WEBVTT 00:00:00.000 --> 00:00:02.450 align:middle line:90% [SQUEAKING] 00:00:02.450 --> 00:00:04.410 align:middle line:90% [RUSTLING] 00:00:04.410 --> 00:00:11.780 align:middle line:90% [CLICKING] 00:00:11.780 --> 00:00:14.960 align:middle line:84% PROFESSOR: In this video, we'll look at a beautiful result that 00:00:14.960 --> 00:00:17.690 align:middle line:84% provides us with the existence of graphs 00:00:17.690 --> 00:00:20.460 align:middle line:84% with high girth and high chromatic number. 00:00:20.460 --> 00:00:23.090 align:middle line:84% And it's a very nice demonstration 00:00:23.090 --> 00:00:26.360 align:middle line:84% of the power of the probabilistic method. 00:00:26.360 --> 00:00:28.430 align:middle line:90% We'll recall some definitions. 00:00:28.430 --> 00:00:31.640 align:middle line:84% Given a graph, G, its chromatic number 00:00:31.640 --> 00:00:33.560 align:middle line:84% is the minimum number of colors we 00:00:33.560 --> 00:00:36.170 align:middle line:84% need to color the vertices of the graph G 00:00:36.170 --> 00:00:39.560 align:middle line:84% so that no edge has both endpoints receiving 00:00:39.560 --> 00:00:41.420 align:middle line:90% the same color. 00:00:41.420 --> 00:00:45.920 align:middle line:84% The girth of a graph G is the length of the shortest cycle. 00:00:45.920 --> 00:00:52.290 align:middle line:84% For example, the following graph has 00:00:52.290 --> 00:00:58.550 align:middle line:84% girth 4 because the shortest cycle in this graph 00:00:58.550 --> 00:00:59.660 align:middle line:90% has length 4. 00:00:59.660 --> 00:01:02.060 align:middle line:90% And there are a couple of them. 00:01:02.060 --> 00:01:11.490 align:middle line:84% Another example-- so this graph here, there are many cycles. 00:01:11.490 --> 00:01:13.720 align:middle line:84% But the shortest one has length 3. 00:01:13.720 --> 00:01:15.060 align:middle line:90% It is a triangle up here. 00:01:15.060 --> 00:01:18.520 align:middle line:90% So the girth is 3. 00:01:18.520 --> 00:01:25.810 align:middle line:84% If the graph is a tree, then it has no cycles, in which case 00:01:25.810 --> 00:01:30.100 align:middle line:84% we say that its girth is infinite. 00:01:30.100 --> 00:01:36.100 align:middle line:84% OK, so what can we say about girth and chromatic number? 00:01:36.100 --> 00:01:46.730 align:middle line:84% Well, suppose your graph contains a clique on k vertices. 00:01:46.730 --> 00:01:51.160 align:middle line:90% 00:01:51.160 --> 00:01:55.270 align:middle line:84% Then just to color this clique, we 00:01:55.270 --> 00:01:57.250 align:middle line:84% need at least k different colors. 00:01:57.250 --> 00:02:04.910 align:middle line:84% So then the chromatic number is at least k. 00:02:04.910 --> 00:02:08.820 align:middle line:90% 00:02:08.820 --> 00:02:13.870 align:middle line:84% Can we say something in the reverse direction? 00:02:13.870 --> 00:02:17.175 align:middle line:84% Suppose we have a graph with high chromatic number. 00:02:17.175 --> 00:02:25.360 align:middle line:90% 00:02:25.360 --> 00:02:35.190 align:middle line:84% Would it be possible to deduce some local information 00:02:35.190 --> 00:02:36.990 align:middle line:90% about the graph? 00:02:36.990 --> 00:02:41.250 align:middle line:84% For example, would it be possible to say that if we have 00:02:41.250 --> 00:02:43.290 align:middle line:84% very high chromatic number, then there 00:02:43.290 --> 00:02:46.440 align:middle line:84% must be a triangle somewhere in the graph, 00:02:46.440 --> 00:02:50.040 align:middle line:84% or there's some fairly dense structure that's 00:02:50.040 --> 00:02:53.870 align:middle line:90% very local in this graph. 00:02:53.870 --> 00:02:56.570 align:middle line:84% So the main theorem that we'll present in this video 00:02:56.570 --> 00:03:02.010 align:middle line:84% gives a definitive no to this question. 00:03:02.010 --> 00:03:12.083 align:middle line:84% And it is a classic result due to Paul Erdos from the '50s. 00:03:12.083 --> 00:03:16.710 align:middle line:90% 00:03:16.710 --> 00:03:19.920 align:middle line:84% And the statement of the theorem is the following. 00:03:19.920 --> 00:03:34.320 align:middle line:84% For any positive integers k and l, there exists a graph. 00:03:34.320 --> 00:03:43.060 align:middle line:90% 00:03:43.060 --> 00:03:55.230 align:middle line:84% There exists a graph with girth l, girth bigger than l, 00:03:55.230 --> 00:04:06.000 align:middle line:84% and chromatic number bigger than k. 00:04:06.000 --> 00:04:08.400 align:middle line:84% OK, so let's pause for a moment and think about what 00:04:08.400 --> 00:04:09.850 align:middle line:90% this theorem is saying. 00:04:09.850 --> 00:04:15.250 align:middle line:84% So given any k and l, no matter how large, we can find a graph. 00:04:15.250 --> 00:04:17.850 align:middle line:84% There exists a graph for which there are no short cycles, 00:04:17.850 --> 00:04:19.950 align:middle line:90% no cycles of length at most l. 00:04:19.950 --> 00:04:24.480 align:middle line:84% And furthermore, to properly color the vertices of the graph, 00:04:24.480 --> 00:04:26.370 align:middle line:90% you need at least one. 00:04:26.370 --> 00:04:29.490 align:middle line:84% You need strictly more than k colors. 00:04:29.490 --> 00:04:33.930 align:middle line:84% So even though this graph has very high chromatic number, 00:04:33.930 --> 00:04:38.940 align:middle line:84% around every vertex, if you only look not too far away, 00:04:38.940 --> 00:04:41.340 align:middle line:84% then the graph locally looks like a tree 00:04:41.340 --> 00:04:45.080 align:middle line:84% because there are no short cycles. 00:04:45.080 --> 00:04:47.480 align:middle line:84% So in particular, having high chromatic number 00:04:47.480 --> 00:04:50.420 align:middle line:84% does not give you local information. 00:04:50.420 --> 00:04:55.010 align:middle line:84% And Erdos shows that there always exists counterexamples 00:04:55.010 --> 00:04:59.090 align:middle line:84% to such a statement if we want some local information 00:04:59.090 --> 00:05:03.140 align:middle line:84% from having high chromatic number. 00:05:03.140 --> 00:05:07.070 align:middle line:84% OK, so the rest of this video will concern 00:05:07.070 --> 00:05:10.640 align:middle line:90% the proof of Erdos' theorem. 00:05:10.640 --> 00:05:14.330 align:middle line:84% It is quite a miraculous result. And certainly before Erdos 00:05:14.330 --> 00:05:17.840 align:middle line:84% proved this result, there were other results in graph theory 00:05:17.840 --> 00:05:22.400 align:middle line:84% that constructed very explicit graphs that 00:05:22.400 --> 00:05:28.580 align:middle line:84% have some large chromatic number and girth, but not for all 00:05:28.580 --> 00:05:30.020 align:middle line:90% values of k and l. 00:05:30.020 --> 00:05:33.840 align:middle line:84% And those constructions are, in some sense, very hands on. 00:05:33.840 --> 00:05:36.500 align:middle line:84% It tells you exactly how to construct these graphs. 00:05:36.500 --> 00:05:38.810 align:middle line:84% But they were not able to get arbitrarily 00:05:38.810 --> 00:05:41.900 align:middle line:84% large girth and chromatic number. 00:05:41.900 --> 00:05:45.360 align:middle line:84% Eldredge's insight is to use randomness, use 00:05:45.360 --> 00:05:47.940 align:middle line:84% the probabilistic method, and show 00:05:47.940 --> 00:05:52.090 align:middle line:84% that by appropriately modifying a random graph, 00:05:52.090 --> 00:05:55.260 align:middle line:84% one can achieve the desired outcome. 00:05:55.260 --> 00:05:56.675 align:middle line:90% So let's see how this works. 00:05:56.675 --> 00:06:02.800 align:middle line:90% 00:06:02.800 --> 00:06:08.080 align:middle line:84% We'll begin by taking a random graph, GNP. 00:06:08.080 --> 00:06:14.205 align:middle line:84% So this is the Erdos Renyi random graph. 00:06:14.205 --> 00:06:22.860 align:middle line:90% 00:06:22.860 --> 00:06:26.600 align:middle line:90% And there are n vertices. 00:06:26.600 --> 00:06:29.810 align:middle line:90% 00:06:29.810 --> 00:06:37.280 align:middle line:84% And the edge probability is p, meaning 00:06:37.280 --> 00:06:42.980 align:middle line:84% that we flip a probability p coin for each possible edge 00:06:42.980 --> 00:06:46.430 align:middle line:84% and put down an edge independently 00:06:46.430 --> 00:06:49.660 align:middle line:84% with probability p between every pair of vertices. 00:06:49.660 --> 00:06:55.670 align:middle line:90% 00:06:55.670 --> 00:07:00.300 align:middle line:84% So let G be drawn from this distribution. 00:07:00.300 --> 00:07:01.765 align:middle line:90% So G is this random graph. 00:07:01.765 --> 00:07:05.260 align:middle line:90% 00:07:05.260 --> 00:07:18.030 align:middle line:84% Now let's take a specific choice of p being locked and squared 00:07:18.030 --> 00:07:18.930 align:middle line:90% over n. 00:07:18.930 --> 00:07:21.690 align:middle line:84% Turns out, other choices of p work as long 00:07:21.690 --> 00:07:23.410 align:middle line:90% as p falls in some range. 00:07:23.410 --> 00:07:26.760 align:middle line:84% But for concreteness, we'll take this very specific choice of p. 00:07:26.760 --> 00:07:28.530 align:middle line:90% So this is a random graph. 00:07:28.530 --> 00:07:30.870 align:middle line:84% And let's construct the random variable 00:07:30.870 --> 00:07:43.370 align:middle line:84% x defined to be the number of cycles of length 00:07:43.370 --> 00:07:49.118 align:middle line:84% at most l and G. These are short cycles in G. 00:07:49.118 --> 00:07:50.660 align:middle line:84% And these are the cycles that we want 00:07:50.660 --> 00:07:53.780 align:middle line:84% to avoid because we want to construct a graph with girth 00:07:53.780 --> 00:07:55.270 align:middle line:90% strictly larger than l. 00:07:55.270 --> 00:07:58.760 align:middle line:84% So we really do not like these cycles. 00:07:58.760 --> 00:08:01.410 align:middle line:84% It would be nice if there were not too many such cycles. 00:08:01.410 --> 00:08:05.990 align:middle line:84% So let's compute the expected value of x. 00:08:05.990 --> 00:08:09.710 align:middle line:84% For this computation, we use linearity of expectations. 00:08:09.710 --> 00:08:18.110 align:middle line:84% And look over all possible cycle lengths from 3 to l. 00:08:18.110 --> 00:08:21.380 align:middle line:90% So I is the target cycle length. 00:08:21.380 --> 00:08:25.550 align:middle line:84% And for cycles of length I, there 00:08:25.550 --> 00:08:31.490 align:middle line:84% are n choose I different choices of which I vertices 00:08:31.490 --> 00:08:33.950 align:middle line:90% the cycles can involve. 00:08:33.950 --> 00:08:38.490 align:middle line:84% And once you specify those vertices, 00:08:38.490 --> 00:08:44.169 align:middle line:84% they are I minus 1 factorial, different permutations, 00:08:44.169 --> 00:08:46.920 align:middle line:84% circular permutations of these vertices. 00:08:46.920 --> 00:08:53.430 align:middle line:84% And we divide it by two because a cycle could be counted 00:08:53.430 --> 00:08:55.620 align:middle line:90% in two different directions. 00:08:55.620 --> 00:08:59.740 align:middle line:84% And they will correspond to the same cycle. 00:08:59.740 --> 00:09:03.510 align:middle line:84% So this quantity here is the number 00:09:03.510 --> 00:09:14.510 align:middle line:84% of cycles of length I in the complete graph on n vertices 00:09:14.510 --> 00:09:17.780 align:middle line:90% with all edges present. 00:09:17.780 --> 00:09:21.620 align:middle line:84% For each such possible cycle, the probability 00:09:21.620 --> 00:09:25.580 align:middle line:84% that the cycle appears in G is p to the I. 00:09:25.580 --> 00:09:30.470 align:middle line:84% So the second term here is probability 00:09:30.470 --> 00:09:40.725 align:middle line:90% that this cycle appears in GNP. 00:09:40.725 --> 00:09:46.360 align:middle line:90% 00:09:46.360 --> 00:09:49.000 align:middle line:90% OK, this is some expression. 00:09:49.000 --> 00:09:50.530 align:middle line:84% It's a little complicated, but we 00:09:50.530 --> 00:09:55.000 align:middle line:84% can simplify it and do some approximations 00:09:55.000 --> 00:09:58.480 align:middle line:90% to upper bound this quantity. 00:09:58.480 --> 00:10:01.870 align:middle line:84% So you see that the n choose I, if you expand it 00:10:01.870 --> 00:10:05.470 align:middle line:84% in terms of factorials, one can upper 00:10:05.470 --> 00:10:10.300 align:middle line:84% bound this quantity by n to the I, 00:10:10.300 --> 00:10:14.920 align:middle line:84% so incorporating the first two factors, 00:10:14.920 --> 00:10:18.753 align:middle line:84% and then leaving p to the I factor intact. 00:10:18.753 --> 00:10:21.540 align:middle line:90% 00:10:21.540 --> 00:10:30.270 align:middle line:84% OK, we chose p so that p times n equals 2 log squared of n. 00:10:30.270 --> 00:10:44.180 align:middle line:84% And so you can rewrite this quantity here like this. 00:10:44.180 --> 00:10:46.900 align:middle line:90% 00:10:46.900 --> 00:10:49.880 align:middle line:90% OK, there are l terms here. 00:10:49.880 --> 00:10:57.160 align:middle line:84% So let me furthermore upper bound to this expression by l 00:10:57.160 --> 00:11:02.000 align:middle line:90% times log n to the 2l. 00:11:02.000 --> 00:11:03.860 align:middle line:90% Here, l is a constant. 00:11:03.860 --> 00:11:08.170 align:middle line:84% So this quantity here is some growth 00:11:08.170 --> 00:11:11.300 align:middle line:84% and the rate of some power of a log 00:11:11.300 --> 00:11:16.730 align:middle line:84% and certainly much slower growing 00:11:16.730 --> 00:11:18.680 align:middle line:90% compared to the function n. 00:11:18.680 --> 00:11:20.630 align:middle line:90% So it's little o of n. 00:11:20.630 --> 00:11:22.350 align:middle line:84% It's a fairly crude approximation, 00:11:22.350 --> 00:11:26.370 align:middle line:84% but it will be sufficient for our purposes. 00:11:26.370 --> 00:11:29.250 align:middle line:90% So this is a calculation. 00:11:29.250 --> 00:11:33.120 align:middle line:84% Conclusion can be read as the expected number of short cycles 00:11:33.120 --> 00:11:36.210 align:middle line:90% is fairly small. 00:11:36.210 --> 00:11:38.470 align:middle line:84% Well, if there are few short cycles, 00:11:38.470 --> 00:11:43.170 align:middle line:84% we can get rid of all of them to get 00:11:43.170 --> 00:11:45.040 align:middle line:90% a graph without short cycles. 00:11:45.040 --> 00:11:47.640 align:middle line:84% So this is the basic idea of the alteration method 00:11:47.640 --> 00:11:49.680 align:middle line:90% in the probabilistic method. 00:11:49.680 --> 00:11:53.310 align:middle line:84% So start with this GNP, this random graph. 00:11:53.310 --> 00:11:55.420 align:middle line:84% It's not going to get us what we want right away, 00:11:55.420 --> 00:11:57.480 align:middle line:84% but we're going to do something to fix 00:11:57.480 --> 00:12:00.400 align:middle line:84% the defects in this graph-- in this case, 00:12:00.400 --> 00:12:03.240 align:middle line:90% get rid of short cycles. 00:12:03.240 --> 00:12:08.340 align:middle line:84% To that end, let us know that by Markov's inequality, 00:12:08.340 --> 00:12:12.810 align:middle line:84% the probability that the number of short cycles-- 00:12:12.810 --> 00:12:14.820 align:middle line:84% so short means length at most l-- 00:12:14.820 --> 00:12:20.940 align:middle line:84% the probability that the number of short cycles x is at least n 00:12:20.940 --> 00:12:26.190 align:middle line:84% over two is upper bounded by the expectation 00:12:26.190 --> 00:12:30.570 align:middle line:90% of x divided by n over two. 00:12:30.570 --> 00:12:36.330 align:middle line:84% And because expectation of x is little o of n, 00:12:36.330 --> 00:12:40.590 align:middle line:84% this final quantity is little o of 1 in the case to 0 00:12:40.590 --> 00:12:44.630 align:middle line:90% as n goes to infinity. 00:12:44.630 --> 00:12:49.170 align:middle line:90% OK, so this is a good sign. 00:12:49.170 --> 00:12:52.320 align:middle line:84% So it means that, typically, the graph 00:12:52.320 --> 00:12:57.130 align:middle line:84% does not have very many short cycles, has fewer than n over 2. 00:12:57.130 --> 00:13:00.240 align:middle line:84% And then eventually, we'll be able to get rid of one edge 00:13:00.240 --> 00:13:04.170 align:middle line:84% from each cycle to get us a graph with still fairly 00:13:04.170 --> 00:13:05.590 align:middle line:90% large number of vertices. 00:13:05.590 --> 00:13:07.920 align:middle line:84% But there are no short cycles left 00:13:07.920 --> 00:13:11.250 align:middle line:90% in this after the alteration. 00:13:11.250 --> 00:13:15.270 align:middle line:84% Now, let's think about chromatic number. 00:13:15.270 --> 00:13:17.400 align:middle line:84% So we want to ensure that the graph 00:13:17.400 --> 00:13:22.920 align:middle line:84% that we end up having has high chromatic number. 00:13:22.920 --> 00:13:28.020 align:middle line:84% To that end, we'll recall the following fact. 00:13:28.020 --> 00:13:33.300 align:middle line:84% If we give you a graph, G, then the chromatic number of G-- 00:13:33.300 --> 00:13:36.795 align:middle line:84% so this is chi of G. So that's the chromatic number. 00:13:36.795 --> 00:13:44.300 align:middle line:90% 00:13:44.300 --> 00:13:47.960 align:middle line:84% Well, the chromatic number is the minimum number 00:13:47.960 --> 00:13:52.190 align:middle line:84% of colors that one needs to color the vertices of the graph 00:13:52.190 --> 00:13:55.990 align:middle line:84% so that no edge has the same color on both sides. 00:13:55.990 --> 00:13:58.250 align:middle line:84% In particular every color class-- 00:13:58.250 --> 00:14:00.500 align:middle line:84% so if you look at a single color like red, 00:14:00.500 --> 00:14:05.040 align:middle line:84% and look at the vertices colored by that color, 00:14:05.040 --> 00:14:09.010 align:middle line:84% these vertices have no edges directly between them, 00:14:09.010 --> 00:14:11.580 align:middle line:84% and so they form an independent set. 00:14:11.580 --> 00:14:14.380 align:middle line:84% So each color class is an independent set. 00:14:14.380 --> 00:14:17.730 align:middle line:84% And so the number of color classes you need, 00:14:17.730 --> 00:14:21.330 align:middle line:84% then, should be at least the number of vertices divided 00:14:21.330 --> 00:14:23.290 align:middle line:90% by the independence number. 00:14:23.290 --> 00:14:28.140 align:middle line:84% This is the size of the largest independent set. 00:14:28.140 --> 00:14:30.000 align:middle line:84% So this alpha of G. The independence 00:14:30.000 --> 00:14:37.850 align:middle line:84% number is the size of the largest independent set. 00:14:37.850 --> 00:14:45.560 align:middle line:90% 00:14:45.560 --> 00:14:48.380 align:middle line:84% Indeed, each color class is an independent set. 00:14:48.380 --> 00:14:51.950 align:middle line:84% And each independent set has size at most alpha of G. 00:14:51.950 --> 00:14:53.870 align:middle line:84% And therefore, the number of colors you need 00:14:53.870 --> 00:14:57.110 align:middle line:84% should be at least the number of vertices divided 00:14:57.110 --> 00:14:58.375 align:middle line:90% by the independence number. 00:14:58.375 --> 00:15:01.460 align:middle line:90% 00:15:01.460 --> 00:15:11.370 align:middle line:84% OK, now for every h, the probability that 00:15:11.370 --> 00:15:13.110 align:middle line:90% the independence number-- 00:15:13.110 --> 00:15:15.510 align:middle line:84% so after here, we see that it makes sense 00:15:15.510 --> 00:15:17.648 align:middle line:84% to think about the independence number. 00:15:17.648 --> 00:15:19.440 align:middle line:84% So the probability that independence number 00:15:19.440 --> 00:15:27.190 align:middle line:84% is bigger than h is at most the following. 00:15:27.190 --> 00:15:31.250 align:middle line:90% So here we'll use union bound. 00:15:31.250 --> 00:15:36.380 align:middle line:84% For each h vertex subset, of which there are n 00:15:36.380 --> 00:15:40.970 align:middle line:84% choose h, such h vertex subsets, the probability 00:15:40.970 --> 00:15:46.250 align:middle line:84% that those h vertices form an independent set is 00:15:46.250 --> 00:15:49.160 align:middle line:84% the probability that there are no edges appearing among these h 00:15:49.160 --> 00:15:50.070 align:middle line:90% vertices. 00:15:50.070 --> 00:15:56.490 align:middle line:84% And that's 1 minus p raised to the power h choose 2. 00:15:56.490 --> 00:15:57.750 align:middle line:90% So this is some expression. 00:15:57.750 --> 00:16:02.460 align:middle line:84% And we can do some manipulations to simplify this expression, 00:16:02.460 --> 00:16:08.010 align:middle line:84% yielding an upper bound of n to the h times. 00:16:08.010 --> 00:16:10.530 align:middle line:84% So the latter expression, we can simplify it 00:16:10.530 --> 00:16:20.460 align:middle line:84% as upper bounded by e to the minus ph, h minus 1 over 2. 00:16:20.460 --> 00:16:28.360 align:middle line:84% So 1 minus p is less than e to the minus p. 00:16:28.360 --> 00:16:32.710 align:middle line:90% 00:16:32.710 --> 00:16:36.820 align:middle line:84% Simplifying even further, we see that this expression 00:16:36.820 --> 00:16:41.330 align:middle line:90% can be rewritten this way. 00:16:41.330 --> 00:16:45.250 align:middle line:90% 00:16:45.250 --> 00:16:50.890 align:middle line:84% And now let's make a choice for what age I want to set. 00:16:50.890 --> 00:16:56.950 align:middle line:84% So let's set age to be the quantity, which, for 00:16:56.950 --> 00:16:58.460 align:middle line:90% now seems a little mysterious. 00:16:58.460 --> 00:17:01.270 align:middle line:84% But it turns out to be a natural thing 00:17:01.270 --> 00:17:04.410 align:middle line:84% to set after we see what the calculation looks like. 00:17:04.410 --> 00:17:08.200 align:middle line:84% So we set age to be this quantity, 3 times log n over p. 00:17:08.200 --> 00:17:11.140 align:middle line:84% And knowing what p was from earlier, 00:17:11.140 --> 00:17:15.260 align:middle line:90% this quantity is 3 n over log n. 00:17:15.260 --> 00:17:17.770 align:middle line:90% 00:17:17.770 --> 00:17:21.079 align:middle line:84% If we set h to be this quantity here, 00:17:21.079 --> 00:17:28.620 align:middle line:84% then you see that here, in this expression, 00:17:28.620 --> 00:17:33.180 align:middle line:84% the numerator here can be simplified. 00:17:33.180 --> 00:17:43.550 align:middle line:84% And the result is that this expression decays according 00:17:43.550 --> 00:17:45.770 align:middle line:84% to some rate, which is n to something 00:17:45.770 --> 00:17:49.700 align:middle line:84% that is a minus constant, something on a constant word. 00:17:49.700 --> 00:17:56.320 align:middle line:84% In particular, it goes to 0 as n goes to infinity. 00:17:56.320 --> 00:17:58.680 align:middle line:84% And that's really the only thing that we need, 00:17:58.680 --> 00:18:01.830 align:middle line:84% that we need out of this calculation. 00:18:01.830 --> 00:18:06.150 align:middle line:84% So if we set h to be of an appropriate quantity, then 00:18:06.150 --> 00:18:10.390 align:middle line:84% typically, one does not have independent sets of size, 00:18:10.390 --> 00:18:13.740 align:middle line:90% at least h in the graph. 00:18:13.740 --> 00:18:17.590 align:middle line:90% OK, so let's regroup. 00:18:17.590 --> 00:18:19.890 align:middle line:90% So we proved a couple of things. 00:18:19.890 --> 00:18:22.950 align:middle line:84% First, we showed that typically-- so when 00:18:22.950 --> 00:18:26.850 align:middle line:84% I say typically, I mean with probability approaching 1 00:18:26.850 --> 00:18:29.040 align:middle line:90% as n goes to infinity. 00:18:29.040 --> 00:18:31.350 align:middle line:84% So typically, this graph does not 00:18:31.350 --> 00:18:37.080 align:middle line:84% have short cycles, meaning cycles of length at most l. 00:18:37.080 --> 00:18:41.190 align:middle line:84% And also, typically, this graph does not 00:18:41.190 --> 00:18:45.240 align:middle line:90% have large independent sets. 00:18:45.240 --> 00:18:49.020 align:middle line:84% OK, now let us put everything together. 00:18:49.020 --> 00:19:06.530 align:middle line:84% So by choosing n sufficiently large, 00:19:06.530 --> 00:19:15.530 align:middle line:84% we can simultaneously ensure that the probability of x, 00:19:15.530 --> 00:19:19.700 align:middle line:84% the number of short cycles, exceeds n over 2, 00:19:19.700 --> 00:19:22.810 align:middle line:90% is strictly less than 1/2. 00:19:22.810 --> 00:19:31.450 align:middle line:84% And the probability that the independence number exceeds h 00:19:31.450 --> 00:19:33.910 align:middle line:90% is also strictly less than half. 00:19:33.910 --> 00:19:37.900 align:middle line:84% So both of these are quantities that decay to 0 as n 00:19:37.900 --> 00:19:38.970 align:middle line:90% goes to infinity. 00:19:38.970 --> 00:19:41.360 align:middle line:84% So by choosing sufficiently r, h, n, 00:19:41.360 --> 00:19:44.140 align:middle line:84% we can make sure that both probabilities are strictly less 00:19:44.140 --> 00:19:45.170 align:middle line:90% than 1/2. 00:19:45.170 --> 00:19:48.910 align:middle line:84% And therefore, there is some possibility 00:19:48.910 --> 00:19:50.625 align:middle line:90% that lie outside both events. 00:19:50.625 --> 00:19:53.580 align:middle line:90% 00:19:53.580 --> 00:20:15.120 align:middle line:84% Thus, there exists a graph G with fewer than n 00:20:15.120 --> 00:20:22.740 align:middle line:84% over 2 cycles of length at most l. 00:20:22.740 --> 00:20:26.000 align:middle line:90% 00:20:26.000 --> 00:20:30.140 align:middle line:84% And furthermore, its independence number, 00:20:30.140 --> 00:20:36.210 align:middle line:84% the size of the largest independent set, is, at most, h, 00:20:36.210 --> 00:20:41.800 align:middle line:84% which we set earlier to be 3 n over log n. 00:20:41.800 --> 00:20:48.220 align:middle line:84% OK, so we have such a graph G that has these nice properties. 00:20:48.220 --> 00:20:52.240 align:middle line:84% Well, we're looking for a graph with high girth. 00:20:52.240 --> 00:20:56.060 align:middle line:84% Here, G has not too many short cycles. 00:20:56.060 --> 00:21:01.660 align:middle line:84% So we can get rid of one vertex from each short cycle 00:21:01.660 --> 00:21:07.760 align:middle line:84% to remove all the short cycles, so remove a vertex 00:21:07.760 --> 00:21:19.880 align:middle line:84% from each short cycle, each cycle of length at most l. 00:21:19.880 --> 00:21:23.590 align:middle line:90% 00:21:23.590 --> 00:21:26.800 align:middle line:84% So remove one vertex from each such cycle. 00:21:26.800 --> 00:21:33.750 align:middle line:84% And doing so results in a subgraph in G prime. 00:21:33.750 --> 00:21:37.980 align:middle line:84% And then we know that the girth of G prime 00:21:37.980 --> 00:21:41.880 align:middle line:84% is strictly larger than l, because we got rid 00:21:41.880 --> 00:21:44.550 align:middle line:90% of all the short cycles. 00:21:44.550 --> 00:21:51.150 align:middle line:84% And furthermore, the chromatic number of g prime 00:21:51.150 --> 00:21:53.940 align:middle line:84% is by what we saw earlier, at least 00:21:53.940 --> 00:21:56.880 align:middle line:84% the number of vertices of g prime divided 00:21:56.880 --> 00:22:00.990 align:middle line:90% by its independence number. 00:22:00.990 --> 00:22:05.010 align:middle line:84% While the number of vertices is at least n over 2 00:22:05.010 --> 00:22:08.130 align:middle line:84% because we removed at most n over 2 00:22:08.130 --> 00:22:12.000 align:middle line:84% vertices from G to obtain G prime. 00:22:12.000 --> 00:22:14.430 align:middle line:84% The independence number of G prime-- 00:22:14.430 --> 00:22:18.450 align:middle line:84% well, the independence number of G prime 00:22:18.450 --> 00:22:20.940 align:middle line:84% is, at most, the independence number of G 00:22:20.940 --> 00:22:23.190 align:middle line:84% because any independent set in G prime 00:22:23.190 --> 00:22:25.410 align:middle line:84% is automatically an independent set in G. 00:22:25.410 --> 00:22:30.930 align:middle line:84% And here we saw that the independence number of G prime 00:22:30.930 --> 00:22:36.920 align:middle line:84% is at most h, which is 3 log n over 3 n over log n. 00:22:36.920 --> 00:22:39.730 align:middle line:90% 00:22:39.730 --> 00:22:46.760 align:middle line:84% So we have, in the denominator, 3 n over log n. 00:22:46.760 --> 00:22:49.510 align:middle line:90% 00:22:49.510 --> 00:22:54.300 align:middle line:84% And whole expression simplifies to log n 00:22:54.300 --> 00:23:00.570 align:middle line:84% over 6, which is bigger than the constant 00:23:00.570 --> 00:23:07.775 align:middle line:84% that we were given, k, as long as n is sufficiently large. 00:23:07.775 --> 00:23:13.250 align:middle line:90% 00:23:13.250 --> 00:23:22.836 align:middle line:84% And thus, G prime satisfies the desired requirements. 00:23:22.836 --> 00:23:28.580 align:middle line:90% 00:23:28.580 --> 00:23:32.300 align:middle line:84% This finishes the proof of this theorem of Erdos 00:23:32.300 --> 00:23:35.930 align:middle line:84% that there exists graphs with arbitrarily large girth 00:23:35.930 --> 00:23:39.530 align:middle line:84% and arbitrarily large chromatic number. 00:23:39.530 --> 00:23:41.880 align:middle line:84% To review some of the key ideas in this proof, 00:23:41.880 --> 00:23:45.800 align:middle line:84% we use the probabilistic method with alterations 00:23:45.800 --> 00:23:51.830 align:middle line:84% to construct first a random graph with various parameters 00:23:51.830 --> 00:23:53.300 align:middle line:84% so that this random graph typically 00:23:53.300 --> 00:23:57.710 align:middle line:84% has very few short cycles so that we can then 00:23:57.710 --> 00:24:01.070 align:middle line:84% remove one vertex from each short cycle 00:24:01.070 --> 00:24:03.090 align:middle line:84% and get rid of all the short cycles. 00:24:03.090 --> 00:24:07.220 align:middle line:84% And this is a way to ensure large girth. 00:24:07.220 --> 00:24:11.600 align:middle line:84% To obtain high chromatic number, we 00:24:11.600 --> 00:24:17.600 align:middle line:84% note that this graph typically has-- 00:24:17.600 --> 00:24:22.740 align:middle line:84% typically does not have large independent sets. 00:24:22.740 --> 00:24:25.230 align:middle line:84% And so its independence number is typically small. 00:24:25.230 --> 00:24:27.660 align:middle line:84% And therefore, the independence number of the subgraph 00:24:27.660 --> 00:24:29.620 align:middle line:84% must be typically small, as well. 00:24:29.620 --> 00:24:32.220 align:middle line:84% And having small independent number 00:24:32.220 --> 00:24:36.510 align:middle line:90% implies high chromatic number. 00:24:36.510 --> 00:24:38.400 align:middle line:84% OK, so it's a beautiful illustration 00:24:38.400 --> 00:24:41.190 align:middle line:84% of the probabilistic method with alterations 00:24:41.190 --> 00:24:44.890 align:middle line:84% that allows us to deduce this highly counterintuitive result, 00:24:44.890 --> 00:24:51.660 align:middle line:84% which was quite surprising because previously, 00:24:51.660 --> 00:24:54.450 align:middle line:84% various researchers have tried to construct 00:24:54.450 --> 00:24:59.410 align:middle line:84% such examples of graphs by hand explicitly, but unsuccessfully. 00:24:59.410 --> 00:25:01.980 align:middle line:84% And Erdos brought this beautiful insight 00:25:01.980 --> 00:25:05.200 align:middle line:84% that introducing randomness by looking at random graphs, 00:25:05.200 --> 00:25:09.300 align:middle line:84% one can, indeed, prove the existence of graphs with highly 00:25:09.300 --> 00:25:12.320 align:middle line:90% counterintuitive properties. 00:25:12.320 --> 00:25:18.000 align:middle line:90%