I greatly enjoyed the episode, but my main irk, if it’s worth calling it that, is that the way the concluding theorem is momentarily presented (quickly flashed in jargon and notation to be blindly admired but with the connotation that it would be beyond the grasp of most of the audience) makes it seem like something “For mathematical experts only”, instead of revealing how easily graspable it is. I dislike the pernicious effect this culturally ubiquitous portrayal of mathematics has, and would’ve been thrilled to see it transcended. Alas. Still, good stuff.
Here’s a very simple proof of the theorem in plain English. To fix any finite amount of mind-swapping:
Step 1: Have everybody who’s messed up arrange themselves in circles, each facing the body their mind should land in (e.g., if Fry’s mind is in Zoidberg’s body, then the Zoidberg body should face the Fry body).
Step 2: Go get two “fresh” (as of yet never mind-swapped) people. Let’s call them Helper A and Helper B.
Step 3: Fix the circles one by one as follows:
3.0) Start each time with Helper A and Helper B’s minds in either their own or each other’s bodies
3.1) Pick any circle of messed-up people you like and unwrap it into a line with whoever you like at the front
3.2) Swap the mind at the front of the line into Helper A’s body
3.3) From back to front, have everybody in the line swap minds with Helper B’s body in turn. (This moves each mind in the line, apart from the front one, forward into the right body)
3.4) Swap the mind in Helper A’s body back where it belongs, into the body at the back of the line. Now the circle/line has been completely fixed, and the Helpers’ minds are back in either their own or each other’s bodies. (The one side effect is that the Helpers’ minds switched places. But that’s ok.)
Step 4: At the very end, after all the circles have been fixed, mind-swap the two Helpers if necessary (i.e., in case there was originally an odd number of messed-up circles).
(For those who care, I’ve taken the liberty of slightly departing from/cleaning up the presentation given on the Futurama greenboard: setting the arbitrary i to 1, commuting the final (x, i + 1) and (y, 1) transpositions, and thinking of {2, 3, …, k, 1} as the back-to-front order of the line. Or, rather, I’ve taken the liberty of just writing down the simplest solution I could come up with, and then describing how it relates to the provided solution.)