Let $(P, E)$ be a directed graph with $P$ being the set of nodes and $E \subset P^2$ the set of directed edges. It is well-known that the transitive closure of $E$ is a preorder $\leq$, which is more precisely a partial order if there are no cycles in the graph. For example, consider the graph below.
How do I get the collection of all simple cycles induced by $\leq$ in $P$? For example, in the graph below, this would be $P_{cycle} = \{\{b, d, c\}, \{b, d, h, e, c\}\}$.
Are all simple cycles chains w.r.t. $\leq$?
How do I get the collection of all maximal totally ordered subsets in which are not cycles? For example, in the graph below, this would be $P_{linear} = \{\{a,b,d,h,e,f\}, \{g, b, d, h, e, f\}\}$.
I want something like $$ P_{cycle} = \left\{S \in \mathbb{P}(P) \mid S \text{ is a simple cycle}\right\} $$ and $$ P_{linear} = \left\{S \in \mathbb{P}(P) \mid S \text{ is a maximal chain } \land \text{ not a cycle } \right\}, $$ where $\mathbb{P}$ is the power set. I want the predicates to be replaced with an expression using $\leq$.
Let $\equiv = \leq \cap \geq$ the equivalence relation of the preorder. Then, ${(P,\leq)_/}_\equiv$ is an ordered set. Let $C$ be a maximal chain in ${(P,\leq)_/}_\equiv$. And $c:C\to P$ a choice function which maps each equivalence class to one of its members, i.e. $c([x]) ∈ [x]$. Then $c[C]:= \{ c([x]) | [x]∈ C\}$ is an element of $P_{linear}$.
Now we need some words about the equivalence classes. Obviously, the vertices of a cycle are equivalent with respect to $\leq$. As equivalence classes either are disjoint or coincide, overlapping cycles belong to the same equivalence class. And that's all.
The preorder restricted to an equivalence class considered as a directed graph results in the full directed graph with loops. So every subset in every permutation with an additonal edge from the end to the beginning is a simple cycle. If you want to count them, note that this procedure generates each circle several times: Every edge in the circle can be the one from the end to the beginning.
I'm leaving out the proofs and some formalism here. These are left as an exercise to the interested reader.