WEBVTT 00:00:00.000 --> 00:00:01.928 align:middle line:90% [SQUEAKING] 00:00:01.928 --> 00:00:03.856 align:middle line:90% [RUSTLING] 00:00:03.856 --> 00:00:05.784 align:middle line:90% [CLICKING] 00:00:05.784 --> 00:00:11.570 align:middle line:90% 00:00:11.570 --> 00:00:13.340 align:middle line:84% YUFEI ZHAO: In this video, we'll look 00:00:13.340 --> 00:00:15.950 align:middle line:84% at an application of the probabilistic method 00:00:15.950 --> 00:00:18.230 align:middle line:90% to graph theory. 00:00:18.230 --> 00:00:20.870 align:middle line:84% An independent set in a graph is a subset 00:00:20.870 --> 00:00:23.450 align:middle line:84% of vertices with no two adjacent. 00:00:23.450 --> 00:00:30.050 align:middle line:84% For example, if the graph is this cycle on four vertices, 00:00:30.050 --> 00:00:36.690 align:middle line:84% an example of an independent set would be two vertices like this. 00:00:36.690 --> 00:00:41.000 align:middle line:90% And these two vertices do not-- 00:00:41.000 --> 00:00:44.270 align:middle line:84% these two vertices are not adjacent to each other, 00:00:44.270 --> 00:00:47.630 align:middle line:84% whereas had I chosen two vertices on the same edge, that 00:00:47.630 --> 00:00:50.240 align:middle line:90% would not be an independent set. 00:00:50.240 --> 00:00:54.630 align:middle line:84% So an important question in graph theory is given a graph, 00:00:54.630 --> 00:00:59.840 align:middle line:84% what can you say about the size of its independent sets? 00:00:59.840 --> 00:01:02.900 align:middle line:84% The following theorem, due to Caro-Wei, 00:01:02.900 --> 00:01:08.660 align:middle line:84% says that every graph G contains a large independent set 00:01:08.660 --> 00:01:09.760 align:middle line:90% in the following sense. 00:01:09.760 --> 00:01:13.860 align:middle line:84% It contains an independent set of size at least 00:01:13.860 --> 00:01:20.910 align:middle line:84% the following quantity, summing over all v among vertices G, 1 00:01:20.910 --> 00:01:23.555 align:middle line:90% over the degree of v plus 1. 00:01:23.555 --> 00:01:26.660 align:middle line:90% 00:01:26.660 --> 00:01:30.240 align:middle line:84% So let us prove this theorem first, 00:01:30.240 --> 00:01:32.930 align:middle line:84% and then we'll see an application and some ways 00:01:32.930 --> 00:01:37.600 align:middle line:84% to interpret this result. The proof of this theorem 00:01:37.600 --> 00:01:41.440 align:middle line:84% applies the probabilistic method. 00:01:41.440 --> 00:01:45.660 align:middle line:84% So we are given this graph G. And the first thing we'll do 00:01:45.660 --> 00:01:57.075 align:middle line:84% is order the vertices of G uniformly at random. 00:01:57.075 --> 00:02:03.010 align:middle line:90% 00:02:03.010 --> 00:02:10.690 align:middle line:84% So consider a random permutation of the vertices of G. And let's 00:02:10.690 --> 00:02:18.220 align:middle line:84% consider the set I of vertices defined as follows. 00:02:18.220 --> 00:02:22.330 align:middle line:84% These are the vertices in G such that v 00:02:22.330 --> 00:02:35.160 align:middle line:84% appears before all it's neighbors 00:02:35.160 --> 00:02:38.150 align:middle line:90% in the random ordering. 00:02:38.150 --> 00:02:45.570 align:middle line:90% 00:02:45.570 --> 00:02:48.420 align:middle line:84% For example, from the graph earlier, 00:02:48.420 --> 00:02:55.740 align:middle line:84% had we ordered the vertices in this order-- 00:02:55.740 --> 00:02:57.910 align:middle line:84% so that's, say, some random ordering. 00:02:57.910 --> 00:03:03.840 align:middle line:84% So let's order the vertices 1, 2, 3, and 4. 00:03:03.840 --> 00:03:13.620 align:middle line:84% So the edges are 1 to 2, 2 to 4, 2 to 4, 3 to 4, and 1 to 3. 00:03:13.620 --> 00:03:19.890 align:middle line:84% What I would pick out is a set of vertices such that-- 00:03:19.890 --> 00:03:23.250 align:middle line:84% so let's see if each vertex would belong to I. 00:03:23.250 --> 00:03:26.620 align:middle line:84% The first vertex appears before all of its neighbors. 00:03:26.620 --> 00:03:31.230 align:middle line:84% So we put the first vertex in I. The second vertex does not 00:03:31.230 --> 00:03:33.130 align:middle line:84% appear before one of its neighbors, 1, 00:03:33.130 --> 00:03:39.570 align:middle line:84% so we do not put it in I. And the third vertex here also does 00:03:39.570 --> 00:03:40.960 align:middle line:90% not appear before its neighbor. 00:03:40.960 --> 00:03:44.070 align:middle line:84% And we do not put it in I. And the fourth vertex, likewise, 00:03:44.070 --> 00:03:46.300 align:middle line:84% does not appear before all of its neighbors. 00:03:46.300 --> 00:03:48.180 align:middle line:84% And we do not put it in I as well. 00:03:48.180 --> 00:03:52.390 align:middle line:84% So in this case, this set picks up only the first vertex. 00:03:52.390 --> 00:03:54.430 align:middle line:84% But in more complicated graphs, it 00:03:54.430 --> 00:03:56.040 align:middle line:90% can pick up additional vertices. 00:03:56.040 --> 00:04:01.140 align:middle line:90% 00:04:01.140 --> 00:04:11.430 align:middle line:84% I claim that I is always an independent set. 00:04:11.430 --> 00:04:17.630 align:middle line:90% 00:04:17.630 --> 00:04:22.040 align:middle line:84% Indeed, it will be impossible to have two vertices in I that 00:04:22.040 --> 00:04:24.560 align:middle line:84% are adjacent to each other because one of them 00:04:24.560 --> 00:04:27.290 align:middle line:84% would appear before the other in the order, 00:04:27.290 --> 00:04:35.680 align:middle line:84% and that would violate the condition on how we chose I. 00:04:35.680 --> 00:04:38.680 align:middle line:84% Next, let's think about how large I is. 00:04:38.680 --> 00:04:52.220 align:middle line:84% For every vertex V and G, the probability that V is in I-- 00:04:52.220 --> 00:05:02.660 align:middle line:84% well, this is the probability that the V appears first 00:05:02.660 --> 00:05:07.650 align:middle line:90% among V and its neighbors. 00:05:07.650 --> 00:05:14.830 align:middle line:90% 00:05:14.830 --> 00:05:19.270 align:middle line:84% Well v has D sub V with degree of V neighbors. 00:05:19.270 --> 00:05:22.880 align:middle line:84% And all the ordering are chosen uniformly at random, 00:05:22.880 --> 00:05:28.410 align:middle line:84% so this probability is 1 divided by the degree of V plus 1. 00:05:28.410 --> 00:05:31.510 align:middle line:84% And thus by the linearity of expectations, 00:05:31.510 --> 00:05:37.320 align:middle line:84% the expected size of this set I is 00:05:37.320 --> 00:05:45.030 align:middle line:84% equal to the sum over vertices V in G of the probability 00:05:45.030 --> 00:05:54.860 align:middle line:84% that this V lies in I. And we just computed this probability 00:05:54.860 --> 00:05:55.360 align:middle line:90% as such. 00:05:55.360 --> 00:05:58.260 align:middle line:90% 00:05:58.260 --> 00:05:59.840 align:middle line:90% This is the expectation. 00:05:59.840 --> 00:06:01.685 align:middle line:90% This is what happens on average. 00:06:01.685 --> 00:06:05.490 align:middle line:84% And therefore, there must be some 00:06:05.490 --> 00:06:12.240 align:middle line:84% ordering that induces an I so there 00:06:12.240 --> 00:06:22.065 align:middle line:84% is some I with the size of I at least the quantity that we just 00:06:22.065 --> 00:06:22.565 align:middle line:90% produced. 00:06:22.565 --> 00:06:27.840 align:middle line:90% 00:06:27.840 --> 00:06:35.070 align:middle line:84% So that finishes the proof of the Caro-Wei theorem. 00:06:35.070 --> 00:06:37.890 align:middle line:84% Again, it says-- the way to think about this theorem 00:06:37.890 --> 00:06:41.880 align:middle line:84% is that if you are given a graph g such that 00:06:41.880 --> 00:06:45.720 align:middle line:84% typically one does not have large degrees, 00:06:45.720 --> 00:06:50.475 align:middle line:84% then it must contain a large independent set. 00:06:50.475 --> 00:06:53.040 align:middle line:90% 00:06:53.040 --> 00:06:57.420 align:middle line:84% By considering the complement of this graph G-- 00:06:57.420 --> 00:07:00.020 align:middle line:90% 00:07:00.020 --> 00:07:12.880 align:middle line:84% by considering the graph complement-- 00:07:12.880 --> 00:07:19.030 align:middle line:84% the graph complement means that we switch, flip 00:07:19.030 --> 00:07:20.810 align:middle line:90% the edge and non-edges. 00:07:20.810 --> 00:07:22.660 align:middle line:84% So this is a graph G. And then this 00:07:22.660 --> 00:07:25.690 align:middle line:84% will be the complement of G, So flip the edges and non-edges 00:07:25.690 --> 00:07:33.550 align:middle line:84% on G. Independent sets in G become 00:07:33.550 --> 00:07:36.857 align:middle line:90% cliques in the complement. 00:07:36.857 --> 00:07:38.940 align:middle line:84% The cliques are complete graphs with all the edges 00:07:38.940 --> 00:07:42.930 align:middle line:90% present and vice versa. 00:07:42.930 --> 00:07:48.870 align:middle line:84% So we can read Caro-Wei's theorem for the complement of G 00:07:48.870 --> 00:07:51.130 align:middle line:84% to deduce the following corollary. 00:07:51.130 --> 00:07:57.808 align:middle line:90% 00:07:57.808 --> 00:08:01.830 align:middle line:90% It says that every n-vertex. 00:08:01.830 --> 00:08:06.590 align:middle line:90% 00:08:06.590 --> 00:08:28.913 align:middle line:84% graph G contains a clique on at least this many vertices. 00:08:28.913 --> 00:08:36.059 align:middle line:90% 00:08:36.059 --> 00:08:40.440 align:middle line:84% So the expression inside the summation is the clique of-- 00:08:40.440 --> 00:08:43.870 align:middle line:84% is the complement or the degree in the complement. 00:08:43.870 --> 00:08:48.780 align:middle line:90% 00:08:48.780 --> 00:08:54.190 align:middle line:84% Let us derive a interesting corollary of this corollary. 00:08:54.190 --> 00:08:57.900 align:middle line:84% And this turns out to be a pretty foundational result 00:08:57.900 --> 00:09:00.155 align:middle line:84% in graph theory known as Turan's theorem. 00:09:00.155 --> 00:09:17.610 align:middle line:90% 00:09:17.610 --> 00:09:19.720 align:middle line:84% Turan's theorem says the following. 00:09:19.720 --> 00:09:26.160 align:middle line:90% 00:09:26.160 --> 00:09:43.890 align:middle line:84% If an n-vertex graph has more than the following number, 00:09:43.890 --> 00:09:48.190 align:middle line:84% 1 minus 1 over r n squared over 2-- 00:09:48.190 --> 00:09:51.545 align:middle line:84% so it has more than this many edges-- 00:09:51.545 --> 00:09:55.550 align:middle line:90% 00:09:55.550 --> 00:10:12.635 align:middle line:84% then the graph contains clique on more than r vertices. 00:10:12.635 --> 00:10:23.900 align:middle line:90% 00:10:23.900 --> 00:10:28.310 align:middle line:84% So Turan's theorem gives us some bound 00:10:28.310 --> 00:10:33.073 align:middle line:84% on the maximum number of edges that a graph can have if it 00:10:33.073 --> 00:10:34.240 align:middle line:90% doesn't have a large clique. 00:10:34.240 --> 00:10:37.870 align:middle line:90% 00:10:37.870 --> 00:10:42.130 align:middle line:84% And furthermore, this bound here is 00:10:42.130 --> 00:10:44.205 align:middle line:84% best possible in the following sense. 00:10:44.205 --> 00:10:56.530 align:middle line:90% 00:10:56.530 --> 00:11:01.120 align:middle line:84% And here let us restrict to the case when n is divisible by r. 00:11:01.120 --> 00:11:02.890 align:middle line:84% When n is not divisible by r, you 00:11:02.890 --> 00:11:05.650 align:middle line:84% need to do a similar construction, 00:11:05.650 --> 00:11:08.770 align:middle line:84% turns out to be best possible, and this bound 00:11:08.770 --> 00:11:12.520 align:middle line:84% can be improved slightly, but not by too much. 00:11:12.520 --> 00:11:14.840 align:middle line:90% So here's the example. 00:11:14.840 --> 00:11:21.585 align:middle line:84% Let us take the n vertices and split them into equal parts. 00:11:21.585 --> 00:11:24.780 align:middle line:90% 00:11:24.780 --> 00:11:26.700 align:middle line:90% And there are r parts. 00:11:26.700 --> 00:11:29.990 align:middle line:84% So here in this illustration r equals to 3. 00:11:29.990 --> 00:11:32.870 align:middle line:90% 00:11:32.870 --> 00:11:38.450 align:middle line:84% And let's put in all the edges between parts, 00:11:38.450 --> 00:11:40.570 align:middle line:90% but no edge within a part. 00:11:40.570 --> 00:11:46.480 align:middle line:84% So this is called the complete r partite graph. 00:11:46.480 --> 00:11:51.080 align:middle line:90% 00:11:51.080 --> 00:11:53.750 align:middle line:84% So you see that this graph here, one 00:11:53.750 --> 00:11:55.190 align:middle line:84% can do a quick calculation to see 00:11:55.190 --> 00:12:00.290 align:middle line:84% that it has exactly this number, 1 minus 1 over r times n 00:12:00.290 --> 00:12:03.350 align:middle line:90% squared over 2 edges. 00:12:03.350 --> 00:12:06.290 align:middle line:84% On the other hand, it does not contain a clique 00:12:06.290 --> 00:12:07.910 align:middle line:90% on more than r vertices. 00:12:07.910 --> 00:12:10.130 align:middle line:84% The biggest clique you have comes 00:12:10.130 --> 00:12:13.670 align:middle line:84% from taking one vertex in each part. 00:12:13.670 --> 00:12:16.820 align:middle line:84% So this is an example showing that the statement of Turan's 00:12:16.820 --> 00:12:19.970 align:middle line:90% theorem is best possible. 00:12:19.970 --> 00:12:24.470 align:middle line:84% Let us prove Turan's theorem as a quick corollary of the earlier 00:12:24.470 --> 00:12:34.120 align:middle line:84% result. So by the earlier corollary, 00:12:34.120 --> 00:12:51.220 align:middle line:84% G has a clique of size, at least sum over the vertices in G, 00:12:51.220 --> 00:12:56.140 align:middle line:90% 1 over n minus the degree of v. 00:12:56.140 --> 00:13:00.166 align:middle line:84% And here, let us use the convexity. 00:13:00.166 --> 00:13:02.980 align:middle line:90% 00:13:02.980 --> 00:13:08.380 align:middle line:84% So in this step, let us use the convexity of the function, 00:13:08.380 --> 00:13:11.270 align:middle line:90% sending x to 1 over n minus x. 00:13:11.270 --> 00:13:13.570 align:middle line:84% So this is a convex function, which 00:13:13.570 --> 00:13:18.910 align:middle line:84% allows us to lower bound the sum by n over n 00:13:18.910 --> 00:13:26.550 align:middle line:90% minus the average degree of G. 00:13:26.550 --> 00:13:32.340 align:middle line:84% On the other hand, if the graph has more than this many edges, 00:13:32.340 --> 00:13:34.470 align:middle line:84% then the average degree is bigger 00:13:34.470 --> 00:13:36.720 align:middle line:84% than some corresponding quantity. 00:13:36.720 --> 00:13:39.510 align:middle line:84% And putting this into the expression, 00:13:39.510 --> 00:13:47.130 align:middle line:84% we can have a final expression strictly bigger than n over n 00:13:47.130 --> 00:13:51.810 align:middle line:90% minus 1 minus 1 over r times n. 00:13:51.810 --> 00:13:56.250 align:middle line:84% And this quantity here equals to r. 00:13:56.250 --> 00:13:59.340 align:middle line:84% So there's a strict equality sign here. 00:13:59.340 --> 00:14:04.200 align:middle line:84% In particular, G has a clique of size strictly greater than r, 00:14:04.200 --> 00:14:06.515 align:middle line:84% and that finishes the proof of Turan's theorem. 00:14:06.515 --> 00:14:10.570 align:middle line:90% 00:14:10.570 --> 00:14:14.170 align:middle line:84% Turan's theorem is a important and foundational result 00:14:14.170 --> 00:14:15.560 align:middle line:90% in graph theory. 00:14:15.560 --> 00:14:18.940 align:middle line:84% And it's a start of a subject that we now call extremal graph 00:14:18.940 --> 00:14:22.210 align:middle line:84% theory, concerning what can you say 00:14:22.210 --> 00:14:26.620 align:middle line:84% about a graph that has certain properties, like not having 00:14:26.620 --> 00:14:28.060 align:middle line:90% a large clique? 00:14:28.060 --> 00:14:31.930 align:middle line:84% And this proof that we just saw is a beautiful illustration 00:14:31.930 --> 00:14:34.585 align:middle line:84% of the probabilistic method to graph theory 00:14:34.585 --> 00:14:39.250 align:middle line:84% that allows us to prove this wonderful result, Turan's 00:14:39.250 --> 00:14:41.920 align:middle line:84% theorem, along with many other results 00:14:41.920 --> 00:14:45.060 align:middle line:84% that I hope that you will learn in the future. 00:14:45.060 --> 00:14:51.000 align:middle line:90%