100 Essential Things You Didn't Know You Didn't Know Read online

  The answer is pretty surprising. The shape of the road surface that gives a steady ride on square wheels is created when you hang two ends of a chain from two points at the same height above ground. This is called the catenary and we have met it before in that connection in Chapter 11. If we turn it upside down we obtain the shape that is used for many of the great arches of the world. But if you take a catenary arch and repeat it over and over again along a line you get a sequence of undulations of the same height. This is the shape of the ground surface that gives us the smooth ride on square wheels. We just need the bottom corners of the square to fit into the sides of the successive ‘valleys’ on the surface. The key feature of the catenary is that when one is placed next to another, the angle between the two sides of the adjacent arches as they come into the lowest point of each valley is a right angle, 90 degrees, and that is the angle at each corner of the square. So, the right-angled wheel just keeps rolling along.fn1

  fn1 The square is not the only possible wheel shape that can give a smooth ride. Any polygonally-shaped wheel will work for a different catenary-shaped road. As the number of sides in the polygon gets very large, it begins to look more and more like a circle and the line of catenaries gets flatter and flatter and looks increasingly like a perfectly flat road. The case of a 3-sided polygon, a triangular wheel, is problematic because it runs into the side of the upcoming catenary before it can roll into the corner and hits the trailing side as it rolls out. You need to lay the road surface bit by bit to avoid these collisions happening. For a rolling polygonal wheel with N equal sides (so N = 4 is the case of our squared wheel), the equation of the catenary-shaped road that gives a smooth, straight-line rider for the cyclist is y = −B cosh(x/B), where B = C cot(π/N) with C constant.

  65

  How Many Guards Does an Art Gallery Need?

  Who will guard the guards?

  Juvenal

  Imagine you are head of security at a large art gallery. You have many valuable paintings covering all of the gallery walls. They are also quite low on the walls, so that they can be viewed at eye level and so are vulnerable to theft or vandalism. The gallery is a collection of different rooms with different shapes and sizes. How are you going to make sure that each one of the pictures can be kept under surveillance all of the time. The solution is simple if you have unlimited money: just have one attendant standing on guard next to every picture. But art galleries are rarely awash with money, and wealthy donors don’t tend to earmark their gifts for the provision of guards and their chairs. So, in practice, you have a problem, a mathematical problem: what is the smallest number of attendants that you need to hire, and how should you position them so that all the walls of the gallery are visible at eye level?

  We need to know the minimum number of guards needed to watch N walls. We assume that the walls are straight and a guard at a corner where two walls meet will be assumed to be able to see everything on those two walls, and we will assume that a guard’s view is never obstructed. A square gallery can obviously be watched by just one guard. In fact, if the gallery is shaped like any polygon in which all the sides bulge outwards (a ‘convex’ polygon) then one guard will always suffice.

  Things get more interesting when the walls don’t all bulge outwards. Here is a gallery like that, with 8 walls, which can also be guarded by just one attendant located at the corner O.

  So, this is a rather economical gallery to run.

  Here is another ‘kinkier’ 12-walled gallery that is not so efficient. It needs 4 guards to keep an eye on all the walls.

  To solve the problem in general we just look at how we can divide the gallery up into triangles that don’t overlap. This can always be done. If the polygon has S vertices, then there will be S-2 triangles. Since a triangle is one of those convex shapes (the three-sided one) that needs only a single guard we know that if the gallery can be completely covered by, say, T non-overlapping triangles then it can always be guarded by T attendants. It might, of course, be guarded by fewer. For instance, we can always divide a square into two triangles by joining opposite diagonals, but we don’t need two guards to watch the walls, one will do. In general, the total number of guards that might be necessary to guard a gallery with W walls is the whole number part of W/3. For our 12-sided comb-shaped gallery this maximum is 12/3 = 4, whereas for an 8-sided gallery it is 2. Unfortunately, determining whether you need to use the maximum is not so easy and is a so-called ‘hard’ computer problem (See chapter 27) for which the computing time can double each time you add another wall to the problem.

  Most of the galleries you visit today will not have kinked, jagged wall plans like these examples. They will have walls that are all at right angles like this:

  If there are many corners in a right-angled polygonal gallery like this, then the number of attendents located at the corners that might be necessary and is always sufficient to guard the gallery is the whole number part of ¼ × (Number of Corners). For the 14-cornered gallery shown here this number is 3. This means that it is much more economical on wages to have a gallery design like this, especially when it is large. If you have 150 walls, then the non-right-angled design could need 50 guards, while the right-angled set-up will need at most 37.

  Another traditional type of right-angled gallery will be divided into rooms, like this 10-roomed example:

  In these cases you can always divide the galley up into a collection of non-overlapping rectangles. This is a useful design because if you place an attendent at the opening connecting two different rooms, then he can guard both of them at the same time. But no guard can guard 3 or more rooms at once. So, now the number of guards that is sufficient and occasionally necessary to keep a complete watch on the gallery is the whole number part of ½ × (Number of Rooms), or 5 for the gallery drawn here. This is a very economical use of human resources.

  We have been speaking of people watching the walls, but the things we have said also apply to CCTV cameras or to lights for illuminating the gallery and its rooms. Next time you are planning to steal the Mona Lisa you will have a head start.

  66

  . . . and What About a Prison?

  All my contacts with criminals show that what they’re doing is just a slightly more extreme version of what everybody is doing.

  David Carter

  Art galleries are not the only buildings that have to be guarded. Prisons and castles do too. But they are an inside-out version of art galleries. It is the outside walls that have to be guarded. How many guards do you need to station at the corners of a polygonal fortress in order to observe all of its exterior walls? There is a simple answer: the smallest whole number at least equal to ½ × (Number of corners). So, with 11 corners, you need 6 guards to guard the outside walls. Better still, we know it is the exact number you need. No fewer will be sufficient and no more are necessary. In the inside gallery problem, we only ever knew the largest possible number that we might need.

  We can think about the case of right-angled prison walls again, as we did for the galleries. We might have an outer prison wall like these two right-angled shapes.

  In these right-angled cases we need 1 plus the smallest whole number at least equal to ¼ × (Number of corners) to guard the whole of the outer wall. No fewer are possible and no more are needed. In the two examples shown there are 12 corners and so we need 1 + 3 = 4 guards.

  67

  A Snooker Trick Shot

  Steve is going for the pink ball – and for those of you watching in black and white, the pink is next to the green.

  Ted Lowe

  Some people used to satisfy themselves that if their children spent most of their waking lives playing computer games, it was good for their understanding of maths and computing. I always wondered if they felt that hours at the snooker or pool hall were adding to their knowledge of Newtonian mechanics. Still, a knowledge of simple geometry can certainly provide you with some impressive snooker demonstrations for the uninitiated.

  Suppose that you
want to hit a single ball so that it goes around the table, bounces off three cushions, and returns to the point from which it was struck. Let’s start with the easy case – a square table. Everything is nice and symmetrical, and it should be clear that you should place the ball in the middle of one of the sides of the table and then strike it at an angle of 45 degrees to the side. It will hit the middle of the adjacent side at the same angle and follow a path that is a perfect square, shown dashed in the diagram overleaf.

  You don’t have to start with the ball against one of the cushions, of course; if you strike it from any point on the dashed square path and send it along one of the sides of the dashed square, then the ball will eventually return to where it started (as long as you hit it hard enough). If you want it to stop at exactly the point that you hit it from, then you need to be very skilful – or at least put in a little practice.

  Unfortunately, you won’t very often come across a square snooker table. Modern tables are made from two squares side by side and a full-size table will have dimensions of 3.5 m × 1.75 m. The important general feature is that the table’s length is twice its width. Using this simple fact, you can draw how your trick shot will have to run when the table has these rectangular dimensions. I have drawn in the diagonal as well for reference. Your shot has to go parallel to the diagonal and hit the sides at points that divide each of them in the ratio 2 to 1, the same ratio as the ratio of the length to the width of the table. (In the case of the square table this ratio was equal to 1 and you had to hit the centre of each side of the table.) This means that the angle that the ball’s path makes with the long sides of the table has its tangent equal to ½, or 26.57 degrees, and the angle it makes with the short sides is 90 minus this angle, or 63.43 degrees, since the three interior angles of the right-angled triangles must add up to 180 degrees. The dashed parallelogram marks the only path on the rectangular table that will see the ball return to its starting point.

  If you find yourself playing on a non-standard table, you need to recalculate. In general, the magic angle you need to strike the ball at, relative to the long side of the table, is the angle whose tangent equals the ratio of the width to the length of the table (1:2 for our full-size table and 1:1 for the square table) and the point against the cushion you need to strike from must divide the length of the side in the same ratio as the length to the width of the table.

  68

  Brothers and Sisters

  Sisterhood is powerful.

  Robin Morgan

  One of the strangest things about China is the growing consequences of the ‘one-child’ policy. With the exception of multiple births (typically 1 per cent of the total), every young person in urban areas is an only child.fn1 In this situation, the likelihood of each child being lavished with rather too much parental attention has given rise to the term ‘Little Emperor Syndrome’. In the future, for almost everybody, there will be no brothers and sisters and no aunts or uncles. A concept like ‘fraternity’ will gradually lose all meaning.

  At first, there seems to be a strange general asymmetry about brothers and sisters. If there are 2 children, one boy and one girl, then the boy has a sister but the girl doesn’t. If there are 4 children, 3 girls and one boy, then the boy has got 3 sisters and the girls between them have 3 × 2 = 6 sisters. Each girl only gets to count the other girls as sisters but the boy counts them all. So it looks as if boys should always have more sisters than girls!

  This seems paradoxical. Let’s look more closely. If a family with n children has g girls and n−g boys, then the boys have a total of g(n−g) sisters between them, while the girls have a total of g(g−1) sisters. These numbers can only be equal if g = ½ (n+1). This can never be true when n is an even number because then g would be a fraction.

  The puzzle has been created because there are many ways in which a family of children can be formed. A family of 3 children can have 3 sons, 3 daughters, 2 sons and 1 daughter, or 2 daughters and 1 son. If we assume that there is an equal chance of 1/that a newborn child will be a boy or a girl (this is not quite true in reality) then a family of n children can be made up in 2n different ways. The number of different family make-ups with n children and g daughters is denoted byfn2 nCg and the boys will each have g(n–g) sisters. Given all the 2n different ways in which the family can be divided into sons and daughters, we should be asking what is the average number of sisters that the boys in the n-child families will have. This is the sum of all the answers for the number of sisters they can have for all the possible values of the number g = 0,1,2,. . ., n divided by the total number, 2n. This is

  bn = 2-n ∑g nCg × g(n−g)

  Similarly, the average number of sisters that the girls in the n-child families have is

  gn = 2-n ∑g nCg × g(g−1)

  The solutions to these formulae are much simpler than the formulae would lead us to expect. Remarkably, the average numbers of sisters for the boys and the girls are equal, and bn = gn = ¼ n(n−1). Notice that because it is an average, it doesn’t mean that any given family has to have the average behaviour. When n = 3, the average number of sisters is 1.5 (which no one family could have). When n = 4, the average number is 3. When n gets big, the number gets closer and closer to the square of n/2. Here is the table of the 8 possibilities for families with 3 children:

  We see that the total number of sisters for the boys are the sums from the second and third rows, 3 × 2 + 3 × 2 = 12, and the total number of sisters for the girls is the sum of the third and the fourth rows; it is also 12 = 3 × 2 + 1 × 6. Since there are 8 possible ways to make this family, the average number of sisters for the girls and the boys is equal to 12/8 = 1.5, as predicted by our formula ¼ × n × (n–1) for the case of n = 3, when there are three children.

  fn1 In rural areas a second child is permitted after an interval of 3 years if the first child was disabled or female.

  fn2 nCg is shorthand for n!/g!(n–g)! and is the number of ways of choosing g outcomes from a total of n possibilities.

  69

  Playing Fair with a Biased Coin

  And somewhat surprisingly, Cambridge have won the toss.

  Harry Carpenter

  Sometimes you need a fair coin to toss so that you can make a choice between two options without any bias. At the outset of many sports events, the referee tosses a coin and asks one of the competitors to call ‘heads’ or ‘tails’. You could set up gambling games by scoring sequences of coin tosses. You could use more than one coin simultaneously, so as to create a greater number of possible outcomes. Now suppose that the only coin that you have available is a biased one: it does not have an equal probability (of ½) of falling ‘heads’ or ‘tails’. Or perhaps you just suspect that the coin that your opponent has so thoughtfully provided might not be fair. Is there anything you can do in order to ensure that tossing a biased coin creates two equally likely, unbiased outcomes?

  Suppose that you toss the coin twice and ignore outcomes where both outcomes are the same – that is, toss again if the sequence ‘heads-heads’ (HH) or ‘tails-tails’ (TT) happens. There are two pairs of outcomes that could result: ‘heads’ followed by ‘tails’ (HT), or ‘tails’ followed by ‘heads’ (TH). If the probability of the biased coin coming down ‘heads’ is p, then the probability of getting ‘tails’ is 1−p, and so the probability of getting the sequence HT is p(1−p) and that of getting TH is (1−p)p. These two probabilities are the same, regardless of the probability p of the biased coins. All we have to do to get a fair game is define HEADS by the sequence HT and TAILS by the sequence TH, and the probability of TAILS is the same as the probability of HEADS. And you don’t need to know the bias, p, of the coin.fn1

  fn1 This trick was thought up by the great mathematician, physicist and computer pioneer, John von Neumann. It had wide use in the construction of computer algorithms. One of the questions that was subsequently addressed was whether there were more efficient ways of defining the new HEAD and TAIL states. The way we have done it wastes ‘time’ b
y having to discard all the HH and TT outcomes.

  70

  The Wonders of Tautology

  A good guide to understanding events in the modern world is to assume the opposite of what Lord Rees-Mogg tells you is the case.

  Richard Ingrams

  ‘Tautology’ is a word that gives out bad vibes. It suggests meaninglessness, and my dictionary defines it as ‘a needless repetition of an idea, statement or word’. It is a statement that is true in all eventualities: all red dogs are dogs. But it would be wrong to think that tautologies are of no use. In some sense they may be the only way to really secure knowledge. Here is a situation where your life depends upon finding one.

  Imagine you are locked in a cell that has two doors – a red door and a black door. One of those doors – the red one – leads to certain death and the other – the black one – leads to safety, but you don’t know which leads to which. Each door has a phone next to it from which you can call an adviser who will tell you which door you should take in order to get out safely. The trouble is that one adviser always tells the truth and the other always lies, but you don’t know which one you are talking to. You can ask one question. What question should you ask?

  Take the simplest question you could ask: ‘Which door should I go through?’ The truthful adviser will tell you to go through the black one and the untruthful adviser will tell you to go through the red one. But, since you don’t know which of the advisers you are speaking to is telling the truth, this doesn’t help you. You would do just as well taking a random guess at red or black. ‘Which door should I go through?’ is not therefore a tautology in this situation. It is a question that can have different answers.