Rock, Paper, Scissors, Spider, Man

a year ago
28-06-2023
5 min read
813 words

Have you ever gotten bored of the classic rock, paper, scissors game? Fear no more! Here is a new twist on the classic that will keep you entertained for hours!

The Rules

Spider


The spider wins against:

  • Rock because spiders have been known to withstand the weight of rocks (evolution is amazing)
  • Human as humans are terrified of spiders

The spider loses against:

  • Paper because spiders can be captured by paper
  • Scissors because scissors can cut spider webs (and spiders)

Man


The man wins against:

  • Paper because humans can shoot paper with guns
  • Scissors because humans can shoot scissors with guns

The man loses against:

  • Rock because humans can be crushed by rocks
  • Spider because humans are terrified of spiders

Rock


The rock wins against:

  • Scissors because rocks can crush scissors
  • Human because rocks can crush humans

The rock loses against:

  • Paper because rocks can be covered by paper
  • Spider because spiders can withstand the weight of rocks (evolution is amazing)

Paper


The paper wins against:

  • Rock because paper can cover rocks
  • Spider because spiders can be captured by paper

The paper loses against:

  • Scissors because scissors can cut paper
  • Human because humans can shoot paper with guns

Scissors


The scissors wins against:

  • Paper because scissors can cut paper
  • Spider because scissors can cut spider webs (and spiders)

The scissors loses against:

  • Rock because rocks can crush scissors
  • Human because humans can shoot scissors with guns

Cayley Table

RockPaperScissorsSpiderMan
RockDrawPaperRockSpiderRock
PaperPaperDrawScissorsPaperMan
ScissorsRockScissorsDrawScissorsMan
SpiderSpiderPaperScissorsDrawSpider
ManRockManManSpiderDraw

On the existence of n-dimensional generalizations

Denote the game as G\mathbf{G} with binary operation \circ and identity element II.

Given a set S\mathbf{S} of nn elements (aia_i), we can define a Cayley table for G\mathbf{G} as follows:

a1a2ana1Ia1a2a1ana2a2a1Ia2ananana1ana2I\begin{array}{c|cccc} \circ & a_1 & a_2 & \cdots & a_n \\ \hline a_1 & I & a_1 \circ a_2 & \cdots & a_1 \circ a_n \\ a_2 & a_2 \circ a_1 & I & \cdots & a_2 \circ a_n \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ a_n & a_n \circ a_1 & a_n \circ a_2 & \cdots & I \\ \end{array}

Call the game G\mathbf{G} balanced if {ai  aiaj=ai,  1jn,  aiS}=n12|\{a_i \ | \ a_i \circ a_j = a_i, \ \forall \ 1 \leq j \leq n, \ \forall \ a_i \in \mathbf{S}\}| = \frac{n-1}{2}

Indeed for a G\mathbf{G} to be nontrivial it must satisfy:

  • G\mathbf{G} is closed under \circ (by definition)
  • G\mathbf{G} is balanced

Claim. A balanced game G\mathbf{G} exists if and only if nn is odd.

Proof.


Disincluding, the identity element, there are n1n-1 elements in S\mathbf{S}. For G\mathbf{G} to be balanced, there must be n12\frac{n-1}{2} elements that are their own inverse. This is only possible if n1n-1 is even, i.e. nn is odd.

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