WEBVTT 00:00:39.760 --> 00:00:40.790 YUFEI ZHAO: All right. 00:00:40.790 --> 00:00:44.300 We are going to continue our discussion of extremal graph 00:00:44.300 --> 00:00:45.470 theory. 00:00:45.470 --> 00:00:48.380 Last time, we discussed what happens 00:00:48.380 --> 00:00:53.000 when we exclude a triangle or more generally a clique. 00:00:53.000 --> 00:00:58.740 And we wish to find a graph that maximizes the number of edges. 00:00:58.740 --> 00:01:01.880 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 00:01:24.930 --> 00:01:29.130 number of edges in an n vertex graph that is H-free-- 00:01:29.130 --> 00:01:31.110 so remember this definition, this 00:01:31.110 --> 00:01:44.710 is the maximum number of edges in an n vertex H-free graph. 00:01:44.710 --> 00:01:47.290 So we are going to be looking at this quantity 00:01:47.290 --> 00:01:49.210 in the next few lectures. 00:01:49.210 --> 00:01:51.370 So the Erdos-Stone-Simonovitz theorem 00:01:51.370 --> 00:01:56.680 tells us that, perhaps, quite surprisingly, this quantity 00:01:56.680 --> 00:02:00.040 is largely covered by the chromatic number of H, 00:02:00.040 --> 00:02:05.470 even though H itself might be quite involved. 00:02:05.470 --> 00:02:06.970 So if you knew the chromatic number, 00:02:06.970 --> 00:02:09.639 you already know a lot of information 00:02:09.639 --> 00:02:12.590 about the growth rate of this function. 00:02:12.590 --> 00:02:17.050 And in particular, as long as it's just not bipartite, 00:02:17.050 --> 00:02:19.240 so that the chromatic numbers at least 3, 00:02:19.240 --> 00:02:21.730 we already know the first order asymptotics 00:02:21.730 --> 00:02:25.600 from the Erdos-Stone-Simonovitz theorem. 00:02:25.600 --> 00:02:34.140 However, when H is bipartite so that the chromatic number is 00:02:34.140 --> 00:02:40.260 exactly 2, then the theorem tells us 00:02:40.260 --> 00:02:45.820 only that this quantity is little o of n squared. 00:02:45.820 --> 00:02:48.270 Which is some useful information, 00:02:48.270 --> 00:02:50.660 but it doesn't tell us the whole story. 00:02:50.660 --> 00:02:53.130 And in the next several lecturers, I want to explore 00:02:53.130 --> 00:02:55.980 what more can we say about this quantity 00:02:55.980 --> 00:02:59.610 here for bipartite graphs H. And it turns out 00:02:59.610 --> 00:03:01.800 that there is a lot that we do not know, 00:03:01.800 --> 00:03:04.980 that there are lots of open problems in this area 00:03:04.980 --> 00:03:07.440 having to do with trying to pin down the growth 00:03:07.440 --> 00:03:10.400 rate of this function. 00:03:10.400 --> 00:03:14.180 And, in particular, for bipartite graphs, 00:03:14.180 --> 00:03:17.420 there's a bipartite graph that places 00:03:17.420 --> 00:03:20.470 somewhat special rule, namely the complete bipartite graph. 00:03:26.760 --> 00:03:30.390 So K st, being the complete bipartite graph, 00:03:30.390 --> 00:03:35.480 with s vertices on one side and t vertices on the other side-- 00:03:40.350 --> 00:03:42.470 so this is a very nice bipartite graph. 00:03:42.470 --> 00:03:46.520 And just to understand, the extremal number for this graph 00:03:46.520 --> 00:03:50.390 is a famous open problem in this area, 00:03:50.390 --> 00:04:00.180 and it has the name of Zarankiewicz problem, which 00:04:00.180 --> 00:04:07.800 is to determine or estimate the extremal numbers 00:04:07.800 --> 00:04:12.510 for these complete bipartite graphs. 00:04:12.510 --> 00:04:17.142 So I'll tell you pretty much all we know about this problem. 00:04:17.142 --> 00:04:19.050 And there are some interesting things. 00:04:19.050 --> 00:04:22.990 But we do not know all that much. 00:04:22.990 --> 00:04:28.060 Now, every bipartite graph is a subset. 00:04:28.060 --> 00:04:30.790 It's a subgraph of such a complete bipartite graph. 00:04:33.970 --> 00:04:46.670 So every bipartite H it is a soft graph of some K st. 00:04:46.670 --> 00:04:53.000 And we know that if H is a subgraph of this K st, 00:04:53.000 --> 00:05:01.430 then the extremal number for H if your graph is H-free, 00:05:01.430 --> 00:05:05.180 then automatically it has to be K s,t-free. 00:05:05.180 --> 00:05:11.700 So there is this bound between these two extremal numbers. 00:05:11.700 --> 00:05:14.870 So in particular, if you have some upper bound on K st, 00:05:14.870 --> 00:05:17.390 then you have some upper bound on bipartite graphs. 00:05:17.390 --> 00:05:19.730 Although, for specific by bipartite graphs H, 00:05:19.730 --> 00:05:23.180 maybe we can do better than using this bound. 00:05:23.180 --> 00:05:27.050 And we'll see examples of that later in the course as well. 00:05:27.050 --> 00:05:31.772 So what can we say about the extremal numbers of these K 00:05:31.772 --> 00:05:33.000 s,t's? 00:05:33.000 --> 00:05:36.760 So the most important theorem in this area of this problem 00:05:36.760 --> 00:05:39.880 is the result due to Kovari-Sos-Turan. 00:05:46.060 --> 00:05:54.460 So the Kovari-Sos-Turan theorem tells us that for every fixed 00:05:54.460 --> 00:05:57.370 integers s and t-- 00:05:57.370 --> 00:05:59.320 s at most t-- 00:05:59.320 --> 00:06:10.550 there exists some constant C, such that the extremal number 00:06:10.550 --> 00:06:15.140 is upper-bounded by something which is on the order of n 00:06:15.140 --> 00:06:18.230 to the 2 minus 1 over s. 00:06:20.960 --> 00:06:23.910 So it gives you some upper bound, showing you 00:06:23.910 --> 00:06:30.330 that it's not only subquadratic, but there is a gap from 2 00:06:30.330 --> 00:06:33.350 in the exponent. 00:06:33.350 --> 00:06:36.350 Now, in combinatorics, as is with many fields 00:06:36.350 --> 00:06:38.990 of mathematics, it can be somewhat intimidating 00:06:38.990 --> 00:06:40.950 to newcomers when you see a lot of names. 00:06:40.950 --> 00:06:42.980 So every theorems seems to be named 00:06:42.980 --> 00:06:45.770 after a string of mathematicians, some of whom 00:06:45.770 --> 00:06:48.260 you may have heard of, some you may not. 00:06:48.260 --> 00:06:51.920 And I agree, it may be difficult to remember who is who and what 00:06:51.920 --> 00:06:55.100 is what, but I think this theorem has a very nice way 00:06:55.100 --> 00:06:58.670 to remember, which is that it's a theorem about K st, 00:06:58.670 --> 00:07:00.331 and this is the K s,t theorem. 00:07:00.331 --> 00:07:02.920 [LAUGHTER] 00:07:02.920 --> 00:07:04.250 OK. 00:07:04.250 --> 00:07:05.690 So this is the K s,t theorem. 00:07:08.520 --> 00:07:13.020 I want to begin by showing you how to prove the K s,t The 00:07:13.020 --> 00:07:16.470 proof is via a nice and not too difficult double-counting 00:07:16.470 --> 00:07:17.520 argument. 00:07:17.520 --> 00:07:19.603 And I show you some applications. 00:07:29.970 --> 00:07:32.360 Let's prove this theorem, and it's 00:07:32.360 --> 00:07:33.800 by double-counting argument. 00:07:45.760 --> 00:07:46.900 What are we going to count? 00:07:46.900 --> 00:07:49.790 Well, let's start with the object that we're working with. 00:07:49.790 --> 00:07:57.430 So there's going to be a graph G that has n vertices, m edges, 00:07:57.430 --> 00:07:58.690 and K s,t-free. 00:08:03.400 --> 00:08:13.670 And let's count the number of stars, K s,1 in this graph G. 00:08:13.670 --> 00:08:19.650 So we're counting configurations like that. 00:08:22.362 --> 00:08:24.210 I'll do an upper bound and a lower bound. 00:08:27.270 --> 00:08:31.020 So let's start with the upper bound. 00:08:31.020 --> 00:08:41.240 On one hand, every subset of s vertices in the graph G 00:08:41.240 --> 00:08:46.590 has at most t minus 1 common neighbors. 00:08:51.990 --> 00:09:03.020 Because if they had t common neighbors, then you get a K st. 00:09:03.020 --> 00:09:05.040 So that's one down. 00:09:05.040 --> 00:09:08.870 On the other hand, let's see what 00:09:08.870 --> 00:09:10.850 happens to the number of common neighbors 00:09:10.850 --> 00:09:15.710 if you knew that this graph has a lot of edges. 00:09:15.710 --> 00:09:17.750 So the number of stars-- 00:09:20.870 --> 00:09:23.600 well, I can calculate this quantity explicitly 00:09:23.600 --> 00:09:30.080 by running over all the vertices of G. For each vertex, 00:09:30.080 --> 00:09:33.260 look at its neighborhood and choose s vertices 00:09:33.260 --> 00:09:35.600 from its neighborhood. 00:09:35.600 --> 00:09:41.750 So for each v I need to find a subset of s vertices 00:09:41.750 --> 00:09:44.960 from its neighborhood. 00:09:44.960 --> 00:09:48.860 By convexity, I can lower bound the sum 00:09:48.860 --> 00:09:53.330 by the average, in some sense, because-- 00:09:53.330 --> 00:09:55.386 let me first write down the expression. 00:10:00.050 --> 00:10:04.940 So I can do a convexity argument that gives me this lower bound. 00:10:08.340 --> 00:10:12.450 Here, I'm abusing notation somewhat 00:10:12.450 --> 00:10:15.690 and writing binomial coefficients 00:10:15.690 --> 00:10:23.760 with a real entry on top, where I mean the expression 00:10:23.760 --> 00:10:33.530 as you would expect, treating this guy as a polynomial in x. 00:10:33.530 --> 00:10:36.020 And the key fact we're using here 00:10:36.020 --> 00:10:40.390 is that this function here is convex 00:10:40.390 --> 00:10:47.190 for x at least and minus 1. 00:10:47.190 --> 00:10:53.920 So, In particular, if you think about this function here, 00:10:53.920 --> 00:10:59.070 you have a lot of 0's, and then it becomes convex afterwards. 00:10:59.070 --> 00:11:02.610 So you can even think of extending this function as 0 00:11:02.610 --> 00:11:03.900 to the left of n minus 1. 00:11:03.900 --> 00:11:08.100 So this is a convex function, and you can apply convexity 00:11:08.100 --> 00:11:11.360 to deduce this inequality. 00:11:11.360 --> 00:11:13.010 But you know the sum of the degrees. 00:11:13.010 --> 00:11:14.720 That's just essentially twice-- 00:11:14.720 --> 00:11:17.000 I mean, that is the twice the number of edges. 00:11:17.000 --> 00:11:25.770 So we have this expression right here. 00:11:25.770 --> 00:11:31.320 So you have an upper bound on the number of s stars coming 00:11:31.320 --> 00:11:35.820 from K s,t-freeness and lower bound on the number of s stars 00:11:35.820 --> 00:11:38.880 coming from just having lots of edges and applying convexity. 00:11:41.490 --> 00:11:47.190 Putting these two things together, 00:11:47.190 --> 00:11:54.150 we find that there's this inequality here. 00:12:00.670 --> 00:12:03.430 Here, we are thinking of s and t as fixed, 00:12:03.430 --> 00:12:05.020 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. 00:12:09.310 --> 00:12:12.790 So it will be helpful to use the asymptotics that n choose 00:12:12.790 --> 00:12:25.420 s grows like n to the s divided by s factorial for a fixed s. 00:12:25.420 --> 00:12:28.600 So looking at that expression and applying this asymptotics 00:12:28.600 --> 00:12:41.130 to both sides, we have this inequality here, 00:12:41.130 --> 00:12:44.640 So I've eliminated s factorial from both sides. 00:12:47.490 --> 00:12:50.730 So now rearrange, clean things up. 00:12:50.730 --> 00:12:53.200 We find the following upper bound-- 00:12:53.200 --> 00:13:06.982 m the number of edges in G. And that's the expression. 00:13:11.320 --> 00:13:17.410 So for fixed s and t it grows like n to the 2 minus 1 over s. 00:13:23.100 --> 00:13:25.340 Any questions? 00:13:25.340 --> 00:13:25.860 Yeah. 00:13:25.860 --> 00:13:27.110 AUDIENCE: Where does the right side of the inequality 00:13:27.110 --> 00:13:28.250 come from? 00:13:28.250 --> 00:13:31.700 Are you counting that from the different cycles [INAUDIBLE]?? 00:13:31.700 --> 00:13:33.310 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. 00:13:37.410 --> 00:13:40.860 So here we are counting the number of K s1's. 00:13:40.860 --> 00:13:42.280 AUDIENCE: Oh, OK. 00:13:42.280 --> 00:13:47.290 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. 00:13:52.490 --> 00:13:53.236 Yes. 00:13:53.236 --> 00:13:55.616 AUDIENCE: Do you actually care that s is less than 00:13:55.616 --> 00:13:57.433 or equal t in the argument? 00:13:57.433 --> 00:13:58.850 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.