WEBVTT 00:00:00.500 --> 00:00:02.000 [SQUEAKING] 00:00:02.000 --> 00:00:03.500 [RUSTLING] 00:00:03.500 --> 00:00:05.500 [CLICKING] 00:00:12.443 --> 00:00:13.610 PROFESSOR: All right, folks. 00:00:13.610 --> 00:00:16.079 I think it's about time to get started. 00:00:16.079 --> 00:00:18.270 Thank you guys for making it to our problem session 00:00:18.270 --> 00:00:20.557 and I guess talking to many of you guys 00:00:20.557 --> 00:00:22.640 virtually, given the number of people in the room. 00:00:22.640 --> 00:00:24.680 But that's perfectly fine. 00:00:24.680 --> 00:00:25.180 Right. 00:00:25.180 --> 00:00:28.190 So we're in problem session number two for 6.006. 00:00:28.190 --> 00:00:30.050 Hopefully you're all in the right place. 00:00:30.050 --> 00:00:31.853 I'm Justin. 00:00:31.853 --> 00:00:33.770 I was looking at a little spreadsheet of who's 00:00:33.770 --> 00:00:35.978 teaching what, and I think you guys are stuck with me 00:00:35.978 --> 00:00:38.725 in a lot of these problem sessions, which actually, I 00:00:38.725 --> 00:00:40.100 think in terms of hours of video, 00:00:40.100 --> 00:00:43.430 means that you're stuck listening to my voice the most. 00:00:43.430 --> 00:00:45.350 So sorry in advance. 00:00:45.350 --> 00:00:48.020 But in any event, it's always fun to do this kind of thing. 00:00:48.020 --> 00:00:51.140 And it gave me the opportunity to do homework, which I have 00:00:51.140 --> 00:00:54.230 not done in a very long time. 00:00:54.230 --> 00:00:58.040 So last night, I worked through all these problems and came up 00:00:58.040 --> 00:01:00.247 with my own hacked-together answers for these, 00:01:00.247 --> 00:01:02.330 which thankfully more or less agreed with what was 00:01:02.330 --> 00:01:05.030 in the answer key, which I have and you don't. 00:01:05.030 --> 00:01:06.960 I think that we release on-- 00:01:06.960 --> 00:01:10.160 so there's a bit of a known bug in 6.006, 00:01:10.160 --> 00:01:13.670 which is that this instructor has horrible handwriting. 00:01:13.670 --> 00:01:14.600 I'm working on it. 00:01:14.600 --> 00:01:17.750 I went to the classroom and practiced the other day. 00:01:17.750 --> 00:01:21.860 But in the meantime, there are horrible handwritten notes 00:01:21.860 --> 00:01:24.440 that are posted in the learning module, at least 00:01:24.440 --> 00:01:26.150 for the duration of the talk today. 00:01:26.150 --> 00:01:28.067 And we'll decide whether we want to leave them 00:01:28.067 --> 00:01:30.300 up or not because they're not super exciting. 00:01:30.300 --> 00:01:32.675 Just so you can make sure you can follow along with that, 00:01:32.675 --> 00:01:33.717 I'm writing on the board. 00:01:33.717 --> 00:01:36.253 And as with before, if there's ever anything 00:01:36.253 --> 00:01:37.670 that you can't read, just stop me. 00:01:37.670 --> 00:01:40.140 And I'll gladly fix. 00:01:40.140 --> 00:01:40.640 Cool. 00:01:40.640 --> 00:01:42.890 So right, so hopefully, does everybody 00:01:42.890 --> 00:01:45.603 have a copy of the problems? 00:01:45.603 --> 00:01:47.270 It'll look a little different from this. 00:01:47.270 --> 00:01:48.562 The green stuff is the answers. 00:01:48.562 --> 00:01:50.738 I get those, and you don't. 00:01:50.738 --> 00:01:53.030 Or sorry, I always ask this question when I'm teaching, 00:01:53.030 --> 00:01:54.197 and you never get an answer. 00:01:54.197 --> 00:01:56.720 Does anybody not have a copy of the problems and the problem 00:01:56.720 --> 00:01:58.030 session? 00:01:58.030 --> 00:01:58.700 Fabulous. 00:01:58.700 --> 00:01:59.730 OK. 00:01:59.730 --> 00:02:01.880 If you don't, there's a pile of handouts up here. 00:02:01.880 --> 00:02:03.060 Here, I'll grab one for you. 00:02:06.560 --> 00:02:08.220 Yeah. 00:02:08.220 --> 00:02:08.720 Great. 00:02:08.720 --> 00:02:11.150 So the basic goal of the problem sessions 00:02:11.150 --> 00:02:13.820 is to go over a bunch of example problems, which I believe 00:02:13.820 --> 00:02:17.330 are mostly lifted from past years' homeworks and exams, 00:02:17.330 --> 00:02:20.030 and work through how we went about thinking 00:02:20.030 --> 00:02:23.390 about them and a solution or at least just kind of sketch 00:02:23.390 --> 00:02:25.400 out enough that we're confident you could fill 00:02:25.400 --> 00:02:26.817 in the blanks of a solution, given 00:02:26.817 --> 00:02:28.010 our limited time together. 00:02:30.860 --> 00:02:31.550 Right. 00:02:31.550 --> 00:02:33.177 And of course, one of the pleasures 00:02:33.177 --> 00:02:35.510 of teaching this problem session is this is not normally 00:02:35.510 --> 00:02:36.468 the class that I teach. 00:02:36.468 --> 00:02:38.780 So these really are problems I haven't seen before. 00:02:38.780 --> 00:02:39.980 And indeed, in the middle of the night 00:02:39.980 --> 00:02:41.397 last night, I woke up and realized 00:02:41.397 --> 00:02:44.160 my answer to the last problem was totally wrong. 00:02:44.160 --> 00:02:46.800 So if you feel that way sometimes, 00:02:46.800 --> 00:02:49.310 the feeling is mutual. 00:02:49.310 --> 00:02:52.970 So in any event, there are four problems on your handout 00:02:52.970 --> 00:02:54.830 that roughly, I think, are in correspondence 00:02:54.830 --> 00:02:57.180 with the stuff we've taught over the last week or so. 00:02:57.180 --> 00:02:59.150 So the first problem involves recurrences. 00:02:59.150 --> 00:03:02.390 The second involves Infinity Stones. 00:03:02.390 --> 00:03:07.610 The third involves some kind of a queue stack structure. 00:03:07.610 --> 00:03:10.700 And the fourth involves all kinds 00:03:10.700 --> 00:03:14.310 of things talking to each other in a way that I got wrong. 00:03:14.310 --> 00:03:17.150 So I guess for lack of a more creative thing, 00:03:17.150 --> 00:03:19.165 we'll just go through these guys in order. 00:03:19.165 --> 00:03:20.540 And essentially, this is supposed 00:03:20.540 --> 00:03:22.940 to be an interactive session, although it doesn't really 00:03:22.940 --> 00:03:24.570 look that way. 00:03:24.570 --> 00:03:26.490 But since you guys are the people in the room, 00:03:26.490 --> 00:03:29.120 you have the advantage of being able to stop me the second 00:03:29.120 --> 00:03:31.343 that I, A, make a mistake or, B, say 00:03:31.343 --> 00:03:32.510 something that confuses you. 00:03:32.510 --> 00:03:35.010 And I'm relying on you to do that, OK? 00:03:35.010 --> 00:03:35.510 Nods. 00:03:35.510 --> 00:03:36.893 That's our deal today. 00:03:36.893 --> 00:03:38.060 Some kind of acknowledgment. 00:03:38.060 --> 00:03:40.200 Noses out of the laptops, maybe, for a second. 00:03:40.200 --> 00:03:40.700 Cool. 00:03:40.700 --> 00:03:41.210 Thank you. 00:03:41.210 --> 00:03:41.930 See, thumbs up. 00:03:41.930 --> 00:03:43.230 That's what I'm looking for. 00:03:43.230 --> 00:03:44.090 OK. 00:03:44.090 --> 00:03:46.160 All right, folks. 00:03:46.160 --> 00:03:48.500 Right, so let's start with the problem number one here. 00:03:48.500 --> 00:03:50.630 So problem number one is solving recurrences, 00:03:50.630 --> 00:03:52.490 which is our favorite way to torture 00:03:52.490 --> 00:03:55.830 undergrads in algorithms classes in the first couple of weeks. 00:03:55.830 --> 00:03:59.540 And in particular, applying this master theorem. 00:03:59.540 --> 00:04:02.870 So I thought I'd spent about 82 seconds reviewing master 00:04:02.870 --> 00:04:06.680 theorem because this is sort of the giant sledgehammer 00:04:06.680 --> 00:04:09.680 for solving recurrences without understanding why you 00:04:09.680 --> 00:04:11.300 got the answer that you did. 00:04:11.300 --> 00:04:13.692 And then for each of these problems, 00:04:13.692 --> 00:04:15.150 conveniently, this homework problem 00:04:15.150 --> 00:04:19.010 asks you to do each big O thing twice, right? 00:04:19.010 --> 00:04:21.230 Once using master theorem to kind of get you 00:04:21.230 --> 00:04:24.045 used to just applying this little set of rules, 00:04:24.045 --> 00:04:25.670 which is sort of the most efficient way 00:04:25.670 --> 00:04:26.720 to solve recurrences. 00:04:26.720 --> 00:04:28.920 It just gives you a formula to look at. 00:04:28.920 --> 00:04:32.240 In addition to that, drawing a tree of computation 00:04:32.240 --> 00:04:34.860 and counting the number of computations you do. 00:04:34.860 --> 00:04:38.030 So the first two are basically straightforward applications 00:04:38.030 --> 00:04:39.440 of master theorem. 00:04:39.440 --> 00:04:41.940 The third one requires a little bit of thinking. 00:04:41.940 --> 00:04:44.210 So naturally, I got stuck and wasted an hour yesterday 00:04:44.210 --> 00:04:46.700 trying to convince myself the answer was right. 00:04:46.700 --> 00:04:47.270 OK. 00:04:47.270 --> 00:04:49.970 So for a tiny bit of review, I know that you also 00:04:49.970 --> 00:04:51.660 did this in recitation. 00:04:51.660 --> 00:04:54.230 But it can't hurt to bring it up again. 00:04:54.230 --> 00:04:55.850 Remember that master theorem-- 00:05:02.970 --> 00:05:05.470 I believe that "master" really is just because it's general, 00:05:05.470 --> 00:05:05.690 right? 00:05:05.690 --> 00:05:07.232 There's not a guy named Master, but I 00:05:07.232 --> 00:05:10.390 notice that it was capitalized a lot in the course notes. 00:05:10.390 --> 00:05:12.910 But in any event, the basic idea is 00:05:12.910 --> 00:05:14.830 that it has some recurrence that looks 00:05:14.830 --> 00:05:22.480 like the following, which is T of n equals aT of n/b 00:05:22.480 --> 00:05:27.370 plus f of n. 00:05:27.370 --> 00:05:31.630 So for instance, in merge sort, for a tiny bit of review, 00:05:31.630 --> 00:05:34.040 remember that we kind of split into two pieces. 00:05:34.040 --> 00:05:36.430 So both a and b in merge sort would be 2. 00:05:36.430 --> 00:05:37.120 Yeah. 00:05:37.120 --> 00:05:39.490 And the amount of work that I do in the merge step 00:05:39.490 --> 00:05:40.840 is kind of like n. 00:05:40.840 --> 00:05:43.570 So that's how to fill in these different constants there. 00:05:43.570 --> 00:05:45.280 And the question is, asymptotically, 00:05:45.280 --> 00:05:47.530 what does this function look like? 00:05:47.530 --> 00:05:49.930 And of course, it's not totally obvious from this formula 00:05:49.930 --> 00:05:52.240 because I've defined T in terms of itself, right? 00:05:52.240 --> 00:05:55.600 And that's the sort of annoying part of solving recurrences. 00:05:55.600 --> 00:05:58.180 And so master theorem, and I'll let you guys review 00:05:58.180 --> 00:06:02.020 your section notes about precisely how this gets proved, 00:06:02.020 --> 00:06:05.980 essentially divides into three cases. 00:06:05.980 --> 00:06:08.980 And the relevant numbers here are as follows, right? 00:06:08.980 --> 00:06:10.618 There's an a, a b, and an f. 00:06:10.618 --> 00:06:12.910 And so the a is kind of like a branching factor, right? 00:06:12.910 --> 00:06:14.890 So if you remember your computation tree, 00:06:14.890 --> 00:06:19.640 like in merge sort, you split into two pieces, right? 00:06:19.640 --> 00:06:21.310 So that number is kind of like a. 00:06:21.310 --> 00:06:24.250 It's the number of splits that you make, right? 00:06:24.250 --> 00:06:27.700 b is the amount that your problem size reduces when 00:06:27.700 --> 00:06:29.412 you go to each of those leaves. 00:06:29.412 --> 00:06:31.120 And then finally, f is the amount of work 00:06:31.120 --> 00:06:32.280 that you do at each node. 00:06:32.280 --> 00:06:35.510 Yeah, so hopefully that basic language makes sense. 00:06:35.510 --> 00:06:38.680 And then essentially what you learn from master theorem 00:06:38.680 --> 00:06:40.460 is that there's three cases. 00:06:40.460 --> 00:06:46.012 The first is that f of n looks like-- 00:06:46.012 --> 00:06:48.220 by the way, in the recitation notes, for some reason, 00:06:48.220 --> 00:06:49.470 it's kind of written backward. 00:06:49.470 --> 00:06:52.060 They give the answer to master theorem and then the condition. 00:06:52.060 --> 00:06:54.480 I'm going to write in the other order. 00:06:54.480 --> 00:06:54.980 Right. 00:06:54.980 --> 00:06:58.090 Looks like the following, which is n to the log 00:06:58.090 --> 00:07:06.958 base b of a minus epsilon for some positive epsilon. 00:07:06.958 --> 00:07:09.250 I'm going to try to be very conscientious about reading 00:07:09.250 --> 00:07:12.620 my writing out loud as I write it down. 00:07:12.620 --> 00:07:13.120 All right. 00:07:13.120 --> 00:07:15.338 And in that case, what we find is notice 00:07:15.338 --> 00:07:16.880 there's something kind of magic here, 00:07:16.880 --> 00:07:18.730 which is that this is just an upper bound. 00:07:18.730 --> 00:07:20.450 There's not a theta here. 00:07:20.450 --> 00:07:23.920 So it's OK if f is below this thing. 00:07:23.920 --> 00:07:29.290 But in any event, the conclusion then is that T of n 00:07:29.290 --> 00:07:36.443 is big theta of n to the log base b of a. 00:07:36.443 --> 00:07:38.110 It's kind of cool if you think about it. 00:07:38.110 --> 00:07:39.568 You only have an upper bound for f, 00:07:39.568 --> 00:07:43.013 but you get a nice theta bound for T, which is pretty cool. 00:07:43.013 --> 00:07:44.680 The reason for that is essentially the f 00:07:44.680 --> 00:07:46.570 is sort of insignificant relative to the work 00:07:46.570 --> 00:07:49.980 of just traversing up and down the tree. 00:07:49.980 --> 00:07:51.990 That's case one. 00:07:51.990 --> 00:07:55.710 Case two-- I'm going to try and leave this on the board as we 00:07:55.710 --> 00:07:57.360 actually solve these problems-- 00:07:57.360 --> 00:07:59.950 is the following, which is f of n 00:07:59.950 --> 00:08:04.920 is big theta of n to the log base 00:08:04.920 --> 00:08:13.020 b of a multiplied by log to the k-th power of n. 00:08:13.020 --> 00:08:16.140 This is a super weird form, but for example, it's 00:08:16.140 --> 00:08:18.640 perfectly kosher to take k equals 0, in which case 00:08:18.640 --> 00:08:21.210 it starts to look like case one, right? 00:08:21.210 --> 00:08:27.450 For some positive k or non-negative k, rather, 00:08:27.450 --> 00:08:32.100 in which case what we learn is that T of n 00:08:32.100 --> 00:08:43.919 is theta of n to the log base b of a times log to the k 00:08:43.919 --> 00:08:47.405 plus 1 of n. 00:08:47.405 --> 00:08:49.530 You can see why we don't love applying this theorem 00:08:49.530 --> 00:08:50.947 in this course because I feel just 00:08:50.947 --> 00:08:53.460 staring at these formulas is totally unenlightening. 00:08:53.460 --> 00:08:55.860 But it is a giant sledgehammer for solving recurrences 00:08:55.860 --> 00:08:57.090 quickly. 00:08:57.090 --> 00:09:00.733 And then finally, case three-- 00:09:00.733 --> 00:09:02.400 I'm going to slide the board up, and I'm 00:09:02.400 --> 00:09:03.942 going to slide it back down because I 00:09:03.942 --> 00:09:05.220 don't want to bend over. 00:09:05.220 --> 00:09:12.690 OK-- is the following, which is f of n is omega, right, 00:09:12.690 --> 00:09:18.270 meaning that it's lower bounded by n to the log base 00:09:18.270 --> 00:09:28.380 b of a plus epsilon for some epsilon greater than 0. 00:09:28.380 --> 00:09:30.375 And we needed a second case. 00:09:33.540 --> 00:09:41.550 What's it called? af of n over b is less than cfn 00:09:41.550 --> 00:09:45.900 for some c between 0 and 1. 00:09:49.510 --> 00:09:51.010 And what's the conclusion there? 00:09:51.010 --> 00:09:53.670 Then it turns out that the f term sort of dominates, 00:09:53.670 --> 00:09:59.610 and what we get is that T of n is theta of f of n. 00:10:03.750 --> 00:10:04.860 OK. 00:10:04.860 --> 00:10:08.280 So essentially, this covers three major cases 00:10:08.280 --> 00:10:09.570 that you see in recurrences. 00:10:09.570 --> 00:10:13.005 There are other more generic versions of the master theorem 00:10:13.005 --> 00:10:14.880 out there where essentially, rather than just 00:10:14.880 --> 00:10:18.210 having one term with the T in it, maybe use more than one 00:10:18.210 --> 00:10:19.380 term with a T in it. 00:10:19.380 --> 00:10:21.510 I don't think those are covered in this class. 00:10:21.510 --> 00:10:23.850 I always confuse the name of the more general one. 00:10:23.850 --> 00:10:25.230 I want to say it's Arzela-Ascoli, 00:10:25.230 --> 00:10:27.230 but I know that that's from functional analysis. 00:10:27.230 --> 00:10:31.103 [INAUDIBLE] theorem, if you want to Google that, learn more. 00:10:31.103 --> 00:10:33.270 But in any event, that's going to be enough for most 00:10:33.270 --> 00:10:36.810 of the recurrences we care about in 006. 00:10:36.810 --> 00:10:38.685 So I've at least written our conditions down. 00:10:38.685 --> 00:10:40.768 And now they're going to sit on the left-hand side 00:10:40.768 --> 00:10:42.540 while we do three example problems to show 00:10:42.540 --> 00:10:43.990 how they show up in practice. 00:10:43.990 --> 00:10:46.770 Are there any questions about what this theorem is telling us 00:10:46.770 --> 00:10:48.180 about life or how to apply it? 00:10:50.833 --> 00:10:52.000 Don't all speak at once now. 00:10:52.000 --> 00:10:52.450 Yes. 00:10:52.450 --> 00:10:53.770 AUDIENCE: Sorry, I have a handwriting question. 00:10:53.770 --> 00:10:54.687 PROFESSOR: No problem. 00:10:54.687 --> 00:10:57.100 AUDIENCE: What's after "some," the last line? 00:10:57.100 --> 00:11:02.260 PROFESSOR: Oh, for some c in 0, 1, not inclusive. 00:11:02.260 --> 00:11:04.060 Fabulous question. 00:11:04.060 --> 00:11:04.690 Any others? 00:11:07.370 --> 00:11:08.000 OK. 00:11:08.000 --> 00:11:11.430 So let's do an example problem here. 00:11:11.430 --> 00:11:13.910 So let's do part a. 00:11:13.910 --> 00:11:15.780 So in part a, they give you a recurrence, 00:11:15.780 --> 00:11:17.950 which isn't all that different from merge 00:11:17.950 --> 00:11:19.320 sort or any of the other ones. 00:11:19.320 --> 00:11:20.720 It looks something like this. 00:11:20.720 --> 00:11:29.218 T of n is equal to 2T of n/2 plus-- 00:11:29.218 --> 00:11:31.760 and then they have some other term, which they don't tell you 00:11:31.760 --> 00:11:37.367 anything about beyond that it's big O of square root of n. 00:11:37.367 --> 00:11:39.200 By the way, this might mean that it actually 00:11:39.200 --> 00:11:40.670 does order 1 work, right? 00:11:40.670 --> 00:11:42.920 It just has to be upper bounded by a square root of n. 00:11:42.920 --> 00:11:44.960 That's the only thing that problem's telling you. 00:11:44.960 --> 00:11:46.190 I'm going to keep kind of driving home 00:11:46.190 --> 00:11:48.705 that point because I think everybody, including myself, 00:11:48.705 --> 00:11:50.580 gets really sloppy about, like, when is it O? 00:11:50.580 --> 00:11:51.320 When is it theta? 00:11:51.320 --> 00:11:52.208 When's it omega? 00:11:52.208 --> 00:11:54.500 And so every single time you write one of those letters 00:11:54.500 --> 00:11:56.838 down, you should step back 50 feet and look at it 00:11:56.838 --> 00:11:58.380 and think, like, did I do that right? 00:11:58.380 --> 00:12:00.410 Or did I just write a Greek letter, right? 00:12:00.410 --> 00:12:02.060 And make sure that you're actually 00:12:02.060 --> 00:12:04.610 thinking through it logically. 00:12:04.610 --> 00:12:05.390 OK. 00:12:05.390 --> 00:12:11.240 So when I apply master theorem in my everyday life, what 00:12:11.240 --> 00:12:13.490 I like to do is to say, OK, somehow, 00:12:13.490 --> 00:12:16.490 the really key quantity in master theorem 00:12:16.490 --> 00:12:20.780 is this dude, right, n to the log base b of a. 00:12:20.780 --> 00:12:23.700 He just keeps showing up in all these different cases. 00:12:23.700 --> 00:12:28.250 So I might as well figure what that is for my recurrence, 00:12:28.250 --> 00:12:31.750 yeah, and then plug it in and check which case I'm in. 00:12:31.750 --> 00:12:33.360 Does that make sense? 00:12:33.360 --> 00:12:35.310 So let's do that. 00:12:35.310 --> 00:12:40.160 So first of all, what is a for this particular recurrence? 00:12:40.160 --> 00:12:40.668 AUDIENCE: 2. 00:12:40.668 --> 00:12:41.210 PROFESSOR: 2. 00:12:43.760 --> 00:12:46.060 What's b? 00:12:46.060 --> 00:12:47.060 You're killing me, guys. 00:12:47.060 --> 00:12:47.560 On 3. 00:12:47.560 --> 00:12:48.560 1, 2, 3. 00:12:48.560 --> 00:12:49.430 AUDIENCE: 2. 00:12:49.430 --> 00:12:51.020 PROFESSOR: OK. 00:12:51.020 --> 00:12:53.180 The enthusiasm is overwhelming. 00:12:53.180 --> 00:12:55.190 OK, and what do we know about the function f? 00:12:55.190 --> 00:12:56.660 AUDIENCE: [INAUDIBLE] 00:12:56.660 --> 00:12:58.430 PROFESSOR: Yeah, it's big O of root n. 00:12:58.430 --> 00:13:00.020 We don't know that it equals root n. 00:13:00.020 --> 00:13:02.538 But we do know that it's at most upper bounded by something 00:13:02.538 --> 00:13:03.830 that kind of looks like root n. 00:13:07.380 --> 00:13:09.130 Eric gave me a whole lesson on this chalk, 00:13:09.130 --> 00:13:10.730 and I'm still failing at it. 00:13:10.730 --> 00:13:11.230 OK. 00:13:11.230 --> 00:13:15.475 So now that gives us everything we need to compute n 00:13:15.475 --> 00:13:16.745 to the log base b of a. 00:13:16.745 --> 00:13:18.370 We're going to do this one real slowly, 00:13:18.370 --> 00:13:20.912 and then we're going to do the next one a little more fastly, 00:13:20.912 --> 00:13:22.540 quickly. 00:13:22.540 --> 00:13:26.740 So n to the log base b of a. 00:13:26.740 --> 00:13:34.355 Well, that's n to the log base 2 of 2. 00:13:34.355 --> 00:13:35.230 Anybody have an idea? 00:13:35.230 --> 00:13:38.158 What's the log base 2 of 2? 00:13:38.158 --> 00:13:39.878 AUDIENCE: 1. 00:13:39.878 --> 00:13:40.420 PROFESSOR: 1. 00:13:40.420 --> 00:13:41.920 OK, I'll take it this time. 00:13:41.920 --> 00:13:43.320 OK. 00:13:43.320 --> 00:13:43.820 Right. 00:13:43.820 --> 00:13:47.320 And so then what does master theorem tell you? 00:13:47.320 --> 00:13:49.060 Well, in some sense, I want to know 00:13:49.060 --> 00:13:52.660 what f is, which in this case looks like square root of n, 00:13:52.660 --> 00:13:56.140 compared to what this is, n to the first. 00:13:56.140 --> 00:13:57.010 Yeah? 00:13:57.010 --> 00:14:02.140 Now, first of all, which of these two things grows faster? 00:14:02.140 --> 00:14:03.430 This guy, right? 00:14:03.430 --> 00:14:07.960 So what we know is that f is really upper bounded by this n 00:14:07.960 --> 00:14:08.590 here. 00:14:08.590 --> 00:14:09.910 "Upper bounded" is not quite the right term 00:14:09.910 --> 00:14:11.660 because big O allows for some wiggle room, 00:14:11.660 --> 00:14:12.640 but asymptotically. 00:14:12.640 --> 00:14:14.740 Hopefully you guys get this concept. 00:14:14.740 --> 00:14:19.750 OK, so in particular, let's see. 00:14:19.750 --> 00:14:21.547 So remember that f of-- 00:14:21.547 --> 00:14:22.880 let's maybe just write it again. 00:14:22.880 --> 00:14:26.860 So f of n is big O of square root of n. 00:14:26.860 --> 00:14:32.050 But let's write that in a suggestive way as n to the 1/2. 00:14:32.050 --> 00:14:33.220 Yeah. 00:14:33.220 --> 00:14:38.740 In particular, this is equal to big O of n to the 1 minus 1/2. 00:14:38.740 --> 00:14:42.590 It's the beautiful thing about 1/2. 00:14:42.590 --> 00:14:43.090 Yeah. 00:14:43.090 --> 00:14:45.490 But what is n to the 1? 00:14:45.490 --> 00:14:48.410 Well, that's n to the log base b of a, right? 00:14:48.410 --> 00:14:51.910 That's what we just showed here, right? 00:14:51.910 --> 00:14:57.970 So this is really O of n to the log b 00:14:57.970 --> 00:15:01.180 of a-- that's a complicated way of writing the number 1, 00:15:01.180 --> 00:15:02.260 right-- 00:15:02.260 --> 00:15:03.380 minus 1/2. 00:15:03.380 --> 00:15:05.630 Let's give 1/2 a special name, too, while we're at it. 00:15:05.630 --> 00:15:09.370 Let's call him epsilon or her, yeah, 00:15:09.370 --> 00:15:14.730 where we take epsilon equals-- 00:15:14.730 --> 00:15:17.540 see what I did there? 00:15:17.540 --> 00:15:19.510 Now, what does it tell me? 00:15:19.510 --> 00:15:21.597 There are three different cases of master theorem. 00:15:21.597 --> 00:15:23.180 And take a look at what I just showed. 00:15:23.180 --> 00:15:26.290 I showed that f of n is equal to big O of n to the log base 00:15:26.290 --> 00:15:31.270 b of a minus epsilon for some epsilon that's equal to 1/2. 00:15:31.270 --> 00:15:35.030 The beautiful thing about 1/2 is that it is greater than 0. 00:15:35.030 --> 00:15:35.920 Yeah. 00:15:35.920 --> 00:15:40.000 So I think we have somewhat laboriously checked 00:15:40.000 --> 00:15:41.480 that we are in case one. 00:15:41.480 --> 00:15:43.930 Any dissidents here? 00:15:43.930 --> 00:15:44.780 Fabulous. 00:15:44.780 --> 00:15:45.280 OK. 00:15:45.280 --> 00:15:48.040 So in that case, master theorem, basically, 00:15:48.040 --> 00:15:49.120 you're just done, right? 00:15:49.120 --> 00:15:50.660 So what is the conclusion? 00:15:50.660 --> 00:15:52.648 AUDIENCE: T of n is theta of n. 00:15:52.648 --> 00:15:53.440 PROFESSOR: Exactly. 00:15:53.440 --> 00:15:56.320 So T of n is theta of n to the log base b of a. 00:15:56.320 --> 00:16:00.040 But we already showed that's equal to n. 00:16:00.040 --> 00:16:02.870 So we're done. 00:16:02.870 --> 00:16:03.370 All right. 00:16:03.370 --> 00:16:08.170 Any questions about how we apply master theorem here? 00:16:08.170 --> 00:16:08.720 OK. 00:16:08.720 --> 00:16:10.907 So now here's the thing about master theorem. 00:16:10.907 --> 00:16:13.240 Did we learn anything about what this recursive function 00:16:13.240 --> 00:16:14.095 is doing? 00:16:14.095 --> 00:16:16.720 No, we just did a bunch of work, plugged in some Greek symbols, 00:16:16.720 --> 00:16:18.510 and out came a big theta. 00:16:18.510 --> 00:16:20.110 Yeah. 00:16:20.110 --> 00:16:23.140 So what we might want to do instead 00:16:23.140 --> 00:16:27.040 is to use a method that I'm still learning myself 00:16:27.040 --> 00:16:29.020 because I'm not used to presenting it this way, 00:16:29.020 --> 00:16:31.660 but that's OK, which is to actually draw out 00:16:31.660 --> 00:16:33.030 the tree of computation. 00:16:33.030 --> 00:16:34.097 Yeah. 00:16:34.097 --> 00:16:35.180 So let's actually do that. 00:16:35.180 --> 00:16:39.100 So let's say that I call my function T of n. 00:16:39.100 --> 00:16:40.083 So what does T of n do? 00:16:40.083 --> 00:16:42.250 Well, in some sense, it does work that kind of looks 00:16:42.250 --> 00:16:44.380 like the square root of n, right, 00:16:44.380 --> 00:16:48.490 and then it makes two function calls, each of which 00:16:48.490 --> 00:16:50.860 has n/2 amount of data. 00:16:50.860 --> 00:16:51.550 Yeah. 00:16:51.550 --> 00:16:53.030 So let's draw what this looks like. 00:16:53.030 --> 00:16:55.810 So the first thing that my function might do 00:16:55.810 --> 00:16:59.320 is work square root of n much. 00:16:59.320 --> 00:17:01.540 Going to put a little square root of n there. 00:17:01.540 --> 00:17:04.599 And now it makes two function calls. 00:17:04.599 --> 00:17:06.480 And how much work does each of these guys do? 00:17:09.300 --> 00:17:14.819 So when it calls T, what goes into T? 00:17:14.819 --> 00:17:15.329 n/2. 00:17:15.329 --> 00:17:16.950 AUDIENCE: So square root of n/2. 00:17:16.950 --> 00:17:17.760 PROFESSOR: Exactly. 00:17:17.760 --> 00:17:22.470 So it does square root of n/2 amount of work 00:17:22.470 --> 00:17:26.033 at each of these two nodes. 00:17:26.033 --> 00:17:26.700 That make sense? 00:17:29.912 --> 00:17:32.370 AUDIENCE: That's determined by the branching number, right? 00:17:32.370 --> 00:17:33.280 PROFESSOR: You have to be careful. 00:17:33.280 --> 00:17:34.720 So the branching number, in this case, 00:17:34.720 --> 00:17:35.845 it happened to be the same. 00:17:35.845 --> 00:17:37.920 The next problem we do, they'll be different. 00:17:37.920 --> 00:17:39.330 So they're both 2 here. 00:17:39.330 --> 00:17:42.390 The outside 2 refers to the fact that there's two children. 00:17:42.390 --> 00:17:42.690 AUDIENCE: Yeah, never mind. 00:17:42.690 --> 00:17:44.460 PROFESSOR: And inside 2 refers to the fact 00:17:44.460 --> 00:17:45.357 that I divided by 2. 00:17:45.357 --> 00:17:47.940 It's a fabulous question because it's exactly what I got wrong 00:17:47.940 --> 00:17:49.600 when I did this problem. 00:17:49.600 --> 00:17:50.550 OK. 00:17:50.550 --> 00:17:55.100 So now each of these guys calls one of their children. 00:17:55.100 --> 00:17:55.600 Yeah. 00:17:55.600 --> 00:17:58.920 And so again, the data divides by 2. 00:17:58.920 --> 00:18:08.070 So now I have the square root of n/4, like this. 00:18:08.070 --> 00:18:08.570 Yeah. 00:18:13.002 --> 00:18:15.210 If you're wondering, inside of each of these circles, 00:18:15.210 --> 00:18:16.320 it says square root of n/4. 00:18:16.320 --> 00:18:16.820 OK. 00:18:21.960 --> 00:18:24.380 So let's say that-- 00:18:24.380 --> 00:18:27.750 so now meanwhile, in our function call tree here, 00:18:27.750 --> 00:18:31.710 how many nodes are in just the first level at the very top? 00:18:31.710 --> 00:18:32.670 One, right? 00:18:32.670 --> 00:18:33.630 Just this dude. 00:18:33.630 --> 00:18:34.860 Yeah. 00:18:34.860 --> 00:18:36.630 One node. 00:18:39.590 --> 00:18:42.600 How many nodes are in the second level? 00:18:42.600 --> 00:18:44.220 Two. 00:18:44.220 --> 00:18:49.710 Or if we want, 2 to the 1 nodes. 00:18:49.710 --> 00:18:57.040 Here, there's four, yeah, which is 2 squared nodes, and so on. 00:18:57.040 --> 00:18:57.540 OK. 00:18:57.540 --> 00:19:00.180 So now we have some pictorial representation 00:19:00.180 --> 00:19:03.327 of what's going on inside of our recurrence. 00:19:03.327 --> 00:19:05.160 If you look at the course notes I wrote out, 00:19:05.160 --> 00:19:09.308 I wrote out one more layer just for fun and profit. 00:19:09.308 --> 00:19:11.100 So now let's see how this helps us actually 00:19:11.100 --> 00:19:12.900 solve our recurrence. 00:19:12.900 --> 00:19:16.620 So what do we know? 00:19:16.620 --> 00:19:18.090 How many levels are in this tree? 00:19:18.090 --> 00:19:20.490 Well, our algorithm kind of stops when the input to T 00:19:20.490 --> 00:19:23.220 looks like 1, right? 00:19:23.220 --> 00:19:28.350 So since this divides by 2 each time, 00:19:28.350 --> 00:19:35.920 the tree has log base 2 of n levels in the tree. 00:19:35.920 --> 00:19:37.200 Yeah. 00:19:37.200 --> 00:19:40.410 And each level does how much work? 00:19:43.770 --> 00:19:44.490 So be careful. 00:19:44.490 --> 00:19:51.660 So each level-- let's call that level l, right? 00:19:51.660 --> 00:19:59.063 So we'll say l equals 0, l equals 1, and so on. 00:19:59.063 --> 00:20:00.480 So there's two different things we 00:20:00.480 --> 00:20:01.938 have to account for when we account 00:20:01.938 --> 00:20:03.240 for the work in this level. 00:20:03.240 --> 00:20:05.623 One is the square root of n/2 in a single node. 00:20:05.623 --> 00:20:08.040 And the other is the fact that there's two different nodes 00:20:08.040 --> 00:20:09.490 in the level. 00:20:09.490 --> 00:20:15.660 So in level l, so the amount of work in level 00:20:15.660 --> 00:20:17.770 l is equal to the product of those two things. 00:20:17.770 --> 00:20:20.470 So how many nodes are in level l? 00:20:20.470 --> 00:20:22.330 2 to the l, yeah? 00:20:22.330 --> 00:20:22.830 OK. 00:20:25.470 --> 00:20:30.210 And how much work does each one of those nodes do? 00:20:30.210 --> 00:20:34.008 It's like n/2 to the l. 00:20:34.008 --> 00:20:36.930 So a different way of writing that 00:20:36.930 --> 00:20:42.270 is the square root of n times 2 to the minus l, like that. 00:20:42.270 --> 00:20:44.950 This is just n/2 to the l. 00:20:44.950 --> 00:20:45.450 Fabulous. 00:20:45.450 --> 00:20:47.970 So now we have everything we need to actually work out 00:20:47.970 --> 00:20:50.200 our solution to the recurrence. 00:20:50.200 --> 00:20:50.700 OK. 00:20:50.700 --> 00:20:51.520 Yes? 00:20:51.520 --> 00:20:53.250 AUDIENCE: What does that say? 00:20:53.250 --> 00:20:58.050 PROFESSOR: This is n times 2 to the minus l. 00:20:58.050 --> 00:20:59.370 I appreciate the [INAUDIBLE]. 00:20:59.370 --> 00:21:03.097 I'm going to work on this more. 00:21:03.097 --> 00:21:04.680 And that's why I also share it online. 00:21:04.680 --> 00:21:05.820 The handwritten notes is-- 00:21:05.820 --> 00:21:07.320 this is literally a scan of the page 00:21:07.320 --> 00:21:09.060 that I'm holding in front of me. 00:21:09.060 --> 00:21:10.560 OK. 00:21:10.560 --> 00:21:14.520 So if we want the total amount of work in our tree, 00:21:14.520 --> 00:21:17.280 right, if we want to account for all of T of n, then what 00:21:17.280 --> 00:21:18.280 are we going to do? 00:21:18.280 --> 00:21:22.140 Well, we have to sum starting at level 0. 00:21:22.140 --> 00:21:25.260 I'm using l instead of i because I'm an analysis person, 00:21:25.260 --> 00:21:26.760 and i is the square root of minus 1. 00:21:26.760 --> 00:21:28.840 And I don't like i's in my algorithms. 00:21:28.840 --> 00:21:32.430 But in any event, how many total levels are there in my tree? 00:21:32.430 --> 00:21:38.810 Well, we know that there's log base 2 of n, right? 00:21:38.810 --> 00:21:42.270 And now we have to sum this quantity for all those levels. 00:21:42.270 --> 00:21:48.810 2 to the l, the square root of n times 2 to the minus l, 00:21:48.810 --> 00:21:50.630 like that. 00:21:50.630 --> 00:21:51.130 OK. 00:21:51.130 --> 00:21:53.970 And then what do we know secretly from the master 00:21:53.970 --> 00:21:55.920 theorem is that we're suspecting that this is 00:21:55.920 --> 00:21:57.170 going to be theta of n, right? 00:21:57.170 --> 00:21:58.590 So that's all we've got to check. 00:21:58.590 --> 00:22:02.230 OK, so let's do that real fast. 00:22:02.230 --> 00:22:05.790 So first of all, notice that this n term 00:22:05.790 --> 00:22:07.600 doesn't depend on the summand. 00:22:07.600 --> 00:22:09.690 So I can just pull them out, right? 00:22:09.690 --> 00:22:15.240 So this is really the square root of n times the sum over l 00:22:15.240 --> 00:22:22.425 equals 0 to the log 2 of n of what? 00:22:25.580 --> 00:22:28.300 Oh, and there's a mistake in my notes. 00:22:28.300 --> 00:22:28.800 Oh, man. 00:22:32.940 --> 00:22:34.280 Oh, no, this is-- 00:22:34.280 --> 00:22:35.667 AUDIENCE: [INAUDIBLE] 00:22:35.667 --> 00:22:36.750 PROFESSOR: Oh, right, duh. 00:22:36.750 --> 00:22:37.050 OK. 00:22:37.050 --> 00:22:39.342 So you have 2 to the l, and you have a square root of 2 00:22:39.342 --> 00:22:41.430 to the minus l. 00:22:41.430 --> 00:22:45.340 So this is really 2 to the minus l/2. 00:22:45.340 --> 00:22:47.730 So you have l minus l/2. 00:22:47.730 --> 00:22:52.486 And this quantity is really 2 to the l/2, like that. 00:22:52.486 --> 00:22:53.880 Does that make sense? 00:22:53.880 --> 00:22:56.790 So this is just properties of exponents from high school 00:22:56.790 --> 00:22:58.210 class. 00:22:58.210 --> 00:23:01.680 And in fact, I don't like this because it's l/2, 00:23:01.680 --> 00:23:03.240 and I want to blindly apply a formula 00:23:03.240 --> 00:23:04.800 without thinking about it. 00:23:04.800 --> 00:23:07.860 And so of course, this is really the same 00:23:07.860 --> 00:23:13.915 as the square root of 2 to the l-th power. 00:23:13.915 --> 00:23:16.020 This is just properties of exponents. 00:23:16.020 --> 00:23:16.950 OK. 00:23:16.950 --> 00:23:19.043 So what is this sum called? 00:23:19.043 --> 00:23:19.960 Do you recognize this? 00:23:19.960 --> 00:23:21.210 So there's some constant here. 00:23:21.210 --> 00:23:22.668 You're taking it to the l-th power, 00:23:22.668 --> 00:23:25.200 and then you're summing it over l's. 00:23:25.200 --> 00:23:27.840 This is called a geometric series, right? 00:23:27.840 --> 00:23:30.450 So it would be like 1 plus x plus x squared 00:23:30.450 --> 00:23:32.670 plus x to the third and dot dot dot, x 00:23:32.670 --> 00:23:35.440 to the, I guess in this case, log base 2 of n. 00:23:35.440 --> 00:23:36.630 Yeah? 00:23:36.630 --> 00:23:39.420 So conveniently, there's a geometrics series formula 00:23:39.420 --> 00:23:41.220 out there. 00:23:41.220 --> 00:23:43.188 I wrote it down in the handwritten notes, 00:23:43.188 --> 00:23:45.480 but maybe for now, because I'm moving slowly as always, 00:23:45.480 --> 00:23:47.100 so we'll skip that part. 00:23:47.100 --> 00:23:51.810 And what you could is as follows, 00:23:51.810 --> 00:23:53.820 that this is the square root of n, right-- 00:23:53.820 --> 00:23:55.320 that's just this outer part-- 00:23:55.320 --> 00:24:00.720 times-- and the inside are going to get the square root of 2 00:24:00.720 --> 00:24:11.250 to the log base 2 of n plus 1 minus 1 00:24:11.250 --> 00:24:15.720 divided by the square root of 2 minus 1. 00:24:15.720 --> 00:24:18.602 And this is the geometric series formula. 00:24:18.602 --> 00:24:21.060 So I encourage you guys to go back home and Google that one 00:24:21.060 --> 00:24:21.643 if you forgot. 00:24:21.643 --> 00:24:22.200 Yes. 00:24:22.200 --> 00:24:24.030 AUDIENCE: Wait, why are we calculating 00:24:24.030 --> 00:24:24.947 the sum of the series? 00:24:24.947 --> 00:24:26.447 PROFESSOR: Because what we're trying 00:24:26.447 --> 00:24:29.260 to do is figure out the total amount of work in this tree, 00:24:29.260 --> 00:24:30.330 right? 00:24:30.330 --> 00:24:32.310 So what we've done so far, we know 00:24:32.310 --> 00:24:35.230 that a single level in this tree is this value. 00:24:35.230 --> 00:24:37.410 And now we have to sum over all the levels from 0 00:24:37.410 --> 00:24:39.060 to log base 2 of n. 00:24:39.060 --> 00:24:41.340 AUDIENCE: OK, and what do we use this number for? 00:24:41.340 --> 00:24:43.632 PROFESSOR: Because we're trying to actually approximate 00:24:43.632 --> 00:24:45.390 or at least bound T of n, right? 00:24:45.390 --> 00:24:47.990 And T of n is the sum of all these values. 00:24:47.990 --> 00:24:48.850 AUDIENCE: Oh, OK. 00:24:48.850 --> 00:24:50.760 And theta of n is not a sufficient bound. 00:24:50.760 --> 00:24:52.260 PROFESSOR: It is a sufficient bound. 00:24:52.260 --> 00:24:54.273 We just got it from kind of a boring way. 00:24:54.273 --> 00:24:55.690 This is a second way that we could 00:24:55.690 --> 00:24:56.800 have proved the same formula. 00:24:56.800 --> 00:24:57.250 AUDIENCE: Oh, OK. 00:24:57.250 --> 00:24:58.020 PROFESSOR: Yeah. 00:24:58.020 --> 00:24:59.940 I see a question over here. 00:24:59.940 --> 00:25:02.672 AUDIENCE: Can you go over the root n, 2 minus l part? 00:25:02.672 --> 00:25:04.380 I understand the 2 to the l is the number 00:25:04.380 --> 00:25:06.340 of nodes second part. 00:25:06.340 --> 00:25:07.170 Does that work? 00:25:07.170 --> 00:25:08.790 PROFESSOR: Oh. 00:25:08.790 --> 00:25:10.930 OK, this is the number of nodes. 00:25:10.930 --> 00:25:12.780 This is the quantity inside of the circle. 00:25:12.780 --> 00:25:14.520 So notice that it's square root of n 00:25:14.520 --> 00:25:17.130 and then square root of n/2 and then square root of n/4 00:25:17.130 --> 00:25:19.500 and then square root of n/8, right? 00:25:19.500 --> 00:25:25.140 So 2 to the minus l here is like 1/2 and then 1/4 and then 1/8. 00:25:25.140 --> 00:25:26.490 Yeah. 00:25:26.490 --> 00:25:26.990 Fabulous. 00:25:26.990 --> 00:25:27.823 Any other questions? 00:25:27.823 --> 00:25:28.350 Yes. 00:25:28.350 --> 00:25:30.470 AUDIENCE: So if we have log base 2n levels 00:25:30.470 --> 00:25:33.410 and we started indexing at 0, doesn't it 00:25:33.410 --> 00:25:36.710 go from l equals 0 to l equals 2n minus 1? 00:25:36.710 --> 00:25:37.820 PROFESSOR: Oh, boy. 00:25:37.820 --> 00:25:38.970 I'm bad at this stuff. 00:25:41.580 --> 00:25:42.210 No, right? 00:25:42.210 --> 00:25:45.730 I think this is correct. 00:25:45.730 --> 00:25:46.230 Let's see. 00:25:46.230 --> 00:25:51.660 So let's say that I have n equals 2, right? 00:25:51.660 --> 00:25:52.830 Then what's going to happen? 00:25:55.407 --> 00:25:56.490 Yeah, no, this is correct. 00:25:56.490 --> 00:26:00.130 So as a sanity check, think about n equals 2. 00:26:00.130 --> 00:26:01.570 So how many levels should it be? 00:26:01.570 --> 00:26:04.100 It should go from l equals 0 and then l equals 1. 00:26:04.100 --> 00:26:06.730 Then I'm done because I have T. 00:26:06.730 --> 00:26:09.160 And then I think-- 00:26:09.160 --> 00:26:12.430 yeah, so the log base 2 of 2 is 1. 00:26:12.430 --> 00:26:15.670 So yeah, this formula checks out. 00:26:15.670 --> 00:26:18.500 OK, any other questions? 00:26:18.500 --> 00:26:19.750 These are all great questions. 00:26:19.750 --> 00:26:22.210 You guys are keeping me honest. 00:26:22.210 --> 00:26:23.650 OK. 00:26:23.650 --> 00:26:26.810 So we have this giant ugly expression here. 00:26:26.810 --> 00:26:29.110 And so our final job here is just to simplify it. 00:26:29.110 --> 00:26:29.933 That's it. 00:26:29.933 --> 00:26:32.350 And then what we're going to find is that this is secretly 00:26:32.350 --> 00:26:33.770 just theta of n. 00:26:33.770 --> 00:26:37.760 A little surprising, given that this is kind of ugly. 00:26:37.760 --> 00:26:38.627 OK. 00:26:38.627 --> 00:26:40.460 And by the way, this was just the constants. 00:26:40.460 --> 00:26:42.043 So this term on the denominator is not 00:26:42.043 --> 00:26:45.590 going to end up mattering for our asymptotic calculation. 00:26:45.590 --> 00:26:46.090 OK. 00:26:49.400 --> 00:26:49.900 Cool. 00:26:49.900 --> 00:26:51.260 So let's see here. 00:26:51.260 --> 00:26:55.690 So remember that the square root of 2 is 2 to the 1/2. 00:26:55.690 --> 00:26:56.190 Yeah. 00:26:56.190 --> 00:26:58.482 So I can do a little bit of reshuffling on our formula. 00:26:58.482 --> 00:27:01.600 And what I'm going to find is that this is exactly 00:27:01.600 --> 00:27:05.830 the square root of n times 1 over the square root 00:27:05.830 --> 00:27:07.270 of 2 minus 1. 00:27:07.270 --> 00:27:08.870 That's just this constant here. 00:27:08.870 --> 00:27:12.380 Now I just have to cope with this funny term inside of here. 00:27:12.380 --> 00:27:13.630 I'm going to do the following. 00:27:13.630 --> 00:27:17.110 I'm going to say, well, square root of 2 is the same as 2 00:27:17.110 --> 00:27:18.400 to the 1/2. 00:27:18.400 --> 00:27:22.340 So I'm going to write this by moving that 1/2 upstairs here, 00:27:22.340 --> 00:27:22.840 right? 00:27:22.840 --> 00:27:25.690 So what am I going to get at the end of the day is that this 00:27:25.690 --> 00:27:36.400 is the same as 2 to the log base 2 of n plus 1 and all of that-- 00:27:36.400 --> 00:27:39.195 oops, plus 1/2, right, because there's the 1/2 exponent that 00:27:39.195 --> 00:27:41.140 got multiplied by the 1-- 00:27:41.140 --> 00:27:42.850 minus 1. 00:27:42.850 --> 00:27:45.290 So this is just an identical expression to this one. 00:27:45.290 --> 00:27:46.915 All I've done is move the 1/2 upstairs. 00:27:49.620 --> 00:27:52.890 And finally, now there's some order in the universe, 00:27:52.890 --> 00:27:55.206 because what is 2 to the log base 2 of n? 00:27:55.206 --> 00:27:57.030 AUDIENCE: n. 00:27:57.030 --> 00:27:58.950 PROFESSOR: n, exactly. 00:27:58.950 --> 00:28:00.900 That's a very complicated way of writing n. 00:28:00.900 --> 00:28:04.980 So this is equal to the square root of n times 00:28:04.980 --> 00:28:07.380 1 over the square root of 2 minus 1, 00:28:07.380 --> 00:28:09.000 which is just a constant. 00:28:09.000 --> 00:28:09.780 I see your hand. 00:28:09.780 --> 00:28:12.060 I'm going to write, and then I'll catch up with you. 00:28:12.060 --> 00:28:15.900 And now this whole quantity is n times the square root of 2, 00:28:15.900 --> 00:28:18.430 right? 00:28:18.430 --> 00:28:19.370 Oops, wait. 00:28:23.690 --> 00:28:24.660 AUDIENCE: [INAUDIBLE] 00:28:24.660 --> 00:28:26.130 PROFESSOR: Oh, I'm sorry. 00:28:26.130 --> 00:28:28.140 That should be-- right, because we 00:28:28.140 --> 00:28:32.080 moved the 1/2 into the exponent, and I didn't account for it. 00:28:32.080 --> 00:28:35.850 So this is the square root of n multiplied 00:28:35.850 --> 00:28:37.350 by the square root of 2. 00:28:37.350 --> 00:28:38.822 AUDIENCE: [INAUDIBLE] 00:28:38.822 --> 00:28:39.780 PROFESSOR: What's that? 00:28:39.780 --> 00:28:44.330 AUDIENCE: It'd be n/2 to the square root of n. 00:28:44.330 --> 00:28:46.197 PROFESSOR: n/2 to the square root of n. 00:28:46.197 --> 00:28:47.280 No, I think this is right. 00:28:47.280 --> 00:28:48.113 AUDIENCE: Oh, sorry. 00:28:48.113 --> 00:28:50.340 PROFESSOR: Yeah. 00:28:50.340 --> 00:28:51.650 You guys are making me nervous. 00:28:51.650 --> 00:28:54.520 OK, so finally, now we can start to see the big O come out, 00:28:54.520 --> 00:28:55.020 right? 00:28:55.020 --> 00:28:57.830 Because now we have the square root of n times itself 00:28:57.830 --> 00:29:00.740 plus some other stuff, which is obviously going to grow slower 00:29:00.740 --> 00:29:02.880 than the remaining terms here. 00:29:02.880 --> 00:29:05.968 So we're good. 00:29:05.968 --> 00:29:07.760 If you're wondering, the official solutions 00:29:07.760 --> 00:29:08.718 are actually incorrect. 00:29:08.718 --> 00:29:10.610 They, too, get this a little bit wrong. 00:29:10.610 --> 00:29:12.485 I'm going to fix that tonight, and then we'll 00:29:12.485 --> 00:29:14.878 post them for you guys. 00:29:14.878 --> 00:29:16.670 And this unfortunately is the kind of thing 00:29:16.670 --> 00:29:18.560 you do a lot of, where you're going to take your tree. 00:29:18.560 --> 00:29:20.720 You write down this giant tree formula here. 00:29:20.720 --> 00:29:22.815 You write down a geometric series of some sort 00:29:22.815 --> 00:29:24.440 and then start bounding terms until you 00:29:24.440 --> 00:29:25.610 get to the expression you want. 00:29:25.610 --> 00:29:27.402 So now you can see why master theorem maybe 00:29:27.402 --> 00:29:29.970 is kind of valuable because it saves you a lot of headache. 00:29:29.970 --> 00:29:30.470 Yes. 00:29:30.470 --> 00:29:32.178 AUDIENCE: When we write up our piece sets 00:29:32.178 --> 00:29:34.260 and we have to write up trees like this, 00:29:34.260 --> 00:29:36.020 can we hand draw them and take pictures? 00:29:36.020 --> 00:29:37.562 PROFESSOR: Yeah, I don't see why not. 00:29:37.562 --> 00:29:40.560 They don't have to draw it in Dixie or something. 00:29:40.560 --> 00:29:42.450 AUDIENCE: You can take pictures of diagrams 00:29:42.450 --> 00:29:43.940 as long as the math doesn't [INAUDIBLE].. 00:29:43.940 --> 00:29:44.720 PROFESSOR: Yeah. 00:29:44.720 --> 00:29:46.790 I think if you're just writing out your math, 00:29:46.790 --> 00:29:49.640 taking a photo of it, and then, like, backslash include 00:29:49.640 --> 00:29:51.020 graphics of your photo-- yeah. 00:29:51.020 --> 00:29:52.895 But just for the figure, I think that's fine. 00:29:52.895 --> 00:29:53.900 AUDIENCE: [INAUDIBLE] 00:29:53.900 --> 00:29:54.350 PROFESSOR: Yeah. 00:29:54.350 --> 00:29:55.520 In fact, when I write research papers, 00:29:55.520 --> 00:29:57.320 that's how I do it until the final draft. 00:29:57.320 --> 00:29:58.500 Yeah. 00:29:58.500 --> 00:29:59.120 OK. 00:29:59.120 --> 00:30:00.578 So essentially, what did we just do 00:30:00.578 --> 00:30:03.200 in this problem, which is part a of one problem, 00:30:03.200 --> 00:30:06.110 is essentially just showed two different ways of solving 00:30:06.110 --> 00:30:06.860 recurrence, right? 00:30:06.860 --> 00:30:09.350 One is the sledgehammer that's very efficient but not 00:30:09.350 --> 00:30:10.520 terribly illustrative. 00:30:10.520 --> 00:30:12.020 The other is really working out all 00:30:12.020 --> 00:30:14.420 of the work that happens at every level of our tree, 00:30:14.420 --> 00:30:16.820 doing a giant sum, and then just actually 00:30:16.820 --> 00:30:18.292 deriving the proper formula. 00:30:18.292 --> 00:30:20.000 And both of these are just different ways 00:30:20.000 --> 00:30:21.230 to skin the same cat. 00:30:21.230 --> 00:30:23.030 Is that a weird phrase? 00:30:23.030 --> 00:30:24.390 I don't know. 00:30:24.390 --> 00:30:24.890 OK. 00:30:24.890 --> 00:30:29.300 So right, so actually, part two, I believe, 00:30:29.300 --> 00:30:32.240 ends up being easier by a funny fluke 00:30:32.240 --> 00:30:34.220 even though it looks like it's harder. 00:30:34.220 --> 00:30:36.360 So let's do that next here. 00:30:36.360 --> 00:30:39.033 So I'm going to leave master theorem on the left-hand side. 00:30:39.033 --> 00:30:40.700 I really want to say "master's theorem," 00:30:40.700 --> 00:30:41.570 but I guess that's not right. 00:30:41.570 --> 00:30:44.240 I'm going to leave the master theorem on the left-hand side. 00:30:47.820 --> 00:30:49.590 Just do the problem over here. 00:30:53.148 --> 00:30:55.440 I'm going to do this one a little bit quicker because I 00:30:55.440 --> 00:30:57.990 prefer not to spend the whole session on one problem. 00:30:57.990 --> 00:30:58.800 OK. 00:30:58.800 --> 00:31:01.050 But I do think it's worth spending a few minutes going 00:31:01.050 --> 00:31:03.008 over this theorem and how to apply it carefully 00:31:03.008 --> 00:31:07.480 because, well, otherwise, you won't. 00:31:07.480 --> 00:31:07.980 OK. 00:31:16.300 --> 00:31:18.070 All right, folks. 00:31:18.070 --> 00:31:19.301 It's a workout up here. 00:31:24.220 --> 00:31:27.640 So in part b, we have a version where 00:31:27.640 --> 00:31:31.480 the branching factor and the amount that the work reduces 00:31:31.480 --> 00:31:32.360 are not the same. 00:31:32.360 --> 00:31:35.080 So now we have to be a little bit more careful in the way 00:31:35.080 --> 00:31:36.580 that we apply master theorem, right? 00:31:36.580 --> 00:31:43.510 So now in part b, we have that T of n 00:31:43.510 --> 00:31:53.440 here is equal to 8T of n/4 plus-- 00:31:53.440 --> 00:31:56.290 and now in the problem, they write big O of n times 00:31:56.290 --> 00:31:57.918 the square root of n. 00:31:57.918 --> 00:31:58.960 We're all grown-ups here. 00:31:58.960 --> 00:32:03.557 That's the same as big O of n to the 3/2. 00:32:05.960 --> 00:32:06.460 Yeah. 00:32:06.460 --> 00:32:08.180 That's just arithmetic. 00:32:08.180 --> 00:32:08.680 OK. 00:32:08.680 --> 00:32:12.560 So first, let's apply the master theorem. 00:32:12.560 --> 00:32:13.060 Let's see. 00:32:13.060 --> 00:32:17.710 So what is n to the log base b of a, in this case? 00:32:17.710 --> 00:32:21.725 Well, what is b? 00:32:21.725 --> 00:32:22.225 4. 00:32:25.180 --> 00:32:26.270 What is a? 00:32:26.270 --> 00:32:26.770 8. 00:32:29.560 --> 00:32:31.390 Anybody know log base 4 of 8? 00:32:31.390 --> 00:32:32.593 AUDIENCE: 3/2. 00:32:32.593 --> 00:32:33.760 PROFESSOR: Wow, you're good. 00:32:33.760 --> 00:32:37.081 Yeah, so this is equal to n to the 3/2. 00:32:37.081 --> 00:32:39.860 And notice that these two things agree now. 00:32:39.860 --> 00:32:40.360 Yeah. 00:32:40.360 --> 00:32:45.880 So which of the three cases of the master theorem are we in? 00:32:45.880 --> 00:32:48.610 Are we in the first one, where one of these kind of dwarfs the 00:32:48.610 --> 00:32:50.560 other? 00:32:50.560 --> 00:32:51.070 No. 00:32:51.070 --> 00:32:53.290 We're in case two, right, where they both kind of 00:32:53.290 --> 00:32:54.880 behave similarly. 00:32:54.880 --> 00:32:55.390 Yeah. 00:32:55.390 --> 00:32:58.120 So this is case two. 00:33:00.790 --> 00:33:04.600 And what is the value of k for case two that's relevant here? 00:33:04.600 --> 00:33:06.865 Do I need a log factor? 00:33:06.865 --> 00:33:08.740 No, these two terms are just the same, right? 00:33:08.740 --> 00:33:10.120 They're both n to the 3/2. 00:33:10.120 --> 00:33:14.530 So this is case two with k equals 0. 00:33:14.530 --> 00:33:23.810 And in particular, immediately, we get that T of n is-- 00:33:23.810 --> 00:33:24.310 let's see. 00:33:24.310 --> 00:33:29.100 We're in case two, so we have theta of n to the log base 00:33:29.100 --> 00:33:34.500 b of a times log to the k plus 1 of n. 00:33:34.500 --> 00:33:40.570 So you have to account for that term, so it's like that. 00:33:40.570 --> 00:33:43.480 That make sense, how we applied master theorem in that case? 00:33:43.480 --> 00:33:44.160 Yes. 00:33:44.160 --> 00:33:46.660 AUDIENCE: So how can we apply case two if it's big O and not 00:33:46.660 --> 00:33:47.740 big theta? 00:33:47.740 --> 00:33:52.060 PROFESSOR: How can we apply case two if f of n 00:33:52.060 --> 00:33:55.163 is big O and not big theta? 00:33:55.163 --> 00:33:56.830 Well, that's coming from a grad student. 00:33:56.830 --> 00:33:59.470 So why don't I ping it back to the students in the class 00:33:59.470 --> 00:34:00.977 rather than answering myself? 00:34:00.977 --> 00:34:02.560 AUDIENCE: Can you repeat the question? 00:34:02.560 --> 00:34:03.227 PROFESSOR: Sure. 00:34:03.227 --> 00:34:05.020 So let's say that we just have a big O. We 00:34:05.020 --> 00:34:06.369 don't have a big theta here. 00:34:10.875 --> 00:34:13.000 Or at least a simple answer, which maybe is the one 00:34:13.000 --> 00:34:15.058 that you're looking for. 00:34:15.058 --> 00:34:15.850 So what do we know? 00:34:15.850 --> 00:34:17.225 In that case, we know that f of n 00:34:17.225 --> 00:34:19.097 might grow more slowly than this function. 00:34:19.097 --> 00:34:21.139 But it doesn't grow more quickly than this thing. 00:34:21.139 --> 00:34:21.653 Yeah. 00:34:21.653 --> 00:34:24.070 AUDIENCE: Would T of n just be big O instead of big theta? 00:34:24.070 --> 00:34:24.370 PROFESSOR: That's exactly right. 00:34:24.370 --> 00:34:26.380 So you can just replace big theta with big O, 00:34:26.380 --> 00:34:27.463 and that's perfectly fine. 00:34:27.463 --> 00:34:29.797 So I don't know if that's the answer you're looking for. 00:34:29.797 --> 00:34:32.000 But at least you get a loose bound this way. 00:34:32.000 --> 00:34:32.500 Yeah. 00:34:32.500 --> 00:34:34.309 AUDIENCE: You should probably write it big O in the answer 00:34:34.309 --> 00:34:34.809 then. 00:34:34.809 --> 00:34:36.370 PROFESSOR: Oops. 00:34:36.370 --> 00:34:37.520 Oh, I'm sorry. 00:34:37.520 --> 00:34:38.150 I see. 00:34:38.150 --> 00:34:39.929 We're being pedantic. 00:34:39.929 --> 00:34:40.900 OK. 00:34:40.900 --> 00:34:42.969 You can just tell me if I made a mistake. 00:34:42.969 --> 00:34:43.594 It happens. 00:34:43.594 --> 00:34:44.969 I think it's correct in my notes. 00:34:44.969 --> 00:34:46.150 It is. 00:34:46.150 --> 00:34:46.750 OK. 00:34:46.750 --> 00:34:49.120 So the reason our colleague brings it up in the back 00:34:49.120 --> 00:34:50.703 is that I made a slight mistake, which 00:34:50.703 --> 00:34:55.000 is that this is a big O, not a big theta, right? 00:34:55.000 --> 00:34:57.400 In which case, the only thing I can draw down here 00:34:57.400 --> 00:34:58.990 is that this is also a big O. So I 00:34:58.990 --> 00:35:02.390 made exactly the mistake I told you guys to avoid before. 00:35:02.390 --> 00:35:02.890 Yes. 00:35:02.890 --> 00:35:08.237 AUDIENCE: So then over there, would those also be big O? 00:35:08.237 --> 00:35:08.820 PROFESSOR: No. 00:35:08.820 --> 00:35:10.660 Yeah, now I've managed to confuse you. 00:35:10.660 --> 00:35:11.947 This is correct. 00:35:11.947 --> 00:35:13.780 This entire statement of the master theorem, 00:35:13.780 --> 00:35:16.513 assuming I copied it correctly, is right. 00:35:16.513 --> 00:35:17.930 But there's a bit of a difference. 00:35:17.930 --> 00:35:20.330 Remember, we wanted to apply case two. 00:35:20.330 --> 00:35:21.560 And look at case two. 00:35:21.560 --> 00:35:23.767 So there's a theta here. 00:35:23.767 --> 00:35:25.600 But in the problem, in the homework problem, 00:35:25.600 --> 00:35:26.920 they were sneaky. 00:35:26.920 --> 00:35:29.943 And they only said that f of T is big O of something. 00:35:29.943 --> 00:35:31.360 So remember, what's the difference 00:35:31.360 --> 00:35:34.300 between big O and big theta? 00:35:34.300 --> 00:35:37.030 Well, intuitively, big theta says that my function really 00:35:37.030 --> 00:35:38.420 does look like this guy. 00:35:38.420 --> 00:35:42.280 Somehow, it's bounded above and below as I go far enough out. 00:35:42.280 --> 00:35:45.430 In big O, there's just a bound above, right? 00:35:45.430 --> 00:35:47.350 So this is somehow looser. 00:35:47.350 --> 00:35:49.910 And so the way to apply master theorem in this case is say, 00:35:49.910 --> 00:35:51.535 well, at least f of n is upper bounded. 00:35:51.535 --> 00:35:52.600 It looks like this. 00:35:52.600 --> 00:35:54.940 So the best that I can do is to replace this guy also 00:35:54.940 --> 00:35:55.870 with an upper bound. 00:35:55.870 --> 00:35:56.560 Yeah. 00:35:56.560 --> 00:35:57.280 That's a great question. 00:35:57.280 --> 00:35:58.655 AUDIENCE: So the reason you chose 00:35:58.655 --> 00:36:01.330 two is that [INAUDIBLE] construction of the math 00:36:01.330 --> 00:36:02.418 is similar to [INAUDIBLE]? 00:36:02.418 --> 00:36:03.210 PROFESSOR: Exactly. 00:36:03.210 --> 00:36:05.370 You got to kind of stand back and squint at it a little bit, 00:36:05.370 --> 00:36:06.660 make sure that it fits. 00:36:06.660 --> 00:36:09.450 But certainly, it's the case that the other two parts 00:36:09.450 --> 00:36:11.246 of this theorem don't apply. 00:36:11.246 --> 00:36:13.080 So we might as well try to squeeze this part 00:36:13.080 --> 00:36:14.430 into the right form. 00:36:14.430 --> 00:36:14.930 All right. 00:36:19.160 --> 00:36:20.702 So let's do the tree version of this. 00:36:20.702 --> 00:36:22.077 And then I think what we're going 00:36:22.077 --> 00:36:24.230 to do is skip part three for now of this problem 00:36:24.230 --> 00:36:26.372 because it's mostly a fun kind of vectorial problem 00:36:26.372 --> 00:36:27.830 rather than something that actually 00:36:27.830 --> 00:36:29.540 is going to help your understanding of algorithms. 00:36:29.540 --> 00:36:31.998 And then if we have time at the end, we'll come back to it. 00:36:31.998 --> 00:36:33.090 OK. 00:36:33.090 --> 00:36:33.590 All right. 00:36:33.590 --> 00:36:36.020 So that was the easy way to solve the problem. 00:36:36.020 --> 00:36:38.420 Now let's do the painful one where we draw the tree. 00:36:38.420 --> 00:36:39.560 I shouldn't say that because I think 00:36:39.560 --> 00:36:41.477 it's something we want to encourage-- the more 00:36:41.477 --> 00:36:44.090 enlightening version, yeah? 00:36:44.090 --> 00:36:51.170 So now at the top of my tree, I do n to the 3/2 work. 00:36:51.170 --> 00:36:54.690 And now how many children does this guy have? 00:36:54.690 --> 00:36:55.610 AUDIENCE: [INAUDIBLE] 00:36:55.610 --> 00:36:56.402 PROFESSOR: Careful. 00:36:56.402 --> 00:37:00.740 You have 1 in 2 chances, either 8 or 4. 00:37:00.740 --> 00:37:01.890 8, right? 00:37:01.890 --> 00:37:04.340 So the outer coefficient is the number of children 00:37:04.340 --> 00:37:06.260 because it's like the number of function calls you make, right? 00:37:06.260 --> 00:37:08.240 That's the way to think about it, right? 00:37:08.240 --> 00:37:10.640 So there's eight of these guys. 00:37:15.020 --> 00:37:15.650 OK. 00:37:15.650 --> 00:37:16.935 And each one of these guys-- 00:37:16.935 --> 00:37:19.100 I'm only going to do one level of these. 00:37:19.100 --> 00:37:21.488 Wait, that should not be connected-- 00:37:21.488 --> 00:37:22.280 does how much work? 00:37:22.280 --> 00:37:25.400 Well, remember that it's divided by 4 here, right? 00:37:25.400 --> 00:37:27.275 So it looks like n/4 to the 3/2. 00:37:34.730 --> 00:37:39.650 That make sense, how we got this picture from our recurrence? 00:37:39.650 --> 00:37:41.360 Let me switch to another piece of chalk. 00:37:41.360 --> 00:37:42.900 OK. 00:37:42.900 --> 00:37:43.400 Right. 00:37:43.400 --> 00:37:49.790 So in other words, if I look at level l of my tree, 00:37:49.790 --> 00:37:51.860 assuming we're going to index from 0 at the top, 00:37:51.860 --> 00:37:55.890 how many nodes are in level l? 00:37:55.890 --> 00:37:58.860 Remember, it branches by 8 each time. 00:37:58.860 --> 00:38:05.540 So it has 8 to the l nodes. 00:38:05.540 --> 00:38:06.860 Yeah. 00:38:06.860 --> 00:38:10.550 And how much work does each do? 00:38:18.410 --> 00:38:22.880 Well, in each level, I divide by 4, right? 00:38:22.880 --> 00:38:27.814 So this is going to look like n times 1/4 to the l to the 3/2. 00:38:32.640 --> 00:38:33.140 Right. 00:38:33.140 --> 00:38:33.640 Where am I? 00:38:36.770 --> 00:38:38.060 There we are. 00:38:38.060 --> 00:38:41.600 So that's n times 4 to the minus l. 00:38:41.600 --> 00:38:47.177 That's fancy notation for 1/4 to the l to the 3/2. 00:38:47.177 --> 00:38:49.010 Notice that this is exactly the same pattern 00:38:49.010 --> 00:38:50.160 as the previous problem. 00:38:50.160 --> 00:38:53.150 I'm just plugging in some different constants. 00:38:53.150 --> 00:38:53.690 Yes. 00:38:53.690 --> 00:38:55.607 AUDIENCE: I have a question about the diagram. 00:38:55.607 --> 00:38:58.605 If you have a really large a and you're 00:38:58.605 --> 00:39:00.230 trying to do the recursion tree method, 00:39:00.230 --> 00:39:03.500 is this type of work OK with where you have-- 00:39:03.500 --> 00:39:05.750 PROFESSOR: If I were grading your papers, it would be. 00:39:05.750 --> 00:39:08.340 But you should probably check with your TAs on Piazza 00:39:08.340 --> 00:39:08.840 on that one. 00:39:08.840 --> 00:39:10.910 AUDIENCE: All right. 00:39:10.910 --> 00:39:13.818 PROFESSOR: But I've learned my license. 00:39:13.818 --> 00:39:15.110 I'm not sure you guys have yet. 00:39:15.110 --> 00:39:17.810 And also, I am not sure I have, either. 00:39:17.810 --> 00:39:18.710 All right, folks. 00:39:18.710 --> 00:39:23.150 So now we need to do our total amount of work here, right? 00:39:23.150 --> 00:39:26.750 So how many levels are there in our tree here? 00:39:26.750 --> 00:39:29.990 Well, the amount of data divides by 4 each time. 00:39:29.990 --> 00:39:34.300 And when the amount of data is 1, then I'm done, right? 00:39:34.300 --> 00:39:42.573 So the total number of levels is equal to the log base 4 of n. 00:39:42.573 --> 00:39:44.990 The basic thing to get right in this problem is just where 00:39:44.990 --> 00:39:47.350 there should be 4's and where there should be 8's. 00:39:47.350 --> 00:39:48.140 OK. 00:39:48.140 --> 00:39:50.760 So now what am I going to do? 00:39:50.760 --> 00:39:53.210 Well, again, as our colleague in the back points out, 00:39:53.210 --> 00:39:56.900 we only have big O. So we can only upper bound our work. 00:39:56.900 --> 00:40:03.295 But our work in the tree is less than or equal to-- 00:40:03.295 --> 00:40:04.670 really, there's a proportion here 00:40:04.670 --> 00:40:06.275 because this is just up to a multiple. 00:40:06.275 --> 00:40:07.650 I guess that's why, in the notes, 00:40:07.650 --> 00:40:11.750 they put a little c in front, which I can do. 00:40:11.750 --> 00:40:14.240 Let's do c like that. 00:40:14.240 --> 00:40:17.690 Well, we're going to have the sum from l equals 0 00:40:17.690 --> 00:40:25.190 to log base 4 of n of exactly the quantity, this guy. 00:40:25.190 --> 00:40:27.830 Multiply it by the number of nodes in the row. 00:40:27.830 --> 00:40:28.700 Yes. 00:40:28.700 --> 00:40:30.890 AUDIENCE: Is the reason why you made this work 00:40:30.890 --> 00:40:32.780 an inequality because of the-- 00:40:32.780 --> 00:40:35.660 PROFESSOR: Because of the big O. That's exactly right. 00:40:35.660 --> 00:40:36.960 Yeah. 00:40:36.960 --> 00:40:38.190 OK. 00:40:38.190 --> 00:40:38.690 All right. 00:40:38.690 --> 00:40:40.520 So what do I get here? 00:40:40.520 --> 00:40:43.550 Well, let's write it out the big ugly way first. 00:40:43.550 --> 00:40:46.670 So this is n times 4 to the minus l. 00:40:46.670 --> 00:40:48.990 This is to the 3/2. 00:40:48.990 --> 00:40:51.350 And then this is multiplied by 8 to the l. 00:40:51.350 --> 00:40:54.270 But whoever designed this problem was really sneaky, 00:40:54.270 --> 00:40:54.890 right? 00:40:54.890 --> 00:40:57.920 So what's 4 to the minus l to the 3/2? 00:41:00.660 --> 00:41:01.445 Any guesses? 00:41:01.445 --> 00:41:02.570 AUDIENCE: 8 to the minus l. 00:41:02.570 --> 00:41:04.715 PROFESSOR: Yeah, this is exactly 8 to the minus l 00:41:04.715 --> 00:41:07.470 if we work through all of your exponent arithmetic. 00:41:07.470 --> 00:41:07.970 Yeah. 00:41:07.970 --> 00:41:08.845 Why does that matter? 00:41:08.845 --> 00:41:11.120 Well, it exactly cancels this 8 to the l term here. 00:41:11.120 --> 00:41:13.490 This is actually exactly what case two in master theorem 00:41:13.490 --> 00:41:15.740 is trying to capture. 00:41:15.740 --> 00:41:24.470 So this whole thing is equal to, really, just n to the 3/2 00:41:24.470 --> 00:41:30.110 from l equals 0 to the log base 4 of n. 00:41:30.110 --> 00:41:31.340 All these constants go away. 00:41:31.340 --> 00:41:32.790 Isn't that beautiful? 00:41:32.790 --> 00:41:33.934 Yes. 00:41:33.934 --> 00:41:34.760 AUDIENCE: Quick question about the tree again. 00:41:34.760 --> 00:41:35.270 PROFESSOR: No problem. 00:41:35.270 --> 00:41:36.100 Uh-huh. 00:41:36.100 --> 00:41:40.575 AUDIENCE: So do people ever use these for optimization 00:41:40.575 --> 00:41:43.630 in terms of trying to write a better algorithm, 00:41:43.630 --> 00:41:48.620 trying to minimize the work per level? 00:41:48.620 --> 00:41:51.623 Yeah, I'm just curious if that's an application that's-- 00:41:51.623 --> 00:41:52.790 PROFESSOR: Sure, I guess so. 00:41:52.790 --> 00:41:54.500 I mean, this is somehow a visualization 00:41:54.500 --> 00:41:57.170 of how your algorithm is making function costs, right? 00:41:57.170 --> 00:42:00.050 So every node here kind of looks like a call 00:42:00.050 --> 00:42:02.283 to your piece of code. 00:42:02.283 --> 00:42:04.700 So every time you make a new one, it's like making a call. 00:42:04.700 --> 00:42:07.670 And then the divide here is kind of like how much data 00:42:07.670 --> 00:42:09.193 goes into that function call. 00:42:09.193 --> 00:42:11.360 So I think this is a useful way to kind of visualize 00:42:11.360 --> 00:42:13.010 what's going on inside of your code. 00:42:13.010 --> 00:42:15.010 And then if you're trying to optimize your code, 00:42:15.010 --> 00:42:17.600 you're trying to basically reduce the number of nodes 00:42:17.600 --> 00:42:20.220 and/or the amount of work that each node does, right? 00:42:20.220 --> 00:42:20.720 Yeah. 00:42:20.720 --> 00:42:23.060 So this is just a nice way to kind of visualize 00:42:23.060 --> 00:42:25.310 what's going on in a recursive algorithm. 00:42:25.310 --> 00:42:26.060 Yeah. 00:42:26.060 --> 00:42:27.440 Great question. 00:42:27.440 --> 00:42:29.030 All right. 00:42:29.030 --> 00:42:31.430 So here's a really nice thing. 00:42:31.430 --> 00:42:34.590 Does n to the 3/2 depend on l? 00:42:34.590 --> 00:42:35.370 No, right? 00:42:35.370 --> 00:42:40.848 So this whole thing is equal to c times n to the 3/2. 00:42:40.848 --> 00:42:42.390 And my colleague in the back is going 00:42:42.390 --> 00:42:45.380 to catch me if I don't make the off by 1 error here. 00:42:45.380 --> 00:42:52.810 So this is log base 4 of n plus 1 because my sum started at 0. 00:42:52.810 --> 00:42:54.180 Yeah. 00:42:54.180 --> 00:43:02.940 And notice that this is exactly big O of n to the 3/2 log 00:43:02.940 --> 00:43:06.430 n, which agrees with what we did in a much more easy case 00:43:06.430 --> 00:43:07.430 with the master theorem. 00:43:07.430 --> 00:43:08.305 Does that make sense? 00:43:08.305 --> 00:43:09.010 Yeah. 00:43:09.010 --> 00:43:11.170 AUDIENCE: So if I'm doing the exact bound, 00:43:11.170 --> 00:43:14.240 can we say that it's theta of n to the [INAUDIBLE]?? 00:43:14.240 --> 00:43:16.110 PROFESSOR: Oh, great question. 00:43:16.110 --> 00:43:18.960 This is where this notation is going to be misleading. 00:43:18.960 --> 00:43:24.080 This expression in a vacuum is theta of this value. 00:43:24.080 --> 00:43:25.610 But the reason that I wrote big O 00:43:25.610 --> 00:43:27.985 is because I only have an inequality all the way up here. 00:43:27.985 --> 00:43:30.153 So if I'm worried about bounding this work, 00:43:30.153 --> 00:43:31.820 there's no reason to write a theta there 00:43:31.820 --> 00:43:33.830 because it's just telling me some kind of intermediate piece 00:43:33.830 --> 00:43:34.670 of information. 00:43:34.670 --> 00:43:36.200 Yeah. 00:43:36.200 --> 00:43:37.460 OK, fabulous. 00:43:37.460 --> 00:43:40.144 So yes. 00:43:40.144 --> 00:43:41.894 AUDIENCE: Does the base of the log matter? 00:43:41.894 --> 00:43:43.686 PROFESSOR: Does the base of the log matter? 00:43:43.686 --> 00:43:44.520 Ah. 00:43:44.520 --> 00:43:47.640 This is a nice formula here. 00:43:47.640 --> 00:43:54.740 So remember that log base b of a is the same as log a 00:43:54.740 --> 00:43:58.680 over log b, if I got that right. 00:43:58.680 --> 00:43:59.180 Yeah. 00:43:59.180 --> 00:44:01.055 So the base of the log when it comes to big O 00:44:01.055 --> 00:44:03.380 doesn't matter because it's a constant factor. 00:44:03.380 --> 00:44:04.710 Yeah. 00:44:04.710 --> 00:44:05.210 OK. 00:44:05.210 --> 00:44:06.668 So that's part two of this problem. 00:44:06.668 --> 00:44:08.150 Part three is a little bit annoying 00:44:08.150 --> 00:44:10.160 because it has two branches in it. 00:44:10.160 --> 00:44:12.560 I wrote out a careful solution. 00:44:12.560 --> 00:44:14.090 It does not apply the master theorem 00:44:14.090 --> 00:44:15.965 because it's irrelevant, at least the version 00:44:15.965 --> 00:44:18.088 that we know in this class. 00:44:18.088 --> 00:44:19.880 So we're going to kind of skip that for now 00:44:19.880 --> 00:44:21.338 because I don't think it's terribly 00:44:21.338 --> 00:44:22.970 relevant to most of the algorithms 00:44:22.970 --> 00:44:24.320 that we'll see in 6.006. 00:44:24.320 --> 00:44:26.330 But we'll come back to it if we have time. 00:44:26.330 --> 00:44:26.990 Yes. 00:44:26.990 --> 00:44:29.198 AUDIENCE: Can we make that a lower bound and an upper 00:44:29.198 --> 00:44:33.080 bound using master theorem, knowing that T of n/4 00:44:33.080 --> 00:44:35.377 is smaller than T of n/3? 00:44:35.377 --> 00:44:37.460 PROFESSOR: I'm going to refer you to the solutions 00:44:37.460 --> 00:44:39.502 rather than talking about that problem right now. 00:44:39.502 --> 00:44:40.490 Yeah. 00:44:40.490 --> 00:44:42.640 OK, cool. 00:44:42.640 --> 00:44:43.140 All right. 00:44:43.140 --> 00:44:45.015 So that's how you apply master theorem, which 00:44:45.015 --> 00:44:46.287 is mildly painful. 00:44:46.287 --> 00:44:48.620 The good news is the rest of this problem set, actually, 00:44:48.620 --> 00:44:50.690 I consider to be much easier than the first one, which 00:44:50.690 --> 00:44:53.060 is why I think it's OK to spend a little bit extra time 00:44:53.060 --> 00:44:56.078 here because I think this stuff is confusing to get right. 00:44:56.078 --> 00:44:58.370 Mostly for me, and hopefully I've conveyed my confusion 00:44:58.370 --> 00:44:59.640 to the rest of you. 00:44:59.640 --> 00:45:00.140 OK. 00:45:00.140 --> 00:45:02.167 So let's erase this. 00:45:02.167 --> 00:45:04.250 And while we're doing that, why don't you guys all 00:45:04.250 --> 00:45:06.560 give problem number two a read? 00:45:06.560 --> 00:45:09.840 So one of the big skills that we need to cover in 6.83-- 00:45:09.840 --> 00:45:17.270 not 6.837, 6.006 and 6.837, but I'm not quite as evil about it 00:45:17.270 --> 00:45:19.670 in that class-- 00:45:19.670 --> 00:45:21.680 is as follows, which is you read a problem. 00:45:21.680 --> 00:45:24.680 And for some reason, your instructors 00:45:24.680 --> 00:45:26.300 have some sick sense of humor. 00:45:26.300 --> 00:45:30.110 And they encode it in this totally weird, goofy language 00:45:30.110 --> 00:45:31.610 which somehow, to a theoretician, 00:45:31.610 --> 00:45:34.290 makes your problem feel more practical. 00:45:34.290 --> 00:45:37.700 So in any event, when you read all of this paragraph, 00:45:37.700 --> 00:45:40.610 the very first skill that you have to do is to figure out, 00:45:40.610 --> 00:45:43.538 OK, like, this is cute notation, and it's about Infinity Stones. 00:45:43.538 --> 00:45:45.080 And if I watch Star Wars or whatever, 00:45:45.080 --> 00:45:46.700 I would know what that meant. 00:45:46.700 --> 00:45:50.840 But in any event, what really matters 00:45:50.840 --> 00:45:53.060 is understanding, OK, but algorithmically, 00:45:53.060 --> 00:45:54.450 what are they asking? 00:45:54.450 --> 00:45:54.972 Yeah. 00:45:54.972 --> 00:45:56.930 So I'm going to try and talk about this problem 00:45:56.930 --> 00:45:58.320 as I erase the board. 00:45:58.320 --> 00:46:02.810 So I believe, what, like, Mickey Mouse or whatever has a bunch, 00:46:02.810 --> 00:46:05.510 has some planet that he's looking for. 00:46:05.510 --> 00:46:09.530 I'm going to open the problem now. 00:46:09.530 --> 00:46:11.990 Right, she's a super villain on a quest. 00:46:11.990 --> 00:46:13.970 She's looking for a stone on a planet, 00:46:13.970 --> 00:46:16.640 and the planet has some index k. 00:46:16.640 --> 00:46:20.600 And unfortunately for us, the number of planets 00:46:20.600 --> 00:46:22.290 is quite large. 00:46:22.290 --> 00:46:27.050 It's infinity, in fact, because it's the Infinity Stone. 00:46:27.050 --> 00:46:28.420 Or, sorry, did I say that? 00:46:28.420 --> 00:46:30.300 I think it's the Infinity Phone or something. 00:46:30.300 --> 00:46:31.050 I forget. 00:46:31.050 --> 00:46:33.650 But in any event, the only thing that you 00:46:33.650 --> 00:46:37.820 can do when you land on a planet is ask an oracle, 00:46:37.820 --> 00:46:40.700 is the index of my planet bigger or smaller 00:46:40.700 --> 00:46:44.300 than the index of the planet that I'm standing on, right? 00:46:44.300 --> 00:46:46.910 And then the question is in log k time where 00:46:46.910 --> 00:46:48.648 k is the index of the planet. 00:46:48.648 --> 00:46:50.690 Notice that's already a little weird because it's 00:46:50.690 --> 00:46:54.020 not the size of your data, quite. 00:46:54.020 --> 00:46:56.390 Can you find the planet? 00:46:56.390 --> 00:46:58.820 Now, what does this kind of problem-- 00:46:58.820 --> 00:47:01.370 like, what is it screaming out for you to use? 00:47:01.370 --> 00:47:02.320 You see a log k. 00:47:02.320 --> 00:47:03.740 You're looking for something. 00:47:03.740 --> 00:47:04.613 AUDIENCE: Bisection. 00:47:04.613 --> 00:47:05.780 PROFESSOR: Bisection, right? 00:47:05.780 --> 00:47:06.080 AUDIENCE: Binary search. 00:47:06.080 --> 00:47:07.290 PROFESSOR: Or binary search. 00:47:07.290 --> 00:47:08.300 That's absolutely right. 00:47:08.300 --> 00:47:10.820 Those are both great answers. 00:47:10.820 --> 00:47:13.940 But there's a bit of a problem, which 00:47:13.940 --> 00:47:16.910 is that the number of planets is unbounded. 00:47:16.910 --> 00:47:18.410 We don't know how many planets there 00:47:18.410 --> 00:47:25.840 are in this little universe that problem 1.2 sets up. 00:47:25.840 --> 00:47:26.340 Yeah. 00:47:26.340 --> 00:47:28.640 So our intuition is that we want to use binary search. 00:47:28.640 --> 00:47:29.960 But in order to do binary search, 00:47:29.960 --> 00:47:32.293 I need to have a left- and a right-hand side and divide, 00:47:32.293 --> 00:47:34.710 and I have no right-hand side. 00:47:34.710 --> 00:47:36.010 So what can I do? 00:47:36.010 --> 00:47:38.830 How's a supervillain to solve this problem? 00:47:42.730 --> 00:47:45.660 Well, remember that each planet has an oracle on it telling me. 00:47:45.660 --> 00:47:47.410 Is there something to my left or my right? 00:47:47.410 --> 00:47:48.480 Yeah. 00:47:48.480 --> 00:47:50.130 So I could start at planet number one, 00:47:50.130 --> 00:47:52.740 and I could just start walking from planet one to planet two 00:47:52.740 --> 00:47:56.250 to planet three, planet four, and asking, am I there yet? 00:47:56.250 --> 00:47:57.030 Am I there yet? 00:47:57.030 --> 00:47:58.530 How much time is that going to take? 00:47:58.530 --> 00:48:00.450 AUDIENCE: Infinite, eventually. 00:48:00.450 --> 00:48:02.650 PROFESSOR: Actually, it won't take infinite time. 00:48:02.650 --> 00:48:03.915 I caught you. 00:48:03.915 --> 00:48:05.040 How much time will it take? 00:48:05.040 --> 00:48:07.500 When will I stop? 00:48:07.500 --> 00:48:09.990 When I hit planet k, right, because I know that planet k 00:48:09.990 --> 00:48:11.340 is out there somewhere. 00:48:11.340 --> 00:48:12.690 The truth is out there. 00:48:12.690 --> 00:48:15.360 And when I find it and I step on it, I stop. 00:48:15.360 --> 00:48:18.300 And I took exactly k steps, maybe k minus 1, 00:48:18.300 --> 00:48:19.770 depending on how you count. 00:48:19.770 --> 00:48:22.000 But we need a log k algorithm. 00:48:22.000 --> 00:48:22.500 Yeah. 00:48:22.500 --> 00:48:24.330 So what can I do? 00:48:24.330 --> 00:48:27.215 AUDIENCE: You start at some k, and then if it's to your right, 00:48:27.215 --> 00:48:29.705 you multiply by another k? 00:48:29.705 --> 00:48:31.080 PROFESSOR: Multiply by another k. 00:48:31.080 --> 00:48:32.220 AUDIENCE: [INAUDIBLE] essentially perceived 00:48:32.220 --> 00:48:33.210 by another k planets. 00:48:33.210 --> 00:48:36.900 So it's like you're at index k squared. 00:48:36.900 --> 00:48:38.091 PROFESSOR: Interesting. 00:48:40.737 --> 00:48:42.356 Hmm. 00:48:42.356 --> 00:48:44.370 AUDIENCE: Couldn't you go from i to i squared? 00:48:44.370 --> 00:48:45.288 AUDIENCE: Yeah. 00:48:45.288 --> 00:48:46.830 PROFESSOR: OK, so you got the right-- 00:48:46.830 --> 00:48:48.622 you're in the right church, wrong pew here. 00:48:48.622 --> 00:48:51.030 So the basic intuition here is that stepping 00:48:51.030 --> 00:48:53.713 one planet at a time doesn't step fast enough. 00:48:53.713 --> 00:48:55.380 If you work out the details on that one, 00:48:55.380 --> 00:48:57.005 you're going to find that you'll end up 00:48:57.005 --> 00:49:00.580 with a runtime that still goes a little bit too quickly in k 00:49:00.580 --> 00:49:01.680 here. 00:49:01.680 --> 00:49:04.340 In some sense, if you were to reverse engineer 00:49:04.340 --> 00:49:06.840 this problem, which isn't really a great way to problem set, 00:49:06.840 --> 00:49:08.760 you really do expect there to be powers of 2 00:49:08.760 --> 00:49:10.690 in every step of your algorithm. 00:49:10.690 --> 00:49:11.567 Yes. 00:49:11.567 --> 00:49:13.270 AUDIENCE: Could you do the binary search 00:49:13.270 --> 00:49:15.470 and start in the middle and then ask 00:49:15.470 --> 00:49:19.980 the oracle if it's higher or lower and then [INAUDIBLE]?? 00:49:19.980 --> 00:49:21.480 PROFESSOR: It's the right intuition. 00:49:21.480 --> 00:49:23.313 But I have a philosophical question for you, 00:49:23.313 --> 00:49:26.298 which is, what is the middle of an infinite set of planets? 00:49:26.298 --> 00:49:27.090 AUDIENCE: Infinite. 00:49:27.090 --> 00:49:28.110 PROFESSOR: Infinite, exactly. 00:49:28.110 --> 00:49:28.770 It's a problem. 00:49:28.770 --> 00:49:29.910 Yes. 00:49:29.910 --> 00:49:31.530 AUDIENCE: We could use-- 00:49:31.530 --> 00:49:34.110 and I don't know who it was back there who was 00:49:34.110 --> 00:49:36.480 suggesting doing i squareds-- 00:49:36.480 --> 00:49:39.467 except for doing i squared, replace it with 2 to the i. 00:49:39.467 --> 00:49:40.800 PROFESSOR: That's exactly right. 00:49:40.800 --> 00:49:42.130 That's a great intuition. 00:49:42.130 --> 00:49:44.350 So let's formalize that a little bit, 00:49:44.350 --> 00:49:46.320 which is like, I could do binary search if I 00:49:46.320 --> 00:49:47.237 had a right-hand side. 00:49:47.237 --> 00:49:48.960 I have a left-hand side because it's 1. 00:49:48.960 --> 00:49:49.800 So what I'm going to do, I'm going 00:49:49.800 --> 00:49:51.480 to keep trying right-hand sides. 00:49:51.480 --> 00:49:53.330 And the oracle is going to tell me, 00:49:53.330 --> 00:49:55.595 is this a valid right-hand side, right, 00:49:55.595 --> 00:49:56.970 because the oracle is telling me, 00:49:56.970 --> 00:49:59.120 is there a planet to my left? 00:49:59.120 --> 00:49:59.700 Yeah. 00:49:59.700 --> 00:50:01.440 So here's what I'm going to do. 00:50:01.440 --> 00:50:02.830 Again, I don't like the index i. 00:50:02.830 --> 00:50:04.455 In the official solution, there's an i. 00:50:04.455 --> 00:50:07.950 But I'm going to use the letter m because I can-- 00:50:07.950 --> 00:50:09.517 which is the following. 00:50:09.517 --> 00:50:10.350 I'm going to visit-- 00:50:13.010 --> 00:50:15.727 I hate teaching. 00:50:15.727 --> 00:50:16.560 No, that's not true. 00:50:16.560 --> 00:50:17.227 I like teaching. 00:50:17.227 --> 00:50:19.920 I just don't like chalk. 00:50:19.920 --> 00:50:25.890 I'm going to visit planet 2 to the m 00:50:25.890 --> 00:50:34.680 for each m starting with m equals 0 00:50:34.680 --> 00:50:36.405 and essentially until-- 00:50:39.300 --> 00:50:42.280 remember that k is the index of the planet I'm looking for. 00:50:42.280 --> 00:50:45.120 So eventually, I'm going to reach the point where k is less 00:50:45.120 --> 00:50:46.810 than or equal to 2 to the m. 00:50:46.810 --> 00:50:48.870 And I know that I can query my oracle, 00:50:48.870 --> 00:50:52.600 and they're going to tell me when that's the case. 00:50:52.600 --> 00:50:54.918 So how much time does this take? 00:50:54.918 --> 00:50:57.210 So I'm going to try planet 1 and then planet 2 and then 00:50:57.210 --> 00:51:02.970 planet 4 and then 8 and 16 all the way up until k. 00:51:02.970 --> 00:51:03.830 AUDIENCE: Log k. 00:51:03.830 --> 00:51:06.210 PROFESSOR: So it's going to take log k time. 00:51:06.210 --> 00:51:08.280 AUDIENCE: Can't you also have a stronger or lower 00:51:08.280 --> 00:51:11.380 bound now to the m minus 1? 00:51:11.380 --> 00:51:13.050 PROFESSOR: Yeah, that's sort of right. 00:51:13.050 --> 00:51:14.467 Right, so eventually, what's going 00:51:14.467 --> 00:51:17.820 to happen when you stop here is that 2 to the m minus 1 00:51:17.820 --> 00:51:19.650 is going to be less than or equal to k. 00:51:19.650 --> 00:51:23.490 This one equals the 2 to the m because this 00:51:23.490 --> 00:51:25.650 is the condition for stopping. 00:51:25.650 --> 00:51:27.760 This is a condition for not stopping. 00:51:27.760 --> 00:51:31.540 So when you stop k, this little sandwich is true. 00:51:31.540 --> 00:51:32.040 Yeah. 00:51:32.040 --> 00:51:34.740 So if I take the log, essentially, what I've shown 00:51:34.740 --> 00:51:36.690 is that I've taken, right-- because m 00:51:36.690 --> 00:51:39.265 is the number of steps in this part of my algorithm. 00:51:39.265 --> 00:51:40.890 So if I take the log, I'm going to find 00:51:40.890 --> 00:51:43.530 that I took log k steps. 00:51:43.530 --> 00:51:44.930 Yeah, nods, acknowledgment. 00:51:44.930 --> 00:51:45.430 Yeah. 00:51:45.430 --> 00:51:46.530 OK. 00:51:46.530 --> 00:51:51.210 And then, well, now I have an upper bound and a lower bound. 00:51:51.210 --> 00:51:52.520 So now what can I do? 00:51:52.520 --> 00:51:53.520 AUDIENCE: Binary search. 00:51:53.520 --> 00:51:53.850 PROFESSOR: Yeah. 00:51:53.850 --> 00:51:55.080 Now I can binary search. 00:51:55.080 --> 00:51:56.967 And that also takes log k time. 00:51:56.967 --> 00:52:01.260 AUDIENCE: Is it 2 times log k [INAUDIBLE]?? 00:52:01.260 --> 00:52:03.570 PROFESSOR: That's exactly right. 00:52:03.570 --> 00:52:05.820 Oops. 00:52:05.820 --> 00:52:08.490 So now I have step two is also binary search. 00:52:08.490 --> 00:52:11.790 And it's also a log k time, and our problem is solved. 00:52:11.790 --> 00:52:13.920 So problem number two is not so hard. 00:52:13.920 --> 00:52:14.842 Is everybody with me? 00:52:14.842 --> 00:52:16.050 Any questions about that one? 00:52:16.050 --> 00:52:17.700 That's a quick one. 00:52:17.700 --> 00:52:18.620 OK. 00:52:18.620 --> 00:52:21.120 Now, problem three really spoke to me as a computer graphics 00:52:21.120 --> 00:52:21.960 professor. 00:52:24.340 --> 00:52:24.840 Right. 00:52:24.840 --> 00:52:27.570 So now I am running Fadobe. 00:52:34.440 --> 00:52:37.970 I collaborate a lot with Fadobe Fresearch. 00:52:37.970 --> 00:52:38.470 Right. 00:52:38.470 --> 00:52:41.400 So Fadobe is trying to make a piece of software 00:52:41.400 --> 00:52:42.540 for image editing. 00:52:42.540 --> 00:52:45.120 And what does my piece of software do? 00:52:45.120 --> 00:52:49.973 Well, my image, or I guess my document, 00:52:49.973 --> 00:52:51.390 consists of a bunch of images that 00:52:51.390 --> 00:52:53.500 are overlaid with one another. 00:52:53.500 --> 00:52:54.000 OK. 00:52:54.000 --> 00:52:55.667 And essentially, what's happening inside 00:52:55.667 --> 00:52:58.290 of the software is I want to keep all the images in order 00:52:58.290 --> 00:53:01.050 from top to bottom, OK, because when I 00:53:01.050 --> 00:53:02.665 render my photo, what do I do? 00:53:02.665 --> 00:53:04.540 I render the bottom one and then the next one 00:53:04.540 --> 00:53:06.300 and so on and just layer them on top of each other, 00:53:06.300 --> 00:53:08.467 like if you ever played with PowerPoint or Photoshop 00:53:08.467 --> 00:53:11.470 or I guess whatever fake name they're giving here. 00:53:11.470 --> 00:53:16.760 That's a pretty common user interface to encounter. 00:53:16.760 --> 00:53:19.170 And so what they've asked you to do 00:53:19.170 --> 00:53:21.900 is to come up with essentially a data structure. 00:53:21.900 --> 00:53:23.730 And your data structure has to support 00:53:23.730 --> 00:53:27.150 a few different operations. 00:53:27.150 --> 00:53:29.760 In particular, you have to be able to make a document, 00:53:29.760 --> 00:53:34.140 import an image and then stick it on top, display, 00:53:34.140 --> 00:53:38.770 which returns all the IDs in the order that you've stored them. 00:53:38.770 --> 00:53:40.990 And then there's the real kicker, 00:53:40.990 --> 00:53:45.300 which is that you need to be able to take one of the layers 00:53:45.300 --> 00:53:47.850 and stick it underneath another one. 00:53:47.850 --> 00:53:49.770 But you have to do that in log n time. 00:53:49.770 --> 00:53:51.645 It's that "but" that makes this whole problem 00:53:51.645 --> 00:53:53.020 kind of a pain in the tuchus. 00:53:53.020 --> 00:53:54.490 Yeah. 00:53:54.490 --> 00:53:54.990 Right. 00:53:54.990 --> 00:53:57.010 So again, here our operations. 00:53:57.010 --> 00:54:01.957 We've got to make an empty doc. 00:54:04.880 --> 00:54:07.650 This is supposed to take order 1 time. 00:54:07.650 --> 00:54:15.900 We're going to import, which adds an x on top, 00:54:15.900 --> 00:54:18.750 and that this should take order n time. 00:54:18.750 --> 00:54:21.300 Notice that this is already a little suspicious. 00:54:21.300 --> 00:54:23.640 If I were trying to psychologically diagnose 00:54:23.640 --> 00:54:28.470 my professors, I would look at this order n with some-- 00:54:28.470 --> 00:54:31.860 like, with a raised eyebrow, because probably what would you 00:54:31.860 --> 00:54:34.440 have in mind if you're talking about stacks of photographs? 00:54:34.440 --> 00:54:37.230 It would be a stack or a queue or some data structure 00:54:37.230 --> 00:54:38.490 like that. 00:54:38.490 --> 00:54:40.740 But then insertion would be like order 1 time. 00:54:40.740 --> 00:54:42.990 So there's clearly something a little more complicated 00:54:42.990 --> 00:54:44.670 going on. 00:54:44.670 --> 00:54:46.830 OK. 00:54:46.830 --> 00:54:49.395 What's number three is display. 00:54:54.420 --> 00:54:56.490 This has to happen in order n time. 00:54:56.490 --> 00:54:57.990 This one kind of makes sense, right, 00:54:57.990 --> 00:54:59.550 because in order to display n things, 00:54:59.550 --> 00:55:02.500 you kind of expect to take at least n time. 00:55:02.500 --> 00:55:07.065 And then finally, we have to move below. 00:55:10.950 --> 00:55:13.080 And this has to take order log n time. 00:55:15.822 --> 00:55:18.030 And this is going to be the kicker, right, because we 00:55:18.030 --> 00:55:18.660 really-- 00:55:18.660 --> 00:55:21.180 this is somehow not totally obvious from the way 00:55:21.180 --> 00:55:24.268 that we set up our problem. 00:55:24.268 --> 00:55:26.310 So everybody understand the problem setup so far? 00:55:26.310 --> 00:55:28.450 We just keep adding objects with IDs, 00:55:28.450 --> 00:55:30.450 and we need to be able to insert them on the top 00:55:30.450 --> 00:55:32.010 and then kind of reorder them. 00:55:32.010 --> 00:55:34.710 And the problem has given you runtime for each 00:55:34.710 --> 00:55:37.910 of these different operations. 00:55:37.910 --> 00:55:39.220 OK. 00:55:39.220 --> 00:55:39.720 Right. 00:55:39.720 --> 00:55:41.030 So here's the thing. 00:55:41.030 --> 00:55:45.110 There's kind of a sequence aspect to our problem and kind 00:55:45.110 --> 00:55:48.030 of a set aspect to our problem. 00:55:48.030 --> 00:55:49.100 Does that make sense? 00:55:49.100 --> 00:55:50.600 The sequence aspect to our problem 00:55:50.600 --> 00:55:52.520 is that we're going to have to display stuff in order n 00:55:52.520 --> 00:55:53.210 time, right? 00:55:53.210 --> 00:55:55.668 We've got to iterate over our whole list, put stuff on top, 00:55:55.668 --> 00:55:56.720 and so on. 00:55:56.720 --> 00:55:58.250 And the set aspect is that we'd like 00:55:58.250 --> 00:56:02.290 to move stuff on top of each other in log n time. 00:56:02.290 --> 00:56:05.420 And the reason that I say that this is somehow set aspect 00:56:05.420 --> 00:56:09.260 is that I need to be able to put any x underneath any y, which 00:56:09.260 --> 00:56:11.960 means that I need to be able to quickly find what layer the x 00:56:11.960 --> 00:56:12.770 and the y is in. 00:56:15.423 --> 00:56:17.090 Any ideas how we can solve this problem? 00:56:17.090 --> 00:56:19.215 How about from some folks I haven't heard from yet? 00:56:22.640 --> 00:56:24.398 Yes. 00:56:24.398 --> 00:56:26.947 AUDIENCE: [INAUDIBLE] 00:56:26.947 --> 00:56:28.780 PROFESSOR: Yeah, that's a really great idea. 00:56:28.780 --> 00:56:30.655 So maybe I'm going to maintain-- so remember, 00:56:30.655 --> 00:56:31.700 we've talked about-- 00:56:31.700 --> 00:56:32.200 let's see. 00:56:32.200 --> 00:56:34.340 In my lecture two lectures ago, we 00:56:34.340 --> 00:56:39.840 kind of thought of a sorted list as a set data structure. 00:56:39.840 --> 00:56:41.817 So yeah, that's a fabulous idea. 00:56:41.817 --> 00:56:43.775 Why don't we, for the set part of our problem-- 00:56:48.330 --> 00:56:57.930 we're going to store in particular a sorted array of-- 00:56:57.930 --> 00:57:02.400 and for now, let's think of it like a sorted array of x's. 00:57:02.400 --> 00:57:05.460 And so this is going to be able to help us answer questions 00:57:05.460 --> 00:57:10.105 like, is this ID in my set of images or not? 00:57:10.105 --> 00:57:12.480 But there's also going to be a second part of my problem, 00:57:12.480 --> 00:57:14.325 which is in addition to that, I need 00:57:14.325 --> 00:57:16.200 to be able to keep a different ordering other 00:57:16.200 --> 00:57:18.870 than being just ordered by x, which is the ordering 00:57:18.870 --> 00:57:21.090 that they're being drawn on the screen, right? 00:57:21.090 --> 00:57:24.690 That's different than the ordering of their IDs. 00:57:24.690 --> 00:57:26.274 So for that-- 00:57:26.274 --> 00:57:27.316 AUDIENCE: Quick question. 00:57:27.316 --> 00:57:30.230 We're going to be building a sorted array with nothing in it 00:57:30.230 --> 00:57:32.700 and it's going to be O of n log n time. 00:57:32.700 --> 00:57:34.450 But that's undefined. 00:57:34.450 --> 00:57:37.526 So can we build a sorted array with nothing in it? 00:57:37.526 --> 00:57:38.280 PROFESSOR: Hmm. 00:57:38.280 --> 00:57:38.780 OK. 00:57:41.500 --> 00:57:43.990 No, I'm not sure I quite follow. 00:57:43.990 --> 00:57:46.920 AUDIENCE: It's building an empty sorted array 00:57:46.920 --> 00:57:49.840 O of n log n time, which is O of 0 log 0 time. 00:57:49.840 --> 00:57:51.000 And log 0 time [INAUDIBLE]. 00:57:51.000 --> 00:57:52.120 AUDIENCE: Empty sorted array? 00:57:52.120 --> 00:57:53.912 PROFESSOR: Yeah, building an empty anything 00:57:53.912 --> 00:57:54.730 that takes 1 time. 00:57:54.730 --> 00:57:55.150 AUDIENCE: Oh. 00:57:55.150 --> 00:57:55.817 PROFESSOR: Yeah. 00:57:55.817 --> 00:57:57.153 AUDIENCE: OK, cool. 00:57:57.153 --> 00:57:57.820 PROFESSOR: Cool. 00:57:57.820 --> 00:57:58.598 All right. 00:57:58.598 --> 00:58:00.640 So in addition to this, we have a sequence aspect 00:58:00.640 --> 00:58:01.265 of our problem. 00:58:05.790 --> 00:58:08.880 And for that, maybe we'll use a linked list, right? 00:58:08.880 --> 00:58:13.620 That's a pretty reasonable sequence data structure. 00:58:13.620 --> 00:58:16.030 And in fact, let's think for a second. 00:58:16.030 --> 00:58:18.453 Now, we're going to need this move below operation, right? 00:58:18.453 --> 00:58:20.370 So our linked list is going to store the order 00:58:20.370 --> 00:58:22.420 of the images in our document. 00:58:22.420 --> 00:58:24.420 In order to move something below something else, 00:58:24.420 --> 00:58:26.640 we're going to have to splice an image in between two 00:58:26.640 --> 00:58:28.790 other guys. 00:58:28.790 --> 00:58:31.210 So for convenience, maybe we have a doubly-linked list 00:58:31.210 --> 00:58:33.030 so that we can move backward and forward so 00:58:33.030 --> 00:58:34.380 that we can insert stuff. 00:58:34.380 --> 00:58:35.462 Yeah. 00:58:35.462 --> 00:58:36.920 So we'll have a doubly-linked list. 00:58:45.330 --> 00:58:47.500 OK. 00:58:47.500 --> 00:58:48.000 Right. 00:58:48.000 --> 00:58:50.400 So let's start solving our problem 00:58:50.400 --> 00:58:53.040 and then thinking about where things 00:58:53.040 --> 00:58:56.083 are going to go wrong with our setup here. 00:58:56.083 --> 00:58:57.750 So first of all, I don't think we really 00:58:57.750 --> 00:58:59.970 have to write anything for part one 00:58:59.970 --> 00:59:03.330 because making an empty anything is a pretty easy algorithm. 00:59:03.330 --> 00:59:06.430 On your homework, of course, you should write that and tell us 00:59:06.430 --> 00:59:08.640 that it takes order 1 time. 00:59:08.640 --> 00:59:11.310 Now, how about importing an object? 00:59:11.310 --> 00:59:17.890 Remember, that puts it on the top of our linked list. 00:59:17.890 --> 00:59:22.860 And we have to insert it into our sorted array, right? 00:59:22.860 --> 00:59:33.150 So when we do that, our insert algorithm is super simple, 00:59:33.150 --> 00:59:33.720 right? 00:59:33.720 --> 00:59:40.500 We're going to add x to the set, which we talked about in class 00:59:40.500 --> 00:59:41.670 for a sorted array. 00:59:41.670 --> 00:59:45.144 How long does this take for a sorted array to add something? 00:59:45.144 --> 00:59:46.325 AUDIENCE: Order n time. 00:59:46.325 --> 00:59:47.700 PROFESSOR: Exactly, order n time. 00:59:47.700 --> 00:59:49.710 But that's actually OK because that's 00:59:49.710 --> 00:59:52.980 our criterion for number two. 00:59:52.980 --> 00:59:53.490 Yeah. 00:59:53.490 --> 00:59:57.220 And in addition to that, we'll put 00:59:57.220 --> 01:00:04.140 x on the top of the linked list, right? 01:00:04.140 --> 01:00:05.484 That's order 1 time. 01:00:05.484 --> 01:00:06.870 That's easy. 01:00:06.870 --> 01:00:08.370 I'm adding space because we're going 01:00:08.370 --> 01:00:09.720 to see that we made a slight mistake 01:00:09.720 --> 01:00:12.345 and that we're going to need to modify our algorithm a tiny bit 01:00:12.345 --> 01:00:14.470 to solve this problem. 01:00:14.470 --> 01:00:18.745 OK, so how do we display? 01:00:18.745 --> 01:00:20.370 I think that's the simplest one, right? 01:00:20.370 --> 01:00:22.578 I'm just going to iterate over the entire linked list 01:00:22.578 --> 01:00:25.313 and just, one by one, output the order because, remember, 01:00:25.313 --> 01:00:26.730 the whole point of the linked list 01:00:26.730 --> 01:00:30.870 is to keep track of all of our documents in order. 01:00:30.870 --> 01:00:41.340 So display, all you have to do is loop over the linked list. 01:00:41.340 --> 01:00:41.880 OK. 01:00:41.880 --> 01:00:43.770 And the real kicker is the last part, right, 01:00:43.770 --> 01:00:45.610 which is how do you move something below? 01:00:45.610 --> 01:00:46.920 So let's think for a minute. 01:00:46.920 --> 01:00:51.160 What's moving below going to entail? 01:00:51.160 --> 01:00:53.190 So is it going to affect the set of keys 01:00:53.190 --> 01:00:55.140 that are in my document? 01:00:55.140 --> 01:00:56.260 No, actually, right? 01:00:56.260 --> 01:00:57.677 It's just changing their ordering, 01:00:57.677 --> 01:01:00.030 but both of those keys are already there. 01:01:00.030 --> 01:01:00.640 Yeah. 01:01:00.640 --> 01:01:03.240 So the only thing it's going to do 01:01:03.240 --> 01:01:06.990 is effect where they are in the sequence. 01:01:06.990 --> 01:01:10.320 But there's something really, really annoying. 01:01:10.320 --> 01:01:12.060 How do I find something in this sequence? 01:01:12.060 --> 01:01:14.295 Let's say that I tell you that document 75 has 01:01:14.295 --> 01:01:18.540 to move below document 352. 01:01:18.540 --> 01:01:20.550 Well, so far in my data structure, 01:01:20.550 --> 01:01:23.372 what's my only option? 01:01:23.372 --> 01:01:25.080 Does the set tell me anything about where 01:01:25.080 --> 01:01:27.820 stuff is in the linked list? 01:01:27.820 --> 01:01:28.320 No. 01:01:28.320 --> 01:01:30.800 So how much time is it going to take for me to find a key? 01:01:30.800 --> 01:01:31.560 AUDIENCE: Order n. 01:01:31.560 --> 01:01:32.760 PROFESSOR: Order n, right, because I've 01:01:32.760 --> 01:01:34.800 got to loop over this whole linked list 01:01:34.800 --> 01:01:37.140 and find the item that I'm missing. 01:01:37.140 --> 01:01:38.970 Is that allowed? 01:01:38.970 --> 01:01:40.470 No, the problem tells me that I have 01:01:40.470 --> 01:01:43.000 to do this in order log n time. 01:01:43.000 --> 01:01:45.060 So somehow in our head, we should be thinking, 01:01:45.060 --> 01:01:50.400 OK, well, we want to kind of use this set here 01:01:50.400 --> 01:01:53.600 to help us find stuff in the doubly-linked list. 01:01:53.600 --> 01:01:56.130 Does that make sense, that intuition? 01:01:56.130 --> 01:01:58.060 So here's how we're going to do it. 01:01:58.060 --> 01:01:59.840 Actually, before I do that, any ideas? 01:01:59.840 --> 01:02:01.620 How can we solve that? 01:02:01.620 --> 01:02:02.640 Yes. 01:02:02.640 --> 01:02:04.580 AUDIENCE: Maybe the [INAUDIBLE]. 01:02:07.700 --> 01:02:09.210 PROFESSOR: That's a fabulous idea. 01:02:09.210 --> 01:02:10.550 So here's what I'm going to do. 01:02:10.550 --> 01:02:12.170 Remember that when we make a set, 01:02:12.170 --> 01:02:13.670 we don't just have to sort the keys. 01:02:13.670 --> 01:02:15.250 We can attach data to our keys. 01:02:15.250 --> 01:02:15.750 Yeah. 01:02:15.750 --> 01:02:21.650 And in particular, I am going to attach a pointer into my linked 01:02:21.650 --> 01:02:22.430 list. 01:02:22.430 --> 01:02:24.780 Really sneaky answer. 01:02:24.780 --> 01:02:25.940 So let's say that I have-- 01:02:25.940 --> 01:02:27.188 let's do a quick example. 01:02:27.188 --> 01:02:28.730 So let's say that I have my documents 01:02:28.730 --> 01:02:35.455 are 1, 5, 3, 2, and 7, like that. 01:02:35.455 --> 01:02:37.330 So my linked list is going to be real simple. 01:02:37.330 --> 01:02:38.240 It's doubly linked. 01:02:38.240 --> 01:02:43.040 See, there's two arrows here, just like that, right? 01:02:43.040 --> 01:02:47.192 And now my sorted array is going to store all the keys in order. 01:02:47.192 --> 01:02:48.650 Let's see if I can do this with me. 01:02:48.650 --> 01:02:52.430 1, 2, 3, 5, 7, right? 01:02:52.430 --> 01:02:55.040 So this is my linked list. 01:02:55.040 --> 01:03:02.030 And here's my sorted array is going to be 1, 2, 3, 5, 7. 01:03:02.030 --> 01:03:04.520 But then what I'm going to attach to each of these guys 01:03:04.520 --> 01:03:09.090 is a pointer to the element of the linked list, right? 01:03:09.090 --> 01:03:12.110 So he is going to additionally contain an arrow here. 01:03:12.110 --> 01:03:13.820 The 2 will contain an arrow there. 01:03:13.820 --> 01:03:16.460 The 3 will contain an arrow there. 01:03:16.460 --> 01:03:21.543 The 5-- this is ugly, I'm sorry. 01:03:21.543 --> 01:03:22.460 Now, here's the thing. 01:03:22.460 --> 01:03:25.250 Let's say that I remove this 3 here. 01:03:25.250 --> 01:03:29.210 Does that actually affect 7's pointer? 01:03:29.210 --> 01:03:30.080 No, right? 01:03:30.080 --> 01:03:32.030 So these pointers remain valid even if I 01:03:32.030 --> 01:03:33.650 edit other parts of the list. 01:03:33.650 --> 01:03:35.710 Really sneaky trick. 01:03:35.710 --> 01:03:36.992 OK. 01:03:36.992 --> 01:03:38.450 So that is going to be our solution 01:03:38.450 --> 01:03:41.840 to this problem is to store not just the sorted array of x's 01:03:41.840 --> 01:03:44.070 but x's and pointers. 01:03:44.070 --> 01:03:50.330 So now we have to modify a tiny bit, right? 01:03:50.330 --> 01:03:54.320 And now what can we do to move something below? 01:03:54.320 --> 01:03:57.200 Well, it's going to happen in sort of three steps, right? 01:03:57.200 --> 01:04:02.120 So our first step, well, we've got to first 01:04:02.120 --> 01:04:06.020 find keys x and y, right? 01:04:06.020 --> 01:04:16.740 So we're going to find x, vx, and y, vy in the set. 01:04:16.740 --> 01:04:18.970 How long does this take? 01:04:18.970 --> 01:04:21.680 Now we can do it in binary search, right? 01:04:21.680 --> 01:04:22.250 Exactly. 01:04:22.250 --> 01:04:27.440 So this is O, like that. 01:04:27.440 --> 01:04:30.350 And now what's my next step? 01:04:32.930 --> 01:04:35.000 Well, remember, the move below operator 01:04:35.000 --> 01:04:37.940 sort of removes x from where it currently is and then puts it 01:04:37.940 --> 01:04:40.680 underneath y, right? 01:04:40.680 --> 01:04:43.340 And both of those are like linked list editing operations, 01:04:43.340 --> 01:04:44.150 right? 01:04:44.150 --> 01:04:45.980 So I'm going to kind of-- 01:04:45.980 --> 01:04:53.360 if I have a linked list, like 1, 2, 3, 4, right, 01:04:53.360 --> 01:04:55.395 and let's say I want to erase the 2, 01:04:55.395 --> 01:04:57.020 I think you guys have all coded this up 01:04:57.020 --> 01:04:58.187 at some point in your lives. 01:04:58.187 --> 01:05:01.412 Then what I'll do is add links like that, which 01:05:01.412 --> 01:05:02.870 I can do because I can move forward 01:05:02.870 --> 01:05:05.140 and backward in the list, right? 01:05:05.140 --> 01:05:06.890 And similarly, if I want to add something, 01:05:06.890 --> 01:05:08.682 then I'll erase the links, put it in there, 01:05:08.682 --> 01:05:10.370 and update the links. 01:05:10.370 --> 01:05:12.797 How much time does that take? 01:05:12.797 --> 01:05:14.630 That's just order 1 time, right, because now 01:05:14.630 --> 01:05:17.450 that I have the pointers to these two locations, 01:05:17.450 --> 01:05:20.450 I'm just doing a lot of rewiring in my linked list. 01:05:20.450 --> 01:05:21.950 But it's all kind of local stuff. 01:05:21.950 --> 01:05:24.533 If you're coding in C, this is where your memory leak happens, 01:05:24.533 --> 01:05:25.790 and your company gets hacked. 01:05:25.790 --> 01:05:27.150 OK. 01:05:27.150 --> 01:05:29.960 Right, so that is what we're going to do next 01:05:29.960 --> 01:05:33.050 is update the linked list. 01:05:33.050 --> 01:05:34.463 And that takes order 1 time. 01:05:34.463 --> 01:05:36.630 On your homework, if you're writing out your answer, 01:05:36.630 --> 01:05:39.297 you should have an answer closer to what I've written on my page 01:05:39.297 --> 01:05:42.030 than the two words that I've written on the board here. 01:05:42.030 --> 01:05:44.180 And then finally, in step three-- 01:05:46.877 --> 01:05:49.460 well, actually, there is no step three that I've written here. 01:05:49.460 --> 01:05:52.058 Sorry. 01:05:52.058 --> 01:05:54.350 There's actually two parts to updating the linked list, 01:05:54.350 --> 01:05:54.850 right? 01:05:54.850 --> 01:06:00.440 One is to remove the old x position 01:06:00.440 --> 01:06:05.140 and then insert in the proper position, right? 01:06:05.140 --> 01:06:08.653 And when I do that, my whole update happened. 01:06:08.653 --> 01:06:10.070 Notice that there's something kind 01:06:10.070 --> 01:06:12.920 of interesting about this move below operation, which is that, 01:06:12.920 --> 01:06:15.890 did I actually edit the set at all? 01:06:15.890 --> 01:06:16.550 No, right? 01:06:16.550 --> 01:06:18.550 That kind of makes sense because the set is just 01:06:18.550 --> 01:06:19.760 a set of keys in my document. 01:06:19.760 --> 01:06:21.200 And just by changing the ordering, 01:06:21.200 --> 01:06:22.980 it doesn't affect what's in my document. 01:06:22.980 --> 01:06:25.860 So somehow, it's a sanity check that that works out. 01:06:25.860 --> 01:06:27.890 So this entire algorithm takes how long? 01:06:27.890 --> 01:06:29.570 Well, there's order log n. 01:06:29.570 --> 01:06:30.470 There's order 1. 01:06:30.470 --> 01:06:32.120 And there's another order 1. 01:06:32.120 --> 01:06:33.740 So the whole thing is log n. 01:06:33.740 --> 01:06:35.750 And life is good for our problem here. 01:06:35.750 --> 01:06:36.860 Yes. 01:06:36.860 --> 01:06:39.440 AUDIENCE: So how we can put such pointers in real life? 01:06:39.440 --> 01:06:41.065 PROFESSOR: How do we input such point-- 01:06:41.065 --> 01:06:43.232 AUDIENCE: [INAUDIBLE] like a set with pointers in it 01:06:43.232 --> 01:06:43.790 in real life? 01:06:43.790 --> 01:06:44.540 PROFESSOR: Got it. 01:06:44.540 --> 01:06:46.640 So I only know the C++ version of this. 01:06:46.640 --> 01:06:49.590 I'm going to get it wrong for Python. 01:06:49.590 --> 01:06:51.840 But I'm not sure that I quite understand the question. 01:06:51.840 --> 01:06:54.090 So it's just like pointers anywhere else in real life. 01:06:54.090 --> 01:06:57.170 So I guess in Python, what you'll 01:06:57.170 --> 01:06:58.903 do is you'll create a new, right-- as you 01:06:58.903 --> 01:07:00.320 add these guys to your set, you'll 01:07:00.320 --> 01:07:01.538 create a new object here. 01:07:01.538 --> 01:07:03.080 And in addition to that, you're going 01:07:03.080 --> 01:07:05.840 to create a new vx that you add to the linked list. 01:07:05.840 --> 01:07:08.267 And the pointer is just an address in memory, right? 01:07:08.267 --> 01:07:09.725 So essentially, what you'll store-- 01:07:09.725 --> 01:07:13.310 AUDIENCE: [INAUDIBLE] linked list nodes are the pointers. 01:07:13.310 --> 01:07:14.810 PROFESSOR: The linked list nodes are 01:07:14.810 --> 01:07:20.930 pairs, a pair of an x value and a-- well, oh, I'm sorry. 01:07:20.930 --> 01:07:22.760 Yeah, actually, I guess that's right. 01:07:22.760 --> 01:07:24.680 No. 01:07:24.680 --> 01:07:28.422 The linked list just contains a long list of x's, right? 01:07:28.422 --> 01:07:30.380 But they don't have to be in contiguous memory. 01:07:30.380 --> 01:07:31.270 That's the only difference. 01:07:31.270 --> 01:07:33.320 So you make them one linked list item a time. 01:07:33.320 --> 01:07:34.970 But it's just like constructing any other linked list. 01:07:34.970 --> 01:07:35.470 Yeah. 01:07:35.470 --> 01:07:37.580 In fact, there's pseudocode in the solutions 01:07:37.580 --> 01:07:39.242 that are distributed. 01:07:39.242 --> 01:07:40.450 So you can take a look there. 01:07:40.450 --> 01:07:40.760 Yeah. 01:07:40.760 --> 01:07:42.177 AUDIENCE: Yeah, speaking of which, 01:07:42.177 --> 01:07:45.080 for implementing these database operations, 01:07:45.080 --> 01:07:47.638 can we pseudocode the algorithms for them? 01:07:47.638 --> 01:07:50.180 PROFESSOR: I think the official answer is no, that you really 01:07:50.180 --> 01:07:53.090 are supposed to write out in words what your thing is doing. 01:07:53.090 --> 01:07:54.840 Now, there's a weird gray area, of course, 01:07:54.840 --> 01:07:55.985 which is that the words that you write out 01:07:55.985 --> 01:07:58.460 are going to look an awful lot like pseudocode. 01:07:58.460 --> 01:08:01.120 But you should make an effort to try and write it 01:08:01.120 --> 01:08:03.620 out a little bit in paragraph form, be descriptive about it. 01:08:03.620 --> 01:08:04.730 This is a good skill to learn. 01:08:04.730 --> 01:08:06.147 At the beginning of this course, I 01:08:06.147 --> 01:08:08.720 think it'll feel a little pedantic at points. 01:08:08.720 --> 01:08:10.970 And that's good because it should, 01:08:10.970 --> 01:08:13.190 and you should deal with it. 01:08:13.190 --> 01:08:13.910 Yeah. 01:08:13.910 --> 01:08:15.770 But really, try and write things out 01:08:15.770 --> 01:08:18.920 in a way that captures your logic rather than just 01:08:18.920 --> 01:08:19.880 Python code. 01:08:19.880 --> 01:08:21.779 Yeah. 01:08:21.779 --> 01:08:22.279 Cool. 01:08:22.279 --> 01:08:22.700 All right. 01:08:22.700 --> 01:08:23.270 So let's see. 01:08:23.270 --> 01:08:26.420 We've got approximately 17 minutes 01:08:26.420 --> 01:08:29.271 left in class, which is approximately 1/8 of the time 01:08:29.271 --> 01:08:31.729 it took me to figure out the solution to this last homework 01:08:31.729 --> 01:08:32.460 problem. 01:08:32.460 --> 01:08:33.590 But we'll get started. 01:08:33.590 --> 01:08:37.310 And I've written out a careful answer because essentially, 01:08:37.310 --> 01:08:40.640 in problem four, I found myself getting 01:08:40.640 --> 01:08:41.930 stuck in a few little details. 01:08:41.930 --> 01:08:43.638 And so I figured the way that I write out 01:08:43.638 --> 01:08:45.830 my little hand-scribbled chicken scratch 01:08:45.830 --> 01:08:49.279 answer online was to just basically give you 01:08:49.279 --> 01:08:52.071 an internal dialogue of my brain and how 01:08:52.071 --> 01:08:53.779 I went about solving these things and all 01:08:53.779 --> 01:08:55.760 the stupid missteps that I made. 01:08:55.760 --> 01:08:56.720 Yeah. 01:08:56.720 --> 01:08:58.130 Because I think that's actually quite valuable. 01:08:58.130 --> 01:09:00.529 Essentially, it's often you see in these problems sessions 01:09:00.529 --> 01:09:01.439 just like, oh, here's a problem. 01:09:01.439 --> 01:09:01.740 Here's the answer. 01:09:01.740 --> 01:09:02.100 Here's a problem. 01:09:02.100 --> 01:09:03.015 Here's the answer. 01:09:03.015 --> 01:09:04.640 But the reality is even your professor, 01:09:04.640 --> 01:09:07.279 especially because he's used to thinking about rendering, 01:09:07.279 --> 01:09:09.979 occasionally gets confused about these algorithms 01:09:09.979 --> 01:09:12.439 problems, especially when they're frankly written 01:09:12.439 --> 01:09:14.720 in a bit of a confusing way. 01:09:14.720 --> 01:09:17.192 And I heard some war stories about this problem actually 01:09:17.192 --> 01:09:18.109 being on the homework. 01:09:18.109 --> 01:09:20.819 Apparently, I'm in good company. 01:09:20.819 --> 01:09:23.660 So let's take a look at this last homework problem, 1-4. 01:09:23.660 --> 01:09:26.390 So this is on brick blowing. 01:09:26.390 --> 01:09:29.720 And this is a great exercise in taking a problem that 01:09:29.720 --> 01:09:32.660 is described in really long, nasty, useless language 01:09:32.660 --> 01:09:36.029 and then extracting the two words that matter. 01:09:36.029 --> 01:09:37.850 By the way, I'm making a joke about that. 01:09:37.850 --> 01:09:40.490 I can tell you that about 82% of my time as a professor 01:09:40.490 --> 01:09:41.960 is spent with people from industry 01:09:41.960 --> 01:09:44.300 visiting my office with exactly this kind of scenario 01:09:44.300 --> 01:09:46.670 where it has a really complicated thing. 01:09:46.670 --> 01:09:49.198 They've been thinking about their construction 01:09:49.198 --> 01:09:51.740 problem for years, and they have all these details and tables 01:09:51.740 --> 01:09:52.407 and flow charts. 01:09:52.407 --> 01:09:54.800 And then it turns out that their problem can be captured 01:09:54.800 --> 01:09:56.225 in about two sentences. 01:09:56.225 --> 01:09:58.400 So it's a good skill to have and one 01:09:58.400 --> 01:10:00.508 that can get you a lot of money as a consultant. 01:10:00.508 --> 01:10:02.550 So it's one that's worth practicing in this class 01:10:02.550 --> 01:10:07.700 even if we're doing it for Porkland with a wolf 01:10:07.700 --> 01:10:10.080 that blows only to the east. 01:10:10.080 --> 01:10:10.580 OK. 01:10:10.580 --> 01:10:12.038 So what's going on in this problem? 01:10:12.038 --> 01:10:15.470 There's a wolf, and the wolf can blow on houses. 01:10:15.470 --> 01:10:19.370 But for some reason, the wolf likes to blow to the right. 01:10:19.370 --> 01:10:21.470 And to make matters even more complicated, whoever 01:10:21.470 --> 01:10:23.270 wrote this problem occasionally ordered things 01:10:23.270 --> 01:10:24.650 from east to west and other times 01:10:24.650 --> 01:10:26.480 ordered them from west to east. 01:10:26.480 --> 01:10:28.610 If I were your instructor at the time, that, 01:10:28.610 --> 01:10:32.750 I would have at least gotten rid of because that's just mean. 01:10:32.750 --> 01:10:34.670 But in any event, I believe the story 01:10:34.670 --> 01:10:41.570 goes as the apocryphal house-blowing pork wolf 01:10:41.570 --> 01:10:43.970 story that we all know and love from our childhood 01:10:43.970 --> 01:10:47.540 is there's a row of houses, each of which 01:10:47.540 --> 01:10:49.430 has a different number of bricks. 01:10:49.430 --> 01:10:50.660 Yeah. 01:10:50.660 --> 01:10:55.060 And now there's a wind which is blowing from west to east. 01:10:55.060 --> 01:10:57.560 I remember all this because I was staring at it this morning 01:10:57.560 --> 01:10:58.580 in my office. 01:10:58.580 --> 01:11:03.530 And essentially, the wolf, being the big, bad wolf that he is, 01:11:03.530 --> 01:11:07.010 says, aha, if I, too, blow from west to east, 01:11:07.010 --> 01:11:11.060 I can knock over more houses more efficiently 01:11:11.060 --> 01:11:12.410 because the wind is helping me. 01:11:12.410 --> 01:11:13.952 You don't want to blow into the wind. 01:11:13.952 --> 01:11:15.900 You get your spit in your face. 01:11:15.900 --> 01:11:18.650 OK, so the wolf does that. 01:11:18.650 --> 01:11:22.790 And then what happens is that the wolf not only knocks down 01:11:22.790 --> 01:11:25.640 the house upon which the wolf blows 01:11:25.640 --> 01:11:29.420 but also all the houses to the east of that house that 01:11:29.420 --> 01:11:32.330 have fewer bricks. 01:11:32.330 --> 01:11:33.523 Why, you might ask? 01:11:33.523 --> 01:11:35.690 I don't know, because whoever wrote this problem was 01:11:35.690 --> 01:11:36.560 being a goofball. 01:11:36.560 --> 01:11:38.430 But that's the basic setup of this problem. 01:11:38.430 --> 01:11:40.250 And you can see that, essentially, this 01:11:40.250 --> 01:11:42.500 is just a long, convoluted way of describing something 01:11:42.500 --> 01:11:43.910 pretty straightforward. 01:11:43.910 --> 01:11:46.280 And so at the end of the day, the wolf, 01:11:46.280 --> 01:11:48.710 being kind of an adversarial wolf, 01:11:48.710 --> 01:11:50.840 wants to blow down as many bricks as possible. 01:11:50.840 --> 01:11:53.390 And we are providing the wolf with the analytical tools 01:11:53.390 --> 01:11:54.950 needed to do so. 01:11:54.950 --> 01:11:55.515 Yeah. 01:11:55.515 --> 01:11:57.890 So in case you didn't think we covered anything practical 01:11:57.890 --> 01:12:02.000 in 6.006, now you know that we actually are doing 01:12:02.000 --> 01:12:05.880 state-of-the-art wolf brick-blowing pig analysis. 01:12:05.880 --> 01:12:06.380 OK. 01:12:08.670 --> 01:12:09.170 Right. 01:12:09.170 --> 01:12:13.110 So let's do an example because this is weirdly complicated. 01:12:13.110 --> 01:12:15.217 So in the first part of the problem, 01:12:15.217 --> 01:12:16.800 it asks you to just do an example out. 01:12:16.800 --> 01:12:19.215 And I think if I were psychologically diagnosing 01:12:19.215 --> 01:12:21.590 the person who wrote this problem, probably the reason is 01:12:21.590 --> 01:12:22.860 that they stood back, and they said, 01:12:22.860 --> 01:12:24.990 nobody's going to understand this unless I force 01:12:24.990 --> 01:12:27.370 them to do one example by hand. 01:12:27.370 --> 01:12:27.870 Right? 01:12:27.870 --> 01:12:29.450 And so the basic problem is asking, 01:12:29.450 --> 01:12:32.900 I could choose to blow on each house. 01:12:32.900 --> 01:12:36.092 Which one should I choose to cause the most damage? 01:12:36.092 --> 01:12:38.550 Or in fact, they actually ask a slightly different problem, 01:12:38.550 --> 01:12:41.510 which is if I were to blow on each house, how much damage 01:12:41.510 --> 01:12:43.337 would I cause? 01:12:43.337 --> 01:12:44.920 OK, so let's step through the problem. 01:12:44.920 --> 01:12:56.698 So they give you an example, a set of numbers, 34, 57, 70, 19, 01:12:56.698 --> 01:12:58.177 48, 2. 01:12:58.177 --> 01:12:59.510 Actually, let's just stop there. 01:12:59.510 --> 01:13:00.650 I don't think there's any reason to do 01:13:00.650 --> 01:13:02.650 a list of a million numbers like in the problem. 01:13:02.650 --> 01:13:04.650 That was just like, the more times you do it, 01:13:04.650 --> 01:13:07.580 the more you convince yourself how useless 6.006 is. 01:13:07.580 --> 01:13:08.760 OK. 01:13:08.760 --> 01:13:09.260 Right. 01:13:09.260 --> 01:13:13.727 So how much damage does blowing on house number one cause? 01:13:13.727 --> 01:13:15.060 Well, let's make a little table. 01:13:15.060 --> 01:13:18.170 So if I blow on house number one, just by default, 01:13:18.170 --> 01:13:19.820 house number one falls. 01:13:19.820 --> 01:13:22.190 I'll pull an x there, meaning I blew it over. 01:13:22.190 --> 01:13:24.810 And then the rule is any house to the right 01:13:24.810 --> 01:13:26.638 which has fewer bricks-- 01:13:26.638 --> 01:13:28.430 the physics of this problem drives me nuts. 01:13:28.430 --> 01:13:30.138 I thought about it for quite a long time, 01:13:30.138 --> 01:13:33.320 and I can't think of a wind that actually has this property. 01:13:33.320 --> 01:13:37.160 But any house that has fewer bricks and is to the east 01:13:37.160 --> 01:13:38.120 also gets blown down. 01:13:38.120 --> 01:13:40.100 So 57 does not, right? 01:13:40.100 --> 01:13:41.540 That's bigger than 34. 01:13:41.540 --> 01:13:42.650 Similarly for 70. 01:13:42.650 --> 01:13:46.610 Aha, the 19 goes with the wind. 01:13:46.610 --> 01:13:49.070 48's bigger, and the 2 goes, right? 01:13:49.070 --> 01:13:53.300 So the element number one of my array would be 3, right? 01:13:53.300 --> 01:13:55.640 Three things get blown over. 01:13:55.640 --> 01:13:57.710 OK, for the next guy, right? 01:13:57.710 --> 01:13:59.390 So the 57 gets blown over. 01:13:59.390 --> 01:14:02.450 It does not blow over the 34 because my wolf only 01:14:02.450 --> 01:14:04.964 knows how to blow to the right. 01:14:04.964 --> 01:14:07.460 Uh. 01:14:07.460 --> 01:14:09.560 So it does not blow over the 70. 01:14:09.560 --> 01:14:12.740 It does blow over the 19. 01:14:12.740 --> 01:14:14.100 It does blow over the 48. 01:14:14.100 --> 01:14:15.320 It does, right? 01:14:15.320 --> 01:14:16.980 So here, there'd be a 4. 01:14:16.980 --> 01:14:18.980 These numbers do not match the ones in the notes 01:14:18.980 --> 01:14:21.938 because this is a shorter list of numbers, I should point out. 01:14:21.938 --> 01:14:23.480 OK, let's do one more, and then we're 01:14:23.480 --> 01:14:25.770 going to stop because this is laborious and boring. 01:14:25.770 --> 01:14:28.790 So the 70 blows over everything. 01:14:28.790 --> 01:14:30.320 Yeah. 01:14:30.320 --> 01:14:34.243 But only everything to its right because, as we all know, 01:14:34.243 --> 01:14:35.285 wolves only blow to the-- 01:14:35.285 --> 01:14:36.530 OK, I'll shut up. 01:14:36.530 --> 01:14:40.200 So there's four things that get blown over here. 01:14:40.200 --> 01:14:41.762 But you guys get the point, right? 01:14:41.762 --> 01:14:43.220 And so essentially, this problem is 01:14:43.220 --> 01:14:44.803 asking you to fill in arrays that look 01:14:44.803 --> 01:14:47.600 like this for your problem. 01:14:47.600 --> 01:14:49.550 Everybody on board with the problem? 01:14:49.550 --> 01:14:52.370 Now, notice that just by virtue of doing this example, 01:14:52.370 --> 01:14:53.840 let's think about this abstractly 01:14:53.840 --> 01:14:55.507 because I think this is a problem that's 01:14:55.507 --> 01:14:58.580 just coded in so many words and made-up weird TA garbage. 01:14:58.580 --> 01:15:01.370 But at the end of the day, this problem 01:15:01.370 --> 01:15:03.830 as an algorithmic problem is not so bad. 01:15:03.830 --> 01:15:07.025 Essentially, what I'm asking you to do is I look at that 57, 01:15:07.025 --> 01:15:08.900 and I look to the right, and I count how many 01:15:08.900 --> 01:15:10.760 things are smaller than 57. 01:15:10.760 --> 01:15:11.870 And I add 1 to it. 01:15:11.870 --> 01:15:13.370 And that's the number that should go 01:15:13.370 --> 01:15:14.787 in that element in the array. 01:15:14.787 --> 01:15:16.370 So notice that those three paragraphs, 01:15:16.370 --> 01:15:17.310 that's what they're communicating. 01:15:17.310 --> 01:15:19.250 You didn't need wolves blowing and winds 01:15:19.250 --> 01:15:23.100 and trees and houses and Dorothy and all that kind of stuff. 01:15:23.100 --> 01:15:24.010 Yeah. 01:15:24.010 --> 01:15:24.510 Right. 01:15:24.510 --> 01:15:27.177 So at the end of the day, that's the problem we can think about. 01:15:27.177 --> 01:15:30.330 And we no longer have to think about Porkland ever again. 01:15:30.330 --> 01:15:30.830 OK. 01:15:30.830 --> 01:15:32.690 So maybe in the remaining 10 minutes 01:15:32.690 --> 01:15:35.180 here, I think we can feasibly do part b. 01:15:35.180 --> 01:15:39.170 And then part c is an extension of part b. 01:15:39.170 --> 01:15:42.110 I got myself confused on part c, so the handwritten solutions 01:15:42.110 --> 01:15:44.960 have an incorrect answer, and then an oops, and then 01:15:44.960 --> 01:15:46.200 a correct answer in it. 01:15:46.200 --> 01:15:48.770 So you can see where my logic went wrong. 01:15:48.770 --> 01:15:51.770 To be fair, I already knew the full answer 01:15:51.770 --> 01:15:55.710 and wrote out both because I thought that was useful. 01:15:55.710 --> 01:15:56.210 OK. 01:15:56.210 --> 01:15:59.275 So in case our problem wasn't contrived enough, 01:15:59.275 --> 01:16:01.400 we're going to add a little bit of condition to it. 01:16:01.400 --> 01:16:03.150 And then what we're going to see in part c 01:16:03.150 --> 01:16:06.080 is that's somehow a stepping stone toward solving the bigger 01:16:06.080 --> 01:16:07.920 problem. 01:16:07.920 --> 01:16:12.560 So in part c, they say that a house in Porkland 01:16:12.560 --> 01:16:16.850 is special if it has one of two properties. 01:16:16.850 --> 01:16:20.660 Either it has no easterly neighbor, 01:16:20.660 --> 01:16:24.890 meaning it's just the rightmost house, 01:16:24.890 --> 01:16:28.940 or its adjacent neighbor to the east 01:16:28.940 --> 01:16:33.680 contains at least as many bricks as it does. 01:16:33.680 --> 01:16:34.280 Right. 01:16:34.280 --> 01:16:37.940 So what does that mean in terms of this list of numbers? 01:16:37.940 --> 01:16:40.130 Which one of these guys is special? 01:16:40.130 --> 01:16:42.960 So the easterly guy is special. 01:16:42.960 --> 01:16:45.530 We just know that by default. And other 01:16:45.530 --> 01:16:47.480 than that, this special property is 01:16:47.480 --> 01:16:50.390 saying that the guy immediately to the right 01:16:50.390 --> 01:16:55.850 has a bigger number or at least equal to or larger to the one 01:16:55.850 --> 01:16:56.840 right before it. 01:16:56.840 --> 01:16:57.740 Yeah. 01:16:57.740 --> 01:17:02.492 So 57 is bigger than 34, so this guy is special. 01:17:02.492 --> 01:17:03.950 By the way, the problem doesn't ask 01:17:03.950 --> 01:17:05.158 you to do this as an example. 01:17:05.158 --> 01:17:07.010 But I totally would if I were solving this. 01:17:07.010 --> 01:17:08.270 Yeah. 01:17:08.270 --> 01:17:11.090 70 is bigger than 57, so it's special. 01:17:11.090 --> 01:17:13.740 19 is smaller than 70, so it's not special. 01:17:13.740 --> 01:17:15.110 I'm sorry. 01:17:15.110 --> 01:17:17.810 40 is bigger than 19. 01:17:17.810 --> 01:17:20.930 And I guess 2 is special just because things on the right 01:17:20.930 --> 01:17:22.520 are special. 01:17:22.520 --> 01:17:25.130 OK, so do we all get the definition? 01:17:25.130 --> 01:17:25.850 Fabulous. 01:17:25.850 --> 01:17:32.360 So in this problem, now what we're given is that I guess, 01:17:32.360 --> 01:17:36.860 what, our evil wolf takes a walk down the block 01:17:36.860 --> 01:17:37.780 and takes a look at-- 01:17:37.780 --> 01:17:39.530 counts the number of bricks at each house. 01:17:39.530 --> 01:17:42.510 And he or she makes an observation, 01:17:42.510 --> 01:17:45.215 which is that all but one house is special. 01:17:47.935 --> 01:17:49.560 Now, let's think about what that means. 01:17:49.560 --> 01:17:51.435 Let's think about the structure of our array. 01:17:51.435 --> 01:17:54.232 So what happens as I walk along my right? 01:17:54.232 --> 01:17:56.690 If a house is special, what do I know about the next house? 01:17:56.690 --> 01:17:58.910 AUDIENCE: It won't get blown if you blow a special house down. 01:17:58.910 --> 01:17:59.900 PROFESSOR: That's true, but we're already 01:17:59.900 --> 01:18:00.710 solving the problem there. 01:18:00.710 --> 01:18:03.080 Let's think qualitatively for a second, like just in terms 01:18:03.080 --> 01:18:04.040 of the number of bricks, right? 01:18:04.040 --> 01:18:05.207 I'm walking down the street. 01:18:05.207 --> 01:18:05.720 I'm looking. 01:18:05.720 --> 01:18:07.940 And the number of bricks increases 01:18:07.940 --> 01:18:10.515 until I get to the one house that isn't special. 01:18:10.515 --> 01:18:11.390 AUDIENCE: [INAUDIBLE] 01:18:11.390 --> 01:18:14.180 PROFESSOR: And then it decreases, potentially, 01:18:14.180 --> 01:18:15.080 or it's the last one. 01:18:15.080 --> 01:18:17.160 And then it starts increasing again. 01:18:17.160 --> 01:18:17.660 Yeah. 01:18:17.660 --> 01:18:19.610 So in other words, my array kind of 01:18:19.610 --> 01:18:23.330 looks like the following, which is that it basically 01:18:23.330 --> 01:18:27.000 has two different pieces that are both sorted. 01:18:27.000 --> 01:18:27.500 Yeah. 01:18:27.500 --> 01:18:29.270 That's what this long paragraph is saying. 01:18:29.270 --> 01:18:31.100 It's saying suppose that the list of houses 01:18:31.100 --> 01:18:32.890 is sorted except for one thing. 01:18:32.890 --> 01:18:33.390 Yeah. 01:18:33.390 --> 01:18:34.807 That's how you could have said it. 01:18:34.807 --> 01:18:36.450 OK, so here's an example. 01:18:36.450 --> 01:18:44.390 So maybe we have 1, 2, 3, 4, and then, I don't know, 2, 3, 4, 5. 01:18:44.390 --> 01:18:47.510 So all of these is special with the exception of this guy, 01:18:47.510 --> 01:18:49.160 sadly for him. 01:18:49.160 --> 01:18:53.090 And so there's kind of a little dividing line 01:18:53.090 --> 01:18:55.310 that happens, to the left of which, it's sorted, 01:18:55.310 --> 01:18:56.750 and to the right. 01:18:56.750 --> 01:18:58.160 Yeah. 01:18:58.160 --> 01:18:59.660 OK, and then the question is, can we 01:18:59.660 --> 01:19:03.110 predict the damage for each house 01:19:03.110 --> 01:19:05.400 based on this information? 01:19:05.400 --> 01:19:08.090 So I want as output to my code an array that says, 01:19:08.090 --> 01:19:10.880 this guy causes so much damage, this guy causes so much damage, 01:19:10.880 --> 01:19:12.440 and so on. 01:19:12.440 --> 01:19:16.298 And I want to get it in order n time. 01:19:16.298 --> 01:19:17.340 Let's think for a second. 01:19:17.340 --> 01:19:19.030 So how are we going to do that? 01:19:19.030 --> 01:19:20.670 So order n time means that I can't 01:19:20.670 --> 01:19:24.170 afford to do sorting and all that kind of fancy stuff. 01:19:24.170 --> 01:19:26.650 So basically, all I can do is walk up and down the array. 01:19:26.650 --> 01:19:31.320 So we should think a bit about, what's special about special-- 01:19:31.320 --> 01:19:32.670 oy vey. 01:19:32.670 --> 01:19:34.590 What's special about this configuration 01:19:34.590 --> 01:19:36.360 that we wouldn't normally have? 01:19:36.360 --> 01:19:39.330 AUDIENCE: Anything before the non-special node should have 01:19:39.330 --> 01:19:40.356 the same-- 01:19:40.356 --> 01:19:42.390 or I'm sorry. 01:19:42.390 --> 01:19:43.470 PROFESSOR: Not quite. 01:19:43.470 --> 01:19:45.570 AUDIENCE: Everything after the non-special node 01:19:45.570 --> 01:19:47.682 causes only one damage. 01:19:47.682 --> 01:19:49.890 PROFESSOR: Yeah, so let's think about this carefully. 01:19:49.890 --> 01:19:53.560 I think that's right, but I'm bad at parsing sentences. 01:19:53.560 --> 01:19:55.590 Remember that I always blow to the right, 01:19:55.590 --> 01:19:58.570 and I only blow short stuff. 01:19:58.570 --> 01:20:04.500 So if I blow on this 2 here, will 01:20:04.500 --> 01:20:07.920 anything before that vertical line ever get blown over? 01:20:07.920 --> 01:20:10.170 No, because by property of being special, 01:20:10.170 --> 01:20:11.670 we know that the list is increasing. 01:20:11.670 --> 01:20:14.028 But it has to decrease to blow something over. 01:20:14.028 --> 01:20:16.020 That make sense? 01:20:16.020 --> 01:20:19.050 So for each one of these guys, the only stuff we can blow 01:20:19.050 --> 01:20:21.168 is in the second array. 01:20:21.168 --> 01:20:22.590 Does that make sense? 01:20:22.590 --> 01:20:23.800 So we made an observation. 01:20:23.800 --> 01:20:25.550 Notice that it is special to the structure 01:20:25.550 --> 01:20:26.730 that we gave to our problem. 01:20:26.730 --> 01:20:27.230 Yeah. 01:20:27.230 --> 01:20:30.927 And moreover, so in both parts of this problem, 01:20:30.927 --> 01:20:32.760 we're going to look at two parts of an array 01:20:32.760 --> 01:20:34.990 and then merge them together in a smart way. 01:20:34.990 --> 01:20:37.470 Let's look on the right-hand side. 01:20:37.470 --> 01:20:38.730 Does any of these things-- 01:20:38.730 --> 01:20:40.140 if I blow on any of these houses, 01:20:40.140 --> 01:20:42.490 what's the damage that I do? 01:20:42.490 --> 01:20:43.198 1. 01:20:43.198 --> 01:20:43.740 You see that? 01:20:43.740 --> 01:20:46.320 Because this is an increasing list, I only blow to the right, 01:20:46.320 --> 01:20:47.830 and it has to get shorter. 01:20:47.830 --> 01:20:48.330 Yeah. 01:20:48.330 --> 01:20:51.040 So if I want partial credit on my homework problem-- 01:20:51.040 --> 01:20:53.910 probably very little partial credit, but spiritually, 01:20:53.910 --> 01:20:56.370 it's half of your credit-- 01:20:56.370 --> 01:21:01.470 we know the second half of my array is just a bunch of 1's. 01:21:01.470 --> 01:21:04.200 Incidentally, one detail which I bet would get a minus 1 01:21:04.200 --> 01:21:05.910 on your homework, we didn't tell you 01:21:05.910 --> 01:21:08.070 that-- all we told you is that there existed 01:21:08.070 --> 01:21:09.630 one house that wasn't special, but we 01:21:09.630 --> 01:21:11.220 didn't tell you which one. 01:21:11.220 --> 01:21:13.580 But notice that I can find it in order n time, right? 01:21:13.580 --> 01:21:15.872 I can just walk along my array and find the first place 01:21:15.872 --> 01:21:18.000 where it decreases, and that's a special house. 01:21:18.000 --> 01:21:20.460 So I can kind of assume that I know where this line is. 01:21:20.460 --> 01:21:23.070 That's kosher. 01:21:23.070 --> 01:21:29.850 So then the only question is, how do I fill in my array here? 01:21:29.850 --> 01:21:30.747 How do I count-- 01:21:30.747 --> 01:21:32.580 and so let's think about this mathematically 01:21:32.580 --> 01:21:35.542 because porks and blowing and stuff is kind of confusing. 01:21:35.542 --> 01:21:37.500 So let's think about it in terms of this array. 01:21:37.500 --> 01:21:39.070 Essentially, what am I trying to do? 01:21:39.070 --> 01:21:41.640 So for this 2, I'm trying to look 01:21:41.640 --> 01:21:44.580 in the second half of my array and find all the numbers here 01:21:44.580 --> 01:21:47.050 that are less than 2, right? 01:21:47.050 --> 01:21:49.650 That's essentially what this problem is asking you to do. 01:21:49.650 --> 01:21:50.970 You see that? 01:21:50.970 --> 01:21:52.800 OK. 01:21:52.800 --> 01:21:54.420 I'm going to make one-- so here's 01:21:54.420 --> 01:21:56.640 a simple kind of n squared algorithm, right, 01:21:56.640 --> 01:22:00.120 which is I could iterate over every item here 01:22:00.120 --> 01:22:02.370 and just loop over here and count the number of things 01:22:02.370 --> 01:22:04.405 that are less than that, right? 01:22:04.405 --> 01:22:06.780 That's a double for loop, each of which could be up to n. 01:22:06.780 --> 01:22:08.168 So that's n squared. 01:22:08.168 --> 01:22:09.210 That's against the rules. 01:22:09.210 --> 01:22:11.270 Can't do it. 01:22:11.270 --> 01:22:13.330 So we need one more additional observation. 01:22:13.330 --> 01:22:14.940 OK. 01:22:14.940 --> 01:22:16.600 Let's think about this for a second. 01:22:16.600 --> 01:22:19.830 Let's say I'm the wolf, the pork-blowing wolf. 01:22:19.830 --> 01:22:22.510 And I walk from one house to the next. 01:22:22.510 --> 01:22:27.450 And I keep track of what houses get blown over here. 01:22:27.450 --> 01:22:28.200 Now, what happens? 01:22:28.200 --> 01:22:33.642 As I walk from one house to the next on the left-hand side, 01:22:33.642 --> 01:22:35.100 they're only getting taller, right? 01:22:35.100 --> 01:22:38.210 That's what it means to be special. 01:22:38.210 --> 01:22:42.360 So the houses that get blown over here, the set of those 01:22:42.360 --> 01:22:43.350 only grows larger. 01:22:43.350 --> 01:22:45.060 Does that make sense? 01:22:45.060 --> 01:22:46.560 Because as these guys get taller, 01:22:46.560 --> 01:22:48.630 more stuff falls over on the right-hand side. 01:22:48.630 --> 01:22:51.850 It's like a total goofball talking about this problem. 01:22:51.850 --> 01:22:55.860 Moreover, these guys are in sorted order. 01:22:55.860 --> 01:22:57.300 Yeah. 01:22:57.300 --> 01:23:01.140 So what's going to end up happening 01:23:01.140 --> 01:23:04.142 is that, always, there's just some interval of houses 01:23:04.142 --> 01:23:05.100 that gets knocked over. 01:23:05.100 --> 01:23:08.832 So for example, the 4 blows over the 2 and the 3, I guess. 01:23:08.832 --> 01:23:11.040 Notice that that always starts on the left-hand side. 01:23:11.040 --> 01:23:13.140 And that's because these guys are sorted. 01:23:13.140 --> 01:23:14.760 The 3 just blows over the 2, which 01:23:14.760 --> 01:23:15.968 is kind of a subset of the 4. 01:23:15.968 --> 01:23:18.052 This isn't a great example because those first two 01:23:18.052 --> 01:23:19.410 guys don't blow over anything. 01:23:19.410 --> 01:23:21.730 That make sense? 01:23:21.730 --> 01:23:22.230 Any ideas? 01:23:22.230 --> 01:23:25.110 How could I use that observation to make an order n algorithm 01:23:25.110 --> 01:23:29.310 for counting blowiness damage? 01:23:32.747 --> 01:23:34.110 I've heard a lot from you. 01:23:34.110 --> 01:23:36.193 Let's hear from some of our other colleagues here. 01:23:36.193 --> 01:23:38.480 Another person. 01:23:38.480 --> 01:23:38.980 Yes. 01:23:38.980 --> 01:23:40.320 AUDIENCE: Binary search. 01:23:40.320 --> 01:23:42.578 PROFESSOR: Binary search. 01:23:42.578 --> 01:23:43.120 Tell me more. 01:23:43.120 --> 01:23:44.120 What would I search for? 01:23:44.120 --> 01:23:46.080 AUDIENCE: So let's say you have 4, 01:23:46.080 --> 01:23:48.545 and you move by yourself from the right, 01:23:48.545 --> 01:23:51.402 take the index of that, and then you take the number. 01:23:51.402 --> 01:23:53.110 PROFESSOR: It's an interesting intuition. 01:23:53.110 --> 01:23:54.960 And it's, like, 82% right. 01:23:54.960 --> 01:23:58.327 So to repeat our colleague's suggestion, which is a good one 01:23:58.327 --> 01:23:59.910 and will make your code a lot faster-- 01:23:59.910 --> 01:24:01.470 in fact, in practical terms, it would probably 01:24:01.470 --> 01:24:02.040 be about the same. 01:24:02.040 --> 01:24:03.720 It would be I'm going to move from the left 01:24:03.720 --> 01:24:05.553 to the right-hand side unless there's a very 01:24:05.553 --> 01:24:06.900 large neighborhood of pigs. 01:24:06.900 --> 01:24:08.520 I'm going to move across this array. 01:24:08.520 --> 01:24:11.670 I'm going to binary search this other side 01:24:11.670 --> 01:24:14.265 and find the first house that shouldn't get blown over 01:24:14.265 --> 01:24:15.300 or something. 01:24:15.300 --> 01:24:18.150 And now I know the total number of things that get blown over. 01:24:18.150 --> 01:24:19.710 Now, unfortunately for you, what's 01:24:19.710 --> 01:24:21.720 the runtime of that algorithm? 01:24:21.720 --> 01:24:22.860 n log n, right? 01:24:22.860 --> 01:24:25.080 Because for every single guy here, 01:24:25.080 --> 01:24:27.973 I incur a binary search there, which is log n time. 01:24:27.973 --> 01:24:29.640 So you're in the church, wrong pew here. 01:24:32.022 --> 01:24:33.980 But there's an observation we haven't used yet, 01:24:33.980 --> 01:24:38.270 which is as I move to the right, these numbers increase. 01:24:38.270 --> 01:24:38.900 Yeah. 01:24:38.900 --> 01:24:42.140 So what's going to happen to the index of the stuff that 01:24:42.140 --> 01:24:44.270 gets blown over? 01:24:44.270 --> 01:24:46.190 It's always going to move this way. 01:24:46.190 --> 01:24:48.440 Is it ever going to move to the left? 01:24:48.440 --> 01:24:48.980 No. 01:24:48.980 --> 01:24:52.280 So there is a term that we used in two lectures ago, which 01:24:52.280 --> 01:24:54.530 is a term I never heard before, but I kind of like it. 01:24:54.530 --> 01:24:57.560 It's called a two-finger algorithm. 01:24:57.560 --> 01:24:58.220 Yeah. 01:24:58.220 --> 01:24:59.990 What does the two-finger algorithm do? 01:24:59.990 --> 01:25:00.780 Looks like this. 01:25:00.780 --> 01:25:01.730 Two fingers. 01:25:01.730 --> 01:25:02.930 Yeah. 01:25:02.930 --> 01:25:04.850 So what I'm going to do is I'm going 01:25:04.850 --> 01:25:07.820 to keep my finger on the right pointing 01:25:07.820 --> 01:25:11.100 at the very first thing that doesn't get blown over. 01:25:11.100 --> 01:25:11.690 OK. 01:25:11.690 --> 01:25:14.250 So here, right, the 1 doesn't blow over anything. 01:25:14.250 --> 01:25:15.338 So it stays here. 01:25:15.338 --> 01:25:16.880 I'm going to iterate to the next guy. 01:25:16.880 --> 01:25:18.980 Does the 2 blow over anything? 01:25:18.980 --> 01:25:21.290 No, by the way the problem is written. 01:25:21.290 --> 01:25:23.510 Now I'm going to move this guy here. 01:25:23.510 --> 01:25:25.580 Well, now the 3 blows over the 2 here. 01:25:25.580 --> 01:25:29.420 So I'm going to move my right finger one thing to the right 01:25:29.420 --> 01:25:31.370 and move the guy to 4. 01:25:31.370 --> 01:25:33.320 The 4 also doesn't blow over anything, 01:25:33.320 --> 01:25:34.740 and then I guess I'm done. 01:25:34.740 --> 01:25:36.770 But the basic point here is that I'm always 01:25:36.770 --> 01:25:39.200 going to be incrementing two different pointers 01:25:39.200 --> 01:25:40.752 and moving to the right. 01:25:40.752 --> 01:25:41.960 And so what am I going to do? 01:25:41.960 --> 01:25:52.850 If I call this finger i and this finger j, 01:25:52.850 --> 01:25:55.670 will i or j ever decrease? 01:25:55.670 --> 01:25:56.660 No. 01:25:56.660 --> 01:25:59.090 So if my algorithm is looping over all of these guys 01:25:59.090 --> 01:26:01.310 and they only ever touch each of these indices once, 01:26:01.310 --> 01:26:03.850 I'm going to get an order n technique. 01:26:03.850 --> 01:26:04.685 Yeah. 01:26:04.685 --> 01:26:07.310 And that's going to be the basic trick to solving this problem. 01:26:07.310 --> 01:26:11.130 Now, I've managed to talk myself out of time. 01:26:11.130 --> 01:26:12.240 But basically, that's it. 01:26:12.240 --> 01:26:15.590 So what I'm going to do if we have 10 extra seconds 01:26:15.590 --> 01:26:17.000 is I'm going to initialize. 01:26:22.130 --> 01:26:34.760 Essentially, for each i, I'm going to increment j 01:26:34.760 --> 01:26:43.250 until the number of bricks at j is greater 01:26:43.250 --> 01:26:45.310 than the number of bricks at i. 01:26:45.310 --> 01:26:47.840 I guess that should be greater than or equal to. 01:26:47.840 --> 01:26:48.920 Right? 01:26:48.920 --> 01:26:51.507 And the reason to do that is that now the number of houses 01:26:51.507 --> 01:26:53.090 that get blown over is just this index 01:26:53.090 --> 01:26:56.300 j minus the index of the first guy, right? 01:26:56.300 --> 01:26:59.060 And then I'll increment i and continue. 01:26:59.060 --> 01:27:01.130 That make any sense? 01:27:01.130 --> 01:27:02.090 Fabulous. 01:27:02.090 --> 01:27:04.490 So I managed by some miracle to make it 01:27:04.490 --> 01:27:06.210 to almost the end of this problem set 01:27:06.210 --> 01:27:07.910 minus one part of one problem. 01:27:07.910 --> 01:27:10.885 Of course, that's the hardest part of the whole problem set. 01:27:10.885 --> 01:27:12.260 But there's an answer written out 01:27:12.260 --> 01:27:14.540 where, essentially, now the question is, can you 01:27:14.540 --> 01:27:16.790 do exactly the same algorithm but I don't give you 01:27:16.790 --> 01:27:22.155 this special assumption that it increases and then decreases? 01:27:22.155 --> 01:27:24.530 And it's going to be basically an extension of this idea, 01:27:24.530 --> 01:27:27.580 as we're going to use this plus a little bit of merge sort. 01:27:27.580 --> 01:27:28.080 OK. 01:27:28.080 --> 01:27:30.000 So with that, we'll see you guys next week. 01:27:30.000 --> 01:27:30.958 It's always a pleasure. 01:27:30.958 --> 01:27:34.180 And yeah, have a lovely weekend.