Understanding the notation $N/(N_{r}\bigcup N_{v})$ in graph theory Currently I'm dealing with a graph problem but I don't understand one specific notation. What does the following mean:
$$N/(N_{r}\bigcup N_{v})$$
$N$, $N_{r}$, $N_{v}$ are sets of nodes. $N$ is the full set, and the other two are subsets. I understand the notation when it's written with \, but I don't know the meaning of /.
 A: The paper can be found here. Looking at the text near Proposition 3 it seems as in Brian's guess is correct: for the node sets $N_r, N_v$ of the directed trees rooted at $r, v$ respectively, $N/(N_r\cup N_v)$ seems to denote just what Brian guessed, the set of nodes in the digraph induced by $N$ where the nodes in the subtrees $N_r, N_v$ are collapsed to a single node.
To see how this "collapse" works (at least as I interpret what Marianov intends), consider this picture:

In the left hand picture I've drawn a directed graph with nodes $N$ consisting of $p, q, r, w, v$ and the three unlabeled nodes at the bottom. As in the article, the tree rooted at $r$ and the tree rooted at $v$ consist of nodes $N_r, N_v$ respectively, so there are three nodes in the set $N_r$ and two in the set $N_v$. If we move those five nodes and superimpose them into one, keeping all of the other edges, we get the picture on the right, where $r, v$, and the three other tree nodes have all been collapsed into one node. In other words, as I interpret things, the four nodes in the right-hand picture are $N/(N_r\cup N_v)$.
A: Reinhard Diestel’s text Graph Theory uses the notation $G/e$, where $e$ is an edge $\{x,y\}$ of $G$, to denote the graph obtained by ‘contracting the edge $e$ into a new vertex $v_e$, which becomes adjacent to all of the former neighbours of $x$ and $y$’. My best guess is that you’re starting with a graph $G=\langle N,E\rangle$ and contracting all of $N_r\cup N_v$ to a single vertex $v^*$. For $v\in N\setminus(N_r\cup N_v)$ the new graph would have an edge $\{v,v^*\}$ iff there is a $u\in N_r\cup N_v$ such that $\{v,u\}\in E$, i.e., iff there is an edge in $G$ from $v$ to some vertex $u\in N_r\cup N_v$.
