Why Futurama is awesome...

Say that you have multiple objects that can switch inherent properties. However, once 2 objects have switched, they can not switch again. Is there a way that all objects can regain their original inherent properties? Let's ask Futurama. Read the original here.

Futurama Writer Invented A New Math Theorem Just To Use In The Show

Casey Chan[1] 

A writer for Futurama created a brand new math theorem based on group theory to explain a plot twist in the show. That is like, going way beyond the call of duty, dude.

Ken Keeler, the Futurama writer behind the theorem, actually has a PhD in math, so this was probably just a walk in the park for him. But for the rest of us non math geniuses, his theorem was used to explain a problem with an invention that let characters switch bodies. In the show, you can only switch bodies once with the same pair of people, so they needed an equation to prove that with enough switching bodies around, everyone will eventually end up as who they really are. Insert: funny jokes, robot humor and black comedy and mix accordingly.

Keeler's theory would mathematically put everybody in the right place so that all could be right in the Futurama world. Without Keeler's work, Bender's brain might have ended up in Amy's body forever! Which, to think about, actually might not be such a bad thing. I'm no Keeler so I can't explain his theory but you can check out the proof in full here[2]. [TheInfosphere.org[3] via Geekosystem[4]]

References
^ Click here to read posts written by Casey Chan (gizmodo.com)
^ here (theinfosphere.org)
^ TheInfosphere.org (theinfosphere.org)
^ Geekosystem (www.geekosystem.com)

First let π be some k-cycle on [n] = {1 ... n} WLOG [without loss of generality] write:

π = 1  2  ...  k  k+1  ...  n
    2  3  ...  1  k+1  ...  n

Let represent the transposition that switches the contents of a and b. By hypothesis π is generated by DISTINCT switches on [n]. Introduce two "new bodies" {x,y} and write

π* = 1  2  ...  k  k+1  ...  n  x  y
     2  3  ...  1  k+1  ...  n  x  y

For any i=1 ... k let σ be the (l-to-r) series of switches

σ = (  ... ) (  ... ) () ()

Note each switch exchanges an element of [n] with one of {x,y} so they are all distinct from the switches within [n] that generated π and also from . By routine verification

π* σ = 1  2  ...  n  x  y
       1  2  ...  n  y  x 

i. e. σ reverts the k-cycle and leaves x and y switched (without performing ).

NOW let π be an ARBITRARY permutation on [n]. It consists of disjoint (nontrivial) cycles and each can be inverted as above in sequence after which x and y can be switched if necessary via , as was desired.