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I've been investigating this problem: Given some straight line planar embedding of a simple connected graph with a simple dual graph, does there exist a straight line planar embedding of that dual graph with each vertex lying in its corresponding face in the original embedding? What if we ignore the infinite face?

I believe I have found a few complicated counterexamples to the main problem, but have no clue how I would go about proving the less restrictive case or finding a counterexample. I would also appreciate smaller counterexample to the initial case.

Here is a graph that I believe is a counterexample. enter image description here

Tbw
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1 Answers1

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Here is a plane embedding whose dual, even ignoring the unbounded region, cannot be drawn so that all dual vertices are in their corresponding regions, and all dual edges are straight lines:

enter image description here

Here is the idea of what goes wrong. This plane embedding has a center square (one region) surrounded by an inner border (8 regions) surrounded by an outer border (3 regions). No matter how we draw the dual vertices and edges for the inner border, they will form an octagon that contains the center square. So the center square (highlighted in red below) is off-limits for the edges connecting the 3 regions of the outer border:

enter image description here

The blue L shape highlighted on the left now represents the portion of the bottom region of the outer border that can "see" the top left region of the outer border without the center square getting in the way. If we pick a dual vertex for the bottom region of the outer border, this is where it must lie, in order for there to be a straight edge to the dual vertex for the top left region.

Similarly, the blue backwards L shape is where the dual vertex for the bottom region must lie in order for there to be a straight edge to the dual vertex for the top right region.

These two blue shapes are disjoint. So it's impossible to pick a location for the dual vertices so that all the dual edges can be drawn as non-intersecting straight lines.

Misha Lavrov
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  • Thanks, this makes a lot of sense, and was the kind of thing I was looking for. My question did say "with a simple dual graph" which in retrospect sounds like we consider the dual to be simple by combining multiple edges. But I meant that the graph must have a strictly simple dual. Could this be adapted into such a solution? – Tbw Nov 22 '23 at 04:57
  • I suppose it could be done by changing the small squares into triangles and adding an additional outer layer of 4 trapezoids. – Tbw Nov 22 '23 at 05:00
  • Yes, that sounds like it would do it. – Misha Lavrov Nov 22 '23 at 05:01