 \in \mathcal F, F\subset E} \frac{w(X\setminus F)}{B-c(F)}$ such that
$c(F)\leq b$. $c(F)\leq b$.
\end{problem} \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 Denote by $\tau$ the optimum of \autoref{nknap}. Define a new weight
$w_\tau:\E\to \R$, $w_\tau:\E\to \R$,
@ -39,12 +40,14 @@ w_\tau(e)=\begin{cases}
Let $(X^N,F^N)$ be the optimal solution to \autoref{nknap}. Let $(X^N,F^N)$ be the optimal solution to \autoref{nknap}.
Every element in $F^N$ is heavy. Every element in $F^N$ is heavy.
\end{lemma} \end{lemma}
proof is exactly the same as lemma 1 in \cite{vygen_fptas_2024}. 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} \begin{lemma}\label{lem:lb}
For any $X\in \mathcal F$, $w_\tau(X)\ge \tau(1+b)$. For any $X\in \mathcal F$, $w_\tau(X)\ge \tau B$.
\end{lemma} \end{lemma}
proof is the same.
\begin{lemma} \begin{lemma}
% $w_\tau (X^N)\le \tau(b+1)$. % $w_\tau (X^N)\le \tau(b+1)$.
@ -53,23 +56,32 @@ proof is the same.
\begin{proof} \begin{proof}
\begin{align*} \begin{align*}
w_\tau (X^N) & \le w(X^N\setminus F^N) + w_\tau(F^N)\\ 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 \cdot(B-c(F^N)) + \tau\cdot c(F^N)\\
& = \tau(b+1) & = \tau B
\end{align*} \end{align*}
Thus by \autoref{lem:lb}, $X^N$ gets the minimum. Thus by \autoref{lem:lb}, $X^N$ gets the minimum.
\end{proof} \end{proof}
Let $(X^*,F^*)$ be the optimal solution to \autoref{bfreeknap}. 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} \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} $X^*$ is either an $\alpha$-approximate solution to $\min_{X\in\mathcal F}
w_\tau(X)$ for some $\alpha>1$, or $w(X^*\setminus F^*)\geq w_\tau(X)$ for some $\alpha>1$, or $w(X^*\setminus F^*)\geq
\tau(\alpha+(\alpha-1)b)$. \tau(\alpha B-b)$.
\end{lemma} \end{lemma}
The proof is the same.
In fact, corollary 1 and theorem 5 are also the same as those in % 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: % \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}] \begin{theorem}[Theorem 4 in \cite{vygen_fptas_2024}]
Let $X^{\min}$ be the optimal solution to $\min_{X\in\mathcal F} w_\tau(X)$. 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 The optimal set $X^*$ in \autoref{bfreeknap} is a