Partially Ordered Sets Corresponding to the Partition Problem

TL;DR

This paper introduces two partially ordered sets related to the partition problem, providing a novel structural approach that offers insights into the problem's complexity and potential polynomial-time solutions for specific cases.

Contribution

The paper proposes a new poset-based methodology for analyzing the partition problem, revealing structural properties and polynomially solvable cases.

Findings

Both posets have width $O(2^n / n^{3/2})$ indicating problem hardness.

Candidate solutions correspond to elements of the second poset, with size $2^{n} - 2 inom{n}{loor{n/2}}$.

Polynomially solvable cases are identified by examining minimal elements of the second poset.

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.