A Faster FPTAS for #Knapsack

Abstract

Given a set W = {w_1,..., 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,..., 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}epsilon^{-1.5}log(n epsilon^{-1})log (n epsilon)) time. The previous best deterministic algorithm [FOCS 2011] runs in O(n^3 epsilon^{-1} log(n epsilon^{-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 epsilon^{-1})} + epsilon^{-2} n^2) time. Therefore, for the case of constant epsilon, we close the gap between the O~(n^{2.5}) randomized algorithm and the 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}epsilon^{-1.5}log(n epsilon^{-1} log U)log (n epsilon) log^2 U) time. The previous best deterministic algorithm [TCS 2016] runs in O(n^3 epsilon^{-1}log(n epsilon^{-1} log U) log^2 U) time.