1 00:00:19,088 --> 00:00:20,630 YUFEI ZHAO: So the first topic that I 2 00:00:20,630 --> 00:00:25,280 want to discuss in this course is extremal graph theory. 3 00:00:38,980 --> 00:00:42,220 And in particular, there is a whole class 4 00:00:42,220 --> 00:00:46,690 of problems which have to do with what happens if you forbid 5 00:00:46,690 --> 00:00:47,750 a specific subgraph. 6 00:00:54,780 --> 00:00:56,400 Forbid a specific subgraph. 7 00:00:56,400 --> 00:00:59,400 And I ask you, what's the maximum number of edges 8 00:00:59,400 --> 00:01:03,330 that can appear in your graph? 9 00:01:03,330 --> 00:01:05,129 In particular, and this is the question 10 00:01:05,129 --> 00:01:08,700 that we saw at the end of last lecture, 11 00:01:08,700 --> 00:01:12,070 which, now we're going to pretend that's a theorem. 12 00:01:12,070 --> 00:01:18,510 Mantel's theorem essentially asks, 13 00:01:18,510 --> 00:01:21,720 if you know that your graph has no triangles, what's 14 00:01:21,720 --> 00:01:25,770 the maximum number of edges it can have? 15 00:01:25,770 --> 00:01:29,880 And Mantel's theorem tells us that the extremal example 16 00:01:29,880 --> 00:01:35,940 is when your graph consists of putting 17 00:01:35,940 --> 00:01:38,580 half the vertices on one side, half 18 00:01:38,580 --> 00:01:41,220 the vertices on the other side, and putting 19 00:01:41,220 --> 00:01:49,350 in all the edges between the two sides. 20 00:01:49,350 --> 00:01:53,490 So this is a complete bipartite graph. 21 00:01:53,490 --> 00:01:58,920 For these partitions, we denote complete bipartite graphs 22 00:01:58,920 --> 00:02:00,980 like that. 23 00:02:00,980 --> 00:02:03,870 And Mantel's theorem tells us that this graph, 24 00:02:03,870 --> 00:02:08,820 among triangle-free graphs, has the most number of edges. 25 00:02:08,820 --> 00:02:21,740 Triangle-free graph, n vertices, has at most-- 26 00:02:21,740 --> 00:02:26,745 so the number of edges there is n squared divided by 4. 27 00:02:26,745 --> 00:02:32,340 Round down-- that many edges. 28 00:02:32,340 --> 00:02:35,800 And from this example, this bound is tight. 29 00:02:35,800 --> 00:02:37,260 So Mantel's theorem them gives us 30 00:02:37,260 --> 00:02:40,620 a completely satisfactory answer to the question of, 31 00:02:40,620 --> 00:02:42,630 what's the maximum number of edges 32 00:02:42,630 --> 00:02:44,755 in a graph without triangles? 33 00:02:47,870 --> 00:02:51,500 And I want to begin by showing you a few different proofs 34 00:02:51,500 --> 00:02:52,760 of Mantel's theorem. 35 00:02:52,760 --> 00:02:55,430 So we're illustrating some different techniques 36 00:02:55,430 --> 00:02:57,870 in graph theory. 37 00:02:57,870 --> 00:03:00,890 So we'll see quite a few proofs in today's lecture. 38 00:03:00,890 --> 00:03:04,010 The first one begins-- 39 00:03:04,010 --> 00:03:05,990 well, here's the setup. 40 00:03:05,990 --> 00:03:09,660 I have G, an n vertex graph. 41 00:03:09,660 --> 00:03:17,190 And let me denote the vertices and edges of G by V and E. 42 00:03:17,190 --> 00:03:23,040 If I have an edge in G between vertices x and y, 43 00:03:23,040 --> 00:03:26,922 then note that they cannot have any common neighbors. 44 00:03:35,300 --> 00:03:38,040 Because if they did, I would see a triangle. 45 00:03:38,040 --> 00:03:40,039 So I assume G is triangle-free. 46 00:03:47,380 --> 00:03:51,190 So what can we say about the degrees of the two 47 00:03:51,190 --> 00:03:53,830 endpoints of this edge? 48 00:03:53,830 --> 00:03:55,746 Well, they cannot add up to? 49 00:03:55,746 --> 00:03:57,783 AUDIENCE: They can't have more than n. 50 00:03:57,783 --> 00:03:59,700 YUFEI ZHAO: They cannot add up to more than n. 51 00:03:59,700 --> 00:04:02,750 So in particular-- so exactly. 52 00:04:02,750 --> 00:04:06,050 So the degrees of these two endpoints, 53 00:04:06,050 --> 00:04:12,450 there's at most n whenever xy is an edge. 54 00:04:12,450 --> 00:04:14,240 So here I'm using d to denote degree. 55 00:04:19,450 --> 00:04:19,950 Well, OK. 56 00:04:19,950 --> 00:04:24,060 So now let me consider the quantity which is 57 00:04:24,060 --> 00:04:25,995 the sum of the squared degrees. 58 00:04:30,140 --> 00:04:37,400 On one hand, I claim that the sum is equal to this quantity 59 00:04:37,400 --> 00:04:40,280 here, where I sum over all edges. 60 00:04:43,300 --> 00:04:45,580 And the reason is, well, look at the sum. 61 00:04:45,580 --> 00:04:48,010 Suppose, imagine writing out all the sum n's. 62 00:04:48,010 --> 00:04:51,632 How many times does each dx come up? 63 00:04:51,632 --> 00:04:59,800 So each dx appears one for each edge x is in, 64 00:04:59,800 --> 00:05:01,820 so appears exactly dx times. 65 00:05:09,710 --> 00:05:14,120 But we saw from up here that each sum n is at most n. 66 00:05:14,120 --> 00:05:18,710 So this sum here is that most mn. 67 00:05:21,950 --> 00:05:25,010 On the other hand, let's consider the quantity which 68 00:05:25,010 --> 00:05:26,465 is just the sum of the degrees. 69 00:05:29,690 --> 00:05:32,950 And you sometimes know it as the handshaking lemma, 70 00:05:32,950 --> 00:05:37,210 that the sums of the degrees is just twice the number of edges. 71 00:05:37,210 --> 00:05:40,840 Each edge is considered twice in the sum. 72 00:05:40,840 --> 00:05:44,130 Well, now we apply the Cauchy-Schwarz inequality, 73 00:05:44,130 --> 00:05:47,980 which, as you'll see many times in this course, 74 00:05:47,980 --> 00:05:52,080 although you might think of it as a fairly simple inequality, 75 00:05:52,080 --> 00:05:54,240 it's extremely powerful. 76 00:05:54,240 --> 00:05:57,925 And it will come up pretty much throughout this course. 77 00:05:57,925 --> 00:05:59,300 By the Cauchy-Schwarz inequality, 78 00:05:59,300 --> 00:06:01,020 you compare these two quantities. 79 00:06:01,020 --> 00:06:15,130 I find that we have this inequality over here relating 80 00:06:15,130 --> 00:06:18,490 the sums of the degrees and square 81 00:06:18,490 --> 00:06:21,300 of the sum of the degrees. 82 00:06:21,300 --> 00:06:30,170 But we saw, on one hand, the left-hand side is 4m squared. 83 00:06:30,170 --> 00:06:34,370 And we also saw that the right-hand side 84 00:06:34,370 --> 00:06:39,870 is at most mn squared. 85 00:06:39,870 --> 00:06:44,850 Putting them together, we see that m is at most n squared 86 00:06:44,850 --> 00:06:45,540 over 4. 87 00:06:45,540 --> 00:06:47,510 And of course, because it's an integer, 88 00:06:47,510 --> 00:06:50,920 it's at most the floor of this number. 89 00:06:50,920 --> 00:06:54,400 And that's a proof of Mantel's theorem. 90 00:07:01,550 --> 00:07:05,090 What can you tell me about the equality case in this proof? 91 00:07:11,660 --> 00:07:13,220 So I'll let you think about that. 92 00:07:13,220 --> 00:07:14,930 But let me show you some other proofs. 93 00:07:19,910 --> 00:07:22,970 In other words, are there graphs with the same number 94 00:07:22,970 --> 00:07:25,820 of edges as the graph shown up there that 95 00:07:25,820 --> 00:07:26,840 is also triangle-free? 96 00:07:26,840 --> 00:07:27,980 Is that a unique example? 97 00:07:31,465 --> 00:07:33,840 So let me show you a different proof of Mantel's theorem. 98 00:07:38,400 --> 00:07:40,260 In this proof, so we begin with a step 99 00:07:40,260 --> 00:07:43,090 that seems a little tricky. 100 00:07:43,090 --> 00:07:48,960 Let's let A be a subset of vertices such 101 00:07:48,960 --> 00:08:01,426 that A is a largest independent set of G. 102 00:08:01,426 --> 00:08:04,010 So remember, independent set is a subset of vertices 103 00:08:04,010 --> 00:08:06,890 with no edges inside. 104 00:08:06,890 --> 00:08:09,770 It may have many independent sets all having 105 00:08:09,770 --> 00:08:11,360 the same maximum size. 106 00:08:11,360 --> 00:08:14,120 Take one of them. 107 00:08:14,120 --> 00:08:15,930 So why should you do this step? 108 00:08:15,930 --> 00:08:17,650 Well, you know, sometimes magic happens. 109 00:08:20,340 --> 00:08:25,280 Let's consider some vertex. 110 00:08:25,280 --> 00:08:28,440 So consider some vertex, let's say x, 111 00:08:28,440 --> 00:08:32,809 and look at its neighborhood. 112 00:08:32,809 --> 00:08:36,400 The neighborhood must be an independent set. 113 00:08:36,400 --> 00:08:39,870 Otherwise, I get a triangle. 114 00:08:39,870 --> 00:08:46,780 So every neighborhood is an independent set. 115 00:08:49,910 --> 00:08:53,740 And as a result, the degree of every vertex 116 00:08:53,740 --> 00:08:58,010 is at most the size of the largest independent set. 117 00:09:03,290 --> 00:09:23,950 So now, let B be the complement of A. So I have this A and B. 118 00:09:23,950 --> 00:09:28,180 Every edge has to intersect B. The edge cannot be entirely 119 00:09:28,180 --> 00:09:30,010 containing A, because it has no edges. 120 00:09:39,870 --> 00:09:48,090 So the number of edges of G is at most-- 121 00:09:48,090 --> 00:09:55,140 well, I count over the vertices in B, the degree. 122 00:09:55,140 --> 00:09:56,690 So maybe I overcount. 123 00:09:56,690 --> 00:10:00,020 So the edges containing B, I count twice, but that's OK. 124 00:10:00,020 --> 00:10:03,550 So this is an upper bound on the number of edges. 125 00:10:03,550 --> 00:10:06,920 But I also know that every vertex in the graph 126 00:10:06,920 --> 00:10:10,490 has a degree at most the size of A. 127 00:10:10,490 --> 00:10:15,270 So each sum n is at most the size of A. And I have B terms. 128 00:10:17,810 --> 00:10:19,440 So I have that. 129 00:10:19,440 --> 00:10:31,240 Now, by the AMGM inequality, you'll have that. 130 00:10:31,240 --> 00:10:36,410 And the sizes of A and B add up to the entire vertex 131 00:10:36,410 --> 00:10:42,322 set, n squared over 4. 132 00:10:42,322 --> 00:10:46,180 So that gives you another proof of Mantel's theorem. 133 00:10:48,910 --> 00:10:51,940 What does this proof tell us about the equality case? 134 00:10:55,876 --> 00:10:58,710 Now, something I always want you to keep in mind when 135 00:10:58,710 --> 00:11:00,870 we do proofs is, especially if we 136 00:11:00,870 --> 00:11:03,660 have a tight example like that-- and later on in the course, 137 00:11:03,660 --> 00:11:05,970 we'll almost never have good examples like that. 138 00:11:05,970 --> 00:11:08,100 So this is still early on in the course. 139 00:11:08,100 --> 00:11:10,530 And we're still very clean in the examples-- is to keep 140 00:11:10,530 --> 00:11:12,960 the extremal example in mind. 141 00:11:12,960 --> 00:11:16,140 And every step in your proof, it should 142 00:11:16,140 --> 00:11:18,180 be tight for that example. 143 00:11:18,180 --> 00:11:21,050 Otherwise, something went wrong with your proof. 144 00:11:21,050 --> 00:11:23,860 Let's think about the inequalities. 145 00:11:23,860 --> 00:11:33,690 At equality, we must have that-- 146 00:11:33,690 --> 00:11:38,430 so looking at this inequality. 147 00:11:38,430 --> 00:11:49,595 So there are no edges in B. We also have that-- 148 00:11:52,385 --> 00:12:07,290 so by that, so every vertex in B is complete to A. And finally, 149 00:12:07,290 --> 00:12:09,180 A and B should have the same size. 150 00:12:12,880 --> 00:12:15,270 So it is exactly the configuration shown up there. 151 00:12:15,270 --> 00:12:18,790 Now, when n is odd, we're rounding down by 1. 152 00:12:18,790 --> 00:12:20,580 So you can lose a little bit. 153 00:12:20,580 --> 00:12:22,870 But you can check that actually what 154 00:12:22,870 --> 00:12:25,420 we described up there is also the unique example. 155 00:12:29,080 --> 00:12:31,380 So that graph, the complete bipartite graph 156 00:12:31,380 --> 00:12:35,530 with two equal parts, is the unique maximal example, 157 00:12:35,530 --> 00:12:37,770 number of edges in a triangle-free graph. 158 00:12:42,216 --> 00:12:43,210 Great. 159 00:12:43,210 --> 00:12:45,970 So once we know what the answer is 160 00:12:45,970 --> 00:12:49,240 for triangle-free, of course, we should ask further questions. 161 00:12:49,240 --> 00:12:50,950 Instead of forbidding a triangle, 162 00:12:50,950 --> 00:12:53,520 what if we forbid other graphs? 163 00:12:53,520 --> 00:12:57,570 And what are some natural next steps to take? 164 00:12:57,570 --> 00:13:01,690 Well, say, instead of triangles, we can ask, what about if we 165 00:13:01,690 --> 00:13:06,550 forbid a K4, a clique of four vertices? 166 00:13:06,550 --> 00:13:10,240 Or in general, what if we forbid a clique on the fixed 167 00:13:10,240 --> 00:13:11,170 number of vertices? 168 00:13:11,170 --> 00:13:20,450 So what is the maximum number of edges in a Kr plus 1-free-- 169 00:13:20,450 --> 00:13:24,080 there's a good reason why index is r plus 1-- 170 00:13:24,080 --> 00:13:28,670 graph on n vertices. 171 00:13:28,670 --> 00:13:31,850 So for example, if we're interested in K4-free-- 172 00:13:38,400 --> 00:13:40,830 so what might be a good candidate for a graph with lots 173 00:13:40,830 --> 00:13:42,790 of edges that has no K4's? 174 00:13:42,790 --> 00:13:44,610 I mean, certainly, that example we saw, 175 00:13:44,610 --> 00:13:47,540 it does not have K4's, because it doesn't have any triangles. 176 00:13:47,540 --> 00:13:50,820 But we can do even better. 177 00:13:50,820 --> 00:13:55,860 Instead of taking two parts, you can take more parts. 178 00:13:55,860 --> 00:14:00,450 For K4, if I take three parts, each 179 00:14:00,450 --> 00:14:03,910 with n/3 number of vertices, of course, 180 00:14:03,910 --> 00:14:07,020 if n is not divisible by 3, round up and run down. 181 00:14:07,020 --> 00:14:13,680 And putting all the edges between different parts, 182 00:14:13,680 --> 00:14:15,900 you can see this graph here has no K4. 183 00:14:18,880 --> 00:14:22,210 So that's an example of a K4-free graph. 184 00:14:22,210 --> 00:14:25,670 Well, does it have the maximum possible number of edges? 185 00:14:25,670 --> 00:14:28,270 So is this the best that we can do? 186 00:14:28,270 --> 00:14:30,220 So it turns out the answer is yes. 187 00:14:30,220 --> 00:14:32,820 And that's the next theorem that we'll see. 188 00:14:32,820 --> 00:14:36,130 But just to give it a name, so we're going to call graphs like 189 00:14:36,130 --> 00:14:38,010 these Turán graphs. 190 00:14:38,010 --> 00:14:40,735 So the next theorem is proved by Turán. 191 00:14:40,735 --> 00:14:42,100 It's Turán's theorem. 192 00:14:42,100 --> 00:14:48,940 So a Turán graph, so we'll denote T sub nr. 193 00:14:48,940 --> 00:14:59,450 It is a complete r-partite graph such 194 00:14:59,450 --> 00:15:06,250 that there are n vertices whose part sizes 195 00:15:06,250 --> 00:15:12,289 are all nearly the same, up to at most 1 difference. 196 00:15:18,160 --> 00:15:19,930 So this is an example here. 197 00:15:19,930 --> 00:15:25,770 But in general, maybe you have r parts. 198 00:15:28,460 --> 00:15:38,780 And I put in all the edges between different parts. 199 00:15:43,720 --> 00:15:45,940 So it's not too hard to calculate the number of edges 200 00:15:45,940 --> 00:15:47,470 in such a graph. 201 00:15:47,470 --> 00:15:50,470 And Turán's theorem tells us that that is the extremal 202 00:15:50,470 --> 00:15:51,620 example. 203 00:15:51,620 --> 00:15:55,720 You cannot do better in terms of getting more edges. 204 00:16:05,510 --> 00:16:15,950 So if G is an n vertex, try a K sub r plus 1-free graph. 205 00:16:18,610 --> 00:16:28,170 Then it has at most the number of edges of the Turán graph. 206 00:16:36,020 --> 00:16:38,360 It's a generalization of Mantel's theorem. 207 00:16:38,360 --> 00:16:42,350 And well, you can think about if the proofs that we did 208 00:16:42,350 --> 00:16:46,310 for Mantel's generalizes to Turán's theorem. 209 00:16:46,310 --> 00:16:50,360 And it's actually not entirely clear how to do it. 210 00:16:50,360 --> 00:16:53,390 So let me present for you three different proofs 211 00:16:53,390 --> 00:16:55,598 of Turán's theorem. 212 00:16:55,598 --> 00:16:57,140 So some of them, you can think about, 213 00:16:57,140 --> 00:16:59,240 are they related to the proofs of Mantel's theorem 214 00:16:59,240 --> 00:16:59,740 that we did? 215 00:16:59,740 --> 00:17:02,930 And they all are going to look somewhat different, but maybe 216 00:17:02,930 --> 00:17:06,280 superficially. 217 00:17:06,280 --> 00:17:13,930 The first proof, we will use induction 218 00:17:13,930 --> 00:17:15,299 on the number of vertices. 219 00:17:18,250 --> 00:17:21,089 So actually, this is one of the very few times in this course 220 00:17:21,089 --> 00:17:23,212 where we will see induction. 221 00:17:23,212 --> 00:17:25,170 So of course, induction is a powerful technique 222 00:17:25,170 --> 00:17:25,920 in combinatorics. 223 00:17:25,920 --> 00:17:28,000 But for almost the rest of the course, 224 00:17:28,000 --> 00:17:30,030 we're not going to have clean examples. 225 00:17:30,030 --> 00:17:32,280 And when we do have clean examples to work with, 226 00:17:32,280 --> 00:17:36,640 somehow increasing n by 1 doesn't buy you all that much. 227 00:17:36,640 --> 00:17:40,020 Here, there are very clean examples, very clean answers, 228 00:17:40,020 --> 00:17:42,360 and induction works out quite well. 229 00:17:45,640 --> 00:17:48,130 When n is small, of course, you should always address that. 230 00:17:48,130 --> 00:17:49,600 You can come up with many funny proofs 231 00:17:49,600 --> 00:17:51,142 if you don't address when n is small. 232 00:17:51,142 --> 00:17:55,840 So when n is small, this problem is basically trivial. 233 00:17:55,840 --> 00:17:57,310 n is almost r. 234 00:17:57,310 --> 00:18:01,210 You could have the complete graph on r vertices. 235 00:18:01,210 --> 00:18:03,350 And then we're good. 236 00:18:03,350 --> 00:18:10,090 So let's assume that we're not in this case. 237 00:18:10,090 --> 00:18:11,810 And also, by induction hypothesis, 238 00:18:11,810 --> 00:18:19,345 let's assume that it is true for all graphs fewer than n 239 00:18:19,345 --> 00:18:19,845 vertices. 240 00:18:24,480 --> 00:18:31,830 And let G be a graph that is K sub r 241 00:18:31,830 --> 00:18:36,990 plus 1-free on n vertices. 242 00:18:36,990 --> 00:18:41,800 And also, let's assume a maximum example to begin with. 243 00:18:41,800 --> 00:18:44,610 So let's assume that the G that you chose 244 00:18:44,610 --> 00:18:50,350 has already the maximum possible number of edges. 245 00:18:50,350 --> 00:18:52,310 There are only finite of any such examples. 246 00:18:52,310 --> 00:18:55,230 So pick one that already has the maximum number of edges. 247 00:18:55,230 --> 00:18:59,190 And let's think about what properties this graph G has. 248 00:18:59,190 --> 00:19:08,460 I claim that G must already contain a clique on r vertices. 249 00:19:13,230 --> 00:19:14,390 So think about that. 250 00:19:14,390 --> 00:19:18,290 If G does not contain a clique on r vertices, 251 00:19:18,290 --> 00:19:20,240 then I can add in more edges. 252 00:19:28,600 --> 00:19:34,100 And I can still maintain the property 253 00:19:34,100 --> 00:19:35,960 of being K sub r plus 1-free. 254 00:19:40,480 --> 00:19:44,320 So I can assume that G must contain a K sub r. 255 00:19:44,320 --> 00:19:48,320 So let's look at one of these K sub r's. 256 00:19:48,320 --> 00:20:04,200 So let n be the vertices of some r clique in G. 257 00:20:04,200 --> 00:20:12,420 So we have some A. And the complement to that, 258 00:20:12,420 --> 00:20:21,490 let me call that B. 259 00:20:21,490 --> 00:20:28,940 Look at the vertex in B. How many neighbors can have in A? 260 00:20:28,940 --> 00:20:32,730 It cannot be complete to A, otherwise I would have an r 261 00:20:32,730 --> 00:20:34,610 plus 1 clique. 262 00:20:34,610 --> 00:20:51,200 So every vertex in B has at most r minus 1 neighbors in A. 263 00:20:51,200 --> 00:20:53,700 So let's count all the edges. 264 00:20:56,640 --> 00:21:00,530 The number of edges in G is that most-- 265 00:21:00,530 --> 00:21:04,070 well, first, we should account for the edges inside A. 266 00:21:04,070 --> 00:21:08,730 And there are-- or choose two of them. 267 00:21:08,730 --> 00:21:13,080 And then the edges between A and B, for every vertex 268 00:21:13,080 --> 00:21:19,050 in B, there are at most r minus 1 vertices going 269 00:21:19,050 --> 00:21:32,650 to A for each vertex in B. And finally, the edge set of B. 270 00:21:32,650 --> 00:21:38,060 Well, we can say something more about these quantities. 271 00:21:38,060 --> 00:21:42,740 We know that the size of B is exactly n minus r. 272 00:21:45,560 --> 00:21:49,060 But what can we say about the number of edges in B? 273 00:21:49,060 --> 00:21:51,450 We can use induction hypothesis. 274 00:21:51,450 --> 00:21:55,990 So B is also r plus 1 clique-free. 275 00:21:55,990 --> 00:21:59,080 So the number of edges is at most the number of edges 276 00:21:59,080 --> 00:22:02,920 in a corresponding Turán graph. 277 00:22:02,920 --> 00:22:05,880 Now, at this point, you can do a calculation. 278 00:22:05,880 --> 00:22:08,100 Well, I mean, you should expect that the answer 279 00:22:08,100 --> 00:22:10,680 we're looking for is-- 280 00:22:10,680 --> 00:22:13,350 I mean, this should be equal to the number of edges 281 00:22:13,350 --> 00:22:14,430 in the Turán graph. 282 00:22:14,430 --> 00:22:16,410 So you can either do a calculation 283 00:22:16,410 --> 00:22:18,630 to figure this out, or remember what 284 00:22:18,630 --> 00:22:23,400 I said earlier, that keep the tight example in mind, 285 00:22:23,400 --> 00:22:26,640 and everything should check out for the tight example. 286 00:22:26,640 --> 00:22:30,570 So in particular, if you are in the situation 287 00:22:30,570 --> 00:22:37,750 of a complete multipartite graph with equal size or nearly 288 00:22:37,750 --> 00:22:42,540 equal-size parts, what is A? 289 00:22:49,780 --> 00:22:52,740 So A is one vertex from each part. 290 00:22:56,270 --> 00:22:58,560 So you take out one vertex from each part. 291 00:23:02,960 --> 00:23:07,610 And read off this calculation for this graph over here. 292 00:23:07,610 --> 00:23:10,042 And then you see that that is indeed equality. 293 00:23:10,042 --> 00:23:11,000 So it should check out. 294 00:23:11,000 --> 00:23:13,824 You don't need to do any actual calculations. 295 00:23:19,150 --> 00:23:21,890 Any questions about the proofs we've done so far? 296 00:23:25,285 --> 00:23:28,200 All right. 297 00:23:28,200 --> 00:23:31,140 Let me show you another proof of Turán's theorem. 298 00:23:39,390 --> 00:23:43,970 So this proof has a name. 299 00:23:43,970 --> 00:23:46,552 So it's called Zykov's symmetrization. 300 00:23:55,408 --> 00:23:58,020 So Zykov has the unfortunate honor 301 00:23:58,020 --> 00:24:00,120 of having a name that's hard to beat 302 00:24:00,120 --> 00:24:02,320 in terms of alphabetical order. 303 00:24:02,320 --> 00:24:04,180 I think, if he and I were to write a paper, 304 00:24:04,180 --> 00:24:05,430 I wouldn't be the last author. 305 00:24:09,730 --> 00:24:12,550 So what's this about? 306 00:24:12,550 --> 00:24:16,930 So let G be the graph. 307 00:24:16,930 --> 00:24:20,690 Again, as before, we're going to take a maximal example. 308 00:24:20,690 --> 00:24:30,810 So be the n-vertex graph that is free of cliques 309 00:24:30,810 --> 00:24:35,790 of r plus 1 vertices and has already the maximum number 310 00:24:35,790 --> 00:24:36,400 of edges. 311 00:24:44,410 --> 00:24:46,850 So here's the property I want to prove about this graph. 312 00:24:50,090 --> 00:24:55,240 I claim that if you look at the complement of the graph, 313 00:24:55,240 --> 00:24:57,980 of this extremal example, it must 314 00:24:57,980 --> 00:25:00,230 be an equivalence relation. 315 00:25:00,230 --> 00:25:07,970 So more precisely, claim that if xy is a non-edge, 316 00:25:07,970 --> 00:25:19,650 and yz is a non-edge, then xz must also be a non-edge. 317 00:25:19,650 --> 00:25:27,270 So in other words, non-edges form equivalence relation. 318 00:25:34,973 --> 00:25:36,390 And again, you should always think 319 00:25:36,390 --> 00:25:39,390 about the extremal example. 320 00:25:39,390 --> 00:25:41,330 And it is true for the extremal example. 321 00:25:41,330 --> 00:25:44,005 Because the complement are a bunch of cliques. 322 00:25:44,005 --> 00:25:45,380 So it is an equivalence relation. 323 00:25:51,830 --> 00:25:53,310 So let's prove this claim. 324 00:25:53,310 --> 00:25:57,240 So let's assume that the conclusion is not true. 325 00:25:57,240 --> 00:26:06,240 So suppose, for the contrary, that we have xy and yz being 326 00:26:06,240 --> 00:26:12,000 non-edges, but xz is an edge. 327 00:26:24,480 --> 00:26:26,580 So let's think about what happens 328 00:26:26,580 --> 00:26:32,210 to the degrees of non-adjacent vertices. 329 00:26:32,210 --> 00:26:36,710 So I claim that if the degree of y 330 00:26:36,710 --> 00:26:40,920 were smaller than the degree of x, then 331 00:26:40,920 --> 00:26:42,880 I can do something to the graph that 332 00:26:42,880 --> 00:26:47,150 violates the maximality of the number of edges. 333 00:26:47,150 --> 00:26:59,250 Namely, I can replace y by a clone of x. 334 00:27:01,840 --> 00:27:05,230 So I chop off y from the graph. 335 00:27:05,230 --> 00:27:06,550 And I look at x. 336 00:27:06,550 --> 00:27:07,990 And I clone. 337 00:27:07,990 --> 00:27:12,100 So cloning it means taking some x prime. 338 00:27:12,100 --> 00:27:17,500 And I join x prime to all the same neighbors as x. 339 00:27:17,500 --> 00:27:21,790 So clone x into some other vertex, x prime. 340 00:27:21,790 --> 00:27:24,430 Now, when you do this, I claim that you also 341 00:27:24,430 --> 00:27:30,820 get a graph that is free of K sub r plus 1. 342 00:27:34,740 --> 00:27:41,630 So we also obtain a K sub r plus 1-free graph. 343 00:27:45,230 --> 00:27:46,325 So why is that? 344 00:27:46,325 --> 00:27:50,710 So if you were to have an r plus 1 clique, 345 00:27:50,710 --> 00:27:54,510 well, it shouldn't have both x and x clone. 346 00:27:54,510 --> 00:27:56,410 Because there's no edge between them. 347 00:27:56,410 --> 00:27:59,468 So it only has one of x and x clone. 348 00:27:59,468 --> 00:28:02,010 But then that would have been a clique in the original graph. 349 00:28:07,300 --> 00:28:12,460 However, what about the number of edges 350 00:28:12,460 --> 00:28:16,300 that we obtain after this transformation? 351 00:28:16,300 --> 00:28:20,800 If x had more outgoing edges, a higher degree 352 00:28:20,800 --> 00:28:24,790 than y, then cloning x to replace y 353 00:28:24,790 --> 00:28:26,500 increases the number of edges. 354 00:28:31,840 --> 00:28:35,780 We obtain a graph that is also K sub r plus 1-free 355 00:28:35,780 --> 00:28:45,820 and has more edges than G, which is not possible, 356 00:28:45,820 --> 00:28:49,180 because we assumed that g started with the maximum number 357 00:28:49,180 --> 00:28:51,163 of edges. 358 00:28:51,163 --> 00:29:00,600 So therefore, the degree of y is at most the degree x, basically 359 00:29:00,600 --> 00:29:05,520 for every non-edge xy. 360 00:29:08,780 --> 00:29:12,080 Likewise, we also have that there 361 00:29:12,080 --> 00:29:13,840 is no edge between y and z. 362 00:29:17,610 --> 00:29:20,400 So the degree of y is-- 363 00:29:23,712 --> 00:29:25,920 thus the degree of y is at least the degree of x now. 364 00:29:25,920 --> 00:29:30,748 So degree of y is at least the degree of z. 365 00:29:33,700 --> 00:29:38,910 Now, y has a lot of outgoing edges. 366 00:29:38,910 --> 00:29:44,160 So now let's replace both x and z by clones of y. 367 00:29:56,760 --> 00:30:02,650 As before, we obtain some graph that we'll call G prime. 368 00:30:02,650 --> 00:30:07,600 And G prime, same reason as before, is K sub r plus 1-free. 369 00:30:07,600 --> 00:30:12,100 If you had a K sub r plus 1, then the same clique 370 00:30:12,100 --> 00:30:16,390 should have shown up in G. So G prime does not 371 00:30:16,390 --> 00:30:18,130 have r plus 1 cliques either. 372 00:30:18,130 --> 00:30:20,960 But what about the number of edges in G prime? 373 00:30:25,460 --> 00:30:27,490 So the number of edges in G prime-- 374 00:30:27,490 --> 00:30:34,320 well, we started with G. We deleted x and z. 375 00:30:34,320 --> 00:30:42,170 So we deleted this many edges. 376 00:30:42,170 --> 00:30:43,730 Here, we're crucially using the fact 377 00:30:43,730 --> 00:30:47,260 that there is an edge between x and z. 378 00:30:47,260 --> 00:30:51,160 But now we also added back in a bunch of edges 379 00:30:51,160 --> 00:30:53,320 coming from the clone of y. 380 00:30:53,320 --> 00:30:59,482 And they are 2 times dy edges added back in. 381 00:30:59,482 --> 00:31:02,240 Now, you see, because y has degree at least that 382 00:31:02,240 --> 00:31:06,680 of both x and z, we obtain that this number of edges 383 00:31:06,680 --> 00:31:13,010 is strictly bigger than that of G, which contradicts 384 00:31:13,010 --> 00:31:29,080 the maximality of G. 385 00:31:29,080 --> 00:31:32,410 So at this point, we know that the complement of G 386 00:31:32,410 --> 00:31:35,390 must be an equivalence relation. 387 00:31:35,390 --> 00:31:38,910 So the complement is a bunch of cliques. 388 00:31:38,910 --> 00:31:40,613 And that's a lot of information. 389 00:31:40,613 --> 00:31:42,030 So that's almost the best thing we 390 00:31:42,030 --> 00:31:43,822 can hope for is the structural information. 391 00:31:43,822 --> 00:31:46,510 We're not quite done yet. 392 00:31:46,510 --> 00:31:47,220 And why is that? 393 00:31:50,860 --> 00:31:52,962 Well, we're almost there. 394 00:31:52,962 --> 00:31:54,170 But we're not quite done yet. 395 00:31:57,128 --> 00:32:00,086 AUDIENCE: [INAUDIBLE] you can figure out 396 00:32:00,086 --> 00:32:02,155 the sizes [INAUDIBLE]. 397 00:32:02,155 --> 00:32:02,780 YUFEI ZHAO: OK. 398 00:32:02,780 --> 00:32:08,590 So we need to figure out the sizes of the individual parts. 399 00:32:08,590 --> 00:32:11,390 So we need to figure out the sizes of individual parts. 400 00:32:11,390 --> 00:32:15,610 And also, note that you cannot have too many parts. 401 00:32:15,610 --> 00:32:20,950 So at this point, after the claim, 402 00:32:20,950 --> 00:32:28,210 we know that G is a complete multipartite graph. 403 00:32:38,810 --> 00:32:43,130 It has at most r parts. 404 00:32:43,130 --> 00:32:45,710 If it had r plus 1 parts, I would see a clique 405 00:32:45,710 --> 00:32:46,745 in r plus 1 vertices. 406 00:32:51,080 --> 00:32:53,665 So then, finally, I need to show-- 407 00:32:53,665 --> 00:32:55,390 I mean, the rest is fairly routine. 408 00:32:55,390 --> 00:32:58,540 I need to show what the part sizes have 409 00:32:58,540 --> 00:33:01,180 to be to maximize the number of edges. 410 00:33:01,180 --> 00:33:03,535 And basically, if you had some two parts-- 411 00:33:06,660 --> 00:33:08,690 so consider exactly r parts. 412 00:33:08,690 --> 00:33:10,090 They're all empty parts. 413 00:33:10,090 --> 00:33:12,780 And some two parts have the number 414 00:33:12,780 --> 00:33:23,390 of vertices differing by more than 1. 415 00:33:23,390 --> 00:33:27,350 Then what I can do is-- 416 00:33:27,350 --> 00:33:32,956 so if you had one part much bigger than the other part, 417 00:33:32,956 --> 00:33:36,730 you can imagine moving one vertex from one part 418 00:33:36,730 --> 00:33:38,690 to the other part. 419 00:33:38,690 --> 00:33:41,540 And you should convince yourself that this operation should 420 00:33:41,540 --> 00:33:44,270 strictly increase the number of edges. 421 00:33:50,520 --> 00:34:01,700 And then moving vertices should strictly increase 422 00:34:01,700 --> 00:34:04,960 the number of edges of G. 423 00:34:04,960 --> 00:34:06,720 And putting this all together, we 424 00:34:06,720 --> 00:34:09,840 see that the extremal example necessarily 425 00:34:09,840 --> 00:34:14,070 has to be the current graph, namely 426 00:34:14,070 --> 00:34:19,630 a complete r-partite graph, where 427 00:34:19,630 --> 00:34:22,860 all the parts have the same size, up 428 00:34:22,860 --> 00:34:24,456 to almost 1 difference. 429 00:34:29,126 --> 00:34:30,540 Great. 430 00:34:30,540 --> 00:34:33,030 Any questions? 431 00:34:33,030 --> 00:34:33,530 Yep? 432 00:34:33,530 --> 00:34:35,530 AUDIENCE: Can the Zykov symmetrization technique 433 00:34:35,530 --> 00:34:37,649 be used for any other type of problem? 434 00:34:40,460 --> 00:34:42,960 YUFEI ZHAO: So the question is, can the Zykov symmetrization 435 00:34:42,960 --> 00:34:46,715 technique be used for any other types of problem? 436 00:34:46,715 --> 00:34:48,090 I do have something else in mind. 437 00:34:48,090 --> 00:34:49,839 But I don't want to discuss it now. 438 00:34:52,719 --> 00:34:54,098 Any more questions? 439 00:34:58,082 --> 00:34:59,078 Great. 440 00:34:59,078 --> 00:35:03,180 And you see that in both proofs, if you look at this proof 441 00:35:03,180 --> 00:35:05,100 as well, you see that the Turán graph, 442 00:35:05,100 --> 00:35:09,370 it's the unique extremizer. 443 00:35:09,370 --> 00:35:12,250 I want to give you a third proof that has 444 00:35:12,250 --> 00:35:13,730 a somewhat different flavor. 445 00:35:13,730 --> 00:35:16,000 So these two proofs, they're both somewhat 446 00:35:16,000 --> 00:35:17,265 combinatorial in flavor. 447 00:35:17,265 --> 00:35:22,340 So I'm doing some arguments either looking at, 448 00:35:22,340 --> 00:35:24,730 in this case, a clique and arguing 449 00:35:24,730 --> 00:35:26,050 what happens outside of it. 450 00:35:26,050 --> 00:35:29,230 And over there, again, I'm looking at maximal example. 451 00:35:29,230 --> 00:35:30,910 The third proof I want to show is 452 00:35:30,910 --> 00:35:32,770 more of a probabilistic proof. 453 00:35:32,770 --> 00:35:35,485 So this highlights an important method in combinatorics, 454 00:35:35,485 --> 00:35:39,280 and almost a probabilistic method, 455 00:35:39,280 --> 00:35:43,550 where we start with a problem that comes with no randomness. 456 00:35:43,550 --> 00:35:48,920 But we introduce some randomness to make the problem amenable. 457 00:35:48,920 --> 00:35:50,120 It's a very pretty idea. 458 00:36:02,480 --> 00:36:11,240 So we start, again, with G being a n-uniform, so n-vertex, K sub 459 00:36:11,240 --> 00:36:17,520 r plus 1-free graph on m edges. 460 00:36:20,532 --> 00:36:25,700 And what I want to do is to randomly sort the vertex set. 461 00:36:25,700 --> 00:36:36,400 Consider a random order of the vertices. 462 00:36:42,620 --> 00:36:45,030 So I put all the vertices on the line, 463 00:36:45,030 --> 00:36:47,610 but the vertices are chosen at random order. 464 00:36:47,610 --> 00:36:53,574 And you see some of the edges like that. 465 00:37:00,400 --> 00:37:03,070 Let me show you how to find a clique. 466 00:37:03,070 --> 00:37:07,810 And essentially, we do it in a not-so-smart way. 467 00:37:07,810 --> 00:37:11,620 Namely, I basically pick all the vertices 468 00:37:11,620 --> 00:37:13,870 in some greedy-like manner, where 469 00:37:13,870 --> 00:37:16,540 I include the vertex in my set if it 470 00:37:16,540 --> 00:37:23,720 is adjacent to all earlier vertices. 471 00:37:23,720 --> 00:37:28,730 So if all of its neighbors-- 472 00:37:28,730 --> 00:37:35,090 so I include V, if V is the earliest 473 00:37:35,090 --> 00:37:38,138 vertex among its neighbors. 474 00:37:45,800 --> 00:37:49,590 So earliest means the leftmost, so if I go from left to right. 475 00:37:49,590 --> 00:37:54,680 So in this case up here, so I look at the first vertex. 476 00:37:54,680 --> 00:37:58,335 Well, the first vertex should always be in your set. 477 00:37:58,335 --> 00:38:01,440 So this vertex has one neighbor, just to the right. 478 00:38:01,440 --> 00:38:04,600 So that's OK. 479 00:38:04,600 --> 00:38:09,350 And hold on. 480 00:38:09,350 --> 00:38:11,930 It's-- no. 481 00:38:11,930 --> 00:38:12,640 Sorry. 482 00:38:12,640 --> 00:38:16,250 That's not what I want to do. 483 00:38:19,890 --> 00:38:20,390 Sorry. 484 00:38:20,390 --> 00:38:25,320 So I actually do mean what I was going to write down initially. 485 00:38:25,320 --> 00:38:41,975 So if V is adjacent to all the earlier vertices 486 00:38:41,975 --> 00:38:42,600 in this order-- 487 00:38:54,450 --> 00:39:00,090 so for example, this vertex, it's OK. 488 00:39:00,090 --> 00:39:04,230 And I'm also going to include this vertex here, 489 00:39:04,230 --> 00:39:12,150 because both of its earlier vertices are in-- 490 00:39:12,150 --> 00:39:16,350 both the earlier vertices are adjacent to the third vertex. 491 00:39:16,350 --> 00:39:18,020 And I think that's it. 492 00:39:20,882 --> 00:39:25,160 Now, this set x, I claim two things. 493 00:39:25,160 --> 00:39:29,490 One, that x has to be a clique. 494 00:39:29,490 --> 00:39:34,050 So claim that x has to be a clique. 495 00:39:37,110 --> 00:39:41,200 Because every vertex in the x is adjacent to all 496 00:39:41,200 --> 00:39:43,540 of the [INAUDIBLE] vertices in particular. 497 00:39:43,540 --> 00:39:48,480 All the vertices in x are joined to each other. 498 00:39:48,480 --> 00:39:54,930 On the other hand, how big is x, at least in expectation? 499 00:39:54,930 --> 00:40:05,202 So for every vertex, I want to understand the probability 500 00:40:05,202 --> 00:40:10,630 that this v is included in x. 501 00:40:10,630 --> 00:40:14,790 So v has a bunch of non-neighbors. 502 00:40:14,790 --> 00:40:19,880 And the property of v being included in x 503 00:40:19,880 --> 00:40:24,660 is that all of its non-neighbors appear after v. 504 00:40:24,660 --> 00:40:31,350 So the probability that v, little v, is in the set x 505 00:40:31,350 --> 00:40:37,740 is equal to the probability that v appears 506 00:40:37,740 --> 00:40:42,210 before all its non-neighbors. 507 00:40:52,300 --> 00:40:55,720 All v and its non-neighbors, they're sorted uniformly. 508 00:40:55,720 --> 00:41:04,300 So the probability that this occurs is exactly 1 over 1 509 00:41:04,300 --> 00:41:06,190 plus the number of non-neighbors, 510 00:41:06,190 --> 00:41:08,320 which is n minus the degree of. 511 00:41:26,340 --> 00:41:28,980 Well, we know that this graph G has 512 00:41:28,980 --> 00:41:31,950 no cliques of size r plus 1. 513 00:41:31,950 --> 00:41:40,720 So if we consider the expected number of the size of x, 514 00:41:40,720 --> 00:41:43,600 on one hand, it is at most r. 515 00:41:43,600 --> 00:41:48,610 Because G K sub r plus 1-free. 516 00:41:48,610 --> 00:41:52,810 On the other hand, by linearity of expectations, 517 00:41:52,810 --> 00:41:55,330 each vertex is included with some probability. 518 00:41:55,330 --> 00:41:58,340 So the size of x in expectation is just 519 00:41:58,340 --> 00:42:05,650 a sum of all of these individual inclusion probabilities, which 520 00:42:05,650 --> 00:42:18,480 individually we've computed above like that. 521 00:42:18,480 --> 00:42:23,190 Now, by convexity, I can conclude 522 00:42:23,190 --> 00:42:28,410 that this quantity, this sum here, is at least the quantity 523 00:42:28,410 --> 00:42:31,620 that would have been obtained if all the decrease were 524 00:42:31,620 --> 00:42:32,490 equal to each other. 525 00:42:37,540 --> 00:42:40,700 And if you rearrange this equation, this inequality, 526 00:42:40,700 --> 00:42:46,370 we obtain that an m is at most 1 minus 1 527 00:42:46,370 --> 00:42:51,740 over r times n squared over 2. 528 00:42:51,740 --> 00:42:55,620 And if you compare this number to the number of edges 529 00:42:55,620 --> 00:42:59,610 in the Turán graph, so you see that this is basically 530 00:42:59,610 --> 00:43:07,300 the number of edges, and this gives you a proof if n is 531 00:43:07,300 --> 00:43:13,810 divisible by r. 532 00:43:13,810 --> 00:43:17,710 And in fact, the number of edges in the Turán graph is exactly 533 00:43:17,710 --> 00:43:22,190 this number here, if this divisibility condition is true. 534 00:43:22,190 --> 00:43:24,940 If it's not, you need to do a little bit more work. 535 00:43:24,940 --> 00:43:29,040 Because we were a little bit lost here in this step. 536 00:43:29,040 --> 00:43:31,570 So you should see that this quantity here 537 00:43:31,570 --> 00:43:35,535 is minimized when all the degrees are roughly 538 00:43:35,535 --> 00:43:36,910 equal to each other, and you make 539 00:43:36,910 --> 00:43:39,250 them as close to each other as possible. 540 00:43:41,840 --> 00:43:45,430 So with a little bit more work, you can get the exact version 541 00:43:45,430 --> 00:43:46,740 of Turán's theorem up there. 542 00:43:46,740 --> 00:43:50,160 But at least what we've shown is-- 543 00:43:50,160 --> 00:43:58,000 so from here, we've shown that the number of edges is at most 544 00:43:58,000 --> 00:43:59,950 this quantity, which, for most purposes, 545 00:43:59,950 --> 00:44:01,997 is basically as good as Turán's theorem. 546 00:44:06,650 --> 00:44:08,470 Any questions about this proof here? 547 00:44:08,470 --> 00:44:11,046 And so it's a probabilistic method of proof. 548 00:44:11,046 --> 00:44:13,600 So we're introducing some randomness 549 00:44:13,600 --> 00:44:16,298 into the problem that originally had no randomness. 550 00:44:19,510 --> 00:44:20,010 Great. 551 00:44:20,010 --> 00:44:22,500 So let's take a very quick 2-minute break. 552 00:44:22,500 --> 00:44:26,150 And then when we come back, I want to show you that even 553 00:44:26,150 --> 00:44:27,977 though we've shown so many different proofs 554 00:44:27,977 --> 00:44:30,560 of Turán's theorem and Mantel's theorem-- you might think, OK, 555 00:44:30,560 --> 00:44:32,900 this is a pretty simple thing-- 556 00:44:32,900 --> 00:44:36,075 even if I tweak the problem just a little bit, 557 00:44:36,075 --> 00:44:38,200 there are so many things that we do not understand, 558 00:44:38,200 --> 00:44:41,460 and many important open problems in combinatorics that are 559 00:44:41,460 --> 00:44:44,743 variants of Turán's theorem. 560 00:44:44,743 --> 00:44:45,910 So let's take a quick break. 561 00:44:51,960 --> 00:44:55,930 So so far we've been talking about Turán's theorem, 562 00:44:55,930 --> 00:44:59,830 or generally the problem of, if you forbid a certain structure, 563 00:44:59,830 --> 00:45:05,730 forbid a certain subgraph, what is the maximum number of edges? 564 00:45:05,730 --> 00:45:07,970 We're going to spend the next few lectures discussing 565 00:45:07,970 --> 00:45:09,480 more problems of that form. 566 00:45:09,480 --> 00:45:13,520 And it turns out, for the answers we've just seen, 567 00:45:13,520 --> 00:45:15,300 they are deceptively simple. 568 00:45:15,300 --> 00:45:17,690 And for almost any other situation, 569 00:45:17,690 --> 00:45:20,340 we don't really know the exact answer. 570 00:45:20,340 --> 00:45:22,910 And for many questions, we don't know anything close 571 00:45:22,910 --> 00:45:24,038 to the truth. 572 00:45:24,038 --> 00:45:25,580 And in particular, I want to show you 573 00:45:25,580 --> 00:45:29,000 one variant of this problem, namely 574 00:45:29,000 --> 00:45:31,640 what happens, instead of looking at graphs, 575 00:45:31,640 --> 00:45:33,640 if you look at hypergraphs. 576 00:45:33,640 --> 00:45:36,860 And there, that's a major open problem in combinatorics, 577 00:45:36,860 --> 00:45:39,150 what the truth should be. 578 00:45:39,150 --> 00:45:47,850 So here is an open problem, which is, 579 00:45:47,850 --> 00:45:53,270 what happens to Turán's theorem for 3-uniform hypergraphs? 580 00:45:59,990 --> 00:46:02,570 So don't be scared by the word hypergraph. 581 00:46:02,570 --> 00:46:06,160 So whereas you think of graphs as having edges consisting 582 00:46:06,160 --> 00:46:08,260 of pairs of vertices, a hypergraph 583 00:46:08,260 --> 00:46:12,200 is simply a structure where, for a 3-uniform hypergraph, 584 00:46:12,200 --> 00:46:14,625 the edges are triples, triples of vertices. 585 00:46:17,130 --> 00:46:26,770 So the question is, what is the maximum number of triples 586 00:46:26,770 --> 00:46:37,820 in 3-uniform hypergraph without-- 587 00:46:37,820 --> 00:46:40,040 well, for Mantel's theorem, we asked what 588 00:46:40,040 --> 00:46:41,885 happens without a triangle. 589 00:46:41,885 --> 00:46:45,560 For a 3-uniform, the basic question you can ask 590 00:46:45,560 --> 00:46:47,940 is, what about forbidding a tetrahedral? 591 00:46:53,590 --> 00:47:02,280 So suppose you do have four vertices 592 00:47:02,280 --> 00:47:08,370 such that every triple inside the four vertices is a niche. 593 00:47:08,370 --> 00:47:10,770 What's the maximum number of edges 594 00:47:10,770 --> 00:47:17,600 you can have in an n-vertex, 3-uniform hypergraph. 595 00:47:17,600 --> 00:47:21,990 Already, it's not so easy to come up with good examples. 596 00:47:21,990 --> 00:47:24,710 So Turán's theorem says take a bipartite graph, 597 00:47:24,710 --> 00:47:26,700 complete bipartite graph. 598 00:47:26,700 --> 00:47:30,090 Well, here, it's not so easy to come up with examples, 599 00:47:30,090 --> 00:47:31,890 but there are some examples. 600 00:47:31,890 --> 00:47:35,630 And in fact, Turán suggested the following construction. 601 00:47:35,630 --> 00:47:41,790 Namely, you divide the set of vertices, as before, 602 00:47:41,790 --> 00:47:46,610 to three roughly equal-sized parts. 603 00:47:46,610 --> 00:47:50,960 And let me take all triples that look like one of the following 604 00:47:50,960 --> 00:47:57,150 forms, either three vertices, one in each vertex set, 605 00:47:57,150 --> 00:48:00,720 or a triple, looking like that-- two 606 00:48:00,720 --> 00:48:03,580 vertex in one set, one in the next set, 607 00:48:03,580 --> 00:48:06,990 or going cyclically, like that. 608 00:48:06,990 --> 00:48:11,870 So I include all triples, one of these forms. 609 00:48:11,870 --> 00:48:15,990 And you should check that this construction 610 00:48:15,990 --> 00:48:19,580 has no tetrahedron. 611 00:48:19,580 --> 00:48:21,340 If it had a tetrahedron, you would 612 00:48:21,340 --> 00:48:24,910 have had at least two vertices in one part, also 613 00:48:24,910 --> 00:48:28,566 then where the other two vertices can be. 614 00:48:28,566 --> 00:48:32,640 If you check that, it cannot happen. 615 00:48:32,640 --> 00:48:35,780 So then, how many edges does it have? 616 00:48:35,780 --> 00:48:37,400 The exact number is not so important. 617 00:48:37,400 --> 00:48:40,442 But what's important is that the edge density-- 618 00:48:40,442 --> 00:48:43,550 of all the possible triples, the fraction 619 00:48:43,550 --> 00:48:47,840 of edges that are containing this construction is 5/9. 620 00:48:50,700 --> 00:48:54,790 And it is conjectured that this is optimal. 621 00:48:59,410 --> 00:49:02,500 However, we're quite far from proving this number. 622 00:49:02,500 --> 00:49:09,920 So the best upper bound that is currently available-- 623 00:49:09,920 --> 00:49:13,640 and it's found quite recently using this fairly new method 624 00:49:13,640 --> 00:49:16,730 in graph theory called flag algebras, essentially 625 00:49:16,730 --> 00:49:20,840 a computerized way to try to prove such inequalities. 626 00:49:20,840 --> 00:49:28,690 And the best upper bound is something like 0.562. 627 00:49:28,690 --> 00:49:32,910 And there's a major open problem to either prove or disprove 628 00:49:32,910 --> 00:49:36,780 that this construction here is the optimal one. 629 00:49:36,780 --> 00:49:39,840 So you see, even though I presented so many proofs 630 00:49:39,840 --> 00:49:42,210 of Mantel's theorem and Turán's theorem, at this point, 631 00:49:42,210 --> 00:49:43,660 hopefully they should all be-- 632 00:49:43,660 --> 00:49:46,200 they seem quite simple in retrospect. 633 00:49:49,210 --> 00:49:50,100 It's deceptive. 634 00:49:50,100 --> 00:49:53,320 And even if I changed and tweak the problem just a little bit, 635 00:49:53,320 --> 00:49:57,000 going to 3-uniform hypergraph instead of graphs, 636 00:49:57,000 --> 00:49:59,670 we really have no idea what's going on. 637 00:49:59,670 --> 00:50:00,170 Yes? 638 00:50:00,170 --> 00:50:00,690 Question? 639 00:50:00,690 --> 00:50:02,790 AUDIENCE: So basically, how do you-- 640 00:50:02,790 --> 00:50:07,250 why you say that this is really bad compared to a two-number? 641 00:50:07,250 --> 00:50:09,355 It doesn't seem like that big a difference. 642 00:50:09,355 --> 00:50:09,980 YUFEI ZHAO: OK. 643 00:50:09,980 --> 00:50:10,620 So great. 644 00:50:10,620 --> 00:50:14,380 So the question is, why do I say it's a pretty big gap. 645 00:50:14,380 --> 00:50:19,070 So we know that there has to be some proportion 646 00:50:19,070 --> 00:50:21,570 of the total number of triples. 647 00:50:21,570 --> 00:50:23,810 So, well, you have two numbers. 648 00:50:23,810 --> 00:50:28,290 And well, I mean, to me, they seem pretty far apart. 649 00:50:28,290 --> 00:50:30,200 It's not some lower-order gap. 650 00:50:30,200 --> 00:50:31,600 So this is a first-order gap. 651 00:50:34,150 --> 00:50:35,145 Any more questions? 652 00:50:35,145 --> 00:50:36,770 But it's true that later in the course, 653 00:50:36,770 --> 00:50:38,930 we'll see gaps that are much bigger. 654 00:50:38,930 --> 00:50:42,820 We'll see gaps where there's a polynomial on one side 655 00:50:42,820 --> 00:50:46,350 and a power of exponentials on the other side. 656 00:50:46,350 --> 00:50:48,290 And there, I agree, that's much worse. 657 00:50:48,290 --> 00:50:49,300 The gap is much bigger. 658 00:50:49,300 --> 00:50:50,810 Here, it's just two numbers. 659 00:50:50,810 --> 00:50:55,285 But in the worst case, there can be two numbers anywhere. 660 00:50:55,285 --> 00:50:58,745 Any more questions? 661 00:50:58,745 --> 00:51:00,120 So throughout this course, I will 662 00:51:00,120 --> 00:51:02,340 try to bring out some open problems. 663 00:51:02,340 --> 00:51:03,850 And there are lots. 664 00:51:03,850 --> 00:51:05,970 So I will tell you what we do understand. 665 00:51:05,970 --> 00:51:08,160 But most things, we do not understand. 666 00:51:08,160 --> 00:51:11,850 And hopefully, some of you will go out and try 667 00:51:11,850 --> 00:51:14,640 to understand them better so the next time I teach the course, 668 00:51:14,640 --> 00:51:16,223 I could have something new to present. 669 00:51:25,790 --> 00:51:28,880 So now that we've addressed the question of what's 670 00:51:28,880 --> 00:51:30,920 the maximum number of edges if you forbid 671 00:51:30,920 --> 00:51:35,090 a triangle or a clique, the next natural thing to ask 672 00:51:35,090 --> 00:51:39,510 is, what about if you forbid a general H? 673 00:51:39,510 --> 00:51:45,440 So give me a graph H. And what's the maximum number of edges 674 00:51:45,440 --> 00:51:46,700 if you forbid that H? 675 00:51:56,430 --> 00:51:58,550 It will be helpful to do some notation. 676 00:51:58,550 --> 00:52:02,750 So the extremal number, which we'll denote by ex, 677 00:52:02,750 --> 00:52:06,920 and this is also called the Turán number sometimes 678 00:52:06,920 --> 00:52:11,350 because of Turán's theorem. 679 00:52:11,350 --> 00:52:15,770 So I'll probably call it the extremal number more 680 00:52:15,770 --> 00:52:18,240 frequently. 681 00:52:18,240 --> 00:52:21,560 So this number here is defined to be the maximum number 682 00:52:21,560 --> 00:52:40,430 of edges in an n-vertex graph containing no copy of H 683 00:52:40,430 --> 00:52:41,050 as a subgraph. 684 00:52:51,900 --> 00:52:55,090 I want to just clarify a piece of notation. 685 00:52:55,090 --> 00:53:02,130 So when I say a subgraph, so what do I mean? 686 00:53:02,130 --> 00:53:06,060 So there are several notions of different kinds of subgraphs. 687 00:53:06,060 --> 00:53:09,720 And a couple that come up somewhat frequently, 688 00:53:09,720 --> 00:53:11,280 one is subgraph. 689 00:53:11,280 --> 00:53:13,920 And then there's something called induced subgraph. 690 00:53:21,200 --> 00:53:24,700 So it's probably easiest if I just show you an example. 691 00:53:24,700 --> 00:53:30,170 So suppose my H is the four cycle. 692 00:53:30,170 --> 00:53:34,910 So in the example of H being a subgraph, 693 00:53:34,910 --> 00:53:38,900 suppose I have this graph here. 694 00:53:38,900 --> 00:53:49,770 So H is a subgraph of this graph here in many different ways, 695 00:53:49,770 --> 00:53:53,316 but in particular, like that. 696 00:53:53,316 --> 00:53:56,070 But you see there are some more edges that 697 00:53:56,070 --> 00:53:57,330 are among the vertices. 698 00:53:57,330 --> 00:53:57,960 But that's OK. 699 00:53:57,960 --> 00:54:02,070 So subgraph only requires you to have a subset of vertices 700 00:54:02,070 --> 00:54:07,560 and a subset of edges, whereas induced subgraphs-- 701 00:54:07,560 --> 00:54:09,745 this is not an example of induced subgraph. 702 00:54:09,745 --> 00:54:17,670 And so induced subgraph means that you 703 00:54:17,670 --> 00:54:22,190 take this set of vertices and you 704 00:54:22,190 --> 00:54:26,090 look at all the edges in the big graph 705 00:54:26,090 --> 00:54:29,060 among your set of vertices. 706 00:54:29,060 --> 00:54:30,430 So that's an induced subgraph. 707 00:54:30,430 --> 00:54:35,720 So here, the four cycle is an induced subgraph, but not 708 00:54:35,720 --> 00:54:38,730 induced subgraph over here. 709 00:54:38,730 --> 00:54:40,700 So I just want to make that distinction clear. 710 00:54:40,700 --> 00:54:43,550 And for now, in this chapter, we'll 711 00:54:43,550 --> 00:54:48,205 only talk about subgraphs-- so not induced, necessarily. 712 00:54:51,460 --> 00:54:52,000 All right. 713 00:54:52,000 --> 00:54:54,710 So let's recap what happened for Turán's theorem. 714 00:55:01,090 --> 00:55:09,900 So Turán's theorem told us that the extremal number of these 715 00:55:09,900 --> 00:55:17,620 cliques, well, we know very precisely to be the number 716 00:55:17,620 --> 00:55:21,110 of edges in the Turán graph. 717 00:55:21,110 --> 00:55:24,440 And in particular, we saw from the last proof, but also, 718 00:55:24,440 --> 00:55:26,630 similarly, if you just do a calculation, 719 00:55:26,630 --> 00:55:32,650 that it is at most this quantity over here. 720 00:55:32,650 --> 00:55:36,860 And it's basically that quantity. 721 00:55:36,860 --> 00:55:45,410 So the number of edges in this Turán graph is asymptotically 722 00:55:45,410 --> 00:55:50,000 that quantity up to a lower-order error; 723 00:55:50,000 --> 00:55:53,330 if you like n go to infinity, r fixed. 724 00:55:56,050 --> 00:55:58,570 And the basic question is, what about general H? 725 00:55:58,570 --> 00:56:05,580 And so if I give you some arbitrary graph H, what can you 726 00:56:05,580 --> 00:56:08,430 tell me about the maximum number of edges 727 00:56:08,430 --> 00:56:14,930 in the graph forbidding this H as a subgraph? 728 00:56:14,930 --> 00:56:18,240 And it turns out, for most H's, we have a pretty good 729 00:56:18,240 --> 00:56:20,370 understanding, and perhaps quite surprisingly, 730 00:56:20,370 --> 00:56:21,870 because you can imagine this problem 731 00:56:21,870 --> 00:56:24,030 looks like it might get quite complicated. 732 00:56:24,030 --> 00:56:27,990 All the proofs that we've done are very specific to cliques. 733 00:56:27,990 --> 00:56:31,590 But it turns out that we already understand a lot. 734 00:56:31,590 --> 00:56:34,290 And the critical parameter that governs 735 00:56:34,290 --> 00:56:39,640 how this quantity behaves is the chromatic number of H. 736 00:56:39,640 --> 00:56:41,850 So if you call me the chromatic number of H, 737 00:56:41,850 --> 00:56:44,520 I can already tell you quite a lot. 738 00:56:44,520 --> 00:56:47,910 So just to remind you, the chromatic number of a graph 739 00:56:47,910 --> 00:56:51,150 is the minimum number of colors you need 740 00:56:51,150 --> 00:56:54,390 to properly color this graph. 741 00:56:54,390 --> 00:57:04,060 Chromatic number of H, denoted chi of H, 742 00:57:04,060 --> 00:57:13,970 is the minimum number of colors needed 743 00:57:13,970 --> 00:57:29,060 to color the vertices of H so that no two adjacent vertices 744 00:57:29,060 --> 00:57:29,930 have the same color. 745 00:57:36,710 --> 00:57:44,730 So for instance, if I give you a clique of r plus 1 vertices, 746 00:57:44,730 --> 00:57:46,920 so all the vertices must receive different colors, 747 00:57:46,920 --> 00:57:49,090 because every pair of vertices is adjacent-- 748 00:57:49,090 --> 00:57:52,520 so the chromatic number is r plus 1. 749 00:57:52,520 --> 00:58:00,991 The chromatic number of the Turán graph is what? 750 00:58:05,290 --> 00:58:05,790 It's r. 751 00:58:05,790 --> 00:58:09,030 So the chromatic number of the Turán graph is r. 752 00:58:09,030 --> 00:58:12,130 I can color each part in this complete bipartite 753 00:58:12,130 --> 00:58:15,120 graph using a different color. 754 00:58:15,120 --> 00:58:17,230 And that's the best I can do. 755 00:58:21,470 --> 00:58:28,220 Now, if I have one graph being a subgraph of another graph, what 756 00:58:28,220 --> 00:58:29,960 can you tell me about the relationships 757 00:58:29,960 --> 00:58:31,460 between their chromatic numbers? 758 00:58:38,945 --> 00:58:47,227 So if H is a subgraph of G, so what can 759 00:58:47,227 --> 00:58:48,810 you tell me about relationship between 760 00:58:48,810 --> 00:58:49,830 their chromatic numbers? 761 00:58:52,752 --> 00:58:54,700 AUDIENCE: Chi of H is less than the other chi. 762 00:58:54,700 --> 00:58:55,450 YUFEI ZHAO: So OK. 763 00:58:55,450 --> 00:59:00,790 You're telling me that the chi of H is at most chi of G. 764 00:59:00,790 --> 00:59:03,050 And why is that? 765 00:59:03,050 --> 00:59:04,688 AUDIENCE: Because if G can be colored 766 00:59:04,688 --> 00:59:10,503 with a certain number of colors, then that also has [INAUDIBLE].. 767 00:59:10,503 --> 00:59:11,470 YUFEI ZHAO: Great. 768 00:59:11,470 --> 00:59:15,140 Whatever coloring you do for G, you use the same coloring, 769 00:59:15,140 --> 00:59:17,760 and that's a proper coloring for H as well. 770 00:59:17,760 --> 00:59:20,240 So the chromatic number of H might be smaller, 771 00:59:20,240 --> 00:59:23,380 but certainly cannot be bigger than that of G. 772 00:59:23,380 --> 00:59:34,580 So in particular, if you have a graph H with chromatic number r 773 00:59:34,580 --> 00:59:44,400 plus 1, then this Turán on graph is always H-free. 774 00:59:47,010 --> 00:59:50,310 So if H requires four colors, it cannot be embedded 775 00:59:50,310 --> 00:59:54,260 into the complete multipartite graph of three parts. 776 00:59:54,260 --> 00:59:59,380 So the Turán graph is also an example of an H-free graph with 777 00:59:59,380 --> 01:00:01,160 lots of edges. 778 01:00:01,160 --> 01:00:07,380 And this tells us that the extremal number of H is 779 01:00:07,380 --> 01:00:11,600 at least that of this Turán number, 780 01:00:11,600 --> 01:00:18,316 where r is defined as the chromatic number minus 1. 781 01:00:18,316 --> 01:00:23,128 So that's some lower-bound construction. 782 01:00:23,128 --> 01:00:25,000 So as we saw earlier, and we know 783 01:00:25,000 --> 01:00:29,170 what the asymptotic is like for the number of edges, 784 01:00:29,170 --> 01:00:31,769 this n goes to infinity. 785 01:00:31,769 --> 01:00:34,970 Namely, it's like that. 786 01:00:38,071 --> 01:00:41,340 And now the question is, is this the right answer? 787 01:00:41,340 --> 01:00:43,410 Is it possible that we completely 788 01:00:43,410 --> 01:00:48,680 missed some construction that might produce a lot more edges? 789 01:00:48,680 --> 01:00:49,710 And it turns out-- 790 01:00:49,710 --> 01:00:52,080 and I think this should be somewhat surprising. 791 01:00:52,080 --> 01:00:55,230 Because so far, I feel like I haven't told you anything 792 01:00:55,230 --> 01:00:58,950 all that surprising yet. 793 01:00:58,950 --> 01:01:01,590 This seems like a fairly mysterious problem. 794 01:01:01,590 --> 01:01:06,210 But it turns out that this is more or less the right answer. 795 01:01:06,210 --> 01:01:11,040 So you cannot do much better than the Turán graph. 796 01:01:11,040 --> 01:01:20,090 And there's the theorem of Erdos, Stone, and Simonovits 797 01:01:20,090 --> 01:01:36,240 rich that for every graph H-- 798 01:01:36,240 --> 01:01:42,890 so if I fixed H, then the limit as n 799 01:01:42,890 --> 01:01:49,340 goes to infinity of the extremal number 800 01:01:49,340 --> 01:01:52,380 is a fraction of total number of pairs. 801 01:01:52,380 --> 01:01:55,270 So here, this is the edge density. 802 01:01:55,270 --> 01:02:04,590 So this quantity here is equal to 1 minus chi of H. 803 01:02:04,590 --> 01:02:08,230 So 1 minus 1 over chi of H minus 1. 804 01:02:08,230 --> 01:02:11,410 So the chromatic number, in some sense, 805 01:02:11,410 --> 01:02:18,530 completely determines how big the extremal number should be. 806 01:02:18,530 --> 01:02:22,010 And you see that so far everything we've proved, 807 01:02:22,010 --> 01:02:25,140 Turán's theorem, Mantel's theorem agree with this formula 808 01:02:25,140 --> 01:02:25,640 here. 809 01:02:25,640 --> 01:02:30,160 But if I give you some H, maybe quite complicated, 810 01:02:30,160 --> 01:02:32,560 and those previous proofs don't work, 811 01:02:32,560 --> 01:02:37,410 well, still you know the first order asymptotics. 812 01:02:37,410 --> 01:02:38,660 But there's still more to say. 813 01:02:38,660 --> 01:02:41,030 But first, let me just run through some examples 814 01:02:41,030 --> 01:02:42,370 for a sanity check. 815 01:02:42,370 --> 01:02:46,774 So if H is a triangle-- 816 01:02:52,702 --> 01:02:59,340 so if H, the chromatic number, and this limit-- 817 01:02:59,340 --> 01:03:03,550 so if H is the triangle, chromatic number is 3. 818 01:03:03,550 --> 01:03:06,230 So then this limit is 1/2. 819 01:03:06,230 --> 01:03:07,750 And that's indeed the case. 820 01:03:07,750 --> 01:03:10,240 So that's what we did with Mantel's theorem. 821 01:03:10,240 --> 01:03:12,850 If H is a clique of four vertices, 822 01:03:12,850 --> 01:03:16,510 then the chromatic number is 4. 823 01:03:16,510 --> 01:03:20,670 And the answer is 2/3. 824 01:03:20,670 --> 01:03:24,120 And also, that agrees with Turán's theorem. 825 01:03:24,120 --> 01:03:25,630 It agrees with Turán's theorem. 826 01:03:25,630 --> 01:03:30,100 But I can give you some fairly complicated-looking H. 827 01:03:30,100 --> 01:03:32,920 So for example, H might be the Petersen graph. 828 01:03:39,680 --> 01:03:41,200 So every good graph theory course 829 01:03:41,200 --> 01:03:45,710 should feature a Petersen graph at least once, somewhere. 830 01:03:45,710 --> 01:03:48,660 So that's the Petersen graph. 831 01:03:48,660 --> 01:03:50,946 What's the chromatic number of the Petersen graph? 832 01:03:50,946 --> 01:03:52,738 Actually, that's kind of a tricky question. 833 01:03:52,738 --> 01:03:54,196 The last time I taught this course, 834 01:03:54,196 --> 01:03:55,710 I even got the answer wrong. 835 01:03:55,710 --> 01:04:00,050 So it turns out you can three-color the Petersen graph. 836 01:04:00,050 --> 01:04:03,390 So it's completely not obvious how to do this. 837 01:04:03,390 --> 01:04:06,660 But there are only so many vertices. 838 01:04:06,660 --> 01:04:09,240 If you stare at it long enough, you can see what happens. 839 01:04:09,240 --> 01:04:12,220 But let me just show you a three-coloring of the Petersen 840 01:04:12,220 --> 01:04:12,720 graph. 841 01:04:21,135 --> 01:04:26,640 Then the third color is like that. 842 01:04:26,640 --> 01:04:30,880 So that's a three-coloring of the Petersen graph. 843 01:04:30,880 --> 01:04:32,340 That's chromatic number 3. 844 01:04:32,340 --> 01:04:33,320 It's not 2. 845 01:04:33,320 --> 01:04:35,146 So why is it not 2? 846 01:04:35,146 --> 01:04:36,460 AUDIENCE: It's not bipartite. 847 01:04:36,460 --> 01:04:37,752 YUFEI ZHAO: It's not bipartite. 848 01:04:37,752 --> 01:04:40,770 It's a five cycle, which cannot be two-colored. 849 01:04:40,770 --> 01:04:44,230 So then the limit is 1/2. 850 01:04:44,230 --> 01:04:47,410 And-- no. 851 01:04:47,410 --> 01:04:50,070 That's 3. 852 01:04:50,070 --> 01:04:51,030 Great. 853 01:04:51,030 --> 01:04:52,740 So the limit is 1/2. 854 01:04:52,740 --> 01:04:55,650 And here, I think that we can apply this theorem 855 01:04:55,650 --> 01:04:58,350 of Erdos-Stone-Simonovits. 856 01:04:58,350 --> 01:05:01,420 And I think this should be somewhat surprising. 857 01:05:01,420 --> 01:05:04,120 Because the Petersen graph looks quite complicated. 858 01:05:04,120 --> 01:05:07,560 If you try to forbid this graph in some big G, 859 01:05:07,560 --> 01:05:10,380 it seems like it's kind of hard to do. 860 01:05:10,380 --> 01:05:12,380 But it turns out the chromatic number completely 861 01:05:12,380 --> 01:05:18,120 governs the behavior of the extremal numbers. 862 01:05:18,120 --> 01:05:23,510 But it turns out that's not the entire story. 863 01:05:23,510 --> 01:05:26,280 Because while this is quite a good theorem, and I said, 864 01:05:26,280 --> 01:05:28,700 it gives you the first-order asymptotics, 865 01:05:28,700 --> 01:05:30,000 actually, that's a lie. 866 01:05:30,000 --> 01:05:34,010 It doesn't always give you the first-order asymptotics. 867 01:05:34,010 --> 01:05:37,830 And when is Erdos-Stone-Simonovits 868 01:05:37,830 --> 01:05:39,300 not effective? 869 01:05:39,300 --> 01:05:41,130 Or rather, it's not the complete-- 870 01:05:41,130 --> 01:05:43,490 it's not the final answer. 871 01:05:43,490 --> 01:05:48,280 Well, you can say, well, what about this little o? 872 01:05:48,280 --> 01:05:52,610 So we're still trying to understand, there's a limit. 873 01:05:52,610 --> 01:05:56,860 And you can understand what is the-- 874 01:05:56,860 --> 01:06:00,810 how quickly does it converge to this number here? 875 01:06:00,810 --> 01:06:03,830 So that's certainly a valid question. 876 01:06:03,830 --> 01:06:09,654 But more importantly, though, if your graphic is bipartite, 877 01:06:09,654 --> 01:06:19,103 if chi of H equals to 2, i.e. bipartite, 878 01:06:19,103 --> 01:06:22,190 then all this theorem tells you is 879 01:06:22,190 --> 01:06:26,570 that the limit is equal to 0, which somehow is not 880 01:06:26,570 --> 01:06:28,010 the most satisfying answer. 881 01:06:28,010 --> 01:06:30,590 You want to know the first-order asymptotics. 882 01:06:30,590 --> 01:06:32,240 I mean, this still tells you something. 883 01:06:32,240 --> 01:06:34,790 So the Erdos-Stone-Simonovits theorem 884 01:06:34,790 --> 01:06:38,230 tells you that the extremal number 885 01:06:38,230 --> 01:06:42,268 is little o of n squared. 886 01:06:42,268 --> 01:06:43,180 But of course, no. 887 01:06:43,180 --> 01:06:44,760 You are a curious mathematician. 888 01:06:44,760 --> 01:06:48,950 And you want to know, really, what is the asymptotics? 889 01:06:48,950 --> 01:06:51,620 Is it like n to the 3/2? 890 01:06:51,620 --> 01:06:53,840 Is it like n to the 4/3? 891 01:06:53,840 --> 01:06:56,670 You know, is this is not satisfying. 892 01:06:56,670 --> 01:07:01,810 And it turns out, from most bipartite graphs H, 893 01:07:01,810 --> 01:07:04,040 it is a very difficult problem that we still 894 01:07:04,040 --> 01:07:06,050 do not know the answer to, even what is 895 01:07:06,050 --> 01:07:08,510 the order of the asymptotics. 896 01:07:08,510 --> 01:07:11,240 So the next few lectures, what I want to do 897 01:07:11,240 --> 01:07:13,400 is to show you some techniques that 898 01:07:13,400 --> 01:07:18,550 will allow you to prove some upper bounds 899 01:07:18,550 --> 01:07:21,550 for this extremal number in the case 900 01:07:21,550 --> 01:07:25,090 when H is bipartite that shows you that the exponent can 901 01:07:25,090 --> 01:07:27,500 be less than 2. 902 01:07:27,500 --> 01:07:31,490 And I will also show you some constructions 903 01:07:31,490 --> 01:07:34,940 that sometimes, but in the very few cases, 904 01:07:34,940 --> 01:07:37,020 matches the upper bound. 905 01:07:37,020 --> 01:07:41,630 So there are very few examples of graphs H for which we know 906 01:07:41,630 --> 01:07:43,860 the first-order asymptotics. 907 01:07:43,860 --> 01:07:47,450 And for most graph H, they are major open problems 908 01:07:47,450 --> 01:07:48,465 in combinatorics. 909 01:07:48,465 --> 01:07:50,840 And there are some really old ones for which any solution 910 01:07:50,840 --> 01:07:52,280 may be quite exciting. 911 01:07:56,580 --> 01:07:59,730 I will not show you a proof of Erdos-Stone-Simonovits now. 912 01:07:59,730 --> 01:08:01,290 We will see that later in the term, 913 01:08:01,290 --> 01:08:03,870 once we have developed some more machinery. 914 01:08:03,870 --> 01:08:07,320 Although, later on in the term, we'll see a proof using 915 01:08:07,320 --> 01:08:09,450 the so-called Szemerédi's regularity lemma, 916 01:08:09,450 --> 01:08:12,720 which I've mentioned a few times in the first lecture. 917 01:08:12,720 --> 01:08:16,270 Although, you don't actually need such heavy machinery. 918 01:08:16,270 --> 01:08:19,500 So the original proofs of Erdos and Stone, 919 01:08:19,500 --> 01:08:21,570 and also by Simonovits-- 920 01:08:21,570 --> 01:08:25,290 so Erdos and Stone first proved this result for H 921 01:08:25,290 --> 01:08:27,750 being a complete multipartite graph. 922 01:08:27,750 --> 01:08:30,750 And Simonovits observed that knowing 923 01:08:30,750 --> 01:08:32,760 H for a complete multipartite graph 924 01:08:32,760 --> 01:08:35,109 actually implies this result in general. 925 01:08:35,109 --> 01:08:38,430 So you will not find a paper with all three 926 01:08:38,430 --> 01:08:41,170 of them being authors. 927 01:08:41,170 --> 01:08:45,328 But this is what the theorem is called. 928 01:08:45,328 --> 01:08:47,870 So later on in the course, once we've developed the machinery 929 01:08:47,870 --> 01:08:51,859 of graph regularity lemmas, I will show you how you can 930 01:08:51,859 --> 01:08:56,992 deduce Erdos-Stone-Simonovits from Turán's theorem. 931 01:08:56,992 --> 01:09:00,140 So somehow you use Turán's theorem and bootstrap it 932 01:09:00,140 --> 01:09:04,380 to Erdos-Stone-Simonovits. 933 01:09:04,380 --> 01:09:06,810 So that should seem somewhat magical. 934 01:09:06,810 --> 01:09:08,580 So on one hand, you have cliques. 935 01:09:08,580 --> 01:09:11,399 On the other hand, you have things that somehow 936 01:09:11,399 --> 01:09:12,750 don't have cliques in them. 937 01:09:12,750 --> 01:09:14,250 They don't really look like cliques. 938 01:09:14,250 --> 01:09:16,460 But you can still bootstrap one to the other. 939 01:09:16,460 --> 01:09:19,170 And so after we develop some machinery, we'll do that. 940 01:09:19,170 --> 01:09:21,180 But there is a combinatorial proof 941 01:09:21,180 --> 01:09:24,689 which doesn't use any of this heavy machinery. 942 01:09:24,689 --> 01:09:27,750 I won't discuss it in lecture, but you can look it up. 943 01:09:27,750 --> 01:09:29,200 And it's quite nice. 944 01:09:29,200 --> 01:09:30,930 It has some combinatorial techniques 945 01:09:30,930 --> 01:09:33,004 and some double-counting arguments. 946 01:09:37,840 --> 01:09:42,880 So going forward, I want to show you some questions, mostly 947 01:09:42,880 --> 01:09:44,609 questions, but some answers as well-- 948 01:09:44,609 --> 01:09:49,640 so questions such as, what is the extremal number of-- 949 01:09:49,640 --> 01:09:53,920 well, so what are some of the basic bipartite graphs? 950 01:09:53,920 --> 01:09:56,620 One of them is just a complete bipartite graph. 951 01:09:59,140 --> 01:10:00,000 So this has a name. 952 01:10:00,000 --> 01:10:02,170 It's called the Zarankiewicz problem. 953 01:10:02,170 --> 01:10:06,650 And for some values of s and t, we know the answer. 954 01:10:06,650 --> 01:10:09,700 But for most values, we have no idea what the answer is. 955 01:10:09,700 --> 01:10:16,510 For example for K4,4, we do not know what is the correct 956 01:10:16,510 --> 01:10:19,472 exponent on the n. 957 01:10:19,472 --> 01:10:23,020 So even fairly small cases, it is very much open. 958 01:10:23,020 --> 01:10:25,240 So this is all to show you that these simple proofs 959 01:10:25,240 --> 01:10:29,425 that we did today, they are perhaps too deceptively simple. 960 01:10:29,425 --> 01:10:32,587 Because even if you change the question a little bit, 961 01:10:32,587 --> 01:10:33,670 we don't understand a lot. 962 01:10:36,540 --> 01:10:38,520 So questions like these will occupy 963 01:10:38,520 --> 01:10:39,630 the next several lectures. 964 01:10:39,630 --> 01:10:41,390 And I'll also show you some constructions. 965 01:10:41,390 --> 01:10:42,932 And there are some constructions that 966 01:10:42,932 --> 01:10:45,540 use nice algebraic ideas and some probabilistic ideas. 967 01:10:45,540 --> 01:10:49,730 So we'll see ideas coming from many different sources. 968 01:10:49,730 --> 01:10:53,360 I want to close off by just a cultural remark. 969 01:10:53,360 --> 01:10:57,110 So in this course, especially the first half, 970 01:10:57,110 --> 01:11:00,530 we'll encounter a lot of Hungarian names. 971 01:11:00,530 --> 01:11:02,600 So we already saw a couple of them. 972 01:11:02,600 --> 01:11:05,000 And I just want to give you a very quick tutorial 973 01:11:05,000 --> 01:11:08,490 on how to pronounce Hungarian names, just 974 01:11:08,490 --> 01:11:10,370 for cultural purposes now. 975 01:11:10,370 --> 01:11:12,380 In the past, when I took this CLASS 976 01:11:12,380 --> 01:11:14,000 there were no Hungarian speakers. 977 01:11:14,000 --> 01:11:16,560 But I think we do have at least one Hungarian speaker 978 01:11:16,560 --> 01:11:17,060 in the room. 979 01:11:17,060 --> 01:11:18,940 So I want you-- 980 01:11:18,940 --> 01:11:21,020 but you are a native Hungarian speaker. 981 01:11:21,020 --> 01:11:22,580 So you should tell us. 982 01:11:22,580 --> 01:11:26,288 So can you help us pronounce these names? 983 01:11:26,288 --> 01:11:27,530 AUDIENCE: Erdos. 984 01:11:27,530 --> 01:11:29,150 YUFEI ZHAO: Erdos. 985 01:11:29,150 --> 01:11:30,900 And this one? 986 01:11:30,900 --> 01:11:33,720 AUDIENCE: That doesn't seem like a Hungarian name. 987 01:11:33,720 --> 01:11:35,130 [LAUGHTER] 988 01:11:36,080 --> 01:11:40,790 YUFEI ZHAO: So this, Hungarian, it's Simonovits. 989 01:11:40,790 --> 01:11:45,860 So I'll just tell you two things about Hungarian names. 990 01:11:45,860 --> 01:11:47,637 One of them is that the S-- 991 01:11:47,637 --> 01:11:48,470 AUDIENCE: It's /sh/. 992 01:11:48,470 --> 01:11:52,160 YUFEI ZHAO: --is pronounced like /sh/. 993 01:11:52,160 --> 01:11:55,070 And another thing that comes up is S-Z, 994 01:11:55,070 --> 01:11:56,870 which we saw in Szemerédi. 995 01:11:56,870 --> 01:11:58,560 So this is pronounced like /s/. 996 01:11:58,560 --> 01:12:01,275 Forget the Z. 997 01:12:01,275 --> 01:12:08,050 So Erdos, another thing about Erdos that you should know 998 01:12:08,050 --> 01:12:09,190 is that-- 999 01:12:09,190 --> 01:12:11,530 what is this accent? 1000 01:12:11,530 --> 01:12:13,502 It is not a double dot. 1001 01:12:13,502 --> 01:12:18,730 So in LaTeX you would type it as slash H, 1002 01:12:18,730 --> 01:12:24,460 and in particular, not like that. 1003 01:12:24,460 --> 01:12:31,320 So just a few cultural remarks about names. 1004 01:12:31,320 --> 01:12:31,820 Great. 1005 01:12:31,820 --> 01:12:32,690 So we'll end today. 1006 01:12:32,690 --> 01:12:34,910 And then next time, we'll start looking 1007 01:12:34,910 --> 01:12:40,260 at other extremal numbers for more bipartite graphs.