refine normalized knapsack problem formulation and update related lemmas
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main.tex
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main.tex
@ -21,9 +21,10 @@ 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 $\min \limits_{X
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\in \mathcal F, F\subset E} \frac{w(X\setminus F)}{b+1-c(F)}$ such that
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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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@ -39,12 +40,14 @@ w_\tau(e)=\begin{cases}
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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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proof is exactly the same as lemma 1 in \cite{vygen_fptas_2024}.
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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 $X\in \mathcal F$, $w_\tau(X)\ge \tau(1+b)$.
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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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proof is the same.
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\begin{lemma}
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% $w_\tau (X^N)\le \tau(b+1)$.
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@ -53,23 +56,32 @@ proof is the same.
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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+1-c(F^N)) + \tau\cdot c(F^N)\\
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& = \tau(b+1)
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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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Let $(X^*,F^*)$ be the optimal solution to \autoref{bfreeknap}.
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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}
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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 $w(X^*\setminus F^*)\geq
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\tau(\alpha+(\alpha-1)b)$.
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\tau(\alpha B-b)$.
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\end{lemma}
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The proof is the same.
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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}. Finally we get a knapsack version of Theorem 4:
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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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