Exploring P versus NP

TL;DR

This paper explores the P versus NP problem by attempting to prove P equals NP through a novel reduction from an EXP-complete problem to an NP problem, aiming to derive a contradiction to the widely accepted P ≠ NP.

Contribution

It introduces a unique approach of reducing an EXP-complete problem to NP to argue for P=NP, diverging from traditional methods focused on NP-complete problems.

Findings

Proposes that if an EXP-complete problem reduces to an NP problem in polynomial time, then EXP equals NP.

Uses this reduction to argue that P equals NP, challenging the common belief.

Suggests that proving P=NP would imply EXP=NP, leading to a contradiction with known complexity class separations.

Abstract

In this article, we discuss the question of whether P equals NP, we do not follow the line of research of many researchers, which is to try to find such a problem Q, and the problem Q belongs to the class of NP-complete, if the problem Q is proved to belong to P, then P and NP are the same, if the problem Q is proved not to belong to P, then P and NP are separated. Our research strategy in this article: Select a problem S of EXP-complete and reduce it to a problem of NP in polynomial time, then S belongs to NP, so EXP = NP, and then from the well-known P neq NP, derive P neq NP.