1 00:00:17,325 --> 00:00:19,270 YUFEI ZHAO: For the last few lectures, 2 00:00:19,270 --> 00:00:23,460 we've been talking about the extremal problem of forbidding 3 00:00:23,460 --> 00:00:26,190 a complete bipartite graph. 4 00:00:26,190 --> 00:00:29,790 So today I want to move beyond the complete bipartite graph 5 00:00:29,790 --> 00:00:34,420 and look at other sparser bipartite graphs. 6 00:00:34,420 --> 00:00:38,760 So we'll be looking at what happens to the extremal problem 7 00:00:38,760 --> 00:00:44,590 if you forbid a sparse bipartite graph. 8 00:00:56,130 --> 00:01:00,900 Recall the Kovari-Sos-Turan theorem, 9 00:01:00,900 --> 00:01:07,560 which tells you that the extremal number for KSC 10 00:01:07,560 --> 00:01:15,300 is upper bounded by something on the order of m to the 2 minus 1 11 00:01:15,300 --> 00:01:17,650 over m. 12 00:01:17,650 --> 00:01:24,870 So if I give you some bipartite H, 13 00:01:24,870 --> 00:01:26,910 we know that because it's a bipartite graph, 14 00:01:26,910 --> 00:01:37,500 it is always contained in some KST for some values of S and P. 15 00:01:37,500 --> 00:01:42,450 So you already automatically have an upper bound, 16 00:01:42,450 --> 00:01:46,810 the extremal number for this H from 17 00:01:46,810 --> 00:01:48,220 the Kovari-Sos-Turan theorem. 18 00:01:53,910 --> 00:01:55,910 But as you might expect, we might 19 00:01:55,910 --> 00:01:59,260 be very wasteful in this step. 20 00:01:59,260 --> 00:02:01,180 And the question is, are there situations 21 00:02:01,180 --> 00:02:04,300 where we can do much better than what is just given 22 00:02:04,300 --> 00:02:06,122 by applying Kovari-Sos-Turan? 23 00:02:06,122 --> 00:02:08,289 And today, I want to show you several examples where 24 00:02:08,289 --> 00:02:10,720 you can significantly improve the bound given 25 00:02:10,720 --> 00:02:16,400 by Kovari-Sos-Turan for various sparser bipartite graphs. 26 00:02:16,400 --> 00:02:22,610 The first result is the following theorem, which 27 00:02:22,610 --> 00:02:26,810 is initially due to Faraday. 28 00:02:26,810 --> 00:02:28,670 And then later, a different proof 29 00:02:28,670 --> 00:02:31,280 was given by Alon-Krivelevich-Sudov. 30 00:02:31,280 --> 00:02:33,890 And the latter proof is the one I want to present, 31 00:02:33,890 --> 00:02:37,250 because it presents an important and intricate 32 00:02:37,250 --> 00:02:41,750 probabilistic technique, which is the main reason for showing 33 00:02:41,750 --> 00:02:44,430 you this theorem. 34 00:02:44,430 --> 00:02:46,940 But let me tell you the theorem first. 35 00:02:46,940 --> 00:02:49,130 Here, H will be a bipartite graph. 36 00:02:53,060 --> 00:02:59,310 And it has vertex bipartitions A and B, 37 00:02:59,310 --> 00:03:09,850 such that every vertex in A has degree at most R. The 38 00:03:09,850 --> 00:03:15,430 bipartition is A and B, and the degree in A 39 00:03:15,430 --> 00:03:24,210 is at most R for every vertex on the left side in the set A. 40 00:03:24,210 --> 00:03:26,220 And I want to understand, is there 41 00:03:26,220 --> 00:03:30,450 some upper bound on the extremal number that does better 42 00:03:30,450 --> 00:03:31,920 than Kovari-Sos-Turan theorem? 43 00:03:31,920 --> 00:03:35,170 And the theorem guarantees such a bound. 44 00:03:35,170 --> 00:03:41,560 So then there exists some constant depending on H, such 45 00:03:41,560 --> 00:03:46,630 that the extremal number is upper 46 00:03:46,630 --> 00:03:53,110 bounded by something to the order of n to the 2 minus 1 47 00:03:53,110 --> 00:03:57,500 over R. 48 00:03:57,500 --> 00:04:00,560 Compared with Kovari-Sos-Turan theorem, on one hand, 49 00:04:00,560 --> 00:04:04,130 if your H is the complete bipartite graph KST, 50 00:04:04,130 --> 00:04:08,160 then this is the same bound as Kovari-Sos-Turan theorem. 51 00:04:08,160 --> 00:04:11,790 On the other hand, you might have a lot more vertices in A 52 00:04:11,790 --> 00:04:15,480 and a lot more vertices in B. The hypothesis only 53 00:04:15,480 --> 00:04:22,362 requires what the degrees in A, max degrees, at most are. 54 00:04:22,362 --> 00:04:23,820 So it could be a much bigger graph. 55 00:04:23,820 --> 00:04:25,730 So if you apply Kovari-Sos-Turan, 56 00:04:25,730 --> 00:04:29,090 you would get a much worse bound compared to what 57 00:04:29,090 --> 00:04:30,806 this theorem guarantees. 58 00:04:37,760 --> 00:04:39,800 This 1 of R is optimal. 59 00:04:39,800 --> 00:04:43,460 In this given statement, you cannot improve this 1 of R, 60 00:04:43,460 --> 00:04:48,470 because we know that from the KST example and the lower 61 00:04:48,470 --> 00:04:53,510 bounds I showed you last time, you cannot improve upon this 1 62 00:04:53,510 --> 00:04:56,970 over R. Also, in this form, this theorem is best possible. 63 00:05:08,510 --> 00:05:11,180 I want to show you a probabilistic technique 64 00:05:11,180 --> 00:05:12,710 for proving this theorem. 65 00:05:15,570 --> 00:05:18,510 And this is an important idea called dependent random choice. 66 00:05:33,670 --> 00:05:36,950 So let me first give you an informal interpretation 67 00:05:36,950 --> 00:05:38,170 of what's going on. 68 00:05:38,170 --> 00:05:46,960 So the idea is that if you have a graph G with many edges, 69 00:05:46,960 --> 00:05:53,440 then inside G, I can find a large subset of vertices 70 00:05:53,440 --> 00:06:06,360 U, such that all small subsets of vertices in U 71 00:06:06,360 --> 00:06:11,137 have many common neighbors. 72 00:06:21,560 --> 00:06:24,640 I won't tell you what the small and many are just yet, 73 00:06:24,640 --> 00:06:27,800 but we'll see it through the proof of the theorem, 74 00:06:27,800 --> 00:06:28,880 but that's the idea. 75 00:06:28,880 --> 00:06:29,880 So I give you a graph. 76 00:06:29,880 --> 00:06:30,830 It's not too sparse. 77 00:06:30,830 --> 00:06:32,630 It's relatively dense. 78 00:06:32,630 --> 00:06:36,500 Then I should be able to find some subset that 79 00:06:36,500 --> 00:06:39,650 is fairly large so that, let's say, 80 00:06:39,650 --> 00:06:41,960 every pair of vertices in the subset 81 00:06:41,960 --> 00:06:45,570 has many common neighbors. 82 00:06:45,570 --> 00:06:46,660 So let me write down-- 83 00:06:46,660 --> 00:06:52,236 or at least attempt to write down-- the formal statement 84 00:06:52,236 --> 00:06:53,820 of dependent random choice. 85 00:06:56,870 --> 00:06:58,960 The statement of the theorem, as I will present, 86 00:06:58,960 --> 00:07:01,940 has a lot of parameters, but don't get-- 87 00:07:01,940 --> 00:07:04,120 I don't want you to be scared off by the parameters, 88 00:07:04,120 --> 00:07:06,140 so I won't even tell you what they are, 89 00:07:06,140 --> 00:07:08,890 but first tell you what the statement of the conclusion is. 90 00:07:08,890 --> 00:07:11,080 And then we'll derive what is the dependencies 91 00:07:11,080 --> 00:07:13,480 on the parameters, so the proof or the technique 92 00:07:13,480 --> 00:07:15,330 is much more important than the statement 93 00:07:15,330 --> 00:07:17,550 of the theorem itself. 94 00:07:17,550 --> 00:07:20,230 So I'll leave some space here. 95 00:07:20,230 --> 00:07:30,290 The conclusion is that every graph with n vertices 96 00:07:30,290 --> 00:07:39,600 and at least alpha n squared over 2 edges 97 00:07:39,600 --> 00:07:45,945 contains a subset U of vertices. 98 00:07:49,200 --> 00:07:55,240 U is not too small, so the size of U is at least little u. 99 00:07:55,240 --> 00:08:07,060 And such that for every subset S of U with R elements, the set S 100 00:08:07,060 --> 00:08:12,232 has at least M common neighbors. 101 00:08:21,400 --> 00:08:22,750 So what's the idea here? 102 00:08:22,750 --> 00:08:24,370 I give you this graph, and I want 103 00:08:24,370 --> 00:08:28,980 you to produce the set U that has this property. 104 00:08:28,980 --> 00:08:31,470 How might you go about finding the set U? 105 00:08:31,470 --> 00:08:33,120 Let me give you an analogy. 106 00:08:33,120 --> 00:08:36,390 Let's suppose you have the friendship graph on MIT campus. 107 00:08:36,390 --> 00:08:38,950 And I want to select a large set, 108 00:08:38,950 --> 00:08:42,600 let's say a hundred students, such that every pair of them 109 00:08:42,600 --> 00:08:48,250 or maybe even just most pairs of them have many common friends. 110 00:08:48,250 --> 00:08:50,460 Well, how might you go about doing that? 111 00:08:50,460 --> 00:08:52,850 Well, if you select a hundred students at random, 112 00:08:52,850 --> 00:08:55,330 you're unlikely to achieve that outcome. 113 00:08:55,330 --> 00:09:00,280 They're going to be pretty dispersed across campus. 114 00:09:00,280 --> 00:09:04,320 But if you focus on some commonalities-- 115 00:09:04,320 --> 00:09:07,920 so, for example, you go to someone, 116 00:09:07,920 --> 00:09:10,080 some specific individual, and look 117 00:09:10,080 --> 00:09:13,630 at their circle of close friends, 118 00:09:13,630 --> 00:09:16,390 then it seems more likely you'll be 119 00:09:16,390 --> 00:09:19,353 able to identify a group of people 120 00:09:19,353 --> 00:09:21,520 who are very well-connected in that they, peer-wise, 121 00:09:21,520 --> 00:09:24,920 have lots of common friends. 122 00:09:24,920 --> 00:09:26,310 And that's the idea here. 123 00:09:26,310 --> 00:09:30,030 We're going to make the random choice by picking 124 00:09:30,030 --> 00:09:33,390 that core individual-- 125 00:09:33,390 --> 00:09:34,690 that's the random choice-- 126 00:09:34,690 --> 00:09:37,800 but then make the subsequent dependent random choice 127 00:09:37,800 --> 00:09:40,560 by looking at the group of friends 128 00:09:40,560 --> 00:09:44,120 from that specific random individual 129 00:09:44,120 --> 00:09:47,570 instead of choosing the hundred people uniformly at random, 130 00:09:47,570 --> 00:09:50,460 which is not going to work. 131 00:09:50,460 --> 00:09:52,580 So let's execute that strategy on this graph. 132 00:09:56,740 --> 00:10:03,820 Let me take T to be the random set, so that's the core set. 133 00:10:03,820 --> 00:10:08,380 So let T be-- 134 00:10:08,380 --> 00:10:11,410 for convenience, I'm going to choose with repetition. 135 00:10:11,410 --> 00:10:12,910 Instead of just choosing one person, 136 00:10:12,910 --> 00:10:19,480 I'm going to use two T vertices, so a list of T vertices 137 00:10:19,480 --> 00:10:32,102 chosen uniformly at random from the vertex set. 138 00:10:32,102 --> 00:10:35,210 So G is going to be my graph. 139 00:10:35,210 --> 00:10:41,010 So the graph is G. 140 00:10:41,010 --> 00:10:46,200 And what we are going to do is look at the set A, which 141 00:10:46,200 --> 00:10:56,660 is the set of common neighbors of T, the vertices that 142 00:10:56,660 --> 00:11:03,280 are adjacent to all of T. And that's the set 143 00:11:03,280 --> 00:11:05,080 I want to think about. 144 00:11:05,080 --> 00:11:08,380 So I want to basically argue that this set 145 00:11:08,380 --> 00:11:12,010 A has more or less the properties that we desire, 146 00:11:12,010 --> 00:11:14,980 maybe with some small number of blemishes, which 147 00:11:14,980 --> 00:11:17,080 we will fix by cleaning up. 148 00:11:20,780 --> 00:11:25,370 First, I want to guarantee that A has a big size-- 149 00:11:25,370 --> 00:11:28,970 we are actually choosing a lot of vertices. 150 00:11:28,970 --> 00:11:34,340 So let's evaluate the size of A in expectation. 151 00:11:34,340 --> 00:11:37,130 By linearity of expectation, we need 152 00:11:37,130 --> 00:11:39,620 to compute the sum of the probabilities 153 00:11:39,620 --> 00:11:48,610 that individual vertices fall in this random A. 154 00:11:48,610 --> 00:11:52,370 For each particular vertex V, when is it in A? 155 00:11:52,370 --> 00:11:59,040 Well, it is in A if T is contained 156 00:11:59,040 --> 00:12:04,430 in the neighborhood of this V. So all of your children 157 00:12:04,430 --> 00:12:06,620 T fall into the neighborhood. 158 00:12:06,620 --> 00:12:10,350 Otherwise, V is not going to be contained in A. 159 00:12:10,350 --> 00:12:15,830 So each individual probability can be computed easily 160 00:12:15,830 --> 00:12:20,600 by looking at the size, so the degree of V 161 00:12:20,600 --> 00:12:24,350 divided by the number of vertices raised to the power 162 00:12:24,350 --> 00:12:25,961 T-- 163 00:12:25,961 --> 00:12:31,720 T independent choices chosen with replacements. 164 00:12:31,720 --> 00:12:38,030 And by convexity applied to the final expression, 165 00:12:38,030 --> 00:12:43,600 we find that it is at least this quantity here where essentially 166 00:12:43,600 --> 00:12:47,230 we're taking the averages of the degrees. 167 00:12:51,600 --> 00:12:52,920 So this is by convexity. 168 00:12:56,470 --> 00:13:00,310 And finally, the graph has at least 169 00:13:00,310 --> 00:13:03,310 that many edges, so the final quantity there 170 00:13:03,310 --> 00:13:11,725 is at least N alpha to the T. Yes, question? 171 00:13:11,725 --> 00:13:13,937 AUDIENCE: [INAUDIBLE] 172 00:13:13,937 --> 00:13:16,020 YUFEI ZHAO: The question is, in our original list, 173 00:13:16,020 --> 00:13:17,800 are we allowed to have repeated vertices? 174 00:13:17,800 --> 00:13:18,300 Yes. 175 00:13:18,300 --> 00:13:21,120 So it's a list, and we're choosing every element 176 00:13:21,120 --> 00:13:23,860 independently at random, allowing repetition. 177 00:13:23,860 --> 00:13:27,143 So it's sometimes called choosing with replacement. 178 00:13:27,143 --> 00:13:28,560 You choose a T. You throw it back. 179 00:13:28,560 --> 00:13:29,352 Choose another one. 180 00:13:39,600 --> 00:13:43,140 The property that we're looking for is that for every S subset, 181 00:13:43,140 --> 00:13:47,650 for every r-element subset of U, it has many common neighbors. 182 00:13:47,650 --> 00:13:55,160 So let's look at such an S. So for each S 183 00:13:55,160 --> 00:13:58,960 which is an r-element subset of the vertex set. 184 00:13:58,960 --> 00:14:03,970 So r-elements subset of V, what is the probability 185 00:14:03,970 --> 00:14:09,000 that this set F is contained in A? 186 00:14:12,696 --> 00:14:17,690 I'll give you this set S. 187 00:14:17,690 --> 00:14:20,600 It is contained in A if-- 188 00:14:20,600 --> 00:14:23,960 well, let's think about how A is chosen. 189 00:14:27,750 --> 00:14:29,740 You want this to happen. 190 00:14:29,740 --> 00:14:34,140 An A is chosen as a common neighborhood of this T-- 191 00:14:34,140 --> 00:14:40,350 so assets containing A, if and only if, P 192 00:14:40,350 --> 00:14:48,760 is contained in the common neighborhood of f. 193 00:14:53,000 --> 00:14:55,250 So f is fixed for now. 194 00:14:55,250 --> 00:14:58,130 P is random. 195 00:14:58,130 --> 00:15:01,750 So we draw the elements of T independently, uniformly, 196 00:15:01,750 --> 00:15:02,530 at random. 197 00:15:02,530 --> 00:15:06,940 Therefore, this probability is equal to the number 198 00:15:06,940 --> 00:15:17,400 of common neighbors of S as a fraction of the total number 199 00:15:17,400 --> 00:15:18,390 of vertices. 200 00:15:18,390 --> 00:15:19,800 This fraction raised to the power 201 00:15:19,800 --> 00:15:31,240 T. We want all S subsets of A-- all r-element subsets 202 00:15:31,240 --> 00:15:35,620 of this A-- to have at least m common neighbors. 203 00:15:35,620 --> 00:15:37,570 But maybe we cannot get that. 204 00:15:37,570 --> 00:15:40,930 So let's figure out how many bad S's are there. 205 00:15:40,930 --> 00:15:45,640 How many S's do not satisfy this condition here? 206 00:15:45,640 --> 00:15:56,150 So let's call such a set S bad if it has 207 00:15:56,150 --> 00:15:59,868 fewer than m common neighbors. 208 00:16:07,520 --> 00:16:18,960 And from this equation, we see that for each fixed S that is 209 00:16:18,960 --> 00:16:25,330 in r-element subset of vertices, it is bad with probability-- 210 00:16:28,710 --> 00:16:30,630 it is bad if it has few common neighbors. 211 00:16:30,630 --> 00:16:32,338 Because if you have few common neighbors, 212 00:16:32,338 --> 00:16:34,780 then this probability is small. 213 00:16:34,780 --> 00:16:42,010 So it is bad with probability strictly less than m over n 214 00:16:42,010 --> 00:16:49,320 raised to the power T. 215 00:16:49,320 --> 00:16:55,480 We chose this A in this dependent random way. 216 00:16:55,480 --> 00:16:58,000 And basically, we want all subsets. 217 00:16:58,000 --> 00:17:01,270 All r-element subsets have many common neighbors, 218 00:17:01,270 --> 00:17:03,850 but maybe we cannot get that at the first try. 219 00:17:03,850 --> 00:17:07,150 But we only have a small number of blemishes, 220 00:17:07,150 --> 00:17:11,010 so we can fix those blemishes by getting 221 00:17:11,010 --> 00:17:14,490 rid of the possible bad r subsets. 222 00:17:17,060 --> 00:17:19,700 And that's fine to do, as long as there are not 223 00:17:19,700 --> 00:17:21,800 too many of them. 224 00:17:21,800 --> 00:17:28,040 And indeed, because S is bad with small probability, 225 00:17:28,040 --> 00:17:40,740 the expected number of bad r-element subsets of A is, 226 00:17:40,740 --> 00:17:42,390 at most-- 227 00:17:42,390 --> 00:17:48,690 well, I look over all possible r-element subsets of vertices. 228 00:17:48,690 --> 00:17:53,640 Each one of them is bad with this probability here. 229 00:17:53,640 --> 00:17:55,110 So we have that bound. 230 00:17:59,860 --> 00:18:03,260 And the point now is that if this number is significantly 231 00:18:03,260 --> 00:18:07,520 smaller than the expected size of A, then 232 00:18:07,520 --> 00:18:10,310 I can clean up all the bad subsets 233 00:18:10,310 --> 00:18:13,790 by plucking out one vertex from each bad subset. 234 00:18:17,410 --> 00:18:21,880 So indeed, that is the case, because the expectation 235 00:18:21,880 --> 00:18:25,270 of the size of A minus-- 236 00:18:25,270 --> 00:18:31,676 so let me call this quantity here star-- 237 00:18:31,676 --> 00:18:39,390 so the size of A minus star by what we have shown 238 00:18:39,390 --> 00:18:46,660 is at least n alpha to the t minus the quantity just now. 239 00:18:49,850 --> 00:18:52,890 And I want this number to be somewhat large. 240 00:18:52,890 --> 00:18:57,590 And now let me put that as a hypothesis of the theory. 241 00:18:57,590 --> 00:19:01,770 So dependent random choice-- 242 00:19:01,770 --> 00:19:14,880 let u, n, r, m, t be positive integers, alpha positive real, 243 00:19:14,880 --> 00:19:22,590 and suppose n alpha to the t minus n to the r. 244 00:19:22,590 --> 00:19:26,566 Cancel the n to the t is at least w. 245 00:19:31,560 --> 00:19:34,500 That's where that inequality comes from. 246 00:19:34,500 --> 00:19:41,300 So if this is at least w, then what we can do 247 00:19:41,300 --> 00:19:56,570 is delete one vertex from each bad subset 248 00:19:56,570 --> 00:20:03,580 S. And after deleting, A then becomes some smaller 249 00:20:03,580 --> 00:20:11,630 set A prime with at least u elements in expectation, 250 00:20:11,630 --> 00:20:14,790 but then there exists some new elements-- 251 00:20:17,680 --> 00:20:18,790 Let me put it this way. 252 00:20:18,790 --> 00:20:25,750 We know that this is true in expectation, thus there 253 00:20:25,750 --> 00:20:31,810 exists some T, such that that inequality is true 254 00:20:31,810 --> 00:20:37,390 without the expectation, such that there exists some T, 255 00:20:37,390 --> 00:20:38,050 such that-- 256 00:20:42,860 --> 00:20:45,380 so this quantity is at least that. 257 00:20:45,380 --> 00:20:48,000 Now, to delete a vertex from each bad subset, 258 00:20:48,000 --> 00:20:51,620 we obtain this A prime with at least u elements. 259 00:20:51,620 --> 00:20:55,170 And we have gotten rid of all the possible bad subsets-- 260 00:20:55,170 --> 00:21:00,880 so no bad r-element subsets. 261 00:21:04,630 --> 00:21:10,440 And that finishes off the proof of dependent random choice. 262 00:21:10,440 --> 00:21:12,480 Just to recap, the idea is that if you 263 00:21:12,480 --> 00:21:17,995 have a dense enough graph, then the conclusion 264 00:21:17,995 --> 00:21:25,780 is that you can find a fairly large subset of vertices, 265 00:21:25,780 --> 00:21:30,510 so every pair, every r-element, have many common neighbors. 266 00:21:30,510 --> 00:21:34,060 And the way you do this, instead of choosing your set at random, 267 00:21:34,060 --> 00:21:38,700 which is not going to work, you choose a small set of anchors. 268 00:21:38,700 --> 00:21:41,090 T, you think of that as the anchors. 269 00:21:41,090 --> 00:21:42,980 You choose a bunch of anchors. 270 00:21:42,980 --> 00:21:46,450 And then you look at their common neighborhoods 271 00:21:46,450 --> 00:21:49,610 and use that as a starting point. 272 00:21:49,610 --> 00:21:51,170 That will almost work. 273 00:21:51,170 --> 00:21:55,160 It might not work perfectly, but then you fix things up 274 00:21:55,160 --> 00:21:58,140 by removing the blemishes. 275 00:21:58,140 --> 00:22:02,030 So this is a very tricky probabilistic idea. 276 00:22:02,030 --> 00:22:04,500 It's also a very important one. 277 00:22:04,500 --> 00:22:07,360 It will allow us to prove the theorem over there 278 00:22:07,360 --> 00:22:11,930 about the extremal numbers of standard degree graphs. 279 00:22:11,930 --> 00:22:13,294 Question? 280 00:22:13,294 --> 00:22:17,677 AUDIENCE: Is the definition of bad set for any subset of V 281 00:22:17,677 --> 00:22:19,625 or is it only for subsets of A? 282 00:22:19,625 --> 00:22:22,420 YUFEI ZHAO: The question is, in the definition of bad, 283 00:22:22,420 --> 00:22:25,110 do I use this definition for all subsets 284 00:22:25,110 --> 00:22:29,080 of V, all r-element subsets, or just subsets of A? 285 00:22:29,080 --> 00:22:33,890 So I use it to mean all subsets of V, because A is random. 286 00:22:33,890 --> 00:22:37,520 A is random, so the definition of bad 287 00:22:37,520 --> 00:22:38,960 does not depend on the randomness. 288 00:22:38,960 --> 00:22:40,820 It only depends on the original graph. 289 00:22:44,264 --> 00:22:52,140 AUDIENCE: How is the badness dependent on any probability? 290 00:22:52,140 --> 00:22:53,848 YUFEI ZHAO: The badness does not-- so the 291 00:22:53,848 --> 00:22:55,515 question us, how does the badness depend 292 00:22:55,515 --> 00:22:56,320 on any probability? 293 00:22:56,320 --> 00:22:58,900 The badness does not depend on the probability, 294 00:22:58,900 --> 00:23:01,930 but A is random. 295 00:23:01,930 --> 00:23:04,580 So the number of bad r-element subsets of A 296 00:23:04,580 --> 00:23:06,910 is a random variable. 297 00:23:06,910 --> 00:23:10,020 So each individual-- so you start with a graph. 298 00:23:10,020 --> 00:23:11,580 Some r-element subsets of graphs. 299 00:23:11,580 --> 00:23:12,450 Some are not. 300 00:23:12,450 --> 00:23:15,280 And now I choose this random A in this dependent random 301 00:23:15,280 --> 00:23:16,380 manner. 302 00:23:16,380 --> 00:23:18,960 And A might contain some bad subsets. 303 00:23:18,960 --> 00:23:24,101 I'm trying to calculate how many bad subsets does A have. 304 00:23:26,870 --> 00:23:28,180 Question? 305 00:23:28,180 --> 00:23:32,386 AUDIENCE: [INAUDIBLE],, because an S is neither bad or not bad. 306 00:23:32,386 --> 00:23:35,491 You said it's bad with probability. 307 00:23:38,760 --> 00:23:41,790 YUFEI ZHAO: So your concern is each S 308 00:23:41,790 --> 00:23:44,130 is bad with probability-- 309 00:23:44,130 --> 00:23:45,420 ah. 310 00:23:45,420 --> 00:23:45,920 Sorry. 311 00:23:45,920 --> 00:23:48,082 AUDIENCE: [INAUDIBLE] 312 00:23:48,082 --> 00:23:48,790 YUFEI ZHAO: Fine. 313 00:23:48,790 --> 00:23:49,430 Yeah. 314 00:23:49,430 --> 00:23:49,930 Thank you. 315 00:23:54,710 --> 00:23:56,970 So each bad subset-- 316 00:23:56,970 --> 00:24:00,990 OK, so for each fixed bad subset, 317 00:24:00,990 --> 00:24:09,870 it is contained in A with probability. 318 00:24:09,870 --> 00:24:11,400 Thank you. 319 00:24:11,400 --> 00:24:13,590 I hope this makes it clear. 320 00:24:13,590 --> 00:24:15,810 So the probability of the-- 321 00:24:15,810 --> 00:24:18,030 the property of being bad is not random, 322 00:24:18,030 --> 00:24:21,848 but it containing A is random. 323 00:24:21,848 --> 00:24:22,348 Question? 324 00:24:22,348 --> 00:24:25,520 AUDIENCE: [INAUDIBLE] 325 00:24:30,798 --> 00:24:32,840 YUFEI ZHAO: So the question is, why is this true? 326 00:24:32,840 --> 00:24:36,550 So why are these two events the same? 327 00:24:36,550 --> 00:24:40,780 And it's kind of hard to steer the definition a bit. 328 00:24:40,780 --> 00:24:45,240 So T is chosen as the common neighborhoods-- 329 00:24:45,240 --> 00:24:45,820 sorry. 330 00:24:45,820 --> 00:24:47,290 You choose T at random. 331 00:24:47,290 --> 00:24:51,140 And you choose A to be the common neighborhood of T. 332 00:24:51,140 --> 00:24:57,210 And so how do you characterize subsets of A 333 00:24:57,210 --> 00:25:01,251 if every element is connected to all of T? 334 00:25:05,160 --> 00:25:06,700 You have to think about it. 335 00:25:06,700 --> 00:25:07,510 Any more questions? 336 00:25:07,510 --> 00:25:08,010 Yes? 337 00:25:08,010 --> 00:25:11,634 AUDIENCE: [INAUDIBLE] 338 00:25:11,634 --> 00:25:13,310 YUFEI ZHAO: We pick T at random. 339 00:25:13,310 --> 00:25:15,008 T is uniform. 340 00:25:15,008 --> 00:25:16,716 So the question is, how are we picking T? 341 00:25:16,716 --> 00:25:18,620 T is uniform at random. 342 00:25:18,620 --> 00:25:19,200 Oh, great. 343 00:25:19,200 --> 00:25:21,780 So the question is, how do we pick the little t 344 00:25:21,780 --> 00:25:23,340 in the theorem statement? 345 00:25:23,340 --> 00:25:24,945 It depends on the application. 346 00:25:24,945 --> 00:25:28,110 And that's a little bit weird, because in the statement 347 00:25:28,110 --> 00:25:30,780 of the theorem, the little t shows up in the inequality, 348 00:25:30,780 --> 00:25:31,980 but not in the conclusion. 349 00:25:31,980 --> 00:25:35,880 So you think of t as an auxiliary parameter. 350 00:25:35,880 --> 00:25:38,520 So the little t comes up in the proof, 351 00:25:38,520 --> 00:25:43,290 but not really in the conclusion. 352 00:25:43,290 --> 00:25:46,297 Any more questions? 353 00:25:46,297 --> 00:25:47,130 It's a tricky lemma. 354 00:25:47,130 --> 00:25:47,922 It's a tricky idea. 355 00:25:52,770 --> 00:25:55,440 So now let me use it to prove the statement over there. 356 00:25:55,440 --> 00:25:57,060 And here, it's not so hard. 357 00:25:57,060 --> 00:26:00,780 So it's mostly an application of this dependent random choice 358 00:26:00,780 --> 00:26:01,575 lemma. 359 00:26:01,575 --> 00:26:06,570 So for all H that's in the theorem-- 360 00:26:09,842 --> 00:26:11,550 so let me prove this lemma-- so for all H 361 00:26:11,550 --> 00:26:16,230 that's in the theorem, there exists a constant C, 362 00:26:16,230 --> 00:26:27,840 such that every graph with at least C n to the 2 minus 1 over 363 00:26:27,840 --> 00:26:38,880 r edges contains a vertex U with-- 364 00:26:38,880 --> 00:26:50,870 so vertex subset U with the size of U equal to B-- 365 00:26:50,870 --> 00:26:54,830 so B comes from the vertex by partition of H. 366 00:26:54,830 --> 00:26:56,810 It's a constant-- 367 00:26:56,810 --> 00:27:12,700 such that every r-element subset in B has-- 368 00:27:12,700 --> 00:27:20,060 so r-element subset in U has lots of common neighbors. 369 00:27:20,060 --> 00:27:22,260 That's at least another constant, 370 00:27:22,260 --> 00:27:26,740 which is the number of vertices in H-- 371 00:27:26,740 --> 00:27:31,770 that many common neighbors. 372 00:27:31,770 --> 00:27:34,820 So you see, it's a direct corollary 373 00:27:34,820 --> 00:27:37,445 of the dependent random choice lemma by setting it 374 00:27:37,445 --> 00:27:41,210 in the right parameters and, indeed, 375 00:27:41,210 --> 00:27:46,920 by the dependent random choice lemma where we choose this T-- 376 00:27:46,920 --> 00:27:49,980 this auxiliary variable T in independent random choice 377 00:27:49,980 --> 00:27:51,170 lemma-- 378 00:27:51,170 --> 00:27:52,670 equal to r. 379 00:27:52,670 --> 00:28:04,040 So suffices to check that there exists of C such that-- 380 00:28:04,040 --> 00:28:08,620 plug it into that expression, that inequality, up there. 381 00:28:08,620 --> 00:28:14,880 So n2Cn to the minus 1 over r-- 382 00:28:14,880 --> 00:28:20,210 raised to the power r minus n choose r. 383 00:28:20,210 --> 00:28:21,770 And then this expression here. 384 00:28:24,580 --> 00:28:28,400 So I'm just plugging in the various graph parameters 385 00:28:28,400 --> 00:28:32,880 into the dependent random choice statement. 386 00:28:32,880 --> 00:28:35,690 And I want to show that you can find a constant C such 387 00:28:35,690 --> 00:28:38,320 that this inequality is true. 388 00:28:38,320 --> 00:28:41,440 And indeed, these exponents, they cancel out, 389 00:28:41,440 --> 00:28:49,310 so the first term is simply 2C raised to the r. 390 00:28:49,310 --> 00:28:52,210 And the second one, because, again, 391 00:28:52,210 --> 00:28:55,160 you notice that the exponents work out just fine, 392 00:28:55,160 --> 00:28:57,145 so it is, at most, a constant. 393 00:29:00,210 --> 00:29:02,550 So you can choose C big enough so that this is true. 394 00:29:07,570 --> 00:29:13,183 So it's a direct verification of the hypothesis of the lemma. 395 00:29:13,183 --> 00:29:15,350 And now we're ready to prove the theorem over there. 396 00:29:31,090 --> 00:29:33,100 Yes, question? 397 00:29:33,100 --> 00:29:35,855 AUDIENCE: How do you have size A plus B common neighbors? 398 00:29:35,855 --> 00:29:37,071 Isn't that the entire graph? 399 00:29:37,071 --> 00:29:39,025 Or can I just [INAUDIBLE]? 400 00:29:39,025 --> 00:29:40,650 YUFEI ZHAO: The question is, how do you 401 00:29:40,650 --> 00:29:42,450 have size A plus B commons neighbors? 402 00:29:42,450 --> 00:29:44,910 So H is fixed. 403 00:29:44,910 --> 00:29:47,640 And A and B are constants, so A plus B 404 00:29:47,640 --> 00:29:51,370 is the number of vertices of H. That's a constant. 405 00:29:51,370 --> 00:29:52,230 AUDIENCE: Oh, sorry. 406 00:29:52,230 --> 00:29:52,932 Oh, OK. 407 00:29:52,932 --> 00:29:55,140 YUFEI ZHAO: Yeah, I'm talking about common neighbors, 408 00:29:55,140 --> 00:29:58,380 not in H, but in the big graph G as n vertices. 409 00:30:02,220 --> 00:30:03,820 Now, I like questions. 410 00:30:03,820 --> 00:30:04,320 It's tricky. 411 00:30:04,320 --> 00:30:06,600 This is a tricky argument, so please do ask questions 412 00:30:06,600 --> 00:30:07,973 if you're confused. 413 00:30:07,973 --> 00:30:09,390 And there are times when I may not 414 00:30:09,390 --> 00:30:13,860 have explained it very well, so please do ask questions. 415 00:30:13,860 --> 00:30:15,360 So let's prove the theorem. 416 00:30:15,360 --> 00:30:16,620 And now we're almost there. 417 00:30:16,620 --> 00:30:19,410 The idea is that we embed the vertices of B 418 00:30:19,410 --> 00:30:22,300 into the big graph G one by one. 419 00:30:22,300 --> 00:30:22,800 Sorry. 420 00:30:22,800 --> 00:30:28,045 First, embed B into the vertices of G using U from the lemma-- 421 00:30:31,160 --> 00:30:35,120 so the lemma that we just stated. 422 00:30:35,120 --> 00:30:37,530 And they claim that once you have done that-- 423 00:30:37,530 --> 00:30:38,800 so the vertices of B-- 424 00:30:44,020 --> 00:30:54,860 and now I need to embed the remaining vertices of A. 425 00:30:54,860 --> 00:31:00,600 And I can do this one by one, because if I 426 00:31:00,600 --> 00:31:08,710 need to embed some vertex of A, has, at most, r neighbors to B. 427 00:31:08,710 --> 00:31:11,380 But I have embedded B in such a way 428 00:31:11,380 --> 00:31:27,080 that they have a lot of common neighbors inside G. 429 00:31:27,080 --> 00:31:28,960 So I can always do it. 430 00:31:28,960 --> 00:31:31,700 So I can always embed the vertices of A one 431 00:31:31,700 --> 00:31:37,010 at a time in such a way that I even avoid collisions. 432 00:31:37,010 --> 00:31:41,730 I don't allow vertices to be embedded into the same place. 433 00:31:50,300 --> 00:31:54,690 This is all using that B, embedding a B, which is U, 434 00:31:54,690 --> 00:31:59,040 has many common neighbors in G. So once you put that in. 435 00:31:59,040 --> 00:32:03,360 And the rest, you just make one choice at a time. 436 00:32:03,360 --> 00:32:06,840 And you need to embed a second vertex, 437 00:32:06,840 --> 00:32:09,450 or you can find somewhere in their common neighborhood that 438 00:32:09,450 --> 00:32:11,511 allows you to do it. 439 00:32:11,511 --> 00:32:13,400 So you embed the vertices one at a time, 440 00:32:13,400 --> 00:32:15,317 and then you finish embedding the whole graph. 441 00:32:18,450 --> 00:32:21,090 Any questions? 442 00:32:21,090 --> 00:32:22,680 It's a tricky argument. 443 00:32:22,680 --> 00:32:27,050 So let's take a break, and I want you to think about it. 444 00:32:27,050 --> 00:32:27,830 Any questions? 445 00:32:35,780 --> 00:32:36,724 Yes. 446 00:32:36,724 --> 00:32:37,840 AUDIENCE: [INAUDIBLE]. 447 00:32:42,717 --> 00:32:44,800 YUFEI ZHAO: So the question is, how do you embed A 448 00:32:44,800 --> 00:32:47,270 without having any collisions? 449 00:32:47,270 --> 00:32:51,520 So I put in the vertices of B so that every r of them 450 00:32:51,520 --> 00:32:53,780 have many neighbors. 451 00:32:53,780 --> 00:32:58,230 And now, I want to try to embed the vertices of A one 452 00:32:58,230 --> 00:33:00,120 by one in any order. 453 00:33:00,120 --> 00:33:02,130 Think about-- you pick the first vertex. 454 00:33:02,130 --> 00:33:04,130 Where can it go? 455 00:33:04,130 --> 00:33:07,740 It has-- let's say it's adjacent to the first three 456 00:33:07,740 --> 00:33:10,920 vertices of B. So, in the embedding, 457 00:33:10,920 --> 00:33:13,170 it has to go in the common neighborhood of those three 458 00:33:13,170 --> 00:33:17,010 vertices, which we know is large. 459 00:33:17,010 --> 00:33:19,520 So I put them anywhere. 460 00:33:19,520 --> 00:33:21,260 And I do the same for the second vertex, 461 00:33:21,260 --> 00:33:24,600 do the same for the fourth vertex. 462 00:33:24,600 --> 00:33:27,770 So I just keep on going. 463 00:33:27,770 --> 00:33:30,260 Because the common neighborhood is large, 464 00:33:30,260 --> 00:33:33,080 it may be that some of the potential vertices I 465 00:33:33,080 --> 00:33:36,950 might embed is already used by the previous steps 466 00:33:36,950 --> 00:33:38,690 in the process. 467 00:33:38,690 --> 00:33:40,550 But because I always have at least A 468 00:33:40,550 --> 00:33:43,400 plus B common neighbors, I always 469 00:33:43,400 --> 00:33:45,950 have some possibilities that remain. 470 00:33:52,820 --> 00:33:53,542 Yes, question. 471 00:33:53,542 --> 00:33:57,786 AUDIENCE: So [INAUDIBLE],, like the last line 472 00:33:57,786 --> 00:34:00,455 of the proof to [INAUDIBLE] delete 473 00:34:00,455 --> 00:34:04,600 a vertex from each bad subset, and then 474 00:34:04,600 --> 00:34:06,212 to make that [INAUDIBLE]. 475 00:34:12,775 --> 00:34:13,400 YUFEI ZHAO: Ah. 476 00:34:13,400 --> 00:34:16,710 The question is, how does this bad subset deletion work? 477 00:34:16,710 --> 00:34:20,300 So you have this A which is fairly large. 478 00:34:20,300 --> 00:34:23,880 And you know that there exists some instance-- there's 479 00:34:23,880 --> 00:34:26,100 some incidence of this randomness that 480 00:34:26,100 --> 00:34:31,860 produces for you a situation where A has very few bad r 481 00:34:31,860 --> 00:34:33,429 subsets. 482 00:34:33,429 --> 00:34:37,540 So then I take A and I delete from A one vertex 483 00:34:37,540 --> 00:34:40,020 in each bad subset. 484 00:34:40,020 --> 00:34:42,620 I haven't changed the size of A very much. 485 00:34:42,620 --> 00:34:46,409 A is still quite large after this deletion, 486 00:34:46,409 --> 00:34:49,020 but now A has no bad subsets remaining, 487 00:34:49,020 --> 00:34:52,110 because I've gotten rid of one vertex from each one, 488 00:34:52,110 --> 00:34:54,980 from each bad subset. 489 00:34:54,980 --> 00:34:56,880 So they're very similar to what we've 490 00:34:56,880 --> 00:35:01,820 seen before, the random process for creating an H-free graph. 491 00:35:01,820 --> 00:35:04,560 You generate a random H-free graph 492 00:35:04,560 --> 00:35:07,920 which has very few copies of H relative to the number 493 00:35:07,920 --> 00:35:10,140 of edges, and then you get rid of them 494 00:35:10,140 --> 00:35:16,560 by removing one edge from each copy of H. 495 00:35:16,560 --> 00:35:20,070 So with the theorem that we saw in the first part 496 00:35:20,070 --> 00:35:22,830 of the lecture, we saw how to improve 497 00:35:22,830 --> 00:35:27,270 on the bound of Kovari-Sos-Turan in some circumstances, namely 498 00:35:27,270 --> 00:35:30,960 one where the graph that you're forbidding, this H, 499 00:35:30,960 --> 00:35:32,592 essentially has bounded degree. 500 00:35:32,592 --> 00:35:34,050 We stated something a bit stronger, 501 00:35:34,050 --> 00:35:38,320 namely has bounded degree from one side. 502 00:35:38,320 --> 00:35:41,340 And that's a pretty general result. 503 00:35:41,340 --> 00:35:44,650 And now I want to look at some more specific situations 504 00:35:44,650 --> 00:35:46,680 where you might be able to improve further. 505 00:35:49,460 --> 00:35:54,530 So what are some nice bipartite graphs? 506 00:35:54,530 --> 00:35:59,070 One that comes up is kind of even cycles. 507 00:35:59,070 --> 00:36:03,490 So if you have C4, C6, and so on. 508 00:36:03,490 --> 00:36:06,410 And you see C4 is the same as K2,2, 509 00:36:06,410 --> 00:36:08,180 which we already saw before. 510 00:36:08,180 --> 00:36:11,420 But even C6, the techniques so far 511 00:36:11,420 --> 00:36:14,990 allow us to obtain, with the theorem that we just saw-- 512 00:36:14,990 --> 00:36:18,710 gives us a bound on C6 that's more or less the same as that 513 00:36:18,710 --> 00:36:23,480 of C4, namely n to the 3/2. 514 00:36:23,480 --> 00:36:26,700 So what's the truth for C6? 515 00:36:26,700 --> 00:36:29,500 Well, it turns out that you can do much better. 516 00:36:32,120 --> 00:36:40,540 So this is the theorem of Bondy and Simonovits 517 00:36:40,540 --> 00:36:45,060 that, for all integers k at least 2, 518 00:36:45,060 --> 00:36:50,380 there exists some constant C such that the extremal number-- 519 00:36:50,380 --> 00:36:53,540 well, I'm going to use C too many times. 520 00:36:53,540 --> 00:37:00,820 So I'll just say that the extremal number of C2 sub k 521 00:37:00,820 --> 00:37:08,850 is, at most, on the order of n to the 1 plus 1 over k. 522 00:37:08,850 --> 00:37:12,220 So, in particular, for 6 cycles, the upper bound 523 00:37:12,220 --> 00:37:15,215 is 4/3 in the exponent. 524 00:37:15,215 --> 00:37:16,760 It's better than the 3/2. 525 00:37:20,350 --> 00:37:22,360 So there's another class of graphs where there 526 00:37:22,360 --> 00:37:23,627 are some nice upper bounds. 527 00:37:23,627 --> 00:37:25,960 So you can ask, well, just like Kovari-Sos-Turan theorem 528 00:37:25,960 --> 00:37:28,030 for complete bipartite graphs, do we 529 00:37:28,030 --> 00:37:31,660 know matching lower bound constructions? 530 00:37:31,660 --> 00:37:38,280 And what is known is that it is tight only 531 00:37:38,280 --> 00:37:40,620 for a small number of cases. 532 00:37:40,620 --> 00:37:43,980 And the others, we do not know whether they are tight. 533 00:37:43,980 --> 00:37:51,630 So this Bondy-Simonovits theorem, it is tight for k 534 00:37:51,630 --> 00:37:55,990 being 2, 3, or 5, and open for others. 535 00:37:59,140 --> 00:38:05,550 So there are constructions for C4,3, C6,3 and C10,3, 536 00:38:05,550 --> 00:38:06,972 but not for C8,3. 537 00:38:06,972 --> 00:38:07,930 That's an open problem. 538 00:38:12,440 --> 00:38:14,150 The proof of the Bondy-Simonovits theorem 539 00:38:14,150 --> 00:38:16,520 is slightly involved, but I want to show you 540 00:38:16,520 --> 00:38:19,010 a weaker result that already contains 541 00:38:19,010 --> 00:38:21,410 a lot of interesting ideas. 542 00:38:21,410 --> 00:38:26,420 So a weaker result is this. 543 00:38:29,840 --> 00:38:34,990 That for every integer k at least 2, 544 00:38:34,990 --> 00:38:44,060 there exists a constant C such that every n-vertex graph G 545 00:38:44,060 --> 00:38:49,490 with at least C-- 546 00:38:49,490 --> 00:38:52,940 so this-- the correct-- 547 00:38:52,940 --> 00:38:58,180 the same order of number of edges-- 548 00:38:58,180 --> 00:38:59,720 so we know for Bondy-Simonovits, it 549 00:38:59,720 --> 00:39:03,390 contains an even cycle of length exactly 2k. 550 00:39:03,390 --> 00:39:05,810 So we'll show something slightly weaker 551 00:39:05,810 --> 00:39:14,510 that contains an even cycle of length, at most, 2k. 552 00:39:20,920 --> 00:39:27,510 Which, in other words, says that the extremal number, if you-- 553 00:39:27,510 --> 00:39:29,140 so we haven't introduced this notation, 554 00:39:29,140 --> 00:39:31,795 but, hopefully, you can guess what it means. 555 00:39:31,795 --> 00:39:36,550 Then, if you forbid all of these cycles, 556 00:39:36,550 --> 00:39:43,325 then it is this quantity there. 557 00:39:43,325 --> 00:39:51,460 So I'll show this weaker result. All right. 558 00:39:51,460 --> 00:39:53,820 Let's' do it. 559 00:39:53,820 --> 00:39:56,860 First, I want to show you a couple of easy preparatory 560 00:39:56,860 --> 00:39:57,360 lemmas. 561 00:40:01,380 --> 00:40:16,430 The first-- so every graph G contains a subgraph with min 562 00:40:16,430 --> 00:40:22,680 degree at least the half of-- 563 00:40:22,680 --> 00:40:27,470 at least half of the average degree of G. 564 00:40:27,470 --> 00:40:30,270 So you have a graph G. It has large average degree. 565 00:40:30,270 --> 00:40:31,560 It has lots of edges. 566 00:40:31,560 --> 00:40:34,650 But maybe there are some small degree vertices. 567 00:40:34,650 --> 00:40:38,370 And it will be useful to know that the minimum degree is 568 00:40:38,370 --> 00:40:40,470 actually quite large as well. 569 00:40:40,470 --> 00:40:42,872 So it turns out, by passing to a subgraph, 570 00:40:42,872 --> 00:40:43,830 you can guarantee that. 571 00:40:48,730 --> 00:40:50,560 How do you think we might prove this? 572 00:40:50,560 --> 00:40:51,430 I give you a graph. 573 00:40:51,430 --> 00:40:52,700 I know it has lots of edges. 574 00:40:52,700 --> 00:40:54,450 But maybe some vertices have small degree. 575 00:40:58,250 --> 00:40:58,806 Yes. 576 00:40:58,806 --> 00:41:00,125 AUDIENCE: [INAUDIBLE]. 577 00:41:00,125 --> 00:41:02,250 YUFEI ZHAO: So you suggest dependent random choice. 578 00:41:02,250 --> 00:41:04,333 That's a heavy hammer for such a simple statement. 579 00:41:04,333 --> 00:41:06,590 And so it turns out we can do something even simpler. 580 00:41:06,590 --> 00:41:08,780 AUDIENCE: [INAUDIBLE]. 581 00:41:08,780 --> 00:41:11,280 YUFEI ZHAO: So we'll just throw out the min degree vertices. 582 00:41:11,280 --> 00:41:14,781 So throw out the small degree vertices. 583 00:41:14,781 --> 00:41:24,150 So if the average degree is 2t then the number of edges 584 00:41:24,150 --> 00:41:28,140 is number of vertices times t. 585 00:41:28,140 --> 00:41:34,280 And you see that removing a vertex of degree t, 586 00:41:34,280 --> 00:41:42,950 or at most t, cannot decrease average degree. 587 00:41:46,070 --> 00:41:47,810 So I have this process where I remove 588 00:41:47,810 --> 00:41:51,680 vertices that are less than half of average degree. 589 00:41:51,680 --> 00:41:53,450 And the average degree never goes down, 590 00:41:53,450 --> 00:41:54,540 so I keep on doing this. 591 00:41:54,540 --> 00:41:55,790 Average degree stays the same. 592 00:41:55,790 --> 00:41:59,820 Well, it can go up, but it never goes down. 593 00:41:59,820 --> 00:42:04,250 And when I stop, I don't have any more small degree vertices 594 00:42:04,250 --> 00:42:06,510 to get rid of. 595 00:42:06,510 --> 00:42:08,520 Why does the process even terminate? 596 00:42:08,520 --> 00:42:11,650 Maybe we end up with the empty graph, which is not so useful. 597 00:42:15,142 --> 00:42:15,642 Yes. 598 00:42:15,642 --> 00:42:18,315 AUDIENCE: The average degree is [INAUDIBLE].. 599 00:42:18,315 --> 00:42:20,690 YUFEI ZHAO: He said the average degree is non-decreasing. 600 00:42:20,690 --> 00:42:23,390 But how do I know the graph has at least some number 601 00:42:23,390 --> 00:42:26,430 of vertices when I stop? 602 00:42:26,430 --> 00:42:26,930 Yes. 603 00:42:26,930 --> 00:42:28,334 AUDIENCE: [INAUDIBLE]. 604 00:42:31,970 --> 00:42:36,070 YUFEI ZHAO: So if you-- you must terminate. 605 00:42:36,070 --> 00:42:39,950 You must terminate because every graph, 606 00:42:39,950 --> 00:42:42,800 if you have way too small number of vertices-- 607 00:42:42,800 --> 00:42:44,750 because, for example, if-- 608 00:42:44,750 --> 00:42:51,170 we'll just notice that every graph with, at most, 609 00:42:51,170 --> 00:42:58,060 2t vertices has average degree less than 2t. 610 00:42:58,060 --> 00:43:01,750 So you'll never get below 2t vertices. 611 00:43:01,750 --> 00:43:04,968 You'll run out of room if you go too far. 612 00:43:04,968 --> 00:43:06,760 Because that's the first preparation lemma. 613 00:43:06,760 --> 00:43:13,240 The second one is that every graph G 614 00:43:13,240 --> 00:43:20,110 has a bipartite subgraph with at least half 615 00:43:20,110 --> 00:43:22,590 of the number of edges as the original graph. 616 00:43:27,940 --> 00:43:30,860 This is a very nice and quick exercise 617 00:43:30,860 --> 00:43:32,680 in the probabilistic method. 618 00:43:32,680 --> 00:43:38,530 So we can color every vertex black and white 619 00:43:38,530 --> 00:43:39,580 uniformly at random. 620 00:43:44,740 --> 00:43:50,850 And the expected number of black to white edges 621 00:43:50,850 --> 00:43:55,970 is exactly half of the original number of edges 622 00:43:55,970 --> 00:43:59,210 by linearity of expectations. 623 00:43:59,210 --> 00:44:03,880 So there's some instance with at least that many black-white 624 00:44:03,880 --> 00:44:05,737 edges, and that's a bipartite subgraph. 625 00:44:09,860 --> 00:44:14,132 So now we can prove the theorem about even cycles, the weaker 626 00:44:14,132 --> 00:44:14,840 theorem at least. 627 00:44:22,392 --> 00:44:26,470 I start with a graph which has a lot of edges. 628 00:44:26,470 --> 00:44:31,900 But by these two lemmas, and changing constant somewhat, 629 00:44:31,900 --> 00:44:39,880 I can obtain a subgraph with-- 630 00:44:39,880 --> 00:44:45,530 that's bipartite and has min degree quite large. 631 00:44:45,530 --> 00:44:54,670 So I lose, at most, a constant factor, 632 00:44:54,670 --> 00:44:58,590 and I get basically within a factor of 2 of-- 633 00:44:58,590 --> 00:45:04,160 well, factor of 4 of the average degree. 634 00:45:04,160 --> 00:45:06,870 So let me call this quantity delta, 635 00:45:06,870 --> 00:45:10,130 the min degree in this subgraph, this bipartite subgraph. 636 00:45:13,240 --> 00:45:15,100 Let's think about what happens to this graph 637 00:45:15,100 --> 00:45:17,998 if I start with an arbitrary vertex. 638 00:45:17,998 --> 00:45:19,540 So now I have a min degree condition. 639 00:45:19,540 --> 00:45:24,390 It's, really, all vertices are now kind of the same to me. 640 00:45:24,390 --> 00:45:29,840 So pick an arbitrary vertex, and look at its neighborhood. 641 00:45:29,840 --> 00:45:39,480 It has at least delta edges coming out. 642 00:45:39,480 --> 00:45:45,820 So let me call the first vertex level 0, and the second set 643 00:45:45,820 --> 00:45:47,800 level 1. 644 00:45:47,800 --> 00:45:51,370 It's bipartite, so there are no edges within level 1. 645 00:45:51,370 --> 00:45:52,810 Let's expand out even further. 646 00:46:01,830 --> 00:46:05,863 Can there be some collisions where two of these edges 647 00:46:05,863 --> 00:46:06,780 go to the same vertex? 648 00:46:09,310 --> 00:46:11,710 Well, if there were, then I find a C4. 649 00:46:15,180 --> 00:46:19,350 So if I assume-- 650 00:46:19,350 --> 00:46:25,620 so let's assume that there's no C4, C6, 651 00:46:25,620 --> 00:46:31,060 and so on, C2k, for contradiction. 652 00:46:31,060 --> 00:46:37,960 So because there's no C4, all the endpoints 653 00:46:37,960 --> 00:46:42,090 of this path of length 2 are distinct in level 2. 654 00:46:45,190 --> 00:46:47,410 And you keep going, or you keep expanding further. 655 00:46:50,120 --> 00:46:50,620 And so on. 656 00:46:56,380 --> 00:46:59,520 And all the way to level t. 657 00:46:59,520 --> 00:47:01,900 And all of these guys must be distinct as well. 658 00:47:06,090 --> 00:47:14,382 Because, otherwise, you'll find a cycle of length, at most, 2t. 659 00:47:14,382 --> 00:47:16,740 So when you do this expansion, each step, 660 00:47:16,740 --> 00:47:18,270 you get distinct vertices. 661 00:47:18,270 --> 00:47:21,288 And you also have no edges inside each part 662 00:47:21,288 --> 00:47:23,330 because you are looking in the bipartite setting. 663 00:47:26,390 --> 00:47:31,130 So how many vertices do you get at the end? 664 00:47:31,130 --> 00:47:32,990 Well, I have a min degree condition. 665 00:47:32,990 --> 00:47:34,790 The min degree condition tells me 666 00:47:34,790 --> 00:47:40,200 that I expand by a factor of at least delta minus 1 each time. 667 00:47:40,200 --> 00:47:44,860 So the number of vertices in here 668 00:47:44,860 --> 00:47:50,510 is at least delta minus 1 raised to the power t-- 669 00:47:50,510 --> 00:47:53,648 so raised to the power k. 670 00:47:53,648 --> 00:47:55,624 Here's k. 671 00:47:55,624 --> 00:47:59,630 And expand all the way to the end. 672 00:47:59,630 --> 00:48:04,180 But, you see, that number there is quite large. 673 00:48:04,180 --> 00:48:06,660 So, in particular, if t is large enough, 674 00:48:06,660 --> 00:48:07,946 then it's bigger than n. 675 00:48:12,806 --> 00:48:15,320 And that will be a contradiction, 676 00:48:15,320 --> 00:48:19,040 because you only had n vertices in the graph to begin with. 677 00:48:24,090 --> 00:48:26,190 Therefore, the assumption that there 678 00:48:26,190 --> 00:48:28,800 are no cycles, even cycles of length, at most, 2k, 679 00:48:28,800 --> 00:48:31,923 is incorrect. 680 00:48:31,923 --> 00:48:33,090 And that finishes the proof. 681 00:48:38,640 --> 00:48:39,640 Any questions? 682 00:48:42,570 --> 00:48:43,070 Yes. 683 00:48:43,070 --> 00:48:46,010 AUDIENCE: Are the ideas for the full Bondy-Simonovits theorem 684 00:48:46,010 --> 00:48:47,490 similar? 685 00:48:47,490 --> 00:48:51,060 YUFEI ZHAO: So the question has to do with, in the original, 686 00:48:51,060 --> 00:48:53,932 in the full Bondy-Simonovits theorem, what do we need to do? 687 00:48:53,932 --> 00:48:54,640 Are they similar? 688 00:48:54,640 --> 00:48:57,310 So, certainly, you want to do something like this. 689 00:48:57,310 --> 00:49:01,110 But, then, you also want to think about, 690 00:49:01,110 --> 00:49:08,070 if you do have short cycles, how can you bypass them? 691 00:49:08,070 --> 00:49:10,170 So there's a more careful analysis 692 00:49:10,170 --> 00:49:12,150 of what happens with shorter cycles. 693 00:49:12,150 --> 00:49:14,520 And we will not get into that. 694 00:49:14,520 --> 00:49:15,732 Any more questions? 695 00:49:19,040 --> 00:49:20,240 All right. 696 00:49:20,240 --> 00:49:23,900 So the first thing that we did in today's lecture 697 00:49:23,900 --> 00:49:24,878 has to deal with what-- 698 00:49:24,878 --> 00:49:25,920 AUDIENCE: Quick question. 699 00:49:25,920 --> 00:49:26,587 YUFEI ZHAO: Yes. 700 00:49:26,587 --> 00:49:28,220 AUDIENCE: So why is-- 701 00:49:28,220 --> 00:49:32,773 so why are all vertices distinct for level k? 702 00:49:32,773 --> 00:49:34,662 Or are they just defined that way? 703 00:49:34,662 --> 00:49:36,770 YUFEI ZHAO: So question is, why are the vertices 704 00:49:36,770 --> 00:49:38,000 distinct for level k? 705 00:49:38,000 --> 00:49:41,120 So at level k, if you have some collapse, 706 00:49:41,120 --> 00:49:46,476 then it came from two different paths, therefore forming a C2k. 707 00:49:46,476 --> 00:49:47,018 AUDIENCE: OK. 708 00:49:53,790 --> 00:49:56,880 YUFEI ZHAO: So now I want to revisit the first theorem that 709 00:49:56,880 --> 00:50:00,120 happened in today's lecture, namely 710 00:50:00,120 --> 00:50:06,920 if you have this H, bipartite A and B. 711 00:50:06,920 --> 00:50:10,850 So we saw the hypothesis was that if every vertex in A 712 00:50:10,850 --> 00:50:13,970 had bounded degree, degree, at most, r, then 713 00:50:13,970 --> 00:50:16,040 we got the upper bound on the extremal number. 714 00:50:16,040 --> 00:50:18,840 That was n to the 2 minus 1 over r. 715 00:50:18,840 --> 00:50:22,750 And, in particular, for r equals to 2, 716 00:50:22,750 --> 00:50:26,210 suppose I have that situation there. 717 00:50:28,740 --> 00:50:35,590 Suppose that the degree for every vertex in A 718 00:50:35,590 --> 00:50:38,110 is, at most, 2. 719 00:50:38,110 --> 00:50:44,320 Then the first theorem guaranteed us that the extremal 720 00:50:44,320 --> 00:50:50,560 number is on the order, at most, n the 3/2, 721 00:50:50,560 --> 00:50:55,770 just like the extremal number for 4 cycles, for K2,2's. 722 00:50:55,770 --> 00:50:58,410 And, of course, this statement is tight, 723 00:50:58,410 --> 00:51:02,330 in the sense that if I change this 3/2 to any smaller number, 724 00:51:02,330 --> 00:51:06,330 then, well, just taking H to be K2,2 violates it. 725 00:51:06,330 --> 00:51:11,520 So I cannot replace in this generality 3/2 by any smaller 726 00:51:11,520 --> 00:51:15,330 number, because we know that the extremal number for K2,2 is 727 00:51:15,330 --> 00:51:16,840 this order. 728 00:51:16,840 --> 00:51:20,540 But is that the only obstruction? 729 00:51:20,540 --> 00:51:28,520 So if H is not K2,2, well, you can make some sillier examples 730 00:51:28,520 --> 00:51:32,780 too by taking K2,2 and add some more edges. 731 00:51:32,780 --> 00:51:36,705 So lets forbid H from having a K2,2 subgraph. 732 00:51:39,840 --> 00:51:41,414 Can you do better now? 733 00:51:41,414 --> 00:51:46,560 So in this case, can you improve for the specific H 734 00:51:46,560 --> 00:51:47,960 that exponent? 735 00:51:47,960 --> 00:51:52,610 And we already saw one case where you can do this, namely 736 00:51:52,610 --> 00:51:57,846 in Bondy-Simonovits theorem for cycles. 737 00:51:57,846 --> 00:52:01,180 Or if you only applied this theorem here, you get 3/2, 738 00:52:01,180 --> 00:52:06,260 but Bondy-Simonovits tells you a much better exponent. 739 00:52:06,260 --> 00:52:08,530 So let's explore the situation. 740 00:52:08,530 --> 00:52:12,800 And it turns out, in a very recent theorem that is only 741 00:52:12,800 --> 00:52:19,100 proved the last couple of years by David Conlon and Joonkyung 742 00:52:19,100 --> 00:52:25,820 Lee, they showed that for every H 743 00:52:25,820 --> 00:52:33,640 as above, there exists constants little c 744 00:52:33,640 --> 00:52:41,560 and big C such that the extremal number of H 745 00:52:41,560 --> 00:52:48,860 is upper bounded by something where I can decrease 3/2 746 00:52:48,860 --> 00:52:50,180 to some even smaller number. 747 00:52:52,960 --> 00:52:56,890 So, somehow, this 3/2, now we understand is really 748 00:52:56,890 --> 00:53:00,280 because of the presence of K2,2. 749 00:53:00,280 --> 00:53:03,040 The graph has-- if H has no K2,2, 750 00:53:03,040 --> 00:53:05,530 then some smaller number suffices. 751 00:53:11,510 --> 00:53:13,960 And I want to use the rest of today 752 00:53:13,960 --> 00:53:17,457 to explain how to prove that theorem there. 753 00:53:17,457 --> 00:53:18,040 Yes, question. 754 00:53:18,040 --> 00:53:20,590 AUDIENCE: The C is not independent of H [INAUDIBLE].. 755 00:53:20,590 --> 00:53:21,632 YUFEI ZHAO: That's right. 756 00:53:21,632 --> 00:53:24,370 So the question is, is C independent of H? 757 00:53:24,370 --> 00:53:37,546 So C depends on H. So C, they are dependent on H. Questions? 758 00:53:43,490 --> 00:53:47,930 Let me put this question in a slightly different formulation 759 00:53:47,930 --> 00:53:50,410 that is also equivalent. 760 00:53:50,410 --> 00:53:53,060 So in graph theory, there is a notion of a subdivision. 761 00:53:53,060 --> 00:54:02,060 And, in particular, a one subdivision of graph H 762 00:54:02,060 --> 00:54:05,910 is this operation where you start with a graph-- 763 00:54:05,910 --> 00:54:08,840 let's say this graph here-- 764 00:54:08,840 --> 00:54:17,490 and you add a vertex to the middle 765 00:54:17,490 --> 00:54:18,840 of every edge of this graph. 766 00:54:27,444 --> 00:54:29,590 So, initially, it's 4 vertices. 767 00:54:29,590 --> 00:54:31,890 Now you add a new vertex to every edge. 768 00:54:31,890 --> 00:54:35,680 So you subdivide every edge into a path of two edges. 769 00:54:35,680 --> 00:54:37,780 That's called a subdivision. 770 00:54:37,780 --> 00:54:42,665 So for today's lecture, let me denote subdivisions by a prime. 771 00:54:42,665 --> 00:54:45,237 And, in particular, if this is K4, 772 00:54:45,237 --> 00:54:47,070 then I will denote this graph here K4 prime. 773 00:54:50,230 --> 00:55:00,220 So, for example, K3 prime, that's a triangle subdivided-- 774 00:55:00,220 --> 00:55:03,506 well, that's a C6, [INAUDIBLE]. 775 00:55:09,570 --> 00:55:14,840 So observe that every H that comes up in this theorem 776 00:55:14,840 --> 00:55:33,340 here is a subgraph of some subdivision, some one 777 00:55:33,340 --> 00:55:35,200 subdivision of a clique. 778 00:55:41,680 --> 00:55:48,090 Because the vertices on the left in A are degree 2, 779 00:55:48,090 --> 00:55:53,490 you think of them as midpoints of edges. 780 00:55:53,490 --> 00:55:57,480 But because you are K2,2 free, you-- 781 00:55:57,480 --> 00:56:01,440 if you collapse those path of lengths 2 to single edges, 782 00:56:01,440 --> 00:56:04,880 you do not end up with parallel edges. 783 00:56:04,880 --> 00:56:11,730 So it is the subgraph of this one subdivision of some graph, 784 00:56:11,730 --> 00:56:13,480 which, then, you can complete to a clique. 785 00:56:19,390 --> 00:56:25,390 So this theorem here is equivalent, 786 00:56:25,390 --> 00:56:33,720 at least qualitatively, to the statement that for every t, 787 00:56:33,720 --> 00:56:38,100 there exists some constants, again 788 00:56:38,100 --> 00:56:46,800 depending on t, such that the extremal number of the one 789 00:56:46,800 --> 00:56:53,260 subdivision of a clique is bounded by something where 790 00:56:53,260 --> 00:56:58,340 we can improve upon the exponent in the first theorem in today's 791 00:56:58,340 --> 00:56:58,840 lecture. 792 00:57:05,020 --> 00:57:06,640 So that theorem there. 793 00:57:06,640 --> 00:57:16,875 So these two theorems are equivalent because 794 00:57:16,875 --> 00:57:17,500 of this remark. 795 00:57:20,020 --> 00:57:23,780 Any questions so far about the statements? 796 00:57:23,780 --> 00:57:24,391 Yes. 797 00:57:24,391 --> 00:57:27,758 AUDIENCE: In the remark, how do you deal with, like, 798 00:57:27,758 --> 00:57:28,720 [INAUDIBLE]. 799 00:57:31,070 --> 00:57:32,570 YUFEI ZHAO: Question, in the remark, 800 00:57:32,570 --> 00:57:35,040 how do you deal with vertices with degree less than 2? 801 00:57:35,040 --> 00:57:37,980 Complete it to a vertex degree of-- vertex of degree 2. 802 00:57:37,980 --> 00:57:41,050 Add another edge. 803 00:57:41,050 --> 00:57:43,020 AUDIENCE: [INAUDIBLE]. 804 00:57:43,020 --> 00:57:45,128 YUFEI ZHAO: Add another edge to a new vertex. 805 00:57:45,128 --> 00:57:47,570 AUDIENCE: OK, sure. 806 00:57:47,570 --> 00:57:51,260 YUFEI ZHAO: Any more questions? 807 00:57:51,260 --> 00:57:52,570 All right. 808 00:57:52,570 --> 00:58:03,330 So the proof I want to show you is due to Oliver Janzer, 809 00:58:03,330 --> 00:58:06,360 and for this clique subdivision theorem. 810 00:58:06,360 --> 00:58:10,840 And this proof produces that C sub t 811 00:58:10,840 --> 00:58:15,450 equals to 1 over 4t minus 6. 812 00:58:15,450 --> 00:58:18,570 So if you plug in t equals to 3, you 813 00:58:18,570 --> 00:58:24,950 find that the exponent here is actually right for the 6 cycle. 814 00:58:24,950 --> 00:58:27,990 So it actually agrees with what we know. 815 00:58:40,650 --> 00:58:45,330 So I want to show you some of the main ideas from this proof. 816 00:58:45,330 --> 00:58:47,790 Just like the proof of-- 817 00:58:47,790 --> 00:58:50,717 that we saw for the cycles, even cycles theorem, 818 00:58:50,717 --> 00:58:52,800 it will be helpful to start with some preparation. 819 00:58:52,800 --> 00:58:55,640 You start with a graph that, even though it 820 00:58:55,640 --> 00:58:57,413 has a lot of edges, may have lots 821 00:58:57,413 --> 00:58:58,830 of vertices with high degree, lots 822 00:58:58,830 --> 00:59:00,270 of vertices with low degree. 823 00:59:00,270 --> 00:59:02,650 It's nice to clean it up somewhat. 824 00:59:02,650 --> 00:59:06,240 And so let me state a preparatory lemma, which 825 00:59:06,240 --> 00:59:09,420 we will not prove, but it's of a similar nature 826 00:59:09,420 --> 00:59:12,120 to this very easy lemma that we saw earlier but 827 00:59:12,120 --> 00:59:13,886 with a bit more work. 828 00:59:13,886 --> 00:59:15,365 AUDIENCE: [INAUDIBLE]. 829 00:59:18,813 --> 00:59:19,780 YUFEI ZHAO: Yes. 830 00:59:19,780 --> 00:59:22,590 Originally, I put a C over here, but now it's OK. 831 00:59:22,590 --> 00:59:24,306 So there exists a Ct. 832 00:59:29,490 --> 00:59:33,360 So the preparation is that we're going 833 00:59:33,360 --> 00:59:39,745 to pass through a large, almost regular subgraph. 834 00:59:43,920 --> 00:59:47,690 The lemma-- so don't worry too much about the details, 835 00:59:47,690 --> 00:59:49,620 and I'll tell you what the idea is. 836 00:59:49,620 --> 00:59:58,480 So for every alpha, there exists constants beta and K such that 837 00:59:58,480 --> 01:00:07,090 for every C and n sufficiently large, 838 01:00:07,090 --> 01:00:16,030 every n-vertex graph G with lots of edges-- 839 01:00:16,030 --> 01:00:20,440 so C n to the 1 plus alpha edges-- 840 01:00:23,120 --> 01:00:30,570 has a subgraph G prime such that-- 841 01:00:30,570 --> 01:00:32,330 so I want some properties. 842 01:00:32,330 --> 01:00:35,795 First, G prime has lots of vertices. 843 01:00:39,390 --> 01:00:43,310 So n to the beta, so it's still lots of vertices. 844 01:00:43,310 --> 01:00:46,360 You do some polynomial in n. 845 01:00:46,360 --> 01:00:49,640 And, 2, it has still lots of edges 846 01:00:49,640 --> 01:00:52,890 relative to the number of vertices it has. 847 01:00:52,890 --> 01:00:55,070 So, basically, changing the constants, 848 01:00:55,070 --> 01:00:56,930 if I start with n to the 1 plus alpha, 849 01:00:56,930 --> 01:01:01,010 I still have, roughly, number of vertices to the 1 plus 850 01:01:01,010 --> 01:01:03,980 alpha, number of edges. 851 01:01:03,980 --> 01:01:10,100 It is almost regular, in the sense that the max degree of G 852 01:01:10,100 --> 01:01:13,640 does not differ from its minimum degree 853 01:01:13,640 --> 01:01:15,380 by more than a constant factor. 854 01:01:18,642 --> 01:01:20,980 So you do have vertices that are too small degree. 855 01:01:20,980 --> 01:01:24,690 You don't have vertices that are too large degree. 856 01:01:24,690 --> 01:01:28,890 And, finally, G prime is bipartite, 857 01:01:28,890 --> 01:01:31,850 and the two parts of this bipartition 858 01:01:31,850 --> 01:01:42,330 have sizes differing by a factor, at most, 2. 859 01:01:51,030 --> 01:01:55,930 So if you like, think of G as a regular bipartite graph. 860 01:01:55,930 --> 01:01:57,350 But this is the preparation lemma. 861 01:01:57,350 --> 01:01:58,933 We'll just make our life a bit easier. 862 01:02:01,680 --> 01:02:08,460 So, from now on, let's treat G as a constant 863 01:02:08,460 --> 01:02:15,620 in our asymptotic notation to simplify the notation. 864 01:02:15,620 --> 01:02:18,140 So you have this graph G. It's a bipartite graph. 865 01:02:22,200 --> 01:02:28,710 And for a pair of vertices on one side A-- 866 01:02:28,710 --> 01:02:31,650 so there are no edges in A. But for a pair of vertices, 867 01:02:31,650 --> 01:02:36,940 I say that this u, v is light. 868 01:02:36,940 --> 01:02:40,610 So it's not an edge, but I talk about these pairs. 869 01:02:40,610 --> 01:02:52,830 I say that it is light if the number of common neighbors 870 01:02:52,830 --> 01:02:54,820 between-- 871 01:02:54,820 --> 01:03:04,940 of u and v is at least 1 and less than t choose 2. 872 01:03:04,940 --> 01:03:08,940 So it has some common neighbors, but not too many. 873 01:03:08,940 --> 01:03:13,420 And then we say that this pair is heavy 874 01:03:13,420 --> 01:03:15,370 if the number of common neighbors 875 01:03:15,370 --> 01:03:19,500 is at least t choose 2. 876 01:03:19,500 --> 01:03:22,560 So if a pair u, v has some common neighbors, 877 01:03:22,560 --> 01:03:24,290 then it's either light or heavy. 878 01:03:32,060 --> 01:03:40,170 I claim that if G is a Kt prime-- 879 01:03:40,170 --> 01:03:43,280 so this is the one subdivision of Kt-- 880 01:03:43,280 --> 01:03:52,495 free bipartite graph with vertex bipartition A union B-- 881 01:03:52,495 --> 01:03:57,440 so U. Not A, but U union B. U will eventually 882 01:03:57,440 --> 01:04:04,600 be a subset of A. Such that all the vertices on the left of U 883 01:04:04,600 --> 01:04:22,955 have to be at least delta, and U is not too small. 884 01:04:31,960 --> 01:04:35,212 So it's at least 4Bt over delta. 885 01:04:35,212 --> 01:04:36,420 Don't worry about it for now. 886 01:04:36,420 --> 01:04:37,960 So think of delta-- 887 01:04:37,960 --> 01:04:39,470 this is a min degree. 888 01:04:39,470 --> 01:04:42,100 So it is somewhat smaller than the average-- 889 01:04:42,100 --> 01:04:44,920 I mean, it's basically the average degree of your graph. 890 01:04:48,010 --> 01:04:49,207 And B, think of as n. 891 01:04:49,207 --> 01:04:51,040 It's more or less the whole set of vertices. 892 01:04:54,230 --> 01:04:59,740 Then the conclusion is that there 893 01:04:59,740 --> 01:05:12,160 exists a u in a lot of light pairs in the set 894 01:05:12,160 --> 01:05:17,170 U. There exists a vertex in many light pairs. 895 01:05:43,340 --> 01:05:45,560 It's important that we assume that this graph G 896 01:05:45,560 --> 01:05:48,680 if Kt prime free. 897 01:05:48,680 --> 01:05:51,410 Because, otherwise, you could imagine a situation where, 898 01:05:51,410 --> 01:05:54,470 essentially, you have a complete bipartite graph 899 01:05:54,470 --> 01:05:58,430 and every pair of vertices is heavy. 900 01:05:58,430 --> 01:06:01,490 So you don't have any light pairs at all. 901 01:06:01,490 --> 01:06:07,210 So having Kt prime free somehow allows us to find light pairs. 902 01:06:07,210 --> 01:06:09,050 So let's see the proof of this lemma. 903 01:06:09,050 --> 01:06:11,860 So you combine some nice ideas that we've seen earlier 904 01:06:11,860 --> 01:06:14,030 in the course, namely double counting, 905 01:06:14,030 --> 01:06:15,650 and also uses Turan's theorem. 906 01:06:20,060 --> 01:06:23,930 So, first, let's do a double counting argument similar 907 01:06:23,930 --> 01:06:29,900 to the proof of the Kovari-Sos-Turan theorem, 908 01:06:29,900 --> 01:06:37,050 where let's count the number of K1,2's like that. 909 01:06:37,050 --> 01:06:44,140 So the number of K1,2's like that between U and B. 910 01:06:44,140 --> 01:06:49,670 I claim one way to count this is to look through all 911 01:06:49,670 --> 01:06:54,720 the vertices on the right side, look at how many neighbors it 912 01:06:54,720 --> 01:06:58,590 has, and sum up the degrees choose 2. 913 01:07:08,964 --> 01:07:13,390 So skipping some-- 914 01:07:13,390 --> 01:07:15,980 I mean, I can tell you what comes out. 915 01:07:15,980 --> 01:07:22,800 So, by convexity, we find that it 916 01:07:22,800 --> 01:07:24,880 is at least this quantity here. 917 01:07:24,880 --> 01:07:27,580 And then, assuming the minimum degree condition, 918 01:07:27,580 --> 01:07:30,610 we find that this quantity is quite large. 919 01:07:41,110 --> 01:07:43,380 So this is a calculation very similar to what 920 01:07:43,380 --> 01:07:45,740 we did for the proof of Kovari-Sos-Turan theorem. 921 01:07:50,810 --> 01:07:56,690 The low degree vertices in B do not contribute much to the sum. 922 01:07:56,690 --> 01:08:01,120 So this sum is large, and it sums over all vertices in B, 923 01:08:01,120 --> 01:08:07,450 but the low degree vertices in B, they contribute very little. 924 01:08:13,870 --> 01:08:18,970 Because if we sum over all the vertices of B 925 01:08:18,970 --> 01:08:30,120 with degree less than 2t, then, for each summand, 926 01:08:30,120 --> 01:08:35,470 it's, at most, 2 t squared, which, 927 01:08:35,470 --> 01:08:43,050 again, by the assumption of delta, is less than half 928 01:08:43,050 --> 01:08:45,810 of the total sum. 929 01:08:45,810 --> 01:08:48,990 So the low degree vertices of B do not contribute very much 930 01:08:48,990 --> 01:08:51,021 to the sum. 931 01:08:51,021 --> 01:08:52,854 So let's look at the higher degree vertices. 932 01:08:56,080 --> 01:08:58,990 For the higher degree vertices, they 933 01:08:58,990 --> 01:09:02,500 contribute a substantial chunk. 934 01:09:16,220 --> 01:09:18,050 And the most important thing here 935 01:09:18,050 --> 01:09:30,350 is that, among these vertices, there are no t mutually heavy-- 936 01:09:33,030 --> 01:09:37,080 so not among these vertices, but in U. If you look at U, 937 01:09:37,080 --> 01:09:43,950 there are no t mutually heavy vertices in U. 938 01:09:43,950 --> 01:09:49,290 If you have t mutually heavy vertices in U, then 939 01:09:49,290 --> 01:09:50,430 what happens? 940 01:09:50,430 --> 01:09:53,640 So if you have t mutually heavy vertices. 941 01:09:53,640 --> 01:09:56,680 If you had, let's say, three vertices in U, 942 01:09:56,680 --> 01:10:00,390 they're mutually heavy, each one of them, because they're heavy, 943 01:10:00,390 --> 01:10:03,780 I can find many common neighbors. 944 01:10:03,780 --> 01:10:10,020 So I can build this path of length 2. 945 01:10:10,020 --> 01:10:12,060 I can build another path of length 2. 946 01:10:12,060 --> 01:10:14,640 And I don't run out of vertices because they're all heavy. 947 01:10:14,640 --> 01:10:17,550 They all have at least t choose 2 common neighbors. 948 01:10:17,550 --> 01:10:20,950 So I can build the subdivision of Kt. 949 01:10:20,950 --> 01:10:25,320 So there are no mutually heavy vertices 950 01:10:25,320 --> 01:10:29,250 in U. So where have we seen this before? 951 01:10:31,770 --> 01:10:38,840 So you have-- think about the neighborhood of a vertex 952 01:10:38,840 --> 01:10:43,110 in B. Because it's inside a neighborhood, 953 01:10:43,110 --> 01:10:46,650 all the pairs are either heavy or light, 954 01:10:46,650 --> 01:10:51,160 and there are no t mutually heavy vertices. 955 01:10:51,160 --> 01:10:57,360 So, then, Turan's theorem tells us 956 01:10:57,360 --> 01:10:58,985 that there must be many light vertices. 957 01:11:03,890 --> 01:11:11,900 That there are many light vertices in this neighborhood. 958 01:11:11,900 --> 01:11:22,100 So the number of light pairs in the neighborhood of this v, 959 01:11:22,100 --> 01:11:28,460 if it has at least t neighbors-- or else you run out of room-- 960 01:11:33,680 --> 01:11:36,440 is at least-- so if you think about what Turan's theorem 961 01:11:36,440 --> 01:11:41,710 says, that the number of known edges is at least this quantity 962 01:11:41,710 --> 01:11:49,740 here, which is at least the degree of v squared-- 963 01:11:49,740 --> 01:11:55,856 degree of v. 964 01:11:55,856 --> 01:12:01,150 So Turan's theorem tells us that there cannot be so many heavy 965 01:12:01,150 --> 01:12:05,770 pairs inside a neighborhood of a vertex in B, 966 01:12:05,770 --> 01:12:07,870 so there must be many light pairs. 967 01:12:12,780 --> 01:12:21,800 And now we sum over all vertices in B. 968 01:12:21,800 --> 01:12:30,120 We obtain that U has a lot of light pairs. 969 01:12:33,700 --> 01:12:37,960 We might have overcounted and a little bit, 970 01:12:37,960 --> 01:12:42,130 because each light pair is overcounted only 971 01:12:42,130 --> 01:12:44,290 by a bounded number of times because it's light. 972 01:12:51,970 --> 01:12:58,610 So it's overcounted by less than K choose 2 times. 973 01:12:58,610 --> 01:13:02,110 So that's just a constant factor, and we're OK with that. 974 01:13:05,090 --> 01:13:06,760 So that's the conclusion for now. 975 01:13:06,760 --> 01:13:13,840 This lemma tells us that you have lots of light pairs in U. 976 01:13:13,840 --> 01:13:20,800 And what we're going to do is to keep on shrinking this U. 977 01:13:20,800 --> 01:13:25,610 So U is going to be a subset of A. Initially, 978 01:13:25,610 --> 01:13:29,710 let's let U be the entire set A. It tells us 979 01:13:29,710 --> 01:13:35,380 that there's one vertex in A with lots of light neighbors. 980 01:13:35,380 --> 01:13:38,410 Take that vertex, choose its neighborhood. 981 01:13:38,410 --> 01:13:39,880 Apply the lemma again. 982 01:13:39,880 --> 01:13:44,350 Find another vertex with lots of internal light neighbors. 983 01:13:44,350 --> 01:13:45,430 Keep on going. 984 01:13:45,430 --> 01:13:49,410 And then we build a large light clique. 985 01:13:49,410 --> 01:13:51,010 So that's the idea. 986 01:13:51,010 --> 01:13:52,040 So we'll find that-- 987 01:13:55,250 --> 01:14:02,290 so we'll see that if this delta is bigger 988 01:14:02,290 --> 01:14:06,960 than, basically, the quantity claimed in the theorem-- 989 01:14:06,960 --> 01:14:13,842 so t minus 2 over 2t minus 3-- 990 01:14:13,842 --> 01:14:22,610 and sufficiently large C, then there 991 01:14:22,610 --> 01:14:33,290 exists the sequence U1, U2, U3, so on, all the way to Ut, 992 01:14:33,290 --> 01:14:41,640 and the sequence of vertices v1, v2, 993 01:14:41,640 --> 01:14:53,510 vt, such that, initially, I take A. And the idea is, 994 01:14:53,510 --> 01:14:56,870 initially, I take v1 to be whatever 995 01:14:56,870 --> 01:14:59,930 comes out of that lemma. 996 01:14:59,930 --> 01:15:05,420 And I want the property that all of the vi, vj's are light. 997 01:15:11,660 --> 01:15:21,350 And 2 is that no three of these v's have a common neighbor. 998 01:15:33,060 --> 01:15:35,310 And, I think, once you have these two properties, then 999 01:15:35,310 --> 01:15:38,370 you can find your clique subdivision. 1000 01:15:38,370 --> 01:15:40,620 You find these t light vertices. 1001 01:15:43,530 --> 01:15:51,220 So if you have v1, v2, v3, v4, you have these light vertices, 1002 01:15:51,220 --> 01:15:52,850 and I can build a clique subdivision 1003 01:15:52,850 --> 01:15:54,830 from these light vertices. 1004 01:15:54,830 --> 01:15:58,880 Because they're light, they have at least one common neighbor 1005 01:15:58,880 --> 01:16:01,330 for each pair. 1006 01:16:01,330 --> 01:16:02,850 I just keep building them. 1007 01:16:02,850 --> 01:16:04,900 So I build these common neighbors. 1008 01:16:04,900 --> 01:16:07,170 Well, you should be somewhat worried 1009 01:16:07,170 --> 01:16:11,720 that I end up using the same vertex twice. 1010 01:16:11,720 --> 01:16:13,880 But, of course, that should not be a worry 1011 01:16:13,880 --> 01:16:18,310 if I guarantee that no three of them have a common neighbor. 1012 01:16:18,310 --> 01:16:21,902 They cannot collapse. 1013 01:16:21,902 --> 01:16:28,480 And you cannot have two vertices end up being the same. 1014 01:16:28,480 --> 01:16:32,610 Otherwise, I would violate property b. 1015 01:16:32,610 --> 01:16:35,340 So these two properties alone allow 1016 01:16:35,340 --> 01:16:41,780 you to build a Kt subdivision. 1017 01:16:41,780 --> 01:16:44,935 But how do we find this sequence? 1018 01:16:44,935 --> 01:16:48,100 Well, so we build it iteratively using that lemma. 1019 01:16:48,100 --> 01:16:53,140 You start with one vertex guaranteed by that lemma. 1020 01:16:53,140 --> 01:16:54,566 You look at its neighborhood. 1021 01:16:57,970 --> 01:17:01,690 You pick another vertex guaranteed by the lemma. 1022 01:17:01,690 --> 01:17:05,580 You look at its neighborhood. 1023 01:17:05,580 --> 01:17:06,780 And so on. 1024 01:17:06,780 --> 01:17:09,748 And you build this sequence of light clique. 1025 01:17:09,748 --> 01:17:10,248 Yes. 1026 01:17:10,248 --> 01:17:12,515 AUDIENCE: [INAUDIBLE]. 1027 01:17:12,515 --> 01:17:13,140 YUFEI ZHAO: Ah. 1028 01:17:13,140 --> 01:17:14,307 The light neighborhood, yes. 1029 01:17:14,307 --> 01:17:15,820 So we're not in the graph any more. 1030 01:17:15,820 --> 01:17:24,780 So everything's inside A. Everything's inside A. 1031 01:17:24,780 --> 01:17:27,130 So let me finish off the list of properties, 1032 01:17:27,130 --> 01:17:30,040 and we're almost there. 1033 01:17:30,040 --> 01:17:33,550 Third property I want is that, when I do this operation, 1034 01:17:33,550 --> 01:17:38,050 I do not reduce my space of possibilities by too much. 1035 01:17:38,050 --> 01:17:45,850 Namely, that the size of U does not go down too substantially. 1036 01:17:51,273 --> 01:17:52,880 And that's guaranteed by the lemma. 1037 01:17:57,430 --> 01:18:00,430 And, 4 is that-- 1038 01:18:00,430 --> 01:18:02,080 basically this picture over here-- 1039 01:18:02,080 --> 01:18:09,940 vi is light to all of U i plus 1. 1040 01:18:15,790 --> 01:18:19,410 So I claim that you can find the sequence satisfying 1041 01:18:19,410 --> 01:18:21,788 these properties. 1042 01:18:21,788 --> 01:18:24,330 And the reason is that you are repeatedly applying the lemma. 1043 01:18:38,330 --> 01:18:45,530 So repeatedly apply the lemma. 1044 01:18:50,330 --> 01:18:52,910 The lemma doesn't address this part 1045 01:18:52,910 --> 01:18:56,890 about triple vertices having common neighbors. 1046 01:18:56,890 --> 01:18:59,870 But I claim that's actually not so hard to deal with. 1047 01:18:59,870 --> 01:19:05,570 Because if you think about how many vertices, 1048 01:19:05,570 --> 01:19:09,890 how many possibilities, this restriction eliminates, 1049 01:19:09,890 --> 01:19:22,480 b only eliminates at each step t choose 2, 1050 01:19:22,480 --> 01:19:32,160 because this is coming from the light restriction, times, 1051 01:19:32,160 --> 01:19:33,836 at most-- 1052 01:19:33,836 --> 01:19:36,870 so another t choose 2. 1053 01:19:36,870 --> 01:19:44,290 And this comes-- the first one comes from pairs of v1 1054 01:19:44,290 --> 01:19:46,400 through vt. 1055 01:19:46,400 --> 01:19:49,610 So eliminates, at most, this many-- 1056 01:19:49,610 --> 01:19:53,170 times the max degree.