Here is a minimal working example of the question:
Consider a network 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. (So $n = m + (m-1)md/2$ nodes total.) Each node takes a state of 0 or 1. All nodes not in row $m$ have $1+d$ "parents" in the next row, and these parents are "local". For example, if $v$ is the $k$th node in row $r < m$, then its parents are nodes $k$ through $k+d$ in row $r+1$.
For each node $v$ in a row $r < m$, there is a table $T_v$ that lists all possible combinations of states of the parents of $v$ and, for each combination, assigns a state of 0 or 1 to $v$.
Let $d\text{-RETROSPECT}$ denote the decision problem of determining whether there is an assignment of states to the nodes of the network that assigns state 1 to the node in the first row and is consistent with all $T_v$.
Are there $d$ for which $d\text{-RETROSPECT}$ is NP-hard?
It seems unlikely that I am the first person to contemplate this. (In fact, in a famous paper in 1990, Cooper showed that if the parents are not required to be local, then the problem is indeed NP-hard for at least $d \geq 2$.) Is a direct proof possible in the local case---e.g., by a polynomial-time reduction from a known NP-complete problem? Or, if not, does anyone know of recent relevant literature?
