The following is a distillation of part of a larger research problem. (I have several clunky proofs for special cases, but an elegant method for the general case is somehow escaping me.)
Consider a graph $G$ with nodes arranged in a pyramid: $1$ node in the first row, $1+d$ nodes in the second, $1+2d$ nodes in the third, and so on, for $m$ rows. Each node takes a state of either 0 or 1.
Consider any row $r < m$. Suppose that $v$ is the $k$th node in this row, and call the $k$th through $(k+d)$th nodes in row $r+1$ the parents of $v$. Let $f_v$ denote a $(1+d)$-ary function from the state assignments of these parents to the state assignment of $v$.
Let $S$ be some subset of the nodes in row $m$. Then $f_S$ is an $|S|$-ary function that maps the state assignments of the nodes in $S$ to a state assignment for the node in row $1$.
Say that a subset $S$ of the nodes in row $m$ is free if there is a state assignment for the rest of row $m$ such that for any function $f_S$, there are functions $f_v$ for all $v$ that produce $f_S$ by composition.
What is a largest free subset of $G$?
If it is helpful, let's look at an example. Suppose $d = 1$ and $n = 6$. Then the following table displays an $f_S$ for $S = \{v_{3,1},v_{3,2},v_{3,3}\}$ that no combination of $f_v$ functions (for the nodes $v_{1,1}$, $v_{2,1}$, and $v_{2,2}$) can produce by composition.
$S = \{v_{3,1}, v_{3,3}\}$ is therefore a largest free subset of this graph.