Let us define a new problem with an instance $I = (G = (V, E), K, L)$, whereas:

  • $G$ is an undirected graph
  • $K \le |V|$
  • $L > 0$ is the maximum limit
  • Each vertex $v \in V$ has a weight $W(v)$
  • Each edge $e \in E$ has a length $W(e)$

Let $P(v)$ be a function that returns the minimum length of a path (basically shortest path) from vertex $v$ to a vertex in $V'$.

The decision question is whether there exists a vertex set $V' \subseteq V, |V'| \le K$, such that:

$$ \sum_{v \in V} W(v) \cdot P(v) \le L $$


Consider the following graph $G$, with $K = 1$ and $L = 9$:

enter image description here

Taking the set $\{v_3\}$ as $V'$ would be the solution to the question, because the total cost is:

$$ 3 \cdot 0 + 2 \cdot 2 + 1 \cdot 5 = 9 $$

Therefore, this is a yes-instance.


How do I prove that this problem is in $\mathsf{NPC}$? I tried reducing a $\text{VC}$-instance to this, but that does not seem to work.

What I have tried as well is by converting the above undirected graph to a directed graph with the weighted paths already computed (this is polynomial computable):

$$ \begin{array}{l|l|l} & v_1 & v_2 & v_3 \\ \hline v_1 & 0 & 3 & 5 \\ \hline v_2 & 6 & 0 & 4 \\ \hline v_3 & 15 & 6 & 0 \end{array} $$

However, I'm not entirely sure what $\mathsf{NPC}$ problem to reduce from. I got the tip to use the $\text{CLIQUE}$ problem from this answer, but I do not see how to perform the reduction, so that seems unlikely. To me, this looks more like something that can be reduced from the $\text{KNAPSACK}$-problem or the $\text{SET COVER}$-problem, using the weights from the transformed directed graph, but I fail to see how exactly to this. I am starting to think that the transformed directed graph is of no use here.

What $\mathsf{NPC}$-problem can I use for the reduction?

  • 1
    $\begingroup$ Why doesn't it work? Have you considered reducing from Dominating Set? Set weights =1, $K=k$ and $L=n-k$? $\endgroup$
    – Pål GD
    Mar 18, 2019 at 6:40
  • $\begingroup$ @PålGD Yes, I considered the reduction function $f : \Sigma^* \rightarrow {\Sigma^*}'$ that transforms a Dominating Set ($\text{DS}$) problem instance to the above Weighted Vertices and Edges problem instance ($\text{WVE}$). Showing that $\forall x \in \text{DS} \rightarrow f(x) \in \text{WVE}$ is easy, just by setting the weights of the vertices to $1$ and the lengths of the edges to $1$, just like you said. However, I am having trouble proving that $\forall f(x) \in \text{WVE} \rightarrow x \in \text{DS}$. $\endgroup$
    – Adnan
    Mar 18, 2019 at 10:11
  • $\begingroup$ Oh actually, does that mean that I only have to consider transformed instances for the $\text{WVE}$ problem, which will always have vertex weights $1$ and edge lengths $1$? (so the example that I have given in the OP is actually useless?) $\endgroup$
    – Adnan
    Mar 18, 2019 at 10:13
  • 1
    $\begingroup$ If you can show that WVE is NP-hard even on unweighted graphs, that's a stronger statement, yes. $\endgroup$
    – Pål GD
    Mar 18, 2019 at 11:43

1 Answer 1


This problem is a generalization of Dominating Set, and if you let weights of vertices and edges be 1, then you have exactly Dominating Set.

The problem definition of Dominating Set is:

Given a graph $G = (V, E)$ and an integer $k$, is there a set $V' \subseteq V$ of size at most $k$ such that the distance of any vertex $v \in V \setminus V'$ to $V'$ is 1.

(This is assuming that the distance from $v' \in V'$ to $V'$ is defined as 0.)

It follows that WVE is NP-hard. This is occasionally called proof by generalization.


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