The Algebraic Cost of a Boolean Sum

TL;DR

This paper investigates why the permanent polynomial is VNP-complete while the determinant is not, identifying a key property called the 'cost of a boolean sum' that distinguishes them.

Contribution

The authors isolate three properties needed for VNP-hardness and demonstrate that the permanent satisfies the 'cost of a boolean sum,' unlike the determinant, revealing a fundamental complexity difference.

Findings

Permanent satisfies the 'cost of a boolean sum' property.

Determinant does not satisfy this property.

The differentiation applies in border complexity and counting complexity.

Abstract

It is a well-known fact that the permanent polynomial is complete for the complexity class VNP, and it is largely suspected that the determinant does not share this property, despite its similar expression. We study the question of why the VNP-completeness proof of the permanent fails for the determinant. We isolate three fundamental properties that are sufficient to prove a polynomial sequence is VNP-hard, of which two are shared by both the permanent and the determinant. We proceed to show that the permanent satisfies the third property, which we refer to as the ``cost of a boolean sum," while the determinant does not, showcasing the fundamental difference between the polynomial families. We further note that this differentiation also applies in the border complexity setting and that our results apply for counting complexity.