  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.
