You've probably played rock-paper-scissors at some point in time - two players choose from rock, paper, or scissors, and the winner is the one who is more powerful than their opponent. Rock crushes scissors, scissors cut(s?) paper, and paper covers rock. What's interesting about this game is that it has symmetry. Any one choice is, overall, no more powerful than any other, so one just has to anticipate what their opponent will choose. A graph of rock-paper-scissors (RPS for short) would look something like this (with arrows pointing from winner to loser):
Notice how rock beats scissors and scissors beats paper. That would normally imply that rock beats paper, but it does not. For this reason, RPS is a non-transitive game.
Another interesting way to represent RPS is with rings. Every ring goes over the one it wins against and under the one it loses against. Here is that arrangement of rings.
Another interesting way to represent RPS is with rings. Every ring goes over the one it wins against and under the one it loses against. Here is that arrangement of rings.
These are the Borromean Rings. Red goes over green, green goes over blue, and blue goes over red. Interestingly enough, despite no two rings being linked, the entire set of rings cannot be pulled apart (unless one of them breaks). However, is the normal game of RPS too boring? Oh, I know! Let's play rock-paper-scissors-lizard-Spock (RPS5)! It's very simple. Scissors cuts paper, paper covers rock, rock crushes lizard, lizard poisons Spock, Spock smashes scissors, scissors decapitates lizard, lizard eats paper, paper disproves Spock, Spock vaporizes rock, and as it always has, rock crushes scissors. (Anyone who recognizes that quote, I applaud you). If you didn't catch all of that, here's a graph with what beats what and the game in ring format.
These rings are similar to the Borromean rings in that no two are linked, but the whole thing holds together.
However, what if we changed the rules to RPS-5? In the diagram, every choice beats the one immediately after it (going clockwise), and also the one 3 choices over. What if every choice beat the one after it and the one 2 places after it? It turns out that they're the same. If you take these new rules and switch some of the choices around (or relabel the graph's nodes), the old rules can be produced. In mathematical terms, the two graphs are isomorphic. So there is only one way to play RPS-5. We have to go deeper.
But before we do, a quick explanation of why there is no RPS-4. Every choice has to beat n others and lose to n others. That means that the number of choices must be written as 2n+1 (the 1 being the original choice). 2n+1 is the formula for odd numbers, so RPS can only be played with an odd number of choices.
Back to the point! RPS-7 is the next possible game. There are multiple ways to name the 7 choices, so I will not give them names. There are 2 symmetric ways to play RPS-7, "symmetric" meaning that if every node were shifted one place over, it would be exactly the same. There is actually one more asymmetric way to play RPS-7, but I won't talk about that right now. The two ways are that every choice beats the ones 1, 2, and 3 places after it, and that every choice beats the ones 1, 2, and 4 after it. These two graphs are non-isomorphic, meaning that no amount of node-relabeling will transform one graph into the other. What's interesting about the second game is that, given any two choices, there is always one that will beat both of them. I don't have graph representations of these games, but I did hand-draw the ring representation of the second game. Here it is.
However, what if we changed the rules to RPS-5? In the diagram, every choice beats the one immediately after it (going clockwise), and also the one 3 choices over. What if every choice beat the one after it and the one 2 places after it? It turns out that they're the same. If you take these new rules and switch some of the choices around (or relabel the graph's nodes), the old rules can be produced. In mathematical terms, the two graphs are isomorphic. So there is only one way to play RPS-5. We have to go deeper.
But before we do, a quick explanation of why there is no RPS-4. Every choice has to beat n others and lose to n others. That means that the number of choices must be written as 2n+1 (the 1 being the original choice). 2n+1 is the formula for odd numbers, so RPS can only be played with an odd number of choices.
Back to the point! RPS-7 is the next possible game. There are multiple ways to name the 7 choices, so I will not give them names. There are 2 symmetric ways to play RPS-7, "symmetric" meaning that if every node were shifted one place over, it would be exactly the same. There is actually one more asymmetric way to play RPS-7, but I won't talk about that right now. The two ways are that every choice beats the ones 1, 2, and 3 places after it, and that every choice beats the ones 1, 2, and 4 after it. These two graphs are non-isomorphic, meaning that no amount of node-relabeling will transform one graph into the other. What's interesting about the second game is that, given any two choices, there is always one that will beat both of them. I don't have graph representations of these games, but I did hand-draw the ring representation of the second game. Here it is.
Pretty interesting, right?
For a more in-depth (but slightly more confusing) source on the relation between RPS and Borromean rings, see this paper. However, I mainly got the information from this video.
For a more in-depth (but slightly more confusing) source on the relation between RPS and Borromean rings, see this paper. However, I mainly got the information from this video.