Partially Ordered Sets Corresponding to the Partition Problem
Susumu Kubo
TL;DR
This work reframes the partition problem, an NP-hard optimization task, through a pair of posets $P(n)$ and $Q(n)$ built from sign-vectors in $\{1,-1\}^n$ with a cumulative-sum order. It establishes an order-isomorphism between $P(n)$ and the classic $M(n)$ via a mapping $f$, and proves that the width of these posets scales as $O(2^n / n^{3/2})$ for $n \equiv 0$ or $3$ (mod $4$). A key insight is that all candidate partitions correspond to elements of $Q(n)$, with their number given by $2^{n-1} - \binom{n}{\lfloor n/2 \rfloor}$, and that partial-order relations can reduce the search space, notably through the minimal elements $\mathbf{m}_k$, enabling polynomial-time solvable cases. This provides a structural, order-theoretic perspective on partition and subset-sum problems and offers a new angle on the P vs NP discussion.
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.