[Skip to 10:50 for the actual problems]
The motto for today is "A good sketch is better than a long speech"
which was supposedly said by Napoleon Bonaparte.
Also known as "A picture is worth a thousand words."
I always wanted to
create a talk
where I succeed in getting across complex contents
without using any words (not to mention formulas),
just amazing graphics and animations.
I won't manage to do that today, but I promise
you'll see many pictures and not a lof text.
And no formulas, in case you're afraid of them,
although formulas aren't evil.
And, like last year, I wrote some programs
to create animations for today.
The subject for today is: "Four Problems"
I've selected four unsolved mathematical problems
which I think are all interesting.
But there were also other reasons to pick them.
Unsolved Problems?
When I did my PhD in mathematics
I sometimes talked to people
who asked me what I was doing.
And they often said: "Research in mathematics?"
"What's that about?  Don't we know everything already?"
"Two plus two is four, and so on..."
Of course, that's NOT the case.
If you're sitting in my math lectures
you already know that two plus two is not always four.
And there are many unsolved problems in math!
Some of them are so hard
that you need to study mathematics
just to understand the problem statement.
But there are also some beautiful problems
that you can almost explain to every child.
That's the type of problems I selected for today.
Not that I think you are children... :)
But these problems are easy to explain and they can be visualized.
But before I start, I'd like to philosophize for five minutes
because this is a Christmas lecture...
I'd like to clarify
the setting for today.
The reason is that
I'm suffering a bit here.
I like math, that's why I studied math.
And I'm giving math lectures here
for students of which I'm pretty sure
that they'd rather hear something else.
No matter what, just something else.
But they have to be there as part of their degree.
So I try to make it palatable for them
by showing them applications
so that they see what math is good for.
I try to demonstrate how math might
for example be used in computer science.
But this goes against my grain.
Applications are not the reason mathematicians do math.
They do it for other reasons.
That's why I'm starting this talk with the question:
"What is math about?"
What is mathematics?  What do mathematicians do?
And that'll lead to the problems I mentioned.
So, what is mathematics?
I saw an article about this
in a newspaper recently.
And the author's first question was:
Does math belong to the (natural) sciences?
One could think that.
Scientists use a lot of math,
especially the physicists,
but also chemists, biologists, and so on.
But of course math isn't a (natural) science.
Scientists use empirical methods
to examine natural objects.
That's not what mathematician's do.
Physicists and chemists use math,
and some physicists even helped to develop new mathematics,
when math wasn't yet where they needed it,
but they don't DO math, they use it as a tool.
Math surely doesn't belong to the (natural) sciences.
If you're at an American university, chances are
you'll have a faculty called "Sciences and Math",
which shows that math doesn't belong to the sciences.
One could think math belongs to the
"Geisteswissenschaften".  [see https://en.wikipedia.org/wiki/Geisteswissenschaft]
This is a term used mainly
in German-speaking countries.
If you interpret this term literally ["Geist" = "mind"]
that seems to make sense as mathematics deals with things
which are only happening in our minds.
You only use paper and pencil and the objects
of your research reside in your head, so to say.
But that's not how the term "Geisteswissenschaft" is
typically used.  According to the agreed-upon usage,
mathematics is not a "Geisteswissenschaft".
So, what about engineering?
Certainly not.  Some of you are already shaking their heads.
Engineering is about technology 
and manufacturing and so on.
But engineers, like scientists,
use a lot of math.
What engineers also have in common with mathematicians
is that they're both interested in developing methods,
procedures for doing things in a clever way.
That's where they agree.
But math certainly isn't the same as engineering.
The last suggestion in the newspaper was:
Is math an art?
Now some of you might believe I'm crazy.
But this is not as absurd as you might think.
Pretty much all great achievements in math
were only possible because the persons
who came up with them
had been very creative.
You need a lot of creativity to create new mathematics.
If you're just reproducing existing math,
you don't need creativity.
Unfortunately, that's what's mostly happening in school.
But creating new math or re-creating, by yourself,
existing math is a creative process.
And, by the way, most laypersons are surprised to hear
that mathematicians care a lot about esthetics.
Whether for example a new proof
is considered good or bad math
often depends on esthetical criteria.
So, there are certainly some similarities to the arts.
But somehow math doesn't really belong to one of these four categories.
Now, if you look around a bit
you might find terms like
"formal sciences" or "structural sciences".
But I'm not really convinced by these.
To me the descriptions of these terms sound
like "mathematics in a broader sense".
It's not much more than another word for the same thing.
Maybe it's more interesting to ask
what the topics of mathematics are.
That'll also lead us to the problems this talk is about.
Most people will say: math is about numbers.
But there we already have the first question.
Have you ever SEEN a number?
If I were talking about oranges and I would show you 
a slide with an orange, you would rightly say 
that I didn't show you an orange - only a picture of an orange.
You have of course seen real oranges.
But you've never seen a seven!
You've seen the symbol on the slide here.
That's a Hindu-Arabic numeral which denotes the number seven.
You've probably seen seven as a Roman numeral,
as an English word,
as a tally mark.
Below left is the Japanese numeral.
Below right is seven in Braille.
But these are just symbols for an abstract concept.
There is no such a thing as a seven that you can touch.
So, even one of the most basic parts
of mathematics is
a purely abstract concept.
And it get's worth.  The next thing mathematician's
work with are points.
I guess you don't even see this point
if you're sitting in one of the last rows.
Here it is.
It's so small for a reason.
The points mathematicians think about have neither 
height nor width.  There's no way to draw them.
That's another purely abstract concept.
The points we mathematicians work with don't exist.
But only those are fun to work with.
Otherwise you'd have to deal
with misshapen rocks that
don't behave the way you want.
We could also talk about circles.
Real circles don't exist either.
You can of course fabricate circular objects,
and you might be able to do that very precisely.
But if you measure accurately enough, you'll realize you don't have a circle.
A circle is something where all points have exactly the same distance to the center.
You'll never manage to fabricate that in the real world.
But you can create a perfect circle in your head.
And we can talk about circles and then we hopefully think about the same kind of objects.
But you can't examine circles using empirical methods.
I could go on, of course.
The Greeks more than 2000 years ago already knew
that there are triangles like this one
where you can either measure the long edge
or one of the short edges, but not all three.
And an engineer will of course say:
"Nonsense!  Of course I can measure them all."
But from a mathematical viewpoint you can't
because one of the lengths must be irrational.
And finally we have things like the Klein bottle here
a two-dimensional "surface" which needs
four dimensions in order
to avoid self-intersections.
We can talk and reason about it.
But does it EXIST?  Probably not...
And the high point is infinity.
There are various kinds of infinity
appearing here and there in mathematics.
All the things here are symbols for different kinds
of infinty, like the "tilted eight"
or the C for "continuum"
or "aleph-naught" in the middle, a Hebrew letter,
and so on and so on.
Every physicist will tell you that our universe is finite,
and that there's nothing that's infinite in the real world.
But every mathematician will tell you that math
only begins to get really exciting when
you're thinking about infinite objects.
Maybe that's what math is - the science of infinity.
OK, so much for the setting of what's about to follow.
I mainly wanted to clarify one thing:
Mathematicians sometimes do strange stuff.
And if you're just interested in the "intended purpose",
if you're just asking yourself "How can I USE this",
you'll quickly come to a point where
none of this seems to make sense.
This whole enterprise can only be fun
if you get involved,
if you dive in,
if you fool around
and try to enjoy yourself.
That's what our four problems are about...
And here they are.
I gave them somewhat lurid headlines:
1. Colorful Points in the Plane
2. Avoiding Patterns,
3. Hidden Picture Puzzles with Squares
4. Ambitious Rodents
They don't have a lot in common.
They're from different areas.
The first one is from geometry,
but also related to set theory.
The second one is a
combinatorial problem, or
you could say it's from graph theory.
The third one is also geometrical in nature,
but we'll realize it has connections to topology.
And the fourth one is from theoretical computer science.
They all have in common that one can visualize them.
At least I'll try to.
That they are otherwise unrelated is actually an advantage for you:
Should you get lost in between,
just wait until the next chapter starts,
and everything starts again from scratch... :)
But that shouldn't happen!
If you're really lost in between,
raise your hand and ask.
My hope is that you'll understand all of this.
I'm not trying to impress you with incomprehensible stuff.
I want you to leave after the talk, saying
"I got it!  And now I'll take pencil and paper
and try it myself!!!"
OK, the first problem is about colorful points in the plane.
The problem also has an "official" name.
It's called the Hadwiger-Nelson problem.
It's from 1945 and Nelson is an American mathematician
who created the problem as a puzzle.
It later turned out that
a Swiss mathematican called Hadwiger had done
something similar earlier.  And so it's named after both now.
And what is the problem?
We want to color the plane.
As you see, I already started with this task.
If mathematicians talk about "the plane",
they mean what is sometimes called "R squared".
The important point is that the plane
extends indefinitely in all directions.
You'll see in a minute that the problem only makes sense
if you don't stop somewhere.
If the plane were only the white
rectangle you see on the slide,
the problem would be fairly easy.
It is hard because the plane is infinite.
So, every point of the plane must get a color.
And remember that "in reality" you can't see the points
because they don't have height or width.
What you see here are "markers" for the real points
so that we have something to talk about.
There's only one rule:
If the distance of two points is EXACTLY ONE
(assuming we have agreed on a unit like centimeters),
then the two points must have DIFFERENT COLORS.
So the encircled point shown here
must NOT be red
because the other one's already red.
This rule doesn't apply to distances
smaller than one.
And it also doesn't apply to points
which are farther apart.
It only applies if the distance is EXACTLY ONE.
And the question simply is:
How many colors do we need?
What's the minimal number of colors needed?
Now think about this for a millisecond.
And then I'll be mean and show you a solution.
But that's probably not the best one.
One could do it like this.
You take honeycombs, hexagons.
You distribute them evenly across the plane.
Again, we imagine this goes on forever
and in all directions.
We're using seven colors.
And each hexagon is colored
in such a way that its six neighbors
have the other six colors.
As you can see, that's easy to achieve.
We have a nice, regular pattern.
And if your hexagons have the right size,
so that one hexagon is a bit smaller than a circle with diameter one
(look for example at the gray one),
then two points within the same hexagon
have a distance less than one,
so they're allowed to have the same color.
And all other gray points are far
enough away, so they are also OK.
This demonstrates that seven colors suffice.
So we now have an upper bound.
But the question was: What's the minimum?
We now know that seven colors suffice.
We could now try with six or five or four colors.
But we'll do it the other way around:
Will two colors be enough?
What do you think?
That was quick!
(We didn't rehearse this, did we?)
If you draw an equilateral triangle like this one
where the edges all have length one,
and you color two vertices, say with blue and green,
then the third vertex (with the question mark)
also needs a color.  But it must be neither green nor blue.
So you obviously need at least three colors.
So the answer to the question whether
two colors suffice is: "Nope!"
The next question of course is if three colors suffice.
This is a bit harder.
Coincidentally, I saw exactly this question
a few weeks ago in math competition
for ninth graders in Hamburg.
We'll see that three colors don't suffice.
And the ninth graders were supposed to
proof why that is the case.
We'll start again with our equilateral triangle.
All edges have length one.
We know we need three colors for the vertices.
We now add a second such triangle like this.
Now, the point with the question mark:
It can neither be red nor green.
Now let's ASSUME we only
need three colors.
Then this point must obviously be blue.
We have just learned:
If we have a constellation like this,
if we have two points with this distance,
they MUST have the same color.
How long is this line segment?
You don't have to tell me!  But can you
at least tell me how one would compute this?
(Remember I promised not to incommode you with formulas...)
I was hoping somebody would say "Pythagoras"...
If you're using the Pythagorean theorem, you'll see
that the distance of the two blue points is the square root of three.
We've just learned: If two points have EXACTLY the distance
square root of three, they MUST have the SAME color.
(Of course only under our assumption that three colors suffice!)
Now we're almost done.
Take a compass and construct a circle with radius
square root of three around one of these points.
Then use your compass again to create
a circle with radius one around the other point.
These two circles intersect - see the point with the question mark.
This point has distance one from the right blue point,
so it MUST NOT be blue.
But is also has distance square root of three from
the left blue point and thus MUST be blue.
This is a contradiction!
So, three colors don't suffice.
So we now know:
Seven colors are enough, three colors aren't enough.
I'll introduce the standard term for this
(so you've actually learned something):
The minimum number of colors needed to color the plane
by these rules is called
the chromatic number of the plane
(from the Greek word for "color").
And what do we know about this number?
We know it's at least four
(because three is not enough)
and at most seven.
But it could also be five or six.
That's what we know.
And nobody knows more!
The problem has been around since 1945
and nobody found a better answer so far!
[But see https://arxiv.org/abs/1804.02385v3 for a 2018 update!]
With problems like these
that have been unsolved for such a long time,
you can be pretty sure they're not so easy.
But there can of course be surprises.  I certainly
don't want to discourage you from thinking about this problem.
Maybe you're the one to find
a beautiful solution to this problem!
But here's a warning:
There's a paper from 2003
from the famous Israeli mathematician Saharon Shelah
and the Russian-American Alexander Soifer.
The proved that in similar constellations
the answer depends
on the axioms of set theory.
This COULD mean that
this question is UNDECIDABLE...
In layman's terms:
We don't know enough about the real numbers.
Although mathematicians have been working with them
for a long time, we still can't axiomatize
them precisely enough
to answer all questions about them.
And this problem is essentially about the real numbers.
In case you've heard about things like Lebesgue measurability,
they are related to this problem...
But that would take us too far for today.
Anyway, this was the first example
for a seemingly SIMPLE question
where the answer currently seems out of reach.
But maybe we know more next week
once you've thought about it a bit...
That was our warm up and here's the next problem.
I called this one "Avoiding Patterns"
which is a cheap joke because mathematicians
usually try to FIND patterns.
Now we're trying to avoid them.
The problem is about "Ramsey numbers".
Ramsey was a philosopher and logician,
a contemporary of Wittgenstein.
I think he even was Wittgenstein's tutor or instructor in Cambridge.
He was mainly interested in logic,
and his Ramsey numbers are kind of a "by-product"
belonging to the area of combinatorics.
Unfortunately, Ramsey already died at the age of 26.
But several questions related to
Ramsey numbers are still an area
of active research today, though.
So, what is this about?
This is what mathematicians call a "graph".
The terms aren't so important.
You should only remember:
The black dots we see here are called "vertices".
And the lines connecting them are called "edges".
We now have five vertices.
And each pair of vertices is connected by an edge.
We now want to color the edges.
Imagine having two buckets of paint,
one with red and one with blue paint.
Some edges will be painted red, some blue.
But all edges have to have a color.
You could for example do it like this.
I highlighted some of the red edges.
Because we can now see that
if I color the edges like this,
I have a distinguished sub-pattern - a red triangle.
Of course, I could have colored differently.
For example like this, now with a blue triangle.
Or...
Like this.
And you won't find any monochromatic triangles here.
To clarify: If I say "triangles", I only mean those triangles
the vertices of which are among the five vertices of your pentagon.
So I managed to
avoid a pattern!
That was this problem's headline.
Now we make this a bit harder - six vertices.
And we ask ourselves
if in this case we can also "avoid patterns".
Can we color the edges with red and blue in such a way
that we see neither a red nor a blue triangle?
Why don't you try this yourself?
[After the break:] Some of you conjectured this might be impossible
and that this was the only reason I offered cake for a solution.
And you were right.  This is impossible!
I'm sorry... :)
It's impossible to color this without having triangles.
But of course we have to prove this.  I'll do that now.
This graph here is totally symmetric.
All vertices are equal.
If I pick one, it doesn't matter which one it is.
I have now selected one vertex and five edges emanate from it.
There are five edges starting at each of the six vertices.
I have two colors at my disposal.
If I color these five edges with two colors,
one of the colors must be used at least three times.
Let's say this color is red - it doesn't matter.
So we have at least three red edges connected to our vertex.
So it looks like this.
I'm only interested in this part of the graph now,
only the selected vertex and the three red edges.
I don't care about the color of the other two edges.
Now we concentrate on the three vertices at the ends of the three red edges.
What can happen there?
We have two possibilities.
The first one is:
One of the edges connecting these three vertices is red.
In that case, I'd have a red triangle - which I wanted to avoid.
The second possibility:
None of the connecting edges is red.
But if none of them is red,
then they're all blue
and I have a blue triangle.
So we've seen it's impossible
to "avoid triangle patterns" in this graph.
Let's summarize our results so far.
A long sentence...
If we have a graph with six vertices
and if all pairs of vertices are connected by edges
and if all edges are either red or blue,
then we'll always find - independent of the coloring -
a triangle that's completely red or completely blue.
An we also saw:
This doesn't work with five instead of six vertices.
Remember the first example where we managed
to "avoid triangles".
This means that six is the smallest number having this property.
Six is the smallest number of vertices where triangles can't be avoided.
You've probably already guessed what comes next:
I want to generalize this.  Mathematicians love to generalize!
But I'll need another technical term for this.
The term is "complete graph".
The triangle on the left
has three vertices and
each pair of vertices is connected by an edge.
We'll call this K3 from now on,
the complete graph with three vertices.
With four vertices you'll get the
graph labeled K4 above.
K5 is the graph from our first example.
K6 is the graph we just talked about.
And we can obviously proceed like this.
I can now restate my long sentence from above
using the new terms - like mathematician would say it.
If we have a complete graph with at least six vertices
and we color each edge red or blue,
then the graph always contains
a copy of K3 which is either completely red or complete blue.
And six is the smallest number with this property.
Now the generalization:
The Ramsey number R(a,b)
is the smallest number n
such that the complete graph Kn
always contains a red copy of Ka
or a blue copy of Kb,
no matter how the edges are colored.
We just figured out that R(3,3) is 6.
In other words: 6 is the smallest number n for which we
every coloring of Kn contains a red or a blue K3.
That's what we just proved.
And we'll now prove two other things.  (But no more exercises for you this time.)
We'll first show R(3,4)=9:
If we have nine vertices, we can't avoid
having either a red K3 or a blue K4.
And then R(4,4)=18: With 18 vertices
we can't avoid either a red or a blue K4.
You'll see that this is not extremely complicated.
And of course you'll just a have to watch me coping with this.
But let's first ask a question that Ramsey couldn't have asked:
Can't we just use a computer to do this?
Take for example K6 where you tried it yourself with pencil and paper.
How about letting a computer do this?
Coincidentally, I have a computer right here.
Here's our K6.
And you could now ask your computer
to try out all possible colorings.
First all edges red, then all red except of one, and so on.
And while the program is running, it looks for triangles.
And it either stops if it finds one or it finishes without finding one.
There's your answer.
Why even think about this problem if you have computers?
I made this program slow on purpose.
It checks one coloring per second
so that we can see what's happening.
At this speed it would take hours to finish.
But I could make it real fast so that it finishes in seconds.
I've computed the amount of work the computer has to do.
We need to color all edges.
There are five edges emanating from the first vertex,
four edges from the next vertex (one was already counted),
and so on,
which amounts to a total of 15 edges.
Each edge can be colored red or blue.
So there are 2 to the 15th power possibilities.
The computer would have to try 32,768 colorings.
That's a lot for a human being, but laughing stock for a computer.
So, once again, why do we bother with this problem?
But remember that one of our goals was R(4,4)=18.
The computer would have to run through all colorings with 18 vertices.
So this program
which is still running
would be run with K18 instead of K6.
So we'll do the same computation with 18 vertices.
We have 17+16+15+...+1 = 153 edges.
And the number of possible colorings is now 2 to the 153rd power.
I had to write down the name of this number as I wasn't familiar with it.
We have approximately eleven quattuordecillions
colorings for the computer to check.
Now image you have a super-fast computer.
The NSA gave you one.
Your supercomputer can check one trillion (!) colorings per second.
That's a bit unrealistic with today's hardware,
but let's assume it is possible.
Then you'd need about 300 septillion years...
For comparison: The universe
is "only" 14 billion years old.
You'd have to start your program pretty early... :)
Maybe it is better to THINK than to just TRY.
So, what was the next step again?
We wanted to figure out the Ramsey number R(3,4)
and I already said it would be nine.
So we need to do two things:
We must look at K8, pictured here,
and we must find a coloring of it
which doesn't contain a red copy of K3
or a blue copy of K4.
I'll just show you the coloring, step by step.
We'll start with the red edges.
You won't find any red triangles here.
This probably happens a bit too fast
to check "live",
but you can later
watch the video and stop it
at this point to convince yourself.
Now the blue edges.
And here we don't have any blue K4s.
Remember that K4 is a quadrilateral
including the diagonals!
Here we have the red and the blue edges combined.
And because we have managed to avoid red K3s and blue K4s,
we now know that R(3,4) is greater than eight.
Now we need to show that we can't avoid these patterns
once we have nine vertices.
That's a little bit more work,
but it's not too hard.
So here we have nine vertices.
And I select one of them.
Like before, the graph is symmetric, 
all vertices are equal,
so it doesn't matter which one I pick.
We have two possibilities:
Either we have at least six blue edges emanating from "our" vertex.
Or it is not possible to pick such a vertex.
That would mean there are at most five blue edges attached to each vertex.
But as there are always eight "other" vertices,
we would then have at least three red egdes attached to EACH vertex.
Let me repeat the two cases:
Either we can pick one vertex where at least six blue edges meet,
or each vertex is the endpoint of at least three red edges.
Case 1: Six blue edges have six endpoints.
If we only look at these six points (which are highlighted)
we have a subgraph with six vertices.
And we already know that we can't "avoid triangles" there.
No matter who we color the edges, will find a red or a blue K3.
If we have a blue K3, for example like so,
we actually have a blue K4 if we add the corresponding blue edges.
Or we find a red K3 instead.
But that's what we wanted to find, a red K3 or a blue K4!
Which means we've covered the first case already.
We can't avoid both a red K3 and a blue K4.
Now case 2: at least three red edges at each vertex.
Here's a clever trick:
We have nine vertices.
Imagine you're "walking" around the graph
and at each vertex you count 
the number of edges emanating from it.
Now let's assume there were EXACTLY three red edges at each vertex.
Your walk visits nine vertices
and you count three red edges at each vertex.
That adds up to 27 red edges.
But you've counted each red edge TWICE!
Because you visited its starting point as well as its endpoint.
But if you count each edge twice, you must end up with an even number...
As 27 is not even, it can't be the case that there
are exactly three red edges at each vertex.
Therefore, there must be at least one vertex where at least four red edges meet.
Right!
So, the situation will look like this.
And again we have two possibilities:
Case 1 is that all four endpoints of these four edges
are connected by blue edges.
Then you have K4 in blue.
Case 2 is they are NOT all connected by blue edges.
Then two of them are connected by a red edge.
And there's your red K3.
We have thus proved that R(3,4) is nine.
If you want to guarantee that a red K3 or a blue K4 can't be avoided,
then you need at least nine vertices.
And of course also the other way around
as you can simply swap the colors:
R(3,4) is the same as R(4,3).
We're almost done.
What remains is to show that R(4,4) is 18.
This will be easy now.
Like before we start with one vertex less,
so we have 17 vertices.
And we need to find a coloring
where we don't find certain patterns.
It's about R(4,4), so we want to avoid red K4 and blue K4.
Here are the red edges.
Cute, isn't it?
As this might be confusing, I've picked out one vertex.
These are the edges emanating from one particular vertex.
And here I have connected all the endpoints with each other.
If a red K4 were in there, we would see it here already.
Again, we don't have enough time.
You have to check this
at your leisure later.
Now for the blue edges.  They look like this.
We'll again pick one vertex to make this easier to check.
And again we connect all endpoints.
Like before, you won't see this in two seconds,
but with enough time you'll see that there's no blue K4 here.
If I now combine the red and the blue edges,
it not only looks nice,
but it also proves that R(4,4) must be greater than 17.
Now I need to establish why the pattern can't be avoided with 18 vertices.
So here's K18.
You know how this works by now.
We pick one vertex.
This vertex is the starting point of 17 edges.
One color must appear at least nine times
among these edges as otherwise
we would have colored at most 16 edges.
So let's assume there are at least nine red edges emanating from here.
That's nine endpoints.
But we just talked about the number nine.
Nine is R(3,4).
So, among these nine endpoints
there's either a red K3
(shown in dashed lines)
or we have a blue K4.
We see that the red K3 together with the other edges
gives you a red K4.
Or you have the blue K4.
Both of these are "inherited" from the number nine.
That's it!  No way to avoid a monochromatic K4.
What have we seen by now?
The Ramsey number R(3,3) is six.
R(3,4) is nine and R(4,4) is 18.
There are a few other Ramsey numbers that are known.
We know R(3,5)=14, R(3,7)=23, etc.
And we know that R(5,5) is somewhere between 43 and 49.
But that's all we know!
Nobody knows the exact value of R(5,5).
And people have tried to figure this out for almost a century.
Not to mention something like a formula for R(n,n)
One could at least imagine that there's a formula where you just have to insert n.
But this seems almost impossible.
If you were just able to improve the estimate above,
if you could for example prove that R(5,5) is NOT 44
or that it can't be greater than 48, that would be progress.
But there hasn't been any progress like this in the last decades.
That's your second homework for this weekend...
Here's the third problem.
I called it "Hidden Picture Puzzles with Squares"
Its "official" name is "Toeplitz Conjecture,"
named after the German mathematician Otto Toeplitz
who was one of the many, many scientists who were 
either chased out of Germany or killed by the Nazis.
Mathematics was, by the way, the science that was
affected the most by the Nazi purges.
Before we start with Toeplitz's real conjecture from 1911,
I'll first show you a somewhat simpler problem.
That will make it easier for us to understand the real thing.
We start with an arbitrary picture.
We want to frame this picture
and we want to frame it tightly.
That's pretty easy...
There's no problem in finding
a rectangle that
squeezes the picture.
By which I mean that the rectangle's edges touch the picture.
Can I also do this with a square?
Yes, that's also possible.
Let me state this precisely:
We want the figure to be inside the square
and ALL FOUR edges of the square must touch the figure.
The question is: Does this always work?
Can I take an arbitrary figure and
always find a square which circumscribes the figure?
That's the technical term.
A circumscribed square is one
where all edges touch the figure.
I want to circumscribe a square around this figure now.
And I'll tell you in advance that it'll work.
But I'll mainly want to demonstrate the method which will achieve this.
We start with a rectangle.
And I think we all understand how this works.
You start with two pairs of parallel lines which are orthogonal to each other
and you move them until they touch the figure.
I've also added "coordinate axes"
parallel to rectangle edges.
In my little demo program
you can rotate these axes
together with the rectangle.
Now an important observation:
I hope that most of you see that
this is a continuous motion.
In layman's terms: If I make only small changes with the mouse,
they will result in small changes of the rectangle.
A bit more precisely: If the angle changes only a little bit,
the edge lengths of the rectangle will also change only by a little bit.
We'll now keep track of the difference
between the two edge lengths of the rectangle.
Let me see - my program had a key for that...
There it is.  Look at the function in the lower right corner.
If I now change the angle,
you see how the difference changes.
The angle is on the horizontal axis
and the difference is on the vertical axis.
And now you're probably already seeing where this is heading.
If I rotate this thing by 90 degrees,
the "roles" of the edges (long vs. short) have changed.
That implies there was a sign change, i.e.
I must have crossed the horizontal axis.
But at that point the difference is zero
which in turn means the edges have the same length.
So if rotate carefully...
...I find such a point and that's a square!
This obviously always works.
So the answer to our question is "yes".
Here's a precise mathematical formulation of this,
for the sake of completeness:
You have an arbitrary set of (at least two) points
in the plane (e.g. the Homer Simpson picture).
You can always circumscribe a square
around this set
if the set is compact.
That's a technical term from topology
which means that the set doesn't extend indefinitely
(obviously you couldn't wrap a square around it then)
and also that the set's boundary belongs to the set.
That's a detail I'll skip today.
Let's just say:
If you can draw it, you can circumscribe it with a square.
If you can't draw it, it's some strange abstract
mathematical object we won't worry about now.
So far for the prelude to the real Toeplitz question.
Oh, wait, here's a little remark I almost forgot.
In case you've been wondering about "at least two points:"
How am I supposed to touch the four edges of a square with two points?
If you touch a vertex of the square, you touch two edges.
You can see here how the tip of the arrow touches two edges.
But now finally the real question.
This one is about inscribed squares.
Here, as an example, we have a silhouette of Africa.
I can now put points on this curve.
I hope you can see them.
We have four red points here.
These four points are arranged in such a way
that they form the vertices of a square.
The obvious question now is:
Is that always possible?
What I want is this:
All four vertices of the square are located on the figure's boundary.
Can I start with an arbitrary figure
and always find an inscribed square for it?
(Another technical term.)
Let's begin with simple figures.
With a circle you all see what we can do.
You can have an inscribed square like this.
Or like this, or like this...
There are some figures where
it is easy to find such a square.
An ellipse also looks pretty harmless.
But it ain't that easy anymore.
You can find an inscribed square,
but it is fairly easy to prove that there's only ONE way to do it.
While we had infinitely many inscribed squares for the circle,
there's only one for an ellipse.
So the obvious question is:
Are the cases where it doesn't work at all?
More formally, this is about closed curves without self-intersections,
think of the boundaries of figures.
The questions is:
Can we always place the vertices of a square on such a curve?
Let's start with something simple that looks like an ellipse.
While I'm doing this:
As you can see, I'm composing this out of straight line segments.
This is only for technical reasons.
After all, I had to write the program somehow.
Hmm, this looks more like a potatoe...
I'm gonna tell you in a minute what I'm actually doing here.
So, this is our curve.
And now I let a horizontal secant run up and down.
We have a straight line segment
going back and forth
which is always connected to the curve on both sides.
And in the next step,
we'll mark the midpoint of this moving secant.
That's the red point.
And we can also
draw the trace of this moving midpoint.
You can also do it like this.
A vertical secant instead.
And you can again mark the midpoint
and draw its trace.
And now I show only the two midpoint traces.
I hope one thing is clear (again due to continuity):
Because one trace starts at the top and goes to the bottom
and the other one starts on the left and goes to the right
they have to intersect somewhere.
I've marked the intersection point here.
Now remember how it was constructed.
This intersection point must lie on the midpoints
of two orthogonal secants.
That means...
we have a rhombus.
The two secants are the diagonals of the rhombus.
And now I'll do what I did in the previous problem:
I'll rotate the curve.
That'll of course change the rhombus.
By the way, you might see "wrong" pictures in between,
but that's due to inaccuracies in the program.
It's not a mathematical error,
it's because computers aren't that good with numbers.
There was a bit of rounding involved to make the program work.
But if you're patient and you're rotating long enough...
...this almost looks like a square.
And here we really have one.
This was a very rough sketch
of how it's supposed to work.
I skipped a few details.
A few strange things can happen.
You can for example have more than one intersection
if the curve is complicated.
And you can have intersections which coalesce and then separate again.
It works nevertheless.
I can't really prove why this always works in a Christmas lecture like this one,
but maybe you'll believe me that it does.
But I'll at least give you a hint
what's so difficult about this.
Put let's first return to our Africa example.
This is the silhouette we already had.
And you can now...
...draw the traces.
And then...
...we have two rhombuses.
Now for the rotation...
We just saw several intersection points.
And here we already have a square.
But I wanted to show you that this ain't easy.
Because this is a Christmas lecture, I'll draw a Christmas tree.
It won't be beautiful.
This is more about what's
problematic with this example.
This is what happens if you buy the tree in the last minute...
I'm used to this.
But I'll add a candle here.
The main reason I'm doing this
is to show how much work
went into this program.
When you're programming the secant movement
you have to be pretty careful.
You always have to keep the contact to both sides.
The secant can't simply "move downwards."
That might not be a continuous motion.
It sometimes has to "go back."
You want to start at the top of the tree
and reach the bottom of the tree
without losing contact
with the right and the left side.
You'll now see how the program is doing this.
Maybe you should write such a program yourself, as an exercise.
Now we're almost done.
So, this always works, but it might be
complicated depending on the curve.
And this will of course
also lead to complicated traces.
Like here.
But I'll remove it now.
Let's see where a square might fit.
Here's one.
Did you expect one there?
Here's a very strange curve.  Let's see...
Here's a square.
The method is always the same,
but the curve can be very complicated.
Back to our question.  It was about closed curves
that don't self-intersect.
What I just showed you
was based on a proof from 1916
from an American mathematician called Emch.
He showed that this will always work
if the curve is piecewise analytic.
Another technical term which says
that the curve consists of finitely many pieces
which can be locally parametrized by power series.
In layman's terms these are curves
that are "pretty smooth"
apart from a few corners.
This applies to all the curves we've seen so far.
But the most general case would be
to show this for arbitrary continuous curves.
Every curve you can "draw in one line"
in an abstract sense.
That's still an unsolved problem.
Toeplitz's original question
was about so-called Jordan curves
which don't have to be "smooth" in any way.
Can I still find an inscribed square for such a curve?
As I said, this is still unsolved.
You might now be wondering where the problem is.
Didn't we just see that this works
for every imaginable curve?
Why is this hard?
It's hard because there exists very strange figures.
This is the Mandelbrot set that you might know.
That's an example of a so-called fractal curve.
I'm showing this one because it's well-known and cute.
But we don't know a lot about its boundary.
There are however fractal curves
of which we know that
Emch's method won't work for them.
One example would be the Weierstraß function.
So this is our third unsolved problem:
Can we place squares on arbitrary Jordan curves?
If it's a curve you can actually draw,
you can place a square on it.
The problematic curves are the ones
you can imagine but not draw.
And so we've reached the last chapter.
The one about ambitious rodents.
The original name is "busy beaver".
It's from an Hungarian mathematician
called Tibor Radó.
1962, the youngest of the four problems.
It's a problem of theoretical computer science.
So, why does this appear in a math lecture?
Well, from a mathematician's viewpoint,
theoretical computer science is math...
And Radó was a mathematician.
Radó used beavers,
but I'd like to use railways.
Railways are kind of important in the history of computer science.
The MIT had the Tech Model Railroad Club
which has been a wellspring of hacker culture.
That's why I thought it was fitting
to use a locomotive for this example.
This locomotive can be programmed.
There's a kind of control desk below
where I can enter instructions for the locomotive.
You should imagine infinite tracks here.
If the locomotive reaches the end on the
right, it re-enters on the left
and vice versa.
The screen is finite, the tracks aren't.
The most basic instruction is "R" for "right".
If I now click "Start"
the locomotive drives to the right
and does the same in each step.
But that's a bit boring.
I'll let it continue a bit so you can see
how the tracks continue "on the other side."
New program, just the letter "L" for "left".
Now it drives into the other direction.
I admit this hasn't been
very exciting so far.
But the programs can have more than one line
and you can have something like "GOTO".
You can for example enter
Drive to the left, then goto line 2.
And there: Drive to the left, then goto line 3.
And there, finally:
Drive to the right, then go back to line 1.
So, "L" is for "drive to the left"
and "2" is for "goto line 2 of the program."
And similarly for "L3" and "R1".
Now let's see what the locomotive will do.
So, she's kind of moving to the left,
but...
OK, still not super-exciting.
But you've already seen the words "white" and "black" here.
I'll type another example program and then I'll tell you what it means.
The second character is the one we already had, for left or right.
The third character is the "GOTO" statement.
The first character is "B" or "W" for "black" or "white".
That's because the locomotive can color the square it's on
before it moves to another square.
It can not only drive,
it can also paint.
This program does the same as before,
but now the squares are colored in black.
And now the last thing you need to know
for your locomotive driver's license.
You can also stop the locomotive.
"H" is for "halt".
So this is the same as before,
but we stop in the third line.
Let's see...
OK.
We'll now focus on programs
where the locomotive stops at the end.
It'll turn out that this is more interesting.
We want the locomotive to stop at one point.
This is also what you'd want as a passenger...
To make this more like a real program
our "control desk" has two columns.
The locomotive can do different things
depending on the color of the square it's on.
I prepared some programs in advance to save me some typing.
Let's try this one.
The first three lines mean that color doesn't matter.
First line: paint black, drive right, goto line 2.
Second line: paint white, drive left, goto line 3.
Third line: paint white, drive right, goto line 4.
Fourth line: IF the current square is white, leave it white, drive left, stay at line 4.
Second column: IF the current square is black, paint it white and halt.
Let's see what this does.
Fascinating what locomotives can do!
Here's another program.
Its most important property is:
It is only two lines long.
Let's see what the locomotive does with these two lines.
Now.
That's it.
Let's remember the following:
The locomotive managed to paint four squares black with two lines of code.
That's the world record!
No program with two lines (that halts)
can blacken more than four squares.
What's so funny about this?
What about three lines?
This is also the world record holder.
Yes, it has to halt.  Otherwise you could paint infinitely many squares...
As I said, the best possible three line program.
And that's it.
With three lines you can paint six squares black.
You know what comes next...
...four lines.
I'll let this one run a little faster.
Yes, you were wondering about the checkbox all the time... :)
Any other plans for tonight?
See, we're done already.
We won't wait for the next one to finish,
but I'll start a five-line program now.
There it is.
Reset.
Start.
We'll check back later.
Computer scientists call S(n) the maximal
number of steps a locomotive can do
if programmed with an n-line program.
And Sigma(n) is the maximal number of squares
it can paint black with n lines of code.
What I just showed you
where the best possible results for Sigma(n) for n=2,3,4.
For n=5 there's a program...
...this one, still running...
...which paints 4098 squares black
(I mixed up the rows)
and needs about 47 million steps for that.
But you see the question marks.
Nobody knows if this is the best possible program!
The fourth unsolved problem.
In case you've been sitting in my
theoretical computer science lectures,
these locomotives might seem familiar to you.
These are Turing machines in disguise... :)
But isn't this strange?
We're talking about super-simple programs
with no more than five lines.
Why is this so hard?
Let's briefly return to the Ramsey numbers
which were about coloring graphs.
That was the problem of computing R(n,n)
and even for n=5 we don't know the answer.
I already said that one could at least
imagine a formula or an algorithm
to compute R(n,n) for any given n.
A computer scientist might say:
"We already have an algorithm:
Just try out all possible graph colorings."
We know that would take very, very long.
But from a theoretical point of view
there is indeed an algorithm for R(n,n).
Can we find a beautiful "closed" formula?
We don't know.  Maybe...
How about the "busy beaver" problem,
S(n) and Sigma(n)?
Is it possible to specify
a formula or an algorithm
to compute S(n) or Sigma(n)?
No, this is impossible!
One can prove there's no way to do this.
As I already said, these little locomotives
are Turing machines in disguise.
And if you could compute these numbers somehow,
then you could solve the so-called halting problem.
But this problem is provably unsolvable.
So we might one day find out the value of Sigma(5)
or even of Sigma(6).
But there'll never be an algorithm
to compute Sigma(n).
No way.
While there might be hope for Ramsey numbers.
And that's all for today.
