# Delaunay Triangulation

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?

• Outside the convex hull, to infinity. – Yves Daoust Nov 20 '16 at 11:12
• I don't understand. – cripto Nov 20 '16 at 11:17

The exterior face is the green area.

• how do I derive the formula 2n − 2 − b? – cripto Nov 20 '16 at 11:36
• @cripto: what are $V,F,E$ ? – Yves Daoust Nov 20 '16 at 11:40
• V=vertex, F=faces and E=edges – cripto Nov 20 '16 at 11:43
• @cripto: I was sure you would answer this ! – Yves Daoust Nov 20 '16 at 11:48
• I don't known where to use 1=V-E+F – cripto Nov 20 '16 at 11:56

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.