Professor Stewart's Hoard of Mathematical Treasures (63 page)

In general, using n cuts, the maximum number of pieces is
+ 1, the nth triangular number plus 1.

Like this.

Note that on the third move, coin 5 just slides out from between coins 2 and 4. The arrow doesn’t show the direction of the move, just which coin goes where.

One of the dice stopped rolling with a 6 uppermost. The other hit a rock, split in two, and the two pieces showed a 6 and a 1. So Olaf scored 13, beating the King of Sweden’s feeble 12.

In mathematical circles this kind of thing is called ‘expanding the state space’. That is, extending the range of possible outcomes. It is one of the reasons why mathematical models never match reality perfectly.

In gambling circles it’s called ‘rigging the dice’.

In political circles it’s called ‘politics’.

I learned this story from Ivar Ekeland’s The Broken Dice.

The donkey was carrying five sacks and the mule was carrying seven.

Suppose the donkey carries x sacks and the mule y. Then the mule tells us two things:

The second equation tells us that
y
=
x
+ 2. Now the first equation tells us that
x
+ 3 = 2
x
- 2, which implies that
x
= 5. So
y
= 7.

Each character has a probability of 1/36 of being selected on any given throw, so on average it takes 36 throws to get any given character. To get DEAR SIR, with 8 characters including the space, you need
throws. The complete works of Shakespeare would need 36
^5000000
throws, which is roughly 10
^2385606
. If the monkey typed 10 characters per second, which is faster than a really good typist, it would take roughly 3 × 10
^2385597
years to complete the task.

Dan Oliver ran a computer program in 2004, and after a simulated time of 42 octillion years, the digital monkey typed

VALENTINE. Cease toIdor:eFLP0FRjWK78aXzVOwm)-’;8.t

The first 19 letters appear in The Two Gentlemen of Verona. Similar results are reported at:
en.wikipedia.org/wiki/Infinite_monkey_theorem

If a square board with no missing corner has at least one even side, which is the case here, then the first player can always win. Imagine the board tiled with dominoes: 2×1 rectangles. Any tiling will do - for example, this one on the complete 8×8 board.

Domino grid for winning strategy on a complete board.

Whatever the second player does, other than losing by hitting the edge, the first player can always find a move for which the snake ends in the middle of a domino. That can’t be a losing move, since it hasn’t hit an edge, and eventually the second player runs out of options.

If both sides of the board are odd, the second player wins with a similar strategy, using a tiling that omits the initial square with + on it.

Cutting out the bottom right corner square interferes with these domino strategies. The first player can’t cover the modified board with dominos, since it has an odd number of squares. The second player can’t cover all but the starting square with dominos, either, but that’s less obvious since those squares are even in number. But if you imagine the usual chessboard pattern of alternating black and white squares, there are 30 of one colour and 32 of the other, again omitting the starting square marked with a +. However, any domino covers one square of each colour, so a tiling would have to cover 31 of each.

There must still be a winning strategy for one or other player, because this is a finite game and can’t be drawn. But it’s no longer clear what that strategy might be, or who should win.

If you’ve followed the instructions correctly, the two handkerchiefs will miraculously separate.

If not, try again and be more careful.

The mathematical aspect is topological: when you convert the handkerchiefs to closed loops by clasping their ends, the loops are not linked. They just look as though they are.

She should write down:
to avoid losing her money.

The solution Dudeney asked for is 3
.

The others, including the example I gave when posing the puzzle, are:

This proof isn’t a joke - it’s how you do it in courses on the foundations of mathematics. It looks like the hard part is to prove the associative law, which of course I assumed. Actually the hard part is defining numbers and addition. That’s why Russell and Whitehead needed 379 pages to prove the simpler theorem 1 + 1 = 2 in Principia Mathematica. After that, 2 + 2 = 4 is a doddle.

You can get nine pieces. Here are two possible ways.

Two ways to get nine pieces with three cuts.

When the tippe top turns over, it still spins clockwise, looking down from above.

If you imagine the tippe top spinning in space, unsupported, and then turn it upside down, it would be spinning anticlockwise. But that’s not what the top does. As it starts to tip over, the end of the stalk hits the ground, and itself starts spinning. This changes the behaviour when the top finally ends up standing on its stalk.

The physical point here is angular momentum (see page 30), a quantity associated with a rotating body, roughly equal to its mass times its rate of spin round an appropriate axis. The angular momentum of a moving body is conserved - does not change - unless some force such as friction is acting.

Most of the angular momentum of the tippe top arises from the spherical part, not the stalk. Since angular momentum must be conserved - subject to a small loss through friction - the final direction of spin has to be the same as the initial direction. Friction just slows the spin down a little.

No other opening moves in JG-40 can force a win. There is a similar strategy for JG-100, which I’ll explain in a moment, and Mathophila wins. As for JG-n, let’s save that for a bit.

This game seems first to have appeared in a course on number theory given by the great mathematical physicist Eugene Wigner at Princeton in the late 1930s. More recently it was reinvented independently by Rob Porteous, a schoolteacher, to teach young children about multiplication and division. Porteous’s pupils discovered that Mathophila can always win JG-100 if (and only if) she starts with 58 or 62.