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PROFESSOR: We've been spending
quite a few lectures so

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far discussing Szemeredi's
regularity lemma,

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it's applications in variants
of the regularity Lemma.

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So I want to spend one more
lecture, before moving on

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to a different
topic, to tell you

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about other extensions
and other perspectives

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on Szemeredi's regularity lemma.

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So hopefully, you
all will become

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experts of the regularity
lemma, especially

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the next homework
problem set where

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there will be plenty of
problems for you to practice

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using the regularity lemma.

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So one of the things that I
would like to discuss today

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is a hypergraph extension of
the triangle removal lemma.

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So as we saw, the
triangle removal lemma

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was one of the important
application of Szemeredi's

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graph regularity lemma.

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OK.

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So that works inside graphs,
but now let's go to hypergraphs.

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In particular, even in
the case of three uniform

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hypergraphs where--

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so let's set some terminology.

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So when r uniform hypergraph, or
simply abbreviated as r graph,

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consists of a vertex
set and an edge

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set where the edge set
consists of r couples.

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So the edges are r element
subsets of the vertex set.

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So r equals the 2 corresponds
to the graph case.

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And you can talk about
sub-graphs or density,

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various counts all analogously
to how we did it for graphs.

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And you can imagine what
the hypergraph removal

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lemma might look like.

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So let me write
down a statement.

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That for all r graph h and
epsilon bigger than zero,

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there exists a delta such that--

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so the last time, there was
some complaints about sentences

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going on too long, so let
me try to cut sentences

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into smaller parts.

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So if g is an end vertex r graph
with small number of copies

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of each, so h is sub-graphs,
then g can be made h free.

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So far, everything's still
the same as in the graph case,

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but now in the graph
case, the number of edges

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is quadratic, most quadratic.

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Here, it's a most
n to the r, so I

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want to make this
graph, this r graph,

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h free by removing
less than epsilon

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n to the r edges from g.

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So that's the statement of
the hypergraph removal lemma.

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So it's an extension of
the graph removal lemma.

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Any questions about
the statement?

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So before discussing
the proof, let

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me show you why you might
care about this statement.

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And we used the triangle removal
lemma to deduce Roth's theorem.

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Remember, there was a graph
theoretic set up where

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start with a 3-AP-free subset.

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We set up a graph,
and then that graph

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has some nice
properties that allows

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us to use a corollary of
the triangle removal lemma,

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namely the corollary that says
that if you have a graph, where

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every edge sits on
exactly one triangle,

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then it has a sub
quadratic number of edges.

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So we can do a similar
type of deduction

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showing that the hypergraph
removal lemma implies

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Szemeredi's theorem.

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So that's what we'll do.

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So let's deduce
Szemeredi's theorem

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from the hypergraph
removal lemma.

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So recall Szemeredi's
theorem says

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that for every fixed
positive integer k,

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if a is a subset of 1
through n, that is kp free,

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has no k term
arithmetic progressions,

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then the size of
a is sub-lineal.

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Instead of illustrating how to
do this proof for general k,

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I'm just going to do it for
the case of k close to 4.

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And you can look at the
proof and it will be clear

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how to generalize.

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So we'll just illustrate
the proof for 4-APs.

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Now, before even showing
you what this proof looks

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like, you might wonder,
do we really need

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the hypergraph removal lemma?

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Could it be that with
the graph removal

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lemma and a more clever
choice of a graph,

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you could prove Szemeredi's
theorem using just that.

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So we set up some
graph previously

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where triangles
correspond to 3-APs.

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Maybe you can set
up some other graph

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where some other kinds of
structure correspond to 4-APs.

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And it turns out the
answer is emphatically no.

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So there's a very good
reason for this that for--

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these are things which we might
go into more when we discuss

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additive combinatorics,
but 4-APs

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is a pattern that's sometimes
called complexity two.

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Where as 3-APs is a pattern
which is called complexity one.

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I won't go into the precise
definitions of what this means,

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but the message is that you
cannot prove the 4-AP theorem

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with just graph machinery.

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You really have to use
something stronger.

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That there is a very real sense
in which just a graph removal

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lemma is at least-- or
Szemeredi's theorem,

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Szemeredi's regularity
lemma is not enough.

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And so we really do have
to go to hypergraph.

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And this extra
layer of complexity,

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like in this word sense
of the complexity,

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introduces also additional
difficulties in there.

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So that there makes
significantly harder

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than graph removal lemma.

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So I won't even
show you anything

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that's close to
a complete proof,

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but I will illustrate
some of the ideas

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and highlight some of
the difficulties today.

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But deducing Szemeredi's theorem
from the hypergraph removal

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lemma is actually not so bad.

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So I will show you how
to do that right now.

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So from the hypergraph
removal lemma,

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even just for three
uniform hypergraph, even

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for tetrahedra-- so into the
triangles graph, tetrahedron,

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we can have the following
corollary analogous to all

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we have for triangles.

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So if g is a three graph such
that every edge is contained

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in a unique tetrahedron, then g
has sub-cubic number of edges.

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So completely
analogous to the one

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we have for triangles and
the proof is identical.

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You read the proof
and everything

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works exactly the same way,
once you have the removal lemma.

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By tetrahedron, I mean
the complete graph

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on four vertices.

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Complete three graph
of four vertices.

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OK?

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So you four vertices and you
look at all possible triples

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of vertices.

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All right.

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So now let's prove
Szemeredi's theorem, at least

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the 4-AP case.

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The general case is
completely analogous.

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Of course, you have
to go to higher order.

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Instead of three graphs, you
have to look at r graphs.

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So just as in the proof
of Roth's theorem,

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we're going to set up
some particular three

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graph where let's look
at a certain modulus

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m being 6n plus 1.

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The exact number is
not so important here.

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I really just wanted
this number is bigger

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than 3n and it's co-prime to 6,
as well as t sum divisibility's

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by 2 and 3, that will be useful.

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So let's build a
4-partite three graph

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g where the four vertex
parts x, y, z, w, they all

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have m vertices each.

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And I'll show you
what the edges are.

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So four vertices, little
x and big X and so on.

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So here are the rules
for putting the edges.

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I'll just tell you
exactly what they are.

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So I put in an edge
xyz if and only

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if the following expression
3x plus 2y plus z lies in a.

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I put in the edge xyw if and
only if 2x plus y minus w

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lies in a.

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OK?

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So xzw if and only if x
minus the minus 2w lies in a.

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And finally, yzw if and only
if minus y minus 2z minus 3w

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lies in a.

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So these are the
rules for putting

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in edges of this hypergraph.

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Of course, you have
this hypergraph.

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And you might be wondering, why
did I choose these expressions?

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So if it's not clear
yet, it will soon

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be very clear why we do this.

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Just as in the proof
of Roth's theorem

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using triangle removal lemma,
let's examine the tetrahedra

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in this three graph.

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So what are the tetrahedra?

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Notice that xyzw
is a tetrahedron.

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So if all three
triples are present, if

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and only if all four of
these expressions lie in a.

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Well, just like in a
proof of Roth's theorem,

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these four expressions
form a 4-AP.

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And the common
difference is minus

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x minus y minus z minus w.

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OK?

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So they were chosen to
satisfy this property.

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But furthermore, notice
that I can't just

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put any expressions in.

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I put these expressions in
with the very nice property

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that the i-th linear form does
not use the i-th variable.

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So each expression
really corresponds

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to an edge in this three graph.

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All right.

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But we started with a
set a that is 4-AP free.

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It follows that you don't have
this kind of configurations

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unless the common
difference is zero.

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So the only
tetrahedra correspond

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to these trivial 4-APs.

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And just as in the
proof of Roth's theorem

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that we saw from
triangle removal lemma,

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the conclusion is
that every edge lies

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in exactly one tetrahedron.

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And therefore, by the
corollary, the number of edges

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is equal to little o of m cubed.

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But on the other hand,
how many edges are there?

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So for each of the
four conditions here,

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so for each of these four
parts, the number of edges is,

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well, you get to choose x
and y whatever you want.

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And the last variable
has a choices.

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So this implies that the
size of a is little o of m,

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and m is on the same order as
n, and that proves the theorem.

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OK.

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Any questions so far?

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Yeah.

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AUDIENCE: Where do we use
the m is co-prime to 6?

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PROFESSOR: Great.

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00:16:42,290 --> 00:16:44,990
Question's where do
we use the condition

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that m is co-prime to 6?

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Anyone know the answer?

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AUDIENCE: To solve for that
last variable, divide by 2 or 3.

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PROFESSOR: Great.

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00:16:54,942 --> 00:16:56,710
So to solve for
the last variable,

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we ought to maybe
divide by 2 or 3.

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And to do that, you need to
have co-primality with 6.

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00:17:02,670 --> 00:17:04,420
Yeah, so I'm hiding a
bit of details here.

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OK?

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00:17:09,359 --> 00:17:12,130
But it's a great question.

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So if you work out the details
of this statement here,

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every edge slicing exactly
one tetra tetrahedron

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and also counting
the number of edges,

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but more especially,
the first sentence.

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You need to actually do
just a tiny bit of work.

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Any more questions?

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So this deduction is the
same deduction as the one

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that we did four
triangles, but you

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have to come up with a slightly
different set of linear forms.

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And usually, if you're
given a specific pattern,

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you know you can play by
hand and try to come up

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with a set of linear forms.

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You can think about, also,
how to do this generally.

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And more generally, this
hypergraph removal lemma

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using this type of
ideas, it allows

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00:18:05,450 --> 00:18:10,810
you to deduce multi-dimensional
Szemeredi theorem.

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00:18:19,470 --> 00:18:23,810
So if you give me
some pattern, then

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a subset of the integer
lattice in the fixed dimension

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avoiding that pattern must
have density going to zero.

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So we stated this in
the very first lecture.

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And I won't spell
all the details,

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but you can follow
this kind of framework

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00:18:41,780 --> 00:18:43,650
and gave you that theorem.

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00:18:43,650 --> 00:18:46,070
And I will post
the problem where

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you asked to do this
for a specific pattern,

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namely that of a geometric
square, axis-aligned square.

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00:18:52,010 --> 00:18:58,360
So if you have
that pattern in z2,

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00:18:58,360 --> 00:19:01,530
so it's worth thinking about
how would you run this argument

247
00:19:01,530 --> 00:19:03,900
for that pattern in z2.

248
00:19:03,900 --> 00:19:04,400
Yes.

249
00:19:04,400 --> 00:19:09,007
AUDIENCE: How close
can the small o m be?

250
00:19:09,007 --> 00:19:09,590
PROFESSOR: OK.

251
00:19:09,590 --> 00:19:12,400
The question's how
close can the--

252
00:19:12,400 --> 00:19:15,310
so you're asking what is the
rate, this little o m gives?

253
00:19:18,410 --> 00:19:20,410
Let me address-- OK, so
hold onto this question.

254
00:19:20,410 --> 00:19:24,070
I will address it once
I discuss what is known

255
00:19:24,070 --> 00:19:25,620
about hypergraph removal lemma.

256
00:19:25,620 --> 00:19:28,360
And that's a great
question and there's

257
00:19:28,360 --> 00:19:32,750
a lot of mystery surrounding
what happens there.

258
00:19:32,750 --> 00:19:33,250
OK.

259
00:19:33,250 --> 00:19:33,750
Questions?

260
00:19:33,750 --> 00:19:34,830
Any others?

261
00:19:37,320 --> 00:19:37,820
Great.

262
00:19:37,820 --> 00:19:40,950
So let's discuss this
hypergraph removal lemma.

263
00:19:40,950 --> 00:19:44,420
And as I had warned you already
at the beginning of lecture,

264
00:19:44,420 --> 00:19:48,410
this one is very difficult. So
I mentioned in the very first

265
00:19:48,410 --> 00:19:53,420
lecture that the development
of Szemeredi's regularity lemma

266
00:19:53,420 --> 00:19:55,550
was a stroke of ingenuity.

267
00:19:55,550 --> 00:19:57,100
But this one here--

268
00:19:57,100 --> 00:19:58,710
but we saw that the
proof and we did

269
00:19:58,710 --> 00:20:01,010
the proof of Szemeredi's
graph regularity lemma

270
00:20:01,010 --> 00:20:02,170
in one lecture.

271
00:20:02,170 --> 00:20:04,740
And once you understand
it, it's not so bad.

272
00:20:04,740 --> 00:20:08,010
You do the energy increment,
conceptually it's not so bad.

273
00:20:08,010 --> 00:20:11,120
But that when there is
actually incredibly difficult.

274
00:20:11,120 --> 00:20:14,270
It's incredibly difficult both
conceptually and technically.

275
00:20:14,270 --> 00:20:16,540
But I want to at least
illustrate some of the ideas

276
00:20:16,540 --> 00:20:18,290
and give you some sense
of the difficulty,

277
00:20:18,290 --> 00:20:25,910
like why is this difficult.
So as we imagine,

278
00:20:25,910 --> 00:20:27,440
we have graph regularity.

279
00:20:27,440 --> 00:20:30,200
And to prove
hypergraph removal, one

280
00:20:30,200 --> 00:20:33,260
would develop some kind of a
hypergraph regularity method.

281
00:20:41,130 --> 00:20:44,930
And the basic idea in hypergraph
regularity or just regularity,

282
00:20:44,930 --> 00:20:47,420
in general, is that I give
you some arbitrary graph

283
00:20:47,420 --> 00:20:50,930
or hypergraph, and you want to
find some kind of partitioning,

284
00:20:50,930 --> 00:20:55,910
some kind of regularization into
some bounded number of pieces,

285
00:20:55,910 --> 00:20:57,860
some bound amount of
data so that that's

286
00:20:57,860 --> 00:21:01,170
a good approximation
for the actual graph.

287
00:21:01,170 --> 00:21:04,080
Just like in the
graph regularity case.

288
00:21:04,080 --> 00:21:06,020
So let's try to do this.

289
00:21:06,020 --> 00:21:09,060
So what does this
partition even look like?

290
00:21:09,060 --> 00:21:10,302
So here's an attempt.

291
00:21:10,302 --> 00:21:11,760
And of course, I
call it an attempt

292
00:21:11,760 --> 00:21:13,860
because eventually,
it will not work.

293
00:21:13,860 --> 00:21:17,250
But it's a very natural
first thing to try.

294
00:21:19,790 --> 00:21:24,630
And maybe, I shouldn't call
it naive because it's actually

295
00:21:24,630 --> 00:21:27,570
not a bad idea to begin with.

296
00:21:27,570 --> 00:21:29,430
OK, so let's see.

297
00:21:29,430 --> 00:21:37,470
So suppose you were
given a three graph g.

298
00:21:37,470 --> 00:21:42,120
I'm going to just to
help you remember what's

299
00:21:42,120 --> 00:21:44,820
the uniformity of each graph.

300
00:21:44,820 --> 00:21:48,580
I will denote in parentheses,
in the superscript,

301
00:21:48,580 --> 00:21:51,550
so just help you remember
that this is the three graph.

302
00:21:51,550 --> 00:21:54,120
So in geometry, they also
do this with manifolds.

303
00:21:54,120 --> 00:21:56,270
So if you put an n on
top, it's an n manifold.

304
00:21:56,270 --> 00:21:59,860
But this 3 is for three graph.

305
00:21:59,860 --> 00:22:09,650
Suppose we partition the
vertex set of this three graph

306
00:22:09,650 --> 00:22:16,280
similar to proof of
Szemeredi's regularity lemma.

307
00:22:22,360 --> 00:22:25,060
Think about how the proof of
Szemeredi's regularity lemma

308
00:22:25,060 --> 00:22:25,850
goes.

309
00:22:25,850 --> 00:22:27,880
So you have this partition,
but in the proof,

310
00:22:27,880 --> 00:22:30,130
there is this
iterative refinement.

311
00:22:30,130 --> 00:22:32,980
And each step you say, well, I
have this notion of regularity.

312
00:22:32,980 --> 00:22:34,450
If it doesn't
satisfy regularity,

313
00:22:34,450 --> 00:22:36,610
I can keep cutting
things up further.

314
00:22:36,610 --> 00:22:38,050
So what's the
notion of regularity

315
00:22:38,050 --> 00:22:43,080
that might get you some
kind of vertex partition?

316
00:22:43,080 --> 00:22:46,920
Can anyone think of a
notion of a regularity

317
00:22:46,920 --> 00:22:49,200
for three uniform hypergraphs?

318
00:22:49,200 --> 00:22:50,166
Yep.

319
00:22:50,166 --> 00:22:52,581
AUDIENCE: So the same
sort of thing, [INAUDIBLE]

320
00:22:52,581 --> 00:22:54,996
variations like [INAUDIBLE].

321
00:22:59,360 --> 00:23:02,650
PROFESSOR: So let me try to
rephrase what you're saying.

322
00:23:02,650 --> 00:23:13,320
So if we have a
notion of regularity,

323
00:23:13,320 --> 00:23:19,200
let's say I have three
vertex sets, V1, V2, V3.

324
00:23:19,200 --> 00:23:29,970
And I want that the density
between these three,

325
00:23:29,970 --> 00:23:31,605
they do not differ from--

326
00:23:37,450 --> 00:23:45,250
if I restrict these
vertex sets to subsets

327
00:23:45,250 --> 00:23:46,530
that are not too small.

328
00:23:52,400 --> 00:23:54,210
Does this make sense?

329
00:23:54,210 --> 00:24:08,700
So here, this d is the
fraction of triples that

330
00:24:08,700 --> 00:24:11,980
are edges of the hypergraph.

331
00:24:15,050 --> 00:24:17,780
So this is the
natural extension,

332
00:24:17,780 --> 00:24:20,510
natural generalization of
the notion of regularity

333
00:24:20,510 --> 00:24:22,550
that we saw earlier for graphs.

334
00:24:22,550 --> 00:24:25,330
And indeed, it's a
very natural notion.

335
00:24:25,330 --> 00:24:26,570
it is a nice notion.

336
00:24:26,570 --> 00:24:28,130
And actually, if
you use this notion,

337
00:24:28,130 --> 00:24:30,650
if you use more or less
precisely what I've written,

338
00:24:30,650 --> 00:24:34,550
you can run through the entire
proof of Szemeredi's graph

339
00:24:34,550 --> 00:24:38,990
regularity lemma and produce
a regularity theorem that

340
00:24:38,990 --> 00:24:42,350
tells you given an arbitrary
three uniform hypergraph,

341
00:24:42,350 --> 00:24:46,070
you can decompose the
vertex set in such a way

342
00:24:46,070 --> 00:24:53,280
that most triples of vertex
sets have this property.

343
00:24:53,280 --> 00:25:02,400
So the same proof as Szemeredi's
regularity lemma implies--

344
00:25:02,400 --> 00:25:04,860
so I won't write down
the entire statement,

345
00:25:04,860 --> 00:25:05,880
but you get the idea.

346
00:25:05,880 --> 00:25:08,070
So for every
epsilon, there exists

347
00:25:08,070 --> 00:25:15,600
m such that we can partition
into the vertex set

348
00:25:15,600 --> 00:25:23,470
into at most m parts, even
equitable, if you like,

349
00:25:23,470 --> 00:25:39,220
so that at most an epsilon
fraction of triples or parts

350
00:25:39,220 --> 00:25:45,360
are not epsilon regular in
the sense that I just said.

351
00:25:45,360 --> 00:25:46,920
So it's literally
is the same proof.

352
00:25:46,920 --> 00:25:49,340
So you really look at the
proof and then you get that.

353
00:25:49,340 --> 00:25:49,840
OK?

354
00:25:49,840 --> 00:25:55,340
So far everything seems
pretty good, pretty easy.

355
00:25:55,340 --> 00:25:55,840
OK.

356
00:25:55,840 --> 00:26:00,530
So why did I say,
initially, that actually,

357
00:26:00,530 --> 00:26:02,890
hypergraph regularity
is incredibly difficult?

358
00:26:02,890 --> 00:26:06,260
So what not good about this one?

359
00:26:09,660 --> 00:26:11,940
Also remember, in
our application

360
00:26:11,940 --> 00:26:16,394
of the regularity method,
there were three steps.

361
00:26:16,394 --> 00:26:17,190
What are they?

362
00:26:17,190 --> 00:26:20,901
Partition, clean, and count.

363
00:26:20,901 --> 00:26:21,401
OK.

364
00:26:21,401 --> 00:26:22,950
So partition, OK,
you do partition.

365
00:26:22,950 --> 00:26:26,980
Clean, well, you do
some kind of cleaning.

366
00:26:26,980 --> 00:26:30,800
But counting,
that's a big thing.

367
00:26:30,800 --> 00:26:32,630
And that's something
that wasn't so hard.

368
00:26:32,630 --> 00:26:35,420
We had to do a little bit of
work, but it wasn't so hard.

369
00:26:35,420 --> 00:26:38,690
And you can ask, is there
a counting lemma associated

370
00:26:38,690 --> 00:26:41,670
to this regularity lemma?

371
00:26:41,670 --> 00:26:44,550
And the answer is
emphatically no.

372
00:26:44,550 --> 00:26:47,910
And I want to convince you that
for this notion of regularity,

373
00:26:47,910 --> 00:26:50,104
there's no counting lemma.

374
00:26:50,104 --> 00:26:50,604
OK.

375
00:27:05,020 --> 00:27:06,246
Yes.

376
00:27:06,246 --> 00:27:07,871
AUDIENCE: Is this
version true, though?

377
00:27:07,871 --> 00:27:08,788
PROFESSOR: It is true.

378
00:27:08,788 --> 00:27:10,240
So you ask, is
this version true?

379
00:27:10,240 --> 00:27:14,585
So this statement is true
with this definition.

380
00:27:14,585 --> 00:27:15,960
And you can prove
it by literally

381
00:27:15,960 --> 00:27:19,110
rerunning the entire proof of
Szemeredi's graph regularity

382
00:27:19,110 --> 00:27:20,950
lemma.

383
00:27:20,950 --> 00:27:23,640
So the regularity statement
I've written down is true,

384
00:27:23,640 --> 00:27:25,410
but it is not useful.

385
00:27:25,410 --> 00:27:28,290
For example, it cannot be
used to prove the tetrahedron

386
00:27:28,290 --> 00:27:32,280
removal lemma because if you
try to run the same regularity

387
00:27:32,280 --> 00:27:35,340
proof of the removal lemma, you
run to the issue that you do

388
00:27:35,340 --> 00:27:37,650
not have a counting lemma.

389
00:27:37,650 --> 00:27:38,150
OK.

390
00:27:38,150 --> 00:27:40,785
So why is it that you do
not have a counting lemma?

391
00:27:40,785 --> 00:27:42,035
So let me show you an example.

392
00:27:45,680 --> 00:27:49,370
And keep in mind that the
notions of regularity,

393
00:27:49,370 --> 00:27:58,770
they are supposed to model
the idea of pseudo randomness,

394
00:27:58,770 --> 00:28:01,340
which is the topic we'll
explore in further length

395
00:28:01,340 --> 00:28:02,580
in the next chapter.

396
00:28:02,580 --> 00:28:04,130
But the idea of
pseudo randomness

397
00:28:04,130 --> 00:28:06,800
is that I want some graph
which is not random,

398
00:28:06,800 --> 00:28:09,810
but in some aspects look random.

399
00:28:09,810 --> 00:28:13,310
So this is an important concept
in mathematics and computer

400
00:28:13,310 --> 00:28:16,280
science, and it's a
very important idea.

401
00:28:16,280 --> 00:28:18,530
But of course, you can
generate a pseudo random object

402
00:28:18,530 --> 00:28:22,220
by just taking a random
object, and it should hopefully

403
00:28:22,220 --> 00:28:24,560
satisfy some properties
of pseudo randomness.

404
00:28:24,560 --> 00:28:30,900
So let's see what this
notion of regularity,

405
00:28:30,900 --> 00:28:36,030
how it works even for
random hypergraph.

406
00:28:36,030 --> 00:28:38,280
What's a random hypergraph?

407
00:28:38,280 --> 00:28:42,830
So there are different ways to
generate a random hypergraph.

408
00:28:42,830 --> 00:28:47,640
One way is to have
a bunch of triples

409
00:28:47,640 --> 00:28:50,130
all appearing
uniformly at random,

410
00:28:50,130 --> 00:28:53,630
independently at random.

411
00:28:53,630 --> 00:28:55,810
So I have a bunch
of possible triples

412
00:28:55,810 --> 00:28:58,940
that I can make as edges,
each one I flip a coin.

413
00:28:58,940 --> 00:29:00,440
But there's a
different way, and let

414
00:29:00,440 --> 00:29:03,770
me show you a different way to
generate a random three graph.

415
00:29:11,080 --> 00:29:15,480
Let me give you two
parameters, p and q.

416
00:29:15,480 --> 00:29:17,650
They are constants
between 0 and 1.

417
00:29:23,630 --> 00:29:34,370
So let's build first a
graph, so a random two graph

418
00:29:34,370 --> 00:29:37,630
which I'll call g2.

419
00:29:37,630 --> 00:29:41,560
So this is just the
Erdos-Renyi random graph, gnp.

420
00:29:41,560 --> 00:29:44,900
The usual one where you
flip a coin for each edge

421
00:29:44,900 --> 00:29:50,000
so that each edge appears with
probability p independently.

422
00:29:50,000 --> 00:29:56,900
And now, I make the actual
three graph that want, g3,

423
00:29:56,900 --> 00:30:17,550
by including every triple,
so every triangle of g

424
00:30:17,550 --> 00:30:19,260
has an edge.

425
00:30:19,260 --> 00:30:23,030
So here, an edge
means a triple--

426
00:30:23,030 --> 00:30:25,690
edge is three vertices
in the hypergraph--

427
00:30:25,690 --> 00:30:26,890
with probability q.

428
00:30:30,880 --> 00:30:32,100
So it's a two-step process.

429
00:30:32,100 --> 00:30:35,430
I first generate a
random graph, and then I

430
00:30:35,430 --> 00:30:37,450
look at the triangles
on top of that graph.

431
00:30:37,450 --> 00:30:42,050
And each triangle I include as
a triple with a probability q.

432
00:30:42,050 --> 00:30:43,790
If you like, q be even one.

433
00:30:43,790 --> 00:30:46,700
So I do a random graph
and my hypergraph

434
00:30:46,700 --> 00:30:47,930
is a set of triangles.

435
00:30:52,460 --> 00:30:55,470
And let's compare
this construction

436
00:30:55,470 --> 00:31:01,410
to the more naive version
of a random hypergraph where

437
00:31:01,410 --> 00:31:06,520
we look at this
hypergraph of graph

438
00:31:06,520 --> 00:31:17,780
where I put in each triple
appearing independently

439
00:31:17,780 --> 00:31:22,394
with probability p cubed q.

440
00:31:25,260 --> 00:31:27,540
So these are two
different constructions

441
00:31:27,540 --> 00:31:30,790
of random hypergraph.

442
00:31:30,790 --> 00:31:33,790
And you can check that they
have basically the same edge

443
00:31:33,790 --> 00:31:35,770
density.

444
00:31:35,770 --> 00:31:38,820
So how many edges
appear in the first one?

445
00:31:44,480 --> 00:31:51,450
While the density of
triangles in g2 is p cubed,

446
00:31:51,450 --> 00:31:54,260
and each of those triangles
appears an edge further

447
00:31:54,260 --> 00:31:55,310
with probability q.

448
00:31:58,530 --> 00:32:02,070
So they have similar
edge densities.

449
00:32:02,070 --> 00:32:07,830
And furthermore, you can
check that this condition here

450
00:32:07,830 --> 00:32:11,300
is true for both graphs.

451
00:32:11,300 --> 00:32:21,540
So both graphs satisfy this
notion of epsilon regularity

452
00:32:21,540 --> 00:32:27,777
as justifying with
high probability.

453
00:32:31,360 --> 00:32:32,110
OK.

454
00:32:32,110 --> 00:32:33,470
Great.

455
00:32:33,470 --> 00:32:37,550
So if you have the counting
lemma, it should give you

456
00:32:37,550 --> 00:32:40,340
some prediction as to
the number of tetrahedra

457
00:32:40,340 --> 00:32:43,460
that come directly from the
densities, in particular,

458
00:32:43,460 --> 00:32:47,363
they should be the same for
these two constructions.

459
00:32:47,363 --> 00:32:48,280
But are they the same?

460
00:32:53,070 --> 00:33:08,640
So the density of tetrahedra
in the first case,

461
00:33:08,640 --> 00:33:10,980
actually, let's do
the second case first.

462
00:33:10,980 --> 00:33:15,270
In b, so what's the
density of tetrahedra?

463
00:33:15,270 --> 00:33:19,260
So if I have four vertices,
so each of those three edges

464
00:33:19,260 --> 00:33:22,320
appear uniformly at
random, independently,

465
00:33:22,320 --> 00:33:27,360
so the density of tetrahedra
is just the edge density

466
00:33:27,360 --> 00:33:28,826
raised to the power of 4.

467
00:33:32,038 --> 00:33:33,080
What about the first one?

468
00:33:35,930 --> 00:33:38,310
AUDIENCE: 6.

469
00:33:38,310 --> 00:33:41,750
PROFESSOR: So in the first
one, to get a tetrahedra,

470
00:33:41,750 --> 00:33:45,230
the underlying graph
needs to have a k4.

471
00:33:45,230 --> 00:33:51,720
So p raised to 6, and then
on top of that, I want 4q's.

472
00:33:51,720 --> 00:33:54,170
So p raised to 6, q raise to 4.

473
00:33:54,170 --> 00:33:56,731
And when p is different,
these numbers are different.

474
00:33:59,500 --> 00:34:01,620
So this is an
example showing why

475
00:34:01,620 --> 00:34:04,962
there is no counting
lemma because you

476
00:34:04,962 --> 00:34:06,420
have two different
graphs that have

477
00:34:06,420 --> 00:34:08,699
the same type of
regularity and densities,

478
00:34:08,699 --> 00:34:11,973
but have vastly different
densities of tetrahedra.

479
00:34:15,600 --> 00:34:16,980
Any questions
about this example?

480
00:34:20,429 --> 00:34:24,870
It shows you, at least, why
this naive attempt does not

481
00:34:24,870 --> 00:34:28,719
work, at least if you follow
our regularity recipe.

482
00:34:34,730 --> 00:34:37,500
But in any case, it's
good for something.

483
00:34:37,500 --> 00:34:41,010
So you do not have a counting
lemma for tetrahedra,

484
00:34:41,010 --> 00:34:44,690
but you can still
salvage something.

485
00:34:44,690 --> 00:34:48,610
So it turns out
there is a counting

486
00:34:48,610 --> 00:34:52,650
lemma if your graph h--

487
00:35:00,820 --> 00:35:04,640
if this r graph h is linear.

488
00:35:04,640 --> 00:35:08,880
So linear means every pair of
edges intersecting at most one

489
00:35:08,880 --> 00:35:09,380
vertex.

490
00:35:22,700 --> 00:35:27,270
So for example, if you look at--

491
00:35:32,640 --> 00:35:36,930
so hypergraph or each line
is an edge of triples.

492
00:35:36,930 --> 00:35:40,050
So that's a linear hypergraph
because each pair intersecting

493
00:35:40,050 --> 00:35:41,700
at most one vertex.

494
00:35:41,700 --> 00:35:43,440
Tetrahedron is not linear.

495
00:35:43,440 --> 00:35:47,500
Two faces of a tetrahedron
can intersect in two vertices.

496
00:35:47,500 --> 00:35:48,000
OK.

497
00:35:48,000 --> 00:35:49,800
So we can try to prove
that this is true.

498
00:35:49,800 --> 00:35:51,675
And actually, the proof
is basically the same

499
00:35:51,675 --> 00:35:54,000
as the counting lemma
that we saw for graphs.

500
00:35:56,730 --> 00:35:57,325
Yes.

501
00:35:57,325 --> 00:35:59,875
AUDIENCE: How many edges
can a linear graph have?

502
00:35:59,875 --> 00:36:01,500
PROFESSOR: The
question, how many edges

503
00:36:01,500 --> 00:36:04,140
can a linear hypergraph have?

504
00:36:04,140 --> 00:36:06,720
You mean, given the
bounded number of vertices.

505
00:36:06,720 --> 00:36:08,250
OK.

506
00:36:08,250 --> 00:36:12,060
I'll leave you to
think about it.

507
00:36:12,060 --> 00:36:15,050
Any more questions?

508
00:36:15,050 --> 00:36:16,980
But for the graph that
we really care about,

509
00:36:16,980 --> 00:36:20,180
namely tetrahedra which
relates to Szemeredi's theorem,

510
00:36:20,180 --> 00:36:21,280
this method does not work.

511
00:36:26,180 --> 00:36:29,780
So what should we do instead?

512
00:36:29,780 --> 00:36:33,440
Let's come up with a different
notion of regularity.

513
00:36:33,440 --> 00:36:36,200
And that's somewhat
inspired by that example

514
00:36:36,200 --> 00:36:42,280
up there where we need to look
at not just triple densities

515
00:36:42,280 --> 00:36:45,670
between three vertex
sets, but also

516
00:36:45,670 --> 00:36:50,380
what happens to triples
that sit on top of a graph.

517
00:36:53,960 --> 00:37:03,430
So we should come up with
some notion of an edge density

518
00:37:03,430 --> 00:37:05,990
on top of two graphs.

519
00:37:10,650 --> 00:37:19,310
So given a, b, and c being
edge sets of a complete graphs,

520
00:37:19,310 --> 00:37:21,140
so these are graphs
a, b, and c--

521
00:37:21,140 --> 00:37:23,240
you should think
of them as graphs--

522
00:37:23,240 --> 00:37:36,580
and a three graph g, we can
define this quantity b of abc--

523
00:37:36,580 --> 00:37:39,220
so there's always a hidden
g which I'll usually omit--

524
00:37:41,830 --> 00:37:56,870
to be the fraction
of triples xyz

525
00:37:56,870 --> 00:38:03,790
where xyz are such that
they sit on top of abc.

526
00:38:03,790 --> 00:38:06,670
So yz lie in a.

527
00:38:06,670 --> 00:38:10,285
xz lie in b.

528
00:38:10,285 --> 00:38:11,800
xy lie in c.

529
00:38:11,800 --> 00:38:24,200
So diffraction off such triples
that are actually triples of g.

530
00:38:30,290 --> 00:38:38,300
In the case when
abc are the same,

531
00:38:38,300 --> 00:38:53,510
it's asking what fraction
of triangles are edges of g.

532
00:38:53,510 --> 00:38:57,150
But you are allowed to
use three different sets.

533
00:38:57,150 --> 00:39:00,360
So think of abc as red,
green, blue masking

534
00:39:00,360 --> 00:39:03,690
for fractions of red,
green, blue triangles

535
00:39:03,690 --> 00:39:06,887
that are actually triples
of the hypergraph.

536
00:39:11,000 --> 00:39:12,030
All right.

537
00:39:12,030 --> 00:39:20,157
So now, we can try to come up
with some notion of regularity.

538
00:39:36,080 --> 00:39:39,630
And as you might
expect, at this point,

539
00:39:39,630 --> 00:39:43,310
it's not sufficient to
partition the vertex set.

540
00:39:43,310 --> 00:39:45,150
Instead, we'll go further.

541
00:39:45,150 --> 00:39:50,860
We'll partition the edge
set of the complete graph.

542
00:39:50,860 --> 00:39:54,545
So we'll partition the
set of pairs of vertices.

543
00:39:57,700 --> 00:40:02,650
Let's partition each set
of the complete graph

544
00:40:02,650 --> 00:40:16,850
as a union of
graphs such that we

545
00:40:16,850 --> 00:40:21,970
would like a similar type
of regularity condition

546
00:40:21,970 --> 00:40:25,050
but for those
types of densities.

547
00:40:25,050 --> 00:40:41,430
Such that for most ijk such
that we have lots of triangles

548
00:40:41,430 --> 00:40:49,663
on top of these three
graphs in the partition.

549
00:40:53,450 --> 00:41:00,900
So for most ijk, such that this
is the case, this partition,

550
00:41:00,900 --> 00:41:27,482
so this triple is regular in
the sense that for all subgroups

551
00:41:27,482 --> 00:41:40,700
with not too few copies,
so not too few triangles,

552
00:41:40,700 --> 00:41:42,080
on top of these a's.

553
00:41:50,440 --> 00:41:53,950
One has that the density,
the triple density,

554
00:41:53,950 --> 00:42:04,550
among the g's is similar to
the triple density on top

555
00:42:04,550 --> 00:42:05,150
of the a's.

556
00:42:20,970 --> 00:42:22,980
So I'm doing some
kind of partition

557
00:42:22,980 --> 00:42:26,280
where g1 is like that.

558
00:42:26,280 --> 00:42:35,080
g2 like that, and g3 that.

559
00:42:35,080 --> 00:42:40,580
And what I'm saying is that if
you take subset of g1, g2, g3

560
00:42:40,580 --> 00:42:42,910
so that there are still
lots of triangles,

561
00:42:42,910 --> 00:42:45,290
and that's analogous to
this condition of a i's

562
00:42:45,290 --> 00:42:52,820
now being too small, then
counting the number of triples

563
00:42:52,820 --> 00:42:57,360
to the fraction of those
triangles that are edges of g.

564
00:42:57,360 --> 00:43:00,320
That fraction is
roughly the same when

565
00:43:00,320 --> 00:43:02,080
you pass down to sub-graphs.

566
00:43:08,435 --> 00:43:10,060
Don't worry about
the specific details,

567
00:43:10,060 --> 00:43:12,477
and I'm not going to try to
give you the specific details.

568
00:43:12,477 --> 00:43:14,850
But think about the analogy
to instead of partitioning

569
00:43:14,850 --> 00:43:17,730
the vertex set, we are
partitioning the edge

570
00:43:17,730 --> 00:43:21,510
set of a complete graph.

571
00:43:21,510 --> 00:43:24,360
But actually,
hypergraph regularity

572
00:43:24,360 --> 00:43:27,090
involves one more
step, namely that we

573
00:43:27,090 --> 00:43:48,530
need to further regularize these
g's via partitioning the vertex

574
00:43:48,530 --> 00:43:50,510
set.

575
00:43:50,510 --> 00:43:53,900
Similar to what happens in
Szemeredi's graph regularity

576
00:43:53,900 --> 00:43:57,020
lemma, but actually more similar
to the strong regularity lemma

577
00:43:57,020 --> 00:43:59,580
that we discussed last time.

578
00:43:59,580 --> 00:44:08,010
So the data of
hypergraph regularity

579
00:44:08,010 --> 00:44:14,350
is not simply a partition of the
vertex set, but it's twofold.

580
00:44:14,350 --> 00:44:19,470
One is a partition of the edge
set of the complete graph,

581
00:44:19,470 --> 00:44:21,720
so partition of
the vertex pairs,

582
00:44:21,720 --> 00:44:30,080
into pseudo random
graphs, so in into graphs,

583
00:44:30,080 --> 00:44:38,170
so that the hypergraph g
sits pseudo randomly on top.

584
00:44:46,330 --> 00:44:51,820
And furthermore,
there's also a partition

585
00:44:51,820 --> 00:45:04,580
of the vertex set of g so
that the graphs in part one

586
00:45:04,580 --> 00:45:17,720
are extremely pseudo random
with respect to this partition.

587
00:45:17,720 --> 00:45:19,260
And this idea of
extremely random

588
00:45:19,260 --> 00:45:22,500
we saw in the last lecture,
you have some the sequence

589
00:45:22,500 --> 00:45:25,890
of epsilons that depend
on how many parts

590
00:45:25,890 --> 00:45:28,833
you have in the first
step of the regularity.

591
00:45:31,731 --> 00:45:32,700
OK.

592
00:45:32,700 --> 00:45:33,938
Any questions?

593
00:45:37,110 --> 00:45:38,500
Yes.

594
00:45:38,500 --> 00:45:40,872
AUDIENCE: What happens
with triples g's that don't

595
00:45:40,872 --> 00:45:42,405
have a lot of triangles?

596
00:45:42,405 --> 00:45:45,030
PROFESSOR: The question is, what
happens to triples of g's that

597
00:45:45,030 --> 00:45:47,080
do not have lots of triangles?

598
00:45:47,080 --> 00:45:50,460
So they are similar
to in graphs,

599
00:45:50,460 --> 00:45:54,380
you have these small
sets of vertices.

600
00:45:54,380 --> 00:45:56,990
So you have to deal
with them somehow,

601
00:45:56,990 --> 00:45:59,780
but I'm, again, leaving out
all these technical details.

602
00:45:59,780 --> 00:46:04,560
And in fact, I am writing
down a very sketchy version

603
00:46:04,560 --> 00:46:05,787
of hypergraph regularity.

604
00:46:05,787 --> 00:46:07,620
You could write down a
more precise version.

605
00:46:07,620 --> 00:46:09,150
You can find it in literature.

606
00:46:09,150 --> 00:46:11,670
In fact, you can find
more than one version

607
00:46:11,670 --> 00:46:15,180
of the statement of hypergraph
regularity in literature.

608
00:46:15,180 --> 00:46:16,980
And they're not all
obviously equivalent.

609
00:46:16,980 --> 00:46:18,480
It actually takes
a lot of work even

610
00:46:18,480 --> 00:46:20,480
to show that different
versions of the statement

611
00:46:20,480 --> 00:46:21,950
are equivalent to each other.

612
00:46:21,950 --> 00:46:26,310
And it's still somewhat
mysterious as to what

613
00:46:26,310 --> 00:46:30,030
is the right, the most
natural formulation

614
00:46:30,030 --> 00:46:31,318
of hypergraph regularity.

615
00:46:31,318 --> 00:46:33,360
That's something that I
think we still do not yet

616
00:46:33,360 --> 00:46:36,540
have a satisfactory answer.

617
00:46:36,540 --> 00:46:40,320
There was a question
earlier about bounds.

618
00:46:40,320 --> 00:46:40,820
OK.

619
00:46:40,820 --> 00:46:44,925
So what kind of bounds do you
get for hypergraph regularity?

620
00:46:44,925 --> 00:46:46,300
So let me address
that issue now.

621
00:47:02,130 --> 00:47:03,630
So what kind of
bounds do you get?

622
00:47:06,530 --> 00:47:09,020
Well for Szemeredi's
graph regularity lemma,

623
00:47:09,020 --> 00:47:11,480
the bound is a power
function because we

624
00:47:11,480 --> 00:47:13,850
have to iterate the
exponential which

625
00:47:13,850 --> 00:47:15,920
comes out of the partitioning.

626
00:47:15,920 --> 00:47:20,900
And in hypergraph regularity,
because of this extremely

627
00:47:20,900 --> 00:47:21,622
pseudo random--

628
00:47:21,622 --> 00:47:24,080
so you are doing some kind of
partitioning the first stage,

629
00:47:24,080 --> 00:47:27,740
and then you are iterating that
on top for the second stage.

630
00:47:27,740 --> 00:47:32,970
Similar to how we did strong
regularity in the last lecture.

631
00:47:32,970 --> 00:47:38,060
So the bounds for
hypergraph regularity

632
00:47:38,060 --> 00:47:43,690
is also iterated power
which we saw last time,

633
00:47:43,690 --> 00:47:44,940
and this is known as a Wowzer.

634
00:47:49,730 --> 00:47:53,130
So it's even worse
than graph regularity.

635
00:47:53,130 --> 00:47:56,730
And just like in the
case of graph regularity,

636
00:47:56,730 --> 00:48:00,290
this was Wowzer type
bound is necessary,

637
00:48:00,290 --> 00:48:04,340
at least for most statements,
any of these useful statements

638
00:48:04,340 --> 00:48:08,220
of hypergraph regularity.

639
00:48:08,220 --> 00:48:10,520
What about the applications?

640
00:48:10,520 --> 00:48:17,105
So applications to
multi-dimensional

641
00:48:17,105 --> 00:48:17,980
Szemeredi's theorem--

642
00:48:17,980 --> 00:48:27,950
OK, so first of all to
Szemeredi theorem, well,

643
00:48:27,950 --> 00:48:30,290
you can prove Szemeredi
theorem this way

644
00:48:30,290 --> 00:48:33,840
and you would get inverse
Wowzer type bounds,

645
00:48:33,840 --> 00:48:36,860
which is it's not so great.

646
00:48:36,860 --> 00:48:38,310
But there are better proofs.

647
00:48:38,310 --> 00:48:41,300
So there are more efficient
proofs quantitatively.

648
00:48:41,300 --> 00:48:45,790
So for Szemeredi's theorem,
the best result for general k

649
00:48:45,790 --> 00:48:52,480
is due to Gowers' which tells
you that a must be, at most,

650
00:48:52,480 --> 00:48:57,480
n over something that's
log log n raised to power

651
00:48:57,480 --> 00:49:01,530
some constant c depending on k.

652
00:49:01,530 --> 00:49:04,510
That's for k equals 3 and 4,
you can do somewhat better

653
00:49:04,510 --> 00:49:06,160
before general k.

654
00:49:06,160 --> 00:49:08,290
This is the best bounds so far.

655
00:49:14,590 --> 00:49:21,040
But for multi-dimensions, for
multi-dimensional patterns,

656
00:49:21,040 --> 00:49:22,880
it turns out that--

657
00:49:22,880 --> 00:49:24,640
well, historically,
the first proof

658
00:49:24,640 --> 00:49:26,620
of the multi-dimensional
Szemeredi's theorem

659
00:49:26,620 --> 00:49:29,230
was done using
Ergodic theory which

660
00:49:29,230 --> 00:49:31,760
has even worse bounds compared
to this approach in that

661
00:49:31,760 --> 00:49:35,610
the Ergodic theoretic proof
gives no bounds because it has

662
00:49:35,610 --> 00:49:37,800
to use compactness
arguments, so they actually

663
00:49:37,800 --> 00:49:39,600
give no quantitative bounds.

664
00:49:39,600 --> 00:49:42,510
And one of the motivations
for this hypergraph regularity

665
00:49:42,510 --> 00:49:45,420
method, the removal
lemma, is to produce

666
00:49:45,420 --> 00:49:48,540
quantitative proof of
multi-dimensional Szemeredi's

667
00:49:48,540 --> 00:49:49,120
theorem.

668
00:49:49,120 --> 00:49:56,330
So in general, still
the best bounds

669
00:49:56,330 --> 00:50:05,810
come from this removal lemma,
so hypergraph removal lemma.

670
00:50:05,810 --> 00:50:08,030
Although in special
cases, and really,

671
00:50:08,030 --> 00:50:10,070
not that many special
cases, but really just

672
00:50:10,070 --> 00:50:13,970
the case of a corner
as we saw earlier

673
00:50:13,970 --> 00:50:15,670
you have somewhat better bounds.

674
00:50:15,670 --> 00:50:18,790
So for corner, you
have bounce, which

675
00:50:18,790 --> 00:50:21,680
have density like polylog log.

676
00:50:21,680 --> 00:50:24,860
But even for a
geometric square, we

677
00:50:24,860 --> 00:50:27,650
do not know any Fourier
analytic methods,

678
00:50:27,650 --> 00:50:29,590
we do not know other methods.

679
00:50:29,590 --> 00:50:32,210
And this is basically
the best bound coming out

680
00:50:32,210 --> 00:50:35,203
of hypergraph regularity.

681
00:50:35,203 --> 00:50:36,620
And there are
serious obstructions

682
00:50:36,620 --> 00:50:41,150
for trying to use Fourier
methods to do other patterns

683
00:50:41,150 --> 00:50:45,110
such as geometric square.

684
00:50:45,110 --> 00:50:45,610
OK?

685
00:50:45,610 --> 00:50:47,560
Any questions?

686
00:50:47,560 --> 00:50:48,237
Yes.

687
00:50:48,237 --> 00:50:51,006
AUDIENCE: So what do the bounds
look like for higher degree

688
00:50:51,006 --> 00:50:51,506
uniformity.

689
00:50:51,506 --> 00:50:53,937
Are they still just Wowzers?

690
00:50:53,937 --> 00:50:54,520
PROFESSOR: OK.

691
00:50:54,520 --> 00:50:59,050
So what are the bounds like
for higher degree uniformity?

692
00:50:59,050 --> 00:51:04,480
So this is Wowzers for free
uniform, and for full uniform,

693
00:51:04,480 --> 00:51:07,040
you iterate Wowzer.

694
00:51:07,040 --> 00:51:10,920
So you go up in
Ackermann hierarchy.

695
00:51:10,920 --> 00:51:13,220
You iterate Wowzer,
you get a four uniform

696
00:51:13,220 --> 00:51:17,392
hypergraph regularity
lemma as so.

697
00:51:21,078 --> 00:51:22,120
Let's take a short break.

698
00:51:27,620 --> 00:51:29,900
So the second topic I
want to discuss today

699
00:51:29,900 --> 00:51:33,830
is a different approach to
proving Szemeredi's graph

700
00:51:33,830 --> 00:51:35,120
regularity lemma.

701
00:51:35,120 --> 00:51:38,810
And this is a good segue
into our next topic,

702
00:51:38,810 --> 00:51:41,480
the next lecture, which is
about pseudo random graphs,

703
00:51:41,480 --> 00:51:46,670
in particular, the idea of
the spectrum eigenvalues,

704
00:51:46,670 --> 00:51:50,110
in particular, play
a central role.

705
00:51:50,110 --> 00:51:59,070
So I want to consider a
spectral approach giving

706
00:51:59,070 --> 00:52:03,570
an alternative way to prove
the Szemeredi regularity lemma.

707
00:52:18,290 --> 00:52:20,220
And if you're already
sick of the regularity

708
00:52:20,220 --> 00:52:23,070
lemma at this point, this
will be the last topic

709
00:52:23,070 --> 00:52:25,470
on regularity lemma
for now, although it

710
00:52:25,470 --> 00:52:27,500
will come up again
later in this course

711
00:52:27,500 --> 00:52:29,240
when we discuss graph limits.

712
00:52:29,240 --> 00:52:31,590
But for now, this is the
last thing I want to say.

713
00:52:31,590 --> 00:52:34,443
And just like the discussion
about hypergraph regularity,

714
00:52:34,443 --> 00:52:35,610
it will be somewhat sketchy.

715
00:52:41,490 --> 00:52:45,630
So this idea has appeared
in literature in the past,

716
00:52:45,630 --> 00:52:48,750
but it was popularized by
many good things in life

717
00:52:48,750 --> 00:52:52,133
by Terry Tao's blog.

718
00:52:52,133 --> 00:52:54,300
So it's a good place to
look up a discussion of what

719
00:52:54,300 --> 00:52:56,710
I'm about to say.

720
00:52:56,710 --> 00:52:57,210
OK.

721
00:52:57,210 --> 00:52:59,550
So we saw the proof
of regularity lemma

722
00:52:59,550 --> 00:53:02,370
via this iterated
partitioning and keeping

723
00:53:02,370 --> 00:53:06,990
track of our progress
through the use of an energy.

724
00:53:06,990 --> 00:53:09,030
But here's a
different perspective,

725
00:53:09,030 --> 00:53:13,560
namely if we start
with a graph g,

726
00:53:13,560 --> 00:53:17,280
I can look at the
adjacency matrix a sub g.

727
00:53:17,280 --> 00:53:20,050
So this is the n
by n matrix where

728
00:53:20,050 --> 00:53:34,630
n is a number of vertices
whose i j-th is zero

729
00:53:34,630 --> 00:53:41,430
if i is not adjacent to j,
and 1 if i is adjacent to j.

730
00:53:41,430 --> 00:53:43,230
So this is a pretty
standard thing

731
00:53:43,230 --> 00:53:46,640
to look at to associate
a graph to this matrix.

732
00:53:46,640 --> 00:53:54,140
So this graph here would
be like that and so on.

733
00:53:56,670 --> 00:54:04,360
It's a real symmetric matrix
and that's always pretty nice.

734
00:54:04,360 --> 00:54:07,870
Symmetric matrices have
lots of great properties

735
00:54:07,870 --> 00:54:09,200
that will be convenient to use.

736
00:54:09,200 --> 00:54:11,290
In fact, if you're
like myself, if you're

737
00:54:11,290 --> 00:54:13,270
too used to working
with symmetric matrices,

738
00:54:13,270 --> 00:54:14,937
you forget that some
of these properties

739
00:54:14,937 --> 00:54:18,460
actually do not apply in general
to non-symmetric matrices.

740
00:54:18,460 --> 00:54:21,440
But it is symmetric,
so we're happy.

741
00:54:21,440 --> 00:54:25,630
So for symmetric matrices, we
have a set of real eigenvalues.

742
00:54:29,380 --> 00:54:31,820
We have real eigenvalues
and eigenvectors.

743
00:54:31,820 --> 00:54:42,070
And for now, let me enumerate
the eigenvalues by lambda 1

744
00:54:42,070 --> 00:54:46,270
through lambda n, so
multiplicity included,

745
00:54:46,270 --> 00:54:51,730
and I sort them according to the
size of their absolute value.

746
00:54:55,780 --> 00:55:00,650
So the spectral theorem
tells us a decomposition.

747
00:55:04,500 --> 00:55:07,130
So here again, we're using
the a as real symmetric.

748
00:55:07,130 --> 00:55:09,540
So it tells us
that this matrix a

749
00:55:09,540 --> 00:55:16,710
can be written as the sum
coming from the eigenvalues

750
00:55:16,710 --> 00:55:30,930
and eigenvectors where the
u i's are the eigenvectors,

751
00:55:30,930 --> 00:55:36,380
but I can choose them so that
they form an orthogonal basis

752
00:55:36,380 --> 00:55:38,930
orthonormal basis, so
they're all unit vectors.

753
00:55:46,070 --> 00:55:50,610
So when I say the spectrum,
I mean this data also,

754
00:55:50,610 --> 00:55:52,590
specifically, this
set of eigenvalues.

755
00:55:54,756 --> 00:55:55,256
All right.

756
00:55:55,256 --> 00:55:58,600
So let's go through some basic
properties of the spectrum.

757
00:55:58,600 --> 00:56:02,650
So first, how big
can the lambdas be?

758
00:56:02,650 --> 00:56:05,140
So I claim that--

759
00:56:05,140 --> 00:56:16,623
so first of all, the sum of the
squares of these lambdas is--

760
00:56:16,623 --> 00:56:18,040
let me not even
call this a lemma,

761
00:56:18,040 --> 00:56:19,460
so it's just an observation.

762
00:56:19,460 --> 00:56:22,230
So the sum of the
squares is this one.

763
00:56:22,230 --> 00:56:29,240
So this is the
trace of a squared,

764
00:56:29,240 --> 00:56:34,050
which is also the sum of
the squares of the entries.

765
00:56:34,050 --> 00:56:39,520
So here, I'm always using
that a is real symmetric--

766
00:56:39,520 --> 00:56:42,960
sum of squares of entries of a.

767
00:56:46,730 --> 00:56:50,420
And the case when you have
a being an adjacency matrix,

768
00:56:50,420 --> 00:56:54,920
this is simply twice the number
of edges, which is at most n

769
00:56:54,920 --> 00:56:56,270
squared.

770
00:56:56,270 --> 00:56:58,020
So that's always a
good thing to remember.

771
00:57:00,840 --> 00:57:01,780
OK?

772
00:57:01,780 --> 00:57:10,650
So as a result, the i-th
eigenvalue cannot be bigger

773
00:57:10,650 --> 00:57:12,620
than what?

774
00:57:12,620 --> 00:57:15,040
So you have i eigenvalues.

775
00:57:15,040 --> 00:57:18,390
So they're sorted
in decreasing order.

776
00:57:18,390 --> 00:57:20,790
So the i-th eigenvalue
cannot be too large,

777
00:57:20,790 --> 00:57:27,300
particular it cannot be
larger than n over root i.

778
00:57:27,300 --> 00:57:32,280
Because otherwise, the sum of
the first i eigenvalue squared

779
00:57:32,280 --> 00:57:36,310
would exceed n squared.

780
00:57:36,310 --> 00:57:39,860
So these things, they do decay.

781
00:57:39,860 --> 00:57:47,640
So second observation
is that if you

782
00:57:47,640 --> 00:57:55,120
have some epsilon and
an arbitrary function,

783
00:57:55,120 --> 00:57:58,930
so this is known as
a growth function.

784
00:57:58,930 --> 00:58:00,750
That's just a name,
don't worry about it.

785
00:58:00,750 --> 00:58:03,000
So which we'll call f.

786
00:58:03,000 --> 00:58:05,700
So its function from
the positive integers,

787
00:58:05,700 --> 00:58:08,140
the positive integers,
and for convenience,

788
00:58:08,140 --> 00:58:11,580
I'm going to assume that f
of j is always at least j.

789
00:58:11,580 --> 00:58:19,610
For every j there
exists some c which

790
00:58:19,610 --> 00:58:23,120
depends only on your epsilon
and this growth function.

791
00:58:23,120 --> 00:58:25,160
So this growth function
plays the same role

792
00:58:25,160 --> 00:58:27,740
as the sequence of
decaying epsilons

793
00:58:27,740 --> 00:58:29,410
in these strong
regularity lemma.

794
00:58:32,090 --> 00:58:34,580
So there exists
some constant bound

795
00:58:34,580 --> 00:58:45,670
such that for every graph g
and ag as above, so associated

796
00:58:45,670 --> 00:58:54,390
with the lambdas and u's, there
exists a j less than c such

797
00:58:54,390 --> 00:58:59,550
that if I sum up the
eigenvalues squared

798
00:58:59,550 --> 00:59:07,050
for eigenvalues i index
between j and c of j,

799
00:59:07,050 --> 00:59:09,530
the sum is fairly small.

800
00:59:09,530 --> 00:59:11,595
It's at most epsilon n squared.

801
00:59:16,730 --> 00:59:19,420
I'll let you ponder
that for a second.

802
00:59:19,420 --> 00:59:21,880
So choose your favorite
growth function.

803
00:59:21,880 --> 00:59:23,830
It can be as quickly
growing as you can.

804
00:59:23,830 --> 00:59:26,280
It can be exponential,
power, or whatever.

805
00:59:26,280 --> 00:59:31,730
And it's saying that I can
look up to a bounded point

806
00:59:31,730 --> 00:59:35,990
so that this stretch
of spectrum squared

807
00:59:35,990 --> 00:59:38,490
is, at most, epsilon n squared.

808
00:59:38,490 --> 00:59:39,275
Question.

809
00:59:39,275 --> 00:59:40,640
AUDIENCE: What is c of j?

810
00:59:43,180 --> 00:59:43,930
PROFESSOR: F of j.

811
00:59:43,930 --> 00:59:44,430
Thank you.

812
00:59:58,568 --> 01:00:00,610
Well, the statement
hopefully will become clearer

813
01:00:00,610 --> 01:00:01,693
once I show you the proof.

814
01:00:07,780 --> 01:00:08,280
OK.

815
01:00:08,280 --> 01:00:10,110
So here's how you
would prove it.

816
01:00:13,310 --> 01:00:20,340
So you first let j1
equal to 1, and I

817
01:00:20,340 --> 01:00:26,130
obtained the subsequent
j's by applying f to j.

818
01:00:35,100 --> 01:00:41,890
So I claim that one cannot
have this inequality violated

819
01:00:41,890 --> 01:00:44,100
for too many of these j i's.

820
01:00:44,100 --> 01:00:59,550
So one cannot have the sum
going between jk and jk plus 1

821
01:00:59,550 --> 01:01:09,400
for all k from 1
to 1 over epsilon.

822
01:01:15,550 --> 01:01:17,305
Let's change this to zero.

823
01:01:23,720 --> 01:01:27,650
But you cannot have this because
if you had this then you sum up

824
01:01:27,650 --> 01:01:31,100
all of these inequalities you
would get that the total sum

825
01:01:31,100 --> 01:01:35,960
would exceed and squared, which
would violate the inequality

826
01:01:35,960 --> 01:01:38,780
about sum of the
squares of the spectrum.

827
01:01:43,030 --> 01:01:49,560
And so therefore, so thus
to the claimed inequality,

828
01:01:49,560 --> 01:02:02,600
so this is true
star holds for sum j

829
01:02:02,600 --> 01:02:10,610
equal to ji so jk where k
is less than 1 over epsilon.

830
01:02:10,610 --> 01:02:14,880
And this j, in
particular, is less than--

831
01:02:14,880 --> 01:02:17,670
well, whatever it
is, it's bounded.

832
01:02:17,670 --> 01:02:27,250
So it's less than f
applied to itself at most 1

833
01:02:27,250 --> 01:02:28,510
over epsilon times--

834
01:02:33,210 --> 01:02:34,160
OK.

835
01:02:34,160 --> 01:02:36,360
So this should look
somewhat familiar.

836
01:02:36,360 --> 01:02:39,930
And I'll ask you to
think about, later on,

837
01:02:39,930 --> 01:02:44,310
how this proof of spectral proof
of Szemeredi's graph regularity

838
01:02:44,310 --> 01:02:47,310
lemma compare to the
proof that we saw earlier.

839
01:02:47,310 --> 01:02:50,550
And you should see where
the analogous step is here.

840
01:02:50,550 --> 01:02:51,593
This is that density.

841
01:02:51,593 --> 01:02:53,010
This is the energy
increment step.

842
01:02:57,456 --> 01:02:58,600
All right.

843
01:02:58,600 --> 01:02:59,100
OK.

844
01:02:59,100 --> 01:03:00,850
So what's the regularity
decomposition?

845
01:03:00,850 --> 01:03:01,950
So I give you this graph.

846
01:03:01,950 --> 01:03:05,010
I give you this
adjacency matrix.

847
01:03:05,010 --> 01:03:07,860
And I want be able
to find a partition,

848
01:03:07,860 --> 01:03:10,740
but there's a different
way to view a partition.

849
01:03:10,740 --> 01:03:15,060
So this is, I think,
a important idea

850
01:03:15,060 --> 01:03:18,320
which, again, is
popularized by Terry Tao,

851
01:03:18,320 --> 01:03:20,790
that instead of looking
at things as a regularity

852
01:03:20,790 --> 01:03:24,950
partition, we can view
these ideas as regularity

853
01:03:24,950 --> 01:03:25,920
de-compositions.

854
01:03:34,330 --> 01:03:45,330
Namely, pick j as in the lemma
and I now write my adjacency

855
01:03:45,330 --> 01:03:50,620
matrix a sub g as a sum
of three matrices, which

856
01:03:50,620 --> 01:03:57,910
we'll call a structured plus
a small plus a pseudo random.

857
01:04:01,600 --> 01:04:15,310
Where a structured equals to
the sum for basically that sum,

858
01:04:15,310 --> 01:04:24,180
this sum here, so this
spectral de-competition

859
01:04:24,180 --> 01:04:27,204
but only for the first
j minus 1 eigenvalues.

860
01:04:35,060 --> 01:04:36,790
So those of you
coming from or who

861
01:04:36,790 --> 01:04:39,050
have taken classes in
something like Statistics

862
01:04:39,050 --> 01:04:41,820
might recognize this as a
principal component analysis.

863
01:04:41,820 --> 01:04:42,900
So this has many names.

864
01:04:42,900 --> 01:04:44,390
It's a very powerful idea.

865
01:04:44,390 --> 01:04:46,880
You look at the
top spectral data,

866
01:04:46,880 --> 01:04:49,610
and that should describe
you most of the information

867
01:04:49,610 --> 01:04:54,740
that you care about about a
graph or a matrix, in general.

868
01:04:54,740 --> 01:05:03,920
The small piece is the sum
but only for i between j

869
01:05:03,920 --> 01:05:04,880
and f of j.

870
01:05:07,510 --> 01:05:15,810
And the pseudo random piece
is for i at least f of j.

871
01:05:23,840 --> 01:05:24,340
OK.

872
01:05:24,340 --> 01:05:30,610
So we decompose this adjacency
matrix into these three pieces.

873
01:05:30,610 --> 01:05:32,590
And the question now
is, what does this

874
01:05:32,590 --> 01:05:35,830
have to do with Szemeredi's
graph regularity lemma?

875
01:05:35,830 --> 01:05:37,330
So what do the
individual components

876
01:05:37,330 --> 01:05:40,030
correspond to in the version
of the regularity lemma

877
01:05:40,030 --> 01:05:44,200
that you've seen and
are now familiar with?

878
01:05:44,200 --> 01:05:46,670
So here is what's going on.

879
01:05:46,670 --> 01:05:53,620
So I want to show you that
this structured piece roughly

880
01:05:53,620 --> 01:05:57,510
corresponds to the partition.

881
01:05:57,510 --> 01:06:02,648
So this is the
bounded partition.

882
01:06:06,000 --> 01:06:08,670
And the small piece
roughly corresponds

883
01:06:08,670 --> 01:06:12,495
to the small fraction
of irregular pairs.

884
01:06:15,370 --> 01:06:17,300
And the pseudo
random piece roughly

885
01:06:17,300 --> 01:06:20,510
corresponds to the idea
of pseudo randomness

886
01:06:20,510 --> 01:06:21,380
between pairs.

887
01:06:37,850 --> 01:06:41,810
First, to understand what the
spectral data have anything

888
01:06:41,810 --> 01:06:44,330
to do with partitions,
let me remind

889
01:06:44,330 --> 01:06:50,030
you a basic fact about
how the spectrum, how

890
01:06:50,030 --> 01:06:53,600
the eigenvalues of a
real symmetric matrix

891
01:06:53,600 --> 01:06:57,090
relate to other
properties of this matrix.

892
01:06:57,090 --> 01:07:00,800
And namely, this
notion of a spectral

893
01:07:00,800 --> 01:07:06,370
radius or sometimes
called spectral norm.

894
01:07:09,200 --> 01:07:13,160
So far I'm only going to
discuss real symmetric matrices.

895
01:07:13,160 --> 01:07:15,200
So many of the things
I will say are not

896
01:07:15,200 --> 01:07:18,110
true for if you're not
in a real symmetric case.

897
01:07:18,110 --> 01:07:25,200
So the spectral radius
spectral norm of a

898
01:07:25,200 --> 01:07:32,670
is the largest eigenvalue
of a in absolute value.

899
01:07:38,540 --> 01:07:42,030
And this quantity turns out
to be equal to the operator

900
01:07:42,030 --> 01:07:54,750
norm which is the norm of
this a as a linear operator,

901
01:07:54,750 --> 01:07:59,100
namely it is the max
or super, it turns out

902
01:07:59,100 --> 01:08:06,620
to be a max, of av over--

903
01:08:06,620 --> 01:08:08,450
length of av divided
by length of v.

904
01:08:08,450 --> 01:08:11,240
So if you hit it with a unit
vector, how far can you go?

905
01:08:14,110 --> 01:08:23,130
So it's also equal to
this bi-linear form.

906
01:08:23,130 --> 01:08:25,220
If you hit it from left
and right by unit vectors,

907
01:08:25,220 --> 01:08:28,750
how big can you get?

908
01:08:28,750 --> 01:08:31,350
So for real symmetric
matrices, these quantities

909
01:08:31,350 --> 01:08:32,350
are equal to each other.

910
01:08:32,350 --> 01:08:33,910
And that will be
an essential fact

911
01:08:33,910 --> 01:08:38,915
for relating the spectral data
with combinatorial quantities.

912
01:08:41,590 --> 01:08:42,850
All right.

913
01:08:42,850 --> 01:08:50,960
So if you give me
this de-composition,

914
01:08:50,960 --> 01:08:53,653
how can I produce
for you a partition?

915
01:08:56,970 --> 01:09:04,029
Basically, you can
look at a structure

916
01:09:04,029 --> 01:09:10,450
which has its state in
its data a bounded number

917
01:09:10,450 --> 01:09:13,130
of eigenvectors.

918
01:09:13,130 --> 01:09:24,160
And by rounding, we can
basically round these guys so

919
01:09:24,160 --> 01:09:27,115
when you round the
individual values by rounding

920
01:09:27,115 --> 01:09:28,120
the coordinate values--

921
01:09:31,120 --> 01:09:42,510
So let's pretend that they
take only a bounded, let's see,

922
01:09:42,510 --> 01:09:43,589
a small number of values.

923
01:09:47,609 --> 01:09:50,310
So just to simplify
things in your mind,

924
01:09:50,310 --> 01:09:53,715
pretend for a second--

925
01:09:53,715 --> 01:09:55,590
well, of course, this
is far from the truth--

926
01:09:55,590 --> 01:10:01,860
pretend for a second that these
guys are 0 comma 1 valued.

927
01:10:01,860 --> 01:10:04,140
Of course, that's not going
to be the case but 0 comma

928
01:10:04,140 --> 01:10:07,270
1 or plus/minus 1, if you like.

929
01:10:07,270 --> 01:10:07,770
OK.

930
01:10:07,770 --> 01:10:09,840
So this is definitely not true.

931
01:10:09,840 --> 01:10:13,350
But for the purpose
of exposition,

932
01:10:13,350 --> 01:10:14,820
let's pretend this is the case.

933
01:10:14,820 --> 01:10:16,770
And you can more
or less achieve it

934
01:10:16,770 --> 01:10:22,160
by rounding the individual
values to their nearby closest

935
01:10:22,160 --> 01:10:23,414
multiple of something.

936
01:10:26,020 --> 01:10:36,950
Then the level sets of
these top eigenvectors,

937
01:10:36,950 --> 01:10:44,940
they partition the vertex set
into a bounded number of parts.

938
01:10:54,800 --> 01:10:57,430
So if you, for example, in the
simplified version where you're

939
01:10:57,430 --> 01:11:00,940
only have plus minus 1
values for this eigenvectors,

940
01:11:00,940 --> 01:11:03,970
then you have, at
most, 2 to the j parts.

941
01:11:03,970 --> 01:11:06,640
But you may get some more
because some epsilons,

942
01:11:06,640 --> 01:11:09,190
but for the purpose
of illustration,

943
01:11:09,190 --> 01:11:10,480
let's not worry about that.

944
01:11:10,480 --> 01:11:13,330
And this is basically
the regularity partition.

945
01:11:13,330 --> 01:11:18,160
I want to show that this set
here has very nice properties

946
01:11:18,160 --> 01:11:20,890
that they basically behave
like the regularity partition

947
01:11:20,890 --> 01:11:24,210
we've gotten previously.

948
01:11:24,210 --> 01:11:29,620
So what I would like to show
is that the other two parts,

949
01:11:29,620 --> 01:11:32,920
they do not contribute very much
in the sense of our regularity

950
01:11:32,920 --> 01:11:34,420
partition.

951
01:11:34,420 --> 01:11:42,210
So for example, if you look
at the pseudo random piece,

952
01:11:42,210 --> 01:11:53,230
if I hit it left and right with
indicator vectors of vertex

953
01:11:53,230 --> 01:11:56,770
sets, how big can
this number get?

954
01:12:00,360 --> 01:12:05,420
So this number here is, at
most, while the norm of u times

955
01:12:05,420 --> 01:12:09,630
the norm of this
w which is just--

956
01:12:09,630 --> 01:12:14,930
so let me write down--

957
01:12:14,930 --> 01:12:18,690
so the norm of indicator
of u, norm indicator

958
01:12:18,690 --> 01:12:22,500
of w multiplied by
the operator norm

959
01:12:22,500 --> 01:12:26,990
of the pseudo random part of a.

960
01:12:26,990 --> 01:12:32,520
But these two guys here, so
they're, at most, root n each.

961
01:12:32,520 --> 01:12:38,910
So this number
here is, at most, n

962
01:12:38,910 --> 01:12:46,570
but we know from our hypothesis
on the pseudo random part

963
01:12:46,570 --> 01:12:50,970
of a that the spectral
norm is no more

964
01:12:50,970 --> 01:12:57,360
than this quantity over here.

965
01:12:57,360 --> 01:12:59,690
And by choosing f
appropriately large,

966
01:12:59,690 --> 01:13:02,160
I can make sure that this
number is extremely small.

967
01:13:10,900 --> 01:13:22,020
f to be large compared to
the number of parts in b

968
01:13:22,020 --> 01:13:29,220
partition so this
quantity is small.

969
01:13:36,670 --> 01:13:40,010
And this is basically the
notion of epsilon regularity

970
01:13:40,010 --> 01:13:42,710
that you saw in
the usual version

971
01:13:42,710 --> 01:13:44,880
or version of Szemeredi's
regularity lemma

972
01:13:44,880 --> 01:13:48,170
I presented in
the first version,

973
01:13:48,170 --> 01:13:53,170
in the very first lecture
that we discussed regularity.

974
01:13:53,170 --> 01:13:55,510
This quantity here
is something which

975
01:13:55,510 --> 01:13:58,090
measures the difference
between-- so for now,

976
01:13:58,090 --> 01:14:01,480
if for a second ignore
the middle piece.

977
01:14:01,480 --> 01:14:03,580
If you ignore the
small piece, then this

978
01:14:03,580 --> 01:14:07,690
is precisely the difference
between the actual densities

979
01:14:07,690 --> 01:14:10,570
between u and w and
the predicted density

980
01:14:10,570 --> 01:14:11,853
between u and w.

981
01:14:11,853 --> 01:14:14,070
AUDIENCE: Why is there
a square root here?

982
01:14:14,070 --> 01:14:16,750
PROFESSOR: The question is, why
is there a square root here?

983
01:14:16,750 --> 01:14:18,850
There should not be
a square root here.

984
01:14:18,850 --> 01:14:20,610
Good.

985
01:14:20,610 --> 01:14:25,120
It then becomes a
square root of the--

986
01:14:25,120 --> 01:14:26,650
yeah, so there is
no square root,

987
01:14:26,650 --> 01:14:30,860
but the length of this vector
is the square root of the size

988
01:14:30,860 --> 01:14:32,250
of u which is, at most, n.

989
01:14:35,512 --> 01:14:36,444
Yeah.

990
01:14:36,444 --> 01:14:39,246
AUDIENCE: Did you say to be
small in general or just small

991
01:14:39,246 --> 01:14:40,995
like you compare
it to f squared?

992
01:14:40,995 --> 01:14:42,495
Because I guess,
isn't it like going

993
01:14:42,495 --> 01:14:44,795
to be this constant function
that you choose before

994
01:14:44,795 --> 01:14:46,220
[INAUDIBLE]?

995
01:14:46,220 --> 01:14:47,020
PROFESSOR: OK.

996
01:14:47,020 --> 01:14:53,450
Question is do, how small do
we want this f of j to be?

997
01:14:53,450 --> 01:14:59,210
So I want this quantity to
be quite a bit smaller than,

998
01:14:59,210 --> 01:14:59,780
let's say--

999
01:14:59,780 --> 01:15:02,780
so basically, I want it to
be less than f of n squared,

1000
01:15:02,780 --> 01:15:09,220
but f of n over the
number of parts square.

1001
01:15:09,220 --> 01:15:15,250
Because this quantity is, let's
say, the sizes of each part.

1002
01:15:15,250 --> 01:15:18,730
So let me just be not precise
and say much less than.

1003
01:15:18,730 --> 01:15:21,580
So this quantity here is
the size of each part.

1004
01:15:21,580 --> 01:15:25,030
And I want to think about
the case when u and w

1005
01:15:25,030 --> 01:15:27,730
they lie inside each part.

1006
01:15:27,730 --> 01:15:29,350
In which case, I
want the difference

1007
01:15:29,350 --> 01:15:32,650
to be much less than epsilon
times the size of the part

1008
01:15:32,650 --> 01:15:33,260
squared.

1009
01:15:33,260 --> 01:15:33,760
Yeah.

1010
01:15:33,760 --> 01:15:37,957
AUDIENCE: [INAUDIBLE] j is
different based on the graph?

1011
01:15:37,957 --> 01:15:40,540
PROFESSOR: Question is, is the
j different based on the graph?

1012
01:15:40,540 --> 01:15:41,040
Yes.

1013
01:15:41,040 --> 01:15:43,950
And that's also the case for
Szemeredi's regularity lemma.

1014
01:15:43,950 --> 01:15:48,657
In Szemeredi's regularity lemma,
you don't know when you stop.

1015
01:15:48,657 --> 01:15:50,740
But you know that you stop
before a certain point.

1016
01:15:56,692 --> 01:15:59,670
OK.

1017
01:15:59,670 --> 01:16:04,390
And finally, what's happening
with a small part of a.

1018
01:16:04,390 --> 01:16:14,930
So in a small, the sum of
the squares of entries--

1019
01:16:17,940 --> 01:16:19,448
so this also has
a convenient name.

1020
01:16:19,448 --> 01:16:20,990
It's called the
Hilbert-Schmidt norm.

1021
01:16:24,190 --> 01:16:26,065
So the sum of the
squares of the entries.

1022
01:16:29,010 --> 01:16:31,200
We basically saw this
calculation earlier,

1023
01:16:31,200 --> 01:16:36,190
it's the sum of the squares of
the eigenvalues in which case

1024
01:16:36,190 --> 01:16:38,210
we've truncated all
the other eigenvalues.

1025
01:16:38,210 --> 01:16:44,490
So the only eigenvalues left are
between j and index between j

1026
01:16:44,490 --> 01:16:45,700
and f of j.

1027
01:16:45,700 --> 01:16:48,355
And we chose j so that
this number is small.

1028
01:16:52,600 --> 01:16:57,900
So a small as in
a bunch of noise,

1029
01:16:57,900 --> 01:17:00,570
but no adversarial
noise, if you will,

1030
01:17:00,570 --> 01:17:03,300
into your graph, but
only a very small amount,

1031
01:17:03,300 --> 01:17:05,610
at most, epsilon amount.

1032
01:17:05,610 --> 01:17:18,830
So it might destroy the epsilon
regularity for, let's say,

1033
01:17:18,830 --> 01:17:23,720
around epsilon
fraction of pairs.

1034
01:17:26,420 --> 01:17:29,940
But that's all it could do.

1035
01:17:29,940 --> 01:17:32,430
So all but epsilon
fraction of your pairs

1036
01:17:32,430 --> 01:17:34,620
will still be epsilon regular.

1037
01:17:34,620 --> 01:17:37,530
And that is the consequence of
Szemeredi's graph regularity

1038
01:17:37,530 --> 01:17:39,650
lemma that we saw earlier.

1039
01:17:39,650 --> 01:17:40,330
Yeah.

1040
01:17:40,330 --> 01:17:43,157
AUDIENCE: Doesn't large
F have to be special?

1041
01:17:43,157 --> 01:17:43,740
PROFESSOR: OK.

1042
01:17:43,740 --> 01:17:45,698
So question, does a large
F have to be special?

1043
01:17:45,698 --> 01:17:47,250
The F should be chosen--

1044
01:17:47,250 --> 01:17:49,740
if you want to achieve
Szemeredi's graph regularity

1045
01:17:49,740 --> 01:17:55,260
lemma, you should find this F so
that basically this inequality

1046
01:17:55,260 --> 01:17:56,600
is true.

1047
01:17:56,600 --> 01:17:58,860
So f should be
quite a bit larger

1048
01:17:58,860 --> 01:18:01,850
than the number of parts.

1049
01:18:01,850 --> 01:18:04,940
But if you choose even
bigger values of f,

1050
01:18:04,940 --> 01:18:07,650
you can achieve more regularity.

1051
01:18:07,650 --> 01:18:09,260
And this is akin
to what happened

1052
01:18:09,260 --> 01:18:10,950
with strong regularity.

1053
01:18:10,950 --> 01:18:14,600
So there's this idea if you
iterate one version regularity,

1054
01:18:14,600 --> 01:18:16,400
you can get a strong
version of regularity.

1055
01:18:16,400 --> 01:18:18,450
And there's some iteration
happening over there.

1056
01:18:18,450 --> 01:18:22,520
So if you choose your f to be
a much bigger function of j,

1057
01:18:22,520 --> 01:18:25,610
you can achieve a much
stronger notion irregularity

1058
01:18:25,610 --> 01:18:29,180
which is similar
and, perhaps, even

1059
01:18:29,180 --> 01:18:33,470
equivalent to strong regularity
that we discussed last time.

1060
01:18:33,470 --> 01:18:36,230
So you get to choose what
f you want to put in here.

1061
01:18:36,230 --> 01:18:37,140
Yeah.

1062
01:18:37,140 --> 01:18:41,240
AUDIENCE: How do you
make it equitable?

1063
01:18:41,240 --> 01:18:45,330
PROFESSOR: The question is,
how do you make it equitable?

1064
01:18:45,330 --> 01:18:45,830
OK.

1065
01:18:45,830 --> 01:18:48,570
So let me now discuss that.

1066
01:18:48,570 --> 01:18:52,910
So in this case, you can
also do very similar things

1067
01:18:52,910 --> 01:18:56,960
to what we've done before,
but to massage the partitions.

1068
01:18:56,960 --> 01:19:01,160
It's not entirely clear
from this formulation.

1069
01:19:01,160 --> 01:19:02,680
But the message
here is that there's

1070
01:19:02,680 --> 01:19:06,440
this equivalence between
operator norm on one hand

1071
01:19:06,440 --> 01:19:08,930
and combinatorial discrepancy
on the other hand.

1072
01:19:08,930 --> 01:19:10,700
And we'll explore
this notion further

1073
01:19:10,700 --> 01:19:13,120
in the next several lectures.