Deterministic Algorithms to Solve the $(n,k)$-Complete Hidden Subset Sum Problem
Lixia Luo, Changheng Li, Qiongxiu Li
TL;DR
This paper tackles the $(n,k)$-complete Hidden Subset Sum Problem (HSSP) by introducing two deterministic solvers. The first is a refined brute-force approach that leverages ordering of the unknowns to prune search, with a running time bound $O\left(\binom{n}{k}\left(\log \binom{n}{k}+\prod_{i=1}^{k-1}(\binom{n-k+i}{i}-i)\right)\right)$. The second method uses symmetric polynomials and Vieta's formulas: all elementary symmetric polynomials of the hidden multiset $X$ are derived from the $k$-subset sums, then $X$ is recovered as the roots of a univariate polynomial constructed via Vieta’s relations and Newton’s identities, yielding a bound $O\left(\sum_{u=1}^n p(u,\le k)^3+\binom{n}{k}n\right)$. A central theoretical result shows the determinant of the key transformation equals the Moser polynomial, tying invertibility to a known combinatorial quantity, and enabling exact recovery when invertible. The work also explores homogeneous symmetric polynomial rings, highlighting deeper algebraic structure and potential AI-privacy applications.
Abstract
The Hidden Subset Sum Problem (HSSP) is a significant NP-complete problem in number theory and combinatorics, with applications in cryptography and AI privacy. For the $(n,k)$-complete HSSP, where a target multiset must be recovered from its all $k$-subset sums, existing algorithms face limitations due to high complexity or intractability. This paper proposes two deterministic algorithms: a brute-force approach, and a novel method leveraging symmetric polynomials and Vieta's formulas with $O\left(\sum_{u=1}^n p(u,\leq k)^3+\binom{n}{k}n\right)$ complexity, where $ p(u,\leq k)$ counts the number of partitions of a positive integer $u$ into at most $k$ parts. The latter constructs an $n$-th degree polynomial via Vieta's formulas, whose roots correspond to the hidden multiset elements. Additionally, the discussion about the homogeneous symmetric polynomial rings is of independent interest.
