YUFEI ZHAO: We're about to embark
on a new chapter in this course where
I want to tell you about Szemeredi's graph regularity
lemma.
Szemeredi's graph regularity lemma
is a very powerful tool in modern graph theory, developed
back in the '70s.
Today I want to show you the statement and the proof
of this graph regularity lemma.
And next time, we'll see how to apply the lemma for graph
theoretic applications.
And we'll also use it to give a proof of Roth's theorem.
The idea of Szemeredi's regularity lemma
is that if you are given a very large graph, G.
And it's a fairly robust theorem, so
any large, dense graph.
And here, "dense" means, let's say, positive x density.
Then it is possible to partition the vertex set of this graph G
into a bounded number of pieces So that G looks random-like
between most pairs of parts.
So for instance, I might produce for you
a partition of the vertex set into some number of parts.
I'll draw five here.
So you give me a graph G. I manage
to produce for you this vertex partition
so that if I look at between a typical pair of parts,
you see here maybe the edge density is close to 0.2
but otherwise, the bipartite graph looks
like a random graph in some precise sense I will describe
in a bit.
And if you look at what the graph looks
like between another pair of parts,
maybe now it's a different x density.
Maybe it's around 0.4.
And again, looks like a random graph with that density.
So in some sense, Szemeredi's regularity lemma
is a universal structural description
that allows you to approximate a graph by a bounded amount
of information.
So that's informally the idea.
And you can already sense that this
can be a very powerful tool.
It doesn't matter what graph you input.
You apply this lemma, and you get an approximate structural
or as later on we'll see, it's also, in some sense,
an analytic description of the graph.
So the first part of today's lecture
will develop just a statement of this regularity lemma.
I'll show you what exactly do I mean by "random-like."
Well, first let me give some definitions.
I denote by the letter e if I input
a pair of vertex sets, x and y.
Here I might, later on, draw the subscript G
if it's clear that I'm always talking about some graph
G. So this is basically the number
of edges between x and y.
And I say "basically" because even though I
will draw and depict everything as if x and y are disjoint
sets, and that's the easiest case to think about,
I'm also going to allow x and y to overlap,
and also allow x and y to be the same set, in which case
you should read a definition as to what this means.
But it's fine to think of it as disjoint sets.
So you're looking at a bipartite graph between x and y.
We're also going to look at the edge density between x and y.
And this is simply the number of edges divided
by the product of the sizes of the sets,
so what fraction of the possible pairs are actual edges.
So from now on, I'll refer to this quantity
as "edge density."
So now, here's the definition of what "random-like"
means for the purpose of Szemeredi's regularity lemma.
So we define a notion of an epsilon regular pair
to be as follows--
throughout, and later on, I will omit even saying this--
G will be some graph.
And we're going to be looking at subsets of vertices of G.
And we say that this pair of subsets of vertices
is epsilon regular, again, in G, but later
on I will even drop saying in G if it's clear
which graph we're working with.
So we say x and y is epsilon regular
if for all subsets A of X, all B subsets of Y that are not
too small, so each at least an epsilon proportion of the sets
that they live in, we find that the x density between A and B
differs from the x density between X and X
by no more than epsilon.
Let me draw you a picture.
I have sets A and B. So I have sets X and Y in my graph G.
And I want to say that the edges between X and Y are epsilon
regular, so it's random-like, if the following holds--
that whenever I pick a subset A in the left set
and a subset B of the right set, the edge density
between A and B is approximately the same as the overall edge
density between X and Y.
So in particular, this bipartite graph, for instance,
is not really dense in one part and really sparse
in another part.
Somehow the edges are evenly distributed
in this precise manner.
So that's the definition of epsilon regular.
Yes, question.
AUDIENCE: What is the epsilon for the size of A the same as
epsilon for [INAUDIBLE]?
YUFEI ZHAO: The question is here,
why are we using the same epsilon here, here, and there?
And that's a great question.
So that's mostly out of convenience.
So you could use different parameters.
And they do play somewhat different roles,
but at the end, we'll generally be looking
at one type of epsilons.
So we just make our life easier.
So you could extend the definition
by having an epsilon comma eta, if you like,
but it will not be necessary for us, and mostly
for simplification.
Any more questions?
All right.
Now if you have a pair x, y that is not epsilon regular,
I just want to introduce a piece of terminology.
So you can read from the definition what it means
to be not epsilon regular.
And sometimes I will say "epsilon irregular,"
but to be precise, I'll stick with not epsilon regular.
Then we can exhibit this A and B that witnesses
the irregularity.
So if x, y is not epsilon regular, then
their irregularity as, we say it's "witnessed by" some pair
A in X and B in Y, satisfying--
basically, you read the definition,
and such that the density between A and B
differs quite a bit from the density between X and Y.
So when I say "to exhibit" or "to witness irregularity,"
that's what I mean.
Now, there's a bit of an unfortunate nomenclature
in graph theory, where previously, we
said "irregular graphs" to mean that every vertex is degree D.
And now we say "epsilon regular" to mean this.
Sorry about that.
These are both standard, so usually from context,
it's clear which one is meant.
So this is what it means for a single pair of vertex sets
to be epsilon regular.
But now I give you a graph.
And I give you a partition of the vertex set.
So what does it mean for that partition
to be epsilon regular?
And here's the second definition.
So an epsilon regular partition, we say that a partition--
and generally, I will denote partition
by curly letters such as that, P. So the partition
will divide a vertex set into a bunch of subsets.
So we say that that partition is epsilon regular
if the following is true--
if I sum over all pairs of irregular
or pairs of vertex sets that are not epsilon regular,
so over Vi, Vj not epsilon regular, and sum
over the product of their sizes, then what I would like
is for the sum to be at most epsilon times
the number of pairs of vertices in G. In other words,
a small fraction of pairs of vertices,
not necessarily edges, but just pairs of vertices,
lie between pairs of vertex parts
that are not epsilon regular.
So for instance, if you do not have epsilon--
if all of your pairs are epsilon regular,
then the partition is epsilon regular.
But I do allow a small number of blemishes.
And that will be necessary.
Just to clarify a subtle point here,
here I do allow in the summation i equals to j,
although in practice it doesn't really matter.
You'll see that it's not really going to come up as an issue.
And one of the reasons that it's not
going to come up as an issue is usually
when we apply this lemma, we're going to have a lot of parts.
In fact, we can make sure that there
is a minimum number of parts.
And if none of the parts are too big, then having i equals to j
contributes very little to that sum anyway.
In particular, if all the set sizes
in this partition are roughly the same--
so if they're all roughly 1 over k fraction of the entire vertex
set--
then that statement up there being epsilon regular partition
up to changing this epsilon is basically the same
as saying that fewer than epsilon fraction of the pairs
Vi, Vj are not epsilon regular.
And here, if k is large enough, I
can even let you make i and j different.
It's not going to affect things after small changes in epsilon.
So when it comes to-- so for people who are seeing
Szemeredi's regularity lemma for the first time--
I think that's maybe all of you, or most of you--
I don't want you to focus on the precise statements
so much as the spirit of the lemma.
Because if you get too nitty gritty with
is that the same as that epsilon, you get very confused
very quickly.
So I want you to focus on the spirit of this lemma.
I will state everything precisely,
but the idea is that most pairs are not epsilon regular.
And don't worry too much about if you are allowed
to take i equals to j or not.
So now we're ready to state Szemeredi's regularity lemma.
And it says that for every epsilon,
there exists some constant M depending only on epsilon such
that every graph has an epsilon regular partition
into at most M parts.
You give me the epsilon, for example 1%,
and there exists some constant such
that every graph has a 1% regular partition
into a bounded number of parts.
In particular-- and this is very important,
make sure you understand this part--
that the number of parts does not depend
on the size of the graph.
Now, it's true that for some graphs, maybe
you do need very many parts.
But the number of parts does not get substantially bigger,
or does not exceed this bound, even if you look at graphs
that have unbounded size.
So it is really a universal theorem in the sense
that it's independent of the size of the graph.
Any questions about the statement of this theorem?
Yes.
AUDIENCE: So in the informal statement at the beginning,
you said G was a large, dense graph.
YUFEI ZHAO: That's right.
AUDIENCE: Is the dense condition appropriate anywhere in here?
YUFEI ZHAO: So the question is, why did I say
that G is a large, dense graph?
And that's a great question.
And that's because if G had a sub-linear number of edges,
then I claim that all--
if you look at the definition of epsilon regular pair,
and your epsilon is a constant, and if your edge densities
are sub-linear, then all of these guys,
they are little o of 1.
They go to 0.
So trivially, you will satisfy the epsilon regular condition.
So if your graph is sparse--
sparse in the sense of having sub-quadratic number of edges--
then you trivially obtain epsilon regularity.
And so the theorem is still true.
It's just not meaningful.
It's just not useful.
But there are settings where having sparse graphs--
and we'll come back to this later in the course--
it's important to explore what happens to sparse graphs.
Yeah.
AUDIENCE: So that M is independent of G.
YUFEI ZHAO: Yes, M is independent of G. M depends
only on epsilon.
AUDIENCE: M is really large, but there's no enough vertices
in the graph.
YUFEI ZHAO: OK, question is, what
happens when M is very large, but there are not
enough vertices in the graph?
Well, if your M is a million, and your graph only
has 1,000 vertices, what you can do
is have every vertex be its own part.
Every vertex is its own part, a singleton partition.
And you can check that that partition satisfies
the properties.
Every pair is a single edge and it's epsilon regular.
Yeah.
AUDIENCE: So in the definition, is it
sort of like all or nothing?
You can either [INAUDIBLE] epsilon regularity [INAUDIBLE]..
Do you get anything where if you, like, say,
make this more continuous, so you allow for it to be--
you quantify how irregular it is, and then can
you make [INAUDIBLE]?
YUFEI ZHAO: OK, so my understanding what
you're asking is in the definition up there,
the sum is--
we put the pair in the sum of this epsilon regular,
and otherwise don't put it.
Is there some gradual way to put some measure of irregularity
into that sum?
And there are versions of regularity lemma that do that,
but they are all, in spirit, morally
the same as that one there.
Yeah.
AUDIENCE: In the informal definition,
what does "random-like" mean?
YUFEI ZHAO: So in the informal definition,
what does "random-like" mean?
This is the formal definition of what "random-like" means.
So actually later on in the course, one of the chapters
will explore what pseudo-random graphs are.
So pseudo-random graph, in some sense,
means graphs that are not random,
but behave in some sense like random.
So "random-like" generally just means that in some aspect,
in some property, it looks like a random object.
And this is one way that something can look like random.
So a random graph has this property,
but random graphs also have many other properties
that are not being exhibited in this definition.
But this is one way that graph can look like random.
So that's a great question.
And we'll come back to that topic later in the course.
All of these are great questions.
So Szemeredi's regularity lemma, the first time you see it,
it can look somewhat scary.
But I want you to try to understand
it more conceptually.
So please do ask questions.
Before diving into the proof, I want to make a few more remarks
about a statement.
It is possible to--
we will prove this version of the regularity lemma.
But as I mentioned, it is the spirit of the regularity lemma
that I care more about.
And it's a very robust statement.
You can add on extra declarations that somehow
doesn't change the spirit.
And the proof will be more or less the same,
but for various applications will be slightly more useful.
So in particular, it is possible to make
the partition equitable.
And "equitable partition" sometimes
is also called an "equipartition," meaning
that it has such that all the Ai's, all
the Bi's have sizes differing by at most 1.
So basically, all the parts have the same size up to at most 1,
because of divisibility.
So let me state a version of regularity lemma
for equitable partitions.
So for every epsilon in m, little m0,
there exists a big M such that every graph
has an epsilon regular equitable partition of the vertex set
into k parts, where k is at least little m, so I can
guarantee a minimum number of parts,
and at most some bounded number.
Again this bound may depend on your inputs epsilon and m0,
but it does not depend on the graph itself.
And you see the slightly stronger conclusion
for many applications is more convenient,
to use this formulation.
And I will comment on how you may modify the proof that we'll
see today into one where you can guarantee equitability.
And you see that for this m, little m0 too small,
for example, if it's somewhat larger than 1 over epsilon,
when you look at the definition of epsilon regular partition,
it suffices to check that at most epsilon k squared,
epsilon fraction of the pairs, Vi, Vj
is epsilon regular over i different
from j, again up to changing epsilon,
let's say, by a factor of 2.
So all of these definitions are basically
the same up to small changes in the parameters.
Next time, we'll see how to apply the regularity lemma.
And we will apply it in the first form,
but you see the second form guarantees
you a somewhat stronger conclusion,
and sometimes more convenient to use.
So for example on the homework problems,
if you wish to use the second form, then please go ahead.
Just make your life somewhat easier,
but it essentially captures all the spirit
of Szemeredi's regularity.
Any questions so far?
I want to explain the idea of the proof of the regularity
lemma.
And this is a very important technique
in this area called the "energy increment argument."
Here's the idea.
We start with some partition, so for example,
the trivial partition--
and by that I mean you only have one part.
All the vertices are in one part.
You're not doing anything to the vertex set.
It's one gigantic part.
Or if you're looking at some other variant,
you can easily modify the proof.
So for example, you can also look at an arbitrary partition
into little m0 parts, if you wish to have
that as your starting point.
So or I'm saying is that this proof is fairly robust.
And we're going to do some iterations.
So as long as your partition is not epsilon regular,
we will do something to the partition to move forward.
And what we will do is look at each pair
of parts in your partition that's not epsilon regular.
Well, if they're not epsilon regular,
then I can find a pair of subsets
which are denoted by the A's that witnesses
this non regularity, that witnesses the irregularity.
And we start with some partition.
So now let us refine the partition
into a partition in even more parts
by simultaneously refining the partition using
all of these Ai, j's that we found in the step above.
So you start with some partition.
If it is not regular, I can chop up
the various parts in some way.
So I start with some partition over here.
And what we are going to do is, let's say between these two,
it's not epsilon regular, so I can
find some pairs of vertex sets that exhibits the irregularity.
I chop it up.
And I can keep further chopping up the rest of the parts.
If these two parts are not epsilon regular,
then I chop it up like that.
And I can keep on doing it.
And originally, I have three parts.
Now I have 12 parts.
And this is a refined partition.
And now I repeat until I am done.
I am done when I obtain a partition that
is epsilon regular.
Now, the basic question when it comes to the strategy
is, are you ever going to be done?
When are you going to be done?
And if this process goes on forever
or goes on for a very long time, then you
might have a lot of parts.
But we want to guarantee that there
is a bounded number of parts.
So what we will show is that--
to show that you have a small number of parts,
in other words, why does this process even stop--
and in particular, we want it to stop
after a small number of steps, after a bounded number
of steps.
And to do this, we will define some notion called
an "energy" of a partition.
And this energy will increase.
So first of all, the energy is some quantity
that we'll define that lies between 0 and 1.
It's some real number lying between 0 and 1.
And each step, the energy goes up by some specific quantity.
Therefore, because the energy cannot increase past 1,
this iteration stops after a bounded number of steps.
And once it's done, we end up with
a epsilon regular partition.
So that's the basic strategy.
And what I want to show you is how to execute that strategy.
Any questions so far?
Yes.
AUDIENCE: Just to clarify [INAUDIBLE] a bit,
if some Vi's into non-epsilon regular partitions,
is it possible for Ai,j and Aik to overlap somehow, right?
Just kind of make those into three partitions?
YUFEI ZHAO: So if I understand correctly,
you are worried about between different pairs,
you might have interactions.
AUDIENCE: Yeah.
YUFEI ZHAO: So you have seen the proof,
but I think this is actually a very important and somewhat
subtle point, is that I do not refine at each step,
I find a pair of witnessing sets.
I find all of these witnessing sets all at the same time,
and I refine everything all at once.
AUDIENCE: OK, so it's like if you do
have overlap between two witnessing sets, that's OK?
YUFEI ZHAO: That is OK, because this step doesn't care.
If you have two witnessing sets that overlap, that is OK.
We'll see the proof.
Yes.
AUDIENCE: Do you just find one pair
of witnessing sets for each Vi, Vj,
even though there might be more?
YUFEI ZHAO: Question is, do we find
just one pair of witnessing sets even
though there could be more?
And the answer is, yes.
We just need to find one.
There could be lots.
So if it's not epsilon regular, it
might be very not epsilon regular.
And in fact, being a witnessing set is a fairly robust notion.
If you just take out a small number of vertices,
it's still a witnessing set.
Any more questions?
Great.
So let's take a quick break and then we'll see the proof.
Let's get started with the proof of Szemeredi's regularity
lemma.
And to do the proof, I want to develop
this notion of energy which you saw in the proof sketch.
So what do I mean by "energy?"
First, if I-- let me define some quantities.
If I have two vertex subsets, U and W,
let me define this quantity, q, which is basically
the edge density squared.
But I normalize it somewhat according
to how big U and W are.
I'm going to use the letter and N to denote
the number of vertices in G. So this is some cube.
And for partitions, if I have a pair of partitions,
Pu of U into k parts, and the partition Pw of W into l parts,
I set this q of Pu and Pw to be the quantity where I sum over
basically all pairs, one part from U,
one part from W of this q between Ui and Wj.
So this is the density squared.
And I'm taking some kind of weighted average
of the squared density.
So here is a weighted average.
If you prefer to think about the special case
where this partition is an equipartition,
then it is really the average of these squared densities.
It's a mean square density.
And finally, for a partition P of the vertex set of G
into m parts, we define this q of this partition P
to be q of P with itself according
to the previous definition.
Or in other words, I do this double sum, i from 1 to m,
j from 1 to m, q of Vi, Vj.
And this is the quantity that I will call
the "energy" of the partition.
It is a mean squared density, some weighted mean
of the edge densities between pairs
of parts in the partition.
You might ask, why is it called an energy?
So you might see from this formula here,
it's some kind of a mean square density,
so it's some kind of an average of squares.
So in particular, it's some kind of an L2 quantity.
And there's a general phenomenon in mathematics,
I think borrowed from physical intuitions,
that you can pretty much call anything that's
an L2 quantity an energy.
And so that's, I think, where the name comes from.
So this is the important object for our proof.
And let's see how to execute a strategy, the energy increment
argument outlined on the board over there.
So we want to show that you can refine a partition that
is not epsilon regular in such a way that the energy goes up.
And to do that, let me state a few lemmas regarding
the energy of a partition under refinement.
And the point of the next several lemmas
is that the energy never decreases under refinement,
and it sometimes increases if your partition is not
epsilon regular.
So the first lemma is that if you
look at the energy between a pair of partitions,
it is never less than the energy between the two vertex sets.
So for instance, if you have U and W like that,
and I partition them into Pu and Pw,
and I measure the energy, just basically
the squared density between U and V versus summing up
the individual squared densities after the partition,
the left side is always at least as great as the right side.
So this is really a claim.
It's a fairly simple claim about convexity,
but let me set it up in a way that will
help some of the later proofs.
So let me define a random variable, which
I call Z, in the following way.
So here's a process that I will use
to define this random variable.
I will select x, little x, to be a vertex uniformly chosen
from U, from the left vertex set.
And I will select a vertex y uniformly chosen from W.
x and y, they fall into some part in the partition.
So suppose Ui is the part where x i
falls, and Wi is the set in the partition where y falls.
So Ui is a member of this partition.
Wi is a member of the other partition of W.
Then I define my random variable Z
to be the x density between Ui and Wj.
So it's Wj.
So that's the definition.
So pick x randomly.
Pick y randomly.
Suppose x falls in Ui.
Suppose y falls in Uj.
Then Z is the x density between these two parts.
So Z is some random variable.
Let's look at properties of this random variable.
First, what is this, it's expectation?
It's a discrete random variable.
And you can easily compute all of these quantities
by just summing up according to how Z is generated.
So I look overall, i and j.
What's the probability that x falls in Ui?
It is the size of Ui as a fraction of U.
What's the probability that y falls in Wj?
It's the size of Wj as a fraction of size W.
And then Z is this quantity here.
So this is what I find to be the expectation of Z.
But you see the density multiplied by the product
of the vertex set sizes, that's just the number of edges
between U and W. And you sum over all the i, j's.
So that, which is simply the edge density between U and W.
So that's the expectation of the Z variable.
On the other hand, what's the second moment?
In other words, what's the expectation of the square of Z?
Again, we do the same computation.
First part is the same.
The second part now becomes a d squared.
And look at how we define energy.
This quantity here is basically the energy q
between the partition U and the partition of W,
except there's normalization.
That's not quite the same as the one we used before.
So we will just put in that normalization.
So now you compare the expectation
of Z versus the expectation of Z squared.
And we know by convexity that the expectation of Z squared
is at least as large as the expectation of Z,
that quantity squared.
But if you plug in what values you get for these two guys,
you derive the inequality claimed in Lemma 1.
You have to cancel some normalization factors,
but that's easy to do.
So that's the first lemma.
So the first one is just about a pair of parts,
and identify partition, each part,
what happens to the energy between this pair.
And the second one is a direct corollary of the first one.
It says that if you have a second partition,
P prime that refines P, then the energy of the second partition,
the refinement, is never less than the energy
of the first partition.
And it is a direct consequence of the first lemma,
because we simply apply this lemma to every pair of parts
in P. Between every pair of parts,
the energy can never go down.
So overall, the energy does not go down.
And finally, so far, we've just said that the partitions can
never make the energy go down.
But in order to do this proof, we
need to show that the energy sometimes goes up.
And that's the point of the third lemma.
The third Lemma tells us that you can get an energy boost.
So this is the Red Bull Lemma.
You can get an energy boost if you are feeling irregular.
So if U, W is not epsilon regular,
and this epsilon regularity is witnessed by U1 in U and W1
in W, then I claim that the energy obtained by chopping U
into U1 and its complement against W1,
against the complement of W1 and W, so here again U and W,
I find a witnessing set for their irregularity.
And now I partition left and right according to--
chop each part into two.
So this energy between this partition
into two on both sides is bigger than the original energy
plus something where we can gain.
And this something where we can gain turns out
to be at least epsilon raised to the power
4 times the size of U, size of W, divided by n squared.
Can you prove it?
Let's define Z the same as in the previous proof,
as in the proof of Lemma 1.
In Lemma 1, we just used the fact that the L2 norm of Z,
the expectation of the square, is at least
the square of the expectation.
But actually, there are differences in it.
It's called a "variance."
The variance of Z is the difference
between these two quantities.
I know that it's always non-negative.
So if you look at how we derived the expectation of Z
and expectation of Z squared, you immediately
see that its variance we can write
as, up to a normalizing factor, the difference
between this energy on one hand and the energy between U and W,
namely the mean square of the normalization.
On the other hand, a different way to calculate the variance
is that it is equal to the expectation of the deviation
from the mean squared.
So let's think about the deviation from the mean.
I am choosing a random vector in, on the left, U,
and another random point, random vertex on the left,
and a random point on the right.
In the event that they both lie in the sets
that witness the irregularity, so in the event
where x falls here and y falls here,
which occurs with this probability,
see that this quantity here is equal to the density between U1
and W1 minus the density--
and this expectation of Z is just
the density between U and W. So interpreting this expectation
for what happens when x falls in U1
and when y falls in W1, ignoring all the other events,
because the quantity is always non-negative everywhere.
But now from the definition of epsilon regularity,
or rather the witnessing of epsilon irregularity,
you see that this U1 is at least an epsilon fraction of U. W1's
at least an epsilon fraction of W.
And this final quantity here is at least epsilon inside, so
at least epsilon squared.
So here we're using all the different components
of the definition of epsilon regular.
Yes.
AUDIENCE: What happens if we're dividing with more witnessing
sets?
YUFEI ZHAO: So you're asking what
happens if we divide with more witnessing sets?
So hold onto that thought.
So right now, I'm just showing what happens
if you have one witnessing set.
Any more questions?
So here we have epsilon to the 4th.
And if you're putting the normalization
comparing these two interpretations,
you'll find the inequality claimed by the lemma.
So now we are ready to show the key part of this iteration.
I'll show you precisely how this iteration works, and show
that you always get an energy boost in the overall partition.
So I'll call the next one Lemma 4.
And this says that if you have a partition P of the vertex set
of G into k parts, if this partition is not epsilon
regular, then there exists a refinement called Q where every
part V sub i is partitioned further into at most 2 the k
parts, and such that the partition of Q--
so the energy of the new partition Q--
increases substantially from the previous partition P.
And we'll show that you can increase
by at least epsilon to the 5th power,
some constant in epsilon.
So if you look at the strategy up there,
if you can do this every step, then that
means that the number of iterations
is bounded by 1 over epsilon to the 5th power.
So to prove this lemma here, we will use the three lemmas
up there and put them together.
So for all the pairs i, j such that V sub i,
V sub j is not epsilon regular, as outlined
in the proof in the outline up there,
we will find this A superscript i, j in Vi,
and A superscript j, i in Vj that witness the irregularity.
So do this simultaneously for all pairs i, comma, j,
where the Vi, Vj is not epsilon regular.
Now what we're going to define Q as is the common refinement.
So take all of, just as indicated
in that picture up there, simultaneously
take all of these A's and use them to refine P. Starting
with P, starting with a partition you have,
simultaneously cut everything up using
all of these witnessing sets.
Now, we only have witnessing pairs for pairs
that are not epsilon regular.
If they're epsilon regular, you don't worry about them.
One of the claims in the lemma now is--
this is the Q that we'll end up with.
We'll show that this Q has that property.
So one of the claims in the lemma
is that every Vi is partitioned into at most 2 to the k parts.
So I hope that part is clear, because how are we
doing the refinement?
We're taking Vi.
It's divided into parts using these A i,
j's, one j coming from each pair that is irregular with Vi.
So I'm cutting up Vi using at most k sets,
so one coming from each of the other possible.
Maybe fewer than k-- that's fine-- but at most k sets
are used to cut up each Vi.
So you have at most 2 to the k parts
once you cut everything up.
But the tricky part is to show that you get an energy boost.
So let's do this.
How do we show that you get an energy boost?
We're going to put the top three lemmas together.
First, we want to analyze the energy of Q.
So let's write it out.
So the energy of Q is the sum over this energy of individual
partitions of the Vi's.
And by this P sub Vi, P sub Vj, I mean the partition of Vj
given by Q. So what happens after you cut up the Vi--
that's what I mean by P sub Vj, or rather I should call it
Q sub Vi, Q sub Vj.
By Lemma 2, we find that--
so let me separate them into two cases.
The first case sums over i, j such that Vi,
Vj is epsilon regular.
And by Lemma 1, so here we're using Lemma 1,
we find that this quantity here cannot be less than the Q
of Vi, Vj.
So take those two parts.
Before and after the refinement by Q,
the energy cannot go down.
So I don't worry too much about pairs that are epsilon regular.
But no let me look up here that are not epsilon regular.
So what we will do now is even though--
so let's look at that picture up there.
So let's focus on what I drew in red.
So let's focus between 1 and 2.
So suppose the shaded part is the witnessing sets.
The witnessing sets got cut up further
by other witnessing sets.
But I don't have to worry about them because Lemma 2 or Lemma
1, really, tells me that I can do an inequality where
I go down to just comparing the energy between this partition
of two parts, this single witnessing set
and its complement, versus what happens in its partner.
So in other words, over here, the Q of this pair,
I am saying that it is no less than if I just
look at what happens if you only cut up these two sets using
the red lines.
Let's go on.
Applying Lemma 3, the energy boost lemma,
the first part stays the same.
So this first part stays the same.
And the second part, now, because I'm
looking at witnessing sets for irregularity,
I get this extra boost.
So this goes back to one of the questions asked earlier,
where in Lemma 3, I don't have to now worry about what happens
if you have further cuts, because I only need to worry
about the case where I only have a single cut
between the epsilon irregular pairs.
So putting it together, we see that the previous line
is at least, if you sum over the Q's of all the pairs
plus this extra epsilon to the 4th term for all pairs
that are not epsilon regular.
I'm applying monotonicity of energy for the types,
for pairs that are epsilon regular,
an energy boost for pairs that are not epsilon regular.
And for the latter type, I obtain this boost.
Now remember what's the definition of an "epsilon
regular partition."
Unfortunately, it's no longer on the board,
but it says that this sum over here,
if it is an epsilon regular partition,
it is at most epsilon.
So if it is not epsilon regular, we can lower bound it.
And that's indeed what we will do.
The first sum here is, by definition,
Q of the partition P. And the second sum,
by the definition of epsilon regular,
is at least epsilon to the power 5.
So here we're using the definition
of epsilon regular partition, namely,
that a large fraction, so at least an epsilon
fraction, basically, of pairs of vertex sets
are not epsilon regular, but in this weighted sense.
And that finishes the proof of Lemma 4 up there.
Any questions so far?
All right, so now we are ready to finish everything
off, and prove Szemeredi's regularity lemma.
So let's prove Szemeredi's regularity lemma.
Let's start with the trivial partition,
meaning just one large part.
And we are going to repeatedly apply
Lemma 4 whenever the partition at hand
is not regular, whenever the current partition is not
epsilon regular.
So let's look at its energy.
The energy of this partition--
so this is a weighted mean of the edge density squared,
so it always lies between 0 and 1,
just from the definition of energy.
On the other hand, Lemma 4 tells us
that the energy increases by at least epsilon to the 5th power
at each iteration.
So this process cannot continue forever.
So it must stop after at most epsilon to the minus 5th power
number of steps.
And when we stop, we must result in an epsilon regular
partition, because otherwise, you're
going to continue applying the lemma and push it even further.
And that's it.
So that proves Szemeredi's graph regularity lemma.
Question.
AUDIENCE: It's going to be some really big value of M.
YUFEI ZHAO: OK, let's talk about bounds.
So let's talk about how many parts.
So how many parts does this proof produce?
We can figure it out.
So we have some number of steps.
Each step increases the number of parts by something.
So if P has k parts, so then Lemma 4 refines
P into at most how many parts?
AUDIENCE: 2 to the k
YUFEI ZHAO: Yeah, so k times 2 to the k.
And I have many iterations of this guy.
So some of you are already laughing,
because it's going to be a very large number.
In fact, because it's going to be so large,
it makes my calculations slightly more convenient.
It really doesn't change the answer
so much if I just bound k to the 2 to the k by 2 to the 2
to the k.
So the final number of parts is this function
iterated on itself epsilon to the minus 5 times.
So it's a power of 2 of height at most 2 to the epsilon
to the 5.
It's a finite number, so it depends only on epsilon and not
on the size of your graph.
And this is the most important thing.
It does not depend on the size of your graph.
It is quite large.
In fact, even for reasonable values of epsilon,
like 1% or even 10%, this number is astronomically large.
And you may ask is it really necessary,
because we did this proof, and it came out fairly elegantly,
I would say it, how the proof was set up.
And you arrived at this finite bound.
But maybe there's a better proof.
Maybe you can work harder and obtain somewhat better bounds.
So you can ask, is it possible that the truth is really
somehow much smaller?
And the answer turns out to be no.
So there is a theorem by Tim Gowers
which says that there exists some constant.
The precise statement, again, is not so important,
but based on what I just said, you cannot improve this bound
given by this proof.
So for every epsilon small enough,
there exists a graph whose epsilon regular partition
requires how many parts?
So the number of parts at least this tower of 2 of height
some epsilon to the minus c.
So really it's a tower of exponentials of size,
essentially polynomial in 1 over epsilon.
So maybe you can squeeze the 5 to something less.
Actually, we don't even know if that's the case,
but certainly you cannot do substantially better than what
the proof gives.
So Szemeredi's regularity lemma is an extremely powerful tool.
And we'll see applications that are basically
very difficult to prove.
And for some of these applications,
we don't really know other proofs
except using Szemeredi's regularity lemma.
But on the other hand, it gives terrible quantitative bounds.
So there is a lot of interest in combinatorics
where once you see a proof that requires Szemeredi's regularity
lemma, or that is first proved using this technique,
to ask can it be used using some other technique?
In fact, Szemeredi himself has worked a lot in that direction,
trying to get rid of the uses of his lemma.
Any questions?
AUDIENCE: How could you modify it for equipartitions?
YUFEI ZHAO: OK, great.
Question is, how can we modify it for equipartitions?
So let's talk about that.
So it's a fantastic question.
So look at this proof and see what can we
do if we really want all the parts to have
roughly the same size, let's say differing by at most 1.
So how to make the epsilon regular partition equitable?
Any guesses?
Any attempts on what we can do?
I mean, basically it's going to follow this proof.
As I said, the spirit of Szemeredi's regularity lemma
is what I've shown you.
But the details and executions may vary somewhat
depending on the specific purpose you have in mind.
Yeah.
AUDIENCE: Can we just add--
[INAUDIBLE] add things to the smaller part
because we know that-- by the fact
that it's not [INAUDIBLE] that parts aren't too small?
YUFEI ZHAO: OK, so you're saying we're going to add something
or to massage the partition to make it epsilon--
AUDIENCE: Add vertices to the smaller parts of the partition.
YUFEI ZHAO: Add vertices to the smaller parts of the partition,
now when are you going to do that?
AUDIENCE: When they're-- like so you do the refinement,
then when they're not [INAUDIBLE]
YUFEI ZHAO: So you want to do this
at every stage of the process.
AUDIENCE: Yes.
[INAUDIBLE]
YUFEI ZHAO: I like that idea.
So here's what we're going to do.
So we still run the same process.
So we're going to have this P, which is the current partition.
So I have current partition.
And as before, we initially have it
as either the trivial partition, if you like,
or m arbitrary equitable parts.
Start with something where you don't really
care about anything except for the size.
And you run basically the same proof,
where if your P is not epsilon regular,
then do what we've done before, so basically
exactly the same thing.
We refine P using pairs witnessing regularity,
same as the proof that we just did.
And now we need to do something a little bit more
to obtain equitability.
And what we will do is right after--
so each step in iteration, right after we do this refinement,
so after we cut up our graph where maybe some of the parts
are really tiny, let's massage the partitions somewhat
to make them equitable.
And to make our life a little bit easier,
we can refine the partition somewhat further
to chop it up into somewhat smaller resolution.
And this part, you can really do it
either arbitrarily or randomly.
Some ways may be slightly easier to execute,
but it doesn't really matter how you do it.
It's fairly robust.
You refine it further.
And basically, I want to make it equitable.
Sometimes, you can just do that by refining,
but maybe if you have some really small parts,
then you might need to move some vertices around,
so I call that "rebalancing."
So move and merge some vertices, but only a very small number
of vertices, to make equitable.
So you run this loop until you find that your partition is
epsilon regular.
Then you're done.
Whenever you run this loop, because we're
doing the second step, your partition
is always going to be equitable.
But we now need to control the energy again
to limit the number of steps.
And the point here is that the first part still
is exactly the same as before, where the energy goes up
by at least epsilon to the minus 5.
But the second part, the energy might go down,
because we're no longer refining, just refining.
Because we're doing some rebalancing.
But you can do it in such a way that the amount of rebalancing
that you do is really small.
You're not actually changing the energy by so much.
So I'll just hand wave here, and say
that we can do this in such a way where the energy might
go down, but only a little bit.
So you're only changing a very small number
of vertices, very small fraction of vertices.
So if you change only an epsilon fraction of vertices,
you don't expect the energy, which
is something that comes out of summing pairs of vertex parts,
to change by all that much.
So putting these two together, you
see that the energy still goes up by, let's say, at least 1/2
of epsilon to the 5th power.
And so then, the rest of the proof runs the same as before.
You finish in some bounded number of steps.
And you result in an equitable partition
that's epsilon regular.
I don't want to belabor the details.
I mean, here, there's some things to check,
but it's, I think, fairly routine.
It's is more of an exercise in technical details.
But the thing that actually is somewhat important
is there's a wrong way to do this.
I just want to point out that what's the wrong way
to do this, is that you apply regularity lemma,
and you think now it has something
that's epsilon regular.
Then I massage it to try to make it equitable at the end.
And so if I don't look into the proof,
I just look at a statement of Szemeredi's regularity lemma,
and I get something that's epsilon regular,
I say I'm just going to divide things up a little
bit further, that doesn't work.
Because the property of being epsilon
regular is actually not preserved under refinement.
So look at the definition.
You have something that's epsilon regular.
You refine the partition.
If might fail to be epsilon regular.
So you really have to take into the proof to get equitability.
So just to repeat, a wrong way to try to get equitability
is to apply regularity lemma, and at the end,
try to massage it to get equitable.
That doesn't work.
Next time, I will show you how to apply Szemeredi's regularity
lemma.