concave octagon with the least number of internal diagonals

[Perhaps check it again]

Consider concave octagons in the plane that are non-self-intersecting.

What is the minimum number of diagonals possible for one of these octagons

if it is required that each of the diagonals lie entirely within the octagon?

Edited July 13, 2014 by Perhaps check it again

[nana77]

I think the minimum number has to be 0.

[Perhaps check it again]

nana77, I don't see that number of yours coming up.

Maybe if you shared even some nonspecific thoughts as to why you

think it is the number you stated, it would lead me to give (possibly needed)

clarifications.

I am open to some other possible answers other than the one I have in mind,

but I want to know if someone is making some more/different assumptions

about the problem than the ones I have stated in the original post.

The problem is still open.

Edited July 13, 2014 by Perhaps check it again

[nana77]

minimum means lowest number, hence 0.

Maximum would be more of a challenging question, I think

[bonanova]

[spoiler='Looks like']Five.

A convex octagon has 20 diagonals, all internal.

If alternate vertices are made concave (internal angle > 180^o) some of the diagonals become exterior.
If the concave points are grouped closely into a quadrilateral and the convex points are judiciously repositioned, (changing one concave vertex back to convex in the process) the internal diagonals can be reduced to six: those of the central quadrilateral. First figure, green edges. Red edges are external.

But we can go one step farther by making the central quadrilateral concave, as in the second figure. The change is to move the white quadrilateral vertex into the interior of the other three. The edge that then becomes external is shown dark red. A concave quadrilateral has only five internal diagonals (green edges) and that is also the minimum achievable for an octagon.