WEBVTT 00:00:00.000 --> 00:00:02.435 align:middle line:90% [SQUEAKING] 00:00:02.435 --> 00:00:03.896 align:middle line:90% [RUSTLING] 00:00:03.896 --> 00:00:07.305 align:middle line:90% [CLICKING] 00:00:07.305 --> 00:00:41.480 align:middle line:90% 00:00:41.480 --> 00:00:43.040 align:middle line:90% PETER SHOR: So, if you have a-- 00:00:43.040 --> 00:00:51.560 align:middle line:90% 00:00:51.560 --> 00:00:54.000 align:middle line:84% so, how would you go about choosing 00:00:54.000 --> 00:00:56.620 align:middle line:90% k things from n objects? 00:00:56.620 --> 00:00:57.980 align:middle line:90% What's the easy way to do it? 00:00:57.980 --> 00:01:04.510 align:middle line:90% 00:01:04.510 --> 00:01:06.310 align:middle line:90% Choose them one at a time. 00:01:06.310 --> 00:01:09.356 align:middle line:90% So, n objects. 00:01:09.356 --> 00:01:13.550 align:middle line:90% 00:01:13.550 --> 00:01:20.470 align:middle line:90% n ways to choose first thing. 00:01:20.470 --> 00:01:23.380 align:middle line:90% 00:01:23.380 --> 00:01:28.750 align:middle line:84% n minus 1 ways to choose the second thing because once 00:01:28.750 --> 00:01:31.420 align:middle line:84% you've chosen the first thing, you can't choose again. 00:01:31.420 --> 00:01:37.430 align:middle line:84% n minus 2 ways to choose [INAUDIBLE] third thing. 00:01:37.430 --> 00:01:41.182 align:middle line:90% 00:01:41.182 --> 00:01:51.990 align:middle line:84% And then with n minus k [INAUDIBLE] 00:01:51.990 --> 00:01:59.166 align:middle line:84% So this gives us n, n minus 1, n minus 2, 00:01:59.166 --> 00:02:03.630 align:middle line:90% through n minus k plus 1. 00:02:03.630 --> 00:02:08.190 align:middle line:84% So I guess this should be n minus k plus 1 [INAUDIBLE]. 00:02:08.190 --> 00:02:09.279 align:middle line:90% So, is this correct? 00:02:09.279 --> 00:02:12.150 align:middle line:90% 00:02:12.150 --> 00:02:12.710 align:middle line:90% No. 00:02:12.710 --> 00:02:15.510 align:middle line:90% Why not? 00:02:15.510 --> 00:02:16.110 align:middle line:90% Yeah? 00:02:16.110 --> 00:02:19.150 align:middle line:84% STUDENT: You want to divide their order. 00:02:19.150 --> 00:02:19.950 align:middle line:90% PETER SHOR: Yeah. 00:02:19.950 --> 00:02:26.730 align:middle line:84% It's basically because we've overcounted. 00:02:26.730 --> 00:02:31.710 align:middle line:84% I mean, if you had a set of 1, 2, 3, 4, 5, 00:02:31.710 --> 00:02:36.350 align:middle line:84% you could choose 3 and 4, but we can also choose 4 and 3, 00:02:36.350 --> 00:02:38.670 align:middle line:84% and these two ways will give us the same set. 00:02:38.670 --> 00:02:41.590 align:middle line:84% So what we need to do is we need to divide by the number of ways 00:02:41.590 --> 00:02:45.310 align:middle line:84% of getting each of these objects, 00:02:45.310 --> 00:02:49.390 align:middle line:84% and that's k factorial because we could have gotten them 00:02:49.390 --> 00:02:50.250 align:middle line:90% in any order. 00:02:50.250 --> 00:02:56.020 align:middle line:84% So we divide by n by k factorial. 00:02:56.020 --> 00:02:59.760 align:middle line:84% So this is n choose k is equal to-- 00:02:59.760 --> 00:03:06.260 align:middle line:90% 00:03:06.260 --> 00:03:13.220 align:middle line:84% Now, there's another expression for this, which is actually 00:03:13.220 --> 00:03:16.580 align:middle line:84% maybe a little simpler [INAUDIBLE], 00:03:16.580 --> 00:03:21.580 align:middle line:84% we can multiply the top by n minus k factorial and the bottom 00:03:21.580 --> 00:03:24.940 align:middle line:90% by n minus k factorial. 00:03:24.940 --> 00:03:28.500 align:middle line:84% And we see that this is equal to n factorial 00:03:28.500 --> 00:03:32.120 align:middle line:90% over k factorial n minus k. 00:03:32.120 --> 00:03:40.780 align:middle line:90% 00:03:40.780 --> 00:03:49.380 align:middle line:84% Now, one of the really nice techniques in proving, I guess, 00:03:49.380 --> 00:03:51.660 align:middle line:84% theorems about binomial coefficients 00:03:51.660 --> 00:03:53.900 align:middle line:84% and other kind of things is you count 00:03:53.900 --> 00:03:55.740 align:middle line:90% things in two different ways. 00:03:55.740 --> 00:03:59.060 align:middle line:84% And if you count the same set of things in two different ways, 00:03:59.060 --> 00:04:04.740 align:middle line:84% these two expressions have to be the same. 00:04:04.740 --> 00:04:07.080 align:middle line:90% So what can we do? 00:04:07.080 --> 00:04:20.159 align:middle line:84% Well, let's count ways to choose k things. 00:04:20.159 --> 00:04:24.620 align:middle line:90% 00:04:24.620 --> 00:04:34.060 align:middle line:84% Well, either the last thing is in the set 00:04:34.060 --> 00:04:35.840 align:middle line:90% or it's not in the set. 00:04:35.840 --> 00:04:38.426 align:middle line:90% 00:04:38.426 --> 00:04:41.630 align:middle line:90% If it's in the set, we have-- 00:04:41.630 --> 00:04:46.580 align:middle line:90% 00:04:46.580 --> 00:04:50.100 align:middle line:84% in the set, we have n minus 1 things here. 00:04:50.100 --> 00:04:51.750 align:middle line:84% And the last thing is in this set, 00:04:51.750 --> 00:04:59.190 align:middle line:84% and also, I guess, k minus 1 things. 00:04:59.190 --> 00:05:01.930 align:middle line:84% And the first n minus 1 things in the set. 00:05:01.930 --> 00:05:06.530 align:middle line:84% And if the last thing is not in the set, 00:05:06.530 --> 00:05:09.650 align:middle line:84% then we'll color it white rather than blue, 00:05:09.650 --> 00:05:11.660 align:middle line:90% and you get this here. 00:05:11.660 --> 00:05:16.250 align:middle line:90% 00:05:16.250 --> 00:05:25.970 align:middle line:84% So what's the number of ways of choosing k minus 1 things 00:05:25.970 --> 00:05:27.320 align:middle line:90% out of n minus 1 things? 00:05:27.320 --> 00:05:29.970 align:middle line:90% 00:05:29.970 --> 00:05:30.770 align:middle line:90% Yeah? 00:05:30.770 --> 00:05:32.530 align:middle line:84% STUDENT: N minus 1 choose k works. 00:05:32.530 --> 00:05:34.050 align:middle line:90% PETER SHOR: Yeah, exactly. 00:05:34.050 --> 00:05:37.810 align:middle line:84% So n minus 1 choose k minus 1 ways up here, 00:05:37.810 --> 00:05:42.530 align:middle line:84% and we get n minus 1 choose k ways 00:05:42.530 --> 00:05:44.610 align:middle line:90% of choosing these things here. 00:05:44.610 --> 00:05:50.480 align:middle line:84% So that shows us that n choose k is 00:05:50.480 --> 00:05:58.622 align:middle line:84% equal to n plus 1 choose k minus 1 plus n minus 1 choose k. 00:05:58.622 --> 00:06:01.800 align:middle line:90% 00:06:01.800 --> 00:06:05.560 align:middle line:90% So, let's see. 00:06:05.560 --> 00:06:10.440 align:middle line:84% So we've got this expression by counting 00:06:10.440 --> 00:06:14.320 align:middle line:84% the number of ways of choosing k things out 00:06:14.320 --> 00:06:15.860 align:middle line:90% of n in two different ways. 00:06:15.860 --> 00:06:25.000 align:middle line:84% First, we colored one of them, and then did the-- 00:06:25.000 --> 00:06:26.280 align:middle line:90% used induction. 00:06:26.280 --> 00:06:32.280 align:middle line:84% And the other way is we just prove the binomial coefficient 00:06:32.280 --> 00:06:33.640 align:middle line:90% n choose k. 00:06:33.640 --> 00:06:35.680 align:middle line:84% So counting things in two different ways 00:06:35.680 --> 00:06:38.440 align:middle line:90% can give you a theorem. 00:06:38.440 --> 00:06:44.600 align:middle line:84% And, in fact, it can give you very deep theorems which-- 00:06:44.600 --> 00:06:48.880 align:middle line:84% as well as relatively easy theorems. 00:06:48.880 --> 00:06:52.600 align:middle line:90% 00:06:52.600 --> 00:06:58.200 align:middle line:84% And there's this beautiful proof of Cayley's tree theorem 00:06:58.200 --> 00:07:00.120 align:middle line:84% in the notes, which depends on counting 00:07:00.120 --> 00:07:03.740 align:middle line:84% things in different ways, and I will eventually get to it, 00:07:03.740 --> 00:07:06.560 align:middle line:90% probably not until next class. 00:07:06.560 --> 00:07:10.020 align:middle line:90% So what do I want to say now? 00:07:10.020 --> 00:07:12.720 align:middle line:90% 00:07:12.720 --> 00:07:15.400 align:middle line:90% Yeah. 00:07:15.400 --> 00:07:33.440 align:middle line:84% This is-- this formula gives us Pascal's triangle here. 00:07:33.440 --> 00:07:42.880 align:middle line:84% I'm just saying 1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 6, 4, 1. 00:07:42.880 --> 00:08:01.750 align:middle line:84% And if this row is n, so n equals 1, n equals 0, 00:08:01.750 --> 00:08:03.910 align:middle line:90% n equals 2, et cetera. 00:08:03.910 --> 00:08:11.290 align:middle line:84% And this is k equals 0, k equals 1, k equals 2. 00:08:11.290 --> 00:08:15.550 align:middle line:90% 00:08:15.550 --> 00:08:22.030 align:middle line:84% Well, you get this number from adding the number above it, 00:08:22.030 --> 00:08:26.150 align:middle line:84% and that follows from this identity. 00:08:26.150 --> 00:08:28.070 align:middle line:84% And from Pascal's triangle, you can 00:08:28.070 --> 00:08:34.549 align:middle line:84% see that n choose k equals n choose n minus k 00:08:34.549 --> 00:08:41.190 align:middle line:84% because Pascal's triangle is symmetric. 00:08:41.190 --> 00:08:44.455 align:middle line:84% The way of getting this number is exactly the same-- 00:08:44.455 --> 00:08:45.830 align:middle line:84% the ways of computing this number 00:08:45.830 --> 00:08:49.910 align:middle line:84% is exactly the same way computing the-- 00:08:49.910 --> 00:08:53.910 align:middle line:84% way we compute this number because each row 00:08:53.910 --> 00:08:57.910 align:middle line:90% is a palindrome. 00:08:57.910 --> 00:09:02.550 align:middle line:84% But you can also see this by the formula n factorial over k 00:09:02.550 --> 00:09:06.010 align:middle line:90% factorial n minus k factorial. 00:09:06.010 --> 00:09:25.750 align:middle line:90% 00:09:25.750 --> 00:09:28.770 align:middle line:84% OK, so the multinomial coefficients. 00:09:28.770 --> 00:09:37.437 align:middle line:90% 00:09:37.437 --> 00:09:45.300 align:middle line:84% If a plus b plus c equals n, n choose a, b, 00:09:45.300 --> 00:09:53.280 align:middle line:84% c is equal to n factorial over a factorial, b 00:09:53.280 --> 00:09:57.680 align:middle line:84% factorial, c factorial, and this is the number of ways-- 00:09:57.680 --> 00:10:08.900 align:middle line:84% this is the number of ways of choosing a-- 00:10:08.900 --> 00:10:11.700 align:middle line:90% 00:10:11.700 --> 00:10:22.710 align:middle line:84% sets of size a, b, c from n things. 00:10:22.710 --> 00:10:27.260 align:middle line:90% 00:10:27.260 --> 00:10:36.580 align:middle line:84% And these sets are labeled, but the things within the sets are-- 00:10:36.580 --> 00:10:44.190 align:middle line:84% I mean things within the sets are interchangeable. 00:10:44.190 --> 00:10:51.010 align:middle line:90% So, and how do you prove this? 00:10:51.010 --> 00:10:54.490 align:middle line:84% Well, there's two ways of proving this. 00:10:54.490 --> 00:11:01.630 align:middle line:90% First, you can use this to-- 00:11:01.630 --> 00:11:04.730 align:middle line:90% 00:11:04.730 --> 00:11:11.080 align:middle line:84% and use a, b, c to n minus 1 choose a minus 1, 00:11:11.080 --> 00:11:18.490 align:middle line:84% b, c, n minus 1 choose a, b minus 1, c, 00:11:18.490 --> 00:11:23.470 align:middle line:84% plus n minus 1 choose a, b, c minus 1. 00:11:23.470 --> 00:11:26.130 align:middle line:90% 00:11:26.130 --> 00:11:28.074 align:middle line:90% So it's one big glick. 00:11:28.074 --> 00:11:30.970 align:middle line:90% 00:11:30.970 --> 00:11:35.090 align:middle line:84% Because you just asked, which set was the last thing in? 00:11:35.090 --> 00:11:38.423 align:middle line:84% And then the n minus 1 do many things. 00:11:38.423 --> 00:11:41.810 align:middle line:84% You know what the size of the sets are, 00:11:41.810 --> 00:11:47.222 align:middle line:84% or you can prove it with n factorial-- 00:11:47.222 --> 00:11:51.290 align:middle line:90% 00:11:51.290 --> 00:11:59.170 align:middle line:84% I'm sorry-- n choose a, b, c is 1, 00:11:59.170 --> 00:12:08.210 align:middle line:84% n choose a times n minus a, b, c. 00:12:08.210 --> 00:12:12.330 align:middle line:84% So here, you first choose the first set, which has size a. 00:12:12.330 --> 00:12:13.770 align:middle line:84% And now you choose the second set 00:12:13.770 --> 00:12:16.550 align:middle line:84% for the remaining n minus 1 thing-- a minus things. 00:12:16.550 --> 00:12:18.330 align:middle line:90% And that has size b. 00:12:18.330 --> 00:12:26.770 align:middle line:84% And this is equal to our formula for the binomial coefficient, n 00:12:26.770 --> 00:12:33.010 align:middle line:84% factorial, a factorial, n minus a factorial times i s. 00:12:33.010 --> 00:12:36.800 align:middle line:90% 00:12:36.800 --> 00:12:43.560 align:middle line:84% n minus a is b plus c, so b plus c, factorial, 00:12:43.560 --> 00:12:48.520 align:middle line:84% over b times the real, c times the real. 00:12:48.520 --> 00:12:52.000 align:middle line:84% And now remember that b plus c equals n minus a. 00:12:52.000 --> 00:12:54.340 align:middle line:90% So these two terms cancel. 00:12:54.340 --> 00:12:57.560 align:middle line:84% And you get n factorial of a factorial, 00:12:57.560 --> 00:12:59.720 align:middle line:90% b factorial, c factorial. 00:12:59.720 --> 00:13:04.520 align:middle line:84% And of course, there's no reason that you would 00:13:04.520 --> 00:13:07.590 align:middle line:90% need to stop these three sets. 00:13:07.590 --> 00:13:10.280 align:middle line:90% 00:13:10.280 --> 00:13:15.960 align:middle line:84% So, we get n choose a1, a2, choose 00:13:15.960 --> 00:13:22.800 align:middle line:84% ak is equal to n factorial over the product i jth 1 00:13:22.800 --> 00:13:33.260 align:middle line:84% through k, a sub bar, with a1 plus k2 plus ak equals n. 00:13:33.260 --> 00:13:36.320 align:middle line:90% 00:13:36.320 --> 00:13:40.300 align:middle line:84% So this is multinomial coefficients. 00:13:40.300 --> 00:13:52.900 align:middle line:90% 00:13:52.900 --> 00:13:53.400 align:middle line:90% Oops. 00:13:53.400 --> 00:14:13.640 align:middle line:90% 00:14:13.640 --> 00:14:16.560 align:middle line:84% So now I want to talk about bijection, which is 00:14:16.560 --> 00:14:18.480 align:middle line:90% another way of counting things. 00:14:18.480 --> 00:14:23.840 align:middle line:84% And common, of course, really like 00:14:23.840 --> 00:14:32.310 align:middle line:84% finding bijective proofs of n and a's, although these are not 00:14:32.310 --> 00:14:37.670 align:middle line:84% necessarily the easiest piece, but often they are, 00:14:37.670 --> 00:14:38.470 align:middle line:90% so bijection. 00:14:38.470 --> 00:14:41.750 align:middle line:90% 00:14:41.750 --> 00:14:52.430 align:middle line:84% The function is a bijection f is projective, 00:14:52.430 --> 00:14:54.610 align:middle line:90% is injective and surjective. 00:14:54.610 --> 00:15:06.390 align:middle line:90% 00:15:06.390 --> 00:15:10.540 align:middle line:84% I guess, a to b, or a and b, are sets of things. 00:15:10.540 --> 00:15:16.030 align:middle line:90% 00:15:16.030 --> 00:15:26.950 align:middle line:84% Injective means f of a1 is not equal to f of a2 00:15:26.950 --> 00:15:31.340 align:middle line:90% if a1 is not equal to a2. 00:15:31.340 --> 00:15:40.500 align:middle line:84% So you don't get mapping to things onto the same object. 00:15:40.500 --> 00:15:50.540 align:middle line:84% And surjective is if b is in d, then 00:15:50.540 --> 00:15:56.100 align:middle line:84% there exists an a such that f of a is b. 00:15:56.100 --> 00:16:00.460 align:middle line:84% So surjective means that everything in b 00:16:00.460 --> 00:16:01.800 align:middle line:90% is in the range of f. 00:16:01.800 --> 00:16:10.900 align:middle line:90% 00:16:10.900 --> 00:16:23.630 align:middle line:84% Also, f is a bijection if and only 00:16:23.630 --> 00:16:30.940 align:middle line:90% if f has f, an inverse of fg. 00:16:30.940 --> 00:16:41.960 align:middle line:84% Now for inverse, we need f of g of a whose a and g of f of bc. 00:16:41.960 --> 00:17:07.420 align:middle line:90% 00:17:07.420 --> 00:17:10.640 align:middle line:84% And again, we'll prove one of these directions. 00:17:10.640 --> 00:17:19.334 align:middle line:90% 00:17:19.334 --> 00:17:29.470 align:middle line:84% If f has an inverse, it is a bijection. 00:17:29.470 --> 00:17:38.730 align:middle line:90% 00:17:38.730 --> 00:17:41.330 align:middle line:84% So what we need to prove is that it's 00:17:41.330 --> 00:17:43.420 align:middle line:90% injective and then surjective. 00:17:43.420 --> 00:17:48.130 align:middle line:90% 00:17:48.130 --> 00:18:01.250 align:middle line:84% If f of a1 equals f of A true, then g of f a1 00:18:01.250 --> 00:18:09.410 align:middle line:84% equals g of f of k2, just because g is a function. 00:18:09.410 --> 00:18:15.130 align:middle line:84% But this has to equal a2, and this has to be a1. 00:18:15.130 --> 00:18:21.210 align:middle line:90% So this implies a1. 00:18:21.210 --> 00:18:27.470 align:middle line:90% So f is injective. 00:18:27.470 --> 00:18:31.250 align:middle line:90% 00:18:31.250 --> 00:18:42.890 align:middle line:84% And now if b is in the range of f, f of g of b equals phi. 00:18:42.890 --> 00:18:45.810 align:middle line:84% So there's an element that maps onto b. 00:18:45.810 --> 00:18:47.470 align:middle line:90% So it's subjective. 00:18:47.470 --> 00:18:54.370 align:middle line:90% 00:18:54.370 --> 00:18:59.030 align:middle line:84% And we've proven that if f has an inverse, that is a bijection. 00:18:59.030 --> 00:19:01.570 align:middle line:90% 00:19:01.570 --> 00:19:04.330 align:middle line:84% And the nice thing about bijections 00:19:04.330 --> 00:19:09.750 align:middle line:90% is that f is a bijection. 00:19:09.750 --> 00:19:12.570 align:middle line:90% 00:19:12.570 --> 00:19:16.470 align:middle line:84% The size of a is equal to the size of b. 00:19:16.470 --> 00:19:23.480 align:middle line:90% 00:19:23.480 --> 00:19:26.800 align:middle line:84% And so if we counted the things in a, 00:19:26.800 --> 00:19:31.240 align:middle line:84% then we will also have counted the things in b. 00:19:31.240 --> 00:19:35.920 align:middle line:84% And we will see several examples of this today. 00:19:35.920 --> 00:19:38.800 align:middle line:84% And the first one I want to show you 00:19:38.800 --> 00:19:43.740 align:middle line:84% is another thing with binomial coefficients. 00:19:43.740 --> 00:19:47.760 align:middle line:90% 00:19:47.760 --> 00:19:48.310 align:middle line:90% Let's see. 00:19:48.310 --> 00:20:03.960 align:middle line:90% 00:20:03.960 --> 00:20:22.000 align:middle line:84% How many different ways of coloring are OK, 00:20:22.000 --> 00:20:24.713 align:middle line:90% n balls with j colors. 00:20:24.713 --> 00:20:30.270 align:middle line:90% 00:20:30.270 --> 00:20:33.350 align:middle line:84% And I should tell you what I mean by different here. 00:20:33.350 --> 00:20:42.239 align:middle line:90% 00:20:42.239 --> 00:20:43.730 align:middle line:90% So let's do an example. 00:20:43.730 --> 00:20:49.710 align:middle line:90% 00:20:49.710 --> 00:20:59.190 align:middle line:90% n equals 4 and j is 3. 00:20:59.190 --> 00:21:01.300 align:middle line:84% Well, we could color all the same color. 00:21:01.300 --> 00:21:04.390 align:middle line:90% 00:21:04.390 --> 00:21:08.910 align:middle line:84% I want to claim that there's only one way of doing that. 00:21:08.910 --> 00:21:11.710 align:middle line:84% For each color, there's only one way of doing that. 00:21:11.710 --> 00:21:13.330 align:middle line:90% So there are three ways total. 00:21:13.330 --> 00:21:19.310 align:middle line:90% 00:21:19.310 --> 00:21:42.050 align:middle line:84% We could color one of them one color, 00:21:42.050 --> 00:21:46.150 align:middle line:84% and the other three another color. 00:21:46.150 --> 00:21:52.950 align:middle line:84% And I want to claim there are six ways of doing this, 00:21:52.950 --> 00:21:58.870 align:middle line:84% because you choose one color for the first one and three 00:21:58.870 --> 00:22:01.050 align:middle line:90% colors for the other one. 00:22:01.050 --> 00:22:03.390 align:middle line:84% So there are three ways to choose 00:22:03.390 --> 00:22:06.163 align:middle line:84% the first color and two ways to choose the second color. 00:22:06.163 --> 00:22:07.080 align:middle line:90% So there are six ways. 00:22:07.080 --> 00:22:09.830 align:middle line:90% 00:22:09.830 --> 00:22:12.870 align:middle line:84% So note that we're not distinguishing 00:22:12.870 --> 00:22:15.950 align:middle line:84% the coloring red, blue, blue, blue from the color of blue, 00:22:15.950 --> 00:22:18.080 align:middle line:90% red, blue, blue. 00:22:18.080 --> 00:22:21.580 align:middle line:84% So these, we're just asking for the total number 00:22:21.580 --> 00:22:23.860 align:middle line:90% of balls of each color. 00:22:23.860 --> 00:22:31.420 align:middle line:84% And you could also color two of them one color and two of them 00:22:31.420 --> 00:22:33.380 align:middle line:90% another color. 00:22:33.380 --> 00:22:37.100 align:middle line:84% And I want to claim there's three ways of doing this, 00:22:37.100 --> 00:22:39.760 align:middle line:84% because there are three colors you don't use. 00:22:39.760 --> 00:22:43.380 align:middle line:84% And once you've decided which color you're not going to use, 00:22:43.380 --> 00:22:48.980 align:middle line:90% the rest of them are given. 00:22:48.980 --> 00:22:51.940 align:middle line:90% So that's 12 so far. 00:22:51.940 --> 00:22:56.400 align:middle line:84% And is there another way I've missed? 00:22:56.400 --> 00:23:00.440 align:middle line:90% 00:23:00.440 --> 00:23:00.940 align:middle line:90% Yeah? 00:23:00.940 --> 00:23:02.580 align:middle line:90% STUDENT: It's like 2, 1, 1. 00:23:02.580 --> 00:23:04.420 align:middle line:90% PETER SHOR: Yeah, 2, 1, 1. 00:23:04.420 --> 00:23:13.560 align:middle line:84% OK, so we could have red, blue, green, green. 00:23:13.560 --> 00:23:15.390 align:middle line:84% So how many ways are there of doing this? 00:23:15.390 --> 00:23:23.450 align:middle line:90% 00:23:23.450 --> 00:23:24.220 align:middle line:90% Yeah? 00:23:24.220 --> 00:23:26.410 align:middle line:90% STUDENT: There are three ways. 00:23:26.410 --> 00:23:28.065 align:middle line:84% So just choose which color will be two. 00:23:28.065 --> 00:23:29.690 align:middle line:84% PETER SHOR: Yeah, there are three ways. 00:23:29.690 --> 00:23:32.290 align:middle line:84% Because once you've chosen which color you're 00:23:32.290 --> 00:23:40.350 align:middle line:84% going to make two balls, then the rest is set. 00:23:40.350 --> 00:23:42.890 align:middle line:90% So there are three ways. 00:23:42.890 --> 00:23:46.850 align:middle line:84% And if we add this up, we get 15. 00:23:46.850 --> 00:23:56.210 align:middle line:84% And 15 just happens to be equal to 5 choose 2. 00:23:56.210 --> 00:24:02.510 align:middle line:84% And just from this one example, it might be a coincidence, 00:24:02.510 --> 00:24:05.390 align:middle line:84% but you can try a whole bunch of other examples. 00:24:05.390 --> 00:24:08.230 align:middle line:84% And it always works out to be a binomial coefficient. 00:24:08.230 --> 00:24:11.490 align:middle line:84% So it turns out not to be a coincidence. 00:24:11.490 --> 00:24:15.570 align:middle line:84% It's a theorem that we'd like to prove. 00:24:15.570 --> 00:24:21.010 align:middle line:84% And 5 choose 2 is a number of ways of choosing two things out 00:24:21.010 --> 00:24:23.090 align:middle line:90% of five things. 00:24:23.090 --> 00:24:30.010 align:middle line:84% So how can we take these colorings of balls 00:24:30.010 --> 00:24:34.890 align:middle line:84% and turn it into finding a subset of two things 00:24:34.890 --> 00:24:35.760 align:middle line:90% out of five things? 00:24:35.760 --> 00:24:38.610 align:middle line:90% 00:24:38.610 --> 00:24:45.830 align:middle line:84% And this is actually a fairly clever bijection, 00:24:45.830 --> 00:24:48.310 align:middle line:84% so I'm not going to ask you to figure it out. 00:24:48.310 --> 00:24:57.910 align:middle line:90% 00:24:57.910 --> 00:24:58.410 align:middle line:90% Oh. 00:24:58.410 --> 00:25:02.282 align:middle line:90% 00:25:02.282 --> 00:25:05.106 align:middle line:84% You'll notice I'm going to do this [INAUDIBLE]. 00:25:05.106 --> 00:25:08.090 align:middle line:90% 00:25:08.090 --> 00:25:14.840 align:middle line:90% So let's pick one of these. 00:25:14.840 --> 00:25:16.901 align:middle line:84% Maybe we should pick two of them. 00:25:16.901 --> 00:25:27.700 align:middle line:84% Let's pick red, blue, green, green. 00:25:27.700 --> 00:25:34.160 align:middle line:90% 00:25:34.160 --> 00:25:40.360 align:middle line:90% So first step, step bijection-- 00:25:40.360 --> 00:25:46.200 align:middle line:84% order balls with some fixed ordering. 00:25:46.200 --> 00:25:48.800 align:middle line:84% And in this case, can be red, blue, green. 00:25:48.800 --> 00:25:52.720 align:middle line:90% 00:25:52.720 --> 00:26:04.480 align:middle line:84% OK, second step is on certain dividers, 00:26:04.480 --> 00:26:07.870 align:middle line:84% so we're going to put a divider between-- 00:26:07.870 --> 00:26:12.800 align:middle line:90% 00:26:12.800 --> 00:26:16.120 align:middle line:84% actually, you know what, nobody caught my mistake. 00:26:16.120 --> 00:26:18.680 align:middle line:90% 00:26:18.680 --> 00:26:23.320 align:middle line:84% 15 is 6 choose 2, and not 5 choose 2. 00:26:23.320 --> 00:26:28.800 align:middle line:84% So, because it's 6 times 5 divided by 2, that's 15. 00:26:28.800 --> 00:26:30.700 align:middle line:90% 5 choose 2 is 10. 00:26:30.700 --> 00:26:33.800 align:middle line:90% 00:26:33.800 --> 00:26:37.330 align:middle line:84% So I should warn you that I cannot do elementary arithmetic 00:26:37.330 --> 00:26:39.920 align:middle line:90% when I'm lecturing. 00:26:39.920 --> 00:26:46.320 align:middle line:84% So the second step is to insert dividers. 00:26:46.320 --> 00:26:55.160 align:middle line:84% So what I do is we're going to take this and put 00:26:55.160 --> 00:26:56.470 align:middle line:90% dividers between the colors. 00:26:56.470 --> 00:27:05.440 align:middle line:90% 00:27:05.440 --> 00:27:13.280 align:middle line:84% OK, now I want to claim the third step is in these colors. 00:27:13.280 --> 00:27:27.210 align:middle line:90% 00:27:27.210 --> 00:27:41.150 align:middle line:84% So now we have two dividers and six things, 00:27:41.150 --> 00:27:44.160 align:middle line:84% because we had four balls and two dividers. 00:27:44.160 --> 00:27:50.490 align:middle line:90% 00:27:50.490 --> 00:27:55.010 align:middle line:84% And I want to claim that this is a bijection. 00:27:55.010 --> 00:27:56.470 align:middle line:90% Why? 00:27:56.470 --> 00:27:58.990 align:middle line:90% Because we can reverse it. 00:27:58.990 --> 00:28:01.550 align:middle line:84% If we know where the divisors are, what we do 00:28:01.550 --> 00:28:04.860 align:middle line:84% is we color everything before the first divider red, 00:28:04.860 --> 00:28:07.660 align:middle line:84% everything between the first and second divider 00:28:07.660 --> 00:28:10.620 align:middle line:84% blue, and everything after the second divider green. 00:28:10.620 --> 00:28:15.080 align:middle line:84% And if we don't have any red balls, 00:28:15.080 --> 00:28:18.620 align:middle line:84% we just put the first divider at the beginning. 00:28:18.620 --> 00:28:25.260 align:middle line:84% So this can be reversed and this can be reversed. 00:28:25.260 --> 00:28:32.220 align:middle line:84% So that means this function that maps from colored balls to balls 00:28:32.220 --> 00:28:38.220 align:middle line:90% and dividers is invertible. 00:28:38.220 --> 00:28:41.160 align:middle line:84% So that means that it's a bijection. 00:28:41.160 --> 00:28:41.660 align:middle line:90% OK. 00:28:41.660 --> 00:28:49.920 align:middle line:90% 00:28:49.920 --> 00:28:52.000 align:middle line:84% So that means that the number of-- 00:28:52.000 --> 00:28:54.586 align:middle line:90% 00:28:54.586 --> 00:28:55.830 align:middle line:90% OK, maybe I should-- 00:28:55.830 --> 00:29:17.980 align:middle line:90% 00:29:17.980 --> 00:29:33.542 align:middle line:84% ways of coloring of n balls with j colors. 00:29:33.542 --> 00:29:41.140 align:middle line:90% 00:29:41.140 --> 00:29:44.132 align:middle line:84% So, how many dividers did we use? 00:29:44.132 --> 00:29:47.314 align:middle line:84% So we have j colors, how many dividers did we use? 00:29:47.314 --> 00:29:50.040 align:middle line:90% 00:29:50.040 --> 00:29:50.540 align:middle line:90% Yeah? 00:29:50.540 --> 00:29:51.380 align:middle line:90% STUDENT: j minus 1. 00:29:51.380 --> 00:29:52.840 align:middle line:84% PETER SHOR: j minus 1, because we 00:29:52.840 --> 00:29:56.780 align:middle line:84% put a dividers between all the pairs of colors. 00:29:56.780 --> 00:30:03.650 align:middle line:84% And we have n plus j minus 1 things. 00:30:03.650 --> 00:30:06.450 align:middle line:84% These are the total dividers and colors. 00:30:06.450 --> 00:30:11.030 align:middle line:84% And we'll choosing j minus 1 dividers out of them. 00:30:11.030 --> 00:30:31.010 align:middle line:90% 00:30:31.010 --> 00:30:34.610 align:middle line:84% So this shows proofs of bijection. 00:30:34.610 --> 00:30:37.910 align:middle line:84% And this is not a completely trivial one. 00:30:37.910 --> 00:30:41.654 align:middle line:90% 00:30:41.654 --> 00:30:44.130 align:middle line:84% And I want to claim that there are other ways you 00:30:44.130 --> 00:30:46.770 align:middle line:90% could have done this. 00:30:46.770 --> 00:31:01.810 align:middle line:90% 00:31:01.810 --> 00:31:05.210 align:middle line:84% So now I want to shift gears and talk about Dyck 00:31:05.210 --> 00:31:13.450 align:middle line:84% walks, which are in Catalan numbers, which are really 00:31:13.450 --> 00:31:18.110 align:middle line:90% very nice, a very nice example. 00:31:18.110 --> 00:31:23.870 align:middle line:84% And we will see them again when we do generate functions. 00:31:23.870 --> 00:31:24.370 align:middle line:90% OK. 00:31:24.370 --> 00:31:35.610 align:middle line:90% 00:31:35.610 --> 00:31:36.600 align:middle line:90% First, let's define. 00:31:36.600 --> 00:32:02.080 align:middle line:90% 00:32:02.080 --> 00:32:20.800 align:middle line:84% So a Dyck path is a path from coordinate 0, 0 to 2n, 0. 00:32:20.800 --> 00:32:23.560 align:middle line:90% 00:32:23.560 --> 00:32:25.970 align:middle line:84% And I guess I should call that a path of length 2n. 00:32:25.970 --> 00:32:28.840 align:middle line:90% 00:32:28.840 --> 00:32:42.320 align:middle line:84% And each step is 1, comma, plus/minus 1. 00:32:42.320 --> 00:32:50.600 align:middle line:90% And it's always above x-axis. 00:32:50.600 --> 00:32:57.110 align:middle line:90% 00:32:57.110 --> 00:33:13.790 align:middle line:84% And further, the number of Dyck paths length 2n is c sub n, 00:33:13.790 --> 00:33:16.340 align:middle line:90% the nth Catalan number. 00:33:16.340 --> 00:33:33.895 align:middle line:90% 00:33:33.895 --> 00:33:37.110 align:middle line:84% So we get a better feel for Dyck-- yes? 00:33:37.110 --> 00:33:39.910 align:middle line:84% STUDENT: So the first step is not legal to have, like, 00:33:39.910 --> 00:33:40.860 align:middle line:90% 1 minus 1? 00:33:40.860 --> 00:33:43.180 align:middle line:84% PETER SHOR: The first step always goes up. 00:33:43.180 --> 00:33:43.680 align:middle line:90% Yes. 00:33:43.680 --> 00:33:45.950 align:middle line:84% And the last step always goes down, 00:33:45.950 --> 00:33:49.150 align:middle line:84% because it has to come back to 0. 00:33:49.150 --> 00:33:51.310 align:middle line:84% And if you have a Dyck path, the reverse of it 00:33:51.310 --> 00:33:53.870 align:middle line:90% is also a Dyck path. 00:33:53.870 --> 00:34:02.110 align:middle line:84% But what I'm going to do is I'm going to give you some examples. 00:34:02.110 --> 00:34:05.380 align:middle line:90% c1 is equal to-- 00:34:05.380 --> 00:34:08.469 align:middle line:84% well, how many paths are there of length 2? 00:34:08.469 --> 00:34:11.030 align:middle line:84% Well, the first step has to go up, 00:34:11.030 --> 00:34:13.630 align:middle line:84% and the second step has to go down. 00:34:13.630 --> 00:34:17.750 align:middle line:90% So c1 is what? 00:34:17.750 --> 00:34:21.630 align:middle line:84% Now let's figure out Dyck paths of length 2. 00:34:21.630 --> 00:34:24.949 align:middle line:84% Well, you could either go up, down, up, down, 00:34:24.949 --> 00:34:31.150 align:middle line:84% or you could go up, up, down, down. 00:34:31.150 --> 00:34:35.460 align:middle line:90% So c2 equals 2, c3 equals 5. 00:34:35.460 --> 00:34:45.710 align:middle line:90% 00:34:45.710 --> 00:34:47.870 align:middle line:90% So we have that one. 00:34:47.870 --> 00:34:50.310 align:middle line:90% We could use this one. 00:34:50.310 --> 00:34:57.060 align:middle line:84% So we could go up twice, down twice, and one more. 00:34:57.060 --> 00:34:59.640 align:middle line:90% And we could do the same thing. 00:34:59.640 --> 00:35:05.340 align:middle line:90% 00:35:05.340 --> 00:35:09.960 align:middle line:84% And we could also go up three and down three. 00:35:09.960 --> 00:35:18.340 align:middle line:90% 00:35:18.340 --> 00:35:21.780 align:middle line:90% And there's one I'm missing. 00:35:21.780 --> 00:35:23.400 align:middle line:84% Anyone want to tell me what it is? 00:35:23.400 --> 00:35:33.960 align:middle line:90% 00:35:33.960 --> 00:35:34.460 align:middle line:90% Yeah? 00:35:34.460 --> 00:35:36.780 align:middle line:84% STUDENT: Up, up, down, up, down, down. 00:35:36.780 --> 00:35:39.240 align:middle line:90% PETER SHOR: Yeah, very good. 00:35:39.240 --> 00:35:44.868 align:middle line:90% 00:35:44.868 --> 00:35:48.580 align:middle line:90% Up, down, up, down, down. 00:35:48.580 --> 00:35:56.320 align:middle line:84% So there are five Dyck paths [INAUDIBLE] three-- or six, 00:35:56.320 --> 00:35:56.960 align:middle line:90% I should say. 00:35:56.960 --> 00:36:07.900 align:middle line:90% 00:36:07.900 --> 00:36:10.120 align:middle line:90% OK, I did not write these. 00:36:10.120 --> 00:36:33.040 align:middle line:90% 00:36:33.040 --> 00:36:37.100 align:middle line:84% And let's, this Catalan number's down here. 00:36:37.100 --> 00:37:01.690 align:middle line:84% Catalan numbers-- noted 1, 2, 5, 14, 42, 132. 00:37:01.690 --> 00:37:04.202 align:middle line:84% And we do not have to memorize this, 00:37:04.202 --> 00:37:06.213 align:middle line:84% although maybe you should memorize. 00:37:06.213 --> 00:37:08.130 align:middle line:84% I mean, we're going to be doing a lot of stuff 00:37:08.130 --> 00:37:09.070 align:middle line:90% with Catalan numbers. 00:37:09.070 --> 00:37:10.970 align:middle line:84% So maybe you should memorize the first three or four 00:37:10.970 --> 00:37:11.637 align:middle line:90% in the sequence. 00:37:11.637 --> 00:37:17.450 align:middle line:90% 00:37:17.450 --> 00:37:20.930 align:middle line:84% So Richard Stanley was a professor at MIT 00:37:20.930 --> 00:37:22.030 align:middle line:90% before he retired. 00:37:22.030 --> 00:37:23.690 align:middle line:90% He was a combinatorist. 00:37:23.690 --> 00:37:26.570 align:middle line:84% And he came up with something like 200 things 00:37:26.570 --> 00:37:29.090 align:middle line:84% that are counted by the Catalan numbers. 00:37:29.090 --> 00:37:32.650 align:middle line:90% So there are really a lot of-- 00:37:32.650 --> 00:37:38.980 align:middle line:84% I mean, they really appear all over mathematics, and-- 00:37:38.980 --> 00:37:41.670 align:middle line:84% STUDENT: Is there a pattern relating them to the n? 00:37:41.670 --> 00:37:43.410 align:middle line:90% PETER SHOR: Yes, there is. 00:37:43.410 --> 00:37:46.262 align:middle line:84% And we will get to that eventually. 00:37:46.262 --> 00:37:47.720 align:middle line:84% But the first thing I'm going to do 00:37:47.720 --> 00:37:50.860 align:middle line:84% is tell you two more things that are counted by Catalan numbers. 00:37:50.860 --> 00:37:56.920 align:middle line:90% 00:37:56.920 --> 00:37:57.420 align:middle line:90% OK. 00:37:57.420 --> 00:38:17.327 align:middle line:90% 00:38:17.327 --> 00:38:29.020 align:middle line:84% So there are plane trees, and n equals 1. 00:38:29.020 --> 00:38:34.040 align:middle line:90% 00:38:34.040 --> 00:38:37.150 align:middle line:90% I'm going to say with roots. 00:38:37.150 --> 00:38:40.400 align:middle line:90% n equals 1, we have a tree. 00:38:40.400 --> 00:38:43.920 align:middle line:84% And there's only one way to draw a tree, really. 00:38:43.920 --> 00:38:47.060 align:middle line:84% n equals 2, well, there are two ways 00:38:47.060 --> 00:38:51.840 align:middle line:84% you can draw tree with two nodes. 00:38:51.840 --> 00:38:55.360 align:middle line:90% So that's 1, 2, just like that. 00:38:55.360 --> 00:39:01.480 align:middle line:84% n equals 3, so let's see we can draw a tree. 00:39:01.480 --> 00:39:02.740 align:middle line:90% Yes. 00:39:02.740 --> 00:39:08.110 align:middle line:84% We can draw a tree that looks like this. 00:39:08.110 --> 00:39:13.320 align:middle line:84% We can draw a tree that looks like this. 00:39:13.320 --> 00:39:17.400 align:middle line:84% We can draw a tree that looks like this. 00:39:17.400 --> 00:39:23.764 align:middle line:90% And there's one there. 00:39:23.764 --> 00:39:35.760 align:middle line:84% And you'll notice that these left and right count when you're 00:39:35.760 --> 00:39:37.320 align:middle line:90% talking about plane trees. 00:39:37.320 --> 00:39:40.460 align:middle line:84% So this is the mirror image of this, 00:39:40.460 --> 00:39:43.510 align:middle line:84% but they count as two different trees. 00:39:43.510 --> 00:39:54.310 align:middle line:84% So which leaf, which vertex something is attached 00:39:54.310 --> 00:39:59.390 align:middle line:90% to matters inflections. 00:39:59.390 --> 00:40:02.280 align:middle line:84% And the next one has 14, but we're not going to draw that. 00:40:02.280 --> 00:40:05.030 align:middle line:90% 00:40:05.030 --> 00:40:12.890 align:middle line:90% And let's move out here. 00:40:12.890 --> 00:40:56.230 align:middle line:90% 00:40:56.230 --> 00:40:57.750 align:middle line:84% I know what I'm saying, last thing 00:40:57.750 --> 00:41:10.362 align:middle line:84% is binary trees where every node has two children. 00:41:10.362 --> 00:41:18.950 align:middle line:90% 00:41:18.950 --> 00:41:27.430 align:middle line:84% So n equals 1, there's only one binary tree on, I guess, 00:41:27.430 --> 00:41:30.030 align:middle line:90% three vertices, two leaves. 00:41:30.030 --> 00:41:38.321 align:middle line:84% n equals 2, well, we can either add 2 00:41:38.321 --> 00:41:43.900 align:middle line:84% to either develop this node, or add, develop for this node. 00:41:43.900 --> 00:41:48.500 align:middle line:84% n equals 3, OK, let's see if I can do this 00:41:48.500 --> 00:41:49.960 align:middle line:90% without taking my nodes. 00:41:49.960 --> 00:42:13.580 align:middle line:90% 00:42:13.580 --> 00:42:15.600 align:middle line:84% OK, I have to put them on the other side. 00:42:15.600 --> 00:42:22.980 align:middle line:90% 00:42:22.980 --> 00:42:25.220 align:middle line:84% And that is supposed to be all of them, 00:42:25.220 --> 00:42:28.060 align:middle line:84% although it's not immediately clear. 00:42:28.060 --> 00:42:35.060 align:middle line:84% But you can only get one of these by taking one of these 00:42:35.060 --> 00:42:39.970 align:middle line:84% and adding a pair of leaves to a vertex-- 00:42:39.970 --> 00:42:42.130 align:middle line:90% or a pair of leaves to a leaf. 00:42:42.130 --> 00:42:44.690 align:middle line:90% And there are only-- 00:42:44.690 --> 00:42:46.470 align:middle line:84% well, you could take care of these two. 00:42:46.470 --> 00:42:50.490 align:middle line:84% Anything here occur, we use any of the three vertices here. 00:42:50.490 --> 00:42:54.730 align:middle line:84% But when you do it to this vertex and this vertex, 00:42:54.730 --> 00:42:56.070 align:middle line:90% you get the same tree. 00:42:56.070 --> 00:42:58.510 align:middle line:90% So this is fine. 00:42:58.510 --> 00:42:59.010 align:middle line:90% Yes? 00:42:59.010 --> 00:43:02.596 align:middle line:84% STUDENT: Is n counting the pair of the leaves? 00:43:02.596 --> 00:43:04.370 align:middle line:90% PETER SHOR: n is counting-- 00:43:04.370 --> 00:43:09.796 align:middle line:90% 00:43:09.796 --> 00:43:21.676 align:middle line:84% n is the number of non-leaf nodes, or 1 less 00:43:21.676 --> 00:43:24.780 align:middle line:90% than the number of leaf nodes. 00:43:24.780 --> 00:43:38.890 align:middle line:90% 00:43:38.890 --> 00:43:45.050 align:middle line:84% And you can prove that every tree has one more leaf 00:43:45.050 --> 00:43:49.890 align:middle line:84% node than interior node, because note 00:43:49.890 --> 00:43:51.970 align:middle line:90% this is true for n equals 1. 00:43:51.970 --> 00:43:54.970 align:middle line:84% And n equals 2, you replace the leaf 00:43:54.970 --> 00:43:58.650 align:middle line:84% with an interior node and two more leaves. 00:43:58.650 --> 00:44:06.050 align:middle line:84% So you preserve the number of leaf nodes minus non-leaf nodes. 00:44:06.050 --> 00:44:14.170 align:middle line:84% So for binary trees, every node, the number of non-leaf node 00:44:14.170 --> 00:44:18.270 align:middle line:84% is one fewer than the number of leaf nodes. 00:44:18.270 --> 00:44:20.810 align:middle line:90% 00:44:20.810 --> 00:44:23.235 align:middle line:90% OK. 00:44:23.235 --> 00:44:25.790 align:middle line:84% STUDENT: Well, wasn't n in the plane tree case 00:44:25.790 --> 00:44:27.170 align:middle line:90% the number of edges or-- 00:44:27.170 --> 00:44:29.690 align:middle line:84% PETER SHOR: n is the number of edges, yeah. 00:44:29.690 --> 00:44:30.410 align:middle line:90% STUDENT: OK. 00:44:30.410 --> 00:44:40.360 align:middle line:90% 00:44:40.360 --> 00:44:47.520 align:middle line:84% PETER SHOR: n plus 1 equals the number of leaves, 00:44:47.520 --> 00:44:48.850 align:middle line:90% or no-- number of vertices. 00:44:48.850 --> 00:44:54.588 align:middle line:90% 00:44:54.588 --> 00:44:57.480 align:middle line:84% And you can similarly prove that a tree always 00:44:57.480 --> 00:45:01.260 align:middle line:90% has one more vertex than edge. 00:45:01.260 --> 00:45:06.640 align:middle line:90% 00:45:06.640 --> 00:45:10.900 align:middle line:90% OK, so we have three things. 00:45:10.900 --> 00:45:14.180 align:middle line:90% We counted five Dyck paths. 00:45:14.180 --> 00:45:16.960 align:middle line:84% And I think what I'm going to do is, first, I'm 00:45:16.960 --> 00:45:21.390 align:middle line:84% going to give you the formula for the Catalan number. 00:45:21.390 --> 00:45:24.160 align:middle line:90% 00:45:24.160 --> 00:45:28.760 align:middle line:90% And this is a really nice-- 00:45:28.760 --> 00:45:33.080 align:middle line:84% well, this is a really nice proof 00:45:33.080 --> 00:45:34.600 align:middle line:84% of the formula for Catalan number, 00:45:34.600 --> 00:45:37.880 align:middle line:84% although it's a little bit complicated. 00:45:37.880 --> 00:46:06.280 align:middle line:84% And we will-- this is one of those proofs. 00:46:06.280 --> 00:46:08.940 align:middle line:84% It's not clear how anyone came up with that. 00:46:08.940 --> 00:46:11.720 align:middle line:90% 00:46:11.720 --> 00:46:26.120 align:middle line:84% So, c sub n is equal to 1 over n plus 1, 2n choose n. 00:46:26.120 --> 00:46:29.270 align:middle line:84% So that's the formula for the number of Catalan numbers. 00:46:29.270 --> 00:46:31.620 align:middle line:90% So let's try it for n equals 3. 00:46:31.620 --> 00:46:35.630 align:middle line:90% 00:46:35.630 --> 00:46:42.762 align:middle line:84% n equals 3-- no, let's try for n equals 2. 00:46:42.762 --> 00:46:48.108 align:middle line:90% 00:46:48.108 --> 00:46:50.550 align:middle line:90% Sorry, for n equals 2. 00:46:50.550 --> 00:46:56.815 align:middle line:90% So 6 choose 3 as 1 equals 4. 00:46:56.815 --> 00:47:03.270 align:middle line:84% 6 minus 5 minus 4, divided by 3 times 2 times 1. 00:47:03.270 --> 00:47:07.070 align:middle line:84% And actually, there's another 4 here. 00:47:07.070 --> 00:47:12.710 align:middle line:84% And when we cancel these out, so 4's cancel, 3's cancels the 2, 00:47:12.710 --> 00:47:13.700 align:middle line:90% and we get 5. 00:47:13.700 --> 00:47:17.510 align:middle line:90% 00:47:17.510 --> 00:47:20.570 align:middle line:84% So it works, at least for one case. 00:47:20.570 --> 00:47:22.520 align:middle line:84% And that will prove it in general. 00:47:22.520 --> 00:47:25.230 align:middle line:90% 00:47:25.230 --> 00:47:26.918 align:middle line:90% So how are we going to prove it? 00:47:26.918 --> 00:47:30.340 align:middle line:90% 00:47:30.340 --> 00:47:31.220 align:middle line:90% OK. 00:47:31.220 --> 00:47:47.140 align:middle line:84% Let p sub n be 0, be the number of paths of paths from 0, 00:47:47.140 --> 00:47:53.220 align:middle line:90% 0 to 2n, 0. 00:47:53.220 --> 00:47:59.960 align:middle line:84% So this is all paths from 0, 0 to 2n, 0. 00:47:59.960 --> 00:48:02.580 align:middle line:90% 00:48:02.580 --> 00:48:03.080 align:middle line:90% OK. 00:48:03.080 --> 00:48:06.775 align:middle line:90% 00:48:06.775 --> 00:48:27.100 align:middle line:84% Let P sub n be a set of Dyck walks from 0, 0 to 2n. 00:48:27.100 --> 00:48:30.082 align:middle line:90% 00:48:30.082 --> 00:48:38.780 align:middle line:84% And we'll let Dn bar be P sub n minus Dn. 00:48:38.780 --> 00:48:45.700 align:middle line:84% So everything that's not a Dyck walk from 2n, 0 to 0. 00:48:45.700 --> 00:48:51.180 align:middle line:84% And what we're going to do, we will count 00:48:51.180 --> 00:48:55.240 align:middle line:90% the number of things in Dn bar. 00:48:55.240 --> 00:48:58.540 align:middle line:90% 00:48:58.540 --> 00:49:01.620 align:middle line:84% And now to get the number of Dyck 00:49:01.620 --> 00:49:07.260 align:middle line:84% walks, we just take the number of paths and subtract Dn bar, 00:49:07.260 --> 00:49:08.620 align:middle line:90% and we get Dn. 00:49:08.620 --> 00:49:12.060 align:middle line:84% So can someone tell me what the number of paths is from 0, 00:49:12.060 --> 00:49:13.400 align:middle line:90% 0 to 2n, 0? 00:49:13.400 --> 00:49:21.420 align:middle line:90% 00:49:21.420 --> 00:49:22.971 align:middle line:90% Yeah? 00:49:22.971 --> 00:49:24.330 align:middle line:90% STUDENT: 2n choose n. 00:49:24.330 --> 00:49:31.810 align:middle line:84% PETER SHOR: 2n choose n, because we need n steps up and n steps 00:49:31.810 --> 00:49:32.350 align:middle line:90% down. 00:49:32.350 --> 00:49:36.710 align:middle line:84% And we don't care whether it goes below the x-axis or not. 00:49:36.710 --> 00:49:39.276 align:middle line:90% 00:49:39.276 --> 00:49:42.430 align:middle line:90% So what is Dn [INAUDIBLE]? 00:49:42.430 --> 00:49:55.210 align:middle line:90% 00:49:55.210 --> 00:50:24.750 align:middle line:84% Well, let P minus 2n be paths from 0, 0 to 2n minus 2. 00:50:24.750 --> 00:50:28.650 align:middle line:90% 00:50:28.650 --> 00:50:31.790 align:middle line:84% And we don't care which way they go. 00:50:31.790 --> 00:50:39.610 align:middle line:90% 00:50:39.610 --> 00:50:47.530 align:middle line:84% So a path from 0, 0 to 2n minus 2 is going to have to have two 00:50:47.530 --> 00:50:51.210 align:middle line:90% more down-steps than up-steps. 00:50:51.210 --> 00:50:59.030 align:middle line:84% And so the number of Pn is equal to 2n choose n minus 1. 00:50:59.030 --> 00:51:04.730 align:middle line:90% 00:51:04.730 --> 00:51:12.810 align:middle line:84% And I want to claim that the number of non-Dyck paths is 00:51:12.810 --> 00:51:18.830 align:middle line:84% equal to the number of paths drawn-- 00:51:18.830 --> 00:51:21.860 align:middle line:90% 0, 0 to 2n minus 2. 00:51:21.860 --> 00:51:24.480 align:middle line:90% 00:51:24.480 --> 00:51:30.095 align:middle line:84% And that will give us the Catalan number when we subtract. 00:51:30.095 --> 00:51:32.640 align:middle line:84% And why did I show you that first? 00:51:32.640 --> 00:51:36.060 align:middle line:90% Because that's pretty great. 00:51:36.060 --> 00:51:50.660 align:middle line:90% 00:51:50.660 --> 00:52:02.630 align:middle line:84% So a path from Pn minus 2 is equal to 2n choose n minus 1, 00:52:02.630 --> 00:52:08.750 align:middle line:90% P sub n, 0 equals 2n choose n. 00:52:08.750 --> 00:52:18.150 align:middle line:84% Pn 0 minus Pn minus 2 is equal to-- 00:52:18.150 --> 00:52:24.510 align:middle line:84% what is 2n factorial over n factorial, n factorial, 00:52:24.510 --> 00:52:30.670 align:middle line:84% minus 2n factorial over n plus 1, factorial, 00:52:30.670 --> 00:52:37.190 align:middle line:90% n minus 1 is equal to-- 00:52:37.190 --> 00:52:39.910 align:middle line:90% well, this is just 2n choose n. 00:52:39.910 --> 00:52:43.110 align:middle line:84% And this, if we take 2n choose n, 00:52:43.110 --> 00:52:48.070 align:middle line:84% and multiply in the denominator by n plus 1, 00:52:48.070 --> 00:52:54.950 align:middle line:84% and the numerator by n, we get this thing. 00:52:54.950 --> 00:53:01.130 align:middle line:84% So that's minus n over n plus 1, 2n. 00:53:01.130 --> 00:53:06.355 align:middle line:90% 00:53:06.355 --> 00:53:11.350 align:middle line:84% And you see, when you multiply by n 00:53:11.350 --> 00:53:18.390 align:middle line:84% plus 1 n in the n factorial, the denominator 00:53:18.390 --> 00:53:20.330 align:middle line:90% is n minus 1 factorial. 00:53:20.330 --> 00:53:24.550 align:middle line:84% When we multiply by n plus 1, the denominator 00:53:24.550 --> 00:53:28.950 align:middle line:84% turns an n factorial to an n plus 1 factorial. 00:53:28.950 --> 00:53:33.750 align:middle line:84% So to get from here to here, we multiply by n over plus 1. 00:53:33.750 --> 00:53:43.070 align:middle line:84% And this is just equal to 2n choose n, divided by n plus 1, 00:53:43.070 --> 00:53:46.030 align:middle line:90% which is the Catalan number. 00:53:46.030 --> 00:53:50.670 align:middle line:84% So what we have to do now is show that every non-Dyck path 00:53:50.670 --> 00:53:57.485 align:middle line:84% show a bijection between non-Dyck paths and paths from 2n 00:53:57.485 --> 00:54:03.920 align:middle line:90% to n from 0, 0 to 2n minus 2. 00:54:03.920 --> 00:54:13.900 align:middle line:90% 00:54:13.900 --> 00:54:29.380 align:middle line:84% And this, I think the best way is by drawing a picture, 00:54:29.380 --> 00:54:35.430 align:middle line:84% because I'm not going to be able to describe it in words. 00:54:35.430 --> 00:54:44.020 align:middle line:90% 00:54:44.020 --> 00:54:47.220 align:middle line:90% So we have a path here. 00:54:47.220 --> 00:54:56.260 align:middle line:84% And it starts at 0 and it goes up, down, down, up, down, maybe. 00:54:56.260 --> 00:54:58.980 align:middle line:84% At some point, it has to go below the x-axis, 00:54:58.980 --> 00:55:01.430 align:middle line:84% because it has to end up at minus 2, 00:55:01.430 --> 00:55:08.840 align:middle line:84% and which is why it's not a Dyck path. 00:55:08.840 --> 00:55:16.100 align:middle line:90% 00:55:16.100 --> 00:55:17.680 align:middle line:90% This, I believe, is minus 2. 00:55:17.680 --> 00:55:23.130 align:middle line:90% 00:55:23.130 --> 00:55:28.105 align:middle line:90% So this is in D bar. 00:55:28.105 --> 00:55:37.500 align:middle line:90% 00:55:37.500 --> 00:55:43.900 align:middle line:84% I guess D bar is fixed, if it makes any difference. 00:55:43.900 --> 00:55:45.660 align:middle line:90% What are we going to do? 00:55:45.660 --> 00:55:55.340 align:middle line:84% What we're going to do is we're going to take the line 00:55:55.340 --> 00:55:58.820 align:middle line:90% x equals minus 1-- 00:55:58.820 --> 00:56:01.180 align:middle line:90% so this is x equals 2-- 00:56:01.180 --> 00:56:03.860 align:middle line:84% and we're going to reflect this path, then, 00:56:03.860 --> 00:56:10.690 align:middle line:84% around x equals minus 1 from every point after the first time 00:56:10.690 --> 00:56:31.730 align:middle line:84% it hits minus 1 from, well, let's say, x minus 1 to the end, 00:56:31.730 --> 00:56:42.740 align:middle line:84% which would be, I guess, 2n minus 2 around x equals minus 1 00:56:42.740 --> 00:56:43.240 align:middle line:90% axis. 00:56:43.240 --> 00:56:47.190 align:middle line:90% 00:56:47.190 --> 00:56:51.410 align:middle line:90% OK, so let's do this set. 00:56:51.410 --> 00:56:54.290 align:middle line:84% Well, when we flipped the last thing in the path, 00:56:54.290 --> 00:56:55.770 align:middle line:90% it's going to end up on 0. 00:56:55.770 --> 00:57:09.000 align:middle line:90% 00:57:09.000 --> 00:57:14.480 align:middle line:84% So now, I want to claim this is a non-Dyck path. 00:57:14.480 --> 00:57:17.200 align:middle line:90% So why is it a non-Dyck path? 00:57:17.200 --> 00:57:24.740 align:middle line:84% Well, I mean, we have a point x on it that is x equals minus 1. 00:57:24.740 --> 00:57:28.240 align:middle line:84% So it has to be a non-Dyck path, because it 00:57:28.240 --> 00:57:31.240 align:middle line:90% goes below the x-axis. 00:57:31.240 --> 00:57:40.760 align:middle line:84% And it is a path from 0 to 2n with every step up, 00:57:40.760 --> 00:57:42.040 align:middle line:90% either up or down. 00:57:42.040 --> 00:57:47.230 align:middle line:90% So it is in Pn plus 0. 00:57:47.230 --> 00:57:50.880 align:middle line:90% 00:57:50.880 --> 00:57:55.520 align:middle line:84% And I'm going to claim this is a bijection between non-Dyck paths 00:57:55.520 --> 00:58:01.320 align:middle line:90% and paths from 0 to 2n minus 2. 00:58:01.320 --> 00:58:15.720 align:middle line:90% 00:58:15.720 --> 00:58:16.490 align:middle line:90% P 2n. 00:58:16.490 --> 00:58:19.240 align:middle line:90% 00:58:19.240 --> 00:58:21.600 align:middle line:90% Now why is that? 00:58:21.600 --> 00:58:26.140 align:middle line:84% Well, I mean, if you have a path from Pn, well, 00:58:26.140 --> 00:58:33.860 align:middle line:84% you have a path from 0 to 2n minus 2, you take this point, 00:58:33.860 --> 00:58:37.680 align:middle line:84% which is the first point that hits [INAUDIBLE] x equals minus 00:58:37.680 --> 00:58:43.360 align:middle line:84% 1, so we flip this around this axis and you get a unique path. 00:58:43.360 --> 00:58:50.320 align:middle line:84% On the other hand, if you have a path from x equals 0 to x equals 00:58:50.320 --> 00:58:55.240 align:middle line:84% 2n that goes below the x-axis, you can take the first time it 00:58:55.240 --> 00:59:04.390 align:middle line:84% hits the x equals minus 1 line, and you end up this function. 00:59:04.390 --> 00:59:09.190 align:middle line:84% So it's a function, f, which has an inverse, g. 00:59:09.190 --> 00:59:10.890 align:middle line:90% So it's a bijection. 00:59:10.890 --> 00:59:15.960 align:middle line:90% 00:59:15.960 --> 00:59:24.990 align:middle line:84% And the number of paths from x equals 0 to x equals minus 2 is 00:59:24.990 --> 00:59:31.830 align:middle line:84% indeed 2n factorial of n plus 1 factorial, n minus 1 factorial. 00:59:31.830 --> 00:59:36.870 align:middle line:84% So this shows that the number of non-Dyck paths 00:59:36.870 --> 00:59:38.398 align:middle line:90% is the right number. 00:59:38.398 --> 00:59:40.190 align:middle line:84% So that shows that the number of Dyck paths 00:59:40.190 --> 00:59:42.110 align:middle line:90% is the Catalan number. 00:59:42.110 --> 00:59:45.750 align:middle line:90% So that's our bijection-- 00:59:45.750 --> 00:59:49.430 align:middle line:84% or that's our proof that Dyck paths 00:59:49.430 --> 00:59:51.180 align:middle line:90% are counted by Catalan numbers. 00:59:51.180 --> 00:59:54.670 align:middle line:90% 00:59:54.670 --> 00:59:59.840 align:middle line:84% So I want to say, how would you ever find this proof? 00:59:59.840 --> 01:00:03.660 align:middle line:90% And I really don't know. 01:00:03.660 --> 01:00:08.840 align:middle line:84% However, if you wait a couple of weeks, 01:00:08.840 --> 01:00:10.940 align:middle line:84% we will be doing generating functions. 01:00:10.940 --> 01:00:14.980 align:middle line:84% And generating functions is a completely straightforward way 01:00:14.980 --> 01:00:18.780 align:middle line:84% of figuring out the formula for the Catalan number, 01:00:18.780 --> 01:00:24.360 align:middle line:84% although it's actually not as nice and easy as this. 01:00:24.360 --> 01:00:25.920 align:middle line:90% It involves some calculations. 01:00:25.920 --> 01:00:34.580 align:middle line:90% 01:00:34.580 --> 01:00:36.860 align:middle line:84% And now the next thing I want to show 01:00:36.860 --> 01:00:41.150 align:middle line:84% is a bijection between binary trees and Dyck paths. 01:00:41.150 --> 01:00:47.780 align:middle line:90% 01:00:47.780 --> 01:00:51.860 align:middle line:84% And that will show you that binary trees are also 01:00:51.860 --> 01:00:53.450 align:middle line:90% counted by Catalan numbers. 01:00:53.450 --> 01:00:59.640 align:middle line:90% 01:00:59.640 --> 01:01:00.140 align:middle line:90% Oh. 01:01:00.140 --> 01:01:33.340 align:middle line:90% 01:01:33.340 --> 01:01:33.880 align:middle line:90% OK. 01:01:33.880 --> 01:01:35.700 align:middle line:90% So how does this work? 01:01:35.700 --> 01:01:38.740 align:middle line:90% Well, let's do an example. 01:01:38.740 --> 01:01:44.410 align:middle line:90% So let's draw a tree. 01:01:44.410 --> 01:02:01.770 align:middle line:90% 01:02:01.770 --> 01:02:04.160 align:middle line:90% OK, so this is a binary tree. 01:02:04.160 --> 01:02:09.850 align:middle line:90% 01:02:09.850 --> 01:02:11.410 align:middle line:84% And what we're going to do is we're 01:02:11.410 --> 01:02:14.180 align:middle line:84% going to traverse the nodes in that first order. 01:02:14.180 --> 01:02:20.770 align:middle line:90% 01:02:20.770 --> 01:02:29.336 align:middle line:90% Traverse nodes in that order. 01:02:29.336 --> 01:02:30.830 align:middle line:90% And what does that mean? 01:02:30.830 --> 01:02:34.210 align:middle line:84% Well, what it means, in this case, 01:02:34.210 --> 01:02:38.410 align:middle line:84% is we're going to just run around the outside of the nodes. 01:02:38.410 --> 01:02:41.670 align:middle line:84% And we get the first node we do this one, 01:02:41.670 --> 01:02:46.410 align:middle line:84% we do a second node we visit, this is the third node we visit. 01:02:46.410 --> 01:02:49.670 align:middle line:84% I'm going to go back and go down and go up. 01:02:49.670 --> 01:02:51.170 align:middle line:90% And this is the fourth we visit. 01:02:51.170 --> 01:02:55.010 align:middle line:84% So we're only counting the first time we visit a node. 01:02:55.010 --> 01:02:58.250 align:middle line:90% This is y 6. 01:02:58.250 --> 01:03:01.850 align:middle line:84% And now we go down, back down and up here. 01:03:01.850 --> 01:03:06.197 align:middle line:84% This is the seventh node, eighth node, ninth node, tenth node up. 01:03:06.197 --> 01:03:09.330 align:middle line:90% 01:03:09.330 --> 01:03:11.410 align:middle line:90% OK. 01:03:11.410 --> 01:03:13.200 align:middle line:84% So what do we do with these things? 01:03:13.200 --> 01:03:27.210 align:middle line:90% 01:03:27.210 --> 01:03:30.450 align:middle line:84% Well, what we're going to do is we're 01:03:30.450 --> 01:03:33.580 align:middle line:84% going to associate each of the nodes-- 01:03:33.580 --> 01:03:39.690 align:middle line:90% 01:03:39.690 --> 01:03:46.530 align:middle line:84% interior node, we'll call it plus. 01:03:46.530 --> 01:03:50.310 align:middle line:84% And a leaf, we'll call it a minus 1. 01:03:50.310 --> 01:03:52.970 align:middle line:90% 01:03:52.970 --> 01:03:54.640 align:middle line:90% So what is this? 01:03:54.640 --> 01:04:05.200 align:middle line:84% This is 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11. 01:04:05.200 --> 01:04:10.860 align:middle line:84% And 1 and 2 are both pluses, because they're interior nodes. 01:04:10.860 --> 01:04:20.660 align:middle line:84% 3 is a minus, 4 is a plus, 5 is a minus, and 6 is a minus. 01:04:20.660 --> 01:04:26.750 align:middle line:84% 7 is a plus, 8 is a minus, 9 is a plus, 10 and 11 are minuses. 01:04:26.750 --> 01:04:35.500 align:middle line:90% 01:04:35.500 --> 01:04:39.160 align:middle line:84% And what we do is we use this for our Dyck path. 01:04:39.160 --> 01:04:42.370 align:middle line:84% We go up when it's plus and down when it's a minus. 01:04:42.370 --> 01:04:56.190 align:middle line:90% 01:04:56.190 --> 01:04:58.590 align:middle line:84% So what does our Dyck path look like? 01:04:58.590 --> 01:05:00.340 align:middle line:84% Well, it starts out going up twice. 01:05:00.340 --> 01:05:03.350 align:middle line:90% 01:05:03.350 --> 01:05:11.310 align:middle line:84% And now we go down and up and down and down. 01:05:11.310 --> 01:05:12.610 align:middle line:90% And then we go up-- 01:05:12.610 --> 01:05:16.350 align:middle line:90% 01:05:16.350 --> 01:05:23.060 align:middle line:84% 1 plus, minus, minus, plus, minus, plus, minus, minus. 01:05:23.060 --> 01:05:27.350 align:middle line:90% 01:05:27.350 --> 01:05:35.310 align:middle line:84% So remember, it is always one more leaf than interior node. 01:05:35.310 --> 01:05:43.590 align:middle line:84% So this always ends up at 2n plus 1 and minus 1. 01:05:43.590 --> 01:05:48.190 align:middle line:90% And this starts at 0, 0. 01:05:48.190 --> 01:05:51.370 align:middle line:84% But the last step has to be a leaf. 01:05:51.370 --> 01:05:53.230 align:middle line:90% So the last step has to be down. 01:05:53.230 --> 01:05:56.830 align:middle line:84% So we just ignore this last step. 01:05:56.830 --> 01:05:59.450 align:middle line:90% And the Dyck path is here. 01:05:59.450 --> 01:06:05.390 align:middle line:90% 01:06:05.390 --> 01:06:08.150 align:middle line:84% So we have to prove this is the Dyck path. 01:06:08.150 --> 01:06:16.110 align:middle line:84% And we also have to prove that every Dyck path corresponds 01:06:16.110 --> 01:06:19.420 align:middle line:90% to a tree. 01:06:19.420 --> 01:06:39.740 align:middle line:90% 01:06:39.740 --> 01:06:42.750 align:middle line:90% So why is this a Dyck path? 01:06:42.750 --> 01:06:54.140 align:middle line:84% Well, how many up steps does it have? 01:06:54.140 --> 01:06:55.880 align:middle line:84% And how many down steps does it have? 01:06:55.880 --> 01:07:02.060 align:middle line:90% 01:07:02.060 --> 01:07:06.860 align:middle line:90% So there are 2n plus 1 leaves. 01:07:06.860 --> 01:07:12.620 align:middle line:84% I'll write it n plus 1 leaves and n internal nodes. 01:07:12.620 --> 01:07:15.580 align:middle line:84% But we're ignoring the last step. 01:07:15.580 --> 01:07:21.220 align:middle line:84% So there are n steps up and n steps down. 01:07:21.220 --> 01:07:26.760 align:middle line:84% So there are n steps up, n steps down. 01:07:26.760 --> 01:07:31.260 align:middle line:90% 01:07:31.260 --> 01:07:37.100 align:middle line:84% And why does it never go below the x-axis? 01:07:37.100 --> 01:07:38.940 align:middle line:84% So suppose it went below the x-axis. 01:07:38.940 --> 01:08:12.740 align:middle line:90% 01:08:12.740 --> 01:08:27.339 align:middle line:84% Look at the tree and just nodes-- 01:08:27.339 --> 01:08:31.819 align:middle line:90% 01:08:31.819 --> 01:08:35.740 align:middle line:84% well, just the nodes up to the point where it goes to minus 1. 01:08:35.740 --> 01:08:41.837 align:middle line:90% 01:08:41.837 --> 01:08:47.850 align:middle line:90% I get 2k plus 1, and minus 1. 01:08:47.850 --> 01:09:05.649 align:middle line:84% Well, this subtree must have, well, 01:09:05.649 --> 01:09:14.910 align:middle line:84% one more leaf than interior node. 01:09:14.910 --> 01:09:21.350 align:middle line:90% 01:09:21.350 --> 01:09:23.830 align:middle line:84% So it has one more leaf than interior node. 01:09:23.830 --> 01:09:29.890 align:middle line:90% 01:09:29.890 --> 01:09:34.830 align:middle line:90% And it's a complete-- 01:09:34.830 --> 01:09:39.550 align:middle line:84% [INAUDIBLE] complete because it's a full binary tree. 01:09:39.550 --> 01:09:45.960 align:middle line:90% 01:09:45.960 --> 01:09:51.319 align:middle line:84% Because basically, if you take a tree 01:09:51.319 --> 01:09:58.280 align:middle line:84% and you remove some part of it before-- 01:09:58.280 --> 01:10:00.760 align:middle line:84% let's stop when it gets to this thing-- 01:10:00.760 --> 01:10:05.400 align:middle line:84% remove some part of it, then I want 01:10:05.400 --> 01:10:12.760 align:middle line:84% to claim this part always has at least as many leaves as 01:10:12.760 --> 01:10:14.560 align:middle line:90% interpreted. 01:10:14.560 --> 01:10:15.720 align:middle line:90% Why? 01:10:15.720 --> 01:10:31.400 align:middle line:84% Well, I mean, if you remove some part of it, 01:10:31.400 --> 01:10:39.960 align:middle line:84% there's always node degree 1, because if every node has degree 01:10:39.960 --> 01:10:42.720 align:middle line:84% 2, it's a complete binary tree and you 01:10:42.720 --> 01:10:44.760 align:middle line:90% can't add anything more to it. 01:10:44.760 --> 01:10:47.940 align:middle line:84% So that means that if you traverse the whole tree, 01:10:47.940 --> 01:10:56.720 align:middle line:84% so there's a node of degree 1 somewhere. 01:10:56.720 --> 01:11:07.840 align:middle line:84% And that-- yeah, I mean, well, there's 01:11:07.840 --> 01:11:21.710 align:middle line:90% either an interior node of tree. 01:11:21.710 --> 01:11:25.080 align:middle line:90% 01:11:25.080 --> 01:11:34.680 align:middle line:84% Either have-- but if you stop at this node 01:11:34.680 --> 01:11:41.870 align:middle line:84% and you remove these two leaves, you either have this guy 01:11:41.870 --> 01:11:49.430 align:middle line:84% or you could stop here and notch through this leaf, 01:11:49.430 --> 01:12:01.990 align:middle line:84% then you have this guy, which just has degree 1. 01:12:01.990 --> 01:12:15.670 align:middle line:84% So if you stop early, you will always have more interior nodes 01:12:15.670 --> 01:12:17.130 align:middle line:90% than leaves. 01:12:17.130 --> 01:12:24.230 align:middle line:84% And that means you haven't actually gotten to the end yet. 01:12:24.230 --> 01:12:27.830 align:middle line:90% 01:12:27.830 --> 01:12:36.630 align:middle line:84% So if you take a tree, then it's a Dyck path. 01:12:36.630 --> 01:12:42.200 align:middle line:84% If you take a binary tree and you stop early, 01:12:42.200 --> 01:12:45.378 align:middle line:84% you do not get to the end of this top, 01:12:45.378 --> 01:12:48.410 align:middle line:84% which has limit 2n plus 1 minus n. 01:12:48.410 --> 01:12:51.510 align:middle line:90% 01:12:51.510 --> 01:12:53.330 align:middle line:84% So that was a little bit hand-wavy, 01:12:53.330 --> 01:12:56.390 align:middle line:84% but you can go through that if you want. 01:12:56.390 --> 01:13:06.550 align:middle line:84% And every path will give you a tree, 01:13:06.550 --> 01:13:14.790 align:middle line:84% because you just go through this and you put a plus 1-- 01:13:14.790 --> 01:13:17.270 align:middle line:90% if it's a plus 1, you put 2. 01:13:17.270 --> 01:13:21.070 align:middle line:84% Goes on, if it's a minus 1, you put leaves on. 01:13:21.070 --> 01:13:29.450 align:middle line:84% And eventually, you will get a tree put up. 01:13:29.450 --> 01:13:32.312 align:middle line:90% 01:13:32.312 --> 01:13:40.910 align:middle line:84% I mean, because you have labeled this one, this one, 01:13:40.910 --> 01:13:45.460 align:middle line:84% and this one, well, I mean, this is a plus 1, 01:13:45.460 --> 01:13:47.480 align:middle line:90% in case you put the leaves on. 01:13:47.480 --> 01:13:51.300 align:middle line:84% Or it's a minus 1 in case [INAUDIBLE] and you end. 01:13:51.300 --> 01:13:55.980 align:middle line:84% And there's always something to do. 01:13:55.980 --> 01:14:01.540 align:middle line:84% I mean, as long as this tree does not 01:14:01.540 --> 01:14:05.260 align:middle line:90% have more interior nodes-- 01:14:05.260 --> 01:14:10.020 align:middle line:84% does not have one more than interior node, 01:14:10.020 --> 01:14:11.240 align:middle line:90% you can always add to it. 01:14:11.240 --> 01:14:13.540 align:middle line:84% And as soon as you get to the end of the path 01:14:13.540 --> 01:14:23.980 align:middle line:84% and it has one more leaf than interior node, 01:14:23.980 --> 01:14:27.520 align:middle line:84% then it's a full binary tree, and we're done. 01:14:27.520 --> 01:14:33.356 align:middle line:90% 01:14:33.356 --> 01:14:38.050 align:middle line:84% So that is more or less what I wanted to tell you today. 01:14:38.050 --> 01:14:44.490 align:middle line:84% I also wanted to tell you what Stirling's formula was. 01:14:44.490 --> 01:14:49.830 align:middle line:84% But, well, I'll just give it to you. 01:14:49.830 --> 01:14:53.130 align:middle line:90% 01:14:53.130 --> 01:14:56.210 align:middle line:84% And we can talk about him more next time. 01:14:56.210 --> 01:14:58.930 align:middle line:90% OK, I have time. 01:14:58.930 --> 01:15:12.914 align:middle line:84% So n factorial is equal to approximately root 2 pi n, 01:15:12.914 --> 01:15:14.800 align:middle line:90% n over e theta. 01:15:14.800 --> 01:15:30.530 align:middle line:90% 01:15:30.530 --> 01:15:33.490 align:middle line:84% And from this, you can make 2n choose n. 01:15:33.490 --> 01:15:47.421 align:middle line:84% It's approximately root 4 pi n over root 2 pi n squared. 01:15:47.421 --> 01:15:54.010 align:middle line:90% 01:15:54.010 --> 01:16:01.710 align:middle line:84% 2n over e to the 2n, over n over e to the n, and squared. 01:16:01.710 --> 01:16:05.730 align:middle line:90% 01:16:05.730 --> 01:16:08.210 align:middle line:84% Because we're just taking n factorial. 01:16:08.210 --> 01:16:12.650 align:middle line:84% We're taking this approximation for 2n factorial and dividing it 01:16:12.650 --> 01:16:16.530 align:middle line:84% by the approximation for n factorial squared. 01:16:16.530 --> 01:16:21.490 align:middle line:84% And you can see that most of these things cancel out. 01:16:21.490 --> 01:16:23.410 align:middle line:90% This e cancels this e. 01:16:23.410 --> 01:16:27.370 align:middle line:90% This n cancels this n. 01:16:27.370 --> 01:16:31.670 align:middle line:90% And more of this cancels. 01:16:31.670 --> 01:16:34.800 align:middle line:90% 01:16:34.800 --> 01:16:44.760 align:middle line:84% And we get 2n choose n is approximately 4 to the n over 01:16:44.760 --> 01:16:46.693 align:middle line:90% square root of pi. 01:16:46.693 --> 01:16:51.120 align:middle line:90% 01:16:51.120 --> 01:16:58.612 align:middle line:84% And that tells us that c sub n is going to 1 over n plus 1, 01:16:58.612 --> 01:17:08.630 align:middle line:84% 2n choose n, is 4 to the n over root pi n to 3/2. 01:17:08.630 --> 01:17:14.400 align:middle line:90% 01:17:14.400 --> 01:17:16.140 align:middle line:90% OK. 01:17:16.140 --> 01:17:18.120 align:middle line:90% So we have an approximation. 01:17:18.120 --> 01:17:21.530 align:middle line:84% We know roughly how fast c sub n goes. 01:17:21.530 --> 01:17:27.220 align:middle line:90% 01:17:27.220 --> 01:17:34.120 align:middle line:84% So next time I will do more examples of counting things. 01:17:34.120 --> 01:17:35.100 align:middle line:90% Any questions? 01:17:35.100 --> 01:17:46.680 align:middle line:90% 01:17:46.680 --> 01:17:48.940 align:middle line:84% We're not going to give a proof of Stirling's formula. 01:17:48.940 --> 01:17:51.080 align:middle line:84% But I wanted to say that the only 01:17:51.080 --> 01:17:54.040 align:middle line:84% hard part about proving Stirling's formula 01:17:54.040 --> 01:17:56.840 align:middle line:90% is this square root of pi. 01:17:56.840 --> 01:18:01.960 align:middle line:84% Everything else-- I guess the square root of 2 pi-- everything 01:18:01.960 --> 01:18:07.600 align:middle line:84% else can be done with elementary calculus 01:18:07.600 --> 01:18:10.880 align:middle line:90% or elementary probability. 01:18:10.880 --> 01:18:15.720 align:middle line:84% But the root pi, but knowing that this constant, this root 01:18:15.720 --> 01:18:17.600 align:middle line:90% pi, is tricky. 01:18:17.600 --> 01:18:20.150 align:middle line:90% OK, thanks. 01:18:20.150 --> 01:18:26.000 align:middle line:90%