WEBVTT 00:00:00.000 --> 00:00:01.473 align:middle line:90% [SQUEAKING] 00:00:01.473 --> 00:00:02.946 align:middle line:90% [RUSTLING] 00:00:02.946 --> 00:00:05.401 align:middle line:90% [CLICKING] 00:00:05.401 --> 00:00:11.790 align:middle line:90% 00:00:11.790 --> 00:00:13.470 align:middle line:84% YUFEI ZHAO: In this video, let us 00:00:13.470 --> 00:00:15.780 align:middle line:84% look at an application of the probabilistic method 00:00:15.780 --> 00:00:17.370 align:middle line:90% to graph theory. 00:00:17.370 --> 00:00:21.400 align:middle line:84% We'll prove what is known as the crossing number inequality. 00:00:21.400 --> 00:00:24.120 align:middle line:84% Now what is the crossing number of a graph? 00:00:24.120 --> 00:00:26.730 align:middle line:84% If I give you a graph, sometimes it's 00:00:26.730 --> 00:00:30.170 align:middle line:84% possible to draw the graph on a plane without crossings. 00:00:30.170 --> 00:00:33.600 align:middle line:84% For example, the complete graph on four vertices 00:00:33.600 --> 00:00:37.680 align:middle line:84% can be drawn on the plane without having any pair of edges 00:00:37.680 --> 00:00:38.760 align:middle line:90% crossed. 00:00:38.760 --> 00:00:41.820 align:middle line:84% Such graph is called a planar graph. 00:00:41.820 --> 00:00:45.330 align:middle line:84% But sometimes, it's not possible to draw such a graph 00:00:45.330 --> 00:00:46.280 align:middle line:90% on the plane. 00:00:46.280 --> 00:00:48.420 align:middle line:84% And the classic example of such a graph 00:00:48.420 --> 00:00:52.260 align:middle line:84% is K5, the complete graph on five vertices. 00:00:52.260 --> 00:00:54.240 align:middle line:84% And then in that case, maybe you want 00:00:54.240 --> 00:00:56.940 align:middle line:84% to know what's the minimum number of edge crossings 00:00:56.940 --> 00:01:03.120 align:middle line:84% you can have in any possible drawing of this graph? 00:01:03.120 --> 00:01:05.040 align:middle line:84% So the crossing number of a graph 00:01:05.040 --> 00:01:08.250 align:middle line:84% is defined to be the minimum number of edge crossings 00:01:08.250 --> 00:01:12.160 align:middle line:84% of a drawing of G of this graph G on the plane using 00:01:12.160 --> 00:01:15.490 align:middle line:90% continuous curves as edges. 00:01:15.490 --> 00:01:18.310 align:middle line:84% It is a classic fact that K5, the complete graph 00:01:18.310 --> 00:01:21.080 align:middle line:84% on five vertices, does not have a planar drawing. 00:01:21.080 --> 00:01:25.150 align:middle line:84% But it is possible to draw this graph using only one crossing. 00:01:25.150 --> 00:01:30.100 align:middle line:84% And we can do this by adding one vertex to this K4 drawing 00:01:30.100 --> 00:01:32.440 align:middle line:90% and draw edges like this. 00:01:32.440 --> 00:01:38.490 align:middle line:84% And the final edge, we need to do a crossing over here. 00:01:38.490 --> 00:01:44.800 align:middle line:84% So this example illustrates that the crossing number of K5 is 1. 00:01:44.800 --> 00:01:47.680 align:middle line:84% If you give me a graph with lots and lots of edges, 00:01:47.680 --> 00:01:50.140 align:middle line:84% should I expect that its crossing number 00:01:50.140 --> 00:01:52.392 align:middle line:90% is necessarily high? 00:01:52.392 --> 00:01:54.100 align:middle line:84% Turns out that's what the crossing number 00:01:54.100 --> 00:01:57.430 align:middle line:90% inequality always guarantees. 00:01:57.430 --> 00:02:02.590 align:middle line:84% The theorem says that in a graph, G, with vertex set E, 00:02:02.590 --> 00:02:09.340 align:middle line:84% vertex set V, and edge set E, if E, this edge set, 00:02:09.340 --> 00:02:14.140 align:middle line:84% is at least four times the number of vertices, 00:02:14.140 --> 00:02:18.910 align:middle line:84% then the crossing number of G is at least the following quantity. 00:02:18.910 --> 00:02:21.910 align:middle line:84% So on the order of the number of edges cubed 00:02:21.910 --> 00:02:24.590 align:middle line:84% divided by number of vertices squared. 00:02:24.590 --> 00:02:31.040 align:middle line:84% Here, c is some constant, some absolute constant. 00:02:31.040 --> 00:02:34.120 align:middle line:84% So this is a theorem that we'll prove in this video. 00:02:34.120 --> 00:02:40.900 align:middle line:84% Before proving it, let me illustrate one corollary to-- 00:02:40.900 --> 00:02:43.000 align:middle line:84% let me state one corollary to illustrate 00:02:43.000 --> 00:02:47.030 align:middle line:90% how to think about this result. 00:02:47.030 --> 00:02:53.810 align:middle line:84% If the number of edges in this graph 00:02:53.810 --> 00:02:59.420 align:middle line:84% is on the order of the number of vertices squared. 00:02:59.420 --> 00:03:03.470 align:middle line:84% So this notation here means that it is larger 00:03:03.470 --> 00:03:07.590 align:middle line:90% than some constant V squared. 00:03:07.590 --> 00:03:11.150 align:middle line:84% So here, c doesn't have to be the same as c earlier. 00:03:11.150 --> 00:03:18.350 align:middle line:84% Then plugging this hypothesis into the theorem, 00:03:18.350 --> 00:03:21.560 align:middle line:84% we obtain that the number of crossings of g 00:03:21.560 --> 00:03:29.050 align:middle line:84% is at least on the order of V to the fourth power. 00:03:29.050 --> 00:03:32.180 align:middle line:90% 00:03:32.180 --> 00:03:35.120 align:middle line:84% So if it has a lot of edges, then the graph 00:03:35.120 --> 00:03:36.720 align:middle line:90% must have a lot of crossings. 00:03:36.720 --> 00:03:39.980 align:middle line:84% And in fact, this is the right order of magnitude, 00:03:39.980 --> 00:03:43.700 align:middle line:84% because you can only have, at most, 00:03:43.700 --> 00:03:46.050 align:middle line:90% E squared number of crossings. 00:03:46.050 --> 00:03:49.980 align:middle line:84% Each pair of edges cross at most once, 00:03:49.980 --> 00:03:54.090 align:middle line:84% even if you just lay out all the vertices in general position 00:03:54.090 --> 00:03:56.345 align:middle line:84% on the plane and draw the edges as straight lines. 00:03:56.345 --> 00:03:59.130 align:middle line:90% 00:03:59.130 --> 00:03:59.820 align:middle line:90% OK. 00:03:59.820 --> 00:04:01.470 align:middle line:84% So in the rest of the video, let me 00:04:01.470 --> 00:04:04.630 align:middle line:84% demonstrate the proof of the crossing number inequality. 00:04:04.630 --> 00:04:12.320 align:middle line:90% 00:04:12.320 --> 00:04:15.290 align:middle line:84% I'll split the proof into three steps. 00:04:15.290 --> 00:04:27.170 align:middle line:84% The first step is an analysis of planar graphs, 00:04:27.170 --> 00:04:30.300 align:middle line:90% namely graphs with no crossings. 00:04:30.300 --> 00:04:40.000 align:middle line:90% 00:04:40.000 --> 00:04:41.830 align:middle line:84% The following turns out to be true, 00:04:41.830 --> 00:04:46.990 align:middle line:90% that if G is a planar graph-- 00:04:46.990 --> 00:04:50.650 align:middle line:84% so "planar" means that it is possible to draw G in the plane 00:04:50.650 --> 00:04:53.110 align:middle line:90% without edge crossings. 00:04:53.110 --> 00:04:58.210 align:middle line:84% Then-- so in other words, the crossing number of G 00:04:58.210 --> 00:05:00.190 align:middle line:90% is equal to 0. 00:05:00.190 --> 00:05:04.510 align:middle line:84% So then the conclusion is that the number of edges of G 00:05:04.510 --> 00:05:08.955 align:middle line:84% is, at most, three times the number of vertices. 00:05:08.955 --> 00:05:10.330 align:middle line:84% Another way to say the conclusion 00:05:10.330 --> 00:05:15.660 align:middle line:84% is that the average degree of this graph is at most 6. 00:05:15.660 --> 00:05:27.650 align:middle line:84% This conclusion can be deduced from effect 00:05:27.650 --> 00:05:34.510 align:middle line:84% from topology known as Euler's formula, which 00:05:34.510 --> 00:05:37.520 align:middle line:84% says that if you draw the graph in the plane in some way-- 00:05:37.520 --> 00:05:41.350 align:middle line:84% so for example, this is some drawing, 00:05:41.350 --> 00:05:45.350 align:middle line:84% then the number of vertices minus the number of edges, 00:05:45.350 --> 00:05:48.430 align:middle line:84% plus the number of faces equals to 2. 00:05:48.430 --> 00:05:52.450 align:middle line:84% Here, by face, we count each individual cell as well as 00:05:52.450 --> 00:05:53.390 align:middle line:90% the outside cell. 00:05:53.390 --> 00:05:57.200 align:middle line:90% So this drawing has three faces. 00:05:57.200 --> 00:06:00.540 align:middle line:84% And it has four vertices and five edges. 00:06:00.540 --> 00:06:03.230 align:middle line:84% And you can verify that this Euler's identity 00:06:03.230 --> 00:06:04.710 align:middle line:90% is true in this case. 00:06:04.710 --> 00:06:07.760 align:middle line:90% It is also true in general. 00:06:07.760 --> 00:06:10.280 align:middle line:84% Starting with Euler's formula, we 00:06:10.280 --> 00:06:13.760 align:middle line:84% can then put some bounds on the number of edges, 00:06:13.760 --> 00:06:17.880 align:middle line:84% and faces, and vertices, relating them to each other, 00:06:17.880 --> 00:06:20.960 align:middle line:84% using some inequalities to derive this conclusion. 00:06:20.960 --> 00:06:22.708 align:middle line:90% And so let me not do that step. 00:06:22.708 --> 00:06:25.375 align:middle line:84% I refer you to the lecture notes if you want to see the details. 00:06:25.375 --> 00:06:28.410 align:middle line:90% 00:06:28.410 --> 00:06:35.610 align:middle line:84% The second step of the proof allows 00:06:35.610 --> 00:06:43.090 align:middle line:84% us to go from one crossing to many crossings. 00:06:43.090 --> 00:06:46.470 align:middle line:90% 00:06:46.470 --> 00:06:49.320 align:middle line:90% So here's what happens. 00:06:49.320 --> 00:06:56.400 align:middle line:84% So given a graph, G, it may not be planar. 00:06:56.400 --> 00:06:58.880 align:middle line:84% So maybe it will require some crossings 00:06:58.880 --> 00:07:01.400 align:middle line:90% to draw G in the plane. 00:07:01.400 --> 00:07:03.930 align:middle line:84% But starting with such a crossing, 00:07:03.930 --> 00:07:08.840 align:middle line:84% we can delete some edges to make it planar. 00:07:08.840 --> 00:07:22.680 align:middle line:84% So given G, by deleting the crossing number of G edges, 00:07:22.680 --> 00:07:29.280 align:middle line:90% we can make it planar. 00:07:29.280 --> 00:07:30.360 align:middle line:90% So if you had-- 00:07:30.360 --> 00:07:34.420 align:middle line:84% so in the earlier example of K5 where this is not planar, 00:07:34.420 --> 00:07:36.130 align:middle line:90% but there's only one crossing. 00:07:36.130 --> 00:07:38.945 align:middle line:84% And by deleting one of the edges corresponding to that crossing. 00:07:38.945 --> 00:07:43.230 align:middle line:90% 00:07:43.230 --> 00:07:49.960 align:middle line:84% And we remove one edge and we end up with a planar graph. 00:07:49.960 --> 00:07:55.060 align:middle line:84% Well, in this planar graph, the inequality in step one 00:07:55.060 --> 00:07:56.380 align:middle line:90% must be satisfied. 00:07:56.380 --> 00:08:00.720 align:middle line:84% So the number of edges of g minus the crossing number of g, 00:08:00.720 --> 00:08:03.760 align:middle line:84% so these are the deleted edges, must 00:08:03.760 --> 00:08:07.510 align:middle line:84% be, at most, three times the number of vertices 00:08:07.510 --> 00:08:14.480 align:middle line:84% of G. Rearranging this inequality, 00:08:14.480 --> 00:08:22.190 align:middle line:84% we obtain the lower bound on the crossing number 00:08:22.190 --> 00:08:28.280 align:middle line:84% being the number of edges, minus three times the number 00:08:28.280 --> 00:08:28.825 align:middle line:90% of vertices. 00:08:28.825 --> 00:08:31.750 align:middle line:90% 00:08:31.750 --> 00:08:35.690 align:middle line:84% So the idea here is really to go from one crossing, 00:08:35.690 --> 00:08:40.450 align:middle line:84% namely that if this inequality, E less than equal to 3V is not 00:08:40.450 --> 00:08:42.650 align:middle line:84% satisfied, then we get at least one crossing. 00:08:42.650 --> 00:08:45.230 align:middle line:84% But then keep on deleting these crossings. 00:08:45.230 --> 00:08:49.200 align:middle line:84% We can go from one crossing to many crossings. 00:08:49.200 --> 00:08:51.970 align:middle line:84% Let's pause and examine this bound that we just proved. 00:08:51.970 --> 00:08:55.480 align:middle line:84% It is a valid bound, but it's not that great. 00:08:55.480 --> 00:08:58.290 align:middle line:84% For example, in this corollary, when 00:08:58.290 --> 00:09:03.490 align:middle line:84% E is on the order of quadratic in number of vertices, 00:09:03.490 --> 00:09:06.270 align:middle line:84% the conclusion that we will get at this stage 00:09:06.270 --> 00:09:09.090 align:middle line:84% is also quadratic in the number of vertices, 00:09:09.090 --> 00:09:11.350 align:middle line:84% which is not as high as we would like, 00:09:11.350 --> 00:09:16.250 align:middle line:84% which is the fourth power in the number of vertices. 00:09:16.250 --> 00:09:21.200 align:middle line:84% So this brings us to the third step. 00:09:21.200 --> 00:09:25.010 align:middle line:84% And this is a step where we'll use the probabilistic method. 00:09:25.010 --> 00:09:28.970 align:middle line:84% And this step, I'll call the bootstrapping step, 00:09:28.970 --> 00:09:34.490 align:middle line:84% where we go from a weak bound to a much stronger bound by using 00:09:34.490 --> 00:09:37.040 align:middle line:90% sampling. 00:09:37.040 --> 00:09:43.170 align:middle line:84% In this step, let's consider some p. 00:09:43.170 --> 00:09:47.770 align:middle line:84% We'll decide this p later on, so to be decided. 00:09:47.770 --> 00:09:51.030 align:middle line:84% But it's some number between 0 and 1. 00:09:51.030 --> 00:09:54.720 align:middle line:84% And let's consider a subgraph of g obtained 00:09:54.720 --> 00:10:11.540 align:middle line:84% by keeping every vertex of G with probability p 00:10:11.540 --> 00:10:22.220 align:middle line:84% independently at random and deleting the other vertices. 00:10:22.220 --> 00:10:30.480 align:middle line:90% 00:10:30.480 --> 00:10:32.350 align:middle line:84% When we delete the other vertices, 00:10:32.350 --> 00:10:35.220 align:middle line:84% we also throw away all the edges that 00:10:35.220 --> 00:10:39.210 align:middle line:84% are adjacent to the deleted vertices. 00:10:39.210 --> 00:10:44.580 align:middle line:84% This process then produces a different graph, G prime, 00:10:44.580 --> 00:10:48.430 align:middle line:84% which contains a subset, V, of vertices, the V prime. 00:10:48.430 --> 00:10:54.090 align:middle line:84% So those are the vertices that are kept by this random process. 00:10:54.090 --> 00:10:58.090 align:middle line:84% And then E prime is a set of edges that were kept, 00:10:58.090 --> 00:11:02.310 align:middle line:84% namely the edges of G that fall between the remaining 00:11:02.310 --> 00:11:06.190 align:middle line:90% vertices in V prime. 00:11:06.190 --> 00:11:08.515 align:middle line:84% Let's think about this graph, G prime here. 00:11:08.515 --> 00:11:09.770 align:middle line:90% Well, it is a graph. 00:11:09.770 --> 00:11:13.000 align:middle line:84% So whatever we proved in step two still 00:11:13.000 --> 00:11:17.130 align:middle line:90% applies to this smaller graph. 00:11:17.130 --> 00:11:19.890 align:middle line:84% Namely, the crossing number of G prime 00:11:19.890 --> 00:11:25.930 align:middle line:84% is at least the number of edges in this G prime, 00:11:25.930 --> 00:11:33.140 align:middle line:84% minus 3 times the number of vertices in this G prime. 00:11:33.140 --> 00:11:36.650 align:middle line:84% So this is a true and valid inequality. 00:11:36.650 --> 00:11:39.900 align:middle line:90% 00:11:39.900 --> 00:11:44.100 align:middle line:84% Well, remember that G is given as a graph that 00:11:44.100 --> 00:11:47.040 align:middle line:84% is deterministic, but we introduce some randomness 00:11:47.040 --> 00:11:48.840 align:middle line:90% to produce G prime. 00:11:48.840 --> 00:11:52.020 align:middle line:90% G prime is a random graph. 00:11:52.020 --> 00:11:55.920 align:middle line:84% So this inequality is true for every instance 00:11:55.920 --> 00:12:01.120 align:middle line:84% of this random graph, but it's also true, then, in expectation. 00:12:01.120 --> 00:12:05.290 align:middle line:84% So let's take the expectation of both sides. 00:12:05.290 --> 00:12:08.190 align:middle line:84% The expectation on the left and the expectation 00:12:08.190 --> 00:12:10.750 align:middle line:84% on the right, which, by linearity of expectations, 00:12:10.750 --> 00:12:16.610 align:middle line:84% we can distribute the expectation into each term. 00:12:16.610 --> 00:12:26.880 align:middle line:90% 00:12:26.880 --> 00:12:29.460 align:middle line:84% Let's think about the various terms. 00:12:29.460 --> 00:12:33.540 align:middle line:84% The easiest one to think about is the expected number 00:12:33.540 --> 00:12:36.510 align:middle line:84% of vertices that remain while each vertex is 00:12:36.510 --> 00:12:38.340 align:middle line:90% kept with probability p. 00:12:38.340 --> 00:12:41.820 align:middle line:84% So the expected number of remaining vertices 00:12:41.820 --> 00:12:47.930 align:middle line:84% is p times the original number of vertices. 00:12:47.930 --> 00:12:50.930 align:middle line:84% Next, the expected number of edges. 00:12:50.930 --> 00:12:54.980 align:middle line:84% Well, an edge is kept if both its endpoints remain. 00:12:54.980 --> 00:12:58.430 align:middle line:84% And the probability that both its endpoints are chosen 00:12:58.430 --> 00:12:59.590 align:middle line:90% is p squared. 00:12:59.590 --> 00:13:05.230 align:middle line:84% So this quantity is p squared times the original number 00:13:05.230 --> 00:13:08.360 align:middle line:90% of edges. 00:13:08.360 --> 00:13:11.540 align:middle line:84% And finally, and the slightly tricky one, 00:13:11.540 --> 00:13:15.940 align:middle line:84% is the expected number of crossings. 00:13:15.940 --> 00:13:26.650 align:middle line:84% Now if I have a crossing in my original graph, G, one way 00:13:26.650 --> 00:13:31.170 align:middle line:84% to go about thinking about G prime is to, 00:13:31.170 --> 00:13:34.540 align:middle line:84% well, take the same planar drawing. 00:13:34.540 --> 00:13:37.570 align:middle line:84% Take the same drawing of G on the plane 00:13:37.570 --> 00:13:39.650 align:middle line:84% and use that drawing for G prime. 00:13:39.650 --> 00:13:43.180 align:middle line:84% So G prime is obtained by keeping some of the vertices. 00:13:43.180 --> 00:13:45.410 align:middle line:84% And so if we use the same drawings, 00:13:45.410 --> 00:13:51.255 align:middle line:84% then this crossing is kept with probability p to the fourth. 00:13:51.255 --> 00:14:13.680 align:middle line:90% 00:14:13.680 --> 00:14:17.220 align:middle line:84% Now G prime is a different graph. 00:14:17.220 --> 00:14:18.480 align:middle line:90% It's a smaller graph. 00:14:18.480 --> 00:14:20.790 align:middle line:84% So potentially, there's a different way 00:14:20.790 --> 00:14:25.320 align:middle line:84% to draw G prime that produces even fewer crossings. 00:14:25.320 --> 00:14:28.530 align:middle line:84% But nevertheless, we have an inequality, 00:14:28.530 --> 00:14:31.440 align:middle line:84% because we could have always kept the cross-- 00:14:31.440 --> 00:14:36.170 align:middle line:90% the drawing from G. 00:14:36.170 --> 00:14:39.900 align:middle line:84% And so we do always have this inequality here. 00:14:39.900 --> 00:14:44.670 align:middle line:84% So this inequality comes from keeping the same drawing, 00:14:44.670 --> 00:14:46.320 align:middle line:84% but keeping in mind that it doesn't 00:14:46.320 --> 00:14:47.790 align:middle line:84% have to be an equality, because we 00:14:47.790 --> 00:14:51.650 align:middle line:84% could have chosen a different drawing for G prime. 00:14:51.650 --> 00:14:57.840 align:middle line:84% OK, so we derived this inequality here. 00:14:57.840 --> 00:15:04.010 align:middle line:84% And by bringing this p to the fourth power to the other side 00:15:04.010 --> 00:15:07.850 align:middle line:84% and rearranging, we obtain the inequality 00:15:07.850 --> 00:15:14.420 align:middle line:84% that the crossing number of G is at least p to the minus 2 times 00:15:14.420 --> 00:15:19.700 align:middle line:84% the size of the edge set times 3 to the p to the minus 3 00:15:19.700 --> 00:15:26.300 align:middle line:84% times the size of the vertex set of G. 00:15:26.300 --> 00:15:31.020 align:middle line:84% p was some parameter that we have yet to specify. 00:15:31.020 --> 00:15:33.530 align:middle line:84% And from this expression, then we can 00:15:33.530 --> 00:15:37.400 align:middle line:90% optimize for the value of p. 00:15:37.400 --> 00:15:42.680 align:middle line:84% And one way to do this, and because we only 00:15:42.680 --> 00:15:45.570 align:middle line:84% look for a bound that is up to a constant factor, 00:15:45.570 --> 00:15:51.800 align:middle line:84% is to set p to be 4 times the number of vertices 00:15:51.800 --> 00:15:55.820 align:middle line:84% divided by the number of edges of the graph. 00:15:55.820 --> 00:16:00.840 align:middle line:84% And it is important here that in the hypothesis of the theorem, 00:16:00.840 --> 00:16:03.020 align:middle line:84% the number of edges is at least four times 00:16:03.020 --> 00:16:05.810 align:middle line:84% the number of vertices, so that this p is indeed 00:16:05.810 --> 00:16:07.590 align:middle line:90% between 0 and 1. 00:16:07.590 --> 00:16:10.170 align:middle line:84% If p ended up being some number bigger than 1, 00:16:10.170 --> 00:16:12.090 align:middle line:84% then this proof would not have made sense. 00:16:12.090 --> 00:16:14.840 align:middle line:84% It would not make sense to choose with probability bigger 00:16:14.840 --> 00:16:15.450 align:middle line:90% than 1. 00:16:15.450 --> 00:16:19.280 align:middle line:84% But p is between 0 and 1, so this makes sense. 00:16:19.280 --> 00:16:22.700 align:middle line:84% And with this choice of p, continuing the earlier 00:16:22.700 --> 00:16:31.040 align:middle line:84% expression, we find that it is equal to one 00:16:31.040 --> 00:16:34.310 align:middle line:84% over 64 times the number of edges 00:16:34.310 --> 00:16:38.450 align:middle line:84% cubed divided by the number of vertices squared. 00:16:38.450 --> 00:16:41.190 align:middle line:84% And that finishes the proof of the result, 00:16:41.190 --> 00:16:45.740 align:middle line:84% showing that the crossing number inequality holds with c 00:16:45.740 --> 00:16:48.360 align:middle line:90% being the constant 1 over 64. 00:16:48.360 --> 00:16:52.750 align:middle line:84% Although, the precise value is not so important. 00:16:52.750 --> 00:16:54.790 align:middle line:84% This is a beautiful demonstration 00:16:54.790 --> 00:16:56.180 align:middle line:90% of the probabilistic method. 00:16:56.180 --> 00:17:01.390 align:middle line:84% And let me review what happened in this proof. 00:17:01.390 --> 00:17:06.109 align:middle line:84% First, we looked at what happens to planar graphs, 00:17:06.109 --> 00:17:08.210 align:middle line:90% those with no crossings. 00:17:08.210 --> 00:17:11.920 align:middle line:84% And here, we derived some bound on the number of edges 00:17:11.920 --> 00:17:15.160 align:middle line:90% that such a graph can have. 00:17:15.160 --> 00:17:19.270 align:middle line:84% And then we said that if you start with a graph, 00:17:19.270 --> 00:17:21.490 align:middle line:84% and then remove all its crossings by removing 00:17:21.490 --> 00:17:23.569 align:middle line:84% one edge according to each crossing, 00:17:23.569 --> 00:17:26.770 align:middle line:84% we can then deduce some lower bound on the number of crossings 00:17:26.770 --> 00:17:29.560 align:middle line:90% from the first step. 00:17:29.560 --> 00:17:32.090 align:middle line:84% Then it gives us this fairly weak bound-- 00:17:32.090 --> 00:17:36.850 align:middle line:84% lower bound on the number of crossings. 00:17:36.850 --> 00:17:39.430 align:middle line:84% And the final step, I think, is the most interesting 00:17:39.430 --> 00:17:40.240 align:middle line:90% mathematically. 00:17:40.240 --> 00:17:41.920 align:middle line:90% It's this bootstrapping step. 00:17:41.920 --> 00:17:45.310 align:middle line:84% We're starting from a fairly weak bound. 00:17:45.310 --> 00:17:50.290 align:middle line:84% We then take our graph G and sample it down 00:17:50.290 --> 00:17:53.120 align:middle line:84% to a smaller subgraph and apply the weaker bound 00:17:53.120 --> 00:17:55.400 align:middle line:90% to the smaller subgraph. 00:17:55.400 --> 00:18:00.320 align:middle line:84% This allows us to boost the weak bound into a much stronger bound 00:18:00.320 --> 00:18:02.570 align:middle line:90% that turns out to be the-- 00:18:02.570 --> 00:18:06.920 align:middle line:84% what is claimed in the result in the theorem. 00:18:06.920 --> 00:18:10.460 align:middle line:84% So this finishes the proof of the crossing number inequality. 00:18:10.460 --> 00:18:13.520 align:middle line:84% And it's just one of many beautiful demonstrations 00:18:13.520 --> 00:18:18.460 align:middle line:84% of the probabilistic method in combinatorics and graph theory. 00:18:18.460 --> 00:18:26.000 align:middle line:90%