Generalized polymorphisms
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_{...
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Zusammenfassung: | 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 $g_0 \neq g_1 = \cdots = g_n$. |
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DOI: | 10.48550/arxiv.2305.10073 |