A faster FPTAS for #knapsack

Given a set W = (w₁, . . ., 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 (u₁, . . ., 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^2.5ε^{− 1.5} log(nε^{− 1}) log(nε)) time. The previous best deterministic algorithm [FOCS 2011] runs in O(n³ε^{− 1} log(nε^{− 1})) time (see also [ESA 2014] for a logarithmic factor improvement). The previous best randomized algorithm [STOC 2003] runs in O(n^2.5log(nε^{− 1}) + ε^{− 2}n²) time. Therefore, for the case of constant ε, we close the gap between the O(n^2.5) randomized algorithm and the O(n³) deterministic algorithm. For the integer version with U = max_i (u_i), we present a deterministic FPTAS running in O(n^2.5ε^{− 1.5} log(nε^{− 1} log U) log(nε) log² U) time. The previous best deterministic algorithm [TCS 2016] runs in O(n³ε^{− 1} log(nε^{− 1} log U) log² U) time.