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

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 math.stackexchange.com/questions/83580/… for some information. Commented Jan 6, 2016 at 16:31
• I suspect the answer is $n-3$. Commented Jan 6, 2016 at 17:41
• @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
Commented Jan 6, 2016 at 18:54

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...