While I’m sure many people have their own explanations for why this film didn’t resonate with a larger crowd, I would like to posit my own: that our culture’s math anxiety runs so deep, we instinctively run when there’s even a hint of mathematics afoot.  For if you look closely enough, you can find some mathematics in this film.

As you probably know, the plot of the film involves Scott Pilgrim fighting his way through the seven evil exes of Ramona Flowers, the girl of his dreams.  And don’t let his wiry frame fool you, for Scott Pilgrim has played enough Street Fighter to know how to throw a serious punch.  This, combined with his sharp wits, make Scott a deadly opponent.

That’s not to say that these fights are a breeze, however.  Consider Scott’s fight against Ramona’s third evil ex, Todd Ingram.  Because of his vegan diet, he has certain psychic and telekinetic powers that give him a distinct advantage.  Luckily for Scott, though, he is not the sharpest of the evil exes.

Immediately prior to this clip, in an act of unparalleled physical strength, Todd Ingram executed a most righteous uppercut on Scott Pilgrim that sent him flying through the air.  He landed some time later in a pile of garbage.  When I watched this scene, one thought came to mind, a thought that I could not dispel for the rest of the movie: how high up did Scott Pilgrim go?  Such a question may seem unanswerable, but with some careful measurement and a simple application of the classical equations of motion, an answer is entirely within our grasp.

Consider the setup: Todd and Scott begin in a face off.

Todd then hits Scott and he flies upward with some initial velocity v₀. After a certain time, Scott returns to earth.

In the film, Scott remains airborne for approximately 23.6 seconds.  If we ignore air resistance, since it takes him about as long to go up as it does to come down, this means he reaches the top of his trajectory around 11.8 seconds into his flight.  We can then find the initial velocity, because we know that when Scott Pilgrim reaches the top of his trajectory, his velocity will momentarily be zero.

More specifically, the velocity at any time t is given by

v(t) = v₀ + gt,

where g is the acceleration due to gravity, typically approximated by -9.8 m/s² (since Scott Pilgrim is Canadian, I think it’s only fair that we perform this analysis in the standard units of his homeland).  In particular, v(11.8) = 0, so we have

v₀ = 9.8 x 11.8 = 115.6 m/s,

so Todd sends Scott flying at a speed of approximately 115.6 meters per second, or a whopping 258.6 miles per hour.

Given this initial speed, how high up will Scott travel?  For this, we can use the equation governing the vertical position (which is the integral of our velocity equation).  If we set Scott’s initial height to be 0, then his height at time t is given by

y(t) = v₀ t+ gt²/2.

Since we now know the initial velocity, we can plug in values at t = 11.8 (when Scott reaches the highest point).  This gives us y(11.8) = 115.6 x 11.8 – 9.8 x (11.8)²/2, which is approximately 681.8 meters (if you prefer, around 2,237 feet).

Of course, this number isn’t very meaningful without context.  As you can see from the picture below (courtesy of Wikipedia), Scott flew way above the Empire State building, and until recently would’ve flown above the highest skyscrapers in the world.  In other words, that punch sent him up fairly high.

Click for a larger view

Admittedly, this analysis is oversimplified by the fact that we ignored air resistance.  If we try to take this into account, what sort of answer will we obtain?

One can obtain analogous answers in the case where air resistance is included, but the mathematics involved quickly becomes more complicated.  Let me simply summarize the results, and for those brave souls who want more explanation, I will provide it at the end.

First of all, even though Scott’s trip takes 23.6 seconds, with air resistance included we can no longer assume that he reaches the zenith of his trip in half the time.  This is because air resistance on the way down will slow his acceleration, so it will actually take longer for him to come down than it did for him to go up.  I have calculated that in this case, instead of reaching the peak at 11.8 seconds, he actually reaches it at closer to 8.1 seconds.

As for the peak itself, once you know the time the peak is reached it’s not hard to compute.  In this case, I found that the peak distance was around 649.1 meters.  Lower than in the previous case, but not as much lower as I would’ve thought.

Perhaps the most interesting thing to note is the change in initial velocity.  To keep Scott in the air for 23.6 seconds with no air resistance, we saw that Todd must propel him with an initial velocity of 115.6 m/s.  With air resistance, however, to keep Scott in the air for that long requires a much greater initial velocity – around 455.1 meters per second, which is over 1,000 miles an hour.  This is also well above the speed of sound, so Scott should have produced a sonic boom as he traveled upwards – the fact that he doesn’t in the film shines a light on Canada’s best kept secret: it lies inside a perfect vacuum.

Those interested in the details of the air resistance problem, read on.  But be careful, there is mathematics ahead!

*

To solve the problem in this case, I began with Newton’s second law:

,

where is Scott’s acceleration as a function of time, and v is his velocity.

The drag force provided by air resistance, ignored in our earlier calculations, is directly proportional to the square of the velocity.  Moreover, when Scott is traveling upwards, the drag force points in the same direction as the force of gravity.  This means that the net force on Scott’s body must be , for some constant c.

When Scott reaches the zenith and begins to descend, the drag force points in the OPPOSITE direction of the gravitational force, so on this side of the trip we see that the net force is .  Note that this net force will shrink in magnitude until the object reaches its terminal velocity, which we will denote by .  Since the sum of the forces is 0 when an object reaches terminal velocity, we see that .

With this setup, we can now prove the following claim.

Claim: Suppose an object is thrown vertically upwards with some initial velocity and lands at a later time .  Then the time at which the object reached the highest point on its trajectory, denoted , satisfies the following equation:

Why is this true?  Well, on the upwards trajectory, isolating the acceleration we see that

.

By separation of variables, this gives

Since , this implies that for some constant k.   Solving for v we find that

Moreover, since , we see that .

Integrating once more and using the fact that the initial height is 0, we find that for ,

A similar argument works for the downwards trajectory – the only difference is that now the force of gravity and the drag force are in opposite directions.  This has the effect of turning the velocity into a function of tanh, rather than tan.  The end result is that for , we have

Since , by integrating once more we can also conclude that for ,

Note that at we have two ways of writing the position, depending on which formula for we’d like to use.  Setting the two equations equal at , canceling out the common factors of , and exponentiating each side, we are led to the equality , which proves the claim.

This allows us to find numerically given g, , and  .  As usual, we approximate g by 9.8 meters per second per second.  For the terminal velocity, I consulted this website to find the terminal velocity of a skydiver.  I averaged the first four estimates that were in a similar range, and came up with an estimate of 54.7 m/s as a terminal velocity for Scott.  Once again, we use 23.6 as the value for .  Graphing for these values of the parameters, you will see that there is a root near t = 8.1.  This gives us the time at which Scott achieves his maximum height, and this in turn can be used to find that maximum height, , and the initial velocity, .

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.

What do I mean by “best possible”?  Well, since 1995 it’s been known that the minimum number of moves needed to solve a Rubik’s cube is at least 20 – in other words, it was proven that there exists a configuration of the Rubik’s cube which is solvable in 20, but no fewer than 20, moves.  But it was not clear if this could be improved – after all, once this result has been proven, is it possible to find a cube that needs 21 moves to be solved?  What about 22? or 25?  or 1,000?

In the same year, it was shown that the maximum number of moves could be bounded above by 29.  So, for the past 15 years, we have known that every possible configuration of the Rubik’s cube (of which there are 43,252,003,274,489,856,000) can be solved in somewhere between 20 and 29 moves.  The question, since then, has been to narrow this gap in order to find the smallest number of moves for which every configuration is actually solvable.

So, we now know that number is 20.  The researchers who discovered this fact have created a website in honor of it, in which you can learn more about this history of the problem, as well as see examples of configurations that require exactly 20 moves to solve.

For those of you who may fancy yourselves Rubik’s cube hobbyists, however (yes, Will Smith, I am talking to you), it’s unlikely that you can solve a Rubik’s cube in such a small number of moves.  When people learn techniques to solve Rubik’s cubes, they are necessarily techniques that are simple enough for a human to remember.  This means that they may not always be the most efficient.  But now we know that even with the most efficient approach, there are some times when 20 moves will be needed.

For this reason, 20 is referred to as “God’s number.”  Yes, I thought it sounded a little pretentious too.  The reasoning is that an omniscient being would presumably choose the most efficient algorithm.  Although my feeling is that God has better things to do than solve Rubik’s cubes.

In any event, the Rubik’s cube is a rare piece of pop culture iconography in that there are a substantial number of advanced mathematical questions one can ask about it.  Mostly this is because the transformations of the Rubik’s cube that you perform when you try to solve it (or, if you are a jerk, the transformations that you perform when you try to screw up somebody’s solved one) endow the cube with a group structure; in less technical terms, it means that composing two transformations gives a transformation, there is an identity transformation (the transformation that doesn’t do anything), every transformation is invertible (every move you make can be reversed), and transformations are associative (given transformations a, b, and c, applying a to the composition of b and c is the same as applying the composition of a and b to c).  The curious reader can find more about the Rubik’s cube group here.

The Rubik’s cube also holds the distinction of being the only mathematical object (that I’m aware of) that had a starring role in an 80′s cartoon show.  Who knew that anthropomorphic Rubik’s cubes could be so creepy looking?

While some readers may be excited by this discovery, others are undoubtedly less impressed.  After all, the Rubik’s cube is just a silly game, isn’t it?  Knowing that it can be solved in under 20 moves doesn’t advance our understanding of how the world works, does it?  Furthermore, a brief examination of their argument reveals that much of the work involved was manual checking of billions of separate cases by computer.  This is not an elegant solution by any means – while this certainly gives a proof of God’s number, it not a proof from The Book.

On the other hand, there are ideas going on here that are interesting and have been asked in a general context. For example, the notion of taking a finite group and a set of generators for that group, and asking how many generators you need to get to any element of the group, has been studied before (such a number is referred to as the diameter of that group with respect to the set of generators).  In general, the problem of calculating the diameter of a group is quite hard, so having an example in which the diameter is known is helpful, regardless of where that example comes from.

Besides, in the words of the Joker, why so serious?  Not all results have to shatter the way we perceive the world.  Math should be recreational, too.  Having said that, I have mixed feelings about this result being used as representative of what mathematicians do, for reasons I’ve stated before.  Mathematics is more about ideas than it is about brute force calculation (although the latter is sometimes necessary).  Of course, there are ideas involved in reducing the number of cubes to consider, so that the calculations can be carried out in a reasonable time frame.  Also, I understand why a result like this can capture the public’s imagination: it’s easy to state for a general audience.

The result raises other questions as well: what if we consider an n by n by n cube, for general 3 x 3 x 3?  What’s the minimum number of moves needed to solve every  2 x 2 x 2 cube, for example, or every 4 x 4 x 4 cube?  Intuition tells me that the amount of computation involved may grow quite rapidly as n increases, so we may not be able to answer this question for many values of n, but since the 3 x 3 x 3 case is known, it shouldn’t be hard to solve the 2 x 2 x 2 case (in fact, the solution is known: in this case, God’s number is 11).  What about general Rubik’s blocks, where the shapes may not be cubes?

Another question: we know that in the 3 x 3 x 3 case, God’s number is 20.  But how frequent are the positions that require 20 cubes to solve?  Are they rare, and in most cases do we expect that 20 can be improved?  In other words, what is the proportion of worst-case scenarios?  This question is addressed on the new Rubik’s Cube website mentioned earlier.   Of course, analogous questions can be formulated for more general cubes.

And as is frequently the case in mathematics, this result is not without controversy.  For even though these researchers claim to have shown that you can’t do better than 20, the video below highlights a method that can be used to solve any Rubik’s cube in just one move!  (Spoiler alert: yes, a Rubik’s cube will blend).

Of course, I’m not sure how many protractors were actually taken.  Unfortunately, most people didn’t seem interested.  Their loss, I suppose.

Now, a show such as this might seem to have very little to do with mathematics.  But look, and ye shall find.  In the second episode of this past season, the chefs were paired up for one of the challenges.  There were 16 chefs at the time, combining to make 8 pairs.  The pairing was determined by drawing knives: 16 knives were presented in a knife block, and each had a number on it from 1 to 8.  The number was printed on the blade, so each chef would walk to the block, draw a knife, and read the number.  The knives were not replaced afterwards.  Pairs were formed by people who drew the same number.

This dude loves the number 3.

In this particular episode, the first six numbers drawn were 2, 1, 3, 6, 7, and 7.  In particular, the first pair was formed on the 6th draw.  This leads to a natural question: how long would you expect it to take before the first pair is formed?  Six draws seemed a bit long to me (I would have expected the first pair to have been formed sooner), so I immediately set about trying to understand the answer to this question.

To ease ourselves into it, let’s simplify things.  Instead of 8 pairs, suppose there were only 3.  And instead of knives, which are dangerous and pointy, let’s suppose people were choosing balls from a bag.  Rather than differentiating the balls by writing numbers on them, let’s differentiate them by color.  So suppose you have a bag with 3 pairs of balls: one pair red, one pair green, and one pair blue.

The game is this: you draw a ball from a bag and put it aside.  You keep doing this until you have drawn a pair.  The question is how long it will take before the first pair is drawn.

Right away we see that you will get a pair some time between your second and your fourth draw.  Obviously you can’t get a pair after only drawing one ball, so you need a minimum of two draws.  On the other hand, if you don’t have a pair after drawing three, you must have one ball of each color, which means your fourth draw MUST give you a pair of some color.

Given this observation, we can now start to calculate probabilities.  What is the probability that you will have a pair after two draws?  Well, this happens precisely when your first and second draw are the same color.  The probability of this happening is equal to 1/5, since there is no restriction on the first ball you draw, but then there is a 1 in 5 chance that the second one you draw will be the other ball with the same color.

You have a 1 in 5 chance of picking the second green ball after picking the first one, for example.

What about the probability that you’ll have a pair after exactly three draws?  In order for this to happen, your second draw must be a different color than the first, and your third draw must be the same color as either your first or second draw.  Of the 5 balls remaining after your first draw, 4 will have a different color from the first, meaning that the probability of drawing a second ball which is a different color than the first is 4/5.  Similarly, the probability of drawing a third ball which is the same color as either the first or the second ball is 1/2 (see the picture below).  Thus, by the laws of conditional probability, the odds that you will have a pair after your third draw is 4/5 x 1/2 = 2/5.

You have a 2 in 4 (i.e. 1 in 2) chance of pulling a blue or green ball given that the results of your first two draws were blue and green, for example.

The same argument works when calculating the odds that the pair will come on the fourth draw.  There is no restriction on the first draw, there is a 4/5 chance that your second draw will be a different color from the first, there is a 1/2 chance that the third draw will be a different color from the second, and there is then a 100% chance that your fourth draw will be the same color as one of your earlier draws.  This again gives a probability of 2/5.  We see that the probabilities add up to one, as they should.

Given these probabilities we can also calculate the expected value: on average, how many draws will you need before you get a pair?  Since the probability of two draws is 1/5, the probability of three draws is 2/5, and the probability of four draws is 4/5, we see that the average is

2 x 1/5 + 3 x 2/5 + 4 x 2/5 = 16/5 = 3.2.

In other words, on average you will need 3.2 draws before you come up with a pair.

It’s more interesting, of course, to deal with c different colors, rather than just 3.  We can still perform this analysis, and try to find probabilities and expectations. Suppose we have c pairs of balls, each pair of a different color. We draw the balls without replacement from a bag until we find a pair of the same color, then we stop. We can define a random variable to be the draw on which we complete our first pair. For example, in the case c = 3 above, we saw that with probability 1/5, with probability 2/5, and with probability 2/5.

As before, notice that must take a value between 2 and c + 1.  This is because we can’t draw a pair before our 2nd draw, and after c draws, the worst case scenario is for us to have each ball of a different color.  Since we have exhausted all color possibilities, the c + 1st draw must give us a pair (in essence we are applying the Pigeonhole Principle).  So, to describe the behavior of we need to calculate the probability that $Y_c = k$ for k between 2 and c + 1.

The same sort of argument as in the simple case c = 3 works here.  Suppose you want to calculate .  In order to find your first pair on the kth draw, you need to NOT draw a pair on your 2nd, 3rd, 4th, …, or k – 1st draw, and then have the color on the kth draw match one of the colors you have already drawn.  Since there are a total of 2c balls in the bag to begin with, we see that the odds of not drawing a pair on the 2nd draw is , the odds of not drawing a pair on the 3rd draw is (since there are 2c – 2 balls remaining, and you want to avoid 2 that are colors you’ve already drawn, leaving you with 2c – 2 – 2 = 2c – 4 options), and so on, so that the odds of not getting a pair on the k – 1st draw is .  Meanwhile, in order to draw a pair on your kth draw, you must pull one of the k – 1 colors that have already been pulled.  Since there are 2c – k + 1 balls remaining, the probability that this happens is .

Combining these, we see that

(Recall that the binomial coefficient is given by , and as usual n! = n x (n-1) x … 3 x 2 x 1 is the product of all integers from 1 to n.) With this formula, we can now see how likely it was for the first pairing on Top Chef to have occurred on or after the 6th draw.  In this case there are 8 pairs, so c = 8, and we see that

which comes out to 16/39, or roughly 41.03%.

Of course, there’s still the question of expectation: approximately how large do we expect to be (remember we saw that )?  I’ll spare you the details, but one can show that for general c, the expected value is given by

In particular, in the case c = 8 from Top Chef, we find that , which is approximately 5.09.  So on average, for c = 8 we expect to find a pair after a little more than 5 draws.

For the more advanced reader, one final question: what happens to the expected value as c grows large?  As it turns out, we can write the expected value in a very nice form in terms of the Gamma function (which one can think of as a generalization of the factorial to the entire real line).  Using the doubling formula , the interested reader can show that

If one then uses Stirling’s formula to approximate the Gamma function, it follows that

in other words as   What a wonderful asymptotic!  This tells us that the number of draws we will need from a bag of c pairs before obtaining our first pair grows like the square root of c times a factor of .  We can compare the estimate given here for with the exact value computed above – in doing so, we find that .  So indeed, the approximation is fairly close to the true value (and the approximation will only get better as c grows).

There are many related questions one could ask.  For example, what if instead of pairs, we look at collections of triplets, or quadruplets?  What if we consider formation of the 2nd pair or 3rd pair instead of only considering the 1st pair?  What if we allow for different numbers of balls of each color (e.g. 2 red balls and 3 green balls)?  But I’ve already gone on too long, so I will leave these questions for another time.  I don’t know if these questions go by a certain name or not – I couldn’t find this particular problem anywhere.  If anyone knows of a paper or book where these problems are discussed, I would be much obliged.

In the mean time, I will close with a picture of Tom Colicchio looking like a badass.  Clearly the worlds of chefs and rock stars have collided – will mathematicians be next?

(Kudos to Dr. Moore for some helpful commentary.)

]]> http://www.mathgoespop.com/2010/07/top-chef-mathematics.html/feed 0 http://www.mathgoespop.com/2010/07/the-housekeeper-and-the-professor.html http://www.mathgoespop.com/2010/07/the-housekeeper-and-the-professor.html#comments Thu, 22 Jul 2010 04:18:01 +0000 Matt http://www.mathgoespop.com/?p=322 Some time ago, I heard about a book from Japan called The Housekeeper and the Professor, written by Yoko Ogawa in 2003 and translated by Stephen Snyder last year.  As the title suggests, the book centers on the relationship between a housekeeper, her son, and a math professor.  The main conceit of the book is that . . . → Read More: The Housekeeper and the Professor]]> Some time ago, I heard about a book from Japan called The Housekeeper and the Professor, written by Yoko Ogawa in 2003 and translated by Stephen Snyder last year.  As the title suggests, the book centers on the relationship between a housekeeper, her son, and a math professor.  The main conceit of the book is that the Professor suffered an accident some years before that impaired his memory, so that his short term memory only lasts around 80 minutes.  In other words, every day the housekeeper and her son come to visit the professor, it is as if they are meeting him for the first time.  He copes by clipping small notes to his clothing, and in spite of his disability he still dabbles in mathematics. 

One part Memento, one part A Beautiful Mind, the book was named a New York Times Book Review Editors’ Choice, and was popular enough in Japan to warrant a film adaptation (the Japanese language trailer for which can be found below).  Clearly, then, the book has resonated with people regardless of language.  But how does the book stack up from a mathematical perspective?

At first glance, it may sound like there’s a lot of fodder for me to complain about here.  After all, how many times have we seen mathematicians in popular culture with some sort of mental handicap?  Granted, memory loss is a new twist – usually insanity is the preferred condition.  Still, though, I don’t think it’s too much to ask for a mathematician who’s just a normal dude (or even, gasp, a normal lady).

Unfortunately, he is also but the latest entry in a long line of mathematicians in popular culture who are socially maladjusted.  He’s also incredibly reclusive – he has a strong aversion to crowds, and when he accompanies the housekeeper and her son to a baseball game midway through the novel, it’s fairly clear that he hasn’t been on an outing in some time.  One could explain these traits as a byproduct of his mental condition, of course: it’s natural for him to be shy around people when he is always meeting them for the first time, and there’s a danger in taking him out for too long lest he should forget what he’s doing out in the first place.  But part of me feels like these are convenient excuses for rehashing familiar tropes about people who study mathematics.

It’s not all bad, though.  In fact, I found myself able to forgive much of what I didn’t like about this portrayal of mathematicians, because there are many positive features about the professor as well.  For starters, the professor is able to form a close relationship to the housekeeper’s son (who he nicknames Root, because the child’s flat head of hear reminds him of the square root sign).  Even though the professor can’t remember who Root is from day to day, every time the boy comes to the professor’s house, the professor dotes on him like a father.  He obsesses over the safety of Root more than the housekeeper, and keeps Root in his mind as much as he can, given his circumstances.

Of course, his concern about Root wouldn’t be complete if it didn’t include concern over his mathematics education.  Here’s another thing Ogawa does quite well – she is able to not only show the professor’s love of mathematics, but she is also able to illustrate how that passion can inspire others.  The professor is a number theorist, and he is always looking for meaning behind numbers (something which, in the hands of a poorer story-teller, would no doubt incite my rage).  What makes the professor’s interest significant is that he never discusses numbers for the sake of random numerological connections – instead, he is able to take small observations and use them to hint at larger mathematical ideas.  Here’s one such example, from the day the three of them went to a baseball game (this may be my favorite passage from the book):

And when he noticed that his seat number was 714 and Root’s was 715, he began to lecture again and completely forgot to sit down.

“The home run record Babe Ruth set in 1935 is 714.  On April 8, 1974, Hank Aaron broke that record, hitting his 715th off of Al Downing of the Dodgers.  The product of 714 and 715 is equal to the product of the first seven prime numbers: 714 x 715 = 2 x 3 x 5 x 7 x 11 x 13 x 17 = 510510.  And, the sum of the prime factors of 714 equals the sum of the prime factors of 715: 714 = 2 x 3 x 7 x 17, 715 = 5 x 11 x 13; 2 + 3 + 7 + 17 = 5 + 11 + 13 = 29.  A pair of consecutive whole numbers with these properties is quite rare.  There are only 26 such pairs up to 20,000.  This one is the Ruth-Aaron pair.  Just like prime numbers, they are more rare as the numbers get larger.  And 5 and 6 are the smallest pair.  But the proof to show that those pairs are infinite in number is quite difficult*. . . . The important thing is that I’m sitting in 714 and you’re in 715, instead of the opposite.  It’s the young who have to break the old records.  That’s the way the world works, don’t you think?” (90)

Happy to have broken the home run record, or happy to be part of an interesting number phenomenon?

There are vignettes like this peppered throughout the book, where the professor will link an everyday observation to some kind of mathematics.  In the story, which is told from the housekeeper’s perspective, we see how this inspires the housekeeper to think about simple mathematical problems.  In spite of her lack of formal training, the professor is able to inspire in her a sense of mathematical curiosity (something which all math teachers should aspire to do).  Those of the mathematical persuasion are rarely shown as being able to interest people who are less mathematically inclined.  I’m glad to see that this book bucks that trend.

Moreover, this book does a better than average job of discussing what makes mathematics so appealing to those of us who study it.  Consider the following exchange between the housekeeper and the professor, after she apologizes for sending a proof of his to a journal via regular mail instead of express:

“No, there was no need to send it express.  Of course, it’s important to arrive at the correct answer before anyone else, but it’s just as important that the proof is elegant.”

“I had no idea a proof could be beautiful . . . or ugly.”

“Of course it can,” he said.  Getting up from the table, he came over to the sink where I was washing the dishes and peered at me as he continued.  “The truly correct proof is one that strikes a harmonious balance between strength and flexibility.  There are plenty of proofs that are technically correct but are messy and inelegant or counterintuitive.  But it’s not something you can put into words—explaining why a formula is beautiful is like trying to explain why the stars are beautiful.” (16)

While I don’t necessarily agree with the last part, I think it’s refreshing to find a discussion of what accounts for mathematical beauty in a book like this.  This type of discussion between a mathematical expert and a mathematical amateur is not often present in works that center on mathematics – more frequently, the conversation is between math experts, or is not about mathematics at all.  Providing this type of simple insight to a reader who may not have a mathematical background is certainly a plus.

Despite falling into some tired stereotypes, the professor emerges as a fully realized character.  His memory problems are much more than a gimmick, and while they enable certain stereotypes to persist, Ogawa also uses his disability to showcase a degree of empathy for other people that is not often found in portrayals of those who study mathematics.  Overall I found that I quite enjoyed the book – if you’ve got a lazy Sunday coming up (the book is short, so you could easily finish it in a weekend), I’d certainly recommend giving this story a shot.

*Actually, the question as to whether or not there are infinitely many such pairs is actually still open.  See here for more information.

At least, that is what the media would have us believe.  For the past several weeks, Paul the Octopus has captured the hearts, minds, and stomachs of people around the world.  He’s a charming octopus, to be sure, but it isn’t his good looks that have gotten him this far.  Instead, it’s his seeming ability to correctly predict the outcome of soccer matches.  As time has gone on and Paul’s predictions have continued to prove themselves accurate, the amount of press he has received has only increased.  Articles about him are everywhere on the internet: here‘s one discussing public outrage after he correctly predicted Spain to defeat Germany in the semifinals, and here‘s an article discussing his preference for Spain over the Netherlands in the finals.  Search for “Paul the Octopus” in Google News and you will find thousands of results.

This is one popular mollusk.

Of course, I suppose it’s possible that Paul really can see into the future, at least as far as soccer is concerned.  After all, he did correctly predict the winner of every game asked of him; an impressive feat, seeing as how his advice was requested a total of 8 times.  However, Occam’s Razor suggests that we should look for a simpler explanation.

A natural first choice is to guess that Paul was simply guessing randomly, and is very lucky.  The odds of this happening are small – assuming he has a 50% chance of picking correctly, the odds of him being right each one of the 8 times he predicted a winner in this World cup would be 1/2⁸, which is only approximately .39%. Very low odds indeed.

This analysis ignores biases that may be present – in particular, Paul’s octopus vision may bias him towards certain flag designs (which, given the fact that he frequently chooses Germany, seems plausible).  The Wikipedia article I linked to discusses other sorts of possible bias.  However, these biases would only influence which box he selected – they wouldn’t necessarily affect the odds that his selection would correspond to the winning team (although it is possible that he is being persuaded to throw his chips in for the favored team, which would increase the likelihood of his success).  Either way, questions concerning how Paul makes his selection are more interesting to me, so let me focus for the moment on that.

First of all, one could easily argue that unlike flipping a coin, the trials here (i.e. Paul’s selections) are NOT independent. Indeed, what’s going on here may be very similar to an article in the New York Times a couple of years ago that discussed the lurking presence of the Monty Hall Problem in a classic experiment from psychology (which I discussed here).

The idea is quite simple.  Paul chooses between two opponents in the World Cup by selecting a piece of food from one of two boxes.  Each box is labeled with a country’s flag, and this is the most obvious distinction between the boxes.  Suppose, for the sake of argument, that Paul has in his mind a ranking of his preferences for the flags, starting with the one he likes the most, and ending with the one he likes the least.  Assuming between any pair of flags Paul prefers one to another, one could theoretically determine his preferences by giving him sufficiently many pairings of different flags.  Moreover, assuming he does have preferences, the game selections are no longer independent, because each game gives us some information about his preferences.

Let’s dig deeper and look at his selections throughout the World Cup.  Paul gave predictions for 8 games, and those 8 games involved 9 separate teams (only one game did not involve Germany).  In the first game, Germany versus Australia, Paul selected Germany.  Let us note that as: Now, already this gives us information about Paul’s preferences.  There are 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 9! = 362,880 possible ways to order these 9 teams, since you can choose any one of the 9 teams to be your favorite, any one of the 8 remaining to be your second favorite, and so on.  But given the above information, we already know that any choice with Australia ranked higher than Germany can not match Paul’s preferences.  This eliminates a surprising number of possible outcomes – half, in fact, since for every list in which Germany is ranked higher than Australia, we can flip these two countries to obtain a ranking in which Australia is higher than Germany.

This then affects the probability of all subsequent pairings!  Let’s take a look at the next game.  In the second game, Paul correctly predicted Serbia to defeat Germany.  We can write this as: This gives us even more information!  Now, not only do we know that Paul prefers Germany to Australia, we also know he prefers Serbia to Germany.

Moreover, suppose (as seems reasonable) we assume the probability that he prefers Germany to Australia is 50%.  Now, after the first match, we know he prefers Germany to Australia – the right followup question to ask is, what is the probability he prefers Serbia to Germany GIVEN that he prefers Germany to Australia?  In particular, we have a conditional probability question here – these two facts are not independent!  For example, the fact that Paul prefers Germany to Australia means that Germany cannot be last in his rankings – this should decrease the probability that he prefers Serbia to Germany.  And indeed it does: supposing that each of the 362,880 rankings is equally likely, the probability that someone will prefer Serbia to Germany given that they prefer Germany to Australia is not 50%, but is only 33%!

To see why, consider an arbitrary ranking of the 9 teams.  If we only consider the relative placement of Serbia (S), Germany (Ge), and Australia (A), there are 6 possible rankings among these three:

1. S > Ge > A
2. S > A > Ge
3. Ge > S > A
4. Ge > A > S
5. A > Ge > S
6. A > S > Ge.

However, if we also know that Ge > A, then rankings 2, 5, and 6 are eliminated, leaving us only with

1. S > Ge > A
2. Ge > S > A
3. Ge > A > S.

In particular, of these three remaining, Serbia is ranked higher than Germany only once.  Therefore, the probability that S > G GIVEN that G > A is only 1/3, or 33%.  (If you prefer, you can use a counting argument to show this as well.)

Of course, just as the first match gave us information, so did the second.  Therefore, when it comes to the third match, we have even more information at our disposal.  The third match was between Germany and Ghana, and Paul correctly identified Germany.  In other words: Now the appropriate question to ask, of course, is: what is the probability that Paul prefers Germany to Ghana, given that he prefers Serbia to Germany and Germany to Australia?  Well, we know that Germany can’t be his first or his last choice, because it must be preceded by Serbia and followed by Australia.  Therefore, among these four countries, Germany must rank second or third.

If Germany ranks second, Serbia must be first, but we are free to make Australia third or fourth.  Similarly, if Germany is third, then Australia must be last, and we are free to make Serbia first or second.  In other words, we have four outcomes:

1. S > Ge > Gh > A
2. S > Ge > A > Gh
3. S > Gh > Ge > A
4. Gh > S > Ge > A.

Among these four choices, only 2 have Germany preferred over Ghana.  Thus the conditional probability that one would prefer Germany to Ghana is again 50%.

The interested reader can easily continue on in this fashion.  If you’re impatient, however, you can calculate the probability that Paul would have made the selections he did more directly.  All we need to know is who Paul selected in each match.  I’ll tell you that in the subsequent matches, Paul picked Germany over England, Germany over Argentina, Spain over Germany, Germany over Uruguay, and Spain over the Netherlands.

Given this information, suppose you want to know the probability that Paul selects Germany over Australia, Argentina, Uruguay, Ghana, and England, while he selected Spain over Germany and the Netherlands and Serbia over Germany.  As stated before, there are 9! possible lists of preferences.  In this case, it’s not hard to determine how many would lead to the behavior seen in this year’s World Cup.  Since Paul picked Germany over 5 teams, but behind 2 teams, we know that Germany can only be ranked 3rd or 4th (any higher and there wouldn’t be room for Spain and Serbia above, any lower and there wouldn’t be room for the 5 teams below).

If Germany is ranked 3rd, then we can choose to put either Spain or Serbia in 1st place.  Whichever one we don’t put in first place will then need to go in 2nd place, so that both countries are ranked higher than Germany.  After that, we are free to order the remaining countries however we like.  In other words, we see that there are 2 x 6! = 1,440 possible lists of preferences if Germany is ranked 3rd:

1. 2 choices
2. 1 choice
3. Germany
4. 6 choices
5. 5 choices
6. 4 choices
7. 3 choices
8. 2 choices
9. 1 choice.

Meanwhile, if Germany is in 4th place, we need to figure out how many ways there are to choose the three teams above it.  Notice that since Australia, Argentina, Uruguay, Ghana and England must all be ranked lower than Germany, the three countries ranked above it must be Spain, Serbia, and the Netherlands.  However, since we also need the Netherlands to be ranked below Spain, this only gives us three possibilities for the ranking of the first three teams: Spain, Serbia, Netherlands; Spain, Netherlands, Serbia; and Serbia, Spain, Netherlands.  Once we have made that selection, however, we are free to choose the 5 teams below Germany however we please.  In other words, if Germany is 4th, there are 3 x 5! = 360 possible lists of preferences.

Combining these, we see that there are 1,800 possible lists of preferences that would lead to the behavior shown by Paul.  Since the total number of outcomes is 9!, this gives a probability of only 1,800/9!, or roughly .49%.  This is the same value you will get if you calculate the remaining conditional probabilities and multiply them together.

Of course, if one wants a more impressive number, one can always try to correct for bias in Paul’s selection.  For example, suppose we assume that Paul will never choose England or Australia if given the option of a different flag – this seems reasonable, given experiments on how his species sees (the other flags have more contrast and are more focused on horizontal shapes, which apparently his species is drawn to).  If we make this assumption, the number of potential preference lists drops to 7! x 2 = 10,080, in which case the probability that Paul would choose as he did jumps up to 17.9%!

There are valid concerns about this model, though.  For instance, given the choice between two flags, why should we assume that Paul will always choose the same one over the other?  Equivalently, why should we believe that Paul’s decisions follow a prescribed preference list?  Indeed, when Paul was used to make predictions in 2008, he selected Germany over Spain, unlike his selection of Spain over Germany in 2010.  In 2008, it was Germany who was the victor, and so Paul guessed incorrectly – perhaps he learns from his mistakes, after all.

Whatever the case, I doubt this octopus has any special ability.  And even if he does, I don’t know that that would necessarily be a good thing.  For we all know that when 8 limbs are combined with super powers, nothing good can come of it.

Is this the future that Paul's powers portend? I believe it is.

Forget about the Tuesday fact for a moment – if you have a friend with two children, one of whom is a boy, what is the probability that the other child is a boy?  You might expect that the answer should be 50%, since the sex of one child shouldn’t affect the sex of the other.  But this is not quite right, because you’re not told whether the boy is the older or younger child.

There are only four possibilities when one has two children, so the situation is easy to analyze.  With two kids, the four possibilities are boy boy, boy girl, girl boy, and girl girl.  If you know that one of the kids is a boy, this eliminates girl girl from the list of potential combinations, leaving us with the three outcomes boy boy, boy girl, and girl boy.  Of these three outcomes, we see that only the first has two boys, and so we conclude that the probability of the second child being a boy is 1/3, NOT 1/2!

There’s another way to answer this question, one that generalizes nicely to the more complicated question asked by Foshee.  There are two cases to consider: either the boy you know about is the younger child, or the older child.  If the boy you know about is the younger child, there are two possibilities for the older child (girl or boy).  Similarly, if the boy you know about is the older child, there are two possibilities for the younger child (girl or boy).  This gives four outcomes, but you have counted the boy boy outcome twice.  In other words, we see there are only three distinct outcomes, and only one of them has two boys, so again we see that the probability is 1/3.

Now let’s return to the original question.  Again, it seems like the fact that the boy was born on a Tuesday shouldn’t matter, but if we do the same analysis as above, we’ll see that this is not the case.  The information about the day does make the number of outcomes larger, however, so it’s easier to get mixed up.

As before, let’s split into two cases, depending on whether the Tuesday boy is younger or older.  If the younger child is the Tuesday boy, then there are 14 possible outcomes for the older child, since there are 2 choices for the sex of the child and 7 choices for the day of the week on which the child was born.  Similarly, if the older child is the Tuesday boy, there are once again 14 possible outcomes for the younger child.  However, notice that we have counted the outcome of two boys both born on Tuesday twice, just as we counted the outcome of two boys twice in the simpler problem.  Correcting for this double counting, we see that there are 14 + 14 – 1 = 27 possible outcomes.  Of these outcomes, 13 of them correspond to having a two boys – if the younger child is the Tuesday boy, there are 7 possible outcomes that will give us two boys, and if the older child is the Tuesday boy, there are again 7 possible outcomes.  As before, though, we’ve counted both children being Tuesday boys twice, so we subtract 1 to correct for this double-counting, which leaves us with a total of 7 + 7 – 1 = 13 desired outcomes.  This means that the probability of the second child being a boy is 13/27.  While still not equal to 1/2, this is much closer to 1/2 than 1/3.

Much like the Monty Hall problem, however, one can understand the mathematical reasoning behind this problem and still have trouble with the intuition.  After all, why should the day on which a child was born have any bearing on the sex of the other child?  On the face of it, the solution to this problem doesn’t make any sense.  One way to try and marry this solution to our intuition is to assign some numbers and explore the data, seeing where our intuition diverges from this picture.

Suppose we take a survey of 19,600 families with two children (you will understand why I’ve chosen this seemingly random number in a moment).  Of those families, suppose they are evenly divided among boy boy, boy girl, girl boy, and girl girl households.  In particular, we see that there are 4,900 families with two boys, 4,900 families with two girls, and 9,800 families with a boy and a girl (in half of these families the boy is older, and in half the boy is younger).

Now suppose we further subdivide the data according to the day of the week in which the child was born.  Suppose that a child is equally likely to have been born on any day of the week.  This means that, for example, among the 4,900 families with two boys, 4,900 ÷ 49 = 100 will have boys who were both born on Monday, 100 will have boys for which the oldest was born on Monday and the youngest Tuesday, 100 will have boys for which the oldest was born on Monday and the youngest Wednesday, and so on.  In other words, for every possible combination of sexes and days of birth, there will be 100 families that match that combination.

Suppose, now, that you take a random family in which one of the children is a boy born on Tuesday.  There will be 1,400 families for which the younger child is a born boy on Tuesday, and 1,400 families for which the older child is a boy born on Tuesday, which means there are 2,700 families total families under consideration (note that we have counted the 100 families with two boys born on Tuesday twice).  Of those 2,700 families, 1,300 will have two boys, for the same reason as above (you can also look at the diagram above – imagine each square as representing 100 families in the survey).  So, if we pick a FAMILY at random with a boy born on Tuesday, the probability that the other child is a boy is 1,300 out of 2,700, or 13/27.

Now, instead of picking a family at random, suppose we pick a BOY at random who was born on Tuesday, and ask for the probability that the boy’s sibling is also a boy.  In this case, note that there are 2,800 boys in our sample who were born on Tuesday – 1,400 who are the younger sibling and 1,400 who are the older sibling.  In particular, note that we don’t subtract out the families with two boys both born on Tuesday for double counting in this case; this is because we are choosing the boy, not the family, and choosing the younger boy born on Tuesday is different from choosing the older boy born on Tuesday.  Because there’s nothing to subtract out, we see that of these 2,800 boys, 1,400 have sisters and 1,400 have brothers.  Therefore the probability that the BOY has a brother is 1/2, the answer our intuition gave us from the beginning!

In other words, you can try to understand this seeming paradox as a difference in perspective.  If we look from the perspective of the family unit, the probability that a two-child FAMILY with one son born on Tuesday will have two sons is 13/27, slightly less than one half.  However, from the perspective of the children, the probability that a BOY born on Tuesday in a two-child family will have a brother as opposed to a sister is 1/2.  In this latter case, the day of the week really is irrelevant.

For some, however, this explanation still may not seem like enough.  I’d encourage anyone with a different approach to sound off below!