Given a weighted directed acyclic graph $G=(V,D,W)$ and a set of arcs $D'$ of $D$, where the weights of $W$ are on the vertices. The problem is to partition $G$ into a minimum number of vertex-disjoint paths that cover all the vertices of $G$ subject to the constraints that:

  1. the weight of each path is at most $k$.
  2. each path should include at least one edge of $D'$.

What is the complexity of this problem?

  • $\begingroup$ Why do you tag hamiltonian-path? $\endgroup$
    – xskxzr
    Commented Sep 11, 2019 at 15:00
  • $\begingroup$ because the minimum path cover problem is a generalization of the hamiltonian path $\endgroup$
    – Farah Mind
    Commented Sep 11, 2019 at 15:10

1 Answer 1


This answer shows this problem is NP-hard. Further research is needed to determine whether it belongs to APX or it is APX-hard.

Let $G$ be a complete directed acyclic graph, i.e., you can name the vertices $1,2,\ldots,n$ and there is an edge $(i,j)$ for all $i<j$. Let $D'=D$, and $k$ is equal to half of the sum of the weights of all vertices. Now there exist 2 vertex-disjoint paths satisfying your conditions if and only if the weights can be partitioned into two subsets both with sum $k$, which is exactly the partition problem. Hence, your problem is NP-hard.

  • $\begingroup$ if $k$ is a constant, the problem becomes polynomial? because searching a minimum number of vertex disjoint paths that cover an acyclic graph can be solved in polynomial time by transforming it in a matching problem. $\endgroup$
    – Farah Mind
    Commented Sep 14, 2019 at 12:43
  • $\begingroup$ I don't know. Maybe. It requires further thinking. And your problem is not only the problem of searching vertex disjoint path. $\endgroup$
    – xskxzr
    Commented Sep 14, 2019 at 12:52
  • $\begingroup$ yes, each path should also pass by at least one of the edges of $D'$ $\endgroup$
    – Farah Mind
    Commented Sep 14, 2019 at 12:59

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