WEBVTT 00:00:00.070 --> 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 --> 00:00:12.700 To make a donation or to view additional materials 00:00:12.700 --> 00:00:16.600 from hundreds of MIT courses, visit MIT OpenCourseWare 00:00:16.600 --> 00:00:17.263 at ocw.mit.edu. 00:00:25.846 --> 00:00:26.720 PROFESSOR: All right. 00:00:26.720 --> 00:00:31.300 Today we continue our theme of approximation, lower bounds 00:00:31.300 --> 00:00:33.020 inapproximability. 00:00:33.020 --> 00:00:35.470 Quick recap of last time. 00:00:35.470 --> 00:00:39.530 We talked about lots of different reductions. 00:00:39.530 --> 00:00:44.900 We, I guess, in particular talked about P-tests, AP and L. 00:00:44.900 --> 00:00:48.720 And in particular we'll be using L-reductions almost exclusively 00:00:48.720 --> 00:00:51.880 today, except the occasional strict reduction, which 00:00:51.880 --> 00:00:55.160 is even stronger, in a sense. 00:00:55.160 --> 00:00:57.117 So what's an L-reduction? 00:00:57.117 --> 00:00:59.450 We're trying to go from one problem A to another problem 00:00:59.450 --> 00:01:05.530 B. We're given an instance x of A. We convert it via function f 00:01:05.530 --> 00:01:10.790 to an instance x prime of B. Then we imagine that somehow we 00:01:10.790 --> 00:01:12.290 obtain a solution. 00:01:12.290 --> 00:01:16.030 We don't know anything about it. y prime to x prime. 00:01:16.030 --> 00:01:17.540 That's in B space. 00:01:17.540 --> 00:01:19.210 And then, in the reduction, we're 00:01:19.210 --> 00:01:21.160 supposed to be able to map any such solution 00:01:21.160 --> 00:01:28.040 y prime to x prime via g into solution y of x in A problem-- 00:01:28.040 --> 00:01:31.560 so that's given by the function g-- such that two things hold. 00:01:31.560 --> 00:01:36.070 The first one is that for f, optimal solution 00:01:36.070 --> 00:01:39.520 of x prime should be at most some constant times 00:01:39.520 --> 00:01:41.930 the optimal solution to x. 00:01:41.930 --> 00:01:44.190 So we don't blow up OPTs too much. 00:01:44.190 --> 00:01:47.660 And secondly the absolute difference 00:01:47.660 --> 00:01:51.600 between the cost of y versus the optimal solution 00:01:51.600 --> 00:01:56.150 for x should be within a constant factor of this kind 00:01:56.150 --> 00:01:59.650 of gap-- additive gap between the cost of y 00:01:59.650 --> 00:02:04.210 prime versus the optimal solution to x prime, 00:02:04.210 --> 00:02:06.430 meaning that if we were given a y prime that's 00:02:06.430 --> 00:02:08.430 very close to optimal for x prime, then the y 00:02:08.430 --> 00:02:10.949 we produce is very close to optimal for x. 00:02:10.949 --> 00:02:14.446 And we want that in an additive sense 00:02:14.446 --> 00:02:17.070 that will imply that things are good in a multiplicative sense. 00:02:17.070 --> 00:02:20.180 Last time we proved that for the min case, 00:02:20.180 --> 00:02:21.490 for minimization problems. 00:02:21.490 --> 00:02:24.110 If you're curious, I worked out the details 00:02:24.110 --> 00:02:26.840 for maximization problems. 00:02:26.840 --> 00:02:29.670 It's a little bit uglier in terms of the arithmetic. 00:02:29.670 --> 00:02:32.920 But you again get that if you had a constant factor 00:02:32.920 --> 00:02:35.970 approximation over here, you preserve a constant factor 00:02:35.970 --> 00:02:39.630 approximation over here, and you only-- you 00:02:39.630 --> 00:02:41.500 lose a reasonable factor. 00:02:44.530 --> 00:02:47.119 We also have that if you can get a PTAS over here, 00:02:47.119 --> 00:02:49.160 so you can get an arbitrarily good approximation, 00:02:49.160 --> 00:02:50.550 you also get a PTAS over here. 00:02:50.550 --> 00:02:52.380 That was the PTAS reduction. 00:02:52.380 --> 00:02:55.430 And it turns out the constant in the end is roughly 00:02:55.430 --> 00:02:59.710 epsilon over alpha beta, where alpha was this constant, 00:02:59.710 --> 00:03:02.270 and beta was this constant. 00:03:02.270 --> 00:03:03.564 That's what we had before. 00:03:03.564 --> 00:03:04.730 It's a little bit different. 00:03:04.730 --> 00:03:06.380 For small epsilon, it's about the same. 00:03:06.380 --> 00:03:09.310 But for large epsilon, it does make a difference. 00:03:09.310 --> 00:03:13.470 And this is why, in case you were confused, 00:03:13.470 --> 00:03:17.230 an L-reduction does not imply in the maximization 00:03:17.230 --> 00:03:21.030 case an AP-reduction, because you have this non-linear term. 00:03:21.030 --> 00:03:23.680 Here, everything was linear in epsilon. 00:03:23.680 --> 00:03:26.320 With minimization, that's true. 00:03:26.320 --> 00:03:27.780 The L implies AP. 00:03:27.780 --> 00:03:30.350 But for maximization it's not quite true. 00:03:30.350 --> 00:03:32.560 It's close. 00:03:32.560 --> 00:03:34.270 So there's some funny. 00:03:34.270 --> 00:03:36.270 What I said didn't quite match this picture. 00:03:36.270 --> 00:03:38.920 That's an explanation. 00:03:38.920 --> 00:03:42.280 And then we did a few reductions. 00:03:42.280 --> 00:03:47.680 I claimed that Max E3SAT-E5, this was exactly 00:03:47.680 --> 00:03:49.650 three distinct literals per clause, 00:03:49.650 --> 00:03:54.790 exactly five occurrences of each variable in five 00:03:54.790 --> 00:03:57.150 different clauses. 00:03:57.150 --> 00:03:58.490 I claimed that was APX-complete. 00:03:58.490 --> 00:04:00.130 We didn't prove it. 00:04:00.130 --> 00:04:02.810 What we did prove is that assuming Max 3SAT 00:04:02.810 --> 00:04:06.940 is APX-complete, we reduce that to Max 3SAT3, 00:04:06.940 --> 00:04:10.240 which is at most three occurrences, each thing, 00:04:10.240 --> 00:04:12.200 first by using expander, and then 00:04:12.200 --> 00:04:14.620 splitting the constant size-- constant occurrence 00:04:14.620 --> 00:04:18.190 variables-- with the cycle of implications trick. 00:04:18.190 --> 00:04:21.019 And then we reduced from that to bounded degree. 00:04:21.019 --> 00:04:23.800 I think we did like max degree 4. 00:04:23.800 --> 00:04:27.320 But all of these can be done in max degree 3. 00:04:27.320 --> 00:04:30.951 Independent set, vertex cover, and dominating set. 00:04:30.951 --> 00:04:32.200 Vertex cover we've seen a lot. 00:04:32.200 --> 00:04:33.940 You want to cover all the edges by choosing vertices. 00:04:33.940 --> 00:04:35.650 Dominating set, you want to cover all the vertices 00:04:35.650 --> 00:04:36.630 by choosing vertices. 00:04:36.630 --> 00:04:38.610 Each vertex covers its neighbor set. 00:04:38.610 --> 00:04:43.556 And independent set, for general graphs this is super hard. 00:04:43.556 --> 00:04:45.055 But for bounded degree graphs, there 00:04:45.055 --> 00:04:46.950 is a constant factor approximation. 00:04:46.950 --> 00:04:50.750 This was choosing vertices that induced no edges. 00:04:50.750 --> 00:04:55.850 So with that in mind, let's do some more APX-reductions, 00:04:55.850 --> 00:04:58.030 APX-hardness, using L-reductions. 00:05:00.600 --> 00:05:06.450 So the next problem we're going to do is Max 2SAT. 00:05:12.110 --> 00:05:15.470 So because we're in the world of optimization, in some sense 00:05:15.470 --> 00:05:19.420 the distinction between 2SAT and 3SAT is not so important. 00:05:19.420 --> 00:05:22.340 It turns out Max 2SAT will be APX-complete 00:05:22.340 --> 00:05:24.230 just like Max 3SAT was. 00:05:24.230 --> 00:05:25.877 So when we didn't have Max, of course 00:05:25.877 --> 00:05:27.460 the complexities were quite different. 00:05:27.460 --> 00:05:29.847 3SAT was hard, 2SAT was easy. 00:05:29.847 --> 00:05:31.430 With maximization, they're going to be 00:05:31.430 --> 00:05:34.460 equivalent in this perspective. 00:05:34.460 --> 00:05:43.150 So I'm going to do an L-reduction 00:05:43.150 --> 00:05:52.250 from independent set of, let's say a degree 3. 00:05:55.275 --> 00:05:56.900 So it'll work with any constant degree, 00:05:56.900 --> 00:05:59.260 but we'll get a different number of occurrences. 00:05:59.260 --> 00:06:01.710 And the reduction is the following. 00:06:01.710 --> 00:06:05.240 There are two types of gadgets for every vertex. 00:06:05.240 --> 00:06:07.280 So I'm given an independent set instance. 00:06:07.280 --> 00:06:11.240 For every vertex v, we're going to convert that 00:06:11.240 --> 00:06:19.484 into a clause-- namely v. I want v to be true, if possible. 00:06:19.484 --> 00:06:21.150 It's a funny way of thinking when you're 00:06:21.150 --> 00:06:23.608 maximizing a number of causes, because a lot of the clauses 00:06:23.608 --> 00:06:24.490 won't be satisfied. 00:06:24.490 --> 00:06:27.240 But you're going to try to put v in the independent set 00:06:27.240 --> 00:06:28.040 if you can. 00:06:28.040 --> 00:06:30.060 That's the meaning of that clause. 00:06:30.060 --> 00:06:34.420 Then for every edge-- let's say connecting v 00:06:34.420 --> 00:06:38.070 to w-- we're going to convert that into a clause which 00:06:38.070 --> 00:06:43.220 is not v or not w. 00:06:43.220 --> 00:06:46.200 We don't want them both to be in the independent set. 00:06:46.200 --> 00:06:48.794 That's the meaning of-- yeah. 00:06:48.794 --> 00:06:50.460 I'm trying to simulate independent sets. 00:06:50.460 --> 00:06:52.200 So I don't want these both to be in. 00:06:52.200 --> 00:06:55.220 This is a 2SAT clause. 00:06:55.220 --> 00:06:56.950 So what's the claim here? 00:06:56.950 --> 00:07:00.800 Suppose you have some assignment to the variable. 00:07:00.800 --> 00:07:03.360 So there's one variable per vertex over here. 00:07:03.360 --> 00:07:07.900 The idea is that variable should indicate whether the vertex is 00:07:07.900 --> 00:07:10.450 in the independent set. 00:07:10.450 --> 00:07:13.290 And the claim is that we will never 00:07:13.290 --> 00:07:18.315 violate an edge constraint, or it's never useful to violate. 00:07:18.315 --> 00:07:21.050 The claim is that there exists an OPT-- optimal 00:07:21.050 --> 00:07:29.075 solution-- satisfying all of these edge constraints. 00:07:32.021 --> 00:07:33.020 So we're doing Max 2SAT. 00:07:33.020 --> 00:07:35.610 So we get a point for every one of these things 00:07:35.610 --> 00:07:37.670 that we satisfy. 00:07:37.670 --> 00:07:40.790 And so in particular, if you didn't 00:07:40.790 --> 00:07:45.530 get this point-- not v or not w-- the converse of this 00:07:45.530 --> 00:07:48.740 is that they are both in. 00:07:48.740 --> 00:07:50.570 Then the idea is that you instead 00:07:50.570 --> 00:07:54.510 take one of those vertices out of the independent set, 00:07:54.510 --> 00:07:57.484 and that will be better for you. 00:07:57.484 --> 00:07:59.900 In general, when you put a variable in an independent set, 00:07:59.900 --> 00:08:02.462 it only helps you for one clause. 00:08:02.462 --> 00:08:04.170 There's only one occurrence of positive v 00:08:04.170 --> 00:08:05.500 in all of these things. 00:08:05.500 --> 00:08:08.710 You might have many edges coming into a vertex, 00:08:08.710 --> 00:08:12.350 and they all prefer the case that v is false. 00:08:12.350 --> 00:08:15.220 So things are going to be easier if you set v to false. 00:08:15.220 --> 00:08:18.420 So if you discover a clause like this, which is currently false, 00:08:18.420 --> 00:08:20.880 meaning both v and w are true, you're 00:08:20.880 --> 00:08:24.170 going to gain a point by setting v to false. 00:08:24.170 --> 00:08:26.564 You'll also lose a point, but you'll only lose one point. 00:08:26.564 --> 00:08:27.980 Potentially, you gain many points, 00:08:27.980 --> 00:08:31.620 but you gain at least one point and lose at most one point 00:08:31.620 --> 00:08:36.240 by switching from both v and w true into just one 00:08:36.240 --> 00:08:37.490 of them true. 00:08:37.490 --> 00:08:41.840 So you can always convert without losing anything in OPT 00:08:41.840 --> 00:08:44.900 into a solution that satisfies all edge constraints. 00:08:44.900 --> 00:08:48.000 And then we know we have an independent set. 00:08:48.000 --> 00:08:50.070 That's what the edge constraints say. 00:08:50.070 --> 00:08:53.006 And therefore the remaining problem 00:08:53.006 --> 00:08:54.630 is to maximize the number vertices that 00:08:54.630 --> 00:08:55.853 are in the independent set. 00:09:04.350 --> 00:09:08.820 So that means if we're given any solution y prime to this Max 00:09:08.820 --> 00:09:11.915 2SAT instance, we can convert it back to an independent set. 00:09:11.915 --> 00:09:14.970 Now it's not quite of the same value. 00:09:14.970 --> 00:09:20.700 In general, the optimal solution here for the 2SAT instance 00:09:20.700 --> 00:09:23.080 is going to be the optimal solution 00:09:23.080 --> 00:09:28.197 for the independent set instance plus the total number of edges, 00:09:28.197 --> 00:09:30.030 because we're going to satisfy all of these. 00:09:30.030 --> 00:09:32.030 That's what we just showed. 00:09:32.030 --> 00:09:36.410 So this is where we get a kind of additive behavior, 00:09:36.410 --> 00:09:39.060 like in this L-reduction. 00:09:39.060 --> 00:09:40.800 The gap is an additive thing. 00:09:40.800 --> 00:09:42.310 But here it's a nice fixed thing. 00:09:42.310 --> 00:09:46.080 And so these are pretty much the same. 00:09:46.080 --> 00:09:48.160 There's just this additive offset. 00:09:48.160 --> 00:09:51.110 So that's going to be fine in terms of the second property. 00:09:51.110 --> 00:09:54.310 The additive difference between one of these solutions and OPT 00:09:54.310 --> 00:09:55.460 will be exactly the same. 00:09:55.460 --> 00:09:58.017 The beta here at this constant will be 1. 00:09:58.017 --> 00:10:00.100 But we do have to worry about the first condition. 00:10:00.100 --> 00:10:02.641 We need to make sure OPT doesn't blow up too much, because we 00:10:02.641 --> 00:10:04.490 did make it bigger. 00:10:04.490 --> 00:10:08.790 So for that, all we need is this is 00:10:08.790 --> 00:10:12.200 omega, the number of vertices. 00:10:12.200 --> 00:10:16.840 And that's because we assumed our graph had bounded degree, 00:10:16.840 --> 00:10:19.610 and so we can always find an independent set of size 00:10:19.610 --> 00:10:23.190 something like n over constant. 00:10:23.190 --> 00:10:24.610 So because that's already linear, 00:10:24.610 --> 00:10:26.630 we only added another linear thing. 00:10:26.630 --> 00:10:32.120 Again, also this is order, number of vertices. 00:10:32.120 --> 00:10:35.200 So we're not adding too much relative to this, 00:10:35.200 --> 00:10:37.110 because bounded degree. 00:10:37.110 --> 00:10:37.900 Cool? 00:10:37.900 --> 00:10:40.460 So that's Max 2SAT, APX-hardness. 00:10:48.110 --> 00:10:53.190 Fun fact which I won't prove. 00:10:53.190 --> 00:11:03.050 Max E2SAT-E3 is also APX-complete. 00:11:03.050 --> 00:11:07.030 So here we got some bounded number of occurrences. 00:11:07.030 --> 00:11:10.680 I guess each variable is going to appear in one 00:11:10.680 --> 00:11:13.060 plus three, four clauses. 00:11:13.060 --> 00:11:16.090 You can get that down to three clauses per variable. 00:11:21.150 --> 00:11:21.650 OK. 00:11:27.810 --> 00:11:29.630 Now that we have Max 2SAT, we can 00:11:29.630 --> 00:11:36.105 do another one, which is Max not all equal 3SAT. 00:11:40.580 --> 00:11:48.060 So from SAT-land, we have 3SAT, not all equal 3SAT, 00:11:48.060 --> 00:11:49.710 and 1 and 3SAT. 00:11:49.710 --> 00:11:51.230 We're going to get all of those. 00:11:51.230 --> 00:11:53.320 Actually, we can even get 1 and 2SAT. 00:11:53.320 --> 00:11:55.160 Little bit stronger. 00:11:55.160 --> 00:11:58.260 But let's do not all equal 3SAT. 00:11:58.260 --> 00:12:02.870 So here we are going to do, I believe, 00:12:02.870 --> 00:12:15.920 a strict reduction from Max 2SAT which we just proved, 00:12:15.920 --> 00:12:18.510 APX-complete. 00:12:18.510 --> 00:12:20.230 Yeah. 00:12:20.230 --> 00:12:22.655 It's again in APX, because you can, say, take 00:12:22.655 --> 00:12:24.330 your random assignment, and you'll 00:12:24.330 --> 00:12:28.100 satisfy some constant fraction of the clauses. 00:12:28.100 --> 00:12:30.710 And OK. 00:12:30.710 --> 00:12:32.630 So here's the reduction. 00:12:32.630 --> 00:12:34.370 Again, very easy. 00:12:34.370 --> 00:12:36.294 Suppose we're starting from Max 2SAT, 00:12:36.294 --> 00:12:37.710 so all our clauses look like this. 00:12:37.710 --> 00:12:40.380 These may be negated or not. 00:12:40.380 --> 00:12:50.700 And we're going to convert it into not all equal of x, y, 00:12:50.700 --> 00:12:52.056 and a. 00:12:52.056 --> 00:12:57.250 a is a new variable, and it appears in every single clause. 00:12:57.250 --> 00:12:57.750 OK? 00:12:57.750 --> 00:12:58.791 So this is kind of funny. 00:13:01.980 --> 00:13:04.880 So a appears everywhere. 00:13:04.880 --> 00:13:06.990 And not all equal has this nice symmetry, right? 00:13:06.990 --> 00:13:08.240 There wasn't really a zero or one. 00:13:08.240 --> 00:13:09.990 You can think of them as red, as blue. 00:13:09.990 --> 00:13:12.910 Doesn't matter whether red is true or blue is true. 00:13:12.910 --> 00:13:15.130 So in particular, we can use that symmetry 00:13:15.130 --> 00:13:18.530 to make a consider it as false. 00:13:18.530 --> 00:13:21.210 So by a possible flipping everything, 00:13:21.210 --> 00:13:25.080 we can imagine that a equals zero. 00:13:25.080 --> 00:13:29.242 If not, flip all the bits, and you'll still be not all equal. 00:13:29.242 --> 00:13:31.700 Or all the things that were not all equal before will still 00:13:31.700 --> 00:13:32.408 be not all equal. 00:13:32.408 --> 00:13:34.350 You'll preserve OPT. 00:13:34.350 --> 00:13:38.540 Now once you think of a is false, then not all equal 00:13:38.540 --> 00:13:41.790 is saying that these are not both 0, which is 00:13:41.790 --> 00:13:44.010 the same thing as saying 2SAT. 00:13:44.010 --> 00:13:45.890 Duh. 00:13:45.890 --> 00:13:46.510 OK. 00:13:46.510 --> 00:13:49.370 Again, I mean this is saying OPT is preserved. 00:13:49.370 --> 00:13:51.560 But if you take any solution to this problem, 00:13:51.560 --> 00:13:54.100 you first possibly flip it so that a is zero, 00:13:54.100 --> 00:13:57.910 and then convert the xy is just exactly the xy's over here, 00:13:57.910 --> 00:14:00.450 and you'll preserve the size of the solution. 00:14:00.450 --> 00:14:02.420 You won't get any scale here, and you also 00:14:02.420 --> 00:14:04.840 preserved OPT exactly. 00:14:04.840 --> 00:14:06.750 So it's in particular an L-reduction, 00:14:06.750 --> 00:14:08.750 but it's even a strict reduction. 00:14:08.750 --> 00:14:10.820 Didn't lose anything. 00:14:10.820 --> 00:14:14.110 No additive slop or whatever. 00:14:14.110 --> 00:14:14.750 OK. 00:14:14.750 --> 00:14:16.430 That's nice. 00:14:16.430 --> 00:14:21.625 Next is usually called Max-Cut. 00:14:24.690 --> 00:14:25.760 You're given a graph. 00:14:25.760 --> 00:14:27.550 You want to split it into two parts 00:14:27.550 --> 00:14:31.660 to maximize the number of edges between the two parts. 00:14:31.660 --> 00:14:43.100 But this is the same thing as max positive 1 and 2SAT, 00:14:43.100 --> 00:14:46.360 which is simpler than 1 and 3SAT. 00:14:46.360 --> 00:14:51.860 You have, I mean, in a cut, again, you have two sides. 00:14:51.860 --> 00:14:54.320 Call them true or false, or red and blue, or whatever. 00:14:54.320 --> 00:14:57.800 You would like to assign exactly one of these to be true. 00:14:57.800 --> 00:15:00.010 Then that edge will be in the cut. 00:15:00.010 --> 00:15:01.630 So it's the same problem. 00:15:01.630 --> 00:15:10.440 And you can also think of it as max positive XOR-SAT. 00:15:10.440 --> 00:15:12.320 Maybe actually call it 2XOR-SAT. 00:15:15.320 --> 00:15:15.890 Same thing. 00:15:15.890 --> 00:15:19.100 It's just every constraint is of the form this x or this. 00:15:19.100 --> 00:15:21.480 You want to maximize the number of those constraints. 00:15:21.480 --> 00:15:23.510 So a lot of these problems have different formulations 00:15:23.510 --> 00:15:25.551 depending on whether you're thinking about logic, 00:15:25.551 --> 00:15:27.970 or thinking about a graph problem. 00:15:27.970 --> 00:15:31.890 So we're going to get all of these four with one reduction. 00:15:31.890 --> 00:15:35.010 And it's going to be from probably this one. 00:15:35.010 --> 00:15:36.430 Yes. 00:15:36.430 --> 00:15:38.260 The great chain of reductions here. 00:15:47.320 --> 00:15:51.450 So we're going to reduce from Max not all equal 3SAT. 00:15:54.060 --> 00:15:58.130 I should mention, all of the reductions we've been seeing, 00:15:58.130 --> 00:16:01.260 including this initial batch where we started from 3SAT, 00:16:01.260 --> 00:16:02.902 converted into 3SAT 3, converted it 00:16:02.902 --> 00:16:04.610 into an independent set, to vertex cover, 00:16:04.610 --> 00:16:09.460 to dominating set to Max 2SAT, to Max not equal 3SAT 00:16:09.460 --> 00:16:12.870 to Max-Cut, are all in this seminal paper by Papadimitriou 00:16:12.870 --> 00:16:15.350 and Yannakakis, 1991. 00:16:15.350 --> 00:16:18.826 This is before APX was really a thing. 00:16:18.826 --> 00:16:20.450 It had a different name at that point-- 00:16:20.450 --> 00:16:24.530 Max SMP-- which later is proved to be essentially equal to APX, 00:16:24.530 --> 00:16:26.710 or the completeness version is the same. 00:16:26.710 --> 00:16:29.019 You don't need to know about that. 00:16:29.019 --> 00:16:31.310 It comes from a different world, but all the reductions 00:16:31.310 --> 00:16:32.570 apply here. 00:16:32.570 --> 00:16:36.230 So here is the reduction for a Max-Cut. 00:16:36.230 --> 00:16:39.040 So again we're trying to simulate Max 00:16:39.040 --> 00:16:40.720 not all equal 3SAT. 00:16:40.720 --> 00:16:44.850 Now we actually saw in the planar lecture, planar 3SAT, 00:16:44.850 --> 00:16:49.650 that you can reduce planar not all equal 3SAT to planar 00:16:49.650 --> 00:16:53.470 Max-Cut, and that we use that to get a polynomial time algorithm 00:16:53.470 --> 00:16:55.340 for planar not all equal 3SAT. 00:16:55.340 --> 00:16:57.190 We're just going to do the reverse. 00:16:57.190 --> 00:17:00.930 And if you recall, this was the heart of that reduction. 00:17:00.930 --> 00:17:03.810 The point is that you can represent 00:17:03.810 --> 00:17:08.800 a not all equal clause as a cut, as a Max-Cut problem 00:17:08.800 --> 00:17:09.599 on a triangle. 00:17:09.599 --> 00:17:11.864 Because in a triangle, either they're all equal, 00:17:11.864 --> 00:17:14.849 and then there's no cut edges, or they're not all equal, 00:17:14.849 --> 00:17:17.520 and then there's exactly two cut edges. 00:17:17.520 --> 00:17:19.079 So that's for a cause of size 3. 00:17:19.079 --> 00:17:22.050 We also need to handle the case of a cause of size 2. 00:17:22.050 --> 00:17:24.905 But that's a two-gon, I guess, instead of a triangle. 00:17:24.905 --> 00:17:26.030 It works the same way here. 00:17:26.030 --> 00:17:29.950 You get 1 if they're not all equal, and zero otherwise. 00:17:29.950 --> 00:17:32.120 This is shown as the zero case. 00:17:32.120 --> 00:17:32.620 OK. 00:17:32.620 --> 00:17:35.910 Now the one thing we need, because not all equal 3SAT 00:17:35.910 --> 00:17:39.300 here, we need negation. 00:17:39.300 --> 00:17:45.480 So we're going to build each variable and its negation 00:17:45.480 --> 00:17:46.350 with this gadget. 00:17:46.350 --> 00:17:48.220 This is a new gadget, variable gadget. 00:17:48.220 --> 00:17:52.790 It's just a whole bunch of edges connecting xi and xi bar. 00:17:52.790 --> 00:17:53.900 And you can make this. 00:17:53.900 --> 00:17:56.570 You can avoid the multigraph aspect here. 00:17:56.570 --> 00:17:59.530 But let's not worry about it here. 00:17:59.530 --> 00:18:03.950 So in general, if there are k occurrences of this variable, 00:18:03.950 --> 00:18:07.370 then we're going to have 2k parallel edges, 00:18:07.370 --> 00:18:11.480 because the cost over here, the potential benefit here is 2. 00:18:11.480 --> 00:18:14.730 Again, we want to argue that if we take an optimal solution, 00:18:14.730 --> 00:18:18.550 we can make it another optimal solution where xi and xi 00:18:18.550 --> 00:18:21.706 bar are on opposite sides of the cut. 00:18:21.706 --> 00:18:23.830 And the reason is, if they're both on the same side 00:18:23.830 --> 00:18:27.690 of the cut, you're not getting this benefit. 00:18:27.690 --> 00:18:29.930 If you flip one of the sides, you 00:18:29.930 --> 00:18:32.150 get this huge benefit, which is 2k. 00:18:32.150 --> 00:18:33.960 And you say, well, how much do I lose 00:18:33.960 --> 00:18:37.470 if I flip this from one side of the cut to the other. 00:18:37.470 --> 00:18:41.590 Well, it appears in at most k different clauses, each of them 00:18:41.590 --> 00:18:43.480 gives me at most two points. 00:18:43.480 --> 00:18:45.970 So I'm losing, at most, 2k points 00:18:45.970 --> 00:18:47.330 by making these opposite. 00:18:47.330 --> 00:18:48.410 But I gain 2k points. 00:18:48.410 --> 00:18:50.990 So it never hurts me to do that switch. 00:18:50.990 --> 00:18:53.560 So I can assume these two guys are on opposite sides, 00:18:53.560 --> 00:18:56.540 and therefore I can assume it's sort of validly doing 00:18:56.540 --> 00:18:57.740 the negation part. 00:18:57.740 --> 00:19:01.810 And then it just reduces to not all equal 3SAT. 00:19:01.810 --> 00:19:04.770 There's a difference between this one, where we only 00:19:04.770 --> 00:19:07.250 get one point, and this one we only get two points. 00:19:07.250 --> 00:19:09.212 AUDIENCE: You get two points. 00:19:09.212 --> 00:19:10.670 PROFESSOR: You get two points here? 00:19:10.670 --> 00:19:10.910 Oh yeah. 00:19:10.910 --> 00:19:11.701 You get two points. 00:19:11.701 --> 00:19:14.820 That's why we doubled the edge. 00:19:14.820 --> 00:19:16.747 So that's cool. 00:19:16.747 --> 00:19:17.830 I think you would be fine. 00:19:17.830 --> 00:19:20.121 It'd still be an L-reduction even if you have one edge. 00:19:20.121 --> 00:19:21.690 But this is nicer. 00:19:21.690 --> 00:19:23.410 And yeah. 00:19:23.410 --> 00:19:24.750 That's it. 00:19:24.750 --> 00:19:25.250 Cool. 00:19:25.250 --> 00:19:28.050 This is Max-Cut. 00:19:28.050 --> 00:19:32.020 It will be a bounded degree based 00:19:32.020 --> 00:19:34.470 on the number of occurrences we got, which was like four. 00:19:34.470 --> 00:19:37.600 I mean, we can use three, and then we'll multiply. 00:19:37.600 --> 00:19:42.270 In general you can prove Max-Cut remains APX-complete 00:19:42.270 --> 00:19:45.440 for degree three graphs. 00:19:45.440 --> 00:19:47.630 So we're not going to prove it here. 00:19:47.630 --> 00:19:51.600 So another kind of reduction trick to reduce degrees, just 00:19:51.600 --> 00:19:55.680 say degree 3 is possible. 00:19:55.680 --> 00:20:03.690 It's also Max Cut in degree 3 graphs is APX-complete. 00:20:03.690 --> 00:20:10.100 So you could call that max positive 1 and 2SAT, hyphen 3. 00:20:10.100 --> 00:20:10.740 Maybe even E3. 00:20:13.580 --> 00:20:14.175 All right. 00:20:16.690 --> 00:20:18.587 So this gives you a flavor. 00:20:18.587 --> 00:20:20.420 This is a fun series of reductions, each one 00:20:20.420 --> 00:20:22.150 building on the previous one. 00:20:22.150 --> 00:20:24.730 But it gives you kind of starting point. 00:20:24.730 --> 00:20:27.310 A lot of the problems we're familiar with in NP 00:20:27.310 --> 00:20:29.480 completeness land, if you just add "Max" in front, 00:20:29.480 --> 00:20:32.930 they become hard. 00:20:32.930 --> 00:20:35.960 I mean I guess Max-Cut always had a Max in front. 00:20:35.960 --> 00:20:38.850 Max 2SAT for NP completeness, we also had a Max in front. 00:20:38.850 --> 00:20:41.341 So those are familiar, and they're APX-complete. 00:20:41.341 --> 00:20:42.840 All of the problems, I've described, 00:20:42.840 --> 00:20:44.298 at least for bounded degree graphs, 00:20:44.298 --> 00:20:46.340 have constant factor approximations. 00:20:46.340 --> 00:20:47.730 So this is the right level. 00:20:47.730 --> 00:20:49.350 They are APX-complete. 00:20:49.350 --> 00:20:51.650 And that determines their approximability. 00:20:51.650 --> 00:20:52.710 Constant factor, no PTAS. 00:20:55.890 --> 00:21:03.060 Now it would be nice to know which problems are hard. 00:21:03.060 --> 00:21:06.790 With NP-completeness, and in the SAT universe, 00:21:06.790 --> 00:21:09.170 we had Schaefer's dichotomy theorem that 00:21:09.170 --> 00:21:12.380 said-- let me cheat and look at my notes from, 00:21:12.380 --> 00:21:17.390 I think, lecture four-- that SAT is polynomial if 00:21:17.390 --> 00:21:18.980 and only if the clauses that you're 00:21:18.980 --> 00:21:21.270 allowed to do-- the operations you're allowed 00:21:21.270 --> 00:21:25.491 to do with variables-- are either have 00:21:25.491 --> 00:21:27.740 the property that when you set all the variables true, 00:21:27.740 --> 00:21:28.810 everything's satisfied. 00:21:28.810 --> 00:21:31.730 Or you set all the variables false, everything satisfied. 00:21:31.730 --> 00:21:37.080 Or every single clause is a conjunction of Horn causes. 00:21:37.080 --> 00:21:43.200 Horn clauses were a few variables, and at most one 00:21:43.200 --> 00:21:45.200 of them is positive. 00:21:45.200 --> 00:21:48.520 Or all the causes you have are conjunctions of Dual-Horn, 00:21:48.520 --> 00:21:54.300 which was, in every clause at most one of them is negated, 00:21:54.300 --> 00:21:58.900 or all of the clauses are conjunctions of 2CNF, 00:21:58.900 --> 00:22:00.660 only like 2SAT. 00:22:00.660 --> 00:22:05.010 Or what I didn't give a name at the time, 00:22:05.010 --> 00:22:10.140 but is essentially a slight generalization of XOR-SAT. 00:22:10.140 --> 00:22:11.580 Let me give it a name here. 00:22:11.580 --> 00:22:13.040 I'm going to call it X(N)OR-SAT. 00:22:19.350 --> 00:22:23.190 You can also phrase them as linear equations over Z2. 00:22:32.390 --> 00:22:34.290 So this is zero and one. 00:22:34.290 --> 00:22:38.120 And it's either X OR, meaning you take the X OR of all 00:22:38.120 --> 00:22:40.340 the things-- that's like the summation of all things, 00:22:40.340 --> 00:22:42.370 or it's X(N)OR, meaning when you take that sum, 00:22:42.370 --> 00:22:44.420 it should equal zero. 00:22:44.420 --> 00:22:46.300 And such systems of linear equations 00:22:46.300 --> 00:22:52.250 can be solved in polynomial time using Gaussian elimination 00:22:52.250 --> 00:22:53.920 over Z2. 00:22:53.920 --> 00:22:56.060 And all of the things I just mentioned 00:22:56.060 --> 00:22:59.420 are all the situations where SAT is polynomial. 00:22:59.420 --> 00:23:03.810 Every other type of clause, SAT is NP-complete-- 00:23:03.810 --> 00:23:05.607 or set of classes. 00:23:05.607 --> 00:23:06.690 Now why do I mention this? 00:23:06.690 --> 00:23:11.520 Because there is an analogous theorem for it's 00:23:11.520 --> 00:23:15.690 not quite SAT, because we need something like this Max. 00:23:15.690 --> 00:23:17.690 We need to turn it into an optimization problem. 00:23:17.690 --> 00:23:21.050 SAT is not normally an optimization problem by itself. 00:23:21.050 --> 00:23:25.270 And characterizing how approximal those problems are. 00:23:25.270 --> 00:23:32.750 Now it is a complicated theorem-- so complicated, 00:23:32.750 --> 00:23:35.200 that I don't want to write it on the board, 00:23:35.200 --> 00:23:36.670 because there's a lot of cases. 00:23:36.670 --> 00:23:39.140 But the point is, it's exhaustive. 00:23:39.140 --> 00:23:41.166 It will tell you if you have anything 00:23:41.166 --> 00:23:42.540 of the type we had with Schaefer, 00:23:42.540 --> 00:23:44.515 which was you define a kind of clause function. 00:23:44.515 --> 00:23:46.190 It's either satisfied or not. 00:23:46.190 --> 00:23:48.120 It applies to some number of variables. 00:23:48.120 --> 00:23:51.150 And then, once you've defined that clause type, 00:23:51.150 --> 00:23:52.830 you can apply it to any combination 00:23:52.830 --> 00:23:54.450 of variables you want. 00:23:54.450 --> 00:23:57.400 That family of problems with no other restrictions 00:23:57.400 --> 00:23:58.505 is what we get. 00:23:58.505 --> 00:24:03.590 And I will just tell you what the problems are. 00:24:03.590 --> 00:24:04.547 There's four of them. 00:24:04.547 --> 00:24:06.380 This is part of what makes the theorem long, 00:24:06.380 --> 00:24:08.750 but also extremely powerful. 00:24:08.750 --> 00:24:12.340 The first dichotomy is max verses min. 00:24:12.340 --> 00:24:15.580 And then the second dichotomy is they 00:24:15.580 --> 00:24:18.162 call it CSP for constraint satisfaction problem. 00:24:18.162 --> 00:24:19.620 So you have a bunch of constraints. 00:24:19.620 --> 00:24:21.970 You want to satisfy as many as possible. 00:24:21.970 --> 00:24:26.750 So this would be the number of satisfied constraints 00:24:26.750 --> 00:24:29.940 is your objective, or your cost function. 00:24:33.240 --> 00:24:37.870 Or the other version is what's called the ones problem, or max 00:24:37.870 --> 00:24:39.530 ones, or min ones. 00:24:39.530 --> 00:24:42.560 This is the number of true variables. 00:24:48.010 --> 00:24:52.060 So again, we have a Schaefer-like SAT style 00:24:52.060 --> 00:24:53.132 of set of clauses. 00:24:53.132 --> 00:24:55.590 Either we want to maximize the number of satisfied clauses, 00:24:55.590 --> 00:24:58.170 or we want to minimize the number satisfied clauses, 00:24:58.170 --> 00:25:02.360 or we want to maximize the number of true variables 00:25:02.360 --> 00:25:03.980 and satisfy everything. 00:25:03.980 --> 00:25:06.360 Or we want to minimize the number of true variables 00:25:06.360 --> 00:25:09.040 and satisfy everything. 00:25:09.040 --> 00:25:09.540 OK. 00:25:09.540 --> 00:25:11.930 Now obviously, if the SAT problem is hard, 00:25:11.930 --> 00:25:13.840 it's going to be hard to do this. 00:25:13.840 --> 00:25:15.710 But it's still interesting. 00:25:15.710 --> 00:25:17.000 You can still think about it. 00:25:17.000 --> 00:25:23.260 And even when the SAT problem is easy, Max ones can be hard. 00:25:23.260 --> 00:25:25.650 So I am going to-- I wrote it all down, 00:25:25.650 --> 00:25:27.280 and then I realized how long it was. 00:25:27.280 --> 00:25:29.060 And so I will just show you. 00:25:29.060 --> 00:25:32.460 Imagine I just hand-wrote this. 00:25:32.460 --> 00:25:35.310 So this is the easy case. 00:25:35.310 --> 00:25:36.401 Max CSP. 00:25:36.401 --> 00:25:38.400 So we want to maximize the number of constraints 00:25:38.400 --> 00:25:40.990 that we satisfy. 00:25:40.990 --> 00:25:45.430 And I'm going to characterize when it is polynomial. 00:25:45.430 --> 00:25:47.710 Now here, PO I haven't defined, but that's 00:25:47.710 --> 00:25:49.737 the analog of P for optimization problems. 00:25:49.737 --> 00:25:51.570 So it's the set of all optimization problems 00:25:51.570 --> 00:25:55.340 that are in P that have a polynomial timed algorithm 00:25:55.340 --> 00:25:57.110 to solve them exactly. 00:25:57.110 --> 00:25:58.520 So it turns out in this situation 00:25:58.520 --> 00:26:01.330 you are either polynomial or APX-complete. 00:26:01.330 --> 00:26:04.940 So it's only about constant factor verses perfect. 00:26:04.940 --> 00:26:08.310 There's never a PTAS, unless there's a polynomial time 00:26:08.310 --> 00:26:09.012 algorithm. 00:26:09.012 --> 00:26:10.470 And the cases should look familiar. 00:26:10.470 --> 00:26:13.170 It's either when you set all the variables true 00:26:13.170 --> 00:26:15.860 or all the variables false, that satisfies everything. 00:26:15.860 --> 00:26:17.690 In that case, Max CSP is, of course, easy. 00:26:17.690 --> 00:26:19.790 You can satisfy everything. 00:26:19.790 --> 00:26:23.150 Another case is if you write the clauses 00:26:23.150 --> 00:26:26.510 in disjunctive normal form-- this is a new type 00:26:26.510 --> 00:26:29.360 that we hadn't seen before, all your causes are-- 00:26:29.360 --> 00:26:32.450 when you write them in DNF, they have exactly two terms. 00:26:32.450 --> 00:26:36.265 So it's the OR of two things that are anded together. 00:26:36.265 --> 00:26:36.765 Sorry. 00:26:36.765 --> 00:26:38.050 There's an "or" in the middle. 00:26:38.050 --> 00:26:40.340 And you have a bunch of things anded together 00:26:40.340 --> 00:26:41.630 in each of my hands. 00:26:41.630 --> 00:26:44.760 And all the ones in here and positive, and all the ones 00:26:44.760 --> 00:26:46.110 in here are negative. 00:26:46.110 --> 00:26:49.090 If every clause looks like that, then you 00:26:49.090 --> 00:26:51.670 can solve this in polynomial time. 00:26:51.670 --> 00:26:56.180 And in all other cases, this problem is APX-complete. 00:26:56.180 --> 00:26:59.582 So that's a nice, very clean characterization. 00:26:59.582 --> 00:27:01.998 AUDIENCE: Wait. [INAUDIBLE] that we learned about earlier. 00:27:01.998 --> 00:27:03.390 Is this the [INAUDIBLE]? 00:27:03.390 --> 00:27:04.015 PROFESSOR: Yes. 00:27:04.015 --> 00:27:05.530 This is disjunctive normal form. 00:27:05.530 --> 00:27:09.390 So it's the or of ands. 00:27:09.390 --> 00:27:12.590 We usually, we deal with CNF ands of ors. 00:27:12.590 --> 00:27:17.530 But for this characterization, every clause 00:27:17.530 --> 00:27:19.810 can be uniquely converted into a DNF, 00:27:19.810 --> 00:27:21.150 and uniquely converted into CNF. 00:27:21.150 --> 00:27:23.990 So that's a well-defined thing to say. 00:27:26.405 --> 00:27:28.530 With Schaefer, we just had to look at the CNF form. 00:27:28.530 --> 00:27:31.990 But here we get a new set of things. 00:27:31.990 --> 00:27:33.130 All right. 00:27:33.130 --> 00:27:35.350 That was one out of four. 00:27:35.350 --> 00:27:37.240 Max Min CSP Ones. 00:27:37.240 --> 00:27:40.486 Next one is Max Ones. 00:27:40.486 --> 00:27:41.860 This is not the most complicated. 00:27:44.540 --> 00:27:46.390 But let's go through them. 00:27:46.390 --> 00:27:49.862 So again, we want to maximize the number of true variables. 00:27:49.862 --> 00:27:51.945 So of course, if we set all the variables to true, 00:27:51.945 --> 00:27:55.570 and everything is satisfied, yay, a polynomial, OK? 00:27:55.570 --> 00:27:58.180 But curiously, if you settle the variables to false, 00:27:58.180 --> 00:28:02.910 and that satisfies everything, that's going to be here. 00:28:02.910 --> 00:28:05.230 That's Poly-APX-complete. 00:28:05.230 --> 00:28:08.050 Poly-APX-complete, you can translate to something like n 00:28:08.050 --> 00:28:10.160 to the 1 minus epsilon, approximable, 00:28:10.160 --> 00:28:12.850 and that's the best you can do. 00:28:12.850 --> 00:28:15.620 Or there's a lower bound of n to the 1 minus epsilon. 00:28:15.620 --> 00:28:18.231 Upper bound might be n or something. 00:28:18.231 --> 00:28:18.730 OK. 00:28:18.730 --> 00:28:23.180 So because maximizing ones, when setting things all at false, 00:28:23.180 --> 00:28:24.450 does not necessarily help you. 00:28:24.450 --> 00:28:26.800 There are some more positive cases. 00:28:26.800 --> 00:28:28.730 If you have a Dual-Horn set up. 00:28:28.730 --> 00:28:31.270 So this is another one of the Schaefer situations. 00:28:31.270 --> 00:28:34.675 If every clause when you write it in CNF every subclause 00:28:34.675 --> 00:28:37.780 is Dual-Horn, at most, one negated thing, 00:28:37.780 --> 00:28:40.070 that is a good situation for maximizing ones, 00:28:40.070 --> 00:28:44.170 because only one of them has to be negative. 00:28:44.170 --> 00:28:48.646 But with Horn, for example, you get Poly-APX-complete, 00:28:48.646 --> 00:28:51.020 because we have an asymmetry here between ones and zeros. 00:28:51.020 --> 00:28:51.968 Question? 00:28:51.968 --> 00:28:53.160 AUDIENCE: In this list, do we just read down it 00:28:53.160 --> 00:28:54.210 until we hit the thing? 00:28:54.210 --> 00:28:55.010 PROFESSOR: Yes. 00:28:55.010 --> 00:28:55.860 Good question. 00:28:55.860 --> 00:29:01.290 This is a sequential algorithm for determining what you have. 00:29:01.290 --> 00:29:03.430 If any of these says, oh, you're in PO, 00:29:03.430 --> 00:29:05.760 then you should stop reading the rest of the theorem. 00:29:05.760 --> 00:29:09.640 The way they write the theorem is less is probably clearer. 00:29:09.640 --> 00:29:11.386 They write an else if for each one, 00:29:11.386 --> 00:29:13.260 but I wrote it backwards, so it's hard for me 00:29:13.260 --> 00:29:14.730 to write else if. 00:29:14.730 --> 00:29:15.410 Yeah. 00:29:15.410 --> 00:29:18.530 Occasionally I'll mention that the previous things 00:29:18.530 --> 00:29:19.030 don't apply. 00:29:19.030 --> 00:29:20.860 But you should read this sequentially. 00:29:24.100 --> 00:29:24.600 OK. 00:29:24.600 --> 00:29:25.870 So it was Dual-Horn. 00:29:25.870 --> 00:29:31.300 Another polynomial case is what I call 2-X(N)OR-SAT, 00:29:31.300 --> 00:29:32.590 where the N is in parentheses. 00:29:32.590 --> 00:29:35.110 So in other words, you have linear equations. 00:29:35.110 --> 00:29:39.300 Each equation only has two terms, sort of like 2SAT. 00:29:39.300 --> 00:29:41.330 And you have equations that say equal zero 00:29:41.330 --> 00:29:44.120 or equal one on those two terms. 00:29:44.120 --> 00:29:45.870 That is also polynomially solvable. 00:29:45.870 --> 00:29:47.490 This is a special case. 00:29:47.490 --> 00:29:49.650 We didn't need the 2 for Schaefer. 00:29:49.650 --> 00:29:54.490 Here we need the 2, because if you have X(N)OR-SAT in general. 00:29:54.490 --> 00:29:57.760 And when I say this, I mean that all constraints 00:29:57.760 --> 00:29:58.940 fall into this category. 00:29:58.940 --> 00:30:00.990 If all constraints are of this form, 00:30:00.990 --> 00:30:03.080 all clauses are of this form, then you're good. 00:30:03.080 --> 00:30:06.420 If all clauses are of the form X(N)OR-SAT, 00:30:06.420 --> 00:30:10.450 but they're not in this class, they're not all of length 2, 00:30:10.450 --> 00:30:12.800 then the problem becomes APX-complete, 00:30:12.800 --> 00:30:16.630 by contrast to Schaefer, where, I mean, 00:30:16.630 --> 00:30:19.370 deciding whether you can satisfy all those things is easy-- 00:30:19.370 --> 00:30:22.670 maximizing the number of ones when you do it is APX-complete. 00:30:22.670 --> 00:30:25.950 So that's particularly interesting. 00:30:25.950 --> 00:30:27.700 AUDIENCE: Not all equal 3SAT fall in that? 00:30:27.700 --> 00:30:28.610 Is that? 00:30:32.620 --> 00:30:35.330 PROFESSOR: Not all equal 3SAT. 00:30:35.330 --> 00:30:37.527 AUDIENCE: Those are X(N)OR clauses, right? 00:30:37.527 --> 00:30:38.110 PROFESSOR: No. 00:30:38.110 --> 00:30:39.526 They should not be X(N)OR clauses, 00:30:39.526 --> 00:30:40.930 because it's NP-complete. 00:30:40.930 --> 00:30:42.800 And when you have X(N)OR clauses, 00:30:42.800 --> 00:30:45.650 it's always polynomial to decide whether you can satisfy 00:30:45.650 --> 00:30:47.040 everything. 00:30:47.040 --> 00:30:49.645 So it's in the other case. 00:30:52.570 --> 00:30:54.070 But good question, because we should 00:30:54.070 --> 00:30:56.920 be getting APX-completeness. 00:30:56.920 --> 00:30:58.837 Yeah, but Max not all equal 3SAT is different. 00:30:58.837 --> 00:31:01.128 Here we're trying to maximize the number of clause that 00:31:01.128 --> 00:31:01.810 were satisfied. 00:31:01.810 --> 00:31:04.309 So if you have not all equal 3SAT, 00:31:04.309 --> 00:31:06.350 and you want to maximize the number of ones, that 00:31:06.350 --> 00:31:08.724 means first you have to satisfy not all equal 3SAT, which 00:31:08.724 --> 00:31:09.610 is hard. 00:31:09.610 --> 00:31:11.760 So that's going to fall into this. 00:31:11.760 --> 00:31:13.800 The bottom one is feasibility. 00:31:13.800 --> 00:31:15.930 Just finding a feasible solution is NP hard. 00:31:18.590 --> 00:31:24.630 The X(N)OR-SAT is this thing-- linear equations over Z2. 00:31:24.630 --> 00:31:27.139 And it could be equal to 0, or equal to 1. 00:31:27.139 --> 00:31:28.930 This is what you might call an X OR clause, 00:31:28.930 --> 00:31:32.940 or this is an X OR clause, this is an X(N)OR clause. 00:31:32.940 --> 00:31:36.890 So if they don't all have size two, then you're APX-complete. 00:31:36.890 --> 00:31:41.400 But you can find a solution by Schaefer's theorem. 00:31:41.400 --> 00:31:42.280 OK. 00:31:42.280 --> 00:31:45.390 So as I mentioned, Horn clauses and 2AT clauses 00:31:45.390 --> 00:31:46.570 are actually really hard. 00:31:46.570 --> 00:31:49.320 They're Poly-APX-complete, n to the 1 minus epsilon. 00:31:49.320 --> 00:31:51.350 Also these are all situations where 00:31:51.350 --> 00:31:54.724 you can find feasible solutions easily by Schaefer, like when 00:31:54.724 --> 00:31:57.140 you can set them all false, and that satisfies everything. 00:31:57.140 --> 00:31:58.020 It doesn't help you when you're trying 00:31:58.020 --> 00:31:59.311 to maximize the number of ones. 00:31:59.311 --> 00:32:01.916 It just gets you to zero. 00:32:01.916 --> 00:32:03.040 Then you want to do better. 00:32:03.040 --> 00:32:06.680 And it's really hard to get any better factor. 00:32:06.680 --> 00:32:08.630 One more situation. 00:32:08.630 --> 00:32:09.130 Sorry. 00:32:11.934 --> 00:32:13.350 There's a slight distinction here. 00:32:13.350 --> 00:32:15.800 So suppose you have the feature that you 00:32:15.800 --> 00:32:20.290 can set one variable true, and the rest false. 00:32:20.290 --> 00:32:22.650 If that satisfies all your constraints, than great, 00:32:22.650 --> 00:32:24.467 you found the value 1. 00:32:24.467 --> 00:32:26.300 And there's a big difference between 0 and 1 00:32:26.300 --> 00:32:28.216 when you're looking at relative approximation, 00:32:28.216 --> 00:32:30.950 because anything divided by 0 is huge. 00:32:30.950 --> 00:32:32.880 So it's really hard to get a good factor. 00:32:32.880 --> 00:32:33.760 That's the situation. 00:32:33.760 --> 00:32:35.260 Distinguishing between 0 and greater 00:32:35.260 --> 00:32:39.150 than 0, which is an infinite ratio, it could be NP-hard. 00:32:39.150 --> 00:32:41.470 That's when you, in this situation, 00:32:41.470 --> 00:32:42.980 we set all the variables false. 00:32:42.980 --> 00:32:43.680 You get zero. 00:32:43.680 --> 00:32:46.690 But finding any other solution is going to be NP-hard. 00:32:46.690 --> 00:32:48.280 Here, if you can at least get 1, you 00:32:48.280 --> 00:32:50.930 can get an N approximation, whereas here you 00:32:50.930 --> 00:32:52.320 can't get an N approximation. 00:32:52.320 --> 00:32:55.290 Here you can get Poly approximation. 00:32:55.290 --> 00:32:57.700 And finally, if you have none of this above situations, 00:32:57.700 --> 00:33:01.950 then testing feasibility is NP-hard by Schaefer's theorem. 00:33:01.950 --> 00:33:04.310 So it's like Schaefer theorem, but some of the cases 00:33:04.310 --> 00:33:08.200 split up into parts. 00:33:08.200 --> 00:33:09.660 Now, that was maximization. 00:33:09.660 --> 00:33:10.510 Question? 00:33:10.510 --> 00:33:12.510 AUDIENCE: So, what's special about 1 here? 00:33:12.510 --> 00:33:15.977 It seems to me if you replace that 1 by K 00:33:15.977 --> 00:33:17.310 it should still be in that case. 00:33:17.310 --> 00:33:18.390 PROFESSOR: This case. 00:33:18.390 --> 00:33:19.330 AUDIENCE: Yeah. 00:33:19.330 --> 00:33:22.620 If I just replace that one with a fixed K. Like 2. 00:33:22.620 --> 00:33:23.880 PROFESSOR: Yes. 00:33:23.880 --> 00:33:27.290 So that problem will still be-- so if you 00:33:27.290 --> 00:33:30.000 can set all but K of them true, I 00:33:30.000 --> 00:33:32.000 think you can also set all but one of them true, 00:33:32.000 --> 00:33:33.430 and still satisfy. 00:33:33.430 --> 00:33:34.190 Yeah. 00:33:34.190 --> 00:33:35.310 So here's the thing. 00:33:35.310 --> 00:33:36.680 This is all variables, right? 00:33:36.680 --> 00:33:39.440 So the idea is you have tons of variables, 00:33:39.440 --> 00:33:41.857 and let's say two of them are set to true. 00:33:41.857 --> 00:33:43.440 So if you look at a clause, the clause 00:33:43.440 --> 00:33:46.685 might just apply to these guys-- all the false guys-- 00:33:46.685 --> 00:33:49.060 or it might apply to false guys and one of the true guys, 00:33:49.060 --> 00:33:52.595 or it might apply to false guys and two of the true guys. 00:33:52.595 --> 00:33:54.220 All of those would have to be satisfied 00:33:54.220 --> 00:33:56.050 in your hypothetical situation. 00:33:56.050 --> 00:33:58.810 If that's true, that implies that all the clauses are 00:33:58.810 --> 00:34:00.950 satisfied when only one of them is set true, 00:34:00.950 --> 00:34:02.400 and the rest are false. 00:34:02.400 --> 00:34:04.980 So your case would fall into this case as well, 00:34:04.980 --> 00:34:07.260 and you'd get Poly-APX-completeness again. 00:34:07.260 --> 00:34:10.040 So it's not totally obvious when these things apply. 00:34:10.040 --> 00:34:14.256 But this is the complete list of different cases. 00:34:14.256 --> 00:34:14.839 Any questions? 00:34:17.480 --> 00:34:19.530 OK. 00:34:19.530 --> 00:34:21.440 Two out of four. 00:34:21.440 --> 00:34:25.460 Next one, this is the longest one, is Min CSP. 00:34:25.460 --> 00:34:28.639 Now here we don't get as nice a characterization, 00:34:28.639 --> 00:34:31.159 because there are some open problems left. 00:34:31.159 --> 00:34:33.420 I haven't checked whether all of these open problems 00:34:33.420 --> 00:34:36.610 remain open, but as of 2001 they were open, 00:34:36.610 --> 00:34:38.639 which was a while ago. 00:34:38.639 --> 00:34:41.800 And we can check whether there's more explicit status. 00:34:41.800 --> 00:34:45.310 But I have the status as of this paper here. 00:34:45.310 --> 00:34:47.150 So Min CSP. 00:34:47.150 --> 00:34:51.130 This is, you want to minimize the number of constraints 00:34:51.130 --> 00:34:54.122 that are satisfied, whereas before we 00:34:54.122 --> 00:34:55.080 looked at maximization. 00:34:55.080 --> 00:34:58.740 There are only three cases which were something like this. 00:34:58.740 --> 00:35:02.270 Again, if setting all the variables false or true 00:35:02.270 --> 00:35:08.810 satisfies all the clauses, this is good, apparently. 00:35:08.810 --> 00:35:10.830 That's less obvious in this case. 00:35:10.830 --> 00:35:12.240 In general, minimization problems 00:35:12.240 --> 00:35:14.365 behave quite differently from maximization problems 00:35:14.365 --> 00:35:16.110 in terms of approximability. 00:35:16.110 --> 00:35:17.970 Maximization is generally easier to 00:35:17.970 --> 00:35:22.130 approximate, because your solutions tend to be big, 00:35:22.130 --> 00:35:24.370 and it's easier to approximate big things. 00:35:24.370 --> 00:35:27.830 Minimization-- small-- is hard. 00:35:27.830 --> 00:35:31.380 Also we had the situation from Max CSP, 00:35:31.380 --> 00:35:33.540 if when you write it in DNF, is exactly 00:35:33.540 --> 00:35:35.107 two terms for every clause. 00:35:35.107 --> 00:35:36.690 One of them is all positive variables, 00:35:36.690 --> 00:35:38.356 and the other is all negative variables. 00:35:38.356 --> 00:35:40.470 That's also easy. 00:35:40.470 --> 00:35:46.270 And here's a new case of APX-completeness. 00:35:46.270 --> 00:35:48.610 So if the problem you're trying to solve 00:35:48.610 --> 00:35:51.290 is exactly this problem, they call this, 00:35:51.290 --> 00:35:54.190 I think, implication hitting set. 00:35:54.190 --> 00:35:57.910 So you have a clause which lets you say x1 implies 00:35:57.910 --> 00:36:01.620 x2 for any two variables. 00:36:01.620 --> 00:36:06.010 And you have some set of clauses like this, where you 00:36:06.010 --> 00:36:08.720 can say here's five variables. 00:36:08.720 --> 00:36:10.680 The OR of them is true. 00:36:10.680 --> 00:36:13.479 No negation here. 00:36:13.479 --> 00:36:15.520 So this is called hitting set, meaning I give you 00:36:15.520 --> 00:36:19.370 a set of vertices and a graph, and I want at least one of them 00:36:19.370 --> 00:36:22.320 to be hit, to be included, to be true. 00:36:22.320 --> 00:36:24.700 And we're trying to minimize the number of such things 00:36:24.700 --> 00:36:26.533 that we satisfy. 00:36:26.533 --> 00:36:31.490 So this turns out to be hard, but only there's no PTAS, 00:36:31.490 --> 00:36:35.600 but there's a constant factor approximation. 00:36:35.600 --> 00:36:38.360 And then we have these four cases 00:36:38.360 --> 00:36:41.770 which show that they are equivalent to known studied 00:36:41.770 --> 00:36:42.860 problems. 00:36:42.860 --> 00:36:44.720 So there are these special cases. 00:36:44.720 --> 00:36:48.414 Other than these getting any approximation 00:36:48.414 --> 00:36:49.830 factor of less than infinity would 00:36:49.830 --> 00:36:52.430 require you to distinguish between zeros OPT, 00:36:52.430 --> 00:36:55.400 and OPT is greater than zero, and it's NP-complete, 00:36:55.400 --> 00:36:57.980 unless you have these. 00:36:57.980 --> 00:37:00.970 So there are some special cases like Min Uncut. 00:37:00.970 --> 00:37:03.150 This is the reverse of Max Cut. 00:37:03.150 --> 00:37:05.880 You want to minimize the number of uncut edges. 00:37:05.880 --> 00:37:10.320 So that plus Max Cut should be equal to the number of edges. 00:37:10.320 --> 00:37:12.920 But the approximability of the two sides is quite different. 00:37:12.920 --> 00:37:16.480 And here are the best results of our APX-hardness, 00:37:16.480 --> 00:37:19.900 and log and upper bound for approximation. 00:37:19.900 --> 00:37:21.870 So that's a little bit harder maybe. 00:37:21.870 --> 00:37:25.110 It's at least as hard as this. 00:37:25.110 --> 00:37:30.480 And that happens when you are in the 2x (N)OR-SAT situation, 00:37:30.480 --> 00:37:33.320 something we saw from the last slide. 00:37:33.320 --> 00:37:35.820 So here it reduces to this other problem. 00:37:35.820 --> 00:37:39.025 Basically the same, but the X(N)ORs don't buy you anything 00:37:39.025 --> 00:37:39.525 new. 00:37:42.580 --> 00:37:44.860 In the case of 2SAT, you get a problem 00:37:44.860 --> 00:37:47.950 known as Min 2CNF deletion. 00:37:47.950 --> 00:37:51.780 And it's similar-- APX-hard, and best approximation 00:37:51.780 --> 00:37:54.680 is log times log log. 00:37:54.680 --> 00:37:57.880 If in the case where you have X(N)OR-SAT in general, 00:37:57.880 --> 00:38:01.330 but it's not all of the linear equations have only two terms-- 00:38:01.330 --> 00:38:05.110 so we have some larger ones-- then it turns out to be 00:38:05.110 --> 00:38:07.000 equivalent to nearest Codeword. 00:38:07.000 --> 00:38:10.120 So it turns out you can write all such equations using 00:38:10.120 --> 00:38:13.260 either equations of length, by using equations of length 3 00:38:13.260 --> 00:38:13.760 always. 00:38:13.760 --> 00:38:15.750 So this is linear equation. 00:38:15.750 --> 00:38:20.820 This should equal 1, or this says equals zero. 00:38:20.820 --> 00:38:23.276 And from that, you can construct all such things. 00:38:23.276 --> 00:38:24.525 This is a really hard problem. 00:38:27.610 --> 00:38:29.800 Poly-APX-hardness is not known. 00:38:29.800 --> 00:38:31.680 Current lower best lower bound is this 2 00:38:31.680 --> 00:38:33.460 to the log to the 1 minus epsilon, which 00:38:33.460 --> 00:38:37.440 we saw in the table of various inapproximability results 00:38:37.440 --> 00:38:37.940 last time. 00:38:37.940 --> 00:38:42.620 So this is a little bit smaller than n to the epsilon, 00:38:42.620 --> 00:38:43.890 but it's kind of close-ish. 00:38:47.150 --> 00:38:50.300 And finally, in the-- I didn't write it. 00:38:50.300 --> 00:38:52.810 If you're in CNF form, and all of the subclauses 00:38:52.810 --> 00:38:55.960 are either Horn, or all of the subclauses are Dual-Horn, 00:38:55.960 --> 00:39:00.350 then you get something called Min Horn Deletion. 00:39:00.350 --> 00:39:02.170 And this has the same inapproximability. 00:39:04.730 --> 00:39:06.070 Here it's known. 00:39:06.070 --> 00:39:07.580 So up here, the best approximation 00:39:07.580 --> 00:39:11.770 is n-- nothing, basically. 00:39:11.770 --> 00:39:13.110 Put them all in. 00:39:13.110 --> 00:39:16.990 And here there's a slightly better approximation known , 00:39:16.990 --> 00:39:18.990 I think, n to the 1 minus epsilon, or something. 00:39:18.990 --> 00:39:20.804 But these are all super hard. 00:39:20.804 --> 00:39:22.470 The main point of this is so that you're 00:39:22.470 --> 00:39:23.820 aware of these problems. 00:39:23.820 --> 00:39:26.640 If you ever encounter a problem that looks anything like this, 00:39:26.640 --> 00:39:29.740 or it looks like some kind of CSP problem, 00:39:29.740 --> 00:39:31.900 you should go to this list and check it out. 00:39:31.900 --> 00:39:35.430 So don't memorize these, but look at the notes. 00:39:35.430 --> 00:39:36.772 Definitely memorize these guys. 00:39:36.772 --> 00:39:37.730 These are good to know. 00:39:37.730 --> 00:39:42.140 But there's a few obscure problems here. 00:39:42.140 --> 00:39:42.640 OK. 00:39:42.640 --> 00:39:47.560 Last one is minimizing the number of ones. 00:39:47.560 --> 00:39:49.990 So this is like the hardest of two worlds. 00:39:49.990 --> 00:39:51.760 Minimization is kind of harder. 00:39:51.760 --> 00:39:54.460 And here you have to satisfy everything, but minimize 00:39:54.460 --> 00:39:56.390 the number of true variables. 00:39:59.530 --> 00:40:03.250 So this is easy if you can set them all false. 00:40:03.250 --> 00:40:04.820 And then you win. 00:40:04.820 --> 00:40:07.120 This is easy in the Horn case. 00:40:07.120 --> 00:40:09.170 The Horn case is when at most one is positive, 00:40:09.170 --> 00:40:11.900 so most of them can be set to zero. 00:40:11.900 --> 00:40:15.990 This is easy in the 2X(N)OR case. 00:40:15.990 --> 00:40:19.060 So if you have linear equations, two terms each, equal to 0 00:40:19.060 --> 00:40:21.320 or equals 1, that's also. 00:40:21.320 --> 00:40:24.100 And you want to minimize the number of true variables. 00:40:24.100 --> 00:40:25.410 That's good. 00:40:25.410 --> 00:40:28.060 If you're in 2CNF form, there's a constant factor 00:40:28.060 --> 00:40:28.780 approximation. 00:40:28.780 --> 00:40:30.240 That's the best you can do. 00:40:30.240 --> 00:40:30.781 APX-complete. 00:40:33.090 --> 00:40:36.300 This is a case from the last slide. 00:40:36.300 --> 00:40:39.290 If you have the hitting set constraints on constant number 00:40:39.290 --> 00:40:41.830 of constant size vertex sets, and you 00:40:41.830 --> 00:40:44.230 have implication constraints, then your problem 00:40:44.230 --> 00:40:45.535 is APX-complete again. 00:40:48.380 --> 00:40:50.300 And then we have these guys appearing, again 00:40:50.300 --> 00:40:51.070 nearest Codeword. 00:40:51.070 --> 00:40:52.980 N Min Horn deletion. 00:40:52.980 --> 00:40:55.020 This one we get in the Dual-Horn case. 00:40:55.020 --> 00:40:56.490 The Horn case is good. 00:40:56.490 --> 00:40:59.880 Dual-Horn, we get this thing, which was like log N 00:40:59.880 --> 00:41:00.380 approximal. 00:41:00.380 --> 00:41:01.490 Or no. 00:41:01.490 --> 00:41:05.880 This was the 2 to the log N to the 1 minus epsilon. 00:41:05.880 --> 00:41:10.380 And this is X(N)OR-SAT when they're not all binary. 00:41:10.380 --> 00:41:12.870 Then we get nearest Codeword-complete. 00:41:12.870 --> 00:41:16.590 And finally, oh, two more. 00:41:16.590 --> 00:41:19.450 The dual to this, if all the variables being set true 00:41:19.450 --> 00:41:22.590 satisfies your constraint, that gives you a solution, 00:41:22.590 --> 00:41:27.780 but it's like the worst solution possible, because you get N. 00:41:27.780 --> 00:41:32.320 And so in that case, you can get probably a poly approximation. 00:41:32.320 --> 00:41:34.740 Not very impressive. 00:41:34.740 --> 00:41:37.380 And that's actually the best you can do, at some N 00:41:37.380 --> 00:41:39.180 to the 1 minus epsilon. 00:41:39.180 --> 00:41:42.250 And in all other cases, by Schaefer's theorem, 00:41:42.250 --> 00:41:45.250 deciding whether even finding a feasible solution is NP-hard. 00:41:45.250 --> 00:41:47.960 So, good luck approximating. 00:41:47.960 --> 00:41:49.360 Cool? 00:41:49.360 --> 00:41:54.275 This is the Khanna, Sudan, Trevisan, Williamson 00:41:54.275 --> 00:41:55.150 multichotomy theorem. 00:41:59.100 --> 00:41:59.600 All right. 00:42:03.860 --> 00:42:11.280 So let's do some more reductions. 00:42:38.260 --> 00:42:42.740 My goal on this page is to get to our good friend 00:42:42.740 --> 00:42:46.180 from one of the first lectures, edge-matching-puzzles. 00:42:46.180 --> 00:42:50.480 You have little square tiles, colors on the edges. 00:42:50.480 --> 00:42:52.910 Normally we want to satisfy all of the edge constraints. 00:42:52.910 --> 00:42:57.480 Only equal colors match, are adjacent to each other. 00:42:57.480 --> 00:43:00.040 Now the problem is going to be maximize the number 00:43:00.040 --> 00:43:03.335 of satisfied edge constraints. 00:43:03.335 --> 00:43:05.160 But before I show you that reduction, 00:43:05.160 --> 00:43:08.020 I need another problem, which is APX-complete. 00:43:08.020 --> 00:43:10.330 So that problem is APX-complete. 00:43:10.330 --> 00:43:14.540 So I need two more problems. 00:43:14.540 --> 00:43:28.996 One is Max independent set in 3-regular 3-edge colorable 00:43:28.996 --> 00:43:29.495 graphs. 00:43:32.790 --> 00:43:33.290 OK. 00:43:33.290 --> 00:43:35.415 I'm not going to prove this one, because we already 00:43:35.415 --> 00:43:37.030 did a version of independent set, 00:43:37.030 --> 00:43:39.340 and it's just tedious to make it-- first, 00:43:39.340 --> 00:43:42.210 to make it exactly degree three everywhere, 00:43:42.210 --> 00:43:45.260 and secondly make it 3-edge colorable. 00:43:45.260 --> 00:43:48.630 With 3 regular 3-edge color is a nice kind of graph, 00:43:48.630 --> 00:43:55.370 because every vertex, you've got one edge of each class. 00:43:55.370 --> 00:43:56.930 So that's kind of cool. 00:43:56.930 --> 00:43:57.990 And we can use this. 00:43:57.990 --> 00:44:00.310 This problem is basically equivalent 00:44:00.310 --> 00:44:03.720 to the actual problem I want, which 00:44:03.720 --> 00:44:07.610 is a variation of three-dimensional matching. 00:44:07.610 --> 00:44:09.980 So remember three-dimensional matching, 00:44:09.980 --> 00:44:16.310 you have three sets-- A, B, and C. You 00:44:16.310 --> 00:44:19.080 look at the triples on A, B, and C. 00:44:19.080 --> 00:44:23.140 And you're given some set of interesting triples 00:44:23.140 --> 00:44:24.960 among those. 00:44:24.960 --> 00:44:32.350 And with 3DM, what we wanted was to choose a set of such triples 00:44:32.350 --> 00:44:36.080 that covers all the vertices, and no two of them intersect. 00:44:36.080 --> 00:44:38.500 That's the matching aspect. 00:44:38.500 --> 00:44:40.740 In this problem, we want to choose as many triples 00:44:40.740 --> 00:44:43.700 as we can that don't intersect each other. 00:44:43.700 --> 00:44:55.530 So the problem is choose max subset S prime of S 00:44:55.530 --> 00:44:59.750 with no duplicate coordinates, I'll say. 00:45:03.720 --> 00:45:05.900 So let's assume A, B, and C are disjoint. 00:45:05.900 --> 00:45:09.020 Then I don't want any element in A union B union C 00:45:09.020 --> 00:45:13.800 to appear twice in this chosen set S prime. 00:45:13.800 --> 00:45:15.710 So that's the problem. 00:45:15.710 --> 00:45:19.521 Now I'm going to prove that that's hard. 00:45:19.521 --> 00:45:24.990 It is basically the same as Max independent set, 00:45:24.990 --> 00:45:29.830 and three regular 3-edge colored graphs, 00:45:29.830 --> 00:45:33.760 because what I do is I take such a graph, 00:45:33.760 --> 00:45:43.490 and for each edge color class-- there are three of them-- 00:45:43.490 --> 00:45:46.040 those are going to be A, B, and C. 00:45:46.040 --> 00:45:47.800 So if I have red, green, and blue, 00:45:47.800 --> 00:45:49.910 all the red edges are going to be elements of A, 00:45:49.910 --> 00:45:52.220 all the green edges are going to be the elements 00:45:52.220 --> 00:45:54.720 of B-- B for green. 00:45:54.720 --> 00:45:58.090 And then all the blue elements are elements of C. 00:45:58.090 --> 00:45:58.710 OK. 00:45:58.710 --> 00:46:06.380 Then a vertex, as I said, has exactly one of each class. 00:46:06.380 --> 00:46:07.790 So that's going to be my triple. 00:46:11.410 --> 00:46:13.540 And that's it. 00:46:13.540 --> 00:46:16.150 So now, if I want to solve three-dimensional matching 00:46:16.150 --> 00:46:17.930 among those triples, that's going 00:46:17.930 --> 00:46:22.735 to correspond to choosing a set of vertices in here, no two 00:46:22.735 --> 00:46:25.760 of which share a color. 00:46:25.760 --> 00:46:30.105 No two of which share the same item of A. Let's say A 00:46:30.105 --> 00:46:32.360 is this color of edge. 00:46:32.360 --> 00:46:35.720 So that means that the vertices over here 00:46:35.720 --> 00:46:37.890 are not connected by an edge. 00:46:37.890 --> 00:46:40.920 So the cool thing here is that each element of A, B, and C 00:46:40.920 --> 00:46:49.000 only appears in two different triples. 00:46:49.000 --> 00:46:51.800 Corresponding to the two ends of the edge. 00:46:51.800 --> 00:46:54.540 So now we have max three-dimensional matching 00:46:54.540 --> 00:46:58.670 where every element in ABC appears in exactly two triples. 00:46:58.670 --> 00:47:03.188 So I guess I can even write E2 if I want to. 00:47:03.188 --> 00:47:05.060 OK. 00:47:05.060 --> 00:47:08.130 That was our sort of homework. 00:47:08.130 --> 00:47:13.370 Now we have max edge matching puzzles. 00:47:13.370 --> 00:47:17.287 Again, we're given square tiles. 00:47:17.287 --> 00:47:18.870 There's different colors on the tiles. 00:47:18.870 --> 00:47:20.780 Any number of colors. 00:47:20.780 --> 00:47:23.950 And we would like to lay things out. 00:47:23.950 --> 00:47:26.880 And I'll tell you the instance here is going to be 2 by N. 00:47:26.880 --> 00:47:29.760 So it's fairly narrow, unlike the construction 00:47:29.760 --> 00:47:32.240 we saw in class. 00:47:32.240 --> 00:47:36.330 And we're reducing from Max 3D M2. 00:47:36.330 --> 00:47:38.156 That's why I introduced it. 00:47:38.156 --> 00:47:43.090 And this is a four years ago result. 00:47:43.090 --> 00:47:47.640 So the idea is the triple is represented by these three 00:47:47.640 --> 00:47:49.210 tiles, and some more. 00:47:49.210 --> 00:47:52.090 But for starters, these three tiles. 00:47:52.090 --> 00:47:54.870 The u glue is unique-- global unique. 00:47:54.870 --> 00:47:57.090 So it wants to be on the boundary. 00:47:57.090 --> 00:47:58.890 And here tiles are not allowed to rotate, 00:47:58.890 --> 00:48:01.490 so it wants to be on the bottom boundary. 00:48:01.490 --> 00:48:08.676 So this ab glues only appear as a single pairs. 00:48:08.676 --> 00:48:10.300 I guess they'll also appear over there. 00:48:10.300 --> 00:48:11.383 But not very many of them. 00:48:11.383 --> 00:48:13.800 So basically a, b, and c have to glue together 00:48:13.800 --> 00:48:14.720 in sequence like that. 00:48:14.720 --> 00:48:15.980 And the percent signs are going to be 00:48:15.980 --> 00:48:17.140 the same on the bottom row. 00:48:17.140 --> 00:48:19.130 So nothing else. 00:48:19.130 --> 00:48:20.832 This is basically forced to do this. 00:48:20.832 --> 00:48:22.540 We'll actually have to do it a few times, 00:48:22.540 --> 00:48:24.920 but you have to build this bottom structure. 00:48:24.920 --> 00:48:28.210 And then the question is what do you build on top. 00:48:28.210 --> 00:48:32.900 And the idea is there are exactly one each of these three 00:48:32.900 --> 00:48:37.110 tiles which just communicate dollar sign left to right, 00:48:37.110 --> 00:48:39.550 and have a, b, c on the bottom. 00:48:39.550 --> 00:48:40.432 So those are cool. 00:48:40.432 --> 00:48:42.890 And if you want to put a triple into your three-dimensional 00:48:42.890 --> 00:48:46.950 matching, then you put those in sequence. 00:48:46.950 --> 00:48:48.020 No mismatches. 00:48:48.020 --> 00:48:48.680 This is great. 00:48:48.680 --> 00:48:49.820 You can take a whole bunch of these, 00:48:49.820 --> 00:48:52.028 stick them next to each other, everything will match. 00:48:52.028 --> 00:48:53.000 No errors. 00:48:53.000 --> 00:48:54.990 So you're getting some constant number 00:48:54.990 --> 00:48:58.230 of points for each of these. 00:48:58.230 --> 00:49:03.240 But you will have to build more-- at least two copies 00:49:03.240 --> 00:49:04.930 of this bottom structure. 00:49:04.930 --> 00:49:07.540 And there's only one copy of this top thing. 00:49:07.540 --> 00:49:09.110 So that's the annoying part. 00:49:09.110 --> 00:49:11.820 But there are some variations of these tiles which 00:49:11.820 --> 00:49:13.570 look like something like this-- I'll 00:49:13.570 --> 00:49:16.930 show you all of them in a moment-- which have exactly one 00:49:16.930 --> 00:49:18.450 mismatch. 00:49:18.450 --> 00:49:20.842 So you don't get quite as many points. 00:49:20.842 --> 00:49:22.800 You get, I don't know, 15 instead of 16 points, 00:49:22.800 --> 00:49:24.740 or whatever. 00:49:24.740 --> 00:49:26.750 Bottom structure looks the same. 00:49:26.750 --> 00:49:31.571 And the point of this is we know a appears 00:49:31.571 --> 00:49:32.570 in two different places. 00:49:32.570 --> 00:49:35.870 So we need two versions of the a tile. 00:49:35.870 --> 00:49:39.015 But we only want one of them to be happy and give you 00:49:39.015 --> 00:49:40.640 all the points, because you should only 00:49:40.640 --> 00:49:44.400 be able to choose the a thing once. 00:49:44.400 --> 00:49:46.520 So yet this triple will still exist. 00:49:46.520 --> 00:49:48.400 adc will still be floating around there. 00:49:48.400 --> 00:49:52.410 You want to still be buildable, but at a cost of negative 1. 00:49:52.410 --> 00:49:54.880 So this part's still built. 00:49:54.880 --> 00:49:57.030 Then you have these sort of filler tiles. 00:49:57.030 --> 00:49:59.000 Your goal is then just get rid of all the stuff 00:49:59.000 --> 00:50:00.770 and pay a penalty. 00:50:00.770 --> 00:50:03.410 But you want to minimize the number of times you do this, 00:50:03.410 --> 00:50:05.850 or maximize the number of times you do this, 00:50:05.850 --> 00:50:09.200 and then it will be simulating Max 3DM. 00:50:09.200 --> 00:50:12.640 There'll be some additive consistent cost, 00:50:12.640 --> 00:50:16.690 which is the cost of all the unpicked triples. 00:50:16.690 --> 00:50:20.525 And then this will be an L-reduction. 00:50:20.525 --> 00:50:21.650 So I have some more slides. 00:50:21.650 --> 00:50:24.070 It's a bit complicated to do all of the details, 00:50:24.070 --> 00:50:28.020 but this is a fully worked-out example with two triples. 00:50:28.020 --> 00:50:30.680 We have a, b, c and a, d, c. 00:50:30.680 --> 00:50:32.264 And because they share a, we don't 00:50:32.264 --> 00:50:33.430 want them both to be picked. 00:50:33.430 --> 00:50:36.380 So the same as what I showed you just in the previous slide. 00:50:36.380 --> 00:50:38.500 But then there are all these other tiles 00:50:38.500 --> 00:50:41.420 that are floating around in order to make 00:50:41.420 --> 00:50:43.320 all the combinations possible. 00:50:43.320 --> 00:50:45.730 And there's all these tiles to basically allow 00:50:45.730 --> 00:50:47.390 them to get thrown away. 00:50:47.390 --> 00:50:50.710 And so that's not so clear. 00:50:50.710 --> 00:50:54.104 This is the overall construction. 00:50:54.104 --> 00:50:56.520 For every triple, you're going to have exactly these three 00:50:56.520 --> 00:50:59.310 tiles that we saw. 00:50:59.310 --> 00:51:01.310 It got rotated relative to the previous picture. 00:51:01.310 --> 00:51:03.560 Maybe rotations are allowed. 00:51:03.560 --> 00:51:05.890 And then for every variable, here 00:51:05.890 --> 00:51:08.150 they're called x, y, z instead of a, b, c. 00:51:08.150 --> 00:51:09.290 But the same thing. 00:51:09.290 --> 00:51:13.100 For every a thing we'll have some constant set of tiles that 00:51:13.100 --> 00:51:15.250 includes the really good one. 00:51:15.250 --> 00:51:15.750 Sorry. 00:51:15.750 --> 00:51:17.270 The good one has two dollar signs. 00:51:17.270 --> 00:51:19.465 This is the one you really like. 00:51:19.465 --> 00:51:21.090 And then there's all this stuff to make 00:51:21.090 --> 00:51:23.350 sure things can get consumed. 00:51:23.350 --> 00:51:24.880 And you can get rid of the triples 00:51:24.880 --> 00:51:27.950 and pay exactly one per unpicked triple. 00:51:27.950 --> 00:51:29.700 So I don't want to go through the details, 00:51:29.700 --> 00:51:34.711 but once you have that, you get an L-reduction from Max 3DN2. 00:51:34.711 --> 00:51:35.210 Questions? 00:51:38.508 --> 00:51:39.494 All right. 00:51:44.960 --> 00:51:50.590 So I want to go up the hierarchy. 00:51:50.590 --> 00:51:55.130 We've been focusing on constant factor, approximable problems 00:51:55.130 --> 00:51:56.270 that have no PTASses. 00:51:59.080 --> 00:52:00.870 I will mention there before we go on 00:52:00.870 --> 00:52:04.050 that there are some constant factor approximable 00:52:04.050 --> 00:52:08.020 problems that are not, that have no PTAS, 00:52:08.020 --> 00:52:10.600 and yet are not APX-complete. 00:52:10.600 --> 00:52:17.520 So APX-complete is not all of APX minus PTAS. 00:52:17.520 --> 00:52:22.460 So there are APX minus PTAS problems 00:52:22.460 --> 00:52:23.620 that are not APX-complete. 00:52:26.380 --> 00:52:29.140 So these are still useful from a reduction standpoint. 00:52:29.140 --> 00:52:33.910 You can use them to show that your problem has no PTAS. 00:52:33.910 --> 00:52:36.450 But you have to state them differently. 00:52:40.690 --> 00:52:43.190 And they're somewhat familiar problems. 00:52:43.190 --> 00:52:46.230 One of them is bin packing. 00:52:46.230 --> 00:52:48.950 This is you're moving out of your house. 00:52:48.950 --> 00:52:50.950 You have a bunch of objects. 00:52:50.950 --> 00:52:52.700 You live in a one-dimensional universe. 00:52:52.700 --> 00:52:55.620 So each box is exactly the same size. 00:52:55.620 --> 00:52:57.240 It's one-dimensional in size. 00:52:57.240 --> 00:52:58.920 And you have a bunch of items which are one-dimensional. 00:52:58.920 --> 00:53:01.211 And you want to pack as many as you can into each box-- 00:53:01.211 --> 00:53:03.190 but overall use the minimum number of boxes. 00:53:03.190 --> 00:53:05.690 It's a minimization problem. 00:53:05.690 --> 00:53:08.770 This has no constant factor approximation. 00:53:08.770 --> 00:53:14.740 But you can find what's called a asymptotic PTAS, where 00:53:14.740 --> 00:53:17.950 you can get a PTAS-style result-- 1 plus epsilon 00:53:17.950 --> 00:53:21.822 times OPT plus 1. 00:53:21.822 --> 00:53:24.881 So an additive error. 00:53:24.881 --> 00:53:26.380 And so in particular, distinguishing 00:53:26.380 --> 00:53:29.930 between two bins and three bins is weakly NP-complete. 00:53:29.930 --> 00:53:36.325 That's like partition, right, between two bins 00:53:36.325 --> 00:53:37.570 and three bins. 00:53:37.570 --> 00:53:39.280 So you need this sort of additive one. 00:53:39.280 --> 00:53:42.060 You can't get a PTAS without the additive one. 00:53:42.060 --> 00:53:45.360 So it's not as hard as all constant factor inapproximable 00:53:45.360 --> 00:53:49.300 problems, but somewhere in between. 00:53:49.300 --> 00:53:52.440 APX-intermediate is the technical term. 00:53:52.440 --> 00:53:56.501 Some other ones are minimum. 00:53:56.501 --> 00:53:58.432 AUDIENCE: [INAUDIBLE]. 00:53:58.432 --> 00:54:00.765 PROFESSOR: Oh, this is all assuming P does not equal NP. 00:54:00.765 --> 00:54:01.120 Yes. 00:54:01.120 --> 00:54:03.530 If P equals NP, then I think all these things are equal. 00:54:03.530 --> 00:54:05.300 So, thank you. 00:54:08.400 --> 00:54:10.670 Another problem I've seen in some situations 00:54:10.670 --> 00:54:15.260 is you want to find the spanning tree in a graph that 00:54:15.260 --> 00:54:16.840 minimizes the maximum degree. 00:54:16.840 --> 00:54:19.070 This is also APX-intermediate. 00:54:19.070 --> 00:54:21.220 There's a constant factor approximation. 00:54:21.220 --> 00:54:26.130 No PTAS, but not as hard as all of APX. 00:54:26.130 --> 00:54:28.440 And another one is min edge coloring, 00:54:28.440 --> 00:54:33.120 which is quite a bit easier than vertex coloring. 00:54:33.120 --> 00:54:34.864 So these are problems to watch out for. 00:54:34.864 --> 00:54:37.280 They're the only ones I know of that are APX-intermediate. 00:54:37.280 --> 00:54:38.280 There may be more known. 00:54:41.330 --> 00:54:42.190 OK. 00:54:42.190 --> 00:54:44.900 So unless there are questions, I want to go up 00:54:44.900 --> 00:54:46.795 to log factor approximation. 00:54:54.180 --> 00:54:56.050 Surprisingly, in the CSP universe, 00:54:56.050 --> 00:54:59.970 we didn't get any log approximation 00:54:59.970 --> 00:55:00.970 as the right answer. 00:55:00.970 --> 00:55:03.350 But there are problems where log is the right answer. 00:55:07.774 --> 00:55:09.690 Again, there's probably intermediate problems. 00:55:09.690 --> 00:55:11.720 But here are some problems that are actually 00:55:11.720 --> 00:55:14.880 complete over all log approximable problems. 00:55:14.880 --> 00:55:16.930 So there's a log lower-bound and upper-bound 00:55:16.930 --> 00:55:19.390 on their approximability. 00:55:19.390 --> 00:55:25.090 I've mentioned two of them-- set cover and dominating set. 00:55:29.859 --> 00:55:32.150 First thing I'd like to show is that these two problems 00:55:32.150 --> 00:55:33.390 are the same. 00:55:33.390 --> 00:55:35.810 I'm not going to try to prove lower bounds on them-- 00:55:35.810 --> 00:55:37.240 at least for now. 00:55:37.240 --> 00:55:40.720 But let me show that you could L-reduce one to the other. 00:55:40.720 --> 00:55:44.080 So the easy direction is L-reducing dominating 00:55:44.080 --> 00:55:47.260 set to set cover, because dominating set 00:55:47.260 --> 00:55:49.100 says, well, if I choose this vertex, 00:55:49.100 --> 00:55:52.550 then I cover these vertices. 00:55:52.550 --> 00:55:53.050 OK. 00:55:53.050 --> 00:55:57.920 So let's call this vertex V, and then maybe a, b, c, d. 00:55:57.920 --> 00:56:04.450 I can represent that by a set-- namely v, a, b, c, d. 00:56:04.450 --> 00:56:06.502 If I choose that set, it covers those elements, 00:56:06.502 --> 00:56:07.960 just like when I choose this vertex 00:56:07.960 --> 00:56:09.450 it covers those vertices. 00:56:09.450 --> 00:56:09.950 OK. 00:56:09.950 --> 00:56:12.490 So that's a strict reduction from dominating 00:56:12.490 --> 00:56:14.865 set to set cover. 00:56:14.865 --> 00:56:18.320 In some sense, the bipartite version gives you more control. 00:56:18.320 --> 00:56:18.820 OK. 00:56:18.820 --> 00:56:22.500 This is the non-bipartite version of set cover. 00:56:22.500 --> 00:56:24.110 So what about the other reduction-- 00:56:24.110 --> 00:56:27.500 reducing set cover to dominating set? 00:56:30.090 --> 00:56:33.170 So this is a little more fun. 00:56:33.170 --> 00:56:35.710 We need to build a graph dominating 00:56:35.710 --> 00:56:39.000 set that somehow has two very different types of vertices. 00:56:39.000 --> 00:56:42.810 We want to represent sets, and we want to represent elements. 00:56:42.810 --> 00:56:44.400 So here's what we're going to do. 00:56:44.400 --> 00:56:49.040 We build a clique representing the sets. 00:56:49.040 --> 00:56:53.560 So there are nodes in this clique-- one for every set. 00:56:53.560 --> 00:56:57.240 And then we're going to have an independent set over here that 00:56:57.240 --> 00:56:59.710 will represent the elements. 00:56:59.710 --> 00:57:01.910 And then whenever a set over here 00:57:01.910 --> 00:57:05.940 contains an element over there, we will add an edge. 00:57:05.940 --> 00:57:08.970 So in general, an element may appear in several sets, 00:57:08.970 --> 00:57:12.200 and the set is going to consist of many elements. 00:57:12.200 --> 00:57:14.440 But over here, there's not going to be any edges 00:57:14.440 --> 00:57:15.410 between these elements. 00:57:15.410 --> 00:57:18.390 These are independent. 00:57:18.390 --> 00:57:22.370 And over here, all of the edges exist. 00:57:22.370 --> 00:57:25.540 So the intent is you choose a set of these vertices 00:57:25.540 --> 00:57:29.620 corresponding to sets in order to cover those vertices. 00:57:29.620 --> 00:57:31.870 And that's going to work, because these vertices 00:57:31.870 --> 00:57:33.820 are super easy to cover in the dominating set. 00:57:33.820 --> 00:57:36.880 You choose any of them, you cover all of them. 00:57:36.880 --> 00:57:40.800 These guys, you never want to put them in a dominating set. 00:57:40.800 --> 00:57:42.800 Why would you put this in a dominating set, when 00:57:42.800 --> 00:57:44.466 you could just follow one of these edges 00:57:44.466 --> 00:57:45.780 and put this in instead? 00:57:45.780 --> 00:57:49.960 That vertex will cover this one, and it will cover all of these. 00:57:49.960 --> 00:57:52.680 And the only edges from here are to over here. 00:57:52.680 --> 00:57:56.451 So if you choose a set, you'll cover all the sets and that one 00:57:56.451 --> 00:57:56.950 element. 00:57:56.950 --> 00:57:58.324 If you choose the element, you'll 00:57:58.324 --> 00:58:01.390 cover the element and some of the sets. 00:58:01.390 --> 00:58:04.100 So in any optimal solution, if this ever appears, 00:58:04.100 --> 00:58:06.530 you can keep it optimal and move over here. 00:58:06.530 --> 00:58:09.170 That is sort of arguments we've been doing over and over. 00:58:09.170 --> 00:58:11.240 So there is an optimal solution where you only 00:58:11.240 --> 00:58:16.810 choose vertices on the left, and then that is a set cover. 00:58:16.810 --> 00:58:19.570 Again, it's a strict reduction. 00:58:19.570 --> 00:58:21.170 No loss. 00:58:21.170 --> 00:58:21.670 Cool? 00:58:21.670 --> 00:58:24.425 So that is why these two problems are equivalent. 00:58:24.425 --> 00:58:26.300 Now we're just going to take on faith for now 00:58:26.300 --> 00:58:29.290 that they are log inapproximable. 00:58:29.290 --> 00:58:32.044 And you've probably seen that this one is log approximable. 00:58:32.044 --> 00:58:33.960 So now you know that this is log approximable. 00:58:39.540 --> 00:58:45.170 I would say most of the literature 00:58:45.170 --> 00:58:50.040 I see for inapproximability is either APX hardness, 00:58:50.040 --> 00:58:52.465 or what people usually call set cover hardness. 00:58:55.140 --> 00:58:57.440 I mean, the fact that set covers log APX-complete, 00:58:57.440 --> 00:58:58.814 that is complete for that class-- 00:58:58.814 --> 00:59:01.230 not just a log lower-bound-- is fairly recent. 00:59:01.230 --> 00:59:03.760 So people usually have called it set cover hardness. 00:59:03.760 --> 00:59:07.000 Now you can call it log APX-hardness. 00:59:07.000 --> 00:59:10.120 So let me show you one example. 00:59:10.120 --> 00:59:11.880 There are a lot of both out there, 00:59:11.880 --> 00:59:15.852 and I'm actually just showing you sort of a small sampling, 00:59:15.852 --> 00:59:17.900 because there's so much. 00:59:17.900 --> 00:59:20.120 So here's a fun problem. 00:59:20.120 --> 00:59:23.167 It's called token reconfiguration. 00:59:23.167 --> 00:59:24.750 And the idea is you're doing some kind 00:59:24.750 --> 00:59:27.410 of motion planning in a graph. 00:59:27.410 --> 00:59:29.380 So something like pushing blocks, 00:59:29.380 --> 00:59:33.300 except you have a bunch of robots, 00:59:33.300 --> 00:59:37.100 which here are represented-- well, you have a graph. 00:59:37.100 --> 00:59:40.760 And each vertex can either have a robot or not. 00:59:40.760 --> 00:59:43.580 In some, you're given an initial configuration 00:59:43.580 --> 00:59:45.320 of how the robots are placed, and you're 00:59:45.320 --> 00:59:46.903 given a final configuration of how you 00:59:46.903 --> 00:59:48.160 want the robots to be placed. 00:59:48.160 --> 00:59:49.826 And they have the same number of robots, 00:59:49.826 --> 00:59:53.220 because you can't eat robots, or create them yet. 00:59:53.220 --> 00:59:55.470 So when robots can create robots, 00:59:55.470 --> 00:59:57.990 that will be another problem. 00:59:57.990 --> 00:59:59.490 So here you have robot conservation. 01:00:03.200 --> 01:00:05.370 So in a configuration, there are three types 01:00:05.370 --> 01:00:08.350 of vertices in that situation. 01:00:08.350 --> 01:00:10.760 It could be you have a vertex that currently 01:00:10.760 --> 01:00:12.580 has a robot-- here they're called tokens, 01:00:12.580 --> 01:00:16.210 to be a little more generic. 01:00:16.210 --> 01:00:19.480 It could have a robot, but not be a place 01:00:19.480 --> 01:00:20.690 that should have a robot. 01:00:20.690 --> 01:00:22.690 So in the initial configuration, it has a robot, 01:00:22.690 --> 01:00:24.910 but in the final configuration it does not. 01:00:24.910 --> 01:00:28.750 It could be you have some robots that are basically 01:00:28.750 --> 01:00:29.940 where they want to be. 01:00:29.940 --> 01:00:33.240 They are robot and also in the target configuration, 01:00:33.240 --> 01:00:34.780 there's a robot there. 01:00:34.780 --> 01:00:36.870 Or I guess there's four cases, but in this case 01:00:36.870 --> 01:00:38.040 we'll only have three. 01:00:38.040 --> 01:00:40.260 Or it could be that you want to have robot there, 01:00:40.260 --> 01:00:42.240 but currently you do not. 01:00:42.240 --> 01:00:46.817 So this is an instance that simulates set cover. 01:00:46.817 --> 01:00:48.650 And this is a situation where robots are all 01:00:48.650 --> 01:00:49.520 treated identically. 01:00:49.520 --> 01:00:52.400 So you don't care which robot goes where. 01:00:52.400 --> 01:00:54.030 So you've got these robots over here, 01:00:54.030 --> 01:00:55.350 which don't want to be here. 01:00:55.350 --> 01:00:56.860 They want to be over there. 01:00:56.860 --> 01:00:58.450 I mean, if you measure this length, 01:00:58.450 --> 01:01:01.900 it's the same as this length. 01:01:01.900 --> 01:01:03.540 And these robots don't want to move, 01:01:03.540 --> 01:01:05.930 but they're going to have to, because they're in the way. 01:01:05.930 --> 01:01:08.590 In this tripartite graph, they're in the way from here 01:01:08.590 --> 01:01:09.950 to there. 01:01:09.950 --> 01:01:12.840 I didn't tell you a move in this scenario 01:01:12.840 --> 01:01:18.220 is that you can take a robot and follow any empty path, OK 01:01:18.220 --> 01:01:21.410 So you can make a sequence of moves all at a cost of one, 01:01:21.410 --> 01:01:23.610 as long as it doesn't hit any other robots. 01:01:23.610 --> 01:01:25.570 So, a collision-free path. 01:01:25.570 --> 01:01:27.850 You follow it, then you can pick up another robot, 01:01:27.850 --> 01:01:29.349 move it along a collision-free path, 01:01:29.349 --> 01:01:32.720 pick up another robot, and so on. 01:01:32.720 --> 01:01:34.884 So if you want to move all these guys over here, 01:01:34.884 --> 01:01:37.300 you're going to have to move some of these out of the way. 01:01:37.300 --> 01:01:38.300 How many? 01:01:38.300 --> 01:01:39.570 Set cover many. 01:01:39.570 --> 01:01:42.330 Here's the set cover instance in this bipartite graph. 01:01:42.330 --> 01:01:45.710 So what you can do is take this robot, move it out of the way, 01:01:45.710 --> 01:01:47.240 move it to one of these elements, 01:01:47.240 --> 01:01:49.200 and then for the remainder of this set, which 01:01:49.200 --> 01:01:51.597 are these two nodes, you can take this guy 01:01:51.597 --> 01:01:53.430 and move it there in one step, take this guy 01:01:53.430 --> 01:01:54.800 and move it there in one step. 01:01:54.800 --> 01:01:56.200 The length of this doesn't matter, because you 01:01:56.200 --> 01:01:57.480 can follow a long path. 01:01:57.480 --> 01:02:01.900 And you just drain out this thing one at a time-- 01:02:01.900 --> 01:02:05.490 except for this guy, who you moved out of the way. 01:02:05.490 --> 01:02:08.260 You move one of these to fill his spot. 01:02:08.260 --> 01:02:10.690 And if you can cover all the elements over here 01:02:10.690 --> 01:02:13.640 with only k of these guys moving, 01:02:13.640 --> 01:02:20.215 then the number of moves will be k plus A. So 01:02:20.215 --> 01:02:21.340 that's what's written here. 01:02:21.340 --> 01:02:26.940 OPT is, this is a fixed added of cost plus the set cover. 01:02:26.940 --> 01:02:30.600 And this is going to be an L-reduction, provided 01:02:30.600 --> 01:02:36.990 this is a linear in A, which is easy enough to arrange. 01:02:36.990 --> 01:02:38.590 So that's the unlabeled case. 01:02:38.590 --> 01:02:40.860 You can also solve the labeled case. 01:02:40.860 --> 01:02:44.170 Maybe you want robot one to go to position one, 01:02:44.170 --> 01:02:47.190 and you want robot two to go to position two. 01:02:47.190 --> 01:02:48.901 Same thing, but here these robots 01:02:48.901 --> 01:02:50.900 are going to have to go back where they started. 01:02:50.900 --> 01:02:53.525 So you just add a little vertex so they can get out of the way. 01:02:53.525 --> 01:02:55.590 Everything can move where they want to. 01:02:55.590 --> 01:02:58.710 Again, choose a set cover, move those over, 01:02:58.710 --> 01:02:59.970 and then move them back. 01:02:59.970 --> 01:03:02.130 So you end up paying two times the set cover. 01:03:02.130 --> 01:03:03.840 But just a constant factor loss. 01:03:03.840 --> 01:03:05.960 Still an L-reduction. 01:03:05.960 --> 01:03:07.960 And this problem is motivated, it's 01:03:07.960 --> 01:03:10.290 sort of a generalization of the 15 puzzle. 01:03:10.290 --> 01:03:12.750 You have a little 4 by 4 grid. 01:03:12.750 --> 01:03:13.990 You've got movable tiles. 01:03:13.990 --> 01:03:16.300 You can only move one at a time in that case, 01:03:16.300 --> 01:03:18.420 because there's only a single gap. 01:03:18.420 --> 01:03:20.890 This is sort of a generalized form of that, 01:03:20.890 --> 01:03:22.770 where you have various tiles. 01:03:22.770 --> 01:03:25.230 You want to get them into the right spots, 01:03:25.230 --> 01:03:28.300 but you can't have collisions during that motion. 01:03:28.300 --> 01:03:31.470 So that's where this problem came from. 01:03:31.470 --> 01:03:34.320 15 puzzle, by the way, in the generalized n by n form 01:03:34.320 --> 01:03:37.077 is NP-hard and in APX, but I think it's open 01:03:37.077 --> 01:03:38.160 whether it's APX-complete. 01:03:40.700 --> 01:03:44.820 I would show the proof, but it's very complicated, so, I won't. 01:03:48.450 --> 01:03:50.140 Cool. 01:03:50.140 --> 01:03:53.170 Well, in the last little bit, I wanted to tell you 01:03:53.170 --> 01:03:56.230 about the super high end. 01:03:56.230 --> 01:03:57.835 So we went to log approximation. 01:04:00.640 --> 01:04:03.720 There are other things known, but not 01:04:03.720 --> 01:04:05.110 a lot of completeness results. 01:04:05.110 --> 01:04:06.610 So we're going to get to other kinds 01:04:06.610 --> 01:04:09.370 of interapproximability next class. 01:04:09.370 --> 01:04:13.430 For now, I want to stick to something APX-complete. 01:04:13.430 --> 01:04:15.790 And the most studied class above log 01:04:15.790 --> 01:04:19.740 is poly, which is like n to the 1 minus epsilon. 01:04:34.860 --> 01:04:38.360 And my main goal here is to tell you about some problems 01:04:38.360 --> 01:04:40.880 that you should, if you think your problem is 01:04:40.880 --> 01:04:44.730 like Poly-APX-hard, these are the standard problems 01:04:44.730 --> 01:04:46.390 to start from. 01:04:46.390 --> 01:04:47.629 There are two of them. 01:04:47.629 --> 01:04:49.920 And I've mentioned them, but not quite in this context. 01:04:57.920 --> 01:05:03.194 They are clique and independent set. 01:05:03.194 --> 01:05:04.610 These are really the same problem. 01:05:04.610 --> 01:05:08.670 One is the complement graph of the other. 01:05:08.670 --> 01:05:09.975 Both maximization problems. 01:05:12.930 --> 01:05:14.480 And those are the standard ones. 01:05:14.480 --> 01:05:16.690 I'll leave it at that. 01:05:16.690 --> 01:05:18.490 I'm going to keep going up. 01:05:18.490 --> 01:05:22.173 The next level most studied is Exp-APX-complete. 01:05:25.136 --> 01:05:27.010 So for these problems, the best approximation 01:05:27.010 --> 01:05:29.960 is n divided by log squared n. 01:05:29.960 --> 01:05:32.234 And there's a lower bound of n to the 1 minus epsilon. 01:05:32.234 --> 01:05:34.400 So there is a gap in terms of their approximability. 01:05:34.400 --> 01:05:35.775 But what we know is that they are 01:05:35.775 --> 01:05:39.930 the hardest problems that have any n to the ce approximation. 01:05:39.930 --> 01:05:44.380 They're all reducible to each other via PTAS reductions. 01:05:44.380 --> 01:05:45.705 So, fairly preserving. 01:05:48.680 --> 01:05:52.010 So our next class up is APX-complete, 01:05:52.010 --> 01:05:59.980 things, problems approximable in exponential and n approximation 01:05:59.980 --> 01:06:00.480 factors. 01:06:00.480 --> 01:06:02.850 How would that happen? 01:06:02.850 --> 01:06:04.420 This is kind of funny. 01:06:04.420 --> 01:06:09.350 And the canonical problem here is the basic reason is numbers. 01:06:12.190 --> 01:06:14.410 We take the traveling salesman problem. 01:06:14.410 --> 01:06:16.950 And every edge can have a weight. 01:06:16.950 --> 01:06:18.730 Let's say it's integer weights. 01:06:18.730 --> 01:06:21.960 But any integer weight that can be expressible in n bits 01:06:21.960 --> 01:06:25.740 is fair game, which means the actual value of that edge 01:06:25.740 --> 01:06:28.500 is going to be exponential in n. 01:06:28.500 --> 01:06:31.210 And from that, you can get a very easy lower bound. 01:06:31.210 --> 01:06:33.480 And in fact, all problems that are 01:06:33.480 --> 01:06:38.430 approximable in exponential APX can be reduced to general TSP, 01:06:38.430 --> 01:06:40.267 where you're just given a bunch of distances 01:06:40.267 --> 01:06:41.350 between pairs of vertices. 01:06:41.350 --> 01:06:43.040 It doesn't satisfy triangle inequality. 01:06:43.040 --> 01:06:44.930 That's the non-metric aspect. 01:06:44.930 --> 01:06:48.100 The triangle inequality TSP, which is what normally happens, 01:06:48.100 --> 01:06:49.260 there is a constant factor. 01:06:49.260 --> 01:06:51.430 It's APX complete. 01:06:51.430 --> 01:06:57.030 But for general waits between pairs of vertices, 01:06:57.030 --> 01:06:59.370 non-metric, it's Exp-APX-complete, 01:06:59.370 --> 01:07:03.220 because you can basically make a graph 01:07:03.220 --> 01:07:05.240 and solve Hamiltonicity by saying 01:07:05.240 --> 01:07:09.050 all the edges in the graph have weight one or zero, 01:07:09.050 --> 01:07:12.790 and all of the edges-- I guess one would be a little bit more 01:07:12.790 --> 01:07:14.240 legitimate. 01:07:14.240 --> 01:07:16.350 And all the non-edges in the graph 01:07:16.350 --> 01:07:17.890 are going to give weight infinity. 01:07:17.890 --> 01:07:19.920 Infinity is the largest expressible number which 01:07:19.920 --> 01:07:22.360 is 1, 1, 1, 1, n bits long. 01:07:22.360 --> 01:07:24.610 And so either you use one of those edges or you don't. 01:07:24.610 --> 01:07:27.540 And there's an exponential gap between them. 01:07:27.540 --> 01:07:29.590 So even if we disallow zeros being an output, 01:07:29.590 --> 01:07:33.446 then we get exponential separation. 01:07:33.446 --> 01:07:35.070 That doesn't prove completeness, but it 01:07:35.070 --> 01:07:38.070 proves that you can't hope for better than exponential 01:07:38.070 --> 01:07:40.910 approximation there. 01:07:40.910 --> 01:07:42.240 OK. 01:07:42.240 --> 01:07:46.620 Two more even crazier classes. 01:07:46.620 --> 01:07:48.420 Now we did see these classes come up 01:07:48.420 --> 01:07:52.580 with the characterization theorem. 01:07:52.580 --> 01:07:54.990 But these are probably how these results were proved. 01:08:17.750 --> 01:08:20.778 So you might think, well, double the exponential. 01:08:20.778 --> 01:08:21.319 I don't know. 01:08:21.319 --> 01:08:22.376 What's next? 01:08:22.376 --> 01:08:24.189 Next, you could define that. 01:08:24.189 --> 01:08:27.550 But what seems to appear most often 01:08:27.550 --> 01:08:33.040 is this is the ultimate class among all NP optimization 01:08:33.040 --> 01:08:34.810 problems, you could imagine being complete 01:08:34.810 --> 01:08:36.060 against all of them. 01:08:36.060 --> 01:08:40.270 And this is with respect to AP-reductions, 01:08:40.270 --> 01:08:41.279 one of the ones we saw. 01:08:44.090 --> 01:08:47.490 And I'm going to define a very closely related class, which 01:08:47.490 --> 01:08:51.560 is NPO PB, NPO polynomially bounded. 01:08:57.700 --> 01:08:58.992 OK. 01:08:58.992 --> 01:09:02.220 So these are the hardest problems to approximate. 01:09:02.220 --> 01:09:04.740 This is basically the problems that have numbers in them, 01:09:04.740 --> 01:09:06.810 and this is the problem that have no numbers, 01:09:06.810 --> 01:09:10.180 or if they have numbers they are polynomially bounded, 01:09:10.180 --> 01:09:12.660 like the polynomial situation. 01:09:12.660 --> 01:09:16.160 So non-metric TSP, well, it's not as hard as NPO-complete, 01:09:16.160 --> 01:09:18.021 but it's more in this category. 01:09:18.021 --> 01:09:20.645 AUDIENCE: Is there a notion of strongness, weakness 01:09:20.645 --> 01:09:22.450 in these kind of things? 01:09:22.450 --> 01:09:23.620 PROFESSOR: That's funny. 01:09:23.620 --> 01:09:25.090 This is a stronger result. 01:09:25.090 --> 01:09:26.560 So there's not quite an analog. 01:09:26.560 --> 01:09:29.569 But you can do exponential tricks 01:09:29.569 --> 01:09:33.140 and give yourself a hard time over here. 01:09:33.140 --> 01:09:36.080 And here you're just not allowed to use. 01:09:36.080 --> 01:09:37.760 Everything's polynomial. 01:09:37.760 --> 01:09:41.870 So a three-partition is sort of more in this universe. 01:09:41.870 --> 01:09:45.490 But in this situation, if you sort of have three partitions, 01:09:45.490 --> 01:09:50.410 but with exponential numbers, then you get this harder class. 01:09:50.410 --> 01:09:53.040 So this is not the analog of weak. 01:09:53.040 --> 01:09:57.724 You could maybe imagine-- well, in some sense, 01:09:57.724 --> 01:09:59.390 weak is a modifier in the problem, where 01:09:59.390 --> 01:10:01.139 you say I want to restrict all the numbers 01:10:01.139 --> 01:10:02.700 to a polynomial size. 01:10:02.700 --> 01:10:05.900 So when you do something like three partition, 01:10:05.900 --> 01:10:10.160 it's sort of a weak problem, or it's 01:10:10.160 --> 01:10:12.270 a polynomially bounded problem. 01:10:12.270 --> 01:10:15.850 Strong NP hardness means that that is NP-complete. 01:10:15.850 --> 01:10:19.012 Anyway vague analog, but not quite. 01:10:19.012 --> 01:10:21.470 It's possible some of these, you could add a weak modifier, 01:10:21.470 --> 01:10:24.590 and it would mean something, but I don't know. 01:10:24.590 --> 01:10:25.090 All right. 01:10:25.090 --> 01:10:27.230 So I just want to give you some sample problems 01:10:27.230 --> 01:10:29.290 on both of these sides. 01:10:29.290 --> 01:10:31.930 Maybe let's start with this side, which 01:10:31.930 --> 01:10:35.117 is a little more interesting, because you 01:10:35.117 --> 01:10:36.575 get some kind of familiar problems, 01:10:36.575 --> 01:10:37.533 and they're super hard. 01:10:40.520 --> 01:10:46.150 Minimum independent dominating set. 01:10:46.150 --> 01:10:47.300 We've seen independent set. 01:10:47.300 --> 01:10:48.383 We've seen dominating set. 01:10:48.383 --> 01:10:51.390 Independent set is already hard to approximate. 01:10:51.390 --> 01:10:56.360 But this problem is worse, because even 01:10:56.360 --> 01:10:58.080 finding an independent dominating set 01:10:58.080 --> 01:11:02.020 is NP-complete, whereas finding an independent set, 01:11:02.020 --> 01:11:04.210 I can choose nothing. 01:11:04.210 --> 01:11:06.990 But if I want to simultaneously be dominating an independent, 01:11:06.990 --> 01:11:07.870 that's NP. 01:11:07.870 --> 01:11:09.570 Hard to find any solution. 01:11:09.570 --> 01:11:17.680 In general in NPO PB problems, NPO PB-complete problems, 01:11:17.680 --> 01:11:20.920 it's always NP-complete to find a feasible solution. 01:11:20.920 --> 01:11:22.543 But it's worse than that. 01:11:22.543 --> 01:11:25.210 So the first level would be to find a feasible solution. 01:11:25.210 --> 01:11:26.910 And this is saying on top of that you 01:11:26.910 --> 01:11:28.490 want to minimize the size. 01:11:28.490 --> 01:11:30.276 I think Max would also be hard. 01:11:30.276 --> 01:11:32.080 But I think there's a general theorem, 01:11:32.080 --> 01:11:33.640 that if you're hard in the min case, 01:11:33.640 --> 01:11:35.570 you're also hard in the max case. 01:11:35.570 --> 01:11:38.960 But it depends on the exact set-up. 01:11:38.960 --> 01:11:41.540 So this is sort of an optimization version 01:11:41.540 --> 01:11:44.110 that makes it even harder than NP-complete. 01:11:44.110 --> 01:11:49.330 So I think this is NP-complete, and this is kind of even worse. 01:11:49.330 --> 01:11:52.910 It's sort of stating the stronger thing about when 01:11:52.910 --> 01:11:55.380 you're trying to optimize over a space of solutions, 01:11:55.380 --> 01:11:57.130 that it's NP-complete to decide. 01:11:57.130 --> 01:11:59.320 Notice that's still an NPO problem. 01:11:59.320 --> 01:12:01.490 We define that solutions need to be 01:12:01.490 --> 01:12:03.244 recognizable in polynomial time. 01:12:03.244 --> 01:12:05.035 But we didn't say that you can generate one 01:12:05.035 --> 01:12:06.200 in polynomial time. 01:12:06.200 --> 01:12:09.030 So it could be NP-complete to find a single solution, 01:12:09.030 --> 01:12:09.744 like here. 01:12:09.744 --> 01:12:11.660 All of these problems will have that property. 01:12:15.990 --> 01:12:21.250 Another fun problem is shortest computation. 01:12:21.250 --> 01:12:23.189 This is sort of the most intuitive one 01:12:23.189 --> 01:12:23.980 at a certain level. 01:12:23.980 --> 01:12:25.480 If you know Turing machines, and you 01:12:25.480 --> 01:12:27.396 have a non-deterministic Turing machine, which 01:12:27.396 --> 01:12:29.020 could take non-deterministic branches, 01:12:29.020 --> 01:12:31.630 you want to find the computation in such a machine that 01:12:31.630 --> 01:12:34.720 terminates the earliest using the fewest steps. 01:12:34.720 --> 01:12:39.080 So you might think of that as canonical NPO PB problem. 01:12:39.080 --> 01:12:41.690 There's no numbers in it, but as you can imagine, 01:12:41.690 --> 01:12:44.440 that's super hard to do. 01:12:44.440 --> 01:12:46.840 Here's some more graph theoretic ones. 01:12:46.840 --> 01:12:50.920 Quite natural problems, but super hard. 01:12:50.920 --> 01:12:52.510 Longest induced path. 01:12:52.510 --> 01:12:55.030 Induced means, there are no other edges 01:12:55.030 --> 01:12:57.320 between the chosen vertices. 01:12:57.320 --> 01:13:00.810 So this is sort of longest path is one thing. 01:13:00.810 --> 01:13:03.030 That's quite hard to approximate-- like, I think, 01:13:03.030 --> 01:13:05.070 n to the 1 minus epsilon. 01:13:05.070 --> 01:13:07.070 That's sort of the analog of Hamiltonicity. 01:13:07.070 --> 01:13:09.740 Along this induced path is worse. 01:13:09.740 --> 01:13:12.190 Even finding an induced path of length k, 01:13:12.190 --> 01:13:16.550 finding a feasible solution, finding an induced path 01:13:16.550 --> 01:13:17.150 is hard. 01:13:24.310 --> 01:13:33.749 Another fun one is longest path with forbidden pairs. 01:13:33.749 --> 01:13:35.540 So there are pairs of edges that you're not 01:13:35.540 --> 01:13:38.160 allowed to choose together, and subject to those constraints 01:13:38.160 --> 01:13:40.000 you want to find the longest path. 01:13:40.000 --> 01:13:42.600 So these are all NPO PB complete. 01:13:42.600 --> 01:13:44.437 No numbers in any of them. 01:13:44.437 --> 01:13:46.145 Now let me give you some number problems. 01:13:58.930 --> 01:14:03.230 So Ones was you want to maximize the number of true variables. 01:14:03.230 --> 01:14:05.710 Now we're going to add weights. 01:14:05.710 --> 01:14:09.330 So we want to maximize the sum of the weights 01:14:09.330 --> 01:14:12.370 of the true variables-- and while 01:14:12.370 --> 01:14:15.830 satisfying a Boolean formula. 01:14:15.830 --> 01:14:17.970 So again, finding a feasible solution is hard. 01:14:17.970 --> 01:14:19.800 That's not surprising. 01:14:19.800 --> 01:14:22.440 Here, the weights can be exponential in value, 01:14:22.440 --> 01:14:24.500 because we allow n bits for the weights. 01:14:24.500 --> 01:14:28.450 And that pushes you into NPO completeness. 01:14:28.450 --> 01:14:31.210 If you say the weights have to be polynomially bounded, 01:14:31.210 --> 01:14:33.276 then this problem is NPO PB complete. 01:14:33.276 --> 01:14:34.900 And that's sort of the starting problem 01:14:34.900 --> 01:14:36.820 that they used to prove all of these are hard. 01:14:36.820 --> 01:14:39.420 So they're reductions from this with polynomial weights 01:14:39.420 --> 01:14:40.100 to these guys. 01:14:44.438 --> 01:14:47.330 AUDIENCE: [INAUDIBLE]? 01:14:47.330 --> 01:14:49.220 PROFESSOR: 3SAT. 01:14:49.220 --> 01:14:54.080 I don't know whether you could go down to 2SAT is interesting. 01:14:54.080 --> 01:14:57.960 Here they say, I think, probably 3SAT or CNFSAT. 01:14:57.960 --> 01:15:00.040 Those reductions definitely still work. 01:15:00.040 --> 01:15:03.050 Whether you could put the 2SAT into the Max aspect, 01:15:03.050 --> 01:15:03.630 I don't know. 01:15:03.630 --> 01:15:06.550 But this could be fun to look at. 01:15:06.550 --> 01:15:09.000 There aren't a ton of papers about these two classes, 01:15:09.000 --> 01:15:11.050 but there are a few before they nailed down 01:15:11.050 --> 01:15:12.860 any interesting problems. 01:15:12.860 --> 01:15:14.740 Here's another interesting problem. 01:15:20.600 --> 01:15:24.830 Suppose you want to do integer linear programming. 01:15:24.830 --> 01:15:28.710 To keep it simple, we'll assume that the variables are 01:15:28.710 --> 01:15:33.452 zero or one, and then that is equally hard. 01:15:33.452 --> 01:15:34.910 Here it's a little, unless you know 01:15:34.910 --> 01:15:37.034 a lot about linear programming, it's not so obvious 01:15:37.034 --> 01:15:39.290 that finding a feasible solution here is hard. 01:15:39.290 --> 01:15:41.589 But in general, linear programing-- at least 01:15:41.589 --> 01:15:43.880 in the non-integer case-- you could reduce optimization 01:15:43.880 --> 01:15:45.660 to feasibility. 01:15:45.660 --> 01:15:47.992 So I think the same thing applies here. 01:15:47.992 --> 01:15:49.950 If you're not familiar with linear programming, 01:15:49.950 --> 01:15:53.450 it's basically a bunch of inequality constraints, 01:15:53.450 --> 01:15:55.260 linear inequality constraints. 01:15:55.260 --> 01:15:58.330 And now this is a bunch of integers. 01:15:58.330 --> 01:16:01.840 These are both given integer matrices and vectors. 01:16:01.840 --> 01:16:05.400 And they can have exponential value. 01:16:05.400 --> 01:16:06.310 Question? 01:16:06.310 --> 01:16:08.750 AUDIENCE: For the max/min weighted ones, 01:16:08.750 --> 01:16:12.320 for polynomial bounded, is it still hard 01:16:12.320 --> 01:16:15.460 if you just do ones and minus ones? 01:16:15.460 --> 01:16:19.560 PROFESSOR: I think min or max ones without weights 01:16:19.560 --> 01:16:21.600 is NPO PB-complete. 01:16:21.600 --> 01:16:23.600 I should double-check. 01:16:23.600 --> 01:16:27.100 I didn't actually mention, but this characterization theorem 01:16:27.100 --> 01:16:30.230 works for weighted problems also. 01:16:30.230 --> 01:16:33.540 For every single case, they show that weighted and unweighted 01:16:33.540 --> 01:16:38.640 are the same complexity, except for this one. 01:16:38.640 --> 01:16:42.740 In the min ones case, if all the variables' true, satisfy it, 01:16:42.740 --> 01:16:45.330 you get Poly-APX-completeness if you're unweighted. 01:16:45.330 --> 01:16:50.300 If you're weighted, then you can't find any approximation. 01:16:50.300 --> 01:16:55.390 It's NP-hard to find any factor, which I think, this is, I 01:16:55.390 --> 01:16:58.037 think, before the introduction or popularization 01:16:58.037 --> 01:16:58.745 of these classes. 01:16:58.745 --> 01:17:03.549 So that may be distinguishing between Poly-APX-complete, 01:17:03.549 --> 01:17:05.590 which is definitely smaller than NPO PB-complete. 01:17:05.590 --> 01:17:08.430 This might be NPO PB-completeness. 01:17:08.430 --> 01:17:08.930 Unclear. 01:17:08.930 --> 01:17:12.120 But it's definitely worse than Poly-APX. 01:17:12.120 --> 01:17:13.150 Yeah? 01:17:13.150 --> 01:17:15.150 AUDIENCE: How is it that distinguished from PXP? 01:17:15.150 --> 01:17:17.550 Because I'm just confused how you would ever get anything 01:17:17.550 --> 01:17:19.950 worse than this, because, that's like the biggest 01:17:19.950 --> 01:17:22.370 that you [INAUDIBLE]. 01:17:22.370 --> 01:17:25.470 PROFESSOR: So this problem is exponential APX-hard 01:17:25.470 --> 01:17:26.795 if you forbid zero. 01:17:26.795 --> 01:17:30.030 If you allow zero, then you can't get any approximation. 01:17:30.030 --> 01:17:32.500 Here, I think even when you allow zero, 01:17:32.500 --> 01:17:34.820 or even when you forbid zero, you still 01:17:34.820 --> 01:17:36.000 can't get an approximation. 01:17:36.000 --> 01:17:39.370 I think that's the idea here. 01:17:39.370 --> 01:17:42.020 Here, these problems generally you 01:17:42.020 --> 01:17:44.812 can get, depending on your set-up, 01:17:44.812 --> 01:17:47.395 these problems you can all get like a factor, n approximation. 01:17:49.900 --> 01:17:52.360 Well, maybe not in polynomial time. 01:17:52.360 --> 01:17:54.090 This is hard to find. 01:17:54.090 --> 01:17:55.320 Some of these you can. 01:17:55.320 --> 01:17:58.750 Longest induced path, just have a path of length 1. 01:17:58.750 --> 01:18:00.220 That will be induced. 01:18:00.220 --> 01:18:02.040 So that gives you a factor n approximation. 01:18:02.040 --> 01:18:05.239 There is a lower bound on this situation, 01:18:05.239 --> 01:18:07.030 n to the 1 minus epsilon inapproximability. 01:18:09.640 --> 01:18:12.810 I think morally it should be a factor n, 01:18:12.810 --> 01:18:15.190 but this is the best result I found. 01:18:15.190 --> 01:18:17.290 So it's funny. 01:18:17.290 --> 01:18:19.800 This is only for number problems. 01:18:19.800 --> 01:18:21.520 So I presented this is as in between. 01:18:21.520 --> 01:18:23.832 But this is actually in some sense lower 01:18:23.832 --> 01:18:24.915 than Exp-APX-completeness. 01:18:27.716 --> 01:18:29.465 It's sort of a harder version of Poly-APX. 01:18:32.130 --> 01:18:34.740 This is a slightly harder version of Exp-APX. 01:18:37.320 --> 01:18:39.920 I think it's a small difference, but it's 01:18:39.920 --> 01:18:43.410 good to know there is this difference. 01:18:43.410 --> 01:18:46.160 Other questions? 01:18:46.160 --> 01:18:46.660 All right. 01:18:46.660 --> 01:18:54.460 So this ends what I plan to say about L-reduction-style proofs, 01:18:54.460 --> 01:18:57.562 which are all about preserving approximability. 01:18:57.562 --> 01:18:59.020 The next class, we're going to look 01:18:59.020 --> 01:19:01.980 at a different take on inapproximability, which 01:19:01.980 --> 01:19:06.180 is called gaps, and gap preserving reductions, 01:19:06.180 --> 01:19:08.320 where you can set up a problem that either 01:19:08.320 --> 01:19:10.980 it has a great solution, or the next solution 01:19:10.980 --> 01:19:12.110 below that is way lower. 01:19:12.110 --> 01:19:15.105 And there's a gap between the best and the next to best. 01:19:15.105 --> 01:19:16.480 And whenever you have such a gap, 01:19:16.480 --> 01:19:18.249 you also have an inapproximability gap, 01:19:18.249 --> 01:19:20.290 because you know there's this solution out there, 01:19:20.290 --> 01:19:24.510 but finding it, if it's NP-complete to find this, 01:19:24.510 --> 01:19:27.240 to solve it exactly, and so the next level down you 01:19:27.240 --> 01:19:28.110 lose some factor. 01:19:28.110 --> 01:19:30.600 And whatever that gap is is your inapproximability bound. 01:19:30.600 --> 01:19:33.280 It doesn't give you completeness results like this 01:19:33.280 --> 01:19:35.030 in general-- not always. 01:19:35.030 --> 01:19:37.732 But it tends to give you really get inapproximability bounds. 01:19:37.732 --> 01:19:40.190 Here I've completely ignored what the constant factors are. 01:19:40.190 --> 01:19:42.860 Most of them are not so great. 01:19:42.860 --> 01:19:44.650 Like when you prove APX-hardness, 01:19:44.650 --> 01:19:48.770 usually you get a 1 plus 1 over 1,000 kind of lower bound 01:19:48.770 --> 01:19:50.540 on the possibility factor. 01:19:50.540 --> 01:19:53.750 But the best upper bound is like 2, or 1.5. 01:19:53.750 --> 01:19:55.290 And what we'll talk about next time, 01:19:55.290 --> 01:19:58.290 you can get much closer-- sometimes exact bounds 01:19:58.290 --> 01:20:00.380 between upper and lower. 01:20:00.380 --> 01:20:03.130 But that will be next week.