Given $\Gamma$ a directed graph. We denote $M$ the set of ordered pairs $(a, b)$, for which we have a straight oriented line between the two vertices, $\vec{ab}$. We say that $M$ is cyclic, if and only if $\Gamma$ contains a cycle, meaning that there exists a sequence of pairs in $M$, $(a_i, b_i)_{1 \leq i \leq n}$, for which $a_2 = b_1, a_3 = b_2, \dots, a_n = b_{n - 1}, a_1 = b_n$. Find the minimum number of subsets in which we can partition $M$, such that none of the sets will be cyclic, but $M$ contains at least a cycle.
If $M$ contains a pair $(a, a)$, we will have a cycle in one of the subsets $(a, a)$, which is the shortest cycle and completely indecomposable in non-cyclic chains. Similarly, $(a, b) \to (b, a)$ is a simple cycle. There are enough $2$ subsets to separate the chain. In general case, we have $\omega_1, \omega_2, \dots, \omega_k$ the simple or composite, disjoint cycles in our graph, each one composed of $\lambda_1, \lambda_2, \dots, \lambda_k$ subcycles. (consider that two cycles are disjoint if there is no cycle composed only of edges from the two cycles, but necessarily at least one from each cycle and that a composite cycle has $\lambda$ subcycles if there are $\lambda$ ways to return to the same point, by passing only through the edges which compose the cycle). Therefore, for $\omega_i$ we need $\lambda_i$ "cuts" into subsets and the result is expressed as $R = \prod_{i = 1}^k \lambda_i$, where $k$ is the number of composite disjoint cycles. However, how could I express this result to be acceptable in terms of a problem (without new definitions of variables, I guess as an expression between $n = |V|$, $m = |E|$, maybe using Turan's theorem (generalized form) - about the condition for $m$ such that there is no $K_r$ a clique - complete graph - in $r$ vertices or any other directed graph application (I am more familiar with simple/undirected graph theorems and lemmas regarding edges vertices, cliques, chains, cycles etc.).