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I invented this little game the other day: First I draw a cloud of as equally as it gets distributed points on a piece of paper, then I try to find a triangulation with the points as vertices such that each vertex meets exactly $6$ edges if it is surrounded by triangles and up to $5$ if it lies on the boundary (see image).

Now I ask myself, given a finite set of points in ${\bf R}^2$ such that no three points lie on the same line, can one always find such a triangulation? What about infinite sets?

[EDIT] User Symlic pointed out that there exists always a rather trivial solution for finite point clouds, you can read it in their comment. So I thought the best way to ask the question would be to consider infinite point sets only (and then ask for any vertex to meet $6$ edges), but there are also some restrictions which should hold, so that e.g. not all points lie in some finite area or on the upper half plane. One idea would be to demand for any ray $X$ and $\epsilon>0$ that there exists some point within $\epsilon$-distance of $X$ (so there are points in any direction if you look far enough) and that the point set is topologically discrete.

Maybe the question is stupid after all, but while drawing these patterns I felt that there should be some underlying result affecting this.

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    "Delaunay triangulation" is the term you want to search for here: https://en.wikipedia.org/wiki/Delaunay_triangulation – 2017-02-02
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    Thanks for the hint, I didn't know that term. But it's a little bit different from my problem, I only make constraints about the number of edges meeting in a single vertex, Delanay makes only constraints about the number of vertices sitting inside the circumcircle of a triangle. – 2017-02-02
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    I guess there should be some more (hidden) constraints. For current version of problem we can do the following. 1. Sort all points by x-coordinate. 2. Connect each point with previous and next one (in case it presents). 3. Triangulate each valley (convex polygon) with zig-zag. Now each point lies on the boundary and has at most 5 incident segments. – 2017-02-03
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    Made an edit... – 2017-02-03
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    Looking at you pictures I propose the following update of finite case: all points laying at some distance $d$ or above away from bounding box should be internal and have degree 6. Also this condition can be two-way: all points at distance at most $d$ from bounding box and only these points should be on the border. The finite case should be more easy to think about and it is even algorithmically solvable unlike the infinite one. – 2017-02-04
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    @Smylic this seems plausible, but you'd have to put in some further restraints, so that it's not happening that e.g. there is a point which should meet $6$ edges (far enough from the boundary) but there are no further points to the right. – 2017-02-04
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    Another formulation which seems plausible to me the following: Given two finite point sets such that no three points in the combined set lie on a straight line and a triangulation of the first set. Then one can find a isomorphic triangulation in the commbined set (with which I mean a triangulation of some subset such that no points left out lie inside any triangle). – 2017-02-04

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