Inevitable patterns in mathematics. Some topics in Ramsey theory

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1 Inevitable patterns in mathematics Some topics in Ramsey theory These notes cover an ambitious guess at what we might get through in the course 1. It is quite likely that we will not manage it all in which case you may be interested in reading the remainder in your own time. 1 Misleading patterns These notes introduce some of the ideas of the branch of mathematics known as Ramsey theory. Ramsey theory is sometimes described by the slogan complete disorder is impossible or (less catchily) as the study of unavoidable regularity in large structures. Like much of mathematics it is concerned with the idea of explaining patterns. Before we get down to some serious mathematics let s look at a few examples which in some sense motivate what follows. Although this is a pure mathematical subject I have chosen concrete real-world examples. How might you explain the following patterns? Birthday coincidences: You notice that of the 23 people in your class there are two who share a birthday. When you check the same thing in other classes of the same size the same thing happens in more than half of them. In classes of 50 people there always seems to be a duplicated birthday. 2. Torah codes: In 1994 Witztum, Rips, and Rosenberg published a claim that the Hebrew text of the book of Genesis contains coded information about mediaeval rabbis. Specifically they found equidistant letter 1 The course was a one-day taster course for sixth form students held at QM on 3 July 1

2 sequences (that is words formed by taking for example every 20th letter starting from a particular point) which apparently spelt out the names and dates of birth of famous rabbis who lived long after the text was written. 3. Friends and strangers: You notice that in every group of 6 people it is possible to find either 3 mutual friends or 3 mutual strangers. In every group of 18 people it is possible to find either 4 mutual friends or 4 mutual strangers The first of these is so surprising that it is tempting to conclude you must have made a mistake or there has been an amazing coincidence. Various claims have been made for the origin of the Torah codes. Your first guess for the third might be something about the dynamics of social interactions. In fact our surprise at the first of these patterns is more to do with our poor intuition for probability. It is easy to check that with 23 people the chance of there being a repeated birthday is ( ) ( ) ( ) ( ) ( ) So it is completely believable that this happens in more than half of our classes. With 50 people the probability of there being a repeated birthday is around 0.97 so it is no surprise that this keeps happening. For the Torah codes a similar (but more subtle) probabilistic phenomenon is at work. In 1999 McKay, Bar-Natan, Bar-Hillel, and Kalai showed that the appearance of these messages is not statistically significant. Amusingly they demonstrated this by finding equidistant letter sequences predicting famous assassinations in Moby Dick. Brendan McKay also found his own name, date of birth and place of residence hidden in the text of War and Peace and remarked The lesson to be drawn from this paper is clear enough. Anyone with the skill and the perseverance can make ELS [equidistant letter sequence] experiments that seem to show remarkable results. In this paper we found a significance level well below 1/1000 from a single name and a single date. Did it happen by chance? Yes! 2

3 A mathematical approach to the Torah code claims is described on Brendan McKay s webpage at The third example is slightly different. Unlike the birthday coincidences and the Torah codes this pattern is not just more likely than we would guess but inevitable. Exploring this sort of phenomenon is what today s course (and these notes) is about. We will investigate the friends and strangers example in the next section. 2 Towards Ramsey s theorem Firstly let s formalise the problem a little as follows: represent the people by points ( blobs ), join two points with a red line if they are friends, join two points with a blue line if they are strangers. The observation becomes: Theorem 2.1 (The Party theorem). If we colour the lines between all pairs from 6 points with red and blue then we always get either a red triangle or a blue triangle. Let s prove this. Proof. Pick one of the points and call it a. Point a must have either 3 red or 3 blue lines coming out of it (if not then it has at most 2 of each colour so 4 lines in total but there are 5 other points). Suppose that a has 3 red lines joining it to points x, y and z. If any of the lines xy, xz and yz are red then we have a red triangle. The only other possibility is that they are all blue and so xyz is a blue triangle. The other possibility is that a has 3 blue lines joining it to points x, y and z. If any of the lines xy, xz and yz are blue then we have a blue triangle. The only other possibility is that they are all red and so xyz is a red triangle. 3

4 More generally what can we say if we have more people? If the number of people is large enough can we always find a large set of mutual friends or mutual strangers? Let s define R(k, l) to be the smallest integer n such that whenever we have a group of n people there must be either k mutual friends or l mutual strangers. 2 Equivalently, in the colouring language. Let R(k, l) be the smallest integer n such that whenever the lines between all pairs from n points are coloured with red and blue then we always get either a set of k points any two of which are joined by a red line or a set of l points any two of which are joined by a blue line. It is not at all obvious that R(k, l) exists for all k and l but we have proved that R(3, 3) 6. This definition takes some getting your head round. The first few problems should help you to digest it. 3 Ramsey s theorem Our task in this section is to prove the following: Theorem 3.1 (Ramsey 1930). The number R(k, l) exists for all k and l. Recall that this means that if I want to guarantee the existence of either a set of k points all connected by red lines or a set of l points all connected by blue lines then I can do this simply by starting with enough points. This is an example of the sort of inevitable patterns that the course title refers to. I do not need to insist on a particular pattern or structure to the way I colour the lines: simply having enough points guarantees the existence of a nice pattern (a bunch of points joined by lines of the same colour) somewhere. Proof. Let s start gently by showing that R(3, 4) exists. I claim that having 10 points is sufficient to give either 3 points connected by red lines or 4 connected by blue lines. 2 The R is in honour of Ramsey. 4

5 Pick a point a. It has either 4 red lines or 6 blue lines coming out of it (if not then there are at most 3 reds and 5 blues but there are 9 other points). If there are 4 red lines, joining a to w, x, y, z, then either one of wx, wy, wz, xy, xz, yz is red giving a red triangle with a or they are all blue giving 4 points joined by blue lines. If there are 6 blue lines joining a to u, v, w, x, y, z then since R(3, 3) 6 there must be either a red triangle in u, v, w, x, y, z (and then we are done) or a blue triangle in u, v, w, x, y, z which together with a makes 4 points joined by blue lines. We conclude that R(3, 4) 10 and (by an exercise) R(4, 3) 10 as well. Essentially the argument shows that R(3, 4) exists provided that R(2, 4) and R(3, 3) exist. What about R(4, 4)? I claim that 20 points suffice. Pick a point a. It has either 10 red lines or 10 blue lines coming out of it (if not then there are at most 9 reds and 9 blues but there are 19 other points). If there are 10 red lines then look among the points joined to a by red lines. Since there are at least 10 of these and R(3, 4) is at most 10 we must have either a red triangle or 4 points joined by blue lines among these 10. In the first case we have 4 points joined by red lines (the red triangle together with a); in the second case we are done immediately. If there are 10 blue lines then the same argument with red and blue exchanged does the trick. Specifically, since there are at least 10 points joined to a with blue lines and R(4, 3) is at most 10 we must have either 4 points joined by red lines or a blue triangle among these 10. In the first case we are done immediately; in the first case we have 4 points joined by blue lines (the blue triangle together with a); We conclude that R(4, 4) 20. 5

6 We have shown that R(4, 4) exists using the fact that R(3, 4) and R(4, 3) exist. Similarly to show that R(5, 5) exists we would need the fact that exist. Which follows from the fact that exist. Which follows from the fact that R(5, 4), R(4, 5) R(5, 3), R(4, 4), R(3, 5) R(5, 2), R(4, 3), R(3, 4), R(2, 5) exist. By an exercise R(5, 2) and R(2, 5) certainly exist. The existence of the others follows from the fact that R(4, 2), R(3, 3), R(2, 4) exist. By an exercise R(4, 2) and R(2, 4) certainly exist. The existence of R(3, 3) follows from the fact that R(3, 2), R(2, 3) exist. By the same exercise these certainly exist and so we are done. Hopefully you can see how this same argument repeated sufficiently often shows that R(k, l) exists for any k and l. So we have proved Ramsey s theorem. Those of you who know what is meant by a proof by induction will be able to write the previous argument out more slickly. If you don t know what induction means then why not ask one of your teachers? 6

7 4 Huge numbers ridiculous bounds We have shown that R(k, l) always exists but what about determining its exact value? The values we have discussed (or looked at in the exercises) are: R(2, n) = n R(3, 3) = 6 R(3, 4) = 9 R(4, 4) = 18 In addition to this it is known that: R(3, 5) = 14 R(3, 6) = 17 R(3, 7) = 23 R(3, 8) = 28 R(3, 9) = 36 R(4, 5) = 25 The last three of these required substantial use of a computer. Given this we turn our attention to giving bounds on these numbers. It is known for example that: 43 R(5, 5) R(6, 6) R(7, 7) 540 At first sight it is quite ridiculous that R(5, 5) is not known exactly. However Exercise 1.5 does gives some feel for why R(5, 5) would take a lot of work to determine. Analysing the proof of Theorem 3.1 gives the following 7

8 Theorem 4.1. From which it follows that Theorem 4.2. R(k, l) R(k 1, l) + R(k, l 1) R(k, l) 2 k+l Proof of Theorem 4.2. We will use induction on k + l to prove this. If you haven t come across this then you may want to skip the proof. Firstly, it is clear that R(k, 2) = k 2 k+2 and R(2, l) = l 2 l+2. Now by Theorem 4.1 R(k, l) R(k 1, l) + R(k, l 1) 2 k 1+l + 2 k+l 1 = 2 2 k+l 1 = 2 k+l Where the second inequality holds since by the induction hypothesis we may assume the result holds for the two terms on the righthand side. So in the most natural case that k = l we have R(k, k) 4 k. To give a lower bound for R(k, l) the obvious thing to do is to find some clever colouring which has no large sets of points joined by lines of the same colour. In fact the best known lower bound takes a completely different approach. Imagine that you wanted to show that R(4, 4) 7. Of course you would do this by finding a colouring of the lines between 6 points with no 4 points all joined by lines of the same colour but imagine (for a moment) that you are not able to find one. What other approach might work? There are 15 pairs of points involved so 15 lines to colour. In total this gives 2 15 possible colourings. There are 15 possible sets of 4 points from the 6 (count them!). If I choose one of these there are 2 10 ways to colour the lines in such a way that I only use a single colour on my chosen set of 4 points (2 choices for the colour on that set and 9 more lines to colour). 8

9 So the number of colourings which do contain 4 points joined by lines of the same colour is at most (If I list the sets of 4 points then 2 10 colourings use only one colour on the first one, 2 10 use only one colour on the second one and so on for all 15 sets of 4 points.) This number is less that the total number of colourings so there must be some colourings which do not contain 4 points all joined by lines of the same colour. This argument gives a rather feeble result for R(4, 4) but in general it is much more effective than the best known constructions. Theorem 4.3 (Erdős 1947). If k 5 then R(k, k) 2 k. Proof. There are ( n k) different sets of k points 3. Let s list them as S 1, S 2,..., S ( n k). Given a colouring let s call a set S monochromatic if all points in it are joined by lines of the same colour. Plainly the number of colourings in which there is a monochromatic set of k points is less than or equal to (The number of colourings in which S 1 is monochromatic) + (The number of colourings in which S 2 is monochromatic) + (The number of colourings in which S 3 is monochromatic) (The number of colourings in which S ( n is monochromatic). k) (If I have such a colouring then I have certainly counted it in this sum. We have an inequality because if the colouring contains more than one monochromatic set of k points then it is overcounted.) Each term in the above sum is the number of colourings in which a fixed set of k points is monochromatic. There are 2 (n 2) ( k 2)+1 such colourings. So the number of colourings containing a monochromatic set of k points is at most ( ) n 2 (n 2) ( k 2)+1. k 3 See the appendix on counting subsets if you don t know what this means. 9

10 If we choose n and k so that this is less than the total number of colourings then we will be done. Indeed, since the number of colourings with a monochromatic set of k points will then be less than the total number of colourings there must be at least one colouring without a monochromatic set of k points. This is precisely what we need to prove the lower bound. The total number of colourings of the lines between n points is So we need Equivalently, Now, ( ) n 2 (k 2)+1 k 2 (n 2). ( ) n 2 (n 2) ( k 2)+1 < 2 (n 2). k ( ) n 2 (k 2)+1 < 1. k n(n 1)(n 2)... (n k + 1) 2 k(k 1) k nk k k 2 2k 2. So if n 2 k = 2 k 2 we have ( ) n 2 (k 2)+1 2 k2 2 k+2 k2 2 + k 2 k 2 2 k 2 < 1 since k 5. Hence the result. This argument is usually phrased in terms of picking a random colouring and showing that the probability that it has a monochromatic set of k points is strictly less than 1. As such it was one of the starting points for the study of random graphs, an important area of mathematics both in its own right and because of its applications to other parts of mathematics and beyond. 10

11 In fact there are many examples where ideas and techniques developed in the context of Ramsey theory have gone on to influence other parts of mathematics. Referring to the problem of improving the bounds for R(k, k), the Fields medal winner (a Fields medal is sometimes considered the mathematical equivalent of a Nobel prize) Timothy Gowers remarked 4 I consider this to be one of the major problems in combinatorics and have devoted many months of my life unsuccessfully trying to solve it. And yet I feel almost embarrassed to write this, conscious as I am that many mathematicians would regard the question as more of a puzzle than a serious mathematical problem.[...] A better bound seems to demand a more global argument, involving the whole graph, and there is simply no adequate model for such an argument in graph theory. Therefore, a solution to this problem is almost bound to introduce a major new technique. In fact the bounds: 2 k R(k, k) 4 k are close to being the best known. Very recently Conlon has improved the upper bound slightly. The exact result is rather hard to write down and the proof is long and difficult. Remarkably the best lower bound is proved by a probabilistic argument (a slightly refined version of the one we saw). So although we know that colourings of 2 k points with no monochromatic set of k points exist it is not known how to construct such a colouring. 4 The quote is from The two cultures of mathematics available at wtg10/papers.html 11

12 5 Inevitable patterns in sets of numbers In this section we look at the idea of inevitable patterns in a different context. As was mentioned earlier, the general philosophy of Ramsey theory is sometimes expressed as Complete disorder is impossible or more precisely in any large enough piece of disorder you can find a small piece of order. In the previous sections the disorder is the colouring and the piece of order is a monochromatic set of k points. Our result was that if we have enough points then any colouring of the lines between then is forced to contain a monochromatic set of k points. The integers form another natural setting for results of this type. If I take the integers {1, 2, 3,..., n} and colour each red or blue then what patterns must inevitably appear? 5.1 Towards van der Waerden s theorem One possible pattern that we might look for is 3 equally spaced integers (for example 12, 14, 16 or 25, 32, 39) all of the same colour. We call such a pattern an arithmetic progression (AP) of length 3. Let s denote by W (3) the smallest number n such that whenever {1, 2,..., n} is coloured with red and blue then there is always an AP of length 3 in one of the colours. Theorem 5.1. W (3) 325 That is to say, whenever {1, 2,..., 325} is coloured with red and blue then there is always an AP of length 3 in one of the colours Proof. We will assume that we have a colouring without an AP of length 3 in either of the colours and show that this is impossible. Split {1, 2, 3,..., 325} into 65 blocks each of 5 consecutive integers: {1,..., 5}, {6,..., 10}, {11,..., 15},..., {321,..., 325}. There are 2 5 = 32 ways to colour a block and so among the first 33 blocks there must be 2 which are identically coloured. That is we have {a, a + 1,..., a + 4} and {b, b + 1,..., b + 4} identically coloured. Take c so that a, b and c are equally spaced (form an AP). Note that the block {c, c+1,..., c+4} is still within {1, 2,..., 325}. 12

13 Now let s look at the colouring of the block containing a (I ll call this block A from now on). In a block of 5 the first 3 integers must include 2 of the same colour, red say. Having 2 red integers but no red AP of length 3 forces a further integer to be blue (for example if 1 and 3 are red then 5 must be blue). The possibilities are Case 1: r r b?? Case 2: r? r? b Case 3:? r r b? Now looking at block B which is identically coloured we have in the first case r} r {{ b??} } r r {{ b??}......?}? {{??} Block A Block B Block C If c + 2 (marked above) is red then a, b + 1, c + 2 is a red AP; if it is blue then a + 2, b + 2, c + 2 is a blue AP. So either way we have a monochromatic AP of length 3. In the second case we have r }? {{ r? } b } r? {{ r? } b......?}? {{?? } Block A Block B Block C and in the third case we have?} r {{ r b?}......?} r {{ r b?}......?}? {{??} Block A Block B Block C Either way the integer marked completes a monochromatic AP of length 3 whichever colour is used on it. If we had only set out to show that an AP of length 3 in one colour must exist then we could have come up with an argument that works for a much smaller set (see problems 2.1 and 2.2). However the method shown is probably shorter and has the advantage that it will generalise to prove much more. Extending the result to show that an AP of length 4 (that is 4 equally spaced integers) exists is harder. However it is not too difficult to extend it 13

14 to show that if 3 colours are used then an AP of length 3 inevitably appears provided that we colour {1, 2,..., n} for some huge n. Let s denote by W 3 (3) the smallest number n such that whenever {1, 2,..., n} is coloured with red, blue and green then there is always an AP of length 3 in one of the colours. Theorem 5.2. W 3 (3) 7( )(2 3 7( ) + 1). If {1, 2, 3,..., 7( )(2 3 7( ) + 1)} is coloured with red, blue and green then there is always an AP of length 3 in one of the colours. The proof is based on the proof of the previous theorem. Make sure that you are completely comfortable with that before tackling it. Proof. Again we will assume that we have a colouring without an AP of length 3 in any of the colours and show that this is impossible. Let n = 7( )(2 3 7( ) + 1) and m = 7( ). Split {1, 2, 3,..., n} into 2 3 7( ) +1 blocks each of 7( ) = m consecutive integers. There are 3 7( ) ways to colour a block and so among the first 3 7( ) + 1 blocks there must be 2 which are identically coloured. That is we have {a, a + 1,..., a + m 1} and {b, b + 1,..., b + m 1} identically coloured. Take c so that a, b and c are equally spaced (form an AP). Note that the block {c, c + 1,..., c + m 1} is still within {1, 2,..., n}. Again we will call these block A, block B and block C. Now let s look at how block A is coloured. Split it up into blocks each of 7 consecutive integers. There are 3 7 ways to colour a block and so among the first blocks there must be 2 which are identically coloured. That is we have {a+x, a+x+1,..., a+x+6} and {a+y, a+y+1,..., a+y+6} identically coloured. Take z so that a + x, a + y and a + z are equally spaced (form an AP). Note that the block {a + z, a + z + 1,..., a + z + 6} is still within block A. We will call these block A+X, block A+Y and block A+Z. Among the first 4 integers in block A + X there must be 2 of the same colour (red say). There are several possibilities but we will just look at the one where a + x and a + x + 3 are both red (the other cases come out in a similar way). Now a + x + 6 must be a colour other than red (blue say). If it were red than a + x, a + x + 3, a + x + 6 would be a red AP. 14

15 Also a + z + 6 must take the third colour green. If it were red then a + x, a + y + 3, a + z + 6 would be a red AP (remembering that blocks A + X and A + Y are identically coloured so a + y + 3 is red because a + x + 3 is). If it were blue then a + x + 6, a + y + 6, a + z + 6 would be a blue AP. Now let s look ahead to block C. What colour can c + z + 6 be? If it is red then a + x, b + y + 3, c + z + 6 would be a red AP (remembering that blocks A and B are coloured identically so b + y + 3 is red because a + y + 3 is). If it is blue then a + x + 6, b + y + 6, c + z + 6 would be a blue AP. If it is green then a + z + 6, b + z + 6, c + z + 6 would be a green AP. We conclude that however {1, 2, 3,..., n} is coloured there must be an AP of length 3 in one of the colours. As before the bound is extremely bad. The true answer is that W 3 (3) = 27. However, for a larger number of colours it is a challenge to show any bound on W r (3) (the smallest n which guarantees an AP of length 3 in one colour when {1, 2, 3,..., n} is coloured with r colours) so we should not be too put off by the huge numbers. The argument you have seen contains all the ideas to prove such a bound although the notation gets tricky and the bounds are horrendous. In fact the generalisation to arbitrarily many colours and length of AP is also true. This is a result of van der Waerden. Theorem 5.3 (van der Waerden 1927). Given r and k there is an n such that whenever {1, 2, 3,..., n} is coloured with r colours we are guaranteed to be able to find an AP of length k. We have shown is that this result is true for r = 2, k = 3 and for r = 3, k = 3. The best known bounds for W (k) (the smallest number needed to guarantee an AP of length k when 2 colours are used) are roughly: 2 k 8k W (k) k. The upper bound (due to Gowers) looks ridiculous but it was a major improvement on the best previously known bound. 15

16 5.2 Density Versions In fact W (3) = 9 so however {1, 2, 3,..., 9} is coloured with red and blue we can always find an AP of length 3 in one of the colours. Which colour could it be in? A tempting guess would be that the colour which appears most often always contains an AP of length 3 but this is false. Suppose that 1, 2, 6, 7, 9 are coloured red and 3, 4, 5, 8 are blue then there are more reds than blues but the red numbers do not contain an AP of length 3. However, our intuition is right in that if n is large enough then one colour being used many times does guarantee a long AP. This is a deep result of Szemerédi. Before stating Szemerédi s theorem let s give a specific example. If I take a tiny number (1/10000 say) and a long length (30 say) then there is some n such that every set of n/10000 integers chosen from {1, 2, 3,..., n} contains an AP of length 30. Theorem 5.4 (Szemerédi 1975). For any δ > 0 and integer k there is an n such that every subset of δn integers from {1, 2, 3,..., n} contains an AP of length k. This theorem can be thought of as a density version of van der Waerden. It has several proofs, all of them long and difficult. The question of how large n must be is not well understood. A related beautiful result of Green and Tao from 2004 is that the set of prime numbers contain APs of any length k. 6 Conclusion Hopefully I have convinced you that a simple starting point like the Party theorem can lead to some deep mathematics. Most of this theory was developed in the past hundred years and the most recent results I have mentioned are only a few years old. There are many open problems: determining R(5, 5), improving the bounds on R(k, k) and W (k), giving a constructive lower bound on R(k, k) which is as good as the probabilistic one, and generalising and sharpening Szemeredi s theorem which although simple to state are hard and important. If progress is made on any of these it is likely to have a profound effect on the subject. Ramsey theory is usually taught in the second or third year of undergraduate maths courses as part of modules called things like Discrete Mathemat- 16

17 ics, Combinatorics or Graph Theory. It is also an important research area. If you have found this interesting then two articles at a similar level which are available on the web are given below: Ben Green Ramsey Theory at the IMO available at bjg23/papers/gazette.ps Imre Leader Friends and Strangers : available at ( Plus is an online maths magazine which is well worth looking at.) Anyone interested in the people behind the maths should take a look at: history/biographies/ramsey.html history/biographies/erdos.html If you have any questions or comments on the course or these notes then I would be happy to hear from you. My address is r.johnson@qmul.ac.uk For more information on maths at Queen Mary take a look at and in particular 17

18 7 Appendix: Counting Subsets If I have n people how many lines do we need to join all pairs? For small n you could just count (3 when n = 2, 6 when n = 4, 10 when n = 5) but in general it would be useful to have a formula. Each line joins a pair of points. There are n ways of picking the first of the pair. For each of these choices there are n 1 ways of picking the second of the pair (we cannot choose the first point again but each of the others could be chosen). This makes n (n 1) choices overall. However we have counted each pair twice (once as (a, b) and once as (b, a) say) and so the number of pairs is n(n 1) 2 This number is usually written as ( n 2) (pronounced n choose 2 ). We also needed a formula for the number of sets of size k from n points. In a similar way to the above we can choose such a set one point at a time. There are n choices for the first point in it. For each of these there are n 1 choices for the second, n 2 for the third, and so on down to n k + 1 for the kth. This makes n (n 1) (n 2) (n k + 1) choices overall. However we have counted each set many times. In fact you should be able to convince yourself that each has been counted k (k 1) (k 2) 2 1. (For example the set {1, 2, 3} appears as (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1) 6 ways in all.) It follows that the number of sets of size k from n points is n (n 1) (n k + 1). k (k 1) 1 This number is usually written as ( n k) (pronounced n choose k ). 18

19 Inevitable patterns in mathematics Problems There are probably too many problems to think about in the time you have. I encourage you to spend a serious amount of time thinking about a few problems rather than taking a superficial look at all of them. Some of the problems are quite open-ended so there is not necessarily a single correct answer or correct approach. If you look at them again after today and have any comments or questions then feel free to get in touch (my address is r.johnson@qmul.ac.uk). You are welcome to work individually, in pairs, or in larger groups. Helpers will be circulating to provide guidance and suggestions but they will not just tell you how to do the problems! 8 Points and Lines To do these problems you will need the definition of Ramsey numbers that I talked about. That is, R(k, l) is the smallest integer n such that whenever the lines between n points are coloured red and blue there is always either a set of k points joined by red lines or a set of l points joined by blue lines. Problem 8.1. What is the value of R(3, 2)? R(4, 2)? R(n, 2)? Problem 8.2. If I tell you that R(a, b) exists and give you its value what can you say about R(b, a)? Problem 8.3. We showed that R(3, 3) 6 (convince yourself that this is indeed the content of the Party theorem). What would you need to do to show that R(3, 3) = 6? Show that R(3, 3) = 6. 19

20 Problem 8.4. Play the following game with the person next to you. Start with 6 blobs on the page. Take it in turns to join two unconnected blobs with one of you always using a red pen and the other always using a blue pen. The first person to make a triangle of their own colour loses the game. What does the fact that R(3, 3) = 6 tell you about this game? Can you come up with some good strategies for playing the game? If you enjoyed problem 1.4 but do not have a ready supply of friends then take a look at: A variation is to allow each player to draw more than one line at each turn. Problem 8.5. A (bad) way to prove that R(3, 3) 6 would be to list all possible red/blue colourings for 6 points and check that each contains either a red or a blue triangle. How many colourings would you need to write down to do this? How many colourings would you need to write down to prove that R(5, 5) 48 by this method? Problem 8.6. Find the exact value of R(3, 4) (or at least find the best bounds you can). What about R(4, 4)? Problem 8.7. Show that if the lines between R(3, 6) points are coloured with red, blue and green there must be a triangle in one of the three colours. Having done this can you formulate and prove an extension of Ramsey s theorem to three colours? What about r colours? Problem 8.8. Show that in any list of 6 distinct integers it is possible to find 3 in increasing order or 3 in decreasing order. The 3 do not have to be consecutive so (1, 4, 2, 5, 3, 6) contains the increasing sequence (1, 2, 3). Is the same thing true for a list of 5 distinct integers? Is the same thing true for a list of 4 distinct integers? What if you want to be able to find a longer increasing/decreasing sequence? 20

21 9 Integers Problem 9.1. What do you need to do to find a lower bound for W (3)? What is the best lower bound for W (3) you can find? Remember that W (3) is the smallest n for which any colouring of {1, 2, 3,..., n} with red and blue is guaranteed to contain an AP of length 3. Problem 9.2. Improve the upper bound of 325 on W (3) as much as you can. Problem 9.3. Denote by W (k) the smallest integer n for which any colouring of {1, 2, 3,..., n} with red and blue is guaranteed to contain an AP of length k. Give some lower bound for W (4), W (5), W (k). Problem 9.4. What is the largest subset of {1, 2, 3,..., 9} which contains no AP of length 3. Problem 9.5. What is the largest subset of {1, 2, 3,..., 10} which contains no AP of length 3. Problem 9.6. Complete the proof of Theorem 5.2 by dealing with the other possibilities for having 2 red numbers among the first 4 of a group of 7. [Hint: If possible do this without listing all the cases. As a warm up you could try rewriting the proof of Theorem 5.1 without splitting into 3 cases.] Problem 9.7. Find a set of the lines between n points which contains more than half of all possible lines but contains no triangle. Explain why this shows that there is no density analogue of Ramsey s theorem. Problem 9.8. If you re feeling bold try to extend the proofs of the existence of W (3) and W 3 (3) to show that W r (3) exists. That is for every r there is some n such that whenever {1, 2, 3,..., n} is coloured with r colours there is an AP of length 3 in one of the colours. 21