1 00:00:18,675 --> 00:00:20,050 YUFEI ZHAO: We're about to embark 2 00:00:20,050 --> 00:00:22,900 on a new chapter in this course where 3 00:00:22,900 --> 00:00:25,520 I want to tell you about Szemeredi's graph regularity 4 00:00:25,520 --> 00:00:26,020 lemma. 5 00:00:46,820 --> 00:00:48,730 Szemeredi's graph regularity lemma 6 00:00:48,730 --> 00:00:52,400 is a very powerful tool in modern graph theory, developed 7 00:00:52,400 --> 00:00:54,440 back in the '70s. 8 00:00:54,440 --> 00:00:58,310 Today I want to show you the statement and the proof 9 00:00:58,310 --> 00:01:00,410 of this graph regularity lemma. 10 00:01:00,410 --> 00:01:05,120 And next time, we'll see how to apply the lemma for graph 11 00:01:05,120 --> 00:01:06,700 theoretic applications. 12 00:01:06,700 --> 00:01:09,050 And we'll also use it to give a proof of Roth's theorem. 13 00:01:11,850 --> 00:01:14,940 The idea of Szemeredi's regularity lemma 14 00:01:14,940 --> 00:01:24,460 is that if you are given a very large graph, G. 15 00:01:24,460 --> 00:01:28,090 And it's a fairly robust theorem, so 16 00:01:28,090 --> 00:01:33,370 any large, dense graph. 17 00:01:33,370 --> 00:01:38,640 And here, "dense" means, let's say, positive x density. 18 00:01:38,640 --> 00:01:50,050 Then it is possible to partition the vertex set of this graph G 19 00:01:50,050 --> 00:02:10,710 into a bounded number of pieces So that G looks random-like 20 00:02:10,710 --> 00:02:13,020 between most pairs of parts. 21 00:02:22,050 --> 00:02:24,930 So for instance, I might produce for you 22 00:02:24,930 --> 00:02:29,840 a partition of the vertex set into some number of parts. 23 00:02:29,840 --> 00:02:32,430 I'll draw five here. 24 00:02:32,430 --> 00:02:34,550 So you give me a graph G. I manage 25 00:02:34,550 --> 00:02:37,690 to produce for you this vertex partition 26 00:02:37,690 --> 00:02:42,130 so that if I look at between a typical pair of parts, 27 00:02:42,130 --> 00:02:46,870 you see here maybe the edge density is close to 0.2 28 00:02:46,870 --> 00:02:50,320 but otherwise, the bipartite graph looks 29 00:02:50,320 --> 00:02:53,580 like a random graph in some precise sense I will describe 30 00:02:53,580 --> 00:02:55,060 in a bit. 31 00:02:55,060 --> 00:02:57,150 And if you look at what the graph looks 32 00:02:57,150 --> 00:02:59,610 like between another pair of parts, 33 00:02:59,610 --> 00:03:01,770 maybe now it's a different x density. 34 00:03:01,770 --> 00:03:04,040 Maybe it's around 0.4. 35 00:03:04,040 --> 00:03:08,890 And again, looks like a random graph with that density. 36 00:03:08,890 --> 00:03:11,700 So in some sense, Szemeredi's regularity lemma 37 00:03:11,700 --> 00:03:14,760 is a universal structural description 38 00:03:14,760 --> 00:03:19,890 that allows you to approximate a graph by a bounded amount 39 00:03:19,890 --> 00:03:22,620 of information. 40 00:03:22,620 --> 00:03:24,300 So that's informally the idea. 41 00:03:24,300 --> 00:03:25,980 And you can already sense that this 42 00:03:25,980 --> 00:03:27,810 can be a very powerful tool. 43 00:03:27,810 --> 00:03:30,780 It doesn't matter what graph you input. 44 00:03:30,780 --> 00:03:36,630 You apply this lemma, and you get an approximate structural 45 00:03:36,630 --> 00:03:39,390 or as later on we'll see, it's also, in some sense, 46 00:03:39,390 --> 00:03:41,310 an analytic description of the graph. 47 00:03:43,840 --> 00:03:48,430 So the first part of today's lecture 48 00:03:48,430 --> 00:03:52,210 will develop just a statement of this regularity lemma. 49 00:03:52,210 --> 00:03:58,030 I'll show you what exactly do I mean by "random-like." 50 00:03:58,030 --> 00:04:03,294 Well, first let me give some definitions. 51 00:04:03,294 --> 00:04:06,350 I denote by the letter e if I input 52 00:04:06,350 --> 00:04:10,750 a pair of vertex sets, x and y. 53 00:04:13,900 --> 00:04:20,050 Here I might, later on, draw the subscript G 54 00:04:20,050 --> 00:04:23,530 if it's clear that I'm always talking about some graph 55 00:04:23,530 --> 00:04:27,220 G. So this is basically the number 56 00:04:27,220 --> 00:04:30,460 of edges between x and y. 57 00:04:41,710 --> 00:04:44,090 And I say "basically" because even though I 58 00:04:44,090 --> 00:04:47,720 will draw and depict everything as if x and y are disjoint 59 00:04:47,720 --> 00:04:51,320 sets, and that's the easiest case to think about, 60 00:04:51,320 --> 00:04:54,470 I'm also going to allow x and y to overlap, 61 00:04:54,470 --> 00:04:57,620 and also allow x and y to be the same set, in which case 62 00:04:57,620 --> 00:05:01,270 you should read a definition as to what this means. 63 00:05:01,270 --> 00:05:03,860 But it's fine to think of it as disjoint sets. 64 00:05:03,860 --> 00:05:07,760 So you're looking at a bipartite graph between x and y. 65 00:05:07,760 --> 00:05:13,100 We're also going to look at the edge density between x and y. 66 00:05:13,100 --> 00:05:17,300 And this is simply the number of edges divided 67 00:05:17,300 --> 00:05:23,570 by the product of the sizes of the sets, 68 00:05:23,570 --> 00:05:28,250 so what fraction of the possible pairs are actual edges. 69 00:05:28,250 --> 00:05:30,440 So from now on, I'll refer to this quantity 70 00:05:30,440 --> 00:05:31,720 as "edge density." 71 00:05:36,840 --> 00:05:41,370 So now, here's the definition of what "random-like" 72 00:05:41,370 --> 00:05:44,480 means for the purpose of Szemeredi's regularity lemma. 73 00:05:44,480 --> 00:05:50,880 So we define a notion of an epsilon regular pair 74 00:05:50,880 --> 00:05:52,140 to be as follows-- 75 00:05:52,140 --> 00:05:55,950 throughout, and later on, I will omit even saying this-- 76 00:05:55,950 --> 00:05:59,340 G will be some graph. 77 00:05:59,340 --> 00:06:05,220 And we're going to be looking at subsets of vertices of G. 78 00:06:05,220 --> 00:06:10,290 And we say that this pair of subsets of vertices 79 00:06:10,290 --> 00:06:18,250 is epsilon regular, again, in G, but later 80 00:06:18,250 --> 00:06:20,560 on I will even drop saying in G if it's clear 81 00:06:20,560 --> 00:06:22,710 which graph we're working with. 82 00:06:22,710 --> 00:06:25,540 So we say x and y is epsilon regular 83 00:06:25,540 --> 00:06:36,360 if for all subsets A of X, all B subsets of Y that are not 84 00:06:36,360 --> 00:06:48,720 too small, so each at least an epsilon proportion of the sets 85 00:06:48,720 --> 00:06:55,470 that they live in, we find that the x density between A and B 86 00:06:55,470 --> 00:06:59,160 differs from the x density between X and X 87 00:06:59,160 --> 00:07:03,010 by no more than epsilon. 88 00:07:03,010 --> 00:07:06,040 Let me draw you a picture. 89 00:07:06,040 --> 00:07:13,780 I have sets A and B. So I have sets X and Y in my graph G. 90 00:07:13,780 --> 00:07:21,000 And I want to say that the edges between X and Y are epsilon 91 00:07:21,000 --> 00:07:25,770 regular, so it's random-like, if the following holds-- 92 00:07:25,770 --> 00:07:31,700 that whenever I pick a subset A in the left set 93 00:07:31,700 --> 00:07:38,640 and a subset B of the right set, the edge density 94 00:07:38,640 --> 00:07:44,130 between A and B is approximately the same as the overall edge 95 00:07:44,130 --> 00:07:47,080 density between X and Y. 96 00:07:47,080 --> 00:07:51,100 So in particular, this bipartite graph, for instance, 97 00:07:51,100 --> 00:07:54,290 is not really dense in one part and really sparse 98 00:07:54,290 --> 00:07:55,150 in another part. 99 00:07:55,150 --> 00:07:58,390 Somehow the edges are evenly distributed 100 00:07:58,390 --> 00:08:00,810 in this precise manner. 101 00:08:00,810 --> 00:08:02,890 So that's the definition of epsilon regular. 102 00:08:02,890 --> 00:08:04,002 Yes, question. 103 00:08:04,002 --> 00:08:08,260 AUDIENCE: What is the epsilon for the size of A the same as 104 00:08:08,260 --> 00:08:11,345 epsilon for [INAUDIBLE]? 105 00:08:11,345 --> 00:08:12,720 YUFEI ZHAO: The question is here, 106 00:08:12,720 --> 00:08:15,763 why are we using the same epsilon here, here, and there? 107 00:08:15,763 --> 00:08:16,930 And that's a great question. 108 00:08:16,930 --> 00:08:19,070 So that's mostly out of convenience. 109 00:08:19,070 --> 00:08:21,780 So you could use different parameters. 110 00:08:21,780 --> 00:08:24,120 And they do play somewhat different roles, 111 00:08:24,120 --> 00:08:27,510 but at the end, we'll generally be looking 112 00:08:27,510 --> 00:08:29,740 at one type of epsilons. 113 00:08:29,740 --> 00:08:31,390 So we just make our life easier. 114 00:08:31,390 --> 00:08:33,539 So you could extend the definition 115 00:08:33,539 --> 00:08:37,600 by having an epsilon comma eta, if you like, 116 00:08:37,600 --> 00:08:40,020 but it will not be necessary for us, and mostly 117 00:08:40,020 --> 00:08:41,780 for simplification. 118 00:08:41,780 --> 00:08:43,144 Any more questions? 119 00:08:46,539 --> 00:08:47,509 All right. 120 00:08:51,890 --> 00:08:58,670 Now if you have a pair x, y that is not epsilon regular, 121 00:08:58,670 --> 00:09:00,690 I just want to introduce a piece of terminology. 122 00:09:00,690 --> 00:09:02,732 So you can read from the definition what it means 123 00:09:02,732 --> 00:09:04,160 to be not epsilon regular. 124 00:09:04,160 --> 00:09:06,370 And sometimes I will say "epsilon irregular," 125 00:09:06,370 --> 00:09:11,600 but to be precise, I'll stick with not epsilon regular. 126 00:09:16,680 --> 00:09:20,910 Then we can exhibit this A and B that witnesses 127 00:09:20,910 --> 00:09:23,080 the irregularity. 128 00:09:23,080 --> 00:09:29,700 So if x, y is not epsilon regular, then 129 00:09:29,700 --> 00:09:45,980 their irregularity as, we say it's "witnessed by" some pair 130 00:09:45,980 --> 00:09:52,380 A in X and B in Y, satisfying-- 131 00:09:52,380 --> 00:09:57,760 basically, you read the definition, 132 00:09:57,760 --> 00:10:01,200 and such that the density between A and B 133 00:10:01,200 --> 00:10:06,150 differs quite a bit from the density between X and Y. 134 00:10:06,150 --> 00:10:09,450 So when I say "to exhibit" or "to witness irregularity," 135 00:10:09,450 --> 00:10:11,640 that's what I mean. 136 00:10:11,640 --> 00:10:13,690 Now, there's a bit of an unfortunate nomenclature 137 00:10:13,690 --> 00:10:15,880 in graph theory, where previously, we 138 00:10:15,880 --> 00:10:19,960 said "irregular graphs" to mean that every vertex is degree D. 139 00:10:19,960 --> 00:10:23,710 And now we say "epsilon regular" to mean this. 140 00:10:23,710 --> 00:10:24,460 Sorry about that. 141 00:10:24,460 --> 00:10:27,070 These are both standard, so usually from context, 142 00:10:27,070 --> 00:10:28,900 it's clear which one is meant. 143 00:10:34,170 --> 00:10:37,050 So this is what it means for a single pair of vertex sets 144 00:10:37,050 --> 00:10:38,670 to be epsilon regular. 145 00:10:38,670 --> 00:10:40,680 But now I give you a graph. 146 00:10:40,680 --> 00:10:44,110 And I give you a partition of the vertex set. 147 00:10:44,110 --> 00:10:45,900 So what does it mean for that partition 148 00:10:45,900 --> 00:10:47,955 to be epsilon regular? 149 00:10:47,955 --> 00:10:49,818 And here's the second definition. 150 00:10:53,940 --> 00:11:05,980 So an epsilon regular partition, we say that a partition-- 151 00:11:05,980 --> 00:11:08,120 and generally, I will denote partition 152 00:11:08,120 --> 00:11:12,860 by curly letters such as that, P. So the partition 153 00:11:12,860 --> 00:11:23,030 will divide a vertex set into a bunch of subsets. 154 00:11:23,030 --> 00:11:30,880 So we say that that partition is epsilon regular 155 00:11:30,880 --> 00:11:33,430 if the following is true-- 156 00:11:33,430 --> 00:11:45,590 if I sum over all pairs of irregular 157 00:11:45,590 --> 00:11:55,100 or pairs of vertex sets that are not epsilon regular, 158 00:11:55,100 --> 00:11:59,370 so over Vi, Vj not epsilon regular, and sum 159 00:11:59,370 --> 00:12:07,020 over the product of their sizes, then what I would like 160 00:12:07,020 --> 00:12:12,540 is for the sum to be at most epsilon times 161 00:12:12,540 --> 00:12:18,320 the number of pairs of vertices in G. In other words, 162 00:12:18,320 --> 00:12:22,060 a small fraction of pairs of vertices, 163 00:12:22,060 --> 00:12:24,790 not necessarily edges, but just pairs of vertices, 164 00:12:24,790 --> 00:12:30,050 lie between pairs of vertex parts 165 00:12:30,050 --> 00:12:33,060 that are not epsilon regular. 166 00:12:33,060 --> 00:12:37,350 So for instance, if you do not have epsilon-- 167 00:12:37,350 --> 00:12:39,360 if all of your pairs are epsilon regular, 168 00:12:39,360 --> 00:12:41,850 then the partition is epsilon regular. 169 00:12:41,850 --> 00:12:46,610 But I do allow a small number of blemishes. 170 00:12:46,610 --> 00:12:48,003 And that will be necessary. 171 00:12:53,680 --> 00:12:56,340 Just to clarify a subtle point here, 172 00:12:56,340 --> 00:13:03,310 here I do allow in the summation i equals to j, 173 00:13:03,310 --> 00:13:05,990 although in practice it doesn't really matter. 174 00:13:05,990 --> 00:13:08,690 You'll see that it's not really going to come up as an issue. 175 00:13:14,680 --> 00:13:16,430 And one of the reasons that it's not 176 00:13:16,430 --> 00:13:19,280 going to come up as an issue is usually 177 00:13:19,280 --> 00:13:22,220 when we apply this lemma, we're going to have a lot of parts. 178 00:13:22,220 --> 00:13:24,330 In fact, we can make sure that there 179 00:13:24,330 --> 00:13:26,700 is a minimum number of parts. 180 00:13:26,700 --> 00:13:34,010 And if none of the parts are too big, then having i equals to j 181 00:13:34,010 --> 00:13:36,030 contributes very little to that sum anyway. 182 00:13:41,410 --> 00:13:49,530 In particular, if all the set sizes 183 00:13:49,530 --> 00:13:51,870 in this partition are roughly the same-- 184 00:13:55,730 --> 00:14:01,950 so if they're all roughly 1 over k fraction of the entire vertex 185 00:14:01,950 --> 00:14:02,970 set-- 186 00:14:02,970 --> 00:14:13,410 then that statement up there being epsilon regular partition 187 00:14:13,410 --> 00:14:16,500 up to changing this epsilon is basically the same 188 00:14:16,500 --> 00:14:23,030 as saying that fewer than epsilon fraction of the pairs 189 00:14:23,030 --> 00:14:29,670 Vi, Vj are not epsilon regular. 190 00:14:29,670 --> 00:14:32,840 And here, if k is large enough, I 191 00:14:32,840 --> 00:14:35,910 can even let you make i and j different. 192 00:14:35,910 --> 00:14:39,110 It's not going to affect things after small changes in epsilon. 193 00:14:42,450 --> 00:14:45,140 So when it comes to-- so for people who are seeing 194 00:14:45,140 --> 00:14:48,040 Szemeredi's regularity lemma for the first time-- 195 00:14:48,040 --> 00:14:51,370 I think that's maybe all of you, or most of you-- 196 00:14:51,370 --> 00:14:54,820 I don't want you to focus on the precise statements 197 00:14:54,820 --> 00:14:56,710 so much as the spirit of the lemma. 198 00:14:56,710 --> 00:14:59,220 Because if you get too nitty gritty with 199 00:14:59,220 --> 00:15:02,350 is that the same as that epsilon, you get very confused 200 00:15:02,350 --> 00:15:03,860 very quickly. 201 00:15:03,860 --> 00:15:06,820 So I want you to focus on the spirit of this lemma. 202 00:15:06,820 --> 00:15:09,190 I will state everything precisely, 203 00:15:09,190 --> 00:15:13,588 but the idea is that most pairs are not epsilon regular. 204 00:15:13,588 --> 00:15:15,630 And don't worry too much about if you are allowed 205 00:15:15,630 --> 00:15:17,110 to take i equals to j or not. 206 00:15:19,960 --> 00:15:22,490 So now we're ready to state Szemeredi's regularity lemma. 207 00:15:38,450 --> 00:15:42,530 And it says that for every epsilon, 208 00:15:42,530 --> 00:15:50,170 there exists some constant M depending only on epsilon such 209 00:15:50,170 --> 00:16:05,520 that every graph has an epsilon regular partition 210 00:16:05,520 --> 00:16:09,140 into at most M parts. 211 00:16:14,330 --> 00:16:17,390 You give me the epsilon, for example 1%, 212 00:16:17,390 --> 00:16:20,240 and there exists some constant such 213 00:16:20,240 --> 00:16:25,280 that every graph has a 1% regular partition 214 00:16:25,280 --> 00:16:28,838 into a bounded number of parts. 215 00:16:28,838 --> 00:16:30,630 In particular-- and this is very important, 216 00:16:30,630 --> 00:16:32,130 make sure you understand this part-- 217 00:16:32,130 --> 00:16:35,840 that the number of parts does not depend 218 00:16:35,840 --> 00:16:36,890 on the size of the graph. 219 00:16:48,200 --> 00:16:50,700 Now, it's true that for some graphs, maybe 220 00:16:50,700 --> 00:16:53,600 you do need very many parts. 221 00:16:53,600 --> 00:16:58,190 But the number of parts does not get substantially bigger, 222 00:16:58,190 --> 00:17:02,480 or does not exceed this bound, even if you look at graphs 223 00:17:02,480 --> 00:17:04,630 that have unbounded size. 224 00:17:04,630 --> 00:17:07,310 So it is really a universal theorem in the sense 225 00:17:07,310 --> 00:17:10,290 that it's independent of the size of the graph. 226 00:17:13,839 --> 00:17:16,200 Any questions about the statement of this theorem? 227 00:17:16,200 --> 00:17:16,876 Yes. 228 00:17:16,876 --> 00:17:19,209 AUDIENCE: So in the informal statement at the beginning, 229 00:17:19,209 --> 00:17:20,759 you said G was a large, dense graph. 230 00:17:20,759 --> 00:17:21,801 YUFEI ZHAO: That's right. 231 00:17:21,801 --> 00:17:23,859 AUDIENCE: Is the dense condition appropriate anywhere in here? 232 00:17:23,859 --> 00:17:26,220 YUFEI ZHAO: So the question is, why did I say 233 00:17:26,220 --> 00:17:29,220 that G is a large, dense graph? 234 00:17:29,220 --> 00:17:30,900 And that's a great question. 235 00:17:30,900 --> 00:17:35,370 And that's because if G had a sub-linear number of edges, 236 00:17:35,370 --> 00:17:39,270 then I claim that all-- 237 00:17:39,270 --> 00:17:46,090 if you look at the definition of epsilon regular pair, 238 00:17:46,090 --> 00:17:52,180 and your epsilon is a constant, and if your edge densities 239 00:17:52,180 --> 00:17:54,250 are sub-linear, then all of these guys, 240 00:17:54,250 --> 00:17:56,170 they are little o of 1. 241 00:17:56,170 --> 00:17:57,520 They go to 0. 242 00:17:57,520 --> 00:18:03,170 So trivially, you will satisfy the epsilon regular condition. 243 00:18:03,170 --> 00:18:04,940 So if your graph is sparse-- 244 00:18:04,940 --> 00:18:09,410 sparse in the sense of having sub-quadratic number of edges-- 245 00:18:09,410 --> 00:18:15,070 then you trivially obtain epsilon regularity. 246 00:18:15,070 --> 00:18:17,620 And so the theorem is still true. 247 00:18:17,620 --> 00:18:19,000 It's just not meaningful. 248 00:18:19,000 --> 00:18:20,050 It's just not useful. 249 00:18:20,050 --> 00:18:22,720 But there are settings where having sparse graphs-- 250 00:18:22,720 --> 00:18:25,410 and we'll come back to this later in the course-- 251 00:18:25,410 --> 00:18:27,757 it's important to explore what happens to sparse graphs. 252 00:18:27,757 --> 00:18:28,257 Yeah. 253 00:18:28,257 --> 00:18:31,670 AUDIENCE: So that M is independent of G. 254 00:18:31,670 --> 00:18:35,030 YUFEI ZHAO: Yes, M is independent of G. M depends 255 00:18:35,030 --> 00:18:35,900 only on epsilon. 256 00:18:39,108 --> 00:18:41,566 AUDIENCE: M is really large, but there's no enough vertices 257 00:18:41,566 --> 00:18:42,290 in the graph. 258 00:18:42,290 --> 00:18:42,990 YUFEI ZHAO: OK, question is, what 259 00:18:42,990 --> 00:18:44,948 happens when M is very large, but there are not 260 00:18:44,948 --> 00:18:47,170 enough vertices in the graph? 261 00:18:47,170 --> 00:18:50,620 Well, if your M is a million, and your graph only 262 00:18:50,620 --> 00:18:52,690 has 1,000 vertices, what you can do 263 00:18:52,690 --> 00:18:54,610 is have every vertex be its own part. 264 00:18:58,730 --> 00:19:02,060 Every vertex is its own part, a singleton partition. 265 00:19:02,060 --> 00:19:05,240 And you can check that that partition satisfies 266 00:19:05,240 --> 00:19:06,980 the properties. 267 00:19:06,980 --> 00:19:12,152 Every pair is a single edge and it's epsilon regular. 268 00:19:12,152 --> 00:19:13,090 Yeah. 269 00:19:13,090 --> 00:19:15,163 AUDIENCE: So in the definition, is it 270 00:19:15,163 --> 00:19:16,330 sort of like all or nothing? 271 00:19:16,330 --> 00:19:18,788 You can either [INAUDIBLE] epsilon regularity [INAUDIBLE].. 272 00:19:18,788 --> 00:19:21,110 Do you get anything where if you, like, say, 273 00:19:21,110 --> 00:19:24,838 make this more continuous, so you allow for it to be-- 274 00:19:24,838 --> 00:19:26,974 you quantify how irregular it is, and then can 275 00:19:26,974 --> 00:19:28,790 you make [INAUDIBLE]? 276 00:19:28,790 --> 00:19:30,660 YUFEI ZHAO: OK, so my understanding what 277 00:19:30,660 --> 00:19:33,360 you're asking is in the definition up there, 278 00:19:33,360 --> 00:19:36,300 the sum is-- 279 00:19:36,300 --> 00:19:39,360 we put the pair in the sum of this epsilon regular, 280 00:19:39,360 --> 00:19:40,680 and otherwise don't put it. 281 00:19:40,680 --> 00:19:44,550 Is there some gradual way to put some measure of irregularity 282 00:19:44,550 --> 00:19:45,450 into that sum? 283 00:19:45,450 --> 00:19:48,410 And there are versions of regularity lemma that do that, 284 00:19:48,410 --> 00:19:51,030 but they are all, in spirit, morally 285 00:19:51,030 --> 00:19:53,885 the same as that one there. 286 00:19:53,885 --> 00:19:54,385 Yeah. 287 00:19:54,385 --> 00:19:55,927 AUDIENCE: In the informal definition, 288 00:19:55,927 --> 00:19:57,200 what does "random-like" mean? 289 00:19:57,200 --> 00:19:58,950 YUFEI ZHAO: So in the informal definition, 290 00:19:58,950 --> 00:20:00,510 what does "random-like" mean? 291 00:20:00,510 --> 00:20:03,330 This is the formal definition of what "random-like" means. 292 00:20:03,330 --> 00:20:06,240 So actually later on in the course, one of the chapters 293 00:20:06,240 --> 00:20:09,390 will explore what pseudo-random graphs are. 294 00:20:09,390 --> 00:20:10,980 So pseudo-random graph, in some sense, 295 00:20:10,980 --> 00:20:12,480 means graphs that are not random, 296 00:20:12,480 --> 00:20:15,030 but behave in some sense like random. 297 00:20:15,030 --> 00:20:19,550 So "random-like" generally just means that in some aspect, 298 00:20:19,550 --> 00:20:23,970 in some property, it looks like a random object. 299 00:20:23,970 --> 00:20:27,830 And this is one way that something can look like random. 300 00:20:27,830 --> 00:20:31,910 So a random graph has this property, 301 00:20:31,910 --> 00:20:34,580 but random graphs also have many other properties 302 00:20:34,580 --> 00:20:37,790 that are not being exhibited in this definition. 303 00:20:37,790 --> 00:20:42,260 But this is one way that graph can look like random. 304 00:20:42,260 --> 00:20:43,430 So that's a great question. 305 00:20:43,430 --> 00:20:45,680 And we'll come back to that topic later in the course. 306 00:20:47,655 --> 00:20:49,030 All of these are great questions. 307 00:20:49,030 --> 00:20:51,870 So Szemeredi's regularity lemma, the first time you see it, 308 00:20:51,870 --> 00:20:54,310 it can look somewhat scary. 309 00:20:54,310 --> 00:20:55,910 But I want you to try to understand 310 00:20:55,910 --> 00:20:56,900 it more conceptually. 311 00:20:56,900 --> 00:20:58,550 So please do ask questions. 312 00:21:05,900 --> 00:21:09,350 Before diving into the proof, I want to make a few more remarks 313 00:21:09,350 --> 00:21:10,709 about a statement. 314 00:21:15,300 --> 00:21:18,130 It is possible to-- 315 00:21:18,130 --> 00:21:21,840 we will prove this version of the regularity lemma. 316 00:21:21,840 --> 00:21:25,930 But as I mentioned, it is the spirit of the regularity lemma 317 00:21:25,930 --> 00:21:27,360 that I care more about. 318 00:21:27,360 --> 00:21:29,610 And it's a very robust statement. 319 00:21:29,610 --> 00:21:32,697 You can add on extra declarations that somehow 320 00:21:32,697 --> 00:21:33,780 doesn't change the spirit. 321 00:21:33,780 --> 00:21:35,613 And the proof will be more or less the same, 322 00:21:35,613 --> 00:21:38,770 but for various applications will be slightly more useful. 323 00:21:38,770 --> 00:21:41,300 So in particular, it is possible to make 324 00:21:41,300 --> 00:21:43,329 the partition equitable. 325 00:21:56,060 --> 00:21:58,000 And "equitable partition" sometimes 326 00:21:58,000 --> 00:22:04,540 is also called an "equipartition," meaning 327 00:22:04,540 --> 00:22:09,490 that it has such that all the Ai's, all 328 00:22:09,490 --> 00:22:13,240 the Bi's have sizes differing by at most 1. 329 00:22:19,090 --> 00:22:22,910 So basically, all the parts have the same size up to at most 1, 330 00:22:22,910 --> 00:22:25,700 because of divisibility. 331 00:22:25,700 --> 00:22:29,570 So let me state a version of regularity lemma 332 00:22:29,570 --> 00:22:31,690 for equitable partitions. 333 00:22:31,690 --> 00:22:37,010 So for every epsilon in m, little m0, 334 00:22:37,010 --> 00:22:48,560 there exists a big M such that every graph 335 00:22:48,560 --> 00:23:05,250 has an epsilon regular equitable partition of the vertex set 336 00:23:05,250 --> 00:23:13,440 into k parts, where k is at least little m, so I can 337 00:23:13,440 --> 00:23:15,420 guarantee a minimum number of parts, 338 00:23:15,420 --> 00:23:17,940 and at most some bounded number. 339 00:23:17,940 --> 00:23:24,750 Again this bound may depend on your inputs epsilon and m0, 340 00:23:24,750 --> 00:23:28,740 but it does not depend on the graph itself. 341 00:23:28,740 --> 00:23:30,950 And you see the slightly stronger conclusion 342 00:23:30,950 --> 00:23:33,170 for many applications is more convenient, 343 00:23:33,170 --> 00:23:37,120 to use this formulation. 344 00:23:37,120 --> 00:23:41,890 And I will comment on how you may modify the proof that we'll 345 00:23:41,890 --> 00:23:46,820 see today into one where you can guarantee equitability. 346 00:23:46,820 --> 00:23:57,750 And you see that for this m, little m0 too small, 347 00:23:57,750 --> 00:24:02,040 for example, if it's somewhat larger than 1 over epsilon, 348 00:24:02,040 --> 00:24:06,260 when you look at the definition of epsilon regular partition, 349 00:24:06,260 --> 00:24:10,080 it suffices to check that at most epsilon k squared, 350 00:24:10,080 --> 00:24:16,225 epsilon fraction of the pairs, Vi, Vj 351 00:24:16,225 --> 00:24:20,710 is epsilon regular over i different 352 00:24:20,710 --> 00:24:23,950 from j, again up to changing epsilon, 353 00:24:23,950 --> 00:24:25,580 let's say, by a factor of 2. 354 00:24:25,580 --> 00:24:28,540 So all of these definitions are basically 355 00:24:28,540 --> 00:24:30,940 the same up to small changes in the parameters. 356 00:24:42,420 --> 00:24:45,080 Next time, we'll see how to apply the regularity lemma. 357 00:24:45,080 --> 00:24:47,960 And we will apply it in the first form, 358 00:24:47,960 --> 00:24:49,682 but you see the second form guarantees 359 00:24:49,682 --> 00:24:51,140 you a somewhat stronger conclusion, 360 00:24:51,140 --> 00:24:53,370 and sometimes more convenient to use. 361 00:24:53,370 --> 00:24:55,430 So for example on the homework problems, 362 00:24:55,430 --> 00:24:59,900 if you wish to use the second form, then please go ahead. 363 00:24:59,900 --> 00:25:01,760 Just make your life somewhat easier, 364 00:25:01,760 --> 00:25:04,040 but it essentially captures all the spirit 365 00:25:04,040 --> 00:25:05,720 of Szemeredi's regularity. 366 00:25:09,815 --> 00:25:10,690 Any questions so far? 367 00:25:13,610 --> 00:25:17,600 I want to explain the idea of the proof of the regularity 368 00:25:17,600 --> 00:25:19,220 lemma. 369 00:25:19,220 --> 00:25:21,980 And this is a very important technique 370 00:25:21,980 --> 00:25:25,493 in this area called the "energy increment argument." 371 00:25:35,850 --> 00:25:38,110 Here's the idea. 372 00:25:38,110 --> 00:25:43,710 We start with some partition, so for example, 373 00:25:43,710 --> 00:25:46,250 the trivial partition-- 374 00:25:46,250 --> 00:25:49,680 and by that I mean you only have one part. 375 00:25:52,295 --> 00:25:53,670 All the vertices are in one part. 376 00:25:53,670 --> 00:25:55,830 You're not doing anything to the vertex set. 377 00:25:55,830 --> 00:25:58,770 It's one gigantic part. 378 00:25:58,770 --> 00:26:01,350 Or if you're looking at some other variant, 379 00:26:01,350 --> 00:26:03,120 you can easily modify the proof. 380 00:26:03,120 --> 00:26:09,050 So for example, you can also look at an arbitrary partition 381 00:26:09,050 --> 00:26:13,523 into little m0 parts, if you wish to have 382 00:26:13,523 --> 00:26:14,690 that as your starting point. 383 00:26:14,690 --> 00:26:16,898 So or I'm saying is that this proof is fairly robust. 384 00:26:19,880 --> 00:26:23,050 And we're going to do some iterations. 385 00:26:26,680 --> 00:26:37,970 So as long as your partition is not epsilon regular, 386 00:26:37,970 --> 00:26:44,660 we will do something to the partition to move forward. 387 00:26:44,660 --> 00:26:54,590 And what we will do is look at each pair 388 00:26:54,590 --> 00:26:59,600 of parts in your partition that's not epsilon regular. 389 00:27:04,170 --> 00:27:05,960 Well, if they're not epsilon regular, 390 00:27:05,960 --> 00:27:08,390 then I can find a pair of subsets 391 00:27:08,390 --> 00:27:17,520 which are denoted by the A's that witnesses 392 00:27:17,520 --> 00:27:26,880 this non regularity, that witnesses the irregularity. 393 00:27:26,880 --> 00:27:29,370 And we start with some partition. 394 00:27:29,370 --> 00:27:33,210 So now let us refine the partition 395 00:27:33,210 --> 00:27:36,720 into a partition in even more parts 396 00:27:36,720 --> 00:27:53,090 by simultaneously refining the partition using 397 00:27:53,090 --> 00:28:00,360 all of these Ai, j's that we found in the step above. 398 00:28:00,360 --> 00:28:01,930 So you start with some partition. 399 00:28:01,930 --> 00:28:05,560 If it is not regular, I can chop up 400 00:28:05,560 --> 00:28:08,120 the various parts in some way. 401 00:28:08,120 --> 00:28:12,090 So I start with some partition over here. 402 00:28:12,090 --> 00:28:15,930 And what we are going to do is, let's say between these two, 403 00:28:15,930 --> 00:28:18,180 it's not epsilon regular, so I can 404 00:28:18,180 --> 00:28:23,260 find some pairs of vertex sets that exhibits the irregularity. 405 00:28:23,260 --> 00:28:24,650 I chop it up. 406 00:28:24,650 --> 00:28:31,230 And I can keep further chopping up the rest of the parts. 407 00:28:31,230 --> 00:28:33,420 If these two parts are not epsilon regular, 408 00:28:33,420 --> 00:28:35,580 then I chop it up like that. 409 00:28:35,580 --> 00:28:38,802 And I can keep on doing it. 410 00:28:38,802 --> 00:28:40,260 And originally, I have three parts. 411 00:28:40,260 --> 00:28:43,040 Now I have 12 parts. 412 00:28:43,040 --> 00:28:45,580 And this is a refined partition. 413 00:28:45,580 --> 00:28:49,130 And now I repeat until I am done. 414 00:28:49,130 --> 00:28:52,490 I am done when I obtain a partition that 415 00:28:52,490 --> 00:28:55,750 is epsilon regular. 416 00:28:55,750 --> 00:28:58,260 Now, the basic question when it comes to the strategy 417 00:28:58,260 --> 00:29:00,270 is, are you ever going to be done? 418 00:29:00,270 --> 00:29:02,570 When are you going to be done? 419 00:29:02,570 --> 00:29:04,650 And if this process goes on forever 420 00:29:04,650 --> 00:29:07,047 or goes on for a very long time, then you 421 00:29:07,047 --> 00:29:08,130 might have a lot of parts. 422 00:29:08,130 --> 00:29:09,600 But we want to guarantee that there 423 00:29:09,600 --> 00:29:12,680 is a bounded number of parts. 424 00:29:12,680 --> 00:29:18,470 So what we will show is that-- 425 00:29:18,470 --> 00:29:22,360 to show that you have a small number of parts, 426 00:29:22,360 --> 00:29:24,740 in other words, why does this process even stop-- 427 00:29:30,915 --> 00:29:33,080 and in particular, we want it to stop 428 00:29:33,080 --> 00:29:38,320 after a small number of steps, after a bounded number 429 00:29:38,320 --> 00:29:40,750 of steps. 430 00:29:40,750 --> 00:29:43,560 And to do this, we will define some notion called 431 00:29:43,560 --> 00:29:47,627 an "energy" of a partition. 432 00:29:51,370 --> 00:29:55,280 And this energy will increase. 433 00:29:55,280 --> 00:29:58,300 So first of all, the energy is some quantity 434 00:29:58,300 --> 00:30:02,980 that we'll define that lies between 0 and 1. 435 00:30:02,980 --> 00:30:05,860 It's some real number lying between 0 and 1. 436 00:30:05,860 --> 00:30:17,800 And each step, the energy goes up by some specific quantity. 437 00:30:22,920 --> 00:30:26,520 Therefore, because the energy cannot increase past 1, 438 00:30:26,520 --> 00:30:30,200 this iteration stops after a bounded number of steps. 439 00:30:33,160 --> 00:30:36,800 And once it's done, we end up with 440 00:30:36,800 --> 00:30:42,000 a epsilon regular partition. 441 00:30:42,000 --> 00:30:44,800 So that's the basic strategy. 442 00:30:44,800 --> 00:30:49,620 And what I want to show you is how to execute that strategy. 443 00:30:49,620 --> 00:30:50,986 Any questions so far? 444 00:30:50,986 --> 00:30:51,486 Yes. 445 00:30:51,486 --> 00:30:55,760 AUDIENCE: Just to clarify [INAUDIBLE] a bit, 446 00:30:55,760 --> 00:30:59,574 if some Vi's into non-epsilon regular partitions, 447 00:30:59,574 --> 00:31:04,347 is it possible for Ai,j and Aik to overlap somehow, right? 448 00:31:04,347 --> 00:31:07,502 Just kind of make those into three partitions? 449 00:31:07,502 --> 00:31:09,210 YUFEI ZHAO: So if I understand correctly, 450 00:31:09,210 --> 00:31:13,880 you are worried about between different pairs, 451 00:31:13,880 --> 00:31:15,110 you might have interactions. 452 00:31:15,110 --> 00:31:15,875 AUDIENCE: Yeah. 453 00:31:15,875 --> 00:31:17,500 YUFEI ZHAO: So you have seen the proof, 454 00:31:17,500 --> 00:31:19,917 but I think this is actually a very important and somewhat 455 00:31:19,917 --> 00:31:25,330 subtle point, is that I do not refine at each step, 456 00:31:25,330 --> 00:31:27,520 I find a pair of witnessing sets. 457 00:31:27,520 --> 00:31:31,860 I find all of these witnessing sets all at the same time, 458 00:31:31,860 --> 00:31:35,758 and I refine everything all at once. 459 00:31:35,758 --> 00:31:37,258 AUDIENCE: OK, so it's like if you do 460 00:31:37,258 --> 00:31:39,780 have overlap between two witnessing sets, that's OK? 461 00:31:39,780 --> 00:31:42,900 YUFEI ZHAO: That is OK, because this step doesn't care. 462 00:31:42,900 --> 00:31:45,720 If you have two witnessing sets that overlap, that is OK. 463 00:31:48,280 --> 00:31:49,210 We'll see the proof. 464 00:31:49,210 --> 00:31:49,815 Yes. 465 00:31:49,815 --> 00:31:51,273 AUDIENCE: Do you just find one pair 466 00:31:51,273 --> 00:31:53,652 of witnessing sets for each Vi, Vj, 467 00:31:53,652 --> 00:31:55,030 even though there might be more? 468 00:31:55,030 --> 00:31:56,488 YUFEI ZHAO: Question is, do we find 469 00:31:56,488 --> 00:31:58,235 just one pair of witnessing sets even 470 00:31:58,235 --> 00:31:59,360 though there could be more? 471 00:31:59,360 --> 00:32:00,100 And the answer is, yes. 472 00:32:00,100 --> 00:32:01,360 We just need to find one. 473 00:32:01,360 --> 00:32:02,430 There could be lots. 474 00:32:02,430 --> 00:32:04,600 So if it's not epsilon regular, it 475 00:32:04,600 --> 00:32:07,600 might be very not epsilon regular. 476 00:32:07,600 --> 00:32:11,450 And in fact, being a witnessing set is a fairly robust notion. 477 00:32:11,450 --> 00:32:13,700 If you just take out a small number of vertices, 478 00:32:13,700 --> 00:32:17,206 it's still a witnessing set. 479 00:32:17,206 --> 00:32:20,160 Any more questions? 480 00:32:20,160 --> 00:32:20,660 Great. 481 00:32:20,660 --> 00:32:25,090 So let's take a quick break and then we'll see the proof. 482 00:32:25,090 --> 00:32:28,260 Let's get started with the proof of Szemeredi's regularity 483 00:32:28,260 --> 00:32:29,050 lemma. 484 00:32:29,050 --> 00:32:31,420 And to do the proof, I want to develop 485 00:32:31,420 --> 00:32:34,510 this notion of energy which you saw in the proof sketch. 486 00:32:37,290 --> 00:32:39,830 So what do I mean by "energy?" 487 00:32:39,830 --> 00:32:46,640 First, if I-- let me define some quantities. 488 00:32:46,640 --> 00:32:53,870 If I have two vertex subsets, U and W, 489 00:32:53,870 --> 00:33:00,810 let me define this quantity, q, which is basically 490 00:33:00,810 --> 00:33:05,470 the edge density squared. 491 00:33:05,470 --> 00:33:08,580 But I normalize it somewhat according 492 00:33:08,580 --> 00:33:11,220 to how big U and W are. 493 00:33:14,560 --> 00:33:18,180 I'm going to use the letter and N to denote 494 00:33:18,180 --> 00:33:26,880 the number of vertices in G. So this is some cube. 495 00:33:26,880 --> 00:33:36,980 And for partitions, if I have a pair of partitions, 496 00:33:36,980 --> 00:33:56,230 Pu of U into k parts, and the partition Pw of W into l parts, 497 00:33:56,230 --> 00:34:14,770 I set this q of Pu and Pw to be the quantity where I sum over 498 00:34:14,770 --> 00:34:18,370 basically all pairs, one part from U, 499 00:34:18,370 --> 00:34:27,409 one part from W of this q between Ui and Wj. 500 00:34:30,150 --> 00:34:32,010 So this is the density squared. 501 00:34:32,010 --> 00:34:35,909 And I'm taking some kind of weighted average 502 00:34:35,909 --> 00:34:39,280 of the squared density. 503 00:34:39,280 --> 00:34:41,370 So here is a weighted average. 504 00:34:41,370 --> 00:34:44,280 If you prefer to think about the special case 505 00:34:44,280 --> 00:34:47,310 where this partition is an equipartition, 506 00:34:47,310 --> 00:34:51,750 then it is really the average of these squared densities. 507 00:34:51,750 --> 00:34:53,070 It's a mean square density. 508 00:34:55,860 --> 00:35:07,450 And finally, for a partition P of the vertex set of G 509 00:35:07,450 --> 00:35:18,560 into m parts, we define this q of this partition P 510 00:35:18,560 --> 00:35:23,480 to be q of P with itself according 511 00:35:23,480 --> 00:35:25,730 to the previous definition. 512 00:35:25,730 --> 00:35:32,000 Or in other words, I do this double sum, i from 1 to m, 513 00:35:32,000 --> 00:35:37,460 j from 1 to m, q of Vi, Vj. 514 00:35:40,550 --> 00:35:42,660 And this is the quantity that I will call 515 00:35:42,660 --> 00:35:53,300 the "energy" of the partition. 516 00:35:53,300 --> 00:36:01,440 It is a mean squared density, some weighted mean 517 00:36:01,440 --> 00:36:05,220 of the edge densities between pairs 518 00:36:05,220 --> 00:36:07,072 of parts in the partition. 519 00:36:09,970 --> 00:36:12,800 You might ask, why is it called an energy? 520 00:36:12,800 --> 00:36:15,370 So you might see from this formula here, 521 00:36:15,370 --> 00:36:19,250 it's some kind of a mean square density, 522 00:36:19,250 --> 00:36:22,600 so it's some kind of an average of squares. 523 00:36:22,600 --> 00:36:27,640 So in particular, it's some kind of an L2 quantity. 524 00:36:27,640 --> 00:36:31,900 And there's a general phenomenon in mathematics, 525 00:36:31,900 --> 00:36:35,110 I think borrowed from physical intuitions, 526 00:36:35,110 --> 00:36:37,390 that you can pretty much call anything that's 527 00:36:37,390 --> 00:36:40,900 an L2 quantity an energy. 528 00:36:40,900 --> 00:36:43,460 And so that's, I think, where the name comes from. 529 00:36:46,260 --> 00:36:48,920 So this is the important object for our proof. 530 00:36:48,920 --> 00:36:53,090 And let's see how to execute a strategy, the energy increment 531 00:36:53,090 --> 00:36:56,410 argument outlined on the board over there. 532 00:36:56,410 --> 00:37:01,060 So we want to show that you can refine a partition that 533 00:37:01,060 --> 00:37:06,100 is not epsilon regular in such a way that the energy goes up. 534 00:37:08,690 --> 00:37:11,960 And to do that, let me state a few lemmas regarding 535 00:37:11,960 --> 00:37:14,930 the energy of a partition under refinement. 536 00:37:24,230 --> 00:37:26,560 And the point of the next several lemmas 537 00:37:26,560 --> 00:37:42,170 is that the energy never decreases under refinement, 538 00:37:42,170 --> 00:37:48,020 and it sometimes increases if your partition is not 539 00:37:48,020 --> 00:37:48,860 epsilon regular. 540 00:37:51,460 --> 00:37:54,380 So the first lemma is that if you 541 00:37:54,380 --> 00:38:03,340 look at the energy between a pair of partitions, 542 00:38:03,340 --> 00:38:11,730 it is never less than the energy between the two vertex sets. 543 00:38:11,730 --> 00:38:22,540 So for instance, if you have U and W like that, 544 00:38:22,540 --> 00:38:28,380 and I partition them into Pu and Pw, 545 00:38:28,380 --> 00:38:31,000 and I measure the energy, just basically 546 00:38:31,000 --> 00:38:35,620 the squared density between U and V versus summing up 547 00:38:35,620 --> 00:38:38,700 the individual squared densities after the partition, 548 00:38:38,700 --> 00:38:41,200 the left side is always at least as great as the right side. 549 00:38:45,110 --> 00:38:47,080 So this is really a claim. 550 00:38:47,080 --> 00:38:49,925 It's a fairly simple claim about convexity, 551 00:38:49,925 --> 00:38:51,550 but let me set it up in a way that will 552 00:38:51,550 --> 00:38:54,440 help some of the later proofs. 553 00:38:54,440 --> 00:39:00,900 So let me define a random variable, which 554 00:39:00,900 --> 00:39:02,772 I call Z, in the following way. 555 00:39:02,772 --> 00:39:04,230 So here's a process that I will use 556 00:39:04,230 --> 00:39:06,510 to define this random variable. 557 00:39:06,510 --> 00:39:12,540 I will select x, little x, to be a vertex uniformly chosen 558 00:39:12,540 --> 00:39:16,160 from U, from the left vertex set. 559 00:39:16,160 --> 00:39:27,670 And I will select a vertex y uniformly chosen from W. 560 00:39:27,670 --> 00:39:34,540 x and y, they fall into some part in the partition. 561 00:39:34,540 --> 00:39:42,630 So suppose Ui is the part where x i 562 00:39:42,630 --> 00:39:52,450 falls, and Wi is the set in the partition where y falls. 563 00:39:52,450 --> 00:39:55,840 So Ui is a member of this partition. 564 00:39:55,840 --> 00:40:00,500 Wi is a member of the other partition of W. 565 00:40:00,500 --> 00:40:03,980 Then I define my random variable Z 566 00:40:03,980 --> 00:40:11,870 to be the x density between Ui and Wj. 567 00:40:11,870 --> 00:40:12,500 So it's Wj. 568 00:40:17,030 --> 00:40:19,170 So that's the definition. 569 00:40:19,170 --> 00:40:21,500 So pick x randomly. 570 00:40:21,500 --> 00:40:22,850 Pick y randomly. 571 00:40:22,850 --> 00:40:25,010 Suppose x falls in Ui. 572 00:40:25,010 --> 00:40:27,140 Suppose y falls in Uj. 573 00:40:27,140 --> 00:40:31,610 Then Z is the x density between these two parts. 574 00:40:34,170 --> 00:40:37,120 So Z is some random variable. 575 00:40:37,120 --> 00:40:41,700 Let's look at properties of this random variable. 576 00:40:41,700 --> 00:40:45,947 First, what is this, it's expectation? 577 00:40:45,947 --> 00:40:47,280 It's a discrete random variable. 578 00:40:47,280 --> 00:40:49,363 And you can easily compute all of these quantities 579 00:40:49,363 --> 00:40:54,670 by just summing up according to how Z is generated. 580 00:40:54,670 --> 00:40:57,540 So I look overall, i and j. 581 00:40:57,540 --> 00:41:01,430 What's the probability that x falls in Ui? 582 00:41:01,430 --> 00:41:04,680 It is the size of Ui as a fraction of U. 583 00:41:04,680 --> 00:41:07,110 What's the probability that y falls in Wj? 584 00:41:07,110 --> 00:41:13,080 It's the size of Wj as a fraction of size W. 585 00:41:13,080 --> 00:41:17,460 And then Z is this quantity here. 586 00:41:17,460 --> 00:41:24,390 So this is what I find to be the expectation of Z. 587 00:41:24,390 --> 00:41:27,580 But you see the density multiplied by the product 588 00:41:27,580 --> 00:41:31,090 of the vertex set sizes, that's just the number of edges 589 00:41:31,090 --> 00:41:38,980 between U and W. And you sum over all the i, j's. 590 00:41:38,980 --> 00:41:50,080 So that, which is simply the edge density between U and W. 591 00:41:50,080 --> 00:41:53,570 So that's the expectation of the Z variable. 592 00:41:53,570 --> 00:41:56,520 On the other hand, what's the second moment? 593 00:41:56,520 --> 00:42:01,040 In other words, what's the expectation of the square of Z? 594 00:42:01,040 --> 00:42:05,080 Again, we do the same computation. 595 00:42:05,080 --> 00:42:06,280 First part is the same. 596 00:42:11,610 --> 00:42:13,760 The second part now becomes a d squared. 597 00:42:19,610 --> 00:42:27,070 And look at how we define energy. 598 00:42:27,070 --> 00:42:32,960 This quantity here is basically the energy q 599 00:42:32,960 --> 00:42:37,460 between the partition U and the partition of W, 600 00:42:37,460 --> 00:42:40,630 except there's normalization. 601 00:42:40,630 --> 00:42:43,140 That's not quite the same as the one we used before. 602 00:42:43,140 --> 00:42:45,620 So we will just put in that normalization. 603 00:42:54,740 --> 00:42:57,440 So now you compare the expectation 604 00:42:57,440 --> 00:43:01,540 of Z versus the expectation of Z squared. 605 00:43:01,540 --> 00:43:09,030 And we know by convexity that the expectation of Z squared 606 00:43:09,030 --> 00:43:13,530 is at least as large as the expectation of Z, 607 00:43:13,530 --> 00:43:14,640 that quantity squared. 608 00:43:17,400 --> 00:43:22,710 But if you plug in what values you get for these two guys, 609 00:43:22,710 --> 00:43:29,140 you derive the inequality claimed in Lemma 1. 610 00:43:32,260 --> 00:43:34,300 You have to cancel some normalization factors, 611 00:43:34,300 --> 00:43:35,217 but that's easy to do. 612 00:43:37,717 --> 00:43:38,800 So that's the first lemma. 613 00:43:43,060 --> 00:43:46,230 So the first one is just about a pair of parts, 614 00:43:46,230 --> 00:43:49,330 and identify partition, each part, 615 00:43:49,330 --> 00:43:52,030 what happens to the energy between this pair. 616 00:43:52,030 --> 00:43:54,820 And the second one is a direct corollary of the first one. 617 00:43:54,820 --> 00:43:57,760 It says that if you have a second partition, 618 00:43:57,760 --> 00:44:08,660 P prime that refines P, then the energy of the second partition, 619 00:44:08,660 --> 00:44:11,510 the refinement, is never less than the energy 620 00:44:11,510 --> 00:44:12,887 of the first partition. 621 00:44:15,570 --> 00:44:18,660 And it is a direct consequence of the first lemma, 622 00:44:18,660 --> 00:44:32,790 because we simply apply this lemma to every pair of parts 623 00:44:32,790 --> 00:44:39,890 in P. Between every pair of parts, 624 00:44:39,890 --> 00:44:41,520 the energy can never go down. 625 00:44:41,520 --> 00:44:43,600 So overall, the energy does not go down. 626 00:44:48,150 --> 00:44:52,170 And finally, so far, we've just said that the partitions can 627 00:44:52,170 --> 00:44:54,550 never make the energy go down. 628 00:44:54,550 --> 00:44:56,190 But in order to do this proof, we 629 00:44:56,190 --> 00:44:58,740 need to show that the energy sometimes goes up. 630 00:45:01,203 --> 00:45:02,870 And that's the point of the third lemma. 631 00:45:02,870 --> 00:45:10,400 The third Lemma tells us that you can get an energy boost. 632 00:45:10,400 --> 00:45:12,020 So this is the Red Bull Lemma. 633 00:45:12,020 --> 00:45:16,962 You can get an energy boost if you are feeling irregular. 634 00:45:20,820 --> 00:45:28,360 So if U, W is not epsilon regular, 635 00:45:28,360 --> 00:45:37,270 and this epsilon regularity is witnessed by U1 in U and W1 636 00:45:37,270 --> 00:45:51,070 in W, then I claim that the energy obtained by chopping U 637 00:45:51,070 --> 00:46:00,260 into U1 and its complement against W1, 638 00:46:00,260 --> 00:46:10,260 against the complement of W1 and W, so here again U and W, 639 00:46:10,260 --> 00:46:15,830 I find a witnessing set for their irregularity. 640 00:46:15,830 --> 00:46:18,980 And now I partition left and right according to-- 641 00:46:18,980 --> 00:46:22,190 chop each part into two. 642 00:46:22,190 --> 00:46:25,900 So this energy between this partition 643 00:46:25,900 --> 00:46:33,670 into two on both sides is bigger than the original energy 644 00:46:33,670 --> 00:46:38,280 plus something where we can gain. 645 00:46:38,280 --> 00:46:41,660 And this something where we can gain turns out 646 00:46:41,660 --> 00:46:46,050 to be at least epsilon raised to the power 647 00:46:46,050 --> 00:46:52,095 4 times the size of U, size of W, divided by n squared. 648 00:47:02,860 --> 00:47:03,700 Can you prove it? 649 00:47:08,190 --> 00:47:15,250 Let's define Z the same as in the previous proof, 650 00:47:15,250 --> 00:47:17,550 as in the proof of Lemma 1. 651 00:47:22,180 --> 00:47:26,070 In Lemma 1, we just used the fact that the L2 norm of Z, 652 00:47:26,070 --> 00:47:28,985 the expectation of the square, is at least 653 00:47:28,985 --> 00:47:30,610 the square of the expectation. 654 00:47:30,610 --> 00:47:32,360 But actually, there are differences in it. 655 00:47:32,360 --> 00:47:35,108 It's called a "variance." 656 00:47:35,108 --> 00:47:38,780 The variance of Z is the difference 657 00:47:38,780 --> 00:47:41,340 between these two quantities. 658 00:47:41,340 --> 00:47:44,852 I know that it's always non-negative. 659 00:47:47,810 --> 00:47:51,115 So if you look at how we derived the expectation of Z 660 00:47:51,115 --> 00:47:53,320 and expectation of Z squared, you immediately 661 00:47:53,320 --> 00:48:01,080 see that its variance we can write 662 00:48:01,080 --> 00:48:07,230 as, up to a normalizing factor, the difference 663 00:48:07,230 --> 00:48:20,090 between this energy on one hand and the energy between U and W, 664 00:48:20,090 --> 00:48:26,380 namely the mean square of the normalization. 665 00:48:26,380 --> 00:48:31,810 On the other hand, a different way to calculate the variance 666 00:48:31,810 --> 00:48:38,500 is that it is equal to the expectation of the deviation 667 00:48:38,500 --> 00:48:40,150 from the mean squared. 668 00:48:44,310 --> 00:48:47,320 So let's think about the deviation from the mean. 669 00:48:47,320 --> 00:48:53,940 I am choosing a random vector in, on the left, U, 670 00:48:53,940 --> 00:48:57,240 and another random point, random vertex on the left, 671 00:48:57,240 --> 00:49:00,570 and a random point on the right. 672 00:49:00,570 --> 00:49:05,220 In the event that they both lie in the sets 673 00:49:05,220 --> 00:49:17,110 that witness the irregularity, so in the event 674 00:49:17,110 --> 00:49:21,430 where x falls here and y falls here, 675 00:49:21,430 --> 00:49:24,930 which occurs with this probability, 676 00:49:24,930 --> 00:49:36,310 see that this quantity here is equal to the density between U1 677 00:49:36,310 --> 00:49:40,730 and W1 minus the density-- 678 00:49:40,730 --> 00:49:44,470 and this expectation of Z is just 679 00:49:44,470 --> 00:49:50,710 the density between U and W. So interpreting this expectation 680 00:49:50,710 --> 00:49:55,750 for what happens when x falls in U1 681 00:49:55,750 --> 00:50:02,210 and when y falls in W1, ignoring all the other events, 682 00:50:02,210 --> 00:50:06,230 because the quantity is always non-negative everywhere. 683 00:50:06,230 --> 00:50:14,660 But now from the definition of epsilon regularity, 684 00:50:14,660 --> 00:50:19,010 or rather the witnessing of epsilon irregularity, 685 00:50:19,010 --> 00:50:26,560 you see that this U1 is at least an epsilon fraction of U. W1's 686 00:50:26,560 --> 00:50:29,800 at least an epsilon fraction of W. 687 00:50:29,800 --> 00:50:35,500 And this final quantity here is at least epsilon inside, so 688 00:50:35,500 --> 00:50:38,600 at least epsilon squared. 689 00:50:38,600 --> 00:50:41,110 So here we're using all the different components 690 00:50:41,110 --> 00:50:43,270 of the definition of epsilon regular. 691 00:50:43,270 --> 00:50:44,506 Yes. 692 00:50:44,506 --> 00:50:48,988 AUDIENCE: What happens if we're dividing with more witnessing 693 00:50:48,988 --> 00:50:50,705 sets? 694 00:50:50,705 --> 00:50:52,080 YUFEI ZHAO: So you're asking what 695 00:50:52,080 --> 00:50:54,150 happens if we divide with more witnessing sets? 696 00:50:54,150 --> 00:50:55,650 So hold onto that thought. 697 00:50:55,650 --> 00:50:58,050 So right now, I'm just showing what happens 698 00:50:58,050 --> 00:50:59,640 if you have one witnessing set. 699 00:51:02,250 --> 00:51:04,590 Any more questions? 700 00:51:04,590 --> 00:51:08,070 So here we have epsilon to the 4th. 701 00:51:08,070 --> 00:51:11,010 And if you're putting the normalization 702 00:51:11,010 --> 00:51:14,180 comparing these two interpretations, 703 00:51:14,180 --> 00:51:16,490 you'll find the inequality claimed by the lemma. 704 00:51:23,180 --> 00:51:28,470 So now we are ready to show the key part of this iteration. 705 00:51:31,080 --> 00:51:33,720 I'll show you precisely how this iteration works, and show 706 00:51:33,720 --> 00:51:39,270 that you always get an energy boost in the overall partition. 707 00:51:39,270 --> 00:51:43,110 So I'll call the next one Lemma 4. 708 00:51:43,110 --> 00:51:52,350 And this says that if you have a partition P of the vertex set 709 00:51:52,350 --> 00:52:01,040 of G into k parts, if this partition is not epsilon 710 00:52:01,040 --> 00:52:21,900 regular, then there exists a refinement called Q where every 711 00:52:21,900 --> 00:52:35,720 part V sub i is partitioned further into at most 2 the k 712 00:52:35,720 --> 00:52:46,180 parts, and such that the partition of Q-- 713 00:52:46,180 --> 00:52:49,070 so the energy of the new partition Q-- 714 00:52:49,070 --> 00:52:54,800 increases substantially from the previous partition P. 715 00:52:54,800 --> 00:52:56,300 And we'll show that you can increase 716 00:52:56,300 --> 00:53:02,690 by at least epsilon to the 5th power, 717 00:53:02,690 --> 00:53:05,133 some constant in epsilon. 718 00:53:05,133 --> 00:53:06,800 So if you look at the strategy up there, 719 00:53:06,800 --> 00:53:08,960 if you can do this every step, then that 720 00:53:08,960 --> 00:53:10,610 means that the number of iterations 721 00:53:10,610 --> 00:53:13,670 is bounded by 1 over epsilon to the 5th power. 722 00:53:22,580 --> 00:53:27,480 So to prove this lemma here, we will use the three lemmas 723 00:53:27,480 --> 00:53:29,580 up there and put them together. 724 00:53:35,900 --> 00:53:43,710 So for all the pairs i, j such that V sub i, 725 00:53:43,710 --> 00:53:51,310 V sub j is not epsilon regular, as outlined 726 00:53:51,310 --> 00:53:53,800 in the proof in the outline up there, 727 00:53:53,800 --> 00:54:02,350 we will find this A superscript i, j in Vi, 728 00:54:02,350 --> 00:54:11,869 and A superscript j, i in Vj that witness the irregularity. 729 00:54:17,250 --> 00:54:23,180 So do this simultaneously for all pairs i, comma, j, 730 00:54:23,180 --> 00:54:25,550 where the Vi, Vj is not epsilon regular. 731 00:54:28,390 --> 00:54:34,900 Now what we're going to define Q as is the common refinement. 732 00:54:34,900 --> 00:54:37,390 So take all of, just as indicated 733 00:54:37,390 --> 00:54:39,550 in that picture up there, simultaneously 734 00:54:39,550 --> 00:54:55,040 take all of these A's and use them to refine P. Starting 735 00:54:55,040 --> 00:54:57,080 with P, starting with a partition you have, 736 00:54:57,080 --> 00:54:59,750 simultaneously cut everything up using 737 00:54:59,750 --> 00:55:01,240 all of these witnessing sets. 738 00:55:05,980 --> 00:55:09,560 Now, we only have witnessing pairs for pairs 739 00:55:09,560 --> 00:55:10,970 that are not epsilon regular. 740 00:55:10,970 --> 00:55:13,262 If they're epsilon regular, you don't worry about them. 741 00:55:19,570 --> 00:55:21,390 One of the claims in the lemma now is-- 742 00:55:21,390 --> 00:55:23,160 this is the Q that we'll end up with. 743 00:55:23,160 --> 00:55:25,470 We'll show that this Q has that property. 744 00:55:25,470 --> 00:55:26,880 So one of the claims in the lemma 745 00:55:26,880 --> 00:55:33,390 is that every Vi is partitioned into at most 2 to the k parts. 746 00:55:33,390 --> 00:55:37,710 So I hope that part is clear, because how are we 747 00:55:37,710 --> 00:55:39,360 doing the refinement? 748 00:55:39,360 --> 00:55:40,520 We're taking Vi. 749 00:55:44,810 --> 00:55:57,610 It's divided into parts using these A i, 750 00:55:57,610 --> 00:56:06,850 j's, one j coming from each pair that is irregular with Vi. 751 00:56:06,850 --> 00:56:12,500 So I'm cutting up Vi using at most k sets, 752 00:56:12,500 --> 00:56:15,090 so one coming from each of the other possible. 753 00:56:15,090 --> 00:56:18,090 Maybe fewer than k-- that's fine-- but at most k sets 754 00:56:18,090 --> 00:56:20,630 are used to cut up each Vi. 755 00:56:20,630 --> 00:56:24,355 So you have at most 2 to the k parts 756 00:56:24,355 --> 00:56:25,480 once you cut everything up. 757 00:56:30,820 --> 00:56:39,570 But the tricky part is to show that you get an energy boost. 758 00:56:39,570 --> 00:56:40,320 So let's do this. 759 00:56:40,320 --> 00:56:42,153 How do we show that you get an energy boost? 760 00:56:49,720 --> 00:56:52,020 We're going to put the top three lemmas together. 761 00:56:54,864 --> 00:57:01,510 First, we want to analyze the energy of Q. 762 00:57:01,510 --> 00:57:03,320 So let's write it out. 763 00:57:03,320 --> 00:57:12,790 So the energy of Q is the sum over this energy of individual 764 00:57:12,790 --> 00:57:14,020 partitions of the Vi's. 765 00:57:17,810 --> 00:57:28,280 And by this P sub Vi, P sub Vj, I mean the partition of Vj 766 00:57:28,280 --> 00:57:36,060 given by Q. So what happens after you cut up the Vi-- 767 00:57:36,060 --> 00:57:43,170 that's what I mean by P sub Vj, or rather I should call it 768 00:57:43,170 --> 00:57:45,070 Q sub Vi, Q sub Vj. 769 00:57:52,260 --> 00:58:01,830 By Lemma 2, we find that-- 770 00:58:01,830 --> 00:58:04,490 so let me separate them into two cases. 771 00:58:04,490 --> 00:58:10,650 The first case sums over i, j such that Vi, 772 00:58:10,650 --> 00:58:16,140 Vj is epsilon regular. 773 00:58:16,140 --> 00:58:22,620 And by Lemma 1, so here we're using Lemma 1, 774 00:58:22,620 --> 00:58:29,130 we find that this quantity here cannot be less than the Q 775 00:58:29,130 --> 00:58:31,580 of Vi, Vj. 776 00:58:31,580 --> 00:58:35,510 So take those two parts. 777 00:58:35,510 --> 00:58:38,680 Before and after the refinement by Q, 778 00:58:38,680 --> 00:58:39,830 the energy cannot go down. 779 00:58:43,660 --> 00:58:47,780 So I don't worry too much about pairs that are epsilon regular. 780 00:58:47,780 --> 00:58:51,900 But no let me look up here that are not epsilon regular. 781 00:59:05,630 --> 00:59:09,685 So what we will do now is even though-- 782 00:59:09,685 --> 00:59:11,310 so let's look at that picture up there. 783 00:59:15,750 --> 00:59:20,970 So let's focus on what I drew in red. 784 00:59:20,970 --> 00:59:24,670 So let's focus between 1 and 2. 785 00:59:24,670 --> 00:59:32,280 So suppose the shaded part is the witnessing sets. 786 00:59:32,280 --> 00:59:34,890 The witnessing sets got cut up further 787 00:59:34,890 --> 00:59:39,490 by other witnessing sets. 788 00:59:39,490 --> 00:59:46,810 But I don't have to worry about them because Lemma 2 or Lemma 789 00:59:46,810 --> 00:59:53,110 1, really, tells me that I can do an inequality where 790 00:59:53,110 --> 01:00:00,940 I go down to just comparing the energy between this partition 791 01:00:00,940 --> 01:00:05,030 of two parts, this single witnessing set 792 01:00:05,030 --> 01:00:16,250 and its complement, versus what happens in its partner. 793 01:00:24,620 --> 01:00:32,600 So in other words, over here, the Q of this pair, 794 01:00:32,600 --> 01:00:40,670 I am saying that it is no less than if I just 795 01:00:40,670 --> 01:00:47,280 look at what happens if you only cut up these two sets using 796 01:00:47,280 --> 01:00:47,950 the red lines. 797 01:00:55,790 --> 01:00:57,960 Let's go on. 798 01:00:57,960 --> 01:01:02,400 Applying Lemma 3, the energy boost lemma, 799 01:01:02,400 --> 01:01:04,960 the first part stays the same. 800 01:01:04,960 --> 01:01:06,820 So this first part stays the same. 801 01:01:06,820 --> 01:01:08,640 And the second part, now, because I'm 802 01:01:08,640 --> 01:01:11,160 looking at witnessing sets for irregularity, 803 01:01:11,160 --> 01:01:12,630 I get this extra boost. 804 01:01:15,950 --> 01:01:19,240 So this goes back to one of the questions asked earlier, 805 01:01:19,240 --> 01:01:23,260 where in Lemma 3, I don't have to now worry about what happens 806 01:01:23,260 --> 01:01:28,780 if you have further cuts, because I only need to worry 807 01:01:28,780 --> 01:01:31,240 about the case where I only have a single cut 808 01:01:31,240 --> 01:01:34,740 between the epsilon irregular pairs. 809 01:01:34,740 --> 01:01:39,900 So putting it together, we see that the previous line 810 01:01:39,900 --> 01:01:48,740 is at least, if you sum over the Q's of all the pairs 811 01:01:48,740 --> 01:01:59,420 plus this extra epsilon to the 4th term for all pairs 812 01:01:59,420 --> 01:02:01,540 that are not epsilon regular. 813 01:02:17,890 --> 01:02:23,840 I'm applying monotonicity of energy for the types, 814 01:02:23,840 --> 01:02:25,760 for pairs that are epsilon regular, 815 01:02:25,760 --> 01:02:29,080 an energy boost for pairs that are not epsilon regular. 816 01:02:29,080 --> 01:02:33,118 And for the latter type, I obtain this boost. 817 01:02:33,118 --> 01:02:35,160 Now remember what's the definition of an "epsilon 818 01:02:35,160 --> 01:02:36,270 regular partition." 819 01:02:36,270 --> 01:02:38,100 Unfortunately, it's no longer on the board, 820 01:02:38,100 --> 01:02:42,890 but it says that this sum over here, 821 01:02:42,890 --> 01:02:44,990 if it is an epsilon regular partition, 822 01:02:44,990 --> 01:02:47,660 it is at most epsilon. 823 01:02:47,660 --> 01:02:51,980 So if it is not epsilon regular, we can lower bound it. 824 01:02:51,980 --> 01:02:55,680 And that's indeed what we will do. 825 01:02:55,680 --> 01:02:58,050 The first sum here is, by definition, 826 01:02:58,050 --> 01:03:04,020 Q of the partition P. And the second sum, 827 01:03:04,020 --> 01:03:07,890 by the definition of epsilon regular, 828 01:03:07,890 --> 01:03:12,840 is at least epsilon to the power 5. 829 01:03:12,840 --> 01:03:14,480 So here we're using the definition 830 01:03:14,480 --> 01:03:17,590 of epsilon regular partition, namely, 831 01:03:17,590 --> 01:03:22,330 that a large fraction, so at least an epsilon 832 01:03:22,330 --> 01:03:28,240 fraction, basically, of pairs of vertex sets 833 01:03:28,240 --> 01:03:34,940 are not epsilon regular, but in this weighted sense. 834 01:03:34,940 --> 01:03:38,106 And that finishes the proof of Lemma 4 up there. 835 01:03:41,760 --> 01:03:42,930 Any questions so far? 836 01:03:47,650 --> 01:03:50,890 All right, so now we are ready to finish everything 837 01:03:50,890 --> 01:03:53,750 off, and prove Szemeredi's regularity lemma. 838 01:04:21,530 --> 01:04:28,280 So let's prove Szemeredi's regularity lemma. 839 01:04:33,480 --> 01:04:43,160 Let's start with the trivial partition, 840 01:04:43,160 --> 01:04:46,250 meaning just one large part. 841 01:04:46,250 --> 01:04:53,880 And we are going to repeatedly apply 842 01:04:53,880 --> 01:05:01,060 Lemma 4 whenever the partition at hand 843 01:05:01,060 --> 01:05:14,290 is not regular, whenever the current partition is not 844 01:05:14,290 --> 01:05:15,190 epsilon regular. 845 01:05:20,600 --> 01:05:22,930 So let's look at its energy. 846 01:05:22,930 --> 01:05:28,590 The energy of this partition-- 847 01:05:28,590 --> 01:05:34,860 so this is a weighted mean of the edge density squared, 848 01:05:34,860 --> 01:05:40,020 so it always lies between 0 and 1, 849 01:05:40,020 --> 01:05:44,760 just from the definition of energy. 850 01:05:44,760 --> 01:05:51,990 On the other hand, Lemma 4 tells us 851 01:05:51,990 --> 01:06:00,840 that the energy increases by at least epsilon to the 5th power 852 01:06:00,840 --> 01:06:03,304 at each iteration. 853 01:06:10,570 --> 01:06:14,370 So this process cannot continue forever. 854 01:06:14,370 --> 01:06:23,408 So it must stop after at most epsilon to the minus 5th power 855 01:06:23,408 --> 01:06:24,075 number of steps. 856 01:06:27,080 --> 01:06:32,590 And when we stop, we must result in an epsilon regular 857 01:06:32,590 --> 01:06:39,220 partition, because otherwise, you're 858 01:06:39,220 --> 01:06:43,156 going to continue applying the lemma and push it even further. 859 01:06:43,156 --> 01:06:46,240 And that's it. 860 01:06:46,240 --> 01:06:50,500 So that proves Szemeredi's graph regularity lemma. 861 01:06:53,440 --> 01:06:54,910 Question. 862 01:06:54,910 --> 01:06:57,360 AUDIENCE: It's going to be some really big value of M. 863 01:06:57,360 --> 01:06:59,030 YUFEI ZHAO: OK, let's talk about bounds. 864 01:06:59,030 --> 01:07:00,490 So let's talk about how many parts. 865 01:07:06,580 --> 01:07:11,410 So how many parts does this proof produce? 866 01:07:11,410 --> 01:07:12,350 We can figure it out. 867 01:07:12,350 --> 01:07:14,470 So we have some number of steps. 868 01:07:14,470 --> 01:07:18,440 Each step increases the number of parts by something. 869 01:07:18,440 --> 01:07:30,840 So if P has k parts, so then Lemma 4 refines 870 01:07:30,840 --> 01:07:37,910 P into at most how many parts? 871 01:07:37,910 --> 01:07:38,823 AUDIENCE: 2 to the k 872 01:07:38,823 --> 01:07:40,490 YUFEI ZHAO: Yeah, so k times 2 to the k. 873 01:07:44,330 --> 01:07:48,720 And I have many iterations of this guy. 874 01:07:48,720 --> 01:07:50,220 So some of you are already laughing, 875 01:07:50,220 --> 01:07:53,280 because it's going to be a very large number. 876 01:07:53,280 --> 01:07:55,310 In fact, because it's going to be so large, 877 01:07:55,310 --> 01:07:57,460 it makes my calculations slightly more convenient. 878 01:07:57,460 --> 01:07:58,918 It really doesn't change the answer 879 01:07:58,918 --> 01:08:01,820 so much if I just bound k to the 2 to the k by 2 to the 2 880 01:08:01,820 --> 01:08:03,050 to the k. 881 01:08:03,050 --> 01:08:10,440 So the final number of parts is this function 882 01:08:10,440 --> 01:08:15,030 iterated on itself epsilon to the minus 5 times. 883 01:08:15,030 --> 01:08:22,590 So it's a power of 2 of height at most 2 to the epsilon 884 01:08:22,590 --> 01:08:23,090 to the 5. 885 01:08:26,560 --> 01:08:29,477 It's a finite number, so it depends only on epsilon and not 886 01:08:29,477 --> 01:08:30,560 on the size of your graph. 887 01:08:30,560 --> 01:08:32,102 And this is the most important thing. 888 01:08:32,102 --> 01:08:35,290 It does not depend on the size of your graph. 889 01:08:35,290 --> 01:08:38,020 It is quite large. 890 01:08:38,020 --> 01:08:42,880 In fact, even for reasonable values of epsilon, 891 01:08:42,880 --> 01:08:48,850 like 1% or even 10%, this number is astronomically large. 892 01:08:48,850 --> 01:08:51,490 And you may ask is it really necessary, 893 01:08:51,490 --> 01:08:54,372 because we did this proof, and it came out fairly elegantly, 894 01:08:54,372 --> 01:08:56,080 I would say it, how the proof was set up. 895 01:08:56,080 --> 01:08:58,779 And you arrived at this finite bound. 896 01:08:58,779 --> 01:09:00,319 But maybe there's a better proof. 897 01:09:00,319 --> 01:09:04,170 Maybe you can work harder and obtain somewhat better bounds. 898 01:09:04,170 --> 01:09:08,250 So you can ask, is it possible that the truth is really 899 01:09:08,250 --> 01:09:11,103 somehow much smaller? 900 01:09:11,103 --> 01:09:12,520 And the answer turns out to be no. 901 01:09:15,189 --> 01:09:21,100 So there is a theorem by Tim Gowers 902 01:09:21,100 --> 01:09:24,087 which says that there exists some constant. 903 01:09:24,087 --> 01:09:26,170 The precise statement, again, is not so important, 904 01:09:26,170 --> 01:09:28,899 but based on what I just said, you cannot improve this bound 905 01:09:28,899 --> 01:09:31,180 given by this proof. 906 01:09:31,180 --> 01:09:38,910 So for every epsilon small enough, 907 01:09:38,910 --> 01:09:48,470 there exists a graph whose epsilon regular partition 908 01:09:48,470 --> 01:09:54,120 requires how many parts? 909 01:09:54,120 --> 01:10:02,950 So the number of parts at least this tower of 2 of height 910 01:10:02,950 --> 01:10:06,010 some epsilon to the minus c. 911 01:10:09,630 --> 01:10:15,200 So really it's a tower of exponentials of size, 912 01:10:15,200 --> 01:10:18,800 essentially polynomial in 1 over epsilon. 913 01:10:18,800 --> 01:10:22,250 So maybe you can squeeze the 5 to something less. 914 01:10:22,250 --> 01:10:24,250 Actually, we don't even know if that's the case, 915 01:10:24,250 --> 01:10:28,700 but certainly you cannot do substantially better than what 916 01:10:28,700 --> 01:10:31,530 the proof gives. 917 01:10:31,530 --> 01:10:36,600 So Szemeredi's regularity lemma is an extremely powerful tool. 918 01:10:36,600 --> 01:10:41,550 And we'll see applications that are basically 919 01:10:41,550 --> 01:10:42,560 very difficult to prove. 920 01:10:42,560 --> 01:10:44,260 And for some of these applications, 921 01:10:44,260 --> 01:10:45,960 we don't really know other proofs 922 01:10:45,960 --> 01:10:48,760 except using Szemeredi's regularity lemma. 923 01:10:48,760 --> 01:10:52,680 But on the other hand, it gives terrible quantitative bounds. 924 01:10:52,680 --> 01:10:57,450 So there is a lot of interest in combinatorics 925 01:10:57,450 --> 01:11:01,400 where once you see a proof that requires Szemeredi's regularity 926 01:11:01,400 --> 01:11:04,020 lemma, or that is first proved using this technique, 927 01:11:04,020 --> 01:11:07,870 to ask can it be used using some other technique? 928 01:11:07,870 --> 01:11:12,270 In fact, Szemeredi himself has worked a lot in that direction, 929 01:11:12,270 --> 01:11:14,854 trying to get rid of the uses of his lemma. 930 01:11:19,500 --> 01:11:22,410 Any questions? 931 01:11:22,410 --> 01:11:25,120 AUDIENCE: How could you modify it for equipartitions? 932 01:11:25,120 --> 01:11:26,170 YUFEI ZHAO: OK, great. 933 01:11:26,170 --> 01:11:29,038 Question is, how can we modify it for equipartitions? 934 01:11:29,038 --> 01:11:30,080 So let's talk about that. 935 01:11:30,080 --> 01:11:32,370 So it's a fantastic question. 936 01:11:32,370 --> 01:11:34,420 So look at this proof and see what can we 937 01:11:34,420 --> 01:11:37,720 do if we really want all the parts to have 938 01:11:37,720 --> 01:11:41,280 roughly the same size, let's say differing by at most 1. 939 01:12:00,800 --> 01:12:09,929 So how to make the epsilon regular partition equitable? 940 01:12:14,920 --> 01:12:16,220 Any guesses? 941 01:12:16,220 --> 01:12:18,810 Any attempts on what we can do? 942 01:12:18,810 --> 01:12:21,920 I mean, basically it's going to follow this proof. 943 01:12:21,920 --> 01:12:24,410 As I said, the spirit of Szemeredi's regularity lemma 944 01:12:24,410 --> 01:12:25,630 is what I've shown you. 945 01:12:25,630 --> 01:12:28,640 But the details and executions may vary somewhat 946 01:12:28,640 --> 01:12:31,210 depending on the specific purpose you have in mind. 947 01:12:31,210 --> 01:12:31,792 Yeah. 948 01:12:31,792 --> 01:12:33,238 AUDIENCE: Can we just add-- 949 01:12:33,238 --> 01:12:35,166 [INAUDIBLE] add things to the smaller part 950 01:12:35,166 --> 01:12:38,058 because we know that-- by the fact 951 01:12:38,058 --> 01:12:41,440 that it's not [INAUDIBLE] that parts aren't too small? 952 01:12:41,440 --> 01:12:44,510 YUFEI ZHAO: OK, so you're saying we're going to add something 953 01:12:44,510 --> 01:12:46,790 or to massage the partition to make it epsilon-- 954 01:12:46,790 --> 01:12:48,870 AUDIENCE: Add vertices to the smaller parts of the partition. 955 01:12:48,870 --> 01:12:51,495 YUFEI ZHAO: Add vertices to the smaller parts of the partition, 956 01:12:51,495 --> 01:12:53,828 now when are you going to do that? 957 01:12:53,828 --> 01:12:56,323 AUDIENCE: When they're-- like so you do the refinement, 958 01:12:56,323 --> 01:12:58,183 then when they're not [INAUDIBLE] 959 01:12:58,183 --> 01:12:59,600 YUFEI ZHAO: So you want to do this 960 01:12:59,600 --> 01:13:01,337 at every stage of the process. 961 01:13:01,337 --> 01:13:01,920 AUDIENCE: Yes. 962 01:13:01,920 --> 01:13:04,500 [INAUDIBLE] 963 01:13:04,500 --> 01:13:06,100 YUFEI ZHAO: I like that idea. 964 01:13:06,100 --> 01:13:09,920 So here's what we're going to do. 965 01:13:09,920 --> 01:13:12,410 So we still run the same process. 966 01:13:12,410 --> 01:13:15,760 So we're going to have this P, which is the current partition. 967 01:13:15,760 --> 01:13:20,980 So I have current partition. 968 01:13:20,980 --> 01:13:24,730 And as before, we initially have it 969 01:13:24,730 --> 01:13:27,420 as either the trivial partition, if you like, 970 01:13:27,420 --> 01:13:32,160 or m arbitrary equitable parts. 971 01:13:37,418 --> 01:13:39,210 Start with something where you don't really 972 01:13:39,210 --> 01:13:41,940 care about anything except for the size. 973 01:13:41,940 --> 01:13:46,080 And you run basically the same proof, 974 01:13:46,080 --> 01:13:53,000 where if your P is not epsilon regular, 975 01:13:53,000 --> 01:13:56,750 then do what we've done before, so basically 976 01:13:56,750 --> 01:13:58,160 exactly the same thing. 977 01:13:58,160 --> 01:14:13,590 We refine P using pairs witnessing regularity, 978 01:14:13,590 --> 01:14:15,750 same as the proof that we just did. 979 01:14:15,750 --> 01:14:17,910 And now we need to do something a little bit more 980 01:14:17,910 --> 01:14:21,480 to obtain equitability. 981 01:14:21,480 --> 01:14:24,660 And what we will do is right after-- 982 01:14:24,660 --> 01:14:29,180 so each step in iteration, right after we do this refinement, 983 01:14:29,180 --> 01:14:36,180 so after we cut up our graph where maybe some of the parts 984 01:14:36,180 --> 01:14:45,560 are really tiny, let's massage the partitions somewhat 985 01:14:45,560 --> 01:14:47,210 to make them equitable. 986 01:14:47,210 --> 01:14:51,000 And to make our life a little bit easier, 987 01:14:51,000 --> 01:14:56,130 we can refine the partition somewhat further 988 01:14:56,130 --> 01:15:00,720 to chop it up into somewhat smaller resolution. 989 01:15:00,720 --> 01:15:03,050 And this part, you can really do it 990 01:15:03,050 --> 01:15:04,370 either arbitrarily or randomly. 991 01:15:06,950 --> 01:15:10,888 Some ways may be slightly easier to execute, 992 01:15:10,888 --> 01:15:12,680 but it doesn't really matter how you do it. 993 01:15:12,680 --> 01:15:13,970 It's fairly robust. 994 01:15:13,970 --> 01:15:16,460 You refine it further. 995 01:15:16,460 --> 01:15:20,500 And basically, I want to make it equitable. 996 01:15:20,500 --> 01:15:23,260 Sometimes, you can just do that by refining, 997 01:15:23,260 --> 01:15:25,810 but maybe if you have some really small parts, 998 01:15:25,810 --> 01:15:28,300 then you might need to move some vertices around, 999 01:15:28,300 --> 01:15:31,150 so I call that "rebalancing." 1000 01:15:34,040 --> 01:15:41,600 So move and merge some vertices, but only a very small number 1001 01:15:41,600 --> 01:15:45,578 of vertices, to make equitable. 1002 01:15:50,960 --> 01:15:55,713 So you run this loop until you find that your partition is 1003 01:15:55,713 --> 01:15:56,380 epsilon regular. 1004 01:15:56,380 --> 01:15:58,000 Then you're done. 1005 01:15:58,000 --> 01:15:59,950 Whenever you run this loop, because we're 1006 01:15:59,950 --> 01:16:02,080 doing the second step, your partition 1007 01:16:02,080 --> 01:16:05,900 is always going to be equitable. 1008 01:16:05,900 --> 01:16:09,110 But we now need to control the energy again 1009 01:16:09,110 --> 01:16:12,260 to limit the number of steps. 1010 01:16:12,260 --> 01:16:15,330 And the point here is that the first part still 1011 01:16:15,330 --> 01:16:20,850 is exactly the same as before, where the energy goes up 1012 01:16:20,850 --> 01:16:25,720 by at least epsilon to the minus 5. 1013 01:16:25,720 --> 01:16:28,360 But the second part, the energy might go down, 1014 01:16:28,360 --> 01:16:30,850 because we're no longer refining, just refining. 1015 01:16:30,850 --> 01:16:34,270 Because we're doing some rebalancing. 1016 01:16:34,270 --> 01:16:36,880 But you can do it in such a way that the amount of rebalancing 1017 01:16:36,880 --> 01:16:39,220 that you do is really small. 1018 01:16:39,220 --> 01:16:42,100 You're not actually changing the energy by so much. 1019 01:16:42,100 --> 01:16:45,980 So I'll just hand wave here, and say 1020 01:16:45,980 --> 01:16:52,980 that we can do this in such a way where the energy might 1021 01:16:52,980 --> 01:16:56,550 go down, but only a little bit. 1022 01:17:02,240 --> 01:17:04,040 So you're only changing a very small number 1023 01:17:04,040 --> 01:17:07,430 of vertices, very small fraction of vertices. 1024 01:17:07,430 --> 01:17:09,950 So if you change only an epsilon fraction of vertices, 1025 01:17:09,950 --> 01:17:12,200 you don't expect the energy, which 1026 01:17:12,200 --> 01:17:17,000 is something that comes out of summing pairs of vertex parts, 1027 01:17:17,000 --> 01:17:18,290 to change by all that much. 1028 01:17:21,540 --> 01:17:24,110 So putting these two together, you 1029 01:17:24,110 --> 01:17:30,650 see that the energy still goes up by, let's say, at least 1/2 1030 01:17:30,650 --> 01:17:35,780 of epsilon to the 5th power. 1031 01:17:35,780 --> 01:17:39,800 And so then, the rest of the proof runs the same as before. 1032 01:17:39,800 --> 01:17:42,590 You finish in some bounded number of steps. 1033 01:17:42,590 --> 01:17:44,930 And you result in an equitable partition 1034 01:17:44,930 --> 01:17:46,990 that's epsilon regular. 1035 01:17:46,990 --> 01:17:48,790 I don't want to belabor the details. 1036 01:17:48,790 --> 01:17:50,582 I mean, here, there's some things to check, 1037 01:17:50,582 --> 01:17:52,960 but it's, I think, fairly routine. 1038 01:17:52,960 --> 01:17:56,960 It's is more of an exercise in technical details. 1039 01:17:56,960 --> 01:18:01,070 But the thing that actually is somewhat important 1040 01:18:01,070 --> 01:18:04,277 is there's a wrong way to do this. 1041 01:18:04,277 --> 01:18:06,360 I just want to point out that what's the wrong way 1042 01:18:06,360 --> 01:18:09,610 to do this, is that you apply regularity lemma, 1043 01:18:09,610 --> 01:18:11,150 and you think now it has something 1044 01:18:11,150 --> 01:18:12,650 that's epsilon regular. 1045 01:18:12,650 --> 01:18:18,350 Then I massage it to try to make it equitable at the end. 1046 01:18:18,350 --> 01:18:21,040 And so if I don't look into the proof, 1047 01:18:21,040 --> 01:18:23,500 I just look at a statement of Szemeredi's regularity lemma, 1048 01:18:23,500 --> 01:18:25,490 and I get something that's epsilon regular, 1049 01:18:25,490 --> 01:18:27,532 I say I'm just going to divide things up a little 1050 01:18:27,532 --> 01:18:30,000 bit further, that doesn't work. 1051 01:18:30,000 --> 01:18:32,280 Because the property of being epsilon 1052 01:18:32,280 --> 01:18:37,340 regular is actually not preserved under refinement. 1053 01:18:37,340 --> 01:18:38,560 So look at the definition. 1054 01:18:38,560 --> 01:18:40,310 You have something that's epsilon regular. 1055 01:18:40,310 --> 01:18:41,450 You refine the partition. 1056 01:18:41,450 --> 01:18:45,960 If might fail to be epsilon regular. 1057 01:18:45,960 --> 01:18:51,290 So you really have to take into the proof to get equitability. 1058 01:18:51,290 --> 01:18:54,220 So just to repeat, a wrong way to try to get equitability 1059 01:18:54,220 --> 01:18:56,930 is to apply regularity lemma, and at the end, 1060 01:18:56,930 --> 01:18:59,420 try to massage it to get equitable. 1061 01:18:59,420 --> 01:19:01,820 That doesn't work. 1062 01:19:01,820 --> 01:19:05,690 Next time, I will show you how to apply Szemeredi's regularity 1063 01:19:05,690 --> 01:19:07,240 lemma.