Given a directed graph $G = (V, E)$ and a vertex $v \in V$, how to find all vertices $v'$ such that exists a simple cycle $v \to ... \to v' \to ... \to v$? That is, to find the set of vertices $$V' = \{v' : \exists c, \text{c is a simple cycle in G}, v \in c, v' \in c \}$$
I found this question related: Find all cycles through a given vertex. I can first find all cycles and then union them to get $V'$. However, is there a more efficient way to do so?
I believe the problem is NP-hard, by reduction from the two disjoint paths problem.
The two disjoint paths problem is as follows:
Given a directed graph $G$ and $(s_1,t_1),(s_2,t_2)$, find two paths $s_1 \leadsto t_1$ and $s_2 \leadsto t_2$ that are vertex-disjoint (they have no vertex in common).
This problem is NP-hard.
We can reduce two disjoint paths to your problem. In particular, given $G$ and $(s_1,t_1),(s_2,t_2)$, we construct a new graph $G'$ by adding new vertices $v,v'$ and adding edges $v \to s_1$, $t_1 \to v'$, $v' \to s_2$, $t_2 \to v$.
Notice that any pair of vertex-disjoint paths $s_1 \leadsto t_1$, $s_2 \leadsto t_2$ will yield a simple cycle
$$v \to s_1 \leadsto t_1 \to v' \to s_2 \leadsto t_2 \to v.$$
Also, within any simple cycle $v \leadsto v' \leadsto v$ we can find simple paths $s_1 \leadsto t_1$, $s_2 \leadsto t_2$ that are vertex-disjoint. It follows that there exists a simple cycle $v \leadsto v' \leadsto v$ in $G'$ iff there exist two vertex-disjoint paths $s_1 \leadsto t_1$, $s_2 \leadsto t_2$ in $G$. So, any algorithm for solving your problem would immediately yield an algorithm for solving the two disjoint paths problem.
