WEBVTT 00:00:00.000 --> 00:00:05.335 align:middle line:84% [SQUEAKING] [RUSTLING] [CLICKING] 00:00:05.335 --> 00:00:12.607 align:middle line:90% 00:00:12.607 --> 00:00:14.440 align:middle line:84% PETER SHOR: But the first thing I want to do 00:00:14.440 --> 00:00:19.120 align:middle line:84% is explain the Cayley tree theorem. 00:00:19.120 --> 00:00:45.160 align:middle line:84% And so the question is, how many labelings of a tree on n 00:00:45.160 --> 00:00:47.440 align:middle line:90% vertices? 00:00:47.440 --> 00:00:50.840 align:middle line:84% And what I want to do is I want to give you 00:00:50.840 --> 00:00:53.240 align:middle line:84% the first few elements-- the first few numbers 00:00:53.240 --> 00:00:58.280 align:middle line:84% in this sequence and let you guess the answer. 00:00:58.280 --> 00:01:03.270 align:middle line:84% So the first one is a tree on two vertices. 00:01:03.270 --> 00:01:11.090 align:middle line:84% And if you want to label it with 1 and 2, 00:01:11.090 --> 00:01:14.630 align:middle line:84% there's only really one way of doing it, which is that. 00:01:14.630 --> 00:01:22.558 align:middle line:84% And so that's n equals 2, n equals 3, 1, 2, 00:01:22.558 --> 00:01:33.230 align:middle line:84% 3, which is clearly the same as 1, 3, 2, because these trees-- 00:01:33.230 --> 00:01:43.950 align:middle line:90% actually, 2, 1, 3, and 3, 1, 2. 00:01:43.950 --> 00:01:49.310 align:middle line:84% So there's one way of doing this, three ways of doing this. 00:01:49.310 --> 00:01:50.490 align:middle line:90% N equals 4. 00:01:50.490 --> 00:01:54.790 align:middle line:90% 00:01:54.790 --> 00:01:59.890 align:middle line:84% OK, this time I'm not going to list them all out because there 00:01:59.890 --> 00:02:01.030 align:middle line:90% are too many to count. 00:02:01.030 --> 00:02:10.190 align:middle line:84% But I'm going to list them in the form of the tree. 00:02:10.190 --> 00:02:13.130 align:middle line:84% So how many ways are they doing it? 00:02:13.130 --> 00:02:16.090 align:middle line:84% Are there labeling a tree which is a path? 00:02:16.090 --> 00:02:17.190 align:middle line:90% Can anyone tell me? 00:02:17.190 --> 00:02:23.770 align:middle line:90% 00:02:23.770 --> 00:02:25.690 align:middle line:90% Yeah? 00:02:25.690 --> 00:02:28.350 align:middle line:84% AUDIENCE: Maybe like 4 choose 2 to choose the two middle ones. 00:02:28.350 --> 00:02:30.530 align:middle line:90% And then 2 times 2. 00:02:30.530 --> 00:02:32.910 align:middle line:84% PETER SHOR: Yeah, 4 choose 2 times 2, 00:02:32.910 --> 00:02:37.010 align:middle line:84% which is 12, because you get to choose 00:02:37.010 --> 00:02:39.450 align:middle line:90% which two are the middle ones. 00:02:39.450 --> 00:02:47.570 align:middle line:84% And then there are two ways of completing that. 00:02:47.570 --> 00:02:52.890 align:middle line:84% And how about a tree that looks like this? 00:02:52.890 --> 00:02:58.960 align:middle line:84% Wait, I drew one too many leaves on that. 00:02:58.960 --> 00:03:00.740 align:middle line:84% How many ways are there of doing this? 00:03:00.740 --> 00:03:08.640 align:middle line:90% 00:03:08.640 --> 00:03:11.200 align:middle line:84% 4, because the only thing that matters is the middle one. 00:03:11.200 --> 00:03:15.320 align:middle line:90% So 4 and this becomes 16. 00:03:15.320 --> 00:03:19.520 align:middle line:90% And let's do n equals 5. 00:03:19.520 --> 00:03:23.960 align:middle line:90% And there's one of these-- 00:03:23.960 --> 00:03:27.120 align:middle line:84% oops, I drew an extra leaf on that one, too. 00:03:27.120 --> 00:03:28.160 align:middle line:90% Sorry. 00:03:28.160 --> 00:03:31.080 align:middle line:90% There are five of these. 00:03:31.080 --> 00:03:41.600 align:middle line:84% And there are 5 factorial over 2 ways, 00:03:41.600 --> 00:03:45.680 align:middle line:84% because there are 5 factorial ways of labeling the vertices 00:03:45.680 --> 00:03:47.010 align:middle line:90% from left to right. 00:03:47.010 --> 00:03:50.600 align:middle line:90% 00:03:50.600 --> 00:03:54.040 align:middle line:84% But this tree is symmetric in left and right, 00:03:54.040 --> 00:03:59.500 align:middle line:84% so you're double counting if you do that, which is 60. 00:03:59.500 --> 00:04:06.300 align:middle line:84% And then there are again 5 factorial over 2 00:04:06.300 --> 00:04:17.579 align:middle line:84% because you can label these 1, 3, 2, 4, 5. 00:04:17.579 --> 00:04:20.579 align:middle line:84% But you can swap 5 and 4 and get the same tree. 00:04:20.579 --> 00:04:23.140 align:middle line:90% So there's an extra factor of 2. 00:04:23.140 --> 00:04:26.620 align:middle line:90% And this is 60. 00:04:26.620 --> 00:04:30.100 align:middle line:84% And you add these up, you get 125. 00:04:30.100 --> 00:04:31.790 align:middle line:90% So does anybody see a pattern? 00:04:31.790 --> 00:04:39.300 align:middle line:90% 00:04:39.300 --> 00:04:40.540 align:middle line:90% Yeah. 00:04:40.540 --> 00:04:44.780 align:middle line:84% AUDIENCE: Like you multiply by 3 then by 4 then by 5. 00:04:44.780 --> 00:04:47.592 align:middle line:84% Like I guess within the sequence. 00:04:47.592 --> 00:04:48.300 align:middle line:90% PETER SHOR: Yeah. 00:04:48.300 --> 00:04:50.260 align:middle line:90% Well, I mean yes. 00:04:50.260 --> 00:04:52.210 align:middle line:84% That's essentially the right idea. 00:04:52.210 --> 00:04:55.050 align:middle line:90% This is 3 to the 1. 00:04:55.050 --> 00:04:56.890 align:middle line:90% This is 2 to the 0. 00:04:56.890 --> 00:04:59.250 align:middle line:90% This is 4 squared. 00:04:59.250 --> 00:05:03.330 align:middle line:90% And this one is 5 cubed. 00:05:03.330 --> 00:05:06.970 align:middle line:84% So the Cayley tree theorem says there 00:05:06.970 --> 00:05:10.070 align:middle line:84% are n to the n minus 2 ways of doing this. 00:05:10.070 --> 00:05:20.570 align:middle line:90% 00:05:20.570 --> 00:05:24.650 align:middle line:84% So that's the statement of the Cayley tree theorem. 00:05:24.650 --> 00:05:29.250 align:middle line:84% And there are lots and lots of proofs. 00:05:29.250 --> 00:05:32.530 align:middle line:84% So one way you might think of proving 00:05:32.530 --> 00:05:36.250 align:middle line:84% it is to find a bijection between labeled trees 00:05:36.250 --> 00:05:42.090 align:middle line:84% and sequences of the numbers 1 through n of length n minus 2. 00:05:42.090 --> 00:05:43.550 align:middle line:90% And you can do it that way. 00:05:43.550 --> 00:05:47.250 align:middle line:84% But this bijection is really incredibly complicated 00:05:47.250 --> 00:05:50.150 align:middle line:90% and completely non-intuitive. 00:05:50.150 --> 00:05:53.270 align:middle line:84% So that's not really the right way to do it. 00:05:53.270 --> 00:05:58.070 align:middle line:84% So what I'm going to do is try to explain 00:05:58.070 --> 00:06:03.190 align:middle line:84% a way of doing this by counting these certain objects twice, 00:06:03.190 --> 00:06:06.150 align:middle line:84% or certain object in two different ways. 00:06:06.150 --> 00:06:07.820 align:middle line:90% And how do we do this? 00:06:07.820 --> 00:06:10.350 align:middle line:90% 00:06:10.350 --> 00:06:23.320 align:middle line:84% So let T sub n be number of labeled trees on n vertices. 00:06:23.320 --> 00:06:30.470 align:middle line:90% 00:06:30.470 --> 00:06:32.150 align:middle line:84% And the first thing we're going to do 00:06:32.150 --> 00:06:39.430 align:middle line:84% is change the thing to number of labeled rooted trees 00:06:39.430 --> 00:06:40.240 align:middle line:90% on n vertices. 00:06:40.240 --> 00:06:55.700 align:middle line:90% 00:06:55.700 --> 00:07:00.260 align:middle line:84% OK, so what do you do to get a rooted tree? 00:07:00.260 --> 00:07:09.940 align:middle line:90% 00:07:09.940 --> 00:07:20.080 align:middle line:84% Is a tree with a specified vertex, called the root. 00:07:20.080 --> 00:07:34.220 align:middle line:90% 00:07:34.220 --> 00:07:43.220 align:middle line:84% OK, so if T sub n is the number of labeled trees 00:07:43.220 --> 00:07:46.480 align:middle line:84% on n vertices, what is the number of labeled 00:07:46.480 --> 00:07:47.990 align:middle line:90% rooted trees on n vertices? 00:07:47.990 --> 00:07:56.520 align:middle line:90% 00:07:56.520 --> 00:08:00.720 align:middle line:84% OK, well, what you do is you start with a tree. 00:08:00.720 --> 00:08:04.960 align:middle line:90% Let's do this tree, 1, 3, 2. 00:08:04.960 --> 00:08:09.720 align:middle line:84% And now you pick a vertex to call the root. 00:08:09.720 --> 00:08:16.880 align:middle line:84% So in this case, you can do this is the root. 00:08:16.880 --> 00:08:21.720 align:middle line:90% And 3, 1, 2, this is the root. 00:08:21.720 --> 00:08:27.960 align:middle line:84% Or you could do 2, 3, 1, and this is the root. 00:08:27.960 --> 00:08:32.280 align:middle line:84% So to go from a labeled tree to a rooted tree, labeled 00:08:32.280 --> 00:08:35.120 align:middle line:84% rooted tree, we just pick one vertex. 00:08:35.120 --> 00:08:39.419 align:middle line:84% And so there are n times T sub n of these. 00:08:39.419 --> 00:08:44.270 align:middle line:90% 00:08:44.270 --> 00:08:55.600 align:middle line:84% And now what we're going to count-- 00:08:55.600 --> 00:08:59.350 align:middle line:90% 00:08:59.350 --> 00:09:11.550 align:middle line:84% what we count in two ways is labeled rooted trees 00:09:11.550 --> 00:09:16.820 align:middle line:90% with ordered edges. 00:09:16.820 --> 00:09:23.110 align:middle line:90% 00:09:23.110 --> 00:09:31.930 align:middle line:84% So let's pick the middle one there, 1, 3, 2. 00:09:31.930 --> 00:09:34.790 align:middle line:90% 00:09:34.790 --> 00:09:36.730 align:middle line:90% So we need to order the edges. 00:09:36.730 --> 00:09:40.230 align:middle line:84% And we can either order this one and then this one, and then 00:09:40.230 --> 00:09:42.210 align:middle line:90% this one and then this one. 00:09:42.210 --> 00:09:45.490 align:middle line:84% So I'm going to use letters to order 00:09:45.490 --> 00:09:49.010 align:middle line:84% the edges because otherwise we have too many numbers floating 00:09:49.010 --> 00:10:03.610 align:middle line:84% around and it gets confusing, so 3, 2, 1, A, B. 00:10:03.610 --> 00:10:08.170 align:middle line:90% So how many labeled trees-- 00:10:08.170 --> 00:10:12.530 align:middle line:84% in terms of T sub n, the number of labeled trees, 00:10:12.530 --> 00:10:14.210 align:middle line:84% how many labeled rooted trees would 00:10:14.210 --> 00:10:15.910 align:middle line:84% ordered edges are there going to be? 00:10:15.910 --> 00:10:20.730 align:middle line:90% 00:10:20.730 --> 00:10:22.550 align:middle line:90% So there are n minus 1 edges. 00:10:22.550 --> 00:10:25.230 align:middle line:84% How many ways are there of ordering n minus 1 edges? 00:10:25.230 --> 00:10:28.955 align:middle line:90% 00:10:28.955 --> 00:10:30.590 align:middle line:90% n minus 1 factorial. 00:10:30.590 --> 00:10:31.690 align:middle line:90% Very good. 00:10:31.690 --> 00:10:39.930 align:middle line:84% So this is T sub n times n times n minus 1 factorial. 00:10:39.930 --> 00:10:42.355 align:middle line:90% OK. 00:10:42.355 --> 00:10:49.920 align:middle line:84% So if we can count this in a way that tells us 00:10:49.920 --> 00:10:55.240 align:middle line:84% the answer is n to the n minus 2 times n times n 00:10:55.240 --> 00:10:59.390 align:middle line:84% minus 1 factorial, then we have proven the theorem. 00:10:59.390 --> 00:11:06.297 align:middle line:90% 00:11:06.297 --> 00:11:06.797 align:middle line:90% OK. 00:11:06.797 --> 00:11:10.576 align:middle line:90% 00:11:10.576 --> 00:11:19.040 align:middle line:84% So the second way of counting, lemma, 00:11:19.040 --> 00:11:35.440 align:middle line:84% there are n to the n minus 2 n factorial labeled rooted trees 00:11:35.440 --> 00:11:36.270 align:middle line:90% with ordered edges. 00:11:36.270 --> 00:11:54.460 align:middle line:90% 00:11:54.460 --> 00:11:56.700 align:middle line:90% Good? 00:11:56.700 --> 00:11:58.210 align:middle line:90% So how do we prove this lemma? 00:11:58.210 --> 00:12:00.780 align:middle line:90% 00:12:00.780 --> 00:12:03.860 align:middle line:90% What we will do-- 00:12:03.860 --> 00:12:04.570 align:middle line:90% proof method. 00:12:04.570 --> 00:12:12.340 align:middle line:90% 00:12:12.340 --> 00:12:24.020 align:middle line:84% Construct a labeled rooted tree, I'm 00:12:24.020 --> 00:12:36.770 align:middle line:84% going to say, via labeled rooted forests, 00:12:36.770 --> 00:12:38.470 align:middle line:90% by adding one edge at a time. 00:12:38.470 --> 00:12:55.170 align:middle line:90% 00:12:55.170 --> 00:13:01.330 align:middle line:84% So back here, we have this n minus 1 factorial, which 00:13:01.330 --> 00:13:03.050 align:middle line:90% is the order of the edges. 00:13:03.050 --> 00:13:05.010 align:middle line:84% And this order of the edges is just 00:13:05.010 --> 00:13:06.730 align:middle line:84% going to be the order that we add 00:13:06.730 --> 00:13:15.490 align:middle line:84% the edges to get to labeled rooted forests to get 00:13:15.490 --> 00:13:16.990 align:middle line:90% to the label rooted tree. 00:13:16.990 --> 00:13:24.998 align:middle line:90% 00:13:24.998 --> 00:13:27.190 align:middle line:84% And I'm going to do it by example. 00:13:27.190 --> 00:13:31.450 align:middle line:84% But-- well, basically because it's much easier 00:13:31.450 --> 00:13:33.530 align:middle line:84% to see by example than if I try to describe 00:13:33.530 --> 00:13:35.710 align:middle line:90% the process in words. 00:13:35.710 --> 00:13:42.750 align:middle line:90% So we start with a forest. 00:13:42.750 --> 00:13:44.570 align:middle line:84% I never told you what a forest is. 00:13:44.570 --> 00:13:47.630 align:middle line:84% A forest is a collection of trees, 00:13:47.630 --> 00:13:55.110 align:middle line:84% which I'm going to say that this makes perfect sense. 00:13:55.110 --> 00:14:06.650 align:middle line:84% So start with a forest with n trees with one vertex each. 00:14:06.650 --> 00:14:20.710 align:middle line:90% 00:14:20.710 --> 00:14:23.720 align:middle line:90% 1, 2, 3, 4, 5, 6. 00:14:23.720 --> 00:14:28.630 align:middle line:90% 00:14:28.630 --> 00:14:32.000 align:middle line:90% Choose one of these trees. 00:14:32.000 --> 00:14:41.140 align:middle line:90% 00:14:41.140 --> 00:14:42.980 align:middle line:84% And what we're going to do is we're 00:14:42.980 --> 00:14:54.620 align:middle line:84% going to add it to a vertex and add it on top 00:14:54.620 --> 00:15:03.250 align:middle line:90% of a vertex in another tree. 00:15:03.250 --> 00:15:08.900 align:middle line:90% 00:15:08.900 --> 00:15:09.960 align:middle line:90% And let's see-- 00:15:09.960 --> 00:15:45.720 align:middle line:90% 00:15:45.720 --> 00:15:59.600 align:middle line:84% So let's take one of these trees and stick it 00:15:59.600 --> 00:16:02.900 align:middle line:84% on top of a vertex in another tree. 00:16:02.900 --> 00:16:05.640 align:middle line:90% 00:16:05.640 --> 00:16:18.840 align:middle line:84% So this gives us 1, 2, 4, 5, 3, 6. 00:16:18.840 --> 00:16:22.440 align:middle line:84% And we want to label this with the letter A 00:16:22.440 --> 00:16:25.720 align:middle line:84% because it was the first thing we did. 00:16:25.720 --> 00:16:34.150 align:middle line:84% Now let's take a different tree and stick it 00:16:34.150 --> 00:16:40.270 align:middle line:84% on top of another vertex in a tree. 00:16:40.270 --> 00:16:51.150 align:middle line:90% And we get 1, 4, 5, 6, 2. 00:16:51.150 --> 00:16:53.470 align:middle line:90% And here this was 3. 00:16:53.470 --> 00:16:55.630 align:middle line:90% And this was a first step. 00:16:55.630 --> 00:16:59.150 align:middle line:90% And this was our second step. 00:16:59.150 --> 00:17:01.510 align:middle line:90% And we keep on doing this. 00:17:01.510 --> 00:17:07.109 align:middle line:84% And maybe the next step is we pick this tree 00:17:07.109 --> 00:17:12.790 align:middle line:84% and stick it on top of this vertex. 00:17:12.790 --> 00:17:31.850 align:middle line:84% And we have 1, 4, 5, 6, 2, 3, and B, A, and C. 00:17:31.850 --> 00:17:35.050 align:middle line:84% And we keep on doing this until we get-- 00:17:35.050 --> 00:17:38.670 align:middle line:84% until we used up all the trees except one. 00:17:38.670 --> 00:17:41.490 align:middle line:84% And we have a labeled rooted tree 00:17:41.490 --> 00:17:42.920 align:middle line:90% with an ordering on the edges. 00:17:42.920 --> 00:17:47.090 align:middle line:90% 00:17:47.090 --> 00:18:13.550 align:middle line:84% So how many ways are there of doing this? 00:18:13.550 --> 00:18:22.450 align:middle line:90% 00:18:22.450 --> 00:18:23.650 align:middle line:90% Oh gosh. 00:18:23.650 --> 00:18:29.630 align:middle line:84% This is one of these hard to erase chunks. 00:18:29.630 --> 00:18:37.580 align:middle line:90% 00:18:37.580 --> 00:18:38.080 align:middle line:90% OK. 00:18:38.080 --> 00:18:40.760 align:middle line:90% 00:18:40.760 --> 00:18:45.940 align:middle line:84% Well, how many configurations are there to start with? 00:18:45.940 --> 00:19:01.360 align:middle line:90% 00:19:01.360 --> 00:19:03.560 align:middle line:90% Anybody? 00:19:03.560 --> 00:19:10.760 align:middle line:84% How many labeled-- how many labeled forests with n tree 00:19:10.760 --> 00:19:12.580 align:middle line:90% with one vertex each are there? 00:19:12.580 --> 00:19:17.440 align:middle line:90% 00:19:17.440 --> 00:19:19.840 align:middle line:84% I can see nobody's answering it because this question is 00:19:19.840 --> 00:19:22.560 align:middle line:90% too easy. 00:19:22.560 --> 00:19:24.550 align:middle line:90% There's only one of them. 00:19:24.550 --> 00:19:27.630 align:middle line:90% There are n vertices. 00:19:27.630 --> 00:19:29.790 align:middle line:84% And they have the labels 1 through n. 00:19:29.790 --> 00:19:40.350 align:middle line:84% And since order doesn't matter, there's one of them, so one 00:19:40.350 --> 00:19:48.970 align:middle line:90% after K steps have-- 00:19:48.970 --> 00:19:51.750 align:middle line:90% 00:19:51.750 --> 00:19:58.790 align:middle line:90% well, each step we have n-- 00:19:58.790 --> 00:20:02.950 align:middle line:90% each step we reduce one tree. 00:20:02.950 --> 00:20:07.500 align:middle line:90% So we have n minus k trees. 00:20:07.500 --> 00:20:16.110 align:middle line:90% 00:20:16.110 --> 00:20:17.980 align:middle line:90% Take one node. 00:20:17.980 --> 00:20:24.130 align:middle line:90% 00:20:24.130 --> 00:20:26.900 align:middle line:90% Add a different tree. 00:20:26.900 --> 00:20:31.930 align:middle line:90% 00:20:31.930 --> 00:20:32.950 align:middle line:90% We take one node. 00:20:32.950 --> 00:20:35.420 align:middle line:84% We add a different tree on that node-- 00:20:35.420 --> 00:20:39.050 align:middle line:90% 00:20:39.050 --> 00:20:40.280 align:middle line:90% to that node. 00:20:40.280 --> 00:20:43.250 align:middle line:90% 00:20:43.250 --> 00:20:57.290 align:middle line:84% So, for example, here, we chose node number 5. 00:20:57.290 --> 00:21:01.050 align:middle line:84% And we added the fourth tree to that node. 00:21:01.050 --> 00:21:03.200 align:middle line:84% So how many ways are there of doing this step? 00:21:03.200 --> 00:21:13.370 align:middle line:90% 00:21:13.370 --> 00:21:16.010 align:middle line:84% How many ways are there of choosing a node? 00:21:16.010 --> 00:21:16.650 align:middle line:90% Anybody? 00:21:16.650 --> 00:21:24.230 align:middle line:90% 00:21:24.230 --> 00:21:24.730 align:middle line:90% Yes. 00:21:24.730 --> 00:21:30.000 align:middle line:90% 00:21:30.000 --> 00:21:32.800 align:middle line:90% How many nodes are there? 00:21:32.800 --> 00:21:33.640 align:middle line:90% n. 00:21:33.640 --> 00:21:35.480 align:middle line:84% And we can choose any one of them. 00:21:35.480 --> 00:21:38.360 align:middle line:84% So there are n ways of choosing the node. 00:21:38.360 --> 00:21:42.520 align:middle line:90% And now we have n minus k trees. 00:21:42.520 --> 00:21:45.080 align:middle line:84% And we want to choose a different tree 00:21:45.080 --> 00:21:47.040 align:middle line:90% to add on top of the node. 00:21:47.040 --> 00:21:51.940 align:middle line:84% So there are n minus k minus 1 trees we can add. 00:21:51.940 --> 00:21:55.600 align:middle line:90% 00:21:55.600 --> 00:22:04.860 align:middle line:84% So there's n times n minus k minus 1 ways to do this. 00:22:04.860 --> 00:22:09.720 align:middle line:90% 00:22:09.720 --> 00:22:12.030 align:middle line:90% So number of-- 00:22:12.030 --> 00:22:15.560 align:middle line:90% 00:22:15.560 --> 00:22:17.200 align:middle line:90% OK, what was it? 00:22:17.200 --> 00:22:33.060 align:middle line:84% Labeled rooted trees-- labeled rooted trees with edge ordering 00:22:33.060 --> 00:22:35.900 align:middle line:90% is the product. 00:22:35.900 --> 00:22:40.600 align:middle line:90% Well, so after k steps-- 00:22:40.600 --> 00:22:44.320 align:middle line:84% we want to start with how many ways there are after 0 steps. 00:22:44.320 --> 00:22:50.840 align:middle line:84% So 0 to-- well, there are n minus 1 steps altogether. 00:22:50.840 --> 00:22:53.080 align:middle line:90% So n minus 2, I believe. 00:22:53.080 --> 00:22:57.596 align:middle line:90% 00:22:57.596 --> 00:23:03.520 align:middle line:90% n, n minus k minus 1. 00:23:03.520 --> 00:23:22.770 align:middle line:90% 00:23:22.770 --> 00:23:30.890 align:middle line:84% So this is equal to pi from k equals 0 00:23:30.890 --> 00:23:38.770 align:middle line:84% to n minus 2 n times pi from k equals 0 to n minus 2 00:23:38.770 --> 00:23:45.290 align:middle line:84% of n minus k minus 1, which is equal to-- 00:23:45.290 --> 00:23:46.630 align:middle line:90% well, there are n-- 00:23:46.630 --> 00:23:54.210 align:middle line:90% 00:23:54.210 --> 00:23:57.130 align:middle line:90% there are n minus 1 steps. 00:23:57.130 --> 00:23:58.950 align:middle line:84% And each of these, you have a factor of n. 00:23:58.950 --> 00:24:02.130 align:middle line:84% So this is just n times n minus 1. 00:24:02.130 --> 00:24:17.450 align:middle line:84% And this should be n minus 1 factorial, which-- 00:24:17.450 --> 00:24:20.190 align:middle line:90% 00:24:20.190 --> 00:24:21.810 align:middle line:90% if I haven't erased it-- 00:24:21.810 --> 00:24:24.470 align:middle line:90% 00:24:24.470 --> 00:24:28.830 align:middle line:84% yeah, we want to show that there are n to the n minus 2n 00:24:28.830 --> 00:24:29.570 align:middle line:90% factorial. 00:24:29.570 --> 00:24:32.070 align:middle line:84% But that's the same as n to the n minus 1 times 00:24:32.070 --> 00:24:33.990 align:middle line:90% n minus 1 factorial. 00:24:33.990 --> 00:24:38.470 align:middle line:84% So this is the number of labeled rooted trees 00:24:38.470 --> 00:24:41.110 align:middle line:90% with an edge ordering. 00:24:41.110 --> 00:24:46.790 align:middle line:84% So this implies that T sub n times 00:24:46.790 --> 00:24:58.310 align:middle line:90% n times n minus 1 factorial-- 00:24:58.310 --> 00:25:02.670 align:middle line:84% so this is the number of labeled rooted trees on n node. 00:25:02.670 --> 00:25:05.910 align:middle line:84% And this is the number of edge orderings-- 00:25:05.910 --> 00:25:10.310 align:middle line:84% is equal to n to the n minus 1 times n 00:25:10.310 --> 00:25:19.580 align:middle line:84% minus 1 factorial, which shows you that T sub n equals minus 2 00:25:19.580 --> 00:25:20.980 align:middle line:90% to the n-- 00:25:20.980 --> 00:25:38.240 align:middle line:84% n to the n minus 2, sorry, which is what we wanted to prove. 00:25:38.240 --> 00:25:50.780 align:middle line:90% 00:25:50.780 --> 00:25:52.260 align:middle line:90% OK. 00:25:52.260 --> 00:25:54.420 align:middle line:90% Yes? 00:25:54.420 --> 00:25:58.340 align:middle line:84% AUDIENCE: When you do the n times n minus k minus 1 thing 00:25:58.340 --> 00:26:00.900 align:middle line:84% is if you pick a node and then isn't the second one just 00:26:00.900 --> 00:26:05.298 align:middle line:84% picking a tree that's not one of the trees you picked? 00:26:05.298 --> 00:26:06.340 align:middle line:90% PETER SHOR: That's right. 00:26:06.340 --> 00:26:08.382 align:middle line:90% Well, it's picking a tree-- 00:26:08.382 --> 00:26:10.840 align:middle line:84% no, because you can pick a tree that you've already picked. 00:26:10.840 --> 00:26:12.600 align:middle line:90% You just cannot pick a tree-- 00:26:12.600 --> 00:26:15.440 align:middle line:84% you cannot pick the tree that node is in. 00:26:15.440 --> 00:26:17.320 align:middle line:90% Yeah. 00:26:17.320 --> 00:26:22.600 align:middle line:84% So we pick a node here, and then we pick a different tree 00:26:22.600 --> 00:26:24.080 align:middle line:90% to put on top of it. 00:26:24.080 --> 00:26:26.310 align:middle line:90% And that's 5, 6, 2. 00:26:26.310 --> 00:26:29.880 align:middle line:84% AUDIENCE: But doesn't it matter how you connect them 00:26:29.880 --> 00:26:30.882 align:middle line:90% with the other tree? 00:26:30.882 --> 00:26:32.840 align:middle line:84% Because you're just picking like another tree-- 00:26:32.840 --> 00:26:34.257 align:middle line:84% PETER SHOR: Oh, we're always going 00:26:34.257 --> 00:26:38.960 align:middle line:84% to put the root of this tree on top of this node. 00:26:38.960 --> 00:26:43.880 align:middle line:84% And you can see that the process can 00:26:43.880 --> 00:26:48.080 align:middle line:84% be undone because if we have A, B, C, the way 00:26:48.080 --> 00:26:51.560 align:middle line:84% we undo this step is we find the last edge. 00:26:51.560 --> 00:26:53.360 align:middle line:90% And we cut it. 00:26:53.360 --> 00:26:57.240 align:middle line:84% And we have a tree on top of that edge. 00:26:57.240 --> 00:27:01.440 align:middle line:84% And we have the tree that's left when you remove this edge. 00:27:01.440 --> 00:27:04.240 align:middle line:90% So that undoes the step. 00:27:04.240 --> 00:27:07.920 align:middle line:84% So the whole process is, I want to say 00:27:07.920 --> 00:27:20.990 align:middle line:84% that there is a this function that takes you from trees on k-- 00:27:20.990 --> 00:27:23.590 align:middle line:90% n minus k nodes to trees-- 00:27:23.590 --> 00:27:27.670 align:middle line:84% or to forests of n minus k trees to forests of n 00:27:27.670 --> 00:27:30.050 align:middle line:84% minus k minus 1 trees is reversible. 00:27:30.050 --> 00:27:32.920 align:middle line:90% So it's a bijection. 00:27:32.920 --> 00:27:34.420 align:middle line:84% But, yeah, that was a good question. 00:27:34.420 --> 00:27:52.910 align:middle line:90% 00:27:52.910 --> 00:27:59.270 align:middle line:84% And last year, we actually used this theorem 00:27:59.270 --> 00:28:03.790 align:middle line:84% for one of our term paper projects. 00:28:03.790 --> 00:28:07.530 align:middle line:90% This year we're not. 00:28:07.530 --> 00:28:10.670 align:middle line:84% I wanted to tell you this at the beginning of the class, 00:28:10.670 --> 00:28:11.510 align:middle line:90% but I forgot. 00:28:11.510 --> 00:28:15.630 align:middle line:84% So what we're going to do is tonight, assuming all goes well, 00:28:15.630 --> 00:28:20.050 align:middle line:84% I will put up the term paper projects and their assignments 00:28:20.050 --> 00:28:21.250 align:middle line:90% to students. 00:28:21.250 --> 00:28:31.610 align:middle line:84% And tomorrow in recitation, the recitation leaders 00:28:31.610 --> 00:28:35.250 align:middle line:84% will discuss them a little bit, although most of the recitation 00:28:35.250 --> 00:28:39.130 align:middle line:84% tomorrow is going to be devoted to writing. 00:28:39.130 --> 00:28:46.720 align:middle line:84% So the term paper projects are hopefully coming out tonight. 00:28:46.720 --> 00:28:51.210 align:middle line:90% 00:28:51.210 --> 00:28:57.170 align:middle line:84% So now, we are going to switch topics and go 00:28:57.170 --> 00:28:59.150 align:middle line:90% to generating functions. 00:28:59.150 --> 00:29:03.870 align:middle line:90% 00:29:03.870 --> 00:29:04.370 align:middle line:90% OK. 00:29:04.370 --> 00:29:25.800 align:middle line:90% 00:29:25.800 --> 00:29:29.640 align:middle line:84% And generating functions are a very useful concept 00:29:29.640 --> 00:29:31.080 align:middle line:90% in mathematics. 00:29:31.080 --> 00:29:33.230 align:middle line:84% And you can use them to count things. 00:29:33.230 --> 00:29:35.840 align:middle line:90% 00:29:35.840 --> 00:29:38.360 align:middle line:84% I mean, there are proofs of the Cayley tree theorem 00:29:38.360 --> 00:29:40.080 align:middle line:90% that use generating functions. 00:29:40.080 --> 00:29:45.280 align:middle line:84% And there are proofs of probably nearly everything we've 00:29:45.280 --> 00:29:48.440 align:middle line:84% discussed in terms of counting that use generating functions, 00:29:48.440 --> 00:29:51.560 align:middle line:84% although often, the generating functions proofs 00:29:51.560 --> 00:29:55.383 align:middle line:84% are more difficult than the non-generating functions proofs. 00:29:55.383 --> 00:29:56.800 align:middle line:84% On the other hand, there are cases 00:29:56.800 --> 00:30:00.720 align:middle line:84% where generating functions are really the only way to do it. 00:30:00.720 --> 00:30:02.680 align:middle line:84% And the other advantage of generating functions 00:30:02.680 --> 00:30:04.180 align:middle line:90% is that they're mechanical. 00:30:04.180 --> 00:30:09.300 align:middle line:84% So given, say, combinatorial identity and you look at it 00:30:09.300 --> 00:30:15.060 align:middle line:84% and you say, what things can I count in two different ways 00:30:15.060 --> 00:30:20.440 align:middle line:84% to prove it, or something else, well, it's-- 00:30:20.440 --> 00:30:23.300 align:middle line:90% 00:30:23.300 --> 00:30:26.460 align:middle line:84% you know, you have to use some creativity to find this. 00:30:26.460 --> 00:30:32.860 align:middle line:84% But if you have a generating combinatorial identity 00:30:32.860 --> 00:30:36.260 align:middle line:84% with generating functions, often it's 00:30:36.260 --> 00:30:39.520 align:middle line:84% fairly straightforward to prove it using generating functions, 00:30:39.520 --> 00:30:43.600 align:middle line:84% although it involves actual algebra. 00:30:43.600 --> 00:30:51.000 align:middle line:84% So it's maybe more difficult. So why am I? 00:30:51.000 --> 00:31:00.700 align:middle line:90% 00:31:00.700 --> 00:31:07.570 align:middle line:84% Let a1-- well, a0, a1, a2, et cetera, 00:31:07.570 --> 00:31:08.660 align:middle line:90% be a sequence of numbers. 00:31:08.660 --> 00:31:21.850 align:middle line:90% 00:31:21.850 --> 00:31:44.210 align:middle line:84% The generating function associated with the sequence 00:31:44.210 --> 00:31:56.930 align:middle line:84% is F of x is equal to sum a sub i x to the i, i 00:31:56.930 --> 00:31:59.030 align:middle line:90% equals 0 to infinity. 00:31:59.030 --> 00:32:12.350 align:middle line:90% 00:32:12.350 --> 00:32:18.210 align:middle line:84% So this is an algebraic function. 00:32:18.210 --> 00:32:22.830 align:middle line:84% And that means you can use all the tools of algebraic functions 00:32:22.830 --> 00:32:24.550 align:middle line:90% to manipulate it. 00:32:24.550 --> 00:32:28.630 align:middle line:84% And this, we will find, is going to be very useful. 00:32:28.630 --> 00:32:34.750 align:middle line:84% Now, you can ask what does x mean here? 00:32:34.750 --> 00:32:36.770 align:middle line:84% And x doesn't really mean anything. 00:32:36.770 --> 00:32:44.350 align:middle line:84% It's just a tool for keeping track of what 00:32:44.350 --> 00:32:48.310 align:middle line:90% the i is on ai x to the i. 00:32:48.310 --> 00:32:58.500 align:middle line:84% So it's a way of keeping the i-th ai 00:32:58.500 --> 00:33:02.740 align:middle line:90% distinct from the j-th ai. 00:33:02.740 --> 00:33:05.090 align:middle line:84% So it usually doesn't really mean anything. 00:33:05.090 --> 00:33:11.580 align:middle line:90% 00:33:11.580 --> 00:33:24.300 align:middle line:84% So the example-- well, one example-- 00:33:24.300 --> 00:33:28.520 align:middle line:84% is we'll take the sequence of binomial coefficients. 00:33:28.520 --> 00:33:32.020 align:middle line:90% 00:33:32.020 --> 00:33:33.620 align:middle line:90% So what is this? 00:33:33.620 --> 00:33:36.640 align:middle line:90% f sub n-- oh, right-- 00:33:36.640 --> 00:33:46.340 align:middle line:90% 00:33:46.340 --> 00:33:50.120 align:middle line:90% a sub n is equal to k choose n. 00:33:50.120 --> 00:33:53.660 align:middle line:90% 00:33:53.660 --> 00:34:02.840 align:middle line:84% So the generating function for this, F of x is equal to sum. 00:34:02.840 --> 00:34:04.880 align:middle line:84% And here n is what we're summing over. 00:34:04.880 --> 00:34:08.400 align:middle line:90% n equals 0 through infinity. 00:34:08.400 --> 00:34:13.739 align:middle line:84% And-- no, this is 0 if n is bigger than k. 00:34:13.739 --> 00:34:26.520 align:middle line:84% So that's sum n equals 0 through K, a k choose n. 00:34:26.520 --> 00:34:34.719 align:middle line:84% But let me write an x to the n, k n choose n, x to the n. 00:34:34.719 --> 00:34:37.780 align:middle line:84% And you should know what this is. 00:34:37.780 --> 00:34:40.400 align:middle line:90% 00:34:40.400 --> 00:34:42.199 align:middle line:90% This is the binomial theorem. 00:34:42.199 --> 00:34:46.620 align:middle line:90% And it's just 1 plus x to the k. 00:34:46.620 --> 00:34:49.719 align:middle line:90% 00:34:49.719 --> 00:34:52.120 align:middle line:84% And one way of thinking about this 00:34:52.120 --> 00:35:05.140 align:middle line:84% is F of x equals 1 plus x, 1 plus x, and 1 plus x. 00:35:05.140 --> 00:35:09.402 align:middle line:90% 00:35:09.402 --> 00:35:10.980 align:middle line:90% So for the k-th term-- 00:35:10.980 --> 00:35:18.830 align:middle line:90% 00:35:18.830 --> 00:35:37.650 align:middle line:84% no, for n-th term, I'm sorry, choose n x's, k minus n 1's. 00:35:37.650 --> 00:35:38.150 align:middle line:90% Good? 00:35:38.150 --> 00:35:42.110 align:middle line:84% So we need to choose whether we'll use the x from this one, 00:35:42.110 --> 00:35:46.250 align:middle line:84% the x from this one, the x from this one, et cetera. 00:35:46.250 --> 00:35:48.590 align:middle line:90% So we're choosing k x's-- 00:35:48.590 --> 00:35:55.060 align:middle line:84% we're choosing n x's out of k things, which is just 00:35:55.060 --> 00:35:56.880 align:middle line:90% n choose k-- or k choose n. 00:35:56.880 --> 00:35:57.860 align:middle line:90% I'm sorry. 00:35:57.860 --> 00:36:01.560 align:middle line:84% And so that's why a sub n is k choose n. 00:36:01.560 --> 00:36:20.420 align:middle line:90% 00:36:20.420 --> 00:36:20.920 align:middle line:90% OK. 00:36:20.920 --> 00:36:24.920 align:middle line:90% 00:36:24.920 --> 00:36:58.680 align:middle line:84% So example 2, the a sub n is the probability of n heads 00:36:58.680 --> 00:37:12.640 align:middle line:84% when you flip a coin, a biased coin, k times. 00:37:12.640 --> 00:37:16.570 align:middle line:84% And we'll let p equal probability of heads. 00:37:16.570 --> 00:37:24.960 align:middle line:90% 00:37:24.960 --> 00:37:29.060 align:middle line:90% And q equals 1 minus p. 00:37:29.060 --> 00:37:38.720 align:middle line:90% 00:37:38.720 --> 00:37:50.725 align:middle line:84% So A of x is equal to summation a sub n x to the n is equal to k 00:37:50.725 --> 00:38:00.122 align:middle line:84% choose n, p to the k, q to the n minus k times x to the k-- 00:38:00.122 --> 00:38:01.150 align:middle line:90% x to the n. 00:38:01.150 --> 00:38:04.990 align:middle line:90% 00:38:04.990 --> 00:38:08.240 align:middle line:84% I keep mixing up my k's and my n's. 00:38:08.240 --> 00:38:13.350 align:middle line:90% 00:38:13.350 --> 00:38:15.750 align:middle line:90% Let me apologize for that. 00:38:15.750 --> 00:38:16.490 align:middle line:90% There we go. 00:38:16.490 --> 00:38:30.670 align:middle line:90% 00:38:30.670 --> 00:38:33.470 align:middle line:90% And this is equal to-- 00:38:33.470 --> 00:38:35.910 align:middle line:90% I forgot the sum here. 00:38:35.910 --> 00:38:47.890 align:middle line:84% This is equal to q plus px to the n, 00:38:47.890 --> 00:38:49.400 align:middle line:90% again, by the binomial theorem. 00:38:49.400 --> 00:38:55.210 align:middle line:90% 00:38:55.210 --> 00:39:00.690 align:middle line:84% So hopefully, I will have time to do a real example that 00:39:00.690 --> 00:39:02.610 align:middle line:84% shows you why generating functions are 00:39:02.610 --> 00:39:06.750 align:middle line:90% useful before the end of class. 00:39:06.750 --> 00:39:11.170 align:middle line:84% But if not, we'll get to it tomorrow, or not tomorrow, 00:39:11.170 --> 00:39:12.620 align:middle line:90% Thursday. 00:39:12.620 --> 00:39:13.120 align:middle line:90% OK. 00:39:13.120 --> 00:39:16.250 align:middle line:90% 00:39:16.250 --> 00:39:19.010 align:middle line:84% I want to say a few more general things 00:39:19.010 --> 00:39:22.150 align:middle line:90% about generating functions. 00:39:22.150 --> 00:39:24.970 align:middle line:90% 00:39:24.970 --> 00:39:31.380 align:middle line:84% Let A, B be disjoint classes of objects. 00:39:31.380 --> 00:39:44.920 align:middle line:90% 00:39:44.920 --> 00:39:54.210 align:middle line:84% And C equals A union B, where U is disjoint union. 00:39:54.210 --> 00:39:58.880 align:middle line:90% 00:39:58.880 --> 00:40:05.120 align:middle line:84% So that just says that A intersection B is the empty set. 00:40:05.120 --> 00:40:07.780 align:middle line:84% So I want to say the generating function for C of x. 00:40:07.780 --> 00:40:11.017 align:middle line:90% 00:40:11.017 --> 00:40:13.600 align:middle line:84% So what we're going to do is we want to express the generating 00:40:13.600 --> 00:40:16.680 align:middle line:84% function for C of x in terms of A 00:40:16.680 --> 00:40:20.800 align:middle line:84% of x and B of x is equal to A of-- 00:40:20.800 --> 00:40:30.720 align:middle line:84% well, it's going to be sum C sub n, x to the n. 00:40:30.720 --> 00:40:35.120 align:middle line:84% Well, C sub n is just A sub n plus B sub 00:40:35.120 --> 00:40:45.060 align:middle line:84% n is equal to sum A sub n plus B sub n, x to the n, 00:40:45.060 --> 00:40:46.700 align:middle line:84% which you can see is the generating 00:40:46.700 --> 00:40:49.680 align:middle line:90% function of A of x plus B of x. 00:40:49.680 --> 00:40:55.460 align:middle line:90% 00:40:55.460 --> 00:40:59.540 align:middle line:84% So the next one, we want to talk about multiplication 00:40:59.540 --> 00:41:01.140 align:middle line:90% of generating functions. 00:41:01.140 --> 00:41:04.640 align:middle line:84% This is actually, first, pretty straightforward. 00:41:04.640 --> 00:41:08.900 align:middle line:84% And second, is not really that useful 00:41:08.900 --> 00:41:10.720 align:middle line:84% when manipulating generating functions. 00:41:10.720 --> 00:41:14.860 align:middle line:84% Multiplying generating functions is incredibly useful. 00:41:14.860 --> 00:41:17.350 align:middle line:90% And we'll do an example. 00:41:17.350 --> 00:41:26.860 align:middle line:90% 00:41:26.860 --> 00:41:29.300 align:middle line:90% So let's do the example first. 00:41:29.300 --> 00:41:46.590 align:middle line:84% And then we will do the formal explanation. 00:41:46.590 --> 00:41:55.130 align:middle line:90% 00:41:55.130 --> 00:41:59.880 align:middle line:84% So let's say we have a 6-sided dice and an 8-sided dice. 00:41:59.880 --> 00:42:10.210 align:middle line:90% 00:42:10.210 --> 00:42:11.280 align:middle line:90% 8-sided dice. 00:42:11.280 --> 00:42:16.490 align:middle line:90% 00:42:16.490 --> 00:42:29.490 align:middle line:84% And we want to know what is the probability of rolling, 00:42:29.490 --> 00:42:37.630 align:middle line:84% OK, let's call it n, when we add up the numbers from the two die. 00:42:37.630 --> 00:42:44.470 align:middle line:90% 00:42:44.470 --> 00:42:49.950 align:middle line:84% And I should specify that the 6-sided dice is numbered 1 00:42:49.950 --> 00:42:53.330 align:middle line:84% through 6, and the 8-sided dice is numbered 1 through 8. 00:42:53.330 --> 00:43:04.310 align:middle line:90% 00:43:04.310 --> 00:43:16.730 align:middle line:84% So let C sub n is equal to number of ways of rolling n. 00:43:16.730 --> 00:43:24.870 align:middle line:90% 00:43:24.870 --> 00:43:30.110 align:middle line:90% Claim-- C sub n is equal to-- 00:43:30.110 --> 00:43:34.230 align:middle line:90% or rather C sub n-- 00:43:34.230 --> 00:43:39.940 align:middle line:84% C of x is equal to sum C sub n x to the n 00:43:39.940 --> 00:43:47.740 align:middle line:84% is equal to x plus x squared plus x cubed plus-- 00:43:47.740 --> 00:43:55.760 align:middle line:84% plus x to the 6th, x plus x squared plus x cubed plus dot, 00:43:55.760 --> 00:44:07.940 align:middle line:84% dot, dot plus x to the 8th equals x plus 2x squared plus 3x 00:44:07.940 --> 00:44:16.260 align:middle line:84% cubed plus dot, dot, dot, plus 5x to the 5th plus 6x 00:44:16.260 --> 00:44:20.540 align:middle line:90% to the 6th, plus 7x to the-- 00:44:20.540 --> 00:44:34.900 align:middle line:84% plus, sorry, 6x to the 7th, plus 6x to the 8th plus 6x to the 9th 00:44:34.900 --> 00:44:39.020 align:middle line:84% plus 5x to the 10th, plus dot, dot, dot. 00:44:39.020 --> 00:44:44.120 align:middle line:90% 00:44:44.120 --> 00:44:46.500 align:middle line:90% Plus x to the 14th. 00:44:46.500 --> 00:45:00.640 align:middle line:90% 00:45:00.640 --> 00:45:02.030 align:middle line:90% So how do we get that? 00:45:02.030 --> 00:45:23.680 align:middle line:90% 00:45:23.680 --> 00:45:25.360 align:middle line:84% So you can see from this that there 00:45:25.360 --> 00:45:27.960 align:middle line:90% are five ways of rolling a 10. 00:45:27.960 --> 00:45:33.860 align:middle line:90% You can roll a 6 and 4. 00:45:33.860 --> 00:45:43.830 align:middle line:84% Or you could roll a 5 and 5, a 4 and 6, a 3 and 7, 00:45:43.830 --> 00:45:46.710 align:middle line:90% and 2 and an 8. 00:45:46.710 --> 00:45:55.470 align:middle line:84% And this works because C sub n is 00:45:55.470 --> 00:46:02.780 align:middle line:90% equal to summation j equals 0. 00:46:02.780 --> 00:46:07.310 align:middle line:90% 00:46:07.310 --> 00:46:09.890 align:middle line:90% j equals 1 through n. 00:46:09.890 --> 00:46:27.810 align:middle line:90% 00:46:27.810 --> 00:46:28.310 align:middle line:90% OK. 00:46:28.310 --> 00:46:31.450 align:middle line:90% 00:46:31.450 --> 00:46:35.710 align:middle line:90% A sub j, B sub n minus j. 00:46:35.710 --> 00:46:38.570 align:middle line:90% 00:46:38.570 --> 00:46:42.450 align:middle line:84% So if you want to get an n, the first place has to be j. 00:46:42.450 --> 00:46:46.090 align:middle line:84% And the second one has to be n minus j. 00:46:46.090 --> 00:46:56.250 align:middle line:84% And A j equals 1, if 1 is less than j is less than 6, 00:46:56.250 --> 00:46:57.870 align:middle line:90% 0 otherwise. 00:46:57.870 --> 00:47:05.890 align:middle line:84% And B sub j equals 1 if 1 is less than j is less than 8, 00:47:05.890 --> 00:47:07.470 align:middle line:90% and 0 otherwise. 00:47:07.470 --> 00:47:16.490 align:middle line:90% 00:47:16.490 --> 00:47:18.490 align:middle line:90% Is this clear? 00:47:18.490 --> 00:47:22.010 align:middle line:84% I mean to get an n, you have to roll a j on the first dice 00:47:22.010 --> 00:47:24.070 align:middle line:84% and an n minus j on the second dice. 00:47:24.070 --> 00:47:26.170 align:middle line:84% And you can only roll a j on the first dice 00:47:26.170 --> 00:47:29.690 align:middle line:90% if j is between 1 and 6. 00:47:29.690 --> 00:47:33.320 align:middle line:84% And you can only roll it n minus j on the second die 00:47:33.320 --> 00:47:35.620 align:middle line:90% if j is between 1 and 8. 00:47:35.620 --> 00:47:39.240 align:middle line:90% 00:47:39.240 --> 00:47:53.000 align:middle line:84% So if A of x is equal to summation, 00:47:53.000 --> 00:47:58.720 align:middle line:90% i equals 1 through 6 x to the i. 00:47:58.720 --> 00:48:09.480 align:middle line:84% And B of x is equal to summation i equals 1 through 8 x to the i. 00:48:09.480 --> 00:48:10.750 align:middle line:90% That's not a very good 8. 00:48:10.750 --> 00:48:15.640 align:middle line:90% 00:48:15.640 --> 00:48:26.280 align:middle line:84% Then C of x equals of A of x, B of x, 00:48:26.280 --> 00:48:30.440 align:middle line:84% because this is the formula for multiplying polynomials. 00:48:30.440 --> 00:48:43.620 align:middle line:90% 00:48:43.620 --> 00:48:46.140 align:middle line:84% So we have the generating function for the first die 00:48:46.140 --> 00:48:48.940 align:middle line:84% and the generating function for the second die. 00:48:48.940 --> 00:48:52.820 align:middle line:84% And we get the generating function for the sum of the two 00:48:52.820 --> 00:48:54.630 align:middle line:90% dies by just multiplying them. 00:48:54.630 --> 00:49:20.460 align:middle line:90% 00:49:20.460 --> 00:49:23.820 align:middle line:90% And let's make this rigorous. 00:49:23.820 --> 00:49:31.710 align:middle line:84% Let A, B be two classes of objects. 00:49:31.710 --> 00:49:50.159 align:middle line:90% 00:49:50.159 --> 00:49:56.350 align:middle line:84% A of x and B of x be their generating functions. 00:49:56.350 --> 00:50:08.890 align:middle line:90% 00:50:08.890 --> 00:50:12.990 align:middle line:90% Then-- oh, well-- 00:50:12.990 --> 00:50:17.010 align:middle line:90% 00:50:17.010 --> 00:50:33.118 align:middle line:84% the Cartesian product, Product A cross B, 00:50:33.118 --> 00:50:42.870 align:middle line:84% is A, B, A and A, B and B. And we 00:50:42.870 --> 00:51:08.670 align:middle line:84% want to claim that size of A, B equals size of A plus size of B. 00:51:08.670 --> 00:51:27.540 align:middle line:84% Then there are the Cartesian product, A cross B, 00:51:27.540 --> 00:51:48.280 align:middle line:84% has generating function A of x, B of x equals C of x. 00:51:48.280 --> 00:52:25.880 align:middle line:90% 00:52:25.880 --> 00:52:46.280 align:middle line:84% Proof-- OK, C sub n equals summation k equals 0 through n, 00:52:46.280 --> 00:52:53.200 align:middle line:90% a sub k, b sub n minus k. 00:52:53.200 --> 00:52:57.440 align:middle line:84% So we want to look at the things of size n 00:52:57.440 --> 00:52:59.680 align:middle line:90% in the Cartesian product. 00:52:59.680 --> 00:53:09.480 align:middle line:84% Well, they're the things of size k in the first generating 00:53:09.480 --> 00:53:12.720 align:middle line:84% function and times the things of size n 00:53:12.720 --> 00:53:14.880 align:middle line:84% minus k in the second generating function. 00:53:14.880 --> 00:53:19.110 align:middle line:84% And we want to sum this over all k. 00:53:19.110 --> 00:53:25.150 align:middle line:84% And now, we just need to check that summation C 00:53:25.150 --> 00:53:32.270 align:middle line:84% sub n x to the n is equal to summation k equals 0 00:53:32.270 --> 00:53:38.750 align:middle line:84% through n, a sub k, b n minus k x to the n is 00:53:38.750 --> 00:53:43.310 align:middle line:90% equal to sum k equals 0-- 00:53:43.310 --> 00:53:50.710 align:middle line:84% or rather, n equals 0 to infinity. 00:53:50.710 --> 00:53:53.630 align:middle line:84% I forgot the second sum here-- n equals 0 00:53:53.630 --> 00:53:59.990 align:middle line:84% to infinity, k equals 0 to n, which is just sum k equals 0 00:53:59.990 --> 00:54:04.390 align:middle line:84% to infinity, a sub k, x to the k, 00:54:04.390 --> 00:54:16.580 align:middle line:84% summation j equals 0 to infinity, b sub j, x to the j, 00:54:16.580 --> 00:54:22.140 align:middle line:84% because when you multiply these, you group the things 00:54:22.140 --> 00:54:26.580 align:middle line:84% that have the same k plus j equals n. 00:54:26.580 --> 00:54:30.420 align:middle line:84% And that gives you this term, sum k 00:54:30.420 --> 00:54:34.580 align:middle line:84% equals 0 to n, A sub k, B sub n minus k. 00:54:34.580 --> 00:54:35.260 align:middle line:90% Yeah. 00:54:35.260 --> 00:54:35.927 align:middle line:90% AUDIENCE: Sorry. 00:54:35.927 --> 00:54:37.380 align:middle line:90% What does size mean? 00:54:37.380 --> 00:54:44.220 align:middle line:84% PETER SHOR: Size is just an arbitrary function 00:54:44.220 --> 00:54:56.580 align:middle line:84% on these objects that correspond to the n here in summation 00:54:56.580 --> 00:54:58.980 align:middle line:90% a sub k, x to the k. 00:54:58.980 --> 00:55:04.580 align:middle line:84% So when you have generating functions, 00:55:04.580 --> 00:55:10.340 align:middle line:84% you're grouping objects into things that go with x to the k. 00:55:10.340 --> 00:55:17.520 align:middle line:84% And those are what we're calling size. 00:55:17.520 --> 00:55:19.920 align:middle line:84% AUDIENCE: So is it the power of x? 00:55:19.920 --> 00:55:22.640 align:middle line:84% PETER SHOR: It's the power of x that 00:55:22.640 --> 00:55:24.020 align:middle line:90% is associated with the object. 00:55:24.020 --> 00:55:29.560 align:middle line:84% And I think we should look at an actual example. 00:55:29.560 --> 00:55:30.890 align:middle line:90% So this is a simple example. 00:55:30.890 --> 00:56:04.960 align:middle line:90% 00:56:04.960 --> 00:56:20.110 align:middle line:84% Example-- OK, suppose you have one penny-- 00:56:20.110 --> 00:56:27.320 align:middle line:84% no, 6 pennies, 1 nickel, 2 dimes. 00:56:27.320 --> 00:56:31.870 align:middle line:90% 00:56:31.870 --> 00:56:48.880 align:middle line:84% How many different ways can you make n cents? 00:56:48.880 --> 00:56:58.510 align:middle line:90% 00:56:58.510 --> 00:57:07.670 align:middle line:84% Claim-- the generating function, let's 00:57:07.670 --> 00:57:18.770 align:middle line:84% call it C of x is equal to sum c n, x to the n. 00:57:18.770 --> 00:57:24.290 align:middle line:84% So we'll let c sub n be the number of different ways, 00:57:24.290 --> 00:57:33.050 align:middle line:84% cn be number of ways to make n cents with 6 00:57:33.050 --> 00:57:36.930 align:middle line:90% pennies, a nickel, and 2 dimes. 00:57:36.930 --> 00:57:45.050 align:middle line:84% C of x is equal to 1 plus x plus x squared plus x cubed plus x 00:57:45.050 --> 00:57:53.410 align:middle line:84% 4th plus x 5th plus x 6th, 1 plus x to the fifth, 00:57:53.410 --> 00:57:58.100 align:middle line:84% and 1 plus x to the 10th plus x to the 20th. 00:57:58.100 --> 00:58:07.930 align:middle line:90% 00:58:07.930 --> 00:58:14.960 align:middle line:84% OK, so when you multiply these things together-- 00:58:14.960 --> 00:58:17.000 align:middle line:90% let's look at the x to the 20-- 00:58:17.000 --> 00:58:24.210 align:middle line:84% let's look at the x to the, I'm thinking, x to the 21st term. 00:58:24.210 --> 00:58:31.360 align:middle line:90% 00:58:31.360 --> 00:58:38.080 align:middle line:84% So it can be got from x to the 6th times x to the 5th times x 00:58:38.080 --> 00:58:38.690 align:middle line:90% to the 10th. 00:58:38.690 --> 00:58:45.120 align:middle line:90% 00:58:45.120 --> 00:58:50.830 align:middle line:84% And we can also get it from x times x to the 20th. 00:58:50.830 --> 00:58:54.840 align:middle line:90% 00:58:54.840 --> 00:58:57.600 align:middle line:84% And those are the only two ways you can get it. 00:58:57.600 --> 00:59:01.360 align:middle line:84% So there are two ways to make change-- 00:59:01.360 --> 00:59:15.420 align:middle line:90% two ways to pay $0.21. 00:59:15.420 --> 00:59:21.130 align:middle line:84% So here the size is just the number of cents. 00:59:21.130 --> 00:59:37.260 align:middle line:90% 00:59:37.260 --> 00:59:40.160 align:middle line:84% OK, let's do something a little bit more complicated. 00:59:40.160 --> 00:59:56.180 align:middle line:90% 00:59:56.180 --> 01:00:07.650 align:middle line:84% How many ways are there to pay, let's say, 01:00:07.650 --> 01:00:19.760 align:middle line:84% n cents in an arbitrary number of pennies, nickels, dimes. 01:00:19.760 --> 01:00:25.490 align:middle line:90% 01:00:25.490 --> 01:00:30.930 align:middle line:84% Well, I want to claim that it's 1 plus x plus x squared plus x 01:00:30.930 --> 01:00:36.210 align:middle line:84% cubed plus dot, dot, dot, times 1 plus x to the 5th plus x 01:00:36.210 --> 01:00:42.850 align:middle line:84% to the 10th plus x to the 15th, plus dot, dot, dot, 01:00:42.850 --> 01:00:47.530 align:middle line:84% times 1 plus x to the 10th, plus x to the 20th, 01:00:47.530 --> 01:00:53.530 align:middle line:84% plus x to the 30th, plus dot, dot, dot, 01:00:53.530 --> 01:01:00.170 align:middle line:84% which is equal to 1 over 1 minus x, 1 over 1 minus x to the 5th, 01:01:00.170 --> 01:01:02.720 align:middle line:84% times 1 over 1 minus x to the 10th. 01:01:02.720 --> 01:01:07.750 align:middle line:90% 01:01:07.750 --> 01:01:11.290 align:middle line:84% So cn x to the n is equal to that. 01:01:11.290 --> 01:01:26.190 align:middle line:90% 01:01:26.190 --> 01:01:33.190 align:middle line:84% OK, so now I want to give an example 01:01:33.190 --> 01:01:37.010 align:middle line:84% for how you would actually use a generating function. 01:01:37.010 --> 01:01:39.270 align:middle line:84% I mean, we have not gotten to the end 01:01:39.270 --> 01:01:44.295 align:middle line:84% of our overall description of generating functions. 01:01:44.295 --> 01:01:45.670 align:middle line:84% But I want to give you an example 01:01:45.670 --> 01:01:50.210 align:middle line:84% first of how you would actually calculate something with them. 01:01:50.210 --> 01:01:56.950 align:middle line:90% 01:01:56.950 --> 01:02:01.690 align:middle line:84% And I want to claim this is really pretty neat. 01:02:01.690 --> 01:02:13.460 align:middle line:90% 01:02:13.460 --> 01:02:23.895 align:middle line:84% What is sum k equals 0 through n, k n choose k. 01:02:23.895 --> 01:02:27.980 align:middle line:90% 01:02:27.980 --> 01:02:31.540 align:middle line:84% So the first thing is you can do it 01:02:31.540 --> 01:02:34.340 align:middle line:84% by counting something two different ways. 01:02:34.340 --> 01:02:38.240 align:middle line:84% And I want to do it with generating functions first. 01:02:38.240 --> 01:02:40.640 align:middle line:84% And if I have any time at the end of the class, 01:02:40.640 --> 01:02:44.500 align:middle line:84% I'll check it by using the technique of counting 01:02:44.500 --> 01:02:45.680 align:middle line:90% in two different ways. 01:02:45.680 --> 01:02:48.060 align:middle line:90% But I'm not sure I will. 01:02:48.060 --> 01:02:51.420 align:middle line:90% So what is it? 01:02:51.420 --> 01:03:02.360 align:middle line:84% k equals 0 through n, n choose k times x to the k is equal to-- 01:03:02.360 --> 01:03:04.220 align:middle line:84% does anybody want to tell me what this is? 01:03:04.220 --> 01:03:11.560 align:middle line:90% 01:03:11.560 --> 01:03:12.440 align:middle line:90% Yeah. 01:03:12.440 --> 01:03:15.820 align:middle line:84% AUDIENCE: Is it n times 2 to the n minus 1? 01:03:15.820 --> 01:03:18.720 align:middle line:90% 01:03:18.720 --> 01:03:22.080 align:middle line:84% PETER SHOR: No, because we want this as a function of something 01:03:22.080 --> 01:03:23.273 align:middle line:90% with x's in it. 01:03:23.273 --> 01:03:24.440 align:middle line:90% AUDIENCE: 1 plus x to the n. 01:03:24.440 --> 01:03:26.960 align:middle line:84% PETER SHOR: Yeah, 1 plus x to the n, that's right. 01:03:26.960 --> 01:03:29.460 align:middle line:90% 1 plus x to the n. 01:03:29.460 --> 01:03:31.360 align:middle line:90% Excellent. 01:03:31.360 --> 01:03:32.640 align:middle line:90% OK. 01:03:32.640 --> 01:03:36.040 align:middle line:90% Now-- and this is-- 01:03:36.040 --> 01:03:40.640 align:middle line:90% we should call this F of x. 01:03:40.640 --> 01:03:45.200 align:middle line:84% Now, how can we get this out of that? 01:03:45.200 --> 01:03:47.920 align:middle line:90% Well, let's differentiate. 01:03:47.920 --> 01:03:59.470 align:middle line:84% d/dx F of x is equal to sum k equals 0 through n, 01:03:59.470 --> 01:04:05.340 align:middle line:84% k n choose k, x to the k minus 1. 01:04:05.340 --> 01:04:09.430 align:middle line:90% 01:04:09.430 --> 01:04:11.810 align:middle line:90% Right? 01:04:11.810 --> 01:04:12.870 align:middle line:90% OK. 01:04:12.870 --> 01:04:15.550 align:middle line:84% And now, let's differentiate this. 01:04:15.550 --> 01:04:20.170 align:middle line:84% Well, that's just n times 1 plus x to the n minus 1. 01:04:20.170 --> 01:04:24.270 align:middle line:90% 01:04:24.270 --> 01:04:32.710 align:middle line:84% And now let's substitute x equals 1. 01:04:32.710 --> 01:04:40.590 align:middle line:84% Well, in this, we get sum k equals 0 through n, k n choose 01:04:40.590 --> 01:04:43.230 align:middle line:90% k. 01:04:43.230 --> 01:04:51.570 align:middle line:84% And in this, we get n times 2 to the n minus 1. 01:04:51.570 --> 01:04:56.083 align:middle line:90% 01:04:56.083 --> 01:04:56.583 align:middle line:90% OK. 01:04:56.583 --> 01:04:59.754 align:middle line:90% 01:04:59.754 --> 01:05:01.020 align:middle line:90% We have our answer. 01:05:01.020 --> 01:05:13.610 align:middle line:90% 01:05:13.610 --> 01:05:26.440 align:middle line:84% And I want to say that we kind of skipped a step here. 01:05:26.440 --> 01:05:33.650 align:middle line:90% 01:05:33.650 --> 01:05:54.680 align:middle line:84% Generating function, for k n choose k is sum k n 01:05:54.680 --> 01:05:56.920 align:middle line:90% choose k, x to the k. 01:05:56.920 --> 01:05:59.626 align:middle line:90% 01:05:59.626 --> 01:06:01.620 align:middle line:90% k equals 0 through n. 01:06:01.620 --> 01:06:04.560 align:middle line:90% 01:06:04.560 --> 01:06:12.040 align:middle line:84% And what we calculated here was just 01:06:12.040 --> 01:06:15.160 align:middle line:84% k n choose k, x to the k minus 1. 01:06:15.160 --> 01:06:22.620 align:middle line:84% So this is actually n x 1 minus x to the n minus 1. 01:06:22.620 --> 01:06:28.320 align:middle line:90% 01:06:28.320 --> 01:06:31.640 align:middle line:84% But since x, if we substitute in x equals 1, 01:06:31.640 --> 01:06:35.520 align:middle line:84% it doesn't matter if we have x to the k or x 01:06:35.520 --> 01:06:39.430 align:middle line:84% to the k minus 1 here, or if we have this extra-- 01:06:39.430 --> 01:06:42.360 align:middle line:90% 01:06:42.360 --> 01:06:47.600 align:middle line:84% sorry, plus-- or if we have this extra x here. 01:06:47.600 --> 01:06:52.980 align:middle line:84% It will-- if x is 1, they're the same. 01:06:52.980 --> 01:07:04.700 align:middle line:84% So to multiply the n-th term in our-- 01:07:04.700 --> 01:07:09.380 align:middle line:84% to multiply the k-th term in a generating function by k, 01:07:09.380 --> 01:07:14.500 align:middle line:84% we just differentiate and multiply by x. 01:07:14.500 --> 01:07:17.140 align:middle line:84% And this is a very useful trick, playing 01:07:17.140 --> 01:07:20.020 align:middle line:90% with generating functions. 01:07:20.020 --> 01:07:24.660 align:middle line:84% So how would you do this with counting two different ways? 01:07:24.660 --> 01:07:25.340 align:middle line:90% Anybody? 01:07:25.340 --> 01:07:33.860 align:middle line:90% 01:07:33.860 --> 01:07:34.950 align:middle line:90% Yes. 01:07:34.950 --> 01:07:37.280 align:middle line:84% AUDIENCE: Is that-- I'm just like, 01:07:37.280 --> 01:07:42.420 align:middle line:84% what's the question is it the line above it or the-- 01:07:42.420 --> 01:07:43.700 align:middle line:90% PETER SHOR: You mean this? 01:07:43.700 --> 01:07:44.820 align:middle line:90% AUDIENCE: Yeah. 01:07:44.820 --> 01:07:48.060 align:middle line:90% PETER SHOR: It's k n choose k-- 01:07:48.060 --> 01:07:52.210 align:middle line:84% AUDIENCE: Oh, you ask like what's like-- 01:07:52.210 --> 01:07:56.570 align:middle line:84% sorry, I want to ask, what question are you answering? 01:07:56.570 --> 01:08:00.290 align:middle line:84% PETER SHOR: Equals n to the n minus 1. 01:08:00.290 --> 01:08:04.388 align:middle line:84% Well, we're answering what is summation k n choose k. 01:08:04.388 --> 01:08:05.930 align:middle line:84% AUDIENCE: I mean like you just asked, 01:08:05.930 --> 01:08:08.192 align:middle line:90% counting two ways, like which-- 01:08:08.192 --> 01:08:10.650 align:middle line:84% PETER SHOR: Oh, we want to answer this question by counting 01:08:10.650 --> 01:08:11.970 align:middle line:90% in two different ways. 01:08:11.970 --> 01:08:16.410 align:middle line:84% AUDIENCE: So it's like you have a set of n elements 01:08:16.410 --> 01:08:18.149 align:middle line:84% and you pick k elements out of it. 01:08:18.149 --> 01:08:20.821 align:middle line:84% And then from the k elements you pick one special element? 01:08:20.821 --> 01:08:21.529 align:middle line:90% PETER SHOR: Yeah. 01:08:21.529 --> 01:08:23.370 align:middle line:90% Exactly. 01:08:23.370 --> 01:08:29.480 align:middle line:90% So we have a set of n elements. 01:08:29.480 --> 01:08:33.250 align:middle line:90% 01:08:33.250 --> 01:08:36.300 align:middle line:90% Choose k elements. 01:08:36.300 --> 01:08:40.529 align:middle line:90% 01:08:40.529 --> 01:08:45.699 align:middle line:90% Choose one of the k elements. 01:08:45.699 --> 01:08:49.590 align:middle line:90% 01:08:49.590 --> 01:08:54.590 align:middle line:90% So we have a special element. 01:08:54.590 --> 01:09:00.310 align:middle line:84% And then we have a subset of k and a subset 01:09:00.310 --> 01:09:03.210 align:middle line:90% of n minus k minus 1. 01:09:03.210 --> 01:09:06.670 align:middle line:90% 01:09:06.670 --> 01:09:25.130 align:middle line:84% And sum over k, we just have a special element of k n choose k. 01:09:25.130 --> 01:09:32.550 align:middle line:90% 01:09:32.550 --> 01:09:42.630 align:middle line:84% We have a special element and a subset of the remaining subset 01:09:42.630 --> 01:09:45.660 align:middle line:90% of n minus 1 other elements. 01:09:45.660 --> 01:09:53.180 align:middle line:90% 01:09:53.180 --> 01:09:56.480 align:middle line:84% And you can pick the special element n ways. 01:09:56.480 --> 01:10:10.220 align:middle line:90% 01:10:10.220 --> 01:10:17.780 align:middle line:84% Other elements-- or the subset, sorry-- 01:10:17.780 --> 01:10:20.340 align:middle line:90% 01:10:20.340 --> 01:10:33.910 align:middle line:84% the subset of other elements is 2 to the n minus 1 ways. 01:10:33.910 --> 01:10:36.620 align:middle line:90% 01:10:36.620 --> 01:10:43.380 align:middle line:84% And this gives n times 2 to the n minus 1, 01:10:43.380 --> 01:10:48.560 align:middle line:84% which is what we wanted in the first place. 01:10:48.560 --> 01:10:51.510 align:middle line:84% OK, so that's two different ways to solve this problem. 01:10:51.510 --> 01:11:01.320 align:middle line:90% 01:11:01.320 --> 01:11:04.840 align:middle line:84% And I want to say that so far we're 01:11:04.840 --> 01:11:06.880 align:middle line:84% only doing things with generating functions 01:11:06.880 --> 01:11:10.260 align:middle line:84% that we could have done more easily some other way. 01:11:10.260 --> 01:11:14.198 align:middle line:84% But there are things that you can 01:11:14.198 --> 01:11:16.240 align:middle line:84% do with generating functions that are much harder 01:11:16.240 --> 01:11:17.300 align:middle line:90% to do other ways. 01:11:17.300 --> 01:11:20.200 align:middle line:84% And we will eventually get to those 01:11:20.200 --> 01:11:23.880 align:middle line:84% when we start talking about Fibonacci numbers. 01:11:23.880 --> 01:11:32.160 align:middle line:84% And I think the next thing I want to do 01:11:32.160 --> 01:11:34.140 align:middle line:84% is too long to fit into 3 minutes. 01:11:34.140 --> 01:11:38.350 align:middle line:84% So I will let you go and see everyone on Thursday. 01:11:38.350 --> 01:11:45.000 align:middle line:90%