Suppose a directed graph $G=(V,E)$ and an initial subset of colored nodes $R_{0} \subseteq V$. We then produce the following subsets $R_{1} \subseteq R_{2} \subseteq ..\subseteq R_{n}$ with the following iterative procedure:

$$ R_{1} = R_{0} \cup \{v\;|\; \forall (w,v) \in E: w \in R_{0}\}$$

and generally

$$ R_{i+1} = R_{i} \cup \{v\;|\; \forall (w,v) \in E: w \in R_{i}\}$$

Essentially, at each step of the iteration, we color a node if all of it's incoming edges originate from nodes that have already been colored.

This procedure can have two outcomes, (a) all nodes eventually get colored at a step $j$ of the iteration such that $R_{j}=V$ or (b) an iteration is reached where no other nodes can get colored and some remain uncolored ($R_{j} = R_{j+1} \subset V$).

Now, the problem is to compute the smallest possible inital set $R_{0}$ such that eventually all nodes will be colored. My questions are:

a) Does this problem have a name in the literature of graph theory?

b) If so, are there any efficient algorithmic solutions?

Thank you!


If the graph is non-oriented (that is, $(u,v) \in E \Leftrightarrow (v,u) \in E$), then you're looking for a vertex cover of the graph. Indeed, if an edge $(u,v)$ has none of its ends in $R_0$, this will be a deadlock preventing both $u$ and $v$ from ever getting in one of the $R_i$ ; and if you have a vertex cover for $R_0$ then $R_1 = V$.

Even in this subclass of graphs, the problem already is NP-hard, so I guess the answer to b) is no. But for the vertex cover problem there are approximation algorithms, so maybe you could look there to find something similar for your problem...

  • $\begingroup$ 1) So the problem as I defined it is at least as hard as vertex cover? 2) The question "does a solution $R_{0}$ with less than k nodes exist?" is an NP problem because if you are given a solution $R_{0}$ with less than k nodes you can verify it fast by expanding to $R_{1}, ..., R_{k}$. From (1) & (2) is it correct to say that the problem is NP-hard? $\endgroup$
    – Enk9456
    Jan 26 at 21:13
  • $\begingroup$ I meant to say that the problem is NP-complete, because (1) implies it is in NP-hard and (2) it is in NP? $\endgroup$
    – Enk9456
    Jan 26 at 21:24
  • $\begingroup$ What About the constraint that the graph is directed and the source of the edges are in $R_i$ and the targets in $R_{i+1}$? $\endgroup$ Jan 27 at 7:47
  • $\begingroup$ @Enk9456 yes, for both (1) and (2), it is indeed NP-complete. $\endgroup$
    – Hugo Manet
    Jan 27 at 9:55
  • $\begingroup$ @DavidScholz a non-directed graph can be interpreted as a directed graph with two oriented edges for each non-oriented edge (in this case, it's OK to do so because we don't examinate properties on the edges themselves). My argument is : if you have a non-oriented edge (u,v), that is both u->v and v->u, if ($u \notin R_i$ and $v \notin R_i$) then ($u \notin R_{i+1}$ and $v \notin R_{i+1}$) so both won't appear anytime in the process. $\endgroup$
    – Hugo Manet
    Jan 27 at 10:02

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