Generating cyclical dependency graphs from k-way partitions of DAGs representing boolean networks My question stems from something mentioned in the following paper*:

Acyclic Multi-Way Partitioning of Boolean Networks by Jason Cong, Zheng Li, and Rajive Bagrodia

Given a DAG representing a Boolean network, $N$, where each node represents a logic gate, define a k-way partitioning solution as $S = (A_1, A_2, ..., A_k)$ satisfying $$A_i \cap A_j = \emptyset \quad{\text{and,}}$$ $$\bigcup\limits_{i=1}^{k} A_i = N$$

Then define a dependency graph $D(S)$. This is a directed graph with $k$ nodes where each node represents a block, $A_i$, and has an edge $(A_i,A_j)$ if and only if there exists an edge $(x,y)$ in $N$ such that $x \in{A_i}$ and $y \in A_j$ (in other words, the edges are the edges removed when forming the partition). 

The authors note that it is possible to get dependency graphs that are cyclic after some k-way partition. My question is, how is this possible? I can't seem to come up with an example where partitioning a DAG would ever produce a cyclic dependency graph. As I see it, a cyclic dependency graph would require a an edge to be "reversed" somehow, which would produce a cycle in the original DAG. Perhaps it has something to do with the fact these graphs are representing boolean networks?

*Link to pdf may be paywalled

**Note the authors also have a balance constraint on the partitions which I didn't describe but I don't think that's relevant to my question.
Consider the following graph on three vertices:

$$u \to v \to w$$

This is acyclic.  If you set $A_1 = \{u,w\}$ and $A_2 = \{v\}$, you end up with a dependency graph that is cyclic: it has an edge $A_1 \to A_2$ (due to the edge $u \to v$) and an edge $A_2 \to A_1$ (due to the edge $v \to w$).