WEBVTT 00:00:00.090 --> 00:00:02.490 The following content is provided under a Creative 00:00:02.490 --> 00:00:04.030 Commons license. 00:00:04.030 --> 00:00:06.360 Your support will help MIT OpenCourseWare 00:00:06.360 --> 00:00:10.720 continue to offer high quality educational resources for free. 00:00:10.720 --> 00:00:13.320 To make a donation or view additional materials 00:00:13.320 --> 00:00:17.280 from hundreds of MIT courses, visit MIT OpenCourseWare 00:00:17.280 --> 00:00:18.450 at ocw.mit.edu. 00:00:22.100 --> 00:00:24.680 ERIK DEMAINE: All right, welcome back to Dynamic Optimality. 00:00:24.680 --> 00:00:27.020 This is the second of two lectures. 00:00:27.020 --> 00:00:29.770 And today we're going to focus mainly on lower bounds. 00:00:29.770 --> 00:00:32.479 So last time we saw this geometric connection 00:00:32.479 --> 00:00:34.020 to binary search trees. 00:00:34.020 --> 00:00:37.760 So again, this is about is there one best binary search tree. 00:00:37.760 --> 00:00:40.070 And we represented binary search trees, or at least 00:00:40.070 --> 00:00:42.080 the execution of those algorithms, 00:00:42.080 --> 00:00:45.920 as point sets in time space. 00:00:45.920 --> 00:00:49.340 And of course a point set corresponded 00:00:49.340 --> 00:00:52.410 to a valid execution of a BST tree where each of these points 00:00:52.410 --> 00:00:55.970 represented which nodes got touched during an access. 00:00:55.970 --> 00:00:59.300 If and only if the point set was arborally satisfied, 00:00:59.300 --> 00:01:01.910 meaning you take any two points in the point set, 00:01:01.910 --> 00:01:03.530 if they span a rectangle that is not 00:01:03.530 --> 00:01:05.720 just a horizontal or vertical line segment, 00:01:05.720 --> 00:01:07.640 there must be a third point somewhere 00:01:07.640 --> 00:01:09.410 inside that rectangle, which in the end 00:01:09.410 --> 00:01:14.310 implies that there's a monotone path between those two points. 00:01:14.310 --> 00:01:16.190 And then we saw, on the upper bound side, 00:01:16.190 --> 00:01:19.100 we saw a greedy algorithm, which was the obvious offline thing 00:01:19.100 --> 00:01:22.010 to do, which is as these points come along, 00:01:22.010 --> 00:01:24.170 as you do the accesses, the white dots, 00:01:24.170 --> 00:01:26.540 you add the necessary red dots in order 00:01:26.540 --> 00:01:30.867 to make it arborally satisfied row by row. 00:01:30.867 --> 00:01:33.200 And so that seemed like the obvious offline thing to do. 00:01:33.200 --> 00:01:36.140 Turns out it could be done online up to constant factors. 00:01:36.140 --> 00:01:37.790 I sketched that last time. 00:01:37.790 --> 00:01:40.100 And this is conjectured to be within a constant factor 00:01:40.100 --> 00:01:41.060 of optimal. 00:01:41.060 --> 00:01:41.810 We can't prove it. 00:01:41.810 --> 00:01:43.851 What I'm going to show today is our best attempts 00:01:43.851 --> 00:01:45.166 at proving this is optimal. 00:01:45.166 --> 00:01:46.790 In particular, there's something called 00:01:46.790 --> 00:01:49.730 the signed greedy algorithm, which is almost the same as 00:01:49.730 --> 00:01:52.100 greedy, but it's a lower bound. 00:01:52.100 --> 00:01:54.007 And greedy is an upper bound. 00:01:54.007 --> 00:01:56.090 So all you need to do is show these two things are 00:01:56.090 --> 00:01:59.564 within a constant factor of each other and we're done. 00:01:59.564 --> 00:02:01.230 We're not going to get there, obviously, 00:02:01.230 --> 00:02:02.830 because we haven't solved that yet. 00:02:02.830 --> 00:02:05.900 But along the way, we're going to see 00:02:05.900 --> 00:02:10.639 tango trees, which achieve this log log n competitive bound. 00:02:10.639 --> 00:02:13.120 So we think greedy is constant competitive. 00:02:13.120 --> 00:02:14.840 The best we know right now is log log n, 00:02:14.840 --> 00:02:17.330 this is an improvement over red black trees, 00:02:17.330 --> 00:02:19.340 which achieve log n. 00:02:19.340 --> 00:02:21.710 Any balance binary search tree is within a log n 00:02:21.710 --> 00:02:23.030 factor of optimal. 00:02:23.030 --> 00:02:27.350 So it's all between constant and log n that we're trying to do. 00:02:27.350 --> 00:02:29.450 Another fun consequence of lower bounds 00:02:29.450 --> 00:02:31.450 is a particular sense in which log n 00:02:31.450 --> 00:02:34.350 is necessary for some access sequences. 00:02:34.350 --> 00:02:37.340 So I argued last time that if you take, like, 00:02:37.340 --> 00:02:39.650 a random access sequence, or-- 00:02:39.650 --> 00:02:42.487 for example, if you look at a binary search tree 00:02:42.487 --> 00:02:44.570 and you say, oh, I'll just access the thing that's 00:02:44.570 --> 00:02:46.160 deepest in the tree, there's always 00:02:46.160 --> 00:02:48.950 something that's deep in the tree of depth, at least log n. 00:02:48.950 --> 00:02:50.450 And so for any binary search tree 00:02:50.450 --> 00:02:51.710 there is an access sequence. 00:02:51.710 --> 00:02:53.543 No matter what that binary search tree does, 00:02:53.543 --> 00:02:56.300 I can choose the next access to force you 00:02:56.300 --> 00:02:58.520 to take log n per operation. 00:02:58.520 --> 00:03:01.370 What we're going to see today in these lower bounds 00:03:01.370 --> 00:03:05.600 is one access sequence that for all binary search trees, 00:03:05.600 --> 00:03:08.780 they must spend log n time. 00:03:08.780 --> 00:03:11.240 Just changing the quantifiers around. 00:03:11.240 --> 00:03:13.970 So instead of for every binary search tree 00:03:13.970 --> 00:03:15.470 there is an access sequence, there's 00:03:15.470 --> 00:03:17.480 going to be there is an access sequence 00:03:17.480 --> 00:03:19.070 such that for every binary search tree 00:03:19.070 --> 00:03:20.330 you need log n time. 00:03:20.330 --> 00:03:22.810 That's something we'll get easily out of these lower 00:03:22.810 --> 00:03:24.470 bounds. 00:03:24.470 --> 00:03:26.660 So let's jump into the lower bounds. 00:03:26.660 --> 00:03:29.620 And we're going to cover you could say three different lower 00:03:29.620 --> 00:03:30.120 bounds. 00:03:30.120 --> 00:03:32.930 The independent rectangles is kind of a generic class 00:03:32.930 --> 00:03:34.250 of lower bounds. 00:03:34.250 --> 00:03:36.260 Then we're going to see two specific choices 00:03:36.260 --> 00:03:38.810 of these independent rectangles, which are actually 00:03:38.810 --> 00:03:41.480 older than this result. So this is 00:03:41.480 --> 00:03:44.450 sort of a modern interpretation of two older results 00:03:44.450 --> 00:03:47.060 and a more general result. 00:03:47.060 --> 00:03:50.210 Signed greedy is going to turn out to be the best lower bound. 00:03:50.210 --> 00:03:52.790 Is It's better than all the ones that we will cover, 00:03:52.790 --> 00:03:56.630 but each of them has their own uses for analysis. 00:03:56.630 --> 00:03:59.300 Each of them is going to let us analyze an algorithm 00:03:59.300 --> 00:04:01.340 that we couldn't or that we don't otherwise 00:04:01.340 --> 00:04:04.160 know how to analyze. 00:04:04.160 --> 00:04:08.600 So let's do the independent rectangle lower bound. 00:04:08.600 --> 00:04:09.700 The sort of generic one. 00:04:25.540 --> 00:04:28.690 So these lower bounds are all going 00:04:28.690 --> 00:04:31.780 to refer to the original point set, 00:04:31.780 --> 00:04:34.720 the white dots, the accesses. 00:04:34.720 --> 00:04:37.330 The idea is you're given an access sequence, a sequence x 00:04:37.330 --> 00:04:41.170 i-- x 1 up to x n, and you want to know some lower bound 00:04:41.170 --> 00:04:43.690 that every binary search tree requires 00:04:43.690 --> 00:04:46.930 a certain number of accesses, a certain number of node touches 00:04:46.930 --> 00:04:47.980 for that access sequence. 00:04:47.980 --> 00:04:49.021 You know it's at least n. 00:04:49.021 --> 00:04:50.910 You want something bigger than n. 00:04:50.910 --> 00:04:54.475 We've got to at least touch the nodes that are being accessed. 00:04:58.983 --> 00:05:00.370 I'm going to drop this. 00:05:15.520 --> 00:05:19.090 So I want the notion of independent rectangles. 00:05:19.090 --> 00:05:24.400 And general idea of dependent rectangles 00:05:24.400 --> 00:05:27.340 would be something like this. 00:05:30.328 --> 00:05:31.300 Ah, I see. 00:05:42.610 --> 00:05:44.170 So these are two rectangles. 00:05:44.170 --> 00:05:46.570 I consider them dependent because one of the corners 00:05:46.570 --> 00:05:48.900 is inside the other rectangle. 00:05:48.900 --> 00:05:51.320 This is true no matter where the points are. 00:05:51.320 --> 00:05:59.750 So, for example, if I take two points, they span a rectangle. 00:05:59.750 --> 00:06:04.660 If I take these two points, for example, they span a rectangle. 00:06:04.660 --> 00:06:06.180 This corner is inside that one. 00:06:06.180 --> 00:06:08.440 So these are considered dependent rectangles 00:06:08.440 --> 00:06:11.127 in either case. 00:06:11.127 --> 00:06:13.210 So corner here does not necessarily mean a point-- 00:06:13.210 --> 00:06:14.782 any of the four corners. 00:06:14.782 --> 00:06:16.240 Rectangle is defined by two points, 00:06:16.240 --> 00:06:19.960 but it has all four corners. 00:06:19.960 --> 00:06:23.380 And so, in particular, independent rectangles-- 00:06:23.380 --> 00:06:26.080 for example, they might be completely disjoint. 00:06:26.080 --> 00:06:28.700 Those are going to be independent. 00:06:28.700 --> 00:06:31.700 Something like that is independent. 00:06:31.700 --> 00:06:33.140 But there are some other cases. 00:06:33.140 --> 00:06:37.180 You can have rectangles that look like this. 00:06:37.180 --> 00:06:38.110 OK? 00:06:38.110 --> 00:06:39.901 And it doesn't matter where the points are. 00:06:39.901 --> 00:06:42.000 Maybe here, here, here, and here. 00:06:42.000 --> 00:06:44.210 Or the other way. 00:06:44.210 --> 00:06:46.990 These are independent. 00:06:46.990 --> 00:06:52.960 And there's one other kind of special case, which maybe I'll 00:06:52.960 --> 00:06:56.680 use color to draw the other one because they're 00:06:56.680 --> 00:06:57.970 right on top of each other. 00:07:03.405 --> 00:07:10.000 So I've got a point here, a point here, and a point here. 00:07:10.000 --> 00:07:13.810 These are two rectangles defined on three points. 00:07:13.810 --> 00:07:16.990 So they both use this point. 00:07:16.990 --> 00:07:19.930 And if you check, it does satisfy this condition. 00:07:19.930 --> 00:07:23.260 So no corner strictly inside the other. 00:07:23.260 --> 00:07:25.660 But we also need that the rectangles are unsatisfied. 00:07:25.660 --> 00:07:28.660 So this is saying that there's no other point even 00:07:28.660 --> 00:07:30.410 on the boundary of the rectangle. 00:07:30.410 --> 00:07:34.360 So this part says, OK, there's nothing strictly inside. 00:07:34.360 --> 00:07:36.610 But we also need that on the boundary 00:07:36.610 --> 00:07:38.060 there's no other points. 00:07:38.060 --> 00:07:40.060 So this is the only sort of situation other 00:07:40.060 --> 00:07:44.200 than reflections where you get this working 00:07:44.200 --> 00:07:45.840 out as independent. 00:07:50.632 --> 00:07:52.090 AUDIENCE: Last case is independent? 00:07:52.090 --> 00:07:53.730 ERIK DEMAINE: Last case is independent. 00:07:56.730 --> 00:07:58.800 All right? 00:07:58.800 --> 00:08:00.470 So this is a definition. 00:08:00.470 --> 00:08:02.470 If I give you a set of rectangles, 00:08:02.470 --> 00:08:05.070 they're independent. 00:08:05.070 --> 00:08:06.930 I mean, I was looking at pairwise. 00:08:06.930 --> 00:08:08.640 But if they're are pairwise independent, 00:08:08.640 --> 00:08:10.710 then they will be independent. 00:08:10.710 --> 00:08:12.270 No corner of any rectangle strictly 00:08:12.270 --> 00:08:14.590 inside any other rectangle. 00:08:14.590 --> 00:08:17.940 And there's no points of those rectangles 00:08:17.940 --> 00:08:21.360 that are inside others. 00:08:21.360 --> 00:08:22.166 OK. 00:08:22.166 --> 00:08:22.665 Cool. 00:08:25.390 --> 00:08:27.790 So what? 00:08:32.350 --> 00:08:36.640 Lower bound says the optimal offline binary search 00:08:36.640 --> 00:08:41.799 tree, or the optimal way to add dots to satisfy your point set, 00:08:41.799 --> 00:08:45.490 is going to be at least the size of the input-- 00:08:45.490 --> 00:08:48.340 meaning the number of initial points you have-- 00:08:48.340 --> 00:08:53.170 plus half the maximum number of independent rectangles. 00:09:02.591 --> 00:09:03.090 OK. 00:09:03.090 --> 00:09:05.460 So this is a max independence set problem. 00:09:05.460 --> 00:09:07.350 In general, that's NP-complete. 00:09:07.350 --> 00:09:10.260 Turns out we'll be able to at least approximate the number 00:09:10.260 --> 00:09:12.990 of independent rectangles within a constant factor 00:09:12.990 --> 00:09:14.560 by the end of class. 00:09:14.560 --> 00:09:16.220 That's going to be signed greedy. 00:09:16.220 --> 00:09:18.180 So signed greedy is going to be the best 00:09:18.180 --> 00:09:21.270 way up to constant factors to choose independent rectangles. 00:09:21.270 --> 00:09:24.780 For now, someone magically tells you 00:09:24.780 --> 00:09:28.770 what's the best way or you just choose some reasonable-- 00:09:28.770 --> 00:09:30.534 any choice of independent rectangles 00:09:30.534 --> 00:09:31.450 will be a lower bound. 00:09:31.450 --> 00:09:34.600 But you get the best lower bound by choosing the max. 00:09:34.600 --> 00:09:35.100 OK? 00:09:35.100 --> 00:09:38.400 So we're going to prove this theorem, 00:09:38.400 --> 00:09:43.432 and then we're going to see three different ways to choose 00:09:43.432 --> 00:09:44.640 those independent rectangles. 00:09:44.640 --> 00:09:46.770 And we'll use them for various things. 00:09:46.770 --> 00:09:48.780 Wilber 1, Wilber, 2 and signed greedy 00:09:48.780 --> 00:09:51.090 are going to be the three choices 00:09:51.090 --> 00:09:53.530 for independent rectangles. 00:09:53.530 --> 00:09:54.030 All right. 00:09:54.030 --> 00:09:57.660 To prove this theorem, we're going to change it a little bit 00:09:57.660 --> 00:09:59.850 first. 00:09:59.850 --> 00:10:02.700 And this is kind of the focus of today-- 00:10:02.700 --> 00:10:04.890 is the idea of signed rectangles. 00:10:09.090 --> 00:10:11.580 If you look at the rectangles in the world spanned 00:10:11.580 --> 00:10:14.800 by two points, there are two different kinds. 00:10:14.800 --> 00:10:20.460 There's the top right, lower left kind. 00:10:20.460 --> 00:10:23.910 And then there's the top left, lower right kind. 00:10:23.910 --> 00:10:26.610 These are positive slope or negative slope. 00:10:29.340 --> 00:10:31.590 Those are the two kinds of rectangles. 00:10:31.590 --> 00:10:36.630 And it's helpful to think about just the positive rectangles 00:10:36.630 --> 00:10:39.690 or the slash rectangles and just the backslash rectangles 00:10:39.690 --> 00:10:41.380 separately. 00:10:41.380 --> 00:10:43.920 So we're going to call a point set-- 00:10:48.090 --> 00:10:50.400 it's a little hard to pronounce-- 00:10:50.400 --> 00:10:53.820 we used to call this plus satisfied. 00:10:53.820 --> 00:10:55.620 So maybe it's easiest to pronounce it 00:10:55.620 --> 00:11:00.990 that way, the symbol formerly known as plus satisfied, 00:11:00.990 --> 00:11:11.370 if all plus rectangles that are not 00:11:11.370 --> 00:11:15.000 on a horizontal or vertical line contain another point. 00:11:23.360 --> 00:11:27.390 So a point set is arborally satisfied 00:11:27.390 --> 00:11:31.310 if and only if it is plus satisfied and minus satisfied-- 00:11:31.310 --> 00:11:33.787 just breaking apart that definition into two parts. 00:11:33.787 --> 00:11:36.120 But now, we're going to look at point sets that are just 00:11:36.120 --> 00:11:39.300 plus satisfied or point sets that are just minus satisfied. 00:11:39.300 --> 00:11:42.630 And then we can look at the optimal solution 00:11:42.630 --> 00:11:46.960 if you only care about plus rectangles. 00:11:46.960 --> 00:12:03.840 So this is the smallest plus satisfied point set 00:12:03.840 --> 00:12:08.776 containing all the access points, all the given points. 00:12:08.776 --> 00:12:12.470 So we'll call that the input. 00:12:12.470 --> 00:12:15.090 OPT was the smallest arborally satisfied. 00:12:15.090 --> 00:12:17.340 OPT plus is the-- 00:12:17.340 --> 00:12:19.483 you just look at plus rectangles. 00:12:24.351 --> 00:12:24.850 OK. 00:12:24.850 --> 00:12:26.570 Why are we doing this? 00:12:26.570 --> 00:12:29.030 Well, for now, we're going to do it to prove this theorem. 00:12:29.030 --> 00:12:35.830 So lemma, which was what we're actually going to prove, 00:12:35.830 --> 00:12:38.650 is if you look at this OPT plus thing, 00:12:38.650 --> 00:12:41.010 it's got to be at least the size of the input-- 00:12:41.010 --> 00:12:44.620 everything has to at least contain the input-- 00:12:44.620 --> 00:12:54.460 plus maximum number of independent plus rectangles. 00:13:02.260 --> 00:13:04.680 So this is where we're actually going to prove. 00:13:04.680 --> 00:13:06.180 If you want to get plus satisfied 00:13:06.180 --> 00:13:09.240 and you've got k independent plus rectangles, 00:13:09.240 --> 00:13:11.050 you need to add at least that many points-- 00:13:11.050 --> 00:13:14.850 so at least one point per plus rectangle. 00:13:14.850 --> 00:13:16.830 If you can prove this, you prove the theorem 00:13:16.830 --> 00:13:18.900 because this holds for minus just 00:13:18.900 --> 00:13:21.120 as well as plus by symmetry. 00:13:21.120 --> 00:13:26.400 And so you take your maximum independent set of rectangles. 00:13:26.400 --> 00:13:28.830 At least half of them are plus or at least half of them 00:13:28.830 --> 00:13:30.000 are minus. 00:13:30.000 --> 00:13:34.290 You apply this bound, and that's where you get the 1/2 here. 00:13:34.290 --> 00:13:38.182 So this is stronger, I guess, than the theorem, 00:13:38.182 --> 00:13:40.140 and this is what we're actually going to prove. 00:13:40.140 --> 00:13:41.850 And so, in this world, we just are 00:13:41.850 --> 00:13:44.790 thinking about plus rectangles, which is a little weird. 00:13:44.790 --> 00:13:47.340 But it works. 00:13:51.120 --> 00:13:54.366 And the proof is going to be in three steps. 00:13:54.366 --> 00:13:56.760 I'm first going to give you an overview of the steps, 00:13:56.760 --> 00:13:58.470 and then we'll actually do them. 00:13:58.470 --> 00:14:03.030 So this is like a two-level proof. 00:14:03.030 --> 00:14:07.470 First thing we do, the top level, 00:14:07.470 --> 00:14:13.590 is we're going to find a rectangle 00:14:13.590 --> 00:14:20.370 in the independent set, and we're 00:14:20.370 --> 00:14:27.540 going to find a vertical line that hits only that rectangle. 00:14:33.730 --> 00:14:37.160 So we're going to have some rectangle 00:14:37.160 --> 00:14:43.380 in the independent set, and we want a vertical line stabbing 00:14:43.380 --> 00:14:46.470 it such that no other rectangle is 00:14:46.470 --> 00:14:48.490 stabbed by this vertical line. 00:14:48.490 --> 00:14:50.910 So all other rectangles-- that's independence, 00:14:50.910 --> 00:14:54.260 so maybe they look something like this-- 00:14:54.260 --> 00:14:55.690 but nothing like this. 00:14:59.297 --> 00:15:01.380 Not obvious that such a thing exists, but it does. 00:15:01.380 --> 00:15:03.240 Actually, not that hard to find. 00:15:03.240 --> 00:15:05.510 We just need some rectangle with some line. 00:15:09.540 --> 00:15:16.230 Then, using that property, we're going 00:15:16.230 --> 00:15:31.200 to be able to find some points in that rectangle that are also 00:15:31.200 --> 00:15:43.530 in the optimal plus solution in the rectangle 00:15:43.530 --> 00:15:44.490 crossing the line. 00:15:49.950 --> 00:15:52.770 Let me get another color. 00:15:52.770 --> 00:15:56.490 So we're going to find a point on the left 00:15:56.490 --> 00:15:59.964 of the line and a point on the right of the line. 00:15:59.964 --> 00:16:01.380 And they're horizontally adjacent, 00:16:01.380 --> 00:16:03.790 meaning there's no other point between them. 00:16:03.790 --> 00:16:07.164 So we know there's some point in this box. 00:16:07.164 --> 00:16:08.580 Because this is a plus box, it has 00:16:08.580 --> 00:16:09.750 got to be satisfied somehow. 00:16:09.750 --> 00:16:11.375 And I claim there's actually two points 00:16:11.375 --> 00:16:12.600 on either side of the line. 00:16:12.600 --> 00:16:14.992 One of them could be equal to this or this, but not both 00:16:14.992 --> 00:16:16.950 obviously because they're horizontally aligned. 00:16:21.270 --> 00:16:29.230 And then what we're going to do is charge the rectangle 00:16:29.230 --> 00:16:29.970 to those points. 00:16:34.346 --> 00:16:36.720 And then, basically, we're going to remove that rectangle 00:16:36.720 --> 00:16:38.340 and repeat. 00:16:38.340 --> 00:16:40.830 And the claim is this charging sort of only happens once 00:16:40.830 --> 00:16:41.910 per point. 00:16:41.910 --> 00:16:44.700 And therefore, the number of points in the optimal solution 00:16:44.700 --> 00:16:48.150 has to be at least the number of rectangles-- number 00:16:48.150 --> 00:16:50.940 of plus rectangles in the independent set. 00:16:50.940 --> 00:16:53.740 So, basically, this is a way of ordering the rectangles. 00:16:53.740 --> 00:16:56.640 We're going to take one that has one of these vertical lines, 00:16:56.640 --> 00:17:00.630 find two points that pay for that rectangle, 00:17:00.630 --> 00:17:03.360 and therefore argue that OPT has to be at least the number 00:17:03.360 --> 00:17:04.394 of rectangles. 00:17:04.394 --> 00:17:08.250 So we have to argue that at least one of these points 00:17:08.250 --> 00:17:11.470 is not one of the original points. 00:17:11.470 --> 00:17:15.644 And that's where we're getting the input plus this. 00:17:15.644 --> 00:17:17.310 So there's lots of things to check here. 00:17:17.310 --> 00:17:18.476 Let's do them one at a time. 00:17:21.560 --> 00:17:23.930 And throughout, I'm going to assume-- 00:17:23.930 --> 00:17:27.150 let me write that the bottom-- 00:17:27.150 --> 00:17:31.700 assume all x- and y-coordinates are unique. 00:17:39.860 --> 00:17:43.490 This is an idea I mentioned last time as well. 00:17:43.490 --> 00:17:46.890 If you have lots of accesses to the same key, 00:17:46.890 --> 00:17:50.870 imagine them being accesses to slightly different keys. 00:17:50.870 --> 00:17:53.330 Just skew them a little bit, and it doesn't 00:17:53.330 --> 00:17:56.420 change any of the bounds much. 00:17:56.420 --> 00:18:00.290 I won't to argue that here, but at the least think of this 00:18:00.290 --> 00:18:03.530 as just a simplifying assumption to make the proofs cleaner. 00:18:07.010 --> 00:18:09.230 So how are we going to do step one? 00:18:09.230 --> 00:18:12.710 I need to find some rectangle and some vertical line that 00:18:12.710 --> 00:18:16.580 only stabs that rectangle. 00:18:16.580 --> 00:18:18.560 And the way we're going to do that is just 00:18:18.560 --> 00:18:30.190 take the widest rectangle that just has the maximum x extent. 00:18:30.190 --> 00:18:34.810 There might be more than one, but just take one of them. 00:18:34.810 --> 00:18:38.534 So it's very wide. 00:18:38.534 --> 00:18:39.950 What this tells us is that there's 00:18:39.950 --> 00:18:42.710 no other rectangle like this. 00:18:42.710 --> 00:18:45.110 This would be independent, but it would be wider. 00:18:45.110 --> 00:18:46.040 So that's not allowed. 00:18:51.365 --> 00:18:53.490 Now, we have to think about all sorts of scenarios. 00:18:53.490 --> 00:18:55.990 So we've got a point here and a point here. 00:18:55.990 --> 00:18:58.960 It could still be that we have rectangles like this. 00:18:58.960 --> 00:19:00.970 They just can't go farther to the right. 00:19:00.970 --> 00:19:03.670 It could be we have rectangles that go like this-- 00:19:03.670 --> 00:19:07.354 just can't go too far to the left. 00:19:07.354 --> 00:19:08.770 These rectangles that are anchored 00:19:08.770 --> 00:19:10.330 in the lower left and these rectangles that 00:19:10.330 --> 00:19:12.610 are anchored in the upper right can't touch each other 00:19:12.610 --> 00:19:16.369 because then one of them would be satisfied. 00:19:16.369 --> 00:19:18.160 This one's going to have a point down here. 00:19:18.160 --> 00:19:20.230 This one is going to have a point here. 00:19:20.230 --> 00:19:23.200 I guess-- yeah, let's see, how would it 00:19:23.200 --> 00:19:24.560 go if they were touching? 00:19:31.110 --> 00:19:33.465 We'd have a corner-- 00:19:33.465 --> 00:19:34.910 hmm, touching is a little weird. 00:19:51.000 --> 00:19:52.000 Ah, I see. 00:19:52.000 --> 00:19:52.500 Good. 00:19:52.500 --> 00:19:53.958 This can't happen because we assume 00:19:53.958 --> 00:19:56.790 the x-coordinates are distinct. 00:19:56.790 --> 00:19:59.360 So that's why I did this. 00:19:59.360 --> 00:20:01.860 That's the reason. 00:20:01.860 --> 00:20:03.580 So this can't happen. 00:20:03.580 --> 00:20:07.410 And I also can't have them go like this because then there's 00:20:07.410 --> 00:20:12.120 a corner in the strict interior of the other rectangle. 00:20:12.120 --> 00:20:12.912 Is that clear? 00:20:12.912 --> 00:20:14.370 This rectangle can't come over here 00:20:14.370 --> 00:20:18.090 because then that would be not independent. 00:20:18.090 --> 00:20:20.160 Rectangle can't come right to the same spot 00:20:20.160 --> 00:20:21.460 because there is no same spot. 00:20:21.460 --> 00:20:24.190 That would be two points on the same vertical line. 00:20:24.190 --> 00:20:26.460 And so what we must have is a picture more 00:20:26.460 --> 00:20:30.810 like this where there's an empty region in between. 00:20:30.810 --> 00:20:34.680 that not hit by-- there can be many of these rectangles, many 00:20:34.680 --> 00:20:35.911 of these rectangles. 00:20:35.911 --> 00:20:37.410 They're independent from each other. 00:20:37.410 --> 00:20:40.620 That's like this case here. 00:20:40.620 --> 00:20:43.560 There can also be some rectangles like this. 00:20:43.560 --> 00:20:47.100 But by the same argument, these guys can't touch each other 00:20:47.100 --> 00:20:49.380 and they can't overlap horizontally 00:20:49.380 --> 00:20:52.770 because then one of the corners would be inside the other. 00:20:52.770 --> 00:20:54.080 Question? 00:20:54.080 --> 00:20:56.455 AUDIENCE: For that picture, you drew a rectangle 00:20:56.455 --> 00:20:59.282 under the other one. 00:20:59.282 --> 00:21:00.740 ERIK DEMAINE: This one or this one? 00:21:00.740 --> 00:21:01.531 AUDIENCE: That one. 00:21:01.531 --> 00:21:03.480 ERIK DEMAINE: Yeah, this one cannot happen. 00:21:03.480 --> 00:21:04.575 That's what we claim-- 00:21:04.575 --> 00:21:05.250 haha, right. 00:21:05.250 --> 00:21:07.140 So you're right. 00:21:07.140 --> 00:21:10.590 So we worry about-- 00:21:10.590 --> 00:21:11.612 interesting. 00:21:11.612 --> 00:21:13.320 Well, we worry about something like this. 00:21:16.194 --> 00:21:19.425 AUDIENCE: Sorry, why can't that happen? 00:21:19.425 --> 00:21:20.800 ERIK DEMAINE: Yeah, you're right. 00:21:20.800 --> 00:21:22.216 I actually drew the wrong picture. 00:21:22.216 --> 00:21:23.215 Sorry. 00:21:23.215 --> 00:21:23.715 Kidding. 00:21:31.080 --> 00:21:31.580 Yeah. 00:21:31.580 --> 00:21:33.250 I really meant line segment here. 00:21:33.250 --> 00:21:34.305 I'm sorry. 00:21:39.640 --> 00:21:42.020 Poor choice of wording. 00:21:42.020 --> 00:21:43.720 So vertical line is actually just going 00:21:43.720 --> 00:21:47.770 to go the extent of the rectangle-- 00:21:47.770 --> 00:21:49.510 something like this. 00:21:49.510 --> 00:21:50.230 Sorry. 00:21:50.230 --> 00:21:53.910 We can't forbid rectangles like this. 00:21:53.910 --> 00:21:57.260 What we can forbid our rectangles like this 00:21:57.260 --> 00:22:01.632 that also try to cross that segment. 00:22:01.632 --> 00:22:03.340 We'll see why this is enough in a moment. 00:22:03.340 --> 00:22:04.060 Sorry about that. 00:22:08.350 --> 00:22:12.070 I really only care about the interior of this rectangle. 00:22:12.070 --> 00:22:15.370 I'm trying to get a vertical line that only stabs 00:22:15.370 --> 00:22:17.530 this rectangle, nothing else-- 00:22:17.530 --> 00:22:19.600 inside the rectangle. 00:22:19.600 --> 00:22:20.740 Sorry, poor wording. 00:22:20.740 --> 00:22:22.674 I don't care about these guys outside 00:22:22.674 --> 00:22:24.340 because I can't say anything about them. 00:22:24.340 --> 00:22:29.275 They could be all over the place in an independent set. 00:22:29.275 --> 00:22:31.150 I mean, relative to what hits this rectangle, 00:22:31.150 --> 00:22:32.050 there's stuff on the left. 00:22:32.050 --> 00:22:33.050 There's stuff on the right. 00:22:33.050 --> 00:22:33.924 There are these guys. 00:22:33.924 --> 00:22:37.340 There can also be things like this. 00:22:37.340 --> 00:22:40.600 But still remaining are these regions 00:22:40.600 --> 00:22:42.959 which are not hit by any rectangles, 00:22:42.959 --> 00:22:44.500 and that's because what I was saying. 00:22:44.500 --> 00:22:46.050 These guys can't touch each other because then there 00:22:46.050 --> 00:22:47.680 would be equal x-coordinates. 00:22:47.680 --> 00:22:49.660 They can't overlap because then one of them 00:22:49.660 --> 00:22:52.180 would not be independent from the other. 00:22:52.180 --> 00:22:58.540 So I get my vertical lines. 00:22:58.540 --> 00:23:01.510 I just need one, but it could be any of these. 00:23:01.510 --> 00:23:06.400 In general, for example, if you take all of these lower 00:23:06.400 --> 00:23:08.470 left anchored rectangles and take just 00:23:08.470 --> 00:23:10.540 to the right of the rightmost one, 00:23:10.540 --> 00:23:13.210 that will be a valid choice for your line. 00:23:13.210 --> 00:23:17.060 Because you can argue none of these can overlap it. 00:23:17.060 --> 00:23:18.550 So that's step one. 00:23:18.550 --> 00:23:20.686 We just take a widest rectangle. 00:23:20.686 --> 00:23:22.060 The one thing we needed to forbid 00:23:22.060 --> 00:23:25.060 was something going like this all the way across. 00:23:28.120 --> 00:23:28.870 Step two. 00:23:31.822 --> 00:23:33.560 Step two is actually pretty easy. 00:23:33.560 --> 00:23:37.490 Once you've identified this red line-- 00:23:37.490 --> 00:23:40.370 inside the rectangle, you know there are some points. 00:23:40.370 --> 00:23:43.714 And I'm going to take the rightmost. 00:23:43.714 --> 00:23:45.380 And then among all the rightmost points, 00:23:45.380 --> 00:23:47.560 I'm going to take the topmost point that 00:23:47.560 --> 00:23:50.420 is to the left of the line and inside the rectangle. 00:23:50.420 --> 00:23:59.340 So let p be the topmost, leftmost point-- 00:23:59.340 --> 00:24:14.020 sorry, rightmost-- that is both in the rectangle 00:24:14.020 --> 00:24:16.210 and left of the line. 00:24:22.240 --> 00:24:28.420 Let me erase this one for a little bit more room. 00:24:28.420 --> 00:24:31.150 So I'm looking at all of this region 00:24:31.150 --> 00:24:33.370 to the left of the line in the rectangle. 00:24:33.370 --> 00:24:39.130 I want to take the rightmost and then topmost 00:24:39.130 --> 00:24:42.100 point-- something like this. 00:24:42.100 --> 00:24:43.960 How do I know such a point exists? 00:24:43.960 --> 00:24:46.640 Because this point is such a point. 00:24:46.640 --> 00:24:49.030 And this point is to the left of the line. 00:24:49.030 --> 00:24:52.510 So if there's nothing else in here, that is a valid choice. 00:24:52.510 --> 00:24:54.950 But in general, we go to the right as much as possible. 00:24:54.950 --> 00:24:56.390 Then we go up as much as possible. 00:24:56.390 --> 00:24:59.110 So that's a point, which we will call them p. 00:24:59.110 --> 00:25:03.534 AUDIENCE: Couldn't it be on the border of the rectangle? 00:25:03.534 --> 00:25:05.200 ERIK DEMAINE: It could be on the border. 00:25:05.200 --> 00:25:06.550 It could be interior. 00:25:06.550 --> 00:25:08.177 We don't know. 00:25:08.177 --> 00:25:11.890 AUDIENCE: When you said topmost, what is your topmost? 00:25:11.890 --> 00:25:15.448 ERIK DEMAINE: Topmost means of maximum y-coordinate. 00:25:15.448 --> 00:25:16.226 AUDIENCE: Oh, OK. 00:25:16.226 --> 00:25:16.850 Got it. 00:25:16.850 --> 00:25:19.490 ERIK DEMAINE: So it could be up here. 00:25:19.490 --> 00:25:21.380 We don't know. 00:25:21.380 --> 00:25:22.760 First, we go rightmost. 00:25:22.760 --> 00:25:25.019 Then, among all the things in that column, 00:25:25.019 --> 00:25:26.060 we go to the topmost one. 00:25:26.060 --> 00:25:27.140 So it might be on the top. 00:25:27.140 --> 00:25:27.806 It might not be. 00:25:30.760 --> 00:25:36.265 These are points-- sorry, this is a point in OPT plus. 00:25:40.270 --> 00:25:42.966 And then q is going to be a similar thing. 00:25:42.966 --> 00:25:52.780 It's going to be the bottom-most, leftmost point 00:25:52.780 --> 00:26:03.970 in OPT plus that is in the rectangle 00:26:03.970 --> 00:26:05.230 and right of the line. 00:26:08.112 --> 00:26:09.670 Not totally symmetric, though. 00:26:09.670 --> 00:26:12.420 We're also going to say and not below p. 00:26:16.570 --> 00:26:25.450 So now we're looking at this upper region here. 00:26:25.450 --> 00:26:28.290 Among all the things that are not below p-- 00:26:28.290 --> 00:26:30.280 should have drawn this more horizontal-- 00:26:33.430 --> 00:26:35.960 and to the right of the red line-- 00:26:35.960 --> 00:26:38.960 so that's up to here, I guess-- 00:26:38.960 --> 00:26:41.662 I want to take the leftmost column that 00:26:41.662 --> 00:26:43.370 has any points in it and then among those 00:26:43.370 --> 00:26:47.370 take the bottom-most point in the column. 00:26:47.370 --> 00:26:50.539 I claim that's actually going to be on this line. 00:26:50.539 --> 00:26:52.580 First thing to check is that such a point exists. 00:26:52.580 --> 00:26:54.980 Such a point exists because, in particular, this 00:26:54.980 --> 00:26:56.689 is such a point. 00:26:56.689 --> 00:26:57.980 It is to the right of the line. 00:26:57.980 --> 00:27:00.800 It's above the blue line, right of the red line, 00:27:00.800 --> 00:27:02.660 in the rectangle. 00:27:02.660 --> 00:27:05.350 But I claim that if we take the leftmost, bottom-most one, then 00:27:05.350 --> 00:27:06.600 they must actually be aligned. 00:27:06.600 --> 00:27:08.010 Why? 00:27:08.010 --> 00:27:10.520 So if it was somewhere else, like up here 00:27:10.520 --> 00:27:16.590 or like this point, then I claim that is an unsatisfied box. 00:27:16.590 --> 00:27:24.180 Let me draw that picture, make a little clearer. 00:27:24.180 --> 00:27:27.240 So something like this. 00:27:30.760 --> 00:27:39.540 So we've got our red line and we've got this picture. 00:27:43.090 --> 00:27:54.000 This is p, and then actually we don't know anything 00:27:54.000 --> 00:27:55.095 about down here. 00:27:59.770 --> 00:28:01.990 This is q. 00:28:01.990 --> 00:28:06.230 I claim that these black regions cannot have any points in them 00:28:06.230 --> 00:28:09.010 because, by definition, p was in the rightmost column. 00:28:09.010 --> 00:28:11.677 So there's nothing in this strip and between p and the red line. 00:28:11.677 --> 00:28:14.218 And it was the topmost within the column, which means there's 00:28:14.218 --> 00:28:15.520 nothing above p in the column. 00:28:15.520 --> 00:28:17.620 So that's why all the points are confined 00:28:17.620 --> 00:28:19.300 to this region over here. 00:28:19.300 --> 00:28:22.360 Similarly, for q, if you look at the things that are above 00:28:22.360 --> 00:28:24.640 or on this horizontal line, which 00:28:24.640 --> 00:28:28.630 was the blue line over there, then we 00:28:28.630 --> 00:28:33.320 know that there's nothing in this strip in between 00:28:33.320 --> 00:28:34.330 because q is leftmost. 00:28:34.330 --> 00:28:36.163 And then among leftmost, it was bottom-most, 00:28:36.163 --> 00:28:37.630 so there's nothing down here. 00:28:37.630 --> 00:28:40.360 So that means, if these guys are not horizontally aligned, 00:28:40.360 --> 00:28:42.840 there is an unsatisfied box here. 00:28:42.840 --> 00:28:43.940 Contradiction. 00:28:43.940 --> 00:28:45.170 It's a plus box. 00:28:45.170 --> 00:28:48.227 So in OPT plus, there's got to be another point, which 00:28:48.227 --> 00:28:49.060 was a contradiction. 00:28:49.060 --> 00:28:52.360 So in fact, p and q must be horizontally aligned. 00:28:52.360 --> 00:28:54.360 So that was step two. 00:28:57.040 --> 00:28:58.300 Finally, we get step three. 00:29:03.250 --> 00:29:05.290 So the idea with step three-- 00:29:05.290 --> 00:29:07.390 now we're going to do a charging argument. 00:29:07.390 --> 00:29:13.060 We want to say, OK, basically, for every independent 00:29:13.060 --> 00:29:15.370 rectangle, we want to find a point that's 00:29:15.370 --> 00:29:19.720 in OPT that was not in the original input. 00:29:19.720 --> 00:29:22.390 And therefore, then OPT plus has to be 00:29:22.390 --> 00:29:25.120 at least the size of the input plus 1 per each 00:29:25.120 --> 00:29:26.350 of these plus rectangles. 00:29:29.510 --> 00:29:32.530 So the idea is the following. 00:29:32.530 --> 00:29:33.730 Because of all this set-up-- 00:29:33.730 --> 00:29:39.340 because we made pq horizontally aligned-- 00:29:39.340 --> 00:29:40.930 they're inside the rectangle. 00:29:40.930 --> 00:29:43.030 And furthermore, they're adjacent 00:29:43.030 --> 00:29:45.040 and they cross this vertical line. 00:29:45.040 --> 00:29:48.880 And that vertical line is not crossed by any other rectangle. 00:29:48.880 --> 00:29:51.780 When I say line, I mean line segment. 00:29:51.780 --> 00:29:54.880 There's no other rectangle that hits this red thing. 00:29:54.880 --> 00:29:56.470 Therefore, these two points are not 00:29:56.470 --> 00:30:00.640 going to get charged as a pair ever again. 00:30:00.640 --> 00:30:05.260 If you remove this rectangle, repeat this process, 00:30:05.260 --> 00:30:09.070 pq is never going to get charged again. 00:30:09.070 --> 00:30:10.990 So we charge to pq. 00:30:10.990 --> 00:30:26.230 And the pair never charged again, never 00:30:26.230 --> 00:30:33.860 be charged by another rectangle because no rectangle 00:30:33.860 --> 00:30:39.520 hits the red thing. 00:30:46.600 --> 00:30:53.380 So no rectangle contains the segment pq, 00:30:53.380 --> 00:30:54.620 the horizontal segment pq. 00:30:57.600 --> 00:30:59.130 So this is almost what we want. 00:30:59.130 --> 00:31:01.900 We really want a single point which is not in the input. 00:31:01.900 --> 00:31:03.360 So we have p and q. 00:31:03.360 --> 00:31:04.680 They're horizontally aligned. 00:31:04.680 --> 00:31:07.080 Now, if they're horizontally aligned, 00:31:07.080 --> 00:31:12.090 we know that not both of them are in the original input 00:31:12.090 --> 00:31:14.564 because all y-coordinates are distinct. 00:31:14.564 --> 00:31:16.230 This is usually true because you're only 00:31:16.230 --> 00:31:19.800 accessing one point per row, per time step. 00:31:19.800 --> 00:31:23.550 So one of these might be in the input, 00:31:23.550 --> 00:31:25.170 but the other one is not. 00:31:25.170 --> 00:31:26.970 So that's the one I want to hold onto. 00:31:26.970 --> 00:31:32.040 And say, OK, that's a point added to OPT plus that pays 00:31:32.040 --> 00:31:34.295 for this rectangle. 00:31:34.295 --> 00:31:35.670 It's not quite so simple, though, 00:31:35.670 --> 00:31:37.230 because we might have a whole bunch 00:31:37.230 --> 00:31:42.060 of horizontally-aligned things. 00:31:42.060 --> 00:31:44.950 And one rectangle charges to this one. 00:31:44.950 --> 00:31:46.980 One rectangle charges to this one. 00:31:46.980 --> 00:31:48.960 One rectangle charges to this pair. 00:31:51.510 --> 00:31:54.660 That's OK, though, because here we have four points. 00:31:54.660 --> 00:31:57.180 Again, one of them could be in the input. 00:31:57.180 --> 00:31:59.160 The other three have to be added. 00:31:59.160 --> 00:32:01.720 And so you've got three rectangles, three added points, 00:32:01.720 --> 00:32:03.190 and we're happy. 00:32:03.190 --> 00:32:04.149 Question? 00:32:04.149 --> 00:32:05.940 AUDIENCE: Just to make the argument formal, 00:32:05.940 --> 00:32:11.667 wouldn't you want to say that only when your saying assume 00:32:11.667 --> 00:32:13.916 that x and y are always distinct-- but then, 00:32:13.916 --> 00:32:16.050 if you have the same either x or y-- 00:32:16.050 --> 00:32:17.490 ERIK DEMAINE: Ah, good point. 00:32:17.490 --> 00:32:21.960 So this is distinct in the input is what I meant. 00:32:21.960 --> 00:32:23.832 Obviously, in OPT, any satisfied set 00:32:23.832 --> 00:32:25.290 is not going to have this property. 00:32:25.290 --> 00:32:26.159 Yeah, good. 00:32:26.159 --> 00:32:28.200 So I want to assume x- and y-coordinates are only 00:32:28.200 --> 00:32:29.490 distinct in the input. 00:32:29.490 --> 00:32:31.020 OPT will not have that property. 00:32:31.020 --> 00:32:34.422 And that's why p and q can exist and have the same y-coordinate. 00:32:34.422 --> 00:32:35.130 Another question? 00:32:39.397 --> 00:32:40.938 AUDIENCE: Does this still [INAUDIBLE] 00:32:40.938 --> 00:32:43.116 the special case where your two points are 00:32:43.116 --> 00:32:44.932 the points of the rectangle? 00:32:44.932 --> 00:32:45.640 ERIK DEMAINE: OK. 00:32:45.640 --> 00:32:50.560 So the question is can p and q be the points of the rectangle? 00:32:50.560 --> 00:32:52.180 One of them can be. 00:32:52.180 --> 00:32:55.360 Like, p could be here, and then another point is over here. 00:32:55.360 --> 00:32:58.480 So then that will be the segment that you are using, 00:32:58.480 --> 00:33:00.770 between p and q. 00:33:00.770 --> 00:33:03.690 Or it could be q is here, and p is over here. 00:33:03.690 --> 00:33:04.875 Then that's the segment. 00:33:04.875 --> 00:33:07.000 You can't have them both equal because p and q have 00:33:07.000 --> 00:33:10.320 to be horizontally aligned and also because there's got 00:33:10.320 --> 00:33:13.750 to be another point in there. 00:33:13.750 --> 00:33:15.220 Yeah, so that should work. 00:33:15.220 --> 00:33:17.710 You have to check that this boundary case is still OK. 00:33:17.710 --> 00:33:23.350 But the claim is no other rectangle touches this red line 00:33:23.350 --> 00:33:24.825 even on the endpoint. 00:33:24.825 --> 00:33:26.200 And therefore, no other rectangle 00:33:26.200 --> 00:33:28.760 will wholly contain p and q. 00:33:28.760 --> 00:33:32.470 And so that means you're only charging to this pair once. 00:33:32.470 --> 00:33:34.280 And then this pair charging is OK 00:33:34.280 --> 00:33:38.142 because, luckily, there's three edges here, four vertices. 00:33:38.142 --> 00:33:39.850 One of those vertices we can't charge to, 00:33:39.850 --> 00:33:43.840 so there's exactly the right number of things for the edges, 00:33:43.840 --> 00:33:46.210 and we're OK. 00:33:46.210 --> 00:33:48.580 Yeah, this can really happen. 00:33:48.580 --> 00:33:54.420 In fact our favorite example of the pinwheel-- 00:33:54.420 --> 00:33:58.903 if instead of doing the greedy addition, we do this addition-- 00:34:03.740 --> 00:34:07.120 these are supposed to be horizontally aligned. 00:34:07.120 --> 00:34:09.770 A little hard without a grid-- 00:34:09.770 --> 00:34:12.350 a graph blackboard would a great. 00:34:12.350 --> 00:34:14.870 So this is not quite satisfied. 00:34:14.870 --> 00:34:19.969 You've got to add some more points here or something. 00:34:19.969 --> 00:34:22.380 But it has the feature that-- 00:34:22.380 --> 00:34:24.560 here's an independent set of rectangles. 00:34:24.560 --> 00:34:33.199 I can do this one, this one, and this one. 00:34:36.030 --> 00:34:38.330 So this is three independent rectangles. 00:34:38.330 --> 00:34:42.852 As the white points go, they're independent rectangles. 00:34:42.852 --> 00:34:44.810 The corners are not strictly inside each other, 00:34:44.810 --> 00:34:46.393 and none of the white points satisfies 00:34:46.393 --> 00:34:49.590 any of the other rectangles. 00:34:49.590 --> 00:34:53.132 And indeed, if you applied this argument, 00:34:53.132 --> 00:34:54.590 first you take the widest rectangle 00:34:54.590 --> 00:34:57.570 and say, OK, here is my vertical red segment. 00:34:57.570 --> 00:35:00.597 I'm going to charge to these two guys, this segment, 00:35:00.597 --> 00:35:02.930 and then eventually this guy will charge to this segment 00:35:02.930 --> 00:35:05.480 and this guy will charge to this segment. 00:35:05.480 --> 00:35:07.460 And luckily, there are three added points 00:35:07.460 --> 00:35:11.140 for exactly the three segments for the three rectangles. 00:35:11.140 --> 00:35:12.390 There had to be another point. 00:35:15.080 --> 00:35:16.140 So that's a lower bound. 00:35:21.453 --> 00:35:22.930 A lot of work-- 00:35:22.930 --> 00:35:25.060 but in the end, it says, look, just 00:35:25.060 --> 00:35:28.930 find an independent set of plus boxes, plus rectangles. 00:35:28.930 --> 00:35:30.350 That's a lower bound on OPT. 00:35:30.350 --> 00:35:32.620 So now, the question remains, how 00:35:32.620 --> 00:35:36.340 do we find a good independent set of plus boxes? 00:35:36.340 --> 00:35:38.500 And now we'll go through the three different ways 00:35:38.500 --> 00:35:40.230 we know how to do it. 00:35:40.230 --> 00:35:42.580 I'll start actually with Wilber 2. 00:35:42.580 --> 00:35:45.220 It's called Wilber 2 because it was in a paper by Wilber, 00:35:45.220 --> 00:35:49.090 and I think he called it lower bound number 1 and lower bound 00:35:49.090 --> 00:35:50.410 number 2. 00:35:50.410 --> 00:35:53.620 But for pragmatic reasons, I'm going to start with number 2. 00:36:00.670 --> 00:36:03.460 It's from 1989, so it's actually an old paper. 00:36:03.460 --> 00:36:06.410 And it was sort of lost for a long time. 00:36:06.410 --> 00:36:09.540 I don't think Wilber wrote any other papers. 00:36:09.540 --> 00:36:11.755 It was in SICOMP, a big journal. 00:36:14.372 --> 00:36:18.400 so a few years after splay trees and then 00:36:18.400 --> 00:36:21.610 sort of rediscovered in the early 2000s 00:36:21.610 --> 00:36:25.060 and turns out to be really useful for a lot of theorems. 00:36:25.060 --> 00:36:27.970 So here's the lower bound. 00:36:27.970 --> 00:36:29.590 Again, we're looking at the input 00:36:29.590 --> 00:36:33.970 point set-- no added points, just the original points. 00:36:33.970 --> 00:36:37.450 Look at every point, and look at all the points 00:36:37.450 --> 00:36:40.970 that you can see from this point downward. 00:36:40.970 --> 00:36:41.830 What does see mean? 00:36:41.830 --> 00:36:46.210 I'm interested in points below p that when 00:36:46.210 --> 00:36:49.910 I draw the rectangle contain no other points. 00:36:49.910 --> 00:36:53.740 So this is sort of like a lower envelope. 00:36:53.740 --> 00:36:56.770 It's going to look something like this-- 00:36:56.770 --> 00:37:02.180 and maybe some points like this. 00:37:02.180 --> 00:37:05.950 So all of these rectangles have to be empty. 00:37:23.540 --> 00:37:37.670 So these are the downward visible points from p. 00:37:41.330 --> 00:37:45.680 And now, among these points, you can sort them by y-coordinate. 00:37:45.680 --> 00:37:48.590 And I want to see how many times do 00:37:48.590 --> 00:37:51.630 they cross this vertical line. 00:37:51.630 --> 00:37:54.560 So if I order them by y-coordinate-- 00:37:57.590 --> 00:37:59.780 so I start here, and maybe I go to here. 00:37:59.780 --> 00:38:03.470 Then the next one is over here, so that's across. 00:38:03.470 --> 00:38:04.850 Then I go over here. 00:38:04.850 --> 00:38:05.520 Then I cross. 00:38:05.520 --> 00:38:06.710 Then I go here. 00:38:06.710 --> 00:38:08.250 Then I cross. 00:38:08.250 --> 00:38:10.670 Go here, here, cross. 00:38:10.670 --> 00:38:12.350 So if I visit them in order, I want 00:38:12.350 --> 00:38:15.170 to know how many times do I cross this vertical line. 00:38:19.320 --> 00:38:23.330 So this is the past of p, all of the accesses before p. 00:38:23.330 --> 00:38:25.880 Think of this is how many times you alternate 00:38:25.880 --> 00:38:27.985 between accessing on the left of the line 00:38:27.985 --> 00:38:29.610 and accessing on the right of the line. 00:38:33.594 --> 00:38:44.270 So count number of alternations left or right of p. 00:38:44.270 --> 00:38:46.412 And again, if we assume that no key is ever 00:38:46.412 --> 00:38:47.870 accessed more than once, then there 00:38:47.870 --> 00:38:49.910 will always be left or right, never exactly on. 00:38:54.931 --> 00:38:56.555 And then I want to sum over all points. 00:38:59.820 --> 00:39:03.950 And I claim this is a lower bound. 00:39:03.950 --> 00:39:07.550 Why is it a lower bound? 00:39:07.550 --> 00:39:13.580 Essentially, I take each of these red lines that 00:39:13.580 --> 00:39:20.005 cross the p vertical line and I turn them into a box. 00:39:20.005 --> 00:39:28.360 So there's one there, one there, one there, and one there. 00:39:28.360 --> 00:39:33.597 I claim if I do this for all p, all those boxes 00:39:33.597 --> 00:39:34.430 will be independent. 00:39:34.430 --> 00:39:36.805 All those rectangles will be independent from each other. 00:39:36.805 --> 00:39:41.810 I won't prove that formally here, but you can check it. 00:39:41.810 --> 00:39:44.240 So it's obvious for one p because each of these 00:39:44.240 --> 00:39:46.250 has a different vertical span. 00:39:46.250 --> 00:39:49.270 If you do it for all p-- all the points p-- 00:39:49.270 --> 00:39:51.980 these won't conflict. 00:39:51.980 --> 00:39:55.560 So by the independent rectangle lower bound, 00:39:55.560 --> 00:40:02.490 this is a lower bound on OPT up to a factor of 2. 00:40:02.490 --> 00:40:04.950 So what? 00:40:04.950 --> 00:40:07.080 Wilber 2 is quite interesting. 00:40:07.080 --> 00:40:08.520 For a long time, we've conjectured 00:40:08.520 --> 00:40:11.060 that it is the right answer. 00:40:11.060 --> 00:40:16.620 So conjecture-- I know it's a weird lower 00:40:16.620 --> 00:40:18.290 bound to even think of. 00:40:18.290 --> 00:40:20.850 It's a very hard paper to read. 00:40:20.850 --> 00:40:22.950 Without the geometric view, it's even harder 00:40:22.950 --> 00:40:26.800 to imagine the definition of this bound. 00:40:26.800 --> 00:40:28.400 It's sort of an algorithm. 00:40:28.400 --> 00:40:29.955 It's a way to assign boxes. 00:40:29.955 --> 00:40:31.080 It gives you a lower bound. 00:40:31.080 --> 00:40:32.810 It's a little weird. 00:40:32.810 --> 00:40:34.560 We conjecture that it's proportional 00:40:34.560 --> 00:40:37.540 to the optimal solution. 00:40:37.540 --> 00:40:38.550 We can't prove it. 00:40:38.550 --> 00:40:40.440 We've tried many times. 00:40:40.440 --> 00:40:47.630 It's a pain to work with, but it is what it is. 00:40:47.630 --> 00:40:49.340 There's one theorem that uses it, 00:40:49.340 --> 00:40:51.150 so I want to tell you about that theorem. 00:40:51.150 --> 00:40:54.200 But I don't want to go into it in too much detail. 00:40:54.200 --> 00:40:56.660 It's a neat theorem. 00:40:56.660 --> 00:41:02.900 And it's in a paper by Iacono, 2002. 00:41:02.900 --> 00:41:05.900 And it was the first paper to revitalize the Wilber stuff. 00:41:05.900 --> 00:41:08.390 So it's like, hey, there's this Wilber 2 bound. 00:41:08.390 --> 00:41:10.820 We can use it to solve a new problem, which is called 00:41:10.820 --> 00:41:12.905 key independent optimality. 00:41:23.960 --> 00:41:27.100 Briefly, the idea with key independent optimality 00:41:27.100 --> 00:41:29.640 is, suppose you've heard about dynamic optimality. 00:41:29.640 --> 00:41:32.510 You know, it's really cool because splay trees and whatnot 00:41:32.510 --> 00:41:34.940 seem to really adapt to whatever your inputs are. 00:41:34.940 --> 00:41:37.250 But suppose your inputs really don't have keys. 00:41:37.250 --> 00:41:42.140 They're just arbitrary objects labeled however, just randomly. 00:41:42.140 --> 00:41:44.570 In fact, let's assume that they're labeled randomly. 00:41:44.570 --> 00:41:46.340 Suppose the keys are generated randomly 00:41:46.340 --> 00:41:48.797 because they're meaningless or just arbitrary things. 00:41:48.797 --> 00:41:50.630 So you figure, oh, maybe I'll make it better 00:41:50.630 --> 00:41:54.560 and just randomize them completely. 00:41:54.560 --> 00:42:06.380 If keys are random, then dynamic OPT 00:42:06.380 --> 00:42:09.481 is the same thing up to constant factors as the working set 00:42:09.481 --> 00:42:09.980 bound. 00:42:15.830 --> 00:42:17.390 That's the theorem. 00:42:17.390 --> 00:42:20.120 So this is cool because it means splay trees are actually 00:42:20.120 --> 00:42:22.070 optimal in the setting where keys are random. 00:42:25.580 --> 00:42:30.200 This is in expectation over the randomized keys. 00:42:30.200 --> 00:42:34.070 And the way this theorem is proved is basically-- 00:42:34.070 --> 00:42:36.890 so what this is saying is, if we take a point set-- 00:42:36.890 --> 00:42:39.290 arbitrary point set-- but then we re-randomize 00:42:39.290 --> 00:42:43.350 the x-coordinates-- leave the y-coordinates as they are-- 00:42:43.350 --> 00:42:48.106 then you can compute how Wilber 2 behaves. 00:42:48.106 --> 00:42:49.730 Because now you have a bunch of points, 00:42:49.730 --> 00:42:53.940 and you're randomly shifting their x-coordinate. 00:42:53.940 --> 00:42:58.070 So it's like if you're randomly bouncing around an x 00:42:58.070 --> 00:43:00.380 and you're interested in this envelope 00:43:00.380 --> 00:43:02.720 on the left and the right, you want to know basically 00:43:02.720 --> 00:43:03.605 how many times-- 00:43:08.030 --> 00:43:11.950 I guess since I last accessed p, which is here. 00:43:11.950 --> 00:43:14.330 We didn't do that here, but in the working set bound 00:43:14.330 --> 00:43:16.280 that's part of the deal. 00:43:20.960 --> 00:43:23.000 If you look on the left side, it's 00:43:23.000 --> 00:43:26.060 like how many times does the max change. 00:43:26.060 --> 00:43:28.520 And you may know if you n random numbers 00:43:28.520 --> 00:43:33.530 and you want to know how many times does the max changes 00:43:33.530 --> 00:43:35.390 as I go left to right, as I take larger 00:43:35.390 --> 00:43:37.130 and larger prefixes of those n numbers, 00:43:37.130 --> 00:43:40.610 the answer is log n in expectation. 00:43:40.610 --> 00:43:43.040 Because the more points you have, the less and less likely 00:43:43.040 --> 00:43:46.240 it is for the max to change. 00:43:46.240 --> 00:43:51.880 So basically, you show the expected Wilber 00:43:51.880 --> 00:43:57.200 2 of a point over this randomization 00:43:57.200 --> 00:44:03.750 is theta log ti, where ti is the working set bound. 00:44:03.750 --> 00:44:08.260 And so, that gives you the theorem. 00:44:08.260 --> 00:44:10.664 This gives you a lower bound of the working set bound. 00:44:10.664 --> 00:44:12.580 We have upper bounds of the working set bound, 00:44:12.580 --> 00:44:15.190 and therefore that's OPT. 00:44:15.190 --> 00:44:17.911 So that's just a very quick sketch. 00:44:17.911 --> 00:44:19.660 If you're interested, check out the paper. 00:44:22.230 --> 00:44:25.865 That's unfortunately all we know what to do with Wilber 2. 00:44:25.865 --> 00:44:27.490 But there's this other bound, Wilber 1, 00:44:27.490 --> 00:44:33.030 which seems less good yet we can do a lot more with it. 00:44:33.030 --> 00:44:35.140 So let me go to that. 00:44:54.470 --> 00:44:57.220 It's a lot easier to analyze algorithms with respect 00:44:57.220 --> 00:45:00.440 to Wilber 1. 00:45:00.440 --> 00:45:01.230 What's Wilber 1? 00:45:04.080 --> 00:45:09.870 We're going to fix something called a lower bound tree. 00:45:09.870 --> 00:45:12.660 I'm going to call it because it's basically 00:45:12.660 --> 00:45:16.515 going to be a perfect binary tree on my keys. 00:45:19.200 --> 00:45:20.740 This tree never changes. 00:45:20.740 --> 00:45:22.320 That's why I say fix. 00:45:22.320 --> 00:45:25.250 It is not the binary search tree you're looking for. 00:45:25.250 --> 00:45:29.040 It is not the binary search tree that you're interested in. 00:45:29.040 --> 00:45:30.540 It's just a thing to think about. 00:45:33.930 --> 00:45:41.400 Now, for each node of that tree-- 00:45:41.400 --> 00:45:44.700 let's look at this node, I'll give the node a name, y. 00:45:47.880 --> 00:45:48.720 So here's y. 00:45:52.000 --> 00:45:53.960 There's the left subtree of y, and there's 00:45:53.960 --> 00:45:56.360 the right subtree of y. 00:45:56.360 --> 00:45:57.766 These are a bunch of keys. 00:45:57.766 --> 00:45:59.390 There's keys that are to the left of y. 00:45:59.390 --> 00:46:01.100 There's keys to the right of y. 00:46:01.100 --> 00:46:02.840 There's keys outside the subtree. 00:46:02.840 --> 00:46:04.880 We're going to ignore those. 00:46:04.880 --> 00:46:08.030 I want to look at the accesses to these keys and accesses 00:46:08.030 --> 00:46:10.220 to these keys and see how many times do I 00:46:10.220 --> 00:46:13.040 switch between left and right. 00:46:13.040 --> 00:46:19.730 So count the number of alternations-- 00:46:19.730 --> 00:46:22.887 so very similar in spirit to Wilber 2, 00:46:22.887 --> 00:46:24.470 it's just relative to this weird tree, 00:46:24.470 --> 00:46:26.045 which is kind of arbitrary-- 00:46:29.090 --> 00:46:34.880 in the access sequence-- which is x1 up to xn-- 00:46:34.880 --> 00:46:48.020 between left and right subtrees of y 00:46:48.020 --> 00:46:50.240 So we're going to ignore accesses to y itself. 00:46:50.240 --> 00:46:53.240 We're going to ignore accesses to keys outside of y. 00:46:53.240 --> 00:46:56.990 Just look at how many times do I switch between x and y. 00:46:56.990 --> 00:46:58.322 That's a lower bound. 00:46:58.322 --> 00:46:59.030 That's the claim. 00:47:09.550 --> 00:47:11.200 It's a lower bound for the same reason 00:47:11.200 --> 00:47:14.800 we use the independent rectangle lower bound. 00:47:14.800 --> 00:47:17.710 And the claim is, if you look at these alternations, 00:47:17.710 --> 00:47:20.370 draw the corresponding rectangles-- 00:47:26.170 --> 00:47:27.820 so over here, we had a vertical line 00:47:27.820 --> 00:47:30.820 which corresponded to the key, and we see how many times do we 00:47:30.820 --> 00:47:32.020 cross the line. 00:47:32.020 --> 00:47:38.874 Basically, the same thing over here except now 00:47:38.874 --> 00:47:41.290 there's one big vertical line that corresponds to the root 00:47:41.290 --> 00:47:43.330 node, then there's some vertical lines that 00:47:43.330 --> 00:47:45.670 correspond to this node and this node, 00:47:45.670 --> 00:47:48.350 and you're interested in the access sequence. 00:47:48.350 --> 00:47:51.591 How many times-- let's do some kind of access sequence like 00:47:51.591 --> 00:47:52.090 this-- 00:47:55.890 --> 00:47:56.920 these are our points-- 00:48:02.590 --> 00:48:06.220 and you just look at what lines are you crossing. 00:48:08.830 --> 00:48:11.127 So like this crosses the big line. 00:48:11.127 --> 00:48:13.210 So that's going to be one alternation between left 00:48:13.210 --> 00:48:14.157 and right here. 00:48:14.157 --> 00:48:16.240 Here's another alternation between left and right. 00:48:16.240 --> 00:48:18.820 Here is another alternation between left and right. 00:48:18.820 --> 00:48:22.750 Here's another alternation between left and right. 00:48:22.750 --> 00:48:24.880 And one more. 00:48:24.880 --> 00:48:27.280 So for the big line, for the root node, 00:48:27.280 --> 00:48:29.530 that's how many times you cross between left and right 00:48:29.530 --> 00:48:31.750 relative to the root. 00:48:31.750 --> 00:48:34.140 Then, for the left subtree the root, 00:48:34.140 --> 00:48:36.120 there's one crossing here. 00:48:36.120 --> 00:48:40.060 There is one crossing here, one crossing here. 00:48:42.940 --> 00:48:45.330 These are touching, but they're not satisfied. 00:48:45.330 --> 00:48:46.200 So it's OK. 00:48:46.200 --> 00:48:48.540 The claim is all these rectangles will be independent. 00:48:48.540 --> 00:48:52.010 Again, I won't prove that formally, but it's true. 00:48:55.010 --> 00:48:55.940 OK? 00:48:55.940 --> 00:48:58.790 Rough sketch. 00:48:58.790 --> 00:49:00.250 So that's Wilber 1. 00:49:00.250 --> 00:49:02.600 It's, again, an independent rectangle lower bound. 00:49:02.600 --> 00:49:04.944 It's a little weird because it depends on this tree. 00:49:04.944 --> 00:49:06.860 You could choose it to be a nice perfect tree. 00:49:06.860 --> 00:49:07.970 You could choose it to be a different tree. 00:49:07.970 --> 00:49:10.344 You'll get a different lower bound each time. 00:49:10.344 --> 00:49:12.260 So of course, you take the max over all trees. 00:49:12.260 --> 00:49:16.100 That will give you the biggest Wilber 1 lower bound. 00:49:16.100 --> 00:49:21.140 We don't know much about that biggest Wilber 1 lower bound. 00:49:21.140 --> 00:49:26.960 I guess you could ask the following open question. 00:49:26.960 --> 00:49:30.380 Is it true that for every access sequence 00:49:30.380 --> 00:49:37.700 there exists a tree p such that Wilber 1 is theta OPT? 00:49:37.700 --> 00:49:39.920 Or is theta Wilber 2 or something? 00:49:39.920 --> 00:49:41.940 Wilber 2 is a single quantity. 00:49:41.940 --> 00:49:43.005 You compute it. 00:49:43.005 --> 00:49:44.420 It gives you a bound. 00:49:44.420 --> 00:49:46.790 Wilber 1, it depends on this p. 00:49:46.790 --> 00:49:48.950 Maybe if you choose the best p for your sequence 00:49:48.950 --> 00:49:50.190 you get the right answer. 00:49:50.190 --> 00:49:55.550 But it's definitely the case that Wilber 1 for a fixed p 00:49:55.550 --> 00:49:58.850 is not the right answer. 00:49:58.850 --> 00:50:00.740 I recall that's easy to prove. 00:50:06.580 --> 00:50:08.690 Well, maybe we'll come back to that. 00:50:08.690 --> 00:50:10.017 Yeah, question? 00:50:10.017 --> 00:50:11.686 AUDIENCE: So how do you construct 00:50:11.686 --> 00:50:12.879 this lower bound tree? 00:50:12.879 --> 00:50:13.932 Like, is it just-- 00:50:13.932 --> 00:50:16.140 ERIK DEMAINE: I'll tell you what we're going to use-- 00:50:16.140 --> 00:50:17.884 the question is how do we construct p. 00:50:17.884 --> 00:50:19.300 You can make it whatever you want. 00:50:19.300 --> 00:50:21.870 What we're going to use is the perfect tree, 00:50:21.870 --> 00:50:23.730 which is sort of unique. 00:50:23.730 --> 00:50:27.076 It's kind of arbitrary, but it works. 00:50:27.076 --> 00:50:28.950 It has the property that its height is log n. 00:50:28.950 --> 00:50:30.300 That's all we need. 00:50:30.300 --> 00:50:32.430 We're going to use that to get tango trees. 00:50:32.430 --> 00:50:34.500 Other questions? 00:50:34.500 --> 00:50:36.600 All right. 00:50:36.600 --> 00:50:39.270 Let me briefly mention a fun access sequence. 00:50:45.315 --> 00:50:48.990 You may recognize this sequence. 00:50:48.990 --> 00:50:52.080 This would be in-order traversal in binary. 00:50:52.080 --> 00:50:53.760 But if I take these bit sequences 00:50:53.760 --> 00:51:09.311 and read them backwards, then I get 0, 4, 2, 6, 1, 5, 3, 7. 00:51:09.311 --> 00:51:11.310 This is the number 0 through 7 in a funny order. 00:51:11.310 --> 00:51:13.890 It's called the bit reversal sequence. 00:51:13.890 --> 00:51:24.000 If you access 0, 4, 2, 6, 1, 5, 3, 7 in a perfect binary tree, 00:51:24.000 --> 00:51:26.410 it maximizes Wilber 1. 00:51:26.410 --> 00:51:36.960 So in-order traversal-- 0, 1, 2, 3, 4, 5, 6. 00:51:36.960 --> 00:51:38.010 Ignore 7. 00:51:38.010 --> 00:51:39.647 There's not 7 in this tree. 00:51:43.135 --> 00:51:46.050 I do 0, 4-- 00:51:46.050 --> 00:51:48.640 if you look at the root, alternate left, 00:51:48.640 --> 00:51:53.820 right, left, right, left, right. 00:51:53.820 --> 00:51:56.500 Because the high-order bit is switching every time, 00:51:56.500 --> 00:51:58.614 and so whether I go to the left of the tree here 00:51:58.614 --> 00:52:00.780 or the right of the tree, it's switching every time. 00:52:00.780 --> 00:52:02.321 And also, if you look in any subtree, 00:52:02.321 --> 00:52:04.710 like when I'm accessing things within the subtree of one, 00:52:04.710 --> 00:52:06.530 it alternates 0, too. 00:52:06.530 --> 00:52:09.030 It's too small a tree to really see that happening, 00:52:09.030 --> 00:52:10.600 but it's true. 00:52:10.600 --> 00:52:15.380 And so, if you do this for k bits, 00:52:15.380 --> 00:52:18.060 n equals 2 to the k roughly. 00:52:18.060 --> 00:52:21.680 And Wilber 1, the lower bound, is 00:52:21.680 --> 00:52:27.210 log n per [INAUDIBLE] because the every access alternates. 00:52:27.210 --> 00:52:29.622 So if you look at a subtree, whatever 00:52:29.622 --> 00:52:31.080 the size of that subtree is, that's 00:52:31.080 --> 00:52:33.060 how many alternations there are. 00:52:33.060 --> 00:52:38.070 And so, number of alternations is theta n log n 00:52:38.070 --> 00:52:41.780 because it's the sum over all nodes of their subtree sizes. 00:52:41.780 --> 00:52:48.300 And so OPT is theta n log n. 00:52:48.300 --> 00:52:50.420 We know we can achieve n log n-- 00:52:50.420 --> 00:52:53.520 this is to do n accesses-- 00:52:53.520 --> 00:52:55.899 we know we can n log n with a red-black tree or whatever, 00:52:55.899 --> 00:52:57.690 but there's actually a lower bound of n log 00:52:57.690 --> 00:52:59.730 n, meaning all binary search trees-- 00:52:59.730 --> 00:53:01.590 if you're given this access sequence, 00:53:01.590 --> 00:53:04.080 doesn't matter what you're doing-- you have to pay 00:53:04.080 --> 00:53:04.770 n log n. 00:53:04.770 --> 00:53:06.680 It's kind of cool. 00:53:06.680 --> 00:53:07.990 A little side effect-- 00:53:07.990 --> 00:53:09.840 that's Wilber's paper ended. 00:53:09.840 --> 00:53:13.350 It's like, hey, cool, can find one access sequence that 00:53:13.350 --> 00:53:17.340 is bad for everybody. 00:53:17.340 --> 00:53:20.480 But now we're going to use Wilber 1 00:53:20.480 --> 00:53:22.890 to get one binary search tree that's pretty 00:53:22.890 --> 00:53:25.770 good for all access sequences. 00:53:25.770 --> 00:53:28.590 Pretty good meaning within a log log n factor of optimal. 00:53:51.280 --> 00:53:59.152 And this is tango trees, which would 00:53:59.152 --> 00:54:04.400 be log log n competitive online binary search trees. 00:54:09.380 --> 00:54:11.270 Why are they called tango trees? 00:54:11.270 --> 00:54:14.330 People made up all sorts of reasons, but I can tell you-- 00:54:14.330 --> 00:54:16.010 because I was there-- 00:54:16.010 --> 00:54:20.810 they were invented mostly on a flight from New York 00:54:20.810 --> 00:54:24.650 to Buenos Aires, which is the center of tango. 00:54:24.650 --> 00:54:26.900 I bought this T-shirt I think the day after. 00:54:26.900 --> 00:54:29.150 And then that week, we wrote the paper, 00:54:29.150 --> 00:54:30.610 and that was tango trees. 00:54:30.610 --> 00:54:35.630 So no particular reason, but it sounds good. 00:54:35.630 --> 00:54:37.130 Always good to have a cool name. 00:54:37.130 --> 00:54:39.095 So the secret is revealed. 00:54:39.095 --> 00:54:46.100 The true meaning of tango trees is nothing, but you we'll see. 00:54:46.100 --> 00:54:48.630 So how do they work? 00:54:48.630 --> 00:54:49.640 It's very simple. 00:54:49.640 --> 00:54:55.040 Basically, we take Wilber 1 and we simulate it. 00:54:55.040 --> 00:54:59.310 So let me be more precise. 00:54:59.310 --> 00:55:05.870 There's one key idea, which is to look at the preferred 00:55:05.870 --> 00:55:10.510 child of a node. 00:55:15.870 --> 00:55:19.242 I'm going to say the preferred child is left. 00:55:19.242 --> 00:55:23.818 Let's see, node y in p. 00:55:23.818 --> 00:55:38.060 It's left if we accessed some node in the left subtree of y 00:55:38.060 --> 00:55:38.675 most recently. 00:55:43.300 --> 00:55:46.760 It's the right child if we accessed 00:55:46.760 --> 00:55:48.960 something in the right subtree most recently. 00:55:48.960 --> 00:55:52.850 So we're just looking at left and right subtree accesses, 00:55:52.850 --> 00:55:54.240 what was most recent? 00:55:54.240 --> 00:55:56.390 There is a special case in the beginning, which 00:55:56.390 --> 00:55:58.681 is you don't have a preferred child because you haven't 00:55:58.681 --> 00:56:00.710 accessed either left or right yet. 00:56:00.710 --> 00:56:12.365 So this is if no access to the left or right yet. 00:56:12.365 --> 00:56:14.340 So that just happens in the beginning. 00:56:14.340 --> 00:56:17.180 Once you've touched everything, everybody 00:56:17.180 --> 00:56:19.970 will have a left or right preferred child. 00:56:19.970 --> 00:56:23.780 So this is just what was your most recent child. 00:56:23.780 --> 00:56:26.580 This is like a parent with a very short memory. 00:56:26.580 --> 00:56:31.400 Just whichever child I most recently talked to, 00:56:31.400 --> 00:56:34.292 that is my preferred child at the moment. 00:56:34.292 --> 00:56:35.750 It's kind of like I don't know when 00:56:35.750 --> 00:56:38.575 you're going to job interviews. 00:56:38.575 --> 00:56:40.160 You know, the most recent interview 00:56:40.160 --> 00:56:42.702 is the one you remember most fondly and so, ah, 00:56:42.702 --> 00:56:44.660 you like that one the best independent of which 00:56:44.660 --> 00:56:45.570 is the coolest. 00:56:45.570 --> 00:56:48.620 So let me draw a picture. 00:56:48.620 --> 00:56:52.520 And I guess I'm going to draw a big picture-- 00:56:52.520 --> 00:57:00.121 my favorite-- a perfectly balanced binary search tree 00:57:00.121 --> 00:57:02.810 with eight leaves. 00:57:02.810 --> 00:57:10.840 And so now, suppose that every node has a preferred child. 00:57:10.840 --> 00:57:12.710 Let's say they all do just because it makes 00:57:12.710 --> 00:57:13.834 a more interesting picture. 00:57:17.240 --> 00:57:20.780 I'm going to draw that with a big fat arrow. 00:57:20.780 --> 00:57:23.390 And now, what does that do? 00:57:23.390 --> 00:57:24.950 It decomposes our tree. 00:57:24.950 --> 00:57:27.860 This is the perfect tree. p is going to be perfectly balanced, 00:57:27.860 --> 00:57:28.700 log n height. 00:57:28.700 --> 00:57:30.290 It could be any log n height tree, 00:57:30.290 --> 00:57:32.990 but we'll make it perfect. 00:57:32.990 --> 00:57:38.840 And it decomposes that tree into paths. 00:57:38.840 --> 00:57:40.120 And there's a path here. 00:57:40.120 --> 00:57:43.880 You just keep following parent pointers, you get a path-- 00:57:43.880 --> 00:57:45.680 not parent pointers, preferred pointers. 00:57:45.680 --> 00:57:47.600 It's also true if you follow parent pointers you get a path, 00:57:47.600 --> 00:57:49.160 but they'll overlap each other. 00:57:49.160 --> 00:57:50.660 You follow preferred child pointers, 00:57:50.660 --> 00:57:52.050 you get non-overlapping paths. 00:57:54.515 --> 00:57:55.140 There they are. 00:57:55.140 --> 00:57:58.950 We also get these singleton paths at the leaves. 00:57:58.950 --> 00:58:01.170 Some of the leaves are in singleton paths. 00:58:01.170 --> 00:58:02.685 These are called preferred paths. 00:58:12.820 --> 00:58:15.635 Why do I care? 00:58:15.635 --> 00:58:19.860 So this tells me the most recently accessed element 00:58:19.860 --> 00:58:21.940 was somebody on this path. 00:58:21.940 --> 00:58:22.911 I don't quite know who. 00:58:22.911 --> 00:58:24.910 It could have been this one, and that would say, 00:58:24.910 --> 00:58:26.410 OK, this is the most recent direction 00:58:26.410 --> 00:58:27.110 we went through all of them. 00:58:27.110 --> 00:58:28.270 Let's say it's that one. 00:58:28.270 --> 00:58:30.280 Now suppose I access this node. 00:58:30.280 --> 00:58:32.410 What does that tell me? 00:58:32.410 --> 00:58:35.020 Well, if I most recently accessed left here 00:58:35.020 --> 00:58:38.650 and now I'm accessing the right, if you look at this node, 00:58:38.650 --> 00:58:40.840 the Wilber 1 bound goes up by 1. 00:58:40.840 --> 00:58:42.400 Because I just accessed left. 00:58:42.400 --> 00:58:44.170 Now I accessed right. 00:58:44.170 --> 00:58:49.120 Also, if I access this node, this guy, his Wilber 1 bound 00:58:49.120 --> 00:58:51.430 goes up by 1 because now he's going to the right, 00:58:51.430 --> 00:58:53.200 whereas last time he went to the left. 00:58:53.200 --> 00:58:56.170 Also, this node previously went to the right 00:58:56.170 --> 00:58:57.580 and went to the left. 00:58:57.580 --> 00:59:02.200 So Wilber 1 went up because of this edge, 00:59:02.200 --> 00:59:03.920 and it went up because of this edge. 00:59:03.920 --> 00:59:07.240 In general, following non-preferred edges, 00:59:07.240 --> 00:59:10.240 I can pay for because Wilber 1 goes up by 1 00:59:10.240 --> 00:59:12.190 every time I use a non-preferred edge. 00:59:12.190 --> 00:59:15.280 This is another way to state the Wilber 1 bound. 00:59:15.280 --> 00:59:17.320 This is the cool thing. 00:59:17.320 --> 00:59:20.350 As long as I can go through a path quickly-- 00:59:23.419 --> 00:59:25.210 ideally, if I could do it in constant time, 00:59:25.210 --> 00:59:27.460 this would be a dynamically-optimal binary 00:59:27.460 --> 00:59:27.990 search tree. 00:59:27.990 --> 00:59:30.573 If I could instantly transport to where I need to go on a path 00:59:30.573 --> 00:59:32.680 and then jump off the path to the next path, 00:59:32.680 --> 00:59:35.900 that I can pay for-- 00:59:35.900 --> 00:59:38.590 I can spend constant time to do that-- 00:59:38.590 --> 00:59:40.125 then I'd be OK. 00:59:40.125 --> 00:59:42.250 I'm not going to be able to do it in constant time, 00:59:42.250 --> 00:59:44.935 but I'm going to be able to do it log log n time. 00:59:44.935 --> 00:59:47.710 I'm going to be able to jump through a path in log log n 00:59:47.710 --> 00:59:49.270 time, and then jump-- 00:59:49.270 --> 00:59:51.730 figure out where I need to diverge from the path 00:59:51.730 --> 00:59:53.364 because maybe I'm accessing this guy. 00:59:53.364 --> 00:59:54.280 Jump to the next path. 00:59:54.280 --> 00:59:56.099 Do that in log log n time. 00:59:56.099 --> 00:59:57.640 I've got to update the path structure 00:59:57.640 --> 00:59:59.620 because now the preferred child is to the right. 00:59:59.620 --> 01:00:00.703 It used to be to the left. 01:00:00.703 --> 01:00:05.886 So I've got to do something that will only cost log log n time. 01:00:05.886 --> 01:00:08.260 If I can do that, the lower bound is the number of edges. 01:00:08.260 --> 01:00:11.390 The upper bound is the number of non-preferred edges 01:00:11.390 --> 01:00:13.250 times log log n. 01:00:13.250 --> 01:00:21.190 So we get a lower bound Wilber 1, 01:00:21.190 --> 01:00:22.930 which is going to be equal to the number 01:00:22.930 --> 01:00:25.480 of non-preferred edges. 01:00:29.099 --> 01:00:30.640 And we're going to get an upper bound 01:00:30.640 --> 01:00:35.950 through tango trees, which is going 01:00:35.950 --> 01:00:39.430 to be order number of non-preferred edges times 01:00:39.430 --> 01:00:42.080 log log n. 01:00:42.080 --> 01:00:42.580 OK. 01:00:42.580 --> 01:00:44.230 Why is it log log n? 01:00:44.230 --> 01:00:47.710 Because each of these paths has length only log n. 01:00:47.710 --> 01:00:50.500 So put them in a balanced binary search tree, 01:00:50.500 --> 01:00:53.350 and it has height log log n. 01:00:53.350 --> 01:00:57.220 So take these paths, squish them into a tree-- 01:00:57.220 --> 01:01:01.430 it's hard, I don't know which way you're squishing. 01:01:01.430 --> 01:01:02.830 It says log n depth. 01:01:02.830 --> 01:01:03.990 It's a path. 01:01:03.990 --> 01:01:05.680 I'm going to fold it into a tree. 01:01:05.680 --> 01:01:07.450 So it has height only log log n. 01:01:07.450 --> 01:01:09.886 Then I can jump around it in log log n time. 01:01:09.886 --> 01:01:11.260 That's the idea with tango trees. 01:01:11.260 --> 01:01:12.345 You're basically done. 01:01:12.345 --> 01:01:15.550 A few details in how they work. 01:01:15.550 --> 01:01:17.510 I don't want to spend too much time on them, 01:01:17.510 --> 01:01:18.926 but let's go through some of them. 01:01:42.410 --> 01:01:53.930 So we're going to store each preferred path 01:01:53.930 --> 01:01:59.840 as an auxiliary tree, which is just-- 01:01:59.840 --> 01:02:02.650 I don't know-- a red-black tree, say. 01:02:07.800 --> 01:02:10.200 What is the red-black tree sorted by? 01:02:10.200 --> 01:02:11.710 I don't have a choice. 01:02:11.710 --> 01:02:13.500 Whatever I do has to be a binary search 01:02:13.500 --> 01:02:15.610 tree among the original keys. 01:02:15.610 --> 01:02:18.180 So if I take these items and I just throw them 01:02:18.180 --> 01:02:21.854 into a red-black tree, they will be sorted by whatever 01:02:21.854 --> 01:02:22.770 their x-coordinate is. 01:02:22.770 --> 01:02:24.910 So this is the max, this is the min. 01:02:24.910 --> 01:02:26.480 This is somewhere in between. 01:02:26.480 --> 01:02:28.320 This is to the left of that. 01:02:28.320 --> 01:02:29.640 So the order is a little weird. 01:02:29.640 --> 01:02:31.980 I'd really like to store them sorted by depth, 01:02:31.980 --> 01:02:33.390 but I can't do that. 01:02:33.390 --> 01:02:34.965 They are sorted by their key values. 01:02:38.430 --> 01:02:42.740 Now, what do I need to do with these auxiliary trees? 01:02:42.740 --> 01:02:46.110 I mean, the basic thing I do is a search, right? 01:02:46.110 --> 01:02:47.330 I'm searching for a key. 01:02:47.330 --> 01:02:49.580 It's a binary search tree, so I can still do a search. 01:02:49.580 --> 01:02:52.740 I can figure out this tree gets represented 01:02:52.740 --> 01:02:56.040 as something more like this. 01:02:56.040 --> 01:02:59.390 That would be a nicely balanced version of these four nodes. 01:02:59.390 --> 01:03:05.250 So if I called them, I don't know, a, b, c, d. 01:03:05.250 --> 01:03:06.900 That's their sorted order. 01:03:06.900 --> 01:03:10.290 It's going to be a, b, c, d. 01:03:10.290 --> 01:03:12.502 That's also their sorted order over here. 01:03:12.502 --> 01:03:14.460 So if I search for my key, I'll figure out, oh, 01:03:14.460 --> 01:03:18.090 do I fall off here, here, here, here, or here? 01:03:18.090 --> 01:03:20.400 Now, each of those corresponds to another path 01:03:20.400 --> 01:03:21.420 I need to visit. 01:03:21.420 --> 01:03:23.730 So if I fall off the left side of a, 01:03:23.730 --> 01:03:26.940 then I should have a pointer to this structure. 01:03:26.940 --> 01:03:29.486 If I fall off the-- 01:03:29.486 --> 01:03:32.640 I guess these two are empty. 01:03:32.640 --> 01:03:35.820 Those correspond to these two places. 01:03:35.820 --> 01:03:40.920 If I fall off here, the right side of c, which is now here, 01:03:40.920 --> 01:03:45.240 this is going to be a pointer to my new structure 01:03:45.240 --> 01:03:48.570 which corresponds to this one. 01:03:48.570 --> 01:03:51.652 And then this one is going to correspond to all this stuff-- 01:03:51.652 --> 01:03:52.860 well, in particular this one. 01:03:55.960 --> 01:03:59.970 It's a little hard to draw this picture, but you get the idea. 01:03:59.970 --> 01:04:02.220 You just rebalance each of these things. 01:04:02.220 --> 01:04:05.160 Keep that the pointers between the preferred paths 01:04:05.160 --> 01:04:06.774 just as they were. 01:04:06.774 --> 01:04:08.940 This is uniquely defined how to do this because it's 01:04:08.940 --> 01:04:11.340 a binary search tree. 01:04:11.340 --> 01:04:20.450 So leaves point to other-- 01:04:20.450 --> 01:04:24.775 let's call them child auxiliary trees. 01:04:24.775 --> 01:04:26.640 It uniquely defines which ones they 01:04:26.640 --> 01:04:29.160 have to point to in order to still navigate 01:04:29.160 --> 01:04:30.760 the whole structure. 01:04:30.760 --> 01:04:33.810 So it's a weird way of rebalancing your tree. 01:04:33.810 --> 01:04:36.360 And the point is each of these red-black trees has height log 01:04:36.360 --> 01:04:39.750 log n because the number of nodes in it is only log n. 01:04:39.750 --> 01:04:41.626 And that gives us the bound. 01:04:55.620 --> 01:05:03.330 Now, key thing to think about is what happens when you change-- 01:05:03.330 --> 01:05:05.010 I said I have to be able to achieve 01:05:05.010 --> 01:05:07.990 number of non-preferred edges times log log n. 01:05:07.990 --> 01:05:08.490 So fine. 01:05:08.490 --> 01:05:10.920 I do a log log n search in here. 01:05:10.920 --> 01:05:12.850 Maybe I decide I have to go off here. 01:05:12.850 --> 01:05:14.589 Then I do a log log n search in here. 01:05:14.589 --> 01:05:16.130 And then maybe I have to go this way. 01:05:16.130 --> 01:05:18.360 So number of non-preferred edges was 2. 01:05:18.360 --> 01:05:20.440 I did two, maybe three searches. 01:05:20.440 --> 01:05:21.880 Fine. 01:05:21.880 --> 01:05:23.910 It's going to be number of non-preferred edges 01:05:23.910 --> 01:05:25.410 plus 1 time log log n. 01:05:25.410 --> 01:05:25.980 No big deal. 01:05:33.600 --> 01:05:35.120 Now I have to update. 01:05:35.120 --> 01:05:37.710 Now this is the preferred edge from the root, 01:05:37.710 --> 01:05:41.070 and this is the preferred edge from this node. 01:05:41.070 --> 01:05:43.930 How do I update preferred edges? 01:05:43.930 --> 01:05:45.400 That's something to think about. 01:05:45.400 --> 01:05:49.480 So I've got a path represented by a red-black tree. 01:05:49.480 --> 01:05:54.360 And now I fall off here, and there's another path here. 01:05:54.360 --> 01:06:00.960 I need to convert this into a path that goes like this 01:06:00.960 --> 01:06:02.490 and then does this. 01:06:02.490 --> 01:06:05.040 And separately, a path that does this. 01:06:05.040 --> 01:06:07.110 That's the new version. 01:06:07.110 --> 01:06:08.120 How do I do that? 01:06:08.120 --> 01:06:11.170 Conceptually, it's pretty simple. 01:06:11.170 --> 01:06:18.630 I want to cut the path here and then rejoin along there, 01:06:18.630 --> 01:06:20.490 like that. 01:06:20.490 --> 01:06:23.700 So conceptually, if things were stored by depth, 01:06:23.700 --> 01:06:26.060 this is what we'd call a split and a concatenate. 01:06:26.060 --> 01:06:28.680 You should know this from regular binary search trees. 01:06:28.680 --> 01:06:31.530 This is a standard exercise for red-black trees. 01:06:31.530 --> 01:06:35.670 Given a query, x, you can cut this tree into two halves 01:06:35.670 --> 01:06:39.000 and get two red-black trees, which represent everything 01:06:39.000 --> 01:06:42.960 to the left of x and everything to the right of x. 01:06:42.960 --> 01:06:44.499 Similarly, given two trees that are 01:06:44.499 --> 01:06:46.290 sorted like this where all the elements are 01:06:46.290 --> 01:06:48.112 less than all the elements over here, 01:06:48.112 --> 01:06:50.070 I can concatenate them into one red-black tree. 01:06:50.070 --> 01:06:51.809 And all of these take log n time, 01:06:51.809 --> 01:06:53.100 where n is the number of nodes. 01:06:53.100 --> 01:06:56.790 Here, that would be log log n time. 01:06:56.790 --> 01:06:58.620 In this world, it's not quite so simple 01:06:58.620 --> 01:07:00.600 because things are not sorted by depth. 01:07:00.600 --> 01:07:02.670 They're sorted by key value. 01:07:02.670 --> 01:07:04.860 But it's not so bad. 01:07:04.860 --> 01:07:12.730 Because, if you look at some path and you want to say, 01:07:12.730 --> 01:07:21.150 OK, I want everything that's below this key value 01:07:21.150 --> 01:07:24.300 or something, then that's the same as saying, 01:07:24.300 --> 01:07:27.790 well, take everything that is within this interval of keys. 01:07:27.790 --> 01:07:29.860 So it's strictly between here and here. 01:07:32.900 --> 01:07:34.460 Let me redraw this slightly. 01:07:45.020 --> 01:07:54.312 So if you look at the nodes of depth greater than d, 01:07:54.312 --> 01:07:56.350 I want to cut off everybody that's 01:07:56.350 --> 01:07:58.120 deeper than a particular spot in order 01:07:58.120 --> 01:08:01.760 to do this kind of change. 01:08:01.760 --> 01:08:12.970 These are equal to nodes in subtree of that. 01:08:12.970 --> 01:08:16.850 So let me give it a name. 01:08:16.850 --> 01:08:18.830 Let's say I want to cut here. 01:08:18.830 --> 01:08:21.580 So I'm going to look at this node y. 01:08:21.580 --> 01:08:24.580 This is nodes in the subtree of y. 01:08:24.580 --> 01:08:26.290 All of the nodes that are below y 01:08:26.290 --> 01:08:30.790 are obviously going to have smaller depth than that path. 01:08:30.790 --> 01:08:31.960 This is nodes in a path. 01:08:35.229 --> 01:08:41.920 And nodes in a subtree are equal to nodes 01:08:41.920 --> 01:08:50.109 with keys in the min of that subtree 01:08:50.109 --> 01:08:51.600 to the max of that tree. 01:08:51.600 --> 01:08:53.899 It's an interval. 01:08:53.899 --> 01:08:55.090 So what do I do? 01:08:55.090 --> 01:08:56.590 I split at min of y. 01:08:56.590 --> 01:08:58.660 I split at max of y. 01:08:58.660 --> 01:09:00.010 That gives me the interval. 01:09:00.010 --> 01:09:01.210 So here's the picture. 01:09:01.210 --> 01:09:02.290 I have a tree. 01:09:02.290 --> 01:09:04.229 I want to cut out this interval of nodes. 01:09:04.229 --> 01:09:07.300 This is like range queries kind of in 1D. 01:09:07.300 --> 01:09:08.229 So I split here. 01:09:08.229 --> 01:09:09.040 I split here. 01:09:09.040 --> 01:09:10.779 What I will have are the things I 01:09:10.779 --> 01:09:13.160 care about, the things to the left of it 01:09:13.160 --> 01:09:15.040 and the things to the right of it. 01:09:15.040 --> 01:09:17.170 What I wanted was this and everything else. 01:09:17.170 --> 01:09:18.160 How do I do that? 01:09:18.160 --> 01:09:24.010 I concatenate-- this is y. 01:09:24.010 --> 01:09:32.380 This is in the interval min of y to max of y. 01:09:32.380 --> 01:09:33.348 So I wanted those guys. 01:09:33.348 --> 01:09:35.139 Those are the nodes that are deeper than d. 01:09:35.139 --> 01:09:37.240 I also want the nodes all together that 01:09:37.240 --> 01:09:38.870 are less deep than d. 01:09:38.870 --> 01:09:40.939 That's these nodes and these nodes. 01:09:40.939 --> 01:09:43.029 So I concatenate these together, get 01:09:43.029 --> 01:09:45.970 one big tree that represents things with depth less than d. 01:09:45.970 --> 01:09:49.479 These are the things of depth greater than d. 01:09:49.479 --> 01:09:49.979 OK? 01:09:49.979 --> 01:09:52.810 So I do two splits, one concatenate, 01:09:52.810 --> 01:09:56.500 and that simulates this kind of cut operation. 01:09:56.500 --> 01:09:58.780 Similarly, if I want to do a joint operation, 01:09:58.780 --> 01:10:01.447 it's a constant number of splits and concatenates, and I'm done. 01:10:01.447 --> 01:10:03.988 Just dealing with the fact that things are in the wrong order 01:10:03.988 --> 01:10:05.620 here, but it's not so bad. 01:10:10.510 --> 01:10:15.230 One more thing, which is-- 01:10:15.230 --> 01:10:17.100 I basically described the overall structure 01:10:17.100 --> 01:10:19.350 as a tree of auxiliary trees. 01:10:19.350 --> 01:10:22.680 In reality, we're in the binary search tree model. 01:10:22.680 --> 01:10:25.830 We can only have one tree. 01:10:25.830 --> 01:10:26.910 Not so hard, though. 01:10:26.910 --> 01:10:30.060 I mean, basically, you want one tree that 01:10:30.060 --> 01:10:34.030 represents lots of trees that are kind of pasted together. 01:10:34.030 --> 01:10:36.900 So to do that, you just mark each node 01:10:36.900 --> 01:10:39.780 that transitions from one tree to the next. 01:10:39.780 --> 01:10:42.810 So each node will say, I am the root of a new auxiliary tree 01:10:42.810 --> 01:10:45.720 or just say, no, I'm part of the same auxiliary tree 01:10:45.720 --> 01:10:46.470 as my parent. 01:10:49.530 --> 01:10:51.790 And then you have to define these kinds of split 01:10:51.790 --> 01:10:55.080 and concatenate operations in this setting where you have 01:10:55.080 --> 01:10:56.676 a tree embedded inside a tree. 01:10:56.676 --> 01:10:58.050 But you just ignore all the nodes 01:10:58.050 --> 01:10:59.883 that are claimed to be part of another tree. 01:10:59.883 --> 01:11:02.290 Just pretend they weren't there, and it works. 01:11:02.290 --> 01:11:06.819 So a little hand-wavy there, but it's kind of a tedious detail. 01:11:06.819 --> 01:11:08.610 You can stick all these trees into one tree 01:11:08.610 --> 01:11:13.100 just by marking these roots. 01:11:13.100 --> 01:11:14.976 And that's tango trees. 01:11:14.976 --> 01:11:18.970 I already spoiled the climax, which is this log log n thing, 01:11:18.970 --> 01:11:22.220 but it's pretty obvious how to get there. 01:11:22.220 --> 01:11:25.150 It's just a lot of details to actually do it. 01:11:25.150 --> 01:11:27.100 We're just taking the Wilber 1 bound, 01:11:27.100 --> 01:11:30.070 recasting it in terms of this preferred path thing 01:11:30.070 --> 01:11:33.370 where it's just the non-preferred edges. 01:11:33.370 --> 01:11:35.600 Or the non-preferred edges are what Wilber 1 counts, 01:11:35.600 --> 01:11:37.270 and so we can afford to spend log log n 01:11:37.270 --> 01:11:38.860 time for each of them. 01:11:38.860 --> 01:11:41.100 And the paths themselves only have log n nodes, 01:11:41.100 --> 01:11:45.470 so you can search through them in log log n time pretty easy. 01:11:45.470 --> 01:11:47.110 This also shows you why Wilber 1 is not 01:11:47.110 --> 01:11:50.890 a good bound with a fixed tree. 01:11:50.890 --> 01:11:53.730 Because here are log n nodes. 01:11:53.730 --> 01:11:58.180 I can just sit there all day bouncing around all of them 01:11:58.180 --> 01:11:59.322 in random order. 01:11:59.322 --> 01:12:01.780 I'm definitely going to need log log n time to access them, 01:12:01.780 --> 01:12:04.460 but Wilber 1 is not changing at all. 01:12:04.460 --> 01:12:08.480 So Wilber 1 stays constant, like 0. 01:12:08.480 --> 01:12:11.140 I had to warm it up, but after I test everything, 01:12:11.140 --> 01:12:14.710 I can just sit there and bounce around these guys randomly. 01:12:14.710 --> 01:12:16.730 I've got to spend log log n time to do that, 01:12:16.730 --> 01:12:19.090 but Wilber 1 doesn't justify it for me. 01:12:19.090 --> 01:12:22.720 Wilber 2 will go up, but Wilber 1 with this tree? 01:12:22.720 --> 01:12:24.389 It's kind of lame. 01:12:24.389 --> 01:12:25.930 So this is the best tango trees could 01:12:25.930 --> 01:12:28.790 hope to do using Wilber 1. 01:12:31.420 --> 01:12:33.790 I would guess that tango trees are a log log 01:12:33.790 --> 01:12:37.400 factor away from optimal, though we don't know that for sure. 01:12:37.400 --> 01:12:40.110 But greedy we're still pretty sure is good. 01:12:40.110 --> 01:12:43.031 It should be a constant factor away from optimal. 01:12:43.031 --> 01:12:44.780 So I want to talk a little bit about that. 01:12:48.170 --> 01:12:50.260 There's one thing on this outline 01:12:50.260 --> 01:12:51.260 we haven't talked about. 01:12:51.260 --> 01:12:52.450 We did independent rectangles. 01:12:52.450 --> 01:12:53.390 We did Wilber 1 and 2. 01:12:53.390 --> 01:12:55.919 We saw applications of them in particular tango trees. 01:12:55.919 --> 01:12:57.710 One thing we haven't done is Signed Greedy. 01:13:01.617 --> 01:13:02.700 So let's do Signed Greedy. 01:13:06.550 --> 01:13:10.260 Still left here is we have two ways to choose 01:13:10.260 --> 01:13:12.330 rectangles, independent rectangles. 01:13:12.330 --> 01:13:13.290 They're different. 01:13:13.290 --> 01:13:15.690 It would be kind of nice to know what the best 01:13:15.690 --> 01:13:17.520 way to choose rectangles is. 01:13:17.520 --> 01:13:19.140 And we actually know that-- 01:13:25.340 --> 01:13:26.880 Signed Greedy. 01:13:26.880 --> 01:13:28.500 So there's two kinds of Signed Greedy. 01:13:28.500 --> 01:13:31.260 There's the plus sign greedy, and there's 01:13:31.260 --> 01:13:32.420 the minus sign greedy. 01:13:36.640 --> 01:13:38.136 How does plus greedy work? 01:13:38.136 --> 01:13:39.510 It's the same as greedy, you just 01:13:39.510 --> 01:13:41.970 only look at plus rectangles. 01:13:41.970 --> 01:13:44.140 Remember plus rectangles and minus rectangles. 01:13:44.140 --> 01:13:49.320 So let's look at our favorite example here. 01:13:49.320 --> 01:13:52.020 With greedy, I would sweep up, and every rectangle that 01:13:52.020 --> 01:13:54.390 was unsatisfied, I would satisfy it. 01:13:54.390 --> 01:13:57.570 Now I'm going to ignore minus rectangles, 01:13:57.570 --> 01:14:00.220 only look at plus rectangles. 01:14:00.220 --> 01:14:02.160 So I see this rectangle, and I say, oh, I 01:14:02.160 --> 01:14:05.220 don't care because that's a minus rectangle. 01:14:05.220 --> 01:14:10.880 Then I see this one and this one. 01:14:10.880 --> 01:14:13.385 I say, oh, those are plus rectangles. 01:14:13.385 --> 01:14:14.760 So I'm going to add a point here. 01:14:14.760 --> 01:14:17.190 I'm going to add a point here. 01:14:17.190 --> 01:14:20.760 Then I go up to here. 01:14:20.760 --> 01:14:23.130 I see this rectangle, which is a plus rectangle. 01:14:23.130 --> 01:14:24.070 That's bad. 01:14:24.070 --> 01:14:25.290 So I've got to add a point. 01:14:25.290 --> 01:14:29.670 I see this minus rectangle I don't care about. 01:14:29.670 --> 01:14:31.590 This is plus greedy. 01:14:31.590 --> 01:14:33.150 It does not satisfy the set. 01:14:33.150 --> 01:14:35.790 This rectangle never got satisfied. 01:14:35.790 --> 01:14:37.930 But it plus satisfies the set. 01:14:37.930 --> 01:14:42.780 If I do plus greedy, it will be plus satisfied. 01:14:42.780 --> 01:14:45.030 Every rectangle you draw here, if it's plus rectangle, 01:14:45.030 --> 01:14:47.490 it's got another point in it. 01:14:47.490 --> 01:14:50.430 What's kind of nice, also, is if you actually 01:14:50.430 --> 01:14:54.600 draw the rectangles you are satisfying-- 01:14:54.600 --> 01:14:55.830 maybe I'm use another color. 01:14:58.530 --> 01:15:00.715 There was one rectangle here. 01:15:00.715 --> 01:15:04.650 There was one rectangle here. 01:15:04.650 --> 01:15:08.379 And there was one rectangle here. 01:15:08.379 --> 01:15:10.170 That's a little awkward because they're not 01:15:10.170 --> 01:15:12.690 on the original points. 01:15:12.690 --> 01:15:14.490 So I can change them a little bit, 01:15:14.490 --> 01:15:21.052 maybe move this one down to here and move this one over to here. 01:15:21.052 --> 01:15:22.510 You could say that those rectangles 01:15:22.510 --> 01:15:24.700 came from those points. 01:15:24.700 --> 01:15:27.670 Then this is a set of independent rectangles 01:15:27.670 --> 01:15:30.210 on the original points. 01:15:30.210 --> 01:15:34.900 Maybe not totally obvious, but plus greedy 01:15:34.900 --> 01:15:49.220 always gives an independent set of plus rectangles. 01:15:49.220 --> 01:15:50.629 So it's a lower bound. 01:15:50.629 --> 01:15:53.170 It's not an upper bound because it's not satisfying the point 01:15:53.170 --> 01:15:55.550 set, but it's a lower bound. 01:15:55.550 --> 01:15:58.418 I claim it's a very good lower bound. 01:16:07.420 --> 01:16:09.280 It by itself might not be great, but you 01:16:09.280 --> 01:16:11.686 have to consider both of them. 01:16:11.686 --> 01:16:26.340 So theorem is if I take the max of plus greedy and minus 01:16:26.340 --> 01:16:26.840 greedy-- 01:16:30.290 --> 01:16:32.210 each of them is lower bound, so the max 01:16:32.210 --> 01:16:35.090 is a lower bound on optimal-- 01:16:35.090 --> 01:16:37.760 then this is within a constant factor 01:16:37.760 --> 01:16:41.650 of the biggest possible independent rectangle lower 01:16:41.650 --> 01:16:42.150 bound. 01:16:51.860 --> 01:16:53.360 And so this is the way you should 01:16:53.360 --> 01:16:54.680 choose independent rectangles. 01:16:54.680 --> 01:16:55.430 Run plus greedy. 01:16:55.430 --> 01:16:56.150 Run minus greedy. 01:16:56.150 --> 01:16:57.587 Take the best of the two. 01:16:57.587 --> 01:16:59.420 That will always be within a constant factor 01:16:59.420 --> 01:17:04.015 of the best independent set of rectangles, factors like 4 01:17:04.015 --> 01:17:06.540 or something in the worst case. 01:17:06.540 --> 01:17:08.480 So let me prove this to you. 01:17:12.110 --> 01:17:13.610 It's a kind of a weird argument. 01:17:16.307 --> 01:17:17.765 I'm going to define a new quantity. 01:17:20.670 --> 01:17:23.210 Let's call this OPT x, I guess. 01:17:26.300 --> 01:17:31.940 It's sort of like if you consider plus rectangles 01:17:31.940 --> 01:17:33.837 separately from minus rectangles, which 01:17:33.837 --> 01:17:34.670 is what we're doing. 01:17:52.440 --> 01:17:55.220 So I would like a point set-- 01:17:55.220 --> 01:17:57.620 first, I'd like a plus satisfying point set, 01:17:57.620 --> 01:18:01.440 and then I'd also like a minus satisfying point set. 01:18:01.440 --> 01:18:03.020 And then I take their union. 01:18:03.020 --> 01:18:08.340 And I say the cost of that pair of plus satisfying and minus 01:18:08.340 --> 01:18:10.350 satisfying is the size of the union. 01:18:10.350 --> 01:18:13.640 So I get a bonus point if they happen to overlap. 01:18:13.640 --> 01:18:16.040 Not a big deal, just a factor of 2. 01:18:16.040 --> 01:18:18.200 So this is not a core concept, but it turns out 01:18:18.200 --> 01:18:20.510 to be basically what we were doing over here. 01:18:23.300 --> 01:18:27.200 Let me give you a sequence of crazy inequalities. 01:18:27.200 --> 01:18:29.900 First one is that this OPT thing is greater than 01:18:29.900 --> 01:18:33.980 or equal to size of the input. 01:18:33.980 --> 01:18:36.060 Each of these inequalities is totally obvious, 01:18:36.060 --> 01:18:38.056 but the conclusion is kind of crazy. 01:18:43.860 --> 01:18:46.520 The independent rectangle lower bound, which we proved, 01:18:46.520 --> 01:18:49.040 says that if you look at plus satisfying things that's 01:18:49.040 --> 01:18:51.050 going to be at least size of the input 01:18:51.050 --> 01:18:52.967 plus the max number of independent rectangles. 01:18:52.967 --> 01:18:54.758 If you look at the minus satisfying things, 01:18:54.758 --> 01:18:56.990 that's also going to be at least size of the input 01:18:56.990 --> 01:19:00.320 plus maximum number of minus independent rectangles. 01:19:00.320 --> 01:19:01.520 So we already proved this. 01:19:01.520 --> 01:19:03.344 That, if you look at this union, it's 01:19:03.344 --> 01:19:05.510 going to be at least the size of the input plus half 01:19:05.510 --> 01:19:08.060 the overall max. 01:19:08.060 --> 01:19:12.320 So that's what we proved at the beginning a lecture. 01:19:12.320 --> 01:19:17.840 Now, this is the best way to use independent rectangles. 01:19:17.840 --> 01:19:19.970 This kind of Signed Greedy, which 01:19:19.970 --> 01:19:22.749 is the max of the two signs, is a way 01:19:22.749 --> 01:19:24.040 to find independent rectangles. 01:19:24.040 --> 01:19:25.331 So it's only going to be worse. 01:19:25.331 --> 01:19:27.480 It's going to be smaller. 01:19:27.480 --> 01:19:32.000 So we can say is greater than or equal to half 01:19:32.000 --> 01:19:37.630 the max of plus greedy and minus greedy. 01:19:42.650 --> 01:19:44.160 This was the max. 01:19:44.160 --> 01:19:46.700 So this is another way to do it, so it must be smaller. 01:19:49.920 --> 01:19:58.710 Now, greedy computes a plus satisfying assignment. 01:19:58.710 --> 01:20:02.834 So I could say, well, if you looked at the optimal plus 01:20:02.834 --> 01:20:05.000 satisfying assignment-- this is something we defined 01:20:05.000 --> 01:20:06.440 at the beginning of lecture-- 01:20:06.440 --> 01:20:09.590 and the optimal minus satisfying assignment, that's 01:20:09.590 --> 01:20:11.870 only going to be smaller than greedy because greedy 01:20:11.870 --> 01:20:15.290 is an algorithm for solving OPT plus. 01:20:15.290 --> 01:20:19.302 It can't be better than the optimum. 01:20:19.302 --> 01:20:21.260 Greedy again has to be bigger than the optimum. 01:20:24.890 --> 01:20:29.110 Now I just want to turn this max into a plus 01:20:29.110 --> 01:20:32.550 because the max is always at least the average. 01:20:32.550 --> 01:20:37.670 So if I take the average, which turns it into 1/4 OPT plus 01:20:37.670 --> 01:20:39.995 plus OPT minus. 01:20:42.990 --> 01:20:44.090 Then that holds. 01:20:44.090 --> 01:20:46.120 You turn the max into a plus. 01:20:46.120 --> 01:20:48.440 If I look at the optimal plus satisfying 01:20:48.440 --> 01:20:50.690 plus the optimal minus satisfying, 01:20:50.690 --> 01:20:54.770 that's only going to be bigger than this thing 01:20:54.770 --> 01:20:56.900 because this can only save like a factor of 2 01:20:56.900 --> 01:21:00.436 or whatever over just adding them up. 01:21:00.436 --> 01:21:02.060 I don't even need to factor of 2 thing. 01:21:02.060 --> 01:21:05.810 I just need that if you add them up, 01:21:05.810 --> 01:21:08.270 that's only going to be worse than just counting them 01:21:08.270 --> 01:21:10.350 as the union. 01:21:10.350 --> 01:21:13.100 So we get what I call a sandwich. 01:21:13.100 --> 01:21:15.860 On the one side, we have OPT x. 01:21:15.860 --> 01:21:17.489 On the other side, we have 1/4 OPT x. 01:21:17.489 --> 01:21:19.280 I really don't care about OPT x personally. 01:21:19.280 --> 01:21:21.690 I mean, it's kind of interesting to see that it's here. 01:21:21.690 --> 01:21:23.910 But the point is these are within a constant factor. 01:21:23.910 --> 01:21:25.618 Therefore, all of these things in between 01:21:25.618 --> 01:21:27.690 are within a constant factor of each other. 01:21:27.690 --> 01:21:31.970 So in particular, this thing, max of the two greedys, 01:21:31.970 --> 01:21:34.070 is within a constant factor of this thing. 01:21:34.070 --> 01:21:37.070 This is the independent rectangle lower bound, 01:21:37.070 --> 01:21:38.540 the best one. 01:21:38.540 --> 01:21:40.460 It also tells you that OPT x is basically 01:21:40.460 --> 01:21:43.230 what we're computing here. 01:21:43.230 --> 01:21:44.640 So this is weird. 01:21:44.640 --> 01:21:49.190 I'm going to draw one more picture, which 01:21:49.190 --> 01:21:56.630 is greedy versus Signed Greedy. 01:21:56.630 --> 01:21:58.760 Remember greedy from last lecture. 01:21:58.760 --> 01:22:02.930 Greedy says, look, I'm going to fix plus rectangles, 01:22:02.930 --> 01:22:04.640 and I'm going to fix minus rectangles. 01:22:04.640 --> 01:22:06.740 It does them both at the same time. 01:22:06.740 --> 01:22:08.600 Signed Greedy says, look, I'm going 01:22:08.600 --> 01:22:11.270 to do the plus rectangles separately, 01:22:11.270 --> 01:22:14.510 and then I'm going to the minus rectangles separately, 01:22:14.510 --> 01:22:17.605 and then add them up or take the union or take the max. 01:22:17.605 --> 01:22:18.370 It doesn't matter. 01:22:18.370 --> 01:22:19.760 It's a constant factor. 01:22:19.760 --> 01:22:22.670 Just add them separately. 01:22:22.670 --> 01:22:24.920 This one is an upper bound. 01:22:24.920 --> 01:22:27.680 It is a binary search tree. 01:22:27.680 --> 01:22:29.030 This thing is a lower bound. 01:22:29.030 --> 01:22:32.780 All binary search trees must take at least this. 01:22:32.780 --> 01:22:35.720 Are they equal up to constant factors? 01:22:35.720 --> 01:22:36.500 We don't know. 01:22:36.500 --> 01:22:37.970 That's the big question. 01:22:37.970 --> 01:22:40.400 They look almost identical. 01:22:40.400 --> 01:22:43.280 But what greedy has to deal with is sort of the interrelations. 01:22:43.280 --> 01:22:45.980 When I fix some plus rectangles, I 01:22:45.980 --> 01:22:49.305 might get new minus rectangles that I have to fix with greedy. 01:22:49.305 --> 01:22:51.180 Signed Greedy doesn't have to deal with that. 01:22:51.180 --> 01:22:52.989 It's just the plus rectangles. 01:22:52.989 --> 01:22:54.530 They might make more plus rectangles, 01:22:54.530 --> 01:22:56.120 but that's all I have to deal with. 01:22:56.120 --> 01:22:57.620 It doesn't deal with the interaction 01:22:57.620 --> 01:22:59.420 between plus and minus rectangles. 01:22:59.420 --> 01:23:03.290 Seems like the interaction kind of fades away 01:23:03.290 --> 01:23:04.207 as a geometric series. 01:23:04.207 --> 01:23:05.873 And therefore, these things are the same 01:23:05.873 --> 01:23:07.040 up to constant factors. 01:23:07.040 --> 01:23:08.900 But we have no way to prove that. 01:23:08.900 --> 01:23:12.380 It could be the interaction blows you out of the water 01:23:12.380 --> 01:23:14.750 somehow. 01:23:14.750 --> 01:23:19.271 That's the best we know for dynamic optimality. 01:23:19.271 --> 01:23:21.770 Maybe next time I teach this class we'll have a final answer 01:23:21.770 --> 01:23:25.730 and it'll be constant, but that's where we are today.