The Futurama Theorem

Dana C. Ernst

Friday Afternoon Mathematics Undergraduate Seminar (FAMUS)

December 4, 2015

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Here is short clip with Professor Farnsworth and the Globe Trotters.

The Prisoner of Benda

The mind swap problem

Here are the swaps that take place during the episode:

The upshot: \[a \rightarrow h \rightarrow l \rightarrow p \rightarrow b \rightarrow e \rightarrow w \rightarrow a \text{ and } f \leftrightarrow z\]

The mind swap problem (continued)

The question is:

Can the machine be used to put everyone's mind back in the proper body if we are not allowed to use the machine on the same pair of bodies more than once?

We need to use math!

The symmetric group

We can model this problem in terms of permutations (i.e., rearrangements) of numbers. It turns out that the set of all permutations of the numbers 1 through $n$ forms a "group" under composition.

A group is a set with an associative binary operation satisfying:

The group of permutations of $\{1,2,\ldots, n\}$ is called the symmetric group and is denoted by $S_n$.

Permutation diagrams

One way of representing an element of the symmetric group is via permutation diagrams, which we illustrate by way of example.



Permutation diagrams (continued)

Let's try multiplying.


Let $\alpha$ and $\beta$ be as on previous slide. Then


But on the other hand


We see that products of permutations do not necessarily commute (order matters).

Permutation diagrams (continued)

However, sometimes permutations do commute.


Let $\beta$ and $\gamma$ be as before. Then


Notice that these result in the same diagram. So, $\beta$ and $\gamma$ commute.

Some comments

The mind swaps as a diagram

If we let $a=1$, $h=2$, $l=3$, $p=4$, $b=5$, $e=6$, and $w=7$, $f=8$, and $z=9$, then the mind swaps from Futurama can be depicted as:


Rephrasing the problem

The problem can be restated as:

Suppose $p$ is a permutation that results from multiplying distinct transpositions. Can we multiply $p$ by a sequence of transpositions that we have not yet used to obtain the identity permutation?

It turns out that, in general, the answer is yes. Keeler's trick to pulling this off is to add in two more people that have not used the machine.

The Futurama Theorem

Here is the solution presented in the show.


Note:This image is taken from The Infosphere: The Futurama Wiki.

Cycle notation

Before describing the solution, we need to introduce a more efficient way of encoding our permutations. One such method is called cycle notation.



Cycle notation (continued)

Let's try multiplying using cycle notation. Note: We will multiply right to left (like function composition).


Consider $\alpha, \beta$, and $\gamma$ in $S_{5}$ as in the previous examples. Then

\[\alpha \beta=(1\; 2\; 3\; 4\; 5)(2\; 4\; 3)=(1\; 2\; 5).\]


\[\gamma\beta =(2\; 4\; 3)(1\; 5)=(1\; 5)(2\; 4\; 3).\]

We saw earlier that $\beta$ and $\gamma$ commute with each other, which is why it looks like nothing happened in the second example.

Some more comments

Fixing a single 7-cycle

Consider just the 7-cycle $(1\; 2\; 3\; 4\; 5\; 6\; 7)$. Introduce two new bodies that have not had their minds swapped with anyone else, say $x$ and $y$. To return all minds of 1-7 back to their rightful owner, multiply the cycle (on the left) by the following sequence of transpositions:

\[{\color{blue}{(x\; 7)}}{\color{red}{(y\; 1)(y\; 2)(y\; 3)(y\; 4)(y\; 5)(y\; 6)(y\; 7)}}{\color{blue}{(x\; 1)}}\]

Let's verify that this actually works! First, notice that all of these transpositions are distinct and since $x$ and $y$ are new, we never used any of the above transpositions to obtain our original scrambling of minds.

\[{\color{blue}{(x\; 7)}}{\color{red}{(y\; 1)(y\; 2)(y\; 3)(y\; 4)(y\; 5)(y\; 6)(y\; 7)}}{\color{blue}{(x\; 1)}}(1\; 2\; 3\; 4\; 5\; 6\; 7)=???\]

Note: $x$ and $y$ now have their minds swapped with each other, but they never sat in the machine at the same time!

Fixing an arbitrary $k$-cycle

Consider the $k$-cycle $(1\; 2\; \ldots\; k)$. How do we "fix" this cycle?

As before, introduce two new bodies that have not used the machine, say $x$ and $y$. To fix a $k$-cycle, multiply the cycle (on the left) by the following:

\[{\color{blue}{(x\; k)}}{\color{red}{(y\; 1)(y\; 2)\cdots (y\; k-1)(y\; k)}}{\color{blue}{(x\; 1)}}\]

Why does this work?!

Fixing products of disjoint cycles

What do we do when we have multiple disjoint cycles to start with?

Some comments and questions


Natural questions

Okay, that's it. Thank you!

These slides were made using deck.js and MathJax.