1 00:00:17,558 --> 00:00:19,850 PROFESSOR: So we spent the last few lectures discussing 2 00:00:19,850 --> 00:00:21,873 Szemerédi's regularity lemma. 3 00:00:21,873 --> 00:00:23,540 So we saw that this is an important tool 4 00:00:23,540 --> 00:00:26,450 with important applications, allowing 5 00:00:26,450 --> 00:00:30,080 you to do things like a proof of Roth's theorem via graph 6 00:00:30,080 --> 00:00:31,640 theory. 7 00:00:31,640 --> 00:00:34,820 One of the concepts that came up when we were discussing 8 00:00:34,820 --> 00:00:38,210 the statement of Szemerédi's regularity lemma is that 9 00:00:38,210 --> 00:00:39,830 of pseudorandomness. 10 00:00:39,830 --> 00:00:44,480 So the statement of Szemerédi's graph regularity lemma is that 11 00:00:44,480 --> 00:00:48,710 you can partition an arbitrary graph into a bounded number 12 00:00:48,710 --> 00:00:53,630 of pieces so that the graph looks random-like, 13 00:00:53,630 --> 00:00:58,040 as we called it, between most pairs of parts. 14 00:00:58,040 --> 00:01:00,017 So what does random-like mean? 15 00:01:00,017 --> 00:01:02,600 So that's something that I want to discuss for the next couple 16 00:01:02,600 --> 00:01:05,239 of lectures. 17 00:01:05,239 --> 00:01:18,580 And this is the idea of pseudorandomness, 18 00:01:18,580 --> 00:01:23,100 which is a concept that is really 19 00:01:23,100 --> 00:01:26,880 prevalent in combinatorics, in theoretical computer science, 20 00:01:26,880 --> 00:01:28,860 and in many different areas. 21 00:01:28,860 --> 00:01:33,840 And what pseudorandomness tries to capture is, in what ways 22 00:01:33,840 --> 00:01:37,270 can a non-random object look random? 23 00:01:37,270 --> 00:01:40,230 So before diving into some specific mathematics, 24 00:01:40,230 --> 00:01:43,530 I want to offer some philosophical remarks. 25 00:01:43,530 --> 00:01:46,110 So you might know that, on a computer, 26 00:01:46,110 --> 00:01:48,420 you want to generate a random number. 27 00:01:48,420 --> 00:01:51,510 Well, you type in a "rand," and it gives you a random number. 28 00:01:51,510 --> 00:01:56,450 But of course, that's not necessarily true randomness. 29 00:01:56,450 --> 00:02:00,870 It came from some pseudorandom generator. 30 00:02:00,870 --> 00:02:03,930 Probably there's some seed and some complex-looking function 31 00:02:03,930 --> 00:02:06,510 and outputs something that you couldn't 32 00:02:06,510 --> 00:02:09,000 distinguish from random. 33 00:02:09,000 --> 00:02:11,430 But it might not actually be random but just something 34 00:02:11,430 --> 00:02:15,730 that looks, in many different ways, like random. 35 00:02:15,730 --> 00:02:17,560 So there is this concept of random. 36 00:02:17,560 --> 00:02:19,590 You can think about a random graph, 37 00:02:19,590 --> 00:02:21,910 right, generate this Erdos-Renyi random graph. 38 00:02:21,910 --> 00:02:25,360 Every edge occurs independently with some probability. 39 00:02:25,360 --> 00:02:28,660 But I can also show you some graph, some specific graph, 40 00:02:28,660 --> 00:02:32,860 which I say, well, it's, for all intents and purposes, 41 00:02:32,860 --> 00:02:38,160 just as good as a random graph. 42 00:02:38,160 --> 00:02:42,380 So in what ways can we capture that concept? 43 00:02:42,380 --> 00:02:44,100 So that's what I want to discuss. 44 00:02:44,100 --> 00:02:46,520 And that's the topic of pseudorandomness. 45 00:02:46,520 --> 00:02:49,250 And of course, well, this idea extends 46 00:02:49,250 --> 00:02:51,690 to many areas, number theory and whatnot, 47 00:02:51,690 --> 00:02:53,820 but we'll stick with graph theory. 48 00:02:53,820 --> 00:02:58,220 In particular, I want to explore today just one specific notion 49 00:02:58,220 --> 00:02:59,660 of pseudorandomness. 50 00:02:59,660 --> 00:03:04,060 And this comes from an important paper called "Quasi-random 51 00:03:04,060 --> 00:03:04,560 graphs." 52 00:03:08,790 --> 00:03:12,350 And this concept is due to Chung, Graham, and Wilson 53 00:03:12,350 --> 00:03:13,570 back in the late '80s. 54 00:03:23,690 --> 00:03:28,980 So they defined various notions of pseudorandomness, 55 00:03:28,980 --> 00:03:30,930 and I want to state them. 56 00:03:30,930 --> 00:03:33,390 And what it turns out-- and the surprising part 57 00:03:33,390 --> 00:03:36,750 is that these notions, these definitions, 58 00:03:36,750 --> 00:03:39,690 although they look superficially different, 59 00:03:39,690 --> 00:03:43,960 they are actually all equivalent to each other. 60 00:03:43,960 --> 00:03:49,050 So let's see what the theorem says. 61 00:03:49,050 --> 00:03:54,090 So the set-up of this theorem is that you have some fixed 62 00:03:54,090 --> 00:03:56,640 real p between 0 and 1. 63 00:03:56,640 --> 00:04:00,810 And this is going to be your graph edge density. 64 00:04:00,810 --> 00:04:11,070 So for any sequence of graphs, Gn-- 65 00:04:13,660 --> 00:04:19,940 so from now, I'm going to drop the subscript n, 66 00:04:19,940 --> 00:04:22,600 so G will just be Gn-- 67 00:04:22,600 --> 00:04:27,680 such that the number of vertices-- 68 00:04:27,680 --> 00:04:37,170 so G is n vertex with edge density basically p. 69 00:04:42,090 --> 00:04:44,000 So this is your sequence of graphs. 70 00:04:44,000 --> 00:04:49,860 And the claim is that we're going to state 71 00:04:49,860 --> 00:04:51,594 some set of properties. 72 00:04:54,260 --> 00:04:55,750 And these properties are all going 73 00:04:55,750 --> 00:04:57,660 to be equivalent to each other. 74 00:05:03,370 --> 00:05:06,030 So all of these properties capture some notion 75 00:05:06,030 --> 00:05:10,140 of pseudorandomness, so in what ways this is graph G or really 76 00:05:10,140 --> 00:05:11,640 a sequence of graphs. 77 00:05:11,640 --> 00:05:13,860 Or you can talk about a specific graph 78 00:05:13,860 --> 00:05:15,990 and have some error parameters and error balance. 79 00:05:15,990 --> 00:05:19,030 They're all roughly the same ideas. 80 00:05:19,030 --> 00:05:21,030 So in what ways can we talk about this graph G 81 00:05:21,030 --> 00:05:22,600 being random-like? 82 00:05:22,600 --> 00:05:26,570 Well, we already saw one notion when we discussed 83 00:05:26,570 --> 00:05:28,530 Szemerédi's regularity lemma. 84 00:05:28,530 --> 00:05:30,700 And let's see that here. 85 00:05:30,700 --> 00:05:33,930 So this notion is known as discrepancy. 86 00:05:33,930 --> 00:05:39,270 And it says that if I restrict my graph to looking only 87 00:05:39,270 --> 00:05:42,990 at edges between some pair of vertex sets, 88 00:05:42,990 --> 00:05:46,680 then the number of edges should be roughly 89 00:05:46,680 --> 00:05:49,800 what you would expect based on density alone. 90 00:05:57,730 --> 00:05:59,230 So this is basically the notion that 91 00:05:59,230 --> 00:06:02,050 came up in epsilon regularity. 92 00:06:02,050 --> 00:06:04,720 This is essentially the same as saying 93 00:06:04,720 --> 00:06:09,310 that G is epsilon regular with itself 94 00:06:09,310 --> 00:06:15,890 where this epsilon now is hidden in this little o parameter. 95 00:06:15,890 --> 00:06:18,360 So that's one notion of pseudorandomness. 96 00:06:18,360 --> 00:06:23,000 So here's another notion which is very similar. 97 00:06:23,000 --> 00:06:25,310 So it's almost just a semantic difference, 98 00:06:25,310 --> 00:06:27,530 but, OK, so I have to do a little bit of work. 99 00:06:27,530 --> 00:06:29,440 So let me call this DISC prime. 100 00:06:29,440 --> 00:06:35,730 So it says that if you look at only edges within this set-- 101 00:06:35,730 --> 00:06:38,840 so instead of taking two sets, I only look at one set-- 102 00:06:38,840 --> 00:06:41,440 and then look at how many edges are 103 00:06:41,440 --> 00:06:45,170 in there versus how many you should expect based on density 104 00:06:45,170 --> 00:06:50,940 alone, these two numbers are also 105 00:06:50,940 --> 00:06:52,240 very similar to each other. 106 00:07:00,730 --> 00:07:04,890 So let's get to something that looks dramatically different. 107 00:07:04,890 --> 00:07:09,940 The next one, I'm going to call count. 108 00:07:09,940 --> 00:07:14,480 So count says that for every graph 109 00:07:14,480 --> 00:07:28,380 H, the number of labeled copies of H in G-- 110 00:07:28,380 --> 00:07:31,620 OK, so labeled copies, I mean that the vertices of H 111 00:07:31,620 --> 00:07:33,070 are labeled. 112 00:07:33,070 --> 00:07:37,350 So for every triangle, there are six labeled triangles 113 00:07:37,350 --> 00:07:40,530 that correspond to that triangle in the graph. 114 00:07:40,530 --> 00:07:44,790 The number of labeled copies of H is-- 115 00:07:44,790 --> 00:07:48,110 so what should you expect if this graph were truly random? 116 00:07:48,110 --> 00:07:51,900 You would expect p raised to the number of edges of H 117 00:07:51,900 --> 00:08:01,530 plus small error times n raised to number of vertices of H. 118 00:08:01,530 --> 00:08:09,710 And just as a remark, this little o term, little o 1 term, 119 00:08:09,710 --> 00:08:15,600 may depend on H. 120 00:08:15,600 --> 00:08:19,340 So this condition, count, says for every graph H, 121 00:08:19,340 --> 00:08:20,540 this is true. 122 00:08:20,540 --> 00:08:23,420 And by that, I mean for every H, there 123 00:08:23,420 --> 00:08:25,480 is some sequence of decaying errors. 124 00:08:25,480 --> 00:08:27,320 But that sequence of decaying errors 125 00:08:27,320 --> 00:08:34,789 may depend on your graph H. OK. 126 00:08:34,789 --> 00:08:38,809 The next one is almost a special case of count. 127 00:08:38,809 --> 00:08:41,940 It's called C4. 128 00:08:41,940 --> 00:08:46,230 And it says that the number of labeled copies of C4, 129 00:08:46,230 --> 00:08:59,410 so the fourth cycle, is at most p raised to power of 4-- 130 00:08:59,410 --> 00:09:05,810 so again, what you should expect in a random setting just 131 00:09:05,810 --> 00:09:07,000 for cycle count alone. 132 00:09:11,740 --> 00:09:15,600 I see, already, some of you are surprised. 133 00:09:15,600 --> 00:09:19,590 So we'll discuss that this is an important constraint. 134 00:09:19,590 --> 00:09:22,770 It turns out that alone implies everything, just having 135 00:09:22,770 --> 00:09:24,030 the correct C4 count. 136 00:09:27,730 --> 00:09:31,720 The next one, we will call codegree. 137 00:09:31,720 --> 00:09:37,550 And the codegree condition says that if you look at a pair 138 00:09:37,550 --> 00:09:42,400 of vertices and look at their number of common neighbors-- 139 00:09:42,400 --> 00:09:45,460 in other words, their codegree-- 140 00:09:45,460 --> 00:09:49,020 then what should you expect this quantity to be? 141 00:09:49,020 --> 00:09:51,880 So there are n vertices that possibly 142 00:09:51,880 --> 00:09:55,780 could be common neighbors, and each one of them, 143 00:09:55,780 --> 00:09:58,730 if this were a random graph with edge probability p, 144 00:09:58,730 --> 00:10:00,940 then you expect the number of common neighbors 145 00:10:00,940 --> 00:10:03,950 to be around p squared n. 146 00:10:03,950 --> 00:10:10,240 So the codegree condition is that this sum is small. 147 00:10:10,240 --> 00:10:14,800 So most pairs of vertices have roughly the correct number 148 00:10:14,800 --> 00:10:16,477 of common neighbors. 149 00:10:20,213 --> 00:10:27,793 So codegree is number of common neighbors. 150 00:10:31,180 --> 00:10:34,990 Next, and the last one, certainly not the least, 151 00:10:34,990 --> 00:10:39,130 is eigenvalue condition. 152 00:10:39,130 --> 00:10:50,410 So here, we are going to denote by lambda 1 through lambda G 153 00:10:50,410 --> 00:10:58,130 the eigenvalues of the adjacency matrix of G. 154 00:10:58,130 --> 00:11:01,520 So we saw this object in the last lecture. 155 00:11:01,520 --> 00:11:03,940 So I include multiplicities. 156 00:11:03,940 --> 00:11:06,320 If some eigenvalue occurs with multiple times, 157 00:11:06,320 --> 00:11:08,920 I include it multiple times. 158 00:11:08,920 --> 00:11:13,870 So the eigenvalue condition says that the top eigenvalue 159 00:11:13,870 --> 00:11:21,720 is around pn and that, more importantly, 160 00:11:21,720 --> 00:11:30,770 the other eigenvalues are all quite small. 161 00:11:35,810 --> 00:11:39,470 Now, for d regular graph, the top eigenvalue-- 162 00:11:39,470 --> 00:11:41,990 and it's fine to think about d regular graphs 163 00:11:41,990 --> 00:11:44,990 if you want to get some intuition out of this theorem. 164 00:11:44,990 --> 00:11:47,360 For d regular graph, the top eigenvalue 165 00:11:47,360 --> 00:11:52,860 is equal to d, because the top eigenvector is d. 166 00:11:52,860 --> 00:11:56,000 It's the all-one vector. 167 00:11:56,000 --> 00:12:09,180 So top eigenvector is all-one vector, which has eigenvalue d. 168 00:12:09,180 --> 00:12:14,310 And what the eigenvalue condition says 169 00:12:14,310 --> 00:12:27,710 is that all the other eigenvalues are much smaller. 170 00:12:27,710 --> 00:12:33,010 So here, I'm thinking of d as on the same order as n. 171 00:12:33,010 --> 00:12:34,900 OK, so this is the theorem. 172 00:12:34,900 --> 00:12:36,150 So that's what we'll do today. 173 00:12:36,150 --> 00:12:39,090 We'll prove that all of these properties 174 00:12:39,090 --> 00:12:41,065 are equivalent to each other. 175 00:12:41,065 --> 00:12:42,690 And all of these properties, you should 176 00:12:42,690 --> 00:12:45,812 think of as characterizations of pseudorandomness. 177 00:12:45,812 --> 00:12:47,520 And of course, this theorem guarantees us 178 00:12:47,520 --> 00:12:50,070 that it doesn't matter which one you use. 179 00:12:50,070 --> 00:12:51,890 They're all equivalent to each other. 180 00:12:51,890 --> 00:12:54,397 And our proofs are actually going to be-- 181 00:12:54,397 --> 00:12:56,730 I mean, I'm going to try to do everything fairly slowly. 182 00:12:56,730 --> 00:12:58,450 But none of these proofs are difficult. 183 00:12:58,450 --> 00:13:00,390 We're not going to use any fancy tools like 184 00:13:00,390 --> 00:13:02,100 Szemerédi's regularity lemma. 185 00:13:02,100 --> 00:13:06,000 In particular, all of these quantitative errors 186 00:13:06,000 --> 00:13:09,490 are reasonably dependent on each other. 187 00:13:09,490 --> 00:13:12,060 So I've stated this theorem so far 188 00:13:12,060 --> 00:13:15,510 in this form where there is a little 1 error. 189 00:13:15,510 --> 00:13:26,030 But equivalently, so I can equivalently state theorem as-- 190 00:13:28,700 --> 00:13:37,770 for example, have DISC with an epsilon error, which 191 00:13:37,770 --> 00:13:42,630 is that some inequality is true with at most epsilon error 192 00:13:42,630 --> 00:13:44,250 instead of little o. 193 00:13:44,250 --> 00:13:48,850 And you have a different epsilon for each one of them. 194 00:13:48,850 --> 00:13:53,010 And the theorem, it turns out that-- 195 00:13:53,010 --> 00:13:55,270 OK, so the proof of this theorem will 196 00:13:55,270 --> 00:14:00,640 be that these conditions are true, so all equivalent, 197 00:14:00,640 --> 00:14:04,780 up to at most a polynomial change in the epsilons. 198 00:14:13,970 --> 00:14:18,470 In other words, so property one is true 199 00:14:18,470 --> 00:14:22,640 for epsilon implies that property two is 200 00:14:22,640 --> 00:14:26,900 true for some epsilon raised to a constant. 201 00:14:26,900 --> 00:14:30,240 So the changes in parameters are quite reasonable. 202 00:14:30,240 --> 00:14:33,030 And we'll see this from the proof, 203 00:14:33,030 --> 00:14:36,865 but I won't say it again explicitly. 204 00:14:36,865 --> 00:14:39,240 Any questions so far about the statement of this theorem? 205 00:14:46,870 --> 00:14:51,033 So as I mentioned just now, the most surprising part 206 00:14:51,033 --> 00:14:52,450 of this theorem and the one that I 207 00:14:52,450 --> 00:14:54,480 want you to pay the most attention to 208 00:14:54,480 --> 00:14:57,870 is the C4 condition. 209 00:14:57,870 --> 00:14:59,970 This seems, at least at face value, 210 00:14:59,970 --> 00:15:03,660 the weakest condition among all of them. 211 00:15:03,660 --> 00:15:07,510 It just says the correct C4 count. 212 00:15:07,510 --> 00:15:10,433 But it turns out to be equivalent to everything else. 213 00:15:10,433 --> 00:15:12,350 And there's something special about C4, right? 214 00:15:12,350 --> 00:15:16,810 If I replace C4 by C3, by just triangles, then it is not true. 215 00:15:16,810 --> 00:15:20,200 So I want you to think about, where does C4 216 00:15:20,200 --> 00:15:22,300 play this important role? 217 00:15:22,300 --> 00:15:25,530 How does it play this important role? 218 00:15:25,530 --> 00:15:26,140 OK. 219 00:15:26,140 --> 00:15:29,450 So let's get started with a proof. 220 00:15:29,450 --> 00:15:32,340 But before that, let me-- 221 00:15:32,340 --> 00:15:35,288 so in this proof, one recurring theme 222 00:15:35,288 --> 00:15:37,830 is that we're going to be using the Cauchy-Schwarz inequality 223 00:15:37,830 --> 00:15:38,910 many times. 224 00:15:38,910 --> 00:15:41,340 And I want to just begin with an exercise that gives you 225 00:15:41,340 --> 00:15:44,140 some familiarity with applying the Cauchy-Schwarz inequality. 226 00:15:44,140 --> 00:15:47,400 And this is a simple tool, but it's extremely powerful. 227 00:15:47,400 --> 00:15:49,650 And it's worthwhile to master how to use 228 00:15:49,650 --> 00:15:52,230 a Cauchy-Schwarz inequality. 229 00:15:52,230 --> 00:15:54,660 So let's get some practice. 230 00:15:54,660 --> 00:16:01,880 And let me prove a claim which is not directly related 231 00:16:01,880 --> 00:16:03,920 to the proof of the theorem, but it's 232 00:16:03,920 --> 00:16:08,520 indirect in that it explains somewhat the C4 condition 233 00:16:08,520 --> 00:16:11,300 and why we have less than or equal to over there. 234 00:16:14,300 --> 00:16:19,100 So the lemma is that if you have a graph on n 235 00:16:19,100 --> 00:16:27,500 vertices such that the number of edges is at least pn 236 00:16:27,500 --> 00:16:33,950 squared over 2, so edge density basically p, then 237 00:16:33,950 --> 00:16:48,500 the number of labeled copies of C4 is at least p to the 4 minus 238 00:16:48,500 --> 00:16:51,102 little o 1 n to the 4th. 239 00:16:56,020 --> 00:16:59,290 So if you have a graph with each density p-- 240 00:16:59,290 --> 00:17:01,150 p's your constant-- then the number of C4s 241 00:17:01,150 --> 00:17:04,869 is at least roughly what you would expect in a random graph. 242 00:17:07,662 --> 00:17:08,829 So let's see how to do this. 243 00:17:11,540 --> 00:17:14,530 And I want to show this inequality as a-- 244 00:17:14,530 --> 00:17:16,660 well, I'll show you how to prove this inequality. 245 00:17:16,660 --> 00:17:20,470 But I also want to draw a sequence of pictures, at least, 246 00:17:20,470 --> 00:17:22,780 to explain how I think about applications 247 00:17:22,780 --> 00:17:26,540 of the Cauchy-Schwarz inequality. 248 00:17:26,540 --> 00:17:27,040 OK. 249 00:17:27,040 --> 00:17:32,810 So the first thing is that we are 250 00:17:32,810 --> 00:17:35,090 counting labeled copies of C4. 251 00:17:35,090 --> 00:17:38,870 And this is basically but not exactly the same as 252 00:17:38,870 --> 00:17:44,270 number of homomorphic copies of C4 and G. So by this guy here, 253 00:17:44,270 --> 00:17:50,300 I really just mean you are mapping vertices of C4 to G 254 00:17:50,300 --> 00:17:53,150 so that the edges all map to edges. 255 00:17:53,150 --> 00:18:03,950 But we are allowing not necessarily injective maps, 256 00:18:03,950 --> 00:18:08,000 C4 to G. But that's OK. 257 00:18:08,000 --> 00:18:15,260 So the number of non-injective maps is at most cubic. 258 00:18:15,260 --> 00:18:17,770 So we're not really affecting our count. 259 00:18:17,770 --> 00:18:20,635 So it's enough to think about homomorphic copies. 260 00:18:26,460 --> 00:18:27,570 OK. 261 00:18:27,570 --> 00:18:28,600 So what's going on here? 262 00:18:28,600 --> 00:18:30,142 So let me draw a sequence of pictures 263 00:18:30,142 --> 00:18:32,680 illustrating this calculation. 264 00:18:32,680 --> 00:18:38,450 So first, we are thinking about counting C4s. 265 00:18:38,450 --> 00:18:39,760 So that's a C4. 266 00:18:43,030 --> 00:18:51,880 I can rewrite the C4 count as a sum over pairs of vertices of G 267 00:18:51,880 --> 00:18:53,980 as the squared codegree. 268 00:18:59,850 --> 00:19:02,470 And what happens here-- so this is true. 269 00:19:02,470 --> 00:19:04,450 I mean, it's not hard to see why this is true. 270 00:19:04,450 --> 00:19:07,050 But I want to draw this in pictures, 271 00:19:07,050 --> 00:19:09,320 because when you have larger and bigger graphs, 272 00:19:09,320 --> 00:19:12,090 it may be more difficult to think about the algebra 273 00:19:12,090 --> 00:19:14,740 unless you have some visualization. 274 00:19:14,740 --> 00:19:19,820 So what happens here is that I notice that the C4 has 275 00:19:19,820 --> 00:19:21,740 a certain reflection. 276 00:19:21,740 --> 00:19:28,680 Namely, it has a reflection along this horizontal line. 277 00:19:28,680 --> 00:19:35,210 And so if I put these two vertices as u and v, 278 00:19:35,210 --> 00:19:38,030 then this reflection tells you that you 279 00:19:38,030 --> 00:19:42,590 can write this number of homomorphic copies 280 00:19:42,590 --> 00:19:43,580 as the sum of squares. 281 00:19:46,510 --> 00:19:48,850 But once you have this reflection-- and reflections 282 00:19:48,850 --> 00:19:50,410 are super useful, because they allow 283 00:19:50,410 --> 00:19:52,900 us to get something into a square 284 00:19:52,900 --> 00:19:58,260 and then, right after, apply the Cauchy-Schwarz inequality. 285 00:19:58,260 --> 00:20:00,160 So we apply Cauchy-Schwarz here. 286 00:20:00,160 --> 00:20:09,370 And we obtain that this sum is at most where 287 00:20:09,370 --> 00:20:18,060 I can pull the square out. 288 00:20:18,060 --> 00:20:23,210 And I need to think about what is the correct factor 289 00:20:23,210 --> 00:20:25,270 to put out here. 290 00:20:25,270 --> 00:20:26,210 And that should be-- 291 00:20:30,710 --> 00:20:34,835 so what's the correct factor that I should put out there? 292 00:20:34,835 --> 00:20:35,960 AUDIENCE: 1 over n squared. 293 00:20:35,960 --> 00:20:39,190 PROFESSOR: OK, so 1 over n squared. 294 00:20:39,190 --> 00:20:42,700 So I don't actually like doing these kind of calculations 295 00:20:42,700 --> 00:20:45,100 with sums, because then you have to keep track 296 00:20:45,100 --> 00:20:47,110 of these normalizing factors. 297 00:20:47,110 --> 00:20:50,890 One of the upcoming chapters, when we discuss graph limits-- 298 00:20:50,890 --> 00:20:52,390 or in fact, you can even do this. 299 00:20:52,390 --> 00:20:54,805 Instead of taking sums, if you take an average, 300 00:20:54,805 --> 00:20:56,680 if you take an expectation, then it turns out 301 00:20:56,680 --> 00:21:00,100 you never have to worry about these normalizing factors. 302 00:21:00,100 --> 00:21:03,220 So normalizing factors should never bother you 303 00:21:03,220 --> 00:21:04,970 if you do it correctly. 304 00:21:04,970 --> 00:21:08,200 But just to make sure things are correct, please 305 00:21:08,200 --> 00:21:10,550 keep me in check. 306 00:21:10,550 --> 00:21:11,050 All right. 307 00:21:11,050 --> 00:21:13,180 So what happened in this step? 308 00:21:13,180 --> 00:21:15,700 In this step, we pulled out that square. 309 00:21:15,700 --> 00:21:19,030 And pictorially, what happens is that we got rid 310 00:21:19,030 --> 00:21:20,610 of half of this picture. 311 00:21:26,550 --> 00:21:29,220 So we used Cauchy-Schwarz, and we wiped out 312 00:21:29,220 --> 00:21:30,060 half of the picture. 313 00:21:32,850 --> 00:21:35,730 And now what we can do is, well, we're 314 00:21:35,730 --> 00:21:39,540 counting these guys, this path of length 2. 315 00:21:39,540 --> 00:21:43,650 But I can reprioritize this picture 316 00:21:43,650 --> 00:21:47,410 so that it looks like that. 317 00:21:47,410 --> 00:21:51,370 And now I notice that there is one more reflection. 318 00:21:51,370 --> 00:21:52,710 So there's one more reflection. 319 00:21:52,710 --> 00:21:54,835 And that's the reflection around the vertical axis. 320 00:21:57,890 --> 00:22:01,140 So let me call this top vertex x. 321 00:22:01,140 --> 00:22:11,916 And I can rewrite the sum like that. 322 00:22:18,220 --> 00:22:18,720 OK. 323 00:22:18,720 --> 00:22:20,970 So once more, we do Cauchy-Schwarz, 324 00:22:20,970 --> 00:22:24,950 which allows us to get rid of half of the picture. 325 00:22:24,950 --> 00:22:27,740 And now I'm going to draw the picture first, 326 00:22:27,740 --> 00:22:30,600 because then you see that what we should be left with 327 00:22:30,600 --> 00:22:32,860 is just a single edge. 328 00:22:32,860 --> 00:22:37,720 And then you write down the correct sum, 329 00:22:37,720 --> 00:22:42,540 making sure that all the parentheses and normalizations 330 00:22:42,540 --> 00:22:43,580 are correct. 331 00:22:43,580 --> 00:22:46,620 But somehow, that doesn't worry me so much, 332 00:22:46,620 --> 00:22:48,970 because I know this will definitely work out. 333 00:22:53,210 --> 00:22:55,750 But whatever it is, you're just summing the number of edges. 334 00:22:55,750 --> 00:22:59,270 So that's just the number of edges. 335 00:22:59,270 --> 00:23:09,950 And so we put everything in. 336 00:23:09,950 --> 00:23:13,790 And we find that the final quantity is at least p raised 337 00:23:13,790 --> 00:23:17,150 to 4 n to 4. 338 00:23:17,150 --> 00:23:18,650 So I did this quite slowly. 339 00:23:18,650 --> 00:23:23,540 But I'm also emphasizing the sequence of pictures, partly 340 00:23:23,540 --> 00:23:25,520 to tell how I think about these inequalities. 341 00:23:25,520 --> 00:23:29,940 Because for other similar looking inequalities-- in fact, 342 00:23:29,940 --> 00:23:33,260 there is something called Sidorenko's conjecture, which 343 00:23:33,260 --> 00:23:36,350 I may discuss more in a future lecture, that 344 00:23:36,350 --> 00:23:38,990 says that this kind of inequality 345 00:23:38,990 --> 00:23:41,540 should be true whenever you replace C4 346 00:23:41,540 --> 00:23:43,520 by any bipartite graph. 347 00:23:43,520 --> 00:23:47,680 And that's a major open problem in combinatorics. 348 00:23:47,680 --> 00:23:49,990 It's kind of hard to keep track of these calculations 349 00:23:49,990 --> 00:23:51,910 unless you have a visual anchor. 350 00:23:51,910 --> 00:23:55,600 And this is my visual anchor, which I'm trying to explain. 351 00:23:55,600 --> 00:23:57,610 Of course, it's down to earth. 352 00:23:57,610 --> 00:23:59,590 It's just the sequence of inequalities. 353 00:23:59,590 --> 00:24:04,924 And this is also some practice with Cauchy-Schwarz. 354 00:24:04,924 --> 00:24:06,760 All right. 355 00:24:06,760 --> 00:24:07,924 Any questions? 356 00:24:10,782 --> 00:24:12,240 But one thing that this calculation 357 00:24:12,240 --> 00:24:15,300 told us is that if you have edge density p, 358 00:24:15,300 --> 00:24:20,790 then you necessarily have C4 density at least p to the 4th. 359 00:24:20,790 --> 00:24:25,240 So that partly explains why you have at most, then, here. 360 00:24:25,240 --> 00:24:28,030 So you always know that it's at least this quantity. 361 00:24:28,030 --> 00:24:30,680 So the C4 quasi randomness condition 362 00:24:30,680 --> 00:24:34,260 is really the equivalent to replacing this less than 363 00:24:34,260 --> 00:24:35,820 or equal to by an equal sign. 364 00:24:40,010 --> 00:24:41,660 So let's get started with proving 365 00:24:41,660 --> 00:24:44,210 the Chung-Graham-Wilson theorem. 366 00:24:44,210 --> 00:24:46,220 So the first place that we'll look at 367 00:24:46,220 --> 00:24:49,920 is the two versions of DISC. 368 00:24:49,920 --> 00:24:51,610 So DISC stands for discrepancy. 369 00:24:54,630 --> 00:25:00,720 So first, the fact that DISC implies DISC prime, I mean, 370 00:25:00,720 --> 00:25:02,250 this is pretty easy. 371 00:25:02,250 --> 00:25:04,590 You take y to equal to x. 372 00:25:04,590 --> 00:25:10,160 Be slightly careful about the definitions, but you're OK. 373 00:25:10,160 --> 00:25:13,590 So not much to do there. 374 00:25:13,590 --> 00:25:20,050 The other direction, where you only 375 00:25:20,050 --> 00:25:21,898 have discrepancies for a single set 376 00:25:21,898 --> 00:25:23,440 and you want to produce discrepancies 377 00:25:23,440 --> 00:25:25,570 for pairs of sets-- 378 00:25:25,570 --> 00:25:28,440 so this is actually a fairly common technique in algebra 379 00:25:28,440 --> 00:25:31,570 that allows you to go from bilinear forms 380 00:25:31,570 --> 00:25:33,460 to quadratic forms and vice versa. 381 00:25:33,460 --> 00:25:35,090 It's that kind of calculation. 382 00:25:35,090 --> 00:25:37,640 So let me do it here concretely in this setting. 383 00:25:37,640 --> 00:25:39,790 So here, what you should think of 384 00:25:39,790 --> 00:25:45,680 is that you have two sets, x and y, and they might overlap. 385 00:25:45,680 --> 00:25:52,290 And what they correspond to in the-- 386 00:25:52,290 --> 00:25:55,220 when you think about the corresponding Venn diagram, 387 00:25:55,220 --> 00:26:03,130 where I'm looking at ways that a pair of vertices 388 00:26:03,130 --> 00:26:05,290 can fall in x and/or y-- 389 00:26:05,290 --> 00:26:08,410 so if you have x and y. 390 00:26:14,410 --> 00:26:22,360 And so it's useful to keep track of which vertices 391 00:26:22,360 --> 00:26:24,310 are in which set. 392 00:26:24,310 --> 00:26:27,010 But what the thing finally comes down 393 00:26:27,010 --> 00:26:31,240 to is that the number of edges with one vertex in x and one 394 00:26:31,240 --> 00:26:35,430 vertex in y, I can write this bilinear form-type quantity 395 00:26:35,430 --> 00:26:50,540 as an appropriate sum of just number of edges in single sets. 396 00:26:57,040 --> 00:27:00,600 And so there are several ways to check that this is true. 397 00:27:00,600 --> 00:27:04,980 One way is to just tally, keep track of how many edges 398 00:27:04,980 --> 00:27:07,420 are you counting in each step. 399 00:27:07,420 --> 00:27:14,330 So if you are trying to count the number of edges in-- 400 00:27:18,990 --> 00:27:21,140 yeah, so let's say if you're trying to count 401 00:27:21,140 --> 00:27:23,840 the number of edges in-- 402 00:27:26,380 --> 00:27:29,320 with one vertex in x, one vertex in y. 403 00:27:29,320 --> 00:27:34,570 Then what this corresponds to is that count. 404 00:27:34,570 --> 00:27:36,040 But let me do a reflection. 405 00:27:41,760 --> 00:27:45,780 And then you see that you can write this sum 406 00:27:45,780 --> 00:27:49,490 as an alternating sum of principal squares, 407 00:27:49,490 --> 00:27:55,070 so this one big square plus the middle square and minus the two 408 00:27:55,070 --> 00:28:02,080 sides squares, which is what that sum comes to. 409 00:28:02,080 --> 00:28:03,250 All right. 410 00:28:03,250 --> 00:28:06,400 So if we assume DISC prime, then I 411 00:28:06,400 --> 00:28:08,560 know that all of these individual sets 412 00:28:08,560 --> 00:28:33,520 have roughly the correct number of edges up to a little o of n 413 00:28:33,520 --> 00:28:35,120 squared error. 414 00:28:35,120 --> 00:28:37,360 And again, I don't have to do this calculation again, 415 00:28:37,360 --> 00:28:39,440 because it's the same calculation. 416 00:28:39,440 --> 00:28:44,260 So the final thing should be p times the sizes of x and y 417 00:28:44,260 --> 00:28:47,420 together plus this same error. 418 00:28:52,770 --> 00:28:55,070 So that shows you DISC prime implies DISC. 419 00:28:55,070 --> 00:28:57,060 So the self version of discrepancy 420 00:28:57,060 --> 00:28:59,394 implies the pair version of discrepancy. 421 00:29:02,060 --> 00:29:04,850 So let's move on to count. 422 00:29:10,820 --> 00:29:14,930 To show that DISC implies count-- 423 00:29:20,180 --> 00:29:22,540 actually, we already did this. 424 00:29:22,540 --> 00:29:23,908 So this is the counting lemma. 425 00:29:31,240 --> 00:29:34,050 So the counting lemma tells us how 426 00:29:34,050 --> 00:29:38,430 to count labeled copies if you have these epsilon regularity 427 00:29:38,430 --> 00:29:40,770 conditions, which is exactly what DISC is. 428 00:29:44,600 --> 00:29:47,550 So count is good. 429 00:29:47,550 --> 00:29:55,210 Another easy implication is count implies C4. 430 00:29:55,210 --> 00:29:58,560 Well, this is actually just tautological. 431 00:29:58,560 --> 00:30:04,610 C4 condition is a special case of the count hypothesis. 432 00:30:08,460 --> 00:30:09,380 All right. 433 00:30:09,380 --> 00:30:15,090 So let's move on to some additional implications that 434 00:30:15,090 --> 00:30:17,190 require a bit more work. 435 00:30:17,190 --> 00:30:20,030 So what about C4 implies codegree? 436 00:30:26,730 --> 00:30:28,410 So this is where we need to do this kind 437 00:30:28,410 --> 00:30:30,940 of Cauchy-Schwarz exercise. 438 00:30:30,940 --> 00:30:33,780 So let's start with C4. 439 00:30:33,780 --> 00:30:38,740 So assume a C4 condition. 440 00:30:38,740 --> 00:30:42,130 And suppose you have this-- 441 00:30:47,650 --> 00:30:50,840 so I want to deduce that the codegree condition is true. 442 00:30:50,840 --> 00:30:56,500 But first, let's think about just 443 00:30:56,500 --> 00:30:58,640 what is the sum of these codegrees 444 00:30:58,640 --> 00:31:02,050 as I vary u and v over all pairs of vertices. 445 00:31:06,980 --> 00:31:11,000 So this is that picture. 446 00:31:11,000 --> 00:31:18,390 So that is equal to the sums of degrees squared, which now, 447 00:31:18,390 --> 00:31:24,090 by Cauchy-Schwarz, you can deduce to be at least n times 2 448 00:31:24,090 --> 00:31:25,710 raised to number of edges-- namely, 449 00:31:25,710 --> 00:31:28,910 the sum of the degrees-- 450 00:31:28,910 --> 00:31:29,710 that thing squared. 451 00:31:32,760 --> 00:31:37,400 So now we assume the C4 condition-- 452 00:31:37,400 --> 00:31:41,540 actually, no, we assume that G has the density as written up 453 00:31:41,540 --> 00:31:42,260 there. 454 00:31:42,260 --> 00:31:51,890 So this quantity is p squared plus little 1 times 455 00:31:51,890 --> 00:31:54,170 n cubed, which is what you should expect 456 00:31:54,170 --> 00:31:56,480 in a random graph of Gnp. 457 00:31:59,020 --> 00:32:00,890 But that's not quite what we're looking for. 458 00:32:00,890 --> 00:32:03,880 So this is just the sum of the codegrees. 459 00:32:03,880 --> 00:32:08,110 What we actually want is the deviation 460 00:32:08,110 --> 00:32:14,120 of codegrees from its expectations, so to speak. 461 00:32:14,120 --> 00:32:16,360 Now, here's an important technique 462 00:32:16,360 --> 00:32:19,360 from probabilistic combinatorics is 463 00:32:19,360 --> 00:32:25,920 that if you want to control the deviation of a random variable, 464 00:32:25,920 --> 00:32:28,970 one thing you should look at is the variance. 465 00:32:28,970 --> 00:32:30,780 So if you can control the variance, 466 00:32:30,780 --> 00:32:32,442 then you can control the deviation. 467 00:32:32,442 --> 00:32:34,650 And this is a method known as a second moment method. 468 00:32:34,650 --> 00:32:36,275 And that's what we're going to do here. 469 00:32:36,275 --> 00:32:44,470 So what we'll try to show is that the second moment of these 470 00:32:44,470 --> 00:32:45,820 codegrees-- 471 00:32:45,820 --> 00:32:48,030 namely, the sum of their squares-- 472 00:32:48,030 --> 00:32:53,060 is also what you should expect as if the random setting. 473 00:32:53,060 --> 00:32:56,775 And then you can put them together to show what you want. 474 00:32:56,775 --> 00:33:01,370 So this quantity here, well, what is this? 475 00:33:01,370 --> 00:33:08,790 We just saw-- see, up there, it's also codegree squared. 476 00:33:08,790 --> 00:33:12,050 So this quantity is also the number 477 00:33:12,050 --> 00:33:13,650 of labeled copies of C4-- 478 00:33:19,280 --> 00:33:23,520 not quite, because you might have two vertices 479 00:33:23,520 --> 00:33:25,390 and the same vertex. 480 00:33:25,390 --> 00:33:30,430 So I incorporate a small error. 481 00:33:30,430 --> 00:33:37,420 So it's a cubic error, but it's certainly sub n to the 4. 482 00:33:37,420 --> 00:33:42,640 And we assume that the number of labeled copies of C4 by the C4 483 00:33:42,640 --> 00:33:48,520 condition is no more than basically p to the 4 times 484 00:33:48,520 --> 00:33:49,870 n raised to power 4. 485 00:33:55,495 --> 00:33:55,995 OK. 486 00:33:55,995 --> 00:33:57,400 So now you have a first moment. 487 00:33:57,400 --> 00:34:00,300 You have some average, and you have some control. 488 00:34:00,300 --> 00:34:02,920 In the second moment, I can put them together 489 00:34:02,920 --> 00:34:05,950 to bound the deviation using this idea 490 00:34:05,950 --> 00:34:07,035 of controlling variance. 491 00:34:09,921 --> 00:34:21,060 So the codegree deviation is upper bounded by-- 492 00:34:21,060 --> 00:34:23,010 so here, using Cauchy-Schwarz, it's 493 00:34:23,010 --> 00:34:27,739 upper bounded by basically the same sum, except I 494 00:34:27,739 --> 00:34:31,580 want to square the summand. 495 00:34:42,536 --> 00:34:45,060 This also gets rid of the pesky absolute value 496 00:34:45,060 --> 00:34:48,400 side, which is not nicely, algebraically behaved. 497 00:34:48,400 --> 00:34:48,900 OK. 498 00:34:48,900 --> 00:34:51,360 So now I have the square, and I can expand the square. 499 00:35:00,810 --> 00:35:04,860 So I expand the square into these terms. 500 00:35:14,910 --> 00:35:20,225 And the final term here is p to the 4 n to the 6. 501 00:35:24,476 --> 00:35:26,360 No, n to the 4. 502 00:35:29,790 --> 00:35:32,730 All right. 503 00:35:32,730 --> 00:35:35,400 But I have controlled the individual terms 504 00:35:35,400 --> 00:35:38,630 from the calculations above. 505 00:35:38,630 --> 00:35:43,660 So I can upper bound this expression 506 00:35:43,660 --> 00:35:47,700 by what I'm writing down now. 507 00:35:54,540 --> 00:35:57,420 And basically, you should expect that everything 508 00:35:57,420 --> 00:36:01,760 should cancel out, because they do cancel out 509 00:36:01,760 --> 00:36:02,760 in the random case. 510 00:36:02,760 --> 00:36:05,040 Of course, the sanity check, it's 511 00:36:05,040 --> 00:36:07,475 important to write down this calculation. 512 00:36:07,475 --> 00:36:08,850 So if everything works out right, 513 00:36:08,850 --> 00:36:10,058 everything should cancel out. 514 00:36:10,058 --> 00:36:11,880 And indeed, they do cancel out. 515 00:36:11,880 --> 00:36:14,583 And you get that-- 516 00:36:14,583 --> 00:36:20,020 so this is a multiplication. 517 00:36:20,020 --> 00:36:24,010 This is p squared. 518 00:36:24,010 --> 00:36:26,800 Is that OK? 519 00:36:26,800 --> 00:36:28,860 So everything should cancel out. 520 00:36:28,860 --> 00:36:34,320 And you get a little o of n cubed. 521 00:36:34,320 --> 00:36:38,550 To summarize, in this implication from C4 522 00:36:38,550 --> 00:36:43,750 to codegree, what we're doing is we're 523 00:36:43,750 --> 00:36:48,640 controlling the variance of codegrees 524 00:36:48,640 --> 00:36:53,950 using the C4 condition and the second moment bound, 525 00:36:53,950 --> 00:36:56,860 showing that the C4 condition trumps 526 00:36:56,860 --> 00:36:58,741 over the codegree condition. 527 00:37:01,830 --> 00:37:02,850 Any questions so far? 528 00:37:08,100 --> 00:37:10,490 So I'll let you ponder in this calculation. 529 00:37:10,490 --> 00:37:15,390 The next one that we'll do is codegree implies DISC. 530 00:37:15,390 --> 00:37:20,992 And that will be a calculation in a very similar flavor. 531 00:37:20,992 --> 00:37:23,325 But it will be a slightly longer but with similar flavor 532 00:37:23,325 --> 00:37:24,370 of calculation. 533 00:37:24,370 --> 00:37:27,384 So let me do that after the break. 534 00:37:27,384 --> 00:37:29,740 All right. 535 00:37:29,740 --> 00:37:31,200 So what have we done so far? 536 00:37:31,200 --> 00:37:34,080 So let's summarize the chain of implications 537 00:37:34,080 --> 00:37:36,240 that we have already proved. 538 00:37:36,240 --> 00:37:38,850 So first, we started with showing 539 00:37:38,850 --> 00:37:43,545 that the two versions of DISC are equivalent. 540 00:37:48,500 --> 00:37:55,610 And then we also noticed that DISC implies count 541 00:37:55,610 --> 00:37:58,700 through the counting lemma. 542 00:37:58,700 --> 00:38:06,090 So we also observed that count implies C4 tautologically 543 00:38:06,090 --> 00:38:08,650 and C4 implies codegree. 544 00:38:14,670 --> 00:38:18,190 So the next natural thing to do is to complete this circuit 545 00:38:18,190 --> 00:38:21,400 and show that the codegree condition implies 546 00:38:21,400 --> 00:38:24,452 the discrepancy condition. 547 00:38:24,452 --> 00:38:25,660 So that's what we'll do next. 548 00:38:28,760 --> 00:38:31,880 And in some sense, these two steps, 549 00:38:31,880 --> 00:38:35,660 you should think of them as going in this natural chain, 550 00:38:35,660 --> 00:38:37,610 where C4-- 551 00:38:37,610 --> 00:38:43,200 so C4 is like this, C4. 552 00:38:43,200 --> 00:38:46,680 Codegree condition is really about that. 553 00:38:46,680 --> 00:38:49,220 And DISC is really about single edges. 554 00:38:49,220 --> 00:38:50,440 So you can go from-- 555 00:38:50,440 --> 00:38:55,520 so double-- if you half, you get much more power. 556 00:38:55,520 --> 00:38:57,560 So it's going in the right direction, 557 00:38:57,560 --> 00:38:59,940 going downstream, so to speak. 558 00:38:59,940 --> 00:39:03,470 So that's what we're doing now, going downstream. 559 00:39:03,470 --> 00:39:05,769 And then you go upstream via the counting lemma. 560 00:39:08,523 --> 00:39:10,360 All right. 561 00:39:10,360 --> 00:39:13,120 Let's do codegree implies DISC. 562 00:39:24,670 --> 00:39:29,140 So we want to show the discrepancy condition, which 563 00:39:29,140 --> 00:39:30,310 is one written up there. 564 00:39:30,310 --> 00:39:32,290 But before that, let me first show you 565 00:39:32,290 --> 00:39:36,460 that the degrees do not vary too much, 566 00:39:36,460 --> 00:39:39,010 show that the degrees are fairly well distributed, 567 00:39:39,010 --> 00:39:42,580 which is what you should expect in a pseudorandom graph. 568 00:39:42,580 --> 00:39:45,880 So you don't expect the half the vertices, half in degrees, 569 00:39:45,880 --> 00:39:48,500 twice the other half. 570 00:39:48,500 --> 00:39:51,220 So that's the first thing I want to establish. 571 00:39:53,920 --> 00:40:02,910 If you look at degrees, this variance, this deviation, 572 00:40:02,910 --> 00:40:06,550 is not too big. 573 00:40:06,550 --> 00:40:07,050 OK. 574 00:40:07,050 --> 00:40:09,450 So like before, we see an absolute value sign. 575 00:40:09,450 --> 00:40:10,420 We see a sum. 576 00:40:10,420 --> 00:40:12,696 So we'll do Cauchy-Schwarz. 577 00:40:12,696 --> 00:40:16,630 Cauchy-Schwarz allows us to bound this quantity, 578 00:40:16,630 --> 00:40:20,290 replacing the summand by a sum of squared. 579 00:40:33,210 --> 00:40:35,550 I have a square, so I can expand the square. 580 00:40:41,060 --> 00:40:43,310 So let me expand the square. 581 00:40:43,310 --> 00:41:02,020 And I get that, so just expanding this square inside. 582 00:41:04,708 --> 00:41:10,260 And you see this degree squared is that picture, so 583 00:41:10,260 --> 00:41:11,730 that sum of codegrees. 584 00:41:24,280 --> 00:41:27,400 And sum of the degrees is just the number of edges. 585 00:41:39,630 --> 00:41:42,750 But we now assume the codegree condition, 586 00:41:42,750 --> 00:41:46,205 which in particular implies that the sum of the codegrees 587 00:41:46,205 --> 00:41:47,580 is roughly what you would expect. 588 00:41:50,170 --> 00:41:52,170 So the sum of the codegrees should 589 00:41:52,170 --> 00:41:58,650 be p squared n cubed plus a little o n cubed error 590 00:41:58,650 --> 00:42:01,250 at the end. 591 00:42:01,250 --> 00:42:05,630 Likewise, the number of edges is, by assumption, 592 00:42:05,630 --> 00:42:10,130 what you would expect in a random graph. 593 00:42:10,130 --> 00:42:13,480 And then the final term. 594 00:42:13,480 --> 00:42:17,020 And like before-- and of course, it's good to do a sanity 595 00:42:17,020 --> 00:42:17,560 check-- 596 00:42:17,560 --> 00:42:19,750 everything should cancel out. 597 00:42:19,750 --> 00:42:25,310 So what you end up with is little o of n squared, 598 00:42:25,310 --> 00:42:28,070 showing that the degrees do not vary too much. 599 00:42:28,070 --> 00:42:31,790 And once you have that promise, then we 600 00:42:31,790 --> 00:42:35,487 move onto the actual discrepancy condition. 601 00:42:41,570 --> 00:42:47,850 So this discrepancy can be rewritten 602 00:42:47,850 --> 00:42:57,760 as the sum over vertices little x and big X, the degree 603 00:42:57,760 --> 00:43:08,212 from little x to y minus p times the size of y, 604 00:43:08,212 --> 00:43:11,362 so rewriting the sum. 605 00:43:11,362 --> 00:43:16,090 And of course, what should we do next? 606 00:43:16,090 --> 00:43:16,950 Cauchy-Schwarz. 607 00:43:16,950 --> 00:43:17,560 Great. 608 00:43:17,560 --> 00:43:21,010 So we'll do a Cauchy-Schwarz. 609 00:43:21,010 --> 00:43:26,180 OK, so here's an important step or trick, if you will. 610 00:43:26,180 --> 00:43:28,990 So we'll do Cauchy-Schwarz. 611 00:43:28,990 --> 00:43:34,070 And something very nice happens when you do Cauchy-Schwarz 612 00:43:34,070 --> 00:43:35,080 here. 613 00:43:35,080 --> 00:43:35,780 OK. 614 00:43:35,780 --> 00:43:37,340 So you can write down the expression 615 00:43:37,340 --> 00:43:40,070 that you obtain when you do Cauchy-Schwarz. 616 00:43:40,070 --> 00:43:41,420 So let me do that first. 617 00:43:47,490 --> 00:43:47,990 OK. 618 00:43:47,990 --> 00:43:51,690 So here's a step which is very easy to gloss over. 619 00:43:51,690 --> 00:43:53,600 But I want to pause and emphasize this step, 620 00:43:53,600 --> 00:43:57,170 because this is actually really important. 621 00:43:57,170 --> 00:44:00,250 What I'm going to do now is to observe that the summand is 622 00:44:00,250 --> 00:44:01,714 always non-negative. 623 00:44:07,390 --> 00:44:14,740 Therefore, I can enlarge the sum from just little x 624 00:44:14,740 --> 00:44:18,542 and X to the entire vertex set. 625 00:44:18,542 --> 00:44:19,750 And this is important, right? 626 00:44:19,750 --> 00:44:21,850 So it's important that we had to do Cauchy-Schwarz 627 00:44:21,850 --> 00:44:23,942 first to get a non-negative summand. 628 00:44:23,942 --> 00:44:25,525 You couldn't do this in the beginning. 629 00:44:29,800 --> 00:44:32,450 So you do that. 630 00:44:32,450 --> 00:44:35,120 And so I have this sum of squares. 631 00:44:35,120 --> 00:44:36,065 I expand. 632 00:44:43,300 --> 00:44:44,270 I expand. 633 00:44:44,270 --> 00:44:47,800 I write out all these expressions. 634 00:44:57,300 --> 00:45:00,680 And now the little x range over the entire vertex set. 635 00:45:15,715 --> 00:45:17,200 All right. 636 00:45:17,200 --> 00:45:20,780 So what was the point of all of that? 637 00:45:20,780 --> 00:45:25,000 So you see this expression here, the degree 638 00:45:25,000 --> 00:45:31,870 from little x to big Y squared, what is that? 639 00:45:31,870 --> 00:45:34,230 How can we rewrite this expression? 640 00:45:38,550 --> 00:45:41,990 So counting little x and then Y squared-- 641 00:45:45,990 --> 00:45:47,993 AUDIENCE: Sum over u and big Y. 642 00:45:47,993 --> 00:45:48,660 PROFESSOR: Yeah. 643 00:45:48,660 --> 00:45:56,064 So sum of codegree of two vertices in Y, 644 00:45:56,064 --> 00:46:03,570 so Y, Y prime, and Y codegree of little y, little y prime. 645 00:46:09,320 --> 00:46:15,240 And likewise, the next expression 646 00:46:15,240 --> 00:46:23,880 can be written as the sum of the degrees of vertices in Y. 647 00:46:23,880 --> 00:46:27,670 And the third term, I leave unchanged. 648 00:46:27,670 --> 00:46:32,610 So now we've gotten rid of these funny expressions 649 00:46:32,610 --> 00:46:35,810 where it's just degree from the vertex to a set. 650 00:46:35,810 --> 00:46:40,320 And we could do this because of this relaxation up here. 651 00:46:40,320 --> 00:46:41,480 So that was the point. 652 00:46:41,480 --> 00:46:43,550 We had to use this relaxation so that we 653 00:46:43,550 --> 00:46:46,140 get these codegree terms. 654 00:46:46,140 --> 00:46:48,750 But now, because you have the codegree terms 655 00:46:48,750 --> 00:46:51,600 and we assume the codegree hypothesis, 656 00:46:51,600 --> 00:46:55,470 we obtain that this sum is roughly 657 00:46:55,470 --> 00:47:00,480 what you expect as in a random case, because all 658 00:47:00,480 --> 00:47:05,515 the individual deviations do not add up to more than little 659 00:47:05,515 --> 00:47:06,780 o n cubed. 660 00:47:11,150 --> 00:47:15,000 That codegree sum is what you expect. 661 00:47:15,000 --> 00:47:19,040 And the next term, the sum of degrees, 662 00:47:19,040 --> 00:47:21,875 is also, by what we did up there, what you expect. 663 00:47:29,315 --> 00:47:31,380 And finally, the third term. 664 00:47:34,330 --> 00:47:36,640 And as earlier, if you did everything correctly, 665 00:47:36,640 --> 00:47:38,690 everything should cancel. 666 00:47:38,690 --> 00:47:40,780 And they do. 667 00:47:40,780 --> 00:47:43,340 And so what you get at the end is little o of n squared. 668 00:47:52,020 --> 00:47:54,900 This completes this fourth cycle. 669 00:47:59,655 --> 00:48:00,530 Any questions so far? 670 00:48:06,160 --> 00:48:08,610 So we're missing one more condition, 671 00:48:08,610 --> 00:48:11,980 and that's the eigenvalue condition. 672 00:48:11,980 --> 00:48:15,928 So far, everything had to do with counting various things. 673 00:48:15,928 --> 00:48:17,970 So what does eigenvalue have to do with anything? 674 00:48:22,920 --> 00:48:25,360 So the eigenvalue condition is actually 675 00:48:25,360 --> 00:48:26,680 a particularly important one. 676 00:48:26,680 --> 00:48:29,560 And we'll see more of this in the next lecture. 677 00:48:29,560 --> 00:48:32,300 But let me first show you the equivalent implications. 678 00:48:32,300 --> 00:48:35,600 So what we'll show is that the eigenvalue condition is 679 00:48:35,600 --> 00:48:37,660 equivalent to the C4 condition. 680 00:48:37,660 --> 00:48:38,710 So that's the goal. 681 00:48:38,710 --> 00:48:42,640 So I'll show equivalence between EIG and C4. 682 00:48:46,250 --> 00:48:52,050 So first, it implies a C4 condition, because up to-- 683 00:48:52,050 --> 00:48:53,990 so instead of counting C4s, which 684 00:48:53,990 --> 00:48:56,340 is a little bit actually not-- 685 00:48:56,340 --> 00:48:59,780 it's a bit annoying to do actual C4s. 686 00:48:59,780 --> 00:49:04,430 Just like earlier, we want to consider homomorphic copies, 687 00:49:04,430 --> 00:49:09,420 which are also labeled walks, so closed walks of length 4. 688 00:49:09,420 --> 00:49:20,040 So up to a cubic error, the number of labeled C4s 689 00:49:20,040 --> 00:49:36,250 is given by the number of closed walks of length 4, which 690 00:49:36,250 --> 00:49:41,807 is equal to the trace of the 4th power of the adjacency matrix 691 00:49:41,807 --> 00:49:42,390 of this graph. 692 00:49:53,280 --> 00:49:55,320 And the next thing is super important. 693 00:49:55,320 --> 00:49:58,850 So the next thing is sometimes called a trace method. 694 00:49:58,850 --> 00:50:02,150 One important way that the eigenvalue, so the spectrum 695 00:50:02,150 --> 00:50:06,470 of a graph or matrix, relates to other combinatorial quantities 696 00:50:06,470 --> 00:50:07,910 is via this trace. 697 00:50:07,910 --> 00:50:11,440 So we know that the trace of the 4th power 698 00:50:11,440 --> 00:50:14,240 is equal to the fourth moment of the eigenvalues. 699 00:50:19,673 --> 00:50:21,590 So if you haven't seen a proof of this before, 700 00:50:21,590 --> 00:50:23,507 I encourage you to go home and think about it. 701 00:50:23,507 --> 00:50:25,730 So this is an important identity, of course. 702 00:50:25,730 --> 00:50:29,330 4 can be replaced by any number up here. 703 00:50:29,330 --> 00:50:35,420 And now you have the eigenvalue condition. 704 00:50:35,420 --> 00:50:38,720 So I can estimate the sum. 705 00:50:38,720 --> 00:50:41,660 There's a principle term-- namely, lambda 1. 706 00:50:41,660 --> 00:50:42,860 So that's the big term. 707 00:50:42,860 --> 00:50:44,210 Everything else is small. 708 00:50:44,210 --> 00:50:46,950 And the smallness is supposed to capture pseudorandomness. 709 00:50:46,950 --> 00:50:51,150 But the big term, you have to analyze separately. 710 00:50:51,150 --> 00:50:58,380 OK, so let me write it out like that. 711 00:50:58,380 --> 00:51:03,030 So the big term, you know that it is p to the 4 n to the 4 712 00:51:03,030 --> 00:51:06,500 plus little o of n to the 4. 713 00:51:06,500 --> 00:51:07,000 OK. 714 00:51:07,000 --> 00:51:11,220 So next thing is what to do with the little terms. 715 00:51:11,220 --> 00:51:13,650 So we want to show that the contribution in total 716 00:51:13,650 --> 00:51:15,770 is not too big. 717 00:51:15,770 --> 00:51:19,010 So what can we do? 718 00:51:19,010 --> 00:51:22,250 Well, let me first try something. 719 00:51:22,250 --> 00:51:27,360 So first, well, you see that each one of these guys 720 00:51:27,360 --> 00:51:28,590 is not too big. 721 00:51:28,590 --> 00:51:35,250 So maybe let's bound each one of them by little o of n 722 00:51:35,250 --> 00:51:36,930 raised to 4. 723 00:51:36,930 --> 00:51:38,880 But then there are n of them, so you 724 00:51:38,880 --> 00:51:42,760 have to multiply by an extra n. 725 00:51:42,760 --> 00:51:44,840 And that's too much. 726 00:51:44,840 --> 00:51:46,380 That's not good enough. 727 00:51:46,380 --> 00:51:49,430 So you cannot individually bound each one of them. 728 00:51:49,430 --> 00:51:51,793 And this is a novice mistake. 729 00:51:51,793 --> 00:51:53,210 This is something that we actually 730 00:51:53,210 --> 00:51:55,040 will see this type of calculation 731 00:51:55,040 --> 00:51:57,630 later on in the term when we discuss Roth's theorem. 732 00:51:57,630 --> 00:52:00,767 But you're not supposed to bound these terms individually. 733 00:52:00,767 --> 00:52:02,600 The better way to do this or the correct way 734 00:52:02,600 --> 00:52:06,520 to do this is to pull out just a couple-- 735 00:52:06,520 --> 00:52:08,180 some, but not all-- 736 00:52:08,180 --> 00:52:09,750 of these factors. 737 00:52:09,750 --> 00:52:16,740 So it is upper bounded by-- you take max of-- 738 00:52:16,740 --> 00:52:19,560 in this case, you can take out one or two. 739 00:52:19,560 --> 00:52:22,260 But you take out, let's say, two factors. 740 00:52:22,260 --> 00:52:25,610 And then you leave the remaining sum intact. 741 00:52:25,610 --> 00:52:30,970 In fact, I can even put lambda 1 back into the remaining sum. 742 00:52:30,970 --> 00:52:32,130 So that is true. 743 00:52:32,130 --> 00:52:35,020 So what I've written down is just true as an inequality. 744 00:52:35,020 --> 00:52:39,020 And now I apply the hypothesis on the sizes 745 00:52:39,020 --> 00:52:39,980 of the other lambdas. 746 00:52:47,730 --> 00:52:53,370 So the one I pulled out is little o of n squared. 747 00:52:53,370 --> 00:52:57,250 And now what's the second sum? 748 00:52:57,250 --> 00:53:03,360 That sum is the trace of a squared, 749 00:53:03,360 --> 00:53:08,290 which is just twice the number of edges of the graph. 750 00:53:08,290 --> 00:53:13,330 So that's also at most n squared. 751 00:53:13,330 --> 00:53:21,800 So combining everything, you have the desired bound 752 00:53:21,800 --> 00:53:22,650 on the C4 count. 753 00:53:25,352 --> 00:53:27,060 Of course, this gives you an upper bound. 754 00:53:27,060 --> 00:53:28,977 But we also did a calculation before the break 755 00:53:28,977 --> 00:53:32,240 that shows you that the C4 bound has a lower bound, as well. 756 00:53:32,240 --> 00:53:36,240 So really, having the correct eigenvalue-- 757 00:53:36,240 --> 00:53:38,180 actually, no, this already shows you 758 00:53:38,180 --> 00:53:40,580 that the C4 bound is correct in both directions, 759 00:53:40,580 --> 00:53:42,753 because this is the main term. 760 00:53:42,753 --> 00:53:44,170 And then everything else is small. 761 00:53:49,070 --> 00:53:50,330 OK. 762 00:53:50,330 --> 00:53:54,406 The final implication is C4 implies eigenvalue. 763 00:54:01,560 --> 00:54:06,470 For this one, I need to explore the following important 764 00:54:06,470 --> 00:54:08,520 property of the top eigenvalue. 765 00:54:08,520 --> 00:54:10,400 So there's something that we also 766 00:54:10,400 --> 00:54:13,580 saw last time, which is the interpretation 767 00:54:13,580 --> 00:54:19,600 of the top eigenvalue of a matrix interpreted as-- 768 00:54:22,880 --> 00:54:26,130 so this is sometimes called the Courant-Fischer criterion. 769 00:54:26,130 --> 00:54:30,610 Or actually, this is a special case of Courant-Fischer. 770 00:54:30,610 --> 00:54:32,930 This is a basic linear algebra fact. 771 00:54:32,930 --> 00:54:35,620 If you are not familiar with it, I recommend looking it up. 772 00:54:39,110 --> 00:54:49,250 The top eigenvalue of a matrix, of a real, symmetric matrix, 773 00:54:49,250 --> 00:54:56,950 is characterized by the maximum value of this quadratic form. 774 00:54:59,570 --> 00:55:02,070 Let's say if x is a non-zero vector. 775 00:55:05,940 --> 00:55:11,070 So in particular, if I set x to be a specific vector, 776 00:55:11,070 --> 00:55:14,540 I can lower bound lambda 1. 777 00:55:14,540 --> 00:55:28,110 So if we set this boldface 1 to be the all-one vector in R 778 00:55:28,110 --> 00:55:31,680 raised to the number of vertices of G, 779 00:55:31,680 --> 00:55:43,640 then the lambda 1 of the graph is at least 780 00:55:43,640 --> 00:55:46,980 this quantity over here. 781 00:55:46,980 --> 00:55:49,400 The numerator and denominators are all easy things 782 00:55:49,400 --> 00:55:50,300 to evaluate. 783 00:55:50,300 --> 00:55:53,140 The numerator is just twice the number of edges, 784 00:55:53,140 --> 00:55:56,140 because you are summing up all the entries of the matrix. 785 00:55:56,140 --> 00:56:01,180 And the denominator is just n. 786 00:56:01,180 --> 00:56:05,920 So the top eigenvalue is at least roughly pn. 787 00:56:13,650 --> 00:56:15,505 So what about the other eigenvalues? 788 00:56:18,830 --> 00:56:23,420 Well, the other eigenvalues, I can again 789 00:56:23,420 --> 00:56:27,930 refer back to this moment formula relating 790 00:56:27,930 --> 00:56:31,380 the trace and closed walks. 791 00:56:31,380 --> 00:56:37,190 It is at most the trace of the 4th power 792 00:56:37,190 --> 00:56:42,710 minus the top eigenvalue raised to the 4th power. 793 00:56:42,710 --> 00:56:44,370 It's the sum of the other eigenvalue 794 00:56:44,370 --> 00:56:45,860 raised to the 4th power. 795 00:56:45,860 --> 00:56:47,310 And 4 here, we're using the 4. 796 00:56:47,310 --> 00:56:49,220 It's an even number, right? 797 00:56:49,220 --> 00:56:52,710 So you have this over here. 798 00:56:52,710 --> 00:57:08,620 So having a C4 hypothesis and also knowing what lambda 1 is 799 00:57:08,620 --> 00:57:11,230 allows you to control the other lambdas. 800 00:57:22,530 --> 00:57:27,500 See, lambda 1 cannot be much greater than pn. 801 00:57:27,500 --> 00:57:29,270 Also comes out of the same calculation. 802 00:57:29,270 --> 00:57:30,246 Yep. 803 00:57:30,246 --> 00:57:36,102 AUDIENCE: So [INAUDIBLE] number 1 equal to [INAUDIBLE]?? 804 00:57:36,102 --> 00:57:39,255 PROFESSOR: Yeah, thank you. 805 00:57:39,255 --> 00:57:40,640 Yeah, so there's a correction. 806 00:57:40,640 --> 00:57:43,360 So lambda 1 is-- 807 00:57:43,360 --> 00:57:45,160 so in other words, the little o is always 808 00:57:45,160 --> 00:57:47,470 respect to the constant density. 809 00:57:53,288 --> 00:57:53,788 OK, yeah. 810 00:57:53,788 --> 00:57:54,288 Question. 811 00:57:54,288 --> 00:57:58,162 AUDIENCE: You said in the eigenvalue implies C4, 812 00:57:58,162 --> 00:58:01,515 you somewhere also used the lower bound to be proved 813 00:58:01,515 --> 00:58:02,237 [INAUDIBLE]. 814 00:58:02,237 --> 00:58:02,820 PROFESSOR: OK. 815 00:58:02,820 --> 00:58:06,800 So the question is in eigenvalue implies C4, 816 00:58:06,800 --> 00:58:08,480 it says something about the lower bound. 817 00:58:08,480 --> 00:58:09,600 So I'm not saying that. 818 00:58:09,600 --> 00:58:15,760 So as written over here, this is what we have proved. 819 00:58:15,760 --> 00:58:18,220 But when you think about the pseudorandomness condition 820 00:58:18,220 --> 00:58:22,150 for C4, it shouldn't be just that the number of C4 count 821 00:58:22,150 --> 00:58:23,740 is at most something. 822 00:58:23,740 --> 00:58:26,860 It should be that it equals to that, 823 00:58:26,860 --> 00:58:29,770 which would be implied by the C4 condition itself, 824 00:58:29,770 --> 00:58:33,180 because we know, always, it is the case that a C4 count is 825 00:58:33,180 --> 00:58:36,400 at least what it is compared to the random case. 826 00:58:43,380 --> 00:58:47,530 So just one more thing I said was that lambda 1, 827 00:58:47,530 --> 00:58:56,290 you also know that it is at most pn plus little n, because-- 828 00:59:02,286 --> 00:59:04,266 OK. 829 00:59:04,266 --> 00:59:04,766 Yeah. 830 00:59:10,240 --> 00:59:13,140 So this finishes the proof of the Chung-Graham-Wilson theorem 831 00:59:13,140 --> 00:59:14,740 on quasi-random graphs. 832 00:59:14,740 --> 00:59:16,960 We stated all of these hypotheses, 833 00:59:16,960 --> 00:59:18,940 and they are all equivalent to each other. 834 00:59:18,940 --> 00:59:22,030 And I want to emphasize, again, the most surprising one is 835 00:59:22,030 --> 00:59:27,610 that C4 implies everything else, that a fairly seemingly 836 00:59:27,610 --> 00:59:30,040 weak condition, this just having the correct number 837 00:59:30,040 --> 00:59:33,670 of copies of labeled C4s, is enough to guarantee 838 00:59:33,670 --> 00:59:37,060 all of these other much more complicated looking conditions. 839 00:59:37,060 --> 00:59:40,510 And in particular, just having the C4 count correct 840 00:59:40,510 --> 00:59:46,810 implies that the counts of every other graph H is correct. 841 00:59:46,810 --> 00:59:49,830 Now, one thing I want to stress is 842 00:59:49,830 --> 00:59:52,560 that the Chung-Graham-Wilson theorem is really 843 00:59:52,560 --> 00:59:54,540 about dense graphs. 844 01:00:03,230 --> 01:00:05,335 And by dense, here, I mean p constant. 845 01:00:08,940 --> 01:00:11,010 Of course, the theorem as stated is true 846 01:00:11,010 --> 01:00:13,650 if you let p equal to 0. 847 01:00:13,650 --> 01:00:17,140 So there, I said p strictly between 0 and 1. 848 01:00:17,140 --> 01:00:22,710 But it is also OK if you let p be equal to 0. 849 01:00:25,470 --> 01:00:29,370 You don't get such interesting theorems, but it is still true. 850 01:00:32,065 --> 01:00:34,440 But for sparse graphs, what you really want to care about 851 01:00:34,440 --> 01:00:40,230 is approximations of the correct order of magnitude. 852 01:00:40,230 --> 01:00:44,660 So what I mean is that you can write down 853 01:00:44,660 --> 01:00:53,910 some sparse analogs for p going to 0, 854 01:00:53,910 --> 01:00:59,505 so p as a function of n going to 0 as n goes to infinity. 855 01:01:02,912 --> 01:01:04,870 So let me just write down a couple of examples, 856 01:01:04,870 --> 01:01:05,900 but I won't do all of them. 857 01:01:05,900 --> 01:01:07,760 You can imagine what they should look like. 858 01:01:07,760 --> 01:01:13,935 So DISC should say this quantity over here. 859 01:01:13,935 --> 01:01:20,200 And the discrepancy condition is little o 860 01:01:20,200 --> 01:01:26,080 of pn squared, because pn squared is the edge 861 01:01:26,080 --> 01:01:27,080 density overall. 862 01:01:27,080 --> 01:01:29,680 So that's the quantity you should compare against and not 863 01:01:29,680 --> 01:01:30,820 n squared. 864 01:01:30,820 --> 01:01:33,250 If you're comparing n squared, you're 865 01:01:33,250 --> 01:01:36,220 cheating, because n squared is much bigger 866 01:01:36,220 --> 01:01:39,160 than the actual edge density. 867 01:01:39,160 --> 01:01:48,830 Likewise, the number of labeled copies of H is-- 868 01:01:51,550 --> 01:01:54,880 I want to put the little o 1 plus little 869 01:01:54,880 --> 01:02:10,750 in front, so instead of plus little o of n to the H 870 01:02:10,750 --> 01:02:11,250 at the end. 871 01:02:14,230 --> 01:02:15,660 So you understand the difference. 872 01:02:15,660 --> 01:02:19,205 So for sparse, this is the correct normalization 873 01:02:19,205 --> 01:02:20,830 that you should have, when p is allowed 874 01:02:20,830 --> 01:02:23,770 to go to 0 as a function of n. 875 01:02:23,770 --> 01:02:27,180 And you can write down all of these conditions, right? 876 01:02:27,180 --> 01:02:28,710 I'm not saying there's a theorem. 877 01:02:28,710 --> 01:02:30,860 You can write out all these conditions. 878 01:02:30,860 --> 01:02:35,220 And you can ask, is there also some notion of equivalence? 879 01:02:35,220 --> 01:02:37,260 So are these corresponding conditions 880 01:02:37,260 --> 01:02:39,400 also equivalent to each other? 881 01:02:39,400 --> 01:02:42,370 And the answer is emphatically no, absolutely not. 882 01:02:42,370 --> 01:02:50,920 So all of these equivalents fail for sparse. 883 01:02:50,920 --> 01:02:52,198 Some of them are still true. 884 01:02:52,198 --> 01:02:53,740 Some of the easier ones that we did-- 885 01:02:53,740 --> 01:02:56,920 for example, the two versions of DISC are equivalent. 886 01:02:56,920 --> 01:02:58,120 That's still OK. 887 01:02:58,120 --> 01:03:02,860 And some of these calculations involving Cauchy-Schwarz 888 01:03:02,860 --> 01:03:06,250 are mostly still OK. 889 01:03:06,250 --> 01:03:09,370 But the one that really fails is the counting lemma. 890 01:03:21,360 --> 01:03:24,090 And let me explain why with an example. 891 01:03:24,090 --> 01:03:27,450 So I want to give you an example of a graph which 892 01:03:27,450 --> 01:03:36,280 looks pseudorandom in the sense of DISC but has no, 893 01:03:36,280 --> 01:03:38,615 let's say, C3 count. 894 01:03:38,615 --> 01:03:42,180 It also has no C4 count, but it has no-- 895 01:03:42,180 --> 01:03:44,929 has the clean, correct number of triangles. 896 01:03:49,820 --> 01:03:51,460 So what's this example? 897 01:03:51,460 --> 01:03:59,800 So let p be some number which is little o of 1 over root n 898 01:03:59,800 --> 01:04:04,920 so some decaying quantity with n. 899 01:04:04,920 --> 01:04:06,505 And let's consider Gnp. 900 01:04:10,820 --> 01:04:14,120 Well, how many triangles do we expect in Gnp? 901 01:04:14,120 --> 01:04:18,860 So let's think of p as just slightly below 1 over root n. 902 01:04:18,860 --> 01:04:25,430 So the number of triangles in Gnp in expectation is-- 903 01:04:28,520 --> 01:04:30,975 so that's the expected number. 904 01:04:30,975 --> 01:04:32,600 And you should expect the actual number 905 01:04:32,600 --> 01:04:35,970 to be roughly around that. 906 01:04:35,970 --> 01:04:43,600 But on the other hand, the number of edges 907 01:04:43,600 --> 01:04:48,950 is also expected to be this quantity here. 908 01:04:48,950 --> 01:04:51,190 And you expect that the actual number of edges 909 01:04:51,190 --> 01:04:53,930 to be very close to it. 910 01:04:53,930 --> 01:04:59,090 But p is chosen so that the number of triangles 911 01:04:59,090 --> 01:05:04,230 is significantly smaller than the number of edges, 912 01:05:04,230 --> 01:05:09,630 so asymptotically smaller, fewer copies of triangles than edges. 913 01:05:09,630 --> 01:05:17,510 So what we can do now is remove an edge 914 01:05:17,510 --> 01:05:28,270 from each copy of a triangle in this Gnp. 915 01:05:33,080 --> 01:05:46,630 We removed a tiny fraction of edges, 916 01:05:46,630 --> 01:05:48,588 because the number of triangles is much less 917 01:05:48,588 --> 01:05:49,630 than the number of edges. 918 01:05:49,630 --> 01:05:52,370 We removed a tiny fraction of edges. 919 01:05:52,370 --> 01:05:55,560 And as a result, we do not change the discrepancy 920 01:05:55,560 --> 01:06:01,940 condition up to a small error. 921 01:06:01,940 --> 01:06:04,910 So the discrepancy condition still holds. 922 01:06:08,780 --> 01:06:10,850 However, the graph has no more triangles. 923 01:06:27,600 --> 01:06:31,130 So you have this pseudorandom graph in one sense-- 924 01:06:31,130 --> 01:06:32,840 namely, of having a discrepancy-- 925 01:06:32,840 --> 01:06:35,510 but fails to be pseudorandom in a different sense-- 926 01:06:35,510 --> 01:06:38,050 namely, it has no triangles. 927 01:06:38,050 --> 01:06:39,520 Yep. 928 01:06:39,520 --> 01:06:41,480 AUDIENCE: Do the conditions C4 and codegree 929 01:06:41,480 --> 01:06:45,410 also hold here-- so the issue being from DISC to count? 930 01:06:45,410 --> 01:06:47,450 PROFESSOR: Question, do the conditions C4 931 01:06:47,450 --> 01:06:49,130 and codegree still hold here? 932 01:06:49,130 --> 01:06:51,890 Basically, downstream is OK, but upstream is not. 933 01:06:54,550 --> 01:06:56,830 So we can go from C4 to codegree to DISC. 934 01:06:56,830 --> 01:07:00,120 But you can't go upward. 935 01:07:00,120 --> 01:07:07,670 And understanding how to rectify the situation, perhaps adding 936 01:07:07,670 --> 01:07:11,210 additional hypotheses to make this true so 937 01:07:11,210 --> 01:07:14,660 that you could have counting lemmas for triangles 938 01:07:14,660 --> 01:07:16,790 and other graphs and sparser graphs, 939 01:07:16,790 --> 01:07:19,365 that's an important topic. 940 01:07:19,365 --> 01:07:20,990 And this is something that I'll discuss 941 01:07:20,990 --> 01:07:23,870 at greater length in not next lecture, 942 01:07:23,870 --> 01:07:25,280 but the one after that. 943 01:07:25,280 --> 01:07:28,970 And this is, in fact, related to the Green-Tao theorem, 944 01:07:28,970 --> 01:07:32,870 which allows you to approve Szemerédi's theorem among 945 01:07:32,870 --> 01:07:33,490 the primes. 946 01:07:33,490 --> 01:07:37,040 The primes contain arbitrarily long arithmetic progressions, 947 01:07:37,040 --> 01:07:40,400 because the primes are also a sparse set. 948 01:07:40,400 --> 01:07:42,260 So it has density going to 0. 949 01:07:42,260 --> 01:07:44,360 It's density decaying, like, 1 over log n, 950 01:07:44,360 --> 01:07:47,220 according to prime number theorem. 951 01:07:47,220 --> 01:07:49,610 But you want to do regularity method. 952 01:07:49,610 --> 01:07:53,020 So you have to face this kind of issues. 953 01:07:53,020 --> 01:07:57,150 So we'll discuss that more at length in a couple of lectures. 954 01:07:57,150 --> 01:08:00,112 But for now, just a warning that everything here 955 01:08:00,112 --> 01:08:01,320 is really about dense graphs. 956 01:08:04,160 --> 01:08:06,830 The next thing I want to discuss is 957 01:08:06,830 --> 01:08:10,449 an elaboration of what happens to these eigenvalue conditions. 958 01:08:16,939 --> 01:08:19,649 So for dense graphs, in some sense, 959 01:08:19,649 --> 01:08:21,520 everything's very clear from this theorem. 960 01:08:21,520 --> 01:08:23,319 Once you have this, theorem, they're all equivalent. 961 01:08:23,319 --> 01:08:24,640 You can go back and forth. 962 01:08:24,640 --> 01:08:26,950 And you lose a little bit of epsilon here and there, 963 01:08:26,950 --> 01:08:28,617 but everything is more or less the same. 964 01:08:28,617 --> 01:08:30,617 But if you go to sparser world, then you really 965 01:08:30,617 --> 01:08:31,825 need to be much more careful. 966 01:08:31,825 --> 01:08:33,450 And we need to think about other tools. 967 01:08:36,397 --> 01:08:37,939 And so the remainder of today, I want 968 01:08:37,939 --> 01:08:41,990 to just discuss one fairly simple but powerful tool 969 01:08:41,990 --> 01:08:46,069 relating eigenvalues on one hand and the discrepancy condition 970 01:08:46,069 --> 01:08:47,936 on the other hand. 971 01:08:47,936 --> 01:08:48,649 All right. 972 01:08:48,649 --> 01:08:51,109 So you can go from eigenvalue to discrepancy 973 01:08:51,109 --> 01:08:52,370 by going down this chain. 974 01:08:52,370 --> 01:08:56,470 But actually, there's a much quicker route. 975 01:08:56,470 --> 01:08:58,490 And this is known as the expander mixing lemma. 976 01:09:10,950 --> 01:09:15,450 For simplicity and really will make our life much simpler, 977 01:09:15,450 --> 01:09:18,866 we're only going to consider d-regular graphs. 978 01:09:18,866 --> 01:09:22,859 So here, d-regular means every vertex is degree d. 979 01:09:22,859 --> 01:09:25,899 Same word, but different meaning from epsilon regular. 980 01:09:25,899 --> 01:09:28,689 And unfortunately, that's just the way it is. 981 01:09:28,689 --> 01:09:33,590 So d regular, and we're going to have n vertices. 982 01:09:33,590 --> 01:09:43,170 And the adjacency matrix has eigenvalues lambda 1, lambda 2, 983 01:09:43,170 --> 01:09:46,740 and so on, arranged in decreasing order. 984 01:09:50,540 --> 01:09:56,600 Let me write lambda as the maximum 985 01:09:56,600 --> 01:09:59,780 in absolute value of the eigenvalues 986 01:09:59,780 --> 01:10:02,910 except for the top one. 987 01:10:02,910 --> 01:10:05,550 In particular, this is either the absolute value 988 01:10:05,550 --> 01:10:12,030 of the second one or the last one. 989 01:10:12,030 --> 01:10:14,120 As I mentioned earlier, the top eigenvalue 990 01:10:14,120 --> 01:10:19,020 is necessarily d, because you have all-ones vector 991 01:10:19,020 --> 01:10:20,328 as an eigenvector. 992 01:10:22,960 --> 01:10:30,620 So the expander mixing lemma says that if I look at two 993 01:10:30,620 --> 01:10:38,760 vertex subsets, the number of edges between them compared 994 01:10:38,760 --> 01:10:40,810 to what you would expect in a random case-- 995 01:10:40,810 --> 01:10:42,860 so just like in the disc setting, but here, 996 01:10:42,860 --> 01:10:45,620 the correct density I should put is d over n-- 997 01:10:48,460 --> 01:10:55,810 this quantity is upper bounded by lambda times the root 998 01:10:55,810 --> 01:10:57,670 of the product of x and y. 999 01:11:03,400 --> 01:11:07,780 So in particular, if this lambda-- 1000 01:11:07,780 --> 01:11:11,550 so everything except for the top eigenvalue-- is small, 1001 01:11:11,550 --> 01:11:15,250 then this discrepancy should be small. 1002 01:11:15,250 --> 01:11:17,980 And you can verify with what we did, that it's consistent, 1003 01:11:17,980 --> 01:11:19,680 what we just did. 1004 01:11:23,430 --> 01:11:25,640 All right. 1005 01:11:25,640 --> 01:11:27,060 So let's prove the expander mixing 1006 01:11:27,060 --> 01:11:31,820 lemma, which is pretty simple given what we've discussed 1007 01:11:31,820 --> 01:11:35,540 so far, relating-- so there was this spectral characterization 1008 01:11:35,540 --> 01:11:38,067 up there of the top eigenvalue. 1009 01:11:40,810 --> 01:11:45,355 So we can let J be the all-ones matrix. 1010 01:11:50,722 --> 01:11:53,460 So let J be the all-ones matrix. 1011 01:11:53,460 --> 01:11:58,980 And we know that the all-ones vector 1012 01:11:58,980 --> 01:12:06,870 is an eigenvector of the adjacency matrix of G 1013 01:12:06,870 --> 01:12:10,920 with eigenvalue d. 1014 01:12:10,920 --> 01:12:16,860 So the eigendecomposition of J is also the all-ones vector 1015 01:12:16,860 --> 01:12:18,460 and its complement. 1016 01:12:18,460 --> 01:12:28,690 So we now see that A sub G minus d over nJ 1017 01:12:28,690 --> 01:12:39,400 has the same eigenvectors as AG. 1018 01:12:39,400 --> 01:12:41,990 So you can choose the eigenvectors for that. 1019 01:12:41,990 --> 01:12:44,400 It's the same set of eigenvectors. 1020 01:12:44,400 --> 01:12:46,200 Of course, we consider this quantity here, 1021 01:12:46,200 --> 01:12:49,080 because this is exactly the quantity that comes up 1022 01:12:49,080 --> 01:12:51,700 in this expression once we hit it 1023 01:12:51,700 --> 01:12:56,817 by characteristic vectors of subsets from left and right. 1024 01:12:56,817 --> 01:12:57,771 All right. 1025 01:13:02,070 --> 01:13:07,140 So what are the eigenvalues? 1026 01:13:14,700 --> 01:13:18,780 So A previously had eigenvalues lambda 1 through lambda n. 1027 01:13:18,780 --> 01:13:23,100 But now the top one gets chopped down to 0. 1028 01:13:31,110 --> 01:13:32,970 So you can check this explicitly. 1029 01:13:32,970 --> 01:13:37,370 So you can check this explicitly by checking 1030 01:13:37,370 --> 01:13:43,210 that if you take this matrix multiplied 1031 01:13:43,210 --> 01:13:47,170 by the all-ones vector, you get 0. 1032 01:13:47,170 --> 01:13:54,400 And if you have a eigenvector-eigenvalue pair, 1033 01:13:54,400 --> 01:14:00,740 then hitting this by any of the other ones 1034 01:14:00,740 --> 01:14:07,150 gets you the same as in A, because you have 1035 01:14:07,150 --> 01:14:08,530 this orthogonality condition. 1036 01:14:08,530 --> 01:14:10,238 All the other eigenvectors are orthogonal 1037 01:14:10,238 --> 01:14:13,410 to the all-ones vector. 1038 01:14:13,410 --> 01:14:14,190 All right. 1039 01:14:14,190 --> 01:14:20,980 So now we apply the Courant-Fischer criteria, 1040 01:14:20,980 --> 01:14:27,500 which tells us that the number in this discrepancy 1041 01:14:27,500 --> 01:14:52,090 quantity, which we can write in terms of this matrix, 1042 01:14:52,090 --> 01:14:59,440 it is upper bounded by the product of the length 1043 01:14:59,440 --> 01:15:02,920 of these two vectors, x and y, multiplied 1044 01:15:02,920 --> 01:15:04,930 by the spectral norm. 1045 01:15:13,570 --> 01:15:16,270 So I'm not quite using the version up there, 1046 01:15:16,270 --> 01:15:18,760 but I'm using the spectral norm version, 1047 01:15:18,760 --> 01:15:20,060 which we discussed last time. 1048 01:15:20,060 --> 01:15:21,510 It's essentially the one up there, 1049 01:15:21,510 --> 01:15:24,970 but you allow not just single x but x and y. 1050 01:15:24,970 --> 01:15:28,000 And that corresponds to the largest eigenvalue 1051 01:15:28,000 --> 01:15:32,150 in absolute value, which we see that. 1052 01:15:32,150 --> 01:15:34,650 It's at most lambda. 1053 01:15:34,650 --> 01:15:42,992 So at most lambda times size of x, size of y square root. 1054 01:15:45,800 --> 01:15:50,190 And that finishes the proof of the expander mixing lemma. 1055 01:15:54,790 --> 01:15:59,170 So the moral here is that, just like what we saw earlier 1056 01:15:59,170 --> 01:16:03,470 in the dense case but for any parameters-- so here, 1057 01:16:03,470 --> 01:16:05,290 it's a very clean statement. 1058 01:16:05,290 --> 01:16:07,710 You can even have done the degree graphs. 1059 01:16:07,710 --> 01:16:11,110 d could be a constant. 1060 01:16:11,110 --> 01:16:15,710 If lambda is small compared to d, then 1061 01:16:15,710 --> 01:16:18,250 you have this discrepancy condition. 1062 01:16:18,250 --> 01:16:21,080 And the reason why this is called an expander mixing lemma 1063 01:16:21,080 --> 01:16:24,020 is that there's this notion of expanders, 1064 01:16:24,020 --> 01:16:27,380 which is not quite the same but very intimately related 1065 01:16:27,380 --> 01:16:29,300 to pseudorandom graphs. 1066 01:16:29,300 --> 01:16:33,020 So one property of pseudorandom graphs that is quite useful-- 1067 01:16:33,020 --> 01:16:34,692 in particular, in computer science-- 1068 01:16:34,692 --> 01:16:36,650 is that if you take a small subset of vertices, 1069 01:16:36,650 --> 01:16:38,480 it has lots of neighbors. 1070 01:16:38,480 --> 01:16:40,640 So the graph is now somehow clustered 1071 01:16:40,640 --> 01:16:42,830 into a few local pieces. 1072 01:16:42,830 --> 01:16:45,790 So there's lots of expansion. 1073 01:16:45,790 --> 01:16:50,140 And that's something that you can guarantee 1074 01:16:50,140 --> 01:16:54,130 using the expander mixing lemma, that you have lots of-- you 1075 01:16:54,130 --> 01:16:56,080 take a small subset of vertices. 1076 01:16:56,080 --> 01:16:58,510 You can expand outward. 1077 01:16:58,510 --> 01:17:00,990 So graphs with that specific property, 1078 01:17:00,990 --> 01:17:02,490 taking a small subset of vertices 1079 01:17:02,490 --> 01:17:04,440 always gets you lots of neighbors, 1080 01:17:04,440 --> 01:17:06,030 are called expander graphs. 1081 01:17:06,030 --> 01:17:09,960 And these graphs play an important role, in particular, 1082 01:17:09,960 --> 01:17:12,780 in computer science in designing algorithms and proving 1083 01:17:12,780 --> 01:17:15,210 complexity results and so on but also play 1084 01:17:15,210 --> 01:17:17,715 important roles in graph theory and combinatorics. 1085 01:17:20,330 --> 01:17:25,310 Well, next time, we'll address a few questions which are along 1086 01:17:25,310 --> 01:17:33,730 the lines of, one, how small can lambda be as a function of d? 1087 01:17:33,730 --> 01:17:34,570 So here is this. 1088 01:17:34,570 --> 01:17:38,500 If lambda's small compared to d, then you have this discrepancy. 1089 01:17:38,500 --> 01:17:43,480 But if d is, let's say, a million, 1090 01:17:43,480 --> 01:17:46,120 how small can lambda be? 1091 01:17:46,120 --> 01:17:47,980 That's one question. 1092 01:17:47,980 --> 01:17:52,000 Another question is, considering everything 1093 01:17:52,000 --> 01:17:59,720 that we've said so far, what can we say about, 1094 01:17:59,720 --> 01:18:03,110 let's say, the relationship between some 1095 01:18:03,110 --> 01:18:07,580 of these conditions for sparse graphs 1096 01:18:07,580 --> 01:18:11,360 but that are somewhat special-- for example, kd graphs 1097 01:18:11,360 --> 01:18:13,970 or vertex-transitive graphs? 1098 01:18:13,970 --> 01:18:15,950 And it turns out some of these relations 1099 01:18:15,950 --> 01:18:19,360 are also equivalent to each other.