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 $\mathcal{NP}$-complete, if the problem Q is proved to belong to $\mathcal{P}$, then $\mathcal{P}$ and $\mathcal{NP}$ are the same, if the problem Q is proved not to belong to $\mathcal{P}$, then $\mathcal{P}$ and $\mathcal{NP}$ are separated. Our research strategy in this article: Select a problem S of $\mathcal{EXP}$-complete and reduce it to a problem of $\mathcal{NP}$ in polynomial time, then S belongs to $\mathcal{NP}$, so $\mathcal{EXP} = \mathcal{NP}$, and then from the well-known $\mathcal{P} \neq \mathcal{EXP}$, derive $\mathcal{P} \neq \mathcal{NP}$.

翻译：本文探讨了P是否等于NP的问题。我们并未沿袭众多研究者的研究路径——即试图寻找一个属于$\mathcal{NP}$-完全类的问题Q，若证明Q属于$\mathcal{P}$，则$\mathcal{P}$与$\mathcal{NP}$相同；若证明Q不属于$\mathcal{P}$，则$\mathcal{P}$与$\mathcal{NP}$相分离。本文的研究策略为：选取一个$\mathcal{EXP}$-完全问题S，将其在多项式时间内归约为一个$\mathcal{NP}$问题，则S属于$\mathcal{NP}$，故$\mathcal{EXP} = \mathcal{NP}$，进而依据已知结论$\mathcal{P} \neq \mathcal{EXP}$，推导出$\mathcal{P} \neq \mathcal{NP}$。