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.