Professor Stewart's Hoard of Mathematical Treasures (13 page)

In the 1960s the Russian mathematician Vladimir Arnold studied a map (another word for ‘function’ or ‘transformation’) from the torus to itself, defined by
where x and y lie between 0 (included) and 1 (excluded), and (mod 1) means that everything before the decimal point (the integer part) is ignored. So 17.443 (mod 1) = 0.443, for instance. The dynamics of this map are chaotic (Cabinet, page 117); also, it ‘preserves area’, meaning that areas don’t change when it is applied. So it provided a simple model for more complicated area-preserving maps arising naturally in mechanics.

This map quickly became known as Arnold’s cat, because he illustrated its effect by drawing a cat on the torus, and showing how the cat distorts when the map is applied. The same thing is done with a picture of a real cat at:

Author Theoni Pappas wrote a children’s book, The Adventures of Penrose the Mathematical Cat, presumably named after mathematical physicist Roger Penrose.

Arnold’s cat.

In the book
Mathematicians in Love
by Rudy Rucker, two mathematics graduate students prove a theorem characterising all dynamical systems in terms of objects from Dr Seuss’s
The Cat in the Hat.
¹⁶

In his 1964 research text
Abelian Categories,
Peter Freyd included the index entry ‘kittygory’. The page concerned refers to a ‘small category’.

There’s a mathematician named Nicholas Katz - does that count?

Um - Felix Hausdorff?

There’s an old test for divisibility by 11, seldom taught in these days of calculators. Suppose, for example, that the number is 4,375,327. Form the two sums
formed by taking every alternate digit (
4
3
7
5
3
2
7
). Take the difference, 21-10 = 11. If this difference is exactly divisible by 11, so is the original number, and conversely. (The number 0 is exactly divisible by 11, being equal to 11×0.) Here the difference is 11 itself, which is divisible by 11, so the test says that 4,375,327 is divisible by 11. In fact, it is equal to 11×397,757. Initial zeros make no difference, by the way, since they add zero to whichever sum they appear in.

Here are two puzzles and a question; the puzzles are easier if you use this test.

The square array uses each of the nine digits 1-9. The second row 384 is twice the first row 192, and the third row 576 is three times the first row.

There are three other ways to do this. Can you find them?

Answer on page 291

There is an entire genre of puzzles that rests on the counterintuitive properties of ‘common knowledge’ - something that has been made public, so that not only does everyone involved know it, but they know that everyone knows it, and they know that everyone knows that everyone knows it . . . A traditional case concerns the curious habits of the obscure but very polite Glaberine
¹⁷
order of monks.

Not ‘habits’ as in clothing, you appreciate.

Brothers Aelfred, Benedict and Cyril are asleep in their cell, when the novice Legpulla sneaks in and paints a blue blob on the top of each of their shaven heads. When they awake, each notices the blob on the other’s head. Now, the monastery rules are clear: it is impolite to say anything that will cause direct embarrassment to another member of the order, but it is also impolite to conceal anything embarrassing about yourself. And impoliteness is not permitted under any circumstances. So the monks say nothing, and their demeanour gives no hint of what they have seen.

Each vaguely wonders whether he, too, has a blob, but dare not ask, and there are no mirrors in their cell, nothing reflective at all. And so things remain until the Abbot enters, frowns, and informs them (neatly avoiding direct embarrassment) that ‘At least one of you has a blue blob on his head.’

Of course, all three monks know that. So does the information make any difference to them?

If you’ve not met this puzzle before, it helps to start with a simpler version, just two monks, Aelfred and Benedict. Each can see the other’s blob, but has no idea what his own head might bear. After the Abbot’s public announcement, Aelfred starts thinking. ‘I know Benedict has a blob, but he doesn’t, because he can’t see the top of his own head. Dear Lord, do I have a blob? Hmmm . . . Suppose I
don’t
have a blob. Then Benedict will see
that I don’t, so he will immediately deduce from the Abbot’s remark that he must have a blob. But he hasn’t shown any sign of embarrassment. Oh dear, I must have a blob.’ Benedict comes to a similar conclusion.

Without the Abbot’s remark, these deductions don’t work, yet the Abbot tells them nothing - apparently - that they don’t know already. Except ... Each knew that at least one monk (the other one) had a blob, but they didn’t know that the other monk knew that at least one monk had a blob.

Got that? Very well - what happens with three monks? Again, they can all deduce that they have blobs, but only after the Abbot’s announcement (see the answers on page 291). The same goes when there are four, five, or more monks, if all of them have blobs on their heads. Indeed, suppose there are 100 monks. Each bears a blob, each is unaware of that, and each is an amazingly rapid logician. To avoid timing issues, suppose that the Abbot has a bell. ‘Every ten seconds,’ he tells them, ‘I will ring this bell. That will give you time to carry out the necessary logic. Immediately after I ring, all monks who can deduce logically that they have a blob must put their hands up.’ He waits ten minutes, ringing his bell from time to time, but nothing happens. ‘Oh, yes, I forgot,’ he says. ‘Here is one extra piece of information. At least one of you has a blob.’

Now nothing happens for 99 rings, and then all 100 monks simultaneously raise their hands after the 100th ring.

Why? Monk number 100, say, can see that the other 99 all have blobs. ‘If I do not have a blob,’ he thinks, ‘then the other 99 all know this. That takes me out of the reckoning altogether. So they are making whatever series of deductions you get with 99 monks when I don’t have a blob. If I’ve sorted out the 99-monk logic right, then after 99 rings they will all put up their hands.’ He waits for ring 99, and nothing happens. ‘Ah, so my assumption is wrong, and I must have a blob.’ Ring 100, up goes his hand. Ditto for the other 99 monks.

Ah, yes . . . but maybe monk 100 was wrong about the 99-monk logic. Then it all falls apart. However, the 99-monk logic
(with the hypothetical assumption that monk 100 is blobless) is the same. Now monk 99 expects the other 98 to put up their hands at the 98th ring, unless monk 99 has a blob. And so it goes, recursively, until we finally get down to a single hypothetical monk. He sees no blobs anywhere, is startled to discover that somebody has one, immediately deduces it must be him (you don’t need to be an expert logician at that stage) and puts his hand up after the first ring.

Since his 1-monk logic is correct, so is the 2-monk logic, then the 3-monk . . . all the way to the 100-monk logic. So this puzzle is a striking example of the Principle of Mathematical Induction. This says that if some property of whole numbers holds for the number 1, and if its validity for any given number implies its validity for the next number, no matter what those numbers may be, then it must be valid for all numbers.

That’s the usual story, but there’s more. So far I’ve assumed that every monk has a blob. However, very similar reasoning shows that this requirement is not essential. Suppose, for example, that 76 monks out of a total of 100 have blobs. Then, if everyone is logical, nothing happens until just after the 76th ring, when all the monks with blobs put up their hands simultaneously, but none of the others.

At first sight, it’s hard to see how they can work this out. The trick lies in the synchronisation of their deductions by the bell, and the application of common kowledge. Try two or three monks first, with different numbers of blobs, or cheat by peeking at the answers on page 291.

Three weary travellers came to an inn, late in the evening, and asked the landlord to prepare some food.

‘All I got is pickled onions,’ he muttered.

The travellers replied that pickled onions would be fine, thank you very much, since the alternative was no food at all.
The landlord disappeared and eventually came back with a jar of pickled onions. By then, all the travellers had fallen asleep, so he put the jar on the table and went off to bed, leaving his guests to sort themselves out.

The first traveller awoke. Not wishing to make a pig of himself, and not knowing what anyone else had already eaten, he took the lid off the jar, threw away an onion that looked bad, ate one-third of the onions that remained, put the lid back on the jar, and went back to sleep.

The second traveller awoke. Not wishing to make a pig of himself, and not knowing what anyone else had already eaten, he took the lid off the jar, threw away two onions that looked bad, ate one-third of the onions that remained, put the lid back on the jar, and went back to sleep.

The third traveller awoke. Not wishing to make a pig of himself, and not knowing what anyone else had already eaten, he took the lid off the jar, threw away three onions that looked bad, ate one-third of the onions that remained, put the lid back on the jar, and went back to sleep.

At this point the landlord returned and removed the jar, which now contained six pickled onions.

How many were there to start with?

 

Answer on page 292

The Great Whodunni has an endless supply of mathematical card tricks. This one allows him to identify a specific card, chosen from 27 cards taken from a standard pack.

Whodunni shuffles the 27 cards, and lays them out in a fan so that his victim can see all of them.

‘Choose one card, mentally, and remember it,’ he tells him. ‘Turn your back, write down the card, and seal it in this envelope, so that we can verify your choice at the end.’

Now Whodunni deals out the 27 cards, face up, into three piles of 9 cards each, and asks the victim to say which pile the chosen card is in.

He picks up the piles, stacks them together without shuffling, then deals them into three piles and asks for the same information.

Finally, he picks up the piles, stacks them together without shuffling, then deals them into three piles and asks for the same information for a third time.

Then he picks out the chosen card.

How does the trick work?

 

Answer on page 293

Whodunni can do even better. In just two deals, he can correctly identify a card chosen from the full 52-card pack.

First, he deals the cards in 13 rows of 4 cards, and asks which row the card is in.

Then he reassembles the cards into the same order, deals them into 4 rows of 13 cards, and again asks which row the card is in.

After which he unerringly names the chosen card.

How does this trick work?

 

Answer on page 293

Why do mathematicians always confuse Halloween and Christmas?

 

Answer on page 293

Whole numbers are fine for addition and multiplication, but subtraction causes problems because, for instance, 6 - 7 doesn’t work with positive whole numbers. This is why negative numbers were invented. A positive or negative whole number is called an integer.

In the same way, the problem of dividing one number by another, such as 6 ÷ 7,
¹⁸
requires the invention of fractions like
. The number on the top (here 6) is the numerator, the one on the bottom (here 7) is the denominator.