Counting The Number of Variants of a Rock Paper Scissors Game with more than 3 Weapons

Question

The fact that the traditional variant of the game of rock paper scissors has 3 weapons is well known. However with 3 weapons there can be 2 variants:

Variant 1:
    Rock beats Scissors
    Scissors beat Paper
    Paper beats Rock
Variant 2: 
    Rock beats Paper
    Paper beats Scissors
    Scissors beat Rock

Now, if one wants to increase the number of weapons, the number must be increased to an odd number. So variants that satisfy the following criteria are balanced:

• There are $n$ weapons; $\{n|n = 2k+1 \land k \in \mathbb{N} \gt 0\}$
• Each weapon is weak against $k$ weapons, and strong against the other $k$.

My question is: For $n$ weapons, how many variants can there be?

Using an algorithm I've formulated, I found that there are 24 variants for 5 weapons, or in general: $(n-1)!$ variants for $n$ weapons.

Summary of the algorithm:

1. Generate the Strengths and Weaknesses Labelled Adjacency Matrix (SWLAM)
2. Switch the rows and then the columns of the SWLAM
3. Relabel the rows and columns.

For illustrations and examples, please refer to the given link above. What I'm not sure of is whether or not the algorithm can generate all the variants.

Answer 1

When $n\ge 7$, your method will not generate all possible variants. As $n$ gets larger, the proportion of variants you get becomes vanishingly small. This problem is analyzed in the paper Rock-Paper-Scissors meets Borromean Rings by Marc Chamberland and Eugene A. Herman.

When $n=7$, there are three possible isomorphism types for the RPS graph, shown below. The below image is taken from that paper. Your method will generate all permutations of the left graph, but will miss all permutations of the middle and right graphs.