WEBVTT 00:00:00.080 --> 00:00:02.430 The following content is provided under a Creative 00:00:02.430 --> 00:00:03.820 Commons license. 00:00:03.820 --> 00:00:06.050 Your support will help MIT OpenCourseWare 00:00:06.050 --> 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 --> 00:00:17.262 at ocw.mit.edu. 00:00:25.845 --> 00:00:28.880 PROFESSOR: All right, today we are going to learn to count. 00:00:28.880 --> 00:00:34.460 One, two, three-- In a algorithmic sense of course, 00:00:34.460 --> 00:00:37.320 and prove hardness of counting style problems. 00:00:37.320 --> 00:00:41.830 Or more generally, any problem where the output is an integer. 00:00:41.830 --> 00:00:45.370 Like approximation algorithms, we 00:00:45.370 --> 00:00:48.630 need to define a slightly stronger version of our NP 00:00:48.630 --> 00:00:49.215 style problem. 00:00:49.215 --> 00:00:52.200 It's not going to be a decision problem. 00:00:52.200 --> 00:00:54.360 The remaining type is a search problem. 00:01:03.380 --> 00:01:05.379 Search problem you're trying to find a solution. 00:01:05.379 --> 00:01:07.070 This is just like optimization problem, 00:01:07.070 --> 00:01:08.980 but with no objective function. 00:01:08.980 --> 00:01:12.250 Today all solutions are created equal. 00:01:12.250 --> 00:01:16.510 So we just want to know how many there are. 00:01:16.510 --> 00:01:19.320 So we're given an instance, the problem. 00:01:19.320 --> 00:01:21.555 We want to produce a solution. 00:01:24.234 --> 00:01:25.650 Whereas in the decision problem we 00:01:25.650 --> 00:01:29.530 want to know whether a solution exists. 00:01:29.530 --> 00:01:35.234 With a search problem we want to find a solution. 00:01:35.234 --> 00:01:36.150 Almost the same thing. 00:01:56.040 --> 00:01:58.490 So that would be a search problem in general. 00:01:58.490 --> 00:02:02.900 For an NP search problem you can recognize what is an instance. 00:02:02.900 --> 00:02:04.490 And you could recognize solutions 00:02:04.490 --> 00:02:07.290 to instances in polynomial time. 00:02:07.290 --> 00:02:10.050 Given an instance in the solution you can say, 00:02:10.050 --> 00:02:11.295 yes that's a valid solution. 00:02:14.620 --> 00:02:16.687 OK for such a search problem of course, 00:02:16.687 --> 00:02:18.270 is the corresponding decision problem. 00:02:18.270 --> 00:02:21.080 Which is, does there exist a solution? 00:02:21.080 --> 00:02:25.890 If you can solve this, then you can solve that. 00:02:25.890 --> 00:02:29.160 You could solve the decision problem in NP 00:02:29.160 --> 00:02:32.460 by guessing a solution and so on. 00:02:32.460 --> 00:02:37.010 So this is intricately related to the notion of a certificate 00:02:37.010 --> 00:02:38.130 for an NP problem. 00:02:38.130 --> 00:02:40.250 The idea solutions are certificates. 00:02:40.250 --> 00:02:41.890 But when we say problem is in NP, 00:02:41.890 --> 00:02:44.080 we say there is some way to define certificate 00:02:44.080 --> 00:02:47.730 so that this kind of problem can be set up. 00:02:47.730 --> 00:02:49.360 And the goal here-- the point here 00:02:49.360 --> 00:02:51.675 is to solidify a specific notion of certificate. 00:02:51.675 --> 00:02:53.700 We can't just use any one, because we're 00:02:53.700 --> 00:02:55.550 going to count them the-- If you formulate 00:02:55.550 --> 00:02:56.966 the certificates in different ways 00:02:56.966 --> 00:02:58.380 you'll get different counts. 00:02:58.380 --> 00:03:03.350 But in general, every NP problem can 00:03:03.350 --> 00:03:09.210 be converted into an NP search problem in at least one way. 00:03:13.720 --> 00:03:15.880 But each notional certificate gives you 00:03:15.880 --> 00:03:18.510 a notion of the search problem. 00:03:18.510 --> 00:03:22.700 OK, in some complexity context these are called NP relations, 00:03:22.700 --> 00:03:24.450 the way you specify what a certificate is. 00:03:24.450 --> 00:03:27.680 But I think this is the more algorithmic perspective. 00:03:27.680 --> 00:03:29.150 All right, so given such a search 00:03:29.150 --> 00:03:31.145 problem we turn it into a counting problem. 00:03:38.800 --> 00:03:41.860 So that's a search problem, is called A. 00:03:41.860 --> 00:03:46.800 Then counting problem will be sharp A, not hashtag 00:03:46.800 --> 00:03:52.410 A. Some people call it number A. The problem is, 00:03:52.410 --> 00:03:59.305 count the number of solutions for a given instance. 00:04:09.770 --> 00:04:12.780 OK so in particular, you detect whether the number 00:04:12.780 --> 00:04:14.540 is zero, or not zero. 00:04:14.540 --> 00:04:16.850 So this is strictly harder in some sense 00:04:16.850 --> 00:04:20.260 than the decision problem, does there exist a solution. 00:04:20.260 --> 00:04:23.670 And we will see some problems where the search problem is 00:04:23.670 --> 00:04:26.330 polynomial, but the corresponding counting problem 00:04:26.330 --> 00:04:30.200 is actually hard. 00:04:30.200 --> 00:04:31.770 I can't say NP hard, there's going 00:04:31.770 --> 00:04:34.310 to be a new notion of hardness. 00:04:34.310 --> 00:04:36.760 So some examples. 00:04:36.760 --> 00:04:40.630 Pretty much every problem we've defined as a decision problem 00:04:40.630 --> 00:04:44.110 had a search problem in mind. 00:04:44.110 --> 00:04:47.607 So something like SAT, you need to satisfy all of the things. 00:04:47.607 --> 00:04:49.190 So there's no objective function here, 00:04:49.190 --> 00:04:50.750 but you want to know how many different ways, 00:04:50.750 --> 00:04:53.200 how many different variable assignments are there that 00:04:53.200 --> 00:04:55.350 satisfy the given formula? 00:04:55.350 --> 00:05:00.070 Or take your favorite pencil and paper puzzle, 00:05:00.070 --> 00:05:06.720 we'll be looking at Shakashaka today, again. 00:05:06.720 --> 00:05:08.540 How many different solutions are there? 00:05:08.540 --> 00:05:10.190 You'd like, when designing a puzzle, 00:05:10.190 --> 00:05:12.370 usually you want to know that it's unique. 00:05:12.370 --> 00:05:14.920 So it'd be nice if you could count the number of solutions 00:05:14.920 --> 00:05:16.930 and show that it's one. 00:05:16.930 --> 00:05:20.700 These problems are going to turn out to be very hard of course. 00:05:20.700 --> 00:05:24.540 So let's define a notion of hardness. 00:05:24.540 --> 00:05:28.325 Sharp P is going to be the class of all of these counting 00:05:28.325 --> 00:05:28.825 problems. 00:05:42.290 --> 00:05:44.095 This is the sort of certificate. 00:05:44.095 --> 00:05:44.720 Yeah, question? 00:05:44.720 --> 00:05:46.040 AUDIENCE: Just for the puzzle application, 00:05:46.040 --> 00:05:47.965 is it going to turn out that counting if there's 00:05:47.965 --> 00:05:50.048 one solution versus one-to-one solution is as hard 00:05:50.048 --> 00:05:51.898 as just counting total numbers? 00:05:51.898 --> 00:05:56.350 PROFESSOR: It's not quite as hard, 00:05:56.350 --> 00:06:00.320 but we will show that distinguishing one 00:06:00.320 --> 00:06:04.430 for more than one is very hard, is NP complete actually. 00:06:04.430 --> 00:06:07.514 That's a decision problem that could show that's NP complete. 00:06:07.514 --> 00:06:09.180 So normally we think of zero versus one, 00:06:09.180 --> 00:06:11.060 but it turns out one versus two is not-- 00:06:11.060 --> 00:06:15.870 or one versus more than one is about the same difficulty. 00:06:15.870 --> 00:06:19.290 Counting is even harder I would say. 00:06:19.290 --> 00:06:20.730 But it's bad news all around. 00:06:20.730 --> 00:06:22.210 There's different notions of bad. 00:06:24.720 --> 00:06:29.810 Cool, so this is the sort of certificate perspective. 00:06:29.810 --> 00:06:32.190 With NP we-- I had given you two different definitions 00:06:32.190 --> 00:06:35.880 of certificate perspective, and a non-deterministic computation 00:06:35.880 --> 00:06:36.770 perspective. 00:06:36.770 --> 00:06:40.290 You can do the same computational perspective here. 00:06:40.290 --> 00:06:47.290 You could say sharp P is the set of problems, 00:06:47.290 --> 00:06:51.165 solved by polynomial time. 00:06:55.660 --> 00:06:57.995 Call it a non-deterministic counting algorithm. 00:07:03.297 --> 00:07:05.380 We don't need Turing machines for this definition, 00:07:05.380 --> 00:07:08.040 although that was of course the original definition. 00:07:08.040 --> 00:07:11.320 So take your favorite non-deterministic algorithm 00:07:11.320 --> 00:07:14.920 as usual for NP, it makes guesses at various points, 00:07:14.920 --> 00:07:16.580 multiple branches. 00:07:16.580 --> 00:07:20.720 With the NP algorithm the way we'd execute it on an NP style 00:07:20.720 --> 00:07:22.970 machine is that we would see whether there's 00:07:22.970 --> 00:07:24.910 any path that led to a yes. 00:07:24.910 --> 00:07:27.790 Again, it's going to output yes or no at the end. 00:07:27.790 --> 00:07:30.930 In this case, what the computer does, the sharp P style 00:07:30.930 --> 00:07:34.290 computer, is it-- conceptually it runs all the branches. 00:07:34.290 --> 00:07:37.310 It counts the number of yeses and returns that number. 00:07:37.310 --> 00:07:41.210 So even though the algorithm is designed to return yes or no, 00:07:41.210 --> 00:07:44.030 when it executes it actually outputs a number. 00:07:44.030 --> 00:07:47.190 The original paper says magically. 00:07:47.190 --> 00:07:50.014 It's just as magic as an NP machine, but a little-- even 00:07:50.014 --> 00:07:50.930 a little more magical. 00:07:53.630 --> 00:07:57.080 OK so you-- I mean if you're not comfortable with that we just 00:07:57.080 --> 00:08:00.330 use this definition, same thing. 00:08:00.330 --> 00:08:07.727 This is all work done by Les Valiant in the late '70s. 00:08:07.727 --> 00:08:08.310 These notions. 00:08:22.710 --> 00:08:24.320 So it's pretty clear, it's pretty easy 00:08:24.320 --> 00:08:26.410 to show all these problems are in sharp P, 00:08:26.410 --> 00:08:28.951 because they were-- the corresponding decision problems 00:08:28.951 --> 00:08:29.450 were in NP. 00:08:29.450 --> 00:08:34.110 We convert them into sharp P algorithms. 00:08:34.110 --> 00:08:37.210 Now let's think about hardness with respect to sharp P. 00:08:37.210 --> 00:08:38.990 As usual we want it to mean as hard 00:08:38.990 --> 00:08:40.409 as all problems in that class. 00:08:40.409 --> 00:08:45.270 Meaning that we can reduce all those problems to our problem. 00:08:45.270 --> 00:08:48.520 And the question is, by what kind of reductions? 00:08:48.520 --> 00:08:51.785 And here we're going to allow very powerful reductions. 00:08:56.972 --> 00:08:58.430 We've talked about these reductions 00:08:58.430 --> 00:09:01.320 but never actually been allowed to use them. 00:09:01.320 --> 00:09:03.550 Multicall, Cook-style reductions. 00:09:06.080 --> 00:09:08.770 I think in general, this is a common approach 00:09:08.770 --> 00:09:14.530 for FNP, which is Functions NP style functions. 00:09:14.530 --> 00:09:17.520 Which you can also think of this as kind of-- the counting 00:09:17.520 --> 00:09:19.850 version is, the output is a value. 00:09:19.850 --> 00:09:24.050 So you have a function instead of just a decision question. 00:09:24.050 --> 00:09:28.190 When the output is some thing, some number, in this case. 00:09:28.190 --> 00:09:30.790 You might have to manipulate that number at the end. 00:09:30.790 --> 00:09:33.180 And so at the very least, you need to make a call 00:09:33.180 --> 00:09:35.360 and do some stuff before you return your modified 00:09:35.360 --> 00:09:37.290 answer in your reduction. 00:09:37.290 --> 00:09:39.870 But in fact we're going to allow full tilt, 00:09:39.870 --> 00:09:43.790 you could make multiple calls to some hypothetical solution 00:09:43.790 --> 00:09:48.757 to your problem in order to solve all problems in sharp P. 00:09:48.757 --> 00:09:50.840 And we'll actually use multicall a bunch of times. 00:09:50.840 --> 00:09:53.140 We won't always need multicall. 00:09:53.140 --> 00:09:55.850 Often we'll be able to get away with a much simpler kind 00:09:55.850 --> 00:09:56.910 of reduction. 00:09:56.910 --> 00:09:58.570 Let me tell you that kind now. 00:10:01.990 --> 00:10:03.881 But in general we allow arbitrary multicall. 00:10:03.881 --> 00:10:04.380 Yeah? 00:10:04.380 --> 00:10:07.820 AUDIENCE: Your not limited in the number of multicalls? 00:10:07.820 --> 00:10:08.550 PROFESSOR: Right. 00:10:08.550 --> 00:10:10.580 You could do polynomial number of multicalls. 00:10:10.580 --> 00:10:14.030 As before, reduction should be polynomial time. 00:10:14.030 --> 00:10:18.070 But, so you're basically given an algorithm. 00:10:18.070 --> 00:10:24.160 That's usually called an Oracle that solves your problem, 00:10:24.160 --> 00:10:28.960 solves your problem B. And you want to solve-- yeah, 00:10:28.960 --> 00:10:33.030 you want to solve A by multiple calls to B. So 00:10:33.030 --> 00:10:35.370 what, we'll see a bunch of examples of that. 00:10:35.370 --> 00:10:39.370 Here's a more familiar style of reduction. 00:10:39.370 --> 00:10:42.060 And often, for a lot of problems we can get away with this. 00:10:42.060 --> 00:10:43.950 But especially a lot of the early proofs 00:10:43.950 --> 00:10:45.380 needed the multicall. 00:10:45.380 --> 00:10:47.380 And as you'll see you can do lots of cool tricks 00:10:47.380 --> 00:10:50.500 with multicall using number theory. 00:10:53.060 --> 00:10:56.790 So parsimonious reduction. 00:10:56.790 --> 00:10:58.855 This is for NP search problems. 00:11:03.470 --> 00:11:09.080 This is going to be a lot like an NP reduction, regular style. 00:11:09.080 --> 00:11:16.470 So again we convert an instance x of a, five function f, 00:11:16.470 --> 00:11:22.700 into an instance x prime of b. 00:11:26.880 --> 00:11:32.030 And that function should be computable in poly time. 00:11:34.970 --> 00:11:37.386 So far just like an NP reduction. 00:11:40.060 --> 00:11:43.450 And usually we would say, for a search problem, 00:11:43.450 --> 00:11:45.630 we would say there exists a solution 00:11:45.630 --> 00:11:48.740 for x, if and only if, there exists a solution for x prime. 00:11:48.740 --> 00:11:53.670 That would be the direct analog of an NP style reduction. 00:11:53.670 --> 00:11:55.670 But we're going to ask for a stronger condition. 00:11:55.670 --> 00:12:01.040 Which is that the number of solutions to problem 00:12:01.040 --> 00:12:10.890 A to instance x, equals the number of solutions of type B 00:12:10.890 --> 00:12:13.540 to instance x prime. 00:12:13.540 --> 00:12:16.050 OK so in particular, this one's going to equal to one, 00:12:16.050 --> 00:12:18.520 this one will be greater than equal to one, and vice versa. 00:12:18.520 --> 00:12:20.829 So this is stronger than an NP style 00:12:20.829 --> 00:12:22.870 reduction for the corresponding decision problem. 00:12:27.820 --> 00:12:29.780 Yeah, so I even wrote that down. 00:12:29.780 --> 00:12:34.100 This implies that the decision problems have the same answer. 00:12:42.100 --> 00:12:45.435 So in particular, this implies that we have an NP reduction. 00:12:52.020 --> 00:12:59.540 So in particular if A, the decision version of A 00:12:59.540 --> 00:13:02.680 is NP hard, then the decision version of B is NP hard. 00:13:02.680 --> 00:13:10.400 But more interesting is that, if A is sharp P hard, 00:13:10.400 --> 00:13:12.130 then B is sharp P hard. 00:13:17.820 --> 00:13:21.080 But also this holds for NP. 00:13:21.080 --> 00:13:25.130 For the decision versions of A and B. Sorry, 00:13:25.130 --> 00:13:30.505 sharp A and sharp B. OK this is a subtle distinction. 00:13:33.600 --> 00:13:36.710 For sharp P hardness we're talking about the counting 00:13:36.710 --> 00:13:38.630 problems, and we're talking about making calls 00:13:38.630 --> 00:13:40.067 to other counting solutions. 00:13:40.067 --> 00:13:42.400 Then doing things with those numbers and who knows what, 00:13:42.400 --> 00:13:44.290 making many calls. 00:13:44.290 --> 00:13:45.930 With parsimonious reduction we're 00:13:45.930 --> 00:13:48.990 thinking about the non-counting version. 00:13:48.990 --> 00:13:50.830 Just the search problem. 00:13:50.830 --> 00:13:55.050 And so we're not worried about counting solutions directly, 00:13:55.050 --> 00:13:57.300 I mean it-- what's nice about parsimonious reductions 00:13:57.300 --> 00:13:59.040 is they look just like NP reductions 00:13:59.040 --> 00:14:01.250 for the regular old problems. 00:14:01.250 --> 00:14:03.240 We just need this extra property, 00:14:03.240 --> 00:14:07.180 parsimony that the number of solutions to the unit 00:14:07.180 --> 00:14:09.210 is preserved through the transformation. 00:14:09.210 --> 00:14:14.350 And a lot of the proofs that we've covered follow-- 00:14:14.350 --> 00:14:17.160 have this property and will be good for us. 00:14:17.160 --> 00:14:19.410 If we can get our source problems hard, 00:14:19.410 --> 00:14:22.430 then we'll get a lot of target problems hard as well. 00:14:25.390 --> 00:14:34.100 Well let me tell you about one more version which I made up. 00:14:34.100 --> 00:14:35.305 C-monious reductions. 00:14:43.550 --> 00:14:45.060 This is my attempt at understanding 00:14:45.060 --> 00:14:47.760 the entomology of parsimonious. 00:14:47.760 --> 00:14:53.880 Which is something like little money, being very thrifty. 00:14:53.880 --> 00:14:56.030 So this is having a little bit more money. 00:14:56.030 --> 00:14:58.047 You have C dollars. 00:14:58.047 --> 00:14:59.880 But you have to be very consistent about it. 00:14:59.880 --> 00:15:01.920 I should probably add some word that means 00:15:01.920 --> 00:15:03.110 uniform in the middle there. 00:15:03.110 --> 00:15:08.390 But I want the number of solutions of x prime to equal 00:15:08.390 --> 00:15:13.315 c times the number of solutions to x. 00:15:16.110 --> 00:15:19.080 C a fixed constant. 00:15:19.080 --> 00:15:22.510 And it has to be the same for every x. 00:15:25.700 --> 00:15:28.020 This would be just as good from a sharp P perspective. 00:15:28.020 --> 00:15:31.690 Because if I could solve B and I wanted to solve A, 00:15:31.690 --> 00:15:35.680 I would convert A to B, run the thing, then divide by C. 00:15:35.680 --> 00:15:39.830 There will never be a remainder and then I have my answer to A. 00:15:39.830 --> 00:15:41.690 As long as C's not zero. 00:15:41.690 --> 00:15:45.930 C should be an integer here. 00:15:45.930 --> 00:15:51.790 We will see a bunch of C-monious reductions. 00:15:51.790 --> 00:15:58.110 I guess, yeah it doesn't have to be totally independent of x. 00:15:58.110 --> 00:16:00.480 It can depend on things like n. 00:16:00.480 --> 00:16:03.870 Something that we can compute easily I guess. 00:16:03.870 --> 00:16:06.030 Shouldn't be too dependent on x. 00:16:06.030 --> 00:16:10.340 All right, let's do-- let's look at some examples. 00:16:10.340 --> 00:16:23.030 So, going to make a list here of sharp P complete problems. 00:16:25.830 --> 00:16:29.560 And we'll start with versions of SAT because we like SAT. 00:16:29.560 --> 00:16:36.120 So I'm just going to tell you that sharp 3SAT is hard. 00:16:36.120 --> 00:16:40.490 First sharp SAT is hard, and the usual proof of SAT hardness 00:16:40.490 --> 00:16:45.460 shows sharp P completeness for sharp SAT. 00:16:45.460 --> 00:16:49.340 And if you're careful about the conversion from SAT to 3CNF 00:16:49.340 --> 00:16:51.290 you can get sharp three satisfied. 00:16:51.290 --> 00:16:53.240 It's not terribly interesting and tedious. 00:16:53.240 --> 00:16:57.660 So I will skip that one. 00:16:57.660 --> 00:17:01.120 So what about Planar 3SAT? 00:17:03.710 --> 00:17:06.970 I stared at this diagram many times for a while. 00:17:06.970 --> 00:17:12.130 This is lecture seven for replacing a crossing in a 3SAT 00:17:12.130 --> 00:17:14.210 thing with this picture. 00:17:14.210 --> 00:17:17.950 And all of this argument and this table in particular, 00:17:17.950 --> 00:17:20.599 was concluding that the variables 00:17:20.599 --> 00:17:21.900 are forced in this scenario. 00:17:21.900 --> 00:17:25.089 If you know what A and B are-- so 00:17:25.089 --> 00:17:26.650 once you choose what A and B are, 00:17:26.650 --> 00:17:28.650 these two have to be copies of A and B 00:17:28.650 --> 00:17:31.760 and then it ended up that A-- We proved 00:17:31.760 --> 00:17:34.680 that A1 equaled A2 and B1 equaled B2, 00:17:34.680 --> 00:17:36.860 and then these variables were all 00:17:36.860 --> 00:17:38.820 determined by these formula. 00:17:38.820 --> 00:17:41.440 And so once you know A and B all the variable settings 00:17:41.440 --> 00:17:42.220 are forced. 00:17:42.220 --> 00:17:45.180 Which means you preserve the number of solutions. 00:17:45.180 --> 00:17:50.330 So planar sharp 3SAT is sharp P complete. 00:17:53.360 --> 00:17:56.200 I'd like to pretend that there's some debate within the sharp P 00:17:56.200 --> 00:18:00.640 community about whether the sharp P are here or here. 00:18:00.640 --> 00:18:03.580 I kind of prefer it here, but I've seen it over here, 00:18:03.580 --> 00:18:04.240 so you know. 00:18:07.650 --> 00:18:10.140 I don't think I've even seen that mentioned but I'm 00:18:10.140 --> 00:18:12.530 sure it's out there somewhere. 00:18:12.530 --> 00:18:16.550 So let's flip through our other planar hardness proofs. 00:18:16.550 --> 00:18:19.920 This is planar monotone rectilinear 3SAT. 00:18:19.920 --> 00:18:21.790 Mainly the monotone aspect. 00:18:21.790 --> 00:18:24.170 We wanted there to be all the variables 00:18:24.170 --> 00:18:26.440 to be positive or negative in every clause. 00:18:26.440 --> 00:18:29.820 And so we had this trick for forcing this thing 00:18:29.820 --> 00:18:31.070 to be not equal to this thing. 00:18:31.070 --> 00:18:34.790 Basically copying the variable with a flip in its truths. 00:18:34.790 --> 00:18:36.610 But again everything here is forced. 00:18:36.610 --> 00:18:38.662 The variables we make are guaranteed 00:18:38.662 --> 00:18:40.620 to be copies or indications of other variables. 00:18:40.620 --> 00:18:42.550 So we preserve number of solutions. 00:18:42.550 --> 00:18:43.790 So this is hard too. 00:19:00.260 --> 00:19:03.060 OK, what else? 00:19:03.060 --> 00:19:05.210 I think we can-- I just made this up 00:19:05.210 --> 00:19:07.870 --but we can add the dash three as usual 00:19:07.870 --> 00:19:10.000 by replacing each variable with little cycle. 00:19:12.520 --> 00:19:14.530 What about planar 1-in-3 SAT? 00:19:14.530 --> 00:19:17.240 So we had, we actually had two ways to prove this. 00:19:17.240 --> 00:19:23.180 This is one of the reductions and we 00:19:23.180 --> 00:19:26.960 check that this set of SAT clauses, these 00:19:26.960 --> 00:19:29.477 are the SAT clauses implemented this 1-in-3 SAT clause. 00:19:29.477 --> 00:19:32.060 And I stared at this for a while and its kind of hard to tell. 00:19:32.060 --> 00:19:34.420 So I just wrote a program to enumerate all cases 00:19:34.420 --> 00:19:36.970 and found there's exactly one case where 00:19:36.970 --> 00:19:39.790 this is not parsimonious. 00:19:39.790 --> 00:19:45.760 And that's when this is false and these two are true. 00:19:45.760 --> 00:19:47.900 And because of these negations you can either 00:19:47.900 --> 00:19:49.400 solve the internal things like this, 00:19:49.400 --> 00:19:51.380 or you could flip all of the internal nodes 00:19:51.380 --> 00:19:53.550 and that will also be satisfied. 00:19:53.550 --> 00:19:57.600 Now this is bad news, because all the other cases there 00:19:57.600 --> 00:20:00.279 is a unique solution over here. 00:20:00.279 --> 00:20:02.320 But in this case there are exactly two solutions. 00:20:02.320 --> 00:20:03.945 If it was two everywhere we'd be happy, 00:20:03.945 --> 00:20:05.500 that would be twomonious. 00:20:05.500 --> 00:20:07.540 If it was one everywhere we'd be happier, happy 00:20:07.540 --> 00:20:08.910 that would be parsimonious. 00:20:08.910 --> 00:20:11.210 But because it's a mixture of one and two 00:20:11.210 --> 00:20:12.975 we have approximately preserved the counts 00:20:12.975 --> 00:20:13.980 with a factor of two. 00:20:13.980 --> 00:20:15.890 But that's not good enough for sharp P, 00:20:15.890 --> 00:20:19.240 we need exact preservation. 00:20:19.240 --> 00:20:20.960 So this is no good. 00:20:20.960 --> 00:20:23.200 Luckily we had another proof which was actually 00:20:23.200 --> 00:20:27.400 a stronger result, Planar Positive Rectilinear 1-in-3SAT. 00:20:27.400 --> 00:20:29.280 This was a version with no negation, 00:20:29.280 --> 00:20:32.580 and this one does work. 00:20:32.580 --> 00:20:36.200 There's first of all this, the not equal gadget 00:20:36.200 --> 00:20:37.220 and the equal gadget. 00:20:37.220 --> 00:20:40.850 I don't want to go through them, but again A forced to be zero, 00:20:40.850 --> 00:20:43.780 C was forced to be one, which forces B and D to be 00:20:43.780 --> 00:20:45.250 zero in this picture. 00:20:45.250 --> 00:20:48.930 So all is good. 00:20:48.930 --> 00:20:50.550 And again parsimonious there. 00:20:50.550 --> 00:20:53.400 And then this one was too complicated to think about, 00:20:53.400 --> 00:20:55.413 so I again wrote a program to check-- try 00:20:55.413 --> 00:20:59.120 all the cases and every choice of XYZ 00:20:59.120 --> 00:21:02.040 over here that's satisfied, this is a reduction from 3SAT. 00:21:02.040 --> 00:21:03.870 So if at least one of these is true, 00:21:03.870 --> 00:21:06.420 there will be exactly one solution over here. 00:21:06.420 --> 00:21:09.555 And just as before after-- if zero of them are true, 00:21:09.555 --> 00:21:10.930 then they'll be no solution here. 00:21:10.930 --> 00:21:13.400 That we already knew. 00:21:13.400 --> 00:21:14.860 So good news. 00:21:14.860 --> 00:21:17.110 I should check whether that's mentioned in their paper 00:21:17.110 --> 00:21:20.830 but it proves planar positive rectilinear 1-in-3SAT 00:21:20.830 --> 00:21:22.270 is sharp P complete. 00:21:42.460 --> 00:21:43.720 Sharp go here, here. 00:21:46.260 --> 00:21:48.470 OK, so lots of fun results. 00:21:48.470 --> 00:21:50.680 We get a lot of results just from-- by looking 00:21:50.680 --> 00:21:51.360 at old proofs. 00:21:51.360 --> 00:21:54.930 Now they're not all going to work, 00:21:54.930 --> 00:21:57.650 but I have one more example that does work. 00:21:57.650 --> 00:22:00.410 Shakashaka, remember the puzzle? 00:22:00.410 --> 00:22:01.850 It's a Nikoli puzzle. 00:22:01.850 --> 00:22:03.800 Every square, every white squared 00:22:03.800 --> 00:22:07.250 can be filled in with a black thing, but adjacent to it too, 00:22:07.250 --> 00:22:09.870 there should be exactly two of those. 00:22:09.870 --> 00:22:12.050 And you want all of the resulting white regions 00:22:12.050 --> 00:22:14.450 to be rectangles possibly rotated. 00:22:14.450 --> 00:22:19.520 And we had this reduction from planar 3SAT, 00:22:19.520 --> 00:22:22.480 and this basically, this type of wire. 00:22:22.480 --> 00:22:26.674 And there's exactly two ways to solve a wire. 00:22:26.674 --> 00:22:27.840 One for true, one for false. 00:22:27.840 --> 00:22:29.423 So once you know what the variable is, 00:22:29.423 --> 00:22:31.240 you're forced what to do, there is also 00:22:31.240 --> 00:22:33.700 a parity-shifting gadget and splits and turns. 00:22:33.700 --> 00:22:36.530 But again, exactly two ways to solve everything. 00:22:36.530 --> 00:22:39.300 So parsimonious, and then the clause everything 00:22:39.300 --> 00:22:41.260 was basically forced just-- you're 00:22:41.260 --> 00:22:44.100 forced whether to have these square diamonds. 00:22:44.100 --> 00:22:45.700 And you just eliminated the one case 00:22:45.700 --> 00:22:47.440 where the clause is not satisfied. 00:22:47.440 --> 00:22:50.550 So there's really no flexibility here, one way to solve it. 00:22:50.550 --> 00:22:52.310 And so it's a parsimonious reduction 00:22:52.310 --> 00:22:54.450 and indeed in the paper we mentioned 00:22:54.450 --> 00:22:57.330 this implies sharp P completeness of counting 00:22:57.330 --> 00:22:59.770 Shakashaka solutions. 00:22:59.770 --> 00:23:02.010 Cool. 00:23:02.010 --> 00:23:04.900 Here's an example that doesn't work. 00:23:04.900 --> 00:23:07.770 A little different, Hamiltonicity or I 00:23:07.770 --> 00:23:11.130 guess I want to count the number of Hamiltonian cycles. 00:23:11.130 --> 00:23:16.510 The natural counting version of Hamiltonicity. 00:23:16.510 --> 00:23:19.080 So we had two proofs for this, neither of them 00:23:19.080 --> 00:23:22.370 work in a sharp P sense. 00:23:22.370 --> 00:23:24.720 This one, remember the idea was that you 00:23:24.720 --> 00:23:28.980 would traverse back and forth one way or the other 00:23:28.980 --> 00:23:30.295 to get all of these nodes. 00:23:30.295 --> 00:23:31.670 That was a variable, that's fine. 00:23:31.670 --> 00:23:33.425 There're exactly two ways to do that. 00:23:33.425 --> 00:23:35.700 But then the clause, the clause had 00:23:35.700 --> 00:23:39.290 to be satisfied by at least one of the three variables. 00:23:39.290 --> 00:23:41.220 And if it's satisfied for example, 00:23:41.220 --> 00:23:42.920 by all three variables, then it could 00:23:42.920 --> 00:23:44.650 be this node is picked up like this, 00:23:44.650 --> 00:23:46.934 or the node could be picked up this way, 00:23:46.934 --> 00:23:48.600 or the node could be picked up this way. 00:23:48.600 --> 00:23:50.700 So there are three different solutions even 00:23:50.700 --> 00:23:53.130 for one fixed variable assignment. 00:23:53.130 --> 00:23:55.370 So that's bad, we're not allowed to do that. 00:23:55.370 --> 00:23:57.860 It'd be fine if every one was three, but some will be one, 00:23:57.860 --> 00:23:59.870 some will be two, some will be three. 00:23:59.870 --> 00:24:01.786 That's going to be some weird product of those 00:24:01.786 --> 00:24:03.200 over the clauses. 00:24:03.200 --> 00:24:04.770 So that doesn't work. 00:24:04.770 --> 00:24:06.075 We had this other proof. 00:24:09.850 --> 00:24:13.430 This was a notation for this gadget which 00:24:13.430 --> 00:24:16.610 forced either this directed edge or this directed edge 00:24:16.610 --> 00:24:20.910 to be used, but not both. 00:24:20.910 --> 00:24:22.410 It's and x or. 00:24:22.410 --> 00:24:24.390 So that's, remember what these things meant 00:24:24.390 --> 00:24:26.990 and that we have the variable true or false. 00:24:26.990 --> 00:24:28.740 And then we connected them to the clauses, 00:24:28.740 --> 00:24:31.860 then separately we had a crossover. 00:24:31.860 --> 00:24:34.940 But the trouble is in the clauses, 00:24:34.940 --> 00:24:38.810 because again, the idea was that the variable chose this guy, 00:24:38.810 --> 00:24:40.500 this one was forbidden. 00:24:40.500 --> 00:24:42.910 That's actually the good case I think. 00:24:42.910 --> 00:24:46.390 If the variable chose this, chose this one, 00:24:46.390 --> 00:24:48.340 then this one must be included. 00:24:48.340 --> 00:24:52.661 That's bad news, if you followed this, and then this, 00:24:52.661 --> 00:24:54.910 and then this, then you cut off this part of the graph 00:24:54.910 --> 00:24:56.980 and you don't get one Hamiltonian cycle. 00:24:56.980 --> 00:25:02.050 You want at least one variable to allow you to go left here 00:25:02.050 --> 00:25:05.230 and then you can go and grab all this stuff and come back. 00:25:05.230 --> 00:25:08.000 But again if multiple variables satisfy this thing, 00:25:08.000 --> 00:25:10.800 any one of them could grab the left rectangle. 00:25:10.800 --> 00:25:14.115 And so they get multiple solutions, not parsimonious. 00:25:16.630 --> 00:25:18.990 But parts of this proof are useful 00:25:18.990 --> 00:25:22.240 and they are used to make a parsimonious proof. 00:25:22.240 --> 00:25:25.550 So the part that was useful is this x or gadget, and the way 00:25:25.550 --> 00:25:27.200 to implement crossovers. 00:25:27.200 --> 00:25:32.390 So just remember that you can build x ors, 00:25:32.390 --> 00:25:35.670 and that you can cross them over using a bunch of x ors. 00:25:35.670 --> 00:25:38.610 So only x or connections, these notations, 00:25:38.610 --> 00:25:41.010 can be crossed in this view. 00:25:41.010 --> 00:25:42.870 We're going to build more gadgets 00:25:42.870 --> 00:25:46.195 and this is a proof by Sato in 2002. 00:25:46.195 --> 00:25:50.120 It was a bachelor's thesis actually in Japan. 00:25:50.120 --> 00:25:53.199 And so here you see redrawn the x or gadget. 00:25:53.199 --> 00:25:54.990 Here it's going to be for undirected graph, 00:25:54.990 --> 00:25:58.290 same structure works. 00:25:58.290 --> 00:26:01.110 And this is do the crossover using x or. 00:26:01.110 --> 00:26:05.710 So he's denoting x or's as this big X connected to two things. 00:26:05.710 --> 00:26:09.270 Now given that we can build an or gadget, which 00:26:09.270 --> 00:26:11.880 says that either we use this edge or we use this edge, 00:26:11.880 --> 00:26:14.400 or both. 00:26:14.400 --> 00:26:17.450 Here we're not using an x or, but this is the graph. 00:26:17.450 --> 00:26:21.640 And the key is this has to be done uniquely. 00:26:21.640 --> 00:26:24.330 That's in particular the point of these dots. 00:26:24.330 --> 00:26:27.450 This looks asymmetric, it's kind of weird. 00:26:27.450 --> 00:26:33.630 For example, if they're both in, then this guy can do this, 00:26:33.630 --> 00:26:35.890 and this guy can do that. 00:26:35.890 --> 00:26:36.970 But it's not symmetric. 00:26:36.970 --> 00:26:40.594 You couldn't flip-- this guy can't grab these points, that 00:26:40.594 --> 00:26:42.510 would be a second solution which would be bad, 00:26:42.510 --> 00:26:43.850 but we missed these points. 00:26:43.850 --> 00:26:47.370 So this guy has to stay down here if he's in at all. 00:26:47.370 --> 00:26:49.250 And then this guy's the only one who 00:26:49.250 --> 00:26:52.120 can grab those extra points. 00:26:52.120 --> 00:26:57.565 Or if just the top guy's in then you do this. 00:27:00.320 --> 00:27:06.080 And if just the bottom guy's in I think it's symmetric. 00:27:06.080 --> 00:27:10.040 That is symmetric, and it's unique. 00:27:10.040 --> 00:27:11.350 Good. 00:27:11.350 --> 00:27:16.130 If there's an implication-- so this says if this edge is in, 00:27:16.130 --> 00:27:20.210 then that edge must be in the Hamiltonian cycle 00:27:20.210 --> 00:27:23.310 and this is essentially by copying. 00:27:23.310 --> 00:27:25.290 And we just have to grab an extra edge 00:27:25.290 --> 00:27:28.780 and add this little extra thing just for copying value. 00:27:28.780 --> 00:27:33.330 So if this one is in, then this edge must not be used, 00:27:33.330 --> 00:27:36.147 which means this edge if it's used must go straight. 00:27:36.147 --> 00:27:37.480 In particular, this is not used. 00:27:37.480 --> 00:27:39.520 And we have an or that means that this one must 00:27:39.520 --> 00:27:41.930 be forced by a property of or. 00:27:41.930 --> 00:27:47.950 On the other hand, if this is not set, this one must be used. 00:27:47.950 --> 00:27:51.150 So I guess this must be an edge that was already 00:27:51.150 --> 00:27:53.640 going to be used for something. 00:27:53.640 --> 00:27:56.350 So that edge is just going to get diverted down here, 00:27:56.350 --> 00:27:59.390 and then the or doesn't constrain us at all. 00:27:59.390 --> 00:28:03.390 Because zero or one, this one is one so the or is happy. 00:28:03.390 --> 00:28:05.200 So that's an implication. 00:28:05.200 --> 00:28:08.666 It's going to be a little more subtle how we combine these. 00:28:08.666 --> 00:28:12.700 This is the tricky gadget. 00:28:12.700 --> 00:28:15.390 I sort of understand it, but it has 00:28:15.390 --> 00:28:17.110 a lot of details to check, especially 00:28:17.110 --> 00:28:18.770 on the uniqueness front. 00:28:18.770 --> 00:28:20.560 But this is a three-way or which we're 00:28:20.560 --> 00:28:24.480 using for clause, SAT clause, 3SAT clause. 00:28:24.480 --> 00:28:27.420 We want at least one of these three edges 00:28:27.420 --> 00:28:29.230 to be in the Hamiltonian cycle. 00:28:29.230 --> 00:28:34.570 And so here we use x ors, ors, implications, and more x ors. 00:28:34.570 --> 00:28:36.260 I'll show you the intended solution, 00:28:36.260 --> 00:28:39.670 assuming I can remember it. 00:28:39.670 --> 00:28:43.940 So let's say for example, this edge is in the Hamiltonian 00:28:43.940 --> 00:28:45.750 cycle. 00:28:45.750 --> 00:28:54.130 Then we're going to do something like go over here, 00:28:54.130 --> 00:28:57.250 come back around like this. 00:28:57.250 --> 00:28:57.980 And then 00:28:57.980 --> 00:29:00.584 AUDIENCE: Did you just violate the x or already? 00:29:00.584 --> 00:29:01.820 PROFESSOR: This x or? 00:29:01.820 --> 00:29:02.565 AUDIENCE: Yeah. 00:29:02.565 --> 00:29:03.690 PROFESSOR: No, I went here. 00:29:03.690 --> 00:29:04.610 AUDIENCE: And then you went up. 00:29:04.610 --> 00:29:06.150 PROFESSOR: Oh, then I went up, good. 00:29:06.150 --> 00:29:08.970 So actually I have to do this and around. 00:29:08.970 --> 00:29:11.230 Yes, so all these constraints are thrown in, basically 00:29:11.230 --> 00:29:12.271 to force it to be unique. 00:29:12.271 --> 00:29:14.890 Without them you could-- still it would still work, 00:29:14.890 --> 00:29:17.620 but it wouldn't be parsimonious. 00:29:17.620 --> 00:29:19.620 OK, let's see, this-- 00:29:19.620 --> 00:29:20.620 AUDIENCE: In the middle. 00:29:20.620 --> 00:29:22.700 PROFESSOR: --doesn't constrain me. 00:29:22.700 --> 00:29:23.700 This one? 00:29:23.700 --> 00:29:24.846 AUDIENCE: Yeah. 00:29:24.846 --> 00:29:27.270 PROFESSOR: Good, go up there. 00:29:27.270 --> 00:29:32.650 So I did exactly one of those and then I need to grab these. 00:29:32.650 --> 00:29:34.060 OK, so that's one picture. 00:29:34.060 --> 00:29:37.300 I think it's not totally symmetric. 00:29:37.300 --> 00:29:39.390 Again, so you have to check all three of them. 00:29:39.390 --> 00:29:42.580 And you have to check for example, It's not so hard. 00:29:42.580 --> 00:29:45.270 Like if this guy was also in, I could have just gone here 00:29:45.270 --> 00:29:47.610 and this guy would pick up those nodes. 00:29:47.610 --> 00:29:50.900 So as long as, in general, as long as at least one of them 00:29:50.900 --> 00:29:53.450 is on you're OK. 00:29:53.450 --> 00:29:55.496 And furthermore, if they're all on, 00:29:55.496 --> 00:29:57.120 there's still a unique way to solve it. 00:29:57.120 --> 00:29:58.430 And I'm not going to go through that. 00:29:58.430 --> 00:30:00.340 But it's thanks to all of these constraints 00:30:00.340 --> 00:30:04.110 they're cutting out multiple solutions. 00:30:04.110 --> 00:30:08.210 OK, so now we just had to put these together, 00:30:08.210 --> 00:30:09.830 this is pretty easy. 00:30:09.830 --> 00:30:11.410 It's pretty much like the old proof. 00:30:11.410 --> 00:30:14.010 Again, we have-- we represent variables 00:30:14.010 --> 00:30:18.250 by having these double edges in a Hamiltonian cycle. 00:30:18.250 --> 00:30:21.580 You're going to choose one or the other. 00:30:21.580 --> 00:30:24.200 And then we have this exclusive or, forcing y and y bar 00:30:24.200 --> 00:30:25.950 to choose opposite choices. 00:30:25.950 --> 00:30:27.950 And then there are these exclusive words to say, 00:30:27.950 --> 00:30:29.908 if you chose that one, then you can't choose it 00:30:29.908 --> 00:30:31.222 for this clause. 00:30:31.222 --> 00:30:32.930 And then the clauses are just represented 00:30:32.930 --> 00:30:34.190 by those three-way ors. 00:30:34.190 --> 00:30:36.190 So this overall structure is the same. 00:30:36.190 --> 00:30:38.340 The crossovers are done as before 00:30:38.340 --> 00:30:41.150 and it's really these gadgets that have changed. 00:30:41.150 --> 00:30:43.420 And they're complicated, but it's parsimonious. 00:30:43.420 --> 00:30:45.700 So with a little more work. 00:30:45.700 --> 00:30:48.630 We get the number of Hamiltonian cycles 00:30:48.630 --> 00:30:54.400 in planar max degree three graphs is sharp P complete. 00:31:16.920 --> 00:31:20.030 So that's nice. 00:31:20.030 --> 00:31:24.390 And from that, we get Slitherlink. 00:31:24.390 --> 00:31:27.684 So this is-- I've sort of been hiding these facts from you. 00:31:27.684 --> 00:31:29.350 When we originally covered these proofs, 00:31:29.350 --> 00:31:33.240 these papers actually talk about-- This one doesn't quite 00:31:33.240 --> 00:31:35.640 talk about sharp P, but it is also sharp P complete. 00:31:35.640 --> 00:31:37.960 Counting the number of solutions to Slitherlink 00:31:37.960 --> 00:31:40.207 is sharp P complete. 00:31:40.207 --> 00:31:41.040 This was the puzzle. 00:31:41.040 --> 00:31:45.530 Again you have that number of adjacent edges-- each number. 00:31:45.530 --> 00:31:49.740 And the proof was a reduction from planar max 00:31:49.740 --> 00:31:52.810 degree three Hamiltonian cycle. 00:31:52.810 --> 00:31:54.740 And at the time I said, oh you could just 00:31:54.740 --> 00:31:56.420 assume it's a grid graph. 00:31:56.420 --> 00:31:59.880 And then you just need the required gadget, 00:31:59.880 --> 00:32:01.550 which is the b part. 00:32:01.550 --> 00:32:02.970 Just need this gadget. 00:32:02.970 --> 00:32:05.010 This was a gadget because it had these ones. 00:32:05.010 --> 00:32:08.045 It meant it had to be traversed like a regular vertex 00:32:08.045 --> 00:32:10.380 in Hamiltonian cycle and it turns out 00:32:10.380 --> 00:32:13.260 there was a way to traverse it straight or with returns. 00:32:13.260 --> 00:32:15.610 And then we could block off edges wherever there 00:32:15.610 --> 00:32:16.860 wasn't supposed to be an edge. 00:32:16.860 --> 00:32:19.390 And so if you're reducing from Hamiltonicity and grid graphs 00:32:19.390 --> 00:32:22.720 that was the whole proof and we were happy. 00:32:22.720 --> 00:32:25.280 Now we don't know whether Hamiltonicity and grid 00:32:25.280 --> 00:32:26.586 graphs is sharp P complete. 00:32:30.796 --> 00:32:32.170 To prove that we would need to be 00:32:32.170 --> 00:32:34.040 able to put bipartite in here, and I 00:32:34.040 --> 00:32:35.610 don't know of a proof of that. 00:32:35.610 --> 00:32:37.390 Good open problem. 00:32:37.390 --> 00:32:40.450 So this was the reason that they have this other gadget, which 00:32:40.450 --> 00:32:45.140 was to make Hamiltonian-- to make these white vertices that 00:32:45.140 --> 00:32:46.920 don't have to be traversed. 00:32:46.920 --> 00:32:49.390 They're just implementing an edge basically. 00:32:49.390 --> 00:32:51.600 So just the black vertices have to be traversed. 00:32:51.600 --> 00:32:54.660 We needed that for drawing things on the grid. 00:32:54.660 --> 00:32:57.790 But if you just don't put in the ones and add in these zeros 00:32:57.790 --> 00:33:01.320 again you can traverse it or not all. 00:33:01.320 --> 00:33:03.340 And this one, as you can read here, 00:33:03.340 --> 00:33:06.400 says this would be bad to have happen. 00:33:06.400 --> 00:33:08.150 In fact you can rule it out, it will never 00:33:08.150 --> 00:33:10.390 happen in the situations we want. 00:33:10.390 --> 00:33:13.280 Because the white vertices only have two neighbors 00:33:13.280 --> 00:33:15.010 that are traversable. 00:33:18.310 --> 00:33:22.240 Cool, and then furthermore, all of these solutions are unique. 00:33:22.240 --> 00:33:24.540 There is exactly one way to go straight. 00:33:24.540 --> 00:33:26.830 There's exactly one way to turn right, exactly one way 00:33:26.830 --> 00:33:27.980 to turn left. 00:33:27.980 --> 00:33:30.490 That's the subtle thing that was not revealed before. 00:33:30.490 --> 00:33:34.240 But if you stare at it long enough you will be convinced. 00:33:34.240 --> 00:33:36.590 So this is a parsimonious reduction 00:33:36.590 --> 00:33:38.940 from planar max degree three Hamiltonian 00:33:38.940 --> 00:33:40.840 cycle to Slitherlink. 00:33:40.840 --> 00:33:43.870 So counting solutions in Slitherlink is hard. 00:33:43.870 --> 00:33:48.860 That was the fully worked out example. 00:33:48.860 --> 00:33:50.412 Any questions? 00:33:50.412 --> 00:33:50.912 Yeah. 00:33:50.912 --> 00:33:53.200 AUDIENCE: So are there examples of problems in P 00:33:53.200 --> 00:33:56.380 who's counting versions are sharp P complete? 00:33:56.380 --> 00:33:57.060 PROFESSOR: Yes. 00:33:57.060 --> 00:34:00.401 And that will be the next topic. 00:34:00.401 --> 00:34:02.650 Well it's going to be a little while til we get there. 00:34:02.650 --> 00:34:04.440 But I'm going to proof things and then we 00:34:04.440 --> 00:34:06.390 will get to an answer, but the answer is yes. 00:34:10.531 --> 00:34:11.322 AUDIENCE: Question. 00:34:11.322 --> 00:34:12.530 PROFESSOR: Yeah. 00:34:12.530 --> 00:34:14.874 AUDIENCE: So, for if we wanted-- if we're 00:34:14.874 --> 00:34:16.540 like thinking about a problem that we're 00:34:16.540 --> 00:34:18.329 trying to prove something P-hard and we start thinking 00:34:18.329 --> 00:34:20.912 maybe stop, we would just show it's in P by finding algorithm. 00:34:20.912 --> 00:34:25.500 Is there a nice way to show that our problem is not P hard? 00:34:25.500 --> 00:34:28.900 PROFESSOR: Well you usually would say that sharp, 00:34:28.900 --> 00:34:31.300 that problem is NP. 00:34:31.300 --> 00:34:33.830 You find a polynomial counting algorithm. 00:34:33.830 --> 00:34:37.190 And they're lot's of examples of polynomial counting algorithms 00:34:37.190 --> 00:34:38.840 especially on like trees. 00:34:38.840 --> 00:34:40.771 Typical thing is your dynamic program. 00:34:40.771 --> 00:34:42.520 So like maybe you want to know-- let's say 00:34:42.520 --> 00:34:44.179 you have a rooted binary tree and for each node 00:34:44.179 --> 00:34:45.889 you could flip it this way or not. 00:34:45.889 --> 00:34:48.250 How many different ways are there to do that? 00:34:48.250 --> 00:34:50.699 And maybe have some constraints on how that's done. 00:34:50.699 --> 00:34:53.122 Then you just try it flipped, and try it not. 00:34:53.122 --> 00:34:54.580 You do dynamic programming and then 00:34:54.580 --> 00:34:57.937 you multiply the two solution sizes together, 00:34:57.937 --> 00:34:59.520 and you get the overall solution size. 00:34:59.520 --> 00:35:01.380 So you basically do combinatorics 00:35:01.380 --> 00:35:03.130 and if there's independent choices you 00:35:03.130 --> 00:35:06.540 multiply, if they're opposing choices you add, 00:35:06.540 --> 00:35:07.410 that kind of thing. 00:35:07.410 --> 00:35:10.070 And from that you get polynomial time counting algorithms. 00:35:10.070 --> 00:35:14.040 In tree-like things that often works inbound treewidth. 00:35:14.040 --> 00:35:16.562 AUDIENCE: Do you know that NP hard problems who's counting 00:35:16.562 --> 00:35:19.567 problems are not sharp P hard? 00:35:19.567 --> 00:35:21.108 I guess this technique wouldn't work. 00:35:21.108 --> 00:35:23.890 PROFESSOR: I would say generally most problems 00:35:23.890 --> 00:35:26.190 that are hard to decide are hard to count. 00:35:26.190 --> 00:35:29.120 And where NP hard implies sharp P hard. 00:35:29.120 --> 00:35:33.329 I don't think there's a hard theorem in that there's 00:35:33.329 --> 00:35:35.620 nothing that really says-- meta-theorem that says that, 00:35:35.620 --> 00:35:36.578 but that's the feeling. 00:35:39.142 --> 00:35:40.600 It'd be nice, then we wouldn't have 00:35:40.600 --> 00:35:42.070 to do all the parsimonious work. 00:35:45.700 --> 00:35:50.350 All right so it's time for a little bit of linear algebra. 00:35:54.410 --> 00:35:57.270 Let me remind you, I guess linear algebra's not 00:35:57.270 --> 00:35:59.350 a pre-req for this class, but probably you've 00:35:59.350 --> 00:36:01.489 seen the determinant of a matrix and use, 00:36:01.489 --> 00:36:03.530 if it's zero then it's non-invertible blah, blah, 00:36:03.530 --> 00:36:04.784 blah. 00:36:04.784 --> 00:36:06.200 Let me remind you of a definition. 00:36:11.980 --> 00:36:13.640 And we rarely use matrix notation. 00:36:13.640 --> 00:36:17.540 So let me remind you of the usual one. 00:36:17.540 --> 00:36:19.050 N by n, square matrix. 00:36:19.050 --> 00:36:21.130 This is a polynomial time problem. 00:36:21.130 --> 00:36:25.330 It is-- but I'm going to define it in an exponential way. 00:36:25.330 --> 00:36:28.410 But you probably know a polynomial time algorithm. 00:36:28.410 --> 00:36:31.937 This is not an algorithms class so you don't need to know it. 00:36:34.760 --> 00:36:37.010 But it's based on Gaussian elimination, the usual one. 00:36:56.370 --> 00:37:00.790 So you look at all permutation matrices, 00:37:00.790 --> 00:37:02.580 all n by n permutation matrices which 00:37:02.580 --> 00:37:05.860 you can think of as a permutation pi on the numbers 1 00:37:05.860 --> 00:37:06.880 through n. 00:37:06.880 --> 00:37:09.780 And you look at i comma pi event, that 00:37:09.780 --> 00:37:11.410 defines the permutation matrix. 00:37:11.410 --> 00:37:14.500 You take the product of the matrix 00:37:14.500 --> 00:37:20.140 values-- if you superimpose the permutation matrix on that, 00:37:20.140 --> 00:37:23.995 on the given matrix A. 00:37:23.995 --> 00:37:28.040 You take that product you possibly negate it 00:37:28.040 --> 00:37:30.690 if the sign of your permutation was negative, 00:37:30.690 --> 00:37:35.960 if it does an even number-- an odd number of transpositions 00:37:35.960 --> 00:37:37.990 then this will be negative. 00:37:37.990 --> 00:37:40.780 Otherwise, it'd be positive and you add those up. 00:37:40.780 --> 00:37:42.840 So of course, an exponential number permutations 00:37:42.840 --> 00:37:45.730 you wouldn't want to do this as an algorithm but turns out it 00:37:45.730 --> 00:37:49.745 can be done in polynomial time. 00:37:49.745 --> 00:37:51.120 The reason for talking about this 00:37:51.120 --> 00:37:56.690 is by analogy I want the notion of permanent, of an n 00:37:56.690 --> 00:37:57.450 by n matrix. 00:37:57.450 --> 00:37:59.190 The same thing, but with this removed. 00:38:02.370 --> 00:38:06.620 The determinant of A is the sum over all permutations pi, 00:38:06.620 --> 00:38:13.900 of the product from i equals one to n of ai, pi of i. 00:38:13.900 --> 00:38:15.870 Now this may not look like a counting problem. 00:38:15.870 --> 00:38:17.900 It turns out it is a counting problem, sort of, 00:38:17.900 --> 00:38:19.232 a weighted counting problem. 00:38:19.232 --> 00:38:21.315 We will get back to counting problems in a moment. 00:38:21.315 --> 00:38:24.740 This Is related to the number of perfect matchings in a graph. 00:38:24.740 --> 00:38:26.520 But at this point it's just-- it's 00:38:26.520 --> 00:38:27.800 a quantity we want to compute. 00:38:27.800 --> 00:38:29.630 This is a function of a matrix. 00:38:29.630 --> 00:38:33.600 And computing this function is sharp P complete. 00:38:33.600 --> 00:38:34.100 Yeah. 00:38:34.100 --> 00:38:37.260 AUDIENCE: Could it just be sine minus 1 2e sine of pi? 00:38:40.276 --> 00:38:42.310 PROFESSOR: I don't know, it depends. 00:38:42.310 --> 00:38:46.695 If you call the number of transpositions mod two so then 00:38:46.695 --> 00:38:48.045 it's zero one. 00:38:48.045 --> 00:38:48.920 You know what I mean. 00:38:53.210 --> 00:38:56.040 All right. 00:38:56.040 --> 00:38:58.017 So the claim is, permanent is sharp P complete. 00:38:58.017 --> 00:38:59.100 We're going to prove this. 00:38:59.100 --> 00:39:01.970 This was the original problem proved sharp P complete. 00:39:01.970 --> 00:39:05.230 Well other than sharp 3SAT I guess. 00:39:05.230 --> 00:39:05.780 Same paper. 00:39:09.680 --> 00:39:15.130 Great so let me give you a little intuition 00:39:15.130 --> 00:39:16.740 of what the permanent is. 00:39:16.740 --> 00:39:20.496 We'd like a definition that's not so algebraic. 00:39:20.496 --> 00:39:23.980 At least I would like one more graph-theoretic would be nice. 00:39:23.980 --> 00:39:25.810 So here's what we're going to do. 00:39:25.810 --> 00:39:30.120 We're going to convert A, our matrix, into a weighted graph. 00:39:42.130 --> 00:39:49.520 And then let me go to the other board. 00:40:00.580 --> 00:40:02.251 How do we convert into a weighted graph, 00:40:02.251 --> 00:40:03.250 weighted directed graph? 00:40:03.250 --> 00:40:08.420 Well the weight from, vertex I to vertex J is AIJ. 00:40:08.420 --> 00:40:10.950 The obvious transformation if it's zero then 00:40:10.950 --> 00:40:13.450 there's not going to be an edge, although it doesn't matter. 00:40:13.450 --> 00:40:15.325 You could leave the edge in with weight zero, 00:40:15.325 --> 00:40:17.004 it will be the same. 00:40:17.004 --> 00:40:18.920 Because what we're interested in, in the claim 00:40:18.920 --> 00:40:28.430 is the permanent of the matrix equals the sum, 00:40:28.430 --> 00:40:42.217 of the product of edge weights, over all cycle covers 00:40:42.217 --> 00:40:46.040 of the graph. 00:40:46.040 --> 00:40:49.240 OK, this is really just the same thing, but a little bit easier 00:40:49.240 --> 00:40:50.400 to think about. 00:40:50.400 --> 00:40:53.330 So a cycle cover it's kind of like a Hamiltonian cycle 00:40:53.330 --> 00:40:55.440 but there could be multiple cycles. 00:40:55.440 --> 00:40:58.600 So at every vertex you should enter and leave. 00:40:58.600 --> 00:41:01.430 And so you have sort of in degree one and out degree one. 00:41:01.430 --> 00:41:04.980 So you end up decomposing the graph into vertex destroyed 00:41:04.980 --> 00:41:08.940 cycles which hit every vertex. 00:41:08.940 --> 00:41:11.250 That's a cycle cover. 00:41:11.250 --> 00:41:14.190 Important note here, we do have loops in the graph. 00:41:14.190 --> 00:41:17.140 We can't have loops if aii is not zero. 00:41:17.140 --> 00:41:18.370 Then you have a loop. 00:41:18.370 --> 00:41:22.840 So the idea is to look at every cycle cover 00:41:22.840 --> 00:41:25.630 and just take the product of the edge weights 00:41:25.630 --> 00:41:28.550 that are edges in the cycles and then 00:41:28.550 --> 00:41:30.270 add that up overall cycle covers. 00:41:30.270 --> 00:41:32.210 So it's the same thing if you stare at it 00:41:32.210 --> 00:41:34.135 long enough because you're going from I. 00:41:34.135 --> 00:41:35.680 I mean this is basically the cycle 00:41:35.680 --> 00:41:39.410 decomposition of the permutation if you know permutation theory. 00:41:39.410 --> 00:41:40.910 So if you don't, don't worry this is 00:41:40.910 --> 00:41:43.880 your definition of permanent. 00:41:43.880 --> 00:41:46.840 So we're going to prove this problem, is sharp P complete? 00:41:55.360 --> 00:42:05.100 And we're going to prove it by a C-monious reduction 00:42:05.100 --> 00:42:07.380 from sharp 3SAT. 00:42:10.718 --> 00:42:14.742 AUDIENCE: You said this is just the original thing introducing 00:42:14.742 --> 00:42:15.242 that? 00:42:15.242 --> 00:42:16.530 PROFESSOR: Yes. 00:42:16.530 --> 00:42:20.130 AUDIENCE: Did they not call it-- you made up the term C-monious. 00:42:20.130 --> 00:42:22.010 What did they call it? 00:42:22.010 --> 00:42:27.260 PROFESSOR: I think they just called it a reduction. 00:42:27.260 --> 00:42:29.820 I think pretty sure, they just called it reduction. 00:42:29.820 --> 00:42:32.060 And for them-- and they said at the beginning, 00:42:32.060 --> 00:42:34.990 --reduction means multicall reduction. 00:42:34.990 --> 00:42:36.850 So they're thinking about that. 00:42:36.850 --> 00:42:40.190 But it turns out to be a C-monious reduction. 00:42:40.190 --> 00:42:42.120 C will not be literally constant, 00:42:42.120 --> 00:42:46.140 but it will be a function of the problem sets. 00:42:46.140 --> 00:42:48.020 And this is the reduction. 00:42:48.020 --> 00:42:51.910 So as usual, we have a clause gadget, and a variable gadget, 00:42:51.910 --> 00:42:53.770 and then there's this shaded thing 00:42:53.770 --> 00:42:58.776 which is this matrix, which you can think of as a graph. 00:42:58.776 --> 00:43:00.650 But in this case will be easier to just think 00:43:00.650 --> 00:43:03.180 of as a black box matrix. 00:43:03.180 --> 00:43:07.990 OK all of the edges in these pictures have weight one. 00:43:07.990 --> 00:43:09.580 And then these edges are special, 00:43:09.580 --> 00:43:12.730 and here you have some loops in the center. 00:43:12.730 --> 00:43:14.980 No one else has a loop. 00:43:14.980 --> 00:43:18.500 So the high-level idea is if you're 00:43:18.500 --> 00:43:22.680 thinking of a cycle cover in a vertex 00:43:22.680 --> 00:43:25.140 because you-- sorry, in a variable 00:43:25.140 --> 00:43:28.400 because you've got a vertex here and a vertex here you 00:43:28.400 --> 00:43:29.650 have to cover them somehow. 00:43:29.650 --> 00:43:33.290 And the intent is that you either cover this one this way, 00:43:33.290 --> 00:43:35.470 or you cover this one that way, those 00:43:35.470 --> 00:43:38.620 would be the true and the false. 00:43:38.620 --> 00:43:42.100 And then from the clouds perspective 00:43:42.100 --> 00:43:46.031 we need to understand, so then these things are connected. 00:43:46.031 --> 00:43:47.655 This thing would go here and this thing 00:43:47.655 --> 00:43:50.440 would go here, and generally connect variables 00:43:50.440 --> 00:43:53.530 to clauses in the obvious way. 00:43:53.530 --> 00:43:56.515 And in general, for every occurrence of the positive form 00:43:56.515 --> 00:43:59.420 and the negative form, sorry positive or negative form 00:43:59.420 --> 00:44:02.500 you'll have one of these blobs that connects to the clause. 00:44:02.500 --> 00:44:04.920 So overall architecture should be clear. 00:44:04.920 --> 00:44:07.100 What does this gadget do? 00:44:07.100 --> 00:44:13.550 It has some nifty properties let me write them down. 00:44:13.550 --> 00:44:17.790 So this matrix is called x in the paper. 00:44:17.790 --> 00:44:23.299 So first of all, permanent of x equals zero. 00:44:23.299 --> 00:44:24.840 I'm just going to state these, I mean 00:44:24.840 --> 00:44:27.120 you could check them by doing the computation. 00:44:27.120 --> 00:44:32.090 So we're interested in cycle covers whose products are not 00:44:32.090 --> 00:44:32.590 zero. 00:44:32.590 --> 00:44:34.850 Otherwise they don't contribute to the sum. 00:44:34.850 --> 00:44:39.310 So I could also add in non-zero. 00:44:39.310 --> 00:44:41.860 Meaning the product is non-zero. 00:44:41.860 --> 00:44:46.840 OK so if we had a cycle cover that just-- where the cycle 00:44:46.840 --> 00:44:50.040 cover just locally solved this thing by traversing these four 00:44:50.040 --> 00:44:52.170 vertices all by themselves. 00:44:52.170 --> 00:44:56.162 Then that cycle would have permanent zero. 00:44:56.162 --> 00:44:57.870 And then the permanent of the whole cycle 00:44:57.870 --> 00:45:01.090 covers the product of those things. 00:45:01.090 --> 00:45:02.780 And so the overall thing would be zero. 00:45:02.780 --> 00:45:04.220 So if you look at a nonzero cycle 00:45:04.220 --> 00:45:06.400 cover you can't just leave these isolated, 00:45:06.400 --> 00:45:08.180 you have to visit them. 00:45:08.180 --> 00:45:11.384 You have to enter them and leave them. 00:45:11.384 --> 00:45:13.800 Now the question is, where could you enter and leave them? 00:45:13.800 --> 00:45:16.710 This is maybe not totally clear from the drawing 00:45:16.710 --> 00:45:21.230 but, the intent is that the first vertex in-- which 00:45:21.230 --> 00:45:24.980 corresponds to row one, column one here, is the left side. 00:45:24.980 --> 00:45:29.580 And the column four, row four is the right side. 00:45:29.580 --> 00:45:31.940 I claim that you have to enter at one of those 00:45:31.940 --> 00:45:33.960 and leave at the other. 00:45:33.960 --> 00:45:36.040 Why is that true? 00:45:36.040 --> 00:45:37.040 For a couple reasons. 00:45:37.040 --> 00:45:42.950 One is that the permanent of x with row and column 00:45:42.950 --> 00:45:51.552 one removed is zero, and so is the permanent 00:45:51.552 --> 00:45:54.210 of x with row and column four removed. 00:45:58.810 --> 00:46:04.660 OK vertex one and four out of this x matrix 00:46:04.660 --> 00:46:06.650 are the only places you could enter and leave. 00:46:06.650 --> 00:46:08.950 But it's possible you enter and then immediately leave. 00:46:08.950 --> 00:46:10.842 So you just touch the thing and leave. 00:46:10.842 --> 00:46:12.800 That would correspond to leaving behind a three 00:46:12.800 --> 00:46:14.330 by three sub-matrix. 00:46:14.330 --> 00:46:16.070 Either by deleting this row and column, 00:46:16.070 --> 00:46:18.236 or by deleting this row and column if you just visit 00:46:18.236 --> 00:46:19.549 this vertex and leave. 00:46:19.549 --> 00:46:20.840 Those also have permanent zero. 00:46:20.840 --> 00:46:23.210 So if you're looking at a nonzero cycle cover 00:46:23.210 --> 00:46:24.740 you can't do that. 00:46:24.740 --> 00:46:27.360 So together those mean that you enter one of them 00:46:27.360 --> 00:46:29.530 and leave at the other. 00:46:29.530 --> 00:46:43.380 And furthermore, if you look at the permanent of x with rows 00:46:43.380 --> 00:46:49.980 and columns one and four removed, both removed, 00:46:49.980 --> 00:46:51.600 that's also zero. 00:46:51.600 --> 00:46:55.550 So the permanent of this sub-matrix is zero 00:46:55.550 --> 00:46:58.360 and therefore you can't just enter here, jump here, 00:46:58.360 --> 00:46:59.280 and leave. 00:46:59.280 --> 00:47:02.894 Which means finally you have to traverse all four vertices. 00:47:02.894 --> 00:47:04.810 You enter at one of them, traverse everything, 00:47:04.810 --> 00:47:06.470 and leave at the other. 00:47:06.470 --> 00:47:10.960 So basically this is a forced edge. 00:47:10.960 --> 00:47:15.050 If you touch here you have to then traverse and leave there, 00:47:15.050 --> 00:47:17.470 in any cycle cover. 00:47:17.470 --> 00:47:20.200 So we're used to seeing forced edges in Hamiltonian cycle. 00:47:20.200 --> 00:47:22.430 This is sort of a stronger form of it. 00:47:22.430 --> 00:47:26.300 That's cool now one catch. 00:47:26.300 --> 00:47:29.230 So if you do that, if you enter let's say vertex one 00:47:29.230 --> 00:47:33.510 and leave at vertex four, you will end up-- your contribution 00:47:33.510 --> 00:47:38.020 to the cycle will end up being the permanent of x 00:47:38.020 --> 00:47:41.930 minus row one and column four. 00:47:41.930 --> 00:47:44.810 Or symmetrically with four and one. 00:47:44.810 --> 00:47:49.740 You'd like this to be one, but it's four. 00:47:49.740 --> 00:47:54.410 So there are four ways to traverse the forced edge. 00:47:54.410 --> 00:47:56.910 But because it's always four, or zero, 00:47:56.910 --> 00:47:58.520 and then it doesn't contribute at all. 00:47:58.520 --> 00:48:01.410 It's always going to be four so this will be a nice uniform 00:48:01.410 --> 00:48:03.862 blow up in our number of solutions. 00:48:03.862 --> 00:48:05.320 C is not going to be four, but it's 00:48:05.320 --> 00:48:08.400 going to be four to some power. 00:48:08.400 --> 00:48:10.310 It's going to be four to the power, 00:48:10.310 --> 00:48:12.800 the number of those gadgets. 00:48:12.800 --> 00:48:14.247 So because you can predict that, I 00:48:14.247 --> 00:48:16.830 mean in the reduction you know exactly how many there it's not 00:48:16.830 --> 00:48:19.875 depend on the solution, it's only dependent on how 00:48:19.875 --> 00:48:20.750 you built this thing. 00:48:20.750 --> 00:48:22.670 So at the end, we're going to divide by four 00:48:22.670 --> 00:48:31.210 to the power-- it's I think five times the number of clauses. 00:48:31.210 --> 00:48:36.215 So C is going to be four to the power of five times number 00:48:36.215 --> 00:48:37.590 of clauses because there are five 00:48:37.590 --> 00:48:41.054 of these gadgets per clause. 00:48:41.054 --> 00:48:43.220 So at the end we'll just divide by that and be done. 00:48:43.220 --> 00:48:46.503 AUDIENCE: Would it be 10 times because you got it 00:48:46.503 --> 00:48:48.848 on the variable side too? 00:48:48.848 --> 00:48:50.650 PROFESSOR: Yes 10 times, thank you. 00:48:50.650 --> 00:48:52.755 AUDIENCE: Eight-- two of those don't actually 00:48:52.755 --> 00:48:53.630 connect to variables. 00:48:53.630 --> 00:48:55.171 PROFESSOR: Two of them do not connect 00:48:55.171 --> 00:48:57.642 to variables, yeah eight. 00:48:57.642 --> 00:48:59.333 Eight times the number of clauses. 00:49:02.650 --> 00:49:07.090 All right so now it's just a matter of-- now 00:49:07.090 --> 00:49:10.570 that you understand what this is now 00:49:10.570 --> 00:49:12.980 you could sort of see how information is communicated. 00:49:12.980 --> 00:49:15.940 Because if the variable they choose is the true setting 00:49:15.940 --> 00:49:17.780 it must visit these edges. 00:49:17.780 --> 00:49:20.070 And once it touches here it has to leave here 00:49:20.070 --> 00:49:22.430 and this is a edge going the wrong way. 00:49:22.430 --> 00:49:26.480 So you can't try to traverse-- from here you cannot touch 00:49:26.480 --> 00:49:28.651 the clauses down below. 00:49:28.651 --> 00:49:30.400 Once you touch here to you have to go here 00:49:30.400 --> 00:49:33.370 and then you must leave here and so on. 00:49:33.370 --> 00:49:35.350 But you leave this one behind and it 00:49:35.350 --> 00:49:38.070 must be traversed by the clause and vice versa. 00:49:38.070 --> 00:49:42.090 If I choose this one then these must be visited by the clauses. 00:49:42.090 --> 00:49:44.250 So from the clause perspective as he 00:49:44.250 --> 00:49:47.140 said there are five of these gadgets, but only three of them 00:49:47.140 --> 00:49:47.810 are connected. 00:49:47.810 --> 00:49:53.340 So these guys are forced, and there's a bunch of edges here 00:49:53.340 --> 00:49:54.392 and it's a case analysis. 00:49:54.392 --> 00:49:55.600 So it'd be the short version. 00:49:55.600 --> 00:49:58.350 Let's just do a couple of examples. 00:49:58.350 --> 00:50:00.410 If none of these have been traversed 00:50:00.410 --> 00:50:03.300 by the variable then the-- pretty much 00:50:03.300 --> 00:50:05.820 you have to go straight through. 00:50:05.820 --> 00:50:08.030 But then you're in trouble. 00:50:08.030 --> 00:50:11.000 Because there's no pointer back to the beginning. 00:50:11.000 --> 00:50:14.840 You can only go back this far. 00:50:14.840 --> 00:50:17.590 So if none of-- if you have to traverse all these things 00:50:17.590 --> 00:50:20.350 it's not possible with the cycle cover. 00:50:20.350 --> 00:50:22.740 But if any one of them has been traversed. 00:50:22.740 --> 00:50:25.130 So for example, if this one has been traversed then 00:50:25.130 --> 00:50:28.280 we'll jump over it, visit these guys, 00:50:28.280 --> 00:50:31.430 jump back here, visit this guy, and jump back there. 00:50:31.430 --> 00:50:35.340 And if you check that's the only way to do it, it's unique. 00:50:35.340 --> 00:50:38.190 And similarly any one of them has been covered, 00:50:38.190 --> 00:50:40.690 or if all three of them have been covered, or if two of them 00:50:40.690 --> 00:50:42.148 have been covered in all cases this 00:50:42.148 --> 00:50:44.880 is a unique way from the clause perspective 00:50:44.880 --> 00:50:46.155 to connect all the things. 00:50:46.155 --> 00:50:47.530 Now in reality, there's four ways 00:50:47.530 --> 00:50:48.680 to traverse each of these things. 00:50:48.680 --> 00:50:50.346 So the whole thing grows by this product 00:50:50.346 --> 00:50:53.050 but it doesn't matter which cycle they appear in. 00:50:53.050 --> 00:50:56.880 We'll always scale the number of solutions by factor of four. 00:50:56.880 --> 00:51:02.510 So it's nice uniform scaling C-monious. 00:51:02.510 --> 00:51:03.960 And that's permanent. 00:51:03.960 --> 00:51:05.410 Pretty cool. 00:51:05.410 --> 00:51:07.410 Pretty cool proof. 00:51:07.410 --> 00:51:10.340 Now one not so nice thing about this proof 00:51:10.340 --> 00:51:16.640 is it involves numbers negative one, zero, one, two, and three. 00:51:16.640 --> 00:51:18.650 For reasons we will see in a moment 00:51:18.650 --> 00:51:21.260 it'd be really to just use zero and one. 00:51:21.260 --> 00:51:24.520 And it turns out that's possible, 00:51:24.520 --> 00:51:27.240 but it's kind of annoying. 00:51:27.240 --> 00:51:28.000 It's nifty though. 00:51:28.000 --> 00:51:30.416 I think this will demonstrate a bunch of fun number theory 00:51:30.416 --> 00:51:32.120 things you could do. 00:51:32.120 --> 00:51:34.660 So next goal is zero one permanent. 00:51:34.660 --> 00:51:36.830 Now the matrix is just zero's and one's. 00:51:36.830 --> 00:51:41.770 This is sharp P complete. 00:51:41.770 --> 00:51:47.508 I'll tell you the reason that is particularly interesting. 00:51:47.508 --> 00:51:49.300 As it gets rid of the weights it makes 00:51:49.300 --> 00:51:57.710 it more clearly-- makes it more clearly a counting problem. 00:51:57.710 --> 00:52:00.440 So first of all is the number of cycle covers 00:52:00.440 --> 00:52:03.300 in the corresponding graph. 00:52:03.300 --> 00:52:05.190 No more weights it's just-- if it's a zero 00:52:05.190 --> 00:52:06.100 there's not an edge. 00:52:06.100 --> 00:52:07.933 Since you can't traverse there if it's a one 00:52:07.933 --> 00:52:10.120 you can traverse there and then every cycle cover 00:52:10.120 --> 00:52:12.660 will have product one. 00:52:12.660 --> 00:52:14.350 So it's just counting them. 00:52:14.350 --> 00:52:18.750 But it also happens to be the number of perfect matchings 00:52:18.750 --> 00:52:20.190 in a bipartite graph. 00:52:33.380 --> 00:52:36.048 Which bipartite graph? 00:52:36.048 --> 00:52:39.480 The bipartite graph where one side of the bi-partition 00:52:39.480 --> 00:52:43.600 is the rows and the other side is the columns of the matrix. 00:52:47.620 --> 00:52:50.290 And you have an edge I j. 00:52:55.240 --> 00:53:02.770 If and only if a i sorry j is not a good not 00:53:02.770 --> 00:53:05.270 the best terminology for I-- where I here's 00:53:05.270 --> 00:53:09.660 is a vertex of v one and j is a vertex of v two. 00:53:09.660 --> 00:53:12.390 Whenever I j equals 1, if it's zero there's no edge. 00:53:15.170 --> 00:53:19.530 It's a little confusing because the matrix is not symmetric 00:53:19.530 --> 00:53:23.040 so you might say well, how does this make an undirected graph? 00:53:23.040 --> 00:53:25.890 Because you could have and edge from i to j, but not to j to I. 00:53:25.890 --> 00:53:28.640 That's OK because I is interpreted in the left space 00:53:28.640 --> 00:53:31.010 and j is interpreted in the right space. 00:53:31.010 --> 00:53:33.730 So the asymmetric case would be the case 00:53:33.730 --> 00:53:36.831 that there's an edge here to here, but not an edge from here 00:53:36.831 --> 00:53:37.330 to here. 00:53:37.330 --> 00:53:38.850 That's perfectly OK. 00:53:38.850 --> 00:53:41.930 This is one, two, three, for a three by three matrix 00:53:41.930 --> 00:53:45.620 we get a six vertex graph, not three vertex. 00:53:45.620 --> 00:53:49.094 That's what allows you to be asymmetric on a loop, what 00:53:49.094 --> 00:53:51.010 we were normally thinking is a loop just means 00:53:51.010 --> 00:53:53.500 a horizontal edge. 00:53:53.500 --> 00:53:57.000 OK so it turns out if you look at this graph, which 00:53:57.000 --> 00:53:59.750 is a different graph from what we started with, 00:53:59.750 --> 00:54:01.950 the number of perfect matchings in that graph 00:54:01.950 --> 00:54:05.824 is equal to the number of cycle covers in the other graph, 00:54:05.824 --> 00:54:06.740 in the directed graph. 00:54:06.740 --> 00:54:08.880 So this one's undirected and bipartite. 00:54:08.880 --> 00:54:12.090 This one was directed and not bipartite. 00:54:12.090 --> 00:54:15.380 And the rough intuition is that we're just 00:54:15.380 --> 00:54:17.684 kind of pulling the vertices apart 00:54:17.684 --> 00:54:20.100 into two versions, the left version and the right version. 00:54:20.100 --> 00:54:23.010 If you imagine there being a connection from one 00:54:23.010 --> 00:54:26.740 right to one left and similarly from two right to two left 00:54:26.740 --> 00:54:28.157 directed. 00:54:28.157 --> 00:54:30.740 And three right to three left, hey it looks like cycles again. 00:54:30.740 --> 00:54:33.507 You went here, then you went around, then you went here, 00:54:33.507 --> 00:54:34.340 then you went there. 00:54:34.340 --> 00:54:36.920 That was a cycle and similarly this is a cycle. 00:54:36.920 --> 00:54:39.390 So they're the same thing. 00:54:39.390 --> 00:54:41.710 This is cool because perfect matchings 00:54:41.710 --> 00:54:44.150 are interesting in particular perfect matchings 00:54:44.150 --> 00:54:46.470 are easy to find in polynomial time. 00:54:46.470 --> 00:54:49.481 But counting them as Sharp P complete. 00:54:49.481 --> 00:54:50.730 Getting back to that question. 00:54:53.960 --> 00:54:57.990 So that's why we care about zero one permanence or one reason 00:54:57.990 --> 00:54:58.890 to care about it. 00:54:58.890 --> 00:55:00.312 Let's prove that it's hard. 00:55:16.050 --> 00:55:19.350 And here we'll use a number theory, pretty basic number 00:55:19.350 --> 00:55:19.850 theory. 00:55:28.600 --> 00:55:30.320 So first claim is that computing the 00:55:30.320 --> 00:55:35.600 permanent of a matrix, general matrix nonzero one modulo 00:55:35.600 --> 00:55:37.840 given integer r is hard. 00:55:42.260 --> 00:55:44.200 r has to be an input for this problem. 00:55:44.200 --> 00:55:47.055 It's not like computing at Mod three necessarily as hard, 00:55:47.055 --> 00:55:50.977 but computing at modular anything is hard. 00:55:50.977 --> 00:55:53.560 And here we're going to finally use some multi-color reduction 00:55:53.560 --> 00:55:54.910 power. 00:55:54.910 --> 00:55:58.340 So the idea is, suppose you could solve this problem then 00:55:58.340 --> 00:56:00.551 I claim I can solve permanent. 00:56:00.551 --> 00:56:01.700 Anyone know how? 00:56:01.700 --> 00:56:03.200 AUDIENCE: Chinese remainder theorem? 00:56:03.200 --> 00:56:15.927 PROFESSOR: Chinese remainder theorem 00:56:15.927 --> 00:56:18.010 OK if you don't know the Chinese remainder theorem 00:56:18.010 --> 00:56:19.030 but you should know it. 00:56:19.030 --> 00:56:21.490 It's good. 00:56:21.490 --> 00:56:24.740 So the idea is we're going to set r 00:56:24.740 --> 00:56:33.970 to be all primes up to how big? 00:56:33.970 --> 00:56:38.230 Or we can stop when the product of all the primes 00:56:38.230 --> 00:56:44.060 that we've considered is bigger than the potential permanent. 00:56:44.060 --> 00:56:49.190 And the permanent is at most m to the n times 00:56:49.190 --> 00:56:52.350 n factorial, where m is the largest 00:56:52.350 --> 00:56:53.640 absolute value in the matrix. 00:57:00.350 --> 00:57:03.480 That's one upper bound, doesn't really matter. 00:57:03.480 --> 00:57:05.110 But the point is, it is computable 00:57:05.110 --> 00:57:07.560 and if you take the logarithm it's not so big. 00:57:07.560 --> 00:57:09.490 All these numbers are at least two. 00:57:09.490 --> 00:57:11.550 All primes are at least two. 00:57:11.550 --> 00:57:15.900 So the number of primes we'll have to consider 00:57:15.900 --> 00:57:19.480 is at most log base 2 of that. 00:57:19.480 --> 00:57:25.150 So this means the number of primes, 00:57:25.150 --> 00:57:29.000 so that's log of m to the n times n factorial, 00:57:29.000 --> 00:57:32.700 and this is roughly-- I'll put a total here. 00:57:32.700 --> 00:57:36.400 This is like m log n that's exact 00:57:36.400 --> 00:57:42.200 plus n log n that's approximate but it's minus order n 00:57:42.200 --> 00:57:44.530 so it won't hurt us. 00:57:44.530 --> 00:57:46.400 So that's good, that's a polynomial. 00:57:46.400 --> 00:57:50.040 Log m is a reasonable thing, m was given us as a number 00:57:50.040 --> 00:57:52.680 so log n is part of this input. 00:57:52.680 --> 00:57:57.860 So as our input and n is obviously good as the dimension 00:57:57.860 --> 00:57:58.640 of the matrix. 00:57:58.640 --> 00:58:01.274 So this is a polynomial number of calls. 00:58:01.274 --> 00:58:02.940 We're going to compute the permanent mod 00:58:02.940 --> 00:58:05.580 r for all these things and our Chinese remainder theorem 00:58:05.580 --> 00:58:08.850 tells you if you know a number which 00:58:08.850 --> 00:58:12.920 is the permanent module all primes whose products-- 00:58:12.920 --> 00:58:16.770 Well if you know a number of modulo a bunch of primes then 00:58:16.770 --> 00:58:21.550 you can figure out the number of modulo, the product of primes. 00:58:21.550 --> 00:58:24.290 And if we set the product to be bigger than the number could 00:58:24.290 --> 00:58:27.730 be, then we know [? any ?] modulo that is actually 00:58:27.730 --> 00:58:29.500 knowing the number itself. 00:58:29.500 --> 00:58:31.820 So then we can reconstruct the permanent. 00:58:35.540 --> 00:58:37.420 So here we're really using multicall. 00:58:45.984 --> 00:58:47.400 I think that's the first time I've 00:58:47.400 --> 00:58:50.150 used Chinese remainder theorem in a class that I've taught. 00:58:53.101 --> 00:58:53.600 Good stuff. 00:58:57.837 --> 00:59:00.295 AUDIENCE: Do you know why it's called the Chinese remainder 00:59:00.295 --> 00:59:00.860 theorem? 00:59:00.860 --> 00:59:02.830 PROFESSOR: I assume it was proved 00:59:02.830 --> 00:59:06.040 by a Chinese mathematician but I-- anyone know? 00:59:06.040 --> 00:59:07.291 AUDIENCE: Sun Tzu. 00:59:07.291 --> 00:59:08.460 PROFESSOR: Sun Tzu. 00:59:08.460 --> 00:59:12.645 OK I've heard of him. 00:59:12.645 --> 00:59:14.650 AUDIENCE: Not the guy who did Art of War. 00:59:14.650 --> 00:59:15.400 PROFESSOR: Oh, OK. 00:59:15.400 --> 00:59:17.800 I don't know him then. 00:59:17.800 --> 00:59:21.100 Another Sun Tzu. 00:59:21.100 --> 00:59:23.970 Cool, yeah so permanent mod r is hard. 00:59:23.970 --> 00:59:32.400 Next claim is that the zero one permanent mod r is hard. 00:59:34.932 --> 00:59:36.390 The whole reason we're going to mod 00:59:36.390 --> 00:59:38.473 r-- I mean it's interesting to know that computing 00:59:38.473 --> 00:59:40.680 permanent mod r is just as hard as permanent 00:59:40.680 --> 00:59:43.710 but it's kind of standard thing. 00:59:43.710 --> 00:59:45.610 The reason I wanted to go to mod r 00:59:45.610 --> 00:59:47.800 was to get rid of these negative numbers. 00:59:47.800 --> 00:59:49.730 Negative numbers are annoying. 00:59:49.730 --> 00:59:54.960 Once I go to mod any r negative numbers flip around 00:59:54.960 --> 00:59:58.480 and I can figure everything is positive or non-negative. 00:59:58.480 --> 01:00:01.680 Now I can go back to gadget land. 01:00:01.680 --> 01:00:04.220 So suppose I have an edge of weight five, 01:00:04.220 --> 01:00:06.270 this will work for any number greater than one. 01:00:06.270 --> 01:00:08.380 If It's one I don't touch it. 01:00:08.380 --> 01:00:11.680 If it's greater than one I'm going to make this thing. 01:00:14.900 --> 01:00:19.360 And the claim is there are exactly five ways 01:00:19.360 --> 01:00:20.390 to use this edge. 01:00:20.390 --> 01:00:22.470 Either you don't use it and you don't use it. 01:00:22.470 --> 01:00:26.290 But if you do use it, exactly five ways to use it. 01:00:26.290 --> 01:00:29.229 If you make this traversal, there are five ways to do it. 01:00:29.229 --> 01:00:30.770 Remember this has to be a cycle cover 01:00:30.770 --> 01:00:32.860 so we have to visit everything. 01:00:32.860 --> 01:00:36.450 If you come up here, can't go that way gotta go straight. 01:00:36.450 --> 01:00:38.295 Can't go that way, got to go straight. 01:00:38.295 --> 01:00:40.580 Hmm, OK. 01:00:40.580 --> 01:00:42.830 So in fact, the rest of this cycle cover 01:00:42.830 --> 01:00:44.910 must be entirely within this picture. 01:00:44.910 --> 01:00:47.210 And so for example, you could say that's a cycle 01:00:47.210 --> 01:00:49.830 and then that's a cycle, these are loops. 01:00:49.830 --> 01:00:51.950 And then that's a cycle and then that's 01:00:51.950 --> 01:00:55.390 a cycle and then that's a cycle and then that's a cycle. 01:00:55.390 --> 01:00:57.620 And whoops I'm left with one vertex. 01:00:57.620 --> 01:01:00.930 So in fact I have to choose exactly one of the loops, 01:01:00.930 --> 01:01:03.160 put that in and the rest can be covered. 01:01:03.160 --> 01:01:06.202 For example, if I choose this guy then this will be a cycle, 01:01:06.202 --> 01:01:07.910 this'll be a cycle, this will be a cycle. 01:01:07.910 --> 01:01:10.490 Perfect parity this will be a cycle, is will be a cycle, 01:01:10.490 --> 01:01:11.750 and this will be a cycle. 01:01:11.750 --> 01:01:12.730 Can't choose two loops. 01:01:12.730 --> 01:01:14.271 If I chose like these two groups then 01:01:14.271 --> 01:01:15.960 this guy would be isolated. 01:01:15.960 --> 01:01:18.140 So you have to choose exactly one loop 01:01:18.140 --> 01:01:19.710 and there are exactly five of them. 01:01:19.710 --> 01:01:23.890 In general you make-- there'd be k of them if you want weight k, 01:01:23.890 --> 01:01:25.550 and you can simulate. 01:01:25.550 --> 01:01:29.030 If you don't use this edge then there's actually only one way 01:01:29.030 --> 01:01:29.610 to do it. 01:01:29.610 --> 01:01:33.050 You have to go all the way around like this. 01:01:33.050 --> 01:01:33.762 So that's cool. 01:01:33.762 --> 01:01:35.970 If you don't use the edge, didn't mess with anything. 01:01:35.970 --> 01:01:38.097 If you use the edge you get a weight of five. 01:01:38.097 --> 01:01:40.180 Multiplicative, because this is independent of all 01:01:40.180 --> 01:01:42.170 the other such choices. 01:01:42.170 --> 01:01:47.490 So you can simulate a high-weight edge, module r. 01:01:47.490 --> 01:01:49.990 In general you can simulate a non-negative weight edge 01:01:49.990 --> 01:01:51.460 like this. 01:01:51.460 --> 01:01:54.030 Just these are all weight one obviously and absent edges 01:01:54.030 --> 01:01:55.310 are zero. 01:01:55.310 --> 01:01:57.940 So we can convert in the module r 01:01:57.940 --> 01:02:01.840 setting there are no negative numbers essentially. 01:02:01.840 --> 01:02:03.885 So we can just do that simulation, everything 01:02:03.885 --> 01:02:05.820 will work mod r. 01:02:05.820 --> 01:02:07.110 So use gadget. 01:02:16.530 --> 01:02:19.300 Now finally we can prove zero one permanent is hard. 01:02:24.500 --> 01:02:25.010 Why? 01:02:25.010 --> 01:02:27.093 Because if we could compute the zero one permanent 01:02:27.093 --> 01:02:28.730 we can compute at mod r. 01:02:28.730 --> 01:02:32.290 Just compute the permanent, take the answer mod r and you 01:02:32.290 --> 01:02:34.250 solve zero one permanent mod r. 01:02:34.250 --> 01:02:37.410 So this is what I might call a one-call reduction. 01:02:37.410 --> 01:02:39.550 It's not our usual notion of one-call reduction 01:02:39.550 --> 01:02:42.990 because we're doing stuff at the end. 01:02:42.990 --> 01:02:45.520 But I'm going to use one-call in this setting. 01:02:50.700 --> 01:02:53.630 Just call it-- in fact we don't even have to change the input. 01:02:53.630 --> 01:02:55.600 Then we get the output when we computed mod r. 01:02:55.600 --> 01:02:56.100 Question. 01:02:56.100 --> 01:02:59.920 AUDIENCE: So in that the weight you 01:02:59.920 --> 01:03:01.348 can have really high weight. 01:03:01.348 --> 01:03:04.522 But then your [INAUDIBLE] nodes so then 01:03:04.522 --> 01:03:08.387 when you use that wouldn't n? 01:03:08.387 --> 01:03:09.220 PROFESSOR: Oh I see. 01:03:09.220 --> 01:03:10.845 If the weights were exponentially large 01:03:10.845 --> 01:03:12.350 this would be bad news. 01:03:12.350 --> 01:03:13.940 But they're not exponentially large 01:03:13.940 --> 01:03:17.000 because we know they're all at most three. 01:03:17.000 --> 01:03:20.570 They're between negative one and three. 01:03:20.570 --> 01:03:26.760 Oh but negative one-- Yes, OK good 01:03:26.760 --> 01:03:28.300 I need the prime number theorem. 01:03:28.300 --> 01:03:28.800 Thank you. 01:03:28.800 --> 01:03:32.220 So we have numbers negative one then 01:03:32.220 --> 01:03:34.150 we're mapping them modulo some prime. 01:03:34.150 --> 01:03:38.090 Now negative one is actually the prime minus one. 01:03:38.090 --> 01:03:43.440 So indeed, we do get weights that are as large as r. 01:03:43.440 --> 01:03:47.870 So this is-- r here we're assuming is encoded in unary. 01:03:51.610 --> 01:03:55.710 It turns out we can afford to encode the primes in unary 01:03:55.710 --> 01:03:59.130 because the number of primes that we use 01:03:59.130 --> 01:04:04.370 is this polynomial thing, a weakly polynomial thing. 01:04:04.370 --> 01:04:09.670 And by the prime number theorem, the prime with that index 01:04:09.670 --> 01:04:11.000 is roughly that big. 01:04:11.000 --> 01:04:12.950 It's the log factor larger. 01:04:12.950 --> 01:04:16.550 So the primes are going to-- the actual value 01:04:16.550 --> 01:04:21.860 of the prime is going to be something like n log 01:04:21.860 --> 01:04:32.270 m, times log base e, and log m. 01:04:32.270 --> 01:04:35.470 By the prime number theorem and that's again weakly polynomial. 01:04:35.470 --> 01:04:37.520 And so we can assume ours encoded in unary. 01:04:41.450 --> 01:04:45.643 Wow, we get to use the prime number theorem that's fun. 01:04:45.643 --> 01:04:48.226 AUDIENCE: Is that the first time you had to use prime numbers? 01:04:48.226 --> 01:04:50.511 PROFESSOR: Probably. 01:04:50.511 --> 01:04:52.760 Pretty sure I've used prime number theorem if and only 01:04:52.760 --> 01:04:55.630 if I've used Chinese remainder theorem. 01:04:55.630 --> 01:04:56.620 They go hand in hand. 01:04:59.430 --> 01:04:59.930 Clear? 01:04:59.930 --> 01:05:02.564 So this was kind of roundabout way, 01:05:02.564 --> 01:05:04.730 but we ended up getting rid of the negative numbers. 01:05:04.730 --> 01:05:06.230 Luckily it still worked. 01:05:06.230 --> 01:05:08.210 And now it's zero one permanence hard, 01:05:08.210 --> 01:05:10.890 therefore counting the number of perfect matchings in a given 01:05:10.890 --> 01:05:12.890 bipartite graph isn't going to be hard. 01:05:12.890 --> 01:05:15.464 These problems were equivalent, like they're identical. 01:05:15.464 --> 01:05:17.380 So you could-- this was a reduction in one way 01:05:17.380 --> 01:05:20.980 but that's equally reducible both ways. 01:05:20.980 --> 01:05:23.430 So you can reduce this one, reduce this one 01:05:23.430 --> 01:05:27.306 to perfect matchings or vice versa. 01:05:27.306 --> 01:05:27.805 All right. 01:05:31.047 --> 01:05:32.630 Here's some more fun things we can do. 01:05:50.000 --> 01:05:51.480 Guess I can add to my list here. 01:05:51.480 --> 01:05:55.600 But in particular we have zero one permanent. 01:06:00.200 --> 01:06:04.141 So there are other ways you might be interested in counting 01:06:04.141 --> 01:06:04.640 matchings. 01:06:12.010 --> 01:06:15.380 So, so far we know it's hard to count perfect matchings. 01:06:15.380 --> 01:06:20.774 So in a balanced bipartite graph we had n vertices on the left, 01:06:20.774 --> 01:06:21.940 and n vertices on the right. 01:06:21.940 --> 01:06:23.790 We're going to use that. 01:06:23.790 --> 01:06:26.790 Then the hope is that there's matching a size n. 01:06:26.790 --> 01:06:29.129 But in general, you could just count maybe those 01:06:29.129 --> 01:06:29.920 don't exist at all. 01:06:29.920 --> 01:06:32.060 Maybe we just want to count maximal matchings. 01:06:32.060 --> 01:06:34.280 These are not maximum matchings. 01:06:34.280 --> 01:06:37.340 Maximal, meaning you can't add any more edges to them. 01:06:37.340 --> 01:06:40.800 They're locally maximum. 01:06:40.800 --> 01:06:43.140 So that's going to be a bigger number, 01:06:43.140 --> 01:06:45.880 it's going to be always bigger than zero. 01:06:45.880 --> 01:06:47.850 Because the empty matching, well you 01:06:47.850 --> 01:06:49.308 could start with the empty matching 01:06:49.308 --> 01:06:52.000 and add things you'll get at least one maximal matching. 01:06:52.000 --> 01:06:54.150 But this is also sharp P complete, 01:06:54.150 --> 01:06:56.090 and you can prove it using some tricks 01:06:56.090 --> 01:07:01.380 from bipartite perfect matching, counting. 01:07:04.580 --> 01:07:06.070 Don't have a ton of time so I'll go 01:07:06.070 --> 01:07:08.840 through this relatively quick. 01:07:08.840 --> 01:07:12.380 We're going to take each vertex and convert it, basically 01:07:12.380 --> 01:07:14.920 make n copies of it. 01:07:14.920 --> 01:07:18.760 And when we have an edge that's going to turn 01:07:18.760 --> 01:07:23.940 into a biclique between them. 01:07:23.940 --> 01:07:25.800 Why did I address such a big one? 01:07:29.970 --> 01:07:32.650 This was supposed to be n and n. 01:07:32.650 --> 01:07:34.220 OK so just blow up every edge. 01:07:34.220 --> 01:07:37.850 My intent is to make matchings become more plentiful. 01:07:40.480 --> 01:07:47.552 So in general, if I used to have a matching of size i, 01:07:47.552 --> 01:07:54.310 I end up converting it into n factorial to the i-th power, 01:07:54.310 --> 01:08:04.460 distinct matchings of size n times i. 01:08:04.460 --> 01:08:07.480 OK because if I use an edge here, 01:08:07.480 --> 01:08:09.790 now I get to put in a arbitrary perfect matching 01:08:09.790 --> 01:08:14.160 in this biclique and they're n factorial of them. 01:08:14.160 --> 01:08:17.229 Cool, why did I do that? 01:08:17.229 --> 01:08:19.410 Because now I suppose that I knew 01:08:19.410 --> 01:08:22.627 how to count the number of maximum actions. 01:08:22.627 --> 01:08:24.710 So they're going to be matchings of various sizes. 01:08:24.710 --> 01:08:27.740 From that I want to extract the number of perfect matchings. 01:08:27.740 --> 01:08:31.170 Sounds impossible, but when you make things giant 01:08:31.170 --> 01:08:33.785 like this they kind of-- all the matchings kind of separate. 01:08:33.785 --> 01:08:37.029 It's like biology or something. 01:08:37.029 --> 01:08:37.560 Chemistry. 01:08:37.560 --> 01:08:43.260 So let's see, it shows how much I know. 01:08:43.260 --> 01:08:49.790 Number of maximal matchings is going 01:08:49.790 --> 01:08:55.420 to be sum i equals zero to n over 2. 01:08:55.420 --> 01:08:58.800 That's the possible sizes of the matchings. 01:08:58.800 --> 01:09:00.370 Of the old matchings I should say. 01:09:00.370 --> 01:09:05.670 Number of original maximal matchings 01:09:05.670 --> 01:09:15.290 in the input graph of size i, times n factorial to the i. 01:09:15.290 --> 01:09:16.830 OK this is just rewriting. 01:09:16.830 --> 01:09:19.189 I'm just this thing, but I'm summing over all i. 01:09:19.189 --> 01:09:21.939 So I have however many matchings I used to have a size i, 01:09:21.939 --> 01:09:24.010 but then I multiply them by n factorial to the i, 01:09:24.010 --> 01:09:25.426 because these are all independent. 01:09:29.970 --> 01:09:33.319 And this thing, the original number 01:09:33.319 --> 01:09:36.770 of matchings in the worst case, I mean the largest it could be 01:09:36.770 --> 01:09:38.529 is when everything is a biclique. 01:09:38.529 --> 01:09:47.275 And then we have n over two factorial maximal matchings 01:09:47.275 --> 01:09:47.775 originally. 01:09:50.569 --> 01:09:54.650 And this is smaller than that. 01:09:54.650 --> 01:09:58.920 And therefore we can pull this apart as a number modulo 01:09:58.920 --> 01:10:00.650 n factorial. 01:10:00.650 --> 01:10:05.020 And each digit is the number of maximum actions of each size. 01:10:05.020 --> 01:10:07.630 And we look at the last digit, we 01:10:07.630 --> 01:10:11.800 get all the number of maximum matchings of size n 01:10:11.800 --> 01:10:16.140 over two also known as the number of perfect matchings. 01:10:16.140 --> 01:10:16.640 Question? 01:10:16.640 --> 01:10:18.970 AUDIENCE: So that second max is maximal? 01:10:18.970 --> 01:10:20.340 PROFESSOR: Yes this is maximal. 01:10:24.680 --> 01:10:26.360 Other questions? 01:10:26.360 --> 01:10:29.000 So again I kind of numbered [? through ?] trick 01:10:29.000 --> 01:10:33.270 by blowing up-- well by making each of these 01:10:33.270 --> 01:10:36.130 get multiplied by a different huge number 01:10:36.130 --> 01:10:39.070 we can pull them apart, separate them-- In logarithm 01:10:39.070 --> 01:10:42.595 these numbers are not huge so it is actually plausible 01:10:42.595 --> 01:10:43.850 you could compute this thing. 01:10:46.699 --> 01:10:48.490 And a tree graph you definitely could do it 01:10:48.490 --> 01:10:51.950 by various multiplications at each level. 01:10:51.950 --> 01:10:53.300 All right. 01:10:53.300 --> 01:10:58.475 So that's nice, let's do another one of that flavor. 01:11:01.163 --> 01:11:02.704 AUDIENCE: So I was actually wondering 01:11:02.704 --> 01:11:04.745 can you find all those primes in polynomial time? 01:11:04.745 --> 01:11:07.820 PROFESSOR: You could use the Sieve of Eratosthenes. 01:11:07.820 --> 01:11:09.836 AUDIENCE: Is that like going to be-- 01:11:09.836 --> 01:11:12.460 PROFESSOR: We can handle pseudo poly here because as we argued, 01:11:12.460 --> 01:11:14.230 the primes are not that big. 01:11:14.230 --> 01:11:16.650 So we could just spend-- we could just 01:11:16.650 --> 01:11:18.650 do Sieve of Eratosthenes, we can afford 01:11:18.650 --> 01:11:21.620 to spend a linear time on the largest prime. 01:11:21.620 --> 01:11:22.725 Or quadratic in that even. 01:11:22.725 --> 01:11:24.100 You could do the naive algorithm. 01:11:29.510 --> 01:11:33.280 We're not doing real Hogan's [? mid-number ?] theory here. 01:11:33.280 --> 01:11:37.250 I could tell where you have to be more careful. 01:11:37.250 --> 01:11:40.190 All right. . 01:11:40.190 --> 01:11:43.492 So here's another question. 01:11:43.492 --> 01:11:45.700 What if I just want to count the number of matchings? 01:11:45.700 --> 01:11:46.691 No condition? 01:11:50.930 --> 01:11:54.990 This is going to use almost the reverse trick. 01:11:54.990 --> 01:11:57.800 So no maximal constraint just all matchings, 01:11:57.800 --> 01:11:59.050 including the empty matchings. 01:11:59.050 --> 01:12:01.100 Count them all. 01:12:01.100 --> 01:12:04.970 We'll see why in a moment, this is quite natural. 01:12:04.970 --> 01:12:09.660 So I claim I can do a multicall reduction to bipartite 01:12:09.660 --> 01:12:11.780 number of maximal matchings. 01:12:11.780 --> 01:12:15.110 And here are my calls for every graph g, 01:12:15.110 --> 01:12:18.190 I'm going to make a graph g prime. 01:12:18.190 --> 01:12:21.020 Where if I have a vertex I'm going to add 01:12:21.020 --> 01:12:25.130 k extra nodes connected just to that vertex. 01:12:25.130 --> 01:12:28.022 So these are leaves and so if this vertex 01:12:28.022 --> 01:12:30.230 was unmatched over here, then I have k different ways 01:12:30.230 --> 01:12:32.580 to match it over here. 01:12:32.580 --> 01:12:36.840 And my intent is to measure this number of matchings of size n 01:12:36.840 --> 01:12:39.080 over two minus k. 01:12:39.080 --> 01:12:40.060 That would be the hope. 01:12:44.500 --> 01:12:53.110 So if I have m r matchings of size r over here, 01:12:53.110 --> 01:12:58.820 these will get mapped to Mr times k plus one 01:12:58.820 --> 01:13:09.110 to the r, matchings of size no, not of size. 01:13:09.110 --> 01:13:10.040 Matchings over here. 01:13:16.654 --> 01:13:18.820 Because for each one I can either leave it unmatched 01:13:18.820 --> 01:13:20.620 or I could add this edge, or I could add this edge, 01:13:20.620 --> 01:13:22.078 or I could add this edge and that's 01:13:22.078 --> 01:13:24.060 true for every unmatched vertex. 01:13:24.060 --> 01:13:28.062 Sorry, this is not size r, this is size n over two minus r. 01:13:28.062 --> 01:13:30.660 R is going to be the number of leftover guys. 01:13:30.660 --> 01:13:33.610 Then they kind of pops out over here. 01:13:33.610 --> 01:13:36.010 So I'm going to run this algorithm-- I'm 01:13:36.010 --> 01:13:43.040 going to compute the number of matchings in Gk for all k up 01:13:43.040 --> 01:13:49.250 to like n over two plus one. 01:13:49.250 --> 01:14:00.930 So I'm going to do this and what I end up 01:14:00.930 --> 01:14:07.390 computing is number of matchings in each Gk, which is going 01:14:07.390 --> 01:14:10.022 to be the sum over all r. 01:14:10.022 --> 01:14:15.885 r is zero to n over two of Mr the original number 01:14:15.885 --> 01:14:19.830 of matchings the size n over two minus r, 01:14:19.830 --> 01:14:22.510 times k plus one to the r. 01:14:22.510 --> 01:14:28.170 Now this is a polynomial in k plus one. 01:14:28.170 --> 01:14:33.000 And if I evaluate a polynomial at its degree plus one 01:14:33.000 --> 01:14:35.500 different points I can reconstruct the coefficients 01:14:35.500 --> 01:14:36.720 of the polynomial. 01:14:36.720 --> 01:14:39.160 And therefore get all of these and in particular 01:14:39.160 --> 01:14:44.980 get m zero, which is the number of perfect matchings. 01:14:44.980 --> 01:14:45.980 Ta da. 01:14:58.130 --> 01:14:58.990 Having too much fun. 01:15:06.450 --> 01:15:07.190 Enough matchings. 01:15:10.700 --> 01:15:13.720 We have all these versions of SAT, which are hard. 01:15:13.720 --> 01:15:17.230 But in fact there are even funnier versions 01:15:17.230 --> 01:15:19.360 like positive 2SAT. 01:15:22.850 --> 01:15:25.600 Positive 2SAT is really easy to decide, 01:15:25.600 --> 01:15:29.510 but it turns out if I add a sharp it is sharp P complete. 01:15:29.510 --> 01:15:31.402 This is sort of like max 2SAT, but here we 01:15:31.402 --> 01:15:32.860 have to satisfy all the constraints 01:15:32.860 --> 01:15:34.985 but you want to count how many different ways there 01:15:34.985 --> 01:15:37.090 are to solve it. 01:15:37.090 --> 01:15:39.350 This is the same thing as vertex cover remember. 01:15:39.350 --> 01:15:41.550 Vertex cover's the same as positive 2SAT. 01:15:41.550 --> 01:15:43.730 So also sharp vertex cover's hard 01:15:43.730 --> 01:15:46.570 and therefore also sharp clique is hard. 01:15:46.570 --> 01:15:49.300 So this is actually a parsimonious reduction 01:15:49.300 --> 01:15:51.275 from bipartite matching. 01:15:51.275 --> 01:15:53.390 That's why I wanted to get there. 01:15:53.390 --> 01:16:00.710 So for every edge we're going to make a variable, x of e, 01:16:00.710 --> 01:16:05.500 which is true if the edge is not in the perfect matching. 01:16:05.500 --> 01:16:11.230 And so then whenever I have two incident edges in f, 01:16:11.230 --> 01:16:14.440 we're going to have a clause which is either I don't use e, 01:16:14.440 --> 01:16:16.610 or I don't use f. 01:16:16.610 --> 01:16:18.424 That's 2SAT, done. 01:16:23.570 --> 01:16:24.220 Satisfying? 01:16:24.220 --> 01:16:25.050 It's parsimonious. 01:16:25.050 --> 01:16:27.330 So satisfying assignments will be one to one 01:16:27.330 --> 01:16:29.730 with bipartite matchings. 01:16:29.730 --> 01:16:31.660 So this is why you should care. 01:16:31.660 --> 01:16:35.600 And if we instead reduce from-- did I erase it? 01:16:35.600 --> 01:16:40.450 Bipartite maximal matchings up here, 01:16:40.450 --> 01:16:43.220 then I get the counting the number 01:16:43.220 --> 01:16:51.530 of maximally true assignments that satisfy the 2SAT clause 01:16:51.530 --> 01:16:54.430 is also hard. 01:16:54.430 --> 01:16:58.120 For what it's worth, a different way of counting that. 01:16:58.120 --> 01:16:58.820 (BREAK IN VIDEO) 01:16:58.820 --> 01:17:01.710 Maxwell means you can't set any more variables true. 01:17:01.710 --> 01:17:02.210 Yeah. 01:17:02.210 --> 01:17:05.978 AUDIENCE: Since the edges on your positive 2SAT reduction 01:17:05.978 --> 01:17:13.154 are the variables there are true when the edge is not 01:17:13.154 --> 01:17:14.122 in a matching. 01:17:14.122 --> 01:17:15.962 I think it's maximally false. 01:17:15.962 --> 01:17:16.920 And not maximally true. 01:17:16.920 --> 01:17:19.740 Also maximally true, you just said everything was true. 01:17:19.740 --> 01:17:20.810 PROFESSOR: Yes. 01:17:20.810 --> 01:17:22.390 Right, right, right, sorry. 01:17:22.390 --> 01:17:23.223 I see, you're right. 01:17:23.223 --> 01:17:25.450 So it's minimal solutions for a 2SAT, thank you. 01:17:28.700 --> 01:17:40.650 OK in fact, it's known that three regular bipartite planar 01:17:40.650 --> 01:17:41.720 sharp vertex cover. 01:17:46.262 --> 01:17:47.220 I won't prove this one. 01:17:47.220 --> 01:17:50.230 But in particular you can make this planar, 01:17:50.230 --> 01:17:52.420 although it doesn't look like we have positive here. 01:17:52.420 --> 01:17:54.461 And you can also make it bipartite, which doesn't 01:17:54.461 --> 01:17:55.810 mean a lot in 2SAT land. 01:17:55.810 --> 01:17:59.830 But it means-- makes a lot of sense in vertex cover land. 01:17:59.830 --> 01:18:01.620 In my zero remaining minutes I want 01:18:01.620 --> 01:18:03.030 to mention one more concept. 01:18:05.736 --> 01:18:09.560 To go back to the original question 01:18:09.560 --> 01:18:13.400 of uniqueness for puzzles. 01:18:13.400 --> 01:18:16.290 And you'll see this in the literature if you look at it. 01:18:16.290 --> 01:18:18.730 Lot of people won't even mention sharp P, 01:18:18.730 --> 01:18:21.690 but they'll mention ASP. 01:18:21.690 --> 01:18:25.880 ASP is slightly weaker but for most intents and purposes 01:18:25.880 --> 01:18:28.210 essentially identical notion to sharp P, 01:18:28.210 --> 01:18:30.300 from a hardness perspective. 01:18:30.300 --> 01:18:33.270 The goal for-- in general, if I have a problem 01:18:33.270 --> 01:18:35.500 a search problem like we started with. 01:18:35.500 --> 01:18:38.650 The ASP version of that search problem is, 01:18:38.650 --> 01:18:43.180 I give you a solution and I want to know is there another one? 01:18:43.180 --> 01:18:44.930 This is a little different from everything 01:18:44.930 --> 01:18:48.085 we've seen because I actually give you a starting point. 01:18:48.085 --> 01:18:49.960 There's some problems where this helps a lot. 01:18:49.960 --> 01:18:54.110 For example, if the solution is never unique like coloring, 01:18:54.110 --> 01:18:55.890 I give you a k coloring, I can give you 01:18:55.890 --> 01:18:58.310 another one I'll just swap colors one and two. 01:18:58.310 --> 01:19:01.890 Or more subtle if I give you a Hamiltonian cycle in a three 01:19:01.890 --> 01:19:02.610 regular graph. 01:19:02.610 --> 01:19:06.860 There's always a way to switch it and make another one. 01:19:06.860 --> 01:19:09.940 So ASP is sometimes easy. 01:19:09.940 --> 01:19:13.630 But you would basically use the same reductions. 01:19:13.630 --> 01:19:16.700 If you can find a parsimonious reduction that 01:19:16.700 --> 01:19:19.310 also has the property-- so parsimonious reduction 01:19:19.310 --> 01:19:22.490 there is a one to one bijection between solutions 01:19:22.490 --> 01:19:23.900 to x and solutions to x prime. 01:19:23.900 --> 01:19:27.382 If you could also compute that bijection, if you can 01:19:27.382 --> 01:19:29.090 convert a solution from one to the other. 01:19:29.090 --> 01:19:31.080 Which we could do in every single proof we've seen 01:19:31.080 --> 01:19:32.496 and even the ones we haven't seen. 01:19:32.496 --> 01:19:35.000 You can always compute that bijection between solutions 01:19:35.000 --> 01:19:38.110 of A to solutions of B. And so what that means is, 01:19:38.110 --> 01:19:46.550 if I can solve B, consider the ASP version of A 01:19:46.550 --> 01:19:47.890 where I give you a solution. 01:19:47.890 --> 01:19:50.610 If I can convert that solution into a solution for my B 01:19:50.610 --> 01:19:52.800 problem, and then with the parsimonious reduction 01:19:52.800 --> 01:19:58.830 I also get a B instance, then I can solve the ASP problem 01:19:58.830 --> 01:20:00.560 for B. That's a decision problem. 01:20:00.560 --> 01:20:02.000 Is there another solution? 01:20:02.000 --> 01:20:04.480 If yes, then A will also have another solution, 01:20:04.480 --> 01:20:07.420 because it's parsimonious the numbers will be the same. 01:20:07.420 --> 01:20:11.050 We need a weaker version of parsimony we just need the one 01:20:11.050 --> 01:20:13.011 and more than one are kept the same. 01:20:13.011 --> 01:20:15.260 If this one was unique then this one should be unique. 01:20:15.260 --> 01:20:18.040 If this one wasn't unique then this one should be not unique. 01:20:18.040 --> 01:20:20.180 If I can solve B then I can solve A. 01:20:20.180 --> 01:20:24.070 Or if I can solve ASP B then I can solve ASP A. 01:20:24.070 --> 01:20:27.750 A lot of these proofs are done-- so-called ASP reduction 01:20:27.750 --> 01:20:30.660 is usually done by a parsimonious reduction. 01:20:30.660 --> 01:20:33.740 And so in particular this was introduced, 01:20:33.740 --> 01:20:36.410 this concept was introduced in the likes of like Slitherlink, 01:20:36.410 --> 01:20:40.160 I think, was one of the early ASP completeness proofs. 01:20:40.160 --> 01:20:42.310 And they were interested in this because the idea 01:20:42.310 --> 01:20:43.935 is if you're designing a puzzle usually 01:20:43.935 --> 01:20:46.390 you design a puzzle with a solution in mind. 01:20:46.390 --> 01:20:48.780 But then you need to make sure there's no other solution. 01:20:48.780 --> 01:20:52.380 So you have exactly this set up and if you 01:20:52.380 --> 01:20:55.020 want a formal definition of ASP completeness it's in the notes. 01:20:55.020 --> 01:21:00.430 But it's not that difficult. 01:21:00.430 --> 01:21:03.602 The key point here is if you're going to prove sharp P or ASP 01:21:03.602 --> 01:21:06.060 completeness you might as well prove the other one as well. 01:21:06.060 --> 01:21:09.400 Get twice the results for basically the same reduction. 01:21:09.400 --> 01:21:11.460 All the versions of 3SAT and 1-in-3SAT, 01:21:11.460 --> 01:21:14.790 and all those things, those were all parsimonious. 01:21:14.790 --> 01:21:17.730 And all of those are ASP complete as well. 01:21:17.730 --> 01:21:19.330 But once you get into C-monious you're 01:21:19.330 --> 01:21:23.470 no longer preserving one versus more than one. 01:21:23.470 --> 01:21:26.252 So while you preserved-- from a counting perspective 01:21:26.252 --> 01:21:28.460 in multicall reductions where you can divide that off 01:21:28.460 --> 01:21:29.320 that's fine. 01:21:29.320 --> 01:21:32.490 But from an ASP completeness perspective you're not fine. 01:21:32.490 --> 01:21:36.010 So in fact all the stuff that we just 01:21:36.010 --> 01:21:40.870 did with the permanent matchings and sharp 2SAT, 01:21:40.870 --> 01:21:41.880 their sharp P complete. 01:21:41.880 --> 01:21:43.625 They may not be ASP complete. 01:21:46.230 --> 01:21:48.080 But you'll notice the puzzles that we 01:21:48.080 --> 01:21:54.940 reduce from like, whatever, Shakashaka is one of them. 01:21:54.940 --> 01:21:59.210 That was from a version of 3SAT, and the other one we had 01:21:59.210 --> 01:22:00.710 was from a version of Hamiltonicity. 01:22:03.500 --> 01:22:05.130 This guy. 01:22:05.130 --> 01:22:07.090 These are all-- which was also from 3SAT. 01:22:07.090 --> 01:22:10.250 So these are all ASP complete also. 01:22:10.250 --> 01:22:12.850 But if you use the weirder things like 2SAT 01:22:12.850 --> 01:22:14.170 you don't get that. 01:22:14.170 --> 01:22:17.070 So usually in one proof you can get NP hardness, ASP hardness, 01:22:17.070 --> 01:22:19.127 and sharp P hardness. 01:22:19.127 --> 01:22:20.960 But if you're going to go from these weirder 01:22:20.960 --> 01:22:22.777 problems with C-monious reductions then 01:22:22.777 --> 01:22:24.860 you only get sharp P hardness, because they're not 01:22:24.860 --> 01:22:26.900 even NP-hard. 01:22:26.900 --> 01:22:28.230 Cool, all right. 01:22:28.230 --> 01:22:30.000 Thank you.