# Counting non-intersecting double-diagonals in polygons

The purpose of this blog post is to outline some deliberate practice I’ve done on solving a problem. A friend and I were talking about his polygon problem. The original problem statement is in portuguese so I’m posting the original statement and then the english translation:

Translation:

Given a regular N-gon, how many ways can it be cut using two diagonals that don’t intersect (they can’t intersect in the interior and they don’t share a common vertex).

I found the following fact to be useful: In an N-gon the number of diagonals is $\binom{N}{2} - N = \frac{N(N-3)}{2}$ (for proofs see 1, 2, 3, 4).

Sidenote: The title of the first thread is "Number of Diagonals in Regular Polygon Makes me Question my Sanity". That’s a bit too dramatic but anyways.

The previous formula just counts all the ways of picking two vertices but removes the edges of the polygon because those are not diagonals. Another proof would be that after we pick a first vertex, we can pick the second vertex to be different from the first and its adjacent neighbours, and then we need to account for the double-counting.

Returning to the problem at hand, for $N \leq 5$ there are no such layouts because there is no room for the 2nd diagonal.

This is what happens for $N=6$ (hexagons):

And here is $N=7$ (septagons):

Sidenote: I wrote this notebook to generate the layouts seen above.

I wrote the following program to count these layouts:

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N=10
S=0
def L(x):
return (x*(x-3))/2
def C(x):
if x < 5:
return 0
u = L(x) - 2*(x-3)
if u < 0:
return 0
else:
return u
for x in range(N):
y1 = (x-2)%N
y2 = (x+2)%N
while True:
if x<y2:
u=y2-x+1
v=N-u+2
dS=C(u)+C(v)
S+=dS
y2 = (y2+1)%N
if y2 == (y1+1)%N:
break
S/=2
display(S)

This program iterates through all the first diagonals. This diagonal will split the N-gon into a u-gon and a v-gon. Next it needs to find how many second diagonals the u-gon and v-gon can hold. So the second diagonal is part of an u-gon or a v-gon but is not allowed to use any of the two vertices of the first diagonal(which is why we subtract that quantity in function C). It then adds up all these possibilities. As before, we have to handle double-counting because if the first diagonal doesn’t intersect the second, the second doesn’t intersect the first(both (D1,D2) and (D2,D1) will be considered).

A search on OEIS shows that this was already classified as A117662, A005701, A050297, so the problem has a closed-form. Maybe I’ll read more about the closed form at some point.

The program’s output matches the OEIS formula (I’ve only checked up to $N=15$). If I have some time I’ll probably think more about ways to derive that formula. I had some fun writing this but since the problem in its original form was solved I’ll set the problem aside for now.