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Wikipedia on Delaunay Triangulation says:

"In the plane[...], if there are b vertices on the convex hull, then any 
triangulation of the points has at most 2n − 2 − b triangles, plus one 
exterior face."

Here n is the total number of vertices in the plane.

I can't see how they obtained this result when I use Euler's formula 1=V-E+F. So V=n then F=1-n+E. what is face exterior?

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  • $\begingroup$ Outside the convex hull, to infinity. $\endgroup$ – Yves Daoust Nov 20 '16 at 11:12
  • $\begingroup$ I don't understand. $\endgroup$ – cripto Nov 20 '16 at 11:17
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The exterior face is the green area.

enter image description here

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  • $\begingroup$ how do I derive the formula 2n − 2 − b? $\endgroup$ – cripto Nov 20 '16 at 11:36
  • $\begingroup$ @cripto: what are $V,F,E$ ? $\endgroup$ – Yves Daoust Nov 20 '16 at 11:40
  • $\begingroup$ V=vertex, F=faces and E=edges $\endgroup$ – cripto Nov 20 '16 at 11:43
  • $\begingroup$ @cripto: I was sure you would answer this ! $\endgroup$ – Yves Daoust Nov 20 '16 at 11:48
  • $\begingroup$ I don't known where to use 1=V-E+F $\endgroup$ – cripto Nov 20 '16 at 11:56
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Euler's formular for planar graphs with $n$ vertices, $e$ edges and $f$ faces says that $2 = n - e + f$. Note that $f$ includes the outer, unbounded face to infinity, too. So we have $f-1$ triangles and a $b$-gon.

We can assign to each edge two tokens; one for each incident face. The number of tokens can be counted per face, too: $3(f-1)$ for $f-1$ triangles and $b$ for the outer face, so $3 (f-1) + b = 2e$.

Then we have $4 = 2n - 2e + 2f = 2n -f -k + 3$, and hence $f-1 = 2n -2 -b$, which is the number of triangles.

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