The setting of the show naturally lends itself to math and science jokes, and in this department Futurama does not disappoint.  Last week, however, they seriously stepped their game up a notch, by featuring the proof of an original mathematical result as a central feature in the plot of the story.

The mathematics evolves quite organically.  In the show, Amy and Professor Farnsworth have created a mind-switching device, which can swap the minds of any two individuals.  After a brief discussion, they decide it would be neat to swap their own brains:

As you might expect, before long they decide they want to swap back.  Unfortunately, due to the brain’s natural immune response, they discover that once two people have switched minds once, they can never switch back (obviously).
This raises a question: once two people have swapped minds, can they get back to their original bodies by means of a third party?  Bender eagerly switches minds with Amy at this point, for reasons that aren’t important here.  Afterwards, though, Farnsworth and Amy have the following discussion:

Prof. Farnsworth (in Bender’s body): Now then Amy, we’ll simply switch bodies, and then we’ll… no, I’d be back in my body, but then you and Bender would be switched.  And the Amy and Bender bodies can’t trade minds again, since they just did!

Amy (in Prof. Farnsworth’s body): Oh no!  Is it possible to get everyone back to normal using four or more bodies?

Prof. Farnsworth: I’m not sure.  I’m afraid we need to use….MATH!

(dramatic music underscores this last point)

Throughout the rest of the episode, mind swapping occurs quite frequently.  Amy (as the professor) swaps with Leela, Fry swaps with Dr. Zoidberg, Nikolai (the playboy ruler of the Robo-Hungarian empire) swaps with a bucket, and so on.  By the episode’s end, so much mind swapping has been going on that it’s hard to keep track.  Nevertheless, in the end it is discovered by ‘Sweet’ Clyde Dixon that by including no more than 2 new people in this mind swap game, everyone can always return to their original body.  The proof is shown briefly, but for interested parties, here it is in all its glory (the proof was written up by writer Ken Keeler, who earned a Ph.D. in applied math from Harvard University):

Click for a larger version.

The reader with some background will realize that this is  a statement about the symmetric group on n elements (the notation used on the holographic chalk board is explained in the link).  The gist of the argument is as follows: suppose that k people have swapped bodies.  Label these people so that the 1st person is in the 2nd person’s body, the 2nd person is in the 3rd person’s body, and so on, so that the last person (a.k.a. the kth person) must be in the 1st person’s body.

With two additional people (let’s call them Xerxes and Yelena, or x and y for short) we can get everyone back to normal by using the following procedure: have the first body switch with y’s body, then have the kth body switch with x’s body.  Next, have the kth body switch with y’s body, and then, in turn, have x’s body switch with the k-1^st body, k-2^nd body, and so on, until x’s body has switched with the 1st body.  At the end of this procedure, everyone will be back to normal except for x and y, who will be in different bodies.  But since they haven’t switched with each other yet, they can then switch back.

I’ve omitted certain details that keep this from being a complete proof, but this is the heart of the idea (for more advanced readers, I have taken i = k – 1 in the episode’s written argument).  If the procedure seems confusing, let’s take a particular example.  Suppose that Professor Farnsworth and Amy knew of this theorem at the episode’s opening: then they’d know they could switch back with the aid of two other people.  Let’s take Leela and Hermes as the other two players (since they seem to be the most responsible).

The theorem states that to get back to normal, first Amy’s body and Leela’s body should swap (the choice of Amy over Farnsworth or Leela over Hermes is arbitrary).  Recall that Farnsworth is in Amy’s body, so after the swap, he will be in Leela’s body, while Leela will be in Amy’s:

After that, Farnsworth’s body and Hermes’ body should swap.  This gives the following picture:

Now swap Leela’s body and Farnsworth’s body.  Note that after this swap, Farnsworth’s mind and his body have been reunited!

We can also reunite Amy with her body by swapping the bodies of Amy and Hermes:

Now Leela is in Hermes’ body and Hermes is in Leela’s.  But since they haven’t swapped with each other at any point, we can now swap them with each other, thereby returning everyone to their original state!

This shows that we can get everyone back to their original state in only 5 moves.  Moreover, each person only has their mind swapped 3 times.

This gives rise to several questions: in a group of n people, what is the minimum number of swaps needed to return everyone to normal?  What is the maximum number? (The answer depends on the number of cycles in the permutation under consideration.)  What is the maximum number of times an individual’s mind must swap in order for the group to return to normal?  If we place caps on this maximum (say, because the cerebral immune response gets stronger with repeated use), what restrictions does this place on how much fiddling with the mind swapping device before people won’t be able to return to their bodies?

For this episode in particular, everyone returns to their bodies in 13 movies, although the speaker in the video below explains how that number can be reduced to 9.

If you are able to catch this episode when it airs again, I would highly encourage you to do so.  Kudos to the writing staff of Futurama for not shying away from a bit of more advanced mathematics.  While this result doesn’t have far-reaching consequences, the fact that it is an original piece of work inspired by the plot of the episode is a great example of how critical thinking can be used to solve all types of problems.

In closing, I’d like to point out that I’m not the only person to have discussed this problem on the interweb.  The video below offers a great explanation as well.

Some time ago, I wrote an article on the optimal way to select a mate, assuming you know how many eligible partners exist, and that once you’ve dated someone, you can’t go back and date them again (sorry, Drew Barrymore and that dude from the Apple commercials).  This is less romantically known as the secretary problem.  Let me briefly recall the problem and its solution: suppose you have n candidates, from which you want to pick the best one.  This applies to a variety of situations, from hiring a secretary to finding a girlfriend to apartment hunting.  In either case, the outcome is the same: you should look at roughly the first n/e of them (yes, that e), and then select the first one after those n/e which is better than all that you have seen so far.  While this strategy won’t guarantee you get the best choice, it will give you the best choice around 37% of the time.

The major problem with this model is that in many situations, the value of n is unknown.  There are ways to circumvent this problem, which I will not discuss here.  Instead, in the context of finding a mate, I offer the following method to calculate the number of partners you could reasonably expect to find in your area.  This method recently gained some attention when Peter Backus, a Ph.D. candidate from the University of Warwick wrote a paper titled “Why I Don’t Have a Girlfriend.”

The basic technique involves modifying the Drake Equation, an equation used to estimate the number of potential extraterrestrial civilizations in our galaxy.  For those who have never been introduced to this equation, it asserts the following:

N = R · f_p · n_e · f_l · f_i · f_c · L.

According to Wikipedia, these variables represent the following quantities:

R = the average rate of star formation per year in our galaxy,
f_p = the fraction of those stars that have planets,
n_e = the average number of planets that can potentially support life per star that has planets,
f_ℓ = the fraction of the above that actually go on to develop life at some point,
f_i = the fraction of the above that actually go on to develop intelligent life,
f_c = the fraction of civilizations that develop a technology that releases detectable signs of their existence into space,
L = the length of time such civilizations release detectable signals into space.

Given estimates for all of these parameters, one could then estimate the number of civilizations in our galaxy.  Since we don’t know the values of any of these parameters, however, this is more of a thought experiment than anything else.

Nevertheless, the idea can be easily modified to try and find the number of eligible mates in a given area.  Peter Backus’ approach is fairly specific to him, but he links to a more general approach discussed here, in which the following equation is presented:

n = P · f_t · f_o · f_c · A · R

In this case, the parameters are given by

P = the population.  This could be the population of your university, your city, or your country, depending on how ambitious you are.
f_t = the fraction of that population which you would want to mate with, in broad terms.  If you’re a straight male, this would be the fraction of females.  If you’re a gay male, it would be the fraction of males, and so on.
f_o = the fraction of the population you’d want to mate with which wants to mate with you.  For example, if you’re a straight male who wants to mate with females, this will compensate for the fact that some females will be lesbian and therefore unwilling to mate with you.
f_c = author Raymond Francis labels this the out fraction, and describes it as the answer to the question “Of the people in your target gender and orientation, how many of them are open enough about their sexuality to engage in a relationship of the sort you’re hoping for?”  If you are straight, this value is likely 1, or very close to it.  If not, things can be a little bit fuzzier.
A = the fraction of those remaining who fall within your desired age range.  This is, of course, personal to you – if you’d like a socially acceptable age range, you could follow the “half your age plus seven” rule.
R = factors for any remaining filters you wish.  Do you want your partner to have a certain level of education, or a certain income?  Do you need a non-smoker, or demand a euphonium player?  Here’s where you can fold that into the mix.

Yes, Wikipedia has a graph illustrating the half your age plus seven rule. Amazing.

With all these parameters accounted for, N will give you the number of potential mates in your area.

Let’s take this equation for a spin, shall we?  Suppose you are a straight male living in Los Angeles, and looking for a girl to date in Los Angeles.  According to Wikipedia, the estimated population of LA as of 2008 was 3,833,995.  Of those, let’s say that 51% are female, and of the females, let’s posit that 90% are straight or bisexual.  f_c should be high in this case – to be conservative, let’s put it at 95%.

To estimate the age filter, one can obtain some data from this site.  Suppose you are a 25 year old man – then, absent any personal preference, the socially acceptable age range of women for you to date is between 19.5 and 36.  According to census data, in 2000 there were 3,694,820 people in Los Angeles, and of them, 974,004 were between the ages of 20 and 34.  Additionally, there were 251,632 people between the ages of 15 and 19, and 584,036 people between the ages of 35 and 44.  If we make the assumption that ages are roughly uniformly distributed within these brackets, this gives us an additionlal 141,970 people either between the ages of 19.5 and 20, or between the ages of 35 and 36.  Combining this gives a total of 1,115,974 people between the ages of 19.5 and 36 in Los Angeles in 2000, or roughly 30% of the population.  Let’s use this for our value A.

Assuming you have no other restrictions (i.e. taking R = 1), this gives us n = 3,833,995 · .51 · .9 · .95 · .3 = 501,544.  That’s a lot of ladies out there for the taking.  Of course, taking R = 1 is probably unrealistic.  It’s unlikely you want to date women who are married, for example, and everyone has their own personal taste that will decrease the pool even further.   Once you’ve calculated your personal value for R, however, you then know how many eligible mates will be in your area.  Given that, you’ll know how large n/e is, and then you’ll know how many people you should date before you think about settling down.

Although Peter Backus has received a fair amount of buzz for the short paper he has written on this idea, he readily admits that he is not the first to think of applying Drake’s equation in this situation.  I’ve discussed mostly Raymond Francis’ approach here, but Backus has links to many other people that have discussed the idea on his website.  In particular, here’s an exchange from CBS’s The Big Bang Theory (don’t worry, there’s a laugh track so you know when things are supposed to be funny).

In summary, not only can the Drake Equation be used to consider the existence of extraterrestrial life, it can also be used to consider potential mates right here on Earth.  The next step, of course, is obvious: we must combine these two equations to calculate the number of potential extraterrestrial mates.  Undoubtedly the number will be small, but one should never underestimate the power of love.

Interstellar love is a wonder to behold.