Inevitable patterns in mathematics

Some topics in Ramsey theory

These notes cover an ambitious guess at what we might get through inthe course1. It is quite likely that we will not manage it all in which case youmay 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 knownas Ramsey theory. Ramsey theory is sometimes described by the slogan“complete disorder is impossible” or (less catchily) as “the study of unavoid-able regularity in large structures”. Like much of mathematics it is concernedwith the idea of explaining patterns.

Before we get down to some serious mathematics let’s look at a few ex-amples which in some sense motivate what follows. Although this is a puremathematical subject I have chosen concrete real-world examples. How mightyou explain the following patterns?

1. Birthday coincidences: You notice that of the 23 people in your classthere are two who share a birthday. When you check the same thing inother classes of the same size the same thing happens in more than halfof them. In classes of 50 people there always seems to be a duplicatedbirthday.

2. Torah codes: In 1994 Witztum, Rips, and Rosenberg published a claimthat the Hebrew text of the book of Genesis contains coded informa-tion about mediaeval rabbis. Specifically they found equidistant letter

1The course was a one-day taster course for sixth form students held at QM on 3 July2008.

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sequences (that is words formed by taking for example every 20th letterstarting from a particular point) which apparently spelt out the namesand dates of birth of famous rabbis who lived long after the text waswritten.

3. Friends and strangers: You notice that in every group of 6 people it ispossible to find either 3 mutual friends or 3 mutual strangers. In everygroup of 18 people it is possible to find either 4 mutual friends or 4mutual strangers

The first of these is so surprising that it is tempting to conclude you musthave made a mistake or there has been an amazing coincidence. Variousclaims have been made for the origin of the Torah codes. Your first guess forthe 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 ourpoor intuition for probability. It is easy to check that with 23 people thechance of there being a repeated birthday is

1−(

365

365

) (364

365

) (363

365

) (362

365

). . .

(343

365

)≈ 0.507.

So it is completely believable that this happens in more than half of ourclasses. With 50 people the probability of there being a repeated birthday isaround 0.97 so it is no surprise that this keeps happening.

For the Torah codes a similar (but more subtle) probabilistic phenomenonis at work. In 1999 McKay, Bar-Natan, Bar-Hillel, and Kalai showed thatthe appearance of these “messages” is not statistically significant. Amusinglythey demonstrated this by finding equidistant letter sequences predictingfamous assassinations in Moby Dick. Brendan McKay also found his ownname, date of birth and place of residence hidden in the text of War andPeace and remarked

“The lesson to be drawn from this paper is clear enough. Any-one with the skill and the perseverance can make ELS [equidis-tant letter sequence] experiments that seem to show remarkableresults. In this paper we found a significance level well below1/1000 from a single name and a single date. Did it happen bychance? Yes!”

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A mathematical approach to the Torah code claims is described on Bren-dan McKay’s webpage at http://cs.anu.edu.au/~bdm/dilugim/torah.html

The third example is slightly different. Unlike the birthday coincidencesand the Torah codes this pattern is not just more likely than we would guessbut inevitable. Exploring this sort of phenomenon is what today’s course(and these notes) is about. We will investigate the friends and strangersexample 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 allpairs from 6 points with red and blue then we always get either a red triangleor 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 notthen it has at most 2 of each colour so 4 lines in total but there are 5other points).

• Suppose that a has 3 red lines joining it to points x, y and z. If any ofthe lines xy, xz and yz are red then we have a red triangle. The onlyother 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, yand z. If any of the lines xy, xz and yz are blue then we have a bluetriangle. The only other possibility is that they are all red and so xyzis a red triangle.

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More generally what can we say if we have more people? If the numberof people is large enough can we always find a large set of mutual friends ormutual strangers?

Let’s define R(k, l) to be the smallest integer n such that whenever wehave a group of n people there must be either k mutual friends or l mutualstrangers.2

Equivalently, in the colouring language. Let R(k, l) be the smallest integern such that whenever the lines between all pairs from n points are colouredwith 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 haveproved that R(3, 3) ≤ 6.

This definition takes some getting your head round. The first few prob-lems 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 eithera set of k points all connected by red lines or a set of l points all connectedby 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 titlerefers to. I do not need to insist on a particular pattern or structure to theway I colour the lines: simply having enough points guarantees the existenceof 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 having10 points is sufficient to give either 3 points connected by red lines or 4connected by blue lines.

2The R is in honour of Ramsey.

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• 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 otherpoints).

• 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 allblue 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) ≤ 6there must be either a red triangle in u, v, w, x, y, z (and then we aredone) or a blue triangle in u, v, w, x, y, z which together with a makes4 points joined by blue lines.

• We conclude that R(3, 4) ≤ 10 and (by an exercise) R(4, 3) ≤ 10 aswell.

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 outof it (if not then there are at most 9 reds and 9 blues but there are 19other points).

• If there are 10 red lines then look among the points joined to a by redlines. Since there are at least 10 of these and R(3, 4) is at most 10 wemust have either a red triangle or 4 points joined by blue lines amongthese 10. In the first case we have 4 points joined by red lines (the redtriangle together with a); in the second case we are done immediately.

• If there are 10 blue lines then the same argument with red and blueexchanged does the trick. Specifically, since there are at least 10 pointsjoined to a with blue lines and R(4, 3) is at most 10 we must have either4 points joined by red lines or a blue triangle among these 10. In thefirst case we are done immediately; in the first case we have 4 pointsjoined by blue lines (the blue triangle together with a);

• We conclude that R(4, 4) ≤ 20.

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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

R(5, 4), R(4, 5)

exist.Which follows from the fact that

R(5, 3), R(4, 4), R(3, 5)

exist.Which follows from the fact that

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 thatR(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’stheorem.

Those of you who know what is meant by a proof by induction will beable to write the previous argument out more slickly. If you don’t know whatinduction means then why not ask one of your teachers?

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4 Huge numbers ridiculous bounds

We have shown that R(k, l) always exists but what about determining itsexact 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) ≤ 49

102 ≤ R(6, 6) ≤ 165

205 ≤ 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 lotof work to determine.

Analysing the proof of Theorem 3.1 gives the following

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Theorem 4.1.R(k, l) ≤ R(k − 1, l) + R(k, l − 1)

From which it follows that

Theorem 4.2.R(k, l) ≤ 2k+l

Proof of Theorem 4.2. We will use induction on k + l to prove this. If youhaven’t come across this then you may want to skip the proof.

Firstly, it is clear that R(k, 2) = k ≤ 2k+2 and R(2, l) = l ≤ 2l+2. Nowby Theorem 4.1

R(k, l) ≤ R(k − 1, l) + R(k, l − 1)

≤ 2k−1+l + 2k+l−1

= 2× 2k+l−1

= 2k+l

Where the second inequality holds since by the induction hypothesis we mayassume 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) ≤ 4k.

To give a lower bound for R(k, l) the obvious thing to do is to find someclever colouring which has no large sets of points joined by lines of the samecolour. In fact the best known lower bound takes a completely differentapproach.

Imagine that you wanted to show that R(4, 4) ≥ 7. Of course you woulddo this by finding a colouring of the lines between 6 points with no 4 pointsall joined by lines of the same colour but imagine (for a moment) that youare not able to find one. What other approach might work?

• There are 15 pairs of points involved so 15 lines to colour. In total thisgives 215 possible colourings.

• There are 15 possible sets of 4 points from the 6 (count them!). If Ichoose one of these there are 210 ways to colour the lines in such a waythat I only use a single colour on my chosen set of 4 points (2 choicesfor the colour on that set and 9 more lines to colour).

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• So the number of colourings which do contain 4 points joined by linesof the same colour is at most 15 × 210. (If I list the sets of 4 pointsthen 210 colourings use only one colour on the first one, 210 use onlyone 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 mustbe some colourings which do not contain 4 points all joined by lines ofthe same colour.

This argument gives a rather feeble result for R(4, 4) but in general it ismuch more effective than the best known constructions.

Theorem 4.3 (Erdos 1947). If k ≥ 5 then R(k, k) ≥√

2k.

Proof. There are(

nk

)different sets of k points3. Let’s list them as S1, S2, . . . , S(n

k).

Given a colouring let’s call a set S monochromatic if all points in it arejoined by lines of the same colour. Plainly the number of colourings in whichthere is a monochromatic set of k points is less than or equal to

(The number of colourings in which S1 is monochromatic)

+ (The number of colourings in which S2 is monochromatic)

+ (The number of colourings in which S3 is monochromatic)

+ . . .

+ (The number of colourings in which S(nk)

is monochromatic).

(If I have such a colouring then I have certainly counted it in this sum. Wehave an inequality because if the colouring contains more than one monochro-matic set of k points then it is overcounted.)

Each term in the above sum is the number of colourings in which a fixedset of k points is monochromatic. There are

2(n2)−(k

2)+1

such colourings.So the number of colourings containing a monochromatic set of k points

is at most (n

k

)2(n

2)−(k2)+1.

3See the appendix on counting subsets if you don’t know what this means.

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If we choose n and k so that this is less than the total number of colour-ings then we will be done. Indeed, since the number of colourings with amonochromatic set of k points will then be less than the total number ofcolourings there must be at least one colouring without a monochromatic setof 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

2(n2).

So we need (n

k

)2(n

2)−(k2)+1 < 2(n

2).

Equivalently, (n

k

)2−(k

2)+1 < 1.

Now, (n

k

)2−(k

2)+1 ≤ n(n− 1)(n− 2) . . . (n− k + 1)

1× 2× 3× · · · × k2−

k(k−1)2 × 2

≤ nk

2k−22−

k2

2+ k

2 .

So if n ≤√

2k

= 2k2 we have(

n

k

)2−(k

2)+1 ≤ 2k2

2−k+2− k2

2+ k

2

≤ 22− k2

< 1

since k ≥ 5. Hence the result.

This argument is usually phrased in terms of picking a random colouringand showing that the probability that it has a monochromatic set of k pointsis strictly less than 1. As such it was one of the starting points for the studyof random graphs, an important area of mathematics both in its own rightand because of its applications to other parts of mathematics and beyond.

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In fact there are many examples where ideas and techniques developedin the context of Ramsey theory have gone on to influence other parts ofmathematics.

Referring to the problem of improving the bounds for R(k, k), the Fieldsmedal winner (a Fields medal is sometimes considered the mathematicalequivalent of a Nobel prize) Timothy Gowers remarked4

“I consider this to be one of the major problems in combina-torics and have devoted many months of my life unsuccessfullytrying to solve it. And yet I feel almost embarrassed to writethis, conscious as I am that many mathematicians would regardthe question as more of a puzzle than a serious mathematicalproblem.[....] A better bound seems to demand a more globalargument, involving the whole graph, and there is simply no ad-equate model for such an argument in graph theory. Therefore,a solution to this problem is almost bound to introduce a majornew technique.”

In fact the bounds: √2

k≤ R(k, k) ≤ 4k

are close to being the best known. Very recently Conlon has improved theupper bound slightly. The exact result is rather hard to write down and theproof is long and difficult. Remarkably the best lower bound is proved bya probabilistic argument (a slightly refined version of the one we saw). So

although we know that colourings of√

2k

points with no monochromatic setof k points exist it is not known how to construct such a colouring.

4The quote is from “The two cultures of mathematics” available athttp://www.dpmms.cam.ac.uk/˜wtg10/papers.html

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5 Inevitable patterns in sets of numbers

In this section we look at the idea of inevitable patterns in a different con-text. As was mentioned earlier, the general philosophy of Ramsey theory issometimes expressed as “Complete disorder is impossible” or more precisely“in any large enough piece of disorder you can find a small piece of order”. Inthe 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 enoughpoints then any colouring of the lines between then is forced to contain amonochromatic set of k points.

The integers form another natural setting for results of this type. If I takethe integers {1, 2, 3, . . . , n} and colour each red or blue then what patternsmust inevitably appear?

5.1 Towards van der Waerden’s theorem

One possible pattern that we might look for is 3 equally spaced integers (forexample 12, 14, 16 or 25, 32, 39) all of the same colour. We call such a patternan 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 oneof the colours.

Theorem 5.1.W (3) ≤ 325

That is to say, whenever {1, 2, . . . , 325} is coloured with red and blue thenthere 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 3in 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 25 = 32 ways to colour a block and so among the first 33 blocksthere 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, band c are equally spaced (form an AP). Note that the block {c, c+1, . . . , c+4}is still within {1, 2, . . . , 325}.

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Now let’s look at the colouring of the block containing a (I’ll call thisblock A from now on). In a block of 5 the first 3 integers must include 2 ofthe same colour, red say. Having 2 red integers but no red AP of length 3forces a further integer to be blue (for example if 1 and 3 are red then 5 mustbe 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 firstcase

r r b ? ?︸ ︷︷ ︸Block A

. . . . . . r r b ? ?︸ ︷︷ ︸Block B

. . . . . . ? ? ∗ ? ?︸ ︷︷ ︸Block C

If c+2 (marked ∗ above) is red then a, b+1, c+2 is a red AP; if it is bluethen a + 2, b + 2, c + 2 is a blue AP. So either way we have a monochromaticAP of length 3.

In the second case we have

r ? r ? b︸ ︷︷ ︸Block A

. . . . . . r ? r ? b︸ ︷︷ ︸Block B

. . . . . . ? ? ? ? ∗︸ ︷︷ ︸Block C

and in the third case we have

? r r b ?︸ ︷︷ ︸Block A

. . . . . . ? r r b ?︸ ︷︷ ︸Block B

. . . . . . ? ? ? ∗ ?︸ ︷︷ ︸Block C

Either way the integer marked ∗ completes a monochromatic AP of length 3whichever colour is used on it.

If we had only set out to show that an AP of length 3 in one colourmust exist then we could have come up with an argument that works for amuch smaller set (see problems 2.1 and 2.2). However the method shown isprobably shorter and has the advantage that it will generalise to prove muchmore.

Extending the result to show that an AP of length 4 (that is 4 equallyspaced integers) exists is harder. However it is not too difficult to extend it

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to show that if 3 colours are used then an AP of length 3 inevitably appearsprovided that we colour {1, 2, . . . , n} for some huge n.

Let’s denote by W3(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 3in one of the colours.

Theorem 5.2.

W3(3) ≤ 7(2× 37 + 1)(2× 37(2×37+1) + 1).

If {1, 2, 3, . . . , 7(2× 37 + 1)(2× 37(2×37+1) + 1)} is coloured with red, blue andgreen 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 thatyou are completely comfortable with that before tackling it.

Proof. Again we will assume that we have a colouring without an AP oflength 3 in any of the colours and show that this is impossible.

Let n = 7(2× 37 + 1)(2× 37(2×37+1) + 1) and m = 7(2× 37 + 1).Split {1, 2, 3, . . . , n} into 2×37(2×37+1)+1 blocks each of 7(2×37+1) = m

consecutive integers.There are 37(2×37+1) ways to colour a block and so among the first 37(2×37+1)+

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 theblock {c, c + 1, . . . , c + m− 1} is still within {1, 2, . . . , n}. Again we will callthese block A, block B and block C.

Now let’s look at how block A is coloured. Split it up into 2×37+1 blockseach of 7 consecutive integers. There are 37 ways to colour a block and soamong the first 37 + 1 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 stillwithin 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 samecolour (red say). There are several possibilities but we will just look at theone where a + x and a + x + 3 are both red (the other cases come out in asimilar way).

Now a + x + 6 must be a colour other than red (blue say). If it were redthan a + x, a + x + 3, a + x + 6 would be a red AP.

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Also a + z + 6 must take the third colour green. If it were red thena+x, a+y +3, a+z +6 would be a red AP (remembering that blocks A+Xand 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 isred then a + x, b + y + 3, c + z + 6 would be a red AP (remembering thatblocks A and B are coloured identically so b + y + 3 is red because a + y + 3is). If it is blue then a + x + 6, b + y + 6, c + z + 6 would be a blue AP. If itis 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 anAP of length 3 in one of the colours.

As before the bound is extremely bad. The true answer is that W3(3) =27. However, for a larger number of colours it is a challenge to show anybound on Wr(3) (the smallest n which guarantees an AP of length 3 in onecolour when {1, 2, 3, . . . , n} is coloured with r colours) so we should not betoo put off by the huge numbers. The argument you have seen contains allthe ideas to prove such a bound although the notation gets tricky and thebounds are horrendous.

In fact the generalisation to arbitrarily many colours and length of AP isalso true. This is a result of van der Waerden.

Theorem 5.3 (van der Waerden 1927). Given r and k there is an n suchthat whenever {1, 2, 3, . . . , n} is coloured with r colours we are guaranteed tobe 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 guar-antee an AP of length k when 2 colours are used) are roughly:

2k

8k≤ W (k) ≤ 2222

222k

.

The upper bound (due to Gowers) looks ridiculous but it was a major im-provement on the best previously known bound.

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5.2 Density Versions

In fact W (3) = 9 so however {1, 2, 3, . . . , 9} is coloured with red and bluewe can always find an AP of length 3 in one of the colours. Which colourcould it be in? A tempting guess would be that the colour which appearsmost often always contains an AP of length 3 but this is false. Suppose that1, 2, 6, 7, 9 are coloured red and 3, 4, 5, 8 are blue then there are more redsthan 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 usedmany times does guarantee a long AP. This is a deep result of Szemeredi.

Before stating Szemeredi’s theorem let’s give a specific example. If I takea tiny number (1/10000 say) and a long length (30 say) then there is some nsuch that every set of n/10000 integers chosen from {1, 2, 3, . . . , n} containsan AP of length 30.

Theorem 5.4 (Szemeredi 1975). For any δ > 0 and integer k there is ann such that every subset of δn integers from {1, 2, 3, . . . , n} contains an APof 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 Partytheorem can lead to some deep mathematics. Most of this theory was de-veloped in the past hundred years and the most recent results I have men-tioned are only a few years old. There are many open problems: determiningR(5, 5), improving the bounds on R(k, k) and W (k), giving a constructivelower bound on R(k, k) which is as good as the probabilistic one, and gener-alising and sharpening Szemeredi’s theorem which although simple to stateare hard and important. If progress is made on any of these it is likely tohave a profound effect on the subject.

Ramsey theory is usually taught in the second or third year of undergrad-uate maths courses as part of modules called things like “Discrete Mathemat-

16

ics”, “Combinatorics” or “Graph Theory”. It is also an important researcharea.

If you have found this interesting then two articles at a similar level whichare available on the web are given below:

• Ben Green “Ramsey Theory at the IMO”

available at http://www.dpmms.cam.ac.uk/˜bjg23/papers/gazette.ps

• Imre Leader “Friends and Strangers”:

available at http://plus.maths.org/issue16/features/ramsey/index.html

(“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:

• http://www-gap.dcs.st-and.ac.uk/ history/Biographies/Ramsey.html

• http://www-gap.dcs.st-and.ac.uk/ history/Biographies/Erdos.html

If you have any questions or comments on the course or these notes then Iwould be happy to hear from you. My email address is r.johnson@qmul.ac.uk

For more information on maths at Queen Mary take a look athttp://www.maths.qmul.ac.uk/

and in particularhttp://www.maths.qmul.ac.uk/undergraduate/prospective/admissions/

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7 Appendix: Counting Subsets

If I have n people how many lines do we need to join all pairs? For small nyou could just count (3 when n = 2, 6 when n = 4, 10 when n = 5) but ingeneral it would be useful to have a formula.

Each line joins a pair of points. There are n ways of picking the firstof the pair. For each of these choices there are n − 1 ways of picking thesecond of the pair (we cannot choose the first point again but each of theothers could be chosen). This makes n × (n − 1) choices overall. Howeverwe have counted each pair twice (once as (a, b) and once as (b, a) say) and sothe number of pairs is

n(n− 1)

2

This number is usually written as(

n2

)(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− 1choices for the second, n− 2 for the third, and so on down to n− k + 1 forthe kth. This makes

n× (n− 1)× (n− 2)× · · · × (n− k + 1)

choices overall. However we have counted each set many times. In factyou 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 followsthat 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(

nk

)(pronounced “n choose k”).

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Inevitable patterns in mathematics

Problems

There are probably too many problems to think about in the time youhave. I encourage you to spend a serious amount of time thinking about afew problems rather than taking a superficial look at all of them. Some ofthe problems are quite open-ended so there is not necessarily a single correctanswer or correct approach.

If you look at them again after today and have any comments or questionsthen feel free to get in touch (my email address is r.johnson@qmul.ac.uk).You are welcome to work individually, in pairs, or in larger groups. Helperswill be circulating to provide guidance and suggestions but they will not justtell you how to do the problems!

8 Points and Lines

To do these problems you will need the definition of Ramsey numbers thatI talked about. That is, R(k, l) is the smallest integer n such that wheneverthe lines between n points are coloured red and blue there is always either aset 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 whatcan you say about R(b, a)?

Problem 8.3. We showed that R(3, 3) ≤ 6 (convince yourself that this isindeed the content of the Party theorem). What would you need to do to showthat R(3, 3) = 6? Show that R(3, 3) = 6.

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Problem 8.4. Play the following game with the person next to you. Startwith 6 blobs on the page. Take it in turns to join two unconnected blobs withone of you always using a red pen and the other always using a blue pen. Thefirst 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 youcome up with some good strategies for playing the game?

If you enjoyed problem 1.4 but do not have a ready supply of friends thentake a look at:

• http://www.dbai.tuwien.ac.at/proj/ramsey/

A variation is to allow each player to draw more than one line at eachturn.

Problem 8.5. A (bad) way to prove that R(3, 3) ≤ 6 would be to list allpossible red/blue colourings for 6 points and check that each contains eithera red or a blue triangle. How many colourings would you need to write downto do this? How many colourings would you need to write down to prove thatR(5, 5) ≤ 48 by this method?

Problem 8.6. Find the exact value of R(3, 4) (or at least find the best boundsyou can). What about R(4, 4)?

Problem 8.7. Show that if the lines between R(3, 6) points are coloured withred, 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’stheorem to three colours? What about r colours?

Problem 8.8. Show that in any list of 6 distinct integers it is possible tofind 3 in increasing order or 3 in decreasing order. The 3 do not have to beconsecutive 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 thingtrue for a list of 4 distinct integers?

What if you want to be able to find a longer increasing/decreasing se-quence?

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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 blueis guaranteed to contain an AP of length 3.

Problem 9.2. Improve the upper bound of 325 on W (3) as much as youcan.

Problem 9.3. Denote by W (k) the smallest integer n for which any colour-ing of {1, 2, 3, . . . , n} with red and blue is guaranteed to contain an AP oflength 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 containsno AP of length 3.

Problem 9.5. What is the largest subset of {1, 2, 3, . . . , 10} which containsno AP of length 3.

Problem 9.6. Complete the proof of Theorem 5.2 by dealing with the otherpossibilities 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 youcould 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 morethan half of all possible lines but contains no triangle.

Explain why this shows that there is no density analogue of Ramsey’stheorem.

Problem 9.8. If you’re feeling bold try to extend the proofs of the existenceof W (3) and W3(3) to show that Wr(3) exists. That is for every r there issome n such that whenever {1, 2, 3, . . . , n} is coloured with r colours there isan AP of length 3 in one of the colours.

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