Table Tennis Player Grouping Problem

Question

I have changed the description of the problem in a different way.

Q: There are N points and all points must be connected by a line segment. But you can't just draw a line, you have to draw a triangle. You have to connect all the points through multiple triangles, but the sides of two triangles must not touch.

*A point can belong to multiple triangles. But two points must belong to only one triangle.

What are the conditions of N that can satisfy this rule?

Below are some of the conditions I thought of. (Groups: triangle's 3 points)

1. If N is 3

Groups : {A, B, C}

2. If N is 7

Groups : {A, B, C}, {A, D, E}, {B, D, F}, {C, E, F}, {A, G, F}, {B, G, E}, {C, G, D}

case 3, 7 image

3. N = p x q (when p and q can be N)

9 elements in p x q matrix

In the p x q matrix, we can choose 3 columns to match one group of p, 3 rows to match one group of q, so we can choose 9 points.

9 elements

And 9 points can be processed into 12 groups. (So, N can be 9.)

Therefore, p x q can be N.

4. Other numbers?

I want to know the generalized rule, or the case for other numbers.

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What I wrote before.

Q: N players play table tennis. If you make three players into a group, They take turns refereeing each other and play three games. Can we make groups so that all players only play one match against each other? What are the conditions of N where all players can be grouped to play only one game against each other?

*One group must include three players. And players can participate in several groups. However, there must be three matches in the group. Therefore, any two people should have only one overlapping group.

Answer 1

What you are looking for is known as a Steiner triple system. In order for this to be possible, $N$ needs to be of the form $6k+1$ or $6k+3$ for some integer $k$.

Proof: Since there are $\binom{N}2=N(N-1)/2$ edges, we need $N(N-1)$ to be a multiple of $3$, which implies that $N\not\equiv2\pmod 3$.

Furthermore, if you focus on the triples containing the particular point A, and delete A from those triples, then what remains are several disjoint pairs whose union is the complement of {A}, which implies that $N-1$ is even, so $N\equiv 1\pmod 2$.

These two facts imply $N \equiv 1\pmod 6$ or $N\equiv 3\pmod 6$.

It turns out that this necessary condition is also sufficient. One method of construction is described in the following paper by Skolem:

Skolem, T. (1958). Some Remarks on the Triple Systems of Steiner. MATHEMATICA SCANDINAVICA, 6, 273-280. https://doi.org/10.7146/math.scand.a-10551