# How to calculate the number of connecting lines amongst corners given the number of edges

Example:

A shape with 4 edges has 2 connecting lines in the middle.

A shape with 5 edges has 5 connecting lines in the middle.

A shape with 6 edges has 9 connecting lines in the middle.

I can see the increment is

4->5 (2 lines to 5 lines; difference of 3)

5->6 (5 lines to 9 lines difference of 4).

It seems to increment that way. What is the mathematical property of this and what is the formula to calculate this given any number of edges?

Consider a convex $n$-gon for $n \ge 3$. There are ${n \choose 2} = n(n-1)/2$ unordered pairs of vertices. Draw lines joining them. $n$ of these are edges of the $n$-gon, leaving $n(n-3)/2$ lines in the interior.

Your terminology is not standard, but I think you want a formula for the number of diagonals in a convex polygon with $n$ sides. That formula is$$\frac{n(n-3)}2.$$

It does increase that way, but normally one would look at all the edges, so for $3$ it's $3$ and we have $4\to 6$, $5\to 10$ and $6\to 15$. To see why, imagine you have your five-vertex (five-corner) figure, and add the sixth vertex somewhere. Now you have to add five edges to make a complete graph.

In general, when you go from a complete graph (i.e. a set of vertices, with an edge between any two of them) of $n$ vertices and you add another vertex, you need to add $n$ edges to make the graph complete again. So the total number of lines in an $(n+1)$-vertex complete graph is $$0 + 1 + 2 + \cdots + n = \frac{(1+n)n}{2}$$ The numbers that appear in this way are known as the triangular numbers, by the way.

First, determine the number of vertexes: n

Second, determine the number of vertexes that each vertex can be connected with (each vertex has n-3 vertexes left to connect it; the 3 comes from itself, and the two vertexes that are on each side).

Third, since there are n vertexes and each vertex can be connected with n-3 other vertexes, the total number (without considering vice versa) will be n(n-3).

Finally, since each line found above can be connected from point p to point q, and vice versa, the number is doubled. So if we divide by 2, n(n-3)/2 is the number you are finding.

example above: n=5, the number is 5(5-3)/2=5.

This is existentially consequential. My interest is that universal peer-to-peer civic discourse becomes intractably complex as population size increases. Ancient Athens could have a democracy in which each citizen's voice could be heard in the agora, whereas in a country with a population of $$3. 3 \cdot 10 ^ 8$$, citizens can only vote with the voice of each asymptotically approaching zero.

The number of peer communication paths basically is 1/2 the square of the number of communicators. For asymmetrical communication leader and follower communication, the number of communication paths is the same as the number of members of the the group, which is a much smaller number, and, in particular, for each communicatee, it is 1.

Large groups are mathematically inconsistent with peer-to-peer civic discourse. No town hall can realistically accommodate $$(3.3 \cdot 10 ^ 8) ^ 2 / 2$$ communication paths. Is there even a networking super-computer that can multiplex that many communication lines?