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.

翻译：分区问题是一个著名的基本NP完全问题。本文主要考虑其优化版本。该问题长期以来从各个角度得到研究，并在实践中能高效求解。因此，我们可以说仅剩的任务是确定该问题能否在给定整数个数$n$的多项式时间内求解。我们提出了两个偏序集（posets），并给出了一种解决分区问题的新方法。第一个偏序集与一个已知偏序集序同构，其结构与子集和问题的解相关，而第二个是第一个的子偏序集，在本文中起关键作用。我们首先展示了这两个偏序集的若干性质，如大小、高度和宽度（由不可比较元素构成的子集的最大规模）。对于$n$模$4$余$0$或$3$的情况，两者的宽度相同且为$O(2^n / n^{3/2})$，这一事实表明了分区问题的难度。接着我们证明，所有候选解一般对应第二个偏序集的元素，其大小为$2^{n} - 2 \binom{n}{\lfloor n/2 \rfloor}$。由于一个分区对应偏序集中的两个元素，候选分区的数量为该值的一半，即$2^{n-1} - \binom{n}{\lfloor n/2 \rfloor}$。最后我们证明，基于偏序关系可以减少候选解。特别地，通过考虑第二个偏序集的极小元素，我们给出了若干多项式可解的情况。我们的方法为分区问题的结构分析提供了有价值的工具，并为P与NP问题提供了新视角。