> A directed acyclic graph $G$ is *confluent*, if any two vertices
> ${v_1},{v_2}$ in $G$ which have a common ancestor $u$ also have a
> common successor $w$. (I.e. if there are paths from some $u$ to both
> ${v_1}$ and ${v_2}$, then there are also paths from ${v_1}$ and
> ${v_2}$ to some $w$. Vertices $u,{v_1},{v_2}$ and $w$ do not need to
> be distinct, thus e.g. a single "line" of vertices is trivially
> confluent) Design a **linear-time algorithm** that determines whether
> a given DAG $G$ is confluent.

Is the "line" with $3$ vertices $u \to v \to w$ s.t. ${v_1} = {v_2}$ trivially confluent? Or is it nesseccary to have ${v_1} \ne {v_2}$ and thus the smallest number of vertices such line may have is $4$ $u \to {v_1} \to {v_2} \to w$?

Edit: Solved the problem.