THE MATHS OF PYLONS, ART GALLERIES
AND PRISONS UNDER THE SPOTLIGHT
Professor John Barrow
A few years ago, when I was visiting Milan, I spent some time in what was probably the most fascinating museum that I have visited in Europe. It is a museum for the study of art fraud. It is based close to the Adante and it is really part of the Chemistry Department of the University of Milan. It has many floors of exhibits, some real, some fraudulent, and the idea is to show you the scientific techniques that are being used to try and discriminate true from faked works of art. So, for example, on a painting you might study the mathematical structure of the propagation of cracks in paint to see whether the cracking pattern on a modern fake really matches what it ought to look like for something that is hundreds of years old. Another example is, in the African Room, where the key sense to use is smell. This is because most because African artefacts have sat next to village fires for a very long period of time and the smell of wood fires is very hard to fake.
When this museum first opened, it ran a rather advanced seminar series on the sorts of scientific techniques that were being used. But then the organiser, the Professor of Chemistry at the University, was contacted by the Chief of Police, who said that he had been to the last of these lectures and was alarmed that a very large fraction of the audience were suspected or well-known art fraudsters, known to the police force in the north of Italy! Needless to say, the series of talks was hastily curtailed!
Well, today, I too am going to talk quite a bit about art theft and prisons and so forth. So, welcome, whoever you are!
The common factor amongst some of the topics in the title is really straight lines; what you can learn about, or from, the mathematics of straight lines. But since we have all been sleeping soundly, shivering in the cold and doing very little over Christmas, perhaps we all need to get warmed up again to start thinking mathematically, so let us start with a little puzzle to get you thinking.
Suppose that you have got nine points. My challenge to you is that you have got to join up all those points, with four straight lines, without ever taking your pencil off the paper, and without ever reversing:
If we have a go, what sort of thing might we do? You might go from point One to Three, then down to Nine, before going across to Seven and then diagonally to Three. But this would mean that you didn't go to point Four. Alternatively, you might put the lines around the edge - One to Three, Three to Nine, Nine to Seven, Seven to One - but that would leave out point Five in the middle. So can you do it?
The answer is, yes, you can do it, but you have to enlarge your way of thinking. Here is a diagram of how to do it:
Line one goes from point Nine to One. The important move then comes with the second line going out beyond point Three - I didn't say you had to stay in the box. Then you cut back through points Six and Seven with your third line, and then up through points Four and Seven with the final line. So, sometimes there can be interesting mathematical content just in the way that straight lines join up.
If you came down here on the train, like me, and you looked out of the window, you will have seen a lot of electricity pylons, which are also made of straight lines. One of the things you notice about pylons, if you study them carefully, is that they are all made of triangles. But why is this?
There is an interesting geometrical lesson here because there is something special about triangles. Suppose you had metal struts, hinged at the corners, and you made a collection of polygonal shapes - a triangle, with three sides, then the square, and upped the number of sides, heptagon, and so forth, and so on forever - then there is something very special and unique about the triangle: it is the only shape that is rigid. So if you try to apply pressure to it, you could not deform it into another shape without bending the sides and buckling it.
So you can see, for the square, if you just slide it over, it will deform into a parallelogram:
So if you are making something that you want to be rigid, and not to bend, triangles are a good idea. Even in fact, if you were making something that looks a bit rectangular, like a gate, you will probably recall that people like to put bars across gates, and divide your square gate into two triangles or, in this rectangular gate, into a number of triangles:
Again, the reason is because the triangular shapes are rigid - they are not going to break. So this is why you see triangles all over the place in fairly straightforward aspects of structural engineering, such as in pylons.
Because you are mathematicians, you tend to think: 'How can we generalise, the triangle's being the only rigid polygon? What happens if we go to polyhedra?' This leads us into solid shapes. The important distinction to bare in mind is convex and concave polyhedra: the convex ones are solid shapes which are pointed outwards; one where, if you draw a line from any point inside it to any other point, you do not pass outside the object. Well, as was first showed a long time ago, all the convex polyhedra, are rigid. So if there is a great V-shaped slit in the object, you could draw a line from one point of the object to another that would pass outside the object; but if the object is like a triangle, you ca not draw a line between any two interior points that passes outside of the triangle. So all the convex ones are rigid, but what happens if they are not convex? If you create polyhedra like the star-shaped ones that Archimedes found, what can you do with those? And it is the same question if they are not regular or uniform?
Everybody regarded this as an unsolved problem; we did not really know what the situation was. But then, in about 1978, an American mathematician called Robert Connelly found an 18-sided example - each of the faces are triangles - that was not rigid. He did not build it, but he created a picture on-screen, as it were, of what it looks like. Subsequently, a theorem was proved which illustrated why you would fail if you tried to build one of these examples, and that is that almost every non-convex polyhedron is rigid. What does that mean - 'almost every'? It is a technical mathematical phrase to say the only ones excluded are special unstable cases, rather like knitting needles suspended on their points. If you change them only very slightly they will behave differently. So, in order to make one of these, you would have to make it absolutely perfectly true and correct, otherwise you would be in a nearby polyhedron that did not have this property.
Then, a few years later, Steffen found one that was slightly smaller, and here is a perspective picture of it:
So this is another idealised example, in a sense, that you probably would not be able to make unless you had perfect engineering: it is a non-convex, rigid example.
There are a few simple examples of why straight lines can sometimes tell you interesting things, but for the rest of the talk, I want to move onto another problem involving straight lines that has traditionally become known as the Art Gallery Problem or Art Gallery Problems, and there is a collection of problems that sort of spring from them. They really are all problems about surveillance, about watching things, and the geometry of what you can see from particular vantage points.
The reason this problem is also interesting, is that it tells you something about what it is that interests mathematicians. The ordinary person in the street probably thinks that mathematicians spend all their time proving theorems, doing calculations, writing computer programmes. Some mathematicians do some of those things for a fair bit of the time, but what really interests mathematicians is a new type of problem; a problem which is not just a new variant of something that they looked at before, where, by using the same methods that you have used before, you can just sort of turn the handle and get the solution.
So this is a problem which was first posed in the course of a conference in 1976 by Victor Klee. It has a very simple wording. If you imagine that you have an art gallery - that was the scenario he first imagined - you have pictures on the walls, the art gallery can have any shape you like (we assume the walls are straight, for simplicity), then how many guards do you need in order that all the walls can be under observation at any one moment or, going further, the whole of the gallery can be under observation at any one moment?
You can see what the more general problem being suggested is: if we change the shape of the gallery, increase the number of walls, the corners, is there a general rule that tells you how many cameras you might need? If you are in the business of building galleries, this is a big financial question. If you build the gallery with a bad shape, you could end up having to employ twice as many attendants and guards, or buy twice as many cameras, as you would need if you had built a more intelligently geometrically designed gallery. So there are both commercial issues as well as mathematical issues involved. So the Art Gallery Problem is, how many cameras do you need to guard a gallery - how many guards, human or mechanical - and where should you put them?
Here is a very simple type of gallery:
We talked just now about convex polygons and convex polyhedra. So here is a convex polygon - the triangle was the three-sided example - and what you can see, pretty obviously, is that if you have a convex closed polygon like this, then one camera is always going to be enough to see the whole of the interior. So at the moment, there are not going to be obstacles, so one camera suffices for a convex polygon.
But suppose that the gallery looks like something of a Guggenheim type of structure:
If we have n vertices - I will refer to them as corners - how many cameras would you need to deal with that situation, and how would you go about trying to decide?
Let us think about a very simple example and build up some intuition. We can start with a polygon like this:
It has got n sides. The first thing to do with this problem is to divide it up into triangles, like this:
Then we can apply the knowledge that one camera is enough to look after the inside of any triangle. In this case, if you have n sides to your polygon, you can get away with n minus two triangles inside, and so you would just need n-2 cameras. But that is rather a lot because you may suppose that we put the camera on the dividing line between two triangles? Then we could halve the number that we would need. So if we stick the cameras on the dividing lines, we can reduce it by two. But you might do better: you might be able to pick the locations so that they look right into two whole triangles, so maybe locating things at the right corners can help you do even better. So you are on the track to see that there might be a systematic way to look at this, and we have learnt already that inside one triangle, you just need one camera, that seems like a good clue. So the answer, if you are a mathematician or a graph theorist, you should triangulate.
So what does that mean? We know that if we have a general polygon of any shape you like, we can always split it up into a collection of triangles. We do this simply by joining the corners in such a way that the lines never intersect, and the result is a so-called triangulation of the polygon:
You can see that if we put a camera inside each one of those triangles, somewhere, then we will have a surveillance pattern that looks after the whole gallery. But we can do a bit better if we do not just drop the camera in anywhere in the triangle, but learn the lesson about using the corners.
So a mathematical proof is then possible, in which we decide that we will have the colours, red, green, and blue at each of the corners:
Here we created a so-called three-colouring of the gallery, so we give each corner a colouring, so that no two colours are drawn from one line to another; so we do not ever have a red connected to red, or a green to green, or a blue to a blue. This is a process which it is fairly straightforward to do, as we can see from this. If we had a gallery that had 30,000 sides, this would be a computationally long-winded so-called MP problem, so computationally, that is the slow part. But we can always get to a position where we have a colouring where each triangle has got the three different colours at the different corners, and there will be no join within a triangle links corners of the same colour. So now all we have to do is either locate our cameras at all the red points, or at all the green points, or at all the blue points. If we count them up, we will find that if we did it at the red points, we would have four points; if we did it at the green points, again we would have four; and the blue points, it would be three. So we would pick the smaller of these, so we pick the blue locations. So that is the three-colouring and that gives you the smallest number that you could make use of to see the whole gallery, and it shows you where to put them.
There is a general theorem here which can be proven, which is something of a re-statement of what we have found in this exploration, and that is, if you have n corners to the gallery, n sides, then the number of cameras which is sufficient and may sometimes be necessary to survey the whole of the inside of the gallery is given by taking one-third of the number of corners and then taking the integer or the whole number part of that.
[n/3] = x where [x] is the integer part of x
We designate the mathematical function of taking the whole number part by the square brackets, so it is sometimes called the floor function. So the floor of pi is 3, and the floor of 2.5 is 2. So you forget about the fractional part. Therefore, in the case we are dealing with, I think there were eleven sides to this gallery, so we divide that by three, where you get three and a third, so the floor is three, and that is indeed what we found we needed in blue pattern:
[n/3] = x
11 / 3 = 3.333...
[3.333...] = 3
So this is the the theorem, and it was first proved by a Czech computer scientist, called Vasek Chvátal, who is now in the Department of Computer Science at one of the Concordia University in Canada. He heard Klee present this problem in 1976, and a few months? later, he presented this solution. There is a theorem which proves what I said rather rigorously, using the same techniques.
So we know that, for example, the integer part of n/3, is not always necessary. For instance, if our gallery had 100 sides but it was a convex polygon, we would only need one camera. So the floor of n/3 is only the number that you might need; it is the one that is always sufficient to deal with a gallery with n sides.
The worst case scenario is one that has a fairly interesting general characterisation, which can be shown in this diagram:
If we have this comb-like structure for the gallery, where the side rooms are all these long Vs, then you can see that from inside, you can never see any two of these Vs completely, so you have always got to have one camera in each of the Vs. So a gallery that has got n/3 of these shaped rooms is going to be a general shape of gallery that is going to need the full compliment of the integer part of n/3 cameras.
If we specialise our gallery a little bit, then we can arrive at some slightly different results that are rather stronger. Suppose you wanted to build a gallery that was constructionally much cheaper and just used right angles. Such galleries, we might call orthogonal galleries. Here is one:
Everything is at right angles to its neighbour. What happens here is you can get away with rather fewer guards.
There is a similar type of proof. You can divide it up, in a sense, into 4-sided shapes, and you only need the integer part - the whole number part - of a quarter times the number of corners ([n/4]) to guard a gallery like that, in the worst case scenario. This is better than before because then, if you remember, if we had a 100 corners before, it needed 33 guards for that comb-like or zigzag structure, but here you could get away with 25 cameras or guards in the worst case scenario.
Another feature of galleries is they can be divided into rooms:
So this is orthogonal, in the sense that the outer boundary is a rectangle, but if we divide it up into rooms which have openings, then you can see that there is an economical way to carry out the surveillance. If you have someone standing in the openings marked with the dots, then they can look after two of the rooms at the same time. So there is a rather simple pattern here: in this gallery which has got r rooms, then you can get away with the integer part of a half r guards. So if there are 11 rooms, it will be the integer part of 5.5, so you just need 5 guards; if there were 12 rooms, you would need 6. So this is, again, this is rather more economical. This one here has got 8 rooms, and you can see the locations of the 4 guards. So in this case, it is not a recipe which tells you what is the worst case scenario; this is the number that you always need.
This next type of question that we might ask about these galleries is that you might perhaps worry that something like your most valuable work of art is only visible by one guard, who might fall asleep, or the film might be being changed on the camera. Therefore, you could ask what the structure of a gallery is where each point is going to be visible by two or three or any number of guards. This turns out to be a very difficult problem.
Here is a simple example; in general, it is not solved. This is the gallery they are to watch over:
Every point is here visible by a guard sitting at the point on the right, but if you move just arbitrarily close to another point, then it is no longer true that the second guard can see every point. So there is only one special point from which the whole gallery can be viewed. You could try and locate two people at the same point, but there is not another point in this gallery where everything is visible to one guard. Therefore, you would have to have more than one extra to help the person in the special point.
The next type of question that mathematicians started asking about these types of problems is what would happen if you allowed the guards to move around - which is rather more realistic. You can imagine there are two types of movement that the guards might be allowed to take part in: the first type are the so-called edge guards. These are people that just walk along the walls; they march backwards and forwards along particular walls and they survey everything in front of them that it is possible for them to see. The second type is more adventurous and they do not just walk around the edges, we will call them diagonal guards. This type of guard is going to walk across, say from one corner to another, so they have a different beat. You could ask how many of these edge guards would you need, how many diagonal guards would you need, how many guards of any sort would you need, and you can see you would obviously imagine you are going to get away with fewer guards if you allow them to move around than if you have them all sitting on a seat or you just have a stationary camera.
In 1981, a famous conjecture was made by Toussaint, who wanted to try and guess how many, say, edge guards you might need. He had realised that there was no hope of having a theorem which said ?this is the situation for all possible galleries?, because he knew that there were examples like the one we have just seen, which were in effect counter-examples to the conjecture he wanted to make. So he thought that the integer part, the whole number part, of a quarter n was going to be sufficient to guard any polygonal gallery if you allowed edge guards moving around, but he knew that these were counter-examples: there were these Christmas tree shaped types of galleries where if you move around in the 'branches' you cannot see the other parts of the gallery. So his hope was that you could make a little list of a finite number of special types of gallery, which you could just exclude by feat, and for all the rest, it would be the case that the integer part of [n/4], like the orthogonal gallery, would suffice. So far, that is unproven still, so if you want to have a go at that, you can.
One of the experts in this area is someone called Joseph O'Rourke, and he proved a slightly different result, that is quite nice; that the minimum number of mobile guards - be they edge or diagonal - sufficient and necessary to guard the polygon is given by our quantity integer part of [n/4]. So it just tells you that you are never going to be able to do better than that, so there might be a stronger theorem that tells you that there are particular cases where you are going to need more.
He also went on to show what might happen if you have galleries which have holes in them. So this is rather like there really could be a hole, but think of it as there being a large exhibit sitting in the middle of the gallery, which you cannot see through. So the guard located on one side cannot see the things that are immediately in the shadow of that sculpture. So there are regions that you have to take out, and this means you have got to have more guards. He proved a rather pretty result - you do not need to remember it, but the idea is just to get a feel for what types of results are possible.
So in this case, you carry out the triangulation a little differently. Here you are looking at 3 times the number of corners, plus 4, plus 4 times the number of holes, divided by 16, where the whole number part of this tells you how many guards are going to be necessary and sufficient to guard a gallery that is all right angles and it has got h holes in it.
[(3n+4+4h)/16] mobile guards are necessary and sufficient to guard orthogonal polygons with h holes.
So if we have 100 sides again, and we have got no holes, for example, then you are going to need 19 moving guards. This is a lot less than if you have guards which are stationary, when the integer part of n/4 is just 25. So you are able to get away with fewer guards when they are able to move around and do the work of rather more immobile guards.
Here is a picture of that type of case, the worst case if you like, where you need the full quota of guards that the formula tells you:
You have got these large sections taken out in each of the legs of this gallery, because there are where there are some huge opaque exhibits sitting, and you have got this odd, almost swastika-like, shaped gallery. It has got twenty sides, you have got four holes in it, and you are going to need five guards, using the formula we just came across, in order that they can see every point in the gallery, where they move around along the edges or across the diagonals.
Another interesting case is the orthogonal gallery that we mentioned earlier, but with rooms sticking out, so it is not all enclosed in a rectangle:
It is an interesting case: you have got little doorways into each of the rooms. In this case, you are going to need more than the whole number part of n/2. You can see why: that if you take the central room, then you can have somebody located in the doorway to the left, where they can watch the centre room and one of the side rooms. But none of the two side rooms have a common wall, so they are not all enclosed, and so you need an extra guard for each of them. So in this case, you have only got five rooms, but you need four guards. So this is a particularly bad scenario, and an expensive design.
You can work out a rather general result, that if a gallery like this has gotc corners, and it has got r rooms, and it has got h holes in the middle, then this is the number of guards that you are going to need:
[(2r +c - 2h - 4) / 4] guards
In our example, we have got twenty sides, four gaps and five rooms, so you might need four guards, and indeed we do.
So they are all the specific cases of that sort that have been looked at for gallery-type problems, but there is a whole collection of related problems that are inspired by this type of geometrical, mathematical problem.
Suppose you had a giant stage where you are running some pop concert or event, and you want to illuminate the stage, everywhere, and you know what the spread of the searchlight beam is from each of your searchlights. What is the smallest number of spotlights you could get away with to illuminate the whole of your stage? An alternative way of stating the problem is, if you are trying to guard a prison camp, what is the smallest number of cameras or searchlights that you might need to illuminate a particular area?
There are two problems of interest to the criminal fraternity, as it were, following from what we have just been looking at. We talked about mobile guards, but there is a sort of night watchman problem as well, which is rather more difficult computationally.
So suppose, at night, you have a watchman who is going to do his round of the gallery, and you want to find the shortest closed route, so that it ends at the same place that it begins, such that the watchman sees every point in the gallery at least once, so nothing is left unobserved in his beat. This is computationally quite hard.
There is an analogous sort of thief's problem: the thief is interested in the shortest path around the gallery that is not visible from particular points, so these may be the points where the night watchman sits or where a camera is located. Is there a path around the gallery that has that property? If you are designing the gallery, you want to try perhaps to make sure that there is no such path, that there is no path which could be unobserved by your array of cameras.
There are analogous problems which look at safaris and game parks. So if you make a trip around the game park where the animals are moving around what is the path that you should take so that you see all the animals of a particular species that you want to see?
So far, we have been looking at what you might call the inside problem: we are inside a building, and we want to look at all the walls and the interior. But there is an outside problem. Supposing we have a prison with straight walls, the analogous problem is in keeping an eye on the outside walls. How many cameras or guards do you need outside the prison so as to keep a watch on every one of the outside walls?
This problem involves a new mathematical function, which we usually call the ceiling function. If we have a quantity x, then the ceiling of x is the smallest integer that is at least as big as x. So it is like the next integer up if x is not a whole number. So the ceiling of pi is 4, but the ceiling of 3 is 3. So in this problem, what we discover is the number of corner guards - people located actually on the corners of the fortress - that are necessary and sufficient to guard the outside of the fortress is given by the ceiling of half the number of sides, or half the number of corners, the ceiling of n/2.
To take a rather trivial type of fortress, we can imagine a square one, with four sides and four corners. The ceiling of a half multiplied by four is two, so you need two guards to look after the outside of a square prison: the ceiling of 4/2 = 2. You would place them on corners diagonally opposite one another, so that each one would look after two different sides.
But if we have a more exotic-shaped prison, we might ask why we should put the guards always at the corners. Why don't we have the ability to locate them at any point? We will call those point guards. To take this example:
The dots outside of the triangle are the locations of point guards, and in this instance, these two guards are sufficient to see every point on the outside of this fortress. We can see that it is something of an inverse triangulation problem: you are here looking for the ceiling of the number of corners divided by three: the ceiling of n/3. You will remember that we had the integer part of a third times the number of corners for the art gallery problem on the inside; this time, we have got the ceiling of a third times the number of corners, for point guards - that is, when the guards can be anywhere and do not have to be at the corners - but we would need the ceiling of a half n if we located them at the corners. We can apply this to an example:
Here you have got seven sides. This means that if we put it into the ceiling of n/3, to get the ceiling of 7/3, where the ceiling of 2.333 is 3. This means that you would need three point guards, as is shown by the dots in the diagram, which mark the appropriate locations for these guards. But if you did it with corner guards, you would need four: so you would need the ceiling of 7/2, a ceiling of 3.5, which is four. So you do better by allowing your guards to sort of wander off than keep an eye on things from a distance.
We can now look at what I call the Prisoner Cell Block H Problem. If we consider a prison or fortress in the shape of an 'H', we can see that it is an orthogonal shape, just like we had an orthogonal gallery, where all the sides are at right angles. If there are n corners, then the number of corner guards that you are going to need looks like: 1 + the ceiling of n/4.
In this case, we have 12 sides on the H block, so you are going to need 1 plus the ceiling of 12 over 4, which is 1 plus 3 - so you are going to need 4 guards, and the locations are marked on the diagram 1 to 4. They are not the unique positions; there are several other positions where the four guards can operate. But this is a pretty economical way of looking after a structure that is got so many sides.
The last version of these types of problems that springs to mind is that you might be interested in combining keeping an eye on what is going on inside with keeping an eye on what is going on outside. We call this the Prison Yard Problem. You are here pulling together the results from the two sorts of mathematical studies we have looked at, because you want to see if you can locate guards in positions where they can do two jobs at once, so they can look inside and they can look outside at the same time. What we know is that this ceiling of n/2 corner guards is always going to be sufficient, and sometimes it is going to be necessary for a convex polygon with n corners.
If it is non-convex then you are going to have to take the integer part of a n/2. Why is that? Well, if it is convex, you remember, you just need one camera on the inside to keep an eye on everything, so one person at any corner looks after the inside. Therefore, you do not have to worry about the inside and you only have to worry about the outside. So the answer is the same as it is for the outside problem, the ceiling of n/2, and any of those guards on the corners of the outside can also keep an eye on the inside. So the problem is just the same as the outside problem when the fortress is a convex polygon, which means that it has not got any parts of the perimeter that point inwards. But if it is non-convex, then the inside problem is the hard bit, because if you put people on corners, where the non-convex V points inwards, anyone on one of those corners can keep an eye on the whole inside V. So when it is non-convex, it is the inside problem you have to worry about, and the integer part of n/2 guards to keep an eye on the inside will also be able to keep an eye on the outside.
So if we have, for example 101 sides, then you are going to need 51 people to keep an eye on the whole thing if it is convex, and 50 if it is non-convex. There are various other results that one can then prove about this sort of thing which are rather unexpected. So how many corner guards or point guards are going to be necessary if you have got a prison which is all right angles and it has got some holes taken out of it? So if you had 100 sides, for example, then 43 corner guards, or 34 point guards, is always going to be enough, regardless of the number of bits that you have as holes cut out of it.
With this whole class of problems, which are known as surveillance problems, or just Art Gallery Problems, I hope have shown you some of the things that you can learn just by thinking rather simply about straight lines in geometry. The key idea really in all these problems was this idea of triangulation, or dividing up the interior of the gallery into a series of disjoint triangles which cover every part of the gallery. You know what the situation is inside a single triangle, how many cameras that you need - just the one - and then the question is: which of the corners of the triangles do you employ to locate your camera?
There is something of a mini-industry of problems of this sort, which move up into three dimensions, as I indicated with the spotlights. You are then looking in a third dimension: is it possible to see everything in a volume, not just on a flat surface?
The moving guards add a new movable ingredient, and you might be able to think of other generalisations you could make. An obvious one is not to have the sides of the galleries as straight lines. So if the walls are curved, you have a new ingredient. There is also another aspect of the gallery problem that might have occurred to you, and that is, why don't you make some of your walls mirrors, so you can see, by reflection, what is going on in another part of the gallery?
Then you have a sequence of problems which have become known as Illumination Problems in mathematics: that if in one place with barriers and walls of different shapes, suppose they were all mirrors. Suppose all the walls are mirrored, and you turned on a light in one point. Under what conditions would the light illuminate the whole of the inside of the gallery? When do the reflections bounce around and eventually visit every single point of the gallery? Even for galleries with walls which are straight and all mirrored, there is a complicated sequence of results, rather like the ones I have talked about, the conditions under which you can and cannot illuminate the whole of the gallery.
So these are problems in what one might call combinatorial geometry. These problems are a branch of mathematics that is really relatively new - it is about thirty years old. I hope you can see it is rather fascinating because rather simply stated problems turn out to have unexpected, general conclusions, and when you begin thinking in a particular direction, you can really make sense of a very large number of problems.
©Professor John Barrow, Gresham College, 12 January 2008