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Dec 5, 2020 at 17:41 comment added Sylvain JULIEN This would suggest that the implication "tiling of a square implies triangulation of a quadrilateral" is not an equivalence.
Dec 5, 2020 at 15:35 comment added Mohammad Ghomi Well, one could consider the graph obtained by adding a vertex to the middle of a square and connecting that to all the other vertices. This graph is a triangulation of a quadrilateral, but it does not correspond to a tiling of a square.
Dec 5, 2020 at 13:41 comment added Joseph O'Rourke If Schramm is indeed the source, then the graph must be a triangulation of a quadrilateral, and a tetrahedron graph is not a triangulation of a quadrilateral.
Dec 4, 2020 at 20:29 comment added Sylvain JULIEN If you consider the block composed of the squares labeled by 19, 29, 16, and 23, it seems that counting a square sharing only a vertex as "half tangent", each of those 4 squares has 2.5 tangent squares, and 2.5 is precisely the arithmetic mean of the degrees of the corresponding vertices in the subgraph they define.
Dec 4, 2020 at 20:29 comment added Joseph O'Rourke Good question, and point! The text says, "Figure 12 shows a generalization of circle packing by replacing circles by squares to compute the extremal length of a combinatorial quadrilateral. The left frame shows a 3-connected graph, with four corner nodes. The right frame shows the extremal length, where each node is replaced by a square with the same label and color. Two nodes are connected in the graph if and only if their corresponding squares are tangent." So perhaps the 4 corner nodes are special?
Dec 4, 2020 at 20:21 history edited Mohammad Ghomi CC BY-SA 4.0
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Dec 4, 2020 at 20:12 history edited Mohammad Ghomi CC BY-SA 4.0
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Dec 4, 2020 at 20:07 history answered Mohammad Ghomi CC BY-SA 4.0