WEBVTT 00:00:00.070 --> 00:00:02.440 The following content is provided under a Creative 00:00:02.440 --> 00:00:03.820 Commons license. 00:00:03.820 --> 00:00:06.060 Your support will help MIT OpenCourseWare 00:00:06.060 --> 00:00:10.150 continue to offer high quality educational resources for free. 00:00:10.150 --> 00:00:12.690 To make a donation or to view additional materials 00:00:12.690 --> 00:00:16.600 from hundreds of MIT courses, visit MIT OpenCourseWare 00:00:16.600 --> 00:00:17.263 at ocw.mit.edu. 00:00:25.940 --> 00:00:27.440 PROFESSOR: All right, so today we're 00:00:27.440 --> 00:00:30.810 going to think about the-- well, also 00:00:30.810 --> 00:00:34.070 the low end of the polynomial hierarchy. 00:00:34.070 --> 00:00:36.500 So most of this class is about polynomial 00:00:36.500 --> 00:00:39.535 versus not polynomial and various notions of hardness, 00:00:39.535 --> 00:00:41.610 [? NP ?] hardness, and worse. 00:00:41.610 --> 00:00:44.080 We look briefly at P-completeness, which 00:00:44.080 --> 00:00:46.090 is about parallel computing. 00:00:46.090 --> 00:00:48.390 Today we're going to be thinking about regular good old 00:00:48.390 --> 00:00:51.020 sequential computing, but trying to distinguish linear time 00:00:51.020 --> 00:00:53.560 versus non-linear time, and in particular, 00:00:53.560 --> 00:00:56.130 trying to find problems that are quadratic 00:00:56.130 --> 00:01:00.090 or cubic in n for sequential running times. 00:01:00.090 --> 00:01:03.440 And probably the most popular attack on this 00:01:03.440 --> 00:01:09.330 is called 3SUM, which is the following problem. 00:01:09.330 --> 00:01:11.485 You're given n integers. 00:01:14.480 --> 00:01:19.020 Let's say, and you want to know do any three of them sum to 0? 00:01:25.620 --> 00:01:29.250 So of course, you can solve this in cubic time 00:01:29.250 --> 00:01:33.380 by testing all triples, whether they sum to 0. 00:01:33.380 --> 00:01:37.240 But you can also solve it in quadratic time. 00:01:37.240 --> 00:01:39.292 So it's not an algorithms class. 00:01:39.292 --> 00:01:41.000 I won't ask you to come up the algorithm. 00:01:41.000 --> 00:01:42.790 But they're quite easy. 00:01:42.790 --> 00:01:48.040 First order n squared randomized is really easy. 00:01:48.040 --> 00:01:50.620 You take all pairwise sums. 00:01:50.620 --> 00:01:54.460 So first of all you, build a dictionary, a hash table 00:01:54.460 --> 00:01:56.350 of all the integers. 00:01:56.350 --> 00:02:00.450 And then you look at all pairwise sums. 00:02:00.450 --> 00:02:02.640 For each pairwise sum, you see whether the negation 00:02:02.640 --> 00:02:04.970 is in the hash table in constant time. 00:02:04.970 --> 00:02:08.410 So that gives you n squared randomized, 00:02:08.410 --> 00:02:13.820 because if a plus b plus c equals 0 00:02:13.820 --> 00:02:18.100 is the same thing as a plus b equal minus c. 00:02:18.100 --> 00:02:20.080 So you look at all pairwise sums, 00:02:20.080 --> 00:02:23.980 see whether the negation is in the list of integers. 00:02:23.980 --> 00:02:28.130 OK, you can also do order n squared deterministic. 00:02:28.130 --> 00:02:33.000 This a more fun puzzle. 00:02:33.000 --> 00:02:34.594 But I'll spoil the answer for you 00:02:34.594 --> 00:02:37.135 just to give you intuition for why this problem is n squared. 00:02:39.720 --> 00:02:43.070 For every possible target sum minus c-- 00:02:43.070 --> 00:02:44.850 so that's going to happen n times-- 00:02:44.850 --> 00:02:49.490 I'm going to run the following linear time algorithm. 00:02:49.490 --> 00:02:52.540 So this is just two copies of the integers in sorted order. 00:02:52.540 --> 00:02:54.030 I have n log n time to sort. 00:02:54.030 --> 00:02:55.340 So that's no problem. 00:02:55.340 --> 00:02:58.050 I'm going to start from with my left finger 00:02:58.050 --> 00:03:00.950 here and my right finger here and look 00:03:00.950 --> 00:03:02.990 at the sum of those two numbers. 00:03:02.990 --> 00:03:06.500 If it's too big-- I have a particular targets, negative c, 00:03:06.500 --> 00:03:09.020 in mind-- if the sum of these two numbers 00:03:09.020 --> 00:03:13.760 is smaller then negative c, then I'm going to advance this one, 00:03:13.760 --> 00:03:15.540 because if this is in sorted order that 00:03:15.540 --> 00:03:17.190 will make my sum larger. 00:03:17.190 --> 00:03:20.740 If the sum is too big, I will advance this one backwards. 00:03:20.740 --> 00:03:24.260 So in general, I have my right finger advancing 00:03:24.260 --> 00:03:26.590 left or my left finger advancing right. 00:03:26.590 --> 00:03:29.220 In each step, one of the two advances. 00:03:29.220 --> 00:03:31.700 And this will not miss-- and this will tell you 00:03:31.700 --> 00:03:34.520 whether that target sum is among the pairwise sums 00:03:34.520 --> 00:03:35.450 from this thing. 00:03:35.450 --> 00:03:38.040 In linear time, we do that n times, once 00:03:38.040 --> 00:03:39.320 for each target sum. 00:03:39.320 --> 00:03:42.420 So that's the fancy quadratic algorithm. 00:03:42.420 --> 00:03:46.160 No fancy data structures required. 00:03:46.160 --> 00:03:46.660 Cool. 00:03:46.660 --> 00:03:48.150 So 3SUM is quadratic. 00:03:48.150 --> 00:03:51.910 And the big conjecture is that you can't solve it any faster. 00:03:51.910 --> 00:03:54.290 Well, you can. 00:03:54.290 --> 00:03:57.310 So it's not quite the conjecture. 00:03:57.310 --> 00:04:05.110 The conjecture is that there is no n to the 2 minus epsilon 00:04:05.110 --> 00:04:11.500 algorithm in general-- in the worst case. 00:04:11.500 --> 00:04:13.880 Let me tell you before we get to-- and this is 00:04:13.880 --> 00:04:16.640 has led to a whole world of lower bounds of 3SUM hardness. 00:04:16.640 --> 00:04:19.279 If your problem is 3SUM hard, then you 00:04:19.279 --> 00:04:22.720 expect it also has no n to the 2 minus epsilon algorithm, 00:04:22.720 --> 00:04:24.390 because if it did, 3SUM would. 00:04:24.390 --> 00:04:30.920 And people generally believe that's the case for 3SUM. 00:04:30.920 --> 00:04:32.230 But there are some exceptions. 00:04:32.230 --> 00:04:38.380 So one thing is that the numbers have to be fairly large. 00:04:38.380 --> 00:04:43.070 If all of the integers are in the range, let's say, 00:04:43.070 --> 00:04:49.580 minus u to u-- that's the universe size-- then via FFT, 00:04:49.580 --> 00:04:53.990 you can solve the problem in u log u time plus linear time. 00:04:53.990 --> 00:04:58.110 So your numbers better be at least larger than n squared. 00:04:58.110 --> 00:05:00.275 And in fact, we'll see n cube suffices. 00:05:00.275 --> 00:05:01.900 So they don't have to be huge, but they 00:05:01.900 --> 00:05:04.180 do have to be bigger than linear or quadratic. 00:05:07.520 --> 00:05:10.380 More generally, there are a bunch 00:05:10.380 --> 00:05:13.475 of sub-quadratic, but only slightly sub-quadratic, 00:05:13.475 --> 00:05:13.975 algorithms. 00:05:16.710 --> 00:05:22.800 The first one achieves a roughly log squared savings. 00:05:22.800 --> 00:05:27.960 So a little bit less is a log log squared in the denominator. 00:05:27.960 --> 00:05:32.910 This is a randomized algorithm in a model of computation 00:05:32.910 --> 00:05:35.140 called the word RAM. 00:05:35.140 --> 00:05:37.572 So if you're interested in the word RAM, 00:05:37.572 --> 00:05:39.280 you should take advanced data structures. 00:05:39.280 --> 00:05:41.720 But basically you can manipulate, let's say, 00:05:41.720 --> 00:05:44.187 log n bit words in constant time. 00:05:44.187 --> 00:05:45.895 You can add them together, multiply them, 00:05:45.895 --> 00:05:48.397 that sort of thing. 00:05:48.397 --> 00:05:49.980 And you assume that the numbers you're 00:05:49.980 --> 00:05:51.234 dealing with fit in a word. 00:05:51.234 --> 00:05:52.900 Because if you're going to compare them, 00:05:52.900 --> 00:05:55.800 you'd also need that assumption. 00:05:55.800 --> 00:05:58.070 So that's a reasonable model. 00:05:58.070 --> 00:06:01.200 And it essentially affords a kind of logarithmic amount 00:06:01.200 --> 00:06:02.040 of parallelism. 00:06:02.040 --> 00:06:04.690 And so because it's a quadratic problem, roughly speaking 00:06:04.690 --> 00:06:08.470 you get a quadratic amount in the parallelism of the model. 00:06:08.470 --> 00:06:10.190 So it's a bit improved. 00:06:10.190 --> 00:06:15.810 This is by two former MIT students, [? Eli ?] Barron 00:06:15.810 --> 00:06:19.480 and Mihai Petrescu and myself. 00:06:19.480 --> 00:06:24.020 And very recently, this year, there's 00:06:24.020 --> 00:06:26.025 been another nice improvement. 00:06:40.470 --> 00:06:45.550 It's actually three algorithms, depending on your model. 00:06:45.550 --> 00:06:47.220 But all based on a similar idea. 00:06:56.410 --> 00:06:57.940 Did I get these backwards? 00:06:57.940 --> 00:06:59.170 I think so. 00:07:17.380 --> 00:07:20.730 So first, these two results-- so these 00:07:20.730 --> 00:07:23.070 are results by [? Gourmand ?] and Petty. 00:07:23.070 --> 00:07:26.390 So that's Petty, gave a talk here about it recently. 00:07:26.390 --> 00:07:28.440 In a real RAM model of computation-- 00:07:28.440 --> 00:07:30.470 this is a weaker model of computation, 00:07:30.470 --> 00:07:32.950 so the result is stronger. 00:07:32.950 --> 00:07:37.720 In a real RAM, you still assume the numbers you're given, 00:07:37.720 --> 00:07:39.047 you can add them. 00:07:39.047 --> 00:07:40.880 I think actually all it needs is the ability 00:07:40.880 --> 00:07:44.400 to add them and compare them, maybe subtract-- yeah, also 00:07:44.400 --> 00:07:45.440 subtract. 00:07:45.440 --> 00:07:48.550 But no multiplication is really useful 00:07:48.550 --> 00:07:49.970 in that particular model. 00:07:49.970 --> 00:07:52.451 Because you can't extract bits out of the thing, 00:07:52.451 --> 00:07:54.200 so you don't assume that they're integers, 00:07:54.200 --> 00:07:55.699 you just treat them as real numbers. 00:07:55.699 --> 00:07:58.000 And all you know how to do is add a bunch of them 00:07:58.000 --> 00:08:01.050 and compare those additions. 00:08:01.050 --> 00:08:02.760 So that's a weaker model of computation. 00:08:02.760 --> 00:08:05.620 And still they're able to get a roughly logarithmic 00:08:05.620 --> 00:08:10.600 improvement, not quite as strong as the quadratic analog. 00:08:10.600 --> 00:08:13.110 But one advance is that this is the first deterministic 00:08:13.110 --> 00:08:15.330 algorithm to be n squared. 00:08:15.330 --> 00:08:18.880 So randomization isn't necessary to achieve that. 00:08:18.880 --> 00:08:22.005 Though this is 2/3 power. 00:08:22.005 --> 00:08:23.380 But the other advance is that you 00:08:23.380 --> 00:08:25.800 don't need to manipulate the individual bits. 00:08:25.800 --> 00:08:30.060 So even in the randomized model that's nice. 00:08:30.060 --> 00:08:35.620 And then the major thing, and sort of the first thing 00:08:35.620 --> 00:08:38.740 to call into question the 3SUM conjecture, which 00:08:38.740 --> 00:08:44.700 is that conjecture, is I wouldn't really 00:08:44.700 --> 00:08:45.820 call this an algorithm. 00:08:45.820 --> 00:08:50.765 But it's a thing which runs n roughly n to the 1.5 time. 00:08:53.570 --> 00:08:57.120 But here it's a more powerful model, a super powerful model, 00:08:57.120 --> 00:09:00.510 called the decision tree model, where 00:09:00.510 --> 00:09:04.960 the idea is that an algorithm is specified by an entire tree. 00:09:04.960 --> 00:09:07.410 The depth of the tree is this big. 00:09:07.410 --> 00:09:11.120 Each node of the tree says, add these five things, 00:09:11.120 --> 00:09:13.420 compare them to these five things, 00:09:13.420 --> 00:09:16.450 and see which is bigger, and then branch. 00:09:16.450 --> 00:09:18.260 There's a left branch, and a right tree. 00:09:18.260 --> 00:09:20.343 If you've ever seen a comparison tree, same thing. 00:09:20.343 --> 00:09:22.395 But the comparisons are more interesting. 00:09:25.370 --> 00:09:27.600 But this is not an algorithm, because we 00:09:27.600 --> 00:09:30.710 don't know how to compute that tree efficiently. 00:09:30.710 --> 00:09:32.710 We can compute it in probably polynomial time, 00:09:32.710 --> 00:09:36.690 but we don't know how to compute it in sub-quadratic time 00:09:36.690 --> 00:09:38.360 or sub-this time. 00:09:38.360 --> 00:09:40.610 So it's kind of a frustrating situation, 00:09:40.610 --> 00:09:44.910 because we know that this thing is 00:09:44.910 --> 00:09:47.850 sort of out there, but actually finding it is hard. 00:09:47.850 --> 00:09:51.270 And actually several problems in computers since the '80s, 00:09:51.270 --> 00:09:54.240 I think, where we know better decision trees 00:09:54.240 --> 00:09:55.880 than we know algorithms. 00:09:55.880 --> 00:10:00.800 So my sense would be that decision tree model is strictly 00:10:00.800 --> 00:10:01.440 more powerful. 00:10:01.440 --> 00:10:05.910 The 3SUM conjecture is true for regular algorithms 00:10:05.910 --> 00:10:08.207 if you define it this way or this way. 00:10:08.207 --> 00:10:10.040 But in the decision to model, obviously, you 00:10:10.040 --> 00:10:10.921 can do a lot better. 00:10:10.921 --> 00:10:11.420 Question? 00:10:11.420 --> 00:10:14.590 AUDIENCE: What's the reason to believe the decision tree 00:10:14.590 --> 00:10:18.250 model is some bit that we run polynomial algorithm, 00:10:18.250 --> 00:10:20.887 look at the bit and it tells us the incident? 00:10:20.887 --> 00:10:22.220 PROFESSOR: You mean, why would-- 00:10:22.220 --> 00:10:25.500 AUDIENCE: Why would you believe this decision tree is there, 00:10:25.500 --> 00:10:29.740 like a model of computation that we should care about? 00:10:29.740 --> 00:10:32.345 PROFESSOR: Oh, no, you shouldn't. 00:10:32.345 --> 00:10:33.970 Decision tree is not a model you should 00:10:33.970 --> 00:10:36.010 consider a reasonable computer. 00:10:36.010 --> 00:10:39.110 But it's interesting in that it suggests-- it gives you 00:10:39.110 --> 00:10:41.010 this tantalizing feeling that maybe you 00:10:41.010 --> 00:10:42.749 could turn this into a real algorithm, 00:10:42.749 --> 00:10:43.790 you could run a computer. 00:10:43.790 --> 00:10:44.960 But, definitely, yeah, you cannot-- 00:10:44.960 --> 00:10:47.310 if you don't know what decision tree is you can't run it 00:10:47.310 --> 00:10:48.470 on a computer directly. 00:10:48.470 --> 00:10:50.737 So it's not a model of computation 00:10:50.737 --> 00:10:51.570 in the strict sense. 00:10:51.570 --> 00:10:53.110 It's especially interesting-- I mean, 00:10:53.110 --> 00:10:55.060 it's always important to see how the model of competition 00:10:55.060 --> 00:10:56.930 relates to bounds, particularly if you're 00:10:56.930 --> 00:10:58.520 going to try to prove the lower bound. 00:10:58.520 --> 00:11:01.430 There are some lower bounds of n squared in restricted forms 00:11:01.430 --> 00:11:02.810 of the decision tree model. 00:11:02.810 --> 00:11:04.850 This says you can't extend those lower bounds 00:11:04.850 --> 00:11:06.390 to arbitrary decision trees. 00:11:06.390 --> 00:11:09.360 Decision trees are traditionally used as a lower bound model. 00:11:09.360 --> 00:11:10.910 If you can prove a lower bound there, 00:11:10.910 --> 00:11:12.600 because it's a very powerful model, 00:11:12.600 --> 00:11:15.560 that implies lower bounds in something like the real RAM. 00:11:15.560 --> 00:11:17.644 So that's why people care in some sense. 00:11:17.644 --> 00:11:19.310 This says you can't prove a strong lower 00:11:19.310 --> 00:11:20.990 bound in that model, which is annoying. 00:11:20.990 --> 00:11:22.576 Another question? 00:11:22.576 --> 00:11:24.112 AUDIENCE: What is the tree that they 00:11:24.112 --> 00:11:25.649 found the structure and the constants of the nodes, 00:11:25.649 --> 00:11:27.760 does it depend on the values of the integer? 00:11:27.760 --> 00:11:29.660 Only the number? 00:11:29.660 --> 00:11:32.530 PROFESSOR: I'm pretty sure in this model 00:11:32.530 --> 00:11:36.220 you have a constant number of original integers. 00:11:36.220 --> 00:11:37.891 You add them together and compare them 00:11:37.891 --> 00:11:39.640 to a constant number of original integers. 00:11:39.640 --> 00:11:41.990 AUDIENCE: But what was is-- I mean 00:11:41.990 --> 00:11:46.426 I can compute the tree in sum, but what does it depend on? 00:11:46.426 --> 00:11:47.742 Like if I change the value-- 00:11:47.742 --> 00:11:49.950 PROFESSOR: The tree, of course, has exponential size. 00:11:49.950 --> 00:11:52.366 So you never would actually want to compute it explicitly. 00:11:52.366 --> 00:11:54.640 What you want to compute is after I've 00:11:54.640 --> 00:11:57.760 done some number of things, what's the next operation that 00:11:57.760 --> 00:11:58.260 happens. 00:11:58.260 --> 00:11:59.884 AUDIENCE: That's still not my question. 00:11:59.884 --> 00:12:02.131 So my question is what changes the actual structure 00:12:02.131 --> 00:12:03.112 of the tree? 00:12:03.112 --> 00:12:04.070 Is it just the number-- 00:12:04.070 --> 00:12:06.260 PROFESSOR: Oh, yeah, n. 00:12:06.260 --> 00:12:10.210 The decision tree only depends on n. 00:12:10.210 --> 00:12:12.730 I mean in some sense the decision tree's encoding 00:12:12.730 --> 00:12:14.577 an adaptive algorithm, that depending 00:12:14.577 --> 00:12:16.160 on the results of previous comparisons 00:12:16.160 --> 00:12:18.050 tells you what to do next about. 00:12:18.050 --> 00:12:19.660 If you think of it as the entire tree 00:12:19.660 --> 00:12:20.784 that's only depending on n. 00:12:23.690 --> 00:12:28.017 But because it's so big, I mean, that doesn't help us if. 00:12:28.017 --> 00:12:29.600 You could imagine pre-computing for n, 00:12:29.600 --> 00:12:31.270 but there's no way to store it. 00:12:31.270 --> 00:12:34.790 And you couldn't really afford that exponential time. 00:12:34.790 --> 00:12:39.260 OK, so that's a short story about the known upper bounds 00:12:39.260 --> 00:12:41.620 and also the known lower bounds on 3SUM. 00:12:41.620 --> 00:12:44.307 There are some weak lower bounds in a particular version 00:12:44.307 --> 00:12:46.140 of the decision tree model when you can only 00:12:46.140 --> 00:12:50.530 compare I think sums of two items, 00:12:50.530 --> 00:12:52.300 then you can get an n squared lower bound. 00:12:52.300 --> 00:12:53.841 But that's not especially interesting 00:12:53.841 --> 00:12:59.120 given this result anymore. 00:12:59.120 --> 00:13:05.380 Let me tell you briefly about k sum, which 00:13:05.380 --> 00:13:16.430 is the obvious generalization, instead of 3, do any k of them 00:13:16.430 --> 00:13:18.450 sum to 0? 00:13:18.450 --> 00:13:21.120 Here, they're actually stronger and lower bounds. 00:13:21.120 --> 00:13:24.692 So if 3SUM is the most popular thing considered 00:13:24.692 --> 00:13:26.150 for proving quadratic lower bounds, 00:13:26.150 --> 00:13:27.524 because a lot of problems we care 00:13:27.524 --> 00:13:29.230 about are linear or quadratic, so 3SUM 00:13:29.230 --> 00:13:31.020 gets a lot of the attention. 00:13:31.020 --> 00:13:32.830 K sum is a little easier to argue about. 00:13:32.830 --> 00:13:35.950 In particular, it's NP hard in general, right. 00:13:35.950 --> 00:13:40.410 This, in particular, encodes something like partition. 00:13:40.410 --> 00:13:44.640 If you have n integers, and they're overall sum is 0, 00:13:44.640 --> 00:13:46.480 and you want to know whether any n/2 of them 00:13:46.480 --> 00:13:48.025 sum to 0 or something like that, that 00:13:48.025 --> 00:13:50.420 would be roughly partition. 00:13:50.420 --> 00:13:52.500 So this is NP hard. 00:13:52.500 --> 00:13:56.510 So definitely it's got to get hard for some k. 00:13:56.510 --> 00:13:59.390 In fact, you can show fixed parameter hardness, 00:13:59.390 --> 00:14:01.950 W1 hardness, with respect to k. 00:14:05.000 --> 00:14:08.080 So in particular, if you assume the exponential time 00:14:08.080 --> 00:14:11.930 hypothesis, then you get some lower bounds. 00:14:11.930 --> 00:14:16.710 And the best lower bound known so far 00:14:16.710 --> 00:14:24.270 is that there's no n to the little of k algorithm, 00:14:24.270 --> 00:14:27.140 assuming regular ETH. 00:14:27.140 --> 00:14:31.530 For this, we need to assume k is less than or equal to-- is not 00:14:31.530 --> 00:14:38.340 too giant, because, I guess, if k equals n, for example, 00:14:38.340 --> 00:14:39.960 this problem is really easy. 00:14:39.960 --> 00:14:43.150 So it can't go all the way. 00:14:43.150 --> 00:14:46.620 So this says, well, maybe we don't get the constant right, 00:14:46.620 --> 00:14:51.680 but there's some kind of n to the roughly k dependence. 00:14:51.680 --> 00:14:55.760 So there's a reason that there's a number here larger than 1. 00:14:55.760 --> 00:14:59.295 Although we don't know how to prove that, the feeling is 00:14:59.295 --> 00:15:02.670 that 3SUM, k sum, they require roughly n 00:15:02.670 --> 00:15:04.150 to some constant times k. 00:15:04.150 --> 00:15:05.900 You can debate about what the constant is, 00:15:05.900 --> 00:15:08.580 but we have this theorem. 00:15:08.580 --> 00:15:14.930 And on the upper bound side-- and what people believe 00:15:14.930 --> 00:15:21.790 is the right answer-- is k/2 ceiling-- at least randomized. 00:15:21.790 --> 00:15:26.120 If you want deterministic, you might get a log factor. 00:15:26.120 --> 00:15:27.620 But you can definitely achieve this 00:15:27.620 --> 00:15:34.230 by the same kind of do all k/2y sums twice, and then 00:15:34.230 --> 00:15:36.860 look for collisions in the hash table. 00:15:36.860 --> 00:15:40.960 And so the ceiling is what's making 3SUM 00:15:40.960 --> 00:15:43.140 into a quadratic thing. 00:15:43.140 --> 00:15:48.510 But 4 sum is just as easy as 3SUM, 00:15:48.510 --> 00:15:51.590 because you can still solve it in quadratic time. 00:15:51.590 --> 00:15:55.339 But 5 sum, the conjecture is that requires n cube time. 00:15:55.339 --> 00:15:57.130 AUDIENCE: Sorry, I didn't quite catch that. 00:15:57.130 --> 00:15:59.130 Is that a known result? 00:15:59.130 --> 00:16:02.220 PROFESSOR: This is an upper bound is known. 00:16:02.220 --> 00:16:10.380 And then the conjecture is that that's tight. 00:16:10.380 --> 00:16:17.740 There's no end to the ceiling k/2 minus epsilon algorithm. 00:16:17.740 --> 00:16:20.090 That's what we don't know. 00:16:20.090 --> 00:16:23.580 But algorithm is easy. 00:16:23.580 --> 00:16:24.560 So that's k sum. 00:16:24.560 --> 00:16:26.790 And this gives you some more intuition 00:16:26.790 --> 00:16:29.080 for why you should expect these problems are hard. 00:16:29.080 --> 00:16:33.025 In particular, if you could prove k sum requires-- I mean, 00:16:33.025 --> 00:16:34.400 if you can prove this conjecture, 00:16:34.400 --> 00:16:38.190 you prove the exponential time hypothesis. 00:16:38.190 --> 00:16:39.660 So you prove p does not equal NP. 00:16:39.660 --> 00:16:42.118 And you may make $1 million and lots of good things happen. 00:16:42.118 --> 00:16:46.200 So we should try to do this. 00:16:46.200 --> 00:16:48.710 Of course, as I mentioned, 3SUM is the one 00:16:48.710 --> 00:16:53.470 we use the most in this world, because for NP hard problems, 00:16:53.470 --> 00:16:55.990 usually we use NP hardness and all the stuff we did. 00:16:55.990 --> 00:16:58.220 You could use k sum, but that's basically partition. 00:17:03.340 --> 00:17:05.241 But this is sort of motivation for why 00:17:05.241 --> 00:17:06.490 you should think 3SUM is hard. 00:17:06.490 --> 00:17:09.450 I'll leave it at that. 00:17:09.450 --> 00:17:12.185 So let me talk about 3SUM hardness. 00:17:21.530 --> 00:17:29.310 So I'm going to call a problem 3SUM hard if that algorithm has 00:17:29.310 --> 00:17:37.960 an n to the two minus epsilon time algorithm, then so 00:17:37.960 --> 00:17:38.460 does 3SUM. 00:17:42.080 --> 00:17:43.940 OK, this is what we want. 00:17:43.940 --> 00:17:45.630 If I say problem is 3SUM is hard, 00:17:45.630 --> 00:17:48.070 it means it shouldn't be solvable in less 00:17:48.070 --> 00:17:52.280 than quadratic time other than this poly log stuff. 00:17:52.280 --> 00:17:54.840 So this is the formal meaning. 00:17:54.840 --> 00:17:58.000 If you could solve it in sub-quadratic, truly 00:17:58.000 --> 00:18:01.140 sub-quadratic time this is often called the minus epsilon, 00:18:01.140 --> 00:18:03.530 then 3SUM can be solved in truly sub-quadratic time, 00:18:03.530 --> 00:18:05.237 contradicting the 3SUM conjecture. 00:18:05.237 --> 00:18:07.570 So if you believe this is the 3SUM conjecture that means 00:18:07.570 --> 00:18:09.920 this is not possible for your problem. 00:18:09.920 --> 00:18:11.670 And the way we're going to do that usually 00:18:11.670 --> 00:18:14.100 is with a 3SUM reduction. 00:18:20.990 --> 00:18:22.360 There are other ways to do it. 00:18:22.360 --> 00:18:24.890 You don't have to follow this particular style reduction. 00:18:24.890 --> 00:18:28.750 But most of them do. 00:18:28.750 --> 00:18:32.960 So it's going to be a multi call reduction, but in this world, 00:18:32.960 --> 00:18:35.340 we have to be careful about polynomial factors. 00:18:35.340 --> 00:18:37.240 We don't want to call your thing n times 00:18:37.240 --> 00:18:39.550 and say that was a legitimate reduction. 00:18:39.550 --> 00:18:42.990 So let's say you can call-- if you're reducing from a to 00:18:42.990 --> 00:18:47.400 b, then that means you could solve b using a. 00:18:47.400 --> 00:18:52.430 And you're going to make a constant number of calls to a. 00:18:52.430 --> 00:18:53.950 Sorry, other way around. 00:18:53.950 --> 00:18:57.064 That means I can solve a using b. 00:18:57.064 --> 00:18:58.480 Usually, we take an instance here, 00:18:58.480 --> 00:18:59.730 reduce it to an instance here. 00:18:59.730 --> 00:19:00.704 That's OK. 00:19:00.704 --> 00:19:03.120 But because we're going to want to solve not just decision 00:19:03.120 --> 00:19:05.390 problems, we're going to say, OK, 00:19:05.390 --> 00:19:06.950 you take your instance of a. 00:19:06.950 --> 00:19:08.767 You can call an oracle for solving 00:19:08.767 --> 00:19:11.350 b a constant number of times, as long as the thing you call it 00:19:11.350 --> 00:19:13.080 with is also not much bigger. 00:19:16.000 --> 00:19:18.150 So the n prime that you call this thing with 00:19:18.150 --> 00:19:21.690 should be linear in n. 00:19:21.690 --> 00:19:27.930 And the running time of the reduction 00:19:27.930 --> 00:19:28.930 should be sub-quadratic. 00:19:32.940 --> 00:19:35.570 OK, pretty much all reductions, it's like n, 00:19:35.570 --> 00:19:37.500 n log n, maybe n log squared n. 00:19:37.500 --> 00:19:40.280 But it should be strictly less than n squared, 00:19:40.280 --> 00:19:43.600 otherwise the reduction doesn't tell you much about weather 00:19:43.600 --> 00:19:46.130 the problem is quadratic or not. 00:19:46.130 --> 00:19:49.150 So with this much running time, plausibly 00:19:49.150 --> 00:19:50.830 you could construct a larger instance. 00:19:50.830 --> 00:19:53.650 But this constraint says the instance 00:19:53.650 --> 00:19:56.440 of b that you called should be linear in size, 00:19:56.440 --> 00:19:58.290 so the quadratic over here is the same thing 00:19:58.290 --> 00:20:00.140 as quadratic over here. 00:20:00.140 --> 00:20:01.830 OK, so those are the rules of the game. 00:20:01.830 --> 00:20:03.250 We're not going to have to worry about these constraints 00:20:03.250 --> 00:20:04.070 too much. 00:20:04.070 --> 00:20:08.467 Most of our reductions are constant factor blow up and run 00:20:08.467 --> 00:20:10.175 in a reasonable amount of time, But we've 00:20:10.175 --> 00:20:12.180 got to be a little careful here to make 00:20:12.180 --> 00:20:14.140 sure the running time is not huge. 00:20:14.140 --> 00:20:17.280 Usually we're allowed polynomial time. 00:20:17.280 --> 00:20:20.740 OK, so if you have a 3SUM reduction from a to b, 00:20:20.740 --> 00:20:24.474 and you know a is 3SUM hard, then b is 3SUM hard. 00:20:24.474 --> 00:20:25.349 AUDIENCE: [INAUDIBLE] 00:20:30.280 --> 00:20:32.780 PROFESSOR: N prime is the size of the thing that 00:20:32.780 --> 00:20:34.790 you're-- the instance you're calling with. 00:20:34.790 --> 00:20:37.560 So if you have an instance x over here, 00:20:37.560 --> 00:20:39.430 you have some x prime over here. 00:20:39.430 --> 00:20:42.044 N prime is the size of x prime. 00:20:42.044 --> 00:20:43.460 There's a constant number of them. 00:20:43.460 --> 00:20:46.630 But I want all of those instances to be linear size. 00:20:56.430 --> 00:20:59.500 OK, so initially a is going to be 3SUM. 00:20:59.500 --> 00:21:02.290 3SUM is 3SUM hard. 00:21:02.290 --> 00:21:05.870 Because if it as a sub-quadratic algorithm, then so does itself. 00:21:05.870 --> 00:21:08.060 And so that's actually easy. 00:21:08.060 --> 00:21:09.590 NP harnessed, that's not so easy. 00:21:12.420 --> 00:21:15.590 If we let b be some other problem, 00:21:15.590 --> 00:21:17.660 then we'll prove hardness. 00:21:17.660 --> 00:21:21.350 In this world, because we don't have any solid, lower 00:21:21.350 --> 00:21:23.350 bounds to work from, it's also interesting to go 00:21:23.350 --> 00:21:24.821 in both directions. 00:21:24.821 --> 00:21:26.820 I'm not going to define a notion of completeness 00:21:26.820 --> 00:21:29.030 here, just because it hasn't been done. 00:21:29.030 --> 00:21:30.710 But you could define 3SUM completeness 00:21:30.710 --> 00:21:35.890 to mean you can reduce from 3SUM and you can reduce to 3SUM. 00:21:35.890 --> 00:21:39.110 We'll see a few problems in that-- usually people 00:21:39.110 --> 00:21:41.940 call that equivalence, sub-quadratic equivalence. 00:21:44.700 --> 00:21:50.450 OK, so let me start with some base 3SUM hard problems 00:21:50.450 --> 00:21:53.020 to start from. 00:21:53.020 --> 00:21:57.280 A lot of this comes from a paper-- 00:21:57.280 --> 00:22:03.800 this paper-- by Gajentaan and Overmars, 1995. 00:22:03.800 --> 00:22:06.560 And they had been collecting over the years 00:22:06.560 --> 00:22:08.770 a whole bunch of mostly computational geometry 00:22:08.770 --> 00:22:14.130 problems, which are 3SUM hard in that you can reduce 3SUM 00:22:14.130 --> 00:22:16.010 to all of them. 00:22:16.010 --> 00:22:19.330 But in the center here is a bunch of core problems, 3SUM, 00:22:19.330 --> 00:22:23.460 3SUM prime, which I would call ABC version of 3SUM-- 00:22:23.460 --> 00:22:26.210 we've seen a few ABC problems in the past-- 00:22:26.210 --> 00:22:28.300 and a geometric version of 3SUM. 00:22:28.300 --> 00:22:30.520 So let me tell you about those. 00:22:30.520 --> 00:22:38.590 First of all, 3SUM is 3SUM hard, even when u is order n cubed. 00:22:38.590 --> 00:22:39.590 So that's nice to know. 00:22:39.590 --> 00:22:43.120 The integers you're working with don't have to be giant. 00:22:43.120 --> 00:22:45.212 This is based on a hashing argument, 00:22:45.212 --> 00:22:46.170 which we won't go into. 00:22:48.840 --> 00:22:52.160 OK, so what is 3SUM prime? 00:22:52.160 --> 00:23:01.440 3SUM prime you're given three sets of integers, A, B, C-- 00:23:01.440 --> 00:23:02.640 yeah? 00:23:02.640 --> 00:23:04.140 AUDIENCE: For these reductions we're 00:23:04.140 --> 00:23:06.100 using only deterministic reductions? 00:23:09.540 --> 00:23:12.250 PROFESSOR: Let's say we're only using deterministic reductions, 00:23:12.250 --> 00:23:14.720 although randomized would also be interesting, 00:23:14.720 --> 00:23:16.450 You just have to weaken this statement. 00:23:16.450 --> 00:23:19.480 But here I'll say deterministic. 00:23:19.480 --> 00:23:22.844 AUDIENCE: Is the hashing argument for why [INAUDIBLE]? 00:23:22.844 --> 00:23:25.010 PROFESSOR: Yeah, I'm pretty sure you can derandomize 00:23:25.010 --> 00:23:25.940 that hashing scheme. 00:23:28.690 --> 00:23:34.395 I need to double check, but, yeah, that's the claim. 00:23:34.395 --> 00:23:35.670 Yeah, that is good question. 00:23:54.740 --> 00:23:58.590 So this is the ABC version of 3SUM. 00:23:58.590 --> 00:24:00.560 We just want the three items to come 00:24:00.560 --> 00:24:03.117 from three particular sets. 00:24:03.117 --> 00:24:04.700 And traditionally, this one is phrased 00:24:04.700 --> 00:24:07.320 as a plus b equals c, although you could also 00:24:07.320 --> 00:24:09.460 say a plus b plus c equals 0. 00:24:09.460 --> 00:24:10.360 It's the same thing. 00:24:10.360 --> 00:24:12.050 You're just negating all the c's. 00:24:12.050 --> 00:24:16.490 And in this world, because you have one item from each set, 00:24:16.490 --> 00:24:20.930 actually it's really easy to just negate one subset of them. 00:24:20.930 --> 00:24:23.840 So it doesn't matter whether you put a minus sign here or not. 00:24:23.840 --> 00:24:27.280 But I will not, because that's how 3SUM prime is usually 00:24:27.280 --> 00:24:29.700 defined. 00:24:29.700 --> 00:24:33.630 So I claim 3SUM prime is 3SUM hard. 00:24:33.630 --> 00:24:34.850 Why? 00:24:34.850 --> 00:24:40.010 I let capital A-- if I'm given a 3SUM instance, 00:24:40.010 --> 00:24:43.430 let's call it S. S is a set of n integers. 00:24:43.430 --> 00:24:45.080 I'm going to let a equal S. I'm going 00:24:45.080 --> 00:24:48.730 to let b equal S. I'm going to let c equal negative s. 00:24:48.730 --> 00:24:49.530 Done. 00:24:49.530 --> 00:24:52.950 OK, so that's a reduction from 3SUM. 00:24:52.950 --> 00:25:03.571 AUDIENCE: [INAUDIBLE] like if 0 is in your list, 00:25:03.571 --> 00:25:06.430 the way the instance you constructed, 00:25:06.430 --> 00:25:10.142 you might use some element x of the list from a. 00:25:10.142 --> 00:25:14.590 The same element x from b, and then y from c. 00:25:14.590 --> 00:25:21.058 And x plus x might be, I guess, plus y would equal 0. 00:25:21.058 --> 00:25:26.180 In the original list you didn't have x twice. 00:25:26.180 --> 00:25:28.930 I'm just getting the example of choosing 0 three 00:25:28.930 --> 00:25:33.880 times as sort of a convenient example of that. 00:25:33.880 --> 00:25:36.470 PROFESSOR: So that you also detect a linear time. 00:25:36.470 --> 00:25:37.760 AUDIENCE: What do you mean? 00:25:37.760 --> 00:25:39.050 PROFESSOR: You can detect whether there 00:25:39.050 --> 00:25:40.000 are any such triples. 00:25:40.000 --> 00:25:44.280 If you allow repetition, then you answer the question. 00:25:44.280 --> 00:25:46.116 Your goal is to solve 3SUM. 00:25:46.116 --> 00:25:48.440 AUDIENCE: So you're solving 3SUM by making 00:25:48.440 --> 00:25:50.890 oracle calls to 3SUM prime? 00:25:50.890 --> 00:25:53.750 PROFESSOR: Right, so if you can 3SUM ahead of time, 00:25:53.750 --> 00:25:54.980 you're done. 00:25:54.980 --> 00:25:56.995 In linear time, you can check whether there 00:25:56.995 --> 00:26:00.160 are any pairs that allow, with some duplication, 00:26:00.160 --> 00:26:04.330 a solution to 3SUM instance. 00:26:04.330 --> 00:26:06.470 This is not addressed in the paper, which 00:26:06.470 --> 00:26:11.810 makes me think that in this definition of 3SUM, 00:26:11.810 --> 00:26:15.060 we allow the items to-- basically, 00:26:15.060 --> 00:26:18.700 every item could be used three times, up to three times. 00:26:18.700 --> 00:26:20.970 There's no requirement that they're distinct items. 00:26:20.970 --> 00:26:22.627 So then this reduction is fine. 00:26:22.627 --> 00:26:23.960 I don't think that's a big deal. 00:26:23.960 --> 00:26:25.474 And you can get rid of it. 00:26:25.474 --> 00:26:28.260 But that must be how it's normally defined. 00:26:28.260 --> 00:26:32.740 I didn't specify whether it was three distinct items. 00:26:32.740 --> 00:26:35.750 But let's allow multiplicity there. 00:26:35.750 --> 00:26:43.250 And then this reduction is fine, because that's 00:26:43.250 --> 00:26:46.030 what the paper does. 00:26:46.030 --> 00:26:49.800 OK, with more effort, like adding big integers and so on, 00:26:49.800 --> 00:26:51.800 you can reduce-- in the other direction 00:26:51.800 --> 00:26:54.060 reduce from 3SUM prime to 3SUM. 00:26:56.910 --> 00:26:58.436 I won't cover that, because I just 00:26:58.436 --> 00:27:00.310 want to prove the 3SUM hardness about things. 00:27:00.310 --> 00:27:02.830 But in fact, all these three problems 00:27:02.830 --> 00:27:04.681 are identical to each other. 00:27:04.681 --> 00:27:06.180 If any one of them is sub-quadratic, 00:27:06.180 --> 00:27:08.040 then they all are. 00:27:08.040 --> 00:27:10.622 So that's nice, because 3SUM prime is really 00:27:10.622 --> 00:27:11.580 a special case of 3SUM. 00:27:18.200 --> 00:27:21.270 So our next problem is the geometric problem. 00:27:32.230 --> 00:27:33.940 They call it geometric base problem. 00:27:42.190 --> 00:27:44.480 So here we're in 2D. 00:27:44.480 --> 00:27:50.430 And we're given endpoints whose y-coordinates are all 00:27:50.430 --> 00:27:51.990 0, 1, or 2. 00:27:51.990 --> 00:27:56.330 You can imagine why-- because there's three lists. 00:27:56.330 --> 00:28:02.840 And so they all live on three horizontal lines. 00:28:02.840 --> 00:28:05.830 Here are the points. 00:28:05.830 --> 00:28:20.140 And we want to know is there a non-horizontal line that passes 00:28:20.140 --> 00:28:30.095 through three points like this. 00:28:32.610 --> 00:28:37.500 OK, so our claim, GeomBase is 3SUM hard. 00:28:37.500 --> 00:28:40.640 And this is their proof. 00:28:40.640 --> 00:28:44.910 On the first line, we put A. On the last line, we put B. 00:28:44.910 --> 00:28:48.630 And in the middle line, we put every item in C divided by 2. 00:28:51.340 --> 00:28:57.860 So if you look at-- sorry, this is a reduction from 3SUM prime. 00:28:57.860 --> 00:28:59.260 So I have three integers. 00:28:59.260 --> 00:29:03.870 We want to know whether you can every a plus b equals c. 00:29:03.870 --> 00:29:10.660 So the idea is if I have two items, A and B, 00:29:10.660 --> 00:29:13.160 then this point-- I mean, if I just draw the line 00:29:13.160 --> 00:29:15.930 and intersect it with the y' equals 1 line, 00:29:15.930 --> 00:29:18.210 that point will be the average of little a and little 00:29:18.210 --> 00:29:20.970 b-- so a plus b over 2. 00:29:20.970 --> 00:29:25.240 So if there's an item C that matches a plus b, 00:29:25.240 --> 00:29:28.655 then the C/2 will equal the a plus b over 2. 00:29:28.655 --> 00:29:30.780 So there's going to be a line through three points, 00:29:30.780 --> 00:29:34.960 if and only if, 3SUM prime had a yes answer. 00:29:34.960 --> 00:29:40.940 And you can reduce in the reverse direction-- in fact, 00:29:40.940 --> 00:29:42.160 just like this. 00:29:42.160 --> 00:29:44.520 You just multiply all these coordinates by 2. 00:29:44.520 --> 00:29:48.710 That gives you C. 00:29:48.710 --> 00:29:50.250 So those are our starting points. 00:29:50.250 --> 00:29:52.440 And we're going to use all of them. 00:29:52.440 --> 00:29:55.520 And I'm just going to run through a bunch of examples 00:29:55.520 --> 00:29:56.650 of 3SUM hard problems. 00:29:56.650 --> 00:30:00.090 So all of them shouldn't have sub-quadratic time algorithms 00:30:00.090 --> 00:30:03.160 unless 3SUM does. 00:30:03.160 --> 00:30:09.180 So the obvious starting point here 00:30:09.180 --> 00:30:11.510 is what's called degeneracy testing 00:30:11.510 --> 00:30:14.220 in computational geometry. 00:30:14.220 --> 00:30:15.894 So usually, we like to assume that you 00:30:15.894 --> 00:30:17.060 have endpoints in the plane. 00:30:17.060 --> 00:30:20.180 They're in general position, meaning no three are co-linear. 00:30:20.180 --> 00:30:23.300 So the problem is given endpoints, 00:30:23.300 --> 00:30:25.030 are any three of them co-linear? 00:30:25.030 --> 00:30:25.530 Question? 00:30:25.530 --> 00:30:27.790 AUDIENCE: With 3SUM prime, because those are integers, 00:30:27.790 --> 00:30:29.248 how do you deal with them that way? 00:30:29.248 --> 00:30:31.740 PROFESSOR: Oh, integers and rational, same thing, 00:30:31.740 --> 00:30:34.850 you just scale it, multiply. 00:30:34.850 --> 00:30:36.550 So everything here-- because we're 00:30:36.550 --> 00:30:37.966 going to go into geometry quite, I 00:30:37.966 --> 00:30:40.020 will use rationals quite a bit. 00:30:40.020 --> 00:30:45.720 So I can multiply everything by 2 to make it integers again. 00:30:45.720 --> 00:30:47.480 But this problem does not say integers, 00:30:47.480 --> 00:30:49.280 so that's why I'm allowed to do that. 00:30:49.280 --> 00:30:50.320 I start with integers. 00:30:50.320 --> 00:30:52.390 And then I do this. 00:30:52.390 --> 00:30:54.560 But you can also add integers here. 00:30:54.560 --> 00:30:57.580 It wouldn't make a big difference. 00:30:57.580 --> 00:31:12.310 OK, so given endpoints in the plane, are any three co-linear? 00:31:12.310 --> 00:31:14.150 I'm guessing this is the original motivation 00:31:14.150 --> 00:31:14.900 for defining 3SUM. 00:31:18.980 --> 00:31:21.220 This is really a harder version of the problem. 00:31:21.220 --> 00:31:22.640 This is kind of a special case. 00:31:22.640 --> 00:31:26.360 But in particular, it's not exactly the same, 00:31:26.360 --> 00:31:28.370 because we forbid horizontal lines. 00:31:28.370 --> 00:31:30.680 We had to construct a very degenerate instance 00:31:30.680 --> 00:31:33.580 with lots of points on the horizontal lines in order 00:31:33.580 --> 00:31:34.922 for this correspondence to work. 00:31:34.922 --> 00:31:36.630 So the question is can you make something 00:31:36.630 --> 00:31:38.546 that is only degenerate, only has three points 00:31:38.546 --> 00:31:44.130 co-linear, when the 3SUM instance has a solution? 00:31:44.130 --> 00:31:47.205 And this reduction is a little unsatisfying. 00:31:47.205 --> 00:31:49.700 And I don't have a great intuition for it. 00:31:49.700 --> 00:31:51.110 But it's very simple. 00:31:51.110 --> 00:31:54.480 We're going to take-- this is going 00:31:54.480 --> 00:32:01.520 to be a reduction from regular old 3SUM, not 3SUM prime. 00:32:01.520 --> 00:32:05.330 So every number x, we're going to map to the point x comma 00:32:05.330 --> 00:32:08.425 x cubed. 00:32:08.425 --> 00:32:12.420 Cubed because it's odd and not 1 basically. 00:32:12.420 --> 00:32:19.690 And so we take our x values-- probably not a good idea 00:32:19.690 --> 00:32:22.010 to put 0 in there, but whatever. 00:32:22.010 --> 00:32:28.200 And we just project them onto this x cubed curve, x 3 is odd, 00:32:28.200 --> 00:32:31.020 so it has this nice picture. 00:32:31.020 --> 00:32:35.320 And the claim is if you take any two points here-- so 00:32:35.320 --> 00:32:38.997 here's an x-coordinate 1/4 and 3/4, and, of course, 00:32:38.997 --> 00:32:40.330 they would actually be integers. 00:32:40.330 --> 00:32:41.790 That's OK. 00:32:41.790 --> 00:32:43.630 You can scale. 00:32:43.630 --> 00:32:49.280 Then so the sum of those is 1. 00:32:49.280 --> 00:32:55.050 And if you look at negative 1, that will-- 00:32:55.050 --> 00:32:57.560 the cube of negative 1, which is 1, 00:32:57.560 --> 00:33:03.080 is exactly equal to where these two cubed points would 00:33:03.080 --> 00:33:05.755 hit if you extend the line. 00:33:05.755 --> 00:33:06.254 Yes? 00:33:06.254 --> 00:33:07.628 AUDIENCE: Now we have the problem 00:33:07.628 --> 00:33:10.210 that if you want to use the version of 3SUM here, 00:33:10.210 --> 00:33:10.710 [INAUDIBLE]. 00:33:14.090 --> 00:33:15.670 PROFESSOR: Yep, so we definitely need 00:33:15.670 --> 00:33:17.174 that those guys are distinct. 00:33:17.174 --> 00:33:18.840 I'm sure that those two versions of 3SUM 00:33:18.840 --> 00:33:20.870 are equivalent up to sub-quadratic reductions. 00:33:20.870 --> 00:33:24.520 But I don't see how to prove that off hand. 00:33:24.520 --> 00:33:29.390 OK, cool, now why is this true? 00:33:29.390 --> 00:33:30.100 I've checked it. 00:33:30.100 --> 00:33:31.860 It's true. 00:33:31.860 --> 00:33:35.080 Off the page of a algebra and prove it. 00:33:35.080 --> 00:33:37.310 I don't have a great intuition for why this is true. 00:33:37.310 --> 00:33:40.710 But there you go. 00:33:40.710 --> 00:33:43.120 It's easy enough to check where this line should go. 00:33:43.120 --> 00:33:46.900 And it happens to go exactly to the place where the sum goes. 00:33:46.900 --> 00:33:48.170 So sorry. 00:33:50.746 --> 00:33:52.870 I'm guessing it would also work for x to the fifth. 00:33:52.870 --> 00:33:54.220 But I didn't check that. 00:33:58.240 --> 00:34:03.160 OK so those three points on the line. 00:34:23.679 --> 00:34:29.310 OK, so an important life lesson about geometry 00:34:29.310 --> 00:34:31.300 is something called duality. 00:34:31.300 --> 00:34:33.920 So here we're interested whether there was one line that 00:34:33.920 --> 00:34:35.010 goes through three points. 00:34:35.010 --> 00:34:38.239 A complimentary problem is I give you a bunch of lines, 00:34:38.239 --> 00:34:43.030 do any three of them pass through a common point? 00:34:43.030 --> 00:34:47.070 Is there one intersection between three or more lines? 00:34:47.070 --> 00:34:49.810 Is there a point that is on three lines? 00:34:49.810 --> 00:34:52.580 If you know projective geometry, this is totally obvious. 00:34:52.580 --> 00:34:54.949 It's the same problem as this. 00:34:54.949 --> 00:34:57.600 You just apply duality. 00:34:57.600 --> 00:34:59.470 Now, there are many different dualities. 00:34:59.470 --> 00:35:02.750 I'll give you two today. 00:35:02.750 --> 00:35:09.260 First, my favorite is projected duality 00:35:09.260 --> 00:35:11.610 and the sort of most standard, at least, in math. 00:35:11.610 --> 00:35:14.160 If you have a point with x-coordinate a and y-coordinate 00:35:14.160 --> 00:35:22.670 b, you map that to the line ax plus by plus 1 equals 0. 00:35:22.670 --> 00:35:24.420 And vice versa. 00:35:24.420 --> 00:35:26.960 So if I have a line-- almost every line 00:35:26.960 --> 00:35:28.330 can be written this way. 00:35:28.330 --> 00:35:30.480 Everyone line that does not go through the origin 00:35:30.480 --> 00:35:32.060 can be written like this. 00:35:32.060 --> 00:35:34.600 And then you can convert it into the corresponding point. 00:35:34.600 --> 00:35:36.470 So if you give me a bunch of lines, 00:35:36.470 --> 00:35:39.727 just translate so that none of them go through the origin. 00:35:39.727 --> 00:35:41.810 And then convert into corresponding set of points. 00:35:41.810 --> 00:35:43.600 And the nice thing about this duality 00:35:43.600 --> 00:35:46.970 is it preserves incidence, meaning 00:35:46.970 --> 00:35:52.130 if before I apply duality, I have a point and a line that 00:35:52.130 --> 00:35:55.226 are touching, then after apply duality, I will have a line 00:35:55.226 --> 00:35:56.350 and appointed are touching. 00:36:02.690 --> 00:36:07.210 So that's great, because in particular, 00:36:07.210 --> 00:36:09.570 a three-way intersection or a point 00:36:09.570 --> 00:36:12.300 is on three lines will convert into a line 00:36:12.300 --> 00:36:13.870 that it goes through three points. 00:36:13.870 --> 00:36:18.507 And so we get a reduction from here to here. 00:36:18.507 --> 00:36:19.590 That's kind of like magic. 00:36:19.590 --> 00:36:24.220 But it works, essentially because-- well, 00:36:24.220 --> 00:36:27.900 if you think of a line over here as just a pair of coordinates, 00:36:27.900 --> 00:36:30.190 usually written a, b, then we're just 00:36:30.190 --> 00:36:33.090 taking essentially a dot product between those two things. 00:36:33.090 --> 00:36:33.840 It's a plus 1. 00:36:33.840 --> 00:36:35.714 But it doesn't matter which one was the point 00:36:35.714 --> 00:36:37.390 and which one was the line. 00:36:37.390 --> 00:36:39.620 So that's very convenient. 00:36:39.620 --> 00:36:42.017 And you can use that to convert a lot of line problems 00:36:42.017 --> 00:36:42.850 into point problems. 00:36:46.670 --> 00:36:49.650 I won't mention k sum very much. 00:36:49.650 --> 00:36:56.120 But obviously, the d dimensional versions here are d plus 1 sum 00:36:56.120 --> 00:36:56.900 hard. 00:36:56.900 --> 00:37:01.560 So that's these are sort of the more geometric versions of k 00:37:01.560 --> 00:37:02.060 sum. 00:37:09.710 --> 00:37:12.780 So let's do some more problems. 00:37:16.330 --> 00:37:17.670 Next one's called a separator. 00:37:33.370 --> 00:37:40.780 So let's say we're given n line segments in the plane, 00:37:40.780 --> 00:37:43.950 is there a line that separates them 00:37:43.950 --> 00:37:46.310 into any two non-empty groups? 00:38:01.860 --> 00:38:05.190 And that line is not allowed to intersect any of the segments. 00:38:08.850 --> 00:38:13.350 So you're not allowed to split a line segment into two parts. 00:38:13.350 --> 00:38:19.100 You just want to partition the line segments into a left chunk 00:38:19.100 --> 00:38:22.150 and a right chunk. 00:38:22.150 --> 00:38:25.790 So there's actually two versions of this problem. 00:38:25.790 --> 00:38:30.280 The first version allows half infinite rays as segments. 00:38:30.280 --> 00:38:34.000 And then you can assume that all the segments are horizontal. 00:38:34.000 --> 00:38:37.210 So I think that pretty clearly expresses it. 00:38:37.210 --> 00:38:40.020 But we can in particular think that we 00:38:40.020 --> 00:38:42.720 are reducing from GeomBase. 00:38:42.720 --> 00:38:44.460 We had this setup. 00:38:44.460 --> 00:38:46.630 We have points on three lines. 00:38:46.630 --> 00:38:48.677 We want to know whether there's a line that 00:38:48.677 --> 00:38:49.510 passes through them. 00:38:49.510 --> 00:38:52.000 So I'm just going to take the complement essentially 00:38:52.000 --> 00:38:56.600 of those lines, emit tiny intervals wherever I had points 00:38:56.600 --> 00:38:57.470 before. 00:38:57.470 --> 00:38:59.660 And now there will be a separating line, if and only 00:38:59.660 --> 00:39:02.444 if, the original points have a line through them. 00:39:02.444 --> 00:39:03.860 If you make these tiny enough, you 00:39:03.860 --> 00:39:07.415 won't be able to do anything else-- oops, yeah-- 00:39:07.415 --> 00:39:10.030 or you could just split them into 1/3, 2/3, yeah. 00:39:10.030 --> 00:39:13.561 I definitely need here that it's a non-horizontal line. 00:39:13.561 --> 00:39:14.060 Thanks. 00:39:24.450 --> 00:39:27.220 Now this requires having these half infinite rays. 00:39:27.220 --> 00:39:32.280 Otherwise you could make a line like this. 00:39:32.280 --> 00:39:34.650 So you're not allowed to do that if these goes off 00:39:34.650 --> 00:39:37.780 to infinity, then you'd be cutting those rays. 00:39:37.780 --> 00:39:39.530 OK, so maybe you consider that reasonable. 00:39:39.530 --> 00:39:40.030 Maybe not. 00:39:40.030 --> 00:39:41.780 Depends on the application. 00:39:41.780 --> 00:39:44.289 If you don't consider half infinite things reasonable, 00:39:44.289 --> 00:39:46.330 you can replace them with some vertical segments. 00:39:46.330 --> 00:39:49.380 You build this little box, essentially 00:39:49.380 --> 00:39:52.150 a box like a pinwheel pattern. 00:39:52.150 --> 00:39:53.760 So there's no way to cut it up except 00:39:53.760 --> 00:39:56.870 to go through the center. 00:39:56.870 --> 00:40:00.940 So two versions-- version one, we allow half infinite things. 00:40:00.940 --> 00:40:02.630 And every segment is horizontal. 00:40:02.630 --> 00:40:04.860 Version two, horizontal and vertical segments 00:40:04.860 --> 00:40:06.680 are all finite length. 00:40:06.680 --> 00:40:10.560 We'll use this version to reduce from a bunch of times. 00:40:10.560 --> 00:40:15.450 Or we will follow this kind of reduction essentially. 00:40:15.450 --> 00:40:16.740 OK, next problem. 00:40:38.220 --> 00:40:41.520 OK, next problem is called strips cover box. 00:40:47.169 --> 00:40:48.460 I like these names of problems. 00:40:48.460 --> 00:40:49.420 They're pretty clear. 00:40:52.920 --> 00:40:56.450 In fact, they're so clear, here's a figure. 00:40:56.450 --> 00:41:00.470 We have a box, which means axes align rectangle. 00:41:00.470 --> 00:41:01.780 And I have strips. 00:41:01.780 --> 00:41:05.330 Strips are-- I take two parallel lines 00:41:05.330 --> 00:41:07.400 and take the lines in between them. 00:41:07.400 --> 00:41:11.100 All these parallel lines between here and here-- that's a strip. 00:41:11.100 --> 00:41:12.180 So I'm given n strips. 00:41:12.180 --> 00:41:13.030 I'm given a box. 00:41:13.030 --> 00:41:15.010 I want to know whether there's an empty part 00:41:15.010 --> 00:41:16.850 or whether it covers. 00:41:16.850 --> 00:41:19.780 AUDIENCE: Can you [? use ?] angle or just strips? 00:41:19.780 --> 00:41:23.350 PROFESSOR: Oh, yeah, sorry, the strips are placed. 00:41:23.350 --> 00:41:25.050 I give you two lines for each strip. 00:41:25.050 --> 00:41:27.700 And I mean the region in between them. 00:41:27.700 --> 00:41:29.480 They're on the plane. 00:41:29.480 --> 00:41:30.980 You can't slide them around. 00:41:30.980 --> 00:41:33.540 That would be a coverage problem. 00:41:33.540 --> 00:41:34.407 It's pretty easy. 00:41:34.407 --> 00:41:35.990 I just want to compute whether there's 00:41:35.990 --> 00:41:39.320 any point in here is not hit by any of the given strips. 00:41:42.780 --> 00:41:45.935 I should mention, all of these problems, 00:41:45.935 --> 00:41:47.310 except where I say otherwise, can 00:41:47.310 --> 00:41:49.630 be solved in quadratic time. 00:41:49.630 --> 00:41:51.970 And so this is showing that that's essentially tight. 00:41:51.970 --> 00:41:54.490 So you can solve this problem by computing 00:41:54.490 --> 00:41:56.490 the arrangement of these lines in quadratic time 00:41:56.490 --> 00:41:58.990 and checking all the cells, whether they're 00:41:58.990 --> 00:42:00.140 in all the strips. 00:42:00.140 --> 00:42:01.244 So that's not hard. 00:42:01.244 --> 00:42:02.910 But the claim is you can't do any better 00:42:02.910 --> 00:42:05.050 than n squared if you believe the 3SUM conjecture. 00:42:09.140 --> 00:42:11.880 And the reduction is essentially this. 00:42:11.880 --> 00:42:14.600 But I'm going to modify it a little bit. 00:42:22.080 --> 00:42:25.710 So remember, I rotated this 90 degrees for a reason, 00:42:25.710 --> 00:42:29.290 because I wanted to use a particular kind of duality. 00:42:29.290 --> 00:42:32.410 The construction is going to be the dual of this. 00:42:32.410 --> 00:42:36.520 But remember, here, the goal is to find the line that does not 00:42:36.520 --> 00:42:40.115 hit any of these segments. 00:42:40.115 --> 00:42:42.100 And the segments are vertical. 00:42:42.100 --> 00:42:46.020 Or they might be infinite, half infinite rays. 00:42:46.020 --> 00:42:49.100 OK, so I'm going to start from there. 00:42:49.100 --> 00:42:53.170 And then I'm going to dualize, using a different dualization. 00:42:53.170 --> 00:42:54.840 This is probably the most popular one 00:42:54.840 --> 00:42:55.923 in computational geometry. 00:43:02.700 --> 00:43:05.690 I think because everyone remembers y equals mx plus b. 00:43:05.690 --> 00:43:07.520 And so there's the obvious conversion 00:43:07.520 --> 00:43:11.990 between a point, which has b and m, to a line, which 00:43:11.990 --> 00:43:15.169 is mx plus b. 00:43:15.169 --> 00:43:16.710 You could argue about which is which, 00:43:16.710 --> 00:43:18.860 but I think this is the more common. 00:43:18.860 --> 00:43:23.830 So what this means is I start with a point. 00:43:23.830 --> 00:43:27.110 The y-coordinate determines the slope of my line. 00:43:27.110 --> 00:43:31.090 And the x-coordinate determines the y-intercept of my line. 00:43:31.090 --> 00:43:33.420 It's b. 00:43:33.420 --> 00:43:36.821 And you can also convert in the other direction. 00:43:36.821 --> 00:43:38.320 I don't think we'll need to be here. 00:43:38.320 --> 00:43:40.420 And that will work for all non-vertical lines. 00:43:40.420 --> 00:43:42.580 This will not represent vertical lines. 00:43:42.580 --> 00:43:44.880 In case you're curious, if you want vertical lines here 00:43:44.880 --> 00:43:49.180 or if you want lines going through the origin here, 00:43:49.180 --> 00:43:50.450 you need points in infinity. 00:43:50.450 --> 00:43:52.000 That's the projective thing here. 00:43:52.000 --> 00:43:55.000 But we don't need that here. 00:43:55.000 --> 00:43:57.680 Because we're just going to take these points 00:43:57.680 --> 00:44:00.330 and convert each of them to corresponding lines. 00:44:00.330 --> 00:44:02.390 So this is a segment, I'm going to get 00:44:02.390 --> 00:44:04.660 an infinite number of points-- sorry, 00:44:04.660 --> 00:44:06.410 there's an infinite number of points here. 00:44:06.410 --> 00:44:08.785 So I'm going to convert into an infinite number of lines. 00:44:08.785 --> 00:44:10.590 That's actually OK, because these 00:44:10.590 --> 00:44:18.050 points all I have the same-- sorry-- opposite. 00:44:20.800 --> 00:44:23.750 I really want the x-coordinate to m and the y-coordinate 00:44:23.750 --> 00:44:27.290 to be b for this picture to be the right picture. 00:44:27.290 --> 00:44:30.500 So all these points have the same x-coordinate. 00:44:30.500 --> 00:44:32.900 So when I convert them into lines, 00:44:32.900 --> 00:44:34.130 they all have the same slope. 00:44:34.130 --> 00:44:37.270 And they'll also be right next to each other. 00:44:37.270 --> 00:44:39.810 Namely, they will be a strip. 00:44:39.810 --> 00:44:41.850 Isn't that cool? 00:44:41.850 --> 00:44:44.670 So when you do this dualization, a vertical segment 00:44:44.670 --> 00:44:45.470 becomes a strip. 00:44:53.530 --> 00:44:56.030 I think the fancy word would be here 00:44:56.030 --> 00:44:58.390 you have a pencil of points. 00:44:58.390 --> 00:45:01.040 And you convert that into a pencil of parallel lines. 00:45:01.040 --> 00:45:04.960 A pencil of parallel lines is strip. 00:45:04.960 --> 00:45:09.040 Pencil just means like a continuous family. 00:45:09.040 --> 00:45:12.380 Now here, these guys are infinite. 00:45:12.380 --> 00:45:14.530 So it's going to be a strip that it goes off 00:45:14.530 --> 00:45:16.350 to infinity on one end. 00:45:16.350 --> 00:45:18.650 That's a half plane. 00:45:18.650 --> 00:45:22.170 So if we have a ray, vertical ray, 00:45:22.170 --> 00:45:25.500 that's going to convert into half plane. 00:45:25.500 --> 00:45:27.004 Half planes aren't allowed, so we're 00:45:27.004 --> 00:45:28.670 going to have to do something with them. 00:45:28.670 --> 00:45:29.878 But there's only six of them. 00:45:29.878 --> 00:45:34.180 There's three down here and three up here. 00:45:34.180 --> 00:45:38.660 And we also haven't defined what our target rectangle is. 00:45:38.660 --> 00:45:40.750 But at this point what we would like 00:45:40.750 --> 00:45:46.220 to say-- so let's see, what would it correspond to a line 00:45:46.220 --> 00:45:46.960 here? 00:45:46.960 --> 00:45:49.500 Notice, the line will never be vertical. 00:45:49.500 --> 00:45:51.870 What would be a line that happens 00:45:51.870 --> 00:45:53.430 not to hit any of these things? 00:45:53.430 --> 00:45:55.714 In the dual, that line maps to a point. 00:45:55.714 --> 00:45:57.130 And so that's saying that there is 00:45:57.130 --> 00:46:00.410 a point that is not covered by any of these strips or half 00:46:00.410 --> 00:46:02.460 planes. 00:46:02.460 --> 00:46:04.870 So we want to know whether the union of these things 00:46:04.870 --> 00:46:07.340 is the entire plane. 00:46:07.340 --> 00:46:10.420 So it's not quite the problem we wanted to reduce do. 00:46:10.420 --> 00:46:12.620 But it's not hard to fix it. 00:46:12.620 --> 00:46:14.890 We essentially just need to make a retake a really big 00:46:14.890 --> 00:46:15.795 rectangle. 00:46:15.795 --> 00:46:17.670 And then there'll be an empty point in there, 00:46:17.670 --> 00:46:22.017 if and only if, there was a line in this problem. 00:46:22.017 --> 00:46:23.600 How big does the rectangle have to be? 00:46:23.600 --> 00:46:27.970 Well, conveniently, these half planes essentially 00:46:27.970 --> 00:46:30.740 narrows down to a hexagon. 00:46:30.740 --> 00:46:34.800 So you have six of them. 00:46:34.800 --> 00:46:36.442 It might be less than a hexagon. 00:46:36.442 --> 00:46:37.660 But I'll just draw six. 00:46:37.660 --> 00:46:41.220 And we're saying all of this stuff is covered. 00:46:41.220 --> 00:46:43.070 All the things outside the hexagon 00:46:43.070 --> 00:46:44.532 are covered by those half planes. 00:46:44.532 --> 00:46:46.240 So really it's just a matter whether this 00:46:46.240 --> 00:46:47.560 has any empty points. 00:46:47.560 --> 00:46:51.460 So take the bounding box of that hexagon. 00:46:51.460 --> 00:46:52.305 That's my box. 00:46:55.670 --> 00:46:59.590 And now I don't have to worry about half planes anymore. 00:47:06.170 --> 00:47:07.520 I can restrict them. 00:47:10.770 --> 00:47:12.370 I can just say, oh, well, now this 00:47:12.370 --> 00:47:15.510 is a strip, which covers-- in particular, 00:47:15.510 --> 00:47:18.270 I need to cover this part of the rectangle. 00:47:18.270 --> 00:47:20.560 I no longer need to go off to infinity. 00:47:20.560 --> 00:47:24.781 So now, I have a bunch of finite strips and a finite box. 00:47:24.781 --> 00:47:26.780 And it's just a matter of whether that thing has 00:47:26.780 --> 00:47:27.940 any empty parts. 00:47:27.940 --> 00:47:28.440 Yes? 00:47:28.440 --> 00:47:31.000 AUDIENCE: [INAUDIBLE] 00:47:31.000 --> 00:47:33.460 PROFESSOR: Yes, there will be an empty point, if and only 00:47:33.460 --> 00:47:37.460 if the original thing had a yes answer. 00:47:37.460 --> 00:47:40.770 So they will cover, if and only if, if you have a no answer. 00:47:45.290 --> 00:47:45.960 Any questions? 00:47:45.960 --> 00:47:48.803 AUDIENCE: Just nomenclature-- can have word [INAUDIBLE] 00:47:48.803 --> 00:47:50.490 that thing you just put on the right, 00:47:50.490 --> 00:47:52.095 because it's actually a special case of what you call 00:47:52.095 --> 00:47:52.595 [INAUDIBLE]. 00:47:52.595 --> 00:47:54.950 PROFESSOR: Yeah, right, so what they say 00:47:54.950 --> 00:47:57.720 is this a reduction from GeomBase, mimicking 00:47:57.720 --> 00:48:00.370 the proof of separator 1. 00:48:00.370 --> 00:48:02.376 Yeah, I don't have a name for it. 00:48:02.376 --> 00:48:02.875 Sorry. 00:48:05.990 --> 00:48:06.822 Yeah? 00:48:06.822 --> 00:48:07.697 AUDIENCE: [INAUDIBLE] 00:48:14.550 --> 00:48:18.840 PROFESSOR: Yes, this duality also preserves incidence. 00:48:18.840 --> 00:48:20.870 I think it's a slightly perturbed version 00:48:20.870 --> 00:48:22.520 of the regular projector duality. 00:48:22.520 --> 00:48:25.740 It is also a projected duality in a sense, 00:48:25.740 --> 00:48:28.820 but with an extra Mobius transformation thrown in 00:48:28.820 --> 00:48:30.069 or something. 00:48:30.069 --> 00:48:31.485 But those also preserve incidence. 00:48:35.380 --> 00:48:39.000 Yeah, we obviously need that. 00:48:39.000 --> 00:48:44.640 OK so to make this a little more usable-- I mean, 00:48:44.640 --> 00:48:45.787 strips can be nice. 00:48:45.787 --> 00:48:47.120 We'll use it in some situations. 00:48:47.120 --> 00:48:50.500 But geometers tend to like to think about triangles. 00:48:50.500 --> 00:48:54.240 So we can also convert this into whether bunch of triangles 00:48:54.240 --> 00:48:55.790 cover a given triangle. 00:49:06.260 --> 00:49:09.100 Basically, so here we are going to reduce 00:49:09.100 --> 00:49:11.130 from strips covering a box. 00:49:11.130 --> 00:49:13.080 We start with a box. 00:49:13.080 --> 00:49:18.210 And we're going to convert that-- we'll draw on top of it 00:49:18.210 --> 00:49:19.760 so it's a little clearer-- I'm going 00:49:19.760 --> 00:49:27.290 to take a really big triangle, which contains that box. 00:49:27.290 --> 00:49:32.005 But then we'll triangulate the exterior right here. 00:49:35.240 --> 00:49:37.530 And add those triangles to my covering collection. 00:49:37.530 --> 00:49:40.320 So all of this stuff is covered for free. 00:49:40.320 --> 00:49:43.390 And so now what remains-- in order to cover this triangle, 00:49:43.390 --> 00:49:45.090 I just need to cover this box. 00:49:45.090 --> 00:49:48.620 OK so, that's one part of it. 00:49:48.620 --> 00:49:51.490 And then the other thing is that we're given strips. 00:49:51.490 --> 00:49:54.490 So if I have a strip-- actually, do I have a figure for this? 00:49:54.490 --> 00:49:58.520 No-- if I have a strip in the original problem, which 00:49:58.520 --> 00:50:01.560 looks something like this, I really only 00:50:01.560 --> 00:50:06.150 care about the portion of a strip that hits the box here. 00:50:06.150 --> 00:50:11.000 So I will just triangulate that part 00:50:11.000 --> 00:50:14.794 and say those triangles are in my set. 00:50:14.794 --> 00:50:17.210 And then those triangles are going to cover this triangle. 00:50:17.210 --> 00:50:19.300 The red triangles will cover the red triangle, 00:50:19.300 --> 00:50:22.260 the big red triangle, if and only if the white strips cover 00:50:22.260 --> 00:50:24.180 the white rectangle. 00:50:24.180 --> 00:50:26.016 Pretty easy-- all we need is that these 00:50:26.016 --> 00:50:27.890 have constant complexity so we're not blowing 00:50:27.890 --> 00:50:29.230 up more than a constant factor. 00:50:31.830 --> 00:50:34.140 Note here all of these smaller triangles 00:50:34.140 --> 00:50:37.330 are contained inside the big triangle. 00:50:37.330 --> 00:50:39.670 So you can even assume that these guys 00:50:39.670 --> 00:50:41.310 are contained in this guy. 00:50:41.310 --> 00:50:47.950 We'll use that at some point-- possibly very soon. 00:50:53.270 --> 00:50:57.595 So next problem. 00:51:21.620 --> 00:51:24.520 Hole in union-- I give you a bunch of triangles. 00:51:24.520 --> 00:51:25.460 I take their union. 00:51:25.460 --> 00:51:28.170 I want to know whether that's a simply connected polygon 00:51:28.170 --> 00:51:30.100 or whether it has a hole in the center. 00:51:30.100 --> 00:51:30.975 AUDIENCE: [INAUDIBLE] 00:51:34.339 --> 00:51:36.130 PROFESSOR: Almost the same as this problem. 00:51:36.130 --> 00:51:38.310 I do is to do a little bit of work, 00:51:38.310 --> 00:51:42.420 because maybe you'd-- actually, pretty much that reduction will 00:51:42.420 --> 00:51:45.870 work fine, as long as the outer triangle is strictly bigger 00:51:45.870 --> 00:51:49.770 than the strip, then I'll always have these outer red things, 00:51:49.770 --> 00:51:50.779 which make a region. 00:51:50.779 --> 00:51:52.320 And then there'll be a hole in there, 00:51:52.320 --> 00:51:54.940 if and only if the rectangle is not fully covered. 00:51:54.940 --> 00:51:58.280 So I just did to enlarge that outer triangle slightly. 00:51:58.280 --> 00:52:01.800 Then I have proof that this is 3SUM hard. 00:52:01.800 --> 00:52:04.180 Done. 00:52:04.180 --> 00:52:06.750 Here, we're also using that the red triangles are contained 00:52:06.750 --> 00:52:09.050 inside the big red triangle. 00:52:09.050 --> 00:52:13.140 We don't go outside and possibly make a hole in some other way. 00:52:13.140 --> 00:52:16.520 OK, another easy one-- triangle measure. 00:52:21.527 --> 00:52:22.860 I give you a bunch of triangles. 00:52:22.860 --> 00:52:25.960 What is the area of their union? 00:52:25.960 --> 00:52:28.440 Well, it's going to be the area of the big triangle, 00:52:28.440 --> 00:52:30.660 if and only if the big triangle's covered. 00:52:30.660 --> 00:52:35.000 So this is a reduction from triangle covers triangle. 00:52:35.000 --> 00:52:36.045 OK, easy. 00:52:42.430 --> 00:52:45.665 Here's a somewhat different problem-- point covering. 00:52:49.320 --> 00:52:57.890 So here I'm given a bunch of half planes, n of them. 00:52:57.890 --> 00:53:02.536 I want to know is there a k-way intersection? 00:53:09.834 --> 00:53:11.250 You can think of this as a version 00:53:11.250 --> 00:53:13.790 of two-dimensional linear programming. 00:53:13.790 --> 00:53:16.810 So this is all in 2D. 00:53:16.810 --> 00:53:19.040 You're given a bunch of linear inequalities, which 00:53:19.040 --> 00:53:20.320 are half planes. 00:53:20.320 --> 00:53:22.570 You want to know-- not can I satisfy all them, 00:53:22.570 --> 00:53:25.110 maybe that's not possible, but can I satisfy at least k 00:53:25.110 --> 00:53:25.840 of them? 00:53:25.840 --> 00:53:28.090 So this is for approximating linear programming. 00:53:28.090 --> 00:53:29.934 There a lot of algorithms for doing that. 00:53:29.934 --> 00:53:31.350 You can do this in quadratic time. 00:53:31.350 --> 00:53:33.040 But the claim is you can't do it better 00:53:33.040 --> 00:53:37.610 unless if you believe the 3SUM conjecture. 00:53:37.610 --> 00:53:45.720 OK, so this is-- I don't have a figure-- no. 00:53:56.750 --> 00:54:00.075 So we're going to reduce from strips cover box. 00:54:05.910 --> 00:54:08.480 So we're given a bunch-- this figure again-- 00:54:08.480 --> 00:54:09.730 we're given a bunch of strips. 00:54:09.730 --> 00:54:11.250 We're given a rectangle. 00:54:11.250 --> 00:54:15.440 We want to convert that whether the whole thing intersects 00:54:15.440 --> 00:54:18.350 to whether there's a k-way intersection 00:54:18.350 --> 00:54:21.720 between infinite strips on one side, half planes. 00:54:25.839 --> 00:54:27.130 So here's what I'm going to do. 00:54:27.130 --> 00:54:32.060 If I have a strip-- I should maybe really 00:54:32.060 --> 00:54:35.020 draw it this way-- bunch of parallel lines, 00:54:35.020 --> 00:54:37.470 finite segment of them. 00:54:37.470 --> 00:54:39.260 But the lines are infinite in this is 00:54:39.260 --> 00:54:42.340 direction and this direction. 00:54:42.340 --> 00:54:47.260 I'm going to convert that into the complement, a common trick 00:54:47.260 --> 00:54:48.040 here. 00:54:48.040 --> 00:54:50.420 So we have half plane over here. 00:54:50.420 --> 00:54:51.900 And we have a half plane earlier. 00:54:54.450 --> 00:54:56.560 Obviously, you cannot be in both of these at once, 00:54:56.560 --> 00:54:58.050 because they're disjoint. 00:54:58.050 --> 00:55:01.970 So the best you can hope for is to be in one of them, which 00:55:01.970 --> 00:55:03.829 means you're not here. 00:55:03.829 --> 00:55:05.370 And remember, the whole question here 00:55:05.370 --> 00:55:08.600 is whether there's a point that is not in any of the strips. 00:55:08.600 --> 00:55:14.310 So you're going to get one point-- 00:55:14.310 --> 00:55:17.310 you're going to get a score of 1 if you're here or here. 00:55:17.310 --> 00:55:20.570 You're going to get a score of 0 locally if in the region 00:55:20.570 --> 00:55:23.830 that's already covered. 00:55:23.830 --> 00:55:27.650 Now we also need to represent the rectangle in some way. 00:55:27.650 --> 00:55:32.400 So if I'm given a box. 00:55:32.400 --> 00:55:35.740 Draw this line, this line, this line, and this line. 00:55:35.740 --> 00:55:37.826 And the half planes I want are the ones 00:55:37.826 --> 00:55:38.950 that contain the rectangle. 00:55:47.020 --> 00:55:50.240 So the idea is you're going to get four bonus points if you're 00:55:50.240 --> 00:55:52.330 in the rectangle. 00:55:52.330 --> 00:55:58.337 And now the question is-- I'm going to set k to be n plus 4. 00:55:58.337 --> 00:56:00.170 I want to know are there any points that are 00:56:00.170 --> 00:56:03.680 in n plus 4 of the half planes. 00:56:03.680 --> 00:56:05.620 To do that, you have to be in all four 00:56:05.620 --> 00:56:07.900 of these-- in other words, in the box-- 00:56:07.900 --> 00:56:10.780 and in one of these for every strip, 00:56:10.780 --> 00:56:13.100 because you can't be in both. 00:56:13.100 --> 00:56:15.220 It's just to achieve that score of n plus 4. 00:56:15.220 --> 00:56:17.220 So that means you're exterior to all the strips, 00:56:17.220 --> 00:56:18.660 but inside the rectangle. 00:56:18.660 --> 00:56:20.673 So that's the same problem. 00:56:20.673 --> 00:56:21.173 Cool. 00:56:29.850 --> 00:56:42.831 Next problem is a visibility problem. 00:56:42.831 --> 00:56:44.705 So we've got a couple of visibility problems. 00:56:54.020 --> 00:56:55.770 Both of these problems are about something 00:56:55.770 --> 00:56:57.310 called weak visibility. 00:57:00.850 --> 00:57:03.870 If you have two geometric objects, a and b, 00:57:03.870 --> 00:57:05.750 they are strongly visible to each other, 00:57:05.750 --> 00:57:09.320 if every point over here can see every point over here, 00:57:09.320 --> 00:57:17.022 meaning the visibility line doesn't cross anything else. 00:57:17.022 --> 00:57:18.480 But what we're talking about here's 00:57:18.480 --> 00:57:20.570 weak visibility, which says there's 00:57:20.570 --> 00:57:24.010 some point over here which can see some point over here. 00:57:24.010 --> 00:57:26.990 So I want to know, say, given two segments, whether there's 00:57:26.990 --> 00:57:29.230 some point here, some point here, where 00:57:29.230 --> 00:57:31.620 if I draw the connecting visibility line, 00:57:31.620 --> 00:57:34.200 there's no other segment blocking it. 00:57:34.200 --> 00:57:35.805 So this is a. 00:57:35.805 --> 00:57:37.040 And this is b. 00:57:37.040 --> 00:57:39.188 This would be an example where a and b-- well, they 00:57:39.188 --> 00:57:41.354 do weakly see each other, because there's that pair. 00:57:44.320 --> 00:57:46.476 So given n segments, you can construct 00:57:46.476 --> 00:57:48.100 what's called a visibility graph, which 00:57:48.100 --> 00:57:51.380 is all of these things in quadratic time. 00:57:51.380 --> 00:57:54.430 If I want to know whether this segment can 00:57:54.430 --> 00:57:59.082 see another segment in this weak sense, that's 3SUM hard. 00:57:59.082 --> 00:57:59.790 Here's the proof. 00:58:03.281 --> 00:58:07.320 It's exactly the same thing, just add two segments. 00:58:07.320 --> 00:58:09.330 This segment can see that segment, 00:58:09.330 --> 00:58:10.910 if and only if there's a line. 00:58:13.740 --> 00:58:15.719 There are so many fun little things you can do. 00:58:15.719 --> 00:58:17.510 But these are problems that a lot of people 00:58:17.510 --> 00:58:18.390 have thought about. 00:58:18.390 --> 00:58:19.620 And they're always wondering, can we 00:58:19.620 --> 00:58:20.790 do better than quadratic? 00:58:20.790 --> 00:58:25.610 Now we know they're all sort of in the same bucket-- almost. 00:58:25.610 --> 00:58:28.755 OK, here's another problem. 00:58:48.850 --> 00:58:55.440 So the problem is I give you a bunch of triangles in 3D. 00:58:55.440 --> 00:59:03.810 And let's see, and maybe I give you a point up here. 00:59:03.810 --> 00:59:06.380 And I want to know whether that point can weakly 00:59:06.380 --> 00:59:08.500 see a given triangle. 00:59:08.500 --> 00:59:10.461 And here all the triangles are horizontal. 00:59:10.461 --> 00:59:11.460 Make it nice and simple. 00:59:11.460 --> 00:59:13.330 And it is possible to solve this, I think, 00:59:13.330 --> 00:59:16.940 in n squared log n time-- maybe n squared time. 00:59:16.940 --> 00:59:21.640 You can construct the visibility graph here in n squared, 00:59:21.640 --> 00:59:23.270 because they're all horizontal. 00:59:23.270 --> 00:59:27.250 So I'm given this point. 00:59:27.250 --> 00:59:28.910 I want to know can it see any of t. 00:59:28.910 --> 00:59:31.620 Or is it completely blocked by these triangles? 00:59:31.620 --> 00:59:33.510 Sound familiar? 00:59:33.510 --> 00:59:37.100 If we put that point way up near or at infinity, 00:59:37.100 --> 00:59:39.800 depending on what you allow me-- near infinity 00:59:39.800 --> 00:59:41.580 will be enough-- this will essentially 00:59:41.580 --> 00:59:44.720 be orthographic projection of these triangles 00:59:44.720 --> 00:59:45.519 onto this triangle. 00:59:45.519 --> 00:59:47.310 And it's a question whether these triangles 00:59:47.310 --> 00:59:48.870 cover this triangle. 00:59:48.870 --> 00:59:51.680 So this is a reduction from triangles cover triangle 00:59:51.680 --> 00:59:54.179 to visible triangle. 00:59:54.179 --> 00:59:55.970 Pretty easy, just put all these guys really 00:59:55.970 --> 00:59:58.500 close, slightly different z-coordinates 00:59:58.500 --> 01:00:00.850 so they're not overlapping, and put the point far away. 01:00:04.350 --> 01:00:07.420 So visible triangle is 3SUM hard. 01:00:07.420 --> 01:00:10.330 All right, let's do some motion planning, 01:00:10.330 --> 01:00:24.250 robot motion planning-- first in 2D-- so planar motion planning. 01:00:24.250 --> 01:00:26.070 I give you a segment. 01:00:26.070 --> 01:00:28.430 That's the robot. 01:00:28.430 --> 01:00:30.730 And I have-- so this is the special arm, 01:00:30.730 --> 01:00:35.240 the robot-- then I have various obstacles, which the robot is 01:00:35.240 --> 01:00:37.320 not allowed to penetrate. 01:00:37.320 --> 01:00:39.430 I give you a starting position for the robot. 01:00:39.430 --> 01:00:42.010 And I give you a target position for the robot. 01:00:42.010 --> 01:00:46.810 And want to know can I go from here to here by some motion? 01:00:46.810 --> 01:00:50.460 And motion we'd allow rotations and translations. 01:00:50.460 --> 01:00:52.790 So can I slide this robot somehow 01:00:52.790 --> 01:00:56.360 through this obstacle course, in here, just 01:00:56.360 --> 01:00:58.410 some parallel parking, whatever. 01:00:58.410 --> 01:01:00.680 You can do lots of tricks to get from A to B. 01:01:00.680 --> 01:01:02.830 This problem can be solved in quadratic time. 01:01:02.830 --> 01:01:06.070 Most 2D motion planning problems with just rotation 01:01:06.070 --> 01:01:09.270 and translation can be sold in n squared time. 01:01:12.040 --> 01:01:14.890 And that's tight, because of this. 01:01:19.422 --> 01:01:21.630 We build these frames, which are large enough that it 01:01:21.630 --> 01:01:22.747 doesn't concern the robot. 01:01:22.747 --> 01:01:24.330 The robot's big enough that it's going 01:01:24.330 --> 01:01:27.120 to have to simultaneously pierce all three lines here. 01:01:27.120 --> 01:01:29.106 And we're done. 01:01:29.106 --> 01:01:29.770 OK, great. 01:01:32.300 --> 01:01:37.140 All this build up for very simple proofs. 01:01:37.140 --> 01:01:39.064 Cool, one more. 01:01:39.064 --> 01:01:43.150 AUDIENCE: Is that one actually quadratic time? 01:01:43.150 --> 01:01:48.410 PROFESSOR: Yeah, you can solve this in quadratic time. 01:01:48.410 --> 01:01:49.242 Cubic is obvious. 01:01:49.242 --> 01:01:50.700 There are three degrees of freedom. 01:01:50.700 --> 01:01:53.750 But I think, in fact, quadratic time. 01:01:53.750 --> 01:01:54.980 That's claim in the paper. 01:01:54.980 --> 01:01:58.430 I haven't studied motion planning algorithms 01:01:58.430 --> 01:01:58.950 for a while. 01:01:58.950 --> 01:02:01.710 So I don't know exactly how it goes. 01:02:01.710 --> 01:02:04.600 But in general, constant dimensional motion planning 01:02:04.600 --> 01:02:07.510 with constant numbers of objects can solved in polynomial time. 01:02:07.510 --> 01:02:13.776 But you can debate about the constants, 01:02:13.776 --> 01:02:15.000 which do matter here. 01:02:15.000 --> 01:02:17.810 I think the claim is tightness for that one. 01:02:17.810 --> 01:02:20.800 Here's another one which can be solved in-- their claim here, 01:02:20.800 --> 01:02:24.930 they say n squared log n they say. 01:02:24.930 --> 01:02:29.060 We can probably get n squared, but n squared log n is enough. 01:02:29.060 --> 01:02:31.575 It's a particular version of 3D motion planning. 01:02:38.950 --> 01:02:43.020 So we are given a vertical segment. 01:02:43.020 --> 01:02:45.450 That's our robot. 01:02:45.450 --> 01:02:47.030 We're going to have triangles. 01:02:47.030 --> 01:02:48.840 Those are our obstacles, because triangles 01:02:48.840 --> 01:02:49.870 we can make polyhedra. 01:02:49.870 --> 01:02:54.860 In this case, all the triangles will lie in horizontal planes. 01:02:54.860 --> 01:02:58.000 And the segment position will be vertical. 01:02:58.000 --> 01:02:59.920 And here you're only allowed translation. 01:03:04.971 --> 01:03:07.220 With translation and rotation, you could also do this. 01:03:07.220 --> 01:03:09.380 But they restrict to translation only, 01:03:09.380 --> 01:03:11.611 because there they can get a almost quadratic time 01:03:11.611 --> 01:03:12.110 algorithm. 01:03:12.110 --> 01:03:14.980 With rotation you have to add a couple of factors of n 01:03:14.980 --> 01:03:16.000 probably. 01:03:16.000 --> 01:03:19.340 But this version, they can solve in n squared log n. 01:03:19.340 --> 01:03:24.520 And it's 3SUM hard by similar kind of structure 01:03:24.520 --> 01:03:28.670 to the triangles covering triangle. 01:03:28.670 --> 01:03:31.540 Mainly, we build this cage-- the triangles aren't shaded in. 01:03:31.540 --> 01:03:34.750 But there's basically a triangle of triangles here, 01:03:34.750 --> 01:03:35.960 and a bunch of them. 01:03:35.960 --> 01:03:37.290 The segment is a unit length. 01:03:37.290 --> 01:03:41.350 And so there's this many to subdivide the space. 01:03:41.350 --> 01:03:43.190 They call this a cage. 01:03:43.190 --> 01:03:46.260 All of these triangles live in horizontal planes. 01:03:46.260 --> 01:03:48.760 There's no way for this vertical robot with translation 01:03:48.760 --> 01:03:51.060 only to get out. 01:03:51.060 --> 01:03:52.736 It's too long. 01:03:52.736 --> 01:03:54.110 And so the starting configuration 01:03:54.110 --> 01:03:55.270 is going to be up here. 01:03:55.270 --> 01:03:57.140 The destination is going to be down here. 01:03:57.140 --> 01:03:59.517 And in the middle triangle here, we're 01:03:59.517 --> 01:04:01.600 going to put a whole bunch of triangles like this. 01:04:01.600 --> 01:04:03.270 So just squish this down. 01:04:03.270 --> 01:04:04.304 Put them all in here. 01:04:04.304 --> 01:04:06.470 And you're going to be able to penetrate if and only 01:04:06.470 --> 01:04:07.469 if there's a blank spot. 01:04:13.540 --> 01:04:15.630 End of this paper. 01:04:15.630 --> 01:04:19.190 I think we covered almost all of these proofs. 01:04:19.190 --> 01:04:22.410 We started with the base problems, 3SUM, 3SUM prime, 01:04:22.410 --> 01:04:25.189 GeomBase, two versions of separator, 01:04:25.189 --> 01:04:27.480 which we weren't directly reducing from but we mimicked 01:04:27.480 --> 01:04:28.850 a zillion times. 01:04:28.850 --> 01:04:31.914 We had the degeneracy over here, strips 01:04:31.914 --> 01:04:33.330 covering a box, triangles covering 01:04:33.330 --> 01:04:34.980 a triangle, hole in the union, visible triangle. 01:04:34.980 --> 01:04:36.938 Those are actually all identical to each other. 01:04:36.938 --> 01:04:38.830 You could reduce in all directions. 01:04:38.830 --> 01:04:40.210 We talked about point cup. 01:04:40.210 --> 01:04:42.380 This was the linear programming. 01:04:42.380 --> 01:04:44.510 Motion planning, 3D motion planning 01:04:44.510 --> 01:04:50.100 happen to reduce here instead of there, and visibility. 01:04:50.100 --> 01:04:51.570 Cool. 01:04:51.570 --> 01:04:53.700 Lot of fun, little reductions, and it 01:04:53.700 --> 01:04:56.380 gives you a flavor for n squared hard-ish problems. 01:04:56.380 --> 01:04:59.320 AUDIENCE: Are there restrictions at all? 01:04:59.320 --> 01:05:02.260 PROFESSOR: I think all the restrictions are open. 01:05:02.260 --> 01:05:05.150 I mean, you'd have to check all the paper since 1995. 01:05:05.150 --> 01:05:07.420 But definitely in that paper, they're open. 01:05:07.420 --> 01:05:10.417 And I haven't heard of any-- there isn't a ton of work 01:05:10.417 --> 01:05:11.500 going the other direction. 01:05:11.500 --> 01:05:15.600 But it would differently nice to, especially 01:05:15.600 --> 01:05:19.930 if 3SUM-- it builds more if 3SUM is the right problem, 01:05:19.930 --> 01:05:22.300 if you can do reductions in both directions. 01:05:26.910 --> 01:05:30.490 I want to show you a couple more-- I'll show you 01:05:30.490 --> 01:05:34.060 one more reduction, which relates to a problem 01:05:34.060 --> 01:05:38.690 to be proved was strongly NP complete at some point 01:05:38.690 --> 01:05:40.330 way back when. 01:05:40.330 --> 01:05:42.980 This is about fixed angle chains. 01:05:42.980 --> 01:05:44.750 You have this kind of linkage structure. 01:05:44.750 --> 01:05:46.680 These are rigid bars. 01:05:46.680 --> 01:05:48.500 These are rigid angles. 01:05:48.500 --> 01:05:53.740 But you can still twist one segment around another. 01:05:53.740 --> 01:05:56.440 So it preserves the angles and the edge lengths. 01:05:56.440 --> 01:05:58.340 And so if I give you a structure like this, 01:05:58.340 --> 01:06:02.160 I want to know if I spin along this edge, 01:06:02.160 --> 01:06:04.090 if I take all this stuff and rotate it out 01:06:04.090 --> 01:06:07.880 of plane around this edge, does it hit anything? 01:06:07.880 --> 01:06:09.320 Or can it go all the way around? 01:06:09.320 --> 01:06:10.050 What about this edge? 01:06:10.050 --> 01:06:10.760 What about this edge? 01:06:10.760 --> 01:06:11.730 What about this edge? 01:06:11.730 --> 01:06:15.360 This is a polynomial time fixed angle chain problem. 01:06:15.360 --> 01:06:18.750 I want to know for every edge, which one's spinning causes 01:06:18.750 --> 01:06:20.940 a collision. 01:06:20.940 --> 01:06:23.790 In fact, if I just want to know whether these guys cause 01:06:23.790 --> 01:06:28.630 a collision, because if they do, a plus b equals c. 01:06:28.630 --> 01:06:32.320 I think that's maybe clear enough with your negative B/2 01:06:32.320 --> 01:06:33.050 in the middle. 01:06:33.050 --> 01:06:35.735 In this case, we've shifted-- the A, 01:06:35.735 --> 01:06:38.184 B, C is the x-coordinate in this structure. 01:06:38.184 --> 01:06:40.350 And these are candidate foldings here where we miss, 01:06:40.350 --> 01:06:43.390 here we collide. 01:06:43.390 --> 01:06:46.010 We separated things out by taking 01:06:46.010 --> 01:06:48.870 every item of a, subtracting a huge number from it to put it 01:06:48.870 --> 01:06:52.700 over here, and every item of c, adding a huge number. 01:06:52.700 --> 01:06:54.596 And because we're in the 3SUM prime problem, 01:06:54.596 --> 01:06:55.970 we know we get one item for each, 01:06:55.970 --> 01:06:58.600 so adding and subtracting and matching huge number 01:06:58.600 --> 01:07:03.400 will preserve all 3SUM pairs, 3SUM triples. 01:07:03.400 --> 01:07:05.560 So that let's us separate out this picture. 01:07:05.560 --> 01:07:07.520 And then you just have to check this reflection corresponds 01:07:07.520 --> 01:07:09.603 to adding in the right way, because of negation we 01:07:09.603 --> 01:07:12.200 divided by 2. 01:07:12.200 --> 01:07:13.641 That's it. 01:07:13.641 --> 01:07:16.670 So that's another. 01:07:16.670 --> 01:07:18.440 And there are a bunch of other problems, 01:07:18.440 --> 01:07:19.979 like if I give you two polygons, I 01:07:19.979 --> 01:07:22.020 want to know whether I can translate this polygon 01:07:22.020 --> 01:07:23.144 to fit inside that polygon. 01:07:23.144 --> 01:07:24.510 That's also 3SUM hard. 01:07:24.510 --> 01:07:28.730 Proof is a little bit messy, so I don't have it here. 01:07:28.730 --> 01:07:37.390 Let me mention me another more recent use of 3SUM conjecture 01:07:37.390 --> 01:07:39.535 is some non-quadratic lower bounds. 01:07:44.970 --> 01:07:47.940 So we're still going to assume the 3SUM conjecture, 01:07:47.940 --> 01:07:50.040 but we're going to prove that a problem requires 01:07:50.040 --> 01:07:52.490 some time other than n squared. 01:07:52.490 --> 01:07:56.280 So here are two problems where this has been done. 01:08:01.540 --> 01:08:03.682 These are graph problems, which is cool, 01:08:03.682 --> 01:08:05.182 because everything we've seen so far 01:08:05.182 --> 01:08:06.390 has been a geometric problem. 01:08:18.060 --> 01:08:19.540 So the weighted, undirected graph, 01:08:19.540 --> 01:08:21.630 I want to know whether there's a three cycle, 01:08:21.630 --> 01:08:24.297 also called a triangle, of given weight. 01:08:24.297 --> 01:08:26.880 So I want to know, for example, is there triangle weight of 0. 01:08:30.420 --> 01:08:33.630 This can be done in polynomial time, obviously. 01:08:33.630 --> 01:08:44.189 And the lower bound says is the number of edges to the 1.5, 01:08:44.189 --> 01:08:49.100 instead of 2 or however you want to think about this, 01:08:49.100 --> 01:08:52.060 minus epsilon. 01:08:52.060 --> 01:08:53.479 This is Mihai Petrescu. 01:08:53.479 --> 01:08:57.620 What he says is this problem, finding this in this much time, 01:08:57.620 --> 01:09:00.779 is 3SUM hard, meaning if this is possible, 01:09:00.779 --> 01:09:03.660 then 3SUM could be solved in sub-quadratic time. 01:09:03.660 --> 01:09:07.334 So there's a gap between this bound and 3SUM bound, 01:09:07.334 --> 01:09:08.500 introduced by the reduction. 01:09:17.220 --> 01:09:18.370 Here's another fun problem. 01:09:26.620 --> 01:09:29.189 Here we're given an unweighted graph, undirected graph. 01:09:29.189 --> 01:09:32.649 And we just want to find e triangles. 01:09:32.649 --> 01:09:36.370 Just list them form me please or tell me there aren't that many. 01:09:38.990 --> 01:09:42.630 You cannot do that in any better than e to the 4/3 minus epsilon 01:09:42.630 --> 01:09:46.760 times if you believe the 3SUM conjecture. 01:09:46.760 --> 01:09:50.290 So these are both 3SUM hard in a different sense 01:09:50.290 --> 01:09:52.634 from what we were using before. 01:09:52.634 --> 01:09:54.380 Before it was quadratic quadratic. 01:09:54.380 --> 01:09:57.310 With graphs, also there's v verses e. 01:09:57.310 --> 01:09:58.850 But neither of these are quadratic 01:09:58.850 --> 01:10:00.070 no matter how you slice them. 01:10:00.070 --> 01:10:03.046 So there are different. 01:10:03.046 --> 01:10:05.170 So there are a small number of pounds of that form. 01:10:05.170 --> 01:10:10.550 And that's kind of interesting and relatively hot area. 01:10:10.550 --> 01:10:12.880 So a few people are thinking about that. 01:10:12.880 --> 01:10:18.880 In particular, there's been a recent surge of interest 01:10:18.880 --> 01:10:20.650 in thinking about graph problems. 01:10:20.650 --> 01:10:23.860 Now for graph problems, there isn't 01:10:23.860 --> 01:10:25.645 a ton of work relating 3SUM, which 01:10:25.645 --> 01:10:28.120 is a very arithmetic problem to graph problems, 01:10:28.120 --> 01:10:31.140 but this is the beginning of that. 01:10:31.140 --> 01:10:32.530 But there are some other problems 01:10:32.530 --> 01:10:35.280 which people think are hard. 01:10:35.280 --> 01:10:36.850 So let me give you some of them. 01:10:41.630 --> 01:10:48.490 Diameter-- so here we're given a weighted, undirected graph. 01:10:48.490 --> 01:10:54.250 I want to know-- so delta v, w-- this is like CRLS notation. 01:10:54.250 --> 01:10:57.330 This is the weight of the minimum weight 01:10:57.330 --> 01:11:00.010 path from v to w. 01:11:00.010 --> 01:11:01.860 And I want to know the max overall v to w. 01:11:01.860 --> 01:11:04.430 What is the longest shortest path? 01:11:04.430 --> 01:11:05.130 That's diameter. 01:11:07.840 --> 01:11:22.670 Conjecture-- no V to the 3 minus epsilon. 01:11:22.670 --> 01:11:27.730 So here there's a lot of interest around cubic problems, 01:11:27.730 --> 01:11:30.460 because this problem seems cubic. 01:11:30.460 --> 01:11:36.270 Another closely related problem is all pair shortest paths. 01:11:36.270 --> 01:11:41.770 I want to know delta of v, w for all v and w. 01:11:41.770 --> 01:11:44.210 This is, of course, a harder problem than diameter. 01:11:44.210 --> 01:11:48.727 Also, this conjecture implies conjecture over here. 01:11:48.727 --> 01:11:50.060 This one's a little more famous. 01:11:50.060 --> 01:11:53.130 The all pairs shortest path conjecture is that you cannot 01:11:53.130 --> 01:11:56.110 solve this in truly sub-cubic time. 01:11:56.110 --> 01:11:59.070 There are, again, poly log improvements. 01:11:59.070 --> 01:12:04.600 But you cannot beat-- we don't know how to beat by a v 01:12:04.600 --> 01:12:08.880 to the epsilon factor, n to the epsilon factor over 01:12:08.880 --> 01:12:14.330 the standard algorithm, which should be Floyd Warshall, 01:12:14.330 --> 01:12:15.340 the triply nested loop. 01:12:15.340 --> 01:12:20.947 Relax every edge n times is the standard cubed algorithm. 01:12:20.947 --> 01:12:22.530 For sparse graphs you could do better. 01:12:22.530 --> 01:12:26.874 For dense graphs, the claim is that's the best you can do. 01:12:26.874 --> 01:12:29.040 And so there's a bunch of problems that are all pair 01:12:29.040 --> 01:12:30.050 shortest path hard. 01:12:30.050 --> 01:12:34.320 There are some problems that are diameter hard. 01:12:34.320 --> 01:12:37.150 Being diameter hard is a little bit stronger. 01:12:37.150 --> 01:12:41.130 Obviously, diameter can reduce to all pairs 01:12:41.130 --> 01:12:45.190 shortest paths via now we want a sub-cubic reduction, something 01:12:45.190 --> 01:12:47.670 n to the 3 minus epsilon time instead of n 01:12:47.670 --> 01:12:49.000 to the 2 minus epsilon. 01:12:49.000 --> 01:12:51.270 You can reduce diameter to all pairs shortest paths. 01:12:51.270 --> 01:12:53.894 Big open problem is whether you could reduce all pairs shortest 01:12:53.894 --> 01:12:55.620 paths to diameter. 01:12:55.620 --> 01:12:57.360 But you can reduce all pairs shortest 01:12:57.360 --> 01:13:00.050 paths to some cool problems. 01:13:07.500 --> 01:13:12.190 I have an image of some reductions. 01:13:12.190 --> 01:13:18.380 So negative triangle-- is there a triangle of negative weight? 01:13:25.420 --> 01:13:27.910 Over here we wanted a triangle of weight exactly 0. 01:13:27.910 --> 01:13:30.162 That helps you find things faster. 01:13:30.162 --> 01:13:32.370 Over here we just want a triangle of negative weight. 01:13:32.370 --> 01:13:34.090 There's more options for those. 01:13:34.090 --> 01:13:37.090 Finding that is all pairs shortest paths hard. 01:13:37.090 --> 01:13:39.900 You can reduce all pairs shortest path 01:13:39.900 --> 01:13:42.540 that problem-- kind of crazy. 01:13:42.540 --> 01:13:45.080 And here, the reduction is not a single-- 01:13:45.080 --> 01:13:47.150 or a constant number of call reductions 01:13:47.150 --> 01:13:48.630 like we've been doing. 01:13:48.630 --> 01:13:50.260 This has a huge output. 01:13:50.260 --> 01:13:53.040 This only has a yes or no output. 01:13:53.040 --> 01:13:54.830 So this reduction's a little bit crazy. 01:13:54.830 --> 01:14:00.720 But basically you take the sum over all calls 01:14:00.720 --> 01:14:03.060 for this reduction. 01:14:03.060 --> 01:14:07.334 And it just has to work out that if you could solve 01:14:07.334 --> 01:14:09.250 negative triangle and sub-cubic time, then you 01:14:09.250 --> 01:14:11.347 could also solve all pairs shortest path 01:14:11.347 --> 01:14:13.410 in sub-cubic time. 01:14:13.410 --> 01:14:19.090 So if you take the sum of the n primes 01:14:19.090 --> 01:14:22.420 to the power 3 minus epsilon, this 01:14:22.420 --> 01:14:25.570 should work out to n to the 3 minus 01:14:25.570 --> 01:14:29.930 epsilon over the different the different calls you 01:14:29.930 --> 01:14:31.180 make to the negative triangle. 01:14:31.180 --> 01:14:33.834 So it's a weaker notion of reduction, which 01:14:33.834 --> 01:14:35.000 lets you prove these things. 01:14:35.000 --> 01:14:37.177 Kind of a big innovation from just-- this one 01:14:37.177 --> 01:14:38.260 hasn't even published yet. 01:14:38.260 --> 01:14:39.860 It will be in [? Sodo ?] 2015. 01:14:39.860 --> 01:14:42.220 It's on the archive now. 01:14:42.220 --> 01:14:45.080 And there been a few papers over the last few years doing 01:14:45.080 --> 01:14:46.204 these kinds of reductions. 01:14:46.204 --> 01:14:47.620 Actually the negative triangle one 01:14:47.620 --> 01:14:51.420 is from an older paper by Vassilevska Williams 01:14:51.420 --> 01:14:56.540 and Williams. 01:14:56.540 --> 01:15:00.440 It's a husband and wife team, Stanford. 01:15:00.440 --> 01:15:16.350 OK, some more problems, radius-- for vertex v, the radius 01:15:16.350 --> 01:15:20.430 around v is how big a ball do need to grow in order 01:15:20.430 --> 01:15:21.760 to cover all the vertices. 01:15:21.760 --> 01:15:27.240 I want to know what's the farthest vertex w from v. 01:15:27.240 --> 01:15:29.520 And then the radius of the graph is 01:15:29.520 --> 01:15:32.580 what is the best such vertex that minimizes the radius, 01:15:32.580 --> 01:15:34.090 the vertex. 01:15:34.090 --> 01:15:35.270 So closely related to this. 01:15:35.270 --> 01:15:38.720 But it is equivalent to negative triangle. 01:15:38.720 --> 01:15:42.220 Both of these are-- this is a-- well, OK, of course, 01:15:42.220 --> 01:15:44.140 you could reduce negative triangle to-- well, 01:15:44.140 --> 01:15:44.973 it's not so obvious. 01:15:44.973 --> 01:15:47.750 You can reduce radius to all pairs shortest paths, 01:15:47.750 --> 01:15:49.770 because you have all the deltas. 01:15:49.770 --> 01:15:52.960 You can compute that max-min in quadratic time. 01:15:52.960 --> 01:15:55.900 So that means all these problems are equivalent to each other 01:15:55.900 --> 01:15:58.790 up to sub-cubic reductions. 01:15:58.790 --> 01:16:01.950 OK, median is another problem-- very similar. 01:16:01.950 --> 01:16:05.830 I just replace this max with a sum. 01:16:05.830 --> 01:16:10.130 So that's some other kind of central located vertex. 01:16:10.130 --> 01:16:11.890 That's median. 01:16:11.890 --> 01:16:16.350 And that's also equivalent to all these problems. 01:16:16.350 --> 01:16:19.770 So there's a growing list of problems 01:16:19.770 --> 01:16:23.586 that are in this sort of cubic space. 01:16:23.586 --> 01:16:25.460 All the problems are conjectured to be cubic. 01:16:25.460 --> 01:16:27.350 And a lot of them are equivalent to each other, 01:16:27.350 --> 01:16:29.165 though there's this divide between diameter 01:16:29.165 --> 01:16:31.030 and all pairs shortest path. 01:16:31.030 --> 01:16:36.750 There are some bounds, assuming strong ETH. 01:16:36.750 --> 01:16:45.560 So if you have strong ETH, then there's 01:16:45.560 --> 01:16:50.799 no e to the 2 minus epsilon algorithm. 01:16:50.799 --> 01:16:52.090 This is not quite what we want. 01:16:52.090 --> 01:16:54.300 We want to v to the 3 minus epsilon. 01:16:54.300 --> 01:16:59.510 This is a statement about sparse graphs. 01:16:59.510 --> 01:17:01.580 You can beat this for dense graphs. 01:17:01.580 --> 01:17:03.830 For sparse graphs this is interesting. 01:17:03.830 --> 01:17:06.487 This is for the worst case relationship between v and E. 01:17:06.487 --> 01:17:07.820 So you get somethings like this. 01:17:07.820 --> 01:17:10.880 You get this even if you allow some approximatability. 01:17:10.880 --> 01:17:14.550 But we still don't have a way to actually prove this conjecture. 01:17:14.550 --> 01:17:15.195 Yeah. 01:17:15.195 --> 01:17:17.320 AUDIENCE: Since you haven't said anything about it, 01:17:17.320 --> 01:17:20.355 I assume that 3SUM and this world, there 01:17:20.355 --> 01:17:22.149 is no bridge between them. 01:17:22.149 --> 01:17:24.190 PROFESSOR: I believe there's no bridge currently. 01:17:24.190 --> 01:17:25.370 This is the closest thing. 01:17:25.370 --> 01:17:28.680 And they do seem similar. 01:17:28.680 --> 01:17:30.060 And also here we have cubic. 01:17:30.060 --> 01:17:31.840 There we have quadratic. 01:17:31.840 --> 01:17:32.900 So maybe you can do it. 01:17:32.900 --> 01:17:35.180 Maybe with 5 SUM you could show some relation. 01:17:35.180 --> 01:17:35.850 I don't know. 01:17:35.850 --> 01:17:39.700 But there's no such theorem yet. 01:17:39.700 --> 01:17:43.010 These problems sound very similar to these problems, 01:17:43.010 --> 01:17:45.510 in particular listing a bunch of negative triangles 01:17:45.510 --> 01:17:47.160 is just as hard as finding one. 01:17:47.160 --> 01:17:49.749 And so there's some similarity to over here. 01:17:49.749 --> 01:17:51.790 But the constraint on the triangles is different. 01:17:51.790 --> 01:17:53.320 Here we want negative ones. 01:17:53.320 --> 01:17:57.690 Here any triangle or triangle of 0 weight. 01:17:57.690 --> 01:18:00.200 So it's an interesting space. 01:18:00.200 --> 01:18:02.100 It's still very ongoing. 01:18:02.100 --> 01:18:04.770 All of these things-- this stuff, and this stuff-- 01:18:04.770 --> 01:18:07.410 is all within the last four years. 01:18:07.410 --> 01:18:10.270 So it'll be interesting to see how it develops. 01:18:10.270 --> 01:18:12.520 But these are the current approaches 01:18:12.520 --> 01:18:14.710 to understanding n squared, n cubed, 01:18:14.710 --> 01:18:17.890 and sort of the low end of the polynomial spectra. 01:18:17.890 --> 01:18:20.117 I think it's pretty interesting, but also where 01:18:20.117 --> 01:18:22.450 we know the least and have the fewest general techniques 01:18:22.450 --> 01:18:23.670 for proving things. 01:18:23.670 --> 01:18:26.010 3SUM is a little more established. 01:18:26.010 --> 01:18:29.810 And there's a bunch of proofs like the one you've seen-- not 01:18:29.810 --> 01:18:32.470 too many out there actually. 01:18:32.470 --> 01:18:34.700 But they're quite accessible. 01:18:34.700 --> 01:18:37.870 This stuff is still, I think, converging. 01:18:37.870 --> 01:18:39.860 But very exciting things, relating 01:18:39.860 --> 01:18:42.510 all these algorithmic problems to each other and kind 01:18:42.510 --> 01:18:43.995 of a nice way for us to end. 01:18:43.995 --> 01:18:48.190 This is my last lecture for 6890. 01:18:48.190 --> 01:18:52.150 And the next two classes are lectures by [? Acosis ?] about 01:18:52.150 --> 01:18:54.980 how [? arithmic ?] game theory in a class called PPAD and PPAD 01:18:54.980 --> 01:18:57.350 hardness, which is its own universe, 01:18:57.350 --> 01:19:00.260 but around economic game theory and very cool stuff. 01:19:00.260 --> 01:19:02.890 He is the expert on it and he's a professor here, 01:19:02.890 --> 01:19:05.280 so he graciously agreed to do two lectures on it. 01:19:05.280 --> 01:19:05.910 Should be fun. 01:19:05.910 --> 01:19:07.460 I'm looking forward to it. 01:19:07.460 --> 01:19:09.010 Thanks.