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My question is: Given a planar graph defined by its adjacency matrix. Can I always find a set of points, so that the dual graph of the voronoi diagram of that set of points is the same as the planar graph?

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No. The dual graph of a Voronoi diagram is the Delaunay triangulation of its point set so, in particular, every interior face of it is a triangle. But there are plenty of planar graphs (e.g., the $4$-cycle) that have non-triangular interior faces.

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  • $\begingroup$ This is not correct. The Delaunay triangulation may not be unique and this breaks the proof. For example, the points $\{(\pm 1,\pm 1)\}$ have Voronoi regions which have the adjacency of a 4-cycle. $\endgroup$ Aug 26, 2022 at 22:19
  • $\begingroup$ However, this answer is still morally correct, in the sense that I think it highlights the right question to ask. Is every triangulated plane graph a dual graph of some Voronoi diagram? I might ask this on mathoverflow if I can't figure it out. $\endgroup$ Aug 26, 2022 at 22:21

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