WEBVTT
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YUFEI ZHAO: For the
last few lectures,
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we've been talking about the
extremal problem of forbidding
00:00:23.460 --> 00:00:26.190
a complete bipartite graph.
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So today I want to move beyond
the complete bipartite graph
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and look at other
sparser bipartite graphs.
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So we'll be looking at what
happens to the extremal problem
00:00:38.760 --> 00:00:44.590
if you forbid a sparse
bipartite graph.
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Recall the
Kovari-Sos-Turan theorem,
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which tells you that the
extremal number for KSC
00:01:07.560 --> 00:01:15.300
is upper bounded by something on
the order of m to the 2 minus 1
00:01:15.300 --> 00:01:17.650
over m.
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So if I give you
some bipartite H,
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we know that because
it's a bipartite graph,
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it is always contained in some
KST for some values of S and P.
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So you already automatically
have an upper bound,
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the extremal number
for this H from
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the Kovari-Sos-Turan theorem.
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But as you might
expect, we might
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be very wasteful in this step.
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And the question is,
are there situations
00:02:01.180 --> 00:02:04.300
where we can do much better
than what is just given
00:02:04.300 --> 00:02:06.122
by applying Kovari-Sos-Turan?
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And today, I want to show
you several examples where
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you can significantly
improve the bound given
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by Kovari-Sos-Turan for various
sparser bipartite graphs.
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The first result is the
following theorem, which
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is initially due to Faraday.
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And then later,
a different proof
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was given by
Alon-Krivelevich-Sudov.
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And the latter proof is
the one I want to present,
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because it presents an
important and intricate
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probabilistic technique, which
is the main reason for showing
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you this theorem.
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But let me tell you
the theorem first.
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Here, H will be a
bipartite graph.
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And it has vertex
bipartitions A and B,
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such that every vertex in
A has degree at most R. The
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bipartition is A and
B, and the degree in A
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is at most R for every vertex
on the left side in the set A.
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And I want to
understand, is there
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some upper bound on the
extremal number that does better
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than Kovari-Sos-Turan theorem?
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And the theorem
guarantees such a bound.
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So then there exists some
constant depending on H, such
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that the extremal
number is upper
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bounded by something to the
order of n to the 2 minus 1
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over R.
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Compared with Kovari-Sos-Turan
theorem, on one hand,
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if your H is the complete
bipartite graph KST,
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then this is the same bound
as Kovari-Sos-Turan theorem.
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On the other hand, you might
have a lot more vertices in A
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and a lot more vertices
in B. The hypothesis only
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requires what the degrees in
A, max degrees, at most are.
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So it could be a
much bigger graph.
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So if you apply
Kovari-Sos-Turan,
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you would get a much worse
bound compared to what
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this theorem guarantees.
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This 1 of R is optimal.
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In this given statement, you
cannot improve this 1 of R,
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because we know that from
the KST example and the lower
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bounds I showed you last time,
you cannot improve upon this 1
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over R. Also, in this form,
this theorem is best possible.
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I want to show you a
probabilistic technique
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for proving this theorem.
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And this is an important idea
called dependent random choice.
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So let me first give you
an informal interpretation
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of what's going on.
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So the idea is that if you
have a graph G with many edges,
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then inside G, I can find
a large subset of vertices
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U, such that all small
subsets of vertices in U
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have many common neighbors.
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I won't tell you what the
small and many are just yet,
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but we'll see it through
the proof of the theorem,
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but that's the idea.
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So I give you a graph.
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It's not too sparse.
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It's relatively dense.
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Then I should be able
to find some subset that
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is fairly large so
that, let's say,
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every pair of
vertices in the subset
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has many common neighbors.
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So let me write down--
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or at least attempt to write
down-- the formal statement
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of dependent random choice.
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The statement of the
theorem, as I will present,
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has a lot of parameters,
but don't get--
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I don't want you to be
scared off by the parameters,
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so I won't even tell
you what they are,
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but first tell you what the
statement of the conclusion is.
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And then we'll derive
what is the dependencies
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on the parameters, so the
proof or the technique
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is much more important
than the statement
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of the theorem itself.
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So I'll leave some space here.
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The conclusion is that
every graph with n vertices
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and at least alpha n
squared over 2 edges
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contains a subset U of vertices.
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U is not too small, so the
size of U is at least little u.
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And such that for every subset S
of U with R elements, the set S
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has at least M common neighbors.
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So what's the idea here?
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I give you this
graph, and I want
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you to produce the set U
that has this property.
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How might you go about
finding the set U?
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Let me give you an analogy.
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Let's suppose you have the
friendship graph on MIT campus.
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And I want to
select a large set,
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let's say a hundred students,
such that every pair of them
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or maybe even just most pairs of
them have many common friends.
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Well, how might you
go about doing that?
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Well, if you select a
hundred students at random,
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you're unlikely to
achieve that outcome.
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They're going to be pretty
dispersed across campus.
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But if you focus on
some commonalities--
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so, for example,
you go to someone,
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some specific
individual, and look
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at their circle
of close friends,
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then it seems more
likely you'll be
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able to identify
a group of people
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who are very well-connected
in that they, peer-wise,
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have lots of common friends.
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And that's the idea here.
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We're going to make the
random choice by picking
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that core individual--
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that's the random choice--
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but then make the subsequent
dependent random choice
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by looking at the
group of friends
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from that specific
random individual
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instead of choosing the hundred
people uniformly at random,
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which is not going to work.
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So let's execute that
strategy on this graph.
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Let me take T to be the random
set, so that's the core set.
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So let T be--
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for convenience, I'm going
to choose with repetition.
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Instead of just
choosing one person,
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I'm going to use two T vertices,
so a list of T vertices
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chosen uniformly at random
from the vertex set.
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So G is going to be my graph.
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So the graph is G.
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And what we are going to do
is look at the set A, which
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is the set of common neighbors
of T, the vertices that
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are adjacent to all of
T. And that's the set
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I want to think about.
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So I want to basically
argue that this set
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A has more or less the
properties that we desire,
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maybe with some small
number of blemishes, which
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we will fix by cleaning up.
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First, I want to guarantee
that A has a big size--
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we are actually choosing
a lot of vertices.
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So let's evaluate the
size of A in expectation.
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By linearity of
expectation, we need
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to compute the sum
of the probabilities
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that individual vertices
fall in this random A.
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For each particular
vertex V, when is it in A?
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Well, it is in A
if T is contained
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in the neighborhood of this
V. So all of your children
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T fall into the neighborhood.
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Otherwise, V is not going
to be contained in A.
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So each individual probability
can be computed easily
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by looking at the size,
so the degree of V
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divided by the number of
vertices raised to the power
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T--
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T independent choices
chosen with replacements.
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And by convexity applied
to the final expression,
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we find that it is at least this
quantity here where essentially
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we're taking the
averages of the degrees.
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So this is by convexity.
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And finally, the
graph has at least
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that many edges, so the
final quantity there
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is at least N alpha to
the T. Yes, question?
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AUDIENCE: [INAUDIBLE]
00:13:13.937 --> 00:13:16.020
YUFEI ZHAO: The question
is, in our original list,
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are we allowed to have
repeated vertices?
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Yes.
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So it's a list, and we're
choosing every element
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independently at random,
allowing repetition.
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So it's sometimes called
choosing with replacement.
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You choose a T.
You throw it back.
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Choose another one.
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The property that we're looking
for is that for every S subset,
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for every r-element subset of
U, it has many common neighbors.
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So let's look at such
an S. So for each S
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which is an r-element
subset of the vertex set.
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So r-elements subset of
V, what is the probability
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that this set F
is contained in A?
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I'll give you this set S.
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It is contained in A if--
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well, let's think
about how A is chosen.
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You want this to happen.
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An A is chosen as a common
neighborhood of this T--
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so assets containing
A, if and only if, P
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is contained in the
common neighborhood of f.
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So f is fixed for now.
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P is random.
00:14:58.130 --> 00:15:01.750
So we draw the elements of
T independently, uniformly,
00:15:01.750 --> 00:15:02.530
at random.
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Therefore, this probability
is equal to the number
00:15:06.940 --> 00:15:17.400
of common neighbors of S as a
fraction of the total number
00:15:17.400 --> 00:15:18.390
of vertices.
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This fraction
raised to the power
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T. We want all S subsets of
A-- all r-element subsets
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of this A-- to have at
least m common neighbors.
00:15:35.620 --> 00:15:37.570
But maybe we cannot get that.
00:15:37.570 --> 00:15:40.930
So let's figure out how
many bad S's are there.
00:15:40.930 --> 00:15:45.640
How many S's do not satisfy
this condition here?
00:15:45.640 --> 00:15:56.150
So let's call such a
set S bad if it has
00:15:56.150 --> 00:15:59.868
fewer than m common neighbors.
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And from this equation, we see
that for each fixed S that is
00:16:18.960 --> 00:16:25.330
in r-element subset of vertices,
it is bad with probability--
00:16:28.710 --> 00:16:30.630
it is bad if it has
few common neighbors.
00:16:30.630 --> 00:16:32.338
Because if you have
few common neighbors,
00:16:32.338 --> 00:16:34.780
then this probability is small.
00:16:34.780 --> 00:16:42.010
So it is bad with probability
strictly less than m over n
00:16:42.010 --> 00:16:49.320
raised to the power T.
00:16:49.320 --> 00:16:55.480
We chose this A in this
dependent random way.
00:16:55.480 --> 00:16:58.000
And basically, we
want all subsets.
00:16:58.000 --> 00:17:01.270
All r-element subsets have
many common neighbors,
00:17:01.270 --> 00:17:03.850
but maybe we cannot get
that at the first try.
00:17:03.850 --> 00:17:07.150
But we only have a small
number of blemishes,
00:17:07.150 --> 00:17:11.010
so we can fix those
blemishes by getting
00:17:11.010 --> 00:17:14.490
rid of the possible
bad r subsets.
00:17:17.060 --> 00:17:19.700
And that's fine to do,
as long as there are not
00:17:19.700 --> 00:17:21.800
too many of them.
00:17:21.800 --> 00:17:28.040
And indeed, because S is
bad with small probability,
00:17:28.040 --> 00:17:40.740
the expected number of bad
r-element subsets of A is,
00:17:40.740 --> 00:17:42.390
at most--
00:17:42.390 --> 00:17:48.690
well, I look over all possible
r-element subsets of vertices.
00:17:48.690 --> 00:17:53.640
Each one of them is bad
with this probability here.
00:17:53.640 --> 00:17:55.110
So we have that bound.
00:17:59.860 --> 00:18:03.260
And the point now is that if
this number is significantly
00:18:03.260 --> 00:18:07.520
smaller than the
expected size of A, then
00:18:07.520 --> 00:18:10.310
I can clean up all
the bad subsets
00:18:10.310 --> 00:18:13.790
by plucking out one vertex
from each bad subset.
00:18:17.410 --> 00:18:21.880
So indeed, that is the case,
because the expectation
00:18:21.880 --> 00:18:25.270
of the size of A minus--
00:18:25.270 --> 00:18:31.676
so let me call this
quantity here star--
00:18:31.676 --> 00:18:39.390
so the size of A minus
star by what we have shown
00:18:39.390 --> 00:18:46.660
is at least n alpha to the t
minus the quantity just now.
00:18:49.850 --> 00:18:52.890
And I want this number
to be somewhat large.
00:18:52.890 --> 00:18:57.590
And now let me put that as
a hypothesis of the theory.
00:18:57.590 --> 00:19:01.770
So dependent random choice--
00:19:01.770 --> 00:19:14.880
let u, n, r, m, t be positive
integers, alpha positive real,
00:19:14.880 --> 00:19:22.590
and suppose n alpha to
the t minus n to the r.
00:19:22.590 --> 00:19:26.566
Cancel the n to the
t is at least w.
00:19:31.560 --> 00:19:34.500
That's where that
inequality comes from.
00:19:34.500 --> 00:19:41.300
So if this is at least
w, then what we can do
00:19:41.300 --> 00:19:56.570
is delete one vertex
from each bad subset
00:19:56.570 --> 00:20:03.580
S. And after deleting, A
then becomes some smaller
00:20:03.580 --> 00:20:11.630
set A prime with at least
u elements in expectation,
00:20:11.630 --> 00:20:14.790
but then there exists
some new elements--
00:20:17.680 --> 00:20:18.790
Let me put it this way.
00:20:18.790 --> 00:20:25.750
We know that this is true
in expectation, thus there
00:20:25.750 --> 00:20:31.810
exists some T, such that
that inequality is true
00:20:31.810 --> 00:20:37.390
without the expectation, such
that there exists some T,
00:20:37.390 --> 00:20:38.050
such that--
00:20:42.860 --> 00:20:45.380
so this quantity
is at least that.
00:20:45.380 --> 00:20:48.000
Now, to delete a vertex
from each bad subset,
00:20:48.000 --> 00:20:51.620
we obtain this A prime
with at least u elements.
00:20:51.620 --> 00:20:55.170
And we have gotten rid of all
the possible bad subsets--
00:20:55.170 --> 00:21:00.880
so no bad r-element subsets.
00:21:04.630 --> 00:21:10.440
And that finishes off the proof
of dependent random choice.
00:21:10.440 --> 00:21:12.480
Just to recap, the
idea is that if you
00:21:12.480 --> 00:21:17.995
have a dense enough
graph, then the conclusion
00:21:17.995 --> 00:21:25.780
is that you can find a fairly
large subset of vertices,
00:21:25.780 --> 00:21:30.510
so every pair, every r-element,
have many common neighbors.
00:21:30.510 --> 00:21:34.060
And the way you do this, instead
of choosing your set at random,
00:21:34.060 --> 00:21:38.700
which is not going to work, you
choose a small set of anchors.
00:21:38.700 --> 00:21:41.090
T, you think of
that as the anchors.
00:21:41.090 --> 00:21:42.980
You choose a bunch of anchors.
00:21:42.980 --> 00:21:46.450
And then you look at
their common neighborhoods
00:21:46.450 --> 00:21:49.610
and use that as
a starting point.
00:21:49.610 --> 00:21:51.170
That will almost work.
00:21:51.170 --> 00:21:55.160
It might not work perfectly,
but then you fix things up
00:21:55.160 --> 00:21:58.140
by removing the blemishes.
00:21:58.140 --> 00:22:02.030
So this is a very tricky
probabilistic idea.
00:22:02.030 --> 00:22:04.500
It's also a very important one.
00:22:04.500 --> 00:22:07.360
It will allow us to prove
the theorem over there
00:22:07.360 --> 00:22:11.930
about the extremal numbers
of standard degree graphs.
00:22:11.930 --> 00:22:13.294
Question?
00:22:13.294 --> 00:22:17.677
AUDIENCE: Is the definition
of bad set for any subset of V
00:22:17.677 --> 00:22:19.625
or is it only for subsets of A?
00:22:19.625 --> 00:22:22.420
YUFEI ZHAO: The question is,
in the definition of bad,
00:22:22.420 --> 00:22:25.110
do I use this definition
for all subsets
00:22:25.110 --> 00:22:29.080
of V, all r-element subsets,
or just subsets of A?
00:22:29.080 --> 00:22:33.890
So I use it to mean all subsets
of V, because A is random.
00:22:33.890 --> 00:22:37.520
A is random, so the
definition of bad
00:22:37.520 --> 00:22:38.960
does not depend
on the randomness.
00:22:38.960 --> 00:22:40.820
It only depends on
the original graph.
00:22:44.264 --> 00:22:52.140
AUDIENCE: How is the badness
dependent on any probability?
00:22:52.140 --> 00:22:53.848
YUFEI ZHAO: The badness
does not-- so the
00:22:53.848 --> 00:22:55.515
question us, how does
the badness depend
00:22:55.515 --> 00:22:56.320
on any probability?
00:22:56.320 --> 00:22:58.900
The badness does not
depend on the probability,
00:22:58.900 --> 00:23:01.930
but A is random.
00:23:01.930 --> 00:23:04.580
So the number of bad
r-element subsets of A
00:23:04.580 --> 00:23:06.910
is a random variable.
00:23:06.910 --> 00:23:10.020
So each individual-- so
you start with a graph.
00:23:10.020 --> 00:23:11.580
Some r-element
subsets of graphs.
00:23:11.580 --> 00:23:12.450
Some are not.
00:23:12.450 --> 00:23:15.280
And now I choose this random
A in this dependent random
00:23:15.280 --> 00:23:16.380
manner.
00:23:16.380 --> 00:23:18.960
And A might contain
some bad subsets.
00:23:18.960 --> 00:23:24.101
I'm trying to calculate how
many bad subsets does A have.
00:23:26.870 --> 00:23:28.180
Question?
00:23:28.180 --> 00:23:32.386
AUDIENCE: [INAUDIBLE],, because
an S is neither bad or not bad.
00:23:32.386 --> 00:23:35.491
You said it's bad
with probability.
00:23:38.760 --> 00:23:41.790
YUFEI ZHAO: So your
concern is each S
00:23:41.790 --> 00:23:44.130
is bad with probability--
00:23:44.130 --> 00:23:45.420
ah.
00:23:45.420 --> 00:23:45.920
Sorry.
00:23:45.920 --> 00:23:48.082
AUDIENCE: [INAUDIBLE]
00:23:48.082 --> 00:23:48.790
YUFEI ZHAO: Fine.
00:23:48.790 --> 00:23:49.430
Yeah.
00:23:49.430 --> 00:23:49.930
Thank you.
00:23:54.710 --> 00:23:56.970
So each bad subset--
00:23:56.970 --> 00:24:00.990
OK, so for each
fixed bad subset,
00:24:00.990 --> 00:24:09.870
it is contained in
A with probability.
00:24:09.870 --> 00:24:11.400
Thank you.
00:24:11.400 --> 00:24:13.590
I hope this makes it clear.
00:24:13.590 --> 00:24:15.810
So the probability of the--
00:24:15.810 --> 00:24:18.030
the property of being
bad is not random,
00:24:18.030 --> 00:24:21.848
but it containing A is random.
00:24:21.848 --> 00:24:22.348
Question?
00:24:22.348 --> 00:24:25.520
AUDIENCE: [INAUDIBLE]
00:24:30.798 --> 00:24:32.840
YUFEI ZHAO: So the question
is, why is this true?
00:24:32.840 --> 00:24:36.550
So why are these
two events the same?
00:24:36.550 --> 00:24:40.780
And it's kind of hard to
steer the definition a bit.
00:24:40.780 --> 00:24:45.240
So T is chosen as the
common neighborhoods--
00:24:45.240 --> 00:24:45.820
sorry.
00:24:45.820 --> 00:24:47.290
You choose T at random.
00:24:47.290 --> 00:24:51.140
And you choose A to be the
common neighborhood of T.
00:24:51.140 --> 00:24:57.210
And so how do you
characterize subsets of A
00:24:57.210 --> 00:25:01.251
if every element is
connected to all of T?
00:25:05.160 --> 00:25:06.700
You have to think about it.
00:25:06.700 --> 00:25:07.510
Any more questions?
00:25:07.510 --> 00:25:08.010
Yes?
00:25:08.010 --> 00:25:11.634
AUDIENCE: [INAUDIBLE]
00:25:11.634 --> 00:25:13.310
YUFEI ZHAO: We pick T at random.
00:25:13.310 --> 00:25:15.008
T is uniform.
00:25:15.008 --> 00:25:16.716
So the question is,
how are we picking T?
00:25:16.716 --> 00:25:18.620
T is uniform at random.
00:25:18.620 --> 00:25:19.200
Oh, great.
00:25:19.200 --> 00:25:21.780
So the question is, how
do we pick the little t
00:25:21.780 --> 00:25:23.340
in the theorem statement?
00:25:23.340 --> 00:25:24.945
It depends on the application.
00:25:24.945 --> 00:25:28.110
And that's a little bit weird,
because in the statement
00:25:28.110 --> 00:25:30.780
of the theorem, the little t
shows up in the inequality,
00:25:30.780 --> 00:25:31.980
but not in the conclusion.
00:25:31.980 --> 00:25:35.880
So you think of t as
an auxiliary parameter.
00:25:35.880 --> 00:25:38.520
So the little t comes
up in the proof,
00:25:38.520 --> 00:25:43.290
but not really in
the conclusion.
00:25:43.290 --> 00:25:46.297
Any more questions?
00:25:46.297 --> 00:25:47.130
It's a tricky lemma.
00:25:47.130 --> 00:25:47.922
It's a tricky idea.
00:25:52.770 --> 00:25:55.440
So now let me use it to prove
the statement over there.
00:25:55.440 --> 00:25:57.060
And here, it's not so hard.
00:25:57.060 --> 00:26:00.780
So it's mostly an application
of this dependent random choice
00:26:00.780 --> 00:26:01.575
lemma.
00:26:01.575 --> 00:26:06.570
So for all H that's
in the theorem--
00:26:09.842 --> 00:26:11.550
so let me prove this
lemma-- so for all H
00:26:11.550 --> 00:26:16.230
that's in the theorem,
there exists a constant C,
00:26:16.230 --> 00:26:27.840
such that every graph with at
least C n to the 2 minus 1 over
00:26:27.840 --> 00:26:38.880
r edges contains
a vertex U with--
00:26:38.880 --> 00:26:50.870
so vertex subset U with
the size of U equal to B--
00:26:50.870 --> 00:26:54.830
so B comes from the
vertex by partition of H.
00:26:54.830 --> 00:26:56.810
It's a constant--
00:26:56.810 --> 00:27:12.700
such that every r-element
subset in B has--
00:27:12.700 --> 00:27:20.060
so r-element subset in U has
lots of common neighbors.
00:27:20.060 --> 00:27:22.260
That's at least
another constant,
00:27:22.260 --> 00:27:26.740
which is the number
of vertices in H--
00:27:26.740 --> 00:27:31.770
that many common neighbors.
00:27:31.770 --> 00:27:34.820
So you see, it's
a direct corollary
00:27:34.820 --> 00:27:37.445
of the dependent random
choice lemma by setting it
00:27:37.445 --> 00:27:41.210
in the right
parameters and, indeed,
00:27:41.210 --> 00:27:46.920
by the dependent random choice
lemma where we choose this T--
00:27:46.920 --> 00:27:49.980
this auxiliary variable T
in independent random choice
00:27:49.980 --> 00:27:51.170
lemma--
00:27:51.170 --> 00:27:52.670
equal to r.
00:27:52.670 --> 00:28:04.040
So suffices to check that
there exists of C such that--
00:28:04.040 --> 00:28:08.620
plug it into that expression,
that inequality, up there.
00:28:08.620 --> 00:28:14.880
So n2Cn to the minus 1 over r--
00:28:14.880 --> 00:28:20.210
raised to the power
r minus n choose r.
00:28:20.210 --> 00:28:21.770
And then this expression here.
00:28:24.580 --> 00:28:28.400
So I'm just plugging in the
various graph parameters
00:28:28.400 --> 00:28:32.880
into the dependent
random choice statement.
00:28:32.880 --> 00:28:35.690
And I want to show that you
can find a constant C such
00:28:35.690 --> 00:28:38.320
that this inequality is true.
00:28:38.320 --> 00:28:41.440
And indeed, these
exponents, they cancel out,
00:28:41.440 --> 00:28:49.310
so the first term is
simply 2C raised to the r.
00:28:49.310 --> 00:28:52.210
And the second one,
because, again,
00:28:52.210 --> 00:28:55.160
you notice that the
exponents work out just fine,
00:28:55.160 --> 00:28:57.145
so it is, at most, a constant.
00:29:00.210 --> 00:29:02.550
So you can choose C big
enough so that this is true.
00:29:07.570 --> 00:29:13.183
So it's a direct verification
of the hypothesis of the lemma.
00:29:13.183 --> 00:29:15.350
And now we're ready to prove
the theorem over there.
00:29:31.090 --> 00:29:33.100
Yes, question?
00:29:33.100 --> 00:29:35.855
AUDIENCE: How do you have size
A plus B common neighbors?
00:29:35.855 --> 00:29:37.071
Isn't that the entire graph?
00:29:37.071 --> 00:29:39.025
Or can I just [INAUDIBLE]?
00:29:39.025 --> 00:29:40.650
YUFEI ZHAO: The
question is, how do you
00:29:40.650 --> 00:29:42.450
have size A plus B
commons neighbors?
00:29:42.450 --> 00:29:44.910
So H is fixed.
00:29:44.910 --> 00:29:47.640
And A and B are
constants, so A plus B
00:29:47.640 --> 00:29:51.370
is the number of vertices
of H. That's a constant.
00:29:51.370 --> 00:29:52.230
AUDIENCE: Oh, sorry.
00:29:52.230 --> 00:29:52.932
Oh, OK.
00:29:52.932 --> 00:29:55.140
YUFEI ZHAO: Yeah, I'm talking
about common neighbors,
00:29:55.140 --> 00:29:58.380
not in H, but in the big
graph G as n vertices.
00:30:02.220 --> 00:30:03.820
Now, I like questions.
00:30:03.820 --> 00:30:04.320
It's tricky.
00:30:04.320 --> 00:30:06.600
This is a tricky argument,
so please do ask questions
00:30:06.600 --> 00:30:07.973
if you're confused.
00:30:07.973 --> 00:30:09.390
And there are times
when I may not
00:30:09.390 --> 00:30:13.860
have explained it very well,
so please do ask questions.
00:30:13.860 --> 00:30:15.360
So let's prove the theorem.
00:30:15.360 --> 00:30:16.620
And now we're almost there.
00:30:16.620 --> 00:30:19.410
The idea is that we
embed the vertices of B
00:30:19.410 --> 00:30:22.300
into the big graph G one by one.
00:30:22.300 --> 00:30:22.800
Sorry.
00:30:22.800 --> 00:30:28.045
First, embed B into the vertices
of G using U from the lemma--
00:30:31.160 --> 00:30:35.120
so the lemma that
we just stated.
00:30:35.120 --> 00:30:37.530
And they claim that once
you have done that--
00:30:37.530 --> 00:30:38.800
so the vertices of B--
00:30:44.020 --> 00:30:54.860
and now I need to embed the
remaining vertices of A.
00:30:54.860 --> 00:31:00.600
And I can do this one
by one, because if I
00:31:00.600 --> 00:31:08.710
need to embed some vertex of A,
has, at most, r neighbors to B.
00:31:08.710 --> 00:31:11.380
But I have embedded
B in such a way
00:31:11.380 --> 00:31:27.080
that they have a lot of
common neighbors inside G.
00:31:27.080 --> 00:31:28.960
So I can always do it.
00:31:28.960 --> 00:31:31.700
So I can always embed
the vertices of A one
00:31:31.700 --> 00:31:37.010
at a time in such a way that
I even avoid collisions.
00:31:37.010 --> 00:31:41.730
I don't allow vertices to be
embedded into the same place.
00:31:50.300 --> 00:31:54.690
This is all using that B,
embedding a B, which is U,
00:31:54.690 --> 00:31:59.040
has many common neighbors in
G. So once you put that in.
00:31:59.040 --> 00:32:03.360
And the rest, you just
make one choice at a time.
00:32:03.360 --> 00:32:06.840
And you need to embed
a second vertex,
00:32:06.840 --> 00:32:09.450
or you can find somewhere in
their common neighborhood that
00:32:09.450 --> 00:32:11.511
allows you to do it.
00:32:11.511 --> 00:32:13.400
So you embed the
vertices one at a time,
00:32:13.400 --> 00:32:15.317
and then you finish
embedding the whole graph.
00:32:18.450 --> 00:32:21.090
Any questions?
00:32:21.090 --> 00:32:22.680
It's a tricky argument.
00:32:22.680 --> 00:32:27.050
So let's take a break, and I
want you to think about it.
00:32:27.050 --> 00:32:27.830
Any questions?
00:32:35.780 --> 00:32:36.724
Yes.
00:32:36.724 --> 00:32:37.840
AUDIENCE: [INAUDIBLE].
00:32:42.717 --> 00:32:44.800
YUFEI ZHAO: So the question
is, how do you embed A
00:32:44.800 --> 00:32:47.270
without having any collisions?
00:32:47.270 --> 00:32:51.520
So I put in the vertices of
B so that every r of them
00:32:51.520 --> 00:32:53.780
have many neighbors.
00:32:53.780 --> 00:32:58.230
And now, I want to try to
embed the vertices of A one
00:32:58.230 --> 00:33:00.120
by one in any order.
00:33:00.120 --> 00:33:02.130
Think about-- you
pick the first vertex.
00:33:02.130 --> 00:33:04.130
Where can it go?
00:33:04.130 --> 00:33:07.740
It has-- let's say it's
adjacent to the first three
00:33:07.740 --> 00:33:10.920
vertices of B. So,
in the embedding,
00:33:10.920 --> 00:33:13.170
it has to go in the common
neighborhood of those three
00:33:13.170 --> 00:33:17.010
vertices, which
we know is large.
00:33:17.010 --> 00:33:19.520
So I put them anywhere.
00:33:19.520 --> 00:33:21.260
And I do the same for
the second vertex,
00:33:21.260 --> 00:33:24.600
do the same for
the fourth vertex.
00:33:24.600 --> 00:33:27.770
So I just keep on going.
00:33:27.770 --> 00:33:30.260
Because the common
neighborhood is large,
00:33:30.260 --> 00:33:33.080
it may be that some of
the potential vertices I
00:33:33.080 --> 00:33:36.950
might embed is already
used by the previous steps
00:33:36.950 --> 00:33:38.690
in the process.
00:33:38.690 --> 00:33:40.550
But because I always
have at least A
00:33:40.550 --> 00:33:43.400
plus B common
neighbors, I always
00:33:43.400 --> 00:33:45.950
have some possibilities
that remain.
00:33:52.820 --> 00:33:53.542
Yes, question.
00:33:53.542 --> 00:33:57.786
AUDIENCE: So [INAUDIBLE],,
like the last line
00:33:57.786 --> 00:34:00.455
of the proof to
[INAUDIBLE] delete
00:34:00.455 --> 00:34:04.600
a vertex from each
bad subset, and then
00:34:04.600 --> 00:34:06.212
to make that [INAUDIBLE].
00:34:12.775 --> 00:34:13.400
YUFEI ZHAO: Ah.
00:34:13.400 --> 00:34:16.710
The question is, how does
this bad subset deletion work?
00:34:16.710 --> 00:34:20.300
So you have this A
which is fairly large.
00:34:20.300 --> 00:34:23.880
And you know that there
exists some instance-- there's
00:34:23.880 --> 00:34:26.100
some incidence of
this randomness that
00:34:26.100 --> 00:34:31.860
produces for you a situation
where A has very few bad r
00:34:31.860 --> 00:34:33.429
subsets.
00:34:33.429 --> 00:34:37.540
So then I take A and I
delete from A one vertex
00:34:37.540 --> 00:34:40.020
in each bad subset.
00:34:40.020 --> 00:34:42.620
I haven't changed the
size of A very much.
00:34:42.620 --> 00:34:46.409
A is still quite large
after this deletion,
00:34:46.409 --> 00:34:49.020
but now A has no bad
subsets remaining,
00:34:49.020 --> 00:34:52.110
because I've gotten rid of
one vertex from each one,
00:34:52.110 --> 00:34:54.980
from each bad subset.
00:34:54.980 --> 00:34:56.880
So they're very
similar to what we've
00:34:56.880 --> 00:35:01.820
seen before, the random process
for creating an H-free graph.
00:35:01.820 --> 00:35:04.560
You generate a
random H-free graph
00:35:04.560 --> 00:35:07.920
which has very few copies
of H relative to the number
00:35:07.920 --> 00:35:10.140
of edges, and then
you get rid of them
00:35:10.140 --> 00:35:16.560
by removing one edge
from each copy of H.
00:35:16.560 --> 00:35:20.070
So with the theorem that
we saw in the first part
00:35:20.070 --> 00:35:22.830
of the lecture, we
saw how to improve
00:35:22.830 --> 00:35:27.270
on the bound of Kovari-Sos-Turan
in some circumstances, namely
00:35:27.270 --> 00:35:30.960
one where the graph that
you're forbidding, this H,
00:35:30.960 --> 00:35:32.592
essentially has bounded degree.
00:35:32.592 --> 00:35:34.050
We stated something
a bit stronger,
00:35:34.050 --> 00:35:38.320
namely has bounded
degree from one side.
00:35:38.320 --> 00:35:41.340
And that's a pretty
general result.
00:35:41.340 --> 00:35:44.650
And now I want to look at
some more specific situations
00:35:44.650 --> 00:35:46.680
where you might be able
to improve further.
00:35:49.460 --> 00:35:54.530
So what are some nice
bipartite graphs?
00:35:54.530 --> 00:35:59.070
One that comes up is
kind of even cycles.
00:35:59.070 --> 00:36:03.490
So if you have
C4, C6, and so on.
00:36:03.490 --> 00:36:06.410
And you see C4 is
the same as K2,2,
00:36:06.410 --> 00:36:08.180
which we already saw before.
00:36:08.180 --> 00:36:11.420
But even C6, the
techniques so far
00:36:11.420 --> 00:36:14.990
allow us to obtain, with the
theorem that we just saw--
00:36:14.990 --> 00:36:18.710
gives us a bound on C6 that's
more or less the same as that
00:36:18.710 --> 00:36:23.480
of C4, namely n to the 3/2.
00:36:23.480 --> 00:36:26.700
So what's the truth for C6?
00:36:26.700 --> 00:36:29.500
Well, it turns out that
you can do much better.
00:36:32.120 --> 00:36:40.540
So this is the theorem
of Bondy and Simonovits
00:36:40.540 --> 00:36:45.060
that, for all
integers k at least 2,
00:36:45.060 --> 00:36:50.380
there exists some constant C
such that the extremal number--
00:36:50.380 --> 00:36:53.540
well, I'm going to
use C too many times.
00:36:53.540 --> 00:37:00.820
So I'll just say that the
extremal number of C2 sub k
00:37:00.820 --> 00:37:08.850
is, at most, on the order
of n to the 1 plus 1 over k.
00:37:08.850 --> 00:37:12.220
So, in particular, for 6
cycles, the upper bound
00:37:12.220 --> 00:37:15.215
is 4/3 in the exponent.
00:37:15.215 --> 00:37:16.760
It's better than the 3/2.
00:37:20.350 --> 00:37:22.360
So there's another class
of graphs where there
00:37:22.360 --> 00:37:23.627
are some nice upper bounds.
00:37:23.627 --> 00:37:25.960
So you can ask, well, just
like Kovari-Sos-Turan theorem
00:37:25.960 --> 00:37:28.030
for complete bipartite
graphs, do we
00:37:28.030 --> 00:37:31.660
know matching lower
bound constructions?
00:37:31.660 --> 00:37:38.280
And what is known is
that it is tight only
00:37:38.280 --> 00:37:40.620
for a small number of cases.
00:37:40.620 --> 00:37:43.980
And the others, we do not
know whether they are tight.
00:37:43.980 --> 00:37:51.630
So this Bondy-Simonovits
theorem, it is tight for k
00:37:51.630 --> 00:37:55.990
being 2, 3, or 5,
and open for others.
00:37:59.140 --> 00:38:05.550
So there are constructions
for C4,3, C6,3 and C10,3,
00:38:05.550 --> 00:38:06.972
but not for C8,3.
00:38:06.972 --> 00:38:07.930
That's an open problem.
00:38:12.440 --> 00:38:14.150
The proof of the
Bondy-Simonovits theorem
00:38:14.150 --> 00:38:16.520
is slightly involved,
but I want to show you
00:38:16.520 --> 00:38:19.010
a weaker result that
already contains
00:38:19.010 --> 00:38:21.410
a lot of interesting ideas.
00:38:21.410 --> 00:38:26.420
So a weaker result is this.
00:38:29.840 --> 00:38:34.990
That for every
integer k at least 2,
00:38:34.990 --> 00:38:44.060
there exists a constant C such
that every n-vertex graph G
00:38:44.060 --> 00:38:49.490
with at least C--
00:38:49.490 --> 00:38:52.940
so this-- the correct--
00:38:52.940 --> 00:38:58.180
the same order of
number of edges--
00:38:58.180 --> 00:38:59.720
so we know for
Bondy-Simonovits, it
00:38:59.720 --> 00:39:03.390
contains an even cycle
of length exactly 2k.
00:39:03.390 --> 00:39:05.810
So we'll show something
slightly weaker
00:39:05.810 --> 00:39:14.510
that contains an even cycle
of length, at most, 2k.
00:39:20.920 --> 00:39:27.510
Which, in other words, says that
the extremal number, if you--
00:39:27.510 --> 00:39:29.140
so we haven't introduced
this notation,
00:39:29.140 --> 00:39:31.795
but, hopefully, you can
guess what it means.
00:39:31.795 --> 00:39:36.550
Then, if you forbid
all of these cycles,
00:39:36.550 --> 00:39:43.325
then it is this quantity there.
00:39:43.325 --> 00:39:51.460
So I'll show this weaker
result. All right.
00:39:51.460 --> 00:39:53.820
Let's' do it.
00:39:53.820 --> 00:39:56.860
First, I want to show you a
couple of easy preparatory
00:39:56.860 --> 00:39:57.360
lemmas.
00:40:01.380 --> 00:40:16.430
The first-- so every graph G
contains a subgraph with min
00:40:16.430 --> 00:40:22.680
degree at least the half of--
00:40:22.680 --> 00:40:27.470
at least half of the
average degree of G.
00:40:27.470 --> 00:40:30.270
So you have a graph G. It
has large average degree.
00:40:30.270 --> 00:40:31.560
It has lots of edges.
00:40:31.560 --> 00:40:34.650
But maybe there are some
small degree vertices.
00:40:34.650 --> 00:40:38.370
And it will be useful to know
that the minimum degree is
00:40:38.370 --> 00:40:40.470
actually quite large as well.
00:40:40.470 --> 00:40:42.872
So it turns out, by
passing to a subgraph,
00:40:42.872 --> 00:40:43.830
you can guarantee that.
00:40:48.730 --> 00:40:50.560
How do you think we
might prove this?
00:40:50.560 --> 00:40:51.430
I give you a graph.
00:40:51.430 --> 00:40:52.700
I know it has lots of edges.
00:40:52.700 --> 00:40:54.450
But maybe some vertices
have small degree.
00:40:58.250 --> 00:40:58.806
Yes.
00:40:58.806 --> 00:41:00.125
AUDIENCE: [INAUDIBLE].
00:41:00.125 --> 00:41:02.250
YUFEI ZHAO: So you suggest
dependent random choice.
00:41:02.250 --> 00:41:04.333
That's a heavy hammer for
such a simple statement.
00:41:04.333 --> 00:41:06.590
And so it turns out we can
do something even simpler.
00:41:06.590 --> 00:41:08.780
AUDIENCE: [INAUDIBLE].
00:41:08.780 --> 00:41:11.280
YUFEI ZHAO: So we'll just throw
out the min degree vertices.
00:41:11.280 --> 00:41:14.781
So throw out the
small degree vertices.
00:41:14.781 --> 00:41:24.150
So if the average degree is
2t then the number of edges
00:41:24.150 --> 00:41:28.140
is number of vertices times t.
00:41:28.140 --> 00:41:34.280
And you see that removing
a vertex of degree t,
00:41:34.280 --> 00:41:42.950
or at most t, cannot
decrease average degree.
00:41:46.070 --> 00:41:47.810
So I have this
process where I remove
00:41:47.810 --> 00:41:51.680
vertices that are less than
half of average degree.
00:41:51.680 --> 00:41:53.450
And the average degree
never goes down,
00:41:53.450 --> 00:41:54.540
so I keep on doing this.
00:41:54.540 --> 00:41:55.790
Average degree stays the same.
00:41:55.790 --> 00:41:59.820
Well, it can go up,
but it never goes down.
00:41:59.820 --> 00:42:04.250
And when I stop, I don't have
any more small degree vertices
00:42:04.250 --> 00:42:06.510
to get rid of.
00:42:06.510 --> 00:42:08.520
Why does the process
even terminate?
00:42:08.520 --> 00:42:11.650
Maybe we end up with the empty
graph, which is not so useful.
00:42:15.142 --> 00:42:15.642
Yes.
00:42:15.642 --> 00:42:18.315
AUDIENCE: The average
degree is [INAUDIBLE]..
00:42:18.315 --> 00:42:20.690
YUFEI ZHAO: He said the average
degree is non-decreasing.
00:42:20.690 --> 00:42:23.390
But how do I know the graph
has at least some number
00:42:23.390 --> 00:42:26.430
of vertices when I stop?
00:42:26.430 --> 00:42:26.930
Yes.
00:42:26.930 --> 00:42:28.334
AUDIENCE: [INAUDIBLE].
00:42:31.970 --> 00:42:36.070
YUFEI ZHAO: So if you--
you must terminate.
00:42:36.070 --> 00:42:39.950
You must terminate
because every graph,
00:42:39.950 --> 00:42:42.800
if you have way too small
number of vertices--
00:42:42.800 --> 00:42:44.750
because, for example, if--
00:42:44.750 --> 00:42:51.170
we'll just notice that
every graph with, at most,
00:42:51.170 --> 00:42:58.060
2t vertices has average
degree less than 2t.
00:42:58.060 --> 00:43:01.750
So you'll never get
below 2t vertices.
00:43:01.750 --> 00:43:04.968
You'll run out of room
if you go too far.
00:43:04.968 --> 00:43:06.760
Because that's the
first preparation lemma.
00:43:06.760 --> 00:43:13.240
The second one is
that every graph G
00:43:13.240 --> 00:43:20.110
has a bipartite subgraph
with at least half
00:43:20.110 --> 00:43:22.590
of the number of edges
as the original graph.
00:43:27.940 --> 00:43:30.860
This is a very nice
and quick exercise
00:43:30.860 --> 00:43:32.680
in the probabilistic method.
00:43:32.680 --> 00:43:38.530
So we can color every
vertex black and white
00:43:38.530 --> 00:43:39.580
uniformly at random.
00:43:44.740 --> 00:43:50.850
And the expected number
of black to white edges
00:43:50.850 --> 00:43:55.970
is exactly half of the
original number of edges
00:43:55.970 --> 00:43:59.210
by linearity of expectations.
00:43:59.210 --> 00:44:03.880
So there's some instance with
at least that many black-white
00:44:03.880 --> 00:44:05.737
edges, and that's a
bipartite subgraph.
00:44:09.860 --> 00:44:14.132
So now we can prove the theorem
about even cycles, the weaker
00:44:14.132 --> 00:44:14.840
theorem at least.
00:44:22.392 --> 00:44:26.470
I start with a graph
which has a lot of edges.
00:44:26.470 --> 00:44:31.900
But by these two lemmas, and
changing constant somewhat,
00:44:31.900 --> 00:44:39.880
I can obtain a subgraph with--
00:44:39.880 --> 00:44:45.530
that's bipartite and has
min degree quite large.
00:44:45.530 --> 00:44:54.670
So I lose, at most,
a constant factor,
00:44:54.670 --> 00:44:58.590
and I get basically
within a factor of 2 of--
00:44:58.590 --> 00:45:04.160
well, factor of 4 of
the average degree.
00:45:04.160 --> 00:45:06.870
So let me call this
quantity delta,
00:45:06.870 --> 00:45:10.130
the min degree in this subgraph,
this bipartite subgraph.
00:45:13.240 --> 00:45:15.100
Let's think about what
happens to this graph
00:45:15.100 --> 00:45:17.998
if I start with an
arbitrary vertex.
00:45:17.998 --> 00:45:19.540
So now I have a min
degree condition.
00:45:19.540 --> 00:45:24.390
It's, really, all vertices are
now kind of the same to me.
00:45:24.390 --> 00:45:29.840
So pick an arbitrary vertex,
and look at its neighborhood.
00:45:29.840 --> 00:45:39.480
It has at least delta
edges coming out.
00:45:39.480 --> 00:45:45.820
So let me call the first vertex
level 0, and the second set
00:45:45.820 --> 00:45:47.800
level 1.
00:45:47.800 --> 00:45:51.370
It's bipartite, so there
are no edges within level 1.
00:45:51.370 --> 00:45:52.810
Let's expand out even further.
00:46:01.830 --> 00:46:05.863
Can there be some collisions
where two of these edges
00:46:05.863 --> 00:46:06.780
go to the same vertex?
00:46:09.310 --> 00:46:11.710
Well, if there were,
then I find a C4.
00:46:15.180 --> 00:46:19.350
So if I assume--
00:46:19.350 --> 00:46:25.620
so let's assume that
there's no C4, C6,
00:46:25.620 --> 00:46:31.060
and so on, C2k,
for contradiction.
00:46:31.060 --> 00:46:37.960
So because there's no
C4, all the endpoints
00:46:37.960 --> 00:46:42.090
of this path of length 2
are distinct in level 2.
00:46:45.190 --> 00:46:47.410
And you keep going, or you
keep expanding further.
00:46:50.120 --> 00:46:50.620
And so on.
00:46:56.380 --> 00:46:59.520
And all the way to level t.
00:46:59.520 --> 00:47:01.900
And all of these guys
must be distinct as well.
00:47:06.090 --> 00:47:14.382
Because, otherwise, you'll find
a cycle of length, at most, 2t.
00:47:14.382 --> 00:47:16.740
So when you do this
expansion, each step,
00:47:16.740 --> 00:47:18.270
you get distinct vertices.
00:47:18.270 --> 00:47:21.288
And you also have no
edges inside each part
00:47:21.288 --> 00:47:23.330
because you are looking
in the bipartite setting.
00:47:26.390 --> 00:47:31.130
So how many vertices
do you get at the end?
00:47:31.130 --> 00:47:32.990
Well, I have a min
degree condition.
00:47:32.990 --> 00:47:34.790
The min degree
condition tells me
00:47:34.790 --> 00:47:40.200
that I expand by a factor of at
least delta minus 1 each time.
00:47:40.200 --> 00:47:44.860
So the number of
vertices in here
00:47:44.860 --> 00:47:50.510
is at least delta minus
1 raised to the power t--
00:47:50.510 --> 00:47:53.648
so raised to the power k.
00:47:53.648 --> 00:47:55.624
Here's k.
00:47:55.624 --> 00:47:59.630
And expand all the
way to the end.
00:47:59.630 --> 00:48:04.180
But, you see, that number
there is quite large.
00:48:04.180 --> 00:48:06.660
So, in particular,
if t is large enough,
00:48:06.660 --> 00:48:07.946
then it's bigger than n.
00:48:12.806 --> 00:48:15.320
And that will be
a contradiction,
00:48:15.320 --> 00:48:19.040
because you only had n vertices
in the graph to begin with.
00:48:24.090 --> 00:48:26.190
Therefore, the
assumption that there
00:48:26.190 --> 00:48:28.800
are no cycles, even cycles
of length, at most, 2k,
00:48:28.800 --> 00:48:31.923
is incorrect.
00:48:31.923 --> 00:48:33.090
And that finishes the proof.
00:48:38.640 --> 00:48:39.640
Any questions?
00:48:42.570 --> 00:48:43.070
Yes.
00:48:43.070 --> 00:48:46.010
AUDIENCE: Are the ideas for the
full Bondy-Simonovits theorem
00:48:46.010 --> 00:48:47.490
similar?
00:48:47.490 --> 00:48:51.060
YUFEI ZHAO: So the question has
to do with, in the original,
00:48:51.060 --> 00:48:53.932
in the full Bondy-Simonovits
theorem, what do we need to do?
00:48:53.932 --> 00:48:54.640
Are they similar?
00:48:54.640 --> 00:48:57.310
So, certainly, you want
to do something like this.
00:48:57.310 --> 00:49:01.110
But, then, you also
want to think about,
00:49:01.110 --> 00:49:08.070
if you do have short cycles,
how can you bypass them?
00:49:08.070 --> 00:49:10.170
So there's a more
careful analysis
00:49:10.170 --> 00:49:12.150
of what happens
with shorter cycles.
00:49:12.150 --> 00:49:14.520
And we will not get into that.
00:49:14.520 --> 00:49:15.732
Any more questions?
00:49:19.040 --> 00:49:20.240
All right.
00:49:20.240 --> 00:49:23.900
So the first thing that
we did in today's lecture
00:49:23.900 --> 00:49:24.878
has to deal with what--
00:49:24.878 --> 00:49:25.920
AUDIENCE: Quick question.
00:49:25.920 --> 00:49:26.587
YUFEI ZHAO: Yes.
00:49:26.587 --> 00:49:28.220
AUDIENCE: So why is--
00:49:28.220 --> 00:49:32.773
so why are all vertices
distinct for level k?
00:49:32.773 --> 00:49:34.662
Or are they just
defined that way?
00:49:34.662 --> 00:49:36.770
YUFEI ZHAO: So question
is, why are the vertices
00:49:36.770 --> 00:49:38.000
distinct for level k?
00:49:38.000 --> 00:49:41.120
So at level k, if you
have some collapse,
00:49:41.120 --> 00:49:46.476
then it came from two different
paths, therefore forming a C2k.
00:49:46.476 --> 00:49:47.018
AUDIENCE: OK.
00:49:53.790 --> 00:49:56.880
YUFEI ZHAO: So now I want to
revisit the first theorem that
00:49:56.880 --> 00:50:00.120
happened in today's
lecture, namely
00:50:00.120 --> 00:50:06.920
if you have this H,
bipartite A and B.
00:50:06.920 --> 00:50:10.850
So we saw the hypothesis was
that if every vertex in A
00:50:10.850 --> 00:50:13.970
had bounded degree,
degree, at most, r, then
00:50:13.970 --> 00:50:16.040
we got the upper bound
on the extremal number.
00:50:16.040 --> 00:50:18.840
That was n to the
2 minus 1 over r.
00:50:18.840 --> 00:50:22.750
And, in particular,
for r equals to 2,
00:50:22.750 --> 00:50:26.210
suppose I have that
situation there.
00:50:28.740 --> 00:50:35.590
Suppose that the degree
for every vertex in A
00:50:35.590 --> 00:50:38.110
is, at most, 2.
00:50:38.110 --> 00:50:44.320
Then the first theorem
guaranteed us that the extremal
00:50:44.320 --> 00:50:50.560
number is on the order,
at most, n the 3/2,
00:50:50.560 --> 00:50:55.770
just like the extremal number
for 4 cycles, for K2,2's.
00:50:55.770 --> 00:50:58.410
And, of course, this
statement is tight,
00:50:58.410 --> 00:51:02.330
in the sense that if I change
this 3/2 to any smaller number,
00:51:02.330 --> 00:51:06.330
then, well, just taking
H to be K2,2 violates it.
00:51:06.330 --> 00:51:11.520
So I cannot replace in this
generality 3/2 by any smaller
00:51:11.520 --> 00:51:15.330
number, because we know that
the extremal number for K2,2 is
00:51:15.330 --> 00:51:16.840
this order.
00:51:16.840 --> 00:51:20.540
But is that the
only obstruction?
00:51:20.540 --> 00:51:28.520
So if H is not K2,2, well, you
can make some sillier examples
00:51:28.520 --> 00:51:32.780
too by taking K2,2 and
add some more edges.
00:51:32.780 --> 00:51:36.705
So lets forbid H from
having a K2,2 subgraph.
00:51:39.840 --> 00:51:41.414
Can you do better now?
00:51:41.414 --> 00:51:46.560
So in this case, can you
improve for the specific H
00:51:46.560 --> 00:51:47.960
that exponent?
00:51:47.960 --> 00:51:52.610
And we already saw one case
where you can do this, namely
00:51:52.610 --> 00:51:57.846
in Bondy-Simonovits
theorem for cycles.
00:51:57.846 --> 00:52:01.180
Or if you only applied this
theorem here, you get 3/2,
00:52:01.180 --> 00:52:06.260
but Bondy-Simonovits tells
you a much better exponent.
00:52:06.260 --> 00:52:08.530
So let's explore the situation.
00:52:08.530 --> 00:52:12.800
And it turns out, in a very
recent theorem that is only
00:52:12.800 --> 00:52:19.100
proved the last couple of years
by David Conlon and Joonkyung
00:52:19.100 --> 00:52:25.820
Lee, they showed
that for every H
00:52:25.820 --> 00:52:33.640
as above, there exists
constants little c
00:52:33.640 --> 00:52:41.560
and big C such that the
extremal number of H
00:52:41.560 --> 00:52:48.860
is upper bounded by something
where I can decrease 3/2
00:52:48.860 --> 00:52:50.180
to some even smaller number.
00:52:52.960 --> 00:52:56.890
So, somehow, this 3/2, now
we understand is really
00:52:56.890 --> 00:53:00.280
because of the presence of K2,2.
00:53:00.280 --> 00:53:03.040
The graph has--
if H has no K2,2,
00:53:03.040 --> 00:53:05.530
then some smaller
number suffices.
00:53:11.510 --> 00:53:13.960
And I want to use
the rest of today
00:53:13.960 --> 00:53:17.457
to explain how to prove
that theorem there.
00:53:17.457 --> 00:53:18.040
Yes, question.
00:53:18.040 --> 00:53:20.590
AUDIENCE: The C is not
independent of H [INAUDIBLE]..
00:53:20.590 --> 00:53:21.632
YUFEI ZHAO: That's right.
00:53:21.632 --> 00:53:24.370
So the question is,
is C independent of H?
00:53:24.370 --> 00:53:37.546
So C depends on H. So C, they
are dependent on H. Questions?
00:53:43.490 --> 00:53:47.930
Let me put this question in a
slightly different formulation
00:53:47.930 --> 00:53:50.410
that is also equivalent.
00:53:50.410 --> 00:53:53.060
So in graph theory, there is
a notion of a subdivision.
00:53:53.060 --> 00:54:02.060
And, in particular, a one
subdivision of graph H
00:54:02.060 --> 00:54:05.910
is this operation where
you start with a graph--
00:54:05.910 --> 00:54:08.840
let's say this graph here--
00:54:08.840 --> 00:54:17.490
and you add a
vertex to the middle
00:54:17.490 --> 00:54:18.840
of every edge of this graph.
00:54:27.444 --> 00:54:29.590
So, initially, it's 4 vertices.
00:54:29.590 --> 00:54:31.890
Now you add a new
vertex to every edge.
00:54:31.890 --> 00:54:35.680
So you subdivide every edge
into a path of two edges.
00:54:35.680 --> 00:54:37.780
That's called a subdivision.
00:54:37.780 --> 00:54:42.665
So for today's lecture, let me
denote subdivisions by a prime.
00:54:42.665 --> 00:54:45.237
And, in particular,
if this is K4,
00:54:45.237 --> 00:54:47.070
then I will denote this
graph here K4 prime.
00:54:50.230 --> 00:55:00.220
So, for example, K3 prime,
that's a triangle subdivided--
00:55:00.220 --> 00:55:03.506
well, that's a C6, [INAUDIBLE].
00:55:09.570 --> 00:55:14.840
So observe that every H that
comes up in this theorem
00:55:14.840 --> 00:55:33.340
here is a subgraph of
some subdivision, some one
00:55:33.340 --> 00:55:35.200
subdivision of a clique.
00:55:41.680 --> 00:55:48.090
Because the vertices on
the left in A are degree 2,
00:55:48.090 --> 00:55:53.490
you think of them as
midpoints of edges.
00:55:53.490 --> 00:55:57.480
But because you are
K2,2 free, you--
00:55:57.480 --> 00:56:01.440
if you collapse those path
of lengths 2 to single edges,
00:56:01.440 --> 00:56:04.880
you do not end up
with parallel edges.
00:56:04.880 --> 00:56:11.730
So it is the subgraph of this
one subdivision of some graph,
00:56:11.730 --> 00:56:13.480
which, then, you can
complete to a clique.
00:56:19.390 --> 00:56:25.390
So this theorem
here is equivalent,
00:56:25.390 --> 00:56:33.720
at least qualitatively, to the
statement that for every t,
00:56:33.720 --> 00:56:38.100
there exists some
constants, again
00:56:38.100 --> 00:56:46.800
depending on t, such that the
extremal number of the one
00:56:46.800 --> 00:56:53.260
subdivision of a clique is
bounded by something where
00:56:53.260 --> 00:56:58.340
we can improve upon the exponent
in the first theorem in today's
00:56:58.340 --> 00:56:58.840
lecture.
00:57:05.020 --> 00:57:06.640
So that theorem there.
00:57:06.640 --> 00:57:16.875
So these two theorems
are equivalent because
00:57:16.875 --> 00:57:17.500
of this remark.
00:57:20.020 --> 00:57:23.780
Any questions so far
about the statements?
00:57:23.780 --> 00:57:24.391
Yes.
00:57:24.391 --> 00:57:27.758
AUDIENCE: In the remark,
how do you deal with, like,
00:57:27.758 --> 00:57:28.720
[INAUDIBLE].
00:57:31.070 --> 00:57:32.570
YUFEI ZHAO: Question,
in the remark,
00:57:32.570 --> 00:57:35.040
how do you deal with vertices
with degree less than 2?
00:57:35.040 --> 00:57:37.980
Complete it to a vertex degree
of-- vertex of degree 2.
00:57:37.980 --> 00:57:41.050
Add another edge.
00:57:41.050 --> 00:57:43.020
AUDIENCE: [INAUDIBLE].
00:57:43.020 --> 00:57:45.128
YUFEI ZHAO: Add another
edge to a new vertex.
00:57:45.128 --> 00:57:47.570
AUDIENCE: OK, sure.
00:57:47.570 --> 00:57:51.260
YUFEI ZHAO: Any more questions?
00:57:51.260 --> 00:57:52.570
All right.
00:57:52.570 --> 00:58:03.330
So the proof I want to show
you is due to Oliver Janzer,
00:58:03.330 --> 00:58:06.360
and for this clique
subdivision theorem.
00:58:06.360 --> 00:58:10.840
And this proof
produces that C sub t
00:58:10.840 --> 00:58:15.450
equals to 1 over 4t minus 6.
00:58:15.450 --> 00:58:18.570
So if you plug in
t equals to 3, you
00:58:18.570 --> 00:58:24.950
find that the exponent here is
actually right for the 6 cycle.
00:58:24.950 --> 00:58:27.990
So it actually agrees
with what we know.
00:58:40.650 --> 00:58:45.330
So I want to show you some of
the main ideas from this proof.
00:58:45.330 --> 00:58:47.790
Just like the proof of--
00:58:47.790 --> 00:58:50.717
that we saw for the cycles,
even cycles theorem,
00:58:50.717 --> 00:58:52.800
it will be helpful to start
with some preparation.
00:58:52.800 --> 00:58:55.640
You start with a graph
that, even though it
00:58:55.640 --> 00:58:57.413
has a lot of edges,
may have lots
00:58:57.413 --> 00:58:58.830
of vertices with
high degree, lots
00:58:58.830 --> 00:59:00.270
of vertices with low degree.
00:59:00.270 --> 00:59:02.650
It's nice to clean
it up somewhat.
00:59:02.650 --> 00:59:06.240
And so let me state a
preparatory lemma, which
00:59:06.240 --> 00:59:09.420
we will not prove, but
it's of a similar nature
00:59:09.420 --> 00:59:12.120
to this very easy lemma
that we saw earlier but
00:59:12.120 --> 00:59:13.886
with a bit more work.
00:59:13.886 --> 00:59:15.365
AUDIENCE: [INAUDIBLE].
00:59:18.813 --> 00:59:19.780
YUFEI ZHAO: Yes.
00:59:19.780 --> 00:59:22.590
Originally, I put a C over
here, but now it's OK.
00:59:22.590 --> 00:59:24.306
So there exists a Ct.
00:59:29.490 --> 00:59:33.360
So the preparation
is that we're going
00:59:33.360 --> 00:59:39.745
to pass through a large,
almost regular subgraph.
00:59:43.920 --> 00:59:47.690
The lemma-- so don't worry
too much about the details,
00:59:47.690 --> 00:59:49.620
and I'll tell you
what the idea is.
00:59:49.620 --> 00:59:58.480
So for every alpha, there exists
constants beta and K such that
00:59:58.480 --> 01:00:07.090
for every C and n
sufficiently large,
01:00:07.090 --> 01:00:16.030
every n-vertex graph
G with lots of edges--
01:00:16.030 --> 01:00:20.440
so C n to the 1
plus alpha edges--
01:00:23.120 --> 01:00:30.570
has a subgraph G
prime such that--
01:00:30.570 --> 01:00:32.330
so I want some properties.
01:00:32.330 --> 01:00:35.795
First, G prime has
lots of vertices.
01:00:39.390 --> 01:00:43.310
So n to the beta, so it's
still lots of vertices.
01:00:43.310 --> 01:00:46.360
You do some polynomial in n.
01:00:46.360 --> 01:00:49.640
And, 2, it has
still lots of edges
01:00:49.640 --> 01:00:52.890
relative to the number
of vertices it has.
01:00:52.890 --> 01:00:55.070
So, basically,
changing the constants,
01:00:55.070 --> 01:00:56.930
if I start with n
to the 1 plus alpha,
01:00:56.930 --> 01:01:01.010
I still have, roughly, number
of vertices to the 1 plus
01:01:01.010 --> 01:01:03.980
alpha, number of edges.
01:01:03.980 --> 01:01:10.100
It is almost regular, in the
sense that the max degree of G
01:01:10.100 --> 01:01:13.640
does not differ from
its minimum degree
01:01:13.640 --> 01:01:15.380
by more than a constant factor.
01:01:18.642 --> 01:01:20.980
So you do have vertices
that are too small degree.
01:01:20.980 --> 01:01:24.690
You don't have vertices
that are too large degree.
01:01:24.690 --> 01:01:28.890
And, finally, G
prime is bipartite,
01:01:28.890 --> 01:01:31.850
and the two parts
of this bipartition
01:01:31.850 --> 01:01:42.330
have sizes differing by
a factor, at most, 2.
01:01:51.030 --> 01:01:55.930
So if you like, think of G
as a regular bipartite graph.
01:01:55.930 --> 01:01:57.350
But this is the
preparation lemma.
01:01:57.350 --> 01:01:58.933
We'll just make our
life a bit easier.
01:02:01.680 --> 01:02:08.460
So, from now on, let's
treat G as a constant
01:02:08.460 --> 01:02:15.620
in our asymptotic notation
to simplify the notation.
01:02:15.620 --> 01:02:18.140
So you have this graph G.
It's a bipartite graph.
01:02:22.200 --> 01:02:28.710
And for a pair of
vertices on one side A--
01:02:28.710 --> 01:02:31.650
so there are no edges in A.
But for a pair of vertices,
01:02:31.650 --> 01:02:36.940
I say that this u, v is light.
01:02:36.940 --> 01:02:40.610
So it's not an edge, but
I talk about these pairs.
01:02:40.610 --> 01:02:52.830
I say that it is light if the
number of common neighbors
01:02:52.830 --> 01:02:54.820
between--
01:02:54.820 --> 01:03:04.940
of u and v is at least 1
and less than t choose 2.
01:03:04.940 --> 01:03:08.940
So it has some common
neighbors, but not too many.
01:03:08.940 --> 01:03:13.420
And then we say that
this pair is heavy
01:03:13.420 --> 01:03:15.370
if the number of
common neighbors
01:03:15.370 --> 01:03:19.500
is at least t choose 2.
01:03:19.500 --> 01:03:22.560
So if a pair u, v has
some common neighbors,
01:03:22.560 --> 01:03:24.290
then it's either light or heavy.
01:03:32.060 --> 01:03:40.170
I claim that if
G is a Kt prime--
01:03:40.170 --> 01:03:43.280
so this is the one
subdivision of Kt--
01:03:43.280 --> 01:03:52.495
free bipartite graph with
vertex bipartition A union B--
01:03:52.495 --> 01:03:57.440
so U. Not A, but U union
B. U will eventually
01:03:57.440 --> 01:04:04.600
be a subset of A. Such that all
the vertices on the left of U
01:04:04.600 --> 01:04:22.955
have to be at least delta,
and U is not too small.
01:04:31.960 --> 01:04:35.212
So it's at least 4Bt over delta.
01:04:35.212 --> 01:04:36.420
Don't worry about it for now.
01:04:36.420 --> 01:04:37.960
So think of delta--
01:04:37.960 --> 01:04:39.470
this is a min degree.
01:04:39.470 --> 01:04:42.100
So it is somewhat smaller
than the average--
01:04:42.100 --> 01:04:44.920
I mean, it's basically the
average degree of your graph.
01:04:48.010 --> 01:04:49.207
And B, think of as n.
01:04:49.207 --> 01:04:51.040
It's more or less the
whole set of vertices.
01:04:54.230 --> 01:04:59.740
Then the conclusion
is that there
01:04:59.740 --> 01:05:12.160
exists a u in a lot of
light pairs in the set
01:05:12.160 --> 01:05:17.170
U. There exists a vertex
in many light pairs.
01:05:43.340 --> 01:05:45.560
It's important that we
assume that this graph G
01:05:45.560 --> 01:05:48.680
if Kt prime free.
01:05:48.680 --> 01:05:51.410
Because, otherwise, you could
imagine a situation where,
01:05:51.410 --> 01:05:54.470
essentially, you have a
complete bipartite graph
01:05:54.470 --> 01:05:58.430
and every pair of
vertices is heavy.
01:05:58.430 --> 01:06:01.490
So you don't have any
light pairs at all.
01:06:01.490 --> 01:06:07.210
So having Kt prime free somehow
allows us to find light pairs.
01:06:07.210 --> 01:06:09.050
So let's see the
proof of this lemma.
01:06:09.050 --> 01:06:11.860
So you combine some nice
ideas that we've seen earlier
01:06:11.860 --> 01:06:14.030
in the course, namely
double counting,
01:06:14.030 --> 01:06:15.650
and also uses Turan's theorem.
01:06:20.060 --> 01:06:23.930
So, first, let's do a double
counting argument similar
01:06:23.930 --> 01:06:29.900
to the proof of the
Kovari-Sos-Turan theorem,
01:06:29.900 --> 01:06:37.050
where let's count the
number of K1,2's like that.
01:06:37.050 --> 01:06:44.140
So the number of K1,2's
like that between U and B.
01:06:44.140 --> 01:06:49.670
I claim one way to count
this is to look through all
01:06:49.670 --> 01:06:54.720
the vertices on the right side,
look at how many neighbors it
01:06:54.720 --> 01:06:58.590
has, and sum up the
degrees choose 2.
01:07:08.964 --> 01:07:13.390
So skipping some--
01:07:13.390 --> 01:07:15.980
I mean, I can tell
you what comes out.
01:07:15.980 --> 01:07:22.800
So, by convexity,
we find that it
01:07:22.800 --> 01:07:24.880
is at least this quantity here.
01:07:24.880 --> 01:07:27.580
And then, assuming the
minimum degree condition,
01:07:27.580 --> 01:07:30.610
we find that this
quantity is quite large.
01:07:41.110 --> 01:07:43.380
So this is a calculation
very similar to what
01:07:43.380 --> 01:07:45.740
we did for the proof of
Kovari-Sos-Turan theorem.
01:07:50.810 --> 01:07:56.690
The low degree vertices in B do
not contribute much to the sum.
01:07:56.690 --> 01:08:01.120
So this sum is large, and it
sums over all vertices in B,
01:08:01.120 --> 01:08:07.450
but the low degree vertices in
B, they contribute very little.
01:08:13.870 --> 01:08:18.970
Because if we sum over
all the vertices of B
01:08:18.970 --> 01:08:30.120
with degree less than 2t,
then, for each summand,
01:08:30.120 --> 01:08:35.470
it's, at most, 2
t squared, which,
01:08:35.470 --> 01:08:43.050
again, by the assumption
of delta, is less than half
01:08:43.050 --> 01:08:45.810
of the total sum.
01:08:45.810 --> 01:08:48.990
So the low degree vertices of
B do not contribute very much
01:08:48.990 --> 01:08:51.021
to the sum.
01:08:51.021 --> 01:08:52.854
So let's look at the
higher degree vertices.
01:08:56.080 --> 01:08:58.990
For the higher
degree vertices, they
01:08:58.990 --> 01:09:02.500
contribute a substantial chunk.
01:09:16.220 --> 01:09:18.050
And the most
important thing here
01:09:18.050 --> 01:09:30.350
is that, among these vertices,
there are no t mutually heavy--
01:09:33.030 --> 01:09:37.080
so not among these vertices,
but in U. If you look at U,
01:09:37.080 --> 01:09:43.950
there are no t mutually
heavy vertices in U.
01:09:43.950 --> 01:09:49.290
If you have t mutually
heavy vertices in U, then
01:09:49.290 --> 01:09:50.430
what happens?
01:09:50.430 --> 01:09:53.640
So if you have t
mutually heavy vertices.
01:09:53.640 --> 01:09:56.680
If you had, let's say,
three vertices in U,
01:09:56.680 --> 01:10:00.390
they're mutually heavy, each one
of them, because they're heavy,
01:10:00.390 --> 01:10:03.780
I can find many
common neighbors.
01:10:03.780 --> 01:10:10.020
So I can build this
path of length 2.
01:10:10.020 --> 01:10:12.060
I can build another
path of length 2.
01:10:12.060 --> 01:10:14.640
And I don't run out of vertices
because they're all heavy.
01:10:14.640 --> 01:10:17.550
They all have at least t
choose 2 common neighbors.
01:10:17.550 --> 01:10:20.950
So I can build the
subdivision of Kt.
01:10:20.950 --> 01:10:25.320
So there are no
mutually heavy vertices
01:10:25.320 --> 01:10:29.250
in U. So where have
we seen this before?
01:10:31.770 --> 01:10:38.840
So you have-- think about
the neighborhood of a vertex
01:10:38.840 --> 01:10:43.110
in B. Because it's
inside a neighborhood,
01:10:43.110 --> 01:10:46.650
all the pairs are
either heavy or light,
01:10:46.650 --> 01:10:51.160
and there are no t
mutually heavy vertices.
01:10:51.160 --> 01:10:57.360
So, then, Turan's
theorem tells us
01:10:57.360 --> 01:10:58.985
that there must be
many light vertices.
01:11:03.890 --> 01:11:11.900
That there are many light
vertices in this neighborhood.
01:11:11.900 --> 01:11:22.100
So the number of light pairs
in the neighborhood of this v,
01:11:22.100 --> 01:11:28.460
if it has at least t neighbors--
or else you run out of room--
01:11:33.680 --> 01:11:36.440
is at least-- so if you think
about what Turan's theorem
01:11:36.440 --> 01:11:41.710
says, that the number of known
edges is at least this quantity
01:11:41.710 --> 01:11:49.740
here, which is at least
the degree of v squared--
01:11:49.740 --> 01:11:55.856
degree of v.
01:11:55.856 --> 01:12:01.150
So Turan's theorem tells us that
there cannot be so many heavy
01:12:01.150 --> 01:12:05.770
pairs inside a neighborhood
of a vertex in B,
01:12:05.770 --> 01:12:07.870
so there must be
many light pairs.
01:12:12.780 --> 01:12:21.800
And now we sum over
all vertices in B.
01:12:21.800 --> 01:12:30.120
We obtain that U has
a lot of light pairs.
01:12:33.700 --> 01:12:37.960
We might have overcounted
and a little bit,
01:12:37.960 --> 01:12:42.130
because each light pair
is overcounted only
01:12:42.130 --> 01:12:44.290
by a bounded number of
times because it's light.
01:12:51.970 --> 01:12:58.610
So it's overcounted by
less than K choose 2 times.
01:12:58.610 --> 01:13:02.110
So that's just a constant
factor, and we're OK with that.
01:13:05.090 --> 01:13:06.760
So that's the
conclusion for now.
01:13:06.760 --> 01:13:13.840
This lemma tells us that you
have lots of light pairs in U.
01:13:13.840 --> 01:13:20.800
And what we're going to do is
to keep on shrinking this U.
01:13:20.800 --> 01:13:25.610
So U is going to be a
subset of A. Initially,
01:13:25.610 --> 01:13:29.710
let's let U be the
entire set A. It tells us
01:13:29.710 --> 01:13:35.380
that there's one vertex in A
with lots of light neighbors.
01:13:35.380 --> 01:13:38.410
Take that vertex,
choose its neighborhood.
01:13:38.410 --> 01:13:39.880
Apply the lemma again.
01:13:39.880 --> 01:13:44.350
Find another vertex with lots
of internal light neighbors.
01:13:44.350 --> 01:13:45.430
Keep on going.
01:13:45.430 --> 01:13:49.410
And then we build a
large light clique.
01:13:49.410 --> 01:13:51.010
So that's the idea.
01:13:51.010 --> 01:13:52.040
So we'll find that--
01:13:55.250 --> 01:14:02.290
so we'll see that if
this delta is bigger
01:14:02.290 --> 01:14:06.960
than, basically, the quantity
claimed in the theorem--
01:14:06.960 --> 01:14:13.842
so t minus 2 over 2t minus 3--
01:14:13.842 --> 01:14:22.610
and sufficiently
large C, then there
01:14:22.610 --> 01:14:33.290
exists the sequence U1, U2,
U3, so on, all the way to Ut,
01:14:33.290 --> 01:14:41.640
and the sequence
of vertices v1, v2,
01:14:41.640 --> 01:14:53.510
vt, such that, initially,
I take A. And the idea is,
01:14:53.510 --> 01:14:56.870
initially, I take
v1 to be whatever
01:14:56.870 --> 01:14:59.930
comes out of that lemma.
01:14:59.930 --> 01:15:05.420
And I want the property that
all of the vi, vj's are light.
01:15:11.660 --> 01:15:21.350
And 2 is that no three of these
v's have a common neighbor.
01:15:33.060 --> 01:15:35.310
And, I think, once you have
these two properties, then
01:15:35.310 --> 01:15:38.370
you can find your
clique subdivision.
01:15:38.370 --> 01:15:40.620
You find these t light vertices.
01:15:43.530 --> 01:15:51.220
So if you have v1, v2, v3, v4,
you have these light vertices,
01:15:51.220 --> 01:15:52.850
and I can build a
clique subdivision
01:15:52.850 --> 01:15:54.830
from these light vertices.
01:15:54.830 --> 01:15:58.880
Because they're light, they have
at least one common neighbor
01:15:58.880 --> 01:16:01.330
for each pair.
01:16:01.330 --> 01:16:02.850
I just keep building them.
01:16:02.850 --> 01:16:04.900
So I build these
common neighbors.
01:16:04.900 --> 01:16:07.170
Well, you should
be somewhat worried
01:16:07.170 --> 01:16:11.720
that I end up using
the same vertex twice.
01:16:11.720 --> 01:16:13.880
But, of course, that
should not be a worry
01:16:13.880 --> 01:16:18.310
if I guarantee that no three
of them have a common neighbor.
01:16:18.310 --> 01:16:21.902
They cannot collapse.
01:16:21.902 --> 01:16:28.480
And you cannot have two
vertices end up being the same.
01:16:28.480 --> 01:16:32.610
Otherwise, I would
violate property b.
01:16:32.610 --> 01:16:35.340
So these two
properties alone allow
01:16:35.340 --> 01:16:41.780
you to build a Kt subdivision.
01:16:41.780 --> 01:16:44.935
But how do we find
this sequence?
01:16:44.935 --> 01:16:48.100
Well, so we build it
iteratively using that lemma.
01:16:48.100 --> 01:16:53.140
You start with one vertex
guaranteed by that lemma.
01:16:53.140 --> 01:16:54.566
You look at its neighborhood.
01:16:57.970 --> 01:17:01.690
You pick another vertex
guaranteed by the lemma.
01:17:01.690 --> 01:17:05.580
You look at its neighborhood.
01:17:05.580 --> 01:17:06.780
And so on.
01:17:06.780 --> 01:17:09.748
And you build this
sequence of light clique.
01:17:09.748 --> 01:17:10.248
Yes.
01:17:10.248 --> 01:17:12.515
AUDIENCE: [INAUDIBLE].
01:17:12.515 --> 01:17:13.140
YUFEI ZHAO: Ah.
01:17:13.140 --> 01:17:14.307
The light neighborhood, yes.
01:17:14.307 --> 01:17:15.820
So we're not in
the graph any more.
01:17:15.820 --> 01:17:24.780
So everything's inside
A. Everything's inside A.
01:17:24.780 --> 01:17:27.130
So let me finish off
the list of properties,
01:17:27.130 --> 01:17:30.040
and we're almost there.
01:17:30.040 --> 01:17:33.550
Third property I want is that,
when I do this operation,
01:17:33.550 --> 01:17:38.050
I do not reduce my space of
possibilities by too much.
01:17:38.050 --> 01:17:45.850
Namely, that the size of U does
not go down too substantially.
01:17:51.273 --> 01:17:52.880
And that's guaranteed
by the lemma.
01:17:57.430 --> 01:18:00.430
And, 4 is that--
01:18:00.430 --> 01:18:02.080
basically this
picture over here--
01:18:02.080 --> 01:18:09.940
vi is light to
all of U i plus 1.
01:18:15.790 --> 01:18:19.410
So I claim that you can
find the sequence satisfying
01:18:19.410 --> 01:18:21.788
these properties.
01:18:21.788 --> 01:18:24.330
And the reason is that you are
repeatedly applying the lemma.
01:18:38.330 --> 01:18:45.530
So repeatedly apply the lemma.
01:18:50.330 --> 01:18:52.910
The lemma doesn't
address this part
01:18:52.910 --> 01:18:56.890
about triple vertices
having common neighbors.
01:18:56.890 --> 01:18:59.870
But I claim that's actually
not so hard to deal with.
01:18:59.870 --> 01:19:05.570
Because if you think
about how many vertices,
01:19:05.570 --> 01:19:09.890
how many possibilities,
this restriction eliminates,
01:19:09.890 --> 01:19:22.480
b only eliminates at
each step t choose 2,
01:19:22.480 --> 01:19:32.160
because this is coming from
the light restriction, times,
01:19:32.160 --> 01:19:33.836
at most--
01:19:33.836 --> 01:19:36.870
so another t choose 2.
01:19:36.870 --> 01:19:44.290
And this comes-- the first
one comes from pairs of v1
01:19:44.290 --> 01:19:46.400
through vt.
01:19:46.400 --> 01:19:49.610
So eliminates, at
most, this many--
01:19:49.610 --> 01:19:53.170
times the max degree.