TITLE: Geometric dominating set: NP-complete?
QUESTION [6 upvotes]: Let $G=(V,E)$ be a geometric graph, a graph embedded in the plane whose edge lengths are
the Euclidean distance between its endpoint vertices.
Say that a set of vertices $D \subseteq V$ is a geometric dominating set
if for every $v \in V$, the shortest path in the graph
from $v$ to some vertex in $D$ is at most length $1$.
For example, a pentagonal wheel with spokes of length $1$ can be dominated with
one vertex, but if the shortest edge length is greater than $1$, then $D$ must equal $V$:



The traditional dominating set problem has been known to be NP-complete since at least Garey & Johnson's 1979 book.
My question is:

Q. Is finding a smallest geometric dominating set for a given geometric graph also NP-complete?

Conceivably it is not intractable, if the geometric structure can be exploited.
Perhaps plane graphs $G$ would especially allow that exploitation.
If anyone knows of references, I would appreciate pointers. Thanks!

REPLY [7 votes]: I came up with an explicit construction that showed that your problem was NP-hard, even on the real line, but then I realized there's an even simpler argument:
Minimum dominating set is known to be NP-hard even for bipartite graphs, and those embed nicely. For example, you could send one part of the partition to $(0,0)$ and the other to $(1,0)$.
(If you insist that distinct vertices be represented by distinct points, then you may choose different points like $(\varepsilon,0)$ and $(1-\varepsilon,0)$ instead.)