WEBVTT 00:00:00.000 --> 00:00:01.984 [SQUEAKING] 00:00:01.984 --> 00:00:04.464 [RUSTLING] 00:00:04.464 --> 00:00:08.432 [CLICKING] 00:00:25.230 --> 00:00:26.805 MICHAEL SIPSER: OK, everybody. 00:00:26.805 --> 00:00:27.305 Let's begin. 00:00:30.000 --> 00:00:31.850 Welcome back. 00:00:31.850 --> 00:00:36.110 Good to see you all here on Zoom. 00:00:36.110 --> 00:00:38.420 So we're going to pick up with what 00:00:38.420 --> 00:00:43.410 we had been discussing last week, 00:00:43.410 --> 00:00:50.550 which was an introduction to NP-completeness. 00:00:50.550 --> 00:00:56.010 So we're following on our description of time complexity. 00:00:56.010 --> 00:00:58.670 We started talking about the time 00:00:58.670 --> 00:01:04.160 complexity classes, the class P, the nondeterministic classes, 00:01:04.160 --> 00:01:07.130 the class NP, P versus NP problem, and then 00:01:07.130 --> 00:01:10.193 leading to this discussion of NP-completeness. 00:01:10.193 --> 00:01:12.110 And today we're going to prove the big theorem 00:01:12.110 --> 00:01:15.890 in the field, which really kind of got things going back 00:01:15.890 --> 00:01:19.430 in the early 1970s by Cook and Levin 00:01:19.430 --> 00:01:22.850 that there was actually an NP-complete problem, 00:01:22.850 --> 00:01:25.550 that SAT in particular is an NP-complete problem. 00:01:25.550 --> 00:01:29.060 And then we'll also talk about 3SAT, which is a useful tool. 00:01:32.000 --> 00:01:37.760 Just to remember, we had this notion of NP-completeness. 00:01:37.760 --> 00:01:40.910 A language is NP-complete if it's in NP. 00:01:40.910 --> 00:01:43.670 And everything else in NP is polynomial time reducible 00:01:43.670 --> 00:01:45.420 to it. 00:01:45.420 --> 00:01:48.630 And if an NP-complete problem turns out 00:01:48.630 --> 00:01:50.730 to have a polynomial time solution, 00:01:50.730 --> 00:01:54.420 then every NP-problem has a polynomial time solution. 00:01:54.420 --> 00:01:57.420 And that's part of the importance of NP-completeness, 00:01:57.420 --> 00:02:04.620 because since we consider it unlikely that P equals NP, 00:02:04.620 --> 00:02:09.539 and that there probably are some problems that are an NP, 00:02:09.539 --> 00:02:12.660 but are not solvable in polynomial time. 00:02:12.660 --> 00:02:16.110 That would imply that an NP-complete problem 00:02:16.110 --> 00:02:17.680 would have that property. 00:02:17.680 --> 00:02:21.450 And so proving a problem being NP-complete 00:02:21.450 --> 00:02:23.610 is evidence, very strong evidence, 00:02:23.610 --> 00:02:27.010 that it doesn't have a polynomial time solution. 00:02:27.010 --> 00:02:29.140 And so therefore it's called intractable. 00:02:29.140 --> 00:02:30.390 It's a very difficult problem. 00:02:33.590 --> 00:02:39.280 So the way we are going to typically show problems 00:02:39.280 --> 00:02:42.910 NP-complete is by reducing a known-- a previously 00:02:42.910 --> 00:02:46.390 known NP-complete problem to that problem. 00:02:46.390 --> 00:02:50.500 Often it's 3SAT, as we've seen in several examples already, 00:02:50.500 --> 00:02:53.030 or it could be some other example. 00:02:53.030 --> 00:02:57.760 So let's just survey briefly the things that we've already-- 00:02:57.760 --> 00:03:01.940 languages that we've already seen, which are NP-complete. 00:03:01.940 --> 00:03:04.480 So we have the languages SAT, which 00:03:04.480 --> 00:03:07.220 has a direct reduction from every NP problem. 00:03:07.220 --> 00:03:17.630 And so we're going to show that today, that every NP 00:03:17.630 --> 00:03:20.270 language is polynomial time reducible to SAT, which 00:03:20.270 --> 00:03:23.790 is, in turn, reducible to 3SAT. 00:03:23.790 --> 00:03:28.730 And we showed previously that 3SAT 00:03:28.730 --> 00:03:30.530 is reducible to CLIQUE and HAMPATH. 00:03:33.460 --> 00:03:37.330 And in recitation, if you went to that, 00:03:37.330 --> 00:03:40.690 they showed that the SUBSET-SUM problem and the undirected 00:03:40.690 --> 00:03:45.670 HAMPATH problems are also reducible from previously 00:03:45.670 --> 00:03:46.480 shown-- 00:03:46.480 --> 00:03:50.680 well, either from 3SAT, or in the case 00:03:50.680 --> 00:03:52.240 of the undirected HAMPATH problem, 00:03:52.240 --> 00:03:55.630 it's reducible from the HAMPATH problem. 00:03:55.630 --> 00:04:03.190 And the conclusion, once we have these two blue reducibilities 00:04:03.190 --> 00:04:07.450 shown, then we know that all of these problems are NP-complete. 00:04:10.150 --> 00:04:12.810 So the class NP basically breaks down 00:04:12.810 --> 00:04:16.740 into the NP-complete problems, P problems, problems 00:04:16.740 --> 00:04:19.500 that are in P, and then there might be problems 00:04:19.500 --> 00:04:20.605 in between as well. 00:04:20.605 --> 00:04:22.230 So there are some problems that are not 00:04:22.230 --> 00:04:23.880 known to be in either category. 00:04:23.880 --> 00:04:26.070 And in fact, there's an old theorem 00:04:26.070 --> 00:04:28.560 which shows that if P is different from NP, 00:04:28.560 --> 00:04:30.750 then there are actually problems that 00:04:30.750 --> 00:04:32.800 are in the intermediate state. 00:04:32.800 --> 00:04:38.530 And of course, it's possible that P equals NP, 00:04:38.530 --> 00:04:40.660 and then everything collapses down 00:04:40.660 --> 00:04:44.230 to be the same with the tiny exception of the sigma 00:04:44.230 --> 00:04:47.380 star and empty set languages, which can never be complete. 00:04:50.620 --> 00:04:51.850 OK. 00:04:51.850 --> 00:04:54.220 So that's our quick review. 00:04:54.220 --> 00:04:55.960 Here is a check-in that I'm going 00:04:55.960 --> 00:04:59.925 to use to get us ready for the big proof 00:04:59.925 --> 00:05:02.050 that we're going to be spending most of the lecture 00:05:02.050 --> 00:05:06.010 on about showing SAT is NP-complete, 00:05:06.010 --> 00:05:07.930 but just to define a little notation, which 00:05:07.930 --> 00:05:10.990 you might have-- maybe you've seen already. 00:05:10.990 --> 00:05:14.450 But I'm going to do it in the form of a check-in. 00:05:14.450 --> 00:05:18.550 So I'm sure you've all seen the big sum 00:05:18.550 --> 00:05:23.440 notation using a big sigma to represent a sum 00:05:23.440 --> 00:05:29.410 over some set of possibilities. 00:05:29.410 --> 00:05:32.170 Just as there is the big sigma notation, we have-- 00:05:32.170 --> 00:05:35.110 you can have other operations that 00:05:35.110 --> 00:05:39.080 apply to a set of elements. 00:05:39.080 --> 00:05:42.550 So in this case, we're going to be seeing the big end AND 00:05:42.550 --> 00:05:44.920 and the big OR operation-- 00:05:44.920 --> 00:05:47.380 it's going to-- notation, which is 00:05:47.380 --> 00:05:52.540 going to allow us to talk about taking the AND of many things 00:05:52.540 --> 00:05:55.420 or the OR of many things, because we're 00:05:55.420 --> 00:05:57.490 going to be building these Boolean formulas. 00:05:57.490 --> 00:06:00.910 And so ANDs and ORs are going to be the operations that we're 00:06:00.910 --> 00:06:02.510 going to be focusing on. 00:06:02.510 --> 00:06:05.950 And so just as an example just to make sure 00:06:05.950 --> 00:06:08.510 we're all understanding this notation, 00:06:08.510 --> 00:06:15.550 if you have two strings x and y written out in terms 00:06:15.550 --> 00:06:16.723 of their individual symbols. 00:06:16.723 --> 00:06:17.890 So they're both of length n. 00:06:17.890 --> 00:06:19.630 So x is x1 to xn. 00:06:19.630 --> 00:06:22.150 Y is y1 to yn. 00:06:22.150 --> 00:06:26.120 And now I write the following expression. 00:06:26.120 --> 00:06:32.260 The big AND for i ranging between 1 and n of xi 00:06:32.260 --> 00:06:33.670 equal to yi. 00:06:33.670 --> 00:06:37.810 If that big AND is true, what does it tell us about x and y? 00:06:37.810 --> 00:06:40.300 And I'm just going to offer you two possibilities, 00:06:40.300 --> 00:06:44.980 that either x and y agree on some symbol or are equal, 00:06:44.980 --> 00:06:47.590 namely that they agree on every symbol. 00:06:47.590 --> 00:06:52.300 And so let's just pull that up as a quick poll 00:06:52.300 --> 00:06:53.395 to get us going here. 00:06:56.393 --> 00:06:58.560 I just want to make sure you understand the notation 00:06:58.560 --> 00:07:00.630 because we're going to be using that a lot 00:07:00.630 --> 00:07:08.640 in describing the polynomial time reduction from languages 00:07:08.640 --> 00:07:11.920 in NP to the SAT language. 00:07:11.920 --> 00:07:12.420 OK. 00:07:12.420 --> 00:07:15.930 I think most of you have got the idea. 00:07:15.930 --> 00:07:17.310 So let's finish this up quick. 00:07:22.140 --> 00:07:24.870 So another 15 seconds, please, just 00:07:24.870 --> 00:07:26.430 to give you a chance to participate. 00:07:31.680 --> 00:07:32.670 OK. 00:07:32.670 --> 00:07:36.270 We're going to close this poll. 00:07:36.270 --> 00:07:38.430 Last chance, last call. 00:07:38.430 --> 00:07:39.660 All right. 00:07:39.660 --> 00:07:43.470 So yes, the big AND-- 00:07:43.470 --> 00:07:50.160 as most of you can see, the big AND says that x1 equals y1, 00:07:50.160 --> 00:07:57.050 and x2 equals y2, and x3 equals y3 so they're all-- 00:07:57.050 --> 00:08:00.860 every symbol in x is equal to the corresponding symbol in y. 00:08:00.860 --> 00:08:04.220 If we had a big OR instead of a big AND, 00:08:04.220 --> 00:08:05.868 then A would be the correct answer, 00:08:05.868 --> 00:08:07.910 because then just there would be some place where 00:08:07.910 --> 00:08:11.120 they agree instead of every place where they agree. 00:08:14.090 --> 00:08:20.795 So let's then start to launch into this big theorem. 00:08:23.470 --> 00:08:27.880 The proof itself is a mass of details 00:08:27.880 --> 00:08:30.020 with one underlying idea. 00:08:30.020 --> 00:08:33.159 And in fact, it's an idea we've seen before. 00:08:33.159 --> 00:08:35.500 But let's-- before we get ahead of ourselves, 00:08:35.500 --> 00:08:38.809 let's just understand what we're trying to do. 00:08:38.809 --> 00:08:43.900 So we want to show that this language SAT is NP-complete. 00:08:43.900 --> 00:08:45.280 We remember what SAT is. 00:08:45.280 --> 00:08:47.860 It's Boolean formulas that are satisfiable. 00:08:50.500 --> 00:08:54.090 So we already showed that the SAT prob-- so being NP-complete 00:08:54.090 --> 00:08:56.430 means it has these two features. 00:08:56.430 --> 00:09:00.400 It's in NP, and everything in NP is reducible to it. 00:09:00.400 --> 00:09:03.840 So first of all, SAT is in NP, as we've already seen. 00:09:03.840 --> 00:09:07.230 The witness that shows the formula is satisfiable 00:09:07.230 --> 00:09:09.450 is simply the satisfying assignment 00:09:09.450 --> 00:09:12.220 that evaluates to true. 00:09:12.220 --> 00:09:16.650 So now we're going to pick some language in NP, A, 00:09:16.650 --> 00:09:20.490 and show that A is polynomial time reducible to SAT. 00:09:23.350 --> 00:09:25.990 So this is going to apply to any language A in NP. 00:09:25.990 --> 00:09:28.480 So let A be some language in NP. 00:09:28.480 --> 00:09:31.030 It's decided by some nondeterministic Turing machine 00:09:31.030 --> 00:09:33.900 M in time n to the k. 00:09:33.900 --> 00:09:36.810 That's what it means to be an NP. 00:09:36.810 --> 00:09:39.300 I'm going to be ignoring the constant factors when 00:09:39.300 --> 00:09:41.400 we could carry that throughout the proof. 00:09:41.400 --> 00:09:44.220 It just would make the description 00:09:44.220 --> 00:09:46.470 a little more cumbersome and wouldn't 00:09:46.470 --> 00:09:47.640 change any of the ideas. 00:09:47.640 --> 00:09:51.660 So let's just say it's M runs in time n to the k, 00:09:51.660 --> 00:09:57.660 and recognizes or decides this language in A. 00:09:57.660 --> 00:10:02.940 So it's a nondeterministic machine. 00:10:02.940 --> 00:10:04.980 So we've got to give a polynomial time 00:10:04.980 --> 00:10:06.930 reduction from A to SAT. 00:10:06.930 --> 00:10:10.930 That's what I have to demonstrate to you. 00:10:10.930 --> 00:10:12.960 So what does that reduction look like? 00:10:12.960 --> 00:10:16.260 It's going to map strings, which may or may not 00:10:16.260 --> 00:10:21.730 be in A, to formulas, which may or may not be satisfied. 00:10:21.730 --> 00:10:24.460 That's what the reduction has got to do. 00:10:24.460 --> 00:10:31.720 So saying that out more formally is it's 00:10:31.720 --> 00:10:34.030 going to take some string w and map it 00:10:34.030 --> 00:10:36.910 to a formula we'll call phi sub M, w, 00:10:36.910 --> 00:10:45.770 where M is the machine that decides A and w is the input. 00:10:45.770 --> 00:10:49.910 So f is going to map w to this formula, where 00:10:49.910 --> 00:10:53.300 the string is in the language exactly when the formula is 00:10:53.300 --> 00:10:54.930 satisfied. 00:10:54.930 --> 00:10:57.480 And the formula is going to depend upon M and w. 00:10:57.480 --> 00:11:00.310 That's why it just has-- it's written in this way. 00:11:00.310 --> 00:11:02.460 So we've seen that kind of thing before. 00:11:02.460 --> 00:11:05.760 So basically, my job-- 00:11:05.760 --> 00:11:09.330 we have this language A that's in NP. 00:11:09.330 --> 00:11:13.650 And now we have some string w, which might be in A, 00:11:13.650 --> 00:11:19.510 or maybe not is in A. And I have to quickly in polynomial time 00:11:19.510 --> 00:11:24.470 now produce a formula which is going to be satisfiable exactly 00:11:24.470 --> 00:11:30.640 when the string w is in A. 00:11:30.640 --> 00:11:33.270 So how is that formula going to work? 00:11:33.270 --> 00:11:35.350 How can I produce such a formula? 00:11:35.350 --> 00:11:39.190 Which I mean, of course, I don't know whether w is in A or not, 00:11:39.190 --> 00:11:42.400 because I'm just the polynomial time reduction. 00:11:42.400 --> 00:11:45.460 And A is an NP language. 00:11:45.460 --> 00:11:47.800 So polynomial time probably is not 00:11:47.800 --> 00:11:50.740 enough to solve whether w is in A. 00:11:50.740 --> 00:11:57.800 So I've got to do that mapping without knowing the answer. 00:11:57.800 --> 00:11:59.870 And the idea is that-- 00:11:59.870 --> 00:12:03.290 and this is where we've seen things like this before-- 00:12:03.290 --> 00:12:05.780 the formula we're going to construct 00:12:05.780 --> 00:12:08.615 simulates the machine on w. 00:12:11.870 --> 00:12:14.540 So in some way it's going to do that simulation. 00:12:14.540 --> 00:12:18.530 The trick is figuring out how to-- what that means. 00:12:18.530 --> 00:12:22.910 But the interpretation of that formula 00:12:22.910 --> 00:12:28.010 is that in a sense describes, says, in my informal language, 00:12:28.010 --> 00:12:31.740 that M accepts w. 00:12:31.740 --> 00:12:38.820 Very much like-- there's a lot of parallel here between this 00:12:38.820 --> 00:12:41.970 construction and, for example, the construction, 00:12:41.970 --> 00:12:46.590 the PCP construction, where we made an instant-- 00:12:46.590 --> 00:12:51.600 given a machine and an input, we made 00:12:51.600 --> 00:12:54.450 a set of those dominoes, where finding a match 00:12:54.450 --> 00:12:56.930 forced you to simulate the machine. 00:12:56.930 --> 00:12:59.933 Here finding a satisfying assignment 00:12:59.933 --> 00:13:01.850 is going to force you to simulate the machine. 00:13:05.330 --> 00:13:07.340 So the satisfying assignment is going 00:13:07.340 --> 00:13:11.780 to be a computation history for M on w. 00:13:11.780 --> 00:13:13.730 And I'm just going to write that computation 00:13:13.730 --> 00:13:15.813 history in a particular way. 00:13:15.813 --> 00:13:18.230 So it's going to have a somewhat different encoding, which 00:13:18.230 --> 00:13:21.380 is going to help us to visualize what's going on better. 00:13:24.080 --> 00:13:28.210 So that's the approach. 00:13:28.210 --> 00:13:29.960 That's the basic idea of what we're trying 00:13:29.960 --> 00:13:31.127 to-- what we're going to do. 00:13:31.127 --> 00:13:33.920 So I'm happy to take a minute if you have 00:13:33.920 --> 00:13:35.780 any questions about this part. 00:13:35.780 --> 00:13:39.890 But otherwise, I'll just move on to start 00:13:39.890 --> 00:13:43.160 doing the actual construction of what the formula looks like. 00:13:43.160 --> 00:13:44.340 How is it going to do that? 00:13:44.340 --> 00:13:47.190 How is that formula going to work? 00:13:47.190 --> 00:13:47.690 OK. 00:13:47.690 --> 00:13:48.950 So no questions. 00:13:48.950 --> 00:13:54.530 Hopefully, we are all together. 00:13:54.530 --> 00:14:00.820 So first of all, let me describe what my computation 00:14:00.820 --> 00:14:02.620 history is going to look like. 00:14:02.620 --> 00:14:04.420 And the situation is a little bit 00:14:04.420 --> 00:14:08.752 different than what we had before because when 00:14:08.752 --> 00:14:10.960 we were talking about the Post correspondence problem 00:14:10.960 --> 00:14:13.180 we had a deterministic machine. 00:14:13.180 --> 00:14:17.240 And now our machine is nondeterministic. 00:14:17.240 --> 00:14:21.800 So we're going to call the object-- 00:14:21.800 --> 00:14:24.920 instead of an accepting computation or a computation 00:14:24.920 --> 00:14:26.420 history, we're just going to-- we're 00:14:26.420 --> 00:14:29.390 going to call it a tableau, or sometimes an accepting tableau 00:14:29.390 --> 00:14:32.720 if you want to emphasize the accepting nature of it. 00:14:32.720 --> 00:14:35.900 But generally, we're just going to call it a tableau. 00:14:35.900 --> 00:14:40.970 So a tableau is really an accepting computation history 00:14:40.970 --> 00:14:48.520 for the machine on an accepting branch 00:14:48.520 --> 00:14:52.380 of its nondeterministic computation. 00:14:52.380 --> 00:14:59.600 So if M accepts w, it's got to have some accepting branch, 00:14:59.600 --> 00:15:04.790 and the tableau is going to be the sequence of configurations 00:15:04.790 --> 00:15:08.060 that the machine goes through on that accepting branch. 00:15:08.060 --> 00:15:09.920 If there are several accepting branches, 00:15:09.920 --> 00:15:11.780 there may be several tableaus. 00:15:11.780 --> 00:15:13.490 There will be several tableaus. 00:15:13.490 --> 00:15:15.770 So there's going to be one tableau 00:15:15.770 --> 00:15:21.900 for each accepting branch of the machine's computation on w. 00:15:21.900 --> 00:15:26.550 If the machine does not accept w, there won't be any tableaus. 00:15:26.550 --> 00:15:29.490 And so the whole point is that we're 00:15:29.490 --> 00:15:35.780 going to make our formula represent the statement 00:15:35.780 --> 00:15:39.370 that there is a tableau. 00:15:39.370 --> 00:15:41.350 And satisfying that formula is going 00:15:41.350 --> 00:15:46.750 to correspond to filling out the symbols in the tableau to make 00:15:46.750 --> 00:15:48.910 it a tableau. 00:15:48.910 --> 00:15:51.550 So here is a tableau. 00:15:51.550 --> 00:15:55.960 So a tableau is just, again, an accepting computation history 00:15:55.960 --> 00:15:59.440 on some branch, some accepting branch of the machine's 00:15:59.440 --> 00:16:02.290 computation. 00:16:02.290 --> 00:16:03.640 The rows are the-- 00:16:03.640 --> 00:16:07.090 instead of writing the computation history 00:16:07.090 --> 00:16:11.170 out linearly, we're going to represent it in a table form 00:16:11.170 --> 00:16:16.790 where each configuration is going to be on a separate row. 00:16:16.790 --> 00:16:19.220 Now, the dimensions of that table 00:16:19.220 --> 00:16:23.080 are going to be n to the k by n to the k, 00:16:23.080 --> 00:16:30.500 because the machine runs for n to the k steps. 00:16:30.500 --> 00:16:33.700 So there's going to-- if there's an accepting branch, 00:16:33.700 --> 00:16:36.610 it's going to accept within that number of steps. 00:16:36.610 --> 00:16:39.370 And we'll have enough rows here to write down 00:16:39.370 --> 00:16:42.400 all of the configurations that the machine goes through one 00:16:42.400 --> 00:16:45.460 after the next, row by row, each one having 00:16:45.460 --> 00:16:49.020 a configuration in it. 00:16:49.020 --> 00:16:52.690 And then at the bottom, there'll be an accept. 00:16:52.690 --> 00:16:55.990 Minor detail, if the machine accepts earlier, 00:16:55.990 --> 00:16:58.000 we'll just say the machine stays in the-- 00:16:58.000 --> 00:17:01.300 once it enters an accept, the machine 00:17:01.300 --> 00:17:03.470 does not change from that point on. 00:17:03.470 --> 00:17:06.099 So the rule of the machine is nothing changes. 00:17:06.099 --> 00:17:08.829 And it just remains in the same configuration from that point 00:17:08.829 --> 00:17:09.329 on. 00:17:13.470 --> 00:17:19.140 So important to understand what we're-- if you're not following 00:17:19.140 --> 00:17:23.800 what I mean by a tableau, you're doomed for this lecture. 00:17:23.800 --> 00:17:29.430 So it makes sense for you to ask a question 00:17:29.430 --> 00:17:33.810 to understand what we mean by tableau. 00:17:33.810 --> 00:17:36.220 So just a few more elements here. 00:17:36.220 --> 00:17:38.953 So this is going to be the start configuration for M on w. 00:17:38.953 --> 00:17:40.620 This would be an accepting configuration 00:17:40.620 --> 00:17:44.370 here down at the very last row. 00:17:44.370 --> 00:17:46.500 And you might imagine-- 00:17:46.500 --> 00:17:53.270 I think I've filled out some hypothetical first step 00:17:53.270 --> 00:17:58.700 of the machine after the start where maybe the machine was-- 00:17:58.700 --> 00:18:03.060 when it's in-- remember how we encoded our configurations. 00:18:03.060 --> 00:18:05.990 So this is the machine is in state q0, 00:18:05.990 --> 00:18:09.890 looking at the w, the first symbol of the input, w1. 00:18:09.890 --> 00:18:14.510 And maybe when it's in q0 looking at that first symbol, 00:18:14.510 --> 00:18:19.340 it moves to state q7, and goes right, and then 00:18:19.340 --> 00:18:24.960 changes that w1 to an A. And so now here is the head shown 00:18:24.960 --> 00:18:28.680 moved one position to the right in the new state q7. 00:18:28.680 --> 00:18:30.570 I mean, of course, that depends on what 00:18:30.570 --> 00:18:33.730 the machine is designed to do, what the transition function 00:18:33.730 --> 00:18:34.230 is. 00:18:34.230 --> 00:18:39.430 But this possibly is what happens. 00:18:39.430 --> 00:18:40.680 So does the-- 00:18:40.680 --> 00:18:41.640 OK, so good. 00:18:41.640 --> 00:18:44.820 Does the tableau trace all steps of all branches? 00:18:44.820 --> 00:18:45.990 No. 00:18:45.990 --> 00:18:51.660 The tableau corresponds to one accepting branch. 00:18:51.660 --> 00:18:53.610 Each different accepting branch is 00:18:53.610 --> 00:18:54.960 going to have its own tableau. 00:18:54.960 --> 00:18:57.420 So there might be several different ways of filling out 00:18:57.420 --> 00:18:58.440 that second row even. 00:18:58.440 --> 00:19:01.470 Of course, the first row is going to be-- in any tableau, 00:19:01.470 --> 00:19:04.020 it's got to be the same. 00:19:04.020 --> 00:19:06.780 Once I know w-- 00:19:06.780 --> 00:19:08.100 here I've written down-- 00:19:08.100 --> 00:19:10.170 maybe I should unpack this for you. 00:19:10.170 --> 00:19:13.410 It's in the start state here, then here are the first-- 00:19:13.410 --> 00:19:16.680 here are n symbols of w, w's of length n. 00:19:16.680 --> 00:19:19.230 So there's w1, w2, up to wn. 00:19:19.230 --> 00:19:23.180 And then I'm padding out the rest with blanks. 00:19:23.180 --> 00:19:24.770 So I should have said that, too. 00:19:24.770 --> 00:19:26.300 But OK. 00:19:26.300 --> 00:19:29.390 So I want to make my table n to the k by n to the k 00:19:29.390 --> 00:19:32.870 because that's going to be enough to represent 00:19:32.870 --> 00:19:35.600 the entire-- 00:19:35.600 --> 00:19:41.750 all of these configurations of M running for most n to the k 00:19:41.750 --> 00:19:47.660 steps, because the machine, even if it sends its head 00:19:47.660 --> 00:19:51.480 moving to the right as fast as it possibly can, 00:19:51.480 --> 00:19:53.540 it's never going to have-- 00:19:53.540 --> 00:19:59.100 it's never going to go outside this box 00:19:59.100 --> 00:20:01.120 if it only runs for n to the k steps. 00:20:01.120 --> 00:20:02.520 So this is going to be big enough 00:20:02.520 --> 00:20:05.385 to represent the entire computation of M. 00:20:05.385 --> 00:20:08.460 So it means running for n to the k time. 00:20:08.460 --> 00:20:13.010 And the input w is of length n. 00:20:13.010 --> 00:20:14.510 All right. 00:20:14.510 --> 00:20:15.170 What is k? 00:20:15.170 --> 00:20:17.360 So k is the running time of the machine. 00:20:17.360 --> 00:20:21.720 So we assumed from the previous slide that M runs in time n 00:20:21.720 --> 00:20:22.220 to the k. 00:20:25.365 --> 00:20:25.865 OK. 00:20:31.700 --> 00:20:32.930 So a good question here. 00:20:32.930 --> 00:20:35.150 How can the tableau be a square table 00:20:35.150 --> 00:20:37.550 if we have a low number of computation histories, 00:20:37.550 --> 00:20:40.590 but a lot of tape? 00:20:40.590 --> 00:20:44.250 Well, the low number computation history, that's a problem. 00:20:44.250 --> 00:20:45.690 You're not saying that well. 00:20:45.690 --> 00:20:49.500 You know, each computation history 00:20:49.500 --> 00:20:51.700 is in a different table. 00:20:51.700 --> 00:20:54.150 So this, I'm representing a single computation history 00:20:54.150 --> 00:20:54.700 here. 00:20:54.700 --> 00:20:59.250 There may be-- there are going to be many configurations. 00:20:59.250 --> 00:21:01.640 So I can't have a small number of configurations 00:21:01.640 --> 00:21:05.330 using a lot of tape, because I can only 00:21:05.330 --> 00:21:09.020 use one more cell, one additional tape cell each time 00:21:09.020 --> 00:21:10.295 I-- 00:21:10.295 --> 00:21:11.795 each additional step of the machine. 00:21:14.360 --> 00:21:14.900 Yeah. 00:21:14.900 --> 00:21:16.970 So there's really no difference between this. 00:21:16.970 --> 00:21:19.220 If you want to think of this as a computation history, 00:21:19.220 --> 00:21:20.120 that's fine. 00:21:20.120 --> 00:21:22.160 This is really just the standard terminology 00:21:22.160 --> 00:21:26.957 that's used when you're proving this particular theorem. 00:21:26.957 --> 00:21:28.790 Typically people talk about it as a tableau, 00:21:28.790 --> 00:21:30.582 but it's really just a computation history. 00:21:35.450 --> 00:21:36.140 The q state-- 00:21:36.140 --> 00:21:39.830 I mean, the question about what are these states' q, 00:21:39.830 --> 00:21:42.680 this is the way we represent configurations. 00:21:42.680 --> 00:21:46.803 So this means the machine is in a state. 00:21:46.803 --> 00:21:48.470 They're not all going down the diagonal. 00:21:48.470 --> 00:21:52.520 The states are going to zigzag here through this picture 00:21:52.520 --> 00:21:56.290 here, depending upon how the head of the machine moves. 00:21:56.290 --> 00:21:57.700 So you have to go back and review 00:21:57.700 --> 00:22:00.370 how we represent the configurations of the machine. 00:22:00.370 --> 00:22:03.610 Remember the configuration is a snapshot of the machine 00:22:03.610 --> 00:22:04.420 at a given point. 00:22:07.703 --> 00:22:09.620 How do we know that M runs in polynomial time? 00:22:09.620 --> 00:22:11.720 We're assuming M runs in polynomial. 00:22:11.720 --> 00:22:16.040 We started off with a language that's in NP. 00:22:16.040 --> 00:22:19.790 So it's a nondeterministic polynomial time machine, 00:22:19.790 --> 00:22:23.570 and we're picking one branch that accepts 00:22:23.570 --> 00:22:25.708 and writing down all the sequence of configurations 00:22:25.708 --> 00:22:26.750 the machine goes through. 00:22:26.750 --> 00:22:27.770 Let's move on. 00:22:27.770 --> 00:22:31.730 And maybe ask more questions as they come to you, 00:22:31.730 --> 00:22:34.695 and I'll pick out some to answer if that's 00:22:34.695 --> 00:22:35.945 going to be helpful to others. 00:22:41.490 --> 00:22:51.100 So we're going to now construct this formula 00:22:51.100 --> 00:22:52.750 to say that M accepts w. 00:22:52.750 --> 00:22:54.294 Again, that was what we-- 00:22:54.294 --> 00:22:55.990 that's our goal. 00:22:55.990 --> 00:22:58.855 And it says that a tableau for M on w exists. 00:23:02.730 --> 00:23:04.740 And basically what that means is we 00:23:04.740 --> 00:23:08.040 want to say that it starts right, it ends right, 00:23:08.040 --> 00:23:11.130 and everything in between is right, 00:23:11.130 --> 00:23:13.410 and then we're going to need some other stuff to talk 00:23:13.410 --> 00:23:16.620 about how we're going to be encoding those symbols using 00:23:16.620 --> 00:23:18.250 Boolean variables. 00:23:18.250 --> 00:23:20.290 So those are going to be the four parts. 00:23:20.290 --> 00:23:21.040 Here's the start. 00:23:21.040 --> 00:23:22.050 It starts right. 00:23:22.050 --> 00:23:23.340 Here it ends right. 00:23:23.340 --> 00:23:24.660 Here it moves right. 00:23:24.660 --> 00:23:27.390 And here it talks about the encoding of the symbols 00:23:27.390 --> 00:23:30.820 into Boolean variables. 00:23:30.820 --> 00:23:33.900 So those are the four parts of this formula 00:23:33.900 --> 00:23:35.025 that I'm going to describe. 00:23:38.710 --> 00:23:40.960 I hope you got your question answered 00:23:40.960 --> 00:23:43.840 as to why the total number of columns is n to the k, 00:23:43.840 --> 00:23:47.110 because it's just big enough to fit the entire-- 00:23:47.110 --> 00:23:49.300 all configurations that the machine 00:23:49.300 --> 00:23:51.130 is running for at most n to the k 00:23:51.130 --> 00:23:53.420 steps, which is what we're assuming. 00:23:56.970 --> 00:23:57.470 OK. 00:23:57.470 --> 00:24:00.410 So now just getting back to that, 00:24:00.410 --> 00:24:03.398 I'm going to describe these different components now 00:24:03.398 --> 00:24:04.190 on separate slides. 00:24:08.030 --> 00:24:11.390 Let me start out with this component phi cell, which 00:24:11.390 --> 00:24:13.077 is sort of the most fundamental one 00:24:13.077 --> 00:24:14.660 because it talks about how we're going 00:24:14.660 --> 00:24:20.660 to be encoding those symbols of the tableau into the Boolean 00:24:20.660 --> 00:24:21.380 variables. 00:24:25.070 --> 00:24:28.130 So again, here's kind of the picture to have in mind of this 00:24:28.130 --> 00:24:32.080 tableau, this n to the k by n to the k table, 00:24:32.080 --> 00:24:38.680 representing some accepting branch of the machine's 00:24:38.680 --> 00:24:42.830 computation if there is one. 00:24:42.830 --> 00:24:48.950 And so now let me draw one of the cells here. 00:24:48.950 --> 00:24:51.360 I'm going to magnify it. 00:24:51.360 --> 00:24:55.800 So this is the i, j cell here. 00:24:55.800 --> 00:24:58.040 And I'm going to-- 00:24:58.040 --> 00:25:02.520 there are going to be a collection 00:25:02.520 --> 00:25:07.325 of Boolean variables associated with each one of the cells. 00:25:07.325 --> 00:25:08.700 So each one of the cells is going 00:25:08.700 --> 00:25:13.080 to have a bunch of variables all to itself. 00:25:13.080 --> 00:25:16.010 And those are going to be basically indicator variables. 00:25:16.010 --> 00:25:18.950 They're going to indicate which symbol 00:25:18.950 --> 00:25:24.110 that cell gets to have in it. 00:25:24.110 --> 00:25:27.920 So again, picture here-- 00:25:27.920 --> 00:25:29.992 in this tableau, we don't know how 00:25:29.992 --> 00:25:31.200 it's going to get filled out. 00:25:31.200 --> 00:25:34.850 But however it gets filled out, each one of these cells 00:25:34.850 --> 00:25:36.125 gets some symbol. 00:25:38.622 --> 00:25:40.580 And that symbol could either be a tape alphabet 00:25:40.580 --> 00:25:44.530 symbol, or a symbol representing a state. 00:25:44.530 --> 00:25:46.730 That's the way we do our configurations. 00:25:46.730 --> 00:25:48.400 So it could be a tape alphabet symbol 00:25:48.400 --> 00:25:53.080 here showing the magnification. 00:25:53.080 --> 00:25:54.830 Maybe it's the tape alphabet symbol, which 00:25:54.830 --> 00:25:57.470 represents the blank symbol. 00:25:57.470 --> 00:26:00.440 Or maybe it's represents some state. 00:26:03.500 --> 00:26:07.200 Now, how am I going to encode that with variables? 00:26:07.200 --> 00:26:10.390 So this-- let me-- 00:26:10.390 --> 00:26:12.280 this is the collection of variables 00:26:12.280 --> 00:26:16.661 that's going to apply to the entire formula phi sub M, w. 00:26:20.270 --> 00:26:24.050 Each cell is-- so each cell i, j is 00:26:24.050 --> 00:26:27.620 going to have a set of variables, 00:26:27.620 --> 00:26:33.500 one for each possible symbol sigma 00:26:33.500 --> 00:26:39.080 that's in the configuration alphabet, namely a tape 00:26:39.080 --> 00:26:41.450 symbol or a state symbol. 00:26:47.450 --> 00:26:50.080 So I'm going to have-- 00:26:50.080 --> 00:26:52.940 well, maybe this will become clear as I'm writing it. 00:26:52.940 --> 00:27:01.940 So if I turn the variable x i, j, sigma equal to true, 00:27:01.940 --> 00:27:06.550 that's just a way of saying that cell i, j contains a sigma. 00:27:06.550 --> 00:27:12.460 So if I have x i, j, a, that means the symbol contains an a. 00:27:12.460 --> 00:27:15.110 So I'm going to illustrate that now for you. 00:27:15.110 --> 00:27:19.390 So imagine you have lights representing all 00:27:19.390 --> 00:27:25.765 the different x i, j's for the different sigmas. 00:27:25.765 --> 00:27:26.640 Didn't say that well. 00:27:26.640 --> 00:27:30.120 So we're in cell i, j. 00:27:30.120 --> 00:27:32.520 So these are all x i, j variables. 00:27:32.520 --> 00:27:34.470 They're all x i, j sigma variables 00:27:34.470 --> 00:27:40.900 for the different sigmas that can go in this cell. 00:27:40.900 --> 00:27:42.780 So all different possible sigmas. 00:27:42.780 --> 00:27:47.860 So this is gamma union Q. 00:27:47.860 --> 00:27:58.270 So now if I have an A here in that cell, so then 00:27:58.270 --> 00:28:03.520 the variable x i, j, a is true. 00:28:03.520 --> 00:28:06.280 And just helping you visualize that, 00:28:06.280 --> 00:28:08.717 that's going to correspond to turning the light on. 00:28:08.717 --> 00:28:10.300 There's going to be a light associated 00:28:10.300 --> 00:28:11.675 with each one of these variables. 00:28:11.675 --> 00:28:15.660 And it's going to be turned on when that variable is true. 00:28:15.660 --> 00:28:19.340 Similarly, if I have the blank symbol 00:28:19.340 --> 00:28:22.430 is the thing that goes in that cell, 00:28:22.430 --> 00:28:24.193 then that variable gets turned on. 00:28:24.193 --> 00:28:25.110 I hope you can see it. 00:28:25.110 --> 00:28:28.820 Maybe it's a little bit small on your screen. 00:28:28.820 --> 00:28:32.900 The x i, j blank, a variable is true. 00:28:32.900 --> 00:28:39.550 And similarly, if it's q7 here, the x i, j q7 variable is true. 00:28:39.550 --> 00:28:42.460 So that's the way we're going to be encoding 00:28:42.460 --> 00:28:46.720 the contents of these cells using these indicator 00:28:46.720 --> 00:28:48.610 variables. 00:28:48.610 --> 00:28:54.490 And now we have to start making some Boolean logic to make sure 00:28:54.490 --> 00:29:01.140 that those variables reasonably represent the cell con-- 00:29:01.140 --> 00:29:02.685 the contents of these cells. 00:29:02.685 --> 00:29:04.560 So for example, what would be the first thing 00:29:04.560 --> 00:29:07.380 that comes to your mind? 00:29:07.380 --> 00:29:11.593 Well, we better not have two lights going on 00:29:11.593 --> 00:29:13.260 in any one of the cells, because then we 00:29:13.260 --> 00:29:14.520 have two symbols in that cell. 00:29:14.520 --> 00:29:15.823 And that's not allowed. 00:29:15.823 --> 00:29:18.240 Each cell is going-- we want each cell to have exactly one 00:29:18.240 --> 00:29:18.900 symbol. 00:29:18.900 --> 00:29:21.870 And that corresponds to each cell having 00:29:21.870 --> 00:29:23.730 exactly one of its lights turned on, 00:29:23.730 --> 00:29:25.800 or equivalently, each cell should 00:29:25.800 --> 00:29:29.800 have exactly one of the variables be true. 00:29:29.800 --> 00:29:31.740 That's the very first part of the formula 00:29:31.740 --> 00:29:32.970 is just going to say that. 00:29:32.970 --> 00:29:35.580 So let me show you what that looks like. 00:29:38.840 --> 00:29:41.150 So here we're talking about this phi cell. 00:29:41.150 --> 00:29:44.720 It says that there's exactly one light on per cell, 00:29:44.720 --> 00:29:47.367 or in other words, exactly one of the x i, 00:29:47.367 --> 00:29:49.742 j sigmas is true for each i, j. 00:29:56.150 --> 00:29:58.400 So this is how I'm going to actually express 00:29:58.400 --> 00:30:01.130 that using my Boolean formula. 00:30:04.657 --> 00:30:06.490 I'm sort of color coding the different parts 00:30:06.490 --> 00:30:08.532 of the formula, which I'm writing out to you here 00:30:08.532 --> 00:30:09.550 in English. 00:30:09.550 --> 00:30:11.080 So first, I want to say this. 00:30:11.080 --> 00:30:14.770 So in every cell, there's at least one light that's on, 00:30:14.770 --> 00:30:16.600 and there's at most one light that's on. 00:30:19.320 --> 00:30:20.960 So here's the green part. 00:30:20.960 --> 00:30:23.030 This is going to say at least one light is on. 00:30:23.030 --> 00:30:27.290 So I'm going to say that by taking all of the var-- 00:30:27.290 --> 00:30:31.910 all of the symbols that can appear in that cell, 00:30:31.910 --> 00:30:37.040 and taking an OR over all of those different associated 00:30:37.040 --> 00:30:38.370 variables. 00:30:38.370 --> 00:30:40.280 So it's either got the first symbol on, 00:30:40.280 --> 00:30:44.180 or the second symbol is there, or dot dot dot, 00:30:44.180 --> 00:30:45.410 or the last symbol is there. 00:30:45.410 --> 00:30:47.670 One of those has got to be there. 00:30:47.670 --> 00:30:52.790 I'm going to write this using my big OR notation. 00:30:52.790 --> 00:30:58.570 So for sigma appearing in this set of possibilities, 00:30:58.570 --> 00:31:02.580 one of those variables has got to be on at least. 00:31:02.580 --> 00:31:05.410 That's what this big OR tells you. 00:31:05.410 --> 00:31:10.060 Now we want to make sure that there's at most one that's on. 00:31:10.060 --> 00:31:11.380 So that there are not-- 00:31:11.380 --> 00:31:15.880 there's at least one on, but there are not two that are on. 00:31:15.880 --> 00:31:19.660 So I'm going to have an additional part of the formula 00:31:19.660 --> 00:31:21.370 here, which says-- 00:31:21.370 --> 00:31:23.406 and I hope you can read this. 00:31:23.406 --> 00:31:26.860 It's a little small. 00:31:26.860 --> 00:31:33.740 If I have two different symbols, sigma and tau, 00:31:33.740 --> 00:31:36.650 that are configuration-- 00:31:36.650 --> 00:31:39.830 possible configuration symbols, where sigma and tau are not 00:31:39.830 --> 00:31:40.410 equal. 00:31:40.410 --> 00:31:41.870 So that's I'm reading it out to you 00:31:41.870 --> 00:31:44.990 if it's too small for your screen. 00:31:44.990 --> 00:31:47.580 Then I'm going to say it's not possible. 00:31:47.580 --> 00:31:52.490 So I have the negation of x i, j sigma, and x i, j tau. 00:31:52.490 --> 00:31:56.720 So saying it another way, it's not the case 00:31:56.720 --> 00:32:01.700 that that cell contains both sigma and tau for any two 00:32:01.700 --> 00:32:06.790 symbols sigma and tau as long as they're different. 00:32:06.790 --> 00:32:10.280 And we want to take these two formulas and add them together. 00:32:10.280 --> 00:32:15.150 And this tells me in the cell i, j, 00:32:15.150 --> 00:32:18.450 there's exactly one of the variables is true. 00:32:18.450 --> 00:32:20.130 Exactly one of the lights is on. 00:32:20.130 --> 00:32:22.200 And that's going to represent which 00:32:22.200 --> 00:32:24.590 symbol goes into that cell. 00:32:24.590 --> 00:32:28.640 And then I want to take the AND over all possible cells 00:32:28.640 --> 00:32:33.580 to make sure that I'm going to now apply that everywhere. 00:32:33.580 --> 00:32:37.920 And so now I do an AND for i and j ranging between 1 and n 00:32:37.920 --> 00:32:42.405 to the k to apply this logic throughout the picture. 00:32:45.550 --> 00:32:46.050 Yeah. 00:32:46.050 --> 00:32:49.080 Sigma unions-- asking sigma union Q 00:32:49.080 --> 00:32:51.690 contains the input, output, yes. 00:32:51.690 --> 00:32:52.530 This is not a sigma. 00:32:52.530 --> 00:32:54.540 This is a gamma. 00:32:54.540 --> 00:32:56.070 Gamma is the tape alphabet. 00:32:56.070 --> 00:33:00.990 This is any symbol, including an input symbol, from sigma 00:33:00.990 --> 00:33:03.280 is going to be in gamma. 00:33:03.280 --> 00:33:05.990 So this is any symbol can appear on the tape. 00:33:10.350 --> 00:33:17.430 And this expression here, that is the expression phi sub cell. 00:33:17.430 --> 00:33:19.620 So here's a little check-in for you. 00:33:19.620 --> 00:33:21.840 But maybe before we jump into the check-in, 00:33:21.840 --> 00:33:23.430 let's just make sure-- 00:33:23.430 --> 00:33:26.070 it may be better to take some questions, 00:33:26.070 --> 00:33:31.818 and then we can ask the check-in for you. 00:33:31.818 --> 00:33:33.120 Do you understand? 00:33:33.120 --> 00:33:37.390 I mean, if you're not getting this, 00:33:37.390 --> 00:33:42.210 you should try to figure out how to get it, 00:33:42.210 --> 00:33:45.990 because this is really just the foundation. 00:33:45.990 --> 00:33:47.753 It only gets more-- 00:33:47.753 --> 00:33:49.170 it is not a very complicated proof 00:33:49.170 --> 00:33:51.212 once you sort of get the idea of what's going on, 00:33:51.212 --> 00:33:53.010 but if you're not getting this part, 00:33:53.010 --> 00:33:54.720 you won't be able to get the rest. 00:34:01.310 --> 00:34:01.810 OK. 00:34:01.810 --> 00:34:04.660 I have no idea what the-- what that means, but I'll 00:34:04.660 --> 00:34:06.070 read it out to you guys. 00:34:06.070 --> 00:34:09.940 This looks like a one-hot tensor encoding, same from common-- 00:34:09.940 --> 00:34:12.639 same form commonly used in ML. 00:34:12.639 --> 00:34:14.980 OK. 00:34:14.980 --> 00:34:15.947 It's just an indicate-- 00:34:15.947 --> 00:34:17.739 I would just call them indicator variables. 00:34:23.420 --> 00:34:27.409 Why is it sigma union Q for the big AND here? 00:34:30.179 --> 00:34:32.100 Sigma union Q? 00:34:32.100 --> 00:34:34.449 You mean gamma union Q? 00:34:34.449 --> 00:34:34.949 [GASPING] 00:34:34.949 --> 00:34:36.929 This is wrong. 00:34:36.929 --> 00:34:38.020 OK. 00:34:38.020 --> 00:34:40.230 Now I understand why everybody is upset. 00:34:40.230 --> 00:34:44.000 This should be a gamma, not a sigma. 00:34:44.000 --> 00:34:45.699 There's a boo-boo. 00:34:45.699 --> 00:34:46.951 Sorry about that. 00:34:46.951 --> 00:34:48.909 I don't know if I can fix that without breaking 00:34:48.909 --> 00:34:51.460 the whole slide, so I'm not going to even try. 00:34:51.460 --> 00:34:54.880 This symbol here should be gamma, not sigma. 00:34:54.880 --> 00:34:56.548 It's a typo. 00:34:56.548 --> 00:34:57.590 Thanks for catching that. 00:35:04.800 --> 00:35:05.550 Yes. 00:35:05.550 --> 00:35:08.670 So the question is-- 00:35:08.670 --> 00:35:09.240 OK. 00:35:09.240 --> 00:35:17.240 So phi sub cell is just trying to make sure 00:35:17.240 --> 00:35:21.980 that the encoding represents setting a bunch of symbols 00:35:21.980 --> 00:35:23.300 into the tableau. 00:35:23.300 --> 00:35:26.240 Not-- so each cell is going to have one symbol 00:35:26.240 --> 00:35:29.810 exactly, not two, not zero. 00:35:29.810 --> 00:35:30.980 So that's what phi sub-- 00:35:30.980 --> 00:35:34.130 if you've satisfied, if you've set the variables 00:35:34.130 --> 00:35:37.100 to satisfy phi sub cell, then there's 00:35:37.100 --> 00:35:40.070 going to be one symbol on in each-- 00:35:40.070 --> 00:35:43.370 one symbol in each of those cells. 00:35:43.370 --> 00:35:45.980 Now, another question, this is not a CNF. 00:35:45.980 --> 00:35:47.720 No, this is not a CNF. 00:35:47.720 --> 00:35:50.540 That's the second half of-- that's going to be-- 00:35:50.540 --> 00:35:52.010 I hope we don't run out of time. 00:35:52.010 --> 00:35:58.670 But I have a way of converting general SAT formulas to CNFs 00:35:58.670 --> 00:36:00.270 and preserving satisfiability. 00:36:00.270 --> 00:36:02.578 So we're going to do that reduction afterward. 00:36:08.210 --> 00:36:11.510 Here's a little check to see if you understand 00:36:11.510 --> 00:36:13.580 at some level what's going on. 00:36:17.680 --> 00:36:22.020 How many variables does this formula actually have? 00:36:25.000 --> 00:36:26.010 Is order n? 00:36:26.010 --> 00:36:27.370 Order n square? 00:36:27.370 --> 00:36:28.920 n to the k? 00:36:28.920 --> 00:36:32.580 Remember, k is the running time of the machine. 00:36:32.580 --> 00:36:34.880 Or n to the 2k? 00:36:34.880 --> 00:36:36.900 What do you think? 00:36:36.900 --> 00:36:40.620 So I mean, for how many variables. 00:36:40.620 --> 00:36:46.200 I mean that in all of phi sub M. How many 00:36:46.200 --> 00:36:50.218 variables do we have all together in this formula 00:36:50.218 --> 00:36:51.510 if that's what the question is. 00:36:56.680 --> 00:36:58.640 And here are the variables. 00:36:58.640 --> 00:37:02.530 So describing them here-- x i, j, sigma. 00:37:02.530 --> 00:37:03.370 OK. 00:37:03.370 --> 00:37:05.620 I'm going to close this. 00:37:05.620 --> 00:37:08.270 So pick something. 00:37:08.270 --> 00:37:08.960 All right. 00:37:08.960 --> 00:37:13.190 Ending polling, 1, 2, 3. 00:37:13.190 --> 00:37:14.730 OK. 00:37:14.730 --> 00:37:15.230 Yeah. 00:37:21.430 --> 00:37:21.930 OK. 00:37:21.930 --> 00:37:23.800 That's a good question. 00:37:23.800 --> 00:37:25.710 So first of all, the correct answer 00:37:25.710 --> 00:37:29.280 is, in fact, D. It's order n to the 2k. 00:37:29.280 --> 00:37:33.940 Now, I'm getting some questions about, 00:37:33.940 --> 00:37:37.360 what about the size of gamma and Q? 00:37:37.360 --> 00:37:40.660 Well, those are going to be fixed. 00:37:40.660 --> 00:37:45.820 They depend only on the machine, but they don't depend on n. 00:37:45.820 --> 00:37:49.540 So thinking about it functionally in terms of n, 00:37:49.540 --> 00:37:54.320 that's going to be a constant multiplier. 00:37:54.320 --> 00:37:59.560 And so it's going to be absorbed within the big O. 00:37:59.560 --> 00:38:03.874 So that's why we have-- 00:38:03.874 --> 00:38:05.958 these are constant relative to n. 00:38:05.958 --> 00:38:06.625 These are fixed. 00:38:11.380 --> 00:38:13.833 There are not n to the k possible symbols. 00:38:13.833 --> 00:38:15.250 There's a fixed number of symbols. 00:38:15.250 --> 00:38:16.675 It depends only on the machine. 00:38:20.350 --> 00:38:22.410 So we're looking at a particular machine, 00:38:22.410 --> 00:38:24.405 and what happens when you look at large inputs. 00:38:31.250 --> 00:38:34.138 So why is D and not C? 00:38:34.138 --> 00:38:35.930 Well, don't forget-- how big is this table? 00:38:38.670 --> 00:38:41.310 This is n to the k by n to the k. 00:38:41.310 --> 00:38:45.300 So there are n to the 2k cells, n to the k quantity squared, 00:38:45.300 --> 00:38:50.110 or n to the 2k cells here. 00:38:50.110 --> 00:38:51.810 And so there's a collection of variables 00:38:51.810 --> 00:38:54.790 for each cell, some fixed number of variables for each cell. 00:38:54.790 --> 00:38:56.475 So that's why its order n to the 2k. 00:39:00.360 --> 00:39:01.230 Good. 00:39:01.230 --> 00:39:03.930 So let's move on. 00:39:03.930 --> 00:39:07.200 I think we're actually-- hold on. 00:39:07.200 --> 00:39:10.200 So we have one more slide, and I think 00:39:10.200 --> 00:39:12.660 then we have a break after. 00:39:12.660 --> 00:39:20.640 So now let's next talk about constructing two more pieces 00:39:20.640 --> 00:39:24.490 of the phi sub M, w formula. 00:39:24.490 --> 00:39:28.560 So we already got phi sub cell done. 00:39:28.560 --> 00:39:33.310 Let's look at phi sub start and phi sub accept. 00:39:33.310 --> 00:39:36.940 And phi start is going to tell us 00:39:36.940 --> 00:39:42.540 that the start configuration has exactly these symbols. 00:39:42.540 --> 00:39:51.390 And phi accept tells us that the bottom configuration contains 00:39:51.390 --> 00:39:54.320 an accepting state somewhere. 00:39:54.320 --> 00:39:56.350 So how are we going to write that down? 00:39:56.350 --> 00:39:59.890 Well, first of all, I'm going to write these down just cell 00:39:59.890 --> 00:40:01.250 by cell. 00:40:01.250 --> 00:40:04.710 So first of all, phi start is going to say the cell 1, 00:40:04.710 --> 00:40:07.870 1 contains a q0. 00:40:07.870 --> 00:40:12.040 I mean, I know what the start configuration should 00:40:12.040 --> 00:40:19.250 be, because thinking of me, or think of us as the reduction, 00:40:19.250 --> 00:40:21.860 the reduction is given M. It's given w. 00:40:21.860 --> 00:40:23.480 So it knows what the start state is. 00:40:23.480 --> 00:40:28.340 It knows what the symbols of w are. 00:40:28.340 --> 00:40:31.820 So it knows what that start configuration is. 00:40:31.820 --> 00:40:35.630 It's just q0 followed by the n symbols 00:40:35.630 --> 00:40:37.820 of w followed by blanks. 00:40:37.820 --> 00:40:44.360 So it wants the very first cell in the left-hand corner 00:40:44.360 --> 00:40:48.740 here to be a q0, the start state of the machine, which it knows. 00:40:48.740 --> 00:40:55.200 So it's going to say x sub 1, 1, q0, that has to be turned on. 00:40:55.200 --> 00:40:59.030 So it's going to be AND of a bunch of variables here. 00:40:59.030 --> 00:41:03.170 And in order to satisfy phi sub start, all of those variables 00:41:03.170 --> 00:41:05.810 have to be set to true. 00:41:05.810 --> 00:41:09.860 So that means we have to have a q0 in that cell. 00:41:09.860 --> 00:41:13.150 So now we're going to do the next cell here. 00:41:13.150 --> 00:41:17.830 The 1, 2, the next cell of the start configuration. 00:41:17.830 --> 00:41:21.910 So phi start is going to have x 1, 2. 00:41:21.910 --> 00:41:23.665 So that's the next place. 00:41:23.665 --> 00:41:25.930 It contains w1. 00:41:25.930 --> 00:41:27.190 And so on. 00:41:27.190 --> 00:41:32.790 x 1, 3 contains w2, all the way up to wn, 00:41:32.790 --> 00:41:40.010 and then there's going to be a bunch of additional parts 00:41:40.010 --> 00:41:43.460 which say that we have blanks in the rest, 00:41:43.460 --> 00:41:48.750 just spelling out exactly all of the symbols in that top row. 00:41:48.750 --> 00:41:57.170 Because that's what the phi start formula, 00:41:57.170 --> 00:41:59.840 or sub formula of the overall formula 00:41:59.840 --> 00:42:03.860 we're making looks like. 00:42:03.860 --> 00:42:06.710 Now let's take a look at phi accept. 00:42:06.710 --> 00:42:09.590 Phi accept, because I'm just looking 00:42:09.590 --> 00:42:16.350 for q accept to appear somewhere in that bottom row, 00:42:16.350 --> 00:42:19.160 I'm going to do that in terms of an OR. 00:42:19.160 --> 00:42:23.930 So here is-- the variables now, notice, have n to the k 00:42:23.930 --> 00:42:27.150 because it's the last row in the table. 00:42:27.150 --> 00:42:30.460 So row n to the k here. 00:42:30.460 --> 00:42:33.710 And then I'm going to vary j from 1 to n to the k. 00:42:33.710 --> 00:42:35.740 So j, the column number, is going 00:42:35.740 --> 00:42:37.510 to range from 1 to n to the k. 00:42:37.510 --> 00:42:39.250 And I'm looking for that to accept. 00:42:39.250 --> 00:42:44.440 So x n to the k j, where j's vary, and q accept. 00:42:44.440 --> 00:42:47.240 One of those has to be true. 00:42:47.240 --> 00:42:48.870 One of those has to be turned on. 00:42:48.870 --> 00:42:52.560 And so that's why it's a big OR. 00:42:52.560 --> 00:43:00.033 And that's my phi accept piece. 00:43:00.033 --> 00:43:01.450 And now we'll take a little break. 00:43:01.450 --> 00:43:04.690 And feel free to ask me some more questions. 00:43:04.690 --> 00:43:06.135 Let me just start our clock. 00:43:15.370 --> 00:43:20.130 Go grab yourself some coffee, or ask me some questions. 00:43:20.130 --> 00:43:22.760 I'm happy to answer them. 00:43:22.760 --> 00:43:27.920 Why don't we check that q accept only appears once? 00:43:27.920 --> 00:43:30.835 Is it possible for q accept to appear twice? 00:43:30.835 --> 00:43:31.835 That's a great question. 00:43:34.590 --> 00:43:37.940 So that would definitely be a broken configuration 00:43:37.940 --> 00:43:39.740 if that happened, because a configuration 00:43:39.740 --> 00:43:44.960 can have-- must have exactly one state symbol appearing. 00:43:44.960 --> 00:43:50.540 The way we're going to enforce that is with the phi move 00:43:50.540 --> 00:43:52.590 part of the formula, which we haven't seen yet. 00:43:52.590 --> 00:43:54.920 So phi move is going to guarantee 00:43:54.920 --> 00:43:57.500 that the machine is acting correctly, 00:43:57.500 --> 00:44:01.490 so that all of the rows of the tableau 00:44:01.490 --> 00:44:04.460 are all legal configurations, and they all legally follow 00:44:04.460 --> 00:44:06.500 from the previous. 00:44:06.500 --> 00:44:09.020 So really, in a sense, the hard-- heavy lifting 00:44:09.020 --> 00:44:11.960 is coming in phi move, but it's really not that bad. 00:44:18.670 --> 00:44:20.420 Somebody says, out of curiosity, how close 00:44:20.420 --> 00:44:23.390 is this intuition proof to the actual proof? 00:44:23.390 --> 00:44:24.920 This is the actual proof. 00:44:24.920 --> 00:44:28.130 There's no-- I'm not hiding anything. 00:44:28.130 --> 00:44:31.250 I mean, we're being a little loose here, 00:44:31.250 --> 00:44:33.695 but you can turn this-- 00:44:33.695 --> 00:44:35.390 we're not cutting any corners here. 00:44:35.390 --> 00:44:38.570 This is exactly how the proof goes. 00:44:38.570 --> 00:44:44.030 And so not-- you're getting the real deal here. 00:44:44.030 --> 00:44:46.820 Somebody wanted to see the previous slide, so here we are. 00:44:46.820 --> 00:44:49.340 Whoops. 00:44:49.340 --> 00:44:50.885 Is there something you want to ask? 00:44:54.990 --> 00:45:02.240 So phi cell says there's exactly one symbol per cell. 00:45:02.240 --> 00:45:04.820 And the variables are set in a way in terms of thinking 00:45:04.820 --> 00:45:06.500 of them as indicator variables. 00:45:06.500 --> 00:45:11.160 There's exactly one variable set to true in each cell. 00:45:11.160 --> 00:45:13.460 So there's exactly one symbol per cell. 00:45:13.460 --> 00:45:16.460 That's what phi cell tells you. 00:45:16.460 --> 00:45:18.680 If you don't have that, then you have a mess. 00:45:18.680 --> 00:45:20.820 So you've got to start with that. 00:45:20.820 --> 00:45:22.320 And then with the-- other things are 00:45:22.320 --> 00:45:23.880 going to be additional conditions, 00:45:23.880 --> 00:45:27.030 which when satisfied are going to enforce the rest 00:45:27.030 --> 00:45:29.370 of the properties that we want. 00:45:29.370 --> 00:45:32.760 Why would the proof fail if we replace n to the k with 2 00:45:32.760 --> 00:45:33.780 to the n to the k? 00:45:36.920 --> 00:45:39.100 So OK. 00:45:39.100 --> 00:45:43.300 I presume where does-- we'll just use the polynomial running 00:45:43.300 --> 00:45:48.430 time of the machine of M, the nondeterministic machine. 00:45:48.430 --> 00:45:50.350 I mean, if you had an enormous tab-- 00:45:50.350 --> 00:45:53.650 we have to show ultimately that this reduction is 00:45:53.650 --> 00:45:55.960 a polynomial time reduction. 00:45:55.960 --> 00:45:58.210 And that's going to depend on how big the tableau is, 00:45:58.210 --> 00:46:00.585 because that's going to tell us how big the formula we're 00:46:00.585 --> 00:46:02.530 producing is, phi sub M, w. 00:46:02.530 --> 00:46:04.480 If phi sub M, w is exponentially big, 00:46:04.480 --> 00:46:07.750 we don't have a prayer of being able to output that formula 00:46:07.750 --> 00:46:08.740 in polynomial time. 00:46:11.830 --> 00:46:14.650 If there were less than n to the k steps, 00:46:14.650 --> 00:46:16.540 do we repeat the last configuration? 00:46:16.540 --> 00:46:18.250 Yeah, that's what I said. 00:46:18.250 --> 00:46:21.160 If the machine ends early, the last configuration 00:46:21.160 --> 00:46:22.910 just stays there. 00:46:22.910 --> 00:46:25.720 So we're going to modify the definition of the machine 00:46:25.720 --> 00:46:29.600 slightly so it just stays. 00:46:29.600 --> 00:46:30.100 Yeah. 00:46:35.090 --> 00:46:35.590 OK. 00:46:46.880 --> 00:46:47.380 Yeah. 00:46:47.380 --> 00:46:48.893 Let me not take the other-- there's 00:46:48.893 --> 00:46:50.560 a bunch of other questions, some of them 00:46:50.560 --> 00:46:54.160 a little on the technical side. 00:46:54.160 --> 00:46:57.200 Let me-- maybe I'll try to address them 00:46:57.200 --> 00:47:01.340 as they come along if it turns out to work to do that. 00:47:01.340 --> 00:47:06.140 So the break is over. 00:47:06.140 --> 00:47:10.460 Why don't we-- is it possible that the encoding configuration 00:47:10.460 --> 00:47:12.830 will not fit in n to the k? 00:47:15.950 --> 00:47:17.700 So the question is, is there a possibility 00:47:17.700 --> 00:47:19.290 that the encoding of the configuration 00:47:19.290 --> 00:47:20.670 won't fit in n to the k? 00:47:20.670 --> 00:47:23.960 If the machine runs for n to the k steps, 00:47:23.960 --> 00:47:26.720 the configuration has to fit within n to the k, 00:47:26.720 --> 00:47:31.110 because it can't use anything more within n to the k steps. 00:47:31.110 --> 00:47:31.870 So think about it. 00:47:31.870 --> 00:47:32.980 But no. 00:47:32.980 --> 00:47:36.690 The answer is the configurations-- 00:47:36.690 --> 00:47:39.360 if the machine runs in time n to the k, 00:47:39.360 --> 00:47:43.500 that whole tableau is big enough to write down 00:47:43.500 --> 00:47:46.440 the entire computation history. 00:47:46.440 --> 00:47:48.610 All right, so let's continue. 00:47:48.610 --> 00:47:53.860 Phi sub move, this is, in a sense, 00:47:53.860 --> 00:47:57.930 the part which is going to tell us that we 00:47:57.930 --> 00:48:02.080 started right, we ended right. 00:48:02.080 --> 00:48:05.990 Phi cell says every cell contains one symbol. 00:48:05.990 --> 00:48:10.750 And now we have to say that the whole interior is correct. 00:48:10.750 --> 00:48:12.280 How are we going to do that? 00:48:12.280 --> 00:48:15.575 So these are the parts we've already done. 00:48:15.575 --> 00:48:16.950 And the way I'm going to describe 00:48:16.950 --> 00:48:21.310 that is in terms of these kind of little windows 00:48:21.310 --> 00:48:22.870 I'm calling neighborhoods. 00:48:22.870 --> 00:48:28.630 So imagine here we have a 2 by 3 rectangle, which I'm going 00:48:28.630 --> 00:48:31.560 to call a 2 by 3 neighborhood. 00:48:31.560 --> 00:48:35.363 And what I'm going to argue, but I'm not going to prove here. 00:48:35.363 --> 00:48:36.780 I'm just going to really state it, 00:48:36.780 --> 00:48:41.280 but it's really just a sort of more or less obvious fact. 00:48:41.280 --> 00:48:47.070 But the proof-- the book has the formal proof, that if any-- 00:48:47.070 --> 00:48:50.100 every one of these here is legitimate, 00:48:50.100 --> 00:48:52.680 is legal according to the rules of the machine, 00:48:52.680 --> 00:48:56.800 if every single-- imagine you have these-- 00:48:56.800 --> 00:48:57.300 oops. 00:48:57.300 --> 00:48:59.670 Let me put myself back on here so you can see me. 00:49:02.190 --> 00:49:12.040 If you have here every 2 by 3 window, 00:49:12.040 --> 00:49:14.920 you can take this as a window, and you 00:49:14.920 --> 00:49:19.530 slide that over the entire picture of the tableau. 00:49:19.530 --> 00:49:22.530 And everything here looks OK as far 00:49:22.530 --> 00:49:23.850 as the running of the machine. 00:49:23.850 --> 00:49:26.700 So I'll say what that means in a second. 00:49:26.700 --> 00:49:30.690 But if everything looks locally fine everywhere, 00:49:30.690 --> 00:49:34.650 then the whole tableau has to be a valid tableau 00:49:34.650 --> 00:49:39.790 in terms of the rules of M. 00:49:39.790 --> 00:49:42.340 Maybe it's easier if I describe what 00:49:42.340 --> 00:49:44.125 I mean by these being legal. 00:49:47.000 --> 00:49:52.520 So these neighborhoods, these 2 by 3 neighborhoods 00:49:52.520 --> 00:49:56.460 are legal if they're consistent with M's transition function. 00:49:56.460 --> 00:49:58.400 So I'm going to describe rather than-- 00:49:58.400 --> 00:50:00.350 I mean, to do this formally, I would 00:50:00.350 --> 00:50:03.470 have to go through a process that we went-- 00:50:03.470 --> 00:50:07.820 like what we did when we went through the construction 00:50:07.820 --> 00:50:09.470 for the Post correspondence problem, 00:50:09.470 --> 00:50:11.780 and say if the machine moves left, this thing happens. 00:50:11.780 --> 00:50:13.000 If it moves right, that's-- 00:50:13.000 --> 00:50:15.690 I think that it's not really necessary. 00:50:15.690 --> 00:50:18.320 You can kind of get the idea very clearly 00:50:18.320 --> 00:50:20.240 by doing it at a little bit of a higher level. 00:50:23.180 --> 00:50:26.510 So let's look at what I mean by a legal neighborhood. 00:50:26.510 --> 00:50:29.860 So a legal neighborhood is a setting of the values, 00:50:29.860 --> 00:50:34.990 the six values of this 2 by 3 neighborhood, 00:50:34.990 --> 00:50:39.770 in a way which doesn't violate M's rules. 00:50:39.770 --> 00:50:47.520 So for example, if M, when it's in state q7 reading of b, 00:50:47.520 --> 00:50:52.390 goes into state q3 and moves left, 00:50:52.390 --> 00:50:56.680 then this would be a legal neighborhood, because it 00:50:56.680 --> 00:51:02.130 shows the head moving left, the b becoming a c. 00:51:02.130 --> 00:51:03.900 So reading a b, I should also say 00:51:03.900 --> 00:51:07.210 that it converts that b to a c. 00:51:07.210 --> 00:51:09.030 And we just head left into state q3. 00:51:09.030 --> 00:51:11.100 So this would be a legal neighborhood 00:51:11.100 --> 00:51:12.540 if that's the way-- 00:51:12.540 --> 00:51:18.120 so being legal depends upon the transition function 00:51:18.120 --> 00:51:19.720 of the machine. 00:51:19.720 --> 00:51:21.820 So given the transition function, 00:51:21.820 --> 00:51:24.460 that's going to tell you which are the legal neighborhoods. 00:51:24.460 --> 00:51:26.740 So another legal neighborhood-- this would always 00:51:26.740 --> 00:51:29.770 be a legal neighborhood-- is that if nothing changes. 00:51:29.770 --> 00:51:33.260 So that means the head of the machine was somewhere else. 00:51:33.260 --> 00:51:38.890 And so whatever was on the tape in this step 00:51:38.890 --> 00:51:43.240 is going to be the same stuff in those places one step later. 00:51:47.150 --> 00:51:52.210 Here's another possible legal neighborhood. 00:51:52.210 --> 00:52:00.050 If the head suddenly appears on one of the cells 00:52:00.050 --> 00:52:04.040 either in the left or the right, that 00:52:04.040 --> 00:52:08.720 would correspond to the machine moving its head from somewhere 00:52:08.720 --> 00:52:14.460 off the neighborhood into the neighborhood in that step. 00:52:14.460 --> 00:52:16.730 So this could be a legal neighborhood, 00:52:16.730 --> 00:52:18.470 provided the machine actually does 00:52:18.470 --> 00:52:21.020 move its head left into a state q5 00:52:21.020 --> 00:52:26.090 at some point under some conditions. 00:52:26.090 --> 00:52:28.700 And here is another kind of a weird legal neighborhood. 00:52:28.700 --> 00:52:32.810 If you have a, b, c, and then the a changes to a d, 00:52:32.810 --> 00:52:34.910 that could also be a legal neighborhood 00:52:34.910 --> 00:52:39.270 if the machine transition function allows 00:52:39.270 --> 00:52:41.940 an a to get converted to a d when there 00:52:41.940 --> 00:52:44.670 is some machine-- when there is some state reading that a, 00:52:44.670 --> 00:52:46.950 and that state also moves its head left. 00:52:46.950 --> 00:52:49.785 So it doesn't move into this picture. 00:52:49.785 --> 00:52:51.660 So those are examples of legal neighborhoods. 00:52:51.660 --> 00:52:54.488 Let me show you some illegal neighborhoods. 00:52:54.488 --> 00:52:57.030 Just I'm doing this-- this is kind of a proof by example now. 00:52:57.030 --> 00:52:59.490 This is perhaps the most intuitive part. 00:52:59.490 --> 00:53:03.210 But I claim that this is easy to turn this into something 00:53:03.210 --> 00:53:06.260 airtight and formal. 00:53:06.260 --> 00:53:08.330 So this would be clearly illegal. 00:53:08.330 --> 00:53:15.680 If you have a piece of the tape in the previous step where 00:53:15.680 --> 00:53:19.340 it's a, b, c, and then suddenly the b changes to a d. 00:53:19.340 --> 00:53:23.030 The symbol on the tape changes out of nowhere 00:53:23.030 --> 00:53:27.230 without having a head nearby to a different-- 00:53:27.230 --> 00:53:28.370 to something else. 00:53:28.370 --> 00:53:29.730 That could never happen. 00:53:29.730 --> 00:53:30.902 So that would be illegal. 00:53:30.902 --> 00:53:32.360 Another thing that would be illegal 00:53:32.360 --> 00:53:35.200 is if a state appears from nowhere. 00:53:35.200 --> 00:53:37.360 That could never happen. 00:53:37.360 --> 00:53:38.650 Or if it just disappears. 00:53:38.650 --> 00:53:40.640 That could never happen. 00:53:40.640 --> 00:53:43.280 And here's another-- here's an interesting one. 00:53:43.280 --> 00:53:45.005 If a state becomes two states. 00:53:49.060 --> 00:53:51.130 Don't forget the machine is nondeterministic. 00:53:51.130 --> 00:53:53.110 So the machine in principle could 00:53:53.110 --> 00:53:56.710 move its head left on one branch and move its head right 00:53:56.710 --> 00:53:58.750 on a different branch, but those would have 00:53:58.750 --> 00:54:00.070 to be in different tableaus. 00:54:00.070 --> 00:54:02.350 They can't be in the same tableau, 00:54:02.350 --> 00:54:04.240 because that doesn't correspond to any 00:54:04.240 --> 00:54:06.070 of the threads of the computation, those 00:54:06.070 --> 00:54:08.410 with multiple threads. 00:54:08.410 --> 00:54:14.720 And I say this because if you think about my claim, which 00:54:14.720 --> 00:54:18.200 is going to put down over here, that if every 2 00:54:18.200 --> 00:54:21.980 by 3 neighborhood is legal, then the tableau overall corresponds 00:54:21.980 --> 00:54:24.870 to a computation history. 00:54:24.870 --> 00:54:28.890 This illustrates why it's not enough to have a 2 00:54:28.890 --> 00:54:33.300 by 2 neighborhood, where you really need the 2 by 3. 00:54:33.300 --> 00:54:36.210 Because if this was a 2 by 2 neighborhood, 00:54:36.210 --> 00:54:38.970 if you just look at these four-- 00:54:38.970 --> 00:54:46.860 this leftmost 2 by 2, that could be a legal neighborhood if it 00:54:46.860 --> 00:54:51.240 was a 2 by 2, if the rules of the machine allowed for that. 00:54:51.240 --> 00:54:54.000 And the right four box-- 00:54:54.000 --> 00:54:56.970 right four cells could also be a legal neighborhood. 00:54:56.970 --> 00:54:59.490 So you could have something that looks 00:54:59.490 --> 00:55:02.670 OK from the perspective of 2 by 2 neighborhoods, 00:55:02.670 --> 00:55:05.790 but globally, in terms of the overall tableau, 00:55:05.790 --> 00:55:08.295 it's completely nonsensical because it has multiple hits. 00:55:11.370 --> 00:55:13.380 But if you have a 2 by 3 neighborhood, 00:55:13.380 --> 00:55:17.620 it's big enough to prevent this situation from occurring. 00:55:17.620 --> 00:55:22.170 And then you can check the details. 00:55:22.170 --> 00:55:25.620 And I think it's very plausible that it guarantees 00:55:25.620 --> 00:55:29.970 that the overall tableau is legitimate 00:55:29.970 --> 00:55:34.650 if all of the 2 by 3 neighborhoods are legal. 00:55:34.650 --> 00:55:39.870 And so that's what we're going to turn into a Boolean 00:55:39.870 --> 00:55:41.310 expression. 00:55:41.310 --> 00:55:46.380 We're going to say for each cell that the set-- 00:55:46.380 --> 00:55:49.830 for each neighborhood-- so here's a neighborhood at the i, 00:55:49.830 --> 00:55:51.180 j location. 00:55:51.180 --> 00:55:53.790 I'm calling this position here sort 00:55:53.790 --> 00:55:56.985 of the home location for that neighborhood. 00:56:00.066 --> 00:56:04.990 For each neighborhood, it has to be set to one 00:56:04.990 --> 00:56:07.480 of the legal possibilities. 00:56:07.480 --> 00:56:10.180 And there's, again, only a fixed number of those 00:56:10.180 --> 00:56:13.150 because there's a fixed number of possible symbols 00:56:13.150 --> 00:56:16.130 that can appear in those cells. 00:56:16.130 --> 00:56:19.570 So this is that fixed number to the sixth power at most. 00:56:19.570 --> 00:56:24.530 And I'm going to say that the cell in the upper left, 00:56:24.530 --> 00:56:27.880 which would be this one, is in r. 00:56:27.880 --> 00:56:29.320 And this one here is an s. 00:56:29.320 --> 00:56:30.950 And this one is a t. 00:56:30.950 --> 00:56:33.610 And this one is a v, if you just trace down what 00:56:33.610 --> 00:56:36.040 the indices are telling you. 00:56:36.040 --> 00:56:40.120 It says that that piece of the tape, that piece of the tableau 00:56:40.120 --> 00:56:46.550 here is set according to one of the possible legal settings. 00:56:46.550 --> 00:56:50.390 And we're just going to OR over all of those possible fixed 00:56:50.390 --> 00:56:52.580 number of legal settings. 00:56:52.580 --> 00:56:57.890 And then I take an AND over all possible tape cells, 00:56:57.890 --> 00:56:59.897 over all possible neighborhoods. 00:57:02.580 --> 00:57:04.350 And so that's going to be my phi move. 00:57:06.870 --> 00:57:07.950 And that's it. 00:57:10.420 --> 00:57:10.920 OK. 00:57:10.920 --> 00:57:11.760 Let's see. 00:57:11.760 --> 00:57:15.240 Can I explain again the third example of illegal? 00:57:15.240 --> 00:57:19.020 So this one over here, I presume, I'm being asked about. 00:57:19.020 --> 00:57:25.700 Well, if the machine is in a state q7 reading a c, 00:57:25.700 --> 00:57:27.500 the head has to move either left or right. 00:57:27.500 --> 00:57:33.570 So at the next configuration, there's 00:57:33.570 --> 00:57:37.320 got to be a state symbol appearing either in this cell 00:57:37.320 --> 00:57:39.970 or in this cell. 00:57:39.970 --> 00:57:41.020 And here the tape-- 00:57:41.020 --> 00:57:46.120 the head has basically just vanished with nowhere 00:57:46.120 --> 00:57:48.550 to-- it's gone. 00:57:48.550 --> 00:57:52.862 That could not happen according to the rules 00:57:52.862 --> 00:57:55.070 of the machine the way we talk about Turing machines. 00:57:55.070 --> 00:57:56.330 So that's not possible. 00:57:56.330 --> 00:58:02.530 So this would be an illegal neighborhood. 00:58:02.530 --> 00:58:05.140 You want to prevent any of the bad stuff 00:58:05.140 --> 00:58:06.760 from happening anywhere in here. 00:58:06.760 --> 00:58:09.070 So only good stuff can be happening locally. 00:58:09.070 --> 00:58:11.440 And that guarantees the overall picture is OK. 00:58:16.720 --> 00:58:19.383 Do we have to check that the head doesn't 00:58:19.383 --> 00:58:21.550 leave the tableau from the left most to right right? 00:58:21.550 --> 00:58:23.675 Yeah, there are some little details here like that. 00:58:23.675 --> 00:58:27.055 So the question is, do I have to make sure that the-- 00:58:27.055 --> 00:58:29.470 yeah, you probably need to mark. 00:58:29.470 --> 00:58:31.870 I think the book probably does this correctly. 00:58:31.870 --> 00:58:34.480 You may have to mark the left and right ends 00:58:34.480 --> 00:58:35.890 to make sure that-- 00:58:35.890 --> 00:58:37.850 I mean, the right end is not a problem, 00:58:37.850 --> 00:58:45.690 because the machine can never go off the right end. 00:58:45.690 --> 00:58:48.180 And if you design the machine so that it never 00:58:48.180 --> 00:58:51.198 moves its head off the left end either, which you can do, 00:58:51.198 --> 00:58:53.490 then you wouldn't have to worry about that possibility. 00:58:53.490 --> 00:58:56.940 But otherwise, you would have to put some sort of delimiter here 00:58:56.940 --> 00:59:01.110 to enforce the head not moving off the left end. 00:59:01.110 --> 00:59:03.030 So there are some details like that, too. 00:59:05.923 --> 00:59:07.590 There will be two heads in the same row. 00:59:07.590 --> 00:59:10.020 No, this can-- I don't know what you-- somebody says, 00:59:10.020 --> 00:59:12.780 there will be two heads in the same row. 00:59:12.780 --> 00:59:16.710 Please elaborate, because this is designed not to allow 00:59:16.710 --> 00:59:18.000 two heads in the same row. 00:59:21.250 --> 00:59:23.510 Could I go over the OR for legal again? 00:59:23.510 --> 00:59:24.010 OK. 00:59:24.010 --> 00:59:30.560 The OR, the big OR here, what I have in mind is I take-- 00:59:30.560 --> 00:59:33.020 there's going to be-- first of all, I look at the machine 00:59:33.020 --> 00:59:35.580 and I look at the transition function. 00:59:35.580 --> 00:59:41.900 And based on that, I write down the list of all the legal 2 00:59:41.900 --> 00:59:42.980 by 3 neighborhoods. 00:59:42.980 --> 00:59:46.040 So all the settings which correspond to legal 2 00:59:46.040 --> 00:59:47.090 by 3 neighborhoods. 00:59:47.090 --> 00:59:49.460 There's going to be some fixed number of those. 00:59:49.460 --> 00:59:52.400 100. 00:59:52.400 --> 00:59:55.310 There's 100 possible legal neighborhoods of which 00:59:55.310 --> 00:59:57.820 I've written down here four. 00:59:57.820 --> 01:00:00.310 But maybe there's some number, say 100. 01:00:00.310 --> 01:00:03.730 So now there's going to be an OR over those 100 01:00:03.730 --> 01:00:05.360 different possibilities. 01:00:05.360 --> 01:00:07.360 It's either going to be this legal neighborhood, 01:00:07.360 --> 01:00:08.620 or some other legal neighborhood, 01:00:08.620 --> 01:00:10.090 or some other legal neighborhood. 01:00:10.090 --> 01:00:12.580 And for each one of those legal neighborhoods, 01:00:12.580 --> 01:00:16.940 I'm going to say, well, the variables are set according 01:00:16.940 --> 01:00:21.530 to that legal neighborhood, or the variables are set according 01:00:21.530 --> 01:00:23.975 to the next legal neighborhood, or the variables are 01:00:23.975 --> 01:00:25.850 set according to the next legal neighborhood, 01:00:25.850 --> 01:00:28.300 and do that 100 times. 01:00:28.300 --> 01:00:30.530 One of those has got to end up-- 01:00:30.530 --> 01:00:33.170 I mean it's an OR, so one of them has to work. 01:00:33.170 --> 01:00:34.480 Otherwise, the formula fails. 01:00:34.480 --> 01:00:36.520 And it will be false, because you're 01:00:36.520 --> 01:00:39.970 going to AND that over all of the neighborhoods 01:00:39.970 --> 01:00:40.600 in the picture. 01:00:46.630 --> 01:00:48.910 Is it possible to have a head on the far left 01:00:48.910 --> 01:00:51.235 of the configuration and one on the far right? 01:00:55.240 --> 01:00:58.110 You mean a head over here and a head over there? 01:00:58.110 --> 01:01:00.480 I mean, how'd the head get there? 01:01:00.480 --> 01:01:03.730 It can't happen. 01:01:03.730 --> 01:01:08.970 You know, the head has to come from a head above it. 01:01:08.970 --> 01:01:11.670 If you're going to be worrying about the details 01:01:11.670 --> 01:01:16.980 of the boundaries here, all that's fixable. 01:01:16.980 --> 01:01:21.593 So let's not lose sight of the main idea. 01:01:21.593 --> 01:01:23.260 I mean, if you understand the main idea, 01:01:23.260 --> 01:01:25.090 you can fix the little details. 01:01:25.090 --> 01:01:27.465 So I want to make sure you understand the main idea 01:01:27.465 --> 01:01:29.700 of what's happening. 01:01:29.700 --> 01:01:31.395 So let's finish up this proof. 01:01:34.010 --> 01:01:38.950 So in summary, we gave a reduction from 8 to SAT. 01:01:38.950 --> 01:01:41.890 This is what we needed. 01:01:41.890 --> 01:01:43.960 It was in those four pieces. 01:01:43.960 --> 01:01:45.520 And you really just need to argue 01:01:45.520 --> 01:01:49.740 that that formula we're building is not too big. 01:01:49.740 --> 01:01:52.020 And it's going to be basically the size of the tableau 01:01:52.020 --> 01:01:54.960 if you look at what we constructed. 01:01:54.960 --> 01:01:56.760 The number of variables is roughly the size 01:01:56.760 --> 01:01:57.810 of the tableau. 01:01:57.810 --> 01:02:00.540 And the amount of logic that we're putting into the formula 01:02:00.540 --> 01:02:02.310 is also going to be a fixed amount 01:02:02.310 --> 01:02:05.730 of logic independent of n for each 01:02:05.730 --> 01:02:08.430 of the variables in that tableau. 01:02:08.430 --> 01:02:11.850 And somebody asked me about the size-- how big the indices are. 01:02:14.470 --> 01:02:22.060 The indices for the x i value, the i and j values, 01:02:22.060 --> 01:02:24.910 technically, they're going to be numbers between 1 01:02:24.910 --> 01:02:25.990 and n to the k. 01:02:25.990 --> 01:02:27.860 So you're going to have to write those down. 01:02:27.860 --> 01:02:30.235 And so that's going to be a slight additional logarithmic 01:02:30.235 --> 01:02:32.280 cost to write those things down. 01:02:32.280 --> 01:02:34.000 It's not really that interesting a point. 01:02:36.760 --> 01:02:39.490 And so the overall f is going to be 01:02:39.490 --> 01:02:41.980 computable in polynomial time because the output is not 01:02:41.980 --> 01:02:42.800 very big. 01:02:42.800 --> 01:02:46.590 And it's also not complicated to write the output down. 01:02:46.590 --> 01:02:50.670 So that's the end of the proof. 01:02:50.670 --> 01:02:54.580 I can take a couple of questions. 01:02:54.580 --> 01:02:56.455 Why can't we just check that the whole-- this 01:02:56.455 --> 01:02:56.990 is a good question. 01:02:56.990 --> 01:02:59.157 Why can't we just check that the whole row is legal? 01:03:05.200 --> 01:03:08.530 You can check that a row actually is a configuration. 01:03:08.530 --> 01:03:15.390 But to check that the row follows from the previous row, 01:03:15.390 --> 01:03:18.600 ultimately, the operation of a Turing machine 01:03:18.600 --> 01:03:20.820 is a local thing. 01:03:20.820 --> 01:03:26.730 I mean, the way it moves from one configuration to the next 01:03:26.730 --> 01:03:31.230 depends locally on how where the head is. 01:03:31.230 --> 01:03:34.200 And so really, that's just another way of-- 01:03:34.200 --> 01:03:36.450 the way I'm saying it is just really checking 01:03:36.450 --> 01:03:39.482 the whole configuration, but just doing it locally. 01:03:39.482 --> 01:03:41.190 I don't know if that's satisfying to you. 01:03:44.700 --> 01:03:46.278 Why don't I move on because I just 01:03:46.278 --> 01:03:47.820 want to make sure we have enough time 01:03:47.820 --> 01:03:50.400 to get to the very last part, which is a little bit-- 01:03:50.400 --> 01:03:52.710 I'm afraid, a little technical. 01:03:52.710 --> 01:03:56.310 So we're going to kind of shift gears now and talk 01:03:56.310 --> 01:04:00.870 about reducing SAT to 3SAT. 01:04:00.870 --> 01:04:07.010 And let's see how it goes. 01:04:07.010 --> 01:04:08.560 I don't always have the most success 01:04:08.560 --> 01:04:11.380 with presenting this little piece, 01:04:11.380 --> 01:04:13.190 because it's slightly a technical argument. 01:04:13.190 --> 01:04:16.270 But if you don't get it, don't worry. 01:04:16.270 --> 01:04:18.772 Just you have to accept that it's true. 01:04:18.772 --> 01:04:20.230 But I'd like to show it to you just 01:04:20.230 --> 01:04:27.420 to make the whole presentation complete in that sense. 01:04:27.420 --> 01:04:33.240 So I'm going to give a reduction that maps general formulas 01:04:33.240 --> 01:04:35.880 to 3CNF formulas. 01:04:35.880 --> 01:04:40.100 So that's how we map SAT to 3SAT. 01:04:40.100 --> 01:04:42.710 If you remember 3SAT is satisfiability 01:04:42.710 --> 01:04:45.140 but for three CNFs. 01:04:45.140 --> 01:04:48.860 So a conjunctive normal form in the form of those clauses, 01:04:48.860 --> 01:04:50.420 which are ANDed together. 01:04:50.420 --> 01:04:54.530 And each clause is an OR of a bunch of literals, which are 01:04:54.530 --> 01:04:58.080 variables or negated variables. 01:04:58.080 --> 01:05:04.220 So I want to convert phi to phi prime, which is 3CNF formula, 01:05:04.220 --> 01:05:07.070 but preserve the satisfiability. 01:05:07.070 --> 01:05:10.460 And phi prime is not going to be logically equivalent to phi, 01:05:10.460 --> 01:05:12.680 because I could do that, too. 01:05:12.680 --> 01:05:17.480 I can convert any formula to a logically equivalent CNF 01:05:17.480 --> 01:05:18.200 formula. 01:05:18.200 --> 01:05:20.675 Maybe not even a 3CNF, but yeah, you 01:05:20.675 --> 01:05:23.870 won't be able to get a 3CNF, but you can get a CNF. 01:05:23.870 --> 01:05:26.405 But it might be exponentially larger. 01:05:26.405 --> 01:05:27.530 And that's not good enough. 01:05:27.530 --> 01:05:30.270 I have to do the reduction in polynomial time. 01:05:30.270 --> 01:05:32.720 So I can't generate a much larger formula 01:05:32.720 --> 01:05:34.190 that's exponentially larger. 01:05:34.190 --> 01:05:40.650 And so I'm going to do that by adding additional variables. 01:05:40.650 --> 01:05:42.140 So it won't be logically equivalent 01:05:42.140 --> 01:05:44.120 because the new formula is going to have 01:05:44.120 --> 01:05:46.510 additional variables in it. 01:05:46.510 --> 01:05:49.250 I'm going to kind of do it by example. 01:05:49.250 --> 01:05:51.840 And we'll see how that goes. 01:05:51.840 --> 01:05:56.770 So here's phi, which is not in 3CNF. 01:05:56.770 --> 01:05:59.730 It's not even in CNF, because it's 01:05:59.730 --> 01:06:03.840 got ORs of ANDs appearing, which is not 01:06:03.840 --> 01:06:07.600 allowed to happen in a CNF. 01:06:07.600 --> 01:06:10.960 So how are we going to convert that 01:06:10.960 --> 01:06:14.980 into a 3CNF formula preserving the satisfiability? 01:06:14.980 --> 01:06:17.800 And just working it through with this example, 01:06:17.800 --> 01:06:20.140 I hope to at least give you some idea of how you 01:06:20.140 --> 01:06:22.040 do the conversion in general. 01:06:22.040 --> 01:06:24.340 So first of all, I'm going to represent this formula 01:06:24.340 --> 01:06:29.100 as a tree using its natural tree structure. 01:06:29.100 --> 01:06:30.140 So you understand. 01:06:30.140 --> 01:06:34.910 So a AND B becomes a AND b written as a tree. 01:06:34.910 --> 01:06:37.830 And then I OR that with c. 01:06:37.830 --> 01:06:41.310 So I get the tree structure here in sort of the natural way. 01:06:41.310 --> 01:06:45.120 And I'm going to label all of these intermediate nodes, which 01:06:45.120 --> 01:06:49.400 are associated now with operations. 01:06:49.400 --> 01:06:55.760 And I'm assuming also that the formula is fully parenthesized 01:06:55.760 --> 01:06:58.280 so that each operation I'm only thinking about 01:06:58.280 --> 01:07:02.600 is applying just the two-- it's a binary operation. 01:07:02.600 --> 01:07:05.750 And let's ignore negations for the minute, because negations, 01:07:05.750 --> 01:07:08.220 you can always push those through down to the leaves. 01:07:08.220 --> 01:07:11.665 But it's just going to make it too complicated. 01:07:11.665 --> 01:07:13.415 So negations turn out not to be a problem. 01:07:16.140 --> 01:07:17.690 So there's only going to be negations 01:07:17.690 --> 01:07:22.610 at the level of the inputs, not at the negation 01:07:22.610 --> 01:07:25.370 operations in the middle. 01:07:25.370 --> 01:07:27.420 So we have this tree structure here. 01:07:27.420 --> 01:07:31.580 And now I'm going to use these two logical facts. 01:07:31.580 --> 01:07:34.040 And I don't know if-- 01:07:34.040 --> 01:07:36.740 you've probably all seen ANDs and ORs, I hope. 01:07:36.740 --> 01:07:38.990 Otherwise, it's going to be really tough. 01:07:38.990 --> 01:07:40.970 But there's also other logical operators, 01:07:40.970 --> 01:07:43.850 such as the implication operator, 01:07:43.850 --> 01:07:52.500 where you have A implies B sort of as a logical operation. 01:07:52.500 --> 01:07:58.590 And so this requires that if A is true, then B is true. 01:08:02.920 --> 01:08:07.380 However, if A is false, B can be anything. 01:08:07.380 --> 01:08:09.660 And similarly if B is true, A can be anything. 01:08:12.570 --> 01:08:15.710 The only thing that's prohibited is that if A is true 01:08:15.710 --> 01:08:17.180 and B is false. 01:08:17.180 --> 01:08:20.569 That's the only thing that would be invalid. 01:08:20.569 --> 01:08:22.520 And so if you think about it, that's 01:08:22.520 --> 01:08:26.180 going to be equivalent to saying that either A is false 01:08:26.180 --> 01:08:27.770 or B is true. 01:08:27.770 --> 01:08:30.109 One of those has to be. 01:08:30.109 --> 01:08:32.750 And that's going to be logically equivalent to saying 01:08:32.750 --> 01:08:36.649 that A implies B. 01:08:36.649 --> 01:08:39.410 Another logical equivalence may be more familiar to you. 01:08:39.410 --> 01:08:42.200 It's just simply De Morgan's law, 01:08:42.200 --> 01:08:47.210 which says that if you have the not of A AND B, that's 01:08:47.210 --> 01:08:54.750 equivalent to saying the not of A or the not of B. 01:08:54.750 --> 01:08:57.960 I'm going to make use of both of these. 01:08:57.960 --> 01:09:01.660 Now, here, I want to-- 01:09:01.660 --> 01:09:03.080 I ran out of room on this slide. 01:09:03.080 --> 01:09:05.205 So I'm going to take myself out of the picture here 01:09:05.205 --> 01:09:05.950 for a minute. 01:09:08.930 --> 01:09:10.810 I had no place else to put this. 01:09:10.810 --> 01:09:12.729 So here we have-- 01:09:12.729 --> 01:09:14.979 if you're going to think of the AND 01:09:14.979 --> 01:09:19.359 in terms of its truth table, so here's a and b 01:09:19.359 --> 01:09:22.000 in terms of a and b. 01:09:22.000 --> 01:09:27.790 So 1 and 1 is 1, but all other settings of a and b 01:09:27.790 --> 01:09:30.420 yield 0 for the AND. 01:09:30.420 --> 01:09:32.234 And I'm going to represent those-- 01:09:35.300 --> 01:09:38.330 if you imagine a and b is going to be called c. 01:09:38.330 --> 01:09:41.240 I'm going to represent this information 01:09:41.240 --> 01:09:48.620 with four small formulas, which taken together, you AND them 01:09:48.620 --> 01:09:52.490 together, are going to force c to have the correct behavior 01:09:52.490 --> 01:09:54.196 associated with a AND b. 01:10:00.610 --> 01:10:03.970 So if a AND b are both 1, then c is 1. 01:10:03.970 --> 01:10:14.260 If a is 0 and b is 1, then it forces c to be 0. 01:10:14.260 --> 01:10:18.700 And similarly, every other setting besides a AND 01:10:18.700 --> 01:10:24.520 b being true, for C to be false, which is what 01:10:24.520 --> 01:10:26.410 you want when you have AND. 01:10:26.410 --> 01:10:29.170 So I'm going to write this expression here down 01:10:29.170 --> 01:10:35.460 with z1 being in the place of c by just taking those four 01:10:35.460 --> 01:10:38.320 expressions and ANDing them together. 01:10:38.320 --> 01:10:41.190 So this is exactly those same four expressions written out 01:10:41.190 --> 01:10:42.800 linearly. 01:10:42.800 --> 01:10:45.800 Now I want to do the same thing for z2, 01:10:45.800 --> 01:10:48.148 but now that's written in terms of an OR. 01:10:48.148 --> 01:10:49.940 So there's a slightly different truth table 01:10:49.940 --> 01:10:51.515 here up in this corner. 01:10:55.030 --> 01:10:59.420 So now if either one is 1, we get a 1 result. 01:10:59.420 --> 01:11:06.380 And so now if a AND b are true, you get c is true. 01:11:06.380 --> 01:11:08.740 However, if a is true and b is false, 01:11:08.740 --> 01:11:12.210 that still implies c is true. 01:11:12.210 --> 01:11:20.340 So I'm going to write down those rules for specifying 01:11:20.340 --> 01:11:25.320 how z2 must be set. 01:11:25.320 --> 01:11:27.150 And each one of these things is going 01:11:27.150 --> 01:11:31.290 to get converted into clauses, three clauses with three 01:11:31.290 --> 01:11:36.790 literals, using these rules over here. 01:11:36.790 --> 01:11:39.570 So I'm going to do that for each zi. 01:11:39.570 --> 01:11:42.430 And lastly, to make sure the whole thing is satisfied, 01:11:42.430 --> 01:11:44.340 which means there's an output of one here, 01:11:44.340 --> 01:11:47.850 I'm going to have one clause associated, which 01:11:47.850 --> 01:11:52.700 says that z4, the output, is 1. 01:11:52.700 --> 01:11:57.950 Now, I can convert all of those when I have a AND b implies c. 01:11:57.950 --> 01:12:03.147 That's logically equivalent to not a OR not b OR c. 01:12:03.147 --> 01:12:04.730 And the way you can see that is really 01:12:04.730 --> 01:12:08.780 by repeated application of these rules here. 01:12:08.780 --> 01:12:11.070 We're running a little low on time. 01:12:11.070 --> 01:12:14.300 So maybe you'll just have to check this offline. 01:12:14.300 --> 01:12:21.390 But quickly, a AND b implies c using the first equivalence 01:12:21.390 --> 01:12:26.190 is the not of this part, OR C. And then 01:12:26.190 --> 01:12:30.420 I can use De Morgan to convert that not of an AND 01:12:30.420 --> 01:12:33.360 to an OR of the nots. 01:12:33.360 --> 01:12:36.810 And then I can remove the parentheses 01:12:36.810 --> 01:12:39.840 because OR is associative. 01:12:39.840 --> 01:12:43.770 And so I get a clause, which is what I need. 01:12:43.770 --> 01:12:47.130 So each one of these guys is really equivalent to a clause. 01:12:47.130 --> 01:12:49.038 And so I just get a bunch of clauses. 01:12:49.038 --> 01:12:50.580 And actually, technically, this needs 01:12:50.580 --> 01:12:55.410 to be three, a copy of three things here. 01:12:55.410 --> 01:13:02.910 It should be z4 OR z4 OR z4, which is a lot. 01:13:02.910 --> 01:13:05.520 So check-in-- [INAUDIBLE] I realize 01:13:05.520 --> 01:13:08.910 my check-in is broken, because I only realized that last point 01:13:08.910 --> 01:13:10.440 just now as I was talking. 01:13:10.440 --> 01:13:16.140 So the actual value that you get in terms 01:13:16.140 --> 01:13:19.200 of-- oh, no, the number of clauses is correct. 01:13:19.200 --> 01:13:20.640 No, I take it back. 01:13:20.640 --> 01:13:21.820 This is fine. 01:13:21.820 --> 01:13:23.700 So if you understood what I was saying, 01:13:23.700 --> 01:13:27.960 hopefully, you can see how big the formula 01:13:27.960 --> 01:13:33.840 phi prime is in terms of the number of operations in phi. 01:13:33.840 --> 01:13:35.940 So let's see how many people get that. 01:13:35.940 --> 01:13:41.820 I acknowledge this may be a little on the technical side. 01:13:41.820 --> 01:13:42.390 OK. 01:13:42.390 --> 01:13:46.040 I'm going to close it, close it down. 01:13:46.040 --> 01:13:47.040 Please enter your value. 01:13:50.440 --> 01:13:52.670 OK. 01:13:52.670 --> 01:13:56.870 Yeah, the correct answer is 4k plus 1, 01:13:56.870 --> 01:14:02.940 because each one of these operations 01:14:02.940 --> 01:14:04.965 is going to end up being a row in this picture. 01:14:07.520 --> 01:14:10.040 Each operation is going to have a variable associated to it. 01:14:10.040 --> 01:14:12.090 It's going to become a row in this picture. 01:14:12.090 --> 01:14:20.060 And so then each row is going to have four clauses, which 01:14:20.060 --> 01:14:26.090 define what you need-- set what you need in order 01:14:26.090 --> 01:14:30.890 to force that variable to have the right value corresponding 01:14:30.890 --> 01:14:31.850 to that operation. 01:14:34.800 --> 01:14:38.220 And so then you need-- and you need one extra clause here 01:14:38.220 --> 01:14:43.020 for saying that this whole thing evaluates to true. 01:14:45.630 --> 01:14:50.730 So that's all I wanted to do today. 01:14:50.730 --> 01:14:52.860 We proved those two main theorems. 01:14:52.860 --> 01:14:55.750 And now we know that there are NP-complete problems. 01:14:55.750 --> 01:14:59.670 And all of the other problems that we can get from-- 01:14:59.670 --> 01:15:02.550 by reductions from these problems 01:15:02.550 --> 01:15:07.170 are also going to be NP-complete as long as they're in NP. 01:15:07.170 --> 01:15:09.030 So that's it. 01:15:09.030 --> 01:15:13.680 Feel free to put some questions into the chat, 01:15:13.680 --> 01:15:17.580 or move on to whatever else you're going to be doing next. 01:15:17.580 --> 01:15:19.950 So a good question here is, why is phi prime 01:15:19.950 --> 01:15:22.305 not logically equivalent with this construction? 01:15:22.305 --> 01:15:24.780 It can't be logically equivalent. 01:15:24.780 --> 01:15:27.180 Logically equivalent means that it gives you 01:15:27.180 --> 01:15:28.950 exactly the same function. 01:15:28.950 --> 01:15:31.020 If you set the variables in the same way, 01:15:31.020 --> 01:15:34.250 you get the same result coming out. 01:15:34.250 --> 01:15:37.700 Well, phi prime has more variables than phi does. 01:15:37.700 --> 01:15:39.680 So it wouldn't even make sense to talk 01:15:39.680 --> 01:15:41.420 about logical equivalence because they're 01:15:41.420 --> 01:15:44.070 two functions on different numbers of variables. 01:15:44.070 --> 01:15:47.150 So in that sense, it doesn't really make sense. 01:15:47.150 --> 01:15:50.390 What you could say is that for every setting 01:15:50.390 --> 01:15:54.050 of the overlapping variables, so the variables 01:15:54.050 --> 01:15:56.070 that appear in both phi and phi prime-- 01:15:56.070 --> 01:15:58.670 so those are the original variables of phi-- 01:15:58.670 --> 01:16:02.900 there's going to exist some setting of new variables, which 01:16:02.900 --> 01:16:04.960 is going to make the-- 01:16:11.997 --> 01:16:13.580 there's going to be-- there will exist 01:16:13.580 --> 01:16:15.260 some setting of the new variables which 01:16:15.260 --> 01:16:17.660 will make the two formulas agree, 01:16:17.660 --> 01:16:20.914 but that's not the definition of logical equivalence. 01:16:23.970 --> 01:16:28.890 So why-- going back to the proof, the satisfiability proof 01:16:28.890 --> 01:16:30.840 and the legal neighborhoods, could I 01:16:30.840 --> 01:16:35.010 go over why the number of legal neighborhoods is polynomial. 01:16:35.010 --> 01:16:36.390 The number of legal neighborhoods 01:16:36.390 --> 01:16:38.100 is not only polynomial. 01:16:38.100 --> 01:16:40.810 It's constant. 01:16:40.810 --> 01:16:42.610 It depends only on the machine. 01:16:42.610 --> 01:16:49.840 It does not depend on M. Because each cell can have 01:16:49.840 --> 01:16:52.480 at most some fixed number of-- 01:16:52.480 --> 01:16:55.210 can have the number of tape symbols 01:16:55.210 --> 01:16:56.890 plus the number of state symbols. 01:16:56.890 --> 01:16:59.980 That depends on the machine only. 01:16:59.980 --> 01:17:06.340 So now we have six tape cells for the six cells in a 2 01:17:06.340 --> 01:17:07.360 by 3 neighborhood. 01:17:07.360 --> 01:17:09.970 So you're going to have that number to the sixth power. 01:17:09.970 --> 01:17:11.900 But still, it's a constant to the sixth power. 01:17:11.900 --> 01:17:12.817 It's still a constant. 01:17:12.817 --> 01:17:15.340 It doesn't depend on M. So it's not a question 01:17:15.340 --> 01:17:18.340 of even being polynomial. 01:17:18.340 --> 01:17:19.400 It's a constant value. 01:17:19.400 --> 01:17:21.580 It's a constant multiplier if you 01:17:21.580 --> 01:17:26.290 want to think about it in terms of the size of the formula 01:17:26.290 --> 01:17:27.970 that's going to result. 01:17:27.970 --> 01:17:36.310 Don't forget we're trying to make a formula which is-- 01:17:36.310 --> 01:17:40.340 the reduction has to be polynomial. 01:17:40.340 --> 01:17:41.800 It's a polynomial time reduction. 01:17:41.800 --> 01:17:45.410 So that means that as n increases, 01:17:45.410 --> 01:17:47.770 the time to calculate the reduction increases 01:17:47.770 --> 01:17:50.290 as a polynomial. 01:17:50.290 --> 01:17:58.090 But we're fixing M. So M does not change. 01:17:58.090 --> 01:18:00.570 So therefore anything that depends on M only 01:18:00.570 --> 01:18:03.760 is just going to be a constant impact on the formula. 01:18:03.760 --> 01:18:04.830 It's not going to be-- 01:18:04.830 --> 01:18:08.200 it doesn't depend on M. 01:18:08.200 --> 01:18:10.980 OK, everybody. 01:18:10.980 --> 01:18:12.210 Bye bye. 01:18:12.210 --> 01:18:13.940 See you.