Math Goes Pop! » recreational math

RIP Martin Gardner

Matt — Tue, 08 Jun 2010 02:58:53 +0000

Not long ago, I wrote an article in commemoration of Martin Gardner’s 95th birthday. Sadly, it seems this will be my last article in celebration of his birth, as he passed away late last month.

Through his passing, though, his influence has become even more apparent. Perhaps because he published mathematical games in Scientific American for 25 years, the magazine has been the most visible in its veneration of him. There are no less than six articles on Gardner at the SciAm website; while some are reprints of earlier articles, there is also new material from writers and mathematicians who were influenced in some way by Gardner’s unique career. Since I can’t do justice to Gardner the way others already have, let me summarize what you can find if you’re interested in learning more about this stand-up fellow.

If you’d like to learn more about Gardner’s life, SciAm has reprinted to earlier essays on the man: one is a profile written by Philip Yam, originally published in 1995, and one is a tribute to Gardner’s influence written by Douglas Hofstadter, author of Gödel, Escher, Bach: an Eternal Golden Braid. For a look into Gardner’s interest in debunking pseudoscience, there is this article written by Michael Shermer, which was originally published in 2002. Or, if you’d just like to try your hand at some recreational math puzzles, there are a couple of articles that give a sample of the material Gardner published in his column. Here’s one for you to stew over, if you like (taken from the first link above):

Arrange four paper matches on a table as shown at right. They represent a martini glass. A match head goes inside to indicate the onion of a Gibson cocktail. The puzzle is to move just two matches so that the glass is re-formed, but the onion—which must stay where it is—winds up outside the glass. At the finish, the glass may be turned to the left or the right, or even be upside down, but it must be exactly the same shape as before.

Finally, there’s also this collection of musings from a few scientists and mathematicians who were influenced in their careers by Gardner’s work. Hofstadter is quoted here, but for my money, a better quote can be found in his earlier article which I cited above:

There should be, it seems to me, be a prestigious national or international prize for writing about scientific ideas. As everybody knows, human civilization relies on science and technology more than at any time in the past, and that reliance can only increase. Yet the worldwide ignorance of and disdain for science, mathematics and precise thinking in general is appalling. Because of this tragic situation, people like Martin are precious purveyors of precious knowledge … if I dare say so, what Martin Gardner has done is of far greater originality than work that has won many people Nobel Prizes. Simultaneously achieving both depth and breadth is almost unheard of in today’s scientific world, but Martin Gardner is an exception, and it is a delight and a privilege to celebrate here his many achievements. Just as Martin’s writings have inspired me for decades, so they will undoubtedly continue to inspire other people for many decades to come.

RIP, Mr. Gardner.

A Mathematical New Years Game

Matt — Mon, 11 Jan 2010 06:05:00 +0000

First, let me begin by wishing a happy 2010 to you all. If you celebrate the holidays the way I do, then the past few weeks have seen you spending time with friends and family. And if you really celebrate the holidays the way I do, then some of that time with friends and family will have been spent with mathematical puzzles.

Very recently I was with a group of friends, discussing all that would come to pass in this new year. One friend, whose anonymity I will preserve by referring to him only as “Smith,” was in the enviable position of being the only one among us whose age divided the current year (I won’t embarrass him by revealing his age, but given that it’s a divisor of 2010, this certainly restricts the possibilities). Once we realized this, it became natural to ask how common an occurrence this should be. In other words, how often can you expect your age to divide the current year? Of course, implicit in this is a choice of calendar – for our purposes, we will stick to commonly used Gregorian calendar, although the results would be equally valid under a different choice (e.g. the Hebrew calendar or Islamic calendar). For example, if you were 1, 7, 41, or 49 last year, your age divided the year (of 2009). Next year, only the one year olds will win out, since 2011 is a prime number.

Depending on the year you were born, you may find that this happens quite frequently, or not very frequently at all. For example, if you were born in the year 0, you’re in luck, because your age will divide the current year for at least part of the year for every subsequent year. The phrase “part of the year” is important, because in a given year you will be two different ages – the age before your birthday, and the age on and after your birthday. Of course, this isn’t an issue if you were born on January 1st or December 31st, but we will ignore this (simpler) case.

Let’s take a more detailed example. Suppose you were born in 1982. In 1983, after your first birthday, your age will divide the year (since 1 divides everything). Similarly, in 1984, your age will divide the year after your 2nd birthday, since 1984 is even. And in 1986 your age will divide the year until your 4th birthday, since 1986 ÷ 3 = 662. Unfortunately, you will be too young to appreciate this arithmetic coincidence at any of these opportunities, and unless you live to be 661, you’ll never again be able to say that your age divides the year.

However, if you were born just a few years earlier, in 1979, you’ll find that your age divides the year quite frequently. In fact, by the year 2000, the only years in which your age wouldn’t have divided the year at all would have been 1987, 1988, 1993, 1994, 1996, 1997, and 1999.

Why is it that some years allow for one’s age to be divisible by the year quite frequently, while other years do not? The answer is quite simple. Suppose we let b denote the birth year, and we let a denote a person’s age. That person will be a years of age from their birthday in year b + a until their birthday in year b + a + 1. Therefore, your age will divide the year from your birthday until the end of the year if a divides b + a, or from the first of the year until your birthday if divides b + a + 1. So, the question becomes: when does a divide b + a, and when does it divide b + a + 1?

In the first case, since a always divides a, we know that a divides b + a if and only if a divides b. By the second same argument, we see that a divides b + a + 1 if and only if a divides b + 1. In other words, we conclude the following:

Your age will divide the current year if, and only if, either (i) it is between January 1st and your birthday, and your age divides the year after you were born, or (ii) it is between your birthday and December 31st, and your age divides the year you were born. To put it more simply, your age will divide the year for at least part of the time you are at that age if and only if that age divides the year of your birth or the year after your birth.

With this knowledge, it’s easy to see why people born in 1979 will have their age divide the current year more frequently than people born in 1982. In the former case, determining the set of ages which will divide the current year is equivalent to finding the divisors of 1979 and 1980. 1979 is a prime number, so it will never be the case that your age will divide the year between your birthday and December 31st (except after your 1st birthday); on the other hand, 1980 has a prime factorization of 2 x 2 x 3 x 3 x 5 x 11, which gives it a large number of small factors, and consequently a large number of solutions to the problem.

By contrast, if you were born in 1982, you won’t get many factors either way: 1982 factors as 2 x 991, and 1983 factors as 3 x 661. This is why, if you are born in 1982, your age won’t divide the current year after you’re 3.

While it’s not often that mathematics comes up when I’m with my friends at home, I certainly relish every opportunity. I hope that this may serve as an example to all of you who would like to make mathematics more of a part of your everyday life, especially in social circles into which math rarely intrudes. Single guys looking for first date conversation material are especially urged to keep this sentiment in mind.

Martin Gardner and the Three Way Duel

Matt — Thu, 29 Oct 2009 03:38:00 +0000

As you may have heard, last week Martin Gardner celebrated his 95th birthday. Gardner, who authored the “Mathematical Games” column in Scientific American for a quarter of a century, is often credited for introducing generations of young students to the beauty and charm inherent in mathematics. My favorite quote in this vein comes from professor Ron Graham, who is quoted in a recent New York Times article on Gardner as saying that “Martin has turned thousands of children into mathematicians, and thousands of mathematicians into children.”

A warm brain is the key to mathematical dexterity.

Both Scientific American and Wired ran articles on Gardner last week, and each one used a different expression to represent his age. Scientific American congratulated him on reaching an age of 2⁵ x 3 – 1, while Wired proclaimed that Gardner had turned 5! – 2⁵. Upon reflection I think I prefer the latter expression over the former, since the exclamation point in the factorial makes his birthday feel just a little more exciting. No matter your preference, however, I hope this encourages others to write their ages using mathematical expressions – more complicated expressions would be especially useful for ladies who don’t wish to reveal their age.

In any event, I’d like to take the opportunity to celebrate Martin Gardner’s birthday in my own way, by discussing a problem credited to him. There are a few variations on this problem, but all fall under the general name of the Truel. I first encountered this problem during a job interview, and the variant I heard then is the variant I’ll discuss now.

Suppose there are three men: Mr. White, Mr. Gray, and Mr. Black. These three men have a score to settle, and so they decide to engage in a three way duel (or a “Truel,” if you want to be cute). Mr. White is a rather poor marksman who hits his target only 1/3 of the time, In contrast, Mr. Gray is successful 2/3 of the time, and Mr. Black is an expert marksman who never misses. Because of this disparity, they agree that Mr. White shall shoot first, followed by Mr. Gray, followed by Mr. Black, and this sequential rotation shall continue until only one man remains standing.

Setup of the truel.

Although Mr. White isn’t so good with a gun, he is an expert strategist, and surmises that Mr. Black will dispose of Mr. Gray first, since Mr. Gray poses more of a threat than Mr. White. By the same reasoning, Mr. White assumes that Mr. Gray will shoot at Mr. Black before shooting at Mr. White.

Assuming that this analysis holds true, what should Mr. White do on his first turn to maximize the chance that he’ll win?

I’ll discuss the answer below, but you should try this out for yourself if you’ve never seen this problem before. So that you don’t peek at the answer accidentally, let’s pause for a moment and look at some drawings courtesy of M.C. Escher.

Relativity

Drawing Hands

Circle Limit III

So, what should Mr. White do? Should he shoot first at Mr. Gray or at Mr. Black?

Phrased this way, it’s a bit of a trick question, since the answer is that Mr. White shouldn’t shoot at either one. Instead, he should intentionally miss!

Intuitively, this seems reasonable if you reflect on it. The reason is that it’s too risky for Mr. White to shoot at either of his opponents in the first round, because if he actually hits one of them, the remaining opponent will immediately turn and shoot at Mr. White. For example, if Mr. White shoots and hits Mr. Grey, he’s done for, since Mr. Black will shoot at Mr. White and is guaranteed to hit. Similarly, if Mr. White hits Mr. Black, Mr. Grey will turn to shoot Mr. White. If, however, Mr. White shoots neither one, then either Mr. Black or Mr. Grey will be eliminated by the other one, leaving Mr. White with only one opponent to worry about.

For those looking for a more quantitative argument, we can use the odds given in the problem to calculate the probability that Mr. White will win if he a) shoots first at Mr. Black, b) shoots first at Mr. Gray, or c) shoots first at neither one.

As part of the analysis, we’ll need to determine the odds that Mr. White will win when he’s paired one-on-one against either one of his adversaries. For Mr. Black this is easy: if Mr. White faces off against Mr. Black, he has a 1/3 chance of winning, since if he misses Mr. Black will win with certainty.

In a face-off between Mr. White and Mr. Gray, however, things are more complicated. This is because there are many ways for Mr. White to win. He could win by hitting Mr. Gray on his first shot, he could win if both men miss and he hits Mr. Gray on his second shot, he could win if both men miss twice and he hits Mr. Gray on his third shot, and so on. In general, since the probability that both men will miss in a given round is 2/3 x 1/3 = 2/9, the probability that Mr. White wins in the first round is 1/3, the probability that he wins in the second round is (2/9) x 1/3, the probability that he wins in the third round is (2/9)² x 1/3, and so on.

In general, we find that the probability Mr. White wins when facing Mr. Gray is 1/3 x (1 + 2/9 + (2/9)² + (2/9)³ + … ). The value of the geometric series is 9/7, so the probability is just 1/3 x 9/7 = 3/7. That is, when facing off against Mr. Gray directly, Mr. White has a 3/7 chance of winning (assuming Mr. White shoots first).

How are these calculations relevant to the matter at hand? Well, we can analyze the outcome of the truel depending on Mr. White’s initial action by using some tree diagrams. Let’s first see what happens in the simplest case when Mr. White intentionally misses. In this case, Mr. Gray will shoot, and either he will hit Mr. Black or he will miss:

Here, GHB and GMB stand for “Gray hits Black” and “Gray Misses Black,” respectively.

We know that if Gray hits Black, we will now be in a face-off between Gray and White. Similarly, if Gray misses Black, Black will in turn kill Gray, putting us in a face-off between Black and White. Since we calculated the probability of White winning in either one of these face-offs, and since Gray has a 2/3 chance of hitting Black, this diagram shows that the probability White will win must be 2/3 x 3/7 + 1/3 x 1/3 = 25/63, which is roughly 39.7%.

What if White shoots at Gray instead of shooting at Black? In that case, the picture gets a bit more complicated:
In this case, if White hits Gray the truel will end, since Black will hit White with certainty. Therefore, the only way White can win is if White misses Gray – but in this case, the situation is just as it was before, when white shot to miss. Since White has a 2/3 chance of missing, and from above we know that White has a 25/63 chance of winning if he misses, this shows that the probability of white winning is 2/3 x 25/63 = 50/189 in this case, or roughly 26.5%.

Finally, if White shoots at Black, we get an even more complicated tree diagram:

Notice that the probabilities in the bottom half of this diagram are the same as in the earlier diagram, and therefore it suffices to consider what happens in the top half. Here we see that White can only win if he hits Black, if Gray misses White, and then White wins the face-off against Gray. Following the probabilities on the tree, we see that the odds of this happening are 1/3 x 1/3 x 3/7 = 1/21. Therefore the total probability that White will win if he shoots at black is 50/189 + 1/21 = 59/189, which is roughly 31.2%.

From this analysis, we see that White’s odds are significantly increased by shooting to miss: 39.7%, versus 31.2% if White shoots at Black and a paltry 26.5% if White shoots at Gray.

Another interesting feature of this problem is that by shooting to miss, not only does White maximize his chances of winning, but he also becomes the most likely person to win. Using similar arguments, one can show when White shoots to miss in the first round, the probability that Gray will win drops to 38.1%, and the probability that Black will win goes down to just 22.2%. In other words, using this strategy not only makes White the most likely to win, it makes Black the least likely to win, even though Black is the strongest marksman!

As discussed in the Wikipedia entry on truels, there are other variants one can consider as well. One could change the order of the shooters, for example, or toy with the probabilities involved. One could also consider more realistic situations in which the players have only a finite number of bullets, or think about what happens if the players fire simultaneously rather than sequentially. These variants will lead to generalizations of the phenomena observed here.

As with much of mathematics, this recreational problem shows that once you start digging, there’s practically no limit to the questions you can ask and answer with a bit of mathematics. Here’s to Martin Gardner, for providing us with such a delicious taste of the bounty mathematics has to offer.