1 00:00:39,760 --> 00:00:40,790 YUFEI ZHAO: All right. 2 00:00:40,790 --> 00:00:44,300 We are going to continue our discussion of extremal graph 3 00:00:44,300 --> 00:00:45,470 theory. 4 00:00:45,470 --> 00:00:48,380 Last time, we discussed what happens 5 00:00:48,380 --> 00:00:53,000 when we exclude a triangle or more generally a clique. 6 00:00:53,000 --> 00:00:58,740 And we wish to find a graph that maximizes the number of edges. 7 00:00:58,740 --> 00:01:01,880 And at the end of the lecture, I stated the theorem 8 00:01:01,880 --> 00:01:15,760 of Erdos-Stone-Simonovitz, which says, recall, 9 00:01:15,760 --> 00:01:24,930 that if you have a fixed H and I wish to understand the maximum 10 00:01:24,930 --> 00:01:29,130 number of edges in an n vertex graph that is H-free-- 11 00:01:29,130 --> 00:01:31,110 so remember this definition, this 12 00:01:31,110 --> 00:01:44,710 is the maximum number of edges in an n vertex H-free graph. 13 00:01:44,710 --> 00:01:47,290 So we are going to be looking at this quantity 14 00:01:47,290 --> 00:01:49,210 in the next few lectures. 15 00:01:49,210 --> 00:01:51,370 So the Erdos-Stone-Simonovitz theorem 16 00:01:51,370 --> 00:01:56,680 tells us that, perhaps, quite surprisingly, this quantity 17 00:01:56,680 --> 00:02:00,040 is largely covered by the chromatic number of H, 18 00:02:00,040 --> 00:02:05,470 even though H itself might be quite involved. 19 00:02:05,470 --> 00:02:06,970 So if you knew the chromatic number, 20 00:02:06,970 --> 00:02:09,639 you already know a lot of information 21 00:02:09,639 --> 00:02:12,590 about the growth rate of this function. 22 00:02:12,590 --> 00:02:17,050 And in particular, as long as it's just not bipartite, 23 00:02:17,050 --> 00:02:19,240 so that the chromatic numbers at least 3, 24 00:02:19,240 --> 00:02:21,730 we already know the first order asymptotics 25 00:02:21,730 --> 00:02:25,600 from the Erdos-Stone-Simonovitz theorem. 26 00:02:25,600 --> 00:02:34,140 However, when H is bipartite so that the chromatic number is 27 00:02:34,140 --> 00:02:40,260 exactly 2, then the theorem tells us 28 00:02:40,260 --> 00:02:45,820 only that this quantity is little o of n squared. 29 00:02:45,820 --> 00:02:48,270 Which is some useful information, 30 00:02:48,270 --> 00:02:50,660 but it doesn't tell us the whole story. 31 00:02:50,660 --> 00:02:53,130 And in the next several lecturers, I want to explore 32 00:02:53,130 --> 00:02:55,980 what more can we say about this quantity 33 00:02:55,980 --> 00:02:59,610 here for bipartite graphs H. And it turns out 34 00:02:59,610 --> 00:03:01,800 that there is a lot that we do not know, 35 00:03:01,800 --> 00:03:04,980 that there are lots of open problems in this area 36 00:03:04,980 --> 00:03:07,440 having to do with trying to pin down the growth 37 00:03:07,440 --> 00:03:10,400 rate of this function. 38 00:03:10,400 --> 00:03:14,180 And, in particular, for bipartite graphs, 39 00:03:14,180 --> 00:03:17,420 there's a bipartite graph that places 40 00:03:17,420 --> 00:03:20,470 somewhat special rule, namely the complete bipartite graph. 41 00:03:26,760 --> 00:03:30,390 So K st, being the complete bipartite graph, 42 00:03:30,390 --> 00:03:35,480 with s vertices on one side and t vertices on the other side-- 43 00:03:40,350 --> 00:03:42,470 so this is a very nice bipartite graph. 44 00:03:42,470 --> 00:03:46,520 And just to understand, the extremal number for this graph 45 00:03:46,520 --> 00:03:50,390 is a famous open problem in this area, 46 00:03:50,390 --> 00:04:00,180 and it has the name of Zarankiewicz problem, which 47 00:04:00,180 --> 00:04:07,800 is to determine or estimate the extremal numbers 48 00:04:07,800 --> 00:04:12,510 for these complete bipartite graphs. 49 00:04:12,510 --> 00:04:17,142 So I'll tell you pretty much all we know about this problem. 50 00:04:17,142 --> 00:04:19,050 And there are some interesting things. 51 00:04:19,050 --> 00:04:22,990 But we do not know all that much. 52 00:04:22,990 --> 00:04:28,060 Now, every bipartite graph is a subset. 53 00:04:28,060 --> 00:04:30,790 It's a subgraph of such a complete bipartite graph. 54 00:04:33,970 --> 00:04:46,670 So every bipartite H it is a soft graph of some K st. 55 00:04:46,670 --> 00:04:53,000 And we know that if H is a subgraph of this K st, 56 00:04:53,000 --> 00:05:01,430 then the extremal number for H if your graph is H-free, 57 00:05:01,430 --> 00:05:05,180 then automatically it has to be K s,t-free. 58 00:05:05,180 --> 00:05:11,700 So there is this bound between these two extremal numbers. 59 00:05:11,700 --> 00:05:14,870 So in particular, if you have some upper bound on K st, 60 00:05:14,870 --> 00:05:17,390 then you have some upper bound on bipartite graphs. 61 00:05:17,390 --> 00:05:19,730 Although, for specific by bipartite graphs H, 62 00:05:19,730 --> 00:05:23,180 maybe we can do better than using this bound. 63 00:05:23,180 --> 00:05:27,050 And we'll see examples of that later in the course as well. 64 00:05:27,050 --> 00:05:31,772 So what can we say about the extremal numbers of these K 65 00:05:31,772 --> 00:05:33,000 s,t's? 66 00:05:33,000 --> 00:05:36,760 So the most important theorem in this area of this problem 67 00:05:36,760 --> 00:05:39,880 is the result due to Kovari-Sos-Turan. 68 00:05:46,060 --> 00:05:54,460 So the Kovari-Sos-Turan theorem tells us that for every fixed 69 00:05:54,460 --> 00:05:57,370 integers s and t-- 70 00:05:57,370 --> 00:05:59,320 s at most t-- 71 00:05:59,320 --> 00:06:10,550 there exists some constant C, such that the extremal number 72 00:06:10,550 --> 00:06:15,140 is upper-bounded by something which is on the order of n 73 00:06:15,140 --> 00:06:18,230 to the 2 minus 1 over s. 74 00:06:20,960 --> 00:06:23,910 So it gives you some upper bound, showing you 75 00:06:23,910 --> 00:06:30,330 that it's not only subquadratic, but there is a gap from 2 76 00:06:30,330 --> 00:06:33,350 in the exponent. 77 00:06:33,350 --> 00:06:36,350 Now, in combinatorics, as is with many fields 78 00:06:36,350 --> 00:06:38,990 of mathematics, it can be somewhat intimidating 79 00:06:38,990 --> 00:06:40,950 to newcomers when you see a lot of names. 80 00:06:40,950 --> 00:06:42,980 So every theorems seems to be named 81 00:06:42,980 --> 00:06:45,770 after a string of mathematicians, some of whom 82 00:06:45,770 --> 00:06:48,260 you may have heard of, some you may not. 83 00:06:48,260 --> 00:06:51,920 And I agree, it may be difficult to remember who is who and what 84 00:06:51,920 --> 00:06:55,100 is what, but I think this theorem has a very nice way 85 00:06:55,100 --> 00:06:58,670 to remember, which is that it's a theorem about K st, 86 00:06:58,670 --> 00:07:00,331 and this is the K s,t theorem. 87 00:07:00,331 --> 00:07:02,920 [LAUGHTER] 88 00:07:02,920 --> 00:07:04,250 OK. 89 00:07:04,250 --> 00:07:05,690 So this is the K s,t theorem. 90 00:07:08,520 --> 00:07:13,020 I want to begin by showing you how to prove the K s,t The 91 00:07:13,020 --> 00:07:16,470 proof is via a nice and not too difficult double-counting 92 00:07:16,470 --> 00:07:17,520 argument. 93 00:07:17,520 --> 00:07:19,603 And I show you some applications. 94 00:07:29,970 --> 00:07:32,360 Let's prove this theorem, and it's 95 00:07:32,360 --> 00:07:33,800 by double-counting argument. 96 00:07:45,760 --> 00:07:46,900 What are we going to count? 97 00:07:46,900 --> 00:07:49,790 Well, let's start with the object that we're working with. 98 00:07:49,790 --> 00:07:57,430 So there's going to be a graph G that has n vertices, m edges, 99 00:07:57,430 --> 00:07:58,690 and K s,t-free. 100 00:08:03,400 --> 00:08:13,670 And let's count the number of stars, K s,1 in this graph G. 101 00:08:13,670 --> 00:08:19,650 So we're counting configurations like that. 102 00:08:22,362 --> 00:08:24,210 I'll do an upper bound and a lower bound. 103 00:08:27,270 --> 00:08:31,020 So let's start with the upper bound. 104 00:08:31,020 --> 00:08:41,240 On one hand, every subset of s vertices in the graph G 105 00:08:41,240 --> 00:08:46,590 has at most t minus 1 common neighbors. 106 00:08:51,990 --> 00:09:03,020 Because if they had t common neighbors, then you get a K st. 107 00:09:03,020 --> 00:09:05,040 So that's one down. 108 00:09:05,040 --> 00:09:08,870 On the other hand, let's see what 109 00:09:08,870 --> 00:09:10,850 happens to the number of common neighbors 110 00:09:10,850 --> 00:09:15,710 if you knew that this graph has a lot of edges. 111 00:09:15,710 --> 00:09:17,750 So the number of stars-- 112 00:09:20,870 --> 00:09:23,600 well, I can calculate this quantity explicitly 113 00:09:23,600 --> 00:09:30,080 by running over all the vertices of G. For each vertex, 114 00:09:30,080 --> 00:09:33,260 look at its neighborhood and choose s vertices 115 00:09:33,260 --> 00:09:35,600 from its neighborhood. 116 00:09:35,600 --> 00:09:41,750 So for each v I need to find a subset of s vertices 117 00:09:41,750 --> 00:09:44,960 from its neighborhood. 118 00:09:44,960 --> 00:09:48,860 By convexity, I can lower bound the sum 119 00:09:48,860 --> 00:09:53,330 by the average, in some sense, because-- 120 00:09:53,330 --> 00:09:55,386 let me first write down the expression. 121 00:10:00,050 --> 00:10:04,940 So I can do a convexity argument that gives me this lower bound. 122 00:10:08,340 --> 00:10:12,450 Here, I'm abusing notation somewhat 123 00:10:12,450 --> 00:10:15,690 and writing binomial coefficients 124 00:10:15,690 --> 00:10:23,760 with a real entry on top, where I mean the expression 125 00:10:23,760 --> 00:10:33,530 as you would expect, treating this guy as a polynomial in x. 126 00:10:33,530 --> 00:10:36,020 And the key fact we're using here 127 00:10:36,020 --> 00:10:40,390 is that this function here is convex 128 00:10:40,390 --> 00:10:47,190 for x at least and minus 1. 129 00:10:47,190 --> 00:10:53,920 So, In particular, if you think about this function here, 130 00:10:53,920 --> 00:10:59,070 you have a lot of 0's, and then it becomes convex afterwards. 131 00:10:59,070 --> 00:11:02,610 So you can even think of extending this function as 0 132 00:11:02,610 --> 00:11:03,900 to the left of n minus 1. 133 00:11:03,900 --> 00:11:08,100 So this is a convex function, and you can apply convexity 134 00:11:08,100 --> 00:11:11,360 to deduce this inequality. 135 00:11:11,360 --> 00:11:13,010 But you know the sum of the degrees. 136 00:11:13,010 --> 00:11:14,720 That's just essentially twice-- 137 00:11:14,720 --> 00:11:17,000 I mean, that is the twice the number of edges. 138 00:11:17,000 --> 00:11:25,770 So we have this expression right here. 139 00:11:25,770 --> 00:11:31,320 So you have an upper bound on the number of s stars coming 140 00:11:31,320 --> 00:11:35,820 from K s,t-freeness and lower bound on the number of s stars 141 00:11:35,820 --> 00:11:38,880 coming from just having lots of edges and applying convexity. 142 00:11:41,490 --> 00:11:47,190 Putting these two things together, 143 00:11:47,190 --> 00:11:54,150 we find that there's this inequality here. 144 00:12:00,670 --> 00:12:03,430 Here, we are thinking of s and t as fixed, 145 00:12:03,430 --> 00:12:05,020 and we are trying to understand how 146 00:12:05,020 --> 00:12:09,310 m and n depend on each other as they get large. 147 00:12:09,310 --> 00:12:12,790 So it will be helpful to use the asymptotics that n choose 148 00:12:12,790 --> 00:12:25,420 s grows like n to the s divided by s factorial for a fixed s. 149 00:12:25,420 --> 00:12:28,600 So looking at that expression and applying this asymptotics 150 00:12:28,600 --> 00:12:41,130 to both sides, we have this inequality here, 151 00:12:41,130 --> 00:12:44,640 So I've eliminated s factorial from both sides. 152 00:12:47,490 --> 00:12:50,730 So now rearrange, clean things up. 153 00:12:50,730 --> 00:12:53,200 We find the following upper bound-- 154 00:12:53,200 --> 00:13:06,982 m the number of edges in G. And that's the expression. 155 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. 156 00:13:23,100 --> 00:13:25,340 Any questions? 157 00:13:25,340 --> 00:13:25,860 Yeah. 158 00:13:25,860 --> 00:13:27,110 AUDIENCE: Where does the right side of the inequality 159 00:13:27,110 --> 00:13:28,250 come from? 160 00:13:28,250 --> 00:13:31,700 Are you counting that from the different cycles [INAUDIBLE]?? 161 00:13:31,700 --> 00:13:33,310 YUFEI ZHAO: So the question is where 162 00:13:33,310 --> 00:13:35,310 the right side of inequality-- which inequality? 163 00:13:35,310 --> 00:13:35,893 This one here? 164 00:13:35,893 --> 00:13:36,660 AUDIENCE: Yeah. 165 00:13:36,660 --> 00:13:37,410 YUFEI ZHAO: Right. 166 00:13:37,410 --> 00:13:40,860 So here we are counting the number of K s1's. 167 00:13:40,860 --> 00:13:42,280 AUDIENCE: Oh, OK. 168 00:13:42,280 --> 00:13:47,290 YUFEI ZHAO: So here I am upper and lower-bounding 169 00:13:47,290 --> 00:13:52,490 the number of K s1's using what we derived earlier. 170 00:13:52,490 --> 00:13:53,236 Yes. 171 00:13:53,236 --> 00:13:55,616 AUDIENCE: Do you actually care that s is less than 172 00:13:55,616 --> 00:13:57,433 or equal t in the argument? 173 00:13:57,433 --> 00:13:58,850 YUFEI ZHAO: The question is, do we 174 00:13:58,850 --> 00:14:01,460 care that s is less than or equal to t in the argument? 175 00:14:01,460 --> 00:14:02,660 The argument doesn't care. 176 00:14:02,660 --> 00:14:06,000 But, of course, if you want the better asymptotics for fixed s 177 00:14:06,000 --> 00:14:08,910 and t as n gets large, you should take s to be less than 178 00:14:08,910 --> 00:14:10,170 or equal to t. 179 00:14:10,170 --> 00:14:10,670 Question? 180 00:14:10,670 --> 00:14:12,590 AUDIENCE: What happens when t equals 1? 181 00:14:12,590 --> 00:14:14,540 YUFEI ZHAO: What happens when t equals to 1. 182 00:14:14,540 --> 00:14:16,025 Well, that's a great question. 183 00:14:16,025 --> 00:14:17,400 I'll leave you to think about it. 184 00:14:17,400 --> 00:14:24,090 If you know that your graph has maximum degree n most t, 185 00:14:24,090 --> 00:14:26,743 what can you tell me about the number of edges in the graph? 186 00:14:26,743 --> 00:14:28,860 [LAUGHTER] 187 00:14:28,860 --> 00:14:30,480 OK. 188 00:14:30,480 --> 00:14:31,810 Any more questions? 189 00:14:35,037 --> 00:14:36,370 We'll come back to this theorem. 190 00:14:36,370 --> 00:14:39,295 In fact, this will occupy us for at least a couple of lectures. 191 00:14:41,950 --> 00:14:45,000 Basically, is this theorem tight? 192 00:14:45,000 --> 00:14:46,920 And it is conjectured to be, although that 193 00:14:46,920 --> 00:14:49,740 is a major open problem extremal graph theory. 194 00:14:49,740 --> 00:14:52,760 We only know a small number of values. 195 00:14:52,760 --> 00:14:55,890 Well, for most values of s and t, 196 00:14:55,890 --> 00:15:00,390 in particular when s and t are both equal to 4, 197 00:15:00,390 --> 00:15:03,520 we do not know if this bound is tight. 198 00:15:03,520 --> 00:15:05,980 But there are some values of s and t for which 199 00:15:05,980 --> 00:15:07,840 we do know that it is tight. 200 00:15:07,840 --> 00:15:13,691 For example, 2 and 2, 3 and 3, 4 and 7, 201 00:15:13,691 --> 00:15:16,887 s and if t is really, really large. 202 00:15:16,887 --> 00:15:18,720 And I will show you some constructions later 203 00:15:18,720 --> 00:15:25,070 on, creating graphs G that are K s,t-free for those parameters 204 00:15:25,070 --> 00:15:29,780 that matches this K st bound up to a constant factor. 205 00:15:29,780 --> 00:15:30,540 Yes? 206 00:15:30,540 --> 00:15:32,990 AUDIENCE: For a fixed s, what's the bound on t-- 207 00:15:32,990 --> 00:15:35,690 if there a bound on t for t with equality cases? 208 00:15:35,690 --> 00:15:36,440 YUFEI ZHAO: Right. 209 00:15:36,440 --> 00:15:39,500 So the question is, for a fixed value of s, 210 00:15:39,500 --> 00:15:42,770 is there some bound on t for which we get equality cases? 211 00:15:42,770 --> 00:15:46,400 There is a conjecture that this bound is sharp 212 00:15:46,400 --> 00:15:51,240 up to a constant factor for every s and t. 213 00:15:51,240 --> 00:15:53,820 But we only know how to prove that conjecture-- 214 00:15:53,820 --> 00:15:56,670 and I will tell you much more about it later on-- 215 00:15:56,670 --> 00:16:01,860 when t is much larger compared to s. 216 00:16:01,860 --> 00:16:08,430 So there is a lot of unexplored territory on this problem. 217 00:16:08,430 --> 00:16:09,780 Any more questions? 218 00:16:13,420 --> 00:16:17,140 Before diving into the Kovari-Sos-Turan further, 219 00:16:17,140 --> 00:16:19,759 I just want to show you some neat applications. 220 00:16:29,540 --> 00:16:32,685 And I want to begin with a geometric application. 221 00:16:41,720 --> 00:16:44,310 There is a classic problem asked by Erdos, 222 00:16:44,310 --> 00:16:47,090 back in the '40s, called the unit distance problem. 223 00:16:53,150 --> 00:17:12,339 Which asks, what is the maximum number of unit distances 224 00:17:12,339 --> 00:17:18,580 formed by n points in the plane. 225 00:17:23,369 --> 00:17:25,140 Let me give you some examples. 226 00:17:25,140 --> 00:17:26,880 If you have three points, you put them 227 00:17:26,880 --> 00:17:31,730 in equilateral triangle, and all three distances are unit. 228 00:17:31,730 --> 00:17:32,550 Great. 229 00:17:32,550 --> 00:17:36,060 If you have four points, you cannot place them so that all 230 00:17:36,060 --> 00:17:39,800 six distances are units if you're staying in the plane. 231 00:17:39,800 --> 00:17:43,050 So the best thing you can do is something like this-- 232 00:17:46,690 --> 00:17:50,830 so I'm drawing all the edges that are unit distances. 233 00:17:50,830 --> 00:17:54,610 If you have one more point, it turns out 234 00:17:54,610 --> 00:17:58,230 that's the best thing you can do with five points. 235 00:17:58,230 --> 00:18:01,320 With six points there are some more possibilities. 236 00:18:01,320 --> 00:18:05,320 Let me draw for you some possibilities. 237 00:18:05,320 --> 00:18:07,960 You can extend the previous configuration 238 00:18:07,960 --> 00:18:09,285 by adding more triangles. 239 00:18:09,285 --> 00:18:10,660 And there are many different ways 240 00:18:10,660 --> 00:18:12,735 that you can attach an extra triangle. 241 00:18:19,560 --> 00:18:22,720 There's actually one more way to do it with six points. 242 00:18:22,720 --> 00:18:29,120 Namely, I can put them like a projection of a prism. 243 00:18:29,120 --> 00:18:31,870 So all of these configurations have 244 00:18:31,870 --> 00:18:33,730 the maximum number of unit distances 245 00:18:33,730 --> 00:18:37,370 obtained if you draw six points. 246 00:18:37,370 --> 00:18:41,210 If you draw seven points, it turns out 247 00:18:41,210 --> 00:18:43,400 this is the best way to do it. 248 00:18:43,400 --> 00:18:45,090 So you can go on for a while. 249 00:18:45,090 --> 00:18:45,885 And people have. 250 00:18:45,885 --> 00:18:48,230 So you can try to tabulate for every n what's 251 00:18:48,230 --> 00:18:49,910 the maximum number of unit distances 252 00:18:49,910 --> 00:18:53,010 you can generate having n points in the plane. 253 00:18:53,010 --> 00:18:56,342 And the question is, what is the answer for n points? 254 00:18:56,342 --> 00:18:57,800 And it turns out, for this problem, 255 00:18:57,800 --> 00:19:00,410 and for many problems like it in combinatorics, 256 00:19:00,410 --> 00:19:03,350 you can have a lot of fun with playing with small examples, 257 00:19:03,350 --> 00:19:05,850 but they are often misleading. 258 00:19:05,850 --> 00:19:08,660 It doesn't really tell you what the overall structure should 259 00:19:08,660 --> 00:19:10,360 be like. 260 00:19:10,360 --> 00:19:12,310 And it turns out this is the major opium 261 00:19:12,310 --> 00:19:15,070 problem for which we do not understand what the structure 262 00:19:15,070 --> 00:19:16,630 is like for large values of n. 263 00:19:19,300 --> 00:19:22,900 So think n large. 264 00:19:22,900 --> 00:19:27,220 What are some possible ways to generate many unit distances? 265 00:19:27,220 --> 00:19:27,720 Yep. 266 00:19:27,720 --> 00:19:29,360 AUDIENCE: Drawing triangles? 267 00:19:29,360 --> 00:19:30,110 YUFEI ZHAO: Great. 268 00:19:30,110 --> 00:19:33,790 So one way is to draw lots of triangles. 269 00:19:33,790 --> 00:19:38,572 So extend this figure forward. 270 00:19:38,572 --> 00:19:39,280 So let's do that. 271 00:19:44,310 --> 00:19:47,470 Well, actually, let me give you something even simpler 272 00:19:47,470 --> 00:19:48,310 to begin with. 273 00:19:48,310 --> 00:19:53,160 I can just put the n points on a line, equal spaced, OK, 274 00:19:53,160 --> 00:19:55,430 I get n minus 1 distances-- 275 00:19:55,430 --> 00:19:57,198 n minus 1 unit distances. 276 00:19:59,820 --> 00:20:07,870 If I put them like that, how many unit distances do I get? 277 00:20:10,720 --> 00:20:12,145 AUDIENCE: 2 times n minus 1. 278 00:20:12,145 --> 00:20:13,100 YUFEI ZHAO: Good. 279 00:20:13,100 --> 00:20:15,555 2 times n minus 1. 280 00:20:15,555 --> 00:20:19,770 So each new point gives you two extra new distances. 281 00:20:19,770 --> 00:20:23,883 And you can try to think about how to do better. 282 00:20:23,883 --> 00:20:25,800 Well, if you follow that sequence of examples, 283 00:20:25,800 --> 00:20:27,930 maybe you're limited to such ideas. 284 00:20:27,930 --> 00:20:28,560 Yes? 285 00:20:28,560 --> 00:20:31,230 AUDIENCE: Are we allowed to put points in the same place? 286 00:20:31,230 --> 00:20:33,960 YUFEI ZHAO: You are not allowed to put points 287 00:20:33,960 --> 00:20:34,910 in the same place. 288 00:20:34,910 --> 00:20:35,910 That's a great question. 289 00:20:35,910 --> 00:20:38,202 But you're not allowed to put points on the same place. 290 00:20:38,202 --> 00:20:38,747 Yes? 291 00:20:38,747 --> 00:20:41,370 AUDIENCE: Is for something to keep 292 00:20:41,370 --> 00:20:46,870 doubling the number of points, the degrees get big? 293 00:20:46,870 --> 00:20:48,750 YUFEI ZHAO: So I didn't quite understand-- 294 00:20:48,750 --> 00:20:51,160 so you say if you put-- 295 00:20:51,160 --> 00:20:52,632 AUDIENCE: [INAUDIBLE] something. 296 00:20:52,632 --> 00:20:53,340 YUFEI ZHAO: Yeah. 297 00:20:53,340 --> 00:20:56,100 AUDIENCE: Like keep translating in different directions. 298 00:20:56,100 --> 00:20:57,933 YUFEI ZHAO: So you want to take this example 299 00:20:57,933 --> 00:20:59,520 and keep translating in some direction 300 00:20:59,520 --> 00:21:01,437 I think you'll run out of room pretty quickly. 301 00:21:01,437 --> 00:21:07,270 AUDIENCE: No, I mean just copy it and put it over [INAUDIBLE].. 302 00:21:07,270 --> 00:21:08,020 YUFEI ZHAO: I see. 303 00:21:08,020 --> 00:21:09,790 You want to take this configuration 304 00:21:09,790 --> 00:21:12,580 and translate it in some unit direction. 305 00:21:12,580 --> 00:21:14,838 Well, then you double the number of points, 306 00:21:14,838 --> 00:21:17,380 but you don't actually increase the number of units distances 307 00:21:17,380 --> 00:21:20,515 all that much. 308 00:21:20,515 --> 00:21:26,311 AUDIENCE: Don't we get each point having one extra unit 309 00:21:26,311 --> 00:21:28,320 distance added? 310 00:21:28,320 --> 00:21:36,840 YUFEI ZHAO: So the suggestion is we take some configuration, 311 00:21:36,840 --> 00:21:44,740 let's say this graph G, and I form two copies of G 312 00:21:44,740 --> 00:21:48,400 by translating G in some unit directions-- some generic unit 313 00:21:48,400 --> 00:21:49,040 direction. 314 00:21:49,040 --> 00:21:49,540 OK, great. 315 00:21:49,540 --> 00:21:51,625 So what happens to the number of vertices? 316 00:21:51,625 --> 00:21:57,730 So n goes to 2m and the number of edges 317 00:21:57,730 --> 00:22:05,415 goes from m to 2m plus n. 318 00:22:05,415 --> 00:22:06,210 Is that right? 319 00:22:06,210 --> 00:22:06,710 OK, great. 320 00:22:06,710 --> 00:22:09,267 So if you do that, what do you get? 321 00:22:09,267 --> 00:22:10,550 AUDIENCE: Log n. 322 00:22:10,550 --> 00:22:11,300 YUFEI ZHAO: n log. 323 00:22:11,300 --> 00:22:11,950 OK. 324 00:22:11,950 --> 00:22:15,770 That's much better than before. 325 00:22:15,770 --> 00:22:17,440 OK, good. 326 00:22:17,440 --> 00:22:17,940 Good. 327 00:22:17,940 --> 00:22:18,360 Very nice. 328 00:22:18,360 --> 00:22:18,860 Yes? 329 00:22:18,860 --> 00:22:20,520 AUDIENCE: That picture [INAUDIBLE].. 330 00:22:20,520 --> 00:22:23,206 Let's start with the construction for n over 3 331 00:22:23,206 --> 00:22:28,046 and then replace each point with a triangles-- equilateral 332 00:22:28,046 --> 00:22:29,320 triangle. 333 00:22:29,320 --> 00:22:30,070 YUFEI ZHAO: Right. 334 00:22:30,070 --> 00:22:34,290 So the suggestion is to start with some graph 335 00:22:34,290 --> 00:22:37,960 G and then replacing-- 336 00:22:37,960 --> 00:22:44,610 so let's, as an example, look at that one-- 337 00:22:44,610 --> 00:22:47,445 each vertex here by an equilateral triangle. 338 00:22:50,530 --> 00:22:53,060 But I want to maintain the same unit distance. 339 00:22:53,060 --> 00:22:55,108 So the-- uh-huh. 340 00:22:55,108 --> 00:23:00,350 AUDIENCE: So you just choose [INAUDIBLE].. 341 00:23:00,350 --> 00:23:08,290 Or you choose [INAUDIBLE] 342 00:23:08,290 --> 00:23:11,680 YUFEI ZHAO: Is this similar to taking this graph G 343 00:23:11,680 --> 00:23:15,370 and then translating it in two different directions that 344 00:23:15,370 --> 00:23:18,467 form an equilateral triangle? 345 00:23:18,467 --> 00:23:19,300 I think it's-- yeah. 346 00:23:19,300 --> 00:23:20,500 So it's a very similar idea. 347 00:23:20,500 --> 00:23:25,050 And maybe you do a little bit better in terms of constant. 348 00:23:25,050 --> 00:23:26,450 All great suggestions. 349 00:23:26,450 --> 00:23:28,215 Actually, this is really nice. 350 00:23:28,215 --> 00:23:30,292 AUDIENCE: [INAUDIBLE] two times [INAUDIBLE].. 351 00:23:30,292 --> 00:23:31,250 YUFEI ZHAO: Two times-- 352 00:23:31,250 --> 00:23:33,000 you want to make a constant correction? 353 00:23:33,000 --> 00:23:35,674 OK, let me just do that. 354 00:23:35,674 --> 00:23:38,090 [LAUGHTER] 355 00:23:38,090 --> 00:23:41,390 Any more suggestions? 356 00:23:41,390 --> 00:23:41,890 OK. 357 00:23:41,890 --> 00:23:43,420 Yeah, all of this is really nice. 358 00:23:43,420 --> 00:23:47,720 So let me tell you what is the best construction that people 359 00:23:47,720 --> 00:23:51,270 are aware of, and this is A construction due to Erdos. 360 00:23:54,520 --> 00:24:00,410 And the idea is to think big, not build from small examples. 361 00:24:00,410 --> 00:24:08,830 So what Erdos did is to consider a square grid of root n 362 00:24:08,830 --> 00:24:10,090 by root n. 363 00:24:10,090 --> 00:24:12,340 When I see something that's root n that's non-integer, 364 00:24:12,340 --> 00:24:17,260 I just think round down to the nearest integer. 365 00:24:17,260 --> 00:24:18,160 I have a square grid. 366 00:24:22,320 --> 00:24:27,540 Now, if you take these distances as unit distances, 367 00:24:27,540 --> 00:24:29,710 well, you get something which is linear in m. 368 00:24:29,710 --> 00:24:31,660 So you don't gain that much. 369 00:24:31,660 --> 00:24:37,440 But you can take any specific distance as your unit distance. 370 00:24:37,440 --> 00:24:39,990 So what we can do is take a distance 371 00:24:39,990 --> 00:24:44,560 that is represented many times in this grid. 372 00:24:44,560 --> 00:24:58,180 So let's take the unit distance to be a distance root r where 373 00:24:58,180 --> 00:25:05,740 r is some integer that can be represented as a sum of two 374 00:25:05,740 --> 00:25:07,630 squares in many different ways. 375 00:25:27,030 --> 00:25:30,000 So, for example, if we-- 376 00:25:36,340 --> 00:25:41,540 so we can take some r so that it has many appearances. 377 00:25:41,540 --> 00:25:43,840 And if you know some elementary number theory, 378 00:25:43,840 --> 00:25:46,180 then you might know that the best 379 00:25:46,180 --> 00:25:49,660 way to do this is to take r to be a product of primes 380 00:25:49,660 --> 00:25:51,970 that are 1 mod 4. 381 00:25:51,970 --> 00:25:54,000 In any case, you can do this and you 382 00:25:54,000 --> 00:25:55,770 can use some analytic number theory 383 00:25:55,770 --> 00:25:58,920 to calculate if you choose the best possible value of r, 384 00:25:58,920 --> 00:26:00,510 how many distances do you get. 385 00:26:00,510 --> 00:26:03,540 So that calculation was done, and it turns out 386 00:26:03,540 --> 00:26:08,640 you get n raised to the power of 1 plus some constant C 387 00:26:08,640 --> 00:26:12,674 over log log n unit distances. 388 00:26:20,553 --> 00:26:22,220 So this is better than the constructions 389 00:26:22,220 --> 00:26:22,970 we've seen before. 390 00:26:26,730 --> 00:26:28,470 So what can we say about this problem? 391 00:26:28,470 --> 00:26:30,250 So this is a construction. 392 00:26:30,250 --> 00:26:37,530 Well, then you want to understand some upper bounds. 393 00:26:37,530 --> 00:26:41,970 What can we say about the upper bounds to this problems? 394 00:26:41,970 --> 00:26:45,810 And let me show you a fairly easy upper bound, which 395 00:26:45,810 --> 00:26:57,670 can be deduced very quickly from the Kovari-Sos-Turan theorem, 396 00:26:57,670 --> 00:27:08,670 that every set of n points in the plane 397 00:27:08,670 --> 00:27:15,584 has at most on the order off n to the 3/2 unit distances. 398 00:27:23,990 --> 00:27:28,450 So not quite this bound, but it's some bound. 399 00:27:28,450 --> 00:27:30,600 So the trivial upper bound is n choose 2. 400 00:27:30,600 --> 00:27:34,190 So it's much better than the trivial upper bound. 401 00:27:34,190 --> 00:27:36,240 So here's the proof. 402 00:27:36,240 --> 00:27:39,020 So let's consider the unit distance graph, which 403 00:27:39,020 --> 00:27:41,570 is basically the graphs I've been drawing, 404 00:27:41,570 --> 00:27:45,440 where you have the vertices, the points, 405 00:27:45,440 --> 00:27:50,180 and I join an edge between two vertices if 406 00:27:50,180 --> 00:27:53,320 and only if their distance is exactly 1. 407 00:27:56,930 --> 00:28:05,390 I claim this graph is K 2,3 free 408 00:28:05,390 --> 00:28:06,630 Why is that? 409 00:28:06,630 --> 00:28:10,670 Because if you have two points, what 410 00:28:10,670 --> 00:28:12,410 are their common neighbors? 411 00:28:12,410 --> 00:28:14,420 Their common chambers must all be at distance 1 412 00:28:14,420 --> 00:28:17,030 from each of the two points. 413 00:28:17,030 --> 00:28:20,320 So their common neighbors must land 414 00:28:20,320 --> 00:28:22,970 in the intersections of the unit circles centered at these two 415 00:28:22,970 --> 00:28:27,600 points, and they meet at most two points. 416 00:28:27,600 --> 00:28:30,800 So it's K 2,3 free. 417 00:28:30,800 --> 00:28:32,870 Therefore, the Kovari-Sos-Turan theorem 418 00:28:32,870 --> 00:28:39,680 tells us that the number of edges in this G 419 00:28:39,680 --> 00:28:43,838 is upper bounded by n to the 3/2. 420 00:28:43,838 --> 00:28:46,700 And that's it. 421 00:28:46,700 --> 00:28:49,860 That gives you some upper bound. 422 00:28:49,860 --> 00:28:52,940 Any questions? 423 00:28:52,940 --> 00:28:57,060 So it's an application of Kovari-Sos-Turan, 424 00:28:57,060 --> 00:29:01,290 where we design the graph that has to be K 2,3, free 425 00:29:01,290 --> 00:29:04,020 and the extremal number tells us some information 426 00:29:04,020 --> 00:29:06,810 about this graph. 427 00:29:06,810 --> 00:29:09,510 What is the best bound that we know? 428 00:29:09,510 --> 00:29:18,040 So it turns out we do know how to do a little bit better, 429 00:29:18,040 --> 00:29:24,090 but not anywhere close to what we believe to be the truth. 430 00:29:24,090 --> 00:29:28,120 So the current best bound is on the order of n to the 4/3. 431 00:29:28,120 --> 00:29:30,310 And we'll actually see a proof of this later 432 00:29:30,310 --> 00:29:32,890 in the course, once we've developed 433 00:29:32,890 --> 00:29:35,990 something called the crossing numbers inequality. 434 00:29:35,990 --> 00:29:37,730 It's also a very short, very neat proof. 435 00:29:41,310 --> 00:29:43,370 I want to tell you about another problem which 436 00:29:43,370 --> 00:29:48,440 looks superficially similar and also a really 437 00:29:48,440 --> 00:29:49,770 interesting geometric problem. 438 00:29:57,580 --> 00:30:13,270 And this is the Erdos distinct distances problem, 439 00:30:13,270 --> 00:30:21,040 which asks for the maximum number of distinct distances 440 00:30:21,040 --> 00:30:22,750 among n points in the plane. 441 00:30:32,670 --> 00:30:33,670 The minimum-- thank you. 442 00:30:33,670 --> 00:30:35,980 The minimum number of distinct distances. 443 00:30:35,980 --> 00:30:37,390 So maximum would be very easy. 444 00:30:47,920 --> 00:30:50,310 Again, you have lots of examples. 445 00:30:50,310 --> 00:30:53,190 In fact, you can look at basically all the examples 446 00:30:53,190 --> 00:30:55,830 that we've given earlier I mean, these two problems 447 00:30:55,830 --> 00:30:58,680 are very much related to each other. 448 00:30:58,680 --> 00:31:02,790 If you want lots of repeat distances 449 00:31:02,790 --> 00:31:05,770 or if you want very few distinct distances, 450 00:31:05,770 --> 00:31:07,950 whether they seem like they are very much related. 451 00:31:07,950 --> 00:31:10,360 And, of course, you can also write down 452 00:31:10,360 --> 00:31:13,380 the inequality that relates the two of them, 453 00:31:13,380 --> 00:31:20,750 because the number of distinct distances 454 00:31:20,750 --> 00:31:27,780 is at least n choose 2 divided by the maximum number of unit 455 00:31:27,780 --> 00:31:28,868 distances-- 456 00:31:34,016 --> 00:31:36,210 in other words, the number of distances that 457 00:31:36,210 --> 00:31:41,690 can be repeated in any single configuration. 458 00:31:41,690 --> 00:31:45,220 So each of the examples that I just erased-- 459 00:31:45,220 --> 00:31:49,070 so, for example, if you put points on the line, 460 00:31:49,070 --> 00:31:51,440 you have n minus 1 distinct distances. 461 00:31:58,630 --> 00:32:00,090 And we already saw the square grid, 462 00:32:00,090 --> 00:32:02,670 so that might be a good candidate to look at as well. 463 00:32:02,670 --> 00:32:05,160 And there it takes some more number theory 464 00:32:05,160 --> 00:32:09,150 to figure out how many distinct distances there are, 465 00:32:09,150 --> 00:32:11,160 and this roughly corresponds to the number 466 00:32:11,160 --> 00:32:19,910 of integers that can be written as the sum of two squares. 467 00:32:19,910 --> 00:32:24,140 So this has been calculated, and the result 468 00:32:24,140 --> 00:32:29,490 is n divided by square root of log n. 469 00:32:29,490 --> 00:32:30,350 So that's the order. 470 00:32:34,510 --> 00:32:36,760 And Erdos conjectured that this example 471 00:32:36,760 --> 00:32:40,260 should be more or less the best you can do. 472 00:32:40,260 --> 00:32:42,080 So these problems look very similar. 473 00:32:42,080 --> 00:32:44,960 But, actually, this problem here was 474 00:32:44,960 --> 00:32:49,610 solved in a spectacular fashion about a decade ago 475 00:32:49,610 --> 00:32:54,520 by an important paper of Guth and Katz. 476 00:32:59,420 --> 00:33:03,380 So Larry Guth is in our department. 477 00:33:03,380 --> 00:33:14,700 And they showed that every set of n points in the plane 478 00:33:14,700 --> 00:33:19,560 generates at least on the order of n 479 00:33:19,560 --> 00:33:23,546 over log n distinct distances. 480 00:33:33,080 --> 00:33:37,080 So not quite what Erdos conjectured, but nearly there. 481 00:33:37,080 --> 00:33:39,780 And, in fact, all the previous results were much worse. 482 00:33:39,780 --> 00:33:43,670 So they were off in the exponent. 483 00:33:43,670 --> 00:33:47,300 But this one more or less got to the truth. 484 00:33:47,300 --> 00:33:49,640 So this is a very sophisticated result 485 00:33:49,640 --> 00:33:53,060 that used lots of amazing techniques, 486 00:33:53,060 --> 00:33:56,000 ranging from the polynomial method in combinatorics 487 00:33:56,000 --> 00:33:58,630 to some algebraic geometry. 488 00:33:58,630 --> 00:34:03,880 And that's-- I just wanted to mention it for cultural value. 489 00:34:03,880 --> 00:34:04,560 OK. 490 00:34:04,560 --> 00:34:06,210 Any questions? 491 00:34:06,210 --> 00:34:06,710 Yes. 492 00:34:06,710 --> 00:34:08,129 AUDIENCE: For the unit distance problem, 493 00:34:08,129 --> 00:34:09,550 what is believed to be the bound? 494 00:34:09,550 --> 00:34:10,258 YUFEI ZHAO: Yeah. 495 00:34:10,258 --> 00:34:11,840 So for the unit distance problem, 496 00:34:11,840 --> 00:34:14,480 and actually for the distinct distances problem, 497 00:34:14,480 --> 00:34:17,630 the question is, what was is believed to be the bound? 498 00:34:17,630 --> 00:34:19,520 And it's the square grid. 499 00:34:19,520 --> 00:34:21,409 So maybe you can do slightly better, 500 00:34:21,409 --> 00:34:24,230 but it is conjecture that you cannot do much better, 501 00:34:24,230 --> 00:34:26,780 let's say, by more than a constant factor compared 502 00:34:26,780 --> 00:34:29,085 to the square grid. 503 00:34:29,085 --> 00:34:29,730 Yes. 504 00:34:29,730 --> 00:34:32,176 AUDIENCE: What is Katz's first name? 505 00:34:32,176 --> 00:34:32,926 YUFEI ZHAO: Sorry? 506 00:34:32,926 --> 00:34:34,630 AUDIENCE: What is Katz's first name? 507 00:34:34,630 --> 00:34:35,739 YUFEI ZHAO: So this is Nets Katz. 508 00:34:35,739 --> 00:34:36,406 He's at CalTech. 509 00:34:43,300 --> 00:34:46,060 I want to take a short break now. 510 00:34:46,060 --> 00:34:48,429 And when we come back, I want to show you 511 00:34:48,429 --> 00:34:52,360 some ways to generate lower bounds to the extremal number. 512 00:34:56,790 --> 00:35:01,620 In the Kovari-Sos-Turon theorem, we understood some upper bounds 513 00:35:01,620 --> 00:35:03,090 for the extremal number. 514 00:35:03,090 --> 00:35:06,420 So I want to turn our attention now to lower bounds. 515 00:35:06,420 --> 00:35:09,450 Which means constructing, in some sense, 516 00:35:09,450 --> 00:35:14,920 graphs that have lots of edges, at the same time being K 517 00:35:14,920 --> 00:35:15,420 s,t-free. 518 00:35:17,980 --> 00:35:19,650 There are many classes of construction, 519 00:35:19,650 --> 00:35:21,700 so there are several techniques for doing this. 520 00:35:21,700 --> 00:35:24,670 And I want to introduce you to some of these techniques. 521 00:35:24,670 --> 00:35:28,050 And the first one comes from the probabilistic method, 522 00:35:28,050 --> 00:35:35,780 where we use a randomized construction by picking 523 00:35:35,780 --> 00:35:40,400 a graph at random, modifying it a little bit in a way that's 524 00:35:40,400 --> 00:35:41,780 to our desire. 525 00:35:41,780 --> 00:35:43,110 This method is very powerful. 526 00:35:43,110 --> 00:35:43,970 It is very general. 527 00:35:43,970 --> 00:35:46,530 It is applicable in a lot of situations. 528 00:35:46,530 --> 00:35:50,630 But, unfortunately, for the problem of getting tight bound, 529 00:35:50,630 --> 00:35:54,500 it really allows you to get tight bounds. 530 00:35:54,500 --> 00:36:00,200 So it is very robust, but somehow it's often not sharp. 531 00:36:00,200 --> 00:36:05,350 And another class of constructions that 532 00:36:05,350 --> 00:36:06,190 are algebraic-- 533 00:36:11,140 --> 00:36:16,520 all the sharp examples come from algebraic constructions. 534 00:36:16,520 --> 00:36:19,740 And there, we're able to use some nice ideas 535 00:36:19,740 --> 00:36:22,050 from algebra or from algebraic geometry 536 00:36:22,050 --> 00:36:25,308 to get tight constructions. 537 00:36:25,308 --> 00:36:26,850 And you can wonder, is there some way 538 00:36:26,850 --> 00:36:30,260 to combine the best of both worlds? 539 00:36:30,260 --> 00:36:31,380 And it turns out-- 540 00:36:31,380 --> 00:36:33,890 this was an important reason development 541 00:36:33,890 --> 00:36:36,690 that was found just several years ago-- 542 00:36:36,690 --> 00:36:39,800 where one can combine ideas from the two 543 00:36:39,800 --> 00:36:43,007 and obtain a randomized algebraic construction. 544 00:36:52,460 --> 00:36:55,030 And that leads to a new source of constructions 545 00:36:55,030 --> 00:37:00,090 for large H-free graphs with many edges. 546 00:37:00,090 --> 00:37:04,230 So the plan is to show you how these constructions work. 547 00:37:10,270 --> 00:37:12,934 First, let's begin with randomized constructions. 548 00:37:24,550 --> 00:37:26,770 As I mentioned, this construction is very general, 549 00:37:26,770 --> 00:37:30,340 it's very robust, and we can use it to obtain H-free graphs 550 00:37:30,340 --> 00:37:37,120 for every graph H. Of course, when H is not bipartite, 551 00:37:37,120 --> 00:37:41,570 then we saw from the Turon graph that that's pretty much 552 00:37:41,570 --> 00:37:42,920 the right thing to do. 553 00:37:42,920 --> 00:37:46,420 So you should think of bipartite graphs H. So fix a graph 554 00:37:46,420 --> 00:37:50,440 H with at least two edges. 555 00:37:50,440 --> 00:37:53,650 Otherwise, the problem is trivial. 556 00:37:53,650 --> 00:38:00,790 The claim is that there is some constant C such that for every 557 00:38:00,790 --> 00:38:03,520 n-- you should think of and this being large-- 558 00:38:03,520 --> 00:38:13,160 there exists an H-free graph on n vertices 559 00:38:13,160 --> 00:38:16,370 and having lots of edges. 560 00:38:16,370 --> 00:38:21,380 And we will show that you can obtain 561 00:38:21,380 --> 00:38:23,840 the following number of edges. 562 00:38:23,840 --> 00:38:28,130 So 2 minus number of vertices of H minus 2 563 00:38:28,130 --> 00:38:33,294 divided by the number of edges of H minus 1. 564 00:38:33,294 --> 00:38:35,720 Don't worry about this expression in the exponent, 565 00:38:35,720 --> 00:38:38,880 it will come out of the proof. 566 00:38:38,880 --> 00:38:46,100 In other words, the extremal number 567 00:38:46,100 --> 00:38:49,436 is at least this quantity here. 568 00:38:59,910 --> 00:39:02,090 How good is this now? 569 00:39:02,090 --> 00:39:05,050 So it's some scary looking expression in the exponent. 570 00:39:05,050 --> 00:39:09,927 So let me just give you some special cases for comparison 571 00:39:09,927 --> 00:39:11,260 to the Kovari-Sos-Turon theorem. 572 00:39:15,661 --> 00:39:22,200 For K s,t's, the bound, this construction gives you a lower 573 00:39:22,200 --> 00:39:30,510 bound which is of the form m to the 2 minus s plus t minus 2 574 00:39:30,510 --> 00:39:33,860 divided by s,t minus 1. 575 00:39:33,860 --> 00:39:37,320 So I used this symbol, this bigger tilde 576 00:39:37,320 --> 00:39:40,030 to denote dropping constant factors. 577 00:39:42,970 --> 00:39:47,780 In particular, setting s and t to be the same, 578 00:39:47,780 --> 00:40:00,190 we find that the K s,s extremal number is lower-bounded by this 579 00:40:00,190 --> 00:40:02,340 quantity here. 580 00:40:02,340 --> 00:40:05,710 So how does that compare to Kovari-Sos-Turon? 581 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 582 00:40:19,830 --> 00:40:21,510 to the 2 minus 1 over s. 583 00:40:21,510 --> 00:40:27,870 So there's a bit of a gap, and in particular even for 2 and 2. 584 00:40:33,340 --> 00:40:34,930 These results tell you a lower bound 585 00:40:34,930 --> 00:40:38,470 on the order of n to the 4/3 and the upper bound 586 00:40:38,470 --> 00:40:40,095 on the order of 3/2. 587 00:40:40,095 --> 00:40:41,470 So I'm just doing this explicitly 588 00:40:41,470 --> 00:40:45,010 to show you that even in sometimes the simplest case 589 00:40:45,010 --> 00:40:48,160 there is a gap between these two bounds. 590 00:40:48,160 --> 00:40:50,380 Later on, we'll see a construction 591 00:40:50,380 --> 00:40:52,990 that shows that the right-hand side is tight. 592 00:40:52,990 --> 00:40:54,697 So this is the truth. 593 00:40:54,697 --> 00:40:56,530 Randomized construction gives you something, 594 00:40:56,530 --> 00:41:00,390 but it doesn't give you the truth. 595 00:41:00,390 --> 00:41:02,100 On the other hand, I want you to notice 596 00:41:02,100 --> 00:41:09,760 that if t gets very large as a function of s, 597 00:41:09,760 --> 00:41:18,661 this exponent here approaches 1 over s as t goes to infinity. 598 00:41:22,100 --> 00:41:25,690 So for very large values of t, at least in the exponent, 599 00:41:25,690 --> 00:41:27,280 it's not that far off. 600 00:41:27,280 --> 00:41:28,870 Although, you never get the right 601 00:41:28,870 --> 00:41:33,650 exponent for any specific essential. 602 00:41:33,650 --> 00:41:38,310 So this is some limitation of the randomized method, 603 00:41:38,310 --> 00:41:40,070 but it's variable robust. 604 00:41:40,070 --> 00:41:42,430 And, in fact, you can use this result 605 00:41:42,430 --> 00:41:47,960 to bootstrap it to a slightly better one for some graphs H. 606 00:41:47,960 --> 00:41:58,970 So, for example, one might do better 607 00:41:58,970 --> 00:42:04,085 by replacing this H by a subgraph. 608 00:42:07,160 --> 00:42:11,140 Because if you have a subgraph, H prime, 609 00:42:11,140 --> 00:42:13,830 and you construct your G to be H prime-free, 610 00:42:13,830 --> 00:42:16,960 then it is automatically H-free. 611 00:42:16,960 --> 00:42:20,200 But maybe this theorem actually gives you a better construction 612 00:42:20,200 --> 00:42:23,010 when restricted to H prime. 613 00:42:23,010 --> 00:42:24,340 And what can you do better? 614 00:42:24,340 --> 00:42:28,510 Well, you can do better if h prime is such 615 00:42:28,510 --> 00:42:32,270 that whatever the quantity that comes up in the exponent-- 616 00:42:32,270 --> 00:42:33,520 so let me write it like this-- 617 00:42:36,940 --> 00:42:40,150 is superior to the exponent you would 618 00:42:40,150 --> 00:42:47,800 obtain by just looking at H. 619 00:42:47,800 --> 00:42:50,080 So let me give the name to this notion here. 620 00:42:50,080 --> 00:42:53,710 So let me call it the 2-density of the graph 621 00:42:53,710 --> 00:42:58,660 to be whatever quantity here that 622 00:42:58,660 --> 00:43:00,250 would be the maximum if you allow 623 00:43:00,250 --> 00:43:02,050 to pass down to subgraphs. 624 00:43:04,936 --> 00:43:08,880 So the 2-density of a graph H-- 625 00:43:08,880 --> 00:43:13,100 so denoted in literature often as m sub 2-- 626 00:43:13,100 --> 00:43:19,800 to be the maximum where you are allowed to look at subgraphs, 627 00:43:19,800 --> 00:43:21,640 let's say, on at least three vertices 628 00:43:21,640 --> 00:43:25,900 to avoid degeneracies of this ratio that 629 00:43:25,900 --> 00:43:29,425 comes up in the expressions. 630 00:43:34,880 --> 00:43:36,590 So then the theorem-- 631 00:43:39,480 --> 00:43:42,260 that construction there-- implies 632 00:43:42,260 --> 00:43:52,120 that the extremal number is at least not just the expression 633 00:43:52,120 --> 00:43:53,960 I've written, but maybe sometimes you 634 00:43:53,960 --> 00:43:56,870 can do slightly better by passing to a subgraph. 635 00:44:00,170 --> 00:44:01,780 So let me give you a concrete example. 636 00:44:04,430 --> 00:44:15,750 Suppose my H is this graph here. 637 00:44:15,750 --> 00:44:18,030 So you can run the calculation and find 638 00:44:18,030 --> 00:44:19,940 what these numbers are. 639 00:44:19,940 --> 00:44:24,870 So the number of vertices, edges, and this ratio 640 00:44:24,870 --> 00:44:32,870 here to be-- so 5 vertices, 8 edges, and 7/3. 641 00:44:35,620 --> 00:44:38,790 And the idea here is that it's more helpful. 642 00:44:38,790 --> 00:44:42,740 So I want to create something which is H-free, 643 00:44:42,740 --> 00:44:47,550 but dense things are easier to avoid. 644 00:44:47,550 --> 00:44:50,140 So if I have something which has a fairly dense core, 645 00:44:50,140 --> 00:44:51,420 that's easier to avoid. 646 00:44:51,420 --> 00:44:55,360 So maybe it's better to, instead of looking at the whole H, 647 00:44:55,360 --> 00:44:59,560 look at this K 4. 648 00:44:59,560 --> 00:45:02,380 So if you can avoid K 4, of course, you avoid H, 649 00:45:02,380 --> 00:45:04,970 and maybe it's easier to just avoid K 4 650 00:45:04,970 --> 00:45:10,280 and not worry about some of the extra decorations. 651 00:45:10,280 --> 00:45:16,340 So if you look at this H prime and go through the parameters, 652 00:45:16,340 --> 00:45:23,510 you find that it is like that. 653 00:45:23,510 --> 00:45:30,460 And it is somewhat denser in this sense, 654 00:45:30,460 --> 00:45:33,180 compared to looking at the whole graph H. 655 00:45:33,180 --> 00:45:36,250 So you can improve on this theorem for some graphs H 656 00:45:36,250 --> 00:45:37,760 where you can pass to a denser core. 657 00:45:41,290 --> 00:45:42,675 Any questions so far? 658 00:45:42,675 --> 00:45:43,300 Yes. 659 00:45:43,300 --> 00:45:45,383 AUDIENCE: Why is this method called the 2-density? 660 00:45:45,383 --> 00:45:47,425 YUFEI ZHAO: The questions is, why is this measure 661 00:45:47,425 --> 00:45:48,310 called a 2-density? 662 00:45:48,310 --> 00:45:50,260 So that's a name given in literature. 663 00:45:50,260 --> 00:45:53,502 It partly has to do with these extra ratios. 664 00:45:53,502 --> 00:45:55,960 So there are other notions of densities that actually we'll 665 00:45:55,960 --> 00:45:57,350 see later on. 666 00:45:57,350 --> 00:46:00,970 So this term we'll only see today. 667 00:46:00,970 --> 00:46:04,690 So it's more of an ad hoc term for the purpose of this course. 668 00:46:04,690 --> 00:46:06,970 Later on, we'll see notions of density 669 00:46:06,970 --> 00:46:11,710 that are more relevant for our discussions. 670 00:46:11,710 --> 00:46:14,620 Any more questions? 671 00:46:14,620 --> 00:46:16,900 So let me show you how to prove this theorem. 672 00:46:20,820 --> 00:46:22,620 The proof is very intuitive. 673 00:46:22,620 --> 00:46:27,040 The idea is you take something at random and then you fix it. 674 00:46:27,040 --> 00:46:27,540 That's it. 675 00:46:31,410 --> 00:46:33,450 Let's consider a random graph. 676 00:46:36,170 --> 00:46:39,060 The Erdos-Rényi random graph-- so whenever I say random graph, 677 00:46:39,060 --> 00:46:41,720 I almost always refer to this one here. 678 00:46:47,420 --> 00:46:52,930 And the Erdos-Rényi random graph is obtained by considering n 679 00:46:52,930 --> 00:47:00,640 vertices and each possible edge appearing independently 680 00:47:00,640 --> 00:47:03,830 and uniformly with probability p. 681 00:47:03,830 --> 00:47:06,940 And we're going to decide this p later on. 682 00:47:06,940 --> 00:47:11,140 So let me not tell you what p is for now. 683 00:47:11,140 --> 00:47:15,070 I'm interested in avoiding H. So this random graph may 684 00:47:15,070 --> 00:47:17,410 have some copies of H. Let me count 685 00:47:17,410 --> 00:47:27,110 the number of copies of H. 686 00:47:27,110 --> 00:47:34,050 We can compute the expected number of copies of H 687 00:47:34,050 --> 00:47:36,060 by linearity of expectations. 688 00:47:36,060 --> 00:47:38,130 For every possible placement, look 689 00:47:38,130 --> 00:47:43,320 at what's the probability that that placement generates an H. 690 00:47:43,320 --> 00:47:48,120 Namely, look at all different possibilities 691 00:47:48,120 --> 00:47:53,160 for choosing the possible vertices of H. I need 692 00:47:53,160 --> 00:47:55,350 to divide it by a factor that accounts 693 00:47:55,350 --> 00:47:57,580 for the number of automorphisms of H. 694 00:47:57,580 --> 00:48:00,760 But just a constant factor-- don't worry about it. 695 00:48:00,760 --> 00:48:03,210 And for each of these possible placements, 696 00:48:03,210 --> 00:48:06,180 H appears with probability exactly p 697 00:48:06,180 --> 00:48:11,250 to the number of edges of H, just 698 00:48:11,250 --> 00:48:14,520 by linearity of expectations. 699 00:48:14,520 --> 00:48:16,800 And I can upper-bound this quantity 700 00:48:16,800 --> 00:48:20,910 very crudely by e to the number of edges of H times 701 00:48:20,910 --> 00:48:24,990 n, which is the number of vertices of G raised 702 00:48:24,990 --> 00:48:28,770 to the number of vertices of H. 703 00:48:28,770 --> 00:48:31,815 On the other hand, I also want a graph G that has lots of edges, 704 00:48:31,815 --> 00:48:33,440 because that's what we're trying to do. 705 00:48:33,440 --> 00:48:35,190 We're trying to generate a graph with lots 706 00:48:35,190 --> 00:48:36,750 of edges that's H-free. 707 00:48:36,750 --> 00:48:41,580 So the number of edges of G, that's also easy to compute. 708 00:48:41,580 --> 00:48:45,510 It's a binomial distribution, and it has expectation p times 709 00:48:45,510 --> 00:48:48,640 n choose 2. 710 00:48:48,640 --> 00:48:51,380 And, basically, I want this quantity 711 00:48:51,380 --> 00:48:55,140 to be much larger than that quantity. 712 00:48:55,140 --> 00:48:56,850 So I choose an appropriate p. 713 00:49:00,020 --> 00:49:04,325 Namely, by comparing these two quantities, 714 00:49:04,325 --> 00:49:07,110 we can choose p to be, let's say 1/2-- 715 00:49:07,110 --> 00:49:08,930 the 1/2 is not important-- 716 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. 717 00:49:18,830 --> 00:49:22,460 So the exponent comes out of comparing these two 718 00:49:22,460 --> 00:49:24,030 expressions. 719 00:49:24,030 --> 00:49:25,400 So you see a 1 and a 2. 720 00:49:28,050 --> 00:49:34,690 Once you have this p, see that the number which 721 00:49:34,690 --> 00:49:42,260 is the difference-- so take the number of edges of G 722 00:49:42,260 --> 00:49:45,020 minus the number of copies of each. 723 00:49:48,100 --> 00:49:49,930 I know the expectation of both. 724 00:49:49,930 --> 00:49:53,590 I can look at their difference with this value of p. 725 00:49:53,590 --> 00:49:59,500 We find that it is at least 1/2 of the number of edges. 726 00:49:59,500 --> 00:50:06,480 So on expectation, you don't lose too much. 727 00:50:06,480 --> 00:50:09,030 So p is chosen so that this inequality is true. 728 00:50:13,480 --> 00:50:15,170 So you set what p is. 729 00:50:15,170 --> 00:50:19,190 We find that this quantity here is at least some constant times 730 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. 731 00:50:30,390 --> 00:50:33,370 We're still working with a random graph, 732 00:50:33,370 --> 00:50:38,320 and we know that this quantity here is on expectation at least 733 00:50:38,320 --> 00:50:41,290 that number-- so not too small. 734 00:50:41,290 --> 00:50:46,690 Therefore, there exists some instance in this randomness, 735 00:50:46,690 --> 00:50:53,270 some G such that the quantity above for 736 00:50:53,270 --> 00:50:57,200 the specific random instance is at least its expectation. 737 00:51:06,780 --> 00:51:13,420 So this gives us a graph G which on one hand has lots of edges, 738 00:51:13,420 --> 00:51:16,720 but also has very few copies of H 739 00:51:16,720 --> 00:51:19,150 relative to the number of edges. 740 00:51:19,150 --> 00:51:23,530 So we can now get rid of all the copies of H 741 00:51:23,530 --> 00:51:34,700 by removing one edge from each copy of H 742 00:51:34,700 --> 00:51:43,370 in G to remove all copies of H. And now, we 743 00:51:43,370 --> 00:51:45,340 obtain an H-free graph. 744 00:51:50,000 --> 00:51:52,880 How many edges are there in this graph? 745 00:51:52,880 --> 00:51:57,250 Well, we removed at most one edge for each copy of H. 746 00:51:57,250 --> 00:52:11,270 So the number of edges is at least this quantity here, 747 00:52:11,270 --> 00:52:15,160 which is what we wrote just now. 748 00:52:20,470 --> 00:52:21,060 And that's it. 749 00:52:21,060 --> 00:52:24,320 So now we've obtained our graph on n vertices 750 00:52:24,320 --> 00:52:29,030 with lots of edges with the claimed bound that's H-free. 751 00:52:29,030 --> 00:52:31,747 So this is the probabilistic method. 752 00:52:31,747 --> 00:52:33,080 You start with something random. 753 00:52:33,080 --> 00:52:34,140 You try to fix it. 754 00:52:34,140 --> 00:52:36,200 And this method sometimes has the name 755 00:52:36,200 --> 00:52:37,970 of the alteration method. 756 00:52:42,910 --> 00:52:44,950 And this is a very important idea 757 00:52:44,950 --> 00:52:47,830 and one of the key ideas in the probabilistic method, which 758 00:52:47,830 --> 00:52:50,830 I encourage you to go and learn more about. 759 00:52:50,830 --> 00:52:53,440 We'll also see this method later on when 760 00:52:53,440 --> 00:52:56,500 we discuss the randomized algebraic construction. 761 00:52:56,500 --> 00:53:00,340 AUDIENCE: Just to clarify, none of the copies of H [INAUDIBLE].. 762 00:53:03,710 --> 00:53:06,880 YUFEI ZHAO: So the questions is-- yes-- 763 00:53:06,880 --> 00:53:09,460 what do I mean by the number of copies of H? 764 00:53:09,460 --> 00:53:11,500 I mean every instance of H you see. 765 00:53:11,500 --> 00:53:12,910 So there could be intersectings. 766 00:53:12,910 --> 00:53:14,680 I'm not asking for destroying copies. 767 00:53:14,680 --> 00:53:16,740 AUDIENCE: Sure. 768 00:53:16,740 --> 00:53:18,750 YUFEI ZHAO: So a complete graph on n vertices 769 00:53:18,750 --> 00:53:20,655 has n choose 3 triangles. 770 00:53:23,270 --> 00:53:26,470 Any more questions? 771 00:53:26,470 --> 00:53:26,970 OK. 772 00:53:26,970 --> 00:53:29,330 So now we've seen that the probabilistic method 773 00:53:29,330 --> 00:53:31,370 gives you know some bound. 774 00:53:31,370 --> 00:53:34,290 And it's not too hard to apply, but it doesn't give you 775 00:53:34,290 --> 00:53:35,070 the right bound. 776 00:53:35,070 --> 00:53:36,518 It doesn't give you the truth. 777 00:53:36,518 --> 00:53:38,310 So now, I want to show you a different type 778 00:53:38,310 --> 00:53:42,390 of constructions, namely algebraic constructions that 779 00:53:42,390 --> 00:53:45,910 do allow you to get the truth, but they 780 00:53:45,910 --> 00:53:48,740 work in only a small number of cases. 781 00:53:48,740 --> 00:53:50,890 And so it's more magical, but they work 782 00:53:50,890 --> 00:53:52,320 better when the magic happens. 783 00:53:59,330 --> 00:54:02,172 So let's discuss algebraic constructions. 784 00:54:08,450 --> 00:54:11,330 In particular, I want to show you 785 00:54:11,330 --> 00:54:16,250 how to obtain the type bound on the extremal number for K 2,2, 786 00:54:16,250 --> 00:54:17,970 namely a fourth cycle. 787 00:54:17,970 --> 00:54:29,480 So this is a result due to Erdos-Rényi-Sos, 788 00:54:29,480 --> 00:54:37,720 and it tells us that the extremal number for K 2,2 is 789 00:54:37,720 --> 00:54:46,410 at least 1/2 basically up to asymptotics times n to the 3/2. 790 00:54:52,390 --> 00:54:54,700 Actually, if you look at the constant that came out 791 00:54:54,700 --> 00:54:57,170 of the proof of the Kovari-Sos-Turon theorem, 792 00:54:57,170 --> 00:54:59,970 It is also 1/2. 793 00:54:59,970 --> 00:55:14,000 So as a corollary, we see that the extremal number 794 00:55:14,000 --> 00:55:15,200 is like that. 795 00:55:15,200 --> 00:55:17,500 So this is one of extremely few cases 796 00:55:17,500 --> 00:55:21,400 where we know the extremal number so well. 797 00:55:21,400 --> 00:55:24,530 So if you go back to the proof of Kovari-Sos-Turon, 798 00:55:24,530 --> 00:55:26,530 you see that the constant actually there is 1/2. 799 00:55:29,670 --> 00:55:33,960 So I want to construct for you a graph that has no fourth cycles 800 00:55:33,960 --> 00:55:36,670 and has lots of edges. 801 00:55:36,670 --> 00:55:39,030 So that's the name of the game. 802 00:55:39,030 --> 00:55:43,528 And I'll describe this graph for you explicitly. 803 00:55:43,528 --> 00:55:44,570 So this graph has a name. 804 00:55:44,570 --> 00:55:45,778 It's called a polarity graph. 805 00:55:54,300 --> 00:56:03,302 Let's suppose that n is a number such that 1 806 00:56:03,302 --> 00:56:05,260 bigger than this number is a square of a prime. 807 00:56:08,240 --> 00:56:13,870 So our construction will use some finite fields. 808 00:56:13,870 --> 00:56:14,680 I'll explain a bit. 809 00:56:14,680 --> 00:56:16,150 If n is not of this form, then you 810 00:56:16,150 --> 00:56:18,670 can change n to a number of very close to of this form, 811 00:56:18,670 --> 00:56:19,753 and everything will be OK. 812 00:56:22,470 --> 00:56:24,450 The graph will be constructed as follows-- 813 00:56:24,450 --> 00:56:32,410 the vertex set will be the plane over Fp. 814 00:56:32,410 --> 00:56:34,890 Let's remove the origin. 815 00:56:34,890 --> 00:56:37,255 So it has n points exactly. 816 00:56:39,870 --> 00:56:47,250 And the edges are such that I put an edge between x,y 817 00:56:47,250 --> 00:56:54,870 and a,b, if and only if the equation ax plus by equals to 1 818 00:56:54,870 --> 00:56:55,780 holds. 819 00:56:55,780 --> 00:56:58,900 And this equation is meant to be read in Fp. 820 00:57:01,797 --> 00:57:02,630 So that's the graph. 821 00:57:02,630 --> 00:57:05,450 That's an explicit description of this graph. 822 00:57:05,450 --> 00:57:07,200 So I need to show you two things-- 823 00:57:07,200 --> 00:57:11,150 one, that has lots of edges, the claimed number of edges. 824 00:57:11,150 --> 00:57:12,950 And two, it has no fourth cycles. 825 00:57:16,270 --> 00:57:20,590 So let's start with not having fourth cycles. 826 00:57:20,590 --> 00:57:28,020 So y is a K 2,2-free 827 00:57:28,020 --> 00:57:30,150 So what would the K 2,2, be? 828 00:57:30,150 --> 00:57:32,300 So let's consider two points. 829 00:57:41,630 --> 00:57:45,410 And I want to understand the number of common neighbors 830 00:57:45,410 --> 00:57:46,280 of these two points. 831 00:57:54,170 --> 00:57:57,440 Well, look at the description for the edges. 832 00:57:57,440 --> 00:57:59,170 What are the neighbors? 833 00:57:59,170 --> 00:58:01,480 The neighbors correspond to solutions 834 00:58:01,480 --> 00:58:03,498 to this system of equations. 835 00:58:11,810 --> 00:58:14,810 And the basic claim is that there is at most one solution-- 836 00:58:14,810 --> 00:58:15,310 x,y. 837 00:58:28,070 --> 00:58:29,820 So it's basic fact linear algebra. 838 00:58:29,820 --> 00:58:33,210 You have to be slowly careful in case a,b is a multiple 839 00:58:33,210 --> 00:58:34,020 of a prime b prime. 840 00:58:34,020 --> 00:58:36,437 But, actually, in that case, you have no solutions anyway. 841 00:58:41,740 --> 00:58:48,960 The second claim then is that this graph has lots of edges. 842 00:58:48,960 --> 00:58:51,390 Well, actually, that's not too hard to show either. 843 00:58:51,390 --> 00:58:57,090 So I claim that every vertex has degree-- 844 00:58:57,090 --> 00:58:59,640 so how many edges come out of every vertex? 845 00:58:59,640 --> 00:59:03,720 I give you a common b, so how many x comma y 846 00:59:03,720 --> 00:59:07,180 satisfy that equation up there? 847 00:59:07,180 --> 00:59:10,555 Basically, for-- so 1 of x and a and b is non-zero. 848 00:59:13,970 --> 00:59:15,830 So let's say a is non-zero. 849 00:59:15,830 --> 00:59:18,020 Therefore, whatever value of y you set, 850 00:59:18,020 --> 00:59:21,400 I can find a unique x that solves the equation. 851 00:59:21,400 --> 00:59:26,450 I have to be slightly careful, because I don't allow loops 852 00:59:26,450 --> 00:59:28,380 in my graph. 853 00:59:28,380 --> 00:59:31,050 So I might lose one edge because of that. 854 00:59:31,050 --> 00:59:36,365 In any case, every vertex has to agree exactly P or P minus 1. 855 00:59:40,250 --> 00:59:43,320 Just solve that equation in x and y. 856 00:59:48,870 --> 00:59:51,970 So the P minus 1 comes from no loops. 857 00:59:56,770 --> 01:00:06,940 Therefore, the number of edges is equal to the claimed bound. 858 01:00:19,650 --> 01:00:21,410 So this finishes the proof in case 859 01:00:21,410 --> 01:00:25,520 when n has that special number theoretic form. 860 01:00:25,520 --> 01:00:35,000 But we can extend to all values of n like this-- 861 01:00:35,000 --> 01:00:38,540 if n doesn't have that form, then I 862 01:00:38,540 --> 01:00:48,230 can take a prime P. It may not necessarily 863 01:00:48,230 --> 01:00:52,310 be exactly satisfying that inequality, 864 01:00:52,310 --> 01:00:55,070 but I can always take a prime pretty close to it. 865 01:00:55,070 --> 01:00:57,050 So I can always take a prime which 866 01:00:57,050 --> 01:01:04,020 is up to a negligible multiplicative 867 01:01:04,020 --> 01:01:05,490 error what I want. 868 01:01:08,440 --> 01:01:10,765 And then we use-- 869 01:01:10,765 --> 01:01:20,410 and such that P squared minus 1 is at most n. 870 01:01:20,410 --> 01:01:31,000 And use the above construction and add isolated vertices 871 01:01:31,000 --> 01:01:34,828 to finish the job to get exactly n vertices. 872 01:01:34,828 --> 01:01:36,370 And the reason that I can always take 873 01:01:36,370 --> 01:01:38,410 a prime very close to it is because there's 874 01:01:38,410 --> 01:01:42,070 a theorem in number theory that tells us that for n 875 01:01:42,070 --> 01:01:45,370 large enough, I can always find the prime 876 01:01:45,370 --> 01:01:48,670 which is slightly less than n but no more 877 01:01:48,670 --> 01:01:51,790 than a negligible multiplicative factor of n. 878 01:01:51,790 --> 01:01:54,722 The best result of this form is-- 879 01:01:54,722 --> 01:01:56,680 I'm just telling you something in number theory 880 01:01:56,680 --> 01:01:59,290 for cultural reasons-- 881 01:01:59,290 --> 01:02:03,020 due to Baker-Harman-Pintz. 882 01:02:03,020 --> 01:02:05,760 And so, this is the question regarding how large 883 01:02:05,760 --> 01:02:07,720 can gaps between primes be. 884 01:02:07,720 --> 01:02:10,180 So you might know the [Bertrand's postulate]] theorem 885 01:02:10,180 --> 01:02:13,800 that tells you there is always a prime between n and 2n. 886 01:02:13,800 --> 01:02:17,440 So what about between n and n plus root n. 887 01:02:17,440 --> 01:02:18,700 Actually, we don't know that. 888 01:02:18,700 --> 01:02:20,530 So the best result of the form is 889 01:02:20,530 --> 01:02:22,355 that there is always a prime-- 890 01:02:22,355 --> 01:02:27,020 so for n sufficiently large there 891 01:02:27,020 --> 01:02:46,865 exists a prime between n minus n to the exponent 0.525 and n. 892 01:02:46,865 --> 01:02:50,480 In any case, this number here, whatever it is is little n, 893 01:02:50,480 --> 01:02:53,150 and that's enough for our purpose. 894 01:02:55,730 --> 01:03:00,862 So it suffices to look at n of a special number theoretic form 895 01:03:00,862 --> 01:03:02,320 where you're allowed to use primes. 896 01:03:05,000 --> 01:03:06,800 So that's the construction there. 897 01:03:06,800 --> 01:03:09,710 Let me show you a interpretation of that construction which 898 01:03:09,710 --> 01:03:14,730 I think is may be helpful to think about, 899 01:03:14,730 --> 01:03:18,830 and that's that you can view it as the incidence 900 01:03:18,830 --> 01:03:22,760 graph between points and lines in projective space-- 901 01:03:22,760 --> 01:03:24,500 in projective plane. 902 01:03:24,500 --> 01:03:26,690 So I start with a projective plane. 903 01:03:33,640 --> 01:03:36,800 So I can view a bipartite version of that construction. 904 01:03:49,060 --> 01:04:02,870 It can be viewed as the point-line incidence 905 01:04:02,870 --> 01:04:13,930 graph of a projective plane over a finite field. 906 01:04:16,730 --> 01:04:25,980 And by this, I mean put as one vertex set the points 907 01:04:25,980 --> 01:04:29,565 of the projective plane and on the other side the lines. 908 01:04:32,370 --> 01:04:39,780 And I put in an edge between a point and a line if and only 909 01:04:39,780 --> 01:04:44,300 if the point lies on the line. 910 01:04:44,300 --> 01:04:48,290 So you can do this more explicitly in coordinates 911 01:04:48,290 --> 01:04:54,710 if you view points and lines as coordinates. 912 01:04:54,710 --> 01:05:00,940 And so the equation for getting a point to be on the line 913 01:05:00,940 --> 01:05:02,850 is like that. 914 01:05:02,850 --> 01:05:05,000 So now why is there no fourth cycle? 915 01:05:07,700 --> 01:05:12,670 A fourth cycle would correspond to two points lying 916 01:05:12,670 --> 01:05:20,050 on two different lines, which is not possible in this geometry. 917 01:05:20,050 --> 01:05:22,590 So that's the reason for that construction up there. 918 01:05:26,490 --> 01:05:29,750 So no two points in two lines. 919 01:05:34,050 --> 01:05:37,921 Any questions about this polarity constructions? 920 01:05:44,174 --> 01:05:46,299 AUDIENCE: Why is it called a polarity construction. 921 01:05:46,299 --> 01:05:47,882 YUFEI ZHAO: The question is, why is it 922 01:05:47,882 --> 01:05:49,410 called a polarity construction? 923 01:05:49,410 --> 01:05:57,170 So it relates points and their polars, which are lines. 924 01:05:57,170 --> 01:05:58,680 Yeah. 925 01:05:58,680 --> 01:06:02,337 AUDIENCE: Why does this not have-- like, on your two P 926 01:06:02,337 --> 01:06:05,283 squared vertices, looking at one vertex for every [INAUDIBLE],, 927 01:06:05,283 --> 01:06:06,520 one vertex for [INAUDIBLE]? 928 01:06:06,520 --> 01:06:07,145 YUFEI ZHAO: OK. 929 01:06:07,145 --> 01:06:10,300 So the question has to do with the number of vertices here. 930 01:06:10,300 --> 01:06:10,800 It's true. 931 01:06:10,800 --> 01:06:13,540 Here, I double the number of vertices, 932 01:06:13,540 --> 01:06:15,660 and so I don't get that constant there. 933 01:06:15,660 --> 01:06:18,600 But what that graph up there-- that's not a bipartite graph. 934 01:06:18,600 --> 01:06:20,610 It is identifying the points and the lines 935 01:06:20,610 --> 01:06:24,130 and overlaying the two parts into one. 936 01:06:24,130 --> 01:06:25,880 But if you don't care about the constants, 937 01:06:25,880 --> 01:06:27,710 this graph here may be conceptually easier 938 01:06:27,710 --> 01:06:29,090 to think about. 939 01:06:29,090 --> 01:06:29,660 Yes. 940 01:06:29,660 --> 01:06:31,993 AUDIENCE: [INAUDIBLE] to generalize the polarities bound 941 01:06:31,993 --> 01:06:33,690 to K [INAUDIBLE]. 942 01:06:33,690 --> 01:06:34,440 YUFEI ZHAO: Great. 943 01:06:34,440 --> 01:06:36,732 The question is, can you generalize this polarity graph 944 01:06:36,732 --> 01:06:37,837 to K 3,3 and higher? 945 01:06:37,837 --> 01:06:39,420 So that's what we're about to do next. 946 01:06:42,320 --> 01:06:45,350 So for K 3,3, what can you do? 947 01:06:45,350 --> 01:06:49,670 So the main observation here is that two lines 948 01:06:49,670 --> 01:06:51,830 intersecting at most one point. 949 01:06:51,830 --> 01:06:55,170 But there are other geometric facts of that form. 950 01:06:55,170 --> 01:06:59,870 So we're going to use one of them to get K 3,3-free graph. 951 01:06:59,870 --> 01:07:07,570 And this construction is due to Brown, 952 01:07:07,570 --> 01:07:13,270 that the extremal number for K 3,3 953 01:07:13,270 --> 01:07:21,550 is also at least a factor 1/2 minus total 1 times-- 954 01:07:21,550 --> 01:07:23,590 so now, what's the exponent? 955 01:07:23,590 --> 01:07:27,400 What is predicted by Kovari-Sos-Turon 956 01:07:27,400 --> 01:07:33,100 is 2 minus 1/3, and Brown obtains the correct exponent. 957 01:07:36,320 --> 01:07:38,760 It turns out this is also the right constant, 958 01:07:38,760 --> 01:07:41,100 this 1/2, although it doesn't follow 959 01:07:41,100 --> 01:07:44,760 from the Kovari-Sos-Turon theorem I stated. 960 01:07:44,760 --> 01:07:46,570 One needs to do a little bit extra work. 961 01:07:46,570 --> 01:07:48,445 But it turns out it is true that this is also 962 01:07:48,445 --> 01:07:49,640 the correct constant. 963 01:07:49,640 --> 01:07:52,200 And that's actually pretty much all the cases where 964 01:07:52,200 --> 01:07:54,000 we know the correct constant. 965 01:07:54,000 --> 01:07:57,210 And there are other cases where we know the correct exponent, 966 01:07:57,210 --> 01:08:01,442 but these things tend to be hard to come by. 967 01:08:01,442 --> 01:08:03,400 So let me show you how to construct this graph. 968 01:08:03,400 --> 01:08:07,130 So it's based on a similar idea as the polarity graph. 969 01:08:07,130 --> 01:08:08,680 It has some more technicalities. 970 01:08:08,680 --> 01:08:10,802 So I'm not going to do the full proof 971 01:08:10,802 --> 01:08:12,010 and just give you the sketch. 972 01:08:16,359 --> 01:08:19,130 As earlier, I'm using the same trick. 973 01:08:19,130 --> 01:08:22,460 We can assume that n has a special form. 974 01:08:22,460 --> 01:08:25,430 Here, let me assume the n is a cube of a prime. 975 01:08:32,700 --> 01:08:36,100 I'm going to put s edges. 976 01:08:36,100 --> 01:08:38,760 So first, the vertices of my graph 977 01:08:38,760 --> 01:08:44,930 are going to be points in the affine plane over Fp. 978 01:08:47,899 --> 01:08:53,330 Previously, the edges had to do with lines. 979 01:08:53,330 --> 01:08:57,270 And now, let's use spheres. 980 01:08:57,270 --> 01:09:08,600 So the edges of the form where I join two vertices like this 981 01:09:08,600 --> 01:09:11,910 if and only if they're-- 982 01:09:11,910 --> 01:09:13,910 well, it's not really a distance, 983 01:09:13,910 --> 01:09:16,670 but it's something that looks like the equation 984 01:09:16,670 --> 01:09:35,620 of a sphere, where u is some fixed non-zero element of Fp. 985 01:09:35,620 --> 01:09:38,319 You may have to be somewhat careful in choosing this u, 986 01:09:38,319 --> 01:09:41,250 but let me not worry too much about it. 987 01:09:41,250 --> 01:09:45,550 So you fixed some so-called distance, 988 01:09:45,550 --> 01:09:52,770 even though it's not a distance, and I join the vertices 989 01:09:52,770 --> 01:09:56,665 whenever they satisfy that equation having that not 990 01:09:56,665 --> 01:09:57,165 distance. 991 01:09:59,750 --> 01:10:01,750 What's the intuition here? 992 01:10:01,750 --> 01:10:04,200 The intuition is that I want to avoid-- so 993 01:10:04,200 --> 01:10:07,723 how do I know that this graph has no K 3,3? 994 01:10:07,723 --> 01:10:10,140 Well, first, let's think about what happens in real space. 995 01:10:16,280 --> 01:10:21,950 So intuition in the real space-- 996 01:10:21,950 --> 01:10:25,710 well, here, I have this graph that, let's say, 997 01:10:25,710 --> 01:10:29,690 the unit distance graph in r3. 998 01:10:29,690 --> 01:10:33,580 So the neighborhood of each point is a unit sphere. 999 01:10:33,580 --> 01:10:37,040 And what I want to know is that if you have three spheres, 1000 01:10:37,040 --> 01:10:39,950 three unit spheres, how many common intersection points 1001 01:10:39,950 --> 01:10:40,660 can they have? 1002 01:10:43,300 --> 01:10:47,350 Two spheres intersecting a circle. 1003 01:10:47,350 --> 01:10:51,120 And that circle cannot lie on the third circle. 1004 01:10:51,120 --> 01:10:54,090 That you should think about. 1005 01:10:54,090 --> 01:10:56,220 So that circle intersects the third circle 1006 01:10:56,220 --> 01:10:59,370 in at most two points. 1007 01:10:59,370 --> 01:11:07,230 So three unit spheres have at most two common points. 1008 01:11:11,950 --> 01:11:15,460 And so the unit distance graph in r3 is K 3,3-free. 1009 01:11:18,320 --> 01:11:20,720 That entire argument, even though I expressed it 1010 01:11:20,720 --> 01:11:23,660 geometrically, it's an algebraic argument. 1011 01:11:23,660 --> 01:11:26,690 You can write down equations on intersections 1012 01:11:26,690 --> 01:11:28,310 between two spheres. 1013 01:11:28,310 --> 01:11:32,985 It's the intersection of the sphere with its coaxial plane. 1014 01:11:32,985 --> 01:11:34,730 I have a couple of these colossal planes. 1015 01:11:34,730 --> 01:11:36,740 They get me a coaxial line. 1016 01:11:36,740 --> 01:11:41,162 That line has to intersect the sphere in almost two points. 1017 01:11:41,162 --> 01:11:42,620 You should actually do this algebra 1018 01:11:42,620 --> 01:11:43,860 if you want to do the proof, because there 1019 01:11:43,860 --> 01:11:45,950 are funny things that can happen you find fields. 1020 01:11:45,950 --> 01:11:49,500 For example, maybe the sphere contains a line. 1021 01:11:49,500 --> 01:11:51,880 But you choose your parameters correctly, 1022 01:11:51,880 --> 01:11:53,130 and these things don't happen. 1023 01:11:56,340 --> 01:11:58,233 And that's the intuition. 1024 01:11:58,233 --> 01:11:59,650 And if you actually work this out, 1025 01:11:59,650 --> 01:12:02,590 you'll find that this graph here is indeed K 3,3-free. 1026 01:12:09,030 --> 01:12:10,770 So I'm skipping the details. 1027 01:12:10,770 --> 01:12:15,360 But you should do the algebra if you want to have a proof. 1028 01:12:15,360 --> 01:12:17,550 On the other hand, it also has lots of edges. 1029 01:12:17,550 --> 01:12:19,620 And that's basically the same reason as before. 1030 01:12:19,620 --> 01:12:22,830 But I can count the number of edges 1031 01:12:22,830 --> 01:12:31,020 by fixing some x, y, and z, and look at how many abc's 1032 01:12:31,020 --> 01:12:34,260 satisfy that equation. 1033 01:12:34,260 --> 01:12:38,190 And that's, again, something that needs to be checked, 1034 01:12:38,190 --> 01:12:41,090 but the point is that this graph-- 1035 01:12:41,090 --> 01:12:49,490 so every vertex has either a P square-- 1036 01:12:49,490 --> 01:12:53,865 so it has close to P squared degree. 1037 01:13:00,180 --> 01:13:05,150 So lots of edges, and combining with basically the same idea as 1038 01:13:05,150 --> 01:13:07,495 before, you get the construction. 1039 01:13:10,350 --> 01:13:11,452 Any questions? 1040 01:13:15,562 --> 01:13:16,770 So where can we go from here? 1041 01:13:19,620 --> 01:13:21,090 To construct the K 2,2-- 1042 01:13:21,090 --> 01:13:23,950 by the way, if you construct K 2,2-free, 1043 01:13:23,950 --> 01:13:26,815 the same construction works for K 2,3-free. 1044 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. 1045 01:13:31,670 --> 01:13:35,110 Here, likewise, this is also K 3,4-free. 1046 01:13:35,110 --> 01:13:38,080 So now, what about higher K s,t's? 1047 01:13:38,080 --> 01:13:41,950 And you might think, well, let's take these geometric objects 1048 01:13:41,950 --> 01:13:44,480 and try to extend them further. 1049 01:13:44,480 --> 01:13:46,230 But that actually seems kind of difficult. 1050 01:13:46,230 --> 01:13:49,020 We do not really know how to do it. 1051 01:13:49,020 --> 01:13:54,510 We do not know how to obtain a construction which 1052 01:13:54,510 --> 01:13:58,140 is of this form that works for K 4,4. 1053 01:13:58,140 --> 01:14:00,080 In fact, there are even some evidence 1054 01:14:00,080 --> 01:14:02,151 that that might be even impossible. 1055 01:14:04,980 --> 01:14:09,240 As I mentioned, K 4,4 is a major open problem. 1056 01:14:14,980 --> 01:14:16,810 It is an open problem to determine 1057 01:14:16,810 --> 01:14:27,970 the order of the extremal number of K 4,4. 1058 01:14:27,970 --> 01:14:31,790 But in any case, this construction, 1059 01:14:31,790 --> 01:14:34,490 this idea of using algebraic constructions, 1060 01:14:34,490 --> 01:14:39,830 is very enlightening, that we should look at ways to get 1061 01:14:39,830 --> 01:14:43,940 large K s,t-free graphs by coming up with clever algebraic 1062 01:14:43,940 --> 01:14:44,965 constructions. 1063 01:14:44,965 --> 01:14:49,820 And next, time I will show you a couple of very nice ideas 1064 01:14:49,820 --> 01:14:52,130 where you can get-- 1065 01:14:52,130 --> 01:14:55,940 you can come up with a different kind of algebraic construction 1066 01:14:55,940 --> 01:14:58,580 which has some superficial similarities to what we've 1067 01:14:58,580 --> 01:15:04,060 seen today but that's really of a different nature. 1068 01:15:04,060 --> 01:15:10,150 So, next time, we will see the following theorem, 1069 01:15:10,150 --> 01:15:13,600 which is obtained in a sequence of two papers, 1070 01:15:13,600 --> 01:15:24,649 union of authors, Alon, Rényi and Sazbó, 1071 01:15:24,649 --> 01:15:30,300 that shows that if t is much larger than s-- 1072 01:15:33,964 --> 01:15:38,340 so minus 1 factorial plus 1-- 1073 01:15:38,340 --> 01:15:48,060 then the extremal number for K s,t is on the same order 1074 01:15:48,060 --> 01:15:51,990 as the upper bound determined in Kovari-Sos-Turon theorem. 1075 01:15:56,880 --> 01:16:01,650 Just to be more explicit about what these s and t are, 1076 01:16:01,650 --> 01:16:04,890 if you plug in what's the smallest 1077 01:16:04,890 --> 01:16:07,980 t that this theorem gives for various values of s, 1078 01:16:07,980 --> 01:16:10,980 find 2,2, and 3,3. 1079 01:16:10,980 --> 01:16:15,070 And the next one is 4, 7. 1080 01:16:15,070 --> 01:16:16,870 And then, it gets worse from there. 1081 01:16:26,170 --> 01:16:29,080 So these constructions are based on-- 1082 01:16:29,080 --> 01:16:31,240 I mean, they are algebraic constructions. 1083 01:16:31,240 --> 01:16:34,830 So we'll see next time what happens.