Math Goes Pop! » Math Gets Around

Top Chef Mathematics

Matt — Wed, 28 Jul 2010 23:22:55 +0000

If you like food, Washington DC, hubris, or reality television, then chances are you are a fan of Bravo’s cooking competition Top Chef. Every year the show takes a group of aspiring chefs, places them in a house in a new city, and throws weekly challenges their way. Following the Survivor template, every week one chef is voted off, and at the end someone is crowned Top Chef (and given a large check). This season, the action takes place in our nation’s capitol.

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.

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.)

Love and Marriage

Matt — Wed, 16 Jun 2010 20:12:06 +0000

I’ve previously discussed some mathematical approaches to dating. Specifically, we have seen how choosing a partner can be modeled as a type of secretary problem, and, if you like, you can estimate the number of candidates you should consider by using a modified Drake’s equation. However, as you know, building a lasting relationship is about more than choosing the right partner; maintaining a happy relationship takes work. And even though most people go into a relationship believing they will not end up as a statistic, the unfortunate reality is that nearly half of all marriages in this country will end in divorce.

How can it be that despite the best intentions of many couples, such a significant proportion will not endure? As one always should, we can turn to mathematics for possible answers. In fact, José-Manuel Rey of the Department of Economic Analysis at the Universidad Complutense in Madrid has done just that, by proposing a mathematical model to explain the dynamics of long term relationships. His model is based on the following assumptions: first, that the individuals in the couple have similar traits (this is spelled out more precisely in Rey’s paper, but basically this is so that he can consider one utility function for the couple rather than separate utility functions for each individual), and second, he assumes a so-called law of thermodynamics for sentimental relationships. This law can be stated as follows: “There is tendency for the initial feeling for one another to fade away. This kind of inertia must be counteracted by conscious practices.” In other words, the natural deterioration in a relationship must be counteracted by continual effort, such as a romantic evening, a heartfelt conversation, or listening to some slow jams by R. Kelly.

Given these assumptions, Rey constructs a model to try and explain the paradox (which he terms the failure paradox) in the fact that people who believe they will be together forever will, with a high degree of probability, end up separating. And sure enough, his model offers us an explanation. His work is perhaps best summarized by the following picture:

The graph above is of feeling (x axis) versus effort (y axis) – in other words, how satisfied is the couple in their relationship compared to how much effort they are willing to put into it? Rey’s model asserts that there is a minimum happiness (denoted x_min above) below which a relationship cannot survive. In other words, if your happiness crosses the red vertical line, your relationship is doomed. The couple’s initial happiness is given by the value x₀ – note that this is greater than the happiness of the equilibrium point, reflecting the common observation that happiness within a couple frequently decreases from its initial state (you can think of this as the honeymoon period, if you like). This isn’t to say that couples in long term relationships are unhappy, for indeed the happiness at the equilibrium point is still above x_min; all this is saying is that happiness is lower at the equilibrium stage than it was initially.

What about the effort involved? Heuristically speaking, the amount of effort you put into a relationship should increase your own happiness, up to a point, but then should begin to negatively affect your happiness. Driving your partner to the airport every once in a while will probably make you feel good about yourself, but driving your partner to the airport every weekend will probably start to wear on you. On the graph above, c* represents the turning point from when the amount of effort you put in positively affects your happiness to when it negatively affects your happiness.

Here is where an important feature of the model comes into play: the amount of effort at the equilibrium point is greater than c*! In other words, in order to maintain equilibrium in your relationship, you need to be putting in more effort than you would optimally choose to. It is this s0-called “effort gap” that Rey identifies as being the cause of so many problems. Note that if the effort gap is large enough, then maintaining the appropriate level of effort may drive happiness below the minimum required value, and even if this is not the case, maintaining the appropriate level of effort will still require some loss of the couple’s happiness, since the effort required will always be greater than c*. To put it more bluntly, relationships require sacrifice.

I’d encourage you to keep this in mind as you navigate the unpredictable seas of love. When it comes to the effort involved in keeping your relationship afloat, mathematics warns you against complacency. When in doubt, it’s always best to go the extra mile. And if you need help wooing your special someone, you can always call on R. Kelly to help set the mood.

Aziz Ansari does a mighty fine impression of Kells.

How Hard Are Computer Games?

Matt — Tue, 13 Apr 2010 04:01:52 +0000

Every week, on days just like today, millions of Americans are afflicted by the debilitating condition known as The Mondays. Urban Dictionary defines The Mondays as follows:

A day generally created for the purpose of making people wish they were someone else. The day you realize you have 4 days of work ahead of you and that they won’t be going by fast at all. Symptoms generally include feeling like crap, wishing you were dead, or not showing up for work in general.

For sufferers, The Mondays can be quite painful. However, there are remedies that can alleviate some of these symptoms. For one, individuals can spend their day playing games on the internet, in an attempt to push their troubles into the darkest recesses of their minds. Minesweeper is a popular choice, as is Tetris, as any fan of Office Space knows (in fact, I credit this film with teaching me about the terrible condition known as The Mondays).

Or, if your case of The Mondays is so strong that you’ve already exhausted these options, you could try your hand at a newer game, such as Flood-It! In this game, released by LabPixies in 2006, the goal is to make a grid of squares all have the same color, but you start by only being able to change the color of the square in the upper left hand corner. By changing this square’s color to match the squares adjacent to it, however, you can gradually increase the size of your mass in the upper left corner, and, like a spreading virus, try to capture every square in the grid. This explanation is admittedly sparse on details, but as you can see from the video below, the rules are quite simple.

While these are all suitable remedies for a severe case of The Mondays, it’s natural to ask what these games could possible have to do with math. Well, as it turns out, each of these games is related to a problem of such importance, the Clay Mathematics Institute has named it one of the Millennium problems (recall that we briefly discussed the first Millennium problem to be solved). The problem in this instance is the famous P = NP? problem.

As with most things, Wikipedia has a treatment of this problem which you are welcome to read. However, it is a bit technical, so instead I offer you the Clay Mathematics Institute’s introduction to the problem:

Suppose that you are organizing housing accommodations for a group of four hundred university students. Space is limited and only one hundred of the students will receive places in the dormitory. To complicate matters, the Dean has provided you with a list of pairs of incompatible students, and requested that no pair from this list appear in your final choice. This is an example of what computer scientists call an NP-problem, since it is easy to check if a given choice of one hundred students proposed by a coworker is satisfactory (i.e., no pair taken from your coworker’s list also appears on the list from the Dean’s office), however the task of generating such a list from scratch seems to be so hard as to be completely impractical. Indeed, the total number of ways of choosing one hundred students from the four hundred applicants is greater than the number of atoms in the known universe! Thus no future civilization could ever hope to build a supercomputer capable of solving the problem by brute force; that is, by checking every possible combination of 100 students. However, this apparent difficulty may only reflect the lack of ingenuity of your programmer. In fact, one of the outstanding problems in computer science is determining whether questions exist whose answer can be quickly checked, but which require an impossibly long time to solve by any direct procedure. Problems like the one listed above certainly seem to be of this kind, but so far no one has managed to prove that any of them really are so hard as they appear, i.e., that there really is no feasible way to generate an answer with the help of a computer. Stephen Cook and Leonid Levin formulated the P (i.e., easy to find) versus NP (i.e., easy to check) problem independently in 1971.

What do these games have to do with this problem? As it turns out, both Tetris and Minesweeper are NP-complete, while FloodIt! is at least NP-hard (you can learn about these definitions on Wikipedia if you’re curious). Given the above description of the problem, however, it’s natural to ask in what sense these three computer games are hard.

Minesweeper’s NP-completeness was proven by Richard Kaye in 2000 (he discusses some of the ideas involved here). The game is NP-complete in the following sense: suppose you are playing a game of minesweeper, and you want to know whether a certain square has a mine behind it or not. One way you could do this is assume that there is a mine, and see if this leads to a contradiction by checking against the remaining possible configurations of mines. Generating all possible configurations, however, can be an arduous task if we just go case by case. In fact, using this method, most of the time a computer will not be able to solve the puzzle in any reasonable amount of time.

Is the lower right corner mined or not?

In other words, we want to check whether a given configuration of the board is consistent; that is, whether it could have come from a certain arrangement of mines. Given the underlying arrangement of mines for a game, it’s easy to check whether the layout is consistent. But to find the solution to a given board using this idea quickly becomes very complicated.

Tetris, meanwhile, was shown to be NP-complete in 2002 (the paper is available on the arXiv). The Tetris game under study here, however, is a bit different from the one enjoyment by so many. Traditionally the game only ends when the player loses, and the player can only see one block ahead into the string of pieces that will eventually fall. In the version under study by Demain, Hohenberger, and Liben-Nowell, however, the game is played with a predetermined, finite number of blocks, and the player knows the full order in which the blocks will fall. They then discuss a number of NP-complete problems. Let me just highlight one: the problem of maximizing the number of rows cleared.

Given a sequence of plays in a Tetris game of this type, we can easily calculate the number of rows that were cleared. However, finding an algorithm that can optimize the number of cleared rows in a reasonable amount of time (i.e. polynomial time) remains at the heart of the P vs. NP problem.

The related question in FloodIt! is to find a solution to a given board in as few moves as possible. The fact that this problem is NP-hard was only recently proved, which I learned from this link on Slashdot.

So if you have a case of The Mondays, fear not: those games you play on your computer are certainly complex enough to give your brain a workout. And if you spend enough time on them, maybe you can discover algorithms to answer the questions raised here in polynomial time. Is it possible that with enough Tetris, you could play your way to a P=NP solution and a million dollars? Perhaps.

This dude certainly hopes so.

Math Gets Around: Finding a Job and Keeping Your Soul

Matt — Mon, 22 Mar 2010 17:21:00 +0000

Hello friends. My apologies for not writing over the past couple of weeks, but I was away at a conference. Being at math conference has its pluses and minuses (pun intended), but one nice thing about being surrounded by other mathematically inclined individuals is that you never have to explain what it is mathematicians do. You may talk a great deal about your research specifically, but everyone understands what it is to do mathematics.

In general, however, math jobs don’t get much buzz, aside from academic jobs and the oft-mired quants who have received varying degrees of blame for the recent recession. That’s why I’d like to highlight this recent post from the Scientific American blog, which discusses quantitative non-academic job opportunities at start-ups that have nothing to do with finance.

At first glance, it might seem like these companies have nothing to do with one another. Kickstarter aims to connect people with fundraising opportunities for projects they may be interested in, ranging from film, to journalism, to art. OkCupid, arguably the most well known company discussed, is a free online dating sight. Simulmedia data mines information from TV set-top boxes to “address the growing audience fragmentation problem for television networks and simultaneously help viewers discover the shows they might enjoy.” Finally, Snooth is building a database and online community for lovers of fine (and not so fine) wine.

Video games, rap music, and aliens, all in one package? This is just one of the many creative feats on display at Kickstarter.

While the mission statements for these companies are quite different, the problems that each company faces have a common thread: the need for strong and meaningful quantitative analysis. Whether you’re analyzing the effectiveness of a fundraising project, or trying to help people find other people to hook up with, there is undoubtedly a mountain of data running around behind the scenes that is begging for analysis and modeling. Moreover, this disparate collection of companies all beholden to mathematics is not really new – indeed, as our ability to collect reliable data increases, so too will our need to analyze that data to draw accurate conclusions. If anything, this post highlights the growing need for mathematical minds in a variety of fields.

For some, however, simply giving math students a career opportunity outside of finance is the greatest reward. As Columbia mathematician Chris Wiggins points out, “I’ve been watching very talented students go price derivatives and have their souls sucked dry … You can stay on the island [of Manhattan] without inventing weapons of financial mass destruction.” Indeed, whether you have a passion for art, love, television, or booze, this goes to show you that a background in mathematics will serve you well. The same is likely true no matter what your passion, if you can only find a way to bring mathematics into the fold. Be warned, however: if you don’t figure out a way to do it, chances are someone else will.

Finding Love with a Modified Drake’s Equation

Matt — Thu, 25 Feb 2010 20:18:48 +0000

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.

Math Gets Around: Breakfast

Matt — Wed, 16 Dec 2009 17:30:00 +0000

I admire the food blog Serious Eats because, as we’ve seen before, it’s not afraid to get a little mathematical. This month they have upped the ante with a post on the delicious object now known as the Mobius strip bagel.

Named for the classical geometric object of the same name, the Mobius strip bagel (and its cousin, the Mobius strip donut) give an elegant mathematical spin on ordinary edibles. In addition to the aesthetic value, the Mobius strip bagel also has the advantage of added surface area, meaning that one can pile on even more cream cheese before stuffing one’s face.

Mathematician George Hart has step-by-step instructions for the transformation from torus to Mobius strip here. I have yet to try this technique myself, but I can think of no better way to celebrate the holidays than by transforming breakfast food into mathematically themed breakfast food.

Comic (but not Comical) Mathematics

Matt — Wed, 02 Sep 2009 23:00:00 +0000

A few months ago, my girlfriend and I were persuaded to subscribe to the LA Times by a very nice man at a nearby Ralph’s store who offered us $20 in groceries for the exchange. “Just try it out,” he insisted, “because you can always cancel and we’ll simply pro-rate the cost based on how long you were a subscriber.”

Fantastic, we thought. Given the current uncertainty surrounding the future of the newspaper industry, subscribing made us feel like responsible citizens – like giving blood, but with fewer personal questions beforehand.

Unfortunately, once the newspaper began to arrive, we had to face the fact that we never read it. I think I skimmed through it a couple days that first week, but after that the papers would go from our doorstep to the recycling bin. Try as we might, we simply couldn’t fit a morning newspaper routine into our lifestyle. And so, with a heavy heart, we canceled our subscription.

To be honest, I haven’t really noticed a difference. All the news I need I can easily access on the internet. The only part I do sometimes miss is rifling through the comics. Sure, you can always go to a website like comics.com and see what your favorite syndicated cartoon characters are up to, but I still prefer the layout a newspaper provides, with dozens of comics squeezed together, allowing you to read them all rapid-fire without having to wade through a list of comics and be bothered by loading times.

My father remedies this by e-mailing me comics directly. Every day I know that I can look forward to the latest adventures of Liō, but this daily standard is sometimes intermingled with other comics. For example, just the other day my father sent me the following math-themed comic:

This is the work of a man named Dave Coverly, who apparently has been named Cartoonist of the Year by the National Cartoonists Society. More of his work can be found here. Some of his cartoons are quite good. Unfortunately, the one above is not.

The above cartoon has the potential to work on multiple levels, but fails to do so. Certainly anyone can read the comic and get the joke (ha ha, the description on the sign involves math, it’s funny because the books in the bin are math books!), but the sentence on the sign doesn’t make any sense. Forget that the mathematical expression is complicated – replace it by a number, such as 1.8. Then the sign reads “Math Book Sale: Buy 2, Get 1 at 1.8!”

Huh? What is this supposed to mean? If Coverly had written 1.8 instead of (3/4 x 2B) ÷ (7π/100), I would hope that an editor would’ve said something like “Hey man, that doesn’t make any sense, why not try for a joke that isn’t so sloppy?” Instead, we get basically the same punchline, but masked by a more convoluted expression. No doubt the reader is meant to simply see that the sign involves math, chuckle inwardly, and turn the page.

It may sound like I’m just trying to pick a fight, but I don’t think this kind of thing would fly with another subject. For example, suppose the comic had a Spanish theme rather than a Math theme. In such a case, Coverly’s comic might look something like this:

This cartoon, like the last one, works on one level: it’s a Spanish book sale, and the sign describes the sale with some Spanish. It’s not hard to get the joke.

However, anyone with a basic understanding of Spanish would probably find this comic more confusing than funny, because the Spanish phrase doesn’t make sense. The phrase certainly doesn’t explain the sale, but it’s also not particularly funny in it’s own right. Why wouldn’t the sign say “half price” in Spanish? Alternatively, one could write a joke in Spanish – this would be lost on people who don’t speak Spanish, but they would still be able to appreciate the basic joke, namely that the Spanish book sale sign is partially written in Spanish.

The situation with math should be no different than the situation with Spanish, but apparently a comic like this is allowed to slide by. Perhaps our population’s mathematical illiteracy is worse than our Spanish illiteracy. Even so, I think it’s fair to call out this comic for simply being lazy. It’s not hard to get the same joke across, but enhance it so that the sign actually makes sense. Here is just one way Coverly could’ve enriched his original idea:

I feel better already.

Of course, one could argue that Parade Magazine is not the place one should go for a daily dose of mathematical humor. And indeed, you’d be correct. Of the nationally syndicated strips, Foxtrot is the only one that consistently uses math for a punchline, but such examples are few and far between. As always, for math themed comics, one is best advised to search here.

Math Gets Around: Preventing the Zombie Apocalypse

Matt — Wed, 19 Aug 2009 00:30:00 +0000

If pop culture has taught us anything, it is that in the event of a zombie outbreak, we are royally screwed. When faced with an onslaught of classical zombies (of the type first made famous by Romero’s 1968 film Night of the Living Dead), films have shown again and again that we are no match for hordes of cannibalistic undead. With the more recent interpretation of zombies that are faster and smarter, our hopes for survival have diminished even further.

Despite overwhelming odds, however, it is not in our nature to simply roll over in the face of adversity. While the body count is usually high in films chronicling the eventual war between the living and the dead, in most cases there are a few who survive to continue the fight after the credits roll.

But how realistic is this depiction? How prepared are we to defend ourselves from being eaten alive by our deceased ancestors? And what strategies will give us the best chance of survival? You’ll be happy to know that mathematics can answer some of these questions.

Hopefully we are more prepared than Homer Simpson.
Students from the mathematics departments at Carleton University and the University of Ottawa have produced several mathematical models to predict what will happen in the event of a zombie outbreak, and how our response to such an outbreak may affect its outcome. The students, led by Professor Robert J. Smith? (this is not a question, he simply insists on this piece of punctuation after his name) used the theory of differential equations to see what would happen in the event that the dead rise from their graves in search of fresh meat.

As would be expected from a paper with such important pop-culture consequences, this research has already garnered a fair amount of attention. The Globe and Mail ran an article last Friday, as did Wired (and with a much cooler picture, I might add). However, if you really want to get to the heart of the matter, here’s a link to the original paper.

Their model eschews the post-28 Days Later interpretation of zombies, focusing instead on the lumbering, thoughtless monsters that have been the stuff of childhood nightmares for decades. One could argue, however, that the results presented in their work would be even grimmer were we to allow zombies the benefit of intelligence and the ability to run a 6:00 mile.

The underlying ideas are similar to those of the SIR model for the spread of infectious disease, which has been discussed elsewhere on this corner of the internet. One significant difference here, of course, is that when dealing with a zombie infestation, dead people may not necessarily stay dead. This adds complications to the theory, but also allows for a richer analysis.

It’s unclear whether or not the zombies in this model know how to dance.

In the end, their research gives us the following insights into the nature of zombie warfare:

With one exception, human-zombie coexistence is not possible. In order to save humanity from a zombie infestation, we must kill every last one of them.
Often times there is a period of latency from the time a person is bitten until the time they turn into a zombie. Whether or not such latency exists will not have an effect on whether or not the zombie hordes will overwhelm us – the only thing that changes is how long it will take for them to do so.
Quarantining is not an effective way to try and stop a zombie outbreak. All it will do is prolong our extinction.
The only model where coexistence is possible is a model in which a cure exists for zombification. Unfortunately (or fortunately, depending on your preferences), in such a world the zombie population would greatly outnumber the human population. In other words, we would survive, but not by much.
The only way to ensure our survival is to attack with decisive force as frequently as we are able. In particular, the model ignores the impact of birth and death rates, which is fine over a brief period of time, but becomes more significant as the fight continues, since more bodies means more potential zombies. In other words, the longer the fight goes on, the less likely we are to emerge from it.

For the self professed zombie expert, these findings may not be all that surprising. For the rest of us, however, it just goes to show you how far a little mathematics can take you, even when exploring the realms of the highly improbable. You may scoff at this article now, but when World War Z arrives, you’ll be thankful that someone took the time to conduct this preliminary research.

(Hat tip to Patrick for the link to the Globe & Mail article. With a colder climate that no doubt helps to preserve dead bodies, it’s no wonder that Canadians are blazing a trail with this research – they will be on the front lines when the time comes.)

Math Gets Around: On Birthdays and Trading Cards

Matt — Fri, 10 Jul 2009 02:33:00 +0000

Today marks the 1 year anniversary of Math Goes Pop! I started on somewhat of a whim after reading an article about compulsory Algebra I education for all California 8th graders (although what with our finances down the toilet, who knows what the current status is here). When I started writing I wasn’t sure there was enough content out there to sustain a blog with this one’s focus. Once I started digging, though, I found that the rabbit hole went quite deep, and so here I am a year later with plenty left to talk about (the recent obsession with pointless math holidays certainly has helped with my output).

Given the date, it seems fitting to begin by mentioning the birthday problem. This is a standard problem given in any introductory probability course, but many people find the result counter intuitive at first.

The birthday problem asks a simple question: if you have a room full of people, how many people do you need so that there’s a 50% probability that at least two of them share the same birthday? In the simplest setting, we make a few assumptions: for example, we usually ignore the presence of leap years, and we assume that each day of the year is equally likely to be someone’s birthday.

With these assumptions, solving the problem is not difficult. The probability that at least 2 people in a group of n share a birthday is 1 minus the probability that all n people have different birthdays. In other words, the probability is

1 – (1 – 1/365)(1 – 2/365)…(1 – (n-1)/365).

This is because the second person has a 1 – 1/365 = 364/365 chance of having a birthday different from the first person, the third person has a 363/365 chance of having a birthday different from the first two people, and so on.

So, to answer the question, we just need to find the smallest n such that the expression above is greater than 1/2. The punchline is that n only needs to be 23 or more in order for this inequality to hold. In other words, in any room with at least 23 people, there’s a better than 50% chance that two of them have the same birthday.

For many people, this number seems too low – after all, you may have been in several groups of 23 or more and never or rarely met someone with your birthday. The problem with this argument is that it addresses a different question. The reason why we only need 23 people to have a 50% chance of finding a common birthday is that we are placing no restriction on the date of the shared birthday. Requiring that someone share the same birthday as you fixes the date, and this changes the problem.

If instead you want to know how many people you need in a room so that there’s a 50% chance one of them will have the same birthday as you, this probability is given by the new expression

1 – (364/365)ⁿ.

To make this greater than 1/2, we need to take a much larger n: n = 253, to be precise.

As I said, this is a well-known problem in probability theory. A few months ago, however, I was asked about a variant of the birthday problem by my friend Gabe over at Motivated Grammar. Rather than looking for identical birthdays, what if you look for different birthdays? More specifically, how many people do you need in a room to guarantee with, say, a 90% certainty, that every day of the year is someone’s birthday?

Of course, you could always pick poorly so that everyone in the room has the same birthday, but the odds of this happening are quite low. In fact, this problem is more commonly known by another name: the coupon collector’s problem.

For this problem, suppose you are clipping coupons from a newspaper (any newspaper except for USA Today). There are n different coupons you can collect, but each newspaper only has 1 coupon, and you can’t see which coupon the newspaper has until you’ve bought the newspaper. In this context, the question becomes: what’s the probability that you’ll need to buy at least newspapers to collect n coupons?

If you prefer, you can think about this problem in terms of trading cards as well. Each pack of cards is akin to buying a set of newspapers, and you want to know the probability that you’ll need to buy at least a certain number of packs in order to collect all the cards. From baseball players to Pokemon, this same problem governs the distribution (assuming that all cards in the back are equally likely to be in the pack, which may not always be the case).

What’s known about this problem? Well, as I said above, even if you buy hundreds of thousands of cards, or stuff millions of people in a room, there’s no guarantee that you’ll collect every card or every date. However, on average, the number of cards you’ll need to go through to complete a set of size n is about n*logn.

In terms of birthdays, this says that if you want to collect every date, on average you’ll need to pool together around 2,153 people. Why such a large number? It’s not unreasonable to expect something like this – when you first begin collecting people, it won’t be hard to get people with different birthdays. However, as your numbers increase, you’ll get a new birthday less and less frequently. Finding that last birthday could prove to be quite elusive.

The same analysis works for trading cards. Trying to complete your collection of series 2 Teenage Mutant Ninja Turtles Animated Series trading cards? Well, my friend, with 88 cards total and 5 cards per pack, you can expect to buy around 79 packs of cards. Perhaps this box set would be a better investment.

Mondo to the max, indeed.

While the expected values are easy to calculate, it may be that you need to greatly exceed the expected value in order to complete your collection. However, one can use Markov inequalities to get bounds on the probabilities. For example, there’s a 90% chance that you’ll be able to hit every birthday if you take no more than about 21,535 people. To bump those odds up to 95%, take 43,069 people.

So, for parents whose children who have gotta catch ‘em all, you can use these methods to get a rough estimate for how much that completion will cost you. And if you’re trying to get a room full of people together so that every day of the year is someone’s birthday, I’d strongly suggest not picking people at random. What an awkward party that would be.

Math Gets Around in the Big City

Matt — Sat, 13 Jun 2009 02:40:00 +0000

Math has gotten a bit of a visibility boost recently, in the form of posts by professor Steven Strogatz at the New York Times blog. For three weeks, starting at the end of May, Professor Strogatz filled in for usual blogger Olivia Judson, and during that time he used the platform to write some highly readable musings that show the presence of mathematics in unlikely places, and touch on some of the directions math is headed in the 21st century.

Let me highlight the first post, titled “Math and the City.” Professor Strogatz begins this article by describing Zipf’s law, an observation attributed to linguist George Zipf regarding the distribution of words in a language (for a linguistic motivation, you can check the Wikipedia article on Zipf’s law).

One of these things is not like the other.

In the context of cities, the law states the following: in a given country, if you rank the cities within that country by population, the largest city should be about twice as large as the second largest city, about three times as large as the third largest city, and so on. In other words, a city’s population is inversely proportional to its rank.

Using the power of the internet, it’s not too hard to find current population data to try and verify this observation. Here’s a table illustrating Zipf’s law for U.S. cities (I pulled the population data from here):

City	Estimated Population (July 2007)	Zipf Law Ratio	Ranking
New York, NY	8,274,527	1	1
LA, CA	3,834,340	2.158	2
Chicago, IL	2,836,658	2.917	3
Houston, TX	2,208,180	3.747	4
Phoenix, AZ	1,552,259	5.331	5
Philadelphia, PA	1,449,634	5.708	6
San Antonio, TX	1,328,984	6.226	7
San Diego, CA	1,266,731	6.532	8
Dallas, TX	1,240,499	6.670	9
San Jose, CA	939,899	8.804	10

Those with a visual bent can also take a look at a graph of the data:

Professor Strogatz doesn’t provide heuristics for why Zipf’s law should be true, but he does observe that this phenomenon has been around for more than a century, and can be observed in countries throughout the world (with varying degrees of agreement). He then goes on to discuss more recent mathematical observations regarding urban development, including the observation that cities, as with many other things, enjoy economies of scale. For example:

If one city is 10 times as populous as another one, does it need 10 times as many gas stations? No. Bigger cities have more gas stations than smaller ones (of course), but not nearly in direct proportion to their size… the bigger a city is, the fewer gas stations it has per person…

The same pattern holds for other measures of infrastructure. Whether you measure miles of roadway or length of electrical cables, you find that all of these also decrease, per person, as city size increases.

In other words, the distribution of infrastructure is not quite the same as the distribution of the population – as population grows, so too does infrastructure, but it can grow more slowly. Further discussion can be found in the article.

Of course, there are many questions here. First of all, a little digging will show you that this trend is stronger in some countries rather than others. Why is this? Also, why must we break down our analysis by country to look at these trends? Why don’t we see this pattern emerge if we simply rank cities independent of country?

Moreover, this Zipfian trend depends of course on one’s definition of the word “city.” If one extends the notion to municipal areas, the trends become less clear. So, how should we define what it means to be a city?

As I learned from a recent article on The Daily Dish, Tim Gulden of George Mason University has recently tried to answer some of these questions. By using nighttime satellite data, Professor Gulden and his coauthors were able to more rigorously and consistently measure city sizes globally, and were able to use these measurements to compare rankings for population, economic activity, and patented innovations.

In their paper (the abstract of which can be found here), the authors give evidence supporting a Zipf-type distribution for all three statistics, and use this data to argue against the idea that globalization is making the world “flatter,” i.e. more equidistributed with regards to things like population or economic activity. Instead, they argue that the world of the future will feature global ranks that follow more of a Zipf distribution, and that one reason why this hasn’t happened already is because it currently can be difficult to migrate between the barriers of different countries.

For more math in the spotlight, I’d encourage you to read Dr. Strogatz’s other posts (here and here). Happy reading!