A Faster FPTAS for #Knapsack

Given a set W = {w1, . . . , wn} 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, . . . , un} and we are allowed to take up to ui items of weight wi. We present a deterministic FPTAS for #Knapsack running in O(n2.5ε−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 √ log(nε−1) + ε−2n2) time. Therefore, in the natural setting of constant ε, we close the gap between the Õ(n) randomized algorithm and the Õ(n) deterministic algorithm. For the integer version with U = maxi {ui}, we present a deterministic FPTAS running in O(n2.5ε−1.5 log(nε−1 logU) log(nε) log U) time. The previous best deterministic algorithm [APPROX 2016] runs in O(n3ε−1 log(nε−1 logU) log U) time.