The adjacency matrix $A$ of the transitive closure of a directed acyclic graph can have a `checkerboard' pattern like \begin{equation} A = \begin{pmatrix} 0 & ? & ? & ? & {\bf a} & {\bf a} & b & b & b & {\bf c} \\ ? & 0 & ? & ? & {\bf a} & {\bf a} & b & b & b & {\bf c} \\ ? & ? & 0 & ? & {\bf a} & {\bf a} & b & b & b & {\bf c}\\ ? & ? & ? & 0 & {\bf a} & {\bf a} & b & b & b & {\bf c}\\ {\bf a}' &{\bf a}' & {\bf a}' & {\bf a}' & 0 & ? & {\bf d} & {\bf d} & {\bf d} & e \\ {\bf a}' & {\bf a}' & {\bf a}' & {\bf a}' & ? & 0 & {\bf d} & {\bf d} & {\bf d} & e \\ b' & b' & b' & b' & {\bf d}' & {\bf d}' & 0 & ? & ? & {\bf f} \\ b' & b'& b'& b' & {\bf d}' & {\bf d}' & ? & 0 & ? & {\bf f} \\ b' & b'& b' & b' & {\bf d}' & {\bf d}' & ? & ? & 0 & {\bf f} \\ {\bf c}' & {\bf c}' & {\bf c}' & {\bf c}' & e' & e' & {\bf f}' & {\bf f}' & {\bf f}' & 0\end{pmatrix}\ . \end{equation}
The alternate light and heavy fonts are here just to make the `checkerboard' pattern of the lettered constants apparent to the eye.
Each of the square blocks along the diagonal can be viewed as representing a set of nodes that forms in a sense its own collective node; in my work it has been profitable to study the related four-node graph whose adjacency matrix is $$\begin{pmatrix} 0 &{\bf a} & b & {\bf c}\\ {\bf a}' & 0 & {\bf d} & e\\ b' & {\bf d}' & 0 & {\bf f}\\ {\bf c}' & e' & {\bf f}' & 0\\ \end{pmatrix}\ .$$
Question: given any adjacency matrix where the nodes have not already been assigned to sets that make the checkerboard pattern plain, how can this sort be efficiently done?
This problem may belong just to the general mathematics of matrices and not require special knowledge of adjacency matrices at all, just knowledge of their basic properties:
The elements of an adjacency matrix are all $0$'s and $1$'s, and any above-diagonal element $A_{ij}$ and its below-diagonal counterpart $A_{ji}$ must form one of the pairs $(1,0)$, $(0,1)$, and $(0,0)$. The diagonal elements are $0$.
If an adjacency matrix represents a transitive closure, then if $A_{ij}=1$ and $A_{jk}=1$, then $A_{ik}=1$. The $4\times 4$, $2\times 2$, $3\times 3$, and $1\times 1$ blocks along the diagonal of the example above are otherwise arbitrary.
Some problems cannot be reduced to fewer nodes than in the original problem.
One can split the nodes into a set $I$ and a set $J$ and see if the set $I$ forms a collective node by checking if for each $j\in J$ all the values $A_{ij}$ for $i\in I$ match, and all the values $A_{ji}$ also match. If one finds a collective node, finding the next reduces to another solving the same general problem but with an adjacency matrix with fewer nodes, which is helpful; still, running through all the sets is slow and I am looking for a better approach.