Josephus and forty of his fellow soldiers retreated from the siege of Yodfat to a small cave in the hills. Surrounded by the Roman legion with no chance of escape, the soldiers saw no other choice but to commit suicide. Like the people of Masada, they would rather die than face capture.

In order to decide who would die in which order, the soldiers stood in a circle and, starting with the top of the circle and continuing clockwise, counted to three. The third man got the axe and the counting resumed at one. The process continued until there was no one left. Josephus, who didn’t quite agree with the whole “we should all kill ourselves” idea, figured out the perfect way to avoid death: be the last man standing.

Of course, the above illustration has only twelve men, and Josephus somehow managed to figure out very quickly where to stand with forty others.

It was posed as a programming riddle, but I think it’s a very cool mathematical riddle too.

As a programmer, the simplest way to solve it is to write a function that basically goes through the whole process, killing off the virtual “soldiers” until you’re left with the sweet spot. With the speed of today’s computers, that makes perfect sense.

But I was wondering how Josephus solved the problem himself. The programming approach would be akin to him drawing all the forty soldiers on a sheet of paper, and then crossing them off one by one until he figured out which one would be left standing last. Definitely doable, but it might take a while.

So my question to you is… can you think of any mathematical shortcuts to quickly figure out where the safe spot is?

(You have two variables. The number of soldiers, and how many soldiers to skip in each step. Have fun!)

The question is simple: What does…

… tend to as n tends to infinity?

Here are two arguments:

1. The inside of the bracket goes to 1 as n goes to infinity. We”ll end up with 1 to a really big power, which is still 1. Therefore, the limit is 1.

2. The inside of the bracket is always slightly bigger than 1. As n goes to infinity, we’ll have something larger than one to the power of infinity, which is obviously infinity. Therefore, the the answer is infinity.

Which explanation is correct? You choose

Three players enter a room and a red or blue hat is placed on each person’s head. The color of each hat is determined by a coin toss, with the outcome of one coin toss having no effect on the others. Each person can see the other players’ hats but not his own.

No communication of any sort is allowed, except for an initial strategy session before the game begins. Once they have had a chance to look at the other hats, the players must simultaneously guess the color of their own hats or pass. The group shares a hypothetical $3 million prize if at least one player guesses correctly and no players guess incorrectly.

The same game can be played with any number of players. The general problem is to find a strategy for the group that maximizes its chances of winning the prize.

The most obvious solution gives you a 50% chance of winning. But interestingly enough, the best strategy gives you a chance of winning greater than 50%. (at least in the 3-hat case)

I won’t tell you any more details about the problem, because I don’t want to spoil your fun, but let’s just say that if you can solve the 7-person version, you’re well ahead of vast majority of mathematicians.

I solved the general case at 2 o’clock in the morning, when I was trying to get to sleep. I jumped out of my bed (without shouting “Eureka!”, unfortunately :p), turned on my computer and hacked together a quick Python program to check what probability of winning the best strategy gives you for 99 hats. I won’t tell you here what it is though. (If you really want to know, you can e-mail me at vlad@leedsmathgeeks.com).

Happy hat-problem-thingy-solving!

Ok, let us try to derive a probablistic formula for this. We need to consider the total possible outcomes of the random experiment. First we need to create a coordinate system:

Where ‘x’ is the distance OP from the midpoint of the needle to the nearest crack and ‘theta’ is the smallest angle between OP and the needle. Note that half the needle has length 1 inch. Notice that we can mark out a random toss of the needle by constructing a coordinate system with the appropriate variables satisfying certain conditions, i.e. the angle ‘theta’ must be between zero and (pi/2) inclusively and the other variable ‘x’ is between zero and one inclusively.(notice we are using radians here) These conditions cover all possible positions that the needle can fall to. We can further notice that the outcome of interest can be thought of as the variable ‘x’ being strictly less than the cosine of the angle ‘theta’. Now plotting the graph of x = cos (theta) with the intervals just deifned above we have:

And we can see that our outcome of interest describes the region under the graph (above). So now we conclude by considering the ratio: area under graph/area of rectangle. We calculate the area under the graph by integrating our function (cosine) with limits from zero to (pi/2) with respect to ‘theta’. This gives us an answer of 1. And the area of the rectangle is simply (pi/2). Now considering the ratio we have:

area under graph/area of rectangle = 1/(Pi/2) = (2/Pi). This gives us the probability of the needle crossing a crack. Say that I carry out this experiment of tossing the needle and recording how many times it crosses a crack. More precisely say that I toss the needle ‘n’ times and record it ‘k’ times when it crosses a crack. Furthermore consider I do this experiment an infinite amount of times, then mathematically we have:

.  Then assuming that I carry out this experiment an infinite number of times, we can reduce this to:   . And so you see we have found the value of ‘pi’ or a close approximation to it, just using a needle and wooden boards and without no use of geometry i.e. using circles. Note this problem can be done with matchsticks and lined paper as long as it satisfies the conditions of the experiment. This is an illustration of the beauty and mystery of mathematics. By this I mean we used calculus to solve parts of the problem, and then ideas of limits. Beautiful! This problem can also be solved by integral geometry through methods of multivariable calculus.  

Behold the proof!

Let a = 0.111111…
Then 10a = 1.11111…
10a - a = 1.11111… - 0.11111…
9a = 1.

But 9 times 0.11111… is 0.99999….

So 0.999999…. is exactly equal to 1!

This has some fun consequences. Like… decimal expansions are not unique! In other words, some numbers have more than one possible way of being written as a decimal.

Of course, whenever you hear a fancy proof like this, you have to be very careful. Like I showed in Why convergence matters, you can sometimes use seemingly impeccable logic to show complete nonsense.

For more great examples of simple proofs of nonsense with VERY subtle errors, check out the website Find the error.

Okay, so i is the square root of -1. I had come to grips with that concept. But after that, when you start thinking about it… surely the square root of i must be an even more imaginary number, j? But it isn’t! In fact, you can express the square root of any complex number as another complex number!

You might be thinking right now: “Well that’s trivial! You just take the Argand plane, and then use the something-or-other theorem and rotate some dots around the origin and voila! You get the answer!” But I didn’t know that back then.

But I had it on good authority that I didn’t need to invent another super-imaginary number. Complex numbers would do. So I decided to find the square root of i!

And today, I’d like to share that journey with you.

How I found the square root of i

Since the square root of i was a complex number, it could be written as “a + bi” with a and b being real numbers.

SQRT(i) = a + bi

I squared both sides

i = a^2 + 2abi - b^2

And here came a nice logical leap (at least I thought it was nice at the time). When you have two complex numbers that equal each other, the real parts must equal each other, and the complex parts must equal each other. So:

i = 2abi
a^2 - b^2 = 0

And from there it was just a bit of trivial algebra, finally yielding 1 + i over the square root of 2. And by squaring that, I verified that it was indeed equal to the square root of i! Eureka! I found it!

This is a math puzzle book, and quite a nice short sci-fi story as well. It follows good ol’ Dorothy , as an alien (Oz) captures her, her faithful companion Toto and the entirety of Kansas to perform odd experiments. Particularly mathematical experiments on Dorothy who some how for a child a superb mathematician.

These are not easy puzzles, they’re arranged form 1 to 5 stars of difficulty , most peple will find Pickover’s problems seems to be extraterrestrially intelligent they’ll pose a big challenge. The 1 star problems should take you a day maybe if you think about them long enough and if you have a much great ability at solving problems. Then the 5 star ones will sometimes take Phd’s or more to complete. There are 108 problems which train your mind to think, “Damn! There are some hard problems out there to solve.” The problems are very varied with topics from geometry to probability, so you get classic which way to go problems and absolutely scary number problems. Even if you don’t have any talent at mathematics you’ll still enjoy this book, because of the answers…

The answers are the best part of this book with Pickover revealing some of the answers to amazing mathematical problems such as the zebra irrationals, Lissajous’ curves and Legion’s number. Sometimes though he doesn’t even have the answer himself, so you get to hear about some of the unsolved problems in mathematics, which is an extra bonus. The answers are very well written, whether your a mathematician or someone who just likes problems to solve you can understand them. Then as a bonus you learn some history of the problem to give you some extra knowledge!

Pickover’s writing is humorous, educated and enjoyable, because you can instantly tell he has a huge passion for his subject. (plus as already the author of over thirty books you would of thought he would of picked up a few things.) The story has lots of reminiscent memories from watching The Wizard of Oz on TV with quotes coming around that remind you of the good ol’ days.

The first downfall from getting this book is getting bored of problems you can’t solve, for example any problem above 2 stars the average person has no chance of solving, although the answers are great most people with a book like this want over 50% of the book to be something they can solve. If it was written problem, answer, problem answer, it wouldn’t be such an annoyance, but having the answers in the back was irritating. Or another downfall is the very few problems which you just couldn’t care less about. For instance there was way to many “which way should Dorothy go” problems, which the first one is ok but then there should be more interesting and easier problems.

I bought this book on a whim, and I’m very glad I did so, it partly got me to do math at university. So I’m probably going to be pretty biased about it, sorry. But I still would recommend it to anyone who wants to read a short sci-fi, be challenged by puzzles, or even want to learn a little mathematical history.

7/10

I’ve got 4 dice here. We both pick a die, and then roll to see who gets the higher number. And since you’re my guest, I’ll let you pick your die first, and then I’ll pick mine. Sounds good?

Oh, I almost forgot to mention. My dice are numbered a bit strangely…

If we actually played this game a couple of times, you might start to notice something. No matter what die you pick, I tend to win more than my fair share of the time.

Why would that be? Am I master dice roller? Or did I embed an evil high-tech device inside my dice to make them come up with high numbers only when I roll them?

No… my winning strategy is much more trivial, based on simple mathematics.

In fact, no matter what die you pick, I can pick a die that beats yours two thirds of the time. I just need to pick the die that’s to the left of your die in the picture above.

And you thought I was being nice by letting you pick your die first :p

S₁ = 1 + 1 + 1 + 1 + 1 + 1 + 1 + …

S₂ = 1 + 1/2 + 1/4 + 1/8 + 1/16 + …

The first series does NOT add up to a finite limit. (aka goes to infinity). The second DOES add up to a finite limit, 2. That means the second series is convergent, while the first series is divergent.

Convergence vs. Divergence

Here’s the interesting thing. When a series is divergent, we can do almost nothing useful with it. But when a series is convergent, we can do TONS and TONS of cool stuff with it!

In fact, one big branch of mathematics is simply devoted to figuring out which series are convergent and which are divergent. We often don’t even care WHAT the series converges to. It could be 1, or pi, or thirty-seven thousand, but it still converges. And we have a bunch of really clever tests for that.

But today, I won’t focus on that. Instead, I’ll show you a quick example of why fooling around with DIVERGENT series can leads to tons of trouble.

Let:
S = 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + …
From this, we obviously see that S = (1 - 1) + (1 - 1) + (1 - 1) + …, so S = 0 + 0 +0 + …, which means S = 0.
But, also:
S = 1 + (-1 + 1) + (-1 + 1) + (-1 + 1) + …
And no matter how many zeros you tack on, you won’t change the initial number. So S = 1.

And because S = S, therefore 0 = 1.

Wait, there’s more!

How about this:
S = 1 - 1 + 1 - 1 + 1 - 1 + …
S = 1 - (1 - 1 + 1 - 1 + 1 - 1 + …)
S = 1 - S
2S = 1
S = 1/2
Yay!
So 0 = 1 = 1/2! And here I was all along, thinking they’re all different numbers…

Can you see the problem in the above calculations?

That’s the thing. The calculations above are absolutely fine! It’s the ORIGINAL PREMISE (that you can use divergent series in equations) that was wrong. It’s like using infinity in equations and then wondering why you end up with things like 0 = 1.

So from today’s article, I would like you to take away two things:

Firstly, remember to be VERY careful when you meet a divergent series.

Secondly, remember the three ways of summing up S above. They’re fantastic for confusing inexperienced mathematicians

Now that’s kind of obvious. After all, most people who take the bus are going either to or from the city centre. It makes sense the buses are packed full of people there.

But that’s not the only effect at play here. There’s another, much simpler reason for the buses being packed full of people in the city centre. It’s the middle of the bus route.

To show how this works, I made a very simple model, shown below:

In my model, there are 10 bus stops. From every bus stop, one person wants to go to each of the other bus stops. So on the first stop, there will be 9 people getting on, while on the 7th stop, there will be 6 people getting off and only 3 people getting on, because there are only 3 stops left.

The blue bars show how many people arrive at each bus stop, whil the red bars show how many people leave each bus stop.

Actually, after I drew the diagram, I was surprised by how flat the distribution seemed around the centre. It looked like only the first two and last two steps were really empty, and otherwise the bus contained pretty much the same amount of people. So just for the kicks, I decided to build the same model with 30 bus stops:

Again, the distribution is very flat around the “city centre”.

Now that I think of it, that’s pretty much what you see in the real world. The bus is more or less constantly full around the city centre, then it slowly starts emptying, and suddenly on the last 3 or so stops, you see a right old exodus!

Hmm, if I spent a bit more time thinking about this, I could probably come up with a lot more examples of this phenomenon. Like cities having a more or less constant population density around the centre, and then quickly dropping the density at the edges. Or mushrooms having a more or less constant height around the centre and dropping off quickly at the edges. This phenomenon would be named Vlad’s Bus Stop Principle, and I would become rich and famous because of its numerous applications…

Ok, maybe not :).