Yes! Define $$ k = \left[ \begin{array}{} 1 & 1 & 1 \\ 1 & 9 & 1 \\ 1 & 1 & 1 \end{array} \right] $$ and $$ f(u) = \begin{cases} 1 & \text{if } u \in \{3, 11, 12\} \\ 0 & \text{else} \end{cases} $$ This corresponds to the previous parameterization via $(0,3)\to 3$, $(1,2)\to 11$, $(1,3)\to 12$. This works because $k*u_t(x)\geq9$ guarantees $u_t(x)=1$, and likewise $k*u_t(x)<9$ guarantees $u_t(x)=0$, with $k*u_t(x)\bmod 9$ being the neighbor count.
5 of 5
deleted 148 characters in body
user130609