Inevitable patterns in mathematics
Some topics in Ramsey theory
These notes cover an ambitious guess at what we might get through in
the course1 . 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?
1. 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
2008.
1
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
364
363
362
343
365
...
≈ 0.507.
1−
365
365
365
365
365
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
A mathematical approach to the Torah code claims is described on Brendan McKay’s webpage at http://cs.anu.edu.au/~bdm/dilugim/torah.html
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
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
• 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
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 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
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) ≤ 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 lot
of work to determine.
Analysing the proof of Theorem 3.1 gives the following
7
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 you
haven’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 . Now
by 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 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) ≤ 4k .
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 215 possible colourings.
• There are 15 possible sets of 4 points from the 6 (count them!). If I
choose one of these there are 210 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
• So the number of colourings which do contain 4 points joined by lines
of the same colour is at most 15 × 210 . (If I list the sets of 4 points
then 210 colourings use only one colour on the first one, 210 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.
√ k
Theorem 4.3 (Erdős 1947). If k ≥ 5 then R(k, k) ≥ 2 .
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 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 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(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
n
k
2( 2 )−(2)+1
such colourings.
So the number of colourings containing a monochromatic set of k points
is at most
n (n2 )−(k2)+1
2
.
k
3
See the appendix on counting subsets if you don’t know what this means.
9
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
n
2( 2 ) .
So we need
n
n (n2 )−(k2)+1
< 2( 2 ) .
2
k
Equivalently,
n −(k2)+1
2
< 1.
k
Now,
n −(k2)+1 n(n − 1)(n − 2) . . . (n − k + 1) − k(k−1)
2
≤
2 2 ×2
k
1 × 2 × 3 × ··· × k
k2
k
nk
≤ k−2 2− 2 + 2 .
2
√ k
k
So if n ≤ 2 = 2 2 we have
k2
k2
k
n −(k2)+1
2
≤ 2 2 −k+2− 2 + 2
k
k
≤ 22− 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
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 remarked4
“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:
√
k
2 ≤ R(k, k) ≤ 4k
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
√ k
although we know that colourings of 2 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
http://www.dpmms.cam.ac.uk/˜wtg10/papers.html
11
of
mathematics”
available
at
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 25 = 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
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
Case 2: r
? r
Case 3: ? r
b ? ?
r
? b
b ?
Now looking at block B which is identically coloured we have in the first
case
b ? ?} . . . . . . ?| ? {z
∗ ? ?}
r| r {z
b ? ?} . . . . . . |r r {z
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 ? }b . . . . . . ?| ? {z
? ? ∗}
r| ? {z
r ? }b . . . . . . |r ? {z
Block A
Block B
Block C
and in the third case we have
?| r {z
r b ?} . . . . . . ?| r {z
r b ?} . . . . . . ?| ? {z
? ∗ ?}
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
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 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 3
in one of the colours.
Theorem 5.2.
W3 (3) ≤ 7(2 × 37 + 1)(2 × 37(2×3
7 +1)
+ 1).
7
If {1, 2, 3, . . . , 7(2 × 37 + 1)(2 × 37(2×3 +1) + 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.
7
Let n = 7(2 × 37 + 1)(2 × 37(2×3 +1) + 1) and m = 7(2 × 37 + 1).
7
Split {1, 2, 3, . . . , n} into 2×37(2×3 +1) +1 blocks each of 7(2×37 +1) = m
consecutive integers.
7
7
There are 37(2×3 +1) ways to colour a block and so among the first 37(2×3 +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 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 2×37 +1 blocks
each of 7 consecutive integers. There are 37 ways to colour a block and so
among 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 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
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 W3 (3) =
27. However, for a larger number of colours it is a challenge to show any
bound on Wr (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
2k
22
≤ W (k) ≤ 22
8k
22
k
.
The upper bound (due to Gowers) looks ridiculous but it was a major improvement on the best previously known bound.
15
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 Mathemat16
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 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 I
would be happy to hear from you. My email address is [email protected]
For more information on maths at Queen Mary take a look at
http://www.maths.qmul.ac.uk/
and in particular
http://www.maths.qmul.ac.uk/undergraduate/prospective/admissions/
17
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
n
This number is usually written as 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 nk (pronounced “n choose k”).
18
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 email address is [email protected]).
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
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:
• http://www.dbai.tuwien.ac.at/proj/ramsey/
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?
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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 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 W3 (3) to show that Wr (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