A Faster FPTAS for #Knapsack

TL;DR

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Contribution

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Abstract

Given a set $W = \{w_1,\ldots, w_n\}$ of non-negative integer weights and an integer $C$, the #Knapsack problem asks to count the number of distinct subsets of $W$ whose total weight is at most $C$. In the more general integer version of the problem, the subsets are multisets. That is, we are also given a set $ \{u_1,\ldots, u_n\}$ and we are allowed to take up to $u_i$ items of weight $w_i$. We present a deterministic FPTAS for #Knapsack running in $O(n^{2.5}\varepsilon^{-1.5}\log(n \varepsilon^{-1})\log (n \varepsilon))$ time. The previous best deterministic algorithm [FOCS 2011] runs in $O(n^3 \varepsilon^{-1} \log(n\varepsilon^{-1}))$ time (see also [ESA 2014] for a logarithmic factor improvement). The previous best randomized algorithm [STOC 2003] runs in $O(n^{2.5} \sqrt{\log (n\varepsilon^{-1}) } + \varepsilon^{-2} n^2 )$ time. Therefore, in the natural setting of constant $\varepsilon$, we close the gap between the $\tilde O(n^{2.5})$ randomized algorithm and the $\tilde O(n^3)$ deterministic algorithm. For the integer version with $U = \max_i \{u_i\}$, we present a deterministic FPTAS running in $O(n^{2.5}\varepsilon^{-1.5}\log(n\varepsilon^{-1} \log U)\log (n \varepsilon) \log^2 U)$ time. The previous best deterministic algorithm [APPROX 2016] runs in $O(n^3\varepsilon^{-1}\log(n \varepsilon^{-1} \log U) \log^2 U)$ time.