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Classical and Quantum Algorithms for Variants of Subset-S...
Jonathan Allcock, Yassine Hamoudi, Antoine Joux, Felix Klingelhö · 2021-11-13 · via cs.DS updates on arXiv.org

Subset-Sum is an NP-complete problem where one must decide if a multiset of $n$ integers contains a subset whose elements sum to a target value $m$. The best-known classical and quantum algorithms run in time $\tilde{O}(2^{n/2})$ and $\tilde{O}(2^{n/3})$, respectively, based on the well-known meet-in-the-middle technique. Here we introduce a novel classical dynamic-programming-based data structure with applications to Subset-Sum and a number of variants, including Equal-Sums (where one seeks two disjoint subsets with the same sum), 2-Subset-Sum (a relaxed version of Subset-Sum where each item in the input set can be used twice in the summation), and Shifted-Sums, a generalization of both of these variants, where one seeks two disjoint subsets whose sums differ by some specified value. Given any modulus $p$, our data structure can be constructed in time $O(n^2p)$, after which queries can be made in time $O(n^2)$ to the lists of subsets summing to any value modulo $p$. We use this data structure in combination with variable-time amplitude amplification and a new quantum pair finding algorithm, extending the quantum claw finding algorithm to the multiple solutions case, to give an $O(2^{0.504n})$ quantum algorithm for Shifted-Sums, an improvement on the best-known $O(2^{0.773n})$ classical running time. Incidentally, we obtain new $\tilde{O}(2^{n/2})$ and $\tilde{O}(2^{n/3})$ classical and quantum algorithms for Subset-Sum, not based on the seminal meet-in-the-middle method. We also study Pigeonhole Equal-Sums and Pigeonhole Modular Equal-Sums, where the existence of a solution is guaranteed by the pigeonhole principle. For the former problem, we give faster classical and quantum algorithms with running time $\tilde{O}(2^{n/2})$ and $\tilde{O}(2^{2n/5})$, respectively. For the more general modular problem, we give a classical algorithm that also runs in time $\tilde{O}(2^{n/2})$.