WEBVTT 00:00:00.000 --> 00:00:02.898 align:middle line:90% [SQUEAKING] 00:00:02.898 --> 00:00:04.347 align:middle line:90% [RUSTLING] 00:00:04.347 --> 00:00:07.728 align:middle line:90% [CLICKING] 00:00:07.728 --> 00:00:11.610 align:middle line:90% 00:00:11.610 --> 00:00:13.320 align:middle line:84% YUFEI ZHAO: In this video, we'll look 00:00:13.320 --> 00:00:15.840 align:middle line:84% at an application of the probabilistic method 00:00:15.840 --> 00:00:18.540 align:middle line:90% to extremal set theory. 00:00:18.540 --> 00:00:22.410 align:middle line:84% Extremal set theory is an area of combinatorics concerned 00:00:22.410 --> 00:00:25.290 align:middle line:84% with questions such as, what is the largest 00:00:25.290 --> 00:00:29.728 align:middle line:84% collection of sets that satisfy certain desired properties? 00:00:29.728 --> 00:00:31.770 align:middle line:84% And the question that we'll look at in this video 00:00:31.770 --> 00:00:33.630 align:middle line:90% is the following. 00:00:33.630 --> 00:00:36.270 align:middle line:90% What is the largest antichain? 00:00:36.270 --> 00:00:40.020 align:middle line:84% And more precisely, what is the largest collection of subsets 00:00:40.020 --> 00:00:42.550 align:middle line:84% of some ground set numbers 1 through n-- 00:00:42.550 --> 00:00:44.130 align:middle line:90% so this is an n element set. 00:00:44.130 --> 00:00:50.790 align:middle line:84% So I want as many sets as we can such that none of these sets 00:00:50.790 --> 00:00:53.580 align:middle line:90% is a subset of another set. 00:00:53.580 --> 00:00:55.710 align:middle line:84% So this is what we mean by an antichain. 00:00:55.710 --> 00:00:59.820 align:middle line:84% And for a given value of n, what is the largest number of sets 00:00:59.820 --> 00:01:00.500 align:middle line:90% you can have? 00:01:00.500 --> 00:01:05.790 align:middle line:84% What is the largest l that we can have as a function of n? 00:01:05.790 --> 00:01:14.110 align:middle line:84% To give an example, when n equals to 3, 00:01:14.110 --> 00:01:22.360 align:middle line:84% you can have the sets 1, 2; 1, 3; and 2, 3; 00:01:22.360 --> 00:01:29.470 align:middle line:84% and you see that none of these sets is a superset of another. 00:01:29.470 --> 00:01:46.700 align:middle line:84% More generally, when you take all k-element subsets of 1 00:01:46.700 --> 00:01:52.110 align:middle line:84% through n, this collection also has the property 00:01:52.110 --> 00:01:54.600 align:middle line:84% that no set is a subset of another set 00:01:54.600 --> 00:01:56.910 align:middle line:84% because all these k-element subsets have 00:01:56.910 --> 00:02:01.950 align:middle line:84% the same size, so none of them can contain another. 00:02:01.950 --> 00:02:11.840 align:middle line:84% Now this collection of sets have n choose k sets. 00:02:11.840 --> 00:02:18.480 align:middle line:84% And which value of k maximizes this quantity n choose k? 00:02:18.480 --> 00:02:28.830 align:middle line:84% Well, it is maximized at k being n over 2, or when n is odd, 00:02:28.830 --> 00:02:31.880 align:middle line:84% we can either round down or round up, 00:02:31.880 --> 00:02:38.390 align:middle line:84% leading to n choose k over 2, or the floor of k 00:02:38.390 --> 00:02:41.240 align:middle line:84% over 2-- so this is-- so n choose n over 2. 00:02:41.240 --> 00:02:44.390 align:middle line:84% So n over 2 floor, meaning that's rounded 00:02:44.390 --> 00:02:46.190 align:middle line:84% down or, equivalently, rounded up, 00:02:46.190 --> 00:02:47.490 align:middle line:90% will produce the same answer. 00:02:47.490 --> 00:02:52.010 align:middle line:84% So this many sets we can get from an antichain. 00:02:52.010 --> 00:02:54.410 align:middle line:90% Now this is just one example. 00:02:54.410 --> 00:02:57.830 align:middle line:84% And the question is, is this the best example? 00:02:57.830 --> 00:02:58.760 align:middle line:90% Can we do better? 00:02:58.760 --> 00:03:00.260 align:middle line:84% Can we get an even larger collection 00:03:00.260 --> 00:03:03.020 align:middle line:90% of sets that form an antichain? 00:03:03.020 --> 00:03:07.680 align:middle line:84% And the classic result of Sperner's-- 00:03:07.680 --> 00:03:20.830 align:middle line:84% or Sperner's theorem, such that-- 00:03:20.830 --> 00:03:24.700 align:middle line:84% actually, no, this is the best that we can do. 00:03:24.700 --> 00:03:31.510 align:middle line:84% And what that theorem says is that if you have l subsets of 1 00:03:31.510 --> 00:03:46.620 align:middle line:84% through n such that none of the sets is a subset of another, 00:03:46.620 --> 00:03:58.710 align:middle line:84% then, in fact, l is, at most, n choose the floor of n over 2. 00:03:58.710 --> 00:04:01.860 align:middle line:84% So this answers the question that we were asking. 00:04:01.860 --> 00:04:04.880 align:middle line:84% And in this video, we'll use the probabilistic method 00:04:04.880 --> 00:04:06.810 align:middle line:90% to prove Sperner's theorem. 00:04:06.810 --> 00:04:08.490 align:middle line:84% And it's quite a beautiful proof. 00:04:08.490 --> 00:04:12.110 align:middle line:84% And the introduction of randomness into the proof 00:04:12.110 --> 00:04:14.870 align:middle line:84% is quite remarkable for a theorem that really doesn't 00:04:14.870 --> 00:04:18.500 align:middle line:90% involve any randomness at all. 00:04:18.500 --> 00:04:24.590 align:middle line:84% What we will prove is actually the following theorem, 00:04:24.590 --> 00:04:28.280 align:middle line:84% which is slightly stronger than Sperner's theorem. 00:04:28.280 --> 00:04:31.160 align:middle line:84% And we'll prove the following, also known as the LYM 00:04:31.160 --> 00:04:33.920 align:middle line:84% inequality, named after three people who 00:04:33.920 --> 00:04:36.080 align:middle line:90% discovered the theorem. 00:04:36.080 --> 00:04:41.330 align:middle line:84% And it says that in the hypothesis that we just set up, 00:04:41.330 --> 00:04:43.890 align:middle line:84% we furthermore have the following consequence. 00:04:43.890 --> 00:04:46.680 align:middle line:90% 00:04:46.680 --> 00:04:53.898 align:middle line:84% If you take the sum over I from 1i-- 00:04:53.898 --> 00:05:03.180 align:middle line:84% i being 1 to l of the quantity 1 over n choose the size of Ai, 00:05:03.180 --> 00:05:05.735 align:middle line:90% then this sum is, at most, 1. 00:05:05.735 --> 00:05:09.860 align:middle line:90% 00:05:09.860 --> 00:05:15.560 align:middle line:84% So let us first prove this LYM inequality, 00:05:15.560 --> 00:05:18.290 align:middle line:84% and then afterwards, we'll deduce Sperner's theorem 00:05:18.290 --> 00:05:20.670 align:middle line:90% from this LYM inequality. 00:05:20.670 --> 00:05:27.080 align:middle line:90% 00:05:27.080 --> 00:05:30.560 align:middle line:84% Let's now introduce randomness into the problem. 00:05:30.560 --> 00:05:40.040 align:middle line:84% So let sigma 1 through sigma n be a randomly chosen 00:05:40.040 --> 00:05:40.846 align:middle line:90% permutation. 00:05:40.846 --> 00:05:50.080 align:middle line:90% 00:05:50.080 --> 00:05:54.490 align:middle line:90% of numbers 1 through n. 00:05:54.490 --> 00:05:59.290 align:middle line:84% And here, we'll choose this permutation uniformly at random, 00:05:59.290 --> 00:06:01.480 align:middle line:84% meaning that of all the n factorial different 00:06:01.480 --> 00:06:05.470 align:middle line:84% permutations, we'll select one with equal probability 00:06:05.470 --> 00:06:09.880 align:middle line:84% of being any one of those n factorial permutations. 00:06:09.880 --> 00:06:16.720 align:middle line:84% And now let's consider the following chain of sets, 00:06:16.720 --> 00:06:22.030 align:middle line:84% starting with the antiset, and then the set 00:06:22.030 --> 00:06:24.310 align:middle line:84% containing just the first element 00:06:24.310 --> 00:06:29.050 align:middle line:84% of this permutation, sigma 1, and then the set 00:06:29.050 --> 00:06:32.740 align:middle line:84% containing the first two elements of this permutation, 00:06:32.740 --> 00:06:40.660 align:middle line:84% sigma 1, sigma 2, and so on, until we add all the elements 00:06:40.660 --> 00:06:42.730 align:middle line:90% into the set. 00:06:42.730 --> 00:06:46.990 align:middle line:84% So these are the set of prefixes of this permutation where 00:06:46.990 --> 00:06:49.100 align:middle line:90% we add one element at a time. 00:06:49.100 --> 00:06:53.620 align:middle line:84% So this is a chain of subsets of 1 through n. 00:06:53.620 --> 00:06:57.370 align:middle line:84% And let's consider the event for which 00:06:57.370 --> 00:06:59.040 align:middle line:90% we'll evaluate its probability. 00:06:59.040 --> 00:07:01.120 align:middle line:90% So the event that Ai-- 00:07:01.120 --> 00:07:05.620 align:middle line:84% so Ai is one of the sets that was originally given in the 00:07:05.620 --> 00:07:06.900 align:middle line:90% in the theorem statement. 00:07:06.900 --> 00:07:09.190 align:middle line:90% So what's the event? 00:07:09.190 --> 00:07:12.250 align:middle line:84% So the event that we'll consider is the event 00:07:12.250 --> 00:07:16.515 align:middle line:90% that Ai appears in this chain. 00:07:16.515 --> 00:07:19.730 align:middle line:90% 00:07:19.730 --> 00:07:21.930 align:middle line:84% So this chain means this chain over here. 00:07:21.930 --> 00:07:25.460 align:middle line:84% Now, this chain is random because the sigmas 00:07:25.460 --> 00:07:27.680 align:middle line:90% are a random permutation. 00:07:27.680 --> 00:07:30.210 align:middle line:84% Even though the Ai's are deterministic, 00:07:30.210 --> 00:07:33.260 align:middle line:84% the Ai's are given, this permutation is random. 00:07:33.260 --> 00:07:37.440 align:middle line:84% So this is some event which has a probability. 00:07:37.440 --> 00:07:40.830 align:middle line:90% 00:07:40.830 --> 00:07:43.770 align:middle line:84% When do we have Ai appearing in this chain? 00:07:43.770 --> 00:07:50.370 align:middle line:84% Or if all the elements of AI appear in this permutation 00:07:50.370 --> 00:07:53.550 align:middle line:84% before all the non-elements of Ai. 00:07:53.550 --> 00:07:56.830 align:middle line:84% So we can count how many different ways this can happen. 00:07:56.830 --> 00:07:59.710 align:middle line:84% So if all the n factorial are different permutations, 00:07:59.710 --> 00:08:04.950 align:middle line:84% the number of permutations where the elements of Ai appear first 00:08:04.950 --> 00:08:10.620 align:middle line:84% is Ai size factorial, and then all the non-elements of Ai 00:08:10.620 --> 00:08:13.770 align:middle line:90% appear afterwards. 00:08:13.770 --> 00:08:19.080 align:middle line:84% It's n minus the size of Ai factorial. 00:08:19.080 --> 00:08:21.510 align:middle line:84% And this quantity here is equal to 1 00:08:21.510 --> 00:08:25.930 align:middle line:84% over the binomial coefficient n choose the size of Ai. 00:08:25.930 --> 00:08:28.500 align:middle line:90% 00:08:28.500 --> 00:08:29.220 align:middle line:90% OK. 00:08:29.220 --> 00:08:34.110 align:middle line:90% Next, let us note the following. 00:08:34.110 --> 00:08:49.670 align:middle line:84% That no two different Ai's, Aj's can simultaneously 00:08:49.670 --> 00:08:53.595 align:middle line:90% appear in the chain. 00:08:53.595 --> 00:08:56.740 align:middle line:90% 00:08:56.740 --> 00:08:58.000 align:middle line:90% Why is that? 00:08:58.000 --> 00:09:03.130 align:middle line:84% Well, we assumed that no Ai contains 00:09:03.130 --> 00:09:08.440 align:middle line:84% is a subset of another Aj, and if, indeed, you 00:09:08.440 --> 00:09:11.560 align:middle line:84% had Ai and Aj both appearing in the chain, then one of them 00:09:11.560 --> 00:09:14.080 align:middle line:90% would contain another. 00:09:14.080 --> 00:09:20.890 align:middle line:84% So this chain cannot contain two different Ai's at the same time. 00:09:20.890 --> 00:09:36.510 align:middle line:84% And therefore, these events, meaning these events over here, 00:09:36.510 --> 00:09:53.730 align:middle line:84% are disjoint as we run i from 1 through l. 00:09:53.730 --> 00:09:56.580 align:middle line:84% If these events are disjoint and each of them 00:09:56.580 --> 00:10:02.280 align:middle line:84% have this probability, well, the sum of these probabilities 00:10:02.280 --> 00:10:04.720 align:middle line:90% then must add up to, at most, 1. 00:10:04.720 --> 00:10:08.210 align:middle line:84% And that's indeed the conclusion that we're looking for. 00:10:08.210 --> 00:10:18.550 align:middle line:84% So the sum of these probabilities, the probability 00:10:18.550 --> 00:10:23.860 align:middle line:84% that Ai appears in the chain, in this random chain, the sum 00:10:23.860 --> 00:10:25.660 align:middle line:84% of these probabilities add up to, at most, 00:10:25.660 --> 00:10:28.400 align:middle line:84% 1 because these events are disjoint from each other. 00:10:28.400 --> 00:10:33.790 align:middle line:84% And we have calculated earlier that the event probabilities are 00:10:33.790 --> 00:10:40.420 align:middle line:84% given as 1 over n choose the size of Ai. 00:10:40.420 --> 00:10:45.630 align:middle line:84% So this concludes the proof of the LYM inequality. 00:10:45.630 --> 00:10:47.830 align:middle line:84% You see the final line, the final inequality 00:10:47.830 --> 00:10:50.870 align:middle line:84% is precisely what we were trying to establish. 00:10:50.870 --> 00:10:53.830 align:middle line:90% 00:10:53.830 --> 00:11:02.210 align:middle line:84% And finally, let us prove Sperner's theorem 00:11:02.210 --> 00:11:06.340 align:middle line:84% by deducing this theorem from the LYM inequality. 00:11:06.340 --> 00:11:07.640 align:middle line:90% So here is a-- 00:11:07.640 --> 00:11:14.210 align:middle line:84% this is a quick deduction because n choose the size of Ai 00:11:14.210 --> 00:11:24.050 align:middle line:84% is, at most, n choose the floor of n over 2 for all i. 00:11:24.050 --> 00:11:27.560 align:middle line:84% So the binomial coefficients for a given n 00:11:27.560 --> 00:11:31.400 align:middle line:84% maxes out in the middle, we have this inequality. 00:11:31.400 --> 00:11:39.140 align:middle line:84% And thus-- here, I'm just rewriting the LYM inequality. 00:11:39.140 --> 00:11:49.190 align:middle line:84% The sum from-- so i from one to l of 1 over n choose size of Ai. 00:11:49.190 --> 00:11:51.920 align:middle line:84% This is, at most, 1, so that's what we just proved. 00:11:51.920 --> 00:11:55.850 align:middle line:84% And then applying this individual bound to each term, 00:11:55.850 --> 00:12:00.290 align:middle line:84% we get l different terms, and each one of them 00:12:00.290 --> 00:12:06.350 align:middle line:84% in the denominator, we have n choose the floor of n over 2. 00:12:06.350 --> 00:12:08.780 align:middle line:84% And rearranging this inequality, we get 00:12:08.780 --> 00:12:11.550 align:middle line:90% the desired upper bound on l. 00:12:11.550 --> 00:12:14.370 align:middle line:84% So this concludes the proof of Sperner's theorem. 00:12:14.370 --> 00:12:17.610 align:middle line:84% And it's a beautiful result with a beautiful proof that 00:12:17.610 --> 00:12:21.720 align:middle line:84% is a wonderful illustration of the probabilistic method 00:12:21.720 --> 00:12:25.380 align:middle line:84% in combinatorics where, although the theorem statement itself is 00:12:25.380 --> 00:12:27.820 align:middle line:84% deterministic-- doesn't involve any randomness, 00:12:27.820 --> 00:12:34.110 align:middle line:84% the proof works by introducing randomness into the problem. 00:12:34.110 --> 00:12:36.960 align:middle line:84% So this is the application of the probabilistic method 00:12:36.960 --> 00:12:39.120 align:middle line:84% in combinatorics, and I think it's 00:12:39.120 --> 00:12:42.650 align:middle line:84% one of the most beautiful applications. 00:12:42.650 --> 00:12:49.000 align:middle line:90%