We find all functions $f_0,f_1,\dots,f_m\colon \{0,1\}^n \to \{0,1\}$ and $g_0,g_1,\dots,g_n\colon \{0,1\}^m \to \{0,1\}$ satisfying the following identity for all $n \times m$ matrices $(z_{ij}) \in \{0,1\}^{n \times m}$: \[ f_0(g_1(z_{11},\dots,z_{1m}),\dots,g_n(z_{n1},\dots,z_{nm})) = g_0(f_1(z_{11},\dots,z_{n1}),\dots,f_m(z_{1m},\dots,z_{nm})). \] Our results generalize work of Dokow and Holzman (2010), which considered the case $g_0 = g_1 = \cdots = g_n$, and of Chase, Filmus, Minzer, Mossel and Saurabh (2022), which considered the case $g_0 \neq g_1 = \cdots = g_n$.

翻译：我们寻找所有满足如下恒等式的函数 $f_0,f_1,\dots,f_m\colon \{0,1\}^n \to \{0,1\}$ 和 $g_0,g_1,\dots,g_n\colon \{0,1\}^m \to \{0,1\}$，该恒等式对所有 $n \times m$ 矩阵 $(z_{ij}) \in \{0,1\}^{n \times m}$ 成立：\[ f_0(g_1(z_{11},\dots,z_{1m}),\dots,g_n(z_{n1},\dots,z_{nm})) = g_0(f_1(z_{11},\dots,z_{n1}),\dots,f_m(z_{1m},\dots,z_{nm})). \] 我们的结果推广了Dokow和Holzman（2010）的工作（该工作考虑了 $g_0 = g_1 = \cdots = g_n$ 的情形）以及Chase、Filmus、Minzer、Mossel和Saurabh（2022）的工作（该工作考虑了 $g_0 \neq g_1 = \cdots = g_n$ 的情形）。