A Subexponential Reduction from Product Partition to Subset Sum

TL;DR

The paper tackles the Product Partition Problem (PPP), where one seeks a subset $\mathcal{C}$ of $\{1,\dots,n\}$ such that $\prod_{i\in\mathcal{C}} s_i = \prod_{i\notin\mathcal{C}} s_i$. It introduces a subexponential reduction to the Subset Sum Problem (SSP) by factoring each input number into primes, expressing $s_i$ via prime exponents to form an exponent matrix $S_M$, and showing PPP solvability is equivalent to the existence of $x\in\{0,1\}^n$ with $(x-\frac{1}{2}\mathbf{1})^T S_M = 0$. The authors then encode the $q$ simultaneous SSPs into a single SSP using a hat-constructed vector with $\hat{s}_i = \sum_{k=1}^q \alpha_{ik}\,k\,(nm)^{k-1}$ and $\hat{T} = \sum_{k=1}^q T_k\,k\,(nm)^{k-1}$, proving equivalence and bounding $q \le 2mn$, which yields a polynomial-size SSP. A numerical-analysis section proposes refined encodings with $N(q)\in[1,2]$ to potentially reduce growth and discusses a relaxed PPP criterion via a real-valued linear condition $x^T S_M=0$. Overall, the work shows that solving PPP reduces to SSP in subexponential time (via factoring) and to a polynomial-size SSP after exponent extraction, implying that polynomial-time factoring would yield a PPP solver through SSP, with future work aimed at tightening the factorization and improving SSP techniques.

Abstract

In this paper we study the Product Partition Problem (PPP), i.e. we are given a set of $n$ natural numbers represented on $m$ bits each and we are asked if a subset exists such that the product of the numbers in the subset equals the product of the numbers not in the subset. Our approach is to obtain the integer factorization of each number. This is the subexponential step. We then form a matrix with the exponents of the primes and show that the PPP has a solution iff some Subset Sum Problems have a common solution. Finally, using the fact that the exponents are not large we combine all the Subset Sum Problems in a single Subset Sum Problem (SSP) and show that its size is polynomial in $m,n$. We show that the PPP has a solution iff the final SSP has one.

Figures (1)

• Figure 1: In this figure one can see $\frac{\frac{N}{2}\cdot P'(N)}{ q \cdot N^{L \cdot (q-1)}}$ for different values of $q$.

Theorems & Definitions (11)

• Lemma 2.2
• proof
• Remark 2.3
• Remark 2.4
• Lemma 2.5
• proof
• Theorem 2.7
• proof
• Lemma 2.8
• proof