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Wikipedia gives the path cover definition as:

Given a directed graph $G = (V, E)$, a path cover is a set of directed paths such that every vertex $v \in V$ belongs to at least one path.

I'm reading the paper "On k-Path Covers and their Application" (Funke, S., Nusser, A., & Storandt, S. (2014)) and they gave the definition where we select a subset of vertices $C \subseteq V$ such that for every simple path $\pi$ in the graph we have that $C \cap \pi \neq \emptyset$. (There's also a $k$ constraint but it's not particularly relevant to my problem).

Are these definitions equivalent? One uses a set of vertices and the other uses a set of paths.

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The definitions are not equivalent. Clearly, as you point out, one is a set of paths while the other is a set of vertices, but even their sizes are unrelated. In fact, since a single edge is a path, the second definition would be equivalent to saying that $C$ is a vertex cover.* If you want an explicit counterexample, think of a graph consisting of a path of length $2$.

From what I see at a quick glance, the paper studies $k$-Path covers. The $k$ here is critical, since a path $\pi$ needs to be covered by $C$ (i.e., it must contain at least one vertex in $C$) only if $\pi$ is a simple path containing at least $k-1$ edges (i.e., $k$ vertices).

* I'm actually disregarding paths with 0 edges here, otherwise the only feasibly set $C$ would be $V$ itself.

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  • $\begingroup$ Thank you. As a followup, the Wikipedia page also has a note that "path cover may also refer to vertex-disjoint path cover" - can we consider this restriction of exactly one to be equivalent to the more general problem of at least one? I think so, since we can just discard paths that would be covering vertices more than once. $\endgroup$
    – a6623
    Mar 9 at 20:58
  • $\begingroup$ No. You can cover a star with $4$ leaves with $2$ paths that intersect on the star center. If you want to use vertex-disjoint paths, then you need to select at least $3$ paths. $\endgroup$
    – Steven
    Mar 9 at 21:00

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