### Hey, I have this blog thing, let's use it

Jan. 22nd, 2011 09:50 pmSo we've been working our way through the second season's DVDs here the past few weeks. As you may know if you've seen the show, in one episode this season they introduce the game of "Rock, Paper, Scissors, Lizard, Spock" (a game originally described here as "Rock Paper Scissors Spock Lizard"), an extension of "Rock, Paper, Scissors". (I'm going to call this game "RPSSL" in this post)

This got me thinking more generally about extensions to "Rock, Paper, Scissors", (hereafter, RPS) so let me define:

Adegree-, wherenRPS extensionnis an odd number >= 3, is a two-player game in which players secretly select one ofnsymbols and then simultaneously reveal their choices; assuming the player selected differently from each other, the winner of the game is determined by the game's ruleset. The ruleset of an RPS extension must specify that every token wins against (n-1)/2 of the other tokens, and loses against the other (n-1)/2 tokens.

So RPSSL is a degree-5 RPS extension, because for any choice player A makes, exactly two choices for player B will let A win (and exactly two will let B win).

Two RPS extensions are

**equivalent**if you can obtain one from the other merely by renaming tokens.

So here's what I've come up with:

- There's at least one degree-
*n*RPS extension for every odd*n*>= 3. - Up to equivalency, there's only one degree-5 RPS extension.
- Up to equivalency, there are at most two degree-7 RPS extensions.

I need to work more with the two different degree-7 extensions I have to prove that they definitely aren't equivalent. Getting some paper and writing it down would probably help, rather than trying to just keep it in my head.

For the other two points, I'll just note that if you re-arrange the tokens in RPSSL into the order:

Rock, Spock, Paper, Lizard, Scissors

And imagine those tokens in a circle, then every token loses to the two tokens that immediately follow it and beats the two that immediately precede it. This shows a general pattern that can work for any

*n*and working out that with five tokens, there must be such an order proves the second point.