WEBVTT 00:00:00.000 --> 00:00:01.485 align:middle line:90% 00:00:01.485 --> 00:00:02.970 align:middle line:90% [SQUEAKING] 00:00:02.970 --> 00:00:04.455 align:middle line:90% [RUSTLING] 00:00:04.455 --> 00:00:05.445 align:middle line:90% [CLICKING] 00:00:05.445 --> 00:00:12.060 align:middle line:90% 00:00:12.060 --> 00:00:14.120 align:middle line:84% PETER SHOR: So my plan for today is 00:00:14.120 --> 00:00:18.880 align:middle line:84% to start off by showing you how to derive the generating 00:00:18.880 --> 00:00:21.160 align:middle line:90% function for Catalan numbers. 00:00:21.160 --> 00:00:24.680 align:middle line:84% And then I will show you how to find 00:00:24.680 --> 00:00:29.960 align:middle line:84% the formula for Catalan numbers using the generating function. 00:00:29.960 --> 00:00:35.660 align:middle line:84% And, finally, I will answer questions for-- 00:00:35.660 --> 00:00:39.200 align:middle line:90% 00:00:39.200 --> 00:00:43.853 align:middle line:84% because we're having a Thursday, and this Catalan number 00:00:43.853 --> 00:00:45.520 align:middle line:84% generating function is not going to take 00:00:45.520 --> 00:00:49.920 align:middle line:84% the entire hour and a half, I will 00:00:49.920 --> 00:00:54.760 align:middle line:84% throw the floor open to questions about what to-- 00:00:54.760 --> 00:00:58.040 align:middle line:90% about material for the test. 00:00:58.040 --> 00:01:02.360 align:middle line:84% And so the test covers everything we've done so far. 00:01:02.360 --> 00:01:04.160 align:middle line:84% I don't think there are going to be 00:01:04.160 --> 00:01:08.310 align:middle line:84% any Catalan number-generating functions on the test. 00:01:08.310 --> 00:01:11.024 align:middle line:90% 00:01:11.024 --> 00:01:13.520 align:middle line:84% But seeing the material for today 00:01:13.520 --> 00:01:15.800 align:middle line:84% is not going to hurt you on the test 00:01:15.800 --> 00:01:21.528 align:middle line:84% because it's more about generating functions. 00:01:21.528 --> 00:01:25.480 align:middle line:84% So recall that the Catalan numbers counted a whole bunch 00:01:25.480 --> 00:01:27.120 align:middle line:90% of different things. 00:01:27.120 --> 00:01:43.440 align:middle line:84% And Catalan numbers count binary trees, 00:01:43.440 --> 00:01:45.910 align:middle line:84% or maybe I should say rooted binary trees. 00:01:45.910 --> 00:01:48.680 align:middle line:90% 00:01:48.680 --> 00:02:04.670 align:middle line:84% So here are the five rooted binary trees on seven nodes. 00:02:04.670 --> 00:02:13.080 align:middle line:90% 00:02:13.080 --> 00:02:13.820 align:middle line:90% There we go. 00:02:13.820 --> 00:02:17.480 align:middle line:90% 00:02:17.480 --> 00:02:21.970 align:middle line:84% So C sub 3 equals 5, and they count Dyck paths. 00:02:21.970 --> 00:02:28.120 align:middle line:90% 00:02:28.120 --> 00:02:33.690 align:middle line:84% So Dyck walks were walks that never went below 0 and go from 0 00:02:33.690 --> 00:02:36.200 align:middle line:90% to 2n. 00:02:36.200 --> 00:02:45.020 align:middle line:84% So for C3, there should be five of them, and there are. 00:02:45.020 --> 00:02:49.760 align:middle line:90% 00:02:49.760 --> 00:02:50.480 align:middle line:90% Wait. 00:02:50.480 --> 00:02:53.470 align:middle line:84% I want to go up twice, down, and then over. 00:02:53.470 --> 00:02:56.240 align:middle line:90% 00:02:56.240 --> 00:03:01.640 align:middle line:84% And down here, let's do up, down, up, down, down, and up, 00:03:01.640 --> 00:03:04.090 align:middle line:90% up, up, down, down, down. 00:03:04.090 --> 00:03:08.910 align:middle line:90% 00:03:08.910 --> 00:03:11.000 align:middle line:84% So how would you get a generating function 00:03:11.000 --> 00:03:12.070 align:middle line:90% for Catalan numbers? 00:03:12.070 --> 00:03:16.480 align:middle line:90% 00:03:16.480 --> 00:03:23.840 align:middle line:84% Well, you want some kind of recurrence. 00:03:23.840 --> 00:03:33.320 align:middle line:90% 00:03:33.320 --> 00:03:36.810 align:middle line:84% Well, first, let's start out by looking at binary trees. 00:03:36.810 --> 00:03:39.400 align:middle line:90% 00:03:39.400 --> 00:03:46.480 align:middle line:84% So a binary tree, well, unless it's a binary tree on one node, 00:03:46.480 --> 00:03:48.840 align:middle line:90% it starts with a vertex. 00:03:48.840 --> 00:03:57.390 align:middle line:84% And there are two branches because, in our definition 00:03:57.390 --> 00:04:04.110 align:middle line:84% of binary tree, every vertex has two branches coming out of it. 00:04:04.110 --> 00:04:09.230 align:middle line:84% And now, on this branch, we have T let's call it a. 00:04:09.230 --> 00:04:11.070 align:middle line:90% And this branch, we have Tb. 00:04:11.070 --> 00:04:13.990 align:middle line:90% 00:04:13.990 --> 00:04:20.769 align:middle line:84% And suppose we want to, say, count C sub n. 00:04:20.769 --> 00:04:26.030 align:middle line:90% 00:04:26.030 --> 00:04:35.550 align:middle line:84% Let's say C sub n, which is the number of binary trees on 2n 00:04:35.550 --> 00:04:37.730 align:middle line:90% minus 3 nodes, I guess-- 00:04:37.730 --> 00:04:40.710 align:middle line:90% no, 2n plus 1 nodes-- 00:04:40.710 --> 00:04:45.950 align:middle line:84% is going to be equal to sum j equals 0 00:04:45.950 --> 00:04:53.350 align:middle line:84% through n minus 1 C sub j C n minus J minus 1. 00:04:53.350 --> 00:04:55.190 align:middle line:90% Why? 00:04:55.190 --> 00:05:08.150 align:middle line:84% Well, the number of possibilities for this tree 00:05:08.150 --> 00:05:14.790 align:middle line:84% is C sub j because it's just a smaller Catalan number. 00:05:14.790 --> 00:05:22.110 align:middle line:84% And if this side has 2j minus 1 nodes, 00:05:22.110 --> 00:05:27.240 align:middle line:84% this side must have 2n minus j minus 1 nodes. 00:05:27.240 --> 00:05:29.910 align:middle line:90% 00:05:29.910 --> 00:05:41.750 align:middle line:84% And so this is number of possibilities is 2-- 00:05:41.750 --> 00:05:52.470 align:middle line:84% sorry-- is C sub n minus j minus 1. 00:05:52.470 --> 00:05:56.750 align:middle line:84% And when you combine these two trees and add a node, 00:05:56.750 --> 00:06:04.270 align:middle line:84% you get a Catalan number C sub n. 00:06:04.270 --> 00:06:10.790 align:middle line:84% So C sub n equals summation j equals 0 n minus 1 Cj C n 00:06:10.790 --> 00:06:12.910 align:middle line:90% minus j. 00:06:12.910 --> 00:06:18.990 align:middle line:84% And we can check that if you want with the Catalan number 3-- 00:06:18.990 --> 00:06:27.750 align:middle line:90% I'm sorry 5, 13, 14-- 00:06:27.750 --> 00:06:28.710 align:middle line:90% sorry. 00:06:28.710 --> 00:06:32.510 align:middle line:84% I cannot remember my Catalan numbers. 00:06:32.510 --> 00:06:38.310 align:middle line:84% So here, 14 should equal 5 times 1 plus 2 times 1 plus 1 times 00:06:38.310 --> 00:06:40.750 align:middle line:90% 2 plus 1 times 5. 00:06:40.750 --> 00:06:42.310 align:middle line:90% And it does. 00:06:42.310 --> 00:06:46.710 align:middle line:90% So this recurrence checks out. 00:06:46.710 --> 00:06:51.230 align:middle line:90% So now you know this recurrence. 00:06:51.230 --> 00:06:57.470 align:middle line:84% How would you write down a formula, or an equation, 00:06:57.470 --> 00:06:58.650 align:middle line:90% with a C of x? 00:06:58.650 --> 00:07:03.270 align:middle line:90% 00:07:03.270 --> 00:07:16.570 align:middle line:84% i equals 0 to infinity x to the n C sub n. 00:07:16.570 --> 00:07:20.630 align:middle line:90% 00:07:20.630 --> 00:07:25.170 align:middle line:90% So what's a equation for C of x? 00:07:25.170 --> 00:07:30.350 align:middle line:90% 00:07:30.350 --> 00:07:31.030 align:middle line:90% Anybody? 00:07:31.030 --> 00:07:34.990 align:middle line:90% 00:07:34.990 --> 00:07:36.029 align:middle line:90% Yeah? 00:07:36.029 --> 00:07:40.021 align:middle line:90% AUDIENCE: [INAUDIBLE] 00:07:40.021 --> 00:07:45.510 align:middle line:90% 00:07:45.510 --> 00:07:49.070 align:middle line:90% PETER SHOR: Yeah, exactly. 00:07:49.070 --> 00:07:55.090 align:middle line:84% C of x is equal to C of x squared. 00:07:55.090 --> 00:07:58.310 align:middle line:84% Actually, this isn't quite right because if this is j 00:07:58.310 --> 00:08:01.590 align:middle line:84% and this is n minus j minus 1, this is n, 00:08:01.590 --> 00:08:07.470 align:middle line:84% so we need to add another x here to make the degrees right. 00:08:07.470 --> 00:08:12.790 align:middle line:84% And this isn't quite right, either, 00:08:12.790 --> 00:08:22.030 align:middle line:84% because C of x starts 1 plus x plus 2x squared, et cetera. 00:08:22.030 --> 00:08:25.730 align:middle line:84% And this side is divisible by x, so this side has 1 in it. 00:08:25.730 --> 00:08:30.710 align:middle line:84% So we need to add 1 for the initial condition. 00:08:30.710 --> 00:08:33.110 align:middle line:90% And now we have an equation. 00:08:33.110 --> 00:08:34.330 align:middle line:90% It's a quadratic equation. 00:08:34.330 --> 00:08:37.289 align:middle line:84% We know how to solve quadratic equations, so let's do it. 00:08:37.289 --> 00:08:45.230 align:middle line:90% 00:08:45.230 --> 00:08:51.130 align:middle line:90% C of x is equal to-- 00:08:51.130 --> 00:08:55.790 align:middle line:90% 00:08:55.790 --> 00:09:00.510 align:middle line:84% well, let's put this quadratic equation in the standard order-- 00:09:00.510 --> 00:09:09.190 align:middle line:84% C of x squared minus C of x plus 1 equals 0. 00:09:09.190 --> 00:09:14.270 align:middle line:90% And we have 1-- 00:09:14.270 --> 00:09:20.830 align:middle line:84% that's minus b-- plus or minus the square root of 1 minus 4aC-- 00:09:20.830 --> 00:09:26.170 align:middle line:90% that's 4x-- over 2a. 00:09:26.170 --> 00:09:26.740 align:middle line:90% That's 2x. 00:09:26.740 --> 00:09:29.900 align:middle line:90% 00:09:29.900 --> 00:09:32.750 align:middle line:84% So there's two possibilities for this. 00:09:32.750 --> 00:09:34.510 align:middle line:84% How would you figure out which one is 00:09:34.510 --> 00:09:54.510 align:middle line:90% the correct one Any suggestions 00:09:54.510 --> 00:09:58.250 align:middle line:90% OK, so what is C of-- 00:09:58.250 --> 00:10:04.350 align:middle line:90% 00:10:04.350 --> 00:10:10.670 align:middle line:84% what do you get when you substitute x equals 0 nC? 00:10:10.670 --> 00:10:17.590 align:middle line:84% Well, you get 1 plus x plus 2x squared plus dot, dot, dot. 00:10:17.590 --> 00:10:19.230 align:middle line:90% You get 1. 00:10:19.230 --> 00:10:22.390 align:middle line:84% So if you substitute x equals 0 in here 00:10:22.390 --> 00:10:26.830 align:middle line:84% and you have the plus thing, you get 00:10:26.830 --> 00:10:31.230 align:middle line:84% 2 divided by 0, which is infinity, so that's not right. 00:10:31.230 --> 00:10:35.830 align:middle line:84% If you substitute the minus term, you get 1 minus 1 00:10:35.830 --> 00:10:40.110 align:middle line:90% over 0, which might be right. 00:10:40.110 --> 00:10:44.580 align:middle line:84% And you can easily check with calculus that limit as x goes 00:10:44.580 --> 00:10:55.810 align:middle line:84% to 0 1 minus square root of 1 minus 4x over 2x is equal to 1. 00:10:55.810 --> 00:10:58.820 align:middle line:90% 00:10:58.820 --> 00:11:03.810 align:middle line:90% So the right value is minus. 00:11:03.810 --> 00:11:23.900 align:middle line:90% 00:11:23.900 --> 00:11:27.940 align:middle line:90% So how do you evaluate this? 00:11:27.940 --> 00:11:32.080 align:middle line:84% Well, to get the Taylor expansion of this, 00:11:32.080 --> 00:11:33.860 align:middle line:90% you use the binomial theorem-- 00:11:33.860 --> 00:11:34.870 align:middle line:90% binomial theorem. 00:11:34.870 --> 00:11:45.260 align:middle line:90% 00:11:45.260 --> 00:11:46.890 align:middle line:90% Use binomial theorem. 00:11:46.890 --> 00:11:56.140 align:middle line:90% 00:11:56.140 --> 00:12:01.420 align:middle line:84% So what the binomial theorem says is that, I guess, 00:12:01.420 --> 00:12:10.940 align:middle line:84% 1 plus a to the k is equal to 1 plus k 00:12:10.940 --> 00:12:19.180 align:middle line:84% choose 1 a plus k choose 2 a squared plus k choose 00:12:19.180 --> 00:12:25.180 align:middle line:90% 3 a cubed plus dot, dot, dot. 00:12:25.180 --> 00:12:30.340 align:middle line:84% And it's also true for fractional values of k. 00:12:30.340 --> 00:12:40.060 align:middle line:84% So 1 plus a to the 1/2 is equal to, I guess, 00:12:40.060 --> 00:12:56.200 align:middle line:84% 1 plus 1/2 choose 1 a plus 1/2 choose 2 a squared plus dot, 00:12:56.200 --> 00:12:56.800 align:middle line:90% dot, dot. 00:12:56.800 --> 00:13:04.516 align:middle line:90% 00:13:04.516 --> 00:13:10.020 align:middle line:84% And, of course, you have to take this and subtract it from 1 00:13:10.020 --> 00:13:11.170 align:middle line:90% and divide by 2x. 00:13:11.170 --> 00:13:26.844 align:middle line:90% 00:13:26.844 --> 00:13:33.220 align:middle line:84% So I'm going to do 1 minus square root of 1 minus 4x over 00:13:33.220 --> 00:13:34.660 align:middle line:90% 2x-- 00:13:34.660 --> 00:13:38.620 align:middle line:84% so that's the generating function for Catalan numbers-- 00:13:38.620 --> 00:13:43.660 align:middle line:90% is equal to 1 over 2x. 00:13:43.660 --> 00:13:47.120 align:middle line:84% And there's a 1 here as the first term, 00:13:47.120 --> 00:13:49.060 align:middle line:90% but it gets canceled by this 1. 00:13:49.060 --> 00:13:51.720 align:middle line:84% So we don't have a constant term, 00:13:51.720 --> 00:13:54.180 align:middle line:84% which is good because we don't want a 1 over 2x 00:13:54.180 --> 00:13:55.580 align:middle line:90% in our expansion. 00:13:55.580 --> 00:14:14.500 align:middle line:84% So the next term is 1/2 times minus 4x plus 1. 00:14:14.500 --> 00:14:20.620 align:middle line:90% So 1/2 choose 1 equals 1/2. 00:14:20.620 --> 00:14:27.620 align:middle line:84% 1/2 choose 2 equals 1/2 times minus 1/2 divided by 2 00:14:27.620 --> 00:14:29.100 align:middle line:90% factorial. 00:14:29.100 --> 00:14:38.700 align:middle line:84% 1/2 choose 3 is 1/2 times minus 1/2 times minus 3/2 divided by 3 00:14:38.700 --> 00:14:40.190 align:middle line:90% factorial, et cetera. 00:14:40.190 --> 00:14:43.300 align:middle line:90% 00:14:43.300 --> 00:14:52.170 align:middle line:84% So this is 1/2 times minus 1/2 divided by 2 factorial minus 4x 00:14:52.170 --> 00:14:52.670 align:middle line:90% squared-- 00:14:52.670 --> 00:14:57.540 align:middle line:90% 00:14:57.540 --> 00:15:02.140 align:middle line:84% OK, I have forgotten the minus sign here because of the minus 00:15:02.140 --> 00:15:04.380 align:middle line:90% there-- 00:15:04.380 --> 00:15:19.920 align:middle line:84% plus 1/2 minus 1/2 minus 3/2 over 3 factorial. 00:15:19.920 --> 00:15:23.380 align:middle line:90% 00:15:23.380 --> 00:15:30.780 align:middle line:84% And there's a 4X cubed, so let's put that here, 00:15:30.780 --> 00:15:31.690 align:middle line:90% and the minus sign. 00:15:31.690 --> 00:15:34.580 align:middle line:90% 00:15:34.580 --> 00:15:38.220 align:middle line:84% So the first thing you can check is that all of these terms 00:15:38.220 --> 00:15:40.520 align:middle line:84% are positive because this minus cancels this minus. 00:15:40.520 --> 00:15:43.700 align:middle line:90% 00:15:43.700 --> 00:15:47.620 align:middle line:84% This minus is squared, so this minus cancels this minus. 00:15:47.620 --> 00:15:51.860 align:middle line:84% This minus is cubed, but there are two minuses here, 00:15:51.860 --> 00:15:54.980 align:middle line:84% so that one cancels this minus, et cetera. 00:15:54.980 --> 00:15:56.670 align:middle line:84% So we can ignore the minus signs. 00:15:56.670 --> 00:16:08.046 align:middle line:90% 00:16:08.046 --> 00:16:13.410 align:middle line:84% So I would like to take the term for x equals 3. 00:16:13.410 --> 00:16:17.260 align:middle line:90% 00:16:17.260 --> 00:16:36.450 align:middle line:84% Well, I want to claim that that is 7/2 times 5/2 times 3/2 times 00:16:36.450 --> 00:16:37.730 align:middle line:90% 1/2 times 1/2-- 00:16:37.730 --> 00:16:40.570 align:middle line:90% 00:16:40.570 --> 00:16:42.970 align:middle line:90% actually, this is minus 1/2-- 00:16:42.970 --> 00:16:50.150 align:middle line:84% divided by-- so I guess that is 5 factorial. 00:16:50.150 --> 00:16:55.850 align:middle line:90% 00:16:55.850 --> 00:16:57.990 align:middle line:90% There's a 4 to the fourth here. 00:16:57.990 --> 00:17:00.770 align:middle line:90% 00:17:00.770 --> 00:17:11.849 align:middle line:84% There's a 2 here and a 2 to the fourth here. 00:17:11.849 --> 00:17:12.550 align:middle line:90% Well, no. 00:17:12.550 --> 00:17:24.770 align:middle line:90% 00:17:24.770 --> 00:17:30.120 align:middle line:84% Well, I hope I haven't forgotten any terms. 00:17:30.120 --> 00:17:36.190 align:middle line:90% 00:17:36.190 --> 00:17:37.962 align:middle line:90% So what is this? 00:17:37.962 --> 00:17:43.450 align:middle line:84% The top is 7 times 5 times 3 divided by 2 00:17:43.450 --> 00:17:51.170 align:middle line:84% to the fifth times 2 times 5 times 4 times 3 times 2. 00:17:51.170 --> 00:17:53.590 align:middle line:84% And there's a 4 to the fourth up here. 00:17:53.590 --> 00:17:57.090 align:middle line:90% 00:17:57.090 --> 00:18:05.410 align:middle line:84% So there are five, six, seven 2's on the bottom and eight 2's 00:18:05.410 --> 00:18:10.570 align:middle line:84% on the top, so that's 2 to the eighth over 2 to the seventh. 00:18:10.570 --> 00:18:13.450 align:middle line:90% And the 3 and the 5 cancel out. 00:18:13.450 --> 00:18:19.010 align:middle line:90% And you get 7 equals 14. 00:18:19.010 --> 00:18:21.850 align:middle line:90% So we did this right. 00:18:21.850 --> 00:18:26.610 align:middle line:90% And I want to claim just by-- 00:18:26.610 --> 00:18:38.050 align:middle line:84% well, just by generalization, that this is going to be 2 00:18:38.050 --> 00:18:44.290 align:middle line:84% and minus 1 C sub n equals 2n minus 1 double factorial. 00:18:44.290 --> 00:18:48.250 align:middle line:84% So double factorial means 2n minus 1, 2n minus 3, 00:18:48.250 --> 00:18:51.250 align:middle line:84% 2n minus 5 all the way down to 1. 00:18:51.250 --> 00:18:52.850 align:middle line:90% So this is 7 double factorial. 00:18:52.850 --> 00:18:59.410 align:middle line:84% That's 7 times 5 times 3 times 1 4 to the n 00:18:59.410 --> 00:19:11.740 align:middle line:84% plus 1 divided by 2 times 2 to the n plus 1 times-- 00:19:11.740 --> 00:19:22.090 align:middle line:90% 00:19:22.090 --> 00:19:24.270 align:middle line:90% should be n plus 1 factorial. 00:19:24.270 --> 00:19:33.770 align:middle line:90% 00:19:33.770 --> 00:19:36.210 align:middle line:90% And wait. 00:19:36.210 --> 00:19:40.290 align:middle line:84% This is x equals 4, not x equals 3. 00:19:40.290 --> 00:19:54.690 align:middle line:84% And so this is 4 to the fourth times x cubed over 2 times 00:19:54.690 --> 00:19:55.850 align:middle line:90% 5 factorial. 00:19:55.850 --> 00:20:00.170 align:middle line:84% So this is indeed n plus 1 factorial. 00:20:00.170 --> 00:20:03.280 align:middle line:90% And this is 2. 00:20:03.280 --> 00:20:05.650 align:middle line:90% Well, it is 2 to the n plus 1. 00:20:05.650 --> 00:20:08.650 align:middle line:90% 2's here and another 2. 00:20:08.650 --> 00:20:14.590 align:middle line:90% And we have n plus 1 factorial. 00:20:14.590 --> 00:20:28.650 align:middle line:90% 00:20:28.650 --> 00:20:38.418 align:middle line:84% So now, we need to manipulate this into some kind of-- 00:20:38.418 --> 00:20:43.930 align:middle line:84% oh, I should say x to the n here. 00:20:43.930 --> 00:20:48.010 align:middle line:84% We need to manipulate this into a better form. 00:20:48.010 --> 00:20:53.890 align:middle line:84% And the way we do that is we say 2k minus 1 factorial is equal 00:20:53.890 --> 00:21:05.650 align:middle line:84% to 2k factorial over 2k times 2k minus 2 all the way down to 2k 00:21:05.650 --> 00:21:06.510 align:middle line:90% minus 4. 00:21:06.510 --> 00:21:15.290 align:middle line:90% 00:21:15.290 --> 00:21:20.930 align:middle line:84% So this is the odd terms between 1 and 2k minus 1. 00:21:20.930 --> 00:21:23.930 align:middle line:84% So this is all the terms between 1 and 2k. 00:21:23.930 --> 00:21:27.140 align:middle line:84% And we divide by the even terms, and we get the odd terms. 00:21:27.140 --> 00:21:29.890 align:middle line:90% 00:21:29.890 --> 00:21:36.650 align:middle line:84% But this is just 2k factorial over k factorial times 2 00:21:36.650 --> 00:21:37.310 align:middle line:90% to the k. 00:21:37.310 --> 00:21:42.850 align:middle line:90% 00:21:42.850 --> 00:21:53.410 align:middle line:84% So that says that C sub n is equal to 2n factorial divided 00:21:53.410 --> 00:21:57.930 align:middle line:90% by n factorial. 00:21:57.930 --> 00:22:00.750 align:middle line:84% And over here, we have an n plus 1 factorial. 00:22:00.750 --> 00:22:09.650 align:middle line:90% 00:22:09.650 --> 00:22:12.160 align:middle line:84% There's a 4 to the n plus 1 on top. 00:22:12.160 --> 00:22:16.330 align:middle line:90% 00:22:16.330 --> 00:22:29.850 align:middle line:84% And there are 2 times 2 to the n plus 1 times 2 00:22:29.850 --> 00:22:32.170 align:middle line:90% to the n on the bottom. 00:22:32.170 --> 00:22:35.010 align:middle line:84% This 2 to the n comes from this one. 00:22:35.010 --> 00:22:37.450 align:middle line:84% This 2 to the n plus 1 comes from that guy. 00:22:37.450 --> 00:22:41.160 align:middle line:90% And the 2 comes from this 2x. 00:22:41.160 --> 00:22:44.690 align:middle line:90% 00:22:44.690 --> 00:22:47.050 align:middle line:84% And everything cancels out except-- 00:22:47.050 --> 00:22:51.690 align:middle line:84% all the factors of 2 cancel out, and we're left with 2n factorial 00:22:51.690 --> 00:22:56.550 align:middle line:84% over n factorial n plus 1 factorial, 00:22:56.550 --> 00:23:04.450 align:middle line:84% which is 1 over n plus 1 2n choose n, 00:23:04.450 --> 00:23:07.590 align:middle line:84% which you should recognize as the formula for Catalan numbers. 00:23:07.590 --> 00:23:08.410 align:middle line:90% Yes? 00:23:08.410 --> 00:23:10.930 align:middle line:84% AUDIENCE: So I have [INAUDIBLE] question. 00:23:10.930 --> 00:23:15.250 align:middle line:84% How did you get 1 plus x plus 2 [INAUDIBLE]? 00:23:15.250 --> 00:23:16.890 align:middle line:90% PETER SHOR: Oh. 00:23:16.890 --> 00:23:21.780 align:middle line:84% Well, the Catalan numbers are 1, 1, 2, 5, 14. 00:23:21.780 --> 00:23:28.960 align:middle line:84% And this is just summation C sub n x to the n 00:23:28.960 --> 00:23:32.120 align:middle line:90% is a generating function. 00:23:32.120 --> 00:23:38.680 align:middle line:84% And that's 1 plus x plus 2x squared plus 5x to the fourth-- 00:23:38.680 --> 00:23:45.980 align:middle line:84% 5x cubed-- plus 14x to the fourth plus dot, dot, dot. 00:23:45.980 --> 00:23:51.000 align:middle line:90% 00:23:51.000 --> 00:23:53.103 align:middle line:90% So the expansion. 00:23:53.103 --> 00:23:55.020 align:middle line:84% Here are the first few terms of the expansion. 00:23:55.020 --> 00:23:59.580 align:middle line:84% The first term, you have a minus 4x and a minus 1 over 2x 00:23:59.580 --> 00:24:06.600 align:middle line:84% because the minus 1 over 2x term here comes from this minus sign. 00:24:06.600 --> 00:24:14.240 align:middle line:84% The second term, you have 1/2 times minus 1/2 divided by 2 00:24:14.240 --> 00:24:17.600 align:middle line:84% factorial times minus 4x squared. 00:24:17.600 --> 00:24:18.940 align:middle line:90% So this is positive. 00:24:18.940 --> 00:24:20.080 align:middle line:90% This is minus. 00:24:20.080 --> 00:24:22.320 align:middle line:90% And this is minus. 00:24:22.320 --> 00:24:26.080 align:middle line:84% The third term, you have this minus, this minus, 00:24:26.080 --> 00:24:28.160 align:middle line:90% and this minus. 00:24:28.160 --> 00:24:32.560 align:middle line:84% And multiplying them all together, you get a minus sign. 00:24:32.560 --> 00:24:35.680 align:middle line:84% And it cancels out this minus sign, so it's positive. 00:24:35.680 --> 00:24:39.080 align:middle line:90% So all the terms are positive. 00:24:39.080 --> 00:24:50.760 align:middle line:90% And, really-- yeah. 00:24:50.760 --> 00:24:57.760 align:middle line:90% 00:24:57.760 --> 00:25:00.440 align:middle line:84% So now that we know all the terms are positive, 00:25:00.440 --> 00:25:03.040 align:middle line:84% we can ignore the minus sign on this, 00:25:03.040 --> 00:25:07.580 align:middle line:84% although if you're multiplying by-- 00:25:07.580 --> 00:25:08.080 align:middle line:90% yeah. 00:25:08.080 --> 00:25:17.000 align:middle line:90% 00:25:17.000 --> 00:25:25.580 align:middle line:84% So this is another way of getting the formula for C sub n. 00:25:25.580 --> 00:25:43.640 align:middle line:90% 00:25:43.640 --> 00:25:46.840 align:middle line:84% And I want to say that there's a completely different way 00:25:46.840 --> 00:25:56.440 align:middle line:84% of deriving this formula for the Catalan number. 00:25:56.440 --> 00:26:00.560 align:middle line:84% We derive this equation in a completely different way. 00:26:00.560 --> 00:26:04.540 align:middle line:84% And I figured I might as well show it because, well, 00:26:04.540 --> 00:26:11.640 align:middle line:84% because it illustrates a neat technique of generating 00:26:11.640 --> 00:26:14.640 align:middle line:84% functions that I had some questions about in my office 00:26:14.640 --> 00:26:16.020 align:middle line:90% hours yesterday. 00:26:16.020 --> 00:26:20.430 align:middle line:84% So Catalan numbers also correspond to Dyck paths. 00:26:20.430 --> 00:26:57.722 align:middle line:90% 00:26:57.722 --> 00:27:10.640 align:middle line:84% So a Dyck path can be broken into a sequence of paths 00:27:10.640 --> 00:27:26.720 align:middle line:84% with plus and minus 1 steps and which touch 0 only 00:27:26.720 --> 00:27:27.940 align:middle line:90% at their endpoints. 00:27:27.940 --> 00:27:41.760 align:middle line:90% 00:27:41.760 --> 00:27:47.880 align:middle line:84% So let's do it for this Dyck path. 00:27:47.880 --> 00:27:50.560 align:middle line:90% The first Dyck path is this guy. 00:27:50.560 --> 00:27:54.680 align:middle line:90% The second one is this guy-- 00:27:54.680 --> 00:27:55.720 align:middle line:90% 1, 2. 00:27:55.720 --> 00:28:00.920 align:middle line:84% The third one is this guy, and the fourth one will be this guy. 00:28:00.920 --> 00:28:03.920 align:middle line:84% So it's pretty clear that every Dyck path 00:28:03.920 --> 00:28:07.370 align:middle line:84% can be broken into a sequence of this kind of paths. 00:28:07.370 --> 00:28:12.120 align:middle line:90% 00:28:12.120 --> 00:28:31.480 align:middle line:84% So if D of x is the generating function for paths that-- 00:28:31.480 --> 00:28:54.440 align:middle line:84% well, for paths that only touch 0 at their own points, 00:28:54.440 --> 00:29:02.880 align:middle line:84% well, we can use the theorem about generating functions 00:29:02.880 --> 00:29:05.280 align:middle line:84% from sequences to get the generating 00:29:05.280 --> 00:29:09.040 align:middle line:84% function for Catalan numbers, or generating function for Dyck 00:29:09.040 --> 00:29:10.040 align:middle line:90% paths. 00:29:10.040 --> 00:29:29.360 align:middle line:84% So C of x is equal to 1 over 1 minus D of x, right? 00:29:29.360 --> 00:29:43.880 align:middle line:90% 00:29:43.880 --> 00:29:54.680 align:middle line:90% But D of x, well, you have-- 00:29:54.680 --> 00:30:09.110 align:middle line:84% everything in D of x looks like this, where this is a Dyck path, 00:30:09.110 --> 00:30:11.750 align:middle line:84% and you have an upstep on the left 00:30:11.750 --> 00:30:16.470 align:middle line:90% and a downstep on the right. 00:30:16.470 --> 00:30:21.070 align:middle line:84% So every one of this kind of paths, 00:30:21.070 --> 00:30:26.510 align:middle line:84% it just comes from a Dyck path where you add two steps to it. 00:30:26.510 --> 00:30:33.110 align:middle line:84% So this says that D of x is equal to x times C 00:30:33.110 --> 00:30:41.590 align:middle line:90% of x because x to the k was-- 00:30:41.590 --> 00:30:49.430 align:middle line:84% in this generating function, x to the k, k 00:30:49.430 --> 00:30:53.990 align:middle line:84% was twice the number of steps in a Dyck path. 00:30:53.990 --> 00:30:59.210 align:middle line:84% And to get the raised Dyck paths, 00:30:59.210 --> 00:31:05.910 align:middle line:84% you add two steps to a Dyck path, so you get x times C of x. 00:31:05.910 --> 00:31:13.330 align:middle line:84% So we have C of x equals 1 over 1 minus x C of x. 00:31:13.330 --> 00:31:17.230 align:middle line:90% 00:31:17.230 --> 00:31:22.670 align:middle line:84% And you should be able to multiply this equation 00:31:22.670 --> 00:31:27.360 align:middle line:90% on both sides to get-- 00:31:27.360 --> 00:31:33.430 align:middle line:90% 00:31:33.430 --> 00:31:41.390 align:middle line:84% so we have C of x minus x C of x squared minus 1 00:31:41.390 --> 00:31:46.420 align:middle line:84% equals 0, which is the same equation we had earlier. 00:31:46.420 --> 00:31:50.830 align:middle line:90% 00:31:50.830 --> 00:31:54.370 align:middle line:84% And you get the same quadratic equation. 00:31:54.370 --> 00:31:55.990 align:middle line:84% And you can do everything exactly 00:31:55.990 --> 00:31:58.560 align:middle line:90% the same way as we did earlier. 00:31:58.560 --> 00:32:03.090 align:middle line:90% 00:32:03.090 --> 00:32:03.590 align:middle line:90% Yeah? 00:32:03.590 --> 00:32:07.790 align:middle line:84% AUDIENCE: Can you repeat the D of x [INAUDIBLE]? 00:32:07.790 --> 00:32:09.110 align:middle line:90% PETER SHOR: Right, OK. 00:32:09.110 --> 00:32:17.870 align:middle line:84% So D of x are paths that start with an upstep 00:32:17.870 --> 00:32:20.850 align:middle line:84% and with a downstep and never go below 1. 00:32:20.850 --> 00:32:42.390 align:middle line:90% 00:32:42.390 --> 00:32:55.350 align:middle line:84% S equals sum n equals 0 to infinity d sub n x to the n, 00:32:55.350 --> 00:33:13.070 align:middle line:84% d paths of length 2n, which start up, end with a downstep, 00:33:13.070 --> 00:33:19.560 align:middle line:90% and never go below 1. 00:33:19.560 --> 00:33:24.590 align:middle line:90% 00:33:24.590 --> 00:33:30.770 align:middle line:84% So here, we have let's call this 0, call this 1. 00:33:30.770 --> 00:33:33.790 align:middle line:90% 00:33:33.790 --> 00:33:36.630 align:middle line:84% And we have a path up here that never 00:33:36.630 --> 00:33:44.630 align:middle line:84% goes below 1 until the last step, in which case 00:33:44.630 --> 00:33:46.600 align:middle line:90% it goes down to 0. 00:33:46.600 --> 00:33:54.310 align:middle line:90% 00:33:54.310 --> 00:33:59.630 align:middle line:84% Now, if you take off the last step and the first step here, 00:33:59.630 --> 00:34:07.280 align:middle line:84% you get this thing is a Dyck path. 00:34:07.280 --> 00:34:12.989 align:middle line:90% 00:34:12.989 --> 00:34:20.270 align:middle line:84% So d sub n is equal to the number of these raised Dyck 00:34:20.270 --> 00:34:24.150 align:middle line:84% paths is equal to C sub n minus 1 00:34:24.150 --> 00:34:33.270 align:middle line:84% because we have subtracted two edges from d sub n 00:34:33.270 --> 00:34:36.639 align:middle line:90% to get a Dyck path. 00:34:36.639 --> 00:34:52.030 align:middle line:90% 00:34:52.030 --> 00:35:03.670 align:middle line:84% So D of x equals sum d sub n x to the n 00:35:03.670 --> 00:35:10.950 align:middle line:90% is equal to sum C sub n x-- 00:35:10.950 --> 00:35:19.590 align:middle line:84% I'm sorry, C n minus 1 x to the n is equal to sum-- 00:35:19.590 --> 00:35:23.700 align:middle line:90% oh, gosh-- is equal to sum-- 00:35:23.700 --> 00:35:26.270 align:middle line:90% 00:35:26.270 --> 00:35:30.230 align:middle line:84% I'm going to claim this is C sub n minus 1 00:35:30.230 --> 00:35:33.920 align:middle line:90% x to the n minus 1 times x. 00:35:33.920 --> 00:35:44.070 align:middle line:90% 00:35:44.070 --> 00:35:47.790 align:middle line:90% And maybe we should-- 00:35:47.790 --> 00:35:59.700 align:middle line:84% well, there's no d paths of length 0. 00:35:59.700 --> 00:36:03.140 align:middle line:84% The shortest one is of length 2, so this summation starts at n 00:36:03.140 --> 00:36:04.780 align:middle line:90% equals 1. 00:36:04.780 --> 00:36:08.980 align:middle line:84% And we can just say n equals 1 to infinity. 00:36:08.980 --> 00:36:14.380 align:middle line:84% And this is n equals 1 to infinity. 00:36:14.380 --> 00:36:16.300 align:middle line:84% And you can recognize this as just 00:36:16.300 --> 00:36:18.460 align:middle line:84% equal to the generating function for Catalan 00:36:18.460 --> 00:36:21.170 align:middle line:90% numbers, which is x C of x. 00:36:21.170 --> 00:36:28.420 align:middle line:90% 00:36:28.420 --> 00:36:30.860 align:middle line:84% So I said I would hold a question and answer 00:36:30.860 --> 00:36:37.460 align:middle line:84% session in the last, I guess, 30 minutes of this question. 00:36:37.460 --> 00:36:41.020 align:middle line:84% So is there anything that people want to-- 00:36:41.020 --> 00:36:42.740 align:middle line:90% you have a question? 00:36:42.740 --> 00:36:45.220 align:middle line:90% AUDIENCE: I had a question. 00:36:45.220 --> 00:36:46.580 align:middle line:90% How do you get [INAUDIBLE]? 00:36:46.580 --> 00:36:50.300 align:middle line:90% 00:36:50.300 --> 00:36:52.340 align:middle line:90% PETER SHOR: Oh, OK. 00:36:52.340 --> 00:36:59.300 align:middle line:84% Well, that just comes from plugging this equation 00:36:59.300 --> 00:37:02.710 align:middle line:90% into this equation. 00:37:02.710 --> 00:37:04.460 align:middle line:84% AUDIENCE: How do you get the top equation? 00:37:04.460 --> 00:37:09.700 align:middle line:84% PETER SHOR: We got the top equation because of this theorem 00:37:09.700 --> 00:37:13.620 align:middle line:84% about sequences for generating functions, 00:37:13.620 --> 00:37:24.560 align:middle line:84% which maybe you used on your homework yesterday, although-- 00:37:24.560 --> 00:37:43.300 align:middle line:90% 00:37:43.300 --> 00:37:45.800 align:middle line:84% so let's go over this theorem about sequences. 00:37:45.800 --> 00:37:53.380 align:middle line:90% 00:37:53.380 --> 00:37:58.100 align:middle line:84% So suppose a generating function-- 00:37:58.100 --> 00:38:16.210 align:middle line:84% suppose a generating function counts let's call them objects. 00:38:16.210 --> 00:38:21.100 align:middle line:90% 00:38:21.100 --> 00:38:30.740 align:middle line:84% There are a n objects of size n where 00:38:30.740 --> 00:38:33.240 align:middle line:84% size might be the length of a Dyck path, 00:38:33.240 --> 00:38:38.340 align:middle line:84% or, I guess, size might be half the length of a Dyck path, 00:38:38.340 --> 00:38:41.893 align:middle line:84% but it could be anything else, anyway. 00:38:41.893 --> 00:38:43.810 align:middle line:84% You can think of this as an abstract quantity. 00:38:43.810 --> 00:38:47.940 align:middle line:90% 00:38:47.940 --> 00:39:04.060 align:middle line:84% So we ask, how many sequences are there of let's call 00:39:04.060 --> 00:39:06.100 align:middle line:90% these objects alpha-- 00:39:06.100 --> 00:39:19.940 align:middle line:84% of objects alpha 1, alpha 2, alpha 3, through alpha k. 00:39:19.940 --> 00:39:23.180 align:middle line:84% And these sequences can be any length. 00:39:23.180 --> 00:39:33.900 align:middle line:84% And the size of the sequence is equal to sum i 00:39:33.900 --> 00:39:37.000 align:middle line:90% equals 1 through k alpha sub i-- 00:39:37.000 --> 00:39:42.020 align:middle line:90% 00:39:42.020 --> 00:39:45.710 align:middle line:84% I'm sorry, is equal to the sum of the sizes of these objects. 00:39:45.710 --> 00:40:00.940 align:middle line:90% 00:40:00.940 --> 00:40:17.300 align:middle line:84% Then the generating function for the sequences is 1 over 1 00:40:17.300 --> 00:40:19.790 align:middle line:84% minus the generating function for the objects. 00:40:19.790 --> 00:40:22.707 align:middle line:90% 00:40:22.707 --> 00:40:24.540 align:middle line:84% I should write down the generating functions 00:40:24.540 --> 00:40:25.950 align:middle line:90% for the objects. 00:40:25.950 --> 00:40:31.780 align:middle line:84% A of x equals sum n equals, I guess, 00:40:31.780 --> 00:40:38.660 align:middle line:84% 1 to infinity alpha sub n x to the n because you're really not 00:40:38.660 --> 00:40:44.820 align:middle line:84% allowed to have objects of size 0 in this theorem because then 00:40:44.820 --> 00:40:51.060 align:middle line:84% there would be an infinite number of sequences of size k, 00:40:51.060 --> 00:40:54.900 align:middle line:84% which doesn't make sense because if you could add arbitrarily 00:40:54.900 --> 00:41:01.140 align:middle line:84% many objects of size 0, you can get arbitrary-many objects 00:41:01.140 --> 00:41:02.700 align:middle line:90% of any finite size-- 00:41:02.700 --> 00:41:06.340 align:middle line:84% or arbitrary-many sequences of any finite size. 00:41:06.340 --> 00:41:11.120 align:middle line:90% And that gives you infinities. 00:41:11.120 --> 00:41:15.460 align:middle line:90% 00:41:15.460 --> 00:41:19.940 align:middle line:84% So how does this work for Dyck paths? 00:41:19.940 --> 00:41:30.260 align:middle line:84% Well, any Dyck path is just a sequence of raised Dyck paths. 00:41:30.260 --> 00:41:31.860 align:middle line:90% So you have a-- 00:41:31.860 --> 00:41:39.180 align:middle line:84% so here-- actually, here, the Dyck path is the empty path. 00:41:39.180 --> 00:41:44.620 align:middle line:84% So the raised Dyck path has length 2 and size 1. 00:41:44.620 --> 00:41:50.740 align:middle line:84% Here, the Dyck path is up, down, up, down, which has size 2. 00:41:50.740 --> 00:41:54.100 align:middle line:84% And the raised Dyck path has size 1 more-- 00:41:54.100 --> 00:41:57.420 align:middle line:90% up, up, down, up, down, down. 00:41:57.420 --> 00:41:59.600 align:middle line:84% And here, the Dyck path is up, down. 00:41:59.600 --> 00:42:02.300 align:middle line:90% The raised Dyck path has size 2. 00:42:02.300 --> 00:42:08.800 align:middle line:84% And here, we have Dyck path of size 3, 00:42:08.800 --> 00:42:12.580 align:middle line:84% so the raised Dyck path has size 4. 00:42:12.580 --> 00:42:23.140 align:middle line:84% And now what we do is we use the formula for the number of-- 00:42:23.140 --> 00:42:29.440 align:middle line:84% we say a Dyck path was just a sequence of raised Dyck paths. 00:42:29.440 --> 00:42:34.900 align:middle line:84% So that means C of x equals 1 over 1 minus D of x. 00:42:34.900 --> 00:42:43.330 align:middle line:84% But raised Dyck paths are just Dyck paths with two extra edges. 00:42:43.330 --> 00:42:47.690 align:middle line:90% So D of x equals x times C of x. 00:42:47.690 --> 00:42:51.150 align:middle line:84% And you get C of x equals 1 over 1 minus x C of x. 00:42:51.150 --> 00:42:56.450 align:middle line:90% 00:42:56.450 --> 00:42:58.250 align:middle line:84% I believe there was a generating function 00:42:58.250 --> 00:43:02.050 align:middle line:84% problem on the homework with tiling things of length 3 00:43:02.050 --> 00:43:05.570 align:middle line:90% with 2 by 1 tiles. 00:43:05.570 --> 00:43:10.050 align:middle line:84% And you can use the theorem about generating functions 00:43:10.050 --> 00:43:14.450 align:middle line:84% for sequences to solve that one, too. 00:43:14.450 --> 00:43:17.010 align:middle line:84% Well, I have I have prepared another problem, which 00:43:17.010 --> 00:43:21.050 align:middle line:84% is very similar to that problem, so why don't I 00:43:21.050 --> 00:43:22.500 align:middle line:90% present the similar problem? 00:43:22.500 --> 00:43:56.250 align:middle line:90% 00:43:56.250 --> 00:43:58.110 align:middle line:84% And I actually think that I will-- 00:43:58.110 --> 00:44:16.970 align:middle line:90% 00:44:16.970 --> 00:44:32.490 align:middle line:84% how many tilings of 2 by n strip with two shapes of tiles, 00:44:32.490 --> 00:44:38.410 align:middle line:84% which you can rotate and reflect if you want. 00:44:38.410 --> 00:44:39.720 align:middle line:90% So these are the two shapes. 00:44:39.720 --> 00:44:47.050 align:middle line:90% 00:44:47.050 --> 00:44:50.070 align:middle line:84% Use the sequence theorem for generating functions. 00:44:50.070 --> 00:44:57.770 align:middle line:90% 00:44:57.770 --> 00:45:06.003 align:middle line:90% And let A of-- 00:45:06.003 --> 00:45:16.250 align:middle line:90% 00:45:16.250 --> 00:45:18.116 align:middle line:90% let's use a lowercase a-- 00:45:18.116 --> 00:45:36.850 align:middle line:84% a sub n b number of tilings of 2 by n strip which cannot be 00:45:36.850 --> 00:45:37.730 align:middle line:90% separated into-- 00:45:37.730 --> 00:45:42.770 align:middle line:90% 00:45:42.770 --> 00:45:59.020 align:middle line:84% cannot be separated into a 2 by k and 2 by n minus k strip. 00:45:59.020 --> 00:46:03.330 align:middle line:90% 00:46:03.330 --> 00:46:07.130 align:middle line:84% So in some sense, this is the number of, I don't know, 00:46:07.130 --> 00:46:12.970 align:middle line:84% fundamental tilings of a 2 by n strip which cannot be assembled 00:46:12.970 --> 00:46:15.730 align:middle line:90% out of other tilings. 00:46:15.730 --> 00:46:20.330 align:middle line:84% Let's think-- let's try to figure out 00:46:20.330 --> 00:46:23.150 align:middle line:90% what these numbers are. 00:46:23.150 --> 00:46:35.530 align:middle line:90% 00:46:35.530 --> 00:46:43.690 align:middle line:84% So the smallest tiling is clearly this. 00:46:43.690 --> 00:46:54.570 align:middle line:84% And this is-- so this is a sub 3 because this has length 3. 00:46:54.570 --> 00:46:56.010 align:middle line:84% And there are two of these tilings 00:46:56.010 --> 00:46:59.630 align:middle line:84% because you can reflect it along a horizontal line. 00:46:59.630 --> 00:47:01.360 align:middle line:90% So a sub 3 equals 2. 00:47:01.360 --> 00:47:04.890 align:middle line:90% 00:47:04.890 --> 00:47:07.730 align:middle line:90% a sub 4 equals 0. 00:47:07.730 --> 00:47:12.450 align:middle line:84% I want to say there are no ways to tile a 2 by 4 strip 00:47:12.450 --> 00:47:13.280 align:middle line:90% with those tiles. 00:47:13.280 --> 00:47:15.930 align:middle line:90% 00:47:15.930 --> 00:47:21.020 align:middle line:90% a sub 5 equals 2 because-- 00:47:21.020 --> 00:47:33.290 align:middle line:90% 00:47:33.290 --> 00:47:35.040 align:middle line:90% a sub 6 equals 0. 00:47:35.040 --> 00:47:41.290 align:middle line:90% 00:47:41.290 --> 00:47:46.130 align:middle line:84% So let's call the number of tilings 00:47:46.130 --> 00:47:52.100 align:middle line:84% let B sub n be number of tilings. 00:47:52.100 --> 00:47:56.530 align:middle line:90% 00:47:56.530 --> 00:48:02.690 align:middle line:84% So b sub n is going to be 2, 0, 2. 00:48:02.690 --> 00:48:06.990 align:middle line:84% But I want to claim that there is a tiling of b sub n, 00:48:06.990 --> 00:48:09.690 align:middle line:84% but it's not elementary tiling because it's 00:48:09.690 --> 00:48:15.310 align:middle line:90% made up of two subtiles. 00:48:15.310 --> 00:48:20.610 align:middle line:90% 00:48:20.610 --> 00:48:24.890 align:middle line:84% I'm doing a bad job of drawing it. 00:48:24.890 --> 00:48:30.930 align:middle line:84% I want to claim that there are four of these because you could 00:48:30.930 --> 00:48:35.450 align:middle line:84% reflect the first half, or you could reflect the second half. 00:48:35.450 --> 00:48:39.450 align:middle line:84% And that's just this squared because you 00:48:39.450 --> 00:48:44.310 align:middle line:84% take two a3's and put them next to each other. 00:48:44.310 --> 00:48:47.830 align:middle line:84% And if there are two tilings of length 3, 00:48:47.830 --> 00:48:51.280 align:middle line:84% this gives you four tilings of length 6. 00:48:51.280 --> 00:48:57.410 align:middle line:90% 00:48:57.410 --> 00:49:01.530 align:middle line:90% a7 equals 2. 00:49:01.530 --> 00:49:06.425 align:middle line:90% And 00:49:06.425 --> 00:49:07.050 align:middle line:90% AUDIENCE: Wait. 00:49:07.050 --> 00:49:08.450 align:middle line:90% What is bn? 00:49:08.450 --> 00:49:09.810 align:middle line:90% [INAUDIBLE]? 00:49:09.810 --> 00:49:12.010 align:middle line:84% PETER SHOR: b sub n is the number 00:49:12.010 --> 00:49:17.090 align:middle line:84% of tilings where you don't require that they're 00:49:17.090 --> 00:49:18.270 align:middle line:90% fundamental tilings. 00:49:18.270 --> 00:49:22.210 align:middle line:84% So b sub n can be made up of-- the b's can be made up of two 00:49:22.210 --> 00:49:27.410 align:middle line:84% tilings which are just stuck next to each other, 00:49:27.410 --> 00:49:31.560 align:middle line:84% whereas the a's, we want to require that there are no 00:49:31.560 --> 00:49:33.040 align:middle line:90% vertical-- 00:49:33.040 --> 00:49:37.280 align:middle line:84% you cannot find a vertical line that splits this tiling 00:49:37.280 --> 00:49:38.080 align:middle line:90% into two. 00:49:38.080 --> 00:49:39.455 align:middle line:84% AUDIENCE: So the vertical line is 00:49:39.455 --> 00:49:42.308 align:middle line:84% defined by separating into k and minus k? 00:49:42.308 --> 00:49:43.350 align:middle line:90% PETER SHOR: That's right. 00:49:43.350 --> 00:49:49.280 align:middle line:90% 00:49:49.280 --> 00:49:57.520 align:middle line:84% And I want to claim that a sub 8 equals a5 times 00:49:57.520 --> 00:50:09.880 align:middle line:84% a3 plus a3 times a5 is equal to 8 because you can take this tile 00:50:09.880 --> 00:50:13.320 align:middle line:90% and put it next to this tile-- 00:50:13.320 --> 00:50:16.300 align:middle line:84% or you take this tile and put it before this tile, 00:50:16.300 --> 00:50:18.690 align:middle line:84% or you can take this tile and put it before that tile. 00:50:18.690 --> 00:50:24.800 align:middle line:90% 00:50:24.800 --> 00:50:29.800 align:middle line:84% So thinking about this, if you don't want any vertical lines, 00:50:29.800 --> 00:50:42.120 align:middle line:84% the only way to get an a tile is to start with an L and put all 00:50:42.120 --> 00:50:48.280 align:middle line:84% these Z tiles next to it and then end it with an L. 00:50:48.280 --> 00:50:56.800 align:middle line:84% So a sub 2k plus 1 is equal to 2. 00:50:56.800 --> 00:51:04.640 align:middle line:84% So the odd a's, there are two ways 00:51:04.640 --> 00:51:09.840 align:middle line:84% of giving a fundamental tile for odd numbers 00:51:09.840 --> 00:51:13.040 align:middle line:90% and 0 for even numbers. 00:51:13.040 --> 00:51:27.640 align:middle line:84% So that means A of x is equal to summation k equals 0-- 00:51:27.640 --> 00:51:38.600 align:middle line:84% or k equals 1 through infinity x to the 2k plus 1 times 2 equals 00:51:38.600 --> 00:51:42.370 align:middle line:84% x cubed 1 over 1 minus x squared. 00:51:42.370 --> 00:51:51.805 align:middle line:90% 00:51:51.805 --> 00:51:56.680 align:middle line:84% So now, let's compute this generating function. 00:51:56.680 --> 00:52:09.230 align:middle line:84% B of x is equal to 1 over x cubed 1 over 1 minus x squared-- 00:52:09.230 --> 00:52:16.384 align:middle line:90% 00:52:16.384 --> 00:52:18.860 align:middle line:90% also, 1 minus that. 00:52:18.860 --> 00:52:29.120 align:middle line:90% 00:52:29.120 --> 00:52:30.600 align:middle line:90% Oh, wait. 00:52:30.600 --> 00:52:31.300 align:middle line:90% I forgot. 00:52:31.300 --> 00:52:34.560 align:middle line:90% There's a 2 here. 00:52:34.560 --> 00:52:41.840 align:middle line:84% So 1 minus twice that is equal to-- 00:52:41.840 --> 00:52:45.900 align:middle line:84% we can multiply 1 minus x squared on the top and bottom, 00:52:45.900 --> 00:52:52.750 align:middle line:84% and we get 1 minus x squared over 1 minus x squared minus 2x 00:52:52.750 --> 00:52:53.250 align:middle line:90% cubed. 00:52:53.250 --> 00:53:01.104 align:middle line:90% 00:53:01.104 --> 00:53:08.160 align:middle line:84% So we have a generating function for equals sum b sub n x 00:53:08.160 --> 00:53:08.820 align:middle line:90% to the n. 00:53:08.820 --> 00:53:11.560 align:middle line:90% 00:53:11.560 --> 00:53:15.320 align:middle line:84% So we have a generating function for b. 00:53:15.320 --> 00:53:20.640 align:middle line:84% Now, I want to say that you can read off 00:53:20.640 --> 00:53:22.750 align:middle line:84% a recurrence from a generating function. 00:53:22.750 --> 00:53:35.907 align:middle line:90% 00:53:35.907 --> 00:53:36.740 align:middle line:90% AUDIENCE: I'm sorry. 00:53:36.740 --> 00:53:38.320 align:middle line:90% It's kind of a minor thing. 00:53:38.320 --> 00:53:45.000 align:middle line:84% Is that the b 8 of the top right corner on this blackboard? 00:53:45.000 --> 00:53:48.200 align:middle line:90% PETER SHOR: Oh, yeah, right. 00:53:48.200 --> 00:53:53.920 align:middle line:90% b8 equals 8. 00:53:53.920 --> 00:53:54.690 align:middle line:90% Thank you. 00:53:54.690 --> 00:53:56.560 align:middle line:90% AUDIENCE: So it's not a8? 00:53:56.560 --> 00:53:59.520 align:middle line:90% PETER SHOR: a8 equals 0. 00:53:59.520 --> 00:54:07.040 align:middle line:84% OK, so this is b8 equals a5 times a3 plus a3 times a5. 00:54:07.040 --> 00:54:08.300 align:middle line:90% I'm sorry about that. 00:54:08.300 --> 00:54:12.840 align:middle line:90% 00:54:12.840 --> 00:54:13.650 align:middle line:90% So we have-- 00:54:13.650 --> 00:54:16.025 align:middle line:90% 00:54:16.025 --> 00:54:16.900 align:middle line:90% AUDIENCE: A question. 00:54:16.900 --> 00:54:19.520 align:middle line:84% What did you do to get to simplify [INAUDIBLE] 3 sub 00:54:19.520 --> 00:54:24.170 align:middle line:84% ax when you did the [? xq ?] thing before you do the sequence 00:54:24.170 --> 00:54:24.670 align:middle line:90% theorem? 00:54:24.670 --> 00:54:27.440 align:middle line:90% 00:54:27.440 --> 00:54:30.480 align:middle line:84% PETER SHOR: Well, it's just twice times the sum 00:54:30.480 --> 00:54:33.400 align:middle line:90% of x to all odd numbers. 00:54:33.400 --> 00:54:43.840 align:middle line:84% And maybe I should do this more slowly. 00:54:43.840 --> 00:54:54.360 align:middle line:84% This is x cubed times sum 2x to the 2n n equals 0 through 00:54:54.360 --> 00:54:57.360 align:middle line:90% infinity. 00:54:57.360 --> 00:55:01.400 align:middle line:84% Because the smallest one of these is x cubed, 00:55:01.400 --> 00:55:03.120 align:middle line:90% so you can multiply by that. 00:55:03.120 --> 00:55:07.240 align:middle line:84% And then you have all the even ones after that. 00:55:07.240 --> 00:55:13.800 align:middle line:84% And now, this, you should know the formula for a geometric 00:55:13.800 --> 00:55:23.120 align:middle line:84% series is equal to 2x cubed over 1 minus x squared because sum n 00:55:23.120 --> 00:55:27.630 align:middle line:84% equals 0 to infinity of x to the 2n is 1 minus x squared. 00:55:27.630 --> 00:55:30.400 align:middle line:90% 00:55:30.400 --> 00:55:37.360 align:middle line:84% And, actually, that is very easy to see 00:55:37.360 --> 00:55:43.600 align:middle line:84% because 1 plus x squared plus x fourth plus x 00:55:43.600 --> 00:55:50.680 align:middle line:84% sixth plus dot, dot, dot times 1 minus x squared is equal to 1. 00:55:50.680 --> 00:55:52.120 align:middle line:84% And now we have a minus x squared 00:55:52.120 --> 00:55:55.300 align:middle line:84% plus x squared, a minus x to the fourth plus x to the fourth, 00:55:55.300 --> 00:55:58.080 align:middle line:84% a minus x to the sixth plus x to the sixth. 00:55:58.080 --> 00:56:01.080 align:middle line:84% So all the other terms cancel out. 00:56:01.080 --> 00:56:04.950 align:middle line:84% So this is a formula for a geometric series. 00:56:04.950 --> 00:56:09.240 align:middle line:90% 00:56:09.240 --> 00:56:12.300 align:middle line:90% So what did we have? 00:56:12.300 --> 00:56:15.880 align:middle line:90% We have-- yeah. 00:56:15.880 --> 00:56:18.550 align:middle line:84% I want to say that you can read off 00:56:18.550 --> 00:56:21.900 align:middle line:84% a recurrence from a generating function. 00:56:21.900 --> 00:56:25.670 align:middle line:90% 00:56:25.670 --> 00:56:42.198 align:middle line:84% So suppose you have p of x over 1 minus x squared minus 2x cubed 00:56:42.198 --> 00:56:46.930 align:middle line:90% p of x gives initial conditions. 00:56:46.930 --> 00:56:51.110 align:middle line:90% 00:56:51.110 --> 00:56:59.030 align:middle line:84% And 1 minus x squared minus 2x cubed gives recurrence. 00:56:59.030 --> 00:57:03.750 align:middle line:84% So this recurrence says b sub n is 00:57:03.750 --> 00:57:12.220 align:middle line:84% equal to b sub n minus 2 plus 2 b sub n minus 3. 00:57:12.220 --> 00:57:26.990 align:middle line:90% 00:57:26.990 --> 00:57:30.760 align:middle line:90% I guess we can explain this. 00:57:30.760 --> 00:57:36.390 align:middle line:90% 00:57:36.390 --> 00:57:48.830 align:middle line:84% So summation B sub i x to the i is equal to p of x over 1 00:57:48.830 --> 00:57:53.470 align:middle line:90% minus x squared minus x cubed. 00:57:53.470 --> 00:57:55.730 align:middle line:90% So what this means is that-- 00:57:55.730 --> 00:58:12.310 align:middle line:90% 00:58:12.310 --> 00:58:14.550 align:middle line:84% I want to multiply both sides here. 00:58:14.550 --> 00:58:25.530 align:middle line:84% So we're going to get b sub n x to the n minus b 00:58:25.530 --> 00:58:33.350 align:middle line:84% n minus 2 x to the n minus 2 times x squared-- 00:58:33.350 --> 00:58:34.750 align:middle line:90% oh, wait, this had a 2-- 00:58:34.750 --> 00:58:38.310 align:middle line:90% 00:58:38.310 --> 00:58:49.190 align:middle line:84% minus b n minus 3 x to the n minus 3-- 00:58:49.190 --> 00:59:00.166 align:middle line:90% and there's a 2 here-- 00:59:00.166 --> 00:59:10.370 align:middle line:84% equals x to the n term of p of x. 00:59:10.370 --> 00:59:13.350 align:middle line:90% 00:59:13.350 --> 00:59:16.910 align:middle line:84% So multiplying just this by both sides 00:59:16.910 --> 00:59:21.040 align:middle line:84% and pulling out the x to the n term-- 00:59:21.040 --> 00:59:23.550 align:middle line:90% 00:59:23.550 --> 00:59:25.990 align:middle line:90% oh, times x cubed-- 00:59:25.990 --> 00:59:30.270 align:middle line:90% gives you that. 00:59:30.270 --> 00:59:36.590 align:middle line:84% So this says that bn minus bn minus 2 minus bn minus 3 00:59:36.590 --> 00:59:42.430 align:middle line:84% equals 0 unless there is an x to the n term in p of x. 00:59:42.430 --> 00:59:48.010 align:middle line:84% And p of x is a polynomial with only a finite number of terms, 00:59:48.010 --> 00:59:54.870 align:middle line:84% so after this finite number of terms, you have this recurrence. 00:59:54.870 --> 00:59:57.540 align:middle line:90% So bn is bn-- 00:59:57.540 --> 01:00:01.110 align:middle line:90% 01:00:01.110 --> 01:00:06.960 align:middle line:90% OK, let's go back here. 01:00:06.960 --> 01:00:27.790 align:middle line:90% 01:00:27.790 --> 01:00:35.760 align:middle line:84% So we have b sub n equals b n minus 2 plus 2b n minus 3. 01:00:35.760 --> 01:00:40.310 align:middle line:90% 01:00:40.310 --> 01:00:42.830 align:middle line:90% b0 equals 1. 01:00:42.830 --> 01:00:47.030 align:middle line:90% b3 equals 2. 01:00:47.030 --> 01:00:50.429 align:middle line:90% b5 equals 2. 01:00:50.429 --> 01:00:52.500 align:middle line:90% b6 equals 4. 01:00:52.500 --> 01:00:55.710 align:middle line:90% 01:00:55.710 --> 01:00:58.930 align:middle line:90% b7 equals 2. 01:00:58.930 --> 01:01:04.790 align:middle line:90% 01:01:04.790 --> 01:01:09.110 align:middle line:90% b8 equals 8, et cetera. 01:01:09.110 --> 01:01:13.110 align:middle line:90% So let's see if this works. 01:01:13.110 --> 01:01:22.030 align:middle line:84% So b2 should be equal to b0 plus 2b minus 1. 01:01:22.030 --> 01:01:25.550 align:middle line:90% But b2 is 0 and not 1. 01:01:25.550 --> 01:01:41.830 align:middle line:84% But that is because we had the B of x equals 1 minus x squared 01:01:41.830 --> 01:01:49.390 align:middle line:84% over 1 minus x squared minus 2x cubed. 01:01:49.390 --> 01:01:55.190 align:middle line:84% That's because this term was minus 1, 01:01:55.190 --> 01:02:07.460 align:middle line:84% which says that b2 b2 equals b0 minus 1 when you work it 01:02:07.460 --> 01:02:08.780 align:middle line:90% all out. 01:02:08.780 --> 01:02:13.360 align:middle line:84% But after the second term, that recurrence works great. 01:02:13.360 --> 01:02:19.220 align:middle line:90% b3 is equal to 2b0. 01:02:19.220 --> 01:02:25.820 align:middle line:90% b4 equals 2b2 plus-- 01:02:25.820 --> 01:02:28.220 align:middle line:90% I mean, b2 plus 3b1. 01:02:28.220 --> 01:02:33.340 align:middle line:90% b5 is b3. 01:02:33.340 --> 01:02:37.580 align:middle line:90% b6 is 2b3. 01:02:37.580 --> 01:02:43.620 align:middle line:90% b7 is b5. 01:02:43.620 --> 01:02:52.090 align:middle line:90% b8 is 2b6 plus b8 is-- 01:02:52.090 --> 01:02:57.471 align:middle line:90% 01:02:57.471 --> 01:03:04.580 align:middle line:84% yeah, b8 is b6 plus 2b3, which is 8, et cetera. 01:03:04.580 --> 01:03:09.460 align:middle line:84% So you can just read off the recurrence 01:03:09.460 --> 01:03:11.900 align:middle line:84% from the bottom of the generating function, 01:03:11.900 --> 01:03:19.080 align:middle line:84% and the top numerator tells you what the initial conditions are. 01:03:19.080 --> 01:03:24.120 align:middle line:90% 01:03:24.120 --> 01:03:25.320 align:middle line:90% Any other questions? 01:03:25.320 --> 01:03:35.940 align:middle line:90% 01:03:35.940 --> 01:03:39.960 align:middle line:84% I also could say that you could have figured this out, 01:03:39.960 --> 01:03:44.540 align:middle line:84% this recurrence out, without using 01:03:44.540 --> 01:03:51.960 align:middle line:84% generating functions because, basically, if you have a-- 01:03:51.960 --> 01:03:56.100 align:middle line:90% 01:03:56.100 --> 01:04:05.940 align:middle line:84% oh, wait-- tiling that looks like this, 01:04:05.940 --> 01:04:08.260 align:middle line:90% there are two ways to extend it. 01:04:08.260 --> 01:04:11.780 align:middle line:90% You can extend it-- 01:04:11.780 --> 01:04:19.480 align:middle line:84% use the same tiling in the first part 01:04:19.480 --> 01:04:24.900 align:middle line:84% and add one of these Z tiles and add that. 01:04:24.900 --> 01:04:26.440 align:middle line:90% Or you can extend it. 01:04:26.440 --> 01:04:43.500 align:middle line:90% 01:04:43.500 --> 01:04:50.380 align:middle line:84% So this is-- you've extended this by 2. 01:04:50.380 --> 01:04:55.300 align:middle line:84% So that gives you the formula bn minus 2. 01:04:55.300 --> 01:04:57.820 align:middle line:84% Here, you've extended this by 3, but there 01:04:57.820 --> 01:05:01.760 align:middle line:84% were two ways to do it because you could have added this block, 01:05:01.760 --> 01:05:04.220 align:middle line:84% or you could have added this block reflected 01:05:04.220 --> 01:05:08.500 align:middle line:90% around its horizontal axis. 01:05:08.500 --> 01:05:11.820 align:middle line:90% So that gives you 2 bn minus 3. 01:05:11.820 --> 01:05:21.360 align:middle line:84% And you cannot extend the empty tile in this second way, 01:05:21.360 --> 01:05:27.440 align:middle line:84% which explains why b2 is 0 and not 2-- 01:05:27.440 --> 01:05:31.420 align:middle line:90% 01:05:31.420 --> 01:05:35.160 align:middle line:84% which explains why b2 is 0 and not 1. 01:05:35.160 --> 01:05:39.900 align:middle line:90% 01:05:39.900 --> 01:05:43.220 align:middle line:84% So you could have gotten this recurrence 01:05:43.220 --> 01:05:45.160 align:middle line:84% without using generating functions. 01:05:45.160 --> 01:05:58.560 align:middle line:90% 01:05:58.560 --> 01:06:04.720 align:middle line:90% So any other questions? 01:06:04.720 --> 01:06:12.300 align:middle line:90% 01:06:12.300 --> 01:06:12.820 align:middle line:90% Yeah? 01:06:12.820 --> 01:06:15.060 align:middle line:84% AUDIENCE: So this problem, intuitively, 01:06:15.060 --> 01:06:19.000 align:middle line:84% is that a n is like counting the components. 01:06:19.000 --> 01:06:23.147 align:middle line:84% So bn is like a sequence of individual components? 01:06:23.147 --> 01:06:23.980 align:middle line:90% PETER SHOR: Exactly. 01:06:23.980 --> 01:06:27.500 align:middle line:84% bn is a sequence of these a n components. 01:06:27.500 --> 01:06:34.680 align:middle line:84% So you get b of x equals 1 over 1 minus a of x. 01:06:34.680 --> 01:06:39.980 align:middle line:90% 01:06:39.980 --> 01:06:40.540 align:middle line:90% Yes? 01:06:40.540 --> 01:06:43.500 align:middle line:84% AUDIENCE: Can you show how you got the bn recurrence? 01:06:43.500 --> 01:06:47.040 align:middle line:84% PETER SHOR: How I got the bn recurrence? 01:06:47.040 --> 01:06:47.540 align:middle line:90% Oh. 01:06:47.540 --> 01:06:54.580 align:middle line:90% 01:06:54.580 --> 01:06:59.180 align:middle line:84% So it's something-- if you go through our proof about how 01:06:59.180 --> 01:07:11.460 align:middle line:84% to construct generating functions from recurrences, 01:07:11.460 --> 01:07:18.580 align:middle line:84% if a generating function has, I guess, 01:07:18.580 --> 01:07:40.380 align:middle line:84% summation f sub n x to the n equals p of x over 1 minus a1x 01:07:40.380 --> 01:07:46.700 align:middle line:84% minus a2x squared minus a3x cubed-- 01:07:46.700 --> 01:07:51.540 align:middle line:84% and, actually, this is more general than that-- 01:07:51.540 --> 01:08:00.300 align:middle line:90% p of x gives initial conditions. 01:08:00.300 --> 01:08:14.420 align:middle line:84% And we have f sub n equals a1f n minus 1 plus a2 f n minus 2 01:08:14.420 --> 01:08:17.410 align:middle line:90% plus a3f sub n minus 3. 01:08:17.410 --> 01:08:31.140 align:middle line:90% 01:08:31.140 --> 01:08:40.620 align:middle line:84% e of g if p of x equals 1 plus 2x squared, 01:08:40.620 --> 01:08:52.080 align:middle line:84% then I want to say f sub 0 equals 1, f sub 1 equals 0, 01:08:52.080 --> 01:08:54.270 align:middle line:90% and f sub 2 equals 2. 01:08:54.270 --> 01:09:07.609 align:middle line:90% 01:09:07.609 --> 01:09:13.210 align:middle line:84% Well, actually, that's not quite true 01:09:13.210 --> 01:09:20.770 align:middle line:84% because I should probably use this as the example. 01:09:20.770 --> 01:09:23.569 align:middle line:84% We have B of x equals 1 minus x squared over 1 01:09:23.569 --> 01:09:25.440 align:middle line:90% minus x squared minus x cubed. 01:09:25.440 --> 01:09:44.090 align:middle line:90% 01:09:44.090 --> 01:09:53.649 align:middle line:84% So 1 over 1 minus x squared minus 2x cubed would give you f0 01:09:53.649 --> 01:10:07.010 align:middle line:84% equals 1 and fn equals fn minus 2 plus 2f n minus 3. 01:10:07.010 --> 01:10:16.650 align:middle line:84% So that would give you 1, 0, 1, 2, and this is-- 01:10:16.650 --> 01:10:19.170 align:middle line:90% f4 would be f2-- 01:10:19.170 --> 01:10:20.450 align:middle line:90% 1 again. 01:10:20.450 --> 01:10:25.270 align:middle line:84% f5 would be 2 plus 2, which is 4. 01:10:25.270 --> 01:10:30.570 align:middle line:90% 01:10:30.570 --> 01:10:36.010 align:middle line:84% So this is 1 over 1 minus x squared minus 2x cubed. 01:10:36.010 --> 01:10:41.770 align:middle line:84% But we also have minus x squared over 1 minus x squared minus 2x 01:10:41.770 --> 01:10:43.010 align:middle line:90% cubed. 01:10:43.010 --> 01:10:48.450 align:middle line:84% And this gives you, basically, this, except you have to-- 01:10:48.450 --> 01:10:56.810 align:middle line:84% this will give you a minus 1 here and then 0 minus 1. 01:10:56.810 --> 01:11:06.850 align:middle line:84% The next one was minus 2 and minus 1 minus 4. 01:11:06.850 --> 01:11:09.770 align:middle line:84% And then, when you add them up, you actually 01:11:09.770 --> 01:11:13.530 align:middle line:90% get the sequence of f sub i. 01:11:13.530 --> 01:11:25.320 align:middle line:84% So that gives you 1, 0, 0, 2, 0, 2, et cetera. 01:11:25.320 --> 01:11:28.170 align:middle line:90% 01:11:28.170 --> 01:11:38.200 align:middle line:84% And we have 1, 0, 0, 2, 0, 2, et cetera. 01:11:38.200 --> 01:11:49.000 align:middle line:90%