WEBVTT 00:00:00.000 --> 00:00:01.932 align:middle line:90% [SQUEAKING] 00:00:01.932 --> 00:00:03.381 align:middle line:90% [RUSTLING] 00:00:03.381 --> 00:00:04.830 align:middle line:90% [CLICKING] 00:00:04.830 --> 00:00:11.120 align:middle line:90% 00:00:11.120 --> 00:00:12.540 align:middle line:90% ZACHARY ABEL: Good morning. 00:00:12.540 --> 00:00:13.320 align:middle line:90% Good morning. 00:00:13.320 --> 00:00:14.820 align:middle line:90% Welcome. 00:00:14.820 --> 00:00:18.755 align:middle line:84% Welcome to lecture 11 of 6 120, 6 1,200. 00:00:18.755 --> 00:00:20.130 align:middle line:84% I never know how to pronounce it. 00:00:20.130 --> 00:00:22.400 align:middle line:90% It's one of them, probably. 00:00:22.400 --> 00:00:25.320 align:middle line:84% We have finished our number theory unit, 00:00:25.320 --> 00:00:29.510 align:middle line:84% and we're now starting a new unit on graphs, graph theory. 00:00:29.510 --> 00:00:34.790 align:middle line:84% This is a new topic, a new kind of mathematical abstraction 00:00:34.790 --> 00:00:37.640 align:middle line:84% that we'll see really interesting uses 00:00:37.640 --> 00:00:40.500 align:middle line:84% for across this class, across algorithms, 00:00:40.500 --> 00:00:42.860 align:middle line:90% across other contexts as well. 00:00:42.860 --> 00:00:45.420 align:middle line:84% Before we dig into what they're good at, 00:00:45.420 --> 00:00:48.780 align:middle line:84% let's talk a bit about what they are, how they're defined. 00:00:48.780 --> 00:00:51.840 align:middle line:84% So I've written an example of a graph up here. 00:00:51.840 --> 00:00:53.030 align:middle line:90% Totally clear, right? 00:00:53.030 --> 00:00:57.050 align:middle line:84% It's easy to read and comprehend, but let me define it 00:00:57.050 --> 00:00:57.840 align:middle line:90% for you. 00:00:57.840 --> 00:01:02.450 align:middle line:90% So a simple graph. 00:01:02.450 --> 00:01:09.205 align:middle line:84% Here, let's say a simple undirected graph, which 00:01:09.205 --> 00:01:10.580 align:middle line:84% are the only kind of graphs we'll 00:01:10.580 --> 00:01:13.760 align:middle line:84% be talking about for the next couple of lectures, 00:01:13.760 --> 00:01:16.090 align:middle line:90% is defined as a pair. 00:01:16.090 --> 00:01:18.620 align:middle line:90% 00:01:18.620 --> 00:01:25.280 align:middle line:84% So the graph G is a pair of V, E, where 00:01:25.280 --> 00:01:38.900 align:middle line:84% V is a nonempty set of what we call nodes, also known 00:01:38.900 --> 00:01:39.483 align:middle line:90% as vertices. 00:01:39.483 --> 00:01:46.670 align:middle line:90% 00:01:46.670 --> 00:01:59.390 align:middle line:84% And E is a set of two element subsets of V, not all of them, 00:01:59.390 --> 00:02:01.480 align:middle line:84% but some of them, possibly none of them. 00:02:01.480 --> 00:02:04.910 align:middle line:84% Could be all of them as well, any amount that you want. 00:02:04.910 --> 00:02:10.000 align:middle line:84% So a graph is just a set V, where we call 00:02:10.000 --> 00:02:12.850 align:middle line:90% the objects vertices or nodes. 00:02:12.850 --> 00:02:13.635 align:middle line:90% I'm sorry. 00:02:13.635 --> 00:02:17.380 align:middle line:90% 00:02:17.380 --> 00:02:19.493 align:middle line:90% These are edges. 00:02:19.493 --> 00:02:22.180 align:middle line:90% 00:02:22.180 --> 00:02:26.150 align:middle line:84% So it's a set V of nodes and a set E of pairs of nodes. 00:02:26.150 --> 00:02:28.700 align:middle line:84% And those pairs are called edges. 00:02:28.700 --> 00:02:33.350 align:middle line:84% So in its purest set theoretic form, here is a graph. 00:02:33.350 --> 00:02:34.520 align:middle line:90% Here's a set of vertices. 00:02:34.520 --> 00:02:39.320 align:middle line:84% Here's a set of some of the two element subsets of V. 00:02:39.320 --> 00:02:45.550 align:middle line:84% Of course, I don't really think about the graph in this form. 00:02:45.550 --> 00:02:47.470 align:middle line:84% Let me give you an alternative way 00:02:47.470 --> 00:02:51.676 align:middle line:90% to express this graph up here. 00:02:51.676 --> 00:02:55.660 align:middle line:90% 1, 2, 3, 4, 5, 6, 7. 00:02:55.660 --> 00:02:56.960 align:middle line:90% These are going to be my nodes. 00:02:56.960 --> 00:02:58.000 align:middle line:90% This is a. 00:02:58.000 --> 00:03:04.330 align:middle line:90% This is b, c, d, e, f, and g. 00:03:04.330 --> 00:03:07.240 align:middle line:84% And anytime we have an edge, let's connect them 00:03:07.240 --> 00:03:09.500 align:middle line:90% by drawing an arc between them. 00:03:09.500 --> 00:03:12.460 align:middle line:90% So we have ab, ac, and bc. 00:03:12.460 --> 00:03:16.930 align:middle line:84% We have these, these, these, this, this, and this. 00:03:16.930 --> 00:03:20.150 align:middle line:84% This is a visual representation of this graph, 00:03:20.150 --> 00:03:22.610 align:middle line:84% but it's conveying the same information. 00:03:22.610 --> 00:03:25.380 align:middle line:84% This is a lot easier to look at, in my opinion, 00:03:25.380 --> 00:03:27.380 align:middle line:84% but they're both conveying the same information. 00:03:27.380 --> 00:03:29.570 align:middle line:84% This is the general idea of a graph. 00:03:29.570 --> 00:03:31.570 align:middle line:84% It's a set of vertices, which are the dots, 00:03:31.570 --> 00:03:34.540 align:middle line:84% and edges, which are the segments connecting them. 00:03:34.540 --> 00:03:36.880 align:middle line:90% Does that make sense? 00:03:36.880 --> 00:03:39.880 align:middle line:90% Wonderful. 00:03:39.880 --> 00:03:43.660 align:middle line:84% I want to mention that there are some subtle differences 00:03:43.660 --> 00:03:45.780 align:middle line:84% in the definitions in other sources. 00:03:45.780 --> 00:03:52.942 align:middle line:90% 00:03:52.942 --> 00:04:00.880 align:middle line:84% Sometimes when people are talking about undirected graphs, 00:04:00.880 --> 00:04:09.970 align:middle line:84% some people will allow what are called parallel edges, which 00:04:09.970 --> 00:04:13.210 align:middle line:84% is when you have two different edges connecting 00:04:13.210 --> 00:04:23.290 align:middle line:84% the same pair of vertices, or self-loops, which 00:04:23.290 --> 00:04:26.830 align:middle line:84% is when you have a single edge connecting a vertex back 00:04:26.830 --> 00:04:28.540 align:middle line:90% to itself. 00:04:28.540 --> 00:04:31.005 align:middle line:84% And we are disallowing both of those. 00:04:31.005 --> 00:04:36.850 align:middle line:90% 00:04:36.850 --> 00:04:39.400 align:middle line:84% So we will never be working with graphs like this 00:04:39.400 --> 00:04:42.730 align:middle line:84% that have parallel edges or that have self-loops. 00:04:42.730 --> 00:04:46.900 align:middle line:84% So that is what people usually mean by simple. 00:04:46.900 --> 00:04:48.400 align:middle line:84% And it's why all of the undirected 00:04:48.400 --> 00:04:51.540 align:middle line:84% graphs we work with are going to be simple. 00:04:51.540 --> 00:05:01.380 align:middle line:84% Likewise, some sources allow the vertex set to be empty. 00:05:01.380 --> 00:05:04.760 align:middle line:84% I stipulated up there that v should be a nonempty set. 00:05:04.760 --> 00:05:10.320 align:middle line:90% 00:05:10.320 --> 00:05:13.020 align:middle line:84% So we're just disagreeing on a single graph that 00:05:13.020 --> 00:05:15.810 align:middle line:84% has zero vertices and then necessarily zero edges 00:05:15.810 --> 00:05:17.910 align:middle line:84% because there's nothing to connect. 00:05:17.910 --> 00:05:21.840 align:middle line:84% The reason we are not allowing that for our purposes 00:05:21.840 --> 00:05:23.940 align:middle line:84% is that often it's going to be a counterexample 00:05:23.940 --> 00:05:25.780 align:middle line:84% to a lot of the theorems we want to state. 00:05:25.780 --> 00:05:28.930 align:middle line:84% We would always have to say for every nonempty graph, 00:05:28.930 --> 00:05:31.620 align:middle line:84% for every graph with a positive number of vertices, something 00:05:31.620 --> 00:05:33.870 align:middle line:84% like that, and it just gets cumbersome. 00:05:33.870 --> 00:05:35.790 align:middle line:84% So we chose this definition because it 00:05:35.790 --> 00:05:37.330 align:middle line:90% makes that more convenient. 00:05:37.330 --> 00:05:39.240 align:middle line:84% Not really a big deal, not really something 00:05:39.240 --> 00:05:40.750 align:middle line:90% we'll need to think about much. 00:05:40.750 --> 00:05:42.870 align:middle line:84% It's just a part of the definition 00:05:42.870 --> 00:05:45.305 align:middle line:84% to be vaguely aware of in the back of your mind. 00:05:45.305 --> 00:05:48.240 align:middle line:90% 00:05:48.240 --> 00:05:56.370 align:middle line:84% We do allow the set of edges to be empty. 00:05:56.370 --> 00:06:02.310 align:middle line:84% For example, here is a graph with five vertices and zero 00:06:02.310 --> 00:06:03.130 align:middle line:90% edges. 00:06:03.130 --> 00:06:06.930 align:middle line:84% Let's call them vertices 1, 2, 3, 4, and 5. 00:06:06.930 --> 00:06:09.700 align:middle line:84% Five vertices, zero edges, that's totally fine. 00:06:09.700 --> 00:06:11.880 align:middle line:90% That's a valid graph. 00:06:11.880 --> 00:06:15.870 align:middle line:84% So this has an empty set for the set of edges, which is fine. 00:06:15.870 --> 00:06:17.490 align:middle line:90% All right? 00:06:17.490 --> 00:06:19.770 align:middle line:90% So what are graphs good at? 00:06:19.770 --> 00:06:24.030 align:middle line:84% Graphs are great at representing two-way relationships 00:06:24.030 --> 00:06:26.070 align:middle line:90% between pairs of objects. 00:06:26.070 --> 00:06:30.510 align:middle line:84% Let me give some real-world examples, like friendship. 00:06:30.510 --> 00:06:36.300 align:middle line:84% You can imagine the graph of all the friends in the world. 00:06:36.300 --> 00:06:41.160 align:middle line:84% Here, let's see, the friend graph. 00:06:41.160 --> 00:06:45.000 align:middle line:84% The vertices are going to be the set of people. 00:06:45.000 --> 00:06:46.870 align:middle line:90% Each person is a vertex. 00:06:46.870 --> 00:06:50.790 align:middle line:84% The set of edges is pairs of friends. 00:06:50.790 --> 00:06:53.585 align:middle line:90% 00:06:53.585 --> 00:06:55.960 align:middle line:84% Now, I'm making an assumption here that friendship always 00:06:55.960 --> 00:06:57.062 align:middle line:90% goes both ways. 00:06:57.062 --> 00:06:59.270 align:middle line:84% If I'm friends with you, then you're friends with me. 00:06:59.270 --> 00:07:03.160 align:middle line:90% No unrequited friendships here. 00:07:03.160 --> 00:07:07.310 align:middle line:84% For example, Facebook friendships always go both ways. 00:07:07.310 --> 00:07:09.760 align:middle line:84% You're either friends on Facebook, or you're not. 00:07:09.760 --> 00:07:14.140 align:middle line:84% So the graph of Facebook friends is a concrete representation 00:07:14.140 --> 00:07:17.356 align:middle line:84% of this kind of thing, which can be interesting. 00:07:17.356 --> 00:07:20.920 align:middle line:90% 00:07:20.920 --> 00:07:21.500 align:middle line:90% Let's see. 00:07:21.500 --> 00:07:24.130 align:middle line:90% Another example is the brain. 00:07:24.130 --> 00:07:26.690 align:middle line:84% The brain is just a network of neurons. 00:07:26.690 --> 00:07:34.300 align:middle line:84% The neurons are the vertices, neurons and connections. 00:07:34.300 --> 00:07:37.400 align:middle line:84% Some neurons connect to neighboring neurons, 00:07:37.400 --> 00:07:39.850 align:middle line:84% and it's those neurons and their connections 00:07:39.850 --> 00:07:44.240 align:middle line:84% that somehow creates cognition and memory and intelligence. 00:07:44.240 --> 00:07:46.160 align:middle line:90% And I wish we understood how. 00:07:46.160 --> 00:07:47.635 align:middle line:90% But at least it's a graph. 00:07:47.635 --> 00:07:50.210 align:middle line:90% 00:07:50.210 --> 00:07:51.985 align:middle line:90% Also, we have the internet. 00:07:51.985 --> 00:07:55.520 align:middle line:90% 00:07:55.520 --> 00:08:00.590 align:middle line:84% Most network connections on the internet go both ways. 00:08:00.590 --> 00:08:04.340 align:middle line:90% Network connections. 00:08:04.340 --> 00:08:06.170 align:middle line:90% So the servers are the vertices. 00:08:06.170 --> 00:08:08.600 align:middle line:84% And which servers can talk to which 00:08:08.600 --> 00:08:11.210 align:middle line:90% other servers are the edges. 00:08:11.210 --> 00:08:13.400 align:middle line:84% And most communications have to go both ways 00:08:13.400 --> 00:08:15.968 align:middle line:84% because you have to communicate back and forth 00:08:15.968 --> 00:08:17.510 align:middle line:84% and make sure that each of you trusts 00:08:17.510 --> 00:08:19.700 align:middle line:84% the other party before you actually 00:08:19.700 --> 00:08:21.150 align:middle line:90% send any information over. 00:08:21.150 --> 00:08:25.430 align:middle line:84% So you have to go both ways to make that connection first, 00:08:25.430 --> 00:08:26.870 align:middle line:84% and then you can send information 00:08:26.870 --> 00:08:29.900 align:middle line:90% in whatever direction you want. 00:08:29.900 --> 00:08:38.840 align:middle line:84% A non-example is hyperlinks on websites. 00:08:38.840 --> 00:08:41.760 align:middle line:84% If site A has a link that takes you to site B, 00:08:41.760 --> 00:08:44.750 align:middle line:84% there isn't necessarily a link back. 00:08:44.750 --> 00:08:48.830 align:middle line:84% So this would not be an example of an undirected graph. 00:08:48.830 --> 00:08:50.970 align:middle line:84% That would be a directed relationship, 00:08:50.970 --> 00:08:54.200 align:middle line:84% which we're going to talk about later after break. 00:08:54.200 --> 00:08:56.480 align:middle line:84% But that's not an example for today. 00:08:56.480 --> 00:09:00.630 align:middle line:84% Similarly, Twitter followers is a directed relationship. 00:09:00.630 --> 00:09:03.060 align:middle line:84% If I follow you, you don't necessarily follow me. 00:09:03.060 --> 00:09:04.640 align:middle line:84% So that would be a directed graph, 00:09:04.640 --> 00:09:11.390 align:middle line:84% which will fit into lectures after break, not this one. 00:09:11.390 --> 00:09:14.060 align:middle line:90% All right. 00:09:14.060 --> 00:09:17.690 align:middle line:84% And graphs-- I said earlier, graphs 00:09:17.690 --> 00:09:19.920 align:middle line:90% are a really useful abstraction. 00:09:19.920 --> 00:09:23.210 align:middle line:84% And I claim that they're useful because they 00:09:23.210 --> 00:09:26.246 align:middle line:84% strike a nice balance between simplicity and complexity. 00:09:26.246 --> 00:09:32.030 align:middle line:90% 00:09:32.030 --> 00:09:36.120 align:middle line:84% On one hand, graphs are simple to define and describe. 00:09:36.120 --> 00:09:37.350 align:middle line:90% They're simple conceptually. 00:09:37.350 --> 00:09:43.130 align:middle line:84% It's just a set of objects and a set of pairwise relationships. 00:09:43.130 --> 00:09:46.100 align:middle line:90% Easy to define. 00:09:46.100 --> 00:09:50.480 align:middle line:84% So as we just saw, there are lots and lots of real scenarios 00:09:50.480 --> 00:09:53.540 align:middle line:84% that can be modeled faithfully as a graph. 00:09:53.540 --> 00:09:56.720 align:middle line:84% On the other hand, they're complex in that there's 00:09:56.720 --> 00:09:59.693 align:middle line:84% a rich mathematical theory of graphs. 00:09:59.693 --> 00:10:07.460 align:middle line:90% 00:10:07.460 --> 00:10:10.520 align:middle line:84% There are lots of powerful tools and definitions 00:10:10.520 --> 00:10:13.160 align:middle line:84% and theorems that we can bring to bear anytime we're 00:10:13.160 --> 00:10:14.970 align:middle line:90% trying to analyze a graph. 00:10:14.970 --> 00:10:17.090 align:middle line:84% And therefore, we can apply all of that 00:10:17.090 --> 00:10:19.370 align:middle line:84% to all the real-world scenarios that we have 00:10:19.370 --> 00:10:21.390 align:middle line:90% been able to model as a graph. 00:10:21.390 --> 00:10:25.400 align:middle line:84% So it's a really nice balance between the two. 00:10:25.400 --> 00:10:28.700 align:middle line:84% So let's dig into this rich mathematical theory 00:10:28.700 --> 00:10:31.210 align:middle line:84% and see what tools and theorems we can develop. 00:10:31.210 --> 00:10:35.960 align:middle line:90% 00:10:35.960 --> 00:10:38.575 align:middle line:84% So the next definition is vertex degree. 00:10:38.575 --> 00:10:42.440 align:middle line:90% 00:10:42.440 --> 00:10:42.940 align:middle line:90% Sorry. 00:10:42.940 --> 00:10:46.450 align:middle line:84% The next definition is adjacency. 00:10:46.450 --> 00:10:46.950 align:middle line:90% Let's see. 00:10:46.950 --> 00:10:50.800 align:middle line:90% 00:10:50.800 --> 00:11:01.120 align:middle line:84% Two vertices, u and v, are adjacent 00:11:01.120 --> 00:11:16.060 align:middle line:84% if and only if they're connected by an edge, i.e. 00:11:16.060 --> 00:11:19.360 align:middle line:90% u, v is one of our edges. 00:11:19.360 --> 00:11:24.550 align:middle line:84% So their vertices are adjacent if they're connected by an edge. 00:11:24.550 --> 00:11:31.120 align:middle line:90% We'll say that-- let's see-- 00:11:31.120 --> 00:11:43.390 align:middle line:84% an edge u, v has endpoints u and v, just the two endpoints 00:11:43.390 --> 00:11:44.540 align:middle line:90% of the edge. 00:11:44.540 --> 00:11:57.220 align:middle line:84% And we'll say that edge u, v is incident to u 00:11:57.220 --> 00:12:00.490 align:middle line:84% and v. So a vertex and an edge are 00:12:00.490 --> 00:12:05.890 align:middle line:84% incident if that edge has that vertex as one of its endpoints. 00:12:05.890 --> 00:12:07.640 align:middle line:84% These words sound a little alike. 00:12:07.640 --> 00:12:09.460 align:middle line:84% Adjacent is used for two vertices 00:12:09.460 --> 00:12:10.820 align:middle line:90% that have an edge between them. 00:12:10.820 --> 00:12:13.720 align:middle line:84% Incident is used for an edge and a vertex, 00:12:13.720 --> 00:12:17.770 align:middle line:84% where the vertex is an endpoint of the edge. 00:12:17.770 --> 00:12:20.050 align:middle line:90% But useful terminology. 00:12:20.050 --> 00:12:27.940 align:middle line:84% And that lets us define the next thing, which is for a vertex v, 00:12:27.940 --> 00:12:43.690 align:middle line:84% the degree of v is the number of edges incident to v. 00:12:43.690 --> 00:12:47.120 align:middle line:84% So the degree of a vertex is the number of edges it has. 00:12:47.120 --> 00:12:49.690 align:middle line:90% So what's the degree of e? 00:12:49.690 --> 00:12:51.700 align:middle line:90% Just yell it out. 00:12:51.700 --> 00:12:52.520 align:middle line:90% 3. yeah. 00:12:52.520 --> 00:12:53.480 align:middle line:90% 1, 2, 3. 00:12:53.480 --> 00:12:55.690 align:middle line:90% What's the degree of c? 00:12:55.690 --> 00:12:56.530 align:middle line:90% 4. 00:12:56.530 --> 00:12:58.050 align:middle line:90% Degree of f? 00:12:58.050 --> 00:12:59.000 align:middle line:90% 2. 00:12:59.000 --> 00:13:01.870 align:middle line:84% What's the degree of this vertex 2? 00:13:01.870 --> 00:13:02.440 align:middle line:90% 0. 00:13:02.440 --> 00:13:03.530 align:middle line:90% It doesn't have any edges. 00:13:03.530 --> 00:13:04.270 align:middle line:90% Excellent. 00:13:04.270 --> 00:13:07.840 align:middle line:84% Degree is just the number of incident edges. 00:13:07.840 --> 00:13:10.480 align:middle line:90% Wonderful. 00:13:10.480 --> 00:13:13.190 align:middle line:84% You can understand some, not everything, 00:13:13.190 --> 00:13:16.000 align:middle line:84% but some things about a graph just from the degrees 00:13:16.000 --> 00:13:17.740 align:middle line:90% of its vertices. 00:13:17.740 --> 00:13:20.860 align:middle line:84% For example, that graph way over there 00:13:20.860 --> 00:13:23.650 align:middle line:84% on the left, the degrees of its vertices 00:13:23.650 --> 00:13:25.010 align:middle line:90% are-- let's go through them all. 00:13:25.010 --> 00:13:28.030 align:middle line:90% We've got 2, 2, 4. 00:13:28.030 --> 00:13:31.130 align:middle line:90% 3, 3, 2, 2, in that order. 00:13:31.130 --> 00:13:32.670 align:middle line:90% I think I got that right. 00:13:32.670 --> 00:13:36.870 align:middle line:84% Sometimes this is called the degree sequence. 00:13:36.870 --> 00:13:42.640 align:middle line:84% That's less important than the concept of degree itself. 00:13:42.640 --> 00:13:45.310 align:middle line:84% But we can play with this concept a little bit. 00:13:45.310 --> 00:13:47.640 align:middle line:84% For example, does there exist a graph 00:13:47.640 --> 00:13:55.345 align:middle line:90% with degree sequence 2, 2, 1? 00:13:55.345 --> 00:13:57.220 align:middle line:84% Sorry, that's supposed to be a question mark. 00:13:57.220 --> 00:13:57.850 align:middle line:90% There we go. 00:13:57.850 --> 00:14:00.640 align:middle line:84% Does there exist a graph with degree sequence 2, 2, 1? 00:14:00.640 --> 00:14:03.280 align:middle line:84% So three vertices, two have degree 2. 00:14:03.280 --> 00:14:04.250 align:middle line:90% One has degree 1. 00:14:04.250 --> 00:14:07.290 align:middle line:90% 00:14:07.290 --> 00:14:09.980 align:middle line:84% Think about that for a second, and then we'll try it together. 00:14:09.980 --> 00:14:12.750 align:middle line:90% 00:14:12.750 --> 00:14:15.220 align:middle line:84% OK, I see someone violently shaking their head. 00:14:15.220 --> 00:14:18.840 align:middle line:84% I think they're cracking their neck. 00:14:18.840 --> 00:14:20.250 align:middle line:90% Let's try it. 00:14:20.250 --> 00:14:22.950 align:middle line:84% So 2, 2, 1 means we need three vertices-- 00:14:22.950 --> 00:14:25.290 align:middle line:90% 1, 2, 3. 00:14:25.290 --> 00:14:27.610 align:middle line:90% The first vertex has degree 2. 00:14:27.610 --> 00:14:29.500 align:middle line:84% Well, there's really only one way to do that. 00:14:29.500 --> 00:14:31.270 align:middle line:90% It has to connect here and here. 00:14:31.270 --> 00:14:33.803 align:middle line:84% The next vertex also has to have degree 2. 00:14:33.803 --> 00:14:35.720 align:middle line:84% Well, that means it also has to connect there. 00:14:35.720 --> 00:14:37.580 align:middle line:84% The only way for it to get a second edge. 00:14:37.580 --> 00:14:40.520 align:middle line:84% Then the third vertex is supposed to have degree 1, 00:14:40.520 --> 00:14:41.570 align:middle line:90% which it doesn't. 00:14:41.570 --> 00:14:44.800 align:middle line:90% So we're sad it didn't work. 00:14:44.800 --> 00:14:46.220 align:middle line:84% You can think of this, by the way, 00:14:46.220 --> 00:14:48.280 align:middle line:90% as a proof by contradiction. 00:14:48.280 --> 00:14:51.140 align:middle line:84% If there were a graph that satisfied this, 00:14:51.140 --> 00:14:53.060 align:middle line:84% then it would have to look like this. 00:14:53.060 --> 00:14:56.140 align:middle line:84% But this graph doesn't work, so nothing works. 00:14:56.140 --> 00:14:57.670 align:middle line:90% Sound good? 00:14:57.670 --> 00:15:00.490 align:middle line:84% OK, but what if we asked a more complicated version 00:15:00.490 --> 00:15:01.430 align:middle line:90% of the question? 00:15:01.430 --> 00:15:06.170 align:middle line:84% Does there exist a graph with degree sequence 2, 2, 2, 2, 00:15:06.170 --> 00:15:09.100 align:middle line:90% 2, 1? 00:15:09.100 --> 00:15:11.297 align:middle line:84% This time there would need to be six vertices. 00:15:11.297 --> 00:15:12.880 align:middle line:84% And now there are lots of ways that we 00:15:12.880 --> 00:15:16.060 align:middle line:84% could choose two edges for each of those vertices. 00:15:16.060 --> 00:15:18.760 align:middle line:84% So there are lots of cases that we would need to try out 00:15:18.760 --> 00:15:21.640 align:middle line:90% in our proof by contradiction. 00:15:21.640 --> 00:15:24.470 align:middle line:84% So maybe a different strategy would work better. 00:15:24.470 --> 00:15:27.350 align:middle line:84% Does there exist a graph with this degree sequence? 00:15:27.350 --> 00:15:28.535 align:middle line:90% Yes. 00:15:28.535 --> 00:15:30.410 align:middle line:84% AUDIENCE: Does the sum always has to be even? 00:15:30.410 --> 00:15:32.743 align:middle line:84% ZACHARY ABEL: No, because the sum always has to be even. 00:15:32.743 --> 00:15:33.400 align:middle line:90% Interesting. 00:15:33.400 --> 00:15:36.060 align:middle line:84% Why does the sum have to be even? 00:15:36.060 --> 00:15:38.915 align:middle line:84% AUDIENCE: If it's an edge of [INAUDIBLE] another vertex 00:15:38.915 --> 00:15:41.160 align:middle line:90% [INAUDIBLE]. 00:15:41.160 --> 00:15:44.020 align:middle line:84% ZACHARY ABEL: All right, so if an edge connects to one vertex, 00:15:44.020 --> 00:15:46.660 align:middle line:84% then it also connects to another vertex. 00:15:46.660 --> 00:15:49.510 align:middle line:84% So each edge is somehow counted twice. 00:15:49.510 --> 00:15:51.750 align:middle line:84% Yeah, that's exactly the right intuition 00:15:51.750 --> 00:15:53.380 align:middle line:90% for the following theorem. 00:15:53.380 --> 00:15:56.970 align:middle line:84% In fact, that's the correct proof for the following theorem. 00:15:56.970 --> 00:16:02.840 align:middle line:90% We've got the handshake lemma. 00:16:02.840 --> 00:16:06.300 align:middle line:90% 00:16:06.300 --> 00:16:06.980 align:middle line:90% Let's see. 00:16:06.980 --> 00:16:11.190 align:middle line:90% 00:16:11.190 --> 00:16:15.340 align:middle line:84% For a graph G, which is vertices and edges, 00:16:15.340 --> 00:16:21.480 align:middle line:84% we have the sum of vertices in the set of vertices 00:16:21.480 --> 00:16:25.120 align:middle line:84% of the degree of that vertex equals. 00:16:25.120 --> 00:16:28.305 align:middle line:84% So if you take all the degrees and add them all up, 00:16:28.305 --> 00:16:30.680 align:middle line:84% your colleague here claimed that what you're going to get 00:16:30.680 --> 00:16:35.960 align:middle line:84% is two times the number of edges. 00:16:35.960 --> 00:16:39.140 align:middle line:84% And we can see that for this example, if we add this up 4, 00:16:39.140 --> 00:16:44.230 align:middle line:84% 8, 14, 16, 18, and there are 9 edges, that checks out. 00:16:44.230 --> 00:16:47.180 align:middle line:90% 00:16:47.180 --> 00:16:49.850 align:middle line:84% And the reason this is true is also exactly what 00:16:49.850 --> 00:16:51.230 align:middle line:90% your colleague said earlier. 00:16:51.230 --> 00:16:53.060 align:middle line:84% Over here, we're adding the degree 00:16:53.060 --> 00:16:57.650 align:middle line:84% of every vertex, which means we're basically 00:16:57.650 --> 00:17:01.340 align:middle line:84% counting every endpoint of all the edges. 00:17:01.340 --> 00:17:03.210 align:middle line:84% And every edge has two endpoints. 00:17:03.210 --> 00:17:07.819 align:middle line:84% So every edge was counted twice in this count on the left. 00:17:07.819 --> 00:17:09.619 align:middle line:90% So that's the handshake lemma. 00:17:09.619 --> 00:17:11.900 align:middle line:84% It's called handshake lemma because if you imagine 00:17:11.900 --> 00:17:15.030 align:middle line:84% a room full of people and some pairs of them shake hands, 00:17:15.030 --> 00:17:18.555 align:middle line:84% then if each person counts the number of handshakes they had, 00:17:18.555 --> 00:17:20.930 align:middle line:84% and you add that all up, that equals two times the number 00:17:20.930 --> 00:17:23.329 align:middle line:84% of handshakes, because each handshake had 00:17:23.329 --> 00:17:25.430 align:middle line:90% two hands involved. 00:17:25.430 --> 00:17:27.780 align:middle line:84% It's exactly the same explanation and proof. 00:17:27.780 --> 00:17:32.270 align:middle line:84% It's just with hands instead of pairs-- 00:17:32.270 --> 00:17:35.450 align:middle line:90% instead of half edges. 00:17:35.450 --> 00:17:38.455 align:middle line:84% All right, so does this help us with that? 00:17:38.455 --> 00:17:45.590 align:middle line:90% 00:17:45.590 --> 00:17:47.830 align:middle line:84% Can someone tell me why there doesn't exist a graph 00:17:47.830 --> 00:17:48.913 align:middle line:90% with this degree sequence? 00:17:48.913 --> 00:17:51.740 align:middle line:90% 00:17:51.740 --> 00:17:52.430 align:middle line:90% Yes? 00:17:52.430 --> 00:17:53.810 align:middle line:84% AUDIENCE: Because the sum is odd. 00:17:53.810 --> 00:17:55.940 align:middle line:84% ZACHARY ABEL: Yeah, because the sum is odd. 00:17:55.940 --> 00:17:58.940 align:middle line:84% This would have to have, what, 11 divided 00:17:58.940 --> 00:18:04.970 align:middle line:90% by 2 edges, which isn't great. 00:18:04.970 --> 00:18:08.160 align:middle line:84% I don't know how to draw a half edge in these graphs, 00:18:08.160 --> 00:18:10.380 align:middle line:90% so there is no such graph. 00:18:10.380 --> 00:18:11.510 align:middle line:90% Awesome. 00:18:11.510 --> 00:18:15.350 align:middle line:84% Here's one more application of the handshake lemma. 00:18:15.350 --> 00:18:17.540 align:middle line:90% Let's see. 00:18:17.540 --> 00:18:28.463 align:middle line:84% What is the maximum number of edges in an n vertex graph? 00:18:28.463 --> 00:18:33.260 align:middle line:90% 00:18:33.260 --> 00:18:38.360 align:middle line:84% What is the maximum number of edges in an n vertex graph? 00:18:38.360 --> 00:18:47.670 align:middle line:84% Well, each vertex has degree at most n minus 1, 00:18:47.670 --> 00:18:51.000 align:middle line:84% because at most, it connects to all the other vertices. 00:18:51.000 --> 00:18:52.770 align:middle line:84% There are no other vertices to connect to. 00:18:52.770 --> 00:18:54.150 align:middle line:90% You can't connect to yourself. 00:18:54.150 --> 00:18:56.540 align:middle line:84% You can't repeat any connections. 00:18:56.540 --> 00:18:58.650 align:middle line:84% Once again, that's what we meant by simple graph, 00:18:58.650 --> 00:19:01.070 align:middle line:90% and all our graphs are simple. 00:19:01.070 --> 00:19:05.660 align:middle line:84% So every vertex has degree at most n minus 1. 00:19:05.660 --> 00:19:17.150 align:middle line:84% So the sum of the degrees is at most n times n minus 1. 00:19:17.150 --> 00:19:20.180 align:middle line:84% But the sum of the degrees is equal to 2 times the number 00:19:20.180 --> 00:19:22.540 align:middle line:90% of edges. 00:19:22.540 --> 00:19:25.540 align:middle line:84% So the number of edges is at most n times n 00:19:25.540 --> 00:19:28.660 align:middle line:90% minus 1 divided by 2. 00:19:28.660 --> 00:19:30.860 align:middle line:84% Didn't have to do any fancy induction. 00:19:30.860 --> 00:19:33.530 align:middle line:90% Didn't have to compute some sum. 00:19:33.530 --> 00:19:36.105 align:middle line:84% I wonder if this equals a sum that we've seen before. 00:19:36.105 --> 00:19:39.520 align:middle line:90% 00:19:39.520 --> 00:19:41.030 align:middle line:90% So we have an upper bound. 00:19:41.030 --> 00:19:43.010 align:middle line:84% The number of edges is at most that. 00:19:43.010 --> 00:19:45.036 align:middle line:84% And it turns out this is achievable. 00:19:45.036 --> 00:19:50.710 align:middle line:90% 00:19:50.710 --> 00:20:00.160 align:middle line:84% Achieved by the complete graph Kn, 00:20:00.160 --> 00:20:03.495 align:middle line:84% which is the graph with n vertices and all possible edges. 00:20:03.495 --> 00:20:06.760 align:middle line:90% 00:20:06.760 --> 00:20:09.710 align:middle line:84% So this vertex connects to everyone. 00:20:09.710 --> 00:20:12.140 align:middle line:84% This vertex connects to everyone. 00:20:12.140 --> 00:20:15.500 align:middle line:84% This connects to everyone, connects to everyone. 00:20:15.500 --> 00:20:16.700 align:middle line:90% I think I got them all. 00:20:16.700 --> 00:20:20.500 align:middle line:90% So this is a drawing of K6. 00:20:20.500 --> 00:20:23.390 align:middle line:84% There are six vertices, and everyone connects to everyone. 00:20:23.390 --> 00:20:27.740 align:middle line:84% This graph indeed has every vertex with degree n minus 1, 00:20:27.740 --> 00:20:32.110 align:middle line:84% so it has exactly this many edges. 00:20:32.110 --> 00:20:34.220 align:middle line:90% The K for Kn, this is standard. 00:20:34.220 --> 00:20:36.260 align:middle line:90% You'll see Kn everywhere. 00:20:36.260 --> 00:20:38.470 align:middle line:90% I think it comes from German. 00:20:38.470 --> 00:20:42.460 align:middle line:84% It kind of sounds like complete graph, right? 00:20:42.460 --> 00:20:46.000 align:middle line:84% But we'll be using this notation a lot. 00:20:46.000 --> 00:20:46.560 align:middle line:90% All right. 00:20:46.560 --> 00:20:50.080 align:middle line:90% 00:20:50.080 --> 00:20:53.440 align:middle line:90% Let me ask another question. 00:20:53.440 --> 00:20:56.620 align:middle line:84% So most of us here are MIT students. 00:20:56.620 --> 00:20:58.550 align:middle line:84% I know we have some cross-registered students. 00:20:58.550 --> 00:20:59.360 align:middle line:90% Welcome. 00:20:59.360 --> 00:21:00.940 align:middle line:90% I'm glad you're here. 00:21:00.940 --> 00:21:03.080 align:middle line:84% For the MIT students in the room, 00:21:03.080 --> 00:21:05.950 align:middle line:84% do you have any friends at Harvard? 00:21:05.950 --> 00:21:06.880 align:middle line:90% Yeah? 00:21:06.880 --> 00:21:08.890 align:middle line:90% Awesome. 00:21:08.890 --> 00:21:11.300 align:middle line:90% OK, so I got at least one hand. 00:21:11.300 --> 00:21:13.160 align:middle line:90% I got more than one hand. 00:21:13.160 --> 00:21:16.660 align:middle line:84% But let me ask, how many Harvard friends do you think 00:21:16.660 --> 00:21:18.860 align:middle line:90% the typical MIT student has? 00:21:18.860 --> 00:21:19.700 align:middle line:90% Don't answer that. 00:21:19.700 --> 00:21:20.977 align:middle line:90% It's rhetorical. 00:21:20.977 --> 00:21:22.310 align:middle line:90% What about the reverse question? 00:21:22.310 --> 00:21:26.590 align:middle line:84% How many MIT friends does the typical Harvard student have? 00:21:26.590 --> 00:21:30.130 align:middle line:90% Which one's bigger? 00:21:30.130 --> 00:21:30.710 align:middle line:90% Let's see. 00:21:30.710 --> 00:21:33.850 align:middle line:84% So on one hand, we've got the average number 00:21:33.850 --> 00:21:41.740 align:middle line:84% of Harvard friends for the typical MIT student, 00:21:41.740 --> 00:21:47.740 align:middle line:84% or the average number of MIT friends 00:21:47.740 --> 00:21:49.215 align:middle line:90% for the typical Harvard student. 00:21:49.215 --> 00:21:51.850 align:middle line:90% 00:21:51.850 --> 00:21:53.270 align:middle line:90% Which one's bigger? 00:21:53.270 --> 00:21:54.680 align:middle line:90% And how could we know? 00:21:54.680 --> 00:21:57.580 align:middle line:90% How can we decide this? 00:21:57.580 --> 00:22:00.110 align:middle line:90% Maybe I could do a big survey. 00:22:00.110 --> 00:22:02.150 align:middle line:84% I could sample a bunch of MIT students, 00:22:02.150 --> 00:22:04.450 align:middle line:84% sample a bunch of Harvard students. 00:22:04.450 --> 00:22:06.740 align:middle line:84% Take the average of whatever is reported, 00:22:06.740 --> 00:22:09.460 align:middle line:90% and take that as ground truth. 00:22:09.460 --> 00:22:13.485 align:middle line:84% That is how a lot of empirical statistics ends up working. 00:22:13.485 --> 00:22:17.010 align:middle line:90% 00:22:17.010 --> 00:22:19.210 align:middle line:90% Can anyone suggest a better way? 00:22:19.210 --> 00:22:20.010 align:middle line:90% Yes? 00:22:20.010 --> 00:22:22.260 align:middle line:84% AUDIENCE: So we know that every friendship is two-way, 00:22:22.260 --> 00:22:25.270 align:middle line:84% and every [INAUDIBLE] has a bigger undergrad population 00:22:25.270 --> 00:22:26.400 align:middle line:90% at MIT. 00:22:26.400 --> 00:22:31.650 align:middle line:84% So you divide it all by a big number, so [INAUDIBLE]. 00:22:31.650 --> 00:22:32.650 align:middle line:90% ZACHARY ABEL: All right. 00:22:32.650 --> 00:22:34.690 align:middle line:90% So let me rephrase that. 00:22:34.690 --> 00:22:36.970 align:middle line:84% So we know that every friendship goes both ways. 00:22:36.970 --> 00:22:39.190 align:middle line:84% There's one MIT student and one Harvard student. 00:22:39.190 --> 00:22:41.430 align:middle line:84% Again, we're assuming friendship is 00:22:41.430 --> 00:22:44.070 align:middle line:84% two-way, no unrequited friendships. 00:22:44.070 --> 00:22:47.190 align:middle line:84% Can't the two schools just get along? 00:22:47.190 --> 00:22:49.290 align:middle line:84% But because they're the same number of friendships 00:22:49.290 --> 00:22:52.360 align:middle line:84% on both sides, when we're taking the average, 00:22:52.360 --> 00:22:54.360 align:middle line:84% we're dividing by the number of Harvard students 00:22:54.360 --> 00:22:56.860 align:middle line:84% and dividing by the number of MIT students. 00:22:56.860 --> 00:22:59.080 align:middle line:84% But Harvard has more undergrads than MIT, 00:22:59.080 --> 00:23:01.090 align:middle line:84% so we've divided by a bigger number. 00:23:01.090 --> 00:23:03.750 align:middle line:84% So the MIT average should be bigger. 00:23:03.750 --> 00:23:05.880 align:middle line:90% Whew, that was a lot. 00:23:05.880 --> 00:23:07.300 align:middle line:90% Thank you for saying all that. 00:23:07.300 --> 00:23:09.750 align:middle line:84% Let's unpack it a little more slowly. 00:23:09.750 --> 00:23:14.880 align:middle line:84% But you are absolutely correct in every step of your reasoning. 00:23:14.880 --> 00:23:22.620 align:middle line:84% So what I'm going to do is let's draw a graph 00:23:22.620 --> 00:23:28.530 align:middle line:84% where the vertices are going to equal the set of MIT undergrads 00:23:28.530 --> 00:23:32.430 align:middle line:84% and the set of Harvard undergrads. 00:23:32.430 --> 00:23:35.200 align:middle line:84% And I want to emphasize that these are two disjoint sets. 00:23:35.200 --> 00:23:37.770 align:middle line:84% No student is enrolled in both schools. 00:23:37.770 --> 00:23:39.430 align:middle line:84% We're not counting cross-registration. 00:23:39.430 --> 00:23:41.340 align:middle line:84% That's why I mentioned that earlier. 00:23:41.340 --> 00:23:45.700 align:middle line:84% And sometimes to emphasize that these two sets are disjoint, 00:23:45.700 --> 00:23:48.970 align:middle line:84% sometimes you put a little dot inside the union sign. 00:23:48.970 --> 00:23:51.600 align:middle line:84% So our set of vertices is the disjoint union 00:23:51.600 --> 00:23:54.420 align:middle line:84% of MIT students and Harvard students. 00:23:54.420 --> 00:23:59.040 align:middle line:84% And our edge set is going to be the set of-- 00:23:59.040 --> 00:24:03.700 align:middle line:84% let's see-- m, h, where m is an MIT student, 00:24:03.700 --> 00:24:07.085 align:middle line:84% h is a Harvard student, and m and h are friends. 00:24:07.085 --> 00:24:09.660 align:middle line:90% 00:24:09.660 --> 00:24:12.190 align:middle line:84% And we can draw an example of what this might look like. 00:24:12.190 --> 00:24:15.490 align:middle line:84% So here's all of my Harvard vertices on the left. 00:24:15.490 --> 00:24:18.180 align:middle line:84% Here's all of my MIT students on the right. 00:24:18.180 --> 00:24:21.720 align:middle line:84% Here, we have, I don't know, Bill and Helen 00:24:21.720 --> 00:24:28.080 align:middle line:90% and Natalie and Neil and TS. 00:24:28.080 --> 00:24:36.060 align:middle line:84% On the right, maybe we have Buzz and Kenji and Lisa, 00:24:36.060 --> 00:24:38.850 align:middle line:90% and these lists go on. 00:24:38.850 --> 00:24:40.120 align:middle line:90% This is just an example. 00:24:40.120 --> 00:24:41.537 align:middle line:84% I don't know if there are actually 00:24:41.537 --> 00:24:43.080 align:middle line:90% students with these names. 00:24:43.080 --> 00:24:44.740 align:middle line:84% And some pairs of them are friends. 00:24:44.740 --> 00:24:47.980 align:middle line:84% Maybe these are friends and these and these are friends. 00:24:47.980 --> 00:24:51.400 align:middle line:84% And Buzz is just friends with most people. 00:24:51.400 --> 00:24:54.338 align:middle line:84% And TS only likes Lisa among this graph. 00:24:54.338 --> 00:24:55.630 align:middle line:90% Some pairs of them are friends. 00:24:55.630 --> 00:24:56.880 align:middle line:90% Here's our graph. 00:24:56.880 --> 00:24:57.460 align:middle line:90% Again. 00:24:57.460 --> 00:24:58.960 align:middle line:84% there are more vertices, more edges, 00:24:58.960 --> 00:25:01.390 align:middle line:84% but here's an example of our graph. 00:25:01.390 --> 00:25:04.900 align:middle line:84% And this graph has a special property called bipartite. 00:25:04.900 --> 00:25:08.000 align:middle line:90% This is a bipartite graph. 00:25:08.000 --> 00:25:08.500 align:middle line:90% Let's see. 00:25:08.500 --> 00:25:10.970 align:middle line:90% Definition. 00:25:10.970 --> 00:25:26.840 align:middle line:84% G is bipartite when it's possible to partition V 00:25:26.840 --> 00:25:31.117 align:middle line:84% as the disjoint union of two sets. 00:25:31.117 --> 00:25:32.700 align:middle line:84% I'm going to call them left and right. 00:25:32.700 --> 00:25:36.650 align:middle line:84% That's common notation for bipartite graphs, 00:25:36.650 --> 00:25:46.610 align:middle line:84% such that every edge has one endpoint in L 00:25:46.610 --> 00:25:50.450 align:middle line:90% and one endpoint in R. 00:25:50.450 --> 00:25:52.215 align:middle line:84% So you have some vertices on the left, 00:25:52.215 --> 00:25:53.340 align:middle line:90% some vertices on the right. 00:25:53.340 --> 00:25:55.463 align:middle line:84% And every edge crosses between them. 00:25:55.463 --> 00:25:57.630 align:middle line:84% One endpoint on the left, one endpoint on the right. 00:25:57.630 --> 00:25:59.220 align:middle line:84% These graphs are called bipartite. 00:25:59.220 --> 00:26:03.260 align:middle line:84% And the one we just drew certainly has that property. 00:26:03.260 --> 00:26:03.770 align:middle line:90% Sound good? 00:26:03.770 --> 00:26:05.510 align:middle line:90% With me so far? 00:26:05.510 --> 00:26:06.710 align:middle line:90% Awesome. 00:26:06.710 --> 00:26:13.640 align:middle line:84% And now, we have a bipartite version of the handshake lemma. 00:26:13.640 --> 00:26:14.230 align:middle line:90% Let's see. 00:26:14.230 --> 00:26:17.630 align:middle line:90% 00:26:17.630 --> 00:26:25.760 align:middle line:84% Handshake lemma bipartite, which says 00:26:25.760 --> 00:26:27.890 align:middle line:84% that if you add up all the degrees just 00:26:27.890 --> 00:26:37.590 align:middle line:84% on the left, the vertices in the left side degree of V, well, 00:26:37.590 --> 00:26:39.560 align:middle line:84% every edge has one endpoint on the left 00:26:39.560 --> 00:26:40.890 align:middle line:90% and one endpoint on the right. 00:26:40.890 --> 00:26:44.390 align:middle line:84% So this is just equal to the number of edges. 00:26:44.390 --> 00:26:47.035 align:middle line:84% And likewise, you can do the same thing on the right. 00:26:47.035 --> 00:26:51.470 align:middle line:90% 00:26:51.470 --> 00:26:54.930 align:middle line:84% So all the degrees on the left add to the number of edges. 00:26:54.930 --> 00:26:58.670 align:middle line:84% All the degrees on the right add to the number of edges. 00:26:58.670 --> 00:27:01.800 align:middle line:84% But remember, we are interested in the averages. 00:27:01.800 --> 00:27:04.880 align:middle line:84% The average degree on the left versus the average degree 00:27:04.880 --> 00:27:07.640 align:middle line:90% on the right. 00:27:07.640 --> 00:27:09.380 align:middle line:90% All right? 00:27:09.380 --> 00:27:16.850 align:middle line:84% So we've got, let's say, average degree on the MIT side 00:27:16.850 --> 00:27:23.300 align:middle line:84% divided by the average degree on the Harvard side. 00:27:23.300 --> 00:27:28.770 align:middle line:84% Well, this is equal to the total degree on the MIT side. 00:27:28.770 --> 00:27:33.710 align:middle line:84% So just adding up all the degrees of the MIT students 00:27:33.710 --> 00:27:38.930 align:middle line:84% divided by the number of MIT students. 00:27:38.930 --> 00:27:44.780 align:middle line:84% And this denominator is the total degree 00:27:44.780 --> 00:27:49.730 align:middle line:84% on the Harvard side divided by the number of Harvard students. 00:27:49.730 --> 00:27:51.653 align:middle line:90% That's what we mean by average. 00:27:51.653 --> 00:27:53.070 align:middle line:84% Add them up, divide by the number. 00:27:53.070 --> 00:27:55.280 align:middle line:84% Add them up, divide by the number. 00:27:55.280 --> 00:27:58.160 align:middle line:84% But we just said that the total degree on the M side 00:27:58.160 --> 00:28:01.340 align:middle line:84% and the total degree on the H side, both of those 00:28:01.340 --> 00:28:03.970 align:middle line:84% are equal to just the number of edges. 00:28:03.970 --> 00:28:06.700 align:middle line:84% So this is the number of edges divided 00:28:06.700 --> 00:28:11.860 align:middle line:84% by the size of M divided by the number of edges divided 00:28:11.860 --> 00:28:17.330 align:middle line:84% by the size of H. The E's cancel. 00:28:17.330 --> 00:28:23.200 align:middle line:84% This just becomes the size of H divided by the size of M. 00:28:23.200 --> 00:28:25.400 align:middle line:84% All right, we've now caught up to your comment. 00:28:25.400 --> 00:28:27.520 align:middle line:90% Thank you so much. 00:28:27.520 --> 00:28:30.400 align:middle line:84% And now, we can just ask how many students 00:28:30.400 --> 00:28:31.960 align:middle line:90% there are in each school? 00:28:31.960 --> 00:28:34.810 align:middle line:84% This is about-- let's see, I wrote something down-- 00:28:34.810 --> 00:28:40.420 align:middle line:90% about 7,200 divided by 4,600. 00:28:40.420 --> 00:28:41.690 align:middle line:90% Those are close anyway. 00:28:41.690 --> 00:28:43.510 align:middle line:90% They might be a bit out of date. 00:28:43.510 --> 00:28:48.520 align:middle line:90% But this is about 1.6. 00:28:48.520 --> 00:28:54.790 align:middle line:84% So the average MIT student has about 1.6 times more Harvard 00:28:54.790 --> 00:28:57.820 align:middle line:84% friends than the number of MIT friends 00:28:57.820 --> 00:28:59.720 align:middle line:90% the average Harvard student has. 00:28:59.720 --> 00:29:06.340 align:middle line:84% And that's not any judgment about behavior on the two 00:29:06.340 --> 00:29:07.580 align:middle line:90% sides of the schools. 00:29:07.580 --> 00:29:11.380 align:middle line:84% It's just a fact about averages after knowing 00:29:11.380 --> 00:29:13.930 align:middle line:90% a little bit of graph theory. 00:29:13.930 --> 00:29:15.900 align:middle line:84% But our number is bigger, so we're better. 00:29:15.900 --> 00:29:20.470 align:middle line:90% 00:29:20.470 --> 00:29:26.180 align:middle line:84% But, so already, we've just defined a couple of things. 00:29:26.180 --> 00:29:29.950 align:middle line:84% And we've already been able to prove some non-obvious theorems 00:29:29.950 --> 00:29:31.910 align:middle line:90% and facts again. 00:29:31.910 --> 00:29:33.380 align:middle line:84% They still look like toy examples. 00:29:33.380 --> 00:29:36.220 align:middle line:84% But there are real scientific studies 00:29:36.220 --> 00:29:39.010 align:middle line:84% that have studied questions like that 00:29:39.010 --> 00:29:42.550 align:middle line:84% and not known the graph theory, and then came up 00:29:42.550 --> 00:29:46.030 align:middle line:84% with conclusions that were very obviously wrong to anyone 00:29:46.030 --> 00:29:47.980 align:middle line:90% who did know the graph theory. 00:29:47.980 --> 00:29:52.690 align:middle line:84% For example, there's a very famous study out of UChicago 00:29:52.690 --> 00:29:56.980 align:middle line:84% from 1994 called "The Social Organization of Sexuality, 00:29:56.980 --> 00:29:59.200 align:middle line:90% Sexual Practices in the US." 00:29:59.200 --> 00:30:04.150 align:middle line:84% So it was a big sweeping study about sexual practices 00:30:04.150 --> 00:30:07.390 align:middle line:84% and behavior and health across the country. 00:30:07.390 --> 00:30:09.860 align:middle line:84% They had thousands of participants. 00:30:09.860 --> 00:30:13.150 align:middle line:84% They asked lots and lots of questions and came up with lots 00:30:13.150 --> 00:30:16.160 align:middle line:84% of statistics and measurements and trends, 00:30:16.160 --> 00:30:19.880 align:middle line:84% and published all their results in a big 700-page write-up, 00:30:19.880 --> 00:30:22.150 align:middle line:90% a very famous study. 00:30:22.150 --> 00:30:24.050 align:middle line:84% And out of all the questions they asked, 00:30:24.050 --> 00:30:28.510 align:middle line:84% I'm going to single out one of them for the single purpose 00:30:28.510 --> 00:30:33.190 align:middle line:84% that their results are obviously wrong because of exactly 00:30:33.190 --> 00:30:34.220 align:middle line:90% this argument. 00:30:34.220 --> 00:30:36.970 align:middle line:84% So one question they asked specifically 00:30:36.970 --> 00:30:39.820 align:middle line:84% to their male and female participants. 00:30:39.820 --> 00:30:43.820 align:middle line:84% They asked the men how many female partners they've had. 00:30:43.820 --> 00:30:47.050 align:middle line:84% They asked the women how many male partners they've had. 00:30:47.050 --> 00:30:47.750 align:middle line:90% Don't worry. 00:30:47.750 --> 00:30:49.990 align:middle line:84% They had lots of other questions about other kinds 00:30:49.990 --> 00:30:51.490 align:middle line:90% of relationships. 00:30:51.490 --> 00:30:53.950 align:middle line:84% They had lots of other questions for their non-binary 00:30:53.950 --> 00:30:55.250 align:middle line:90% participants as well. 00:30:55.250 --> 00:30:59.910 align:middle line:84% Like I said, I'm just singling out this one of many questions. 00:30:59.910 --> 00:31:03.760 align:middle line:84% So what's the average number of female partners a man has had? 00:31:03.760 --> 00:31:08.010 align:middle line:84% What's the average number of male partners a woman has had? 00:31:08.010 --> 00:31:12.120 align:middle line:84% And according to their reports and surveys, 00:31:12.120 --> 00:31:16.350 align:middle line:84% the men reported about 1.74 times as many partners 00:31:16.350 --> 00:31:17.330 align:middle line:90% as the women. 00:31:17.330 --> 00:31:29.220 align:middle line:90% 00:31:29.220 --> 00:31:30.600 align:middle line:90% Let's see. 00:31:30.600 --> 00:31:37.560 align:middle line:84% So the average degree of the men was 00:31:37.560 --> 00:31:46.140 align:middle line:84% equal to about 1.74 times the average degree of the women. 00:31:46.140 --> 00:31:51.990 align:middle line:84% A different study, this time from ABC News in 2004, 00:31:51.990 --> 00:31:55.690 align:middle line:84% got a similar result, but instead of 1.74-- 00:31:55.690 --> 00:31:56.440 align:middle line:90% so sorry. 00:31:56.440 --> 00:31:58.066 align:middle line:90% This was from UChicago. 00:31:58.066 --> 00:32:02.400 align:middle line:90% 00:32:02.400 --> 00:32:11.070 align:middle line:84% ABC News approximated this number to be about 3.3. 00:32:11.070 --> 00:32:19.470 align:middle line:84% Yet another study from the National Center of Health, 00:32:19.470 --> 00:32:26.910 align:middle line:84% this time from 2007, got this number to be about 1.75. 00:32:26.910 --> 00:32:29.760 align:middle line:84% When, of course, we know the exact answer. 00:32:29.760 --> 00:32:31.770 align:middle line:84% This ratio should be the number of women divided 00:32:31.770 --> 00:32:34.920 align:middle line:84% by the number of men because of the specific question 00:32:34.920 --> 00:32:38.040 align:middle line:84% they were asking about this subset of relationships 00:32:38.040 --> 00:32:40.510 align:middle line:84% among this subset of the population, 00:32:40.510 --> 00:32:43.260 align:middle line:84% they put themselves in a bipartite scenario 00:32:43.260 --> 00:32:44.732 align:middle line:90% where this theorem applies. 00:32:44.732 --> 00:32:46.440 align:middle line:84% And the correct answer is just the number 00:32:46.440 --> 00:32:48.095 align:middle line:84% of women divided by the number of men. 00:32:48.095 --> 00:32:53.020 align:middle line:90% 00:32:53.020 --> 00:32:53.520 align:middle line:90% Let's see. 00:32:53.520 --> 00:32:57.930 align:middle line:84% The number of women divided by the number of men, which-- 00:32:57.930 --> 00:33:00.490 align:middle line:90% according to the 2010 census-- 00:33:00.490 --> 00:33:10.560 align:middle line:84% was about 157 million over 152 million, which is about 1.03. 00:33:10.560 --> 00:33:14.070 align:middle line:90% The correct answer is 1.03. 00:33:14.070 --> 00:33:17.350 align:middle line:84% And it's not a question of behavior. 00:33:17.350 --> 00:33:20.100 align:middle line:84% It's a question of just averages. 00:33:20.100 --> 00:33:22.870 align:middle line:84% There is nothing to learn from this ratio, 00:33:22.870 --> 00:33:25.650 align:middle line:84% except this is how many men and women are in the country. 00:33:25.650 --> 00:33:28.930 align:middle line:84% Not a very interesting example for a behavioral study, 00:33:28.930 --> 00:33:31.320 align:middle line:90% if you ask me. 00:33:31.320 --> 00:33:35.130 align:middle line:84% I'm not going to speculate on why these numbers don't 00:33:35.130 --> 00:33:38.250 align:middle line:90% match reality. 00:33:38.250 --> 00:33:38.860 align:middle line:90% I don't know. 00:33:38.860 --> 00:33:42.075 align:middle line:84% Think about it yourselves if you want. 00:33:42.075 --> 00:33:43.700 align:middle line:84% Are there questions about this example? 00:33:43.700 --> 00:33:47.730 align:middle line:90% 00:33:47.730 --> 00:33:49.655 align:middle line:84% Or anything else we've talked about so far? 00:33:49.655 --> 00:33:53.900 align:middle line:90% 00:33:53.900 --> 00:33:58.050 align:middle line:84% All right, let's move on to another topic in graph theory, 00:33:58.050 --> 00:33:59.606 align:middle line:90% another common application. 00:33:59.606 --> 00:34:04.220 align:middle line:90% 00:34:04.220 --> 00:34:07.820 align:middle line:84% So instead of measuring affinities or friendships 00:34:07.820 --> 00:34:12.099 align:middle line:84% among pairs of objects, let's measure conflicts between them. 00:34:12.099 --> 00:34:17.600 align:middle line:90% 00:34:17.600 --> 00:34:19.304 align:middle line:90% So here's an example graph. 00:34:19.304 --> 00:34:27.739 align:middle line:90% 00:34:27.739 --> 00:34:30.980 align:middle line:84% Let's define the vertex set of our graph 00:34:30.980 --> 00:34:41.300 align:middle line:84% to be the set of all MIT classes with a final exam. 00:34:41.300 --> 00:34:57.920 align:middle line:84% And E is going to be pairs of classes, C1, C2, that 00:34:57.920 --> 00:35:03.100 align:middle line:90% have someone enrolled in both. 00:35:03.100 --> 00:35:06.080 align:middle line:90% 00:35:06.080 --> 00:35:07.770 align:middle line:90% So you're the registrar. 00:35:07.770 --> 00:35:11.000 align:middle line:84% Your goal is to assign a time for the final exam 00:35:11.000 --> 00:35:12.890 align:middle line:90% for every class. 00:35:12.890 --> 00:35:16.610 align:middle line:84% And if some student is taking two different classes 00:35:16.610 --> 00:35:19.040 align:middle line:84% and those classes are scheduled for the same time 00:35:19.040 --> 00:35:23.630 align:middle line:84% slot or overlapping time slots, then that student 00:35:23.630 --> 00:35:26.130 align:middle line:84% can't be in two different places at the same time, 00:35:26.130 --> 00:35:27.920 align:middle line:90% so you have a conflict. 00:35:27.920 --> 00:35:32.030 align:middle line:84% So this graph here is measuring all of the conflicts 00:35:32.030 --> 00:35:34.710 align:middle line:90% among final exam times. 00:35:34.710 --> 00:35:36.980 align:middle line:84% There's an edge precisely when the two 00:35:36.980 --> 00:35:41.330 align:middle line:84% classes can't be scheduled for overlapping times. 00:35:41.330 --> 00:35:43.340 align:middle line:90% Does that make sense so far? 00:35:43.340 --> 00:35:43.880 align:middle line:90% Cool. 00:35:43.880 --> 00:35:53.580 align:middle line:84% So edges are conflicts in exam scheduling. 00:35:53.580 --> 00:36:00.560 align:middle line:90% 00:36:00.560 --> 00:36:03.053 align:middle line:90% And let me draw an example. 00:36:03.053 --> 00:36:09.140 align:middle line:90% 00:36:09.140 --> 00:36:10.790 align:middle line:90% Let's see. 00:36:10.790 --> 00:36:15.620 align:middle line:84% So in this example, MIT only has five courses-- 00:36:15.620 --> 00:36:30.360 align:middle line:84% 6,3700, 6.2000, 6.1200, 6.3000, and 6.4100. 00:36:30.360 --> 00:36:32.750 align:middle line:84% So this is some fake approximation 00:36:32.750 --> 00:36:35.010 align:middle line:84% of what this graph might look like. 00:36:35.010 --> 00:36:37.880 align:middle line:84% So this says that there's someone taking our class 00:36:37.880 --> 00:36:41.940 align:middle line:90% and 6.2000. 00:36:41.940 --> 00:36:44.550 align:middle line:84% There's at least one student taking both of these classes. 00:36:44.550 --> 00:36:49.780 align:middle line:84% There's no one taking 6.3700 and 6.3000. 00:36:49.780 --> 00:36:51.970 align:middle line:84% So for example, these two are allowed 00:36:51.970 --> 00:36:55.120 align:middle line:84% to be scheduled at the same time. 00:36:55.120 --> 00:36:58.810 align:middle line:84% So this is listing all of our conflicts. 00:36:58.810 --> 00:37:01.743 align:middle line:84% And this brings us to the graph coloring problem. 00:37:01.743 --> 00:37:30.100 align:middle line:90% 00:37:30.100 --> 00:37:45.940 align:middle line:84% So we say that a proper K coloring of a graph G 00:37:45.940 --> 00:37:49.390 align:middle line:90% is defined to be-- 00:37:49.390 --> 00:37:51.040 align:middle line:84% I'll write it down mathematically first 00:37:51.040 --> 00:37:52.870 align:middle line:90% and then explain it in words-- 00:37:52.870 --> 00:37:55.900 align:middle line:84% a function f from the set of vertices 00:37:55.900 --> 00:38:04.273 align:middle line:84% to a set C of colors, where there are at most K colors. 00:38:04.273 --> 00:38:08.020 align:middle line:90% 00:38:08.020 --> 00:38:18.430 align:middle line:84% And for all edges, u, v, fu does not equal fv. 00:38:18.430 --> 00:38:21.740 align:middle line:84% So in other words, we have some set of colors available. 00:38:21.740 --> 00:38:26.620 align:middle line:84% And we want to assign a color to every vertex. 00:38:26.620 --> 00:38:29.450 align:middle line:84% And we want to assign a color to every vertex 00:38:29.450 --> 00:38:31.720 align:middle line:84% so that no two adjacent vertices are ever 00:38:31.720 --> 00:38:34.300 align:middle line:90% assigned to the same color. 00:38:34.300 --> 00:38:36.370 align:middle line:84% In this example, you can think of colors 00:38:36.370 --> 00:38:40.180 align:middle line:90% as time slots for final exams. 00:38:40.180 --> 00:38:42.270 align:middle line:84% If two vertices are given the same color, 00:38:42.270 --> 00:38:45.150 align:middle line:84% that means that they'll be administered at the same time. 00:38:45.150 --> 00:38:46.880 align:middle line:84% And this condition that edges have 00:38:46.880 --> 00:38:49.160 align:middle line:84% to have two different colors on their endpoints 00:38:49.160 --> 00:38:52.050 align:middle line:84% means that any time there's a scheduling conflict, 00:38:52.050 --> 00:38:54.570 align:middle line:84% they will always be given different colors, 00:38:54.570 --> 00:38:58.070 align:middle line:84% which means they'll be given different time slots. 00:38:58.070 --> 00:39:00.900 align:middle line:84% So this is what we mean by a proper K coloring. 00:39:00.900 --> 00:39:04.970 align:middle line:84% It's an assignment of a color to every vertex 00:39:04.970 --> 00:39:07.680 align:middle line:84% from a possible choice of K different colors, 00:39:07.680 --> 00:39:09.800 align:middle line:84% where the two endpoints of every edge 00:39:09.800 --> 00:39:11.305 align:middle line:90% are assigned different colors. 00:39:11.305 --> 00:39:14.930 align:middle line:90% 00:39:14.930 --> 00:39:19.280 align:middle line:84% And as an example, so let's try to color this graph. 00:39:19.280 --> 00:39:23.450 align:middle line:90% 00:39:23.450 --> 00:39:25.820 align:middle line:84% Let's say I'm going to color this red 00:39:25.820 --> 00:39:32.120 align:middle line:84% and this green and this blue and this yellow and this purple. 00:39:32.120 --> 00:39:33.775 align:middle line:90% Is that a proper coloring? 00:39:33.775 --> 00:39:37.970 align:middle line:90% 00:39:37.970 --> 00:39:38.560 align:middle line:90% Yeah. 00:39:38.560 --> 00:39:39.060 align:middle line:90% Yeah. 00:39:39.060 --> 00:39:41.227 align:middle line:84% Every edge has two different colors on its endpoints 00:39:41.227 --> 00:39:42.970 align:middle line:84% because all the colors are distinct. 00:39:42.970 --> 00:39:45.960 align:middle line:90% This is a proper five coloring. 00:39:45.960 --> 00:39:48.005 align:middle line:90% Can be five-colored. 00:39:48.005 --> 00:39:51.870 align:middle line:90% 00:39:51.870 --> 00:39:53.160 align:middle line:90% All right? 00:39:53.160 --> 00:39:55.470 align:middle line:84% But that means we would need five different time 00:39:55.470 --> 00:39:57.280 align:middle line:90% slots for the final exams. 00:39:57.280 --> 00:39:59.940 align:middle line:84% And we'd like to be a little more efficient. 00:39:59.940 --> 00:40:01.783 align:middle line:84% Can we color this with just four colors? 00:40:01.783 --> 00:40:04.860 align:middle line:90% 00:40:04.860 --> 00:40:06.400 align:middle line:90% Yes, I claim we can. 00:40:06.400 --> 00:40:09.000 align:middle line:84% Let's make that red instead of purple. 00:40:09.000 --> 00:40:12.870 align:middle line:84% Now, this class and this class are both red, 00:40:12.870 --> 00:40:16.990 align:middle line:84% but because there's no edge between them, that's fine. 00:40:16.990 --> 00:40:19.438 align:middle line:84% All of the edges have two different colors 00:40:19.438 --> 00:40:20.230 align:middle line:90% on their endpoints. 00:40:20.230 --> 00:40:22.110 align:middle line:90% And that's all we need. 00:40:22.110 --> 00:40:22.960 align:middle line:90% All right? 00:40:22.960 --> 00:40:27.450 align:middle line:90% So this can be four-colored. 00:40:27.450 --> 00:40:30.690 align:middle line:90% Can we do better? 00:40:30.690 --> 00:40:32.563 align:middle line:84% Can we get away with just three colors? 00:40:32.563 --> 00:40:35.132 align:middle line:90% 00:40:35.132 --> 00:40:36.090 align:middle line:90% All right, interesting. 00:40:36.090 --> 00:40:40.080 align:middle line:84% I'm seeing one head shake and one head nod. 00:40:40.080 --> 00:40:41.760 align:middle line:90% So yes, please. 00:40:41.760 --> 00:40:44.970 align:middle line:84% AUDIENCE: How about making the yellow red. 00:40:44.970 --> 00:40:46.770 align:middle line:90% ZACHARY ABEL: Make yellow red. 00:40:46.770 --> 00:40:49.730 align:middle line:84% AUDIENCE: And then making that bright red like a blue or green? 00:40:49.730 --> 00:40:51.730 align:middle line:84% ZACHARY ABEL: OK, we'll make this blue or green. 00:40:51.730 --> 00:40:52.440 align:middle line:90% Yeah. 00:40:52.440 --> 00:40:56.140 align:middle line:84% All right, so we've got red, green, blue, red, blue. 00:40:56.140 --> 00:40:59.200 align:middle line:84% So we've got two reds, but there's no edge between them, 00:40:59.200 --> 00:41:00.190 align:middle line:90% so we're fine. 00:41:00.190 --> 00:41:03.100 align:middle line:84% And we've got two blues, but there's no edge between them, 00:41:03.100 --> 00:41:04.290 align:middle line:90% so we're fine. 00:41:04.290 --> 00:41:09.150 align:middle line:84% This is absolutely a valid three-coloring. 00:41:09.150 --> 00:41:11.020 align:middle line:90% Can be three-colored. 00:41:11.020 --> 00:41:14.850 align:middle line:90% 00:41:14.850 --> 00:41:17.970 align:middle line:90% Can this be 17-colored? 00:41:17.970 --> 00:41:21.780 align:middle line:84% Is there a proper 17 coloring for this graph? 00:41:21.780 --> 00:41:23.530 align:middle line:84% This is a bit of a weird edge case, 00:41:23.530 --> 00:41:29.430 align:middle line:84% but yes, this set C says that there have to be 00:41:29.430 --> 00:41:32.940 align:middle line:90% at most 17 colors available. 00:41:32.940 --> 00:41:36.810 align:middle line:84% Now, we don't have to use all of those colors. 00:41:36.810 --> 00:41:41.670 align:middle line:84% So if we had a 17-color palette and we found this proper 00:41:41.670 --> 00:41:45.400 align:middle line:84% coloring, that would be considered a valid 17-coloring. 00:41:45.400 --> 00:41:51.210 align:middle line:84% So this is a proper 17-coloring of the graph. 00:41:51.210 --> 00:41:53.610 align:middle line:84% But more directly, it's a proper three-coloring 00:41:53.610 --> 00:41:57.270 align:middle line:84% because usually we're trying to minimize the number of colors. 00:41:57.270 --> 00:41:57.910 align:middle line:90% All right? 00:41:57.910 --> 00:42:04.350 align:middle line:84% And we can make that notion a concrete definition. 00:42:04.350 --> 00:42:14.610 align:middle line:84% The chromatic number of a graph G, 00:42:14.610 --> 00:42:19.330 align:middle line:84% the chromatic number is denoted chi of G-- 00:42:19.330 --> 00:42:20.963 align:middle line:90% that's a lowercase Greek chi-- 00:42:20.963 --> 00:42:25.470 align:middle line:90% 00:42:25.470 --> 00:42:43.520 align:middle line:84% is the smallest number of colors needed for a proper coloring 00:42:43.520 --> 00:42:47.900 align:middle line:90% of G. All right? 00:42:47.900 --> 00:42:49.730 align:middle line:84% So the chromatic number asks what's 00:42:49.730 --> 00:42:53.005 align:middle line:84% the fewest colors you can get away with. 00:42:53.005 --> 00:42:54.380 align:middle line:84% In our scheduling problem, what's 00:42:54.380 --> 00:42:57.260 align:middle line:84% the fewest number of time slots that we need to cover 00:42:57.260 --> 00:42:59.720 align:middle line:90% all of the final exams? 00:42:59.720 --> 00:43:02.570 align:middle line:84% For this graph, what is its chromatic number? 00:43:02.570 --> 00:43:04.945 align:middle line:90% Have we shown that it's three? 00:43:04.945 --> 00:43:08.090 align:middle line:90% 00:43:08.090 --> 00:43:11.660 align:middle line:90% First of all, is it five? 00:43:11.660 --> 00:43:13.950 align:middle line:84% Is the chromatic number of this graph five? 00:43:13.950 --> 00:43:14.620 align:middle line:90% Why not? 00:43:14.620 --> 00:43:19.030 align:middle line:90% 00:43:19.030 --> 00:43:20.280 align:middle line:90% AUDIENCE: [INAUDIBLE] smaller. 00:43:20.280 --> 00:43:22.613 align:middle line:84% ZACHARY ABEL: Because we can get away with fewer colors. 00:43:22.613 --> 00:43:23.400 align:middle line:90% That's right. 00:43:23.400 --> 00:43:26.030 align:middle line:84% It's not five because we can do fewer. 00:43:26.030 --> 00:43:27.750 align:middle line:84% It's not four because we can do fewer. 00:43:27.750 --> 00:43:28.360 align:middle line:90% Is it three? 00:43:28.360 --> 00:43:31.340 align:middle line:90% 00:43:31.340 --> 00:43:32.280 align:middle line:90% I don't know yet. 00:43:32.280 --> 00:43:35.540 align:middle line:84% Can we get away with fewer colors? 00:43:35.540 --> 00:43:38.030 align:middle line:90% Well, let's see. 00:43:38.030 --> 00:43:39.900 align:middle line:90% Can it be two-colored? 00:43:39.900 --> 00:43:42.970 align:middle line:84% Is there a proper coloring with just two colors for this graph? 00:43:42.970 --> 00:43:45.980 align:middle line:90% 00:43:45.980 --> 00:43:50.420 align:middle line:84% And in this case, there isn't, because this triangle right here 00:43:50.420 --> 00:43:56.120 align:middle line:84% shows us that all three of these vertices need different colors. 00:43:56.120 --> 00:43:58.710 align:middle line:84% You can think of this as a proof by contradiction as well. 00:43:58.710 --> 00:44:02.010 align:middle line:84% Assume it's possible to color with just two colors. 00:44:02.010 --> 00:44:05.390 align:middle line:84% Well, then, well, these three vertices 00:44:05.390 --> 00:44:08.100 align:middle line:84% would have a conflict among them somehow, 00:44:08.100 --> 00:44:10.430 align:middle line:90% which is a contradiction. 00:44:10.430 --> 00:44:16.530 align:middle line:90% So cannot be two-colored. 00:44:16.530 --> 00:44:19.160 align:middle line:90% 00:44:19.160 --> 00:44:21.230 align:middle line:84% And these two statements together 00:44:21.230 --> 00:44:22.920 align:middle line:84% can be colored with three colors. 00:44:22.920 --> 00:44:25.220 align:middle line:84% Cannot be colored with two colors. 00:44:25.220 --> 00:44:28.100 align:middle line:84% This is what we need to confirm that the chromatic number is 00:44:28.100 --> 00:44:29.330 align:middle line:90% three. 00:44:29.330 --> 00:44:32.060 align:middle line:84% Three is the smallest because three works and fewer doesn't 00:44:32.060 --> 00:44:34.220 align:middle line:90% work. 00:44:34.220 --> 00:44:42.050 align:middle line:84% And I want to emphasize that, in general, to show 00:44:42.050 --> 00:44:46.310 align:middle line:84% that the chromatic number of G is some value K, 00:44:46.310 --> 00:44:49.495 align:middle line:84% usually we need both an upper bound and a lower bound. 00:44:49.495 --> 00:44:52.340 align:middle line:90% 00:44:52.340 --> 00:44:59.820 align:middle line:84% We need an upper bound, which is that chi of G is less than 00:44:59.820 --> 00:45:03.590 align:middle line:84% or equal K. And how do we prove that? 00:45:03.590 --> 00:45:06.560 align:middle line:90% Usually by giving a K coloring. 00:45:06.560 --> 00:45:09.360 align:middle line:84% If you know it's possible to color in K colors, 00:45:09.360 --> 00:45:11.830 align:middle line:84% then the chromatic number has to be K or smaller. 00:45:11.830 --> 00:45:14.630 align:middle line:90% 00:45:14.630 --> 00:45:16.083 align:middle line:90% Give a K coloring. 00:45:16.083 --> 00:45:18.980 align:middle line:90% 00:45:18.980 --> 00:45:26.510 align:middle line:84% But you also need a lower bound that chi of G 00:45:26.510 --> 00:45:30.400 align:middle line:84% has to be greater than K minus 1. 00:45:30.400 --> 00:45:32.890 align:middle line:90% And how do you prove that? 00:45:32.890 --> 00:45:40.450 align:middle line:84% Well, you have to prove that it can't be K minus 1 colored, 00:45:40.450 --> 00:45:42.380 align:middle line:90% usually by contradiction. 00:45:42.380 --> 00:45:50.110 align:middle line:90% 00:45:50.110 --> 00:45:52.480 align:middle line:84% So I just wanted to emphasize that the definition 00:45:52.480 --> 00:45:55.810 align:middle line:84% of chromatic number as the smallest thing that works, there 00:45:55.810 --> 00:45:57.080 align:middle line:90% are two parts to that. 00:45:57.080 --> 00:46:00.220 align:middle line:84% It works, and nothing smaller works. 00:46:00.220 --> 00:46:02.590 align:middle line:90% Make sense? 00:46:02.590 --> 00:46:05.200 align:middle line:90% Cool. 00:46:05.200 --> 00:46:06.000 align:middle line:90% All right. 00:46:06.000 --> 00:46:11.080 align:middle line:90% 00:46:11.080 --> 00:46:13.930 align:middle line:90% Yes. 00:46:13.930 --> 00:46:15.020 align:middle line:90% So graph coloring. 00:46:15.020 --> 00:46:17.140 align:middle line:84% We've seen one application, which 00:46:17.140 --> 00:46:20.300 align:middle line:84% is the registrar scheduling final exams. 00:46:20.300 --> 00:46:24.260 align:middle line:84% They actually do something exactly like this, by the way. 00:46:24.260 --> 00:46:26.680 align:middle line:84% This is why final exam schedules don't come out 00:46:26.680 --> 00:46:28.960 align:middle line:84% until a couple of weeks into the semester 00:46:28.960 --> 00:46:30.640 align:middle line:84% because they need to see who's enrolled 00:46:30.640 --> 00:46:35.140 align:middle line:84% in which pairs of classes, so we can try to schedule those apart 00:46:35.140 --> 00:46:35.810 align:middle line:90% from each other. 00:46:35.810 --> 00:46:39.790 align:middle line:84% They're trying to find a coloring with few colors. 00:46:39.790 --> 00:46:41.560 align:middle line:84% In practice, if they were looking 00:46:41.560 --> 00:46:45.370 align:middle line:84% for an actual proper coloring with no conflicts whatsoever, 00:46:45.370 --> 00:46:47.720 align:middle line:84% it probably wouldn't fit into exam week. 00:46:47.720 --> 00:46:49.360 align:middle line:84% So instead, they're looking to minimize 00:46:49.360 --> 00:46:52.640 align:middle line:84% the number of conflicts, which is a slightly different problem, 00:46:52.640 --> 00:46:55.240 align:middle line:90% but still an interesting one. 00:46:55.240 --> 00:46:58.870 align:middle line:84% But there are lots of scenarios where you're 00:46:58.870 --> 00:47:03.550 align:middle line:84% trying to avoid conflicts, just like the registrar is trying 00:47:03.550 --> 00:47:11.706 align:middle line:84% to avoid conflicts, so other coloring applications. 00:47:11.706 --> 00:47:16.690 align:middle line:90% 00:47:16.690 --> 00:47:18.400 align:middle line:84% A really common one, and one that's 00:47:18.400 --> 00:47:20.800 align:middle line:84% appropriate for this computer science class, 00:47:20.800 --> 00:47:22.476 align:middle line:90% is register allocation. 00:47:22.476 --> 00:47:27.520 align:middle line:90% 00:47:27.520 --> 00:47:31.360 align:middle line:84% When designing compilers, you have 00:47:31.360 --> 00:47:33.890 align:middle line:84% multiple values that you're trying to compute, 00:47:33.890 --> 00:47:36.070 align:middle line:84% different operations you need to do with them. 00:47:36.070 --> 00:47:36.570 align:middle line:90% Yes? 00:47:36.570 --> 00:47:39.280 align:middle line:90% 00:47:39.280 --> 00:47:42.260 align:middle line:84% AUDIENCE: Does the elements of C matter or just absolute size? 00:47:42.260 --> 00:47:44.133 align:middle line:84% Because you can put green in the left corner 00:47:44.133 --> 00:47:45.550 align:middle line:84% like there's a bunch of different. 00:47:45.550 --> 00:47:47.133 align:middle line:84% ZACHARY ABEL: That's a great question. 00:47:47.133 --> 00:47:48.760 align:middle line:90% Thank you for asking. 00:47:48.760 --> 00:47:52.490 align:middle line:84% Do the elements of C matter or just the size of C? 00:47:52.490 --> 00:47:55.720 align:middle line:84% C is the set of colors available. 00:47:55.720 --> 00:47:58.142 align:middle line:84% So if you change the set C, then you're 00:47:58.142 --> 00:48:00.100 align:middle line:84% giving a different coloring because your colors 00:48:00.100 --> 00:48:02.530 align:middle line:90% mean different things. 00:48:02.530 --> 00:48:05.718 align:middle line:84% But you're not going to change how many colors are needed. 00:48:05.718 --> 00:48:08.260 align:middle line:84% You're not going to change the chromatic number if you change 00:48:08.260 --> 00:48:11.800 align:middle line:90% the set of colors available. 00:48:11.800 --> 00:48:16.635 align:middle line:84% And you said we could have put green here instead of here? 00:48:16.635 --> 00:48:17.260 align:middle line:90% AUDIENCE: Yeah. 00:48:17.260 --> 00:48:19.150 align:middle line:84% ZACHARY ABEL: Yeah, like we could have made-- 00:48:19.150 --> 00:48:22.090 align:middle line:90% the reds could instead be green. 00:48:22.090 --> 00:48:26.190 align:middle line:84% And this green could be red, or this green could be yellow. 00:48:26.190 --> 00:48:28.720 align:middle line:84% Now, we've got some green, some blues, and a yellow. 00:48:28.720 --> 00:48:31.110 align:middle line:84% Still three colors, different coloring 00:48:31.110 --> 00:48:33.310 align:middle line:84% because we're assigning different colors to them, 00:48:33.310 --> 00:48:36.130 align:middle line:84% but still the same number of colors. 00:48:36.130 --> 00:48:38.650 align:middle line:84% And often, for the applications we're about to talk about, 00:48:38.650 --> 00:48:42.420 align:middle line:84% the number of colors is all that matters, because the colors just 00:48:42.420 --> 00:48:44.370 align:middle line:84% mean these are the different classes that you 00:48:44.370 --> 00:48:46.560 align:middle line:84% need to put them in, and then you do something 00:48:46.560 --> 00:48:49.770 align:middle line:90% with each class. 00:48:49.770 --> 00:48:52.300 align:middle line:84% So this application, register allocation. 00:48:52.300 --> 00:48:55.260 align:middle line:84% When you're computing multiple values, 00:48:55.260 --> 00:49:00.210 align:middle line:84% your processor has a small number, like 16 or 32, 00:49:00.210 --> 00:49:04.380 align:middle line:84% registers, which are really, really fast memory that it 00:49:04.380 --> 00:49:05.470 align:middle line:90% can do operations on. 00:49:05.470 --> 00:49:08.520 align:middle line:84% And in fact, it has to bring data into a register 00:49:08.520 --> 00:49:11.760 align:middle line:84% before it can compute on them usually. 00:49:11.760 --> 00:49:15.900 align:middle line:84% And some values you're computing and you 00:49:15.900 --> 00:49:18.365 align:middle line:84% need lots of different pieces to go into that computation. 00:49:18.365 --> 00:49:19.990 align:middle line:84% Some values, well, you're done with it. 00:49:19.990 --> 00:49:21.790 align:middle line:84% You don't need to remember it anymore. 00:49:21.790 --> 00:49:24.780 align:middle line:84% So let's evict that and put something else in that register. 00:49:24.780 --> 00:49:27.660 align:middle line:84% And ideally, you can figure out a way 00:49:27.660 --> 00:49:30.720 align:middle line:84% to schedule all of your values in the different registers 00:49:30.720 --> 00:49:33.390 align:middle line:84% so that everything fits in the registers and everything's 00:49:33.390 --> 00:49:36.840 align:middle line:84% fast, instead of having to spill them out into RAM. 00:49:36.840 --> 00:49:40.020 align:middle line:84% You don't need to understand compilers 00:49:40.020 --> 00:49:41.710 align:middle line:90% or exactly what I'm saying here. 00:49:41.710 --> 00:49:45.600 align:middle line:84% The point is, your variables, the values you're computing, 00:49:45.600 --> 00:49:48.970 align:middle line:84% have different lifetimes, times when they need to still exist, 00:49:48.970 --> 00:49:52.110 align:middle line:84% and other times when you're allowed to forget them. 00:49:52.110 --> 00:49:59.160 align:middle line:84% And compiler writers often draw exactly the conflict graph 00:49:59.160 --> 00:50:00.730 align:middle line:90% between these lifetimes. 00:50:00.730 --> 00:50:02.820 align:middle line:84% This value has to exist at the same time 00:50:02.820 --> 00:50:04.870 align:middle line:84% that this value also has to exist, 00:50:04.870 --> 00:50:08.070 align:middle line:84% so they can't be in the same register, 00:50:08.070 --> 00:50:10.570 align:middle line:84% because then one would overwrite the other. 00:50:10.570 --> 00:50:12.900 align:middle line:84% And if you can write down your conflict graph 00:50:12.900 --> 00:50:16.230 align:middle line:84% and find a coloring with a small number of colors, 00:50:16.230 --> 00:50:19.032 align:middle line:84% ideally smaller than the number of registers available, 00:50:19.032 --> 00:50:20.490 align:middle line:84% then everything works in registers, 00:50:20.490 --> 00:50:23.580 align:middle line:90% and you're really fast. 00:50:23.580 --> 00:50:29.090 align:middle line:84% Another one is radio tower frequencies. 00:50:29.090 --> 00:50:33.690 align:middle line:90% 00:50:33.690 --> 00:50:35.670 align:middle line:84% If each tower in a radio network needs 00:50:35.670 --> 00:50:38.110 align:middle line:84% to broadcast with some frequencies, 00:50:38.110 --> 00:50:40.830 align:middle line:84% well, if two towers use the same frequency that 00:50:40.830 --> 00:50:43.673 align:middle line:84% can reach the same places, then they're going to conflict. 00:50:43.673 --> 00:50:45.340 align:middle line:84% You're going to get static in the noise. 00:50:45.340 --> 00:50:48.960 align:middle line:84% You're getting noise from two different places at once. 00:50:48.960 --> 00:50:53.740 align:middle line:84% So any conflicting towers need to use different frequencies. 00:50:53.740 --> 00:50:57.060 align:middle line:84% And this is, again, a graph coloring problem. 00:50:57.060 --> 00:50:59.020 align:middle line:90% Draw your conflict graph. 00:50:59.020 --> 00:51:00.790 align:middle line:84% Color it with a small number of colors. 00:51:00.790 --> 00:51:03.310 align:middle line:84% Each color is a different frequency. 00:51:03.310 --> 00:51:06.330 align:middle line:84% And since different towers that are assigned the same color 00:51:06.330 --> 00:51:08.127 align:middle line:84% don't have an edge in the conflict graph, 00:51:08.127 --> 00:51:09.960 align:middle line:84% it means they don't conflict, and so they're 00:51:09.960 --> 00:51:12.485 align:middle line:84% allowed to use the same frequency just fine. 00:51:12.485 --> 00:51:15.180 align:middle line:90% 00:51:15.180 --> 00:51:18.860 align:middle line:84% In general, where did I put this list? 00:51:18.860 --> 00:51:19.710 align:middle line:90% Way over here. 00:51:19.710 --> 00:51:20.575 align:middle line:90% No, come back. 00:51:20.575 --> 00:51:23.505 align:middle line:90% 00:51:23.505 --> 00:51:24.505 align:middle line:90% Let's see, map coloring. 00:51:24.505 --> 00:51:29.150 align:middle line:90% 00:51:29.150 --> 00:51:32.250 align:middle line:84% A while ago, Eric told us about the four-color theorem. 00:51:32.250 --> 00:51:38.490 align:middle line:84% That every map that you draw on a sheet of paper, 00:51:38.490 --> 00:51:41.210 align:middle line:84% you can color all the regions in different colors 00:51:41.210 --> 00:51:44.840 align:middle line:84% so that two neighboring regions never have the same color. 00:51:44.840 --> 00:51:49.190 align:middle line:84% And four colors is always enough. 00:51:49.190 --> 00:51:51.920 align:middle line:84% This is just saying that certain kinds of graphs 00:51:51.920 --> 00:51:53.880 align:middle line:90% can always be four-colored. 00:51:53.880 --> 00:51:55.640 align:middle line:84% Not every graph, but certain kinds 00:51:55.640 --> 00:51:58.250 align:middle line:84% of graphs, the so-called planar graphs 00:51:58.250 --> 00:51:59.850 align:middle line:90% can always be four-colored. 00:51:59.850 --> 00:52:03.220 align:middle line:90% 00:52:03.220 --> 00:52:05.370 align:middle line:84% This is the language that that proof is written in. 00:52:05.370 --> 00:52:07.230 align:middle line:84% It's a theorem about graph theory. 00:52:07.230 --> 00:52:10.640 align:middle line:90% It uses graph theory. 00:52:10.640 --> 00:52:14.280 align:middle line:84% But it's an example, we've already seen, which is nice. 00:52:14.280 --> 00:52:16.020 align:middle line:84% In general, any time you're trying 00:52:16.020 --> 00:52:25.020 align:middle line:84% to avoid pairwise conflicts, graph coloring 00:52:25.020 --> 00:52:28.260 align:middle line:90% is a great tool to reach for. 00:52:28.260 --> 00:52:32.020 align:middle line:84% Because it's such a great tool to reach for in many contexts, 00:52:32.020 --> 00:52:34.440 align:middle line:84% wouldn't it be great if there was 00:52:34.440 --> 00:52:38.860 align:middle line:84% a really fast, simple algorithm that says, here's a graph. 00:52:38.860 --> 00:52:40.570 align:middle line:84% Here's a target number of colors, 00:52:40.570 --> 00:52:45.600 align:middle line:84% please give me a proper coloring that uses this number of colors. 00:52:45.600 --> 00:53:00.870 align:middle line:84% So an algorithm to K color any input graph, or decide it's 00:53:00.870 --> 00:53:01.445 align:middle line:90% not possible. 00:53:01.445 --> 00:53:04.680 align:middle line:90% 00:53:04.680 --> 00:53:06.600 align:middle line:84% If I'm the registrar, and I know I 00:53:06.600 --> 00:53:09.790 align:middle line:84% have-- what is it-- five days of exams, two sessions each. 00:53:09.790 --> 00:53:11.890 align:middle line:84% So I have 10 different exam sessions. 00:53:11.890 --> 00:53:14.215 align:middle line:84% I really want a 10-coloring of my graph. 00:53:14.215 --> 00:53:16.880 align:middle line:90% 00:53:16.880 --> 00:53:19.970 align:middle line:84% And so I want to ask, please give me a 10-coloring of this 00:53:19.970 --> 00:53:20.790 align:middle line:90% conflict graph. 00:53:20.790 --> 00:53:22.370 align:middle line:90% And then, that's my schedule. 00:53:22.370 --> 00:53:25.190 align:middle line:84% Is there an algorithm to do that? 00:53:25.190 --> 00:53:30.830 align:middle line:84% Likewise, instead of asking give me 10 colors, 00:53:30.830 --> 00:53:37.430 align:middle line:84% you can ask the algorithm to compute chi of G. 00:53:37.430 --> 00:53:40.020 align:middle line:84% I don't know in advance how many colors I'm expecting. 00:53:40.020 --> 00:53:41.430 align:middle line:90% Just give me the best coloring. 00:53:41.430 --> 00:53:44.040 align:middle line:84% Give me a coloring that uses the fewest number of colors. 00:53:44.040 --> 00:53:45.900 align:middle line:84% Is there an algorithm to do that? 00:53:45.900 --> 00:53:49.220 align:middle line:84% That would be really great in lots of applications. 00:53:49.220 --> 00:53:51.260 align:middle line:84% Unfortunately, I'm talking in hypotheticals 00:53:51.260 --> 00:53:56.240 align:middle line:84% here because these algorithms very, very likely don't exist. 00:53:56.240 --> 00:54:05.060 align:middle line:84% These problems are what we call NP-- 00:54:05.060 --> 00:54:06.990 align:middle line:84% well, this problem is NP-complete. 00:54:06.990 --> 00:54:08.410 align:middle line:84% This one is possibly even harder. 00:54:08.410 --> 00:54:14.620 align:middle line:90% 00:54:14.620 --> 00:54:16.460 align:middle line:84% At least, this problem is NP-complete. 00:54:16.460 --> 00:54:20.050 align:middle line:84% Like I said, this one might be even harder than NP. 00:54:20.050 --> 00:54:21.410 align:middle line:90% What does this mean? 00:54:21.410 --> 00:54:22.700 align:middle line:90% I don't care so much. 00:54:22.700 --> 00:54:24.520 align:middle line:84% I don't really want to dig into what 00:54:24.520 --> 00:54:28.730 align:middle line:84% NP-complete means right now, or NP-hard, or any of those things. 00:54:28.730 --> 00:54:32.830 align:middle line:84% What I will say is that there are thousands upon thousands 00:54:32.830 --> 00:54:38.230 align:middle line:84% of computational problems, like graph coloring, 00:54:38.230 --> 00:54:40.340 align:middle line:84% like the traveling salesman problem, 00:54:40.340 --> 00:54:42.640 align:middle line:84% if you've heard of that one, it's pretty famous. 00:54:42.640 --> 00:54:48.040 align:middle line:84% Like Sudoku on n squared by n squared boards instead of just 9 00:54:48.040 --> 00:54:49.360 align:middle line:90% by 9. 00:54:49.360 --> 00:54:52.090 align:middle line:84% There are thousands upon thousands of these problems 00:54:52.090 --> 00:54:55.270 align:middle line:84% that computer scientists have decided 00:54:55.270 --> 00:54:57.910 align:middle line:90% are all equally difficult. 00:54:57.910 --> 00:55:02.080 align:middle line:84% So if we were able to find an efficient algorithm to solve 00:55:02.080 --> 00:55:05.478 align:middle line:84% any one of them, like to solve the K coloring problem, then 00:55:05.478 --> 00:55:07.270 align:middle line:84% that would immediately give us an algorithm 00:55:07.270 --> 00:55:10.910 align:middle line:84% to solve all the other ones efficiently as well. 00:55:10.910 --> 00:55:14.900 align:middle line:84% And all the problems that are easier than those. 00:55:14.900 --> 00:55:18.170 align:middle line:84% A notable problem that we've already talked about 00:55:18.170 --> 00:55:25.820 align:middle line:90% is factoring large numbers. 00:55:25.820 --> 00:55:28.370 align:middle line:90% This is known to be in NP-- 00:55:28.370 --> 00:55:30.900 align:middle line:84% it's not known whether it's NP-complete. 00:55:30.900 --> 00:55:34.230 align:middle line:84% It's in fact thought to be even easier than most of these. 00:55:34.230 --> 00:55:39.530 align:middle line:84% So it's thought that factoring numbers is somehow easier than K 00:55:39.530 --> 00:55:40.890 align:middle line:90% coloring your graph. 00:55:40.890 --> 00:55:45.570 align:middle line:84% Also, by the way, even if we're just asking for three colors, 00:55:45.570 --> 00:55:48.320 align:middle line:84% does this graph have a coloring with just three colors. 00:55:48.320 --> 00:55:49.950 align:middle line:90% That's already NP-complete. 00:55:49.950 --> 00:55:52.730 align:middle line:84% That's already too hard, and we don't 00:55:52.730 --> 00:55:55.940 align:middle line:84% believe there's going to be an efficient algorithm for it. 00:55:55.940 --> 00:55:59.330 align:middle line:84% So if you solve this easier problem, factoring 00:55:59.330 --> 00:56:02.640 align:middle line:84% large numbers, if you find an efficient algorithm for that, 00:56:02.640 --> 00:56:06.080 align:middle line:84% well, you'll be able to break all the public RSA keys. 00:56:06.080 --> 00:56:08.060 align:middle line:84% Read everyone's encrypted messages. 00:56:08.060 --> 00:56:12.410 align:middle line:84% Steal all the electronic money, including the real currency, 00:56:12.410 --> 00:56:14.770 align:middle line:90% from electronic banks. 00:56:14.770 --> 00:56:17.530 align:middle line:84% Run from all of the world governments that are trying 00:56:17.530 --> 00:56:19.230 align:middle line:90% to hire and/or silence you. 00:56:19.230 --> 00:56:22.390 align:middle line:90% 00:56:22.390 --> 00:56:25.390 align:middle line:84% But if you solve the even harder problem of three-coloring 00:56:25.390 --> 00:56:29.380 align:middle line:84% a graph, you get all that plus $1 million 00:56:29.380 --> 00:56:32.090 align:middle line:84% prize for solving a famous millennium problem. 00:56:32.090 --> 00:56:35.740 align:middle line:90% So you've got choices. 00:56:35.740 --> 00:56:41.510 align:middle line:84% Point is-- graph coloring is a hard computational problem. 00:56:41.510 --> 00:56:44.920 align:middle line:84% And most computer scientists strongly 00:56:44.920 --> 00:56:47.860 align:middle line:84% believe that there is no efficient algorithm 00:56:47.860 --> 00:56:49.750 align:middle line:90% for graph coloring. 00:56:49.750 --> 00:56:54.080 align:middle line:84% And so, at least for very large graphs, 00:56:54.080 --> 00:56:58.180 align:middle line:84% it's not realistic to ask for a coloring 00:56:58.180 --> 00:57:03.690 align:middle line:84% with a specific number of colors or the best possible coloring, 00:57:03.690 --> 00:57:05.520 align:middle line:84% which means that our applications are 00:57:05.520 --> 00:57:08.670 align:middle line:84% a little bit sad because sometimes we really want 00:57:08.670 --> 00:57:09.970 align:middle line:90% optimal colorings. 00:57:09.970 --> 00:57:13.590 align:middle line:84% Thankfully, in practice, optimal colorings 00:57:13.590 --> 00:57:15.250 align:middle line:90% aren't really necessary. 00:57:15.250 --> 00:57:19.590 align:middle line:84% Sometimes we can get away with good enough. 00:57:19.590 --> 00:57:24.580 align:middle line:84% And so for all of these NP-complete or NP-hard problems, 00:57:24.580 --> 00:57:28.290 align:middle line:84% the best you can hope for is an algorithm that's good enough. 00:57:28.290 --> 00:57:31.120 align:middle line:84% Usually, the answer it gives us a little worse, 00:57:31.120 --> 00:57:33.360 align:middle line:90% but it runs faster. 00:57:33.360 --> 00:57:35.940 align:middle line:84% And let's see an algorithm for that 00:57:35.940 --> 00:57:38.890 align:middle line:84% to give us some colorings of graphs, 00:57:38.890 --> 00:57:41.566 align:middle line:84% even if they're not always optimal colorings. 00:57:41.566 --> 00:57:45.360 align:middle line:90% 00:57:45.360 --> 00:57:46.085 align:middle line:90% All right. 00:57:46.085 --> 00:57:57.630 align:middle line:90% 00:57:57.630 --> 00:58:07.590 align:middle line:84% So we have a greedy coloring algorithm. 00:58:07.590 --> 00:58:10.300 align:middle line:84% And here is the idea of the algorithm. 00:58:10.300 --> 00:58:17.220 align:middle line:84% So first of all, we're going to order the vertices-- 00:58:17.220 --> 00:58:20.650 align:middle line:90% V1, V2, Vn. 00:58:20.650 --> 00:58:24.790 align:middle line:84% So we're going to pick some ordering of the vertices. 00:58:24.790 --> 00:58:26.123 align:middle line:90% We're going to order the colors. 00:58:26.123 --> 00:58:29.310 align:middle line:90% 00:58:29.310 --> 00:58:31.770 align:middle line:84% I'm going to use my favorite color palette-- 00:58:31.770 --> 00:58:34.210 align:middle line:90% 1, 2, 3, and so on. 00:58:34.210 --> 00:58:37.170 align:middle line:90% Those are my colors. 00:58:37.170 --> 00:58:44.460 align:middle line:90% And then, for each Vi in order-- 00:58:44.460 --> 00:58:47.740 align:middle line:84% so we'll look at V1, then V2, then V3 and so on, 00:58:47.740 --> 00:58:50.180 align:middle line:84% and we'll assign each of them a color one at a time. 00:58:50.180 --> 00:58:54.180 align:middle line:90% 00:58:54.180 --> 00:59:08.430 align:middle line:84% Assign Vi the smallest color that 00:59:08.430 --> 00:59:09.916 align:middle line:90% doesn't introduce conflicts. 00:59:09.916 --> 00:59:13.020 align:middle line:90% 00:59:13.020 --> 00:59:16.210 align:middle line:84% So when we get to vertex i, we're going to ask, 00:59:16.210 --> 00:59:17.650 align:middle line:90% does color one work? 00:59:17.650 --> 00:59:18.760 align:middle line:90% Does color two work? 00:59:18.760 --> 00:59:20.170 align:middle line:90% Does color three work? 00:59:20.170 --> 00:59:24.460 align:middle line:84% And the first time we get to a color that is valid, 00:59:24.460 --> 00:59:26.848 align:middle line:84% that it doesn't have a neighbor of that same color, 00:59:26.848 --> 00:59:28.890 align:middle line:84% we're going to assign it that color and then move 00:59:28.890 --> 00:59:29.810 align:middle line:90% on to the next vertex. 00:59:29.810 --> 00:59:33.213 align:middle line:90% 00:59:33.213 --> 00:59:35.130 align:middle line:84% Yeah, it's called greedy because you're always 00:59:35.130 --> 00:59:38.250 align:middle line:84% picking the smallest color you can at any one time. 00:59:38.250 --> 00:59:40.440 align:middle line:84% It doesn't necessarily mean that you're going 00:59:40.440 --> 00:59:42.250 align:middle line:90% to get an optimal coloring. 00:59:42.250 --> 00:59:45.900 align:middle line:84% In fact, usually you won't, at least for big graphs 00:59:45.900 --> 00:59:49.170 align:middle line:84% where it's hard to find optimal colorings. 00:59:49.170 --> 00:59:51.100 align:middle line:84% Let's run through this algorithm. 00:59:51.100 --> 00:59:54.120 align:middle line:84% Let's see it happening just to get a more concrete feel 00:59:54.120 --> 00:59:55.745 align:middle line:90% for what we mean. 00:59:55.745 --> 00:59:58.730 align:middle line:90% 00:59:58.730 --> 01:00:00.050 align:middle line:90% Here we go. 01:00:00.050 --> 01:00:01.290 align:middle line:90% So here's my graph. 01:00:01.290 --> 01:00:02.010 align:middle line:90% No, pause. 01:00:02.010 --> 01:00:03.000 align:middle line:90% Thank you. 01:00:03.000 --> 01:00:07.100 align:middle line:84% Six vertices-- 1, 2, 3, 4, 5, 7 edges. 01:00:07.100 --> 01:00:09.690 align:middle line:84% And this is the vertex order we're going to use. 01:00:09.690 --> 01:00:12.290 align:middle line:84% We could choose any vertex order we want. 01:00:12.290 --> 01:00:15.050 align:middle line:84% But we're going to look at the vertices in this order 01:00:15.050 --> 01:00:17.850 align:middle line:84% and always choose the smallest color that's valid. 01:00:17.850 --> 01:00:20.750 align:middle line:84% So the first color is fine because there 01:00:20.750 --> 01:00:21.960 align:middle line:90% are no possible conflicts. 01:00:21.960 --> 01:00:23.930 align:middle line:90% No one else has colors yet. 01:00:23.930 --> 01:00:27.110 align:middle line:84% Second vertex, the first color is no good 01:00:27.110 --> 01:00:30.350 align:middle line:84% because it already has a neighbor that's 01:00:30.350 --> 01:00:31.920 align:middle line:90% been assigned that color. 01:00:31.920 --> 01:00:34.980 align:middle line:84% So instead of the color triangle, 01:00:34.980 --> 01:00:38.540 align:middle line:84% we have to go up to color square. 01:00:38.540 --> 01:00:40.340 align:middle line:90% All right? 01:00:40.340 --> 01:00:42.470 align:middle line:90% Now, let's go to vertex three. 01:00:42.470 --> 01:00:47.450 align:middle line:84% Our first color is valid this time, so it gets color one. 01:00:47.450 --> 01:00:49.010 align:middle line:90% Still with me? 01:00:49.010 --> 01:00:54.290 align:middle line:84% All right, vertex four can't use color one but can use color two. 01:00:54.290 --> 01:00:56.520 align:middle line:90% Vertex five, let's take a look. 01:00:56.520 --> 01:00:59.550 align:middle line:84% Vertex five can't use the first color. 01:00:59.550 --> 01:01:02.930 align:middle line:84% Can't use the second color, but can use the third color, 01:01:02.930 --> 01:01:05.210 align:middle line:84% so it gets assigned to the third color. 01:01:05.210 --> 01:01:08.580 align:middle line:84% And finally, vertex six gets assigned the first color. 01:01:08.580 --> 01:01:12.260 align:middle line:84% And we have our valid coloring or proper coloring. 01:01:12.260 --> 01:01:14.870 align:middle line:90% Does that algorithm make sense? 01:01:14.870 --> 01:01:16.880 align:middle line:90% Awesome. 01:01:16.880 --> 01:01:26.300 align:middle line:84% One thing I want to mention is that different vertex orderings 01:01:26.300 --> 01:01:34.730 align:middle line:84% will often lead to different colorings 01:01:34.730 --> 01:01:47.303 align:middle line:84% and often lead to colorings with a different number of colors. 01:01:47.303 --> 01:01:50.420 align:middle line:90% 01:01:50.420 --> 01:01:52.820 align:middle line:84% So here with this ordering, we got a coloring 01:01:52.820 --> 01:01:53.670 align:middle line:90% with three colors. 01:01:53.670 --> 01:01:55.430 align:middle line:90% We got a proper three-coloring. 01:01:55.430 --> 01:02:00.260 align:middle line:84% But if instead we use this ordering, where all I've done 01:02:00.260 --> 01:02:03.170 align:middle line:84% is changed the order of vertices one, five, and six, 01:02:03.170 --> 01:02:07.160 align:middle line:84% this is a different vertex order on the same graph, 01:02:07.160 --> 01:02:10.010 align:middle line:84% so let's run the algorithm again in this ordering. 01:02:10.010 --> 01:02:12.150 align:middle line:84% And let's see what coloring it gives us. 01:02:12.150 --> 01:02:13.920 align:middle line:90% So first color is fine here. 01:02:13.920 --> 01:02:15.720 align:middle line:90% First color is fine here. 01:02:15.720 --> 01:02:17.840 align:middle line:84% This one needs the second color because it already 01:02:17.840 --> 01:02:20.240 align:middle line:84% has a neighbor with the first color. 01:02:20.240 --> 01:02:23.180 align:middle line:84% Stop me if I'm going too quickly. 01:02:23.180 --> 01:02:26.280 align:middle line:84% The next one has neighbors with colors one and two, 01:02:26.280 --> 01:02:29.900 align:middle line:90% so it needs the third color. 01:02:29.900 --> 01:02:33.210 align:middle line:84% The fifth vertex has a neighbor with color one, 01:02:33.210 --> 01:02:35.840 align:middle line:90% so it needs the second color. 01:02:35.840 --> 01:02:40.670 align:middle line:84% And then, finally, vertex six has a neighbor with color one 01:02:40.670 --> 01:02:44.910 align:middle line:84% and a neighbor with color two and a neighbor with color three, 01:02:44.910 --> 01:02:48.170 align:middle line:84% so it in fact needs a fourth color. 01:02:48.170 --> 01:02:51.660 align:middle line:84% So with this different vertex ordering on the same graph, 01:02:51.660 --> 01:02:55.750 align:middle line:84% it gave us a four-coloring instead of a three-coloring. 01:02:55.750 --> 01:02:56.390 align:middle line:90% All right? 01:02:56.390 --> 01:02:59.163 align:middle line:84% So some orderings are better than others. 01:02:59.163 --> 01:03:10.360 align:middle line:90% 01:03:10.360 --> 01:03:12.970 align:middle line:90% Yeah? 01:03:12.970 --> 01:03:15.460 align:middle line:90% All right. 01:03:15.460 --> 01:03:17.180 align:middle line:84% Does this algorithm actually work? 01:03:17.180 --> 01:03:19.110 align:middle line:84% Does it always give a proper coloring? 01:03:19.110 --> 01:03:21.940 align:middle line:90% 01:03:21.940 --> 01:03:24.230 align:middle line:90% So yes, in this case, it does. 01:03:24.230 --> 01:03:28.480 align:middle line:84% And an easy way to see that is that the way the algorithm 01:03:28.480 --> 01:03:32.320 align:middle line:84% is designed, it never introduces conflicts. 01:03:32.320 --> 01:03:35.480 align:middle line:84% Always after we've assigned the next vertex, 01:03:35.480 --> 01:03:38.900 align:middle line:84% we never have an edge that has two endpoints of the same color 01:03:38.900 --> 01:03:41.690 align:middle line:84% because that's kind of the point of the algorithm. 01:03:41.690 --> 01:03:45.560 align:middle line:84% So we never introduce a conflict on any edge. 01:03:45.560 --> 01:03:47.950 align:middle line:84% So when we get to the end, we haven't 01:03:47.950 --> 01:03:49.040 align:middle line:90% introduced any conflicts. 01:03:49.040 --> 01:03:50.710 align:middle line:90% So there aren't any conflicts. 01:03:50.710 --> 01:03:55.390 align:middle line:84% That's kind of a state machine style proof of that algorithm. 01:03:55.390 --> 01:03:56.140 align:middle line:90% Yes? 01:03:56.140 --> 01:03:58.688 align:middle line:84% AUDIENCE: Could you try ordering like the middle one 01:03:58.688 --> 01:04:00.230 align:middle line:84% once you get the optimal [INAUDIBLE]? 01:04:00.230 --> 01:04:01.397 align:middle line:90% ZACHARY ABEL: Good question. 01:04:01.397 --> 01:04:04.690 align:middle line:84% If you try all the orderings and then take the best coloring 01:04:04.690 --> 01:04:07.810 align:middle line:90% you get, will that be optimal? 01:04:07.810 --> 01:04:11.950 align:middle line:84% So there are two questions in there. 01:04:11.950 --> 01:04:16.150 align:middle line:84% Can you get all colorings from this algorithm? 01:04:16.150 --> 01:04:17.180 align:middle line:90% And the answer is no. 01:04:17.180 --> 01:04:19.490 align:middle line:84% You can't get all colorings from this algorithm. 01:04:19.490 --> 01:04:22.480 align:middle line:84% For example, colorings that never use color one you 01:04:22.480 --> 01:04:24.670 align:middle line:84% can't get from this algorithm because you're always 01:04:24.670 --> 01:04:28.300 align:middle line:84% going to use color one on the first vertex. 01:04:28.300 --> 01:04:31.270 align:middle line:84% But the other question, can you always get 01:04:31.270 --> 01:04:35.020 align:middle line:84% some coloring that uses the minimal number of colors? 01:04:35.020 --> 01:04:38.210 align:middle line:84% Even if you can't get all different colorings, 01:04:38.210 --> 01:04:40.580 align:middle line:84% can you at least get all different numbers of colors? 01:04:40.580 --> 01:04:43.150 align:middle line:90% And that answer is yes. 01:04:43.150 --> 01:04:46.460 align:middle line:84% It takes a little bit of thought to prove, but the answer is yes. 01:04:46.460 --> 01:04:51.040 align:middle line:84% So if you do try all n factorial vertex orderings 01:04:51.040 --> 01:04:55.280 align:middle line:84% and take the minimal coloring, you will, in fact, 01:04:55.280 --> 01:04:56.540 align:middle line:90% get the chromatic number. 01:04:56.540 --> 01:04:58.900 align:middle line:84% But n factorial is a runtime that 01:04:58.900 --> 01:05:05.140 align:middle line:84% is not considered efficient for things like NP and NP hardness. 01:05:05.140 --> 01:05:07.150 align:middle line:84% So that is absolutely an algorithm 01:05:07.150 --> 01:05:12.580 align:middle line:84% to compute chromatic number, just not an efficient one. 01:05:12.580 --> 01:05:13.740 align:middle line:90% Yeah, great question. 01:05:13.740 --> 01:05:16.480 align:middle line:90% 01:05:16.480 --> 01:05:21.460 align:middle line:84% Let me prove one thing about this greedy algorithm. 01:05:21.460 --> 01:05:22.900 align:middle line:84% As we said, it's not always going 01:05:22.900 --> 01:05:25.510 align:middle line:84% to give the smallest coloring depending 01:05:25.510 --> 01:05:28.480 align:middle line:90% on what order you pick. 01:05:28.480 --> 01:05:34.000 align:middle line:84% But we can prove something about the quality of colorings we get. 01:05:34.000 --> 01:05:36.405 align:middle line:90% And here is that theorem. 01:05:36.405 --> 01:05:44.410 align:middle line:90% 01:05:44.410 --> 01:05:54.090 align:middle line:84% So for all graphs G, for all vertex 01:05:54.090 --> 01:06:09.240 align:middle line:84% orderings, if the max degree of our graph 01:06:09.240 --> 01:06:11.580 align:middle line:90% is less than or equal d-- 01:06:11.580 --> 01:06:13.620 align:middle line:84% by max degree here, I mean, look at the degrees 01:06:13.620 --> 01:06:16.080 align:middle line:84% of all the vertices and take the largest one. 01:06:16.080 --> 01:06:20.610 align:middle line:84% So the maximum of the degrees of the vertices. 01:06:20.610 --> 01:06:23.010 align:middle line:84% So if the max degree is at most d. 01:06:23.010 --> 01:06:25.800 align:middle line:84% Said differently, if all vertices in the graph 01:06:25.800 --> 01:06:37.830 align:middle line:84% have degree at most d, then the greedy algorithm 01:06:37.830 --> 01:06:51.020 align:middle line:84% returns a proper coloring with at most d plus 1 colors. 01:06:51.020 --> 01:06:54.330 align:middle line:90% 01:06:54.330 --> 01:06:58.875 align:middle line:84% So if all the vertices of the graph have degree at most d, 01:06:58.875 --> 01:07:02.195 align:middle line:84% then the greedy algorithm will use at most d plus 1 colors. 01:07:02.195 --> 01:07:04.920 align:middle line:90% 01:07:04.920 --> 01:07:07.710 align:middle line:90% Does this statement make sense? 01:07:07.710 --> 01:07:13.080 align:middle line:84% So in this example, all the degrees are three or less. 01:07:13.080 --> 01:07:15.300 align:middle line:84% So we're guaranteed to produce a coloring that 01:07:15.300 --> 01:07:18.300 align:middle line:90% uses at most four colors. 01:07:18.300 --> 01:07:19.800 align:middle line:84% And I want to emphasize this depends 01:07:19.800 --> 01:07:23.080 align:middle line:84% on the degrees of the vertices, not the number of vertices. 01:07:23.080 --> 01:07:26.530 align:middle line:84% If this graph had thousands and thousands of vertices, 01:07:26.530 --> 01:07:28.810 align:middle line:84% but all the degrees were three or less, 01:07:28.810 --> 01:07:32.730 align:middle line:84% we would still get a coloring with at most four colors. 01:07:32.730 --> 01:07:33.900 align:middle line:90% All right? 01:07:33.900 --> 01:07:35.880 align:middle line:90% It's not necessarily optimal. 01:07:35.880 --> 01:07:38.400 align:middle line:84% In this graph, it wasn't optimal. 01:07:38.400 --> 01:07:40.560 align:middle line:84% We found a three-coloring, whereas this only 01:07:40.560 --> 01:07:44.040 align:middle line:84% promises a four-coloring, but at least it's 01:07:44.040 --> 01:07:46.590 align:middle line:84% some measure of quality of the results 01:07:46.590 --> 01:07:50.040 align:middle line:84% we get out of this greedy algorithm. 01:07:50.040 --> 01:07:51.840 align:middle line:90% Does the claim make sense? 01:07:51.840 --> 01:07:52.680 align:middle line:90% Wonderful. 01:07:52.680 --> 01:07:55.620 align:middle line:90% Let's prove it. 01:07:55.620 --> 01:07:56.740 align:middle line:90% All right. 01:07:56.740 --> 01:08:04.350 align:middle line:84% We're going to prove this by induction. 01:08:04.350 --> 01:08:07.530 align:middle line:90% But induction on what? 01:08:07.530 --> 01:08:10.770 align:middle line:84% We might think, OK, well, there's 01:08:10.770 --> 01:08:15.870 align:middle line:84% a d in the theorem statement, so what if we induct on d? 01:08:15.870 --> 01:08:19.560 align:middle line:84% Well, that means at some point we know something 01:08:19.560 --> 01:08:22.810 align:middle line:84% about all graphs whose degrees are at most eight. 01:08:22.810 --> 01:08:25.080 align:middle line:84% And we need to prove something about all graphs 01:08:25.080 --> 01:08:28.649 align:middle line:90% whose degrees are at most nine. 01:08:28.649 --> 01:08:31.420 align:middle line:84% I don't really see how to get from one to the next. 01:08:31.420 --> 01:08:35.310 align:middle line:84% That doesn't seem easy to me, right? 01:08:35.310 --> 01:08:38.729 align:middle line:84% Graphs with max degree nine look very different 01:08:38.729 --> 01:08:41.120 align:middle line:84% from graphs with max degree eight. 01:08:41.120 --> 01:08:42.960 align:middle line:84% If you have a graph with max degree nine, 01:08:42.960 --> 01:08:46.580 align:middle line:84% there might be many, many, many vertices with degree nine. 01:08:46.580 --> 01:08:49.319 align:middle line:84% But to get anything useful out of our inductive hypothesis, 01:08:49.319 --> 01:08:52.340 align:middle line:84% we would need all degrees to be eight or less. 01:08:52.340 --> 01:08:53.880 align:middle line:90% There is a disconnect there. 01:08:53.880 --> 01:08:57.410 align:middle line:84% I don't know how to make that proof work. 01:08:57.410 --> 01:09:02.130 align:middle line:90% So let's see induction on what? 01:09:02.130 --> 01:09:04.819 align:middle line:90% 01:09:04.819 --> 01:09:09.590 align:middle line:84% We just said induction on d may be not such a good idea. 01:09:09.590 --> 01:09:15.439 align:middle line:84% Often when proving a statement about graphs by induction, 01:09:15.439 --> 01:09:18.390 align:middle line:84% there are two common strategies to try. 01:09:18.390 --> 01:09:21.770 align:middle line:84% You can try inducting on the number of vertices 01:09:21.770 --> 01:09:24.359 align:middle line:84% or inducting on the number of edges. 01:09:24.359 --> 01:09:27.020 align:middle line:84% I'll explain what that means in just a second. 01:09:27.020 --> 01:09:29.000 align:middle line:84% But these are going to be your best friends. 01:09:29.000 --> 01:09:32.390 align:middle line:84% If you're trying to prove a thing about graphs by induction, 01:09:32.390 --> 01:09:34.220 align:middle line:84% first, try proving it by induction 01:09:34.220 --> 01:09:35.640 align:middle line:90% on the number of vertices. 01:09:35.640 --> 01:09:38.580 align:middle line:84% Then try proving it by induction on the number of edges. 01:09:38.580 --> 01:09:41.729 align:middle line:84% If both of those don't work, I don't know. 01:09:41.729 --> 01:09:44.399 align:middle line:90% Cry. 01:09:44.399 --> 01:09:46.760 align:middle line:90% That's what I would do. 01:09:46.760 --> 01:09:49.229 align:middle line:90% But let's try this strategy. 01:09:49.229 --> 01:09:52.140 align:middle line:84% Let's induct on the number of vertices in our graph. 01:09:52.140 --> 01:09:55.010 align:middle line:90% And here's what that means. 01:09:55.010 --> 01:10:02.570 align:middle line:84% So we're going to let p of n be the statement that all 01:10:02.570 --> 01:10:09.860 align:middle line:90% graphs with n vertices-- 01:10:09.860 --> 01:10:12.626 align:middle line:84% so all graphs with exactly n vertices-- 01:10:12.626 --> 01:10:15.140 align:middle line:90% 01:10:15.140 --> 01:10:28.970 align:middle line:84% and max degree at most d result in at most d 01:10:28.970 --> 01:10:32.820 align:middle line:84% plus 1 colors from the greedy algorithm. 01:10:32.820 --> 01:10:40.040 align:middle line:90% 01:10:40.040 --> 01:10:41.930 align:middle line:84% So this is what we mean by we're inducting 01:10:41.930 --> 01:10:43.170 align:middle line:90% on the number of vertices. 01:10:43.170 --> 01:10:45.890 align:middle line:84% Because n, the variable we're going to be inducting on, 01:10:45.890 --> 01:10:49.820 align:middle line:84% is measuring the number of vertices of our graph. 01:10:49.820 --> 01:10:53.630 align:middle line:84% So if we think about induction as proving things 01:10:53.630 --> 01:10:56.990 align:middle line:84% one step at a time, we're going to first prove something 01:10:56.990 --> 01:10:59.250 align:middle line:84% about all graphs with one vertex, 01:10:59.250 --> 01:11:01.290 align:middle line:84% then all graphs with two vertices, 01:11:01.290 --> 01:11:03.210 align:middle line:84% then all graphs with three vertices, 01:11:03.210 --> 01:11:05.130 align:middle line:84% then all graphs with four vertices. 01:11:05.130 --> 01:11:06.240 align:middle line:90% This is fun. 01:11:06.240 --> 01:11:07.920 align:middle line:90% It's 5, 6, 7. 01:11:07.920 --> 01:11:11.090 align:middle line:84% We're going to build up one size at a time. 01:11:11.090 --> 01:11:14.600 align:middle line:84% And that's how our induction is going to go. 01:11:14.600 --> 01:11:18.440 align:middle line:84% All graphs with n vertices is the key phrase here 01:11:18.440 --> 01:11:21.970 align:middle line:84% that makes this inducting on the number of vertices. 01:11:21.970 --> 01:11:23.970 align:middle line:84% If you want to do induct on the number of edges, 01:11:23.970 --> 01:11:28.340 align:middle line:84% you would say all graphs with n edges or something like that. 01:11:28.340 --> 01:11:29.010 align:middle line:90% All right? 01:11:29.010 --> 01:11:30.450 align:middle line:90% Does this setup make sense? 01:11:30.450 --> 01:11:33.680 align:middle line:90% So we've defined Pn here. 01:11:33.680 --> 01:11:53.020 align:middle line:84% Our goal-- we want to prove for all n greater or equal 1 P of n 01:11:53.020 --> 01:11:54.970 align:middle line:90% is true. 01:11:54.970 --> 01:11:56.950 align:middle line:90% Yeah? 01:11:56.950 --> 01:12:02.290 align:middle line:84% So for any number of vertices, all graphs 01:12:02.290 --> 01:12:06.460 align:middle line:84% with that many vertices do the thing. 01:12:06.460 --> 01:12:11.050 align:middle line:84% So together, all of these Pn's, starting at P1 and going upward, 01:12:11.050 --> 01:12:15.640 align:middle line:90% include all finite graphs. 01:12:15.640 --> 01:12:18.220 align:middle line:90% Does that make sense? 01:12:18.220 --> 01:12:20.072 align:middle line:84% I failed to mention earlier, but most 01:12:20.072 --> 01:12:22.030 align:middle line:84% of the graphs we're talking about in this class 01:12:22.030 --> 01:12:23.330 align:middle line:90% are going to be finite. 01:12:23.330 --> 01:12:27.550 align:middle line:84% So if I don't specify, go ahead and assume it's finite. 01:12:27.550 --> 01:12:29.657 align:middle line:84% So this will cover all of the finite graphs. 01:12:29.657 --> 01:12:31.490 align:middle line:84% So it'll cover all the graphs we care about. 01:12:31.490 --> 01:12:33.460 align:middle line:84% And we'll end up proving our theorem. 01:12:33.460 --> 01:12:37.960 align:middle line:84% All right, so let's actually prove our theorem. 01:12:37.960 --> 01:12:42.190 align:middle line:90% Base case P of 1. 01:12:42.190 --> 01:12:44.080 align:middle line:84% Well, if there's only one vertex, 01:12:44.080 --> 01:12:46.095 align:middle line:84% we know what the greedy coloring is going to do. 01:12:46.095 --> 01:12:48.820 align:middle line:90% 01:12:48.820 --> 01:12:53.080 align:middle line:84% Well, let's see, max degree is zero 01:12:53.080 --> 01:12:55.900 align:middle line:84% because there are no edges if there's only one vertex. 01:12:55.900 --> 01:13:04.360 align:middle line:90% Greedy coloring uses one color. 01:13:04.360 --> 01:13:09.370 align:middle line:84% One is 0 plus 1, which is what was promised. 01:13:09.370 --> 01:13:11.560 align:middle line:90% So P of 1 is fine. 01:13:11.560 --> 01:13:21.430 align:middle line:84% Let's do the inductive step, P of n-- so here, assume P of n. 01:13:21.430 --> 01:13:24.340 align:middle line:90% We want to show P of n plus 1. 01:13:24.340 --> 01:13:29.260 align:middle line:84% And I really want to emphasize one point right here. 01:13:29.260 --> 01:13:31.390 align:middle line:84% Very often in this class, students 01:13:31.390 --> 01:13:36.220 align:middle line:84% come away from the next couple of lectures with the belief 01:13:36.220 --> 01:13:39.680 align:middle line:84% that induction on graphs is different. 01:13:39.680 --> 01:13:41.120 align:middle line:90% It follows different rules. 01:13:41.120 --> 01:13:42.200 align:middle line:90% It's backwards. 01:13:42.200 --> 01:13:45.940 align:middle line:84% We're going from n plus 1 to n or something like that. 01:13:45.940 --> 01:13:48.370 align:middle line:84% And I really want to emphasize induction 01:13:48.370 --> 01:13:51.100 align:middle line:84% on graphs is using regular induction in the way we're 01:13:51.100 --> 01:13:52.150 align:middle line:90% used to. 01:13:52.150 --> 01:13:54.790 align:middle line:84% We're not changing the rules here. 01:13:54.790 --> 01:14:00.160 align:middle line:84% What changes is that what it means to prove P of n 01:14:00.160 --> 01:14:02.680 align:middle line:84% implies P of n plus 1 is a little different 01:14:02.680 --> 01:14:03.950 align:middle line:90% from what we're used to. 01:14:03.950 --> 01:14:06.860 align:middle line:84% We're still proving that P of n implies P of n plus 1, 01:14:06.860 --> 01:14:09.340 align:middle line:84% because that's what induction tells us to do. 01:14:09.340 --> 01:14:13.150 align:middle line:84% But if we use our proof outlining techniques 01:14:13.150 --> 01:14:15.100 align:middle line:84% from the first couple lectures, and really 01:14:15.100 --> 01:14:17.350 align:middle line:84% unpack what it means to prove that this implies 01:14:17.350 --> 01:14:20.740 align:middle line:84% that, what it tells you to do is often a little 01:14:20.740 --> 01:14:24.010 align:middle line:84% bit different from what your intuition might tell you 01:14:24.010 --> 01:14:25.720 align:middle line:90% you're supposed to do. 01:14:25.720 --> 01:14:30.990 align:middle line:84% So it's doubly important when inducting on graphs 01:14:30.990 --> 01:14:34.270 align:middle line:84% to lean heavily on our proof outlining skills. 01:14:34.270 --> 01:14:36.870 align:middle line:84% Make sure that we're setting up our proof correctly 01:14:36.870 --> 01:14:40.260 align:middle line:84% to prove Pn implies Pn plus 1, and not something 01:14:40.260 --> 01:14:44.340 align:middle line:84% that we just think means Pn implies Pn plus 1. 01:14:44.340 --> 01:14:45.850 align:middle line:90% So let's do that now. 01:14:45.850 --> 01:14:46.960 align:middle line:90% We're assuming P of n. 01:14:46.960 --> 01:14:55.110 align:middle line:84% In other words, we're assuming all n vertex 01:14:55.110 --> 01:15:05.600 align:middle line:84% graphs with min degree at most d produce at most d plus 1 colors. 01:15:05.600 --> 01:15:08.670 align:middle line:90% 01:15:08.670 --> 01:15:09.580 align:middle line:90% That's P of n. 01:15:09.580 --> 01:15:11.460 align:middle line:90% We're assuming P of n. 01:15:11.460 --> 01:15:20.130 align:middle line:84% And we want to show that all n plus 1 vertex 01:15:20.130 --> 01:15:32.640 align:middle line:84% graphs with min degree at most d produce at most d plus 1 colors. 01:15:32.640 --> 01:15:34.950 align:middle line:84% All I've done is write out what Pn and Pn plus 1 01:15:34.950 --> 01:15:36.330 align:middle line:90% mean in context. 01:15:36.330 --> 01:15:38.340 align:middle line:90% With me so far? 01:15:38.340 --> 01:15:41.280 align:middle line:84% All right, and here's where the confusing bit 01:15:41.280 --> 01:15:44.910 align:middle line:84% happens, or at least the non-intuitive bit. 01:15:44.910 --> 01:15:49.710 align:middle line:84% Our task is we want to prove that all n plus 1 vertex graphs 01:15:49.710 --> 01:15:51.750 align:middle line:90% do some stuff. 01:15:51.750 --> 01:15:55.635 align:middle line:84% So the theorem we're trying to prove is a for all. 01:15:55.635 --> 01:15:58.920 align:middle line:84% All n plus 1 vertex graphs do something. 01:15:58.920 --> 01:16:01.290 align:middle line:84% But we know how to prove a for all. 01:16:01.290 --> 01:16:04.260 align:middle line:84% We start with a generic thing in that set, 01:16:04.260 --> 01:16:07.770 align:middle line:84% and we prove the conclusion for that thing. 01:16:07.770 --> 01:16:10.372 align:middle line:84% So this for all immediately tells us 01:16:10.372 --> 01:16:12.330 align:middle line:84% that the next line of our proof should probably 01:16:12.330 --> 01:16:13.445 align:middle line:90% look something like this. 01:16:13.445 --> 01:16:16.200 align:middle line:90% 01:16:16.200 --> 01:16:29.675 align:middle line:84% Suppose G is an arbitrary n plus 1 vertex graph. 01:16:29.675 --> 01:16:33.990 align:middle line:90% 01:16:33.990 --> 01:16:35.610 align:middle line:90% Yes? 01:16:35.610 --> 01:16:38.310 align:middle line:84% AUDIENCE: For writing out the meanings of P of n 01:16:38.310 --> 01:16:41.460 align:middle line:84% and P of n plus 1, did you mean to say max degree instead 01:16:41.460 --> 01:16:42.278 align:middle line:90% of min degree? 01:16:42.278 --> 01:16:43.320 align:middle line:90% ZACHARY ABEL: Absolutely. 01:16:43.320 --> 01:16:45.610 align:middle line:84% I meant to say max degree instead of min degree. 01:16:45.610 --> 01:16:46.600 align:middle line:90% That is a typo. 01:16:46.600 --> 01:16:48.420 align:middle line:90% Thank you so much. 01:16:48.420 --> 01:16:51.238 align:middle line:84% I guess it's a write-o because I'm not typing. 01:16:51.238 --> 01:16:52.030 align:middle line:90% AUDIENCE: For both? 01:16:52.030 --> 01:16:52.380 align:middle line:90% ZACHARY ABEL: For both, yes. 01:16:52.380 --> 01:16:53.090 align:middle line:90% Thank you. 01:16:53.090 --> 01:16:57.570 align:middle line:90% 01:16:57.570 --> 01:16:59.310 align:middle line:90% Thank you for catching that. 01:16:59.310 --> 01:17:01.470 align:middle line:84% All right, is everyone OK with the fact 01:17:01.470 --> 01:17:04.590 align:middle line:84% that this for all, all n plus 1 vertex graphs 01:17:04.590 --> 01:17:08.220 align:middle line:84% tells us that this next line should start with suppose G is 01:17:08.220 --> 01:17:10.860 align:middle line:90% an n plus 1 vertex graph? 01:17:10.860 --> 01:17:14.910 align:middle line:84% Because this is the part that many students get wrong. 01:17:14.910 --> 01:17:19.540 align:middle line:84% They will often say suppose G is an arbitrary n vertex graph. 01:17:19.540 --> 01:17:21.690 align:middle line:84% And then they go on to try to prove stuff 01:17:21.690 --> 01:17:23.107 align:middle line:84% about some n plus 1 vertex graphs, 01:17:23.107 --> 01:17:24.857 align:middle line:84% but they haven't shown that they're really 01:17:24.857 --> 01:17:26.010 align:middle line:90% getting all of them. 01:17:26.010 --> 01:17:27.770 align:middle line:84% It's this for all that tells us we 01:17:27.770 --> 01:17:31.100 align:middle line:84% should start like this and no other way. 01:17:31.100 --> 01:17:34.400 align:middle line:84% All right, so what do we do next? 01:17:34.400 --> 01:17:44.210 align:middle line:84% So vertices V1, Vn, Vn plus 1, so we have whatever vertex order 01:17:44.210 --> 01:17:45.360 align:middle line:90% we're using. 01:17:45.360 --> 01:17:59.360 align:middle line:84% Now, the greedy algorithm will color V1 through Vn as if Vn 01:17:59.360 --> 01:18:07.840 align:middle line:84% plus 1 and its edges didn't exist. 01:18:07.840 --> 01:18:10.400 align:middle line:90% 01:18:10.400 --> 01:18:12.500 align:middle line:84% The algorithm never looks at edges 01:18:12.500 --> 01:18:16.400 align:middle line:84% that connect to vertices that haven't been colored yet. 01:18:16.400 --> 01:18:18.930 align:middle line:84% So as far as the greedy algorithm is concerned, 01:18:18.930 --> 01:18:22.670 align:middle line:84% coloring vertices V1 through Vn, so all but one 01:18:22.670 --> 01:18:25.370 align:middle line:84% of the vertices in our graph, is going 01:18:25.370 --> 01:18:29.450 align:middle line:84% to look just like coloring an n vertex graph. 01:18:29.450 --> 01:18:30.140 align:middle line:90% Yeah? 01:18:30.140 --> 01:18:41.450 align:middle line:84% So this, what's called a sub-graph on V1 01:18:41.450 --> 01:18:44.150 align:middle line:84% through Vn, if we take just the first n vertices, 01:18:44.150 --> 01:18:49.460 align:middle line:84% make a sub-graph of just its vertices as our vertex set 01:18:49.460 --> 01:18:51.710 align:middle line:84% and the edges between them as our edge 01:18:51.710 --> 01:19:04.435 align:middle line:84% set with all their edges, this sub-graph also has max degree d. 01:19:04.435 --> 01:19:07.700 align:middle line:90% 01:19:07.700 --> 01:19:08.295 align:middle line:90% Oh, sorry. 01:19:08.295 --> 01:19:13.835 align:middle line:90% 01:19:13.835 --> 01:19:15.710 align:middle line:84% We're supposed to prove something about all n 01:19:15.710 --> 01:19:18.260 align:middle line:84% plus 1 vertex graphs with max degree d. 01:19:18.260 --> 01:19:22.862 align:middle line:84% So assume G is some n plus 1 vertex graph with max degree d. 01:19:22.862 --> 01:19:25.070 align:middle line:84% Then, when we look at the sub-graph, where we're only 01:19:25.070 --> 01:19:27.840 align:middle line:84% taking some of the vertices and some of the edges, 01:19:27.840 --> 01:19:31.130 align:middle line:84% well, degree can't go up when you do that. 01:19:31.130 --> 01:19:37.030 align:middle line:84% So this sub-graph also has max degree at most d. 01:19:37.030 --> 01:19:40.608 align:middle line:90% 01:19:40.608 --> 01:19:41.108 align:middle line:90% OK. 01:19:41.108 --> 01:19:44.852 align:middle line:90% 01:19:44.852 --> 01:19:51.320 align:middle line:84% By P of n, because we assumed P of n, 01:19:51.320 --> 01:19:57.325 align:middle line:84% these n vertices will use at most d plus 1 colors. 01:19:57.325 --> 01:20:01.676 align:middle line:90% 01:20:01.676 --> 01:20:04.950 align:middle line:84% We have an n vertex graph that we're talking about. 01:20:04.950 --> 01:20:06.690 align:middle line:84% We know it satisfies the property, 01:20:06.690 --> 01:20:09.120 align:middle line:84% so we get the conclusion because of our assumption. 01:20:09.120 --> 01:20:11.840 align:middle line:84% So it uses at most d plus 1 colors. 01:20:11.840 --> 01:20:13.520 align:middle line:84% So now, all we need to talk about 01:20:13.520 --> 01:20:15.596 align:middle line:84% is the last step of the greedy algorithm. 01:20:15.596 --> 01:20:18.130 align:middle line:90% 01:20:18.130 --> 01:20:22.930 align:middle line:90% Last step for Vn plus 1. 01:20:22.930 --> 01:20:26.230 align:middle line:84% Well Vn plus 1 has degree at most d. 01:20:26.230 --> 01:20:29.890 align:middle line:84% So it has at most d colors that conflict. 01:20:29.890 --> 01:20:36.550 align:middle line:84% So one of the first d plus 1 colors will be sufficient. 01:20:36.550 --> 01:20:43.150 align:middle line:84% Uses 1 through d plus 1 uses one of these colors 01:20:43.150 --> 01:20:46.030 align:middle line:84% because you can only have d conflicts because this 01:20:46.030 --> 01:20:48.250 align:middle line:90% has degree at most d. 01:20:48.250 --> 01:20:52.780 align:middle line:84% Therefore, the last color also uses one of the first d plus 1 01:20:52.780 --> 01:20:53.330 align:middle line:90% colors. 01:20:53.330 --> 01:20:55.490 align:middle line:84% So the whole coloring uses that many colors. 01:20:55.490 --> 01:20:58.810 align:middle line:90% And we're done with the proof. 01:20:58.810 --> 01:21:01.510 align:middle line:90% Did that make sense? 01:21:01.510 --> 01:21:03.375 align:middle line:84% If it didn't, come up, talk to me afterward. 01:21:03.375 --> 01:21:05.000 align:middle line:84% I'm happy to stay and answer questions. 01:21:05.000 --> 01:21:05.870 align:middle line:90% Thank you so much. 01:21:05.870 --> 01:21:07.740 align:middle line:90% We'll see you on Thursday. 01:21:07.740 --> 01:21:15.000 align:middle line:90%