In this article, we discuss the $\mathcal{NP}$ problem, we do not follow the line of research of many researchers, which is to try to find such an instance Q, and the instance Q belongs to the class of $\mathcal{NP}$-complete, if the instance Q is proved to belong to $\mathcal{P}$, then $\mathcal{P}$ and $\mathcal{NP}$ are the same, if the instance Q is proved not to belong to $\mathcal{P}$, then $\mathcal{P}$ and $\mathcal{NP}$ are separated. Our research strategy in this article: Select an instance S of $\mathcal{EXP}$-complete and reduce it to an instance 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：本文讨论$\mathcal{NP}$问题，我们并未沿袭多数研究者所采用的研究路径，即试图寻找一个属于$\mathcal{NP}$完全类的问题实例Q，若证明Q属于$\mathcal{P}$类，则$\mathcal{P}$与$\mathcal{NP}$相同；若证明Q不属于$\mathcal{P}$类，则$\mathcal{P}$与$\mathcal{NP}$相分离。本文的研究策略为：选取一个$\mathcal{EXP}$完全类的问题实例S，将其在多项式时间内归约为一个$\mathcal{NP}$类的问题实例，由此S属于$\mathcal{NP}$类，从而得到$\mathcal{EXP} = \mathcal{NP}$，进而依据已知结论$\mathcal{P} \neq \mathcal{EXP}$，推导出$\mathcal{P} \neq \mathcal{NP}$。