Minimum number of internal diagonals of a simple $n$-gon

Question

What is the least number of internal diagonals a simple $n$–gon may have? (For a fixed $n$)

I know that any simple polygon has at least one internal diagonal.

The main problem is with the concave polygons, how do I generalize for them?

Any help would be truly appreciated.

See http://math.stackexchange.com/questions/83580/every-polygon-has-an-interior-diagonal for some information. — Ethan Bolker, Jan 06 '16 at 16:31
@6005: It's certainly possible to create an $n$-gon with only $n-3$ diagonals. (For $n>3$, create increasingly-"C"-shaped figures.) I don't know if we can do better. — Blue, Jan 06 '16 at 18:54

score 4 · Answer 1 · answered Jan 07 '16 at 22:44

The answer is $n-3$.

If there are $n-3$ consecutive concave vertices and only $3$ convex ones then all internal diagonals must start from the middle convex vertex.

For the lower bound: it is well-known that it is possible to divide the polygon into $n-2$ triangles, using only internal diagonals. To do this we need exaclty $n-3$ diagonals...

Minimum number of internal diagonals of a simple $n$-gon

1 Answers1