WEBVTT 00:00:16.602 --> 00:00:18.810 YUFEI ZHAO: We've been spending the past few lectures 00:00:18.810 --> 00:00:21.750 discussing Szemeredi's Regularity Lemma. 00:00:21.750 --> 00:00:24.060 And one of the first applications 00:00:24.060 --> 00:00:26.730 that we discussed of the Regularity Lemma 00:00:26.730 --> 00:00:29.580 is the triangle removal Lemma. 00:00:29.580 --> 00:00:31.740 So today, I want to revisit this topic 00:00:31.740 --> 00:00:34.860 and show you a strengthening of the Removal Lemma 00:00:34.860 --> 00:00:37.860 for which new regularity techniques are needed. 00:00:42.030 --> 00:00:45.090 But first, recall the graph removal Lemma. 00:00:57.690 --> 00:01:03.590 In the graph removal Lemma, we have that for every graph H 00:01:03.590 --> 00:01:10.040 and epsilon bigger than zero, there exists some delta such 00:01:10.040 --> 00:01:24.730 that if an N vertex graph has fewer than delta 00:01:24.730 --> 00:01:35.270 and to the number of vertices of H, many copies of H, 00:01:35.270 --> 00:01:53.620 then it can be made H-free by removing fewer than epsilon N 00:01:53.620 --> 00:01:54.520 squared edges. 00:01:58.230 --> 00:02:01.065 Even in the case when H is a triangle, when 00:02:01.065 --> 00:02:02.940 this is called a triangle removal Lemma, even 00:02:02.940 --> 00:02:06.330 in that case, basically the regularity method 00:02:06.330 --> 00:02:08.280 is more or less the only way that we currently 00:02:08.280 --> 00:02:10.380 know how to prove this theorem. 00:02:10.380 --> 00:02:14.310 So we saw this a few lectures ago. 00:02:14.310 --> 00:02:16.320 What I would like to discuss today 00:02:16.320 --> 00:02:19.650 is a variant of this result where 00:02:19.650 --> 00:02:24.030 instead of considering copies of H, 00:02:24.030 --> 00:02:29.460 we're now considering induced copies of H. OK? 00:02:29.460 --> 00:02:35.010 So this is the induced graph removal Lemma 00:02:35.010 --> 00:02:39.780 where the only difference is that the hypothesis is now 00:02:39.780 --> 00:02:44.910 going to be changed to induced copies of H. 00:02:44.910 --> 00:02:46.590 And the conclusion is that you can 00:02:46.590 --> 00:02:51.330 make the graph induced H-free. 00:02:51.330 --> 00:02:52.920 So let me remind you, the difference 00:02:52.920 --> 00:02:55.860 between the induced graph subgraph 00:02:55.860 --> 00:02:59.040 and the usual subgraph. 00:02:59.040 --> 00:03:12.220 So we say that H is an induced copy of G, induced subgraph 00:03:12.220 --> 00:03:42.842 of G. If one can obtain H from G by deleting vertices of G. 00:03:42.842 --> 00:03:44.300 You're not allowed to delete edges, 00:03:44.300 --> 00:03:46.910 but only allowed to delete vertices. 00:03:46.910 --> 00:03:55.520 So in other words, the four cycle 00:03:55.520 --> 00:04:05.210 is not an induced subgraph because, well, 00:04:05.210 --> 00:04:08.270 if you select four vertices, you don't generate this four cycle. 00:04:08.270 --> 00:04:09.290 You get extra edges. 00:04:09.290 --> 00:04:12.060 So it is a subgraph, but not an induced subgraph. 00:04:20.050 --> 00:04:23.290 So it is a theorem, the induced graph removal Lemma. 00:04:23.290 --> 00:04:24.790 So it's a theorem, and let's discuss 00:04:24.790 --> 00:04:26.220 how we may prove that theorem. 00:04:26.220 --> 00:04:26.720 Question. 00:04:30.550 --> 00:04:33.000 OK, question is, why is it stronger than the graph removal 00:04:33.000 --> 00:04:34.810 lemma? 00:04:34.810 --> 00:04:40.060 So it's not stronger, but we'll see the relationship 00:04:40.060 --> 00:04:41.270 between the two. 00:04:41.270 --> 00:04:46.330 So I claim that it is more difficult to do this theorem. 00:04:46.330 --> 00:04:48.940 Any more questions? 00:04:48.940 --> 00:04:57.320 So let's pretend for a second that whatever's in here 00:04:57.320 --> 00:04:58.605 is not quite true. 00:04:58.605 --> 00:04:59.480 So here's an example. 00:05:06.110 --> 00:05:14.360 For example, if your H is three isolated vertices. 00:05:14.360 --> 00:05:15.890 So what is that saying? 00:05:15.890 --> 00:05:18.230 We're looking at copies of H which 00:05:18.230 --> 00:05:19.700 are three isolated vertices. 00:05:19.700 --> 00:05:26.330 So really you are looking at triangles in g complement. 00:05:26.330 --> 00:05:30.040 So this is exactly the triangle removal lemma 00:05:30.040 --> 00:05:35.300 in the complement of g, but you can't get rid of these guys 00:05:35.300 --> 00:05:36.680 by removing edges. 00:05:36.680 --> 00:05:38.360 So we need to make the modification 00:05:38.360 --> 00:05:40.580 where instead of removing these edges, 00:05:40.580 --> 00:05:48.080 we need to both remove and add by adding or deleting. 00:05:51.490 --> 00:05:52.758 So maybe at the same time. 00:05:52.758 --> 00:05:55.050 So you're allowed to add some edges, delete some edges. 00:05:55.050 --> 00:05:58.788 But in total, you change no more than epsilon n squared edges. 00:05:58.788 --> 00:06:01.080 So those are sometimes also known as the edit distance. 00:06:05.690 --> 00:06:08.530 You're allowed to change edges. 00:06:08.530 --> 00:06:10.810 So you can add edges and delete edges. 00:06:16.090 --> 00:06:17.940 Any questions about the statement? 00:06:20.880 --> 00:06:24.550 All right, so let's think about how would you 00:06:24.550 --> 00:06:27.760 prove this result following the proof that we did 00:06:27.760 --> 00:06:30.660 for the triangle removal lemma. 00:06:30.660 --> 00:06:33.340 So let's pretend that we go through this proof 00:06:33.340 --> 00:06:35.230 and think about what could go wrong. 00:06:35.230 --> 00:06:37.840 So remember in the application of the removal lemma, 00:06:37.840 --> 00:06:40.250 so the recipe has three steps. 00:06:40.250 --> 00:06:42.454 The first step we do a partition. 00:06:45.420 --> 00:06:49.220 So we partition applying Szemeredi's regularity lemma 00:06:49.220 --> 00:06:50.890 to this partition. 00:06:50.890 --> 00:06:56.400 And the second step is do a cleaning, 00:06:56.400 --> 00:06:59.820 and the two key things that happen in the cleaning 00:06:59.820 --> 00:07:11.130 is we remove low density pairs of parts and irregular pairs. 00:07:13.940 --> 00:07:16.380 And the third step we claim that once we 00:07:16.380 --> 00:07:19.890 do the cleaning, once we remove those edges, 00:07:19.890 --> 00:07:22.230 the resulting graphs should be H3. 00:07:22.230 --> 00:07:26.970 Because if we're not H3, then by considering the vertex parts 00:07:26.970 --> 00:07:29.460 where H lie and applying the counting lemma, 00:07:29.460 --> 00:07:33.430 you can generate many more copies of H. 00:07:33.430 --> 00:07:35.550 So these were the three main steps in the proof 00:07:35.550 --> 00:07:37.530 of the triangle removal lemma. 00:07:37.530 --> 00:07:40.050 So let's see what happens when we 00:07:40.050 --> 00:07:43.530 try to apply this strategy to the induced version. 00:07:43.530 --> 00:07:47.760 I mean, the partition you still do the regularity partition. 00:07:47.760 --> 00:07:50.370 Nothing really changes there. 00:07:50.370 --> 00:07:54.420 So let's see in the cleaning step what happens. 00:07:54.420 --> 00:07:56.460 For low density pairs-- 00:07:56.460 --> 00:07:59.250 well, so now we need to think about not just low density 00:07:59.250 --> 00:08:01.700 pairs, but also high density pairs. 00:08:01.700 --> 00:08:04.770 Because in the induced, we think about edges and non-edges 00:08:04.770 --> 00:08:05.950 at the same time. 00:08:05.950 --> 00:08:09.030 So you might think of a strategy which is like the edge 00:08:09.030 --> 00:08:11.460 density is less than n. 00:08:11.460 --> 00:08:15.480 So less than epsilon, then you remove all those edges. 00:08:15.480 --> 00:08:18.450 And if the edge density is bigger than 1 plus epsilon, 00:08:18.450 --> 00:08:22.100 then you add all of those edges in. 00:08:22.100 --> 00:08:23.860 So this is the natural generalization 00:08:23.860 --> 00:08:25.820 of our strategy for triangle removal 00:08:25.820 --> 00:08:27.400 lemma for the induced setting. 00:08:27.400 --> 00:08:30.570 So so far, everything's still OK. 00:08:30.570 --> 00:08:33.490 But now what would you do for the irregular pairs? 00:08:37.970 --> 00:08:41.030 That's problematic. 00:08:41.030 --> 00:08:43.820 Previously for triangle removal lemma, 00:08:43.820 --> 00:08:47.270 we just said if a pair is irregular, get rid of that pair 00:08:47.270 --> 00:08:51.440 and it will never show up in the counting stage. 00:08:51.440 --> 00:08:54.540 But that strategy no longer works. 00:08:54.540 --> 00:08:59.870 Because for example, if your graph H being counted 00:08:59.870 --> 00:09:08.630 is this here, you do the regularity partition, 00:09:08.630 --> 00:09:12.550 and one of your pairs is irregular. 00:09:12.550 --> 00:09:16.240 So you, let's say, get rid of all those edges in between. 00:09:16.240 --> 00:09:20.680 Then maybe you have some embedding of H 00:09:20.680 --> 00:09:25.905 where you are going to use the removed edges. 00:09:30.010 --> 00:09:34.040 And now you don't have a counting lemma. 00:09:34.040 --> 00:09:41.070 You cannot say, I found this copy of H in my changed graph. 00:09:41.070 --> 00:09:43.650 And by the counting lemma I could get many copies of H 00:09:43.650 --> 00:09:47.620 because you have no control over this irregular pair anymore. 00:09:47.620 --> 00:09:50.880 So the fact that you have to add and remove 00:09:50.880 --> 00:09:52.660 makes it unclear what to do here, 00:09:52.660 --> 00:09:54.810 and this is a big obstacle in the application 00:09:54.810 --> 00:09:59.370 of the regularity lemma to the induced removal lemma 00:09:59.370 --> 00:10:02.310 application. 00:10:02.310 --> 00:10:04.350 Any questions about this obstacle? 00:10:08.550 --> 00:10:10.940 So make sure you understand why this is an issue. 00:10:10.940 --> 00:10:13.840 Otherwise you won't really appreciate 00:10:13.840 --> 00:10:16.550 what will happen next. 00:10:16.550 --> 00:10:21.630 So somehow we need to find some kind of regularity partition 00:10:21.630 --> 00:10:25.170 to get no irregular pairs. 00:10:25.170 --> 00:10:29.780 So the question is, is there a way 00:10:29.780 --> 00:10:36.580 to partition so that there are no irregular pairs? 00:10:42.130 --> 00:10:44.530 For those of you who have started your homework 00:10:44.530 --> 00:10:49.150 problem on time, you realize that the answer is no. 00:10:49.150 --> 00:10:50.770 So one of the homework problems is 00:10:50.770 --> 00:10:53.700 for you to show that for the specific graph known 00:10:53.700 --> 00:10:54.700 as the half graph. 00:10:58.110 --> 00:11:00.588 So there was an example in homework 00:11:00.588 --> 00:11:01.630 that for the half graph-- 00:11:08.665 --> 00:11:11.800 so you'll see in the homework what this graph is-- 00:11:11.800 --> 00:11:15.050 you cannot partition it so that you get rid of all irregular 00:11:15.050 --> 00:11:15.550 pairs. 00:11:15.550 --> 00:11:18.340 Irregular pairs are necessary in the statement 00:11:18.340 --> 00:11:19.650 of regularity lemma. 00:11:22.240 --> 00:11:24.340 So what I want to show you today is a way 00:11:24.340 --> 00:11:29.400 to do what's called a strong regularity lemma in which you 00:11:29.400 --> 00:11:33.000 obtain a somewhat different consequence that will allow 00:11:33.000 --> 00:11:35.438 you to get rid of irregular pairs 00:11:35.438 --> 00:11:36.730 in the more restricted setting. 00:11:39.730 --> 00:11:42.100 So this is the issue, the irregular pairs. 00:11:48.420 --> 00:11:50.790 Before telling you what this regularity lemma is, 00:11:50.790 --> 00:11:55.050 I want to give you a small generalization 00:11:55.050 --> 00:11:58.350 of the induced graph removal lemma, or just a different way 00:11:58.350 --> 00:12:00.540 to think about the statement. 00:12:00.540 --> 00:12:03.720 And you can think of it as a colorful version instead 00:12:03.720 --> 00:12:08.970 of induced where you have edges and no edges. 00:12:08.970 --> 00:12:11.070 You can also have colored edges. 00:12:11.070 --> 00:12:14.190 So colorful removal lemma, although this name 00:12:14.190 --> 00:12:15.300 is not standard. 00:12:21.580 --> 00:12:25.610 So colorful-- so when we talk about graphs, 00:12:25.610 --> 00:12:29.040 it's colorful graph removal lemma. 00:12:29.040 --> 00:12:37.490 So for every k, r, and epsilon, there exists delta such 00:12:37.490 --> 00:12:58.020 that if curly H is a set of r edge of the complete graph 00:12:58.020 --> 00:13:00.590 on little k vertices. 00:13:00.590 --> 00:13:03.800 So edge coloring just means using r colors 00:13:03.800 --> 00:13:04.700 to color the edges. 00:13:04.700 --> 00:13:07.530 So there are no restrictions about what are allowed, 00:13:07.530 --> 00:13:08.450 what are not allowed. 00:13:08.450 --> 00:13:12.293 So just a set of possible r colorings. 00:13:12.293 --> 00:13:22.000 Then if the complete graph-- 00:13:27.823 --> 00:13:28.990 say it slightly differently. 00:13:28.990 --> 00:13:44.650 So then every r edge coloring of the complete graph 00:13:44.650 --> 00:14:02.850 on n vertices with fewer than delta fraction of its k vertex 00:14:02.850 --> 00:14:20.310 subsets, say k vertex subgraphs, belonging to the script H. 00:14:20.310 --> 00:14:34.670 So every such graph can be made curly H free by recoloring, 00:14:34.670 --> 00:14:49.710 so using the same r colors, a fewer than epsilon fraction. 00:14:49.710 --> 00:15:00.710 So less than epsilon fraction of the edges of this kn. 00:15:03.360 --> 00:15:06.540 So in particular, the version that we just stated, 00:15:06.540 --> 00:15:16.260 the induced version, so the induced graph removal lemma, 00:15:16.260 --> 00:15:31.450 is the same as having two colors and H having exactly one 00:15:31.450 --> 00:15:44.620 red-blue coloring of k of the complete graph 00:15:44.620 --> 00:15:48.780 on the same number of vertices as H. 00:15:48.780 --> 00:15:51.930 So you color red the edges and blue the non-edges, 00:15:51.930 --> 00:15:53.650 for instance. 00:15:53.650 --> 00:15:58.680 And you're saying, I want to color the big complete graph 00:15:58.680 --> 00:16:02.760 with red and blue in such a way that there are very few copies 00:16:02.760 --> 00:16:03.900 of that pattern. 00:16:03.900 --> 00:16:05.730 So then I can recolor the red and blue 00:16:05.730 --> 00:16:09.750 in a small number of places to get rid of all such patterns. 00:16:09.750 --> 00:16:11.940 So having a colored pattern somewhere 00:16:11.940 --> 00:16:14.610 in your graph in this complete graph coloring 00:16:14.610 --> 00:16:19.170 is the same as having an induced subgraph. 00:16:19.170 --> 00:16:19.750 Yeah? 00:16:19.750 --> 00:16:21.130 AUDIENCE: So after done-- 00:16:21.130 --> 00:16:23.976 like the statement after done is a really long sentence. 00:16:23.976 --> 00:16:24.560 Can I-- 00:16:24.560 --> 00:16:28.020 YUFEI ZHAO: Yeah, OK. 00:16:28.020 --> 00:16:37.380 So every r edge coloring of kn with a small number of patterns 00:16:37.380 --> 00:16:43.710 can be made h-free by recoloring a small fraction of the edges. 00:16:43.710 --> 00:16:45.630 So like in a triangle removal lemma, 00:16:45.630 --> 00:16:49.170 every graph with a small number of triangles 00:16:49.170 --> 00:16:51.540 can be made triangle-free by removing 00:16:51.540 --> 00:16:52.710 a small number of edges. 00:16:58.020 --> 00:17:01.710 Any other questions? 00:17:01.710 --> 00:17:07.079 So this is a restatement of the induced removal lemma 00:17:07.079 --> 00:17:10.349 with a bit more generality. 00:17:10.349 --> 00:17:12.960 It's OK if you like this one more or less, 00:17:12.960 --> 00:17:15.810 but let's talk about the induced version from now on. 00:17:15.810 --> 00:17:18.450 But the same proofs that I will talk about also 00:17:18.450 --> 00:17:22.710 applies to this version where you have somewhat more colors. 00:17:26.349 --> 00:17:30.940 So the variant of the regularity lemma that we'll need 00:17:30.940 --> 00:17:33.370 is known as a strong regularity lemma. 00:17:47.320 --> 00:17:49.210 To state the strong regularity lemma, 00:17:49.210 --> 00:17:52.420 let me recall a notion that came up in the proof of Szemeredi's 00:17:52.420 --> 00:17:54.820 regularity lemma. 00:17:54.820 --> 00:17:57.740 And this was the notion of an energy. 00:17:57.740 --> 00:18:04.930 So recall that if you have a partition, denoted P. So 00:18:04.930 --> 00:18:11.980 if this is a partition of the vertex set of a graph, G, 00:18:11.980 --> 00:18:17.770 and here n is the number of vertices, 00:18:17.770 --> 00:18:29.390 we defined this notion of energy to be this quantity denoted 00:18:29.390 --> 00:18:36.790 q, which is basically a squared mean of the densities 00:18:36.790 --> 00:18:42.010 between vertex parts appropriately normalized 00:18:42.010 --> 00:18:46.930 if the vertexes do not all have the same size. 00:18:52.860 --> 00:18:58.690 In the proof of Szemeredi's regularity lemma, 00:18:58.690 --> 00:19:00.760 there was an important energy increment 00:19:00.760 --> 00:19:12.250 step which says that if you have some partition p that is not 00:19:12.250 --> 00:19:24.318 epsilon regular, then there exists a refinement, Q. 00:19:24.318 --> 00:19:29.920 And this refinement has the property that Q has 00:19:29.920 --> 00:19:37.440 a small number of pieces, or not too large as a function of P 00:19:37.440 --> 00:19:41.170 So it's bounded at least in terms of P. 00:19:41.170 --> 00:19:48.370 But also if P is not epsilon regular, then the energy of Q 00:19:48.370 --> 00:19:55.160 is significantly larger than the energy of P. So remember, 00:19:55.160 --> 00:19:58.580 this was an important step in the proof of regularity lemma. 00:20:03.130 --> 00:20:07.380 So to state the strong regularity lemma, 00:20:07.380 --> 00:20:09.298 we need that notion of energy. 00:20:09.298 --> 00:20:11.340 And the statement of the strong regularity lemma, 00:20:11.340 --> 00:20:13.298 if you've never seen this kind of thing before, 00:20:13.298 --> 00:20:14.880 will seem a bit intimidating at first 00:20:14.880 --> 00:20:19.230 because it involves a whole sequence of parameters. 00:20:19.230 --> 00:20:20.640 But we'll get used to it. 00:20:23.880 --> 00:20:26.790 So instead of one epsilon parameter, 00:20:26.790 --> 00:20:35.945 now you have a sequence of positive epsilons. 00:20:35.945 --> 00:20:38.030 And part of the strength of this regularity lemma 00:20:38.030 --> 00:20:41.120 is that depending on the application you have in mind, 00:20:41.120 --> 00:20:44.570 you can make the sequence go to zero pretty quickly. 00:20:44.570 --> 00:20:48.810 Thereby increasing the strength of the regularity lemma. 00:20:48.810 --> 00:20:53.750 So there exists some m bound, which depends only 00:20:53.750 --> 00:21:07.960 on your epsilons such that every graph has not just one, 00:21:07.960 --> 00:21:16.170 but now we're going to get a pair of vertex partitions P 00:21:16.170 --> 00:21:22.930 and Q with the following properties. 00:21:22.930 --> 00:21:27.180 So first, P refines-- 00:21:27.180 --> 00:21:36.990 so Q refines P. So it's a pair of partitions, 00:21:36.990 --> 00:21:38.650 one refining the other. 00:21:42.400 --> 00:21:45.250 The number of parts of Q is bounded 00:21:45.250 --> 00:21:47.260 just like in the usual regularity lemma. 00:21:50.070 --> 00:21:54.960 The partition P epsilon 0 regular. 00:21:57.720 --> 00:22:02.350 And here is the new part that's the most important one. 00:22:02.350 --> 00:22:07.220 Q is very epsilon regular. 00:22:07.220 --> 00:22:09.250 So it's not just epsilon 0 regular, 00:22:09.250 --> 00:22:13.410 it's epsilon sub the number of parts of P regular. 00:22:16.330 --> 00:22:23.380 So you should think of this as extremely regular 00:22:23.380 --> 00:22:28.280 because you get to choose what the sequence of epsilon is. 00:22:28.280 --> 00:22:32.720 And finally, the energy difference between P and Q 00:22:32.720 --> 00:22:33.980 is not too big. 00:22:43.690 --> 00:22:46.830 This is the statement of the strong regularity lemma. 00:22:46.830 --> 00:22:48.750 It produces for you not just one partition, 00:22:48.750 --> 00:22:50.780 but a pair of partitions. 00:22:50.780 --> 00:22:52.800 And in this pair of partitions, you 00:22:52.800 --> 00:22:57.480 have one partition, P, which is similar to the one 00:22:57.480 --> 00:22:59.660 that we obtained from Szemeredi's regularity lemma 00:22:59.660 --> 00:23:06.150 is some epsilon 0 regular, but we also get a refinement Q. 00:23:06.150 --> 00:23:11.250 And this Q is extremely regular. 00:23:11.250 --> 00:23:19.510 So you can think that is P, then Q 00:23:19.510 --> 00:23:22.040 is an extremely regular refinement 00:23:22.040 --> 00:23:28.210 of P. Any questions about the statement 00:23:28.210 --> 00:23:29.960 of the strong regularity lemma? 00:23:33.860 --> 00:23:35.610 So the sequence of epsilons gives you 00:23:35.610 --> 00:23:39.810 flexibility on how to apply it, but let's see how to prove it. 00:23:39.810 --> 00:23:43.800 And the proof is once you understand how this works, 00:23:43.800 --> 00:23:47.070 conceptually it's pretty short. 00:23:47.070 --> 00:23:50.370 But let me do it slowly so that we can appreciate 00:23:50.370 --> 00:23:54.970 this sequence of epsilons. 00:23:54.970 --> 00:23:58.270 And the idea is that we will repeatedly apply 00:23:58.270 --> 00:23:59.570 Szemeredi's regularity lemma. 00:24:06.030 --> 00:24:08.510 So start with the regularity lemma. 00:24:08.510 --> 00:24:12.980 We'll apply it repeatedly to generate 00:24:12.980 --> 00:24:16.540 a sequence of partitions. 00:24:16.540 --> 00:24:21.620 So first, let me remind you a statement of Szemeredi's 00:24:21.620 --> 00:24:22.610 regularity lemma. 00:24:22.610 --> 00:24:24.320 This is slightly different from the one 00:24:24.320 --> 00:24:28.330 that we stated, but comes out of the same proof. 00:24:28.330 --> 00:24:33.440 So for every epsilon, there exists some m0 00:24:33.440 --> 00:24:41.500 which depends on epsilon such that for every partition P0, 00:24:41.500 --> 00:24:46.530 so starting with some partition-- 00:24:46.530 --> 00:24:48.510 so actually, let me start with just P. 00:24:48.510 --> 00:24:54.920 So if you start with some partition of the vertex set 00:24:54.920 --> 00:25:10.590 of g, there exists a refinement P prime of P into at most-- 00:25:10.590 --> 00:25:27.520 OK, so the refinement has is such that with each part of P 00:25:27.520 --> 00:25:38.960 refined into at most m0 parts such 00:25:38.960 --> 00:25:44.650 that P prime, the new partition, is epsilon regular. 00:25:52.470 --> 00:25:54.900 So this is a statement of Szemeredi's regularity lemma 00:25:54.900 --> 00:25:56.850 that we will apply repeatedly. 00:25:56.850 --> 00:25:58.870 So in the version that we've seen before, 00:25:58.870 --> 00:26:02.500 we would start with a trivial partition. 00:26:02.500 --> 00:26:05.740 And applying refinements repeatedly 00:26:05.740 --> 00:26:09.880 in the proof to get a partition into a bounded number of parts 00:26:09.880 --> 00:26:14.080 such that the final partition is epsilon regular. 00:26:14.080 --> 00:26:17.020 But instead, in the proof of the regularity lemma 00:26:17.020 --> 00:26:19.180 if you start with not a trivial partition 00:26:19.180 --> 00:26:24.100 but start with a given partition and run this exact same proof, 00:26:24.100 --> 00:26:26.380 you find this consequence. 00:26:26.380 --> 00:26:29.320 Except now you can guarantee that the final partition 00:26:29.320 --> 00:26:31.570 is a refinement of the one that you are given. 00:26:35.880 --> 00:26:42.940 So let's apply the statement, and we obtain 00:26:42.940 --> 00:26:50.640 a sequence of partitions of g-- 00:26:50.640 --> 00:26:52.460 the vertex set of g-- 00:26:52.460 --> 00:27:07.250 starting with P0 being a trivial partition, and so on. 00:27:07.250 --> 00:27:17.330 Such that each partition, each P sub i plus 1 00:27:17.330 --> 00:27:29.940 refines the previous one, and such 00:27:29.940 --> 00:27:39.370 that each P sub i plus 1 is epsilon sub a P sub i regular. 00:27:42.360 --> 00:27:45.460 So you apply the regularity lemma with parameter 00:27:45.460 --> 00:27:49.150 based on the number of parts you currently have. 00:27:49.150 --> 00:27:51.010 Applied to the current partition, 00:27:51.010 --> 00:27:56.820 you get a finer partition that's extremely regular. 00:27:56.820 --> 00:27:58.920 And you also know that the number 00:27:58.920 --> 00:28:04.350 of parts of the new partition is bounded in terms 00:28:04.350 --> 00:28:06.737 of the previous partition. 00:28:23.138 --> 00:28:26.115 All right. 00:28:26.115 --> 00:28:26.990 Any questions so far? 00:28:31.510 --> 00:28:35.790 So now we get this sequence of partitions. 00:28:35.790 --> 00:28:37.320 We can keep on doing this. 00:28:37.320 --> 00:28:43.150 So g could be arbitrarily large, but eventually we 00:28:43.150 --> 00:28:45.980 will be able to obtain the last condition here, 00:28:45.980 --> 00:28:49.170 which is the only thing that is missing so far. 00:28:49.170 --> 00:28:59.510 So since the energy is bounded between 0 and 1, 00:28:59.510 --> 00:29:06.650 there exists some i at most 1 over epsilon 0 00:29:06.650 --> 00:29:15.860 such that the energy goes up by less than epsilon 0. 00:29:26.390 --> 00:29:28.700 Because otherwise your energy would exceed 1. 00:29:33.370 --> 00:29:38.590 So now let's set P to be this Pi, 00:29:38.590 --> 00:29:43.920 and Q to be this, the refinement-- 00:29:43.920 --> 00:29:45.670 the next term in the partition. 00:29:50.470 --> 00:29:54.680 And what we find is that the-- 00:29:54.680 --> 00:29:57.110 so then you have basically all the conditions. 00:29:57.110 --> 00:30:03.080 So p is epsilon 0 regular, because it is epsilon-- 00:30:03.080 --> 00:30:07.870 the previous term, which is at most epsilon 0 regular. 00:30:07.870 --> 00:30:11.970 And you have this one as well, and this one as well. 00:30:11.970 --> 00:30:14.700 And we want to show that the number of parts of Q 00:30:14.700 --> 00:30:16.960 is bounded. 00:30:16.960 --> 00:30:19.310 And that's basically because each time there 00:30:19.310 --> 00:30:22.790 was a bound on the number of parts which depends only 00:30:22.790 --> 00:30:24.890 on the regularity parameters, and you're 00:30:24.890 --> 00:30:27.980 repeating that bound a bounded number of times. 00:30:30.520 --> 00:30:44.610 So Q is-- so it's bounded as a function of the sequence 00:30:44.610 --> 00:30:47.010 of epsilons-- this infinite vector of epsilons, 00:30:47.010 --> 00:30:49.180 but it is a bounded number. 00:30:49.180 --> 00:30:54.323 You're only iterating this bound a bounded number of times. 00:30:54.323 --> 00:30:55.490 And that finishes the proof. 00:31:00.946 --> 00:31:02.363 Any questions? 00:31:09.770 --> 00:31:13.180 It may be somewhat mysterious to you right now why we do this, 00:31:13.180 --> 00:31:14.840 so we'll get that application a second. 00:31:14.840 --> 00:31:18.010 But for now, I just want to comment a bit on the bounds. 00:31:26.090 --> 00:31:32.130 Of course, the bounds depend on what epsilon i's do you use. 00:31:32.130 --> 00:31:33.900 And typically, you want the epsilon i's 00:31:33.900 --> 00:31:37.350 to decrease with more parts that you have. 00:31:37.350 --> 00:31:40.020 And with almost all reasonable applications 00:31:40.020 --> 00:31:44.450 of this regularity lemma, the strong regularity lemma-- 00:31:44.450 --> 00:31:52.960 so for example, with epsilon i being some epsilon divided 00:31:52.960 --> 00:31:58.140 by, let's say, i plus 1 or any polynomial of the i's-- 00:31:58.140 --> 00:32:02.230 or you can even let it decay quicker than that, as well. 00:32:02.230 --> 00:32:06.310 You see, basically what happens is that you 00:32:06.310 --> 00:32:09.310 are applying this m0 bound. 00:32:15.300 --> 00:32:25.360 m0 applied in succession 1 over epsilon times. 00:32:29.250 --> 00:32:33.800 In the regularity lemma, we saw that the m0 that comes out 00:32:33.800 --> 00:32:37.130 of Szemeredi's graph regularity lemma is the tower function. 00:32:39.872 --> 00:32:43.970 So the tower function, that's a tower of i 00:32:43.970 --> 00:32:52.326 is defined to be the exponential function iterated i times. 00:32:52.326 --> 00:32:53.920 So of course, I'm being somewhat loose 00:32:53.920 --> 00:32:57.250 here with the exact dependence, but you get the idea 00:32:57.250 --> 00:33:02.050 that now we want to apply the tower function i times. 00:33:15.220 --> 00:33:17.350 Instead of iterating the exponential i times, 00:33:17.350 --> 00:33:20.380 now you iterate the tower function i times. 00:33:20.380 --> 00:33:23.292 And some of you laughing, this is an incredibly large number. 00:33:23.292 --> 00:33:25.000 It's even larger than the tower function. 00:33:29.490 --> 00:33:33.530 So in literature, especially around the regularity lemma, 00:33:33.530 --> 00:33:37.020 this function where you iterate the tower function i 00:33:37.020 --> 00:33:39.873 times is given the name wowzer. 00:33:39.873 --> 00:33:47.280 [LAUGHTER] As in, wow, this is a huge number. 00:33:47.280 --> 00:33:50.400 So it's a step up in the Ackerman hierarchy. 00:33:50.400 --> 00:33:53.480 So if you repeat the wowzer function i times, 00:33:53.480 --> 00:33:56.640 you move up one ladder in the Ackerman hierarchy 00:33:56.640 --> 00:34:01.060 and this hierarchy of rapidly growing functions. 00:34:01.060 --> 00:34:03.560 But in any case, it's bounded and that's good enough for us. 00:34:10.695 --> 00:34:11.570 Any questions so far? 00:34:16.380 --> 00:34:16.950 Yeah? 00:34:16.950 --> 00:34:19.857 AUDIENCE: What do you call like [INAUDIBLE] 00:34:19.857 --> 00:34:21.690 YUFEI ZHAO: Yes, so question is, what do you 00:34:21.690 --> 00:34:23.920 call wowzer iterated? 00:34:23.920 --> 00:34:27.170 I'm not aware of a standard name for that. 00:34:27.170 --> 00:34:28.980 Actually, even the name wowzer somehow 00:34:28.980 --> 00:34:32.139 is very common in the combinatorics community, 00:34:32.139 --> 00:34:34.199 but I think most people outside this community 00:34:34.199 --> 00:34:35.510 will not recognize this word. 00:34:40.139 --> 00:34:42.389 Any more questions? 00:34:42.389 --> 00:34:46.145 So another way it's a step up in Ackerman hierarchy. 00:34:46.145 --> 00:34:48.270 So it's enumerated one, two, three, four, you know, 00:34:48.270 --> 00:34:49.170 if you keep going up. 00:34:52.994 --> 00:34:56.340 All right. 00:34:56.340 --> 00:35:01.290 Another remark about this strong regularity lemma 00:35:01.290 --> 00:35:04.910 is that it will be convenient for us-- actually, 00:35:04.910 --> 00:35:07.950 some are more essential compared to our previous applications-- 00:35:07.950 --> 00:35:09.912 to make the parts equitable. 00:35:13.150 --> 00:35:18.710 So P and Q equitable. 00:35:18.710 --> 00:35:22.175 And basically, the parts are such that all the-- 00:35:22.175 --> 00:35:25.068 the partitions are such that all the parts have basically 00:35:25.068 --> 00:35:26.235 the same number of vertices. 00:35:26.235 --> 00:35:30.485 So I won't make it precise, but you can do it. 00:35:30.485 --> 00:35:31.610 It's not too hard to do it. 00:35:31.610 --> 00:35:35.180 And you can prove it similar to how 00:35:35.180 --> 00:35:39.030 I described how to modify the proof of the regularity level. 00:35:39.030 --> 00:35:41.240 So I won't belabor that point, but we'll 00:35:41.240 --> 00:35:42.470 use the equitable version. 00:35:45.790 --> 00:35:50.760 All right, so how does one use this regularity lemma? 00:35:50.760 --> 00:35:53.460 Let me state a corollary, and let 00:35:53.460 --> 00:35:55.850 me call this a corollary star because you actually 00:35:55.850 --> 00:35:58.430 need to do some work to get it to follow 00:35:58.430 --> 00:35:59.880 from the strong regularity lemma. 00:35:59.880 --> 00:36:01.547 But the corollary is the version that we 00:36:01.547 --> 00:36:06.420 will apply that if you start with a decreasing 00:36:06.420 --> 00:36:17.215 sequence of this epsilon, then there exists a delta such 00:36:17.215 --> 00:36:18.340 that the following is true. 00:36:22.650 --> 00:36:41.360 Every n vertex graph has an equitable vertex partition, 00:36:41.360 --> 00:36:51.530 call it i through the k, and a subset Wi 00:36:51.530 --> 00:36:59.580 of each Vi such that the following properties hold. 00:36:59.580 --> 00:37:04.960 First, all the W's are fairly large. 00:37:04.960 --> 00:37:07.070 They're at least constant proportion 00:37:07.070 --> 00:37:08.540 of the total vertex set. 00:37:13.040 --> 00:37:22.520 Between every pair of Wi Wj, it is epsilon sub k regular. 00:37:30.890 --> 00:37:32.640 And this is the point I want to emphasize. 00:37:32.640 --> 00:37:35.360 So here there are not you regular pairs anymore. 00:37:35.360 --> 00:37:37.170 So it is every. 00:37:41.274 --> 00:37:44.870 So no irregular pairs between the Wi's, 00:37:44.870 --> 00:37:48.740 and also we need to include the case 00:37:48.740 --> 00:37:50.900 when i equals the j, as well. 00:37:50.900 --> 00:37:53.495 So each Wi is regular with itself. 00:37:57.060 --> 00:38:05.850 And furthermore, the edge densities between the V's are 00:38:05.850 --> 00:38:11.820 similar to the edge densities between the corresponding W's. 00:38:11.820 --> 00:38:18.060 And here it is for most pairs for all 00:38:18.060 --> 00:38:23.550 but at most epsilon k square pairs. 00:38:30.320 --> 00:38:31.745 Epsilon 0, yeah. 00:38:31.745 --> 00:38:32.860 At most epsilon 0. 00:38:40.660 --> 00:38:42.405 Any questions about the statement? 00:38:57.440 --> 00:39:00.950 So let me show you how you could deduce 00:39:00.950 --> 00:39:04.520 the corollary from the strong regularity lemma. 00:39:16.950 --> 00:39:18.720 So first, let me draw your picture. 00:39:23.070 --> 00:39:26.346 So here you have a regularity partition. 00:39:31.010 --> 00:39:38.155 And so these are your V's, and inside each V 00:39:38.155 --> 00:39:50.340 I find a W such that if I look at the edge 00:39:50.340 --> 00:39:53.350 sets between pairwise blue sets, including 00:39:53.350 --> 00:39:58.950 the blue sets with themselves, it is always very regular. 00:39:58.950 --> 00:40:04.620 And also, the edge densities between the blue sets 00:40:04.620 --> 00:40:07.590 is mostly very similar to the edge 00:40:07.590 --> 00:40:10.110 density between their ambient white sets. 00:40:20.040 --> 00:40:21.910 OK, so let me say a few words-- 00:40:21.910 --> 00:40:23.880 I won't go into too many details-- 00:40:23.880 --> 00:40:26.280 about how you might deduce this corollary 00:40:26.280 --> 00:40:28.470 from the strong regularity lemma. 00:40:31.280 --> 00:40:33.320 So first let me do something which 00:40:33.320 --> 00:40:39.830 is slightly simpler, which is to not yet require 00:40:39.830 --> 00:40:44.326 that the blue sets, Wi's, are regular with themselves. 00:40:51.480 --> 00:40:59.630 So without requiring this as regular so we can obtain 00:40:59.630 --> 00:41:13.660 the Wi's by picking a uniform random part 00:41:13.660 --> 00:41:27.870 of the final partition, Q, inside each part of P 00:41:27.870 --> 00:41:29.640 in the strong regularity lemma. 00:41:35.920 --> 00:41:37.840 So you have the strong regularity lemma, 00:41:37.840 --> 00:41:44.080 which produces for you a pair of partitions like that. 00:41:44.080 --> 00:41:47.020 So it produces for you a pair of partitions. 00:41:47.020 --> 00:41:53.560 And what we will do is to pick one of these guys as my W, 00:41:53.560 --> 00:41:55.048 pick one of these guys at random, 00:41:55.048 --> 00:41:56.590 and pick one of those guys at random. 00:42:00.730 --> 00:42:06.620 Because W is so extremely regular, most of these pairs 00:42:06.620 --> 00:42:09.740 will be regular. 00:42:09.740 --> 00:42:13.610 So with high probability, you will not 00:42:13.610 --> 00:42:18.800 encounter any irregular pairs if you 00:42:18.800 --> 00:42:25.570 pick the W's randomly as parts of Q. So that's the key point. 00:42:25.570 --> 00:42:28.840 Here we're using that Q is extremely regular. 00:42:40.930 --> 00:42:47.420 So all the Wi Wj is regular for all i not equal 00:42:47.420 --> 00:42:49.718 to j with high probability. 00:42:53.550 --> 00:42:56.490 But the other thing that we would like is that the edge 00:42:56.490 --> 00:43:00.810 densities between the W's are similar to those between 00:43:00.810 --> 00:43:02.740 the V's. 00:43:02.740 --> 00:43:06.020 And for that, we will use this condition about their energies 00:43:06.020 --> 00:43:07.550 being very similar to each other. 00:43:10.650 --> 00:43:16.800 So the third consequence, C, is-- 00:43:16.800 --> 00:43:22.200 it's a consequence of the energy bound. 00:43:30.820 --> 00:43:34.510 Because recall that in our proof of the Szemeredi regularity 00:43:34.510 --> 00:43:36.730 lemma there was an interpretation 00:43:36.730 --> 00:43:43.860 of the energy as the second moment 00:43:43.860 --> 00:43:47.340 of a certain random variable which we called z. 00:43:51.640 --> 00:43:54.830 And using that interpretation, I can write down 00:43:54.830 --> 00:44:00.320 this expression like that. 00:44:00.320 --> 00:44:03.860 We are here assuming for simplicity 00:44:03.860 --> 00:44:08.030 that Q is completely equitable, so all the parts 00:44:08.030 --> 00:44:09.650 have exactly the same size. 00:44:09.650 --> 00:44:15.740 Z of Q is defined to be the edge density between Vi and Vj 00:44:15.740 --> 00:44:21.010 for random ij. 00:44:21.010 --> 00:44:23.770 So this is a random variable z. 00:44:23.770 --> 00:44:28.410 So you pick pair of parts uniformly, 00:44:28.410 --> 00:44:31.110 or maybe with some weights if they're not exactly equal. 00:44:31.110 --> 00:44:34.480 And you evaluate the edge density. 00:44:34.480 --> 00:44:37.650 So this energy difference is the difference 00:44:37.650 --> 00:44:39.330 between the second moments. 00:44:39.330 --> 00:44:46.710 And because Q is a refinement of P, 00:44:46.710 --> 00:44:55.870 it is the case that this difference of L2 norms 00:44:55.870 --> 00:45:00.860 is equal to the second moment of the difference 00:45:00.860 --> 00:45:02.480 of the random variables. 00:45:02.480 --> 00:45:04.760 So we saw a version of this earlier 00:45:04.760 --> 00:45:07.910 when we were discussing variance in the context 00:45:07.910 --> 00:45:10.520 of the proof of the similar irregularity lemma. 00:45:10.520 --> 00:45:11.810 Here it's basically the same. 00:45:11.810 --> 00:45:16.430 You can either look at this inequality part by part of V, 00:45:16.430 --> 00:45:21.050 or if you like to be a bit more abstract 00:45:21.050 --> 00:45:24.170 then this is actually a case of Pythagorean theorem. 00:45:29.910 --> 00:45:34.350 If you view these as vectors in a certain vector space, 00:45:34.350 --> 00:45:36.100 then you have some orthogonality. 00:45:36.100 --> 00:45:40.378 So you have this sum of squares identity. 00:45:45.860 --> 00:45:47.792 Where does part A come from? 00:45:47.792 --> 00:45:52.340 So part A, we want the parts, that Wi's to be not too small, 00:45:52.340 --> 00:46:15.561 but that comes from a bound on the number of parts of Q. 00:46:15.561 --> 00:46:18.810 So so far this more or less proves the corollary 00:46:18.810 --> 00:46:23.050 except for that we simplified our lives 00:46:23.050 --> 00:46:29.680 by requiring just that the i not equal to j, the Vi Vj's are 00:46:29.680 --> 00:46:31.100 regular. 00:46:31.100 --> 00:46:33.800 But in the statement up there, we also want 00:46:33.800 --> 00:46:37.650 the Vi's-- so the Wi's ice to be regular with themselves, 00:46:37.650 --> 00:46:41.440 which will be important for application. 00:46:41.440 --> 00:46:45.670 So I won't explain how to do that, and part of the reason 00:46:45.670 --> 00:46:49.170 is that this is also one of your homework problems. 00:46:49.170 --> 00:46:51.920 So in one of the homework problems problem set 3, 00:46:51.920 --> 00:46:55.430 you were asked to prove that every graph has 00:46:55.430 --> 00:47:01.100 a subset of vertices that is of least constant proportion such 00:47:01.100 --> 00:47:04.880 that it is regular with itself. 00:47:04.880 --> 00:47:06.770 And the methods you use there will 00:47:06.770 --> 00:47:12.210 be applicable to handle the situation over here, as well. 00:47:12.210 --> 00:47:15.030 So putting all of these ingredients together, 00:47:15.030 --> 00:47:20.020 we get the corollary whereby you have this picture, 00:47:20.020 --> 00:47:21.610 you have this partition. 00:47:21.610 --> 00:47:24.190 I don't even require the Vi's to be regular. 00:47:24.190 --> 00:47:25.640 That doesn't matter anymore. 00:47:25.640 --> 00:47:29.230 All that matters is that between the Wi's they are very regular, 00:47:29.230 --> 00:47:34.450 and that there are no irregular parts between these Wi's. 00:47:34.450 --> 00:47:41.610 And now we'll be able to go back to the induced graph removal 00:47:41.610 --> 00:47:46.920 lemma where previously we had an issue with the existence 00:47:46.920 --> 00:47:51.880 of irregular pairs in the use of Szemeredi regularity partition, 00:47:51.880 --> 00:47:55.250 and now we have a tool to get around that. 00:47:55.250 --> 00:48:00.300 So next we will see how to execute this proof, 00:48:00.300 --> 00:48:03.840 but at this point hopefully you already see an outline. 00:48:03.840 --> 00:48:10.678 Because you no longer need to worry about this thing here. 00:48:10.678 --> 00:48:11.720 Let's take a quick break. 00:48:14.760 --> 00:48:15.830 Any questions so far? 00:48:20.600 --> 00:48:21.480 Yes? 00:48:21.480 --> 00:48:31.540 AUDIENCE: Why are we able to [INAUDIBLE] 00:48:31.540 --> 00:48:33.040 YUFEI ZHAO: OK, so the question was, 00:48:33.040 --> 00:48:37.170 there was a step where we were looking at some expectations 00:48:37.170 --> 00:48:39.300 of squares. 00:48:39.300 --> 00:48:43.670 And so why was that identity true? 00:48:43.670 --> 00:48:46.250 So if you look back to the proof of Szemeredi's regularity 00:48:46.250 --> 00:48:48.932 lemma, we already saw an instance of that inequality 00:48:48.932 --> 00:48:50.390 in the computation of the variance. 00:48:57.860 --> 00:49:01.270 So you know that the variance of x, on one 00:49:01.270 --> 00:49:10.470 hand it is equal to where mu is the mean of x. 00:49:10.470 --> 00:49:15.670 And on the other hand, it is equal to this quantity. 00:49:20.410 --> 00:49:23.360 So you agree with this formula? 00:49:23.360 --> 00:49:28.360 And you can expand it to prove it, and the thing that-- 00:49:28.360 --> 00:49:30.990 the question that you raised basically you 00:49:30.990 --> 00:49:34.680 can prove by looking at this formula part by part. 00:49:39.250 --> 00:49:40.602 Any more questions? 00:49:49.760 --> 00:49:54.350 So let's now prove the induced graph removal lemma. 00:49:54.350 --> 00:49:57.110 And we'll follow the regularity partition, 00:49:57.110 --> 00:49:59.510 but with a small twist that Instead 00:49:59.510 --> 00:50:01.850 of using Szemeredi's regularity lemma, 00:50:01.850 --> 00:50:03.740 we will use that corollary up there. 00:50:11.560 --> 00:50:14.880 So let's prove the induced graph removal lemma. 00:50:20.820 --> 00:50:21.800 So the three steps. 00:50:21.800 --> 00:50:23.451 First, we do partition. 00:50:30.050 --> 00:50:33.370 So let's suppose you have a-- 00:50:36.850 --> 00:50:39.100 so we suppose g is like above. 00:50:39.100 --> 00:50:45.200 You have very few induced copies of H. 00:50:45.200 --> 00:50:51.170 Let's apply the corollary to get a partition of the vertex set 00:50:51.170 --> 00:50:57.660 of g into k parts. 00:50:57.660 --> 00:51:04.690 And inside each part I have a W. Satisfying 00:51:04.690 --> 00:51:11.950 the following properties that each Wi Wj 00:51:11.950 --> 00:51:18.517 is regular with the following parameter which 00:51:18.517 --> 00:51:21.100 will come out of later when we need to use the counting lemma. 00:51:21.100 --> 00:51:23.740 But it's some number, but don't worry too much about it. 00:51:27.050 --> 00:51:30.310 So here I'm going to-- 00:51:30.310 --> 00:51:35.685 so let's say H has little H vertices. 00:51:45.405 --> 00:51:48.060 So between Wi Wj it is this regular. 00:51:48.060 --> 00:51:50.730 So we actually have not yet used the full strength 00:51:50.730 --> 00:51:58.740 of the corollary where I can make the regularity even depend 00:51:58.740 --> 00:51:59.520 on k. 00:51:59.520 --> 00:52:01.650 So we will not need that here, but we'll 00:52:01.650 --> 00:52:04.090 need it in a later application. 00:52:04.090 --> 00:52:11.120 So the exponent is little H. 00:52:11.120 --> 00:52:18.890 OK, so other properties are that the densities between the Vi's 00:52:18.890 --> 00:52:28.910 and the Wi's do not differ by more than epsilon over 2 00:52:28.910 --> 00:52:33.700 for all but a small fraction-- 00:52:33.700 --> 00:52:36.480 so epsilon k squared over 2-- 00:52:36.480 --> 00:52:36.980 pairs. 00:52:45.150 --> 00:52:53.010 And finally, the sizes of the Wi's are at least delta 0 times 00:52:53.010 --> 00:52:57.010 n where delta 0 depends only on epsilon. 00:53:04.310 --> 00:53:07.600 Epsilon and H. 00:53:21.210 --> 00:53:25.320 This is the partition step, so now let's do the cleaning. 00:53:28.210 --> 00:53:34.830 In the cleaning step, basically we're not going to-- 00:53:34.830 --> 00:53:37.620 I mean, there is no longer an issue of irregular pairs if we 00:53:37.620 --> 00:53:40.100 only look at the Wi's. 00:53:40.100 --> 00:53:43.560 So we just need to think about the low density pairs 00:53:43.560 --> 00:53:46.710 or whatever the corresponding analog is. 00:53:46.710 --> 00:53:52.200 And what happens here is that for every i less than j, 00:53:52.200 --> 00:53:57.790 and crucially including when i equals to j, 00:53:57.790 --> 00:54:06.330 if the edge densities between the W's is too small 00:54:06.330 --> 00:54:18.790 then we remove all edges between Vi and Vj. 00:54:23.070 --> 00:54:31.810 And if the edge density between the Wi's is too big, 00:54:31.810 --> 00:54:35.395 then we remove all edges. 00:54:38.280 --> 00:54:41.560 So we add all edges between Vi and Vj. 00:55:00.900 --> 00:55:02.970 How many edges do we end up adding or removing? 00:55:06.560 --> 00:55:21.460 So the total number of edges added or removed from g is-- 00:55:21.460 --> 00:55:26.680 in this case, so if the edges density 00:55:26.680 --> 00:55:32.160 in g between the Vi's and Vj's is also very small, 00:55:32.160 --> 00:55:36.110 then you do not remove very many edges. 00:55:36.110 --> 00:55:41.190 But most pairs of Vi and Vj have that property. 00:55:41.190 --> 00:55:43.650 So you tidy up what kind of errors 00:55:43.650 --> 00:55:48.140 you can get from here and there, and you 00:55:48.140 --> 00:55:51.890 find that the total number of edges that are added or removed 00:55:51.890 --> 00:55:59.150 from g is less than, let's say, epsilon n squared. 00:55:59.150 --> 00:56:00.830 Maybe even get an extra factor of 2, 00:56:00.830 --> 00:56:04.310 but you know, upon changing some constant factors, 00:56:04.310 --> 00:56:08.180 it's less than epsilon n squared. 00:56:08.180 --> 00:56:13.470 So this is some small details you can work out. 00:56:13.470 --> 00:56:15.780 Here we're using-- asking, how is 00:56:15.780 --> 00:56:19.410 the density between Vi and Vj related to Wi and Wj? 00:56:19.410 --> 00:56:23.540 Well, for most pairs of i and j they're very similar. 00:56:23.540 --> 00:56:26.180 And there's a small fraction of them that are not similar, 00:56:26.180 --> 00:56:32.310 but then you lump everything in to this bound over here. 00:56:40.290 --> 00:56:43.212 So maybe I need to-- 00:56:43.212 --> 00:56:44.920 let me just put a 2 here just to be safe. 00:56:48.690 --> 00:56:50.780 All right. 00:56:50.780 --> 00:56:55.590 So we deleted a very small number of edges, 00:56:55.590 --> 00:56:57.750 and now we want to show that the graph that 00:56:57.750 --> 00:57:01.740 has resulted from this modification 00:57:01.740 --> 00:57:05.250 does not have any induced H sub-graphs. 00:57:11.480 --> 00:57:15.110 And the final step is the counting step. 00:57:15.110 --> 00:57:20.960 So suppose there were any induced 00:57:20.960 --> 00:57:26.860 H left after the modification. 00:57:26.860 --> 00:57:30.020 So I want to show that, in fact, there must be a lot of H's-- 00:57:30.020 --> 00:57:32.160 induced H's originally in the graph, 00:57:32.160 --> 00:57:34.336 thereby contradicting the hypothesis. 00:57:41.690 --> 00:57:45.210 So where does this induced H sit? 00:57:45.210 --> 00:57:55.070 Well, you have the V's, and inside the V's you have 00:57:55.070 --> 00:57:55.570 the W's. 00:58:04.170 --> 00:58:13.200 So suppose my H is that graph for illustration. 00:58:13.200 --> 00:58:16.770 And in particular, I have a non-edge. 00:58:16.770 --> 00:58:20.370 So I have an edge, and I also have a non-edge. 00:58:20.370 --> 00:58:22.895 So between these two, that's the non-edge. 00:58:27.950 --> 00:58:34.590 So suppose you find a copy of H in the cleaned-up graph. 00:58:34.590 --> 00:58:36.110 Where can that cleaned up-- 00:58:36.110 --> 00:58:37.940 this copy of H sit? 00:58:37.940 --> 00:58:39.790 Suppose you find it here. 00:58:43.440 --> 00:58:52.130 The claim now is that if this copy of H existed here, then 00:58:52.130 --> 00:58:57.050 I must be able to find many such copies of H 00:58:57.050 --> 00:58:59.090 in the corresponding yellow parts. 00:59:01.940 --> 00:59:10.050 Because between the yellow parts you have regularity, 00:59:10.050 --> 00:59:15.450 and you also have the right kinds of densities. 00:59:15.450 --> 00:59:17.710 Because if they didn't have the right kind of density, 00:59:17.710 --> 00:59:19.210 we would have cleaned it up already. 00:59:22.900 --> 00:59:26.000 So that's the ideal. 00:59:26.000 --> 00:59:29.720 If you had a copy of this H somewhere, 00:59:29.720 --> 00:59:31.820 then I zoom into the yellow parts, 00:59:31.820 --> 00:59:37.160 zoom into these W's, and I find lots of copies of H in between 00:59:37.160 --> 00:59:39.430 the W's. 00:59:39.430 --> 00:59:40.930 So suppose-- let me write this down. 00:59:40.930 --> 00:59:57.690 So suppose the little V's, so the vertices, lies in the-- 00:59:57.690 --> 01:00:00.540 so I'm just indexing where a little v lies. 01:00:00.540 --> 01:00:03.225 The little v lies in big V sub phi 01:00:03.225 --> 01:00:14.600 V for some phi which since the vertices of H2 went through k. 01:00:14.600 --> 01:00:34.650 So now we apply counting lemma to embed induced copies of H 01:00:34.650 --> 01:00:45.710 in g where the vertex V in H is mapped 01:00:45.710 --> 01:00:54.455 to a vertex in the corresponding W. 01:01:00.630 --> 01:01:05.440 And we would like to know that there are lots of such copies. 01:01:05.440 --> 01:01:06.660 And the counting Lemma-- 01:01:06.660 --> 01:01:11.730 or rather, some variant, but I should read the counting lemma 01:01:11.730 --> 01:01:16.940 that we did last time and view it as a multi-partite version. 01:01:16.940 --> 01:01:21.440 Apply this so far part to part. 01:01:21.440 --> 01:01:30.340 So we find that the number of such induced copies 01:01:30.340 --> 01:01:32.180 is within a small error. 01:01:36.630 --> 01:01:47.700 So that regularity parameter multiplied by the number 01:01:47.700 --> 01:01:51.690 of edges of H, which we already canceled out, 01:01:51.690 --> 01:02:01.015 multiplied by the product of these Wi's. 01:02:04.490 --> 01:02:08.660 So it's within this error of what 01:02:08.660 --> 01:02:13.250 you would suspect if you naively multiply the edge 01:02:13.250 --> 01:02:16.953 densities together along with the vertex densities. 01:02:29.280 --> 01:02:35.790 So these factors are for the edges that you want to embed, 01:02:35.790 --> 01:02:39.460 and then I also need to multiply the densities 01:02:39.460 --> 01:02:40.510 for the long edges. 01:02:51.378 --> 01:02:54.542 So 1 minus these edge densities. 01:02:54.542 --> 01:02:56.500 So one way you can think of it is just consider 01:02:56.500 --> 01:02:58.510 the complement in g. 01:02:58.510 --> 01:03:02.500 So consider the complement of g to get this version here. 01:03:02.500 --> 01:03:06.240 And then finally, the product of the vertex set sizes. 01:03:18.000 --> 01:03:22.500 And the point is that this is not a small number. 01:03:22.500 --> 01:03:35.870 So hence the number of induced copies of H in g 01:03:35.870 --> 01:03:41.450 is at least on the order of-- 01:03:41.450 --> 01:03:42.370 well, OK? 01:03:44.910 --> 01:03:50.590 So it's at least some number, which is basically 01:03:50.590 --> 01:03:51.880 this guy over here. 01:03:51.880 --> 01:03:55.820 So epsilon over 4 raised to-- 01:03:55.820 --> 01:03:57.820 all of these are constants, so that's the point. 01:03:57.820 --> 01:03:59.740 All of these guys are constants, minus-- 01:04:05.050 --> 01:04:08.103 so here is the main term, and then the error term. 01:04:13.790 --> 01:04:17.210 And then the product of these vertex set sizes, 01:04:17.210 --> 01:04:19.680 and we saw that each vertex set is not too small. 01:04:25.840 --> 01:04:30.035 So you have lots of induced copies of H in g. 01:04:30.035 --> 01:04:31.430 Yep? 01:04:31.430 --> 01:04:34.600 AUDIENCE: How do you do in the case where 01:04:34.600 --> 01:04:44.740 the density between [INAUDIBLE] 01:04:44.740 --> 01:04:47.360 YUFEI ZHAO: OK, so can you repeat your question? 01:04:47.360 --> 01:04:53.045 AUDIENCE: How are you dealing with the [INAUDIBLE] 01:04:53.045 --> 01:04:53.670 YUFEI ZHAO: OK. 01:04:53.670 --> 01:04:55.860 So question, how do we deal with the all but epsilon 01:04:55.860 --> 01:04:57.080 over two pairs? 01:04:57.080 --> 01:04:59.375 So that comes up in the cleaning step 01:04:59.375 --> 01:05:01.760 in what I wrote in red in dealing 01:05:01.760 --> 01:05:07.780 with the number of total edges that are added or removed. 01:05:07.780 --> 01:05:10.330 So think about how many edges are added or removed. 01:05:10.330 --> 01:05:15.440 In these non-exceptional pairs, the number of edges that are 01:05:15.440 --> 01:05:16.310 added or removed-- 01:05:26.470 --> 01:05:28.670 let's just think about added edges. 01:05:28.670 --> 01:05:45.930 So if the density of V is controlled by that of W, 01:05:45.930 --> 01:05:48.910 then the number of edges added-- 01:05:48.910 --> 01:05:50.730 or removed, in that case-- 01:05:50.730 --> 01:05:56.230 from all such pairs along with-- 01:05:56.230 --> 01:05:56.730 yeah. 01:05:56.730 --> 01:06:01.410 So you have epsilon n squared edges changed. 01:06:06.540 --> 01:06:17.780 On the other hand, if this is not true then you only have 01:06:17.780 --> 01:06:23.000 epsilon k squared such pairs ij for which this cannot be true. 01:06:23.000 --> 01:06:26.380 So you also only have at most epsilon n squared edges 01:06:26.380 --> 01:06:29.850 added or removed in such cases. 01:06:29.850 --> 01:06:32.642 That answers your question? 01:06:32.642 --> 01:06:33.907 Yes? 01:06:33.907 --> 01:06:35.032 AUDIENCE: Is that number 0? 01:06:37.755 --> 01:06:39.624 YUFEI ZHAO: Is which number 0? 01:06:39.624 --> 01:06:45.320 AUDIENCE: The number of induced edges for the [INAUDIBLE] 01:06:45.320 --> 01:06:46.520 YUFEI ZHAO: The-- 01:06:46.520 --> 01:06:47.951 AUDIENCE: Yeah, the top board. 01:06:47.951 --> 01:06:48.868 YUFEI ZHAO: Top board? 01:06:58.840 --> 01:06:59.400 Good. 01:06:59.400 --> 01:07:01.695 So asking about this number. 01:07:01.695 --> 01:07:02.820 So that should have been 2. 01:07:08.580 --> 01:07:10.020 Yes? 01:07:10.020 --> 01:07:12.173 AUDIENCE: I don't see k anywhere. 01:07:12.173 --> 01:07:14.840 YUFEI ZHAO: OK, so question, you don't see k appearing anywhere. 01:07:14.840 --> 01:07:16.893 So the k in the corollary, do you mean? 01:07:16.893 --> 01:07:17.820 AUDIENCE: Yeah. 01:07:17.820 --> 01:07:19.530 YUFEI ZHAO: So that hasn't come up yet. 01:07:19.530 --> 01:07:24.180 So it comes up implicitly because we need to lower bound 01:07:24.180 --> 01:07:26.715 the sizes of these W's. 01:07:31.820 --> 01:07:34.880 So this is partly why we need a bound on the number of parts, 01:07:34.880 --> 01:07:38.150 but it is true that we do not need epsilon k to depend 01:07:38.150 --> 01:07:39.860 on k in this application yet. 01:07:39.860 --> 01:07:42.140 I will mention a different application in the second 01:07:42.140 --> 01:07:43.230 where you do need that k. 01:07:48.810 --> 01:07:52.920 OK, so the number of induced H in g is at least this amount. 01:07:52.920 --> 01:07:54.080 And that's a small lie. 01:07:54.080 --> 01:07:59.780 You need to maybe consider this is the number of homomorphic. 01:07:59.780 --> 01:08:01.730 Well, actually, no, we're OK. 01:08:01.730 --> 01:08:02.514 Never mind. 01:08:11.120 --> 01:08:19.609 So you can set delta to be this quantity here, 01:08:19.609 --> 01:08:21.210 and then that finishes the proof. 01:08:21.210 --> 01:08:23.500 So you have lots of induced copies of H 01:08:23.500 --> 01:08:27.765 in your graph which contradicts the hypothesis. 01:08:27.765 --> 01:08:30.600 So that finishes the proof of the induced removal lemma, 01:08:30.600 --> 01:08:34.109 and basically the proof is the same as the usual graph 01:08:34.109 --> 01:08:36.510 removal lemma except that now we need 01:08:36.510 --> 01:08:40.260 some strengthened regularity lemma which 01:08:40.260 --> 01:08:43.290 allows us to get rid of irregular parts 01:08:43.290 --> 01:08:44.890 but in a more restricted setting. 01:08:44.890 --> 01:08:47.729 Because we saw you cannot completely get rid of irregular 01:08:47.729 --> 01:08:48.229 parts. 01:08:51.720 --> 01:08:52.500 Any questions? 01:08:56.109 --> 01:08:56.849 Yes? 01:08:56.849 --> 01:09:01.473 AUDIENCE: [INAUDIBLE] 01:09:01.473 --> 01:09:03.640 YUFEI ZHAO: So I want to address the question of why 01:09:03.640 --> 01:09:05.950 did I state this corollary in this more 01:09:05.950 --> 01:09:09.520 general form of a decreasing sequence of epsilons? 01:09:09.520 --> 01:09:11.770 So first of all, with strong regularity lemmas, 01:09:11.770 --> 01:09:15.189 the strength is sometimes always nice to-- 01:09:15.189 --> 01:09:17.439 it's always nice to state it with this extra strength. 01:09:17.439 --> 01:09:20.080 Because it's the right way to think 01:09:20.080 --> 01:09:22.770 about these types of theorems. 01:09:22.770 --> 01:09:25.330 That the regularity on the parts depends-- 01:09:25.330 --> 01:09:28.779 you can make it depend on the number of parts 01:09:28.779 --> 01:09:32.170 so that you get much stronger control on the regularity. 01:09:32.170 --> 01:09:33.740 But there are also some applications. 01:09:33.740 --> 01:09:36.970 For example, whether I will state next, 01:09:36.970 --> 01:09:40.210 an application where you do need that kind of strength. 01:09:40.210 --> 01:09:43.899 So here's what's known as the infinite removal lemma. 01:09:47.260 --> 01:09:49.689 Here we have not just a single pattern 01:09:49.689 --> 01:09:52.660 or a finite number of patterns we want to get rid of. 01:09:52.660 --> 01:09:55.330 For now we have infinitely many patterns. 01:09:55.330 --> 01:10:08.880 So for every curly H, which is a possibly infinite set 01:10:08.880 --> 01:10:11.040 of graphs. 01:10:11.040 --> 01:10:13.090 The graphs themselves are always finite, 01:10:13.090 --> 01:10:15.480 but this may be an infinite list. 01:10:15.480 --> 01:10:19.830 And an epsilon parameter. 01:10:19.830 --> 01:10:26.350 There exists an H0 and a delta positive parameter 01:10:26.350 --> 01:10:38.230 such that every n vertex graph with at most delta-- 01:10:38.230 --> 01:10:41.180 so less than delta-- 01:10:41.180 --> 01:10:52.580 V to the H induced copies of H for every H 01:10:52.580 --> 01:11:00.205 in this family with fewer than H0 vertices. 01:11:03.670 --> 01:11:07.670 So every graph with this property 01:11:07.670 --> 01:11:14.910 can be made curly H free. 01:11:14.910 --> 01:11:17.580 So it means free of-- 01:11:17.580 --> 01:11:32.390 induced curly H free by adding or removing fewer 01:11:32.390 --> 01:11:34.380 than epsilon n squared edges. 01:11:38.230 --> 01:11:39.660 So now instead of a single pattern 01:11:39.660 --> 01:11:43.370 you have a possibly infinite set of induced patterns and a 01:11:43.370 --> 01:11:49.700 want to make your graph curly H free-- 01:11:49.700 --> 01:11:51.930 induced curly H free. 01:11:51.930 --> 01:11:55.910 And the theorem is that if there exists 01:11:55.910 --> 01:12:03.320 some finite bound, H0, such that if you have few copies-- 01:12:03.320 --> 01:12:06.460 so for all the patterns up to that point-- 01:12:06.460 --> 01:12:08.050 then you can do what you need to do. 01:12:11.050 --> 01:12:13.390 So take some time to even digest this statement, 01:12:13.390 --> 01:12:16.000 but it's somehow infinite versions-- the correct infinite 01:12:16.000 --> 01:12:18.220 version of the removal lemma if you 01:12:18.220 --> 01:12:21.347 have infinitely many patterns that you need to remove. 01:12:21.347 --> 01:12:22.930 And I claim that the proof is actually 01:12:22.930 --> 01:12:25.640 more or less the same proof as the one that we did here, 01:12:25.640 --> 01:12:27.760 except now you need to take your epsilon 01:12:27.760 --> 01:12:31.900 case, as in this corollary, to depend on k. 01:12:31.900 --> 01:12:36.670 You need to in some way look ahead in this infinite pattern. 01:12:36.670 --> 01:12:50.160 So here in proof, this epsilon k from corollary depends on k. 01:12:50.160 --> 01:13:02.790 And also it depends on your family of patterns H. 01:13:02.790 --> 01:13:05.340 Finally, I want to mention a perspective-- 01:13:05.340 --> 01:13:09.022 a computer science perspective on these removal 01:13:09.022 --> 01:13:10.730 lemmas that we've been discussing so far. 01:13:14.750 --> 01:13:16.510 And that's in the context of something 01:13:16.510 --> 01:13:17.590 called property testing. 01:13:37.830 --> 01:13:43.350 And basically, we would like an efficient-- 01:13:43.350 --> 01:13:51.900 efficient meaning fast-- randomized algorithm 01:13:51.900 --> 01:14:11.820 to distinguish graphs that are triangle-free from those 01:14:11.820 --> 01:14:15.747 that are epsilon far from triangle-free. 01:14:19.330 --> 01:14:22.370 Where being epsilon far from triangle-free means 01:14:22.370 --> 01:14:32.560 that you need to change more than epsilon n squared edges 01:14:32.560 --> 01:14:45.310 here. n is, as usual, the number of vertices to make the graph 01:14:45.310 --> 01:14:45.940 triangle-free. 01:14:45.940 --> 01:14:48.040 So the distance, the [INAUDIBLE] distance 01:14:48.040 --> 01:14:52.750 is more than epsilon away from being triangle-free. 01:14:52.750 --> 01:14:55.450 So somebody gives you a very large graphing. 01:14:55.450 --> 01:14:56.830 n is very large. 01:14:56.830 --> 01:15:00.400 You cannot search through every triple vertices. 01:15:00.400 --> 01:15:01.600 That's too expensive. 01:15:01.600 --> 01:15:06.280 But you want some way to test if a graph is triangle-free 01:15:06.280 --> 01:15:09.551 versus very far away from being triangle-free. 01:15:14.270 --> 01:15:16.670 So there's a very simple randomized algorithm 01:15:16.670 --> 01:15:24.000 to do this, which is to just try randomly 01:15:24.000 --> 01:15:30.850 sample a random triple of vertices 01:15:30.850 --> 01:15:33.090 and check if it's a triangle. 01:15:41.860 --> 01:15:43.710 So you do this. 01:15:43.710 --> 01:15:49.330 And just to make our life a bit more secure, 01:15:49.330 --> 01:15:53.320 let's try it some larger number of times. 01:15:53.320 --> 01:15:59.160 So some c of epsilon some constant number of times. 01:15:59.160 --> 01:16:02.890 And if you find a triangle-- 01:16:02.890 --> 01:16:08.770 so if you don't find a triangle, then we 01:16:08.770 --> 01:16:14.084 return that it's triangle-free. 01:16:18.350 --> 01:16:23.444 Otherwise we return that it is epsilon far from triangle-free. 01:16:31.980 --> 01:16:33.296 So that's the algorithm. 01:16:36.560 --> 01:16:39.470 So it's a very intuitive algorithm, 01:16:39.470 --> 01:16:42.180 but why does it work? 01:16:42.180 --> 01:16:45.050 So we want to know that, indeed, somebody gives you 01:16:45.050 --> 01:16:46.370 one of these two possibilities. 01:16:46.370 --> 01:16:50.340 You run that algorithm, you can succeed with high probability. 01:16:50.340 --> 01:16:51.544 Question? 01:16:51.544 --> 01:16:54.380 AUDIENCE: [INAUDIBLE] 01:16:54.380 --> 01:16:58.500 YUFEI ZHAO: So let's talk about why this works. 01:16:58.500 --> 01:17:02.340 So theorem, for every epsilon, there 01:17:02.340 --> 01:17:08.324 exists a c such that algorithm succeeds 01:17:08.324 --> 01:17:17.760 with probability bigger than 2/3, and 2/3 can be any number. 01:17:17.760 --> 01:17:20.000 So any number that you like because you can always 01:17:20.000 --> 01:17:22.901 repeat it to boost that constant probability. 01:17:26.200 --> 01:17:28.320 So there are two cases. 01:17:28.320 --> 01:17:35.070 If g is triangle-free, then it always succeeds. 01:17:35.070 --> 01:17:36.720 You'll never find this triangle, and it 01:17:36.720 --> 01:17:38.435 would return triangle-free. 01:17:44.260 --> 01:17:55.340 On the other hand, if g is epsilon far from triangle-free, 01:17:55.340 --> 01:18:00.230 then triangle removal lemma tells us 01:18:00.230 --> 01:18:03.913 that g has lots of triangles. 01:18:07.230 --> 01:18:08.850 Delta n cubed triangles. 01:18:12.100 --> 01:18:23.710 So if we sample c being, let's say, 1 over delta times-- 01:18:23.710 --> 01:18:26.770 delta here is a function of epsilon from the triangle 01:18:26.770 --> 01:18:28.640 removal lemma. 01:18:28.640 --> 01:18:34.960 So we find that the probability that the algorithm fails 01:18:34.960 --> 01:18:36.790 is at most-- 01:18:51.710 --> 01:18:53.820 so you have a lot of triangles. 01:18:53.820 --> 01:18:56.580 So very likely you will hit one of these triangles. 01:18:56.580 --> 01:19:00.020 So the probability that the algorithm fails is at most 1 01:19:00.020 --> 01:19:05.780 minus delta n cubed divided by total number of triples 01:19:05.780 --> 01:19:08.240 raised to 1 over delta. 01:19:08.240 --> 01:19:13.090 And this is 1 minus at most 1 minus 6 delta raised 01:19:13.090 --> 01:19:17.900 to 1 over delta, and it's at most e to the minus 6. 01:19:17.900 --> 01:19:21.890 So less than 1/3 in particular. 01:19:21.890 --> 01:19:25.040 So this algorithm succeeds with high probability. 01:19:25.040 --> 01:19:26.850 Now, how big of a c do you need? 01:19:26.850 --> 01:19:30.090 Well, that depends on the triangle removal lemma. 01:19:30.090 --> 01:19:32.290 So it's a constant. 01:19:32.290 --> 01:19:34.410 So it's a constant, does not depend 01:19:34.410 --> 01:19:37.520 on the size of the graph. 01:19:37.520 --> 01:19:39.360 But it's a large constant, because we 01:19:39.360 --> 01:19:41.560 saw in the proof of regularity lemma 01:19:41.560 --> 01:19:42.690 that it can be very large. 01:19:46.800 --> 01:19:49.500 But you know, this theorem here is basically 01:19:49.500 --> 01:19:53.230 the same as a triangle removal lemma. 01:19:53.230 --> 01:19:55.630 So it's highly non-trivial if it's true. 01:19:55.630 --> 01:19:59.470 Even though the algorithm is extremely naive and simple. 01:19:59.470 --> 01:20:01.420 I just want to finish off with one more thing. 01:20:01.420 --> 01:20:03.220 Instead of testing for triangle-freeness, 01:20:03.220 --> 01:20:06.100 you can ask what other properties can you test? 01:20:06.100 --> 01:20:12.590 So which graph properties are testable 01:20:12.590 --> 01:20:13.770 in default in that sense? 01:20:16.950 --> 01:20:19.040 So distinguishing something which 01:20:19.040 --> 01:20:27.070 has the property, so P versus epsilon far from this property 01:20:27.070 --> 01:20:32.120 P. 01:20:32.120 --> 01:20:34.060 And you have this tester which is you 01:20:34.060 --> 01:20:37.020 sample some number of vertices. 01:20:37.020 --> 01:20:38.800 So this is called the oblivious tester. 01:20:42.120 --> 01:20:48.760 So you sample k vertices, and you try 01:20:48.760 --> 01:20:52.710 to see if it has that property. 01:20:52.710 --> 01:20:56.637 So there's a class of properties called hereditary. 01:21:00.050 --> 01:21:02.560 So hereditary properties are properties 01:21:02.560 --> 01:21:05.710 that are closed under vertex deletion. 01:21:11.320 --> 01:21:13.730 And these properties are-- 01:21:13.730 --> 01:21:16.660 lots of properties that you're seeing are of this form. 01:21:16.660 --> 01:21:24.780 So for example, being H3 is this form being planar so this one 01:21:24.780 --> 01:21:31.706 being induced H3, so this one being three-colorable, 01:21:31.706 --> 01:21:33.950 being perfect, they're all examples 01:21:33.950 --> 01:21:35.600 of hereditary properties. 01:21:35.600 --> 01:21:38.210 Properties that if your graph is three-colorable, 01:21:38.210 --> 01:21:41.690 you take out some vertices, it's still three-colorable. 01:21:41.690 --> 01:21:44.120 And all the discussions that we've 01:21:44.120 --> 01:21:47.960 done so far, in particular the infinite removal lemma. 01:21:47.960 --> 01:21:52.010 If you phrase it in the form of property testing given 01:21:52.010 --> 01:22:03.060 the above discussion, it implies that every hereditary property 01:22:03.060 --> 01:22:03.630 is testable. 01:22:06.860 --> 01:22:10.970 In fact, it's testable in the above sense 01:22:10.970 --> 01:22:16.110 with a one-sided error using an oblivious tester. 01:22:16.110 --> 01:22:19.830 One-sided error means that up there if it's triangle-free, 01:22:19.830 --> 01:22:21.260 then it always succeeds. 01:22:21.260 --> 01:22:26.110 So here one of the cases that always succeeds. 01:22:26.110 --> 01:22:28.150 And the reason is that you can characterize 01:22:28.150 --> 01:22:37.350 a hereditary property by a curly H induced H3 01:22:37.350 --> 01:22:41.790 for some curly H. Namely, you're putting everything 01:22:41.790 --> 01:22:45.800 into H that do not have this property. 01:22:53.970 --> 01:22:58.580 This is a possibly infinite set of graphs, 01:22:58.580 --> 01:23:00.500 and that completely characterizes 01:23:00.500 --> 01:23:02.700 this hereditary property. 01:23:02.700 --> 01:23:05.460 And if you read out the infinite removal lemma, 01:23:05.460 --> 01:23:09.950 it says precisely, using above this interpretation, 01:23:09.950 --> 01:23:14.350 that you have a property testing algorithm.