Is realization of unit disk graphs hard? It is known that recognizing a unit disk graph is NP-hard [1].
However, the paper does not mention how hard the realization problem is.
I have looked up several references [2][3][4]. None of the papers answer whether the following problem is NP-hard:

Given a unit disk graph $G = (V,E)$, find a configuration of a set
$\mathcal{D}$ of disks, such that the intersection graph
$G(\mathcal{D})$ of $\mathcal{D}$ is isomorphic to $G$.

The difference between this problem and the recognition problem is that the input of this problem is guaranteed to be a unit disk.
Is there any study that shows the complexity of the above problem? I expect it to be NP-hard, but I am yet to find a full proof.
One of the complications here is that there is no natural &quot;decision variant&quot; of this problem, since this isn't an optimization problem. (and the problem &quot;given an UDG, is there a realization?&quot; is trivial)
Since reductions between decision problems tend to be better studied, one option is to make it an optimisation problem in a chosen parameter. Even though you don't care about that value, you can use binary search on the optimisation problem to solve the realisation problem. Hence, the realisation problem is not much easier than the opt. problem, so showing the opt. problem is hard suffices.
The question then of course is what parameter to choose. Perhaps the problem &quot;Given an UDG, find a realisation with minimum/maximum sum of edge lengths&quot; would be suitable. It seems related to the problem that Aspnes et al. [1] show to be NP-hard.

[1]: ASPNES, James, et al. A theory of network localization. IEEE Transactions on Mobile Computing, 2006, 5.12: 1663-1678.