A recent article on Wired’s website discusses the mathematics of Lego – more specifically, it highlights an article on the complexity of Lego systems.  As any child will tell you, Lego sets can vary from very simple, small sets, to much larger and more complicated ones.  As a simple corollary, smaller sets will have fewer pieces, and larger sets will have more pieces.  But how does the number of types of pieces grow as the size of the set grows?  For example, if a 100 piece set consists of 10 different types of pieces, is it reasonable to guess that a 1000 piece set will consist of 100 different types of pieces?

In a word, no.  Though the number of different types of pieces will grow as the size of a set grows, it will grow slower than the size of the set (in other words, it will grow sub-linearly).  To put it another way, as the size of the Lego set grows, rather than building more and more new types of pieces, the same types of pieces that are present in smaller sets tend to be used in new ways.  The effect is that the proportion of distinct piece types decreases as the size of the set grows.  From a mathematical standpoint, if we let y denote the number of different types of pieces, and x be the number of pieces, then this power law is giving us the following equation:

for some constants A and b, with b between 0 and 1.  While y grows as x grows, it does not grow as quickly as x itself.

Taken in a broader context, though, this should not be surprising.  Examples of similar phenomena are prevalent throughout nature, as well as in made-made phenomena such as urban planning.  One example cited in the Wired article is Kleiber’s Law, which states that the ratio of an animal’s metabolic rate to its mass tends to decrease as mass increases (in other words, larger animals are capable of metabolizing more efficiently).  Here’s an article that discusses an analogue of this power law in the context of brain development, and relates this to the development of cities.

So the next time you give a Lego set to a child, feel free to explain this connection – I’m sure any child will welcome the math lesson (at least, any child worth giving a Lego set to in the first place).  It’s also worth noting that this phenomenon is most likely not unique to Lego sets – I am eagerly awaiting a similar report on the mathematics of Tinkertoys – though unfortunately, in this case the number of piece types seems not to have increased in nearly a century.

Even XKCD is on the Car Talk bandwagon! (Click the image to go to the source)

Most significant to our present discussion, however, is Car Talk’s weekly diversion known as the Puzzler.  Each week, the brothers read a Puzzler (i.e. a brain teaser) to their listeners and request solutions.  The Puzzler’s solution is revealed the following week, and a new Puzzler is then presented; moreover, one of the correct listener submissions is chosen at random, and the winner receives a gift certificate for some Car Talk schwag.  While these puzzles are sometimes car-related, this is not a prerequisite, and indeed the puzzler I would now like to discuss makes no mention of cars.

Here’s the puzzle, with wording modified slightly from Car Talk’s website: Suppose there are 20,000 lights in a row, all turned off.  One person comes through and pulls the cord on every light, turning each of them on.  A second person then comes through and pulls the cord on every second light. A third person then pulls the cord on every third light, and so on. After the 20,000th person has gone through the room, which lights are turned on?

This problem also goes by the name of the Locker Problem, with open and shut lockers substituting for on and off lights.  But no matter how you contextualize the problem, the solution is the same.  To get some idea of what the answer should be, let’s consider what happens just for the first few (say, 10) light bulbs.  Feel free to think about this problem on your own before continuing on!

After the first person walks through the room, all the light bulbs are on.  So, the first 10 light bulbs will look like this:

After every second cord is pulled, half the lights will be on, and half will be off:

Now here’s where the fun begins.  When every third cord is pulled, off lights will turn on, and on lights will turn off.  This means that the arrangement of the first ten lights will look like this:

And, when every fourth cord is pulled, we get the following picture:

You can probably fill in the rest.  Note that when every 6th, 7th, 8th, 9th, or 10th cord is pulled, only one light in first 10 switches, so most of the work is already done.  In the end, the string of the first 10 lights will look like this:

Here we see that three of the first ten lights remain lit: the 1st, 4th, and 9th.  The astute reader will note that 1, 4, and 9 all share a common property; they are all perfect squares (1 = 1², 4 = 2², 9 = 3²).  The question to be asked, of course, is whether this pattern continues, and if it does, why?

To answer these questions we’ll need to think a bit more deeply about what’s going on.  First, when does a given light bulb in our string of 20,000 get its cord pulled?  Extrapolating from the first few cases, we see that if a person goes through the chain and pulls every dth cord, then the light bulbs with their cords pulled are precisely the ones whose numbers are multiples of d.  Therefore, the nth bulb’s cord is pulled once for every divisor of n.  More concretely, we see for example that the 12th light bulb will have its cord pulled 6 times: by the 1st, 2nd, 3rd, 4th, 6th, and 12th person, since 1, 2, 3, 4, 6, and 12 are precisely the divisors of 12.

What does this divisibility information tell us about the status of a given light bulb?  Since all the lights start in the off position, we see that a light will be off at the end if its cord is pulled an even number of times, and it will be on at the end if its cord is pulled an odd number of times.  In other words, the nth light will be on at the end of this process precisely when n has an odd number of divisors.

All we need to do now is convince ourselves that the numbers with an odd number of divisors are precisely the squares.  There are a couple of ways to do this.

Way 1: Pick your favorite whole number n.  For any divisor d of n, n/d is also a divisor (for example, if n = 30, the fact that 5 is a divisor immediately tells us that 30/5 = 6 is a divisor too).  In this way, we can naturally count up all the divisors of n in pairs.  Because of this, the only way we can have an odd number of divisors is if one of the pairs has the same number repeated twice, i.e. if for some divisor d, d and n/d are equal.  But if they are equal, this means that n = d², in other words, n is a square.

Way 2: By the fundamental theorem of arithmetic, Any positive whole number n > 1 can be written in an essentially unique way as a product of prime numbers, i.e. for any n > 1 we can write

,

where the numbers are primes, and the exponents are greater than zero (for example, ).  In particular, any divisor of n must be composed of the same primes as n itself, so that if d is a divisor, we can write

,

where each exponent is no larger than (for example, 12 is a divisor of 180 but 24 isn’t, since 2 goes into 24 three times but into 180 only twice).

How many divisors are there? Well, could divide d as few as 0 times, or as many as times – we have different ways to choose the exponent on , corresponding to the numbers 0, 1, 2, …, up to .  Similarly, we have different ways to choose the exponent on , and so on for each prime, so that there are different ways to choose the exponent on .  By the fundamental counting principle, this means that the number of divisors of n is equal to

Remember that our light will be on only if the number of divisors is odd, in other words, only if the above product is odd.  Notice that if any term in the above product is even, the product itself will be even – so in order for the product to be odd, each term in the product must be odd.  Since each must be odd, this means that each exponent must be even.  But if each is even, then is always a whole number, so

is a whole number whose square is equal to n.  Hence, once again we conclude n is a square.

There are plenty of variants to this problem worth pondering – for example, what if people only come in and pull every dth cord only for d prime?  Or, what if instead of the 2nd person pulling every 2nd cord, and the third person pulling every 3rd cord, the second person pulls every 2nd cord twice, the third person pulls every 3rd cord three times, and so on?  No doubt you can come up with some variants on your own as well.

If you prefer the locker version, here’s an interactive site where you can watch with satisfaction as lockers open and close.  No matter what model you use, though, this is a cute little problem on integers and their divisibility, and the result can be surprising for first time viewers.  So kudos to Car Talk for discussing this problem on a national stage! (Kudos also to Tom for drawing my attention to this particular episode of the program.)

For our own wedding, my bride and I decided it would be natural to incorporate some mathematics into the table numbers.  There is some freedom in how one decides to do this.  For example, we initially toyed with the idea of using numbers for the tables that were somehow significant to us and our relationship, but found it too difficult to come up with examples meeting this criterion.  If one wants intrinsically interesting numbers, there are many examples among the whole numbers (I was particularly fond of using the smallest whole number expressible as the sum of cubes in two different ways).  In the end, though, we decided to expand the realm of p0ssibilities beyond the range of whole numbers.  This turned out to be a good decision, both aesthetically and educationally.

Table number e. Hat tip to Caroline for the shot.

If you are looking for a way to incorporate some math into your celebration, the table numbers are certainly one option.  At each of our tables we had a small placard, with the number on one side and a brief description of the number (and some table exercises!) on the reverse.  I tried to have sympathy for our audience, and give descriptions that a general audience would be able to understand, though I gave myself more flexibility with a table occupied by other math students.  For sake of completeness, here are all the numbers we used, along with their descriptions (see if you can tell which table had the math students!).  In no particular order:

1. (see here for more).

The ratio of a circle’s circumference to its diameter, is perhaps the most famous irrational number. Historically, has also been known as Archimedes’ constant, and Archimedes himself proved that .

More than one trillion digits of the decimal expansion of have been computed, and folks with nothing better to do than recite those digits come together each day (March 14th, naturally) to see who has memorized the longest string of numbers in the decimal expansion. If you’re looking for more interesting properties of , though, here are a few to mull over:

Table exercises!

1. Use geometry to show that 2\sqrt{2}<\pi<4. These bounds are not as good as those of Archimedes, but they are easier to derive.

2. (Harder!) Explain why \pi is irrational, i.e. why it cannot be written as a fraction p/q where p and q are integers.

2. e (see here for more).

e, a.k.a. Euler’s number, a.k.a. Napier’s Constant, is an irrational number of fundamental importance. While it lacks the general public awareness of a number like , I assure you it is no less charming. Typically defined as the limit

e enjoys many other identities, including

and

e also determines the base of the exponential function , unique among all exponential functions in the study of calculus because it is equal to its own derivative.

Table exercises!

1. Use one of the identities above to verify that e < 3.

2. Use one of the identities above to verify that e is irrational, i.e. that it cannot be written as a ratio p/q where p and q are integers.

3. Suppose each of you has brought a hat to this wedding. Everyone leaves his or her hat inside, and when a person leaves, he can’t be bothered to search for the hat he brought, and simply takes one from the hat pile at random. Show that the probability nobody ends up with the hat they came in with tends to 1/e as the number of people increases.

3. (see here for more).

Take all the perfect cubes (, and so on), take the reciprocals of all those perfect cubes, and add them all together. You will end up with a number that is sometimes called Apéry’s constant, and is written

The constant is named in honor of Roger Apéry, who proved in 1978 that this number is irrational. Intuitively, one can interpret as the probability that three randomly chosen whole numbers will have no prime factors in common.

One can consider more general numbers as well. For example, for any whole number k bigger than 1, the sum

will yield some finite value. When k is even, one has nice formulas for the values, for instance , .

In fact, it is possible to let k take on quite a large range of values. The function one gets is called the Riemann zeta function, and lies at the center of one of the most famous unsolved problems in mathematics.

Table exercises!

1. Show that .

2. Given that , show that .

4. (see here for more).

The black portion of the area pictured above is found by drawing rectangles between two integers n and n + 1 with height 1/n (the rectangle between 1 and 2 has height 1, the rectangle between 2 and 3 has height 1/2, and so on), and subtracting the area under the graph of the function y = 1/x.  The total black area, if this picture were to be extented out to infinity, would represent the number .

can be approximated by its decimal expansion, , and while this number comes up quite naturally in number theory and mathematical statistics, surprisingly little is known about it. For example, it is unknown whether or not is a rational number (unlike constants such as or e, which are known to be irrational).

Table exercises!

1. Using geometry and the figure above, show that .

2. Show that the sum on the right hand side of the inequality in the first exercise equals , so that .

5. (see here for more).

is a concept of central importance in mathematics, and ergo, a concept of central importance in all things. While the figure-eight symbol for infinity is known and loved by all, it was not introduced until the year 1655, though many ancient cultures grappled with the idea of the infinite.

Though may seem like a single idea, great minds have shown that not all infinities are created equal. For example, the mathematician Georg Cantor showed that even though there are infinitely many whole numbers, and there are infinitely many real numbers, there are (in a sense that can be made rigorous) infinitely many more real numbers than counting numbers.

On a related note, the love Matt and Meg feel for you all for standing with them on this day is undoubtedly infinite. How this compares to their love for one another, however, is a problem that has yet to be investigated.

Table exercises!

1. Show that there are infinitely many prime numbers.

2. How does the number of even integers compare to the number of integers? Are there more of one type of number?

3. Suppose a set is finite with N elements. Show that the set of subsets of the original set is finite with elements.

6. (see here for more).

Suppose two line segments have length a and b, with a larger than b. If the ratio of a to b is the same as the ratio of a + b to b, this ratio is called the golden ratio, and is written . In other words,

.

This, in turn, implies that , or (by the quadratic formula)

.

The golden ratio has a rich history, both mathematically and artistically. It is also closely related to the Fibonacci sequence, the sequence of numbers whose first two terms are 0 and 1, and where all subsequent terms are found by adding the previous two terms. In other words, the sequence begins . If we let denote the Fibonacci number (so , and so on), then

Table exercises!

1. Show why the above limit formula for is true.

2. Show that .

3. Show that for any n, .

7. (see here for more).

The de Bruijn-Newman constant, the value of which is currently unknown, is intimately connected to the Riemann Hypothesis. There exists a class of functions , one for each real number t. is essentially the Riemann function, and in particular, the Riemann Hypothesis is true if and only if has only real zeros.

Here are some properties of the family of functions :

1. has only real zeros for any .

2. If has only real zeros, then for any , has only real zeros too.

3. There exists a real value such that has at least one non-real zero.

These properties combine to show the existence of a constant , lying somewhere in the range , such that has only real zeroes if and only if . This is how the de Bruijn-Newman constant is defined. Moreover, the Riemann Hypothesis is equivalent to the statement that .

The current best estimates for state that

so if the Riemann Hypothesis is true, it is, in some sense, “just barely” true. In particular, it’s possible that , in which case you are really just sitting at the 0 table. But while your table may be marked as such, you should know that none of you are zeros in our hearts.

Table exercises!

1. Prove or disprove the Riemann Hypothesis.

8. i (see here for more).

i, more formally known as the square root of -1, is defined to be one of two solutions to the equation (the other solution being -i).

While this might seem like an arbitrary construction, in the larger context of history, it makes perfect sense. Just as the whole numbers are perfectly good for solving basic counting problems, but may be insufficient for problems involving debts or losses (where negative numbers play a prominent role), or problems involving rates or ratios (where fractions take the spotlight), the extension of numbers to include i leads to a wide variety of applications. This include (but are not limited to) applications in electrical engineering, signal processing, and fluid dynamics.

i is also one of the key ingredients in Euler’s identity, one of the most popular formulas in mathematics. This formula states that , and is noted for its unification of five constants of fundamental importance in mathematics: e, , i, 1 and 0.

Table exercises!

1. Show that whenever n is divisible by 4.

2. Find all x satisfying the equation .

3. The set of complex numbers is defined as the set of all a + bi, where a and b are real numbers. 1 + i is a complex number, as is . Can you define an addition law on the set of complex numbers? A multiplication law?

9. (see here for more).

The plastic constant can be viewed as a cousin to the golden ratio (see the table for more information). Formally, is equal to the real root of the equation . The value of is

Just as the golden ratio is intimately related to the Fibonacci sequence, the plastic constant is related to a sequence known as the Padovan sequence. The first three numbers in the Padovan sequence are given by

and the nth term is given by adding two earlier terms in the sequence:

For example, the first few terms in the sequence are given by .

One can similarly construct a sequence known as the Perrin sequence. This sequence is similar to the Padovan sequence, but in this case, the equations needed to get started are . In either case,

Table exercises!

1. Show why the limit formulas given above are true.

2. Show that the first few terms of the Perrin sequence are .

3. Show that if p is a prime number, is divisible by p.

10. (see here for more).

Along with , is probably the most well known number on display here. While it may seem mundane, has an interesting mathematical history, notably because it was one of the first examples of an irrational number (i.e. a number that cannot be expressed as a fraction p/q where p and q are both integers). An early proof of this fact is attributed to the Greek thinker Hippasus, a follower of Pythagoras; legend has it that when he discovered was irrational, the result was so controversial that he was thrown out to sea by his colleagues and drowned.

These days, mathematics is (for the most part) less fraught with peril. The following elegant identities involving have been met with much less controversy:

Table exercises!

1. Prove that is irrational (make sure you are removed from any large bodies of water).

2. Try to prove the identities written above.

3. For which whole numbers m is a rational number?

Enjoy the table exercises!

Of all of these shows, though, the one that most piques my mathematical interest is TLC’s Four Weddings.  Based on a British show with the same name, the premise is as follows: four brides-to-be, unknown to one another, meet and attend each others’ weddings.  When one bride gets married, the other three score various aspects of the wedding, and the bride with the highest score among the four wins a honeymoon (contingencies are in place in the event of a tie, though these are not always explained and seem to vary from season to season).  In order to make for good TV, the show frequently manages to bring out the worst aspects of these women, as they nitpick and pass judgment on everyone else’s wedding.  Here’s a short clip to give you a taste for what this show is all about:

How are the weddings scored?  This process is explained in detail during the course of each episode.  The wedding is broken down into four categories: dress, venue, food, and overall experience.  For overall experience, the other three brides in attendance give a score from 1 to 10 (though I’ve never seen a bride give another wedding a 10).  For the rest of the categories, though, the brides can only rank the weddings as being 1st, 2nd, or 3rd in the given category.  1st place gets 10 points, 2nd place gets 6, and 3rd gets 3.  The total possible number of points a wedding can score is therefore 120.  At the end of each episode, the scores for each bride are broken out in detail.  The wedding budget and headcount are provided as well.

After watching a few episodes, it seemed like the best way to ensure a win was to simply outspend your competitors.  Certainly a large budget can help improve the guest experience or score a hip dress, but I was curious as to what overall trend (if any) could be made between, say, money spent on a wedding and the wedding’s overall score.

With this noble goal in mind, I proceeded to DVR 28 episodes of this show.  After recording the scores for 112 weddings, some of my questions were as follows: is the amount a person spends on a wedding correlated to the score they receive from the other competitors?  What about the amount a person spends per guest?  Finally, did the most expensive wedding win more frequently than would be expected by pure chance?

Let’s look at some data.  Here is a scatterplot of each wedding’s budget, vs. the total points earned.

Click to embiggen!

As the dots suggest, there is a slight positive correlation between the amount one spends on a wedding, and the score one receives from one’s fellow competitors.  The coefficient of correlation here is approximately 0.286, though if we discard the two $150,000 weddings, this improves to around 0.348.

Of course, a $10,000 wedding for 10 people might be a much nicer affair than a $10,000 wedding for 1,000 people.  If you spend more money per guest, does this translate into a higher score as well?  The dots don’t lie; here’s another scatterplot:

Click to embiggen!

There is quite clearly an outlier in this set of data – this corresponds to a bride who spent $150,000 for a 120 person wedding, for a whopping $1,250 spent per guest (this particular bride did end up taking first place).  Eliminating this outlier, though, the correlation here is weaker than the correlation for actual cost, at a meager 0.098.  In other words, there may not be a linear relationship between cost per guest and total score.  This may be because certain fixed costs, such as the dress and the venue, won’t necessarily vary much with the guest total, unlike something like food.

This analysis, though, obscures a key point.  In order to win the honeymoon, you don’t necessarily need a high score; you only need a higher score than your three other competitors.  With this in mind, it may be simpler to just look at the scores episode-by-episode, and see how the amount spent on a wedding compares to the wedding’s relative ranking.

In the graph below, the blue bars count the number of times the most expensive wedding was given a certain rank.  The red bars count the number of times the wedding with the highest cost per guest was given a certain rank.  Note that if cost had no bearing on the rank, we would expect an equal number of weddings in each rank (in this case, about 7 per rank).

Click to embiggen!

As you can see, spending the most money seems to bestow an advantage: fully 50% (14 out of 28) of the most expensive weddings were ranked first.  This would be unlikely if cost had no impact on ranking.

The picture is a little murkier for cost per guest -the frequency for each ranking sticks pretty closely to 7, so it’s not clear that spending more per guest gives any advantage.

In conclusion, there is a small positive correlation between amount spent and score received, though this does not transfer to the amount spent per guest.  Compared only to one’s other three competitors, the amount spent appears to confer an even greater advantage, so if you are on this show and want to show your competitors that you are better than them, my advice would be to simply outspend them.  If all you’re interested in is a nice vacation, though, it may be cheaper to just stick to your budget, and plan a vacation on your own.

While combining the arts with the sciences is nothing new, it’s cool to see a program embrace the intersection of these disciplines with such gusto.  Of course, it can be difficult to squeeze educational content out of a song with a science focus, but if School House Rock has taught me anything, it is that education and fly jams need not be mutually exclusive.  If you feel, however, that NIMBioS’s song on sexual selection doesn’t quite make the cut on the education front, here are some other math and science songs to ease you through your hump day.

Here is “The First and Second Law” by Flanders and Swann (this one is highly recommended), a song about thermodynamics:

The University of Tennessee isn’t the only school to join songwriting and science.  Here’s an evolutionary jam courtesy of the University of Chicago:

For the mathematically inclined, there isn’t much that can beat “Finite Simple Group (of Order Two)”:

Though, if those jokes don’t make much sense, one can always listen to muppets singing about math instead. Who knew there was math in tube socks?

(Hat tip to Meg for the NPR link!)

The rules for the patterns are fairly simple, but to make things crystal clear, let me label the dots in the grid as follows:

Here are the rules constraining the types of patterns you can make:

1. The pattern must connect at least 4 dots.
2. No dot may be used more than once.
3. The order in which the dots are connected matters.
4. Two dots which are on opposite sides of the grid (e.g. 1 and 9, 2 and 8, 1 and 3) cannot be connected together directly without going through the dot between them, unless the dot between them has already been used.

To give some examples, the first rule prohibits patterns that are too small, such as 1-2-3.  The second one prohibits patterns like 1-2-3-5-7-4-1, although a pattern like 1-5-2-3-7 is allowed by the fourth rule – in other words, we can jump from 3 to 7 since the 5 has already been used.

Images of the paths described above.

The fourth rule tells us, for example, that 1-2-3-6 and 6-3-2-1 are DIFFERENT passwords, even though the shape they make is the same.  Thus, the password is determined not only by shape, but also by direction.

Same paths, different directions.

I should also point out that it is possible to connect dots that are not immediately adjacent – for example, one can draw a line from 2 to 7.  So, for example, if one starts at 1, it is possible to move to every dot except for 3, 7, and 9.

Given these rules, my friend is interested in determining how many unique passwords one can make.  For comparison’s sake, we can think about the iPhone, which can be secured with a 4-digit passcode.  Since each digit can run from 0 to 9, we see that there are 10,000 = 10⁴ possible iPhone passwords.  How does the number of patterns compare, say, to this number?

Without too much work, we can provide an upper bound for the number of passwords one can make using these pattern rules.  To see how this is done, let’s forget about the pretty pictures, and focus only on the sequence of digits (for example, the paths above can be viewed as the sequences 1236 and 6321).  From this perspective, the second rule tells us that no digit can appear in our sequence more than once, and so in combination with the first rule this tells us that the sequence must be between 4 and 9 digits long.  Suppose for a moment that these we ignore the fourth rule.  Then the number of allowable patterns would be the same as the number of ordered strings of digits of length 4 to 9, where in each string the same digit can appear no more than once.

The fact that the order is important tells us we are dealing with permutations.  These can be counted easily.  For example, the number of ways to make a 4 digit password if no digit from 1 to 9 can appear more than once is 9*8*7*6 = 9!/5!.  Similarly, the number of ways to make a 5 digit password if no digit can appear more than once is 9!/4!.  Adding all these up (for lengths from 4 to 9), we find that under these conditions, the number of passwords equals

9!/5! + 9!/4! + 9!/3! + 9!/2! + 9!/1! + 9!/0! = 985,824.

That’s a nearly 100 fold increase in the number of possible passwords!  Of course, this doesn’t reflect the number of patterns one can make, because we have ignored some of the rules.  This means, for example, that we have counted the string of digits 1397, even though this doesn’t represent an allowable pattern.

How can we modify this argument to eliminate the digit strings that don’t correspond to patterns?  This requires a more careful analysis.  For simplicity, let’s suppose that 3 digit passwords are acceptable, and focus only on them.

In this analysis, there are three types of dots that matter: the dots in the corners, the dots on the sides, and the dot in the middle.  Let’s label these classes of points by C, S, and M.  The first dot in your pattern can be any one of these three types, but the number of options you have depends on the type; in other words, you have 4 choices for a corner dot, 1 choice for a middle dot, and 4 choices for a side dot.

Moreover, the number of options for your second choice depends on what type of dot you start with.  If you choose a corner dot, you have 5 options for your next dot (either the middle, or one of the four sides).  Meanwhile, if you choose the middle dot, the next dot in your pattern can be any of the remaining 8 dots.  Similarly, if you start at a dot on the side, you have 7 options for the next dot, since you can move to any dot except for the one directly opposite your starting dot.

Similarly, the number of options for the third dot depends on your choice for the second dot.  If you choose the corner dot and then the middle dot, you will have 7 choices for the third dot.  Meanwhile, if you choose the center dot and then a side dot, you’ll have 6 choices for the third dot.  And so on – the number of options is summarized in the following chart.

Number of options for a 3 dot pattern (notation explanation below)

In particular, note that if you start at a side dot, the number of patterns you can create depends on whether your choice for the second dot is the middle dot, a corner adjacent to your starting dot (or a “close corner,” labeled cC), a corner not adjacent to the starting dot (or a “far corner,” labeled fC), or another side.  Moving from a side to a close corner gives 5 choices for the third dot, while moving from a side to a far corner gives only 4 choices for the third dot.

There are 5 viable candidates when moving from a side to a close corner, but only 4 when moving from a side to a far corner.

The point is that even in this (relatively) simple example, the number of options for any given point in the pattern is highly dependent upon the choices made up to that point.  In the case of 3 dots, the chart above tells us that the total number of allowable patterns is 4 x 1 x 7 + 4 x 4 x 6 + 1 x 4 x 5 + 1 x 4 x 7 + 4 x 1 x 7 + 4 x 2 x 5 + 4 x 2 x 4 + 4 x 2 x 6 = 320.  This is in contrast to the upper bound of possible permutations of 3 of the 9 digits, which is given by 9 x 8 x 7 = 504.

As you can see, just counting the number of patterns connecting three dots is a little annoying.  Doing the same for patterns of lengths 4 through 9 is even worse.  Although conceptually it’s not difficult to count patterns, technically the task quickly becomes arduous.  Thankfully, we can easily delegate technical calculations to those lovable sidekicks, computers.

After realizing that the calculations would be no fun to do by hand, I went online and discovered that this question has already been answered here.  If you click on the link, you’ll find some savvy folk who wrote programs to count all possible paths.  Final answer: 766,752 (incidentally, this is 7/9ths of the upper bound we calculated above).

So, we still have an over 75 fold increase in the number of possible passwords when compared to a device like the iPhone.  Whether or not this actually makes for a more secure device is a question I can’t answer.  I will point out that many of these phones lock you out for 30 seconds after five failed password attempts – if we assume that a crude hacker could make 5 password attempts per minute, then 766,752 different password options means that it could take over 100 days of continuous guessing before your password is cracked.  Of course, an actual hacker undoubtedly has more sophisticated methods.  Even so, if more celebrities locked their phones with pretty pictures, perhaps hackers would have a harder time stealing all of their naked photographs.

(Big ups to Matt for suggesting this problem!)

Apparently I’m not the only one who thought this a worthwhile problem.  This problem has also made an appearance at the fun-filled blog Mind Your Decisions (it’s an excellent discussion, so if this kind of thing suits you, check it out).  In the comments section, discussion on this problem continues; in particular, one person proposed that the model should be modified to include the possibility of guessing.  This is an entirely reasonable thing to want, and thankfully it can be incorporated into the model without too much added effort.

Let me recall the notations I used when I discussed the problem earlier.  I’ve already mentioned N, M, and K.  Let’s let n represent the number of questions you can answer, and of those n, let X represent the number that actually appear on the exam.  As we’ve seen, X satisfies a hypergeometric distribution, and the probability that X is some value k is given by

Moreover, since you will pass the exam only if X is at least K (in other words, only if the number of questions on the exam that you can answer is at least the minimum number of correct answers need to pass), the probability of passing is

This is all review from the earlier posts, and does not take into account the effect of guessing.  Let us now imagine how we can include this refinement into the model.

Intuitively, if we are allowed to guess, then the probability of our being able to pass should increase.  To make things as simple as possible, let’s assume that if you don’t know the answer to a question, you have a probability p of guessing correctly – in other words, the probability of a correct answer is the same for each question.  Let’s also assume that the probability of guessing correctly is independent of the number of questions on the exam that you can answer without guessing (we’ll use this assumption in a moment).

In this modified situation, you will win if you can answer enough questions, or if you can guess enough correct answers in the event that you can’t answer the minimum number of questions with absolute certainty.  So, if X is at least K, nothing changes – you’ll simply answer the minimum number of questions and call it a day.  What’s new is the situation when X < K, in which case you’ll need to guess in order to try and pass.

Roughly speaking, then, the probability of passing is equal to P(X > K) + P(X < K and you guess correctly enough times).  How many times is “enough”?  Well, X plus the number of correct guesses must be at least K.  Considering the cases X = 0, 1, …, K – 1 separately, we can rewrite this as

Since the value of X is independent of the number of correct guesses you’ll make, this can be written as

Moreover, since we want the number of correct guesses to be between K – i and M – i, we can write the above as

Now, the probability occurring in the sum over j should hopefully look familiar to anyone with a basic background in probability.  The number of successes out of a fixed number of trials given that the probability of success is some number p follows the Binomial distribution, one of the first probability distributions encountered in any course in probability or statistics.  In particular, since we’ve said that the probability of a correct guess is p, knowledge of the binomial distribution tells us that

In summary, the probability of passing is

If we want a more explicit formula, we can also use our knowledge of the probability distribution for X.  Also, notice that P(X = i) is 0 if n < i (the number of questions on the exam for which you know the answer can’t exceed the number of questions in total for which you know the answer), so we can write the probability of success in two ways, depending on whether n < K or n > K.

If n < K, then the probability of passing becomes

In the case that n > K, we have a contribution from the P(X > K) term, and so the total probability is

This is all well and good (and agrees with what commenter Scott derived in the comments of the Mind Your Decisions post), but what does it say about our example from last time (where N = 100, M = 10 and K = 6)?  As before, here are some graphs of the probability of success as a function of how many questions you can answer.  Note that any such graph depends on the probability p.  So, let’s illustrate two examples:

Case 1: p is a fixed value.  Here are the graphs corresponding to p = 0, p = .25, p = .375, p = .5, and p = .75 (i.e. the chances of you guessing correctly are either 25%, 37.5% 50%, or 75% – note that the case of p = 0 corresponds to the previous case where guessing isn’t a factor).

Some highlights – without guessing, you need to know the answers to 55 questions in order to have at least 50% chance of passing.  With a 25% chance of guessing correctly, you only need to know the answers to 40 questions.  At 37.5%, the number of questions decreases to 28, at 50% it drops to 10 questions, and if you have a 75% chance of answering correctly, you have over a 90% chance of passing without knowing any answers at all!  If you want at least an 80% chance of passing, the number of answers becomes 67 (p = 0), 56 (p = .25), 48 (p = .375), and 35 (p = .5).

Case 2: p increases with n.  It seems reasonable to assume that the more answers you know, the better your chances of correctly guessing the answer to a question you don’t know, since you will be more knowledgeable in general.  In this particular example, I’ve taken p to equal n/N (in this case n/100).  Note that with this choice, initially the probability of success will be 0, but as n grows the probability of success should grow relatively rapidly.

The above graph quantifies the above heuristics.  Note that the red line grows very rapidly, so that the probability of success is greater than 50% after memorizing 33 questions, more than 80% after 43 questions, and over 95% after slightly over half of the questions (53).

So there you have it.  If you are feeling lazy the next time you have to prepare for an exam, hopefully this will provide you some guidance as to the minimum amount of work you can do while still being reasonably confident that you won’t fail.

Hope you all have a great weekend!

]]> http://www.mathgoespop.com/2011/01/test-taking-part-3.html/feed 0 http://www.mathgoespop.com/2010/12/humanities2.html http://www.mathgoespop.com/2010/12/humanities2.html#comments Fri, 17 Dec 2010 16:58:16 +0000 Matt http://www.mathgoespop.com/?p=981 Last week we discussed an example of when a mathematical background might prove useful even in the least quantitative of liberal arts courses.  More specifically, we asked the question: if a teacher gives you a list of N questions, tells you that M will be on an exam, and you must answer K of the questions given . . . → Read More: Addendum to Math Gets Around: The Humanities]]> Last week we discussed an example of when a mathematical background might prove useful even in the least quantitative of liberal arts courses.  More specifically, we asked the question: if a teacher gives you a list of N questions, tells you that M will be on an exam, and you must answer K of the questions given on the exam, what’s the minimum number of questions you should prepare to guarantee that you will be able to answer K of the questions on the exam?  (Answer: N + K – M.) We also looked at the question probabilistically – namely, we saw that of the questions appearing on the exam, the number that you’ve prepared for follows a hypergeometric distribution.

As a concrete example I considered the case N = 6, M = 5, K = 3 – in this case, the minimum number of questions you should prepare to guarantee that you can answer 3 of 5 problems on the exam is 4, and we saw that if you only prepare 3 questions, you have a 50% chance of those 3 questions appearing on the list of 5.

Late last week, however, I was made aware of another example, one for which the probabilities might prove more interesting (since there are more cases to consider).  Specifically, let us consider the case of a person studying to become a U.S. citizen.  As part of this process, one must submit to an interview in which one is asked 10 questions, and must answer 6 of those 10 questions correctly.  However, the potential list of questions is made available to people beforehand; there are 100 questions from which the 10 questions can be drawn.  In other words, we have N = 100, M = 10, and K = 6.

In this case, to guarantee that you will be able to answer 6 of the 10 questions presented, our analysis from last time tells you that you should prepare 100 + 6 – 10 = 96 of the questions.  Indeed, this makes sense, since the worst that can happen is that the 4 questions you don’t prepare happen to be precisely 4 of the 10 questions you are asked in the interview.  This also reflects the fact that the closer M is to K, the more questions the test taker will have to prepare (note that if M were closer to N, say M = 90, the test taker would only have to prepare 16 questions).

Still, preparing 96 of the questions may seem like a little much, especially since only 10 questions will come up in the interview.  So, let’s see what happens if someone prepares for fewer than 96 questions.  Obviously one should know how to answer at least 6 of the questions, but what about values between 6 and 96?

Here is a graph showing the probability that one will pass the interview given that one has learned the answer to n questions, for some n between 6 and 96.This graph tells you that, for example, even if one only had time to learn the answers to 73 out of the 100 questions, one’s chances of passing the exam would still be over 90%.  Those are pretty good odds, for only learning the answers to roughly three quarters of the questions.  On the other hand, one needs to learn the answers to 37 questions before one’s odds of passing rise above 10%, so it’s certainly not likely that someone will pass by learning the answers to only a handful of questions (which is probably what the government intends).

If only Apu had known of these findings, perhaps he could have saved himself some trouble.

Anyway, I just wanted to highlight another example where these ideas apply.  If you can think of any others, let me know!  Also, if you are interested in the content of the 100 questions that can be asked of our future citizens, you can find the full list (along with acceptable answers) here.