Partially Ordered Sets Corresponding to the Partition Problem

May 09, 2024 ยท The Ethereal ยท ๐Ÿ› International Symposium on Fundamentals of Computation Theory

๐Ÿ”ฎ THE ETHEREAL: The Ethereal
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Authors Susumu Kubo arXiv ID 2405.05544 Category cs.DM: Discrete Mathematics Cross-listed cs.DS Citations 0 Venue International Symposium on Fundamentals of Computation Theory Last Checked 5 months ago
Abstract
The partition problem is a well-known basic NP-complete problem. We mainly consider the optimization version of it in this paper. The problem has been investigated from various perspectives for a long time and can be solved efficiently in practice. Hence, we might say that the only remaining task is to decide whether the problem can be solved in polynomial time in the number $n$ of given integers. We propose two partially ordered sets (posets) and present a novel methodology for solving the partition problem. The first poset is order-isomorphic to a well-known poset whose structures are related to the solutions of the subset sum problem, while the second is a subposet of the first and plays a crucial role in this paper. We first show several properties of the two posets, such as size, height and width (the largest size of a subset consisting of incomparable elements). Both widths are the same and $O(2^n / n^{3/2})$ for $n$ congruent to $0$ or $3$ (mod $4$). This fact indicates the hardness of the partition problem. We then prove that in general all the candidate solutions correspond to the elements of the second poset, whose size is $2^{n} - 2 \binom{n}{\lfloor n/2 \rfloor}$. Since a partition corresponds to two elements of the poset, the number of the candidate partitions is half of it, that is, $2^{n-1} - \binom{n}{\lfloor n/2 \rfloor}$. We finally prove that the candidate solutions can be reduced based on the partial order. In particular, we give several polynomially solvable cases by considering the minimal elements of the second poset. Our approach offers a valuable tool for structural analysis of the partition problem and provides a new perspective on the P versus NP problem.
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