WEBVTT 00:00:00.000 --> 00:00:01.948 align:middle line:90% [SQUEAKING] 00:00:01.948 --> 00:00:03.409 align:middle line:90% [RUSTLING] 00:00:03.409 --> 00:00:06.331 align:middle line:90% [CLICKING] 00:00:06.331 --> 00:00:11.220 align:middle line:90% 00:00:11.220 --> 00:00:13.560 align:middle line:84% YUFEI ZHAO: The probabilistic method in combinatorics 00:00:13.560 --> 00:00:16.590 align:middle line:84% is a powerful way to demonstrate the existence 00:00:16.590 --> 00:00:18.720 align:middle line:84% of certain special configurations 00:00:18.720 --> 00:00:23.610 align:middle line:84% in combinatorial objects by introducing randomness. 00:00:23.610 --> 00:00:28.050 align:middle line:84% In this video, we will see one of the earliest examples 00:00:28.050 --> 00:00:30.540 align:middle line:84% of an application of this method and its use 00:00:30.540 --> 00:00:32.910 align:middle line:84% to prove lower bounds to Ramsey numbers. 00:00:32.910 --> 00:00:36.930 align:middle line:84% Now, Ramsey numbers are interesting objects 00:00:36.930 --> 00:00:39.640 align:middle line:84% in combinatorics, and they are defined as follows. 00:00:39.640 --> 00:00:44.970 align:middle line:84% The Ramsey number Rkl is defined to be the smallest integer 00:00:44.970 --> 00:00:48.390 align:middle line:84% n, such that no matter how we color 00:00:48.390 --> 00:00:51.890 align:middle line:84% the edges of a complete graph on n vertices, 00:00:51.890 --> 00:00:54.720 align:middle line:90% we'll denote such objects by kn. 00:00:54.720 --> 00:00:58.750 align:middle line:84% So a complete graph is n vertices with all the edges, 00:00:58.750 --> 00:01:02.050 align:middle line:84% all n choose two edges available. 00:01:02.050 --> 00:01:06.000 align:middle line:84% And if we color all such edges, each one of them, 00:01:06.000 --> 00:01:09.450 align:middle line:84% either with red or green, so with one of two colors, then 00:01:09.450 --> 00:01:14.620 align:middle line:84% there always exists a red kl, red clique on k vertices 00:01:14.620 --> 00:01:17.590 align:middle line:90% or green clique on l vertices. 00:01:17.590 --> 00:01:19.660 align:middle line:84% So the Ramsey number is the smallest n such 00:01:19.660 --> 00:01:20.690 align:middle line:90% that this is possible. 00:01:20.690 --> 00:01:28.810 align:middle line:84% So to be more concrete, let me demonstrate an example. 00:01:28.810 --> 00:01:34.480 align:middle line:84% So R 3, 3 equals to 6, and what this means is the following. 00:01:34.480 --> 00:01:40.930 align:middle line:84% First, if we have six vertices and we color 00:01:40.930 --> 00:01:46.780 align:middle line:84% all the possible 6 choose two edges, each one with one 00:01:46.780 --> 00:01:48.590 align:middle line:90% of two colors, red or green. 00:01:48.590 --> 00:01:54.250 align:middle line:84% So for example, maybe these edges are colored green 00:01:54.250 --> 00:02:06.390 align:middle line:84% and the remaining edges are colored red, 00:02:06.390 --> 00:02:08.979 align:middle line:84% then no matter how we do these colorings, 00:02:08.979 --> 00:02:12.870 align:middle line:84% there always exists a triangle, which 00:02:12.870 --> 00:02:16.590 align:middle line:84% is completely red or completely green. 00:02:16.590 --> 00:02:21.490 align:middle line:84% In this example here, you see this. 00:02:21.490 --> 00:02:25.850 align:middle line:84% There's a triangle that is completely red. 00:02:25.850 --> 00:02:32.420 align:middle line:84% On the other hand, if we start with only five vertices, 00:02:32.420 --> 00:02:39.925 align:middle line:84% then it is possible to color the edges with one of two colors. 00:02:39.925 --> 00:02:43.160 align:middle line:90% 00:02:43.160 --> 00:02:45.510 align:middle line:84% So color each edge with one of two colors 00:02:45.510 --> 00:02:50.210 align:middle line:84% so that there is no monochromatic triangle. 00:02:50.210 --> 00:02:55.670 align:middle line:84% So this is what it means for R 3, 3 to be 6. 00:02:55.670 --> 00:03:02.170 align:middle line:84% An important result in combinatorics, due to Ramsey, 00:03:02.170 --> 00:03:09.320 align:middle line:84% so this is known as Ramsey's theorem, 00:03:09.320 --> 00:03:18.060 align:middle line:84% and this was proved back in the 1920s, says that this number, 00:03:18.060 --> 00:03:20.110 align:middle line:90% Rkl is always finite. 00:03:20.110 --> 00:03:22.260 align:middle line:84% It is always a well-defined number. 00:03:22.260 --> 00:03:26.220 align:middle line:84% So as long as n is large enough, no matter 00:03:26.220 --> 00:03:30.310 align:middle line:84% how we color the edges of a complete graph on n vertices, 00:03:30.310 --> 00:03:34.860 align:middle line:84% there always exists a large monochromatic, 00:03:34.860 --> 00:03:40.500 align:middle line:84% either a large red k clique or a large green L clique. 00:03:40.500 --> 00:03:44.970 align:middle line:84% What I want to prove in this video is the following theorem 00:03:44.970 --> 00:03:53.200 align:middle line:90% due to Erdos from the 1940s. 00:03:53.200 --> 00:04:02.290 align:middle line:84% An Erdos showed that Rkk is bigger than 2 to the k over 2 00:04:02.290 --> 00:04:07.090 align:middle line:84% for every integer k bigger than 3. 00:04:07.090 --> 00:04:09.250 align:middle line:84% So this is a lower bound to the Ramsey numbers. 00:04:09.250 --> 00:04:13.300 align:middle line:84% And what this says that if you have too few vertices, namely 00:04:13.300 --> 00:04:15.530 align:middle line:84% if you have 2 to the k over 2 vertices, 00:04:15.530 --> 00:04:20.360 align:middle line:84% then there is always a way to color the edges with two colors 00:04:20.360 --> 00:04:22.600 align:middle line:84% so that there is no monochromatic clique 00:04:22.600 --> 00:04:23.940 align:middle line:90% on k vertices. 00:04:23.940 --> 00:04:28.740 align:middle line:90% 00:04:28.740 --> 00:04:32.130 align:middle line:84% Erdos's theorem is an important foundational result 00:04:32.130 --> 00:04:34.560 align:middle line:84% and is also one of the earliest demonstrations 00:04:34.560 --> 00:04:36.540 align:middle line:90% of the probabilistic method. 00:04:36.540 --> 00:04:41.330 align:middle line:84% So we will see this proof in this lecture, in this video. 00:04:41.330 --> 00:04:44.630 align:middle line:90% 00:04:44.630 --> 00:04:49.190 align:middle line:84% In fact, we will prove a slightly more general statement. 00:04:49.190 --> 00:04:56.456 align:middle line:84% We will show that if two integers, n and k, 00:04:56.456 --> 00:05:00.640 align:middle line:84% satisfy the following inequality, 00:05:00.640 --> 00:05:06.390 align:middle line:84% n choose k times 2 to the 1 minus k choose two, 00:05:06.390 --> 00:05:10.470 align:middle line:90% if this quantity is less than 1. 00:05:10.470 --> 00:05:12.170 align:middle line:84% So right now, this is just some formula. 00:05:12.170 --> 00:05:13.420 align:middle line:90% It will come out of the proof. 00:05:13.420 --> 00:05:16.540 align:middle line:84% So let's not worry too much about it. 00:05:16.540 --> 00:05:26.190 align:middle line:84% Then Rkk is strictly larger than n. 00:05:26.190 --> 00:05:30.340 align:middle line:90% 00:05:30.340 --> 00:05:35.200 align:middle line:84% We will prove the second theorem and, by a calculation that 00:05:35.200 --> 00:05:37.490 align:middle line:84% is fairly straightforward, which I will admit, 00:05:37.490 --> 00:05:44.050 align:middle line:84% one can use the second result to imply the first result. 00:05:44.050 --> 00:05:48.820 align:middle line:84% OK, so let us prove the second theorem over here. 00:05:48.820 --> 00:05:50.110 align:middle line:90% So what is this saying? 00:05:50.110 --> 00:05:52.940 align:middle line:84% If n satisfies certain inequalities, 00:05:52.940 --> 00:05:55.150 align:middle line:84% so the n is not too large, then it 00:05:55.150 --> 00:05:59.140 align:middle line:84% is possible to color the edges of a complete graph 00:05:59.140 --> 00:06:03.445 align:middle line:84% on n vertices so that there is no monochromatic clique on k 00:06:03.445 --> 00:06:03.945 align:middle line:90% vertices. 00:06:03.945 --> 00:06:07.280 align:middle line:90% 00:06:07.280 --> 00:06:09.830 align:middle line:90% How do we find such a coloring? 00:06:09.830 --> 00:06:11.840 align:middle line:90% Well, let's do it randomly. 00:06:11.840 --> 00:06:21.530 align:middle line:84% So let's color the edges of a complete graph on k 00:06:21.530 --> 00:06:36.890 align:middle line:84% vertices using two colors uniformly at random, 00:06:36.890 --> 00:06:40.610 align:middle line:84% so meaning that for each edge, we flip a coin. 00:06:40.610 --> 00:06:42.870 align:middle line:84% If it comes up heads, we color it red. 00:06:42.870 --> 00:06:46.520 align:middle line:84% If it comes up tails, we color it green. 00:06:46.520 --> 00:06:50.510 align:middle line:84% And the goal is to show that we can 00:06:50.510 --> 00:06:52.220 align:middle line:84% do this in such a way that avoids 00:06:52.220 --> 00:06:55.460 align:middle line:84% having a monochromatic clique of size k. 00:06:55.460 --> 00:06:59.240 align:middle line:84% So there are things that we want to avoid, and let us encode 00:06:59.240 --> 00:07:01.460 align:middle line:90% these bad events as follows. 00:07:01.460 --> 00:07:14.260 align:middle line:84% For every k vertex subset s of vertices, 00:07:14.260 --> 00:07:15.770 align:middle line:90% there are n vertices in total. 00:07:15.770 --> 00:07:19.720 align:middle line:84% So for every k vertex subset s, we'll 00:07:19.720 --> 00:07:24.490 align:middle line:84% use A sub s to denote the following bad event that we're 00:07:24.490 --> 00:07:26.320 align:middle line:90% trying to avoid. 00:07:26.320 --> 00:07:44.040 align:middle line:84% This is the event that s induces a monochromatic clique. 00:07:44.040 --> 00:07:46.865 align:middle line:84% OK, it induces a monochromatic clique in this coloring. 00:07:46.865 --> 00:07:50.520 align:middle line:90% 00:07:50.520 --> 00:07:54.690 align:middle line:84% Let's calculate the probability that this occurs. 00:07:54.690 --> 00:07:58.830 align:middle line:84% So what's the probability that in this s, which 00:07:58.830 --> 00:08:03.360 align:middle line:84% let's say k is 4 for purpose of this illustration, 00:08:03.360 --> 00:08:06.550 align:middle line:84% of the six edges, all of them have the same color. 00:08:06.550 --> 00:08:08.850 align:middle line:84% Well, there are two possible colors, 00:08:08.850 --> 00:08:11.700 align:middle line:84% and for each of those colors, the probability that all 00:08:11.700 --> 00:08:15.360 align:middle line:84% of these edges are of that color is this quantity, 00:08:15.360 --> 00:08:17.530 align:middle line:90% 2 to the minus k choose two. 00:08:17.530 --> 00:08:21.270 align:middle line:84% So this is the probability 2 times 2 to the minus k 00:08:21.270 --> 00:08:21.880 align:middle line:90% choose two. 00:08:21.880 --> 00:08:24.960 align:middle line:84% This is the probability that s, all the edges in s 00:08:24.960 --> 00:08:28.670 align:middle line:90% are entirely of the same color. 00:08:28.670 --> 00:08:32.900 align:middle line:84% Well, what about the probability that there 00:08:32.900 --> 00:08:39.470 align:middle line:84% is some clique of size k in the graph that's monochromatic? 00:08:39.470 --> 00:08:41.455 align:middle line:84% That's really the quantity that we care about. 00:08:41.455 --> 00:08:44.480 align:middle line:90% 00:08:44.480 --> 00:08:46.250 align:middle line:84% So let's consider the probability 00:08:46.250 --> 00:09:04.280 align:middle line:84% that there is some monochromatic clique of size k in this graph. 00:09:04.280 --> 00:09:07.750 align:middle line:90% 00:09:07.750 --> 00:09:10.250 align:middle line:84% Well, if there is some monochromatic clique of size k, 00:09:10.250 --> 00:09:12.980 align:middle line:90% then it must be one of the S's. 00:09:12.980 --> 00:09:23.430 align:middle line:84% So we can do what's called a union bound to upper bound 00:09:23.430 --> 00:09:29.070 align:middle line:84% this probability by the sum of the individual event 00:09:29.070 --> 00:09:30.960 align:middle line:90% probabilities. 00:09:30.960 --> 00:09:35.520 align:middle line:84% So summing over all subsets s of the vertices with size exactly 00:09:35.520 --> 00:09:41.630 align:middle line:84% k of the probability that A sub s this event occurs. 00:09:41.630 --> 00:09:47.420 align:middle line:84% In this summation, there are n choose k terms, 00:09:47.420 --> 00:09:49.370 align:middle line:84% and each term is this probability 00:09:49.370 --> 00:09:50.755 align:middle line:90% that we calculated earlier. 00:09:50.755 --> 00:09:53.630 align:middle line:90% 00:09:53.630 --> 00:09:56.030 align:middle line:90% And you see this expression. 00:09:56.030 --> 00:09:59.300 align:middle line:84% It was the expression in the hypothesis of the theorem. 00:09:59.300 --> 00:10:04.310 align:middle line:84% And we assume that it is strictly less than 1. 00:10:04.310 --> 00:10:09.140 align:middle line:84% And this is why we had this expression in the theorem. 00:10:09.140 --> 00:10:10.490 align:middle line:90% OK, great. 00:10:10.490 --> 00:10:14.240 align:middle line:84% So the probability that there is some monochromatic clique 00:10:14.240 --> 00:10:17.840 align:middle line:84% of size k is strictly less than 1, 00:10:17.840 --> 00:10:30.660 align:middle line:84% and thus with positive probability 00:10:30.660 --> 00:10:50.510 align:middle line:84% a random coloring has no monochromatic clique of size k. 00:10:50.510 --> 00:10:55.230 align:middle line:90% 00:10:55.230 --> 00:10:58.310 align:middle line:84% And that's basically, what we are trying to do. 00:10:58.310 --> 00:11:02.040 align:middle line:84% We're trying to demonstrate the existence of a coloring 00:11:02.040 --> 00:11:04.380 align:middle line:84% with no monochromatic clique of size k. 00:11:04.380 --> 00:11:08.340 align:middle line:84% And we show that, in fact, by doing everything at random, 00:11:08.340 --> 00:11:12.740 align:middle line:84% the random coloring succeeds with positive probability. 00:11:12.740 --> 00:11:16.940 align:middle line:84% And therefore, such an object must exist. 00:11:16.940 --> 00:11:20.665 align:middle line:90% Thus such a coloring exists. 00:11:20.665 --> 00:11:24.990 align:middle line:90% 00:11:24.990 --> 00:11:28.830 align:middle line:84% And this finishes the proof of this theorem over here 00:11:28.830 --> 00:11:31.170 align:middle line:84% that we were trying to demonstrate. 00:11:31.170 --> 00:11:35.010 align:middle line:84% And as I mentioned earlier, by a more routine calculation, which 00:11:35.010 --> 00:11:38.400 align:middle line:84% we will not do here, one can deduce Erdos's result 00:11:38.400 --> 00:11:40.170 align:middle line:90% from this theorem. 00:11:40.170 --> 00:11:43.520 align:middle line:84% And in fact, by a more careful calculation that, again, I 00:11:43.520 --> 00:11:48.270 align:middle line:84% will not do, one can show an even more precise lower bound, 00:11:48.270 --> 00:11:53.430 align:middle line:84% that Rkk is bigger than the following asymptotic 1 00:11:53.430 --> 00:11:57.240 align:middle line:84% over the following constant e times root 2. 00:11:57.240 --> 00:12:02.040 align:middle line:84% So here e is the natural constant, around 2.71, 00:12:02.040 --> 00:12:10.460 align:middle line:84% plus a term that goes to 0 as k goes to infinity times k 2 00:12:10.460 --> 00:12:11.400 align:middle line:90% to the k over 2. 00:12:11.400 --> 00:12:14.780 align:middle line:84% So this is a much more precise bound 00:12:14.780 --> 00:12:17.300 align:middle line:84% that one can get out of analyzing 00:12:17.300 --> 00:12:21.610 align:middle line:84% the consequence of this theorem that we just proved. 00:12:21.610 --> 00:12:23.620 align:middle line:84% In the next part of this video, we 00:12:23.620 --> 00:12:28.300 align:middle line:84% will refine or come up with additional techniques that 00:12:28.300 --> 00:12:30.260 align:middle line:84% allows us to improve this lower bound. 00:12:30.260 --> 00:12:35.910 align:middle line:84% But for now, let me make a comment about this technique. 00:12:35.910 --> 00:12:40.830 align:middle line:84% What this theorem shows is that by doing a random coloring 00:12:40.830 --> 00:12:44.820 align:middle line:84% there is a positive probability that this coloring gets us 00:12:44.820 --> 00:12:46.670 align:middle line:90% what we want. 00:12:46.670 --> 00:12:49.970 align:middle line:84% And in fact, the probability that this coloring doesn't 00:12:49.970 --> 00:12:53.100 align:middle line:84% get us what we want is this quantity here, 00:12:53.100 --> 00:12:56.660 align:middle line:84% which is not only less than 1, but it is actually quite small. 00:12:56.660 --> 00:12:58.220 align:middle line:90% It is very, very small. 00:12:58.220 --> 00:13:02.450 align:middle line:84% So if you, for example, on a computer flip these random coins 00:13:02.450 --> 00:13:05.850 align:middle line:84% and come up with this coloring, with very high probability, 00:13:05.850 --> 00:13:07.760 align:middle line:90% it will succeed. 00:13:07.760 --> 00:13:13.070 align:middle line:84% However, it will be very difficult to verify 00:13:13.070 --> 00:13:17.030 align:middle line:84% that you have succeeded because the number of cliques that one 00:13:17.030 --> 00:13:20.810 align:middle line:84% needs to check is n choose k, which grows very 00:13:20.810 --> 00:13:23.570 align:middle line:90% quickly as a function of k. 00:13:23.570 --> 00:13:26.420 align:middle line:84% So this is a phenomenon that is sometimes called 00:13:26.420 --> 00:13:29.905 align:middle line:90% finding hay in a haystack. 00:13:29.905 --> 00:13:38.130 align:middle line:90% 00:13:38.130 --> 00:13:42.150 align:middle line:84% A random coloring works with overwhelming probability. 00:13:42.150 --> 00:13:44.500 align:middle line:84% And so if you just pick something at random, 00:13:44.500 --> 00:13:45.490 align:middle line:90% it will work. 00:13:45.490 --> 00:13:48.000 align:middle line:84% But one, it is very hard to check that what 00:13:48.000 --> 00:13:49.730 align:middle line:90% you found actually works. 00:13:49.730 --> 00:13:53.970 align:middle line:84% And two, it is an active research direction and what 00:13:53.970 --> 00:13:55.470 align:middle line:84% seems to be a very difficult problem 00:13:55.470 --> 00:14:01.680 align:middle line:84% to find other ways that does not involve randomness but maybe 00:14:01.680 --> 00:14:03.900 align:middle line:84% even involving randomness in some other ways 00:14:03.900 --> 00:14:07.200 align:middle line:84% to guarantee that what you found works. 00:14:07.200 --> 00:14:09.600 align:middle line:84% And this finding hay in a haystack 00:14:09.600 --> 00:14:12.550 align:middle line:84% describes this phenomenon, which is quite counterintuitive, 00:14:12.550 --> 00:14:13.110 align:middle line:90% that-- 00:14:13.110 --> 00:14:15.690 align:middle line:84% and also demonstrates the power of the probabilistic method, 00:14:15.690 --> 00:14:17.920 align:middle line:84% that by just doing everything in random, 00:14:17.920 --> 00:14:20.370 align:middle line:90% it works really, really well. 00:14:20.370 --> 00:14:23.460 align:middle line:84% But we don't understand this process as much 00:14:23.460 --> 00:14:25.730 align:middle line:90% as we would really like to. 00:14:25.730 --> 00:14:31.140 align:middle line:84% So now you've seen a proof of Erdos's seminal result 00:14:31.140 --> 00:14:34.150 align:middle line:84% that gives a lower bound to Ramsey numbers. 00:14:34.150 --> 00:14:36.490 align:middle line:84% In the next two segments of this video, 00:14:36.490 --> 00:14:38.850 align:middle line:84% we will see a couple of refinements 00:14:38.850 --> 00:14:41.010 align:middle line:84% to this technique that allows us to get 00:14:41.010 --> 00:14:44.370 align:middle line:84% slightly better lower bounds by introducing new ideas 00:14:44.370 --> 00:14:45.635 align:middle line:90% to the probabilistic method. 00:14:45.635 --> 00:14:49.010 align:middle line:90% 00:14:49.010 --> 00:14:51.350 align:middle line:84% Now, let us introduce an additional idea 00:14:51.350 --> 00:14:54.920 align:middle line:84% to the probabilistic method to prove a slightly better lower 00:14:54.920 --> 00:14:57.500 align:middle line:90% bound on Ramsey numbers. 00:14:57.500 --> 00:14:59.075 align:middle line:84% And this is a method of alterations 00:14:59.075 --> 00:15:02.150 align:middle line:90% called the alteration method. 00:15:02.150 --> 00:15:04.520 align:middle line:84% What we'll prove is the following. 00:15:04.520 --> 00:15:06.360 align:middle line:84% So let me write down the statement. 00:15:06.360 --> 00:15:08.060 align:middle line:90% It's slightly technical. 00:15:08.060 --> 00:15:10.730 align:middle line:84% The statement itself is not as important as the ideas I 00:15:10.730 --> 00:15:13.190 align:middle line:84% will introduce, so bear with me for a second. 00:15:13.190 --> 00:15:20.100 align:middle line:84% So theorem says that for any positive integers k and n, 00:15:20.100 --> 00:15:24.180 align:middle line:84% the Ramsey number, Rkk, is bigger 00:15:24.180 --> 00:15:30.420 align:middle line:84% than the following quantity, n minus n choose k times 2 00:15:30.420 --> 00:15:34.900 align:middle line:90% to the 1 minus k choose two. 00:15:34.900 --> 00:15:36.590 align:middle line:90% OK, so that's the theorem. 00:15:36.590 --> 00:15:39.890 align:middle line:84% It gives you a lower bound on Ramsey numbers. 00:15:39.890 --> 00:15:44.320 align:middle line:84% So, by optimizing over the value of k as a value of n, 00:15:44.320 --> 00:15:49.940 align:middle line:84% as a function of k, we can deduce the following corollary. 00:15:49.940 --> 00:15:52.490 align:middle line:84% And I will not show this deduction here 00:15:52.490 --> 00:15:54.710 align:middle line:84% because it's a more routine calculation. 00:15:54.710 --> 00:16:01.340 align:middle line:84% So the corollary is that Rkk is bigger than the following 00:16:01.340 --> 00:16:06.050 align:middle line:84% quantity, 1 over e-- e is the natural constant 2.71, 00:16:06.050 --> 00:16:08.930 align:middle line:90% and so on-- plus little 1-- 00:16:08.930 --> 00:16:12.260 align:middle line:84% a term that goes to 0 as k goes to infinity-- 00:16:12.260 --> 00:16:16.320 align:middle line:84% times k times 2 to the k over 2, OK? 00:16:16.320 --> 00:16:18.900 align:middle line:84% So that's the asymptotic lower bound 00:16:18.900 --> 00:16:21.570 align:middle line:84% that one obtains by taking this theorem 00:16:21.570 --> 00:16:25.380 align:middle line:84% and optimizing the value of n as a function of k. 00:16:25.380 --> 00:16:27.510 align:middle line:84% We see that it is slightly better than the bound 00:16:27.510 --> 00:16:29.970 align:middle line:84% that we got previously, is slightly better 00:16:29.970 --> 00:16:33.510 align:middle line:90% by a factor of root 2. 00:16:33.510 --> 00:16:36.155 align:middle line:84% OK, so let us prove this theorem here. 00:16:36.155 --> 00:16:38.730 align:middle line:90% 00:16:38.730 --> 00:16:45.630 align:middle line:84% Like before, we will start with a random coloring 00:16:45.630 --> 00:16:47.520 align:middle line:90% but then make some adjustments. 00:16:47.520 --> 00:16:50.870 align:middle line:84% So in fact, this construction will have two steps. 00:16:50.870 --> 00:16:53.975 align:middle line:84% Construct a random coloring in two steps. 00:16:53.975 --> 00:17:02.760 align:middle line:90% 00:17:02.760 --> 00:17:06.000 align:middle line:90% In the first step, We'll. 00:17:06.000 --> 00:17:18.000 align:middle line:84% Do what we did previously just to randomly color each 00:17:18.000 --> 00:17:28.625 align:middle line:84% edge with one of two colors uniformly at random. 00:17:28.625 --> 00:17:34.370 align:middle line:90% 00:17:34.370 --> 00:17:36.410 align:middle line:84% OK, so flip a coin for each edge, 00:17:36.410 --> 00:17:39.830 align:middle line:84% and color it with one of the two colors. 00:17:39.830 --> 00:17:46.150 align:middle line:84% Now, this coloring might have some monochromatic k cliques, 00:17:46.150 --> 00:17:52.210 align:middle line:84% in which case, we will delete a vertex from each such k clique 00:17:52.210 --> 00:17:55.100 align:middle line:84% and to destroy all the monochromatic k cliques. 00:17:55.100 --> 00:17:58.600 align:middle line:84% So the second step is to delete a vertex 00:17:58.600 --> 00:18:07.796 align:middle line:84% from each monochromatic k clique. 00:18:07.796 --> 00:18:10.570 align:middle line:90% 00:18:10.570 --> 00:18:15.280 align:middle line:84% OK, so after this process, see, what we are left with 00:18:15.280 --> 00:18:21.850 align:middle line:84% is an edge coloring of some complete graph where 00:18:21.850 --> 00:18:24.910 align:middle line:84% there are no monochromatic k cliques left because we've 00:18:24.910 --> 00:18:26.710 align:middle line:84% destroyed all such monochromatic k 00:18:26.710 --> 00:18:31.360 align:middle line:84% cliques in the second step of this process. 00:18:31.360 --> 00:18:33.190 align:middle line:90% How many vertices are left? 00:18:33.190 --> 00:18:35.990 align:middle line:84% So we started with initially n vertices, 00:18:35.990 --> 00:18:38.090 align:middle line:84% but in the second step we deleted some. 00:18:38.090 --> 00:18:39.640 align:middle line:84% So we don't have as many vertices 00:18:39.640 --> 00:18:41.900 align:middle line:90% as we started with initially. 00:18:41.900 --> 00:18:50.330 align:middle line:84% So let's ask ourselves, how many vertices are left 00:18:50.330 --> 00:18:51.550 align:middle line:90% at the end of this process? 00:18:51.550 --> 00:18:55.550 align:middle line:90% 00:18:55.550 --> 00:18:59.930 align:middle line:84% Towards this goal, let me introduce a random variable x. 00:18:59.930 --> 00:19:11.602 align:middle line:84% And x is the number of monochromatic k cliques 00:19:11.602 --> 00:19:14.560 align:middle line:90% at the end of the first step. 00:19:14.560 --> 00:19:23.830 align:middle line:90% 00:19:23.830 --> 00:19:30.250 align:middle line:84% Well, the expectation of x can be 00:19:30.250 --> 00:19:33.130 align:middle line:84% computed as follows, which is very similar calculation 00:19:33.130 --> 00:19:35.500 align:middle line:90% to what we did earlier. 00:19:35.500 --> 00:19:39.160 align:middle line:84% There are n choose k possible k cliques, and each one of them 00:19:39.160 --> 00:19:44.200 align:middle line:84% is monochromatic with probability 2 to the 1 minus k 00:19:44.200 --> 00:19:45.370 align:middle line:90% choose two. 00:19:45.370 --> 00:19:49.620 align:middle line:84% So that's the expected number of monochromatic k cliques. 00:19:49.620 --> 00:19:56.370 align:middle line:84% Now, in the second step of this process, 00:19:56.370 --> 00:20:01.240 align:middle line:84% we delete some number of vertices 00:20:01.240 --> 00:20:04.340 align:middle line:84% and one vertex for each monochromatic k clique. 00:20:04.340 --> 00:20:07.330 align:middle line:84% Although, some vertices may be used 00:20:07.330 --> 00:20:11.470 align:middle line:84% to destroy more than one monochromatic k clique. 00:20:11.470 --> 00:20:16.750 align:middle line:84% So the number of vertices that we delete is not necessarily x, 00:20:16.750 --> 00:20:18.520 align:middle line:90% but it is no more than x. 00:20:18.520 --> 00:20:24.895 align:middle line:84% So delete at most the size of x many vertices. 00:20:24.895 --> 00:20:29.550 align:middle line:90% 00:20:29.550 --> 00:20:33.480 align:middle line:84% And thus, the final graph that we 00:20:33.480 --> 00:20:42.770 align:middle line:84% get at the end of this two-step process has at least n 00:20:42.770 --> 00:20:47.420 align:middle line:90% minus the size of x vertices. 00:20:47.420 --> 00:20:50.160 align:middle line:84% Let me remind you that this is a random process. 00:20:50.160 --> 00:20:53.580 align:middle line:84% The final graph and this coloring that we get is random, 00:20:53.580 --> 00:20:57.470 align:middle line:90% and x is also a random quantity. 00:20:57.470 --> 00:21:00.650 align:middle line:84% This is a number of vertices, and this quantity here 00:21:00.650 --> 00:21:11.474 align:middle line:84% has expectation n minus expectation of the size of x. 00:21:11.474 --> 00:21:14.860 align:middle line:90% Oh, so x is already a size. 00:21:14.860 --> 00:21:15.570 align:middle line:90% So let me-- 00:21:15.570 --> 00:21:19.530 align:middle line:84% I don't have to use the absolute value bars. 00:21:19.530 --> 00:21:31.080 align:middle line:84% So the expectation is n minus n choose k 2 to the 1 minus k 00:21:31.080 --> 00:21:31.820 align:middle line:90% choose two. 00:21:31.820 --> 00:21:35.540 align:middle line:90% 00:21:35.540 --> 00:21:38.270 align:middle line:84% OK, so the final graph has at least this many vertices 00:21:38.270 --> 00:21:40.880 align:middle line:90% in expectation. 00:21:40.880 --> 00:21:44.300 align:middle line:84% Well, that's the average number through this random process, 00:21:44.300 --> 00:21:53.720 align:middle line:84% and thus, there must always be, with some positive probability, 00:21:53.720 --> 00:21:56.010 align:middle line:84% some graph that beats this average. 00:21:56.010 --> 00:22:06.990 align:middle line:84% So thus, or therefore, with positive probability 00:22:06.990 --> 00:22:20.643 align:middle line:84% the remaining graph has at least this many vertices. 00:22:20.643 --> 00:22:25.650 align:middle line:90% 00:22:25.650 --> 00:22:37.110 align:middle line:84% And furthermore, it has no monochromatic clique 00:22:37.110 --> 00:22:41.130 align:middle line:84% on k vertices because we destroyed all such cliques 00:22:41.130 --> 00:22:44.800 align:middle line:84% in the second step of this process. 00:22:44.800 --> 00:22:49.690 align:middle line:84% OK, this finishes the proof of this theorem, which, 00:22:49.690 --> 00:22:51.430 align:middle line:84% as you see in the corollary, gives you 00:22:51.430 --> 00:22:53.290 align:middle line:84% a slightly better lower bound compared 00:22:53.290 --> 00:22:55.990 align:middle line:90% to what we saw last time. 00:22:55.990 --> 00:22:57.830 align:middle line:84% In the third segment of this video, 00:22:57.830 --> 00:23:03.160 align:middle line:84% we'll see yet another technique that gets us even further 00:23:03.160 --> 00:23:06.850 align:middle line:84% along to prove an even better lower bound on the Ramsey 00:23:06.850 --> 00:23:09.240 align:middle line:90% numbers. 00:23:09.240 --> 00:23:11.490 align:middle line:84% Earlier, we saw two different approaches 00:23:11.490 --> 00:23:14.340 align:middle line:84% to using the probabilistic method to lower bound Ramsey 00:23:14.340 --> 00:23:15.210 align:middle line:90% numbers. 00:23:15.210 --> 00:23:18.690 align:middle line:84% The first method was by taking a union bound 00:23:18.690 --> 00:23:22.590 align:middle line:84% to show that the sum of all the probabilities of bad events 00:23:22.590 --> 00:23:24.420 align:middle line:90% is fairly small. 00:23:24.420 --> 00:23:27.610 align:middle line:84% And the second method is through alteration, 00:23:27.610 --> 00:23:30.780 align:middle line:84% where we start by constructing some random coloring 00:23:30.780 --> 00:23:34.230 align:middle line:90% and then fix the blemishes. 00:23:34.230 --> 00:23:37.218 align:middle line:84% So we show that by removing the bad parts, 00:23:37.218 --> 00:23:38.760 align:middle line:84% removing the vertices that contribute 00:23:38.760 --> 00:23:41.970 align:middle line:84% to monochromatic cliques, we can get a coloring 00:23:41.970 --> 00:23:45.270 align:middle line:90% with no monochromatic cliques. 00:23:45.270 --> 00:23:48.240 align:middle line:84% Now let me introduce a more advanced method 00:23:48.240 --> 00:23:51.060 align:middle line:90% that can get us even further. 00:23:51.060 --> 00:23:54.750 align:middle line:84% So the general motivation, that we already 00:23:54.750 --> 00:23:56.580 align:middle line:84% saw a couple of times, is that we 00:23:56.580 --> 00:24:07.910 align:middle line:84% wish to avoid a certain collection of bad events, which 00:24:07.910 --> 00:24:13.520 align:middle line:90% we'll denote E1 through EL. 00:24:13.520 --> 00:24:15.170 align:middle line:90% So these are bad events. 00:24:15.170 --> 00:24:16.940 align:middle line:84% In our case, the bad events correspond 00:24:16.940 --> 00:24:20.870 align:middle line:84% to having a monochromatic clique in the random coloring. 00:24:20.870 --> 00:24:22.820 align:middle line:84% There are some extreme situations 00:24:22.820 --> 00:24:25.885 align:middle line:90% that are easier to handle. 00:24:25.885 --> 00:24:34.910 align:middle line:90% 00:24:34.910 --> 00:24:42.590 align:middle line:84% And these extreme situations are one of two types. 00:24:42.590 --> 00:24:48.350 align:middle line:84% The first is when the events collectively 00:24:48.350 --> 00:24:50.870 align:middle line:84% have very small probability, even 00:24:50.870 --> 00:24:54.590 align:middle line:84% when you just sum the probabilities of these events. 00:24:54.590 --> 00:24:57.650 align:middle line:84% So for example, if the sum of the probabilities 00:24:57.650 --> 00:25:00.450 align:middle line:84% of these bad events is less than 1, 00:25:00.450 --> 00:25:07.310 align:middle line:84% then we can apply the union bound 00:25:07.310 --> 00:25:11.910 align:middle line:84% to deduce that with positive probability 00:25:11.910 --> 00:25:13.730 align:middle line:90% none of the bad events occur. 00:25:13.730 --> 00:25:18.140 align:middle line:84% So that's an easy situation to handle. 00:25:18.140 --> 00:25:27.620 align:middle line:84% Another easy situation to handle is if all the events, so all 00:25:27.620 --> 00:25:29.276 align:middle line:90% the bad events are independent. 00:25:29.276 --> 00:25:34.330 align:middle line:90% 00:25:34.330 --> 00:25:36.590 align:middle line:84% So if they are independent events, 00:25:36.590 --> 00:25:44.890 align:middle line:84% then the probability that none of them occur 00:25:44.890 --> 00:25:49.420 align:middle line:84% is equal to the product of the probabilities 00:25:49.420 --> 00:25:52.455 align:middle line:84% that the individual events do not occur. 00:25:52.455 --> 00:26:00.110 align:middle line:90% 00:26:00.110 --> 00:26:04.190 align:middle line:84% And this is positive, provided that all the individual event 00:26:04.190 --> 00:26:08.430 align:middle line:84% probabilities are strictly less than 1. 00:26:08.430 --> 00:26:11.850 align:middle line:84% Now, this situation typically is not 00:26:11.850 --> 00:26:14.170 align:middle line:84% the case in a lot of applications, 00:26:14.170 --> 00:26:16.170 align:middle line:84% including the one that we're considering. 00:26:16.170 --> 00:26:19.320 align:middle line:84% For example, having two triangles, if they 00:26:19.320 --> 00:26:24.930 align:middle line:84% intersect at an edge, if they overlap by an edge, then being-- 00:26:24.930 --> 00:26:30.250 align:middle line:84% or if you have two cliques that intersect in some large portion, 00:26:30.250 --> 00:26:33.930 align:middle line:84% then having them individually being monochromatic 00:26:33.930 --> 00:26:37.680 align:middle line:90% are not independent events. 00:26:37.680 --> 00:26:41.430 align:middle line:84% So what we're often dealing with is 00:26:41.430 --> 00:26:49.440 align:middle line:84% situations where there is some dependencies, 00:26:49.440 --> 00:26:51.536 align:middle line:90% so some event dependencies. 00:26:51.536 --> 00:26:58.140 align:middle line:90% 00:26:58.140 --> 00:27:03.270 align:middle line:84% And often, these are the hardest and most interesting 00:27:03.270 --> 00:27:05.100 align:middle line:84% situations that one has to deal with 00:27:05.100 --> 00:27:08.190 align:middle line:90% in the probabilistic method. 00:27:08.190 --> 00:27:22.298 align:middle line:84% A powerful tool that we will see is the Lovász local lemma, OK? 00:27:22.298 --> 00:27:24.680 align:middle line:84% So let me explain what is a Lovász local lemma 00:27:24.680 --> 00:27:27.530 align:middle line:84% and then apply it to the problem of lower bounds 00:27:27.530 --> 00:27:29.600 align:middle line:90% to Ramsey numbers. 00:27:29.600 --> 00:27:32.660 align:middle line:84% So here is a version of the statement of the Lovász 00:27:32.660 --> 00:27:36.260 align:middle line:84% local lemma and specifically applied 00:27:36.260 --> 00:27:41.075 align:middle line:84% to a setting known as the random variable model, random variables 00:27:41.075 --> 00:27:41.575 align:middle line:90% model. 00:27:41.575 --> 00:27:44.150 align:middle line:90% 00:27:44.150 --> 00:27:46.310 align:middle line:90% The setting is as follows. 00:27:46.310 --> 00:27:53.490 align:middle line:84% Suppose we have x1 through xv, and these are random variables. 00:27:53.490 --> 00:27:58.360 align:middle line:90% 00:27:58.360 --> 00:28:03.900 align:middle line:84% And importantly, they are independent random variables. 00:28:03.900 --> 00:28:05.700 align:middle line:84% So think of them as independent coin 00:28:05.700 --> 00:28:11.610 align:middle line:84% tosses, the outcomes of independent random coin tosses. 00:28:11.610 --> 00:28:20.860 align:middle line:84% We'll have B1 through Bm denote index sets. 00:28:20.860 --> 00:28:33.130 align:middle line:84% So they are subsets of numbers 1 through v. And for each i 00:28:33.130 --> 00:28:54.170 align:middle line:84% ranging from 1 through m, let Ei be some event that depends only 00:28:54.170 --> 00:29:10.470 align:middle line:84% on the variables indexed by Bi, namely, the variables x sub j 00:29:10.470 --> 00:29:13.650 align:middle line:84% as j ranges over the elements of Bi. 00:29:13.650 --> 00:29:17.520 align:middle line:90% 00:29:17.520 --> 00:29:20.430 align:middle line:84% So this is a situation that captures a lot of applications 00:29:20.430 --> 00:29:23.410 align:middle line:84% where you have bad events, for example, 00:29:23.410 --> 00:29:28.200 align:middle line:84% cliques being monochromatic in a coloring of a large graph. 00:29:28.200 --> 00:29:32.280 align:middle line:84% And some of these events may have dependencies 00:29:32.280 --> 00:29:35.700 align:middle line:84% because the events are based on some underlying coin 00:29:35.700 --> 00:29:39.160 align:middle line:84% tosses, underlying independent random variables. 00:29:39.160 --> 00:29:45.060 align:middle line:84% And the different events may involve overlapping variables. 00:29:45.060 --> 00:29:49.360 align:middle line:84% And so this is capturing that kind of setup. 00:29:49.360 --> 00:29:52.660 align:middle line:84% Now, let me introduce some hypotheses that 00:29:52.660 --> 00:29:55.850 align:middle line:84% captures the notion of a weak amount of dependence, 00:29:55.850 --> 00:29:57.610 align:middle line:90% a small amount of dependence. 00:29:57.610 --> 00:30:09.510 align:middle line:84% So suppose for every i from 1 to m, the set, 00:30:09.510 --> 00:30:27.280 align:middle line:84% the index set Bi is disjoint from all but at most d other 00:30:27.280 --> 00:30:29.010 align:middle line:90% Bj's. 00:30:29.010 --> 00:30:33.130 align:middle line:84% So d here is some other parameter in the theorem. 00:30:33.130 --> 00:30:38.770 align:middle line:84% So each Bi has overlapping variables with a small number 00:30:38.770 --> 00:30:42.290 align:middle line:90% of other Bj's. 00:30:42.290 --> 00:30:48.960 align:middle line:84% And furthermore, another hypothesis 00:30:48.960 --> 00:30:57.970 align:middle line:84% is that each bad event occurs with some probability which 00:30:57.970 --> 00:31:05.220 align:middle line:84% is not too large, so 1 over d plus 1 times a constant E. Here, 00:31:05.220 --> 00:31:08.670 align:middle line:84% E is the natural constant, 2.71 so on. 00:31:08.670 --> 00:31:11.880 align:middle line:90% And this is true for every i. 00:31:11.880 --> 00:31:18.550 align:middle line:84% So let me just write it a little bit more clearly like this. 00:31:18.550 --> 00:31:22.660 align:middle line:90% 00:31:22.660 --> 00:31:25.000 align:middle line:84% OK, so these are all the hypotheses. 00:31:25.000 --> 00:31:26.960 align:middle line:84% So there's a lot, but roughly speaking, 00:31:26.960 --> 00:31:30.580 align:middle line:84% it's saying that every bad event is 00:31:30.580 --> 00:31:33.910 align:middle line:84% related to only a small number of other bad events. 00:31:33.910 --> 00:31:38.990 align:middle line:84% And each event occurs with not too high probability, 00:31:38.990 --> 00:31:41.360 align:middle line:84% which depends not on the total number of events, 00:31:41.360 --> 00:31:43.780 align:middle line:84% but only on this d, which is the number of other events 00:31:43.780 --> 00:31:44.985 align:middle line:90% that it is associated with. 00:31:44.985 --> 00:31:47.970 align:middle line:90% 00:31:47.970 --> 00:31:58.190 align:middle line:84% The conclusion is that the probability 00:31:58.190 --> 00:32:01.765 align:middle line:90% that none of the bad events-- 00:32:01.765 --> 00:32:05.710 align:middle line:90% 00:32:05.710 --> 00:32:10.530 align:middle line:84% so the bad events are the events indexed. 00:32:10.530 --> 00:32:13.240 align:middle line:90% 00:32:13.240 --> 00:32:21.460 align:middle line:84% So none of the bad events you went through Em occurs 00:32:21.460 --> 00:32:24.010 align:middle line:90% is strictly positive. 00:32:24.010 --> 00:32:28.380 align:middle line:84% So with positive probability, none of these bad events occur. 00:32:28.380 --> 00:32:30.850 align:middle line:84% OK, so that's the statement of the Lovász local lemma. 00:32:30.850 --> 00:32:33.410 align:middle line:90% 00:32:33.410 --> 00:32:37.040 align:middle line:84% And now let us apply this local lemma 00:32:37.040 --> 00:32:42.800 align:middle line:84% to prove a even better lower bound than what we saw earlier 00:32:42.800 --> 00:32:45.590 align:middle line:90% for Ramsey numbers. 00:32:45.590 --> 00:32:49.340 align:middle line:84% We will not be able to prove the local lemma in this video. 00:32:49.340 --> 00:32:52.730 align:middle line:84% So its proof is, although not long, quite intricate. 00:32:52.730 --> 00:32:56.545 align:middle line:84% And I refer you to the references for a proof. 00:32:56.545 --> 00:32:59.320 align:middle line:90% 00:32:59.320 --> 00:33:10.310 align:middle line:84% So let us now see a lower bound to Ramsey numbers and this is 00:33:10.310 --> 00:33:13.690 align:middle line:84% a result due to Spencer, Joe Spencer from the '70s. 00:33:13.690 --> 00:33:23.280 align:middle line:90% 00:33:23.280 --> 00:33:24.875 align:middle line:90% And such that if-- 00:33:24.875 --> 00:33:33.200 align:middle line:90% 00:33:33.200 --> 00:33:42.870 align:middle line:84% if k choose 2 times n choose 2, or n minus 2 00:33:42.870 --> 00:33:50.150 align:middle line:84% choose k minus 2 plus 1 times 2 to the 1 minus k choose 2 00:33:50.150 --> 00:34:00.420 align:middle line:84% is less than 1 over E, then the Ramsey number, Rkk, is strictly 00:34:00.420 --> 00:34:01.560 align:middle line:90% bigger than n. 00:34:01.560 --> 00:34:07.233 align:middle line:84% So, as with earlier, this is some mysterious-looking formula. 00:34:07.233 --> 00:34:09.150 align:middle line:84% And I want you to not worry too much about it. 00:34:09.150 --> 00:34:12.210 align:middle line:90% It will come out of the proof. 00:34:12.210 --> 00:34:18.300 align:middle line:84% What may be more illuminating is the consequence of this theorem. 00:34:18.300 --> 00:34:27.850 align:middle line:84% So as a corollary, Rkk has the lower bound 1-- 00:34:27.850 --> 00:34:31.449 align:middle line:84% so a lower bound-- the following asymptotic root 2 divided by E 00:34:31.449 --> 00:34:37.270 align:middle line:84% plus little 1 k times 2 to the k over 2. 00:34:37.270 --> 00:34:40.540 align:middle line:84% And this yet beats the bound that we saw earlier 00:34:40.540 --> 00:34:42.820 align:middle line:90% by another factor of root 2. 00:34:42.820 --> 00:34:45.489 align:middle line:84% This may seem like a small improvement, 00:34:45.489 --> 00:34:49.330 align:middle line:84% and quantitatively, indeed, it's only a constant factor more. 00:34:49.330 --> 00:34:52.960 align:middle line:84% Whereas, the remaining asymptotics, 00:34:52.960 --> 00:34:57.400 align:middle line:84% we do not know how far it is away from the actual truth. 00:34:57.400 --> 00:35:02.710 align:middle line:84% But yet, this result proved in the '70s is the best known lower 00:35:02.710 --> 00:35:08.950 align:middle line:84% bound to date for diagonal Ramsey numbers, these Rkk's. 00:35:08.950 --> 00:35:14.530 align:middle line:84% OK, so let us now see the proof of this theorem of Spencer, 00:35:14.530 --> 00:35:17.090 align:middle line:84% which applies the Lovász local lemma. 00:35:17.090 --> 00:35:23.070 align:middle line:90% 00:35:23.070 --> 00:35:26.400 align:middle line:84% As earlier, we will begin by coloring 00:35:26.400 --> 00:35:35.620 align:middle line:84% the edges of the complete graph on n vertices 00:35:35.620 --> 00:35:43.040 align:middle line:84% with two colors uniformly at random. 00:35:43.040 --> 00:35:49.750 align:middle line:90% 00:35:49.750 --> 00:35:54.230 align:middle line:84% And also, similar to the first proof that we saw, 00:35:54.230 --> 00:36:12.210 align:middle line:84% for every k vertex subset s, let us define a bad event, 00:36:12.210 --> 00:36:34.610 align:middle line:84% E sub s to be the event that s induces a monochromatic clique 00:36:34.610 --> 00:36:36.205 align:middle line:90% on the vertices in s. 00:36:36.205 --> 00:36:39.330 align:middle line:90% 00:36:39.330 --> 00:36:43.530 align:middle line:84% And we also saw earlier that the probability of this bad event 00:36:43.530 --> 00:36:48.280 align:middle line:84% is 2 raised to 1 minus k choose two. 00:36:48.280 --> 00:36:51.010 align:middle line:90% 00:36:51.010 --> 00:36:57.660 align:middle line:84% OK, let us see how this setup corresponds 00:36:57.660 --> 00:37:01.140 align:middle line:84% to this random variable setup earlier. 00:37:01.140 --> 00:37:11.420 align:middle line:84% The variables correspond to the coloring of the edges. 00:37:11.420 --> 00:37:16.460 align:middle line:90% 00:37:16.460 --> 00:37:24.980 align:middle line:84% So in other words, these are indexed by the n 00:37:24.980 --> 00:37:30.430 align:middle line:84% choose two edges of the complete graph, Kn. 00:37:30.430 --> 00:37:36.040 align:middle line:84% So there are n choose two such variables. 00:37:36.040 --> 00:37:38.740 align:middle line:84% Well, what about this condition here? 00:37:38.740 --> 00:37:54.180 align:middle line:84% So Bi's, so each Bi is a set of size k 00:37:54.180 --> 00:37:58.100 align:middle line:84% choose two corresponding to the edges of this complete graph 00:37:58.100 --> 00:38:01.560 align:middle line:90% inside this clique of size k. 00:38:01.560 --> 00:38:03.335 align:middle line:90% So when do the Bi's overlap? 00:38:03.335 --> 00:38:06.180 align:middle line:90% 00:38:06.180 --> 00:38:22.070 align:middle line:84% So if s and s prime, so both k vertex subsets of vertices, 00:38:22.070 --> 00:38:25.260 align:middle line:84% so if these are both k vertex subsets, 00:38:25.260 --> 00:38:40.380 align:middle line:84% then their cliques overlap in edges if and only 00:38:40.380 --> 00:38:53.080 align:middle line:84% if the sets of vertices s and s prime 00:38:53.080 --> 00:38:55.670 align:middle line:84% overlapping at least two vertices. 00:38:55.670 --> 00:39:02.460 align:middle line:84% So as an illustration, if you have this clique of size 4 00:39:02.460 --> 00:39:06.870 align:middle line:84% and another clique of size 4, then they 00:39:06.870 --> 00:39:11.850 align:middle line:84% overlap in at least one edge, if and only if the vertex sets 00:39:11.850 --> 00:39:14.045 align:middle line:84% overlap in at least two vertices. 00:39:14.045 --> 00:39:16.890 align:middle line:90% 00:39:16.890 --> 00:39:19.680 align:middle line:84% So to apply the local lemma, we need 00:39:19.680 --> 00:39:25.260 align:middle line:84% to check how many other s primes intersect a given 00:39:25.260 --> 00:39:28.590 align:middle line:90% s in terms of their edge set. 00:39:28.590 --> 00:39:31.950 align:middle line:84% And this is a calculation that we can do as follows. 00:39:31.950 --> 00:39:46.160 align:middle line:84% For each k vertex s, the number of k vertex 00:39:46.160 --> 00:39:55.330 align:middle line:84% s primes that has intersection size at least two with s 00:39:55.330 --> 00:39:57.790 align:middle line:90% is at most following. 00:39:57.790 --> 00:40:02.770 align:middle line:84% Well, I need to pick a pair of vertices of s 00:40:02.770 --> 00:40:08.740 align:middle line:84% where there is this overlap and then choose n choose-- 00:40:08.740 --> 00:40:12.580 align:middle line:84% and then select the remaining vertices of s prime 00:40:12.580 --> 00:40:19.300 align:middle line:84% from the remaining n choose two vertices of the entire clique. 00:40:19.300 --> 00:40:21.190 align:middle line:84% There are some possible overcount 00:40:21.190 --> 00:40:25.150 align:middle line:84% here because some s primes may be counted in more than one way 00:40:25.150 --> 00:40:28.815 align:middle line:84% through this formula here, but it is certainly an upper bound. 00:40:28.815 --> 00:40:31.960 align:middle line:90% 00:40:31.960 --> 00:40:34.640 align:middle line:84% And now we can apply the local lemma. 00:40:34.640 --> 00:40:50.250 align:middle line:90% 00:40:50.250 --> 00:40:52.830 align:middle line:84% By applying the local lemma, one checks 00:40:52.830 --> 00:40:58.860 align:middle line:84% that as long as this hypothesis inequality is satisfied, 00:40:58.860 --> 00:41:02.220 align:middle line:84% this hypothesis in the statement of the local lemma 00:41:02.220 --> 00:41:04.590 align:middle line:90% is satisfied as well. 00:41:04.590 --> 00:41:14.330 align:middle line:84% So this inequality here corresponds to this inequality 00:41:14.330 --> 00:41:15.910 align:middle line:90% in the hypothesis. 00:41:15.910 --> 00:41:19.370 align:middle line:84% So this, it is satisfied, and so we can apply the local lemma 00:41:19.370 --> 00:41:31.570 align:middle line:84% to deduce that with positive probability 00:41:31.570 --> 00:41:39.290 align:middle line:84% none of the bad events, so none of the bad events Es 00:41:39.290 --> 00:41:50.120 align:middle line:84% occur for ranging over all k vertex subsets s in the graph. 00:41:50.120 --> 00:41:54.530 align:middle line:84% And therefore, with positive probability, 00:41:54.530 --> 00:41:59.870 align:middle line:84% none of the k vertex subsets induce a monochromatic clique. 00:41:59.870 --> 00:42:05.010 align:middle line:84% And that's exactly what we're trying to show with this result 00:42:05.010 --> 00:42:06.760 align:middle line:90% over-- to this claim over here. 00:42:06.760 --> 00:42:10.480 align:middle line:84% And this concludes the proof of Spencer's theorem. 00:42:10.480 --> 00:42:12.880 align:middle line:84% So you see that this is a demonstration 00:42:12.880 --> 00:42:16.180 align:middle line:84% of the powerful Lovász local lemma. 00:42:16.180 --> 00:42:20.560 align:middle line:84% And it is a very interesting method 00:42:20.560 --> 00:42:26.500 align:middle line:84% that was pioneered by Lovász and has had important applications 00:42:26.500 --> 00:42:30.130 align:middle line:84% subsequently in the development of the probabilistic method. 00:42:30.130 --> 00:42:32.350 align:middle line:84% Although we did not see the proof of the local lemma 00:42:32.350 --> 00:42:34.150 align:middle line:84% here today, I hope that you can at least 00:42:34.150 --> 00:42:36.530 align:middle line:90% appreciate its application. 00:42:36.530 --> 00:42:40.550 align:middle line:84% And throughout this lecture, throughout this video, 00:42:40.550 --> 00:42:42.280 align:middle line:84% we saw three different approaches 00:42:42.280 --> 00:42:45.910 align:middle line:84% to lower bounding, the Ramsey number, the diagonal Ramsey 00:42:45.910 --> 00:42:49.960 align:middle line:84% numbers using different ideas from the probabilistic method. 00:42:49.960 --> 00:42:54.550 align:middle line:84% It is a powerful method, and it has lots of other applications 00:42:54.550 --> 00:42:57.130 align:middle line:84% and many beautiful applications to various areas 00:42:57.130 --> 00:43:00.180 align:middle line:90% of combinatorics and beyond. 00:43:00.180 --> 00:43:07.000 align:middle line:90%