WEBVTT
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YUFEI ZHAO: All right.
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We are going to continue our
discussion of extremal graph
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theory.
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Last time, we
discussed what happens
00:00:48.380 --> 00:00:53.000
when we exclude a triangle
or more generally a clique.
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And we wish to find a graph that
maximizes the number of edges.
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And at the end of the
lecture, I stated the theorem
00:01:01.880 --> 00:01:15.760
of Erdos-Stone-Simonovitz,
which says, recall,
00:01:15.760 --> 00:01:24.930
that if you have a fixed H and
I wish to understand the maximum
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number of edges in an n
vertex graph that is H-free--
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so remember this
definition, this
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is the maximum number of edges
in an n vertex H-free graph.
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So we are going to be
looking at this quantity
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in the next few lectures.
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So the Erdos-Stone-Simonovitz
theorem
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tells us that, perhaps, quite
surprisingly, this quantity
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is largely covered by the
chromatic number of H,
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even though H itself
might be quite involved.
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So if you knew the
chromatic number,
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you already know a
lot of information
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about the growth rate
of this function.
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And in particular, as long
as it's just not bipartite,
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so that the chromatic
numbers at least 3,
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we already know the
first order asymptotics
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from the
Erdos-Stone-Simonovitz theorem.
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However, when H is bipartite
so that the chromatic number is
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exactly 2, then the
theorem tells us
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only that this quantity
is little o of n squared.
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Which is some
useful information,
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but it doesn't tell
us the whole story.
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And in the next several
lecturers, I want to explore
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what more can we say
about this quantity
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here for bipartite graphs
H. And it turns out
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that there is a lot
that we do not know,
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that there are lots of
open problems in this area
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having to do with trying
to pin down the growth
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rate of this function.
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And, in particular,
for bipartite graphs,
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there's a bipartite
graph that places
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somewhat special rule, namely
the complete bipartite graph.
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So K st, being the
complete bipartite graph,
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with s vertices on one side and
t vertices on the other side--
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so this is a very
nice bipartite graph.
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And just to understand, the
extremal number for this graph
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is a famous open
problem in this area,
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and it has the name of
Zarankiewicz problem, which
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is to determine or estimate
the extremal numbers
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for these complete
bipartite graphs.
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So I'll tell you pretty much
all we know about this problem.
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And there are some
interesting things.
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But we do not know
all that much.
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Now, every bipartite
graph is a subset.
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It's a subgraph of such a
complete bipartite graph.
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So every bipartite H it is
a soft graph of some K st.
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And we know that if H is
a subgraph of this K st,
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then the extremal number for
H if your graph is H-free,
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then automatically it
has to be K s,t-free.
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So there is this bound between
these two extremal numbers.
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So in particular, if you have
some upper bound on K st,
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then you have some upper
bound on bipartite graphs.
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Although, for specific
by bipartite graphs H,
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maybe we can do better
than using this bound.
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And we'll see examples of that
later in the course as well.
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So what can we say about the
extremal numbers of these K
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s,t's?
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So the most important theorem
in this area of this problem
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is the result due
to Kovari-Sos-Turan.
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So the Kovari-Sos-Turan theorem
tells us that for every fixed
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integers s and t--
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s at most t--
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there exists some constant C,
such that the extremal number
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is upper-bounded by something
which is on the order of n
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to the 2 minus 1 over s.
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So it gives you some
upper bound, showing you
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that it's not only subquadratic,
but there is a gap from 2
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in the exponent.
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Now, in combinatorics,
as is with many fields
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of mathematics, it can
be somewhat intimidating
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to newcomers when you
see a lot of names.
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So every theorems
seems to be named
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after a string of
mathematicians, some of whom
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you may have heard
of, some you may not.
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And I agree, it may be difficult
to remember who is who and what
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is what, but I think this
theorem has a very nice way
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to remember, which is that
it's a theorem about K st,
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and this is the K s,t theorem.
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[LAUGHTER]
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OK.
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So this is the K s,t theorem.
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I want to begin by showing
you how to prove the K s,t The
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proof is via a nice and not
too difficult double-counting
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argument.
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And I show you
some applications.
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Let's prove this
theorem, and it's
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by double-counting argument.
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What are we going to count?
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Well, let's start with the
object that we're working with.
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So there's going to be a graph
G that has n vertices, m edges,
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and K s,t-free.
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And let's count the number of
stars, K s,1 in this graph G.
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So we're counting
configurations like that.
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I'll do an upper bound
and a lower bound.
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So let's start with
the upper bound.
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On one hand, every subset
of s vertices in the graph G
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has at most t minus
1 common neighbors.
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Because if they had t common
neighbors, then you get a K st.
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So that's one down.
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On the other hand,
let's see what
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happens to the number
of common neighbors
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if you knew that this
graph has a lot of edges.
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So the number of stars--
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well, I can calculate
this quantity explicitly
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by running over all the
vertices of G. For each vertex,
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look at its neighborhood
and choose s vertices
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from its neighborhood.
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So for each v I need to
find a subset of s vertices
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from its neighborhood.
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By convexity, I can
lower bound the sum
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by the average, in
some sense, because--
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let me first write
down the expression.
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So I can do a convexity argument
that gives me this lower bound.
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Here, I'm abusing
notation somewhat
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and writing binomial
coefficients
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with a real entry on top,
where I mean the expression
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as you would expect, treating
this guy as a polynomial in x.
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And the key fact
we're using here
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is that this function
here is convex
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for x at least and minus 1.
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So, In particular, if you
think about this function here,
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you have a lot of 0's, and then
it becomes convex afterwards.
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So you can even think of
extending this function as 0
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to the left of n minus 1.
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So this is a convex function,
and you can apply convexity
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to deduce this inequality.
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But you know the
sum of the degrees.
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That's just essentially twice--
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I mean, that is the twice
the number of edges.
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So we have this
expression right here.
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So you have an upper bound on
the number of s stars coming
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from K s,t-freeness and lower
bound on the number of s stars
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coming from just having lots of
edges and applying convexity.
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Putting these two
things together,
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we find that there's
this inequality here.
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Here, we are thinking
of s and t as fixed,
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and we are trying
to understand how
00:12:05.020 --> 00:12:09.310
m and n depend on each
other as they get large.
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So it will be helpful to use
the asymptotics that n choose
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s grows like n to the s divided
by s factorial for a fixed s.
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So looking at that expression
and applying this asymptotics
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to both sides, we have
this inequality here,
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So I've eliminated s
factorial from both sides.
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So now rearrange,
clean things up.
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We find the following
upper bound--
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m the number of edges in G.
And that's the expression.
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So for fixed s and t it grows
like n to the 2 minus 1 over s.
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Any questions?
00:13:25.340 --> 00:13:25.860
Yeah.
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AUDIENCE: Where does the
right side of the inequality
00:13:27.110 --> 00:13:28.250
come from?
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Are you counting that from the
different cycles [INAUDIBLE]??
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YUFEI ZHAO: So the
question is where
00:13:33.310 --> 00:13:35.310
the right side of
inequality-- which inequality?
00:13:35.310 --> 00:13:35.893
This one here?
00:13:35.893 --> 00:13:36.660
AUDIENCE: Yeah.
00:13:36.660 --> 00:13:37.410
YUFEI ZHAO: Right.
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So here we are counting
the number of K s1's.
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AUDIENCE: Oh, OK.
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YUFEI ZHAO: So here I am
upper and lower-bounding
00:13:47.290 --> 00:13:52.490
the number of K s1's using
what we derived earlier.
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Yes.
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AUDIENCE: Do you actually
care that s is less than
00:13:55.616 --> 00:13:57.433
or equal t in the argument?
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YUFEI ZHAO: The
question is, do we
00:13:58.850 --> 00:14:01.460
care that s is less than or
equal to t in the argument?
00:14:01.460 --> 00:14:02.660
The argument doesn't care.
00:14:02.660 --> 00:14:06.000
But, of course, if you want the
better asymptotics for fixed s
00:14:06.000 --> 00:14:08.910
and t as n gets large, you
should take s to be less than
00:14:08.910 --> 00:14:10.170
or equal to t.
00:14:10.170 --> 00:14:10.670
Question?
00:14:10.670 --> 00:14:12.590
AUDIENCE: What happens
when t equals 1?
00:14:12.590 --> 00:14:14.540
YUFEI ZHAO: What happens
when t equals to 1.
00:14:14.540 --> 00:14:16.025
Well, that's a great question.
00:14:16.025 --> 00:14:17.400
I'll leave you to
think about it.
00:14:17.400 --> 00:14:24.090
If you know that your graph
has maximum degree n most t,
00:14:24.090 --> 00:14:26.743
what can you tell me about the
number of edges in the graph?
00:14:26.743 --> 00:14:28.860
[LAUGHTER]
00:14:28.860 --> 00:14:30.480
OK.
00:14:30.480 --> 00:14:31.810
Any more questions?
00:14:35.037 --> 00:14:36.370
We'll come back to this theorem.
00:14:36.370 --> 00:14:39.295
In fact, this will occupy us for
at least a couple of lectures.
00:14:41.950 --> 00:14:45.000
Basically, is this
theorem tight?
00:14:45.000 --> 00:14:46.920
And it is conjectured
to be, although that
00:14:46.920 --> 00:14:49.740
is a major open problem
extremal graph theory.
00:14:49.740 --> 00:14:52.760
We only know a small
number of values.
00:14:52.760 --> 00:14:55.890
Well, for most
values of s and t,
00:14:55.890 --> 00:15:00.390
in particular when s and
t are both equal to 4,
00:15:00.390 --> 00:15:03.520
we do not know if
this bound is tight.
00:15:03.520 --> 00:15:05.980
But there are some values
of s and t for which
00:15:05.980 --> 00:15:07.840
we do know that it is tight.
00:15:07.840 --> 00:15:13.691
For example, 2 and
2, 3 and 3, 4 and 7,
00:15:13.691 --> 00:15:16.887
s and if t is
really, really large.
00:15:16.887 --> 00:15:18.720
And I will show you
some constructions later
00:15:18.720 --> 00:15:25.070
on, creating graphs G that are
K s,t-free for those parameters
00:15:25.070 --> 00:15:29.780
that matches this K st bound
up to a constant factor.
00:15:29.780 --> 00:15:30.540
Yes?
00:15:30.540 --> 00:15:32.990
AUDIENCE: For a fixed s,
what's the bound on t--
00:15:32.990 --> 00:15:35.690
if there a bound on t for
t with equality cases?
00:15:35.690 --> 00:15:36.440
YUFEI ZHAO: Right.
00:15:36.440 --> 00:15:39.500
So the question is,
for a fixed value of s,
00:15:39.500 --> 00:15:42.770
is there some bound on t for
which we get equality cases?
00:15:42.770 --> 00:15:46.400
There is a conjecture
that this bound is sharp
00:15:46.400 --> 00:15:51.240
up to a constant factor
for every s and t.
00:15:51.240 --> 00:15:53.820
But we only know how to
prove that conjecture--
00:15:53.820 --> 00:15:56.670
and I will tell you much
more about it later on--
00:15:56.670 --> 00:16:01.860
when t is much
larger compared to s.
00:16:01.860 --> 00:16:08.430
So there is a lot of unexplored
territory on this problem.
00:16:08.430 --> 00:16:09.780
Any more questions?
00:16:13.420 --> 00:16:17.140
Before diving into the
Kovari-Sos-Turan further,
00:16:17.140 --> 00:16:19.759
I just want to show you
some neat applications.
00:16:29.540 --> 00:16:32.685
And I want to begin with
a geometric application.
00:16:41.720 --> 00:16:44.310
There is a classic
problem asked by Erdos,
00:16:44.310 --> 00:16:47.090
back in the '40s, called
the unit distance problem.
00:16:53.150 --> 00:17:12.339
Which asks, what is the maximum
number of unit distances
00:17:12.339 --> 00:17:18.580
formed by n points in the plane.
00:17:23.369 --> 00:17:25.140
Let me give you some examples.
00:17:25.140 --> 00:17:26.880
If you have three
points, you put them
00:17:26.880 --> 00:17:31.730
in equilateral triangle, and
all three distances are unit.
00:17:31.730 --> 00:17:32.550
Great.
00:17:32.550 --> 00:17:36.060
If you have four points, you
cannot place them so that all
00:17:36.060 --> 00:17:39.800
six distances are units if
you're staying in the plane.
00:17:39.800 --> 00:17:43.050
So the best thing you can
do is something like this--
00:17:46.690 --> 00:17:50.830
so I'm drawing all the edges
that are unit distances.
00:17:50.830 --> 00:17:54.610
If you have one more
point, it turns out
00:17:54.610 --> 00:17:58.230
that's the best thing you
can do with five points.
00:17:58.230 --> 00:18:01.320
With six points there are
some more possibilities.
00:18:01.320 --> 00:18:05.320
Let me draw for you
some possibilities.
00:18:05.320 --> 00:18:07.960
You can extend the
previous configuration
00:18:07.960 --> 00:18:09.285
by adding more triangles.
00:18:09.285 --> 00:18:10.660
And there are many
different ways
00:18:10.660 --> 00:18:12.735
that you can attach
an extra triangle.
00:18:19.560 --> 00:18:22.720
There's actually one more
way to do it with six points.
00:18:22.720 --> 00:18:29.120
Namely, I can put them like
a projection of a prism.
00:18:29.120 --> 00:18:31.870
So all of these
configurations have
00:18:31.870 --> 00:18:33.730
the maximum number
of unit distances
00:18:33.730 --> 00:18:37.370
obtained if you draw six points.
00:18:37.370 --> 00:18:41.210
If you draw seven
points, it turns out
00:18:41.210 --> 00:18:43.400
this is the best way to do it.
00:18:43.400 --> 00:18:45.090
So you can go on for a while.
00:18:45.090 --> 00:18:45.885
And people have.
00:18:45.885 --> 00:18:48.230
So you can try to tabulate
for every n what's
00:18:48.230 --> 00:18:49.910
the maximum number
of unit distances
00:18:49.910 --> 00:18:53.010
you can generate having
n points in the plane.
00:18:53.010 --> 00:18:56.342
And the question is, what
is the answer for n points?
00:18:56.342 --> 00:18:57.800
And it turns out,
for this problem,
00:18:57.800 --> 00:19:00.410
and for many problems
like it in combinatorics,
00:19:00.410 --> 00:19:03.350
you can have a lot of fun with
playing with small examples,
00:19:03.350 --> 00:19:05.850
but they are often misleading.
00:19:05.850 --> 00:19:08.660
It doesn't really tell you what
the overall structure should
00:19:08.660 --> 00:19:10.360
be like.
00:19:10.360 --> 00:19:12.310
And it turns out this
is the major opium
00:19:12.310 --> 00:19:15.070
problem for which we do not
understand what the structure
00:19:15.070 --> 00:19:16.630
is like for large values of n.
00:19:19.300 --> 00:19:22.900
So think n large.
00:19:22.900 --> 00:19:27.220
What are some possible ways to
generate many unit distances?
00:19:27.220 --> 00:19:27.720
Yep.
00:19:27.720 --> 00:19:29.360
AUDIENCE: Drawing triangles?
00:19:29.360 --> 00:19:30.110
YUFEI ZHAO: Great.
00:19:30.110 --> 00:19:33.790
So one way is to draw
lots of triangles.
00:19:33.790 --> 00:19:38.572
So extend this figure forward.
00:19:38.572 --> 00:19:39.280
So let's do that.
00:19:44.310 --> 00:19:47.470
Well, actually, let me give
you something even simpler
00:19:47.470 --> 00:19:48.310
to begin with.
00:19:48.310 --> 00:19:53.160
I can just put the n points
on a line, equal spaced, OK,
00:19:53.160 --> 00:19:55.430
I get n minus 1 distances--
00:19:55.430 --> 00:19:57.198
n minus 1 unit distances.
00:19:59.820 --> 00:20:07.870
If I put them like that, how
many unit distances do I get?
00:20:10.720 --> 00:20:12.145
AUDIENCE: 2 times n minus 1.
00:20:12.145 --> 00:20:13.100
YUFEI ZHAO: Good.
00:20:13.100 --> 00:20:15.555
2 times n minus 1.
00:20:15.555 --> 00:20:19.770
So each new point gives you
two extra new distances.
00:20:19.770 --> 00:20:23.883
And you can try to think
about how to do better.
00:20:23.883 --> 00:20:25.800
Well, if you follow that
sequence of examples,
00:20:25.800 --> 00:20:27.930
maybe you're limited
to such ideas.
00:20:27.930 --> 00:20:28.560
Yes?
00:20:28.560 --> 00:20:31.230
AUDIENCE: Are we allowed to
put points in the same place?
00:20:31.230 --> 00:20:33.960
YUFEI ZHAO: You are not
allowed to put points
00:20:33.960 --> 00:20:34.910
in the same place.
00:20:34.910 --> 00:20:35.910
That's a great question.
00:20:35.910 --> 00:20:38.202
But you're not allowed to
put points on the same place.
00:20:38.202 --> 00:20:38.747
Yes?
00:20:38.747 --> 00:20:41.370
AUDIENCE: Is for
something to keep
00:20:41.370 --> 00:20:46.870
doubling the number of
points, the degrees get big?
00:20:46.870 --> 00:20:48.750
YUFEI ZHAO: So I didn't
quite understand--
00:20:48.750 --> 00:20:51.160
so you say if you put--
00:20:51.160 --> 00:20:52.632
AUDIENCE: [INAUDIBLE] something.
00:20:52.632 --> 00:20:53.340
YUFEI ZHAO: Yeah.
00:20:53.340 --> 00:20:56.100
AUDIENCE: Like keep translating
in different directions.
00:20:56.100 --> 00:20:57.933
YUFEI ZHAO: So you want
to take this example
00:20:57.933 --> 00:20:59.520
and keep translating
in some direction
00:20:59.520 --> 00:21:01.437
I think you'll run out
of room pretty quickly.
00:21:01.437 --> 00:21:07.270
AUDIENCE: No, I mean just copy
it and put it over [INAUDIBLE]..
00:21:07.270 --> 00:21:08.020
YUFEI ZHAO: I see.
00:21:08.020 --> 00:21:09.790
You want to take
this configuration
00:21:09.790 --> 00:21:12.580
and translate it in
some unit direction.
00:21:12.580 --> 00:21:14.838
Well, then you double
the number of points,
00:21:14.838 --> 00:21:17.380
but you don't actually increase
the number of units distances
00:21:17.380 --> 00:21:20.515
all that much.
00:21:20.515 --> 00:21:26.311
AUDIENCE: Don't we get each
point having one extra unit
00:21:26.311 --> 00:21:28.320
distance added?
00:21:28.320 --> 00:21:36.840
YUFEI ZHAO: So the suggestion
is we take some configuration,
00:21:36.840 --> 00:21:44.740
let's say this graph G,
and I form two copies of G
00:21:44.740 --> 00:21:48.400
by translating G in some unit
directions-- some generic unit
00:21:48.400 --> 00:21:49.040
direction.
00:21:49.040 --> 00:21:49.540
OK, great.
00:21:49.540 --> 00:21:51.625
So what happens to the
number of vertices?
00:21:51.625 --> 00:21:57.730
So n goes to 2m and
the number of edges
00:21:57.730 --> 00:22:05.415
goes from m to 2m plus n.
00:22:05.415 --> 00:22:06.210
Is that right?
00:22:06.210 --> 00:22:06.710
OK, great.
00:22:06.710 --> 00:22:09.267
So if you do that,
what do you get?
00:22:09.267 --> 00:22:10.550
AUDIENCE: Log n.
00:22:10.550 --> 00:22:11.300
YUFEI ZHAO: n log.
00:22:11.300 --> 00:22:11.950
OK.
00:22:11.950 --> 00:22:15.770
That's much better than before.
00:22:15.770 --> 00:22:17.440
OK, good.
00:22:17.440 --> 00:22:17.940
Good.
00:22:17.940 --> 00:22:18.360
Very nice.
00:22:18.360 --> 00:22:18.860
Yes?
00:22:18.860 --> 00:22:20.520
AUDIENCE: That
picture [INAUDIBLE]..
00:22:20.520 --> 00:22:23.206
Let's start with the
construction for n over 3
00:22:23.206 --> 00:22:28.046
and then replace each point
with a triangles-- equilateral
00:22:28.046 --> 00:22:29.320
triangle.
00:22:29.320 --> 00:22:30.070
YUFEI ZHAO: Right.
00:22:30.070 --> 00:22:34.290
So the suggestion is to
start with some graph
00:22:34.290 --> 00:22:37.960
G and then replacing--
00:22:37.960 --> 00:22:44.610
so let's, as an example,
look at that one--
00:22:44.610 --> 00:22:47.445
each vertex here by an
equilateral triangle.
00:22:50.530 --> 00:22:53.060
But I want to maintain
the same unit distance.
00:22:53.060 --> 00:22:55.108
So the-- uh-huh.
00:22:55.108 --> 00:23:00.350
AUDIENCE: So you just
choose [INAUDIBLE]..
00:23:00.350 --> 00:23:08.290
Or you choose [INAUDIBLE]
00:23:08.290 --> 00:23:11.680
YUFEI ZHAO: Is this similar
to taking this graph G
00:23:11.680 --> 00:23:15.370
and then translating it in
two different directions that
00:23:15.370 --> 00:23:18.467
form an equilateral triangle?
00:23:18.467 --> 00:23:19.300
I think it's-- yeah.
00:23:19.300 --> 00:23:20.500
So it's a very similar idea.
00:23:20.500 --> 00:23:25.050
And maybe you do a little bit
better in terms of constant.
00:23:25.050 --> 00:23:26.450
All great suggestions.
00:23:26.450 --> 00:23:28.215
Actually, this is really nice.
00:23:28.215 --> 00:23:30.292
AUDIENCE: [INAUDIBLE]
two times [INAUDIBLE]..
00:23:30.292 --> 00:23:31.250
YUFEI ZHAO: Two times--
00:23:31.250 --> 00:23:33.000
you want to make a
constant correction?
00:23:33.000 --> 00:23:35.674
OK, let me just do that.
00:23:35.674 --> 00:23:38.090
[LAUGHTER]
00:23:38.090 --> 00:23:41.390
Any more suggestions?
00:23:41.390 --> 00:23:41.890
OK.
00:23:41.890 --> 00:23:43.420
Yeah, all of this
is really nice.
00:23:43.420 --> 00:23:47.720
So let me tell you what is the
best construction that people
00:23:47.720 --> 00:23:51.270
are aware of, and this is A
construction due to Erdos.
00:23:54.520 --> 00:24:00.410
And the idea is to think big,
not build from small examples.
00:24:00.410 --> 00:24:08.830
So what Erdos did is to
consider a square grid of root n
00:24:08.830 --> 00:24:10.090
by root n.
00:24:10.090 --> 00:24:12.340
When I see something that's
root n that's non-integer,
00:24:12.340 --> 00:24:17.260
I just think round down
to the nearest integer.
00:24:17.260 --> 00:24:18.160
I have a square grid.
00:24:22.320 --> 00:24:27.540
Now, if you take these
distances as unit distances,
00:24:27.540 --> 00:24:29.710
well, you get something
which is linear in m.
00:24:29.710 --> 00:24:31.660
So you don't gain that much.
00:24:31.660 --> 00:24:37.440
But you can take any specific
distance as your unit distance.
00:24:37.440 --> 00:24:39.990
So what we can do
is take a distance
00:24:39.990 --> 00:24:44.560
that is represented
many times in this grid.
00:24:44.560 --> 00:24:58.180
So let's take the unit distance
to be a distance root r where
00:24:58.180 --> 00:25:05.740
r is some integer that can be
represented as a sum of two
00:25:05.740 --> 00:25:07.630
squares in many different ways.
00:25:27.030 --> 00:25:30.000
So, for example, if we--
00:25:36.340 --> 00:25:41.540
so we can take some r so
that it has many appearances.
00:25:41.540 --> 00:25:43.840
And if you know some
elementary number theory,
00:25:43.840 --> 00:25:46.180
then you might
know that the best
00:25:46.180 --> 00:25:49.660
way to do this is to take
r to be a product of primes
00:25:49.660 --> 00:25:51.970
that are 1 mod 4.
00:25:51.970 --> 00:25:54.000
In any case, you
can do this and you
00:25:54.000 --> 00:25:55.770
can use some analytic
number theory
00:25:55.770 --> 00:25:58.920
to calculate if you choose
the best possible value of r,
00:25:58.920 --> 00:26:00.510
how many distances do you get.
00:26:00.510 --> 00:26:03.540
So that calculation was
done, and it turns out
00:26:03.540 --> 00:26:08.640
you get n raised to the power
of 1 plus some constant C
00:26:08.640 --> 00:26:12.674
over log log n unit distances.
00:26:20.553 --> 00:26:22.220
So this is better
than the constructions
00:26:22.220 --> 00:26:22.970
we've seen before.
00:26:26.730 --> 00:26:28.470
So what can we say
about this problem?
00:26:28.470 --> 00:26:30.250
So this is a construction.
00:26:30.250 --> 00:26:37.530
Well, then you want to
understand some upper bounds.
00:26:37.530 --> 00:26:41.970
What can we say about the
upper bounds to this problems?
00:26:41.970 --> 00:26:45.810
And let me show you a fairly
easy upper bound, which
00:26:45.810 --> 00:26:57.670
can be deduced very quickly from
the Kovari-Sos-Turan theorem,
00:26:57.670 --> 00:27:08.670
that every set of n
points in the plane
00:27:08.670 --> 00:27:15.584
has at most on the order off
n to the 3/2 unit distances.
00:27:23.990 --> 00:27:28.450
So not quite this bound,
but it's some bound.
00:27:28.450 --> 00:27:30.600
So the trivial upper
bound is n choose 2.
00:27:30.600 --> 00:27:34.190
So it's much better than
the trivial upper bound.
00:27:34.190 --> 00:27:36.240
So here's the proof.
00:27:36.240 --> 00:27:39.020
So let's consider the
unit distance graph, which
00:27:39.020 --> 00:27:41.570
is basically the graphs
I've been drawing,
00:27:41.570 --> 00:27:45.440
where you have the
vertices, the points,
00:27:45.440 --> 00:27:50.180
and I join an edge
between two vertices if
00:27:50.180 --> 00:27:53.320
and only if their
distance is exactly 1.
00:27:56.930 --> 00:28:05.390
I claim this graph is K 2,3 free
00:28:05.390 --> 00:28:06.630
Why is that?
00:28:06.630 --> 00:28:10.670
Because if you have
two points, what
00:28:10.670 --> 00:28:12.410
are their common neighbors?
00:28:12.410 --> 00:28:14.420
Their common chambers
must all be at distance 1
00:28:14.420 --> 00:28:17.030
from each of the two points.
00:28:17.030 --> 00:28:20.320
So their common
neighbors must land
00:28:20.320 --> 00:28:22.970
in the intersections of the unit
circles centered at these two
00:28:22.970 --> 00:28:27.600
points, and they meet
at most two points.
00:28:27.600 --> 00:28:30.800
So it's K 2,3 free.
00:28:30.800 --> 00:28:32.870
Therefore, the
Kovari-Sos-Turan theorem
00:28:32.870 --> 00:28:39.680
tells us that the number
of edges in this G
00:28:39.680 --> 00:28:43.838
is upper bounded
by n to the 3/2.
00:28:43.838 --> 00:28:46.700
And that's it.
00:28:46.700 --> 00:28:49.860
That gives you some upper bound.
00:28:49.860 --> 00:28:52.940
Any questions?
00:28:52.940 --> 00:28:57.060
So it's an application
of Kovari-Sos-Turan,
00:28:57.060 --> 00:29:01.290
where we design the graph
that has to be K 2,3, free
00:29:01.290 --> 00:29:04.020
and the extremal number
tells us some information
00:29:04.020 --> 00:29:06.810
about this graph.
00:29:06.810 --> 00:29:09.510
What is the best
bound that we know?
00:29:09.510 --> 00:29:18.040
So it turns out we do know
how to do a little bit better,
00:29:18.040 --> 00:29:24.090
but not anywhere close to what
we believe to be the truth.
00:29:24.090 --> 00:29:28.120
So the current best bound is
on the order of n to the 4/3.
00:29:28.120 --> 00:29:30.310
And we'll actually see
a proof of this later
00:29:30.310 --> 00:29:32.890
in the course, once
we've developed
00:29:32.890 --> 00:29:35.990
something called the
crossing numbers inequality.
00:29:35.990 --> 00:29:37.730
It's also a very
short, very neat proof.
00:29:41.310 --> 00:29:43.370
I want to tell you about
another problem which
00:29:43.370 --> 00:29:48.440
looks superficially
similar and also a really
00:29:48.440 --> 00:29:49.770
interesting geometric problem.
00:29:57.580 --> 00:30:13.270
And this is the Erdos
distinct distances problem,
00:30:13.270 --> 00:30:21.040
which asks for the maximum
number of distinct distances
00:30:21.040 --> 00:30:22.750
among n points in the plane.
00:30:32.670 --> 00:30:33.670
The minimum-- thank you.
00:30:33.670 --> 00:30:35.980
The minimum number of
distinct distances.
00:30:35.980 --> 00:30:37.390
So maximum would be very easy.
00:30:47.920 --> 00:30:50.310
Again, you have
lots of examples.
00:30:50.310 --> 00:30:53.190
In fact, you can look at
basically all the examples
00:30:53.190 --> 00:30:55.830
that we've given earlier
I mean, these two problems
00:30:55.830 --> 00:30:58.680
are very much related
to each other.
00:30:58.680 --> 00:31:02.790
If you want lots
of repeat distances
00:31:02.790 --> 00:31:05.770
or if you want very
few distinct distances,
00:31:05.770 --> 00:31:07.950
whether they seem like
they are very much related.
00:31:07.950 --> 00:31:10.360
And, of course, you
can also write down
00:31:10.360 --> 00:31:13.380
the inequality that
relates the two of them,
00:31:13.380 --> 00:31:20.750
because the number
of distinct distances
00:31:20.750 --> 00:31:27.780
is at least n choose 2 divided
by the maximum number of unit
00:31:27.780 --> 00:31:28.868
distances--
00:31:34.016 --> 00:31:36.210
in other words, the
number of distances that
00:31:36.210 --> 00:31:41.690
can be repeated in any
single configuration.
00:31:41.690 --> 00:31:45.220
So each of the examples
that I just erased--
00:31:45.220 --> 00:31:49.070
so, for example, if you
put points on the line,
00:31:49.070 --> 00:31:51.440
you have n minus 1
distinct distances.
00:31:58.630 --> 00:32:00.090
And we already saw
the square grid,
00:32:00.090 --> 00:32:02.670
so that might be a good
candidate to look at as well.
00:32:02.670 --> 00:32:05.160
And there it takes
some more number theory
00:32:05.160 --> 00:32:09.150
to figure out how many
distinct distances there are,
00:32:09.150 --> 00:32:11.160
and this roughly
corresponds to the number
00:32:11.160 --> 00:32:19.910
of integers that can be written
as the sum of two squares.
00:32:19.910 --> 00:32:24.140
So this has been
calculated, and the result
00:32:24.140 --> 00:32:29.490
is n divided by
square root of log n.
00:32:29.490 --> 00:32:30.350
So that's the order.
00:32:34.510 --> 00:32:36.760
And Erdos conjectured
that this example
00:32:36.760 --> 00:32:40.260
should be more or less
the best you can do.
00:32:40.260 --> 00:32:42.080
So these problems
look very similar.
00:32:42.080 --> 00:32:44.960
But, actually, this
problem here was
00:32:44.960 --> 00:32:49.610
solved in a spectacular
fashion about a decade ago
00:32:49.610 --> 00:32:54.520
by an important paper
of Guth and Katz.
00:32:59.420 --> 00:33:03.380
So Larry Guth is
in our department.
00:33:03.380 --> 00:33:14.700
And they showed that every
set of n points in the plane
00:33:14.700 --> 00:33:19.560
generates at least
on the order of n
00:33:19.560 --> 00:33:23.546
over log n distinct distances.
00:33:33.080 --> 00:33:37.080
So not quite what Erdos
conjectured, but nearly there.
00:33:37.080 --> 00:33:39.780
And, in fact, all the previous
results were much worse.
00:33:39.780 --> 00:33:43.670
So they were off
in the exponent.
00:33:43.670 --> 00:33:47.300
But this one more or
less got to the truth.
00:33:47.300 --> 00:33:49.640
So this is a very
sophisticated result
00:33:49.640 --> 00:33:53.060
that used lots of
amazing techniques,
00:33:53.060 --> 00:33:56.000
ranging from the polynomial
method in combinatorics
00:33:56.000 --> 00:33:58.630
to some algebraic geometry.
00:33:58.630 --> 00:34:03.880
And that's-- I just wanted to
mention it for cultural value.
00:34:03.880 --> 00:34:04.560
OK.
00:34:04.560 --> 00:34:06.210
Any questions?
00:34:06.210 --> 00:34:06.710
Yes.
00:34:06.710 --> 00:34:08.129
AUDIENCE: For the
unit distance problem,
00:34:08.129 --> 00:34:09.550
what is believed
to be the bound?
00:34:09.550 --> 00:34:10.258
YUFEI ZHAO: Yeah.
00:34:10.258 --> 00:34:11.840
So for the unit
distance problem,
00:34:11.840 --> 00:34:14.480
and actually for the
distinct distances problem,
00:34:14.480 --> 00:34:17.630
the question is, what was
is believed to be the bound?
00:34:17.630 --> 00:34:19.520
And it's the square grid.
00:34:19.520 --> 00:34:21.409
So maybe you can
do slightly better,
00:34:21.409 --> 00:34:24.230
but it is conjecture that
you cannot do much better,
00:34:24.230 --> 00:34:26.780
let's say, by more than a
constant factor compared
00:34:26.780 --> 00:34:29.085
to the square grid.
00:34:29.085 --> 00:34:29.730
Yes.
00:34:29.730 --> 00:34:32.176
AUDIENCE: What is
Katz's first name?
00:34:32.176 --> 00:34:32.926
YUFEI ZHAO: Sorry?
00:34:32.926 --> 00:34:34.630
AUDIENCE: What is
Katz's first name?
00:34:34.630 --> 00:34:35.739
YUFEI ZHAO: So
this is Nets Katz.
00:34:35.739 --> 00:34:36.406
He's at CalTech.
00:34:43.300 --> 00:34:46.060
I want to take a
short break now.
00:34:46.060 --> 00:34:48.429
And when we come back,
I want to show you
00:34:48.429 --> 00:34:52.360
some ways to generate lower
bounds to the extremal number.
00:34:56.790 --> 00:35:01.620
In the Kovari-Sos-Turon theorem,
we understood some upper bounds
00:35:01.620 --> 00:35:03.090
for the extremal number.
00:35:03.090 --> 00:35:06.420
So I want to turn our
attention now to lower bounds.
00:35:06.420 --> 00:35:09.450
Which means constructing,
in some sense,
00:35:09.450 --> 00:35:14.920
graphs that have lots of
edges, at the same time being K
00:35:14.920 --> 00:35:15.420
s,t-free.
00:35:17.980 --> 00:35:19.650
There are many classes
of construction,
00:35:19.650 --> 00:35:21.700
so there are several
techniques for doing this.
00:35:21.700 --> 00:35:24.670
And I want to introduce you
to some of these techniques.
00:35:24.670 --> 00:35:28.050
And the first one comes from
the probabilistic method,
00:35:28.050 --> 00:35:35.780
where we use a randomized
construction by picking
00:35:35.780 --> 00:35:40.400
a graph at random, modifying
it a little bit in a way that's
00:35:40.400 --> 00:35:41.780
to our desire.
00:35:41.780 --> 00:35:43.110
This method is very powerful.
00:35:43.110 --> 00:35:43.970
It is very general.
00:35:43.970 --> 00:35:46.530
It is applicable in
a lot of situations.
00:35:46.530 --> 00:35:50.630
But, unfortunately, for the
problem of getting tight bound,
00:35:50.630 --> 00:35:54.500
it really allows you
to get tight bounds.
00:35:54.500 --> 00:36:00.200
So it is very robust, but
somehow it's often not sharp.
00:36:00.200 --> 00:36:05.350
And another class of
constructions that
00:36:05.350 --> 00:36:06.190
are algebraic--
00:36:11.140 --> 00:36:16.520
all the sharp examples come
from algebraic constructions.
00:36:16.520 --> 00:36:19.740
And there, we're able
to use some nice ideas
00:36:19.740 --> 00:36:22.050
from algebra or from
algebraic geometry
00:36:22.050 --> 00:36:25.308
to get tight constructions.
00:36:25.308 --> 00:36:26.850
And you can wonder,
is there some way
00:36:26.850 --> 00:36:30.260
to combine the best
of both worlds?
00:36:30.260 --> 00:36:31.380
And it turns out--
00:36:31.380 --> 00:36:33.890
this was an important
reason development
00:36:33.890 --> 00:36:36.690
that was found just
several years ago--
00:36:36.690 --> 00:36:39.800
where one can combine
ideas from the two
00:36:39.800 --> 00:36:43.007
and obtain a randomized
algebraic construction.
00:36:52.460 --> 00:36:55.030
And that leads to a new
source of constructions
00:36:55.030 --> 00:37:00.090
for large H-free
graphs with many edges.
00:37:00.090 --> 00:37:04.230
So the plan is to show you
how these constructions work.
00:37:10.270 --> 00:37:12.934
First, let's begin with
randomized constructions.
00:37:24.550 --> 00:37:26.770
As I mentioned, this
construction is very general,
00:37:26.770 --> 00:37:30.340
it's very robust, and we can
use it to obtain H-free graphs
00:37:30.340 --> 00:37:37.120
for every graph H. Of course,
when H is not bipartite,
00:37:37.120 --> 00:37:41.570
then we saw from the Turon
graph that that's pretty much
00:37:41.570 --> 00:37:42.920
the right thing to do.
00:37:42.920 --> 00:37:46.420
So you should think of bipartite
graphs H. So fix a graph
00:37:46.420 --> 00:37:50.440
H with at least two edges.
00:37:50.440 --> 00:37:53.650
Otherwise, the
problem is trivial.
00:37:53.650 --> 00:38:00.790
The claim is that there is some
constant C such that for every
00:38:00.790 --> 00:38:03.520
n-- you should think of
and this being large--
00:38:03.520 --> 00:38:13.160
there exists an H-free
graph on n vertices
00:38:13.160 --> 00:38:16.370
and having lots of edges.
00:38:16.370 --> 00:38:21.380
And we will show
that you can obtain
00:38:21.380 --> 00:38:23.840
the following number of edges.
00:38:23.840 --> 00:38:28.130
So 2 minus number of
vertices of H minus 2
00:38:28.130 --> 00:38:33.294
divided by the number
of edges of H minus 1.
00:38:33.294 --> 00:38:35.720
Don't worry about this
expression in the exponent,
00:38:35.720 --> 00:38:38.880
it will come out of the proof.
00:38:38.880 --> 00:38:46.100
In other words,
the extremal number
00:38:46.100 --> 00:38:49.436
is at least this quantity here.
00:38:59.910 --> 00:39:02.090
How good is this now?
00:39:02.090 --> 00:39:05.050
So it's some scary looking
expression in the exponent.
00:39:05.050 --> 00:39:09.927
So let me just give you some
special cases for comparison
00:39:09.927 --> 00:39:11.260
to the Kovari-Sos-Turon theorem.
00:39:15.661 --> 00:39:22.200
For K s,t's, the bound, this
construction gives you a lower
00:39:22.200 --> 00:39:30.510
bound which is of the form m
to the 2 minus s plus t minus 2
00:39:30.510 --> 00:39:33.860
divided by s,t minus 1.
00:39:33.860 --> 00:39:37.320
So I used this symbol,
this bigger tilde
00:39:37.320 --> 00:39:40.030
to denote dropping
constant factors.
00:39:42.970 --> 00:39:47.780
In particular, setting
s and t to be the same,
00:39:47.780 --> 00:40:00.190
we find that the K s,s extremal
number is lower-bounded by this
00:40:00.190 --> 00:40:02.340
quantity here.
00:40:02.340 --> 00:40:05.710
So how does that compare
to Kovari-Sos-Turon?
00:40:08.330 --> 00:40:19.830
In Kovari-Sos-Turon we saw that
K s,s is at most 2 to the n
00:40:19.830 --> 00:40:21.510
to the 2 minus 1 over s.
00:40:21.510 --> 00:40:27.870
So there's a bit of a gap, and
in particular even for 2 and 2.
00:40:33.340 --> 00:40:34.930
These results tell
you a lower bound
00:40:34.930 --> 00:40:38.470
on the order of n to the
4/3 and the upper bound
00:40:38.470 --> 00:40:40.095
on the order of 3/2.
00:40:40.095 --> 00:40:41.470
So I'm just doing
this explicitly
00:40:41.470 --> 00:40:45.010
to show you that even in
sometimes the simplest case
00:40:45.010 --> 00:40:48.160
there is a gap between
these two bounds.
00:40:48.160 --> 00:40:50.380
Later on, we'll
see a construction
00:40:50.380 --> 00:40:52.990
that shows that the
right-hand side is tight.
00:40:52.990 --> 00:40:54.697
So this is the truth.
00:40:54.697 --> 00:40:56.530
Randomized construction
gives you something,
00:40:56.530 --> 00:41:00.390
but it doesn't
give you the truth.
00:41:00.390 --> 00:41:02.100
On the other hand,
I want you to notice
00:41:02.100 --> 00:41:09.760
that if t gets very
large as a function of s,
00:41:09.760 --> 00:41:18.661
this exponent here approaches
1 over s as t goes to infinity.
00:41:22.100 --> 00:41:25.690
So for very large values of
t, at least in the exponent,
00:41:25.690 --> 00:41:27.280
it's not that far off.
00:41:27.280 --> 00:41:28.870
Although, you
never get the right
00:41:28.870 --> 00:41:33.650
exponent for any
specific essential.
00:41:33.650 --> 00:41:38.310
So this is some limitation
of the randomized method,
00:41:38.310 --> 00:41:40.070
but it's variable robust.
00:41:40.070 --> 00:41:42.430
And, in fact, you
can use this result
00:41:42.430 --> 00:41:47.960
to bootstrap it to a slightly
better one for some graphs H.
00:41:47.960 --> 00:41:58.970
So, for example,
one might do better
00:41:58.970 --> 00:42:04.085
by replacing this
H by a subgraph.
00:42:07.160 --> 00:42:11.140
Because if you have
a subgraph, H prime,
00:42:11.140 --> 00:42:13.830
and you construct your
G to be H prime-free,
00:42:13.830 --> 00:42:16.960
then it is automatically H-free.
00:42:16.960 --> 00:42:20.200
But maybe this theorem actually
gives you a better construction
00:42:20.200 --> 00:42:23.010
when restricted to H prime.
00:42:23.010 --> 00:42:24.340
And what can you do better?
00:42:24.340 --> 00:42:28.510
Well, you can do better
if h prime is such
00:42:28.510 --> 00:42:32.270
that whatever the quantity
that comes up in the exponent--
00:42:32.270 --> 00:42:33.520
so let me write it like this--
00:42:36.940 --> 00:42:40.150
is superior to the
exponent you would
00:42:40.150 --> 00:42:47.800
obtain by just looking at H.
00:42:47.800 --> 00:42:50.080
So let me give the name
to this notion here.
00:42:50.080 --> 00:42:53.710
So let me call it the
2-density of the graph
00:42:53.710 --> 00:42:58.660
to be whatever
quantity here that
00:42:58.660 --> 00:43:00.250
would be the
maximum if you allow
00:43:00.250 --> 00:43:02.050
to pass down to subgraphs.
00:43:04.936 --> 00:43:08.880
So the 2-density of a graph H--
00:43:08.880 --> 00:43:13.100
so denoted in literature
often as m sub 2--
00:43:13.100 --> 00:43:19.800
to be the maximum where you are
allowed to look at subgraphs,
00:43:19.800 --> 00:43:21.640
let's say, on at
least three vertices
00:43:21.640 --> 00:43:25.900
to avoid degeneracies
of this ratio that
00:43:25.900 --> 00:43:29.425
comes up in the expressions.
00:43:34.880 --> 00:43:36.590
So then the theorem--
00:43:39.480 --> 00:43:42.260
that construction
there-- implies
00:43:42.260 --> 00:43:52.120
that the extremal number is at
least not just the expression
00:43:52.120 --> 00:43:53.960
I've written, but
maybe sometimes you
00:43:53.960 --> 00:43:56.870
can do slightly better
by passing to a subgraph.
00:44:00.170 --> 00:44:01.780
So let me give you
a concrete example.
00:44:04.430 --> 00:44:15.750
Suppose my H is this graph here.
00:44:15.750 --> 00:44:18.030
So you can run the
calculation and find
00:44:18.030 --> 00:44:19.940
what these numbers are.
00:44:19.940 --> 00:44:24.870
So the number of vertices,
edges, and this ratio
00:44:24.870 --> 00:44:32.870
here to be-- so 5
vertices, 8 edges, and 7/3.
00:44:35.620 --> 00:44:38.790
And the idea here is
that it's more helpful.
00:44:38.790 --> 00:44:42.740
So I want to create
something which is H-free,
00:44:42.740 --> 00:44:47.550
but dense things
are easier to avoid.
00:44:47.550 --> 00:44:50.140
So if I have something which
has a fairly dense core,
00:44:50.140 --> 00:44:51.420
that's easier to avoid.
00:44:51.420 --> 00:44:55.360
So maybe it's better to, instead
of looking at the whole H,
00:44:55.360 --> 00:44:59.560
look at this K 4.
00:44:59.560 --> 00:45:02.380
So if you can avoid K 4,
of course, you avoid H,
00:45:02.380 --> 00:45:04.970
and maybe it's easier
to just avoid K 4
00:45:04.970 --> 00:45:10.280
and not worry about some
of the extra decorations.
00:45:10.280 --> 00:45:16.340
So if you look at this H prime
and go through the parameters,
00:45:16.340 --> 00:45:23.510
you find that it is like that.
00:45:23.510 --> 00:45:30.460
And it is somewhat
denser in this sense,
00:45:30.460 --> 00:45:33.180
compared to looking
at the whole graph H.
00:45:33.180 --> 00:45:36.250
So you can improve on this
theorem for some graphs H
00:45:36.250 --> 00:45:37.760
where you can pass
to a denser core.
00:45:41.290 --> 00:45:42.675
Any questions so far?
00:45:42.675 --> 00:45:43.300
Yes.
00:45:43.300 --> 00:45:45.383
AUDIENCE: Why is this
method called the 2-density?
00:45:45.383 --> 00:45:47.425
YUFEI ZHAO: The questions
is, why is this measure
00:45:47.425 --> 00:45:48.310
called a 2-density?
00:45:48.310 --> 00:45:50.260
So that's a name
given in literature.
00:45:50.260 --> 00:45:53.502
It partly has to do
with these extra ratios.
00:45:53.502 --> 00:45:55.960
So there are other notions of
densities that actually we'll
00:45:55.960 --> 00:45:57.350
see later on.
00:45:57.350 --> 00:46:00.970
So this term we'll
only see today.
00:46:00.970 --> 00:46:04.690
So it's more of an ad hoc term
for the purpose of this course.
00:46:04.690 --> 00:46:06.970
Later on, we'll see
notions of density
00:46:06.970 --> 00:46:11.710
that are more relevant
for our discussions.
00:46:11.710 --> 00:46:14.620
Any more questions?
00:46:14.620 --> 00:46:16.900
So let me show you how
to prove this theorem.
00:46:20.820 --> 00:46:22.620
The proof is very intuitive.
00:46:22.620 --> 00:46:27.040
The idea is you take something
at random and then you fix it.
00:46:27.040 --> 00:46:27.540
That's it.
00:46:31.410 --> 00:46:33.450
Let's consider a random graph.
00:46:36.170 --> 00:46:39.060
The Erdos-Rényi random graph--
so whenever I say random graph,
00:46:39.060 --> 00:46:41.720
I almost always refer
to this one here.
00:46:47.420 --> 00:46:52.930
And the Erdos-Rényi random graph
is obtained by considering n
00:46:52.930 --> 00:47:00.640
vertices and each possible
edge appearing independently
00:47:00.640 --> 00:47:03.830
and uniformly with
probability p.
00:47:03.830 --> 00:47:06.940
And we're going to
decide this p later on.
00:47:06.940 --> 00:47:11.140
So let me not tell
you what p is for now.
00:47:11.140 --> 00:47:15.070
I'm interested in avoiding
H. So this random graph may
00:47:15.070 --> 00:47:17.410
have some copies
of H. Let me count
00:47:17.410 --> 00:47:27.110
the number of copies of H.
00:47:27.110 --> 00:47:34.050
We can compute the expected
number of copies of H
00:47:34.050 --> 00:47:36.060
by linearity of expectations.
00:47:36.060 --> 00:47:38.130
For every possible
placement, look
00:47:38.130 --> 00:47:43.320
at what's the probability that
that placement generates an H.
00:47:43.320 --> 00:47:48.120
Namely, look at all
different possibilities
00:47:48.120 --> 00:47:53.160
for choosing the possible
vertices of H. I need
00:47:53.160 --> 00:47:55.350
to divide it by a
factor that accounts
00:47:55.350 --> 00:47:57.580
for the number of
automorphisms of H.
00:47:57.580 --> 00:48:00.760
But just a constant factor--
don't worry about it.
00:48:00.760 --> 00:48:03.210
And for each of these
possible placements,
00:48:03.210 --> 00:48:06.180
H appears with
probability exactly p
00:48:06.180 --> 00:48:11.250
to the number of
edges of H, just
00:48:11.250 --> 00:48:14.520
by linearity of expectations.
00:48:14.520 --> 00:48:16.800
And I can upper-bound
this quantity
00:48:16.800 --> 00:48:20.910
very crudely by e to the
number of edges of H times
00:48:20.910 --> 00:48:24.990
n, which is the number
of vertices of G raised
00:48:24.990 --> 00:48:28.770
to the number of vertices of H.
00:48:28.770 --> 00:48:31.815
On the other hand, I also want a
graph G that has lots of edges,
00:48:31.815 --> 00:48:33.440
because that's what
we're trying to do.
00:48:33.440 --> 00:48:35.190
We're trying to generate
a graph with lots
00:48:35.190 --> 00:48:36.750
of edges that's H-free.
00:48:36.750 --> 00:48:41.580
So the number of edges of G,
that's also easy to compute.
00:48:41.580 --> 00:48:45.510
It's a binomial distribution,
and it has expectation p times
00:48:45.510 --> 00:48:48.640
n choose 2.
00:48:48.640 --> 00:48:51.380
And, basically, I
want this quantity
00:48:51.380 --> 00:48:55.140
to be much larger
than that quantity.
00:48:55.140 --> 00:48:56.850
So I choose an appropriate p.
00:49:00.020 --> 00:49:04.325
Namely, by comparing
these two quantities,
00:49:04.325 --> 00:49:07.110
we can choose p to
be, let's say 1/2--
00:49:07.110 --> 00:49:08.930
the 1/2 is not important--
00:49:08.930 --> 00:49:18.830
times n to 2 the v of H minus
2 divided by e of H minus 1.
00:49:18.830 --> 00:49:22.460
So the exponent comes out
of comparing these two
00:49:22.460 --> 00:49:24.030
expressions.
00:49:24.030 --> 00:49:25.400
So you see a 1 and a 2.
00:49:28.050 --> 00:49:34.690
Once you have this p,
see that the number which
00:49:34.690 --> 00:49:42.260
is the difference-- so take
the number of edges of G
00:49:42.260 --> 00:49:45.020
minus the number
of copies of each.
00:49:48.100 --> 00:49:49.930
I know the expectation of both.
00:49:49.930 --> 00:49:53.590
I can look at their difference
with this value of p.
00:49:53.590 --> 00:49:59.500
We find that it is at least
1/2 of the number of edges.
00:49:59.500 --> 00:50:06.480
So on expectation, you
don't lose too much.
00:50:06.480 --> 00:50:09.030
So p is chosen so that
this inequality is true.
00:50:13.480 --> 00:50:15.170
So you set what p is.
00:50:15.170 --> 00:50:19.190
We find that this quantity here
is at least some constant times
00:50:19.190 --> 00:50:27.160
n to the 2 minus v of H minus
2 divided by e the H minus 1.
00:50:30.390 --> 00:50:33.370
We're still working
with a random graph,
00:50:33.370 --> 00:50:38.320
and we know that this quantity
here is on expectation at least
00:50:38.320 --> 00:50:41.290
that number-- so not too small.
00:50:41.290 --> 00:50:46.690
Therefore, there exists some
instance in this randomness,
00:50:46.690 --> 00:50:53.270
some G such that the
quantity above for
00:50:53.270 --> 00:50:57.200
the specific random instance
is at least its expectation.
00:51:06.780 --> 00:51:13.420
So this gives us a graph G which
on one hand has lots of edges,
00:51:13.420 --> 00:51:16.720
but also has very
few copies of H
00:51:16.720 --> 00:51:19.150
relative to the number of edges.
00:51:19.150 --> 00:51:23.530
So we can now get rid
of all the copies of H
00:51:23.530 --> 00:51:34.700
by removing one edge
from each copy of H
00:51:34.700 --> 00:51:43.370
in G to remove all
copies of H. And now, we
00:51:43.370 --> 00:51:45.340
obtain an H-free graph.
00:51:50.000 --> 00:51:52.880
How many edges are
there in this graph?
00:51:52.880 --> 00:51:57.250
Well, we removed at most
one edge for each copy of H.
00:51:57.250 --> 00:52:11.270
So the number of edges is
at least this quantity here,
00:52:11.270 --> 00:52:15.160
which is what we wrote just now.
00:52:20.470 --> 00:52:21.060
And that's it.
00:52:21.060 --> 00:52:24.320
So now we've obtained
our graph on n vertices
00:52:24.320 --> 00:52:29.030
with lots of edges with the
claimed bound that's H-free.
00:52:29.030 --> 00:52:31.747
So this is the
probabilistic method.
00:52:31.747 --> 00:52:33.080
You start with something random.
00:52:33.080 --> 00:52:34.140
You try to fix it.
00:52:34.140 --> 00:52:36.200
And this method
sometimes has the name
00:52:36.200 --> 00:52:37.970
of the alteration method.
00:52:42.910 --> 00:52:44.950
And this is a very
important idea
00:52:44.950 --> 00:52:47.830
and one of the key ideas in
the probabilistic method, which
00:52:47.830 --> 00:52:50.830
I encourage you to go
and learn more about.
00:52:50.830 --> 00:52:53.440
We'll also see this
method later on when
00:52:53.440 --> 00:52:56.500
we discuss the randomized
algebraic construction.
00:52:56.500 --> 00:53:00.340
AUDIENCE: Just to clarify, none
of the copies of H [INAUDIBLE]..
00:53:03.710 --> 00:53:06.880
YUFEI ZHAO: So the
questions is-- yes--
00:53:06.880 --> 00:53:09.460
what do I mean by the
number of copies of H?
00:53:09.460 --> 00:53:11.500
I mean every instance
of H you see.
00:53:11.500 --> 00:53:12.910
So there could be intersectings.
00:53:12.910 --> 00:53:14.680
I'm not asking for
destroying copies.
00:53:14.680 --> 00:53:16.740
AUDIENCE: Sure.
00:53:16.740 --> 00:53:18.750
YUFEI ZHAO: So a complete
graph on n vertices
00:53:18.750 --> 00:53:20.655
has n choose 3 triangles.
00:53:23.270 --> 00:53:26.470
Any more questions?
00:53:26.470 --> 00:53:26.970
OK.
00:53:26.970 --> 00:53:29.330
So now we've seen that
the probabilistic method
00:53:29.330 --> 00:53:31.370
gives you know some bound.
00:53:31.370 --> 00:53:34.290
And it's not too hard to
apply, but it doesn't give you
00:53:34.290 --> 00:53:35.070
the right bound.
00:53:35.070 --> 00:53:36.518
It doesn't give you the truth.
00:53:36.518 --> 00:53:38.310
So now, I want to show
you a different type
00:53:38.310 --> 00:53:42.390
of constructions, namely
algebraic constructions that
00:53:42.390 --> 00:53:45.910
do allow you to get
the truth, but they
00:53:45.910 --> 00:53:48.740
work in only a small
number of cases.
00:53:48.740 --> 00:53:50.890
And so it's more
magical, but they work
00:53:50.890 --> 00:53:52.320
better when the magic happens.
00:53:59.330 --> 00:54:02.172
So let's discuss
algebraic constructions.
00:54:08.450 --> 00:54:11.330
In particular, I
want to show you
00:54:11.330 --> 00:54:16.250
how to obtain the type bound on
the extremal number for K 2,2,
00:54:16.250 --> 00:54:17.970
namely a fourth cycle.
00:54:17.970 --> 00:54:29.480
So this is a result
due to Erdos-Rényi-Sos,
00:54:29.480 --> 00:54:37.720
and it tells us that the
extremal number for K 2,2 is
00:54:37.720 --> 00:54:46.410
at least 1/2 basically up to
asymptotics times n to the 3/2.
00:54:52.390 --> 00:54:54.700
Actually, if you look at
the constant that came out
00:54:54.700 --> 00:54:57.170
of the proof of the
Kovari-Sos-Turon theorem,
00:54:57.170 --> 00:54:59.970
It is also 1/2.
00:54:59.970 --> 00:55:14.000
So as a corollary, we see
that the extremal number
00:55:14.000 --> 00:55:15.200
is like that.
00:55:15.200 --> 00:55:17.500
So this is one of
extremely few cases
00:55:17.500 --> 00:55:21.400
where we know the
extremal number so well.
00:55:21.400 --> 00:55:24.530
So if you go back to the
proof of Kovari-Sos-Turon,
00:55:24.530 --> 00:55:26.530
you see that the constant
actually there is 1/2.
00:55:29.670 --> 00:55:33.960
So I want to construct for you a
graph that has no fourth cycles
00:55:33.960 --> 00:55:36.670
and has lots of edges.
00:55:36.670 --> 00:55:39.030
So that's the name of the game.
00:55:39.030 --> 00:55:43.528
And I'll describe this
graph for you explicitly.
00:55:43.528 --> 00:55:44.570
So this graph has a name.
00:55:44.570 --> 00:55:45.778
It's called a polarity graph.
00:55:54.300 --> 00:56:03.302
Let's suppose that n
is a number such that 1
00:56:03.302 --> 00:56:05.260
bigger than this number
is a square of a prime.
00:56:08.240 --> 00:56:13.870
So our construction will
use some finite fields.
00:56:13.870 --> 00:56:14.680
I'll explain a bit.
00:56:14.680 --> 00:56:16.150
If n is not of
this form, then you
00:56:16.150 --> 00:56:18.670
can change n to a number of
very close to of this form,
00:56:18.670 --> 00:56:19.753
and everything will be OK.
00:56:22.470 --> 00:56:24.450
The graph will be
constructed as follows--
00:56:24.450 --> 00:56:32.410
the vertex set will
be the plane over Fp.
00:56:32.410 --> 00:56:34.890
Let's remove the origin.
00:56:34.890 --> 00:56:37.255
So it has n points exactly.
00:56:39.870 --> 00:56:47.250
And the edges are such that
I put an edge between x,y
00:56:47.250 --> 00:56:54.870
and a,b, if and only if the
equation ax plus by equals to 1
00:56:54.870 --> 00:56:55.780
holds.
00:56:55.780 --> 00:56:58.900
And this equation is
meant to be read in Fp.
00:57:01.797 --> 00:57:02.630
So that's the graph.
00:57:02.630 --> 00:57:05.450
That's an explicit
description of this graph.
00:57:05.450 --> 00:57:07.200
So I need to show
you two things--
00:57:07.200 --> 00:57:11.150
one, that has lots of edges,
the claimed number of edges.
00:57:11.150 --> 00:57:12.950
And two, it has
no fourth cycles.
00:57:16.270 --> 00:57:20.590
So let's start with not
having fourth cycles.
00:57:20.590 --> 00:57:28.020
So y is a K 2,2-free
00:57:28.020 --> 00:57:30.150
So what would the K 2,2, be?
00:57:30.150 --> 00:57:32.300
So let's consider two points.
00:57:41.630 --> 00:57:45.410
And I want to understand the
number of common neighbors
00:57:45.410 --> 00:57:46.280
of these two points.
00:57:54.170 --> 00:57:57.440
Well, look at the
description for the edges.
00:57:57.440 --> 00:57:59.170
What are the neighbors?
00:57:59.170 --> 00:58:01.480
The neighbors
correspond to solutions
00:58:01.480 --> 00:58:03.498
to this system of equations.
00:58:11.810 --> 00:58:14.810
And the basic claim is that
there is at most one solution--
00:58:14.810 --> 00:58:15.310
x,y.
00:58:28.070 --> 00:58:29.820
So it's basic fact
linear algebra.
00:58:29.820 --> 00:58:33.210
You have to be slowly careful
in case a,b is a multiple
00:58:33.210 --> 00:58:34.020
of a prime b prime.
00:58:34.020 --> 00:58:36.437
But, actually, in that case,
you have no solutions anyway.
00:58:41.740 --> 00:58:48.960
The second claim then is that
this graph has lots of edges.
00:58:48.960 --> 00:58:51.390
Well, actually, that's not
too hard to show either.
00:58:51.390 --> 00:58:57.090
So I claim that every
vertex has degree--
00:58:57.090 --> 00:58:59.640
so how many edges come
out of every vertex?
00:58:59.640 --> 00:59:03.720
I give you a common b,
so how many x comma y
00:59:03.720 --> 00:59:07.180
satisfy that equation up there?
00:59:07.180 --> 00:59:10.555
Basically, for-- so 1 of
x and a and b is non-zero.
00:59:13.970 --> 00:59:15.830
So let's say a is non-zero.
00:59:15.830 --> 00:59:18.020
Therefore, whatever
value of y you set,
00:59:18.020 --> 00:59:21.400
I can find a unique x
that solves the equation.
00:59:21.400 --> 00:59:26.450
I have to be slightly careful,
because I don't allow loops
00:59:26.450 --> 00:59:28.380
in my graph.
00:59:28.380 --> 00:59:31.050
So I might lose one
edge because of that.
00:59:31.050 --> 00:59:36.365
In any case, every vertex has
to agree exactly P or P minus 1.
00:59:40.250 --> 00:59:43.320
Just solve that
equation in x and y.
00:59:48.870 --> 00:59:51.970
So the P minus 1
comes from no loops.
00:59:56.770 --> 01:00:06.940
Therefore, the number of edges
is equal to the claimed bound.
01:00:19.650 --> 01:00:21.410
So this finishes
the proof in case
01:00:21.410 --> 01:00:25.520
when n has that special
number theoretic form.
01:00:25.520 --> 01:00:35.000
But we can extend to all
values of n like this--
01:00:35.000 --> 01:00:38.540
if n doesn't have
that form, then I
01:00:38.540 --> 01:00:48.230
can take a prime P.
It may not necessarily
01:00:48.230 --> 01:00:52.310
be exactly satisfying
that inequality,
01:00:52.310 --> 01:00:55.070
but I can always take a
prime pretty close to it.
01:00:55.070 --> 01:00:57.050
So I can always
take a prime which
01:00:57.050 --> 01:01:04.020
is up to a negligible
multiplicative
01:01:04.020 --> 01:01:05.490
error what I want.
01:01:08.440 --> 01:01:10.765
And then we use--
01:01:10.765 --> 01:01:20.410
and such that P squared
minus 1 is at most n.
01:01:20.410 --> 01:01:31.000
And use the above construction
and add isolated vertices
01:01:31.000 --> 01:01:34.828
to finish the job to
get exactly n vertices.
01:01:34.828 --> 01:01:36.370
And the reason that
I can always take
01:01:36.370 --> 01:01:38.410
a prime very close to
it is because there's
01:01:38.410 --> 01:01:42.070
a theorem in number theory
that tells us that for n
01:01:42.070 --> 01:01:45.370
large enough, I can
always find the prime
01:01:45.370 --> 01:01:48.670
which is slightly less
than n but no more
01:01:48.670 --> 01:01:51.790
than a negligible
multiplicative factor of n.
01:01:51.790 --> 01:01:54.722
The best result
of this form is--
01:01:54.722 --> 01:01:56.680
I'm just telling you
something in number theory
01:01:56.680 --> 01:01:59.290
for cultural reasons--
01:01:59.290 --> 01:02:03.020
due to Baker-Harman-Pintz.
01:02:03.020 --> 01:02:05.760
And so, this is the
question regarding how large
01:02:05.760 --> 01:02:07.720
can gaps between primes be.
01:02:07.720 --> 01:02:10.180
So you might know the
[Bertrand's postulate]] theorem
01:02:10.180 --> 01:02:13.800
that tells you there is always
a prime between n and 2n.
01:02:13.800 --> 01:02:17.440
So what about between
n and n plus root n.
01:02:17.440 --> 01:02:18.700
Actually, we don't know that.
01:02:18.700 --> 01:02:20.530
So the best result
of the form is
01:02:20.530 --> 01:02:22.355
that there is always a prime--
01:02:22.355 --> 01:02:27.020
so for n sufficiently
large there
01:02:27.020 --> 01:02:46.865
exists a prime between n minus
n to the exponent 0.525 and n.
01:02:46.865 --> 01:02:50.480
In any case, this number here,
whatever it is is little n,
01:02:50.480 --> 01:02:53.150
and that's enough
for our purpose.
01:02:55.730 --> 01:03:00.862
So it suffices to look at n of
a special number theoretic form
01:03:00.862 --> 01:03:02.320
where you're allowed
to use primes.
01:03:05.000 --> 01:03:06.800
So that's the
construction there.
01:03:06.800 --> 01:03:09.710
Let me show you a interpretation
of that construction which
01:03:09.710 --> 01:03:14.730
I think is may be
helpful to think about,
01:03:14.730 --> 01:03:18.830
and that's that you can
view it as the incidence
01:03:18.830 --> 01:03:22.760
graph between points and
lines in projective space--
01:03:22.760 --> 01:03:24.500
in projective plane.
01:03:24.500 --> 01:03:26.690
So I start with a
projective plane.
01:03:33.640 --> 01:03:36.800
So I can view a bipartite
version of that construction.
01:03:49.060 --> 01:04:02.870
It can be viewed as the
point-line incidence
01:04:02.870 --> 01:04:13.930
graph of a projective
plane over a finite field.
01:04:16.730 --> 01:04:25.980
And by this, I mean put as
one vertex set the points
01:04:25.980 --> 01:04:29.565
of the projective plane and
on the other side the lines.
01:04:32.370 --> 01:04:39.780
And I put in an edge between
a point and a line if and only
01:04:39.780 --> 01:04:44.300
if the point lies on the line.
01:04:44.300 --> 01:04:48.290
So you can do this more
explicitly in coordinates
01:04:48.290 --> 01:04:54.710
if you view points and
lines as coordinates.
01:04:54.710 --> 01:05:00.940
And so the equation for getting
a point to be on the line
01:05:00.940 --> 01:05:02.850
is like that.
01:05:02.850 --> 01:05:05.000
So now why is there
no fourth cycle?
01:05:07.700 --> 01:05:12.670
A fourth cycle would
correspond to two points lying
01:05:12.670 --> 01:05:20.050
on two different lines, which is
not possible in this geometry.
01:05:20.050 --> 01:05:22.590
So that's the reason for
that construction up there.
01:05:26.490 --> 01:05:29.750
So no two points in two lines.
01:05:34.050 --> 01:05:37.921
Any questions about this
polarity constructions?
01:05:44.174 --> 01:05:46.299
AUDIENCE: Why is it called
a polarity construction.
01:05:46.299 --> 01:05:47.882
YUFEI ZHAO: The
question is, why is it
01:05:47.882 --> 01:05:49.410
called a polarity construction?
01:05:49.410 --> 01:05:57.170
So it relates points and
their polars, which are lines.
01:05:57.170 --> 01:05:58.680
Yeah.
01:05:58.680 --> 01:06:02.337
AUDIENCE: Why does this not
have-- like, on your two P
01:06:02.337 --> 01:06:05.283
squared vertices, looking at one
vertex for every [INAUDIBLE],,
01:06:05.283 --> 01:06:06.520
one vertex for [INAUDIBLE]?
01:06:06.520 --> 01:06:07.145
YUFEI ZHAO: OK.
01:06:07.145 --> 01:06:10.300
So the question has to do with
the number of vertices here.
01:06:10.300 --> 01:06:10.800
It's true.
01:06:10.800 --> 01:06:13.540
Here, I double the
number of vertices,
01:06:13.540 --> 01:06:15.660
and so I don't get
that constant there.
01:06:15.660 --> 01:06:18.600
But what that graph up there--
that's not a bipartite graph.
01:06:18.600 --> 01:06:20.610
It is identifying the
points and the lines
01:06:20.610 --> 01:06:24.130
and overlaying the
two parts into one.
01:06:24.130 --> 01:06:25.880
But if you don't care
about the constants,
01:06:25.880 --> 01:06:27.710
this graph here may
be conceptually easier
01:06:27.710 --> 01:06:29.090
to think about.
01:06:29.090 --> 01:06:29.660
Yes.
01:06:29.660 --> 01:06:31.993
AUDIENCE: [INAUDIBLE] to
generalize the polarities bound
01:06:31.993 --> 01:06:33.690
to K [INAUDIBLE].
01:06:33.690 --> 01:06:34.440
YUFEI ZHAO: Great.
01:06:34.440 --> 01:06:36.732
The question is, can you
generalize this polarity graph
01:06:36.732 --> 01:06:37.837
to K 3,3 and higher?
01:06:37.837 --> 01:06:39.420
So that's what we're
about to do next.
01:06:42.320 --> 01:06:45.350
So for K 3,3, what can you do?
01:06:45.350 --> 01:06:49.670
So the main observation
here is that two lines
01:06:49.670 --> 01:06:51.830
intersecting at most one point.
01:06:51.830 --> 01:06:55.170
But there are other
geometric facts of that form.
01:06:55.170 --> 01:06:59.870
So we're going to use one of
them to get K 3,3-free graph.
01:06:59.870 --> 01:07:07.570
And this construction
is due to Brown,
01:07:07.570 --> 01:07:13.270
that the extremal
number for K 3,3
01:07:13.270 --> 01:07:21.550
is also at least a factor
1/2 minus total 1 times--
01:07:21.550 --> 01:07:23.590
so now, what's the exponent?
01:07:23.590 --> 01:07:27.400
What is predicted
by Kovari-Sos-Turon
01:07:27.400 --> 01:07:33.100
is 2 minus 1/3, and Brown
obtains the correct exponent.
01:07:36.320 --> 01:07:38.760
It turns out this is
also the right constant,
01:07:38.760 --> 01:07:41.100
this 1/2, although
it doesn't follow
01:07:41.100 --> 01:07:44.760
from the Kovari-Sos-Turon
theorem I stated.
01:07:44.760 --> 01:07:46.570
One needs to do a
little bit extra work.
01:07:46.570 --> 01:07:48.445
But it turns out it is
true that this is also
01:07:48.445 --> 01:07:49.640
the correct constant.
01:07:49.640 --> 01:07:52.200
And that's actually pretty
much all the cases where
01:07:52.200 --> 01:07:54.000
we know the correct constant.
01:07:54.000 --> 01:07:57.210
And there are other cases where
we know the correct exponent,
01:07:57.210 --> 01:08:01.442
but these things tend
to be hard to come by.
01:08:01.442 --> 01:08:03.400
So let me show you how
to construct this graph.
01:08:03.400 --> 01:08:07.130
So it's based on a similar
idea as the polarity graph.
01:08:07.130 --> 01:08:08.680
It has some more technicalities.
01:08:08.680 --> 01:08:10.802
So I'm not going to
do the full proof
01:08:10.802 --> 01:08:12.010
and just give you the sketch.
01:08:16.359 --> 01:08:19.130
As earlier, I'm
using the same trick.
01:08:19.130 --> 01:08:22.460
We can assume that n
has a special form.
01:08:22.460 --> 01:08:25.430
Here, let me assume the
n is a cube of a prime.
01:08:32.700 --> 01:08:36.100
I'm going to put s edges.
01:08:36.100 --> 01:08:38.760
So first, the
vertices of my graph
01:08:38.760 --> 01:08:44.930
are going to be points in
the affine plane over Fp.
01:08:47.899 --> 01:08:53.330
Previously, the edges
had to do with lines.
01:08:53.330 --> 01:08:57.270
And now, let's use spheres.
01:08:57.270 --> 01:09:08.600
So the edges of the form where
I join two vertices like this
01:09:08.600 --> 01:09:11.910
if and only if they're--
01:09:11.910 --> 01:09:13.910
well, it's not
really a distance,
01:09:13.910 --> 01:09:16.670
but it's something that
looks like the equation
01:09:16.670 --> 01:09:35.620
of a sphere, where u is some
fixed non-zero element of Fp.
01:09:35.620 --> 01:09:38.319
You may have to be somewhat
careful in choosing this u,
01:09:38.319 --> 01:09:41.250
but let me not worry
too much about it.
01:09:41.250 --> 01:09:45.550
So you fixed some
so-called distance,
01:09:45.550 --> 01:09:52.770
even though it's not a distance,
and I join the vertices
01:09:52.770 --> 01:09:56.665
whenever they satisfy that
equation having that not
01:09:56.665 --> 01:09:57.165
distance.
01:09:59.750 --> 01:10:01.750
What's the intuition here?
01:10:01.750 --> 01:10:04.200
The intuition is that
I want to avoid-- so
01:10:04.200 --> 01:10:07.723
how do I know that this
graph has no K 3,3?
01:10:07.723 --> 01:10:10.140
Well, first, let's think about
what happens in real space.
01:10:16.280 --> 01:10:21.950
So intuition in the real space--
01:10:21.950 --> 01:10:25.710
well, here, I have this
graph that, let's say,
01:10:25.710 --> 01:10:29.690
the unit distance graph in r3.
01:10:29.690 --> 01:10:33.580
So the neighborhood of each
point is a unit sphere.
01:10:33.580 --> 01:10:37.040
And what I want to know is
that if you have three spheres,
01:10:37.040 --> 01:10:39.950
three unit spheres, how many
common intersection points
01:10:39.950 --> 01:10:40.660
can they have?
01:10:43.300 --> 01:10:47.350
Two spheres
intersecting a circle.
01:10:47.350 --> 01:10:51.120
And that circle cannot
lie on the third circle.
01:10:51.120 --> 01:10:54.090
That you should think about.
01:10:54.090 --> 01:10:56.220
So that circle intersects
the third circle
01:10:56.220 --> 01:10:59.370
in at most two points.
01:10:59.370 --> 01:11:07.230
So three unit spheres have
at most two common points.
01:11:11.950 --> 01:11:15.460
And so the unit distance
graph in r3 is K 3,3-free.
01:11:18.320 --> 01:11:20.720
That entire argument,
even though I expressed it
01:11:20.720 --> 01:11:23.660
geometrically, it's
an algebraic argument.
01:11:23.660 --> 01:11:26.690
You can write down
equations on intersections
01:11:26.690 --> 01:11:28.310
between two spheres.
01:11:28.310 --> 01:11:32.985
It's the intersection of the
sphere with its coaxial plane.
01:11:32.985 --> 01:11:34.730
I have a couple of
these colossal planes.
01:11:34.730 --> 01:11:36.740
They get me a coaxial line.
01:11:36.740 --> 01:11:41.162
That line has to intersect the
sphere in almost two points.
01:11:41.162 --> 01:11:42.620
You should actually
do this algebra
01:11:42.620 --> 01:11:43.860
if you want to do the
proof, because there
01:11:43.860 --> 01:11:45.950
are funny things that can
happen you find fields.
01:11:45.950 --> 01:11:49.500
For example, maybe the
sphere contains a line.
01:11:49.500 --> 01:11:51.880
But you choose your
parameters correctly,
01:11:51.880 --> 01:11:53.130
and these things don't happen.
01:11:56.340 --> 01:11:58.233
And that's the intuition.
01:11:58.233 --> 01:11:59.650
And if you actually
work this out,
01:11:59.650 --> 01:12:02.590
you'll find that this graph
here is indeed K 3,3-free.
01:12:09.030 --> 01:12:10.770
So I'm skipping the details.
01:12:10.770 --> 01:12:15.360
But you should do the algebra
if you want to have a proof.
01:12:15.360 --> 01:12:17.550
On the other hand, it
also has lots of edges.
01:12:17.550 --> 01:12:19.620
And that's basically the
same reason as before.
01:12:19.620 --> 01:12:22.830
But I can count
the number of edges
01:12:22.830 --> 01:12:31.020
by fixing some x, y, and z,
and look at how many abc's
01:12:31.020 --> 01:12:34.260
satisfy that equation.
01:12:34.260 --> 01:12:38.190
And that's, again, something
that needs to be checked,
01:12:38.190 --> 01:12:41.090
but the point is
that this graph--
01:12:41.090 --> 01:12:49.490
so every vertex has
either a P square--
01:12:49.490 --> 01:12:53.865
so it has close to
P squared degree.
01:13:00.180 --> 01:13:05.150
So lots of edges, and combining
with basically the same idea as
01:13:05.150 --> 01:13:07.495
before, you get
the construction.
01:13:10.350 --> 01:13:11.452
Any questions?
01:13:15.562 --> 01:13:16.770
So where can we go from here?
01:13:19.620 --> 01:13:21.090
To construct the K 2,2--
01:13:21.090 --> 01:13:23.950
by the way, if you
construct K 2,2-free,
01:13:23.950 --> 01:13:26.815
the same construction
works for K 2,3-free.
01:13:26.815 --> 01:13:31.670
If it's K 2,2-free, then, it's
also K 2,3-free or K 2,4-free.
01:13:31.670 --> 01:13:35.110
Here, likewise, this
is also K 3,4-free.
01:13:35.110 --> 01:13:38.080
So now, what about
higher K s,t's?
01:13:38.080 --> 01:13:41.950
And you might think, well, let's
take these geometric objects
01:13:41.950 --> 01:13:44.480
and try to extend them further.
01:13:44.480 --> 01:13:46.230
But that actually seems
kind of difficult.
01:13:46.230 --> 01:13:49.020
We do not really
know how to do it.
01:13:49.020 --> 01:13:54.510
We do not know how to
obtain a construction which
01:13:54.510 --> 01:13:58.140
is of this form that
works for K 4,4.
01:13:58.140 --> 01:14:00.080
In fact, there are
even some evidence
01:14:00.080 --> 01:14:02.151
that that might be
even impossible.
01:14:04.980 --> 01:14:09.240
As I mentioned, K 4,4
is a major open problem.
01:14:14.980 --> 01:14:16.810
It is an open
problem to determine
01:14:16.810 --> 01:14:27.970
the order of the
extremal number of K 4,4.
01:14:27.970 --> 01:14:31.790
But in any case,
this construction,
01:14:31.790 --> 01:14:34.490
this idea of using
algebraic constructions,
01:14:34.490 --> 01:14:39.830
is very enlightening, that
we should look at ways to get
01:14:39.830 --> 01:14:43.940
large K s,t-free graphs by
coming up with clever algebraic
01:14:43.940 --> 01:14:44.965
constructions.
01:14:44.965 --> 01:14:49.820
And next, time I will show you
a couple of very nice ideas
01:14:49.820 --> 01:14:52.130
where you can get--
01:14:52.130 --> 01:14:55.940
you can come up with a different
kind of algebraic construction
01:14:55.940 --> 01:14:58.580
which has some superficial
similarities to what we've
01:14:58.580 --> 01:15:04.060
seen today but that's really
of a different nature.
01:15:04.060 --> 01:15:10.150
So, next time, we will
see the following theorem,
01:15:10.150 --> 01:15:13.600
which is obtained in a
sequence of two papers,
01:15:13.600 --> 01:15:24.649
union of authors,
Alon, Rényi and Sazbó,
01:15:24.649 --> 01:15:30.300
that shows that if t
is much larger than s--
01:15:33.964 --> 01:15:38.340
so minus 1 factorial plus 1--
01:15:38.340 --> 01:15:48.060
then the extremal number for
K s,t is on the same order
01:15:48.060 --> 01:15:51.990
as the upper bound determined
in Kovari-Sos-Turon theorem.
01:15:56.880 --> 01:16:01.650
Just to be more explicit
about what these s and t are,
01:16:01.650 --> 01:16:04.890
if you plug in
what's the smallest
01:16:04.890 --> 01:16:07.980
t that this theorem gives
for various values of s,
01:16:07.980 --> 01:16:10.980
find 2,2, and 3,3.
01:16:10.980 --> 01:16:15.070
And the next one is 4, 7.
01:16:15.070 --> 01:16:16.870
And then, it gets
worse from there.
01:16:26.170 --> 01:16:29.080
So these constructions
are based on--
01:16:29.080 --> 01:16:31.240
I mean, they are
algebraic constructions.
01:16:31.240 --> 01:16:34.830
So we'll see next
time what happens.