WEBVTT 00:00:01.550 --> 00:00:03.920 The following content is provided under a Creative 00:00:03.920 --> 00:00:05.310 Commons license. 00:00:05.310 --> 00:00:07.520 Your support will help MIT OpenCourseWare 00:00:07.520 --> 00:00:11.610 continue to offer high-quality educational resources for free. 00:00:11.610 --> 00:00:14.180 To make a donation or to view additional materials 00:00:14.180 --> 00:00:18.140 from hundreds of MIT courses, visit MIT OpenCourseWare 00:00:18.140 --> 00:00:19.026 at ocw.mit.edu. 00:00:26.090 --> 00:00:27.410 GILBERT STRANG: OK. 00:00:27.410 --> 00:00:31.070 Let me just say our next topic, moving really 00:00:31.070 --> 00:00:35.210 toward machine learning, would be 00:00:35.210 --> 00:00:38.990 the last section of chapter 6 about 00:00:38.990 --> 00:00:40.730 stochastic gradient descent. 00:00:43.280 --> 00:00:49.810 I think Professor Sra from computer science, 00:00:49.810 --> 00:00:54.410 if he is able, he will take the class Friday and tell us 00:00:54.410 --> 00:01:01.130 about SGDs, Stochastic Gradient Descent, critical algorithm. 00:01:01.130 --> 00:01:07.550 I mean that, at a quick look, stochastic gradient descent 00:01:07.550 --> 00:01:11.720 does smaller batches at a time, so it's computationally faster 00:01:11.720 --> 00:01:14.810 than pure gradient descent. 00:01:14.810 --> 00:01:17.150 But also, stochastic gradient descent 00:01:17.150 --> 00:01:20.990 is able to solve probabilistic problems where 00:01:20.990 --> 00:01:24.300 you're trying to minimize an expected value instead 00:01:24.300 --> 00:01:25.310 of a function. 00:01:25.310 --> 00:01:30.220 Anyway, it's an important thing. 00:01:30.220 --> 00:01:35.450 Actually, there is a lecture at 2:00 00:01:35.450 --> 00:01:42.020 in neuroscience about the gradient descent algorithms. 00:01:42.020 --> 00:01:45.110 And then Professor Boyd is speaking here at 4:30 00:01:45.110 --> 00:01:46.040 this afternoon. 00:01:46.040 --> 00:01:49.330 It's overwhelming. 00:01:49.330 --> 00:01:57.350 So I'm going to talk today about several topics that are 00:01:57.350 --> 00:02:01.250 specific parts of optimization. 00:02:01.250 --> 00:02:04.910 Linear programming is a very famous part of optimization. 00:02:04.910 --> 00:02:07.170 I don't know if you've met it. 00:02:07.170 --> 00:02:13.400 I think it's worth knowing what the inputs are 00:02:13.400 --> 00:02:17.270 and what's the key fact about linear programming. 00:02:17.270 --> 00:02:20.870 And, of course, also, what are the algorithms. 00:02:20.870 --> 00:02:23.150 That's usually what we want to know. 00:02:23.150 --> 00:02:25.460 What's the problem? 00:02:25.460 --> 00:02:27.980 What are the results, the mathematical results? 00:02:27.980 --> 00:02:30.910 And what are the computational tools? 00:02:30.910 --> 00:02:33.220 Yes, so that's very established. 00:02:33.220 --> 00:02:37.200 Max-flow min-cut is one specific linear programming 00:02:37.200 --> 00:02:41.780 that I think maybe would serve as the best example. 00:02:41.780 --> 00:02:46.310 And then two-person games, have you met those? 00:02:46.310 --> 00:02:48.830 So would you know what the rules are 00:02:48.830 --> 00:02:53.330 in a two-person zero-sum game? 00:02:53.330 --> 00:02:57.930 In game theory, those are the ones that are clear-- 00:02:57.930 --> 00:03:04.230 everything's well-established there. 00:03:04.230 --> 00:03:07.850 And, in fact, they're equivalent to a linear program. 00:03:07.850 --> 00:03:10.640 So that's why those two are coming in today. 00:03:10.640 --> 00:03:15.860 And then the key fact about two-person games and about 00:03:15.860 --> 00:03:17.870 linear programming is duality. 00:03:17.870 --> 00:03:20.060 So if there's any word on that board, 00:03:20.060 --> 00:03:25.640 it's the one in capital letters that has math content. 00:03:25.640 --> 00:03:28.190 I'm just going to start with linear programming 00:03:28.190 --> 00:03:35.190 and then move on. 00:03:35.190 --> 00:03:37.965 So what's a linear program? 00:03:37.965 --> 00:03:41.610 It's we're optimizing a linear cost function. 00:03:41.610 --> 00:03:46.580 So we're minimizing the cost c transpose x. 00:03:46.580 --> 00:03:50.810 So that vector x is the unknown that we're looking for, 00:03:50.810 --> 00:03:53.210 and this vector c is the cost vector. 00:03:53.210 --> 00:03:59.540 So that is c1 x1 plus, plus cn xn. 00:03:59.540 --> 00:04:02.450 So you can see why it's called linear programming. 00:04:02.450 --> 00:04:04.130 The cost is linear. 00:04:04.130 --> 00:04:06.560 The constraints are also linear. 00:04:06.560 --> 00:04:19.820 So the constraints on x are a set of linear equations. 00:04:19.820 --> 00:04:23.480 Of course, I'm not thinking of A as being a square invertible 00:04:23.480 --> 00:04:24.070 matrix. 00:04:24.070 --> 00:04:25.100 No way. 00:04:25.100 --> 00:04:27.680 If it were, that would tell us what x had to be 00:04:27.680 --> 00:04:29.420 and our problem would be over. 00:04:33.350 --> 00:04:36.630 We have n unknowns. 00:04:36.630 --> 00:04:39.470 This is m by n, of course. 00:04:39.470 --> 00:04:41.060 x is n by 1. 00:04:41.060 --> 00:04:44.000 That's our unknown vector. 00:04:44.000 --> 00:04:49.520 And I'm thinking that m would be smaller than n. 00:04:49.520 --> 00:04:52.250 But we do have constraints, and now comes 00:04:52.250 --> 00:04:56.780 the thing that makes linear programming not actually 00:04:56.780 --> 00:04:58.280 linear. 00:04:58.280 --> 00:05:03.410 And that's the constraint x greater or equal to 0. 00:05:03.410 --> 00:05:05.960 And I've written that as a vector 00:05:05.960 --> 00:05:14.840 but that means x1 greater or equal to 0, on to xn 00:05:14.840 --> 00:05:16.110 greater or equal to 0. 00:05:16.110 --> 00:05:19.010 So it's a vector inequality. 00:05:19.010 --> 00:05:24.080 So we have minimizing a very simple function 00:05:24.080 --> 00:05:28.170 with pretty straightforward constraints but inequality 00:05:28.170 --> 00:05:29.570 constraints. 00:05:29.570 --> 00:05:32.690 So the set of-- 00:05:32.690 --> 00:05:36.210 these two together are the constraints. 00:05:36.210 --> 00:05:38.060 And in linear algebra-- 00:05:38.060 --> 00:05:45.710 in linear programming language this is called a feasible set 00:05:45.710 --> 00:05:48.680 of x's. 00:05:48.680 --> 00:05:54.250 It's the constraint set, capital K in the notes. 00:05:54.250 --> 00:06:01.660 So let me draw a picture to show-- 00:06:01.660 --> 00:06:03.910 or maybe I'll just ask. 00:06:03.910 --> 00:06:06.040 How many of you are already familiar 00:06:06.040 --> 00:06:08.060 with linear programming? 00:06:08.060 --> 00:06:09.700 Oh, quite a lot. 00:06:09.700 --> 00:06:11.900 So I won't belabor the point. 00:06:11.900 --> 00:06:16.600 But let me just draw a picture. 00:06:16.600 --> 00:06:22.340 So my picture will be just here. 00:06:22.340 --> 00:06:25.730 So that's n equals 3. 00:06:25.730 --> 00:06:30.880 So this is the x1, x2, x3 space. 00:06:30.880 --> 00:06:34.450 And x greater or equal to 0 means what? 00:06:34.450 --> 00:06:40.830 It means that I'm in this 1/8 of this space, right? 00:06:40.830 --> 00:06:44.310 It's the non-negative-- well, I would say quadrant 00:06:44.310 --> 00:06:48.190 but I really should say octant because it's 00:06:48.190 --> 00:06:51.630 1/8 of the full 3D space. 00:06:51.630 --> 00:06:53.250 So I'm in here. 00:06:53.250 --> 00:06:59.250 And then maybe I have m as maybe just one constraint. 00:06:59.250 --> 00:07:08.880 So let me take the cost to be, say, x1 plus 2 x2 plus 5 x3, 00:07:08.880 --> 00:07:10.360 say. 00:07:10.360 --> 00:07:14.100 And the constraint, I'm just going to have one equation-- 00:07:14.100 --> 00:07:16.110 x1 plus x2. 00:07:16.110 --> 00:07:22.380 Let me make it easy, make that constraint an easy one to-- 00:07:22.380 --> 00:07:27.960 so this is what we want to minimize. 00:07:27.960 --> 00:07:33.210 And that's what we have to satisfy, as well 00:07:33.210 --> 00:07:35.070 as x greater or equal to 0. 00:07:35.070 --> 00:07:38.150 So that's a plane. 00:07:38.150 --> 00:07:41.410 One equation gives us a plane in R3. 00:07:41.410 --> 00:07:43.085 The plane would go through-- 00:07:47.222 --> 00:07:57.110 would hit the axes at 3, 0, 0. 00:07:57.110 --> 00:07:59.990 And so the min-- 00:07:59.990 --> 00:08:02.840 the point has to lie on that plane. 00:08:02.840 --> 00:08:07.250 And it has to lie in the octant, so that plane is chopped off 00:08:07.250 --> 00:08:08.960 to be a triangle. 00:08:08.960 --> 00:08:13.820 This is a good visualization of linear programming. 00:08:13.820 --> 00:08:15.470 And what's the conclusion? 00:08:15.470 --> 00:08:17.210 Well, our cost is linear. 00:08:19.880 --> 00:08:24.800 So the result is that one of these three corners 00:08:24.800 --> 00:08:27.860 is the winner. 00:08:27.860 --> 00:08:29.040 It could happen. 00:08:29.040 --> 00:08:32.570 It could happen that maybe I have equal values on those two 00:08:32.570 --> 00:08:35.480 corners, and therefore all the way along 00:08:35.480 --> 00:08:37.309 would be also winners. 00:08:37.309 --> 00:08:39.559 But when I have a linear function, 00:08:39.559 --> 00:08:41.960 it's a maximum at the ends. 00:08:41.960 --> 00:08:46.370 And these are the ends, those three corners. 00:08:46.370 --> 00:08:56.330 So 3, 0, 0, 0, 3, 0, or 0, 0, 3 are the three corners, 00:08:56.330 --> 00:08:59.640 and one of those is the winner and the problem is solved. 00:08:59.640 --> 00:09:03.290 And, in fact, for this case, since I'm minimizing, 00:09:03.290 --> 00:09:06.230 I guess it would be that corner that wins. 00:09:06.230 --> 00:09:09.080 So let me give it a star, not an x. 00:09:09.080 --> 00:09:10.480 Yes. 00:09:10.480 --> 00:09:12.980 3, 0, 0 because that-- 00:09:12.980 --> 00:09:13.600 oh. 00:09:13.600 --> 00:09:16.100 No, that's 3. 00:09:16.100 --> 00:09:24.980 So the value turned out to be 3 at the point 3, 0, 0. 00:09:24.980 --> 00:09:26.050 And that's x star. 00:09:30.290 --> 00:09:30.875 Good. 00:09:30.875 --> 00:09:33.770 Good, good, good. 00:09:33.770 --> 00:09:36.290 What more do I want to do, I said? 00:09:36.290 --> 00:09:37.530 Yes. 00:09:37.530 --> 00:09:39.322 AUDIENCE: Shouldn't it be at point 0, 0, 3? 00:09:41.780 --> 00:09:43.650 GILBERT STRANG: Should it be-- 00:09:43.650 --> 00:09:44.870 oh, no. 00:09:44.870 --> 00:09:47.730 0, 0, 3 would have x3 is 3. 00:09:47.730 --> 00:09:48.350 That's fine. 00:09:48.350 --> 00:09:49.630 It gives us a chance, I think. 00:09:49.630 --> 00:09:52.030 That would give me a cost of 15. 00:09:52.030 --> 00:09:53.930 This would give me a cost of 6, and that 00:09:53.930 --> 00:09:55.270 would give me a cost of 3. 00:09:55.270 --> 00:09:55.770 Yes. 00:09:58.690 --> 00:10:02.750 It's obvious that if we could enumerate all the corners, 00:10:02.750 --> 00:10:09.290 we would have a super fast way to get the answer. 00:10:09.290 --> 00:10:14.260 But the trouble is, of course, that for large values of m 00:10:14.260 --> 00:10:17.810 and n, there are exponentially many corners, 00:10:17.810 --> 00:10:21.830 and we don't want to see them all. 00:10:21.830 --> 00:10:30.050 So that's what makes linear programming take time and need 00:10:30.050 --> 00:10:31.200 ideas. 00:10:31.200 --> 00:10:33.800 So, for algorithms, there are two types 00:10:33.800 --> 00:10:39.370 of algorithms, two types of codes that solve these. 00:10:39.370 --> 00:10:45.320 The older, well-established ones are the-- 00:10:45.320 --> 00:10:56.860 so one way is the simplex method, which finds a corner. 00:10:56.860 --> 00:11:00.730 We know the optimum one of the corners. 00:11:00.730 --> 00:11:07.020 So it will find one corner, and it will go to the next corner. 00:11:09.970 --> 00:11:15.280 So if it starts at one of these corners, 00:11:15.280 --> 00:11:21.610 it will travel along an edge that lowers the cost. 00:11:21.610 --> 00:11:25.050 And it has to stop at the end because it's linear. 00:11:25.050 --> 00:11:27.610 The cost will keep going down and going down 00:11:27.610 --> 00:11:32.200 until it bumps into the end point, 00:11:32.200 --> 00:11:33.910 and then it can't go further. 00:11:33.910 --> 00:11:36.940 And then we would, from that next corner, we 00:11:36.940 --> 00:11:38.185 would go to the next corner. 00:11:42.940 --> 00:11:49.090 Each time, it's like steepest descent on the edges. 00:11:49.090 --> 00:11:52.550 From the first corner, we go the steepest way 00:11:52.550 --> 00:11:56.320 till we can't go further, we've hit another corner. 00:11:56.320 --> 00:11:59.020 And we recompute the linear algebra, 00:11:59.020 --> 00:12:03.880 find which direction is steepest from there. 00:12:03.880 --> 00:12:06.490 So that's the idea of the simplex method, 00:12:06.490 --> 00:12:09.190 which was invented by Dantzig. 00:12:15.900 --> 00:12:20.310 And the algebra is not going to be in today's lecture 00:12:20.310 --> 00:12:23.020 but it's straightforward. 00:12:23.020 --> 00:12:25.340 Well, people have to optimize it because that's 00:12:25.340 --> 00:12:28.200 a highly frequently used method. 00:12:28.200 --> 00:12:29.580 Yes. 00:12:29.580 --> 00:12:35.250 But then, about 20 years later maybe, or 30, I 00:12:35.250 --> 00:12:40.980 remember going to a lecture in downtown Boston by Karmarkar. 00:12:40.980 --> 00:12:44.360 So I have to put his name down-- 00:12:44.360 --> 00:12:44.995 Karmarkar. 00:12:50.670 --> 00:12:55.950 And he was in The New York Times, all the newspapers, 00:12:55.950 --> 00:12:58.890 so it was a big deal. 00:12:58.890 --> 00:13:04.190 He had an alternative algorithm. 00:13:04.190 --> 00:13:09.860 And the exact algorithm he proposed 00:13:09.860 --> 00:13:15.200 hasn't survived until today but the idea has. 00:13:15.200 --> 00:13:21.920 And the idea was to go to travel inside the feasible set and not 00:13:21.920 --> 00:13:23.420 around the edges. 00:13:23.420 --> 00:13:27.800 So his idea-- because in here, would maybe 00:13:27.800 --> 00:13:31.670 travel down near that edge, start again, travel again. 00:13:31.670 --> 00:13:38.900 So it's steepest descent in the constraint set, the feasible 00:13:38.900 --> 00:13:39.920 set. 00:13:39.920 --> 00:13:46.310 And you can use calculus and this idea. 00:13:46.310 --> 00:13:51.440 So this is interior point method. 00:13:51.440 --> 00:13:54.440 So I'll just use the word interior. 00:13:54.440 --> 00:14:01.550 That's telling us that we're inside the feasible set. 00:14:01.550 --> 00:14:04.490 We don't hit, come all the way to the boundary 00:14:04.490 --> 00:14:06.920 intentionally, because we want room 00:14:06.920 --> 00:14:10.490 to move, and to find derivatives, and to use 00:14:10.490 --> 00:14:13.780 Newton's method to minimize. 00:14:13.780 --> 00:14:17.930 You choose a search direction, just as all of optimization 00:14:17.930 --> 00:14:18.950 does. 00:14:18.950 --> 00:14:20.780 And in that search direction, you 00:14:20.780 --> 00:14:25.130 track it and you stop before you bump 00:14:25.130 --> 00:14:27.800 into the edge of the feasible set. 00:14:27.800 --> 00:14:30.010 And then you compute derivatives again. 00:14:30.010 --> 00:14:32.270 So you can use calculus. 00:14:32.270 --> 00:14:35.120 And it's a method that-- 00:14:35.120 --> 00:14:37.130 it's an idea that-- 00:14:37.130 --> 00:14:40.760 the interior idea came before Karmarkar. 00:14:40.760 --> 00:14:46.460 But he got super publicity, so it really got attention, 00:14:46.460 --> 00:14:48.020 got new thinking. 00:14:48.020 --> 00:14:52.080 And new ideas came partly from people at MIT. 00:14:56.140 --> 00:15:01.180 And these two are now still locked in competition. 00:15:01.180 --> 00:15:06.710 One hasn't beaten the other in all problems. 00:15:06.710 --> 00:15:11.390 So linear programming is here. 00:15:11.390 --> 00:15:13.760 But then, for nonlinear programming, 00:15:13.760 --> 00:15:17.330 quadratic programming, where the cost is quadratic, 00:15:17.330 --> 00:15:21.320 nonlinear programming, semi-definite programming-- 00:15:21.320 --> 00:15:24.890 that's where you have a matrix unknown and matrix 00:15:24.890 --> 00:15:27.200 constraints-- 00:15:27.200 --> 00:15:31.250 those are all-- the more complicated you get, 00:15:31.250 --> 00:15:35.960 the more it tends to be interior point methods. 00:15:35.960 --> 00:15:39.147 That's a summary of linear programming. 00:15:41.890 --> 00:15:44.500 Now I'd like to give an example. 00:15:44.500 --> 00:15:46.390 And then I haven't told you the main-- 00:15:46.390 --> 00:15:48.525 well, let me tell you the main fact about duality. 00:15:51.430 --> 00:15:55.090 This is really-- and I'll write that maybe here next to it. 00:15:58.550 --> 00:16:05.230 So duality is there's a dual program, a dual LP. 00:16:05.230 --> 00:16:12.770 And that dual is going to do a maximum instead of a minimum 00:16:12.770 --> 00:16:20.030 And the cost is going to involve the b from the primal problem. 00:16:20.030 --> 00:16:24.980 So this is now called the primal problem, and this is its dual. 00:16:24.980 --> 00:16:27.320 So they're twin problems. 00:16:27.320 --> 00:16:30.230 They use the same data but quite differently. 00:16:30.230 --> 00:16:33.030 It's like transposing things. 00:16:33.030 --> 00:16:36.890 So let's call it the unknown y. 00:16:36.890 --> 00:16:47.030 So for y1 to ym, I guess, because b, the right-hand side 00:16:47.030 --> 00:16:49.940 over here, is m by 1. 00:16:53.240 --> 00:16:59.190 So that's maximize that subject to-- what are the constraints? 00:16:59.190 --> 00:17:03.080 Well, the cost function over here, c, the cost vector, 00:17:03.080 --> 00:17:06.260 goes into the constraints. 00:17:06.260 --> 00:17:10.220 And I think the greater or equal sign probably becomes 00:17:10.220 --> 00:17:12.190 a less or equal to sign. 00:17:12.190 --> 00:17:15.650 The A gets transposed. 00:17:15.650 --> 00:17:20.660 I think that's probably the constraint in the dual. 00:17:20.660 --> 00:17:23.020 And it happens it doesn't need y. 00:17:23.020 --> 00:17:26.690 We don't have y greater or equal to 0 over here. 00:17:26.690 --> 00:17:28.820 So that's a dual problem. 00:17:28.820 --> 00:17:31.910 It has a linear cost. 00:17:31.910 --> 00:17:35.720 It has linear inequality constraints. 00:17:35.720 --> 00:17:38.480 It can be solved by the simplex method. 00:17:38.480 --> 00:17:41.570 You could choose whether you solve that one or this one, 00:17:41.570 --> 00:17:45.761 because if you solve one, you solve the other. 00:17:45.761 --> 00:17:48.530 The two are closely connected and that's 00:17:48.530 --> 00:17:51.140 the key idea of duality. 00:17:51.140 --> 00:17:56.745 So maybe I'll put the idea of, first of all, a weak duality. 00:18:03.030 --> 00:18:08.160 Which says that this quantity that we're trying to maximize-- 00:18:08.160 --> 00:18:11.180 we're getting it as large as possible-- 00:18:11.180 --> 00:18:18.630 is always less or equal to any c transpose x. 00:18:18.630 --> 00:18:28.020 This is for any feasible, allowed-- 00:18:28.020 --> 00:18:32.430 any x and y that satisfy the constraints. 00:18:32.430 --> 00:18:35.640 Remember, feasible means satisfies the constraints. 00:18:35.640 --> 00:18:39.300 So, in other words, this problem, this maximization 00:18:39.300 --> 00:18:42.190 problem, you're trying to push this up. 00:18:42.190 --> 00:18:46.020 The minimization problem, you're trying to pull this down. 00:18:46.020 --> 00:18:49.950 And it's easier to show that the one, the b 00:18:49.950 --> 00:18:52.860 transpose-- it's easier to show that inequality. 00:18:52.860 --> 00:18:54.360 Let's do that. 00:18:54.360 --> 00:18:56.150 So b transpose y proof. 00:18:58.730 --> 00:19:01.560 It should be a one-line proof. 00:19:01.560 --> 00:19:04.290 So b transpose y-- 00:19:06.990 --> 00:19:07.830 what do I do? 00:19:07.830 --> 00:19:11.950 So b transpose y well, I look over here. 00:19:11.950 --> 00:19:14.280 That's where b shows up. 00:19:14.280 --> 00:19:24.900 That's x transpose A transpose y, because b is Ax. 00:19:24.900 --> 00:19:29.730 I'm feasible, so my b is Ax. 00:19:29.730 --> 00:19:32.700 And for any x-- this is for this is for any 00:19:32.700 --> 00:19:35.930 x that b is going to be Ax. 00:19:35.930 --> 00:19:38.340 So am I good so far? 00:19:38.340 --> 00:19:40.730 And now what do I do? 00:19:40.730 --> 00:19:43.440 Now I look here. 00:19:43.440 --> 00:19:46.920 I see A transpose y sitting right in front of me. 00:19:46.920 --> 00:19:50.760 So I say, well, OK, less or equal to. 00:19:50.760 --> 00:19:52.560 And I guess I'm done. 00:19:52.560 --> 00:19:54.870 A transpose y is less or equal to c. 00:19:54.870 --> 00:19:59.880 So I have x transpose c, c transpose x. 00:19:59.880 --> 00:20:04.800 So this equals this, less or equal that. 00:20:04.800 --> 00:20:05.780 I've got it. 00:20:05.780 --> 00:20:11.820 Weak duality is just put together the requirements 00:20:11.820 --> 00:20:13.650 on x and y and you have it. 00:20:13.650 --> 00:20:19.600 But was this important? 00:20:22.260 --> 00:20:24.300 If the mathematics is right, everything 00:20:24.300 --> 00:20:27.840 has to have its place, play its role. 00:20:27.840 --> 00:20:32.460 And so what's the role of x greater or equal to 0 there? 00:20:32.460 --> 00:20:34.260 Why do we need x-- 00:20:34.260 --> 00:20:40.160 If we just even think of n equal 1, just a number, 00:20:40.160 --> 00:20:43.970 where do we use the fact that x is greater or equal to 0? 00:20:43.970 --> 00:20:45.020 Do we use it? 00:20:45.020 --> 00:20:52.975 This looks so smooth, because A transpose y-- 00:20:52.975 --> 00:20:58.880 I'll write that as x transpose c for the moment, 00:20:58.880 --> 00:21:03.720 just so your eye says that's the same x transpose, 00:21:03.720 --> 00:21:06.010 and the A transpose y is less or equal to c. 00:21:06.010 --> 00:21:08.926 Where is x greater or equal to 0 coming in? 00:21:08.926 --> 00:21:10.732 AUDIENCE: [INAUDIBLE]. 00:21:10.732 --> 00:21:11.565 GILBERT STRANG: Yes. 00:21:11.565 --> 00:21:12.060 AUDIENCE: Like above. 00:21:12.060 --> 00:21:12.905 It's from the that. 00:21:12.905 --> 00:21:14.280 GILBERT STRANG: Yes, that's true. 00:21:14.280 --> 00:21:17.070 But I want to see-- 00:21:17.070 --> 00:21:19.860 so here's my point. 00:21:19.860 --> 00:21:21.720 The fact that A transpose y is less 00:21:21.720 --> 00:21:23.880 or equal to c, that's fine. 00:21:23.880 --> 00:21:25.420 Highly important. 00:21:25.420 --> 00:21:30.300 But if x was negative, it would be the other way. 00:21:30.300 --> 00:21:31.890 You would go the other way and you 00:21:31.890 --> 00:21:33.780 wouldn't have what you want. 00:21:33.780 --> 00:21:38.220 So we really do use the fact that x is greater or equal to 0 00:21:38.220 --> 00:21:41.250 to say that I have this less or equal this. 00:21:41.250 --> 00:21:44.310 And then I multiply by something positive 00:21:44.310 --> 00:21:46.290 and then I still have less or equal. 00:21:46.290 --> 00:21:47.620 OK, good. 00:21:47.620 --> 00:21:50.800 So the math is right. 00:21:50.800 --> 00:21:53.160 Everything does its part. 00:21:53.160 --> 00:21:56.700 And then, of course, the beautiful result, 00:21:56.700 --> 00:21:59.680 the important result is that there 00:21:59.680 --> 00:22:03.330 is strong duality, just called duality, 00:22:03.330 --> 00:22:06.810 which is that at the maximum-- 00:22:06.810 --> 00:22:10.380 now this is not for any x and y but this is for-- 00:22:10.380 --> 00:22:14.700 this is going to be for x y star, the winner, 00:22:14.700 --> 00:22:16.560 and x star, the winner-- 00:22:16.560 --> 00:22:18.570 equality holds. 00:22:18.570 --> 00:22:19.500 So that's duality. 00:22:22.010 --> 00:22:24.540 The maximum in the dual problem is 00:22:24.540 --> 00:22:27.550 the same as the minimum in the primal problem. 00:22:27.550 --> 00:22:29.640 The two have met. 00:22:29.640 --> 00:22:32.550 There is no duality gap. 00:22:32.550 --> 00:22:36.360 In some cooked up nonlinear problems, 00:22:36.360 --> 00:22:41.620 there could be a gap between the maximum and the minimum, 00:22:41.620 --> 00:22:43.920 but you hope not. 00:22:43.920 --> 00:22:48.010 And here the big theorem is not. 00:22:48.010 --> 00:22:49.390 They're equal. 00:22:49.390 --> 00:22:53.580 You push this up, push that down. 00:22:53.580 --> 00:22:59.100 Another way to say that duality is that this is 00:22:59.100 --> 00:23:01.050 pushing that up at some max-- 00:23:01.050 --> 00:23:06.660 I could write that as a maximum of a minimum equaling a minimum 00:23:06.660 --> 00:23:08.660 of a maximum, if I wanted. 00:23:12.366 --> 00:23:16.270 So the duality in linear programming 00:23:16.270 --> 00:23:20.710 was the same as von Neumann's minimax theorem. 00:23:20.710 --> 00:23:25.300 And his theorem applied to two-person games. 00:23:25.300 --> 00:23:34.090 So the key math result is duality for linear programming, 00:23:34.090 --> 00:23:38.110 and it's going to be-- you'll see the same thing happening 00:23:38.110 --> 00:23:41.200 for two-person games. 00:23:41.200 --> 00:23:44.590 And it's a minimax theorem, or a saddle point. 00:23:44.590 --> 00:23:50.740 Or it's just things come out right. 00:23:50.740 --> 00:23:51.410 Yes. 00:23:51.410 --> 00:23:58.440 So just to mention that mathematical programming, 00:23:58.440 --> 00:24:03.440 of course, includes much more difficult problems. 00:24:03.440 --> 00:24:05.850 This is linear programming. 00:24:05.850 --> 00:24:10.790 That problem, as you see, has a beautiful, simple theory. 00:24:10.790 --> 00:24:16.410 And the paying attention is paying attention 00:24:16.410 --> 00:24:19.410 to the algorithm, because you've got two important choices 00:24:19.410 --> 00:24:21.770 and they both get highly developed. 00:24:24.840 --> 00:24:25.490 Now, OK. 00:24:25.490 --> 00:24:28.740 So for game, now I'm going to turn to-- 00:24:28.740 --> 00:24:33.930 well, I'll do an example of max flow equal min cut, 00:24:33.930 --> 00:24:35.900 just see what-- 00:24:35.900 --> 00:24:38.960 and then go to two-person game. 00:24:38.960 --> 00:24:42.100 So here's an example of a-- 00:24:52.920 --> 00:24:56.200 so I start with a graph. 00:24:56.200 --> 00:24:58.400 Let me just imagine. 00:25:04.330 --> 00:25:06.400 So this is the source. 00:25:09.220 --> 00:25:11.680 This is the sink. 00:25:11.680 --> 00:25:15.770 And I'm sending flow through the network. 00:25:15.770 --> 00:25:19.810 So it's a network that I'm sending flow through. 00:25:19.810 --> 00:25:21.760 So my job is to maximize-- 00:25:24.790 --> 00:25:27.380 I'll set x at the source to be 0. 00:25:32.620 --> 00:25:36.130 And then the flow, the total flow will come into the sink. 00:25:36.130 --> 00:25:42.520 So I want to maximize x at the sink. 00:25:42.520 --> 00:25:45.080 So it's a linear programming and there are constraints. 00:25:45.080 --> 00:25:47.370 So what are the constraints? 00:25:47.370 --> 00:25:50.130 Every edge has a capacity. 00:25:50.130 --> 00:25:51.240 So let's see. 00:25:51.240 --> 00:25:57.460 Suppose that edge has capacity 5. 00:25:57.460 --> 00:26:01.710 Let me put capacities on all these edges. 00:26:01.710 --> 00:26:07.860 2, 1, 3, 4. 00:26:07.860 --> 00:26:09.780 I'm just stabbing around here. 00:26:12.980 --> 00:26:15.710 1,000. 00:26:15.710 --> 00:26:22.310 And let me say 2 and 4. 00:26:22.310 --> 00:26:25.570 I have no idea what's happening here. 00:26:25.570 --> 00:26:31.170 But if we see the problem, we'll probably be able to solve it. 00:26:31.170 --> 00:26:36.710 So these constraints are that the flow variable, 00:26:36.710 --> 00:26:42.520 which would be the y that I'm trying to maximize, cannot-- 00:26:42.520 --> 00:26:45.340 the amount of flow, just in ordinary words, 00:26:45.340 --> 00:26:50.170 the amount of flow on the edges can't go higher 00:26:50.170 --> 00:26:51.520 than the capacity. 00:26:51.520 --> 00:26:54.880 I could send 11 along this edge, but then I've got nowhere 00:26:54.880 --> 00:26:57.870 to send it after that. 00:26:57.870 --> 00:27:00.670 I could send, well, 900 on that edge. 00:27:00.670 --> 00:27:04.980 but, obviously, it would get stuck there. 00:27:04.980 --> 00:27:09.250 So the question is, what's the maximum I 00:27:09.250 --> 00:27:11.620 can send through that network? 00:27:11.620 --> 00:27:15.670 It's a classical problem. 00:27:15.670 --> 00:27:20.400 And, in fact, it's an integer programming problem. 00:27:20.400 --> 00:27:22.950 These are all integer capacities. 00:27:22.950 --> 00:27:27.560 I could insist on integer flow. 00:27:27.560 --> 00:27:32.880 But it's a very remarkable integer problem 00:27:32.880 --> 00:27:37.080 because I could allow fractions and the answer 00:27:37.080 --> 00:27:40.020 would not be a-- would not involve fractions. 00:27:40.020 --> 00:27:44.730 In other words, if I keep it as an integer problem, 00:27:44.730 --> 00:27:49.420 then the mathematics is definitely 00:27:49.420 --> 00:27:51.660 harder for an integer problem. 00:27:51.660 --> 00:27:53.380 What's different? 00:27:53.380 --> 00:28:02.720 So the x's could be integers or they could be real numbers. 00:28:07.480 --> 00:28:11.000 Over here, they were real numbers. 00:28:11.000 --> 00:28:13.040 I happened to get an integer for this, 00:28:13.040 --> 00:28:20.540 but if I had 10 crossing planes in 15 dimensions, 00:28:20.540 --> 00:28:24.330 the integers would be totally lost. 00:28:24.330 --> 00:28:32.460 But the point is, here, that if I allow real numbers, 00:28:32.460 --> 00:28:36.360 it doesn't get me any more flow, that the winning 00:28:36.360 --> 00:28:38.490 flow is an integer anyway. 00:28:38.490 --> 00:28:43.350 So it's an integer problem which can be safely relaxed, 00:28:43.350 --> 00:28:48.240 and you can safely use simplex method, or Karmarkar's method, 00:28:48.240 --> 00:28:53.250 or any interior point method with non-integers, 00:28:53.250 --> 00:28:57.600 because in the interior here you're not starting or ending 00:28:57.600 --> 00:28:59.070 at integers. 00:28:59.070 --> 00:29:04.650 You can do it because the integer answer 00:29:04.650 --> 00:29:07.340 will be, in the end, better. 00:29:07.340 --> 00:29:09.480 And what is that answer? 00:29:09.480 --> 00:29:16.470 I think I've made those capacities too easy to-- 00:29:16.470 --> 00:29:17.940 oh, I didn't do this one. 00:29:24.000 --> 00:29:27.710 So what shall I-- shall I make it large, like 9? 00:29:27.710 --> 00:29:29.280 I don't know. 00:29:29.280 --> 00:29:40.460 What's the best we can do through that network? 00:29:40.460 --> 00:29:42.620 Can you-- I can't see it from here yet? 00:29:45.515 --> 00:29:47.580 What should I do? 00:29:47.580 --> 00:29:49.885 Obviously, in this simple problem 00:29:49.885 --> 00:29:54.360 I can get anything I want pretty much that far. 00:29:54.360 --> 00:29:56.040 But then what can I do? 00:29:58.990 --> 00:30:00.430 Let me even make that 19. 00:30:05.180 --> 00:30:10.710 So I'm not imposing much of a limit on that edge either. 00:30:10.710 --> 00:30:12.920 But these edges are quite tight limits, 00:30:12.920 --> 00:30:15.810 and these are sort of intermediate limits. 00:30:15.810 --> 00:30:22.140 What do you think is the most I can send through? 00:30:22.140 --> 00:30:25.800 And how would you show me that I couldn't send more? 00:30:25.800 --> 00:30:28.020 That's the key question. 00:30:28.020 --> 00:30:31.530 I want to get a bound on the-- 00:30:31.530 --> 00:30:34.250 an upper bound on this maximum. 00:30:34.250 --> 00:30:36.860 This maximum getting into the sink 00:30:36.860 --> 00:30:41.900 is less or equal to-- and what number would you propose? 00:30:41.900 --> 00:30:44.270 Could I get 1,000 through? 00:30:44.270 --> 00:30:49.460 I could start, but of course it would pile up there. 00:30:49.460 --> 00:30:50.810 I couldn't get further. 00:30:54.010 --> 00:30:57.574 What do you think is the best I can do? 00:30:57.574 --> 00:30:58.550 AUDIENCE: 10? 00:30:58.550 --> 00:30:59.860 GILBERT STRANG: 10? 00:30:59.860 --> 00:31:01.380 I can't do more than 10? 00:31:01.380 --> 00:31:04.340 How could I do 10? 00:31:04.340 --> 00:31:07.140 AUDIENCE: So you go 12-- 00:31:07.140 --> 00:31:08.710 GILBERT STRANG: 12 this way? 00:31:08.710 --> 00:31:09.293 AUDIENCE: Yes. 00:31:09.293 --> 00:31:10.242 19. 00:31:10.242 --> 00:31:11.200 GILBERT STRANG: Oh, OK. 00:31:11.200 --> 00:31:13.070 AUDIENCE: And then split the 8 and 2. 00:31:13.070 --> 00:31:14.733 Can you split? 00:31:14.733 --> 00:31:15.650 GILBERT STRANG: I see. 00:31:15.650 --> 00:31:19.280 10 that way, 10 this-- yes, absolutely you can split. 00:31:19.280 --> 00:31:23.650 And then this could go here, and that would be able to go there. 00:31:23.650 --> 00:31:25.910 So I can get 10 through. 00:31:25.910 --> 00:31:27.220 Correct. 00:31:27.220 --> 00:31:29.180 AUDIENCE: [INAUDIBLE]. 00:31:29.180 --> 00:31:31.190 GILBERT STRANG: Can I do anything better? 00:31:31.190 --> 00:31:35.900 Oh, I could be sending some up this way at the same time. 00:31:35.900 --> 00:31:38.950 So I could get 3 along the top. 00:31:38.950 --> 00:31:41.440 So this is like an auction. 00:31:41.440 --> 00:31:43.720 We can get up to 13. 00:31:43.720 --> 00:31:53.300 Can we get-- so that was 3 going this way and 10 going this way. 00:31:53.300 --> 00:31:58.248 Is 13 an-- can I not exceed 13? 00:31:58.248 --> 00:31:58.790 AUDIENCE: 14? 00:31:58.790 --> 00:32:00.670 AUDIENCE: 14. 00:32:00.670 --> 00:32:04.010 GILBERT STRANG: Do I hear 14? 00:32:04.010 --> 00:32:06.470 Oh, I've got a lot of room for one more. 00:32:06.470 --> 00:32:07.670 So, OK. 00:32:07.670 --> 00:32:10.440 So 14 in any case. 00:32:10.440 --> 00:32:10.940 All right. 00:32:10.940 --> 00:32:12.227 Do I hear 15? 00:32:12.227 --> 00:32:14.060 AUDIENCE: Can you do two more on the bottom, 00:32:14.060 --> 00:32:15.710 like 12 instead of 10? 00:32:15.710 --> 00:32:18.350 GILBERT STRANG: If I did 12, then what would I do? 00:32:18.350 --> 00:32:20.380 AUDIENCE: Split the other two up. 00:32:20.380 --> 00:32:23.170 GILBERT STRANG: I'd send two up here. 00:32:23.170 --> 00:32:24.920 What am I going to do with them from here? 00:32:24.920 --> 00:32:26.027 AUDIENCE: [INAUDIBLE]. 00:32:26.027 --> 00:32:27.110 GILBERT STRANG: Send them? 00:32:27.110 --> 00:32:28.070 AUDIENCE: Up again. 00:32:28.070 --> 00:32:30.630 GILBERT STRANG: Up again, and along. 00:32:30.630 --> 00:32:37.850 But then the 3 that I had right now would be cut back to 1. 00:32:41.430 --> 00:32:46.780 It's a lot of fun, this max flow problem. 00:32:46.780 --> 00:32:49.260 And I'm looking for a bound to know 00:32:49.260 --> 00:32:51.990 when to quit, to know when I've optimized. 00:32:51.990 --> 00:32:53.850 That's the whole idea of duality, 00:32:53.850 --> 00:33:00.940 is to find some upper thing that I'm trying to push down 00:33:00.940 --> 00:33:03.350 but I can't go beyond it. 00:33:03.350 --> 00:33:06.540 So I don't think I could get more than-- 00:33:06.540 --> 00:33:09.040 you see, everything has to cross this middle. 00:33:09.040 --> 00:33:10.610 So 3 and 4. 00:33:10.610 --> 00:33:13.540 And I don't think I could beat 23. 00:33:13.540 --> 00:33:18.360 If somebody said more than 23, I would be very doubtful, 00:33:18.360 --> 00:33:21.352 because I couldn't get it across the middle. 00:33:21.352 --> 00:33:24.390 But then, can I get 23? 00:33:24.390 --> 00:33:25.920 I doubt it. 00:33:25.920 --> 00:33:28.840 Maybe 14 is the best possible. 00:33:28.840 --> 00:33:31.330 So how would we show that 14 is the best possible? 00:33:35.330 --> 00:33:37.760 I think if I could find-- 00:33:37.760 --> 00:33:43.610 so this is called a cut, a cut in this network. 00:33:43.610 --> 00:33:45.185 Oh, yes, I see a cut. 00:33:45.185 --> 00:33:46.985 A cut like there. 00:33:50.310 --> 00:33:51.750 You see that? 00:33:51.750 --> 00:33:58.180 Every bit of flow has got to cross that cut. 00:33:58.180 --> 00:34:03.640 And the total capacity crossing is the 3 and the 1 and the 2 00:34:03.640 --> 00:34:06.510 and the 8, which is 14. 00:34:06.510 --> 00:34:08.590 So I can't get more than 14 through. 00:34:08.590 --> 00:34:13.620 And somehow that cut is loaded to capacity. 00:34:13.620 --> 00:34:19.494 Probably those edges all have to be fully up 00:34:19.494 --> 00:34:23.530 to capacity that cross the cut. 00:34:23.530 --> 00:34:27.290 So the cut is a separation of edges 00:34:27.290 --> 00:34:29.800 that go with the source and-- 00:34:29.800 --> 00:34:32.110 sorry-- nodes that go with the source, nodes 00:34:32.110 --> 00:34:33.510 that go with the sink. 00:34:33.510 --> 00:34:34.010 Yes. 00:34:37.030 --> 00:34:39.219 And then it's the edges across the cut. 00:34:42.820 --> 00:34:45.739 Is that OK for an example? 00:34:45.739 --> 00:34:47.679 So that's the duality. 00:34:47.679 --> 00:34:51.679 The maximum flow turned out to be 14, 00:34:51.679 --> 00:34:54.139 and the minimum cut turned out to be 14. 00:34:54.139 --> 00:34:58.110 And when those match, 14 equal 14, I know I'm through. 00:34:58.110 --> 00:35:00.560 I know I'm through because I'm able to get 14 through 00:35:00.560 --> 00:35:02.560 and I could never get more. 00:35:02.560 --> 00:35:03.060 Yes. 00:35:03.060 --> 00:35:04.880 3 and 4 and 6 and 8. 00:35:04.880 --> 00:35:06.770 Yes. 00:35:06.770 --> 00:35:11.180 And, of course, in a big network, 00:35:11.180 --> 00:35:15.110 the maximum cut is not going to be visible. 00:35:15.110 --> 00:35:18.590 Well, you couldn't-- it would have thousands of edges. 00:35:18.590 --> 00:35:21.080 You couldn't see what you were doing, 00:35:21.080 --> 00:35:26.660 But you could solve this problem fast, actually fast. 00:35:26.660 --> 00:35:28.440 And it's an important-- 00:35:28.440 --> 00:35:31.210 in practice, it's an important example. 00:35:31.210 --> 00:35:36.740 A lot of other things fit into the max flow min cut example. 00:35:36.740 --> 00:35:40.310 And therefore solving it in faster than-- oh, 00:35:40.310 --> 00:35:44.150 I didn't even say about speed. 00:35:44.150 --> 00:35:49.760 So the simplex method, almost always it's average case. 00:35:49.760 --> 00:35:51.790 Dan Spielman, who was on the faculty 00:35:51.790 --> 00:35:58.520 here, who's just terrific in this area, 00:35:58.520 --> 00:36:03.260 was maybe among the first to study the average case. 00:36:03.260 --> 00:36:08.960 So for an average choice of A and b and c, 00:36:08.960 --> 00:36:11.180 instead of making the worst choice-- 00:36:11.180 --> 00:36:15.050 for the worst choice, you can create a feasible set 00:36:15.050 --> 00:36:17.030 that it has to go corner, corner, corner, 00:36:17.030 --> 00:36:22.310 corner to get to the end, so the simplex method would 00:36:22.310 --> 00:36:24.890 take exponentially long. 00:36:24.890 --> 00:36:27.020 But that's extremely rare. 00:36:27.020 --> 00:36:30.020 It doesn't happen in practice, I think. 00:36:30.020 --> 00:36:33.590 And the average one is a polynomial, 00:36:33.590 --> 00:36:40.580 so it's a fast method on average. 00:36:40.580 --> 00:36:43.880 And we can show-- actually, that was a famous result that 00:36:43.880 --> 00:36:51.260 came from Russia, that linear programming is in the big P 00:36:51.260 --> 00:36:55.220 versus LP world, linear programming 00:36:55.220 --> 00:36:57.620 goes with P. Linear programming is 00:36:57.620 --> 00:37:01.480 a problem that can be solved in polynomial time. 00:37:01.480 --> 00:37:05.700 The simplex method won't always do it but it can be done. 00:37:05.700 --> 00:37:06.200 Yes. 00:37:06.200 --> 00:37:10.430 There came into the world, the ellipsoid methods, 00:37:10.430 --> 00:37:15.650 and it was an exciting time to decide that linear programming 00:37:15.650 --> 00:37:18.400 was actually P and not LP. 00:37:22.900 --> 00:37:27.130 So this is a case of duality where you can understand 00:37:27.130 --> 00:37:32.620 duality by the statement that the flow cannot exceed 00:37:32.620 --> 00:37:34.720 the capacity of the cut. 00:37:34.720 --> 00:37:42.460 So that's what duality is, that any flow has 00:37:42.460 --> 00:37:51.955 to be less or equal the capacity of any cut, 00:37:51.955 --> 00:37:56.650 the capacity being the sum of the ones. 00:37:56.650 --> 00:37:58.720 So, of course, there are other cuts. 00:37:58.720 --> 00:38:01.810 That cut has to be crossed, but that's 00:38:01.810 --> 00:38:05.710 gone 3, 7, 14, 22 capacity. 00:38:05.710 --> 00:38:11.030 So the 14 capacity was the minimum, 00:38:11.030 --> 00:38:12.620 and that gave the maximum. 00:38:12.620 --> 00:38:13.120 Nice. 00:38:13.120 --> 00:38:14.060 Isn't it nice? 00:38:14.060 --> 00:38:14.560 Yes. 00:38:17.100 --> 00:38:22.860 So there is another world here of two-person games 00:38:22.860 --> 00:38:26.190 that also has duality and could be expressed. 00:38:26.190 --> 00:38:31.680 So let me just talk finally today about two-person games. 00:38:31.680 --> 00:38:35.430 So the two persons are x and y of course. 00:38:35.430 --> 00:38:37.770 What other names could they have? 00:38:37.770 --> 00:38:44.670 And there is a matrix involved. 00:38:44.670 --> 00:38:47.325 This is the payoff matrix. 00:38:51.690 --> 00:38:53.930 So let me take a very simple game first. 00:38:56.690 --> 00:39:01.890 So x is going to choose one of these rows, 00:39:01.890 --> 00:39:05.040 and y is going to choose one of these columns. 00:39:05.040 --> 00:39:07.920 And let's say payoff from x to y. 00:39:11.760 --> 00:39:16.980 I like it that way because, then, x, who's paying, 00:39:16.980 --> 00:39:21.090 is going to be minimizing, as x was up here. 00:39:21.090 --> 00:39:24.170 And y, who's collecting, is going 00:39:24.170 --> 00:39:28.070 to be maximizing, as in the dual. 00:39:28.070 --> 00:39:30.660 So the payoff-- well, let's see. 00:39:30.660 --> 00:39:37.050 I think maybe 1, 4, 2, 8 would be probably a fairly easy game 00:39:37.050 --> 00:39:39.900 to play. 00:39:39.900 --> 00:39:43.290 So it's a two-person game, a zero-sum game. 00:39:43.290 --> 00:39:54.630 Zero sum means-- that means all pay-- 00:39:54.630 --> 00:39:58.026 what x pays goes to y. 00:39:58.026 --> 00:40:03.840 y gets all that x pays. 00:40:03.840 --> 00:40:08.250 There's no third party here. 00:40:08.250 --> 00:40:09.300 No lawyers involved. 00:40:13.310 --> 00:40:16.280 And y is going to choose a column, 00:40:16.280 --> 00:40:18.870 and x is going to choose a row. 00:40:18.870 --> 00:40:23.790 And x wants to make it small, and y wants to make it big. 00:40:23.790 --> 00:40:26.100 So what happens in this game? 00:40:26.100 --> 00:40:28.230 What does y choose to make it big? 00:40:28.230 --> 00:40:29.760 The second column. 00:40:29.760 --> 00:40:31.930 What does x choose to make it small? 00:40:31.930 --> 00:40:33.420 The first row. 00:40:33.420 --> 00:40:37.510 So it's going to be focused in on that 2, 00:40:37.510 --> 00:40:42.920 because if y keeps choosing that column, 2 is the best-- 00:40:42.920 --> 00:40:46.170 the least that-- x is going to have to pay 2, 00:40:46.170 --> 00:40:50.550 and he achieves that by choosing the first row. 00:40:50.550 --> 00:40:52.820 So it's a simple game. 00:40:52.820 --> 00:40:54.890 That's a saddle point. 00:40:54.890 --> 00:40:59.000 It's a minimum for y in its column, 00:40:59.000 --> 00:41:03.090 and it's a maximum for x in its row. 00:41:03.090 --> 00:41:05.400 But, of course, a matrix, another matrix 00:41:05.400 --> 00:41:08.380 might not have such a saddle point. 00:41:08.380 --> 00:41:09.630 So let me-- do you see-- 00:41:09.630 --> 00:41:12.960 OK with this game? 00:41:12.960 --> 00:41:15.090 That would be a sort of straightforward game 00:41:15.090 --> 00:41:17.460 where simple strategy-- 00:41:17.460 --> 00:41:22.320 column 2 every time, row 1 every time is the optimal. 00:41:22.320 --> 00:41:24.570 But now let me just exchange that. 00:41:29.980 --> 00:41:33.440 So, again, x1 and x2. 00:41:33.440 --> 00:41:36.230 y1 and y2 are the two columns. 00:41:36.230 --> 00:41:37.100 What happens now? 00:41:40.740 --> 00:41:43.641 Well, y kind of likes the big number 8 there. 00:41:46.940 --> 00:41:48.550 So it goes for the second. 00:41:48.550 --> 00:41:51.580 The second column still has the bigger numbers. 00:41:51.580 --> 00:41:55.180 So y aims for column 2. 00:41:55.180 --> 00:41:58.440 But then what does x do? 00:41:58.440 --> 00:42:03.143 What row does x choose if y is in the column 2? 00:42:03.143 --> 00:42:04.010 AUDIENCE: Second. 00:42:04.010 --> 00:42:05.520 GILBERT STRANG: The second. 00:42:05.520 --> 00:42:06.160 OK. 00:42:06.160 --> 00:42:09.600 So have I found a saddle point? 00:42:09.600 --> 00:42:14.640 Where y chooses this column, x chooses this row, 00:42:14.640 --> 00:42:16.810 is that a saddle point? 00:42:16.810 --> 00:42:17.310 No. 00:42:17.310 --> 00:42:20.100 Because what will y do now? 00:42:20.100 --> 00:42:24.480 He sees x choosing the second row all the time, 00:42:24.480 --> 00:42:29.280 and y sees a 4, a very tempting 4, in that row. 00:42:29.280 --> 00:42:35.040 So y is going to choose y1, the first column, pretty often. 00:42:35.040 --> 00:42:38.580 Not all the time, because what happens if he just-- 00:42:38.580 --> 00:42:42.300 if y chooses this column all the time, 00:42:42.300 --> 00:42:45.270 then x will choose this row and the payoff 00:42:45.270 --> 00:42:50.880 would only be 1, so that things went downhill for y there. 00:42:50.880 --> 00:42:53.950 So this is not a saddle point. 00:42:53.950 --> 00:42:54.840 And what do we do? 00:43:00.340 --> 00:43:07.390 Mixed strategy, which is y will choose the two columns 00:43:07.390 --> 00:43:10.770 with some probabilities that add to 1. 00:43:10.770 --> 00:43:19.360 So there will be a third possibility, that p x1 and 1-- 00:43:19.360 --> 00:43:22.840 I'm sorry-- p y1. 00:43:22.840 --> 00:43:28.120 p times this column and 1 minus p times the second column. 00:43:28.120 --> 00:43:38.410 So that will be p plus 1 minus p times 8, and 4p plus 1 00:43:38.410 --> 00:43:41.250 minus p times 2. 00:43:41.250 --> 00:43:44.170 That's this mixed column. 00:43:44.170 --> 00:43:51.550 By mixing his strategy, we have a strategy 00:43:51.550 --> 00:43:53.050 like a third person, y3. 00:43:55.630 --> 00:44:03.520 And, of course, the x is going to notice after a while what 00:44:03.520 --> 00:44:06.350 the strategy is. 00:44:06.350 --> 00:44:09.650 This is an open competition. 00:44:09.650 --> 00:44:13.040 You're not hiding-- you're not able to hide anything. 00:44:13.040 --> 00:44:16.220 You might think, well, maybe y will jump around, 00:44:16.220 --> 00:44:19.620 but that's foolishness. 00:44:19.620 --> 00:44:24.470 y is going to end up finding the best 00:44:24.470 --> 00:44:26.880 choice of p between 0 and 1. 00:44:29.510 --> 00:44:32.060 And it will actually be between 0 and 1 00:44:32.060 --> 00:44:38.010 because the extreme strategies of column 1 and column 2 00:44:38.010 --> 00:44:39.760 were not winners. 00:44:39.760 --> 00:44:42.290 But now x is going to do the same thing. 00:44:42.290 --> 00:44:44.660 p x1-- he's going to-- 00:44:44.660 --> 00:44:51.160 he or she is going to combine those rows. 00:44:51.160 --> 00:44:58.420 So this would be p1 for the first row, and 1-- 00:44:58.420 --> 00:45:03.900 sorry-- p-- oh, I better choose another number. 00:45:03.900 --> 00:45:05.510 What's another letter than p? 00:45:05.510 --> 00:45:06.460 AUDIENCE: q. 00:45:06.460 --> 00:45:08.620 GILBERT STRANG: q? 00:45:08.620 --> 00:45:09.810 OK, q. 00:45:14.820 --> 00:45:21.360 So this row is a combination of those rows, so it's q plus 1 00:45:21.360 --> 00:45:32.500 minus q times 4 in the first position, and p times 8 and 1 00:45:32.500 --> 00:45:35.760 minus p-- q times 8. 00:45:35.760 --> 00:45:39.940 q times 8 and 1 minus q times 2. 00:45:39.940 --> 00:45:44.242 I'm sorry that I've written this too small. 00:45:44.242 --> 00:45:46.126 But what's going to happen? 00:45:50.370 --> 00:45:54.200 What's going to determine p and q and solve the game? 00:45:57.840 --> 00:46:04.100 Well, if these are equal, then what happens? 00:46:04.100 --> 00:46:08.180 Then x is OK with either one. 00:46:10.790 --> 00:46:14.330 x has nothing to choose if those are equal 00:46:14.330 --> 00:46:18.740 and y is staying with his mixed strategy, which gives him 00:46:18.740 --> 00:46:22.640 this third combination column. 00:46:22.640 --> 00:46:24.620 Then x is good. 00:46:24.620 --> 00:46:28.730 Now, you might say, well, then x could do what it wanted. 00:46:28.730 --> 00:46:31.850 But x couldn't do what it wanted. 00:46:31.850 --> 00:46:37.850 If x doesn't stick with the optimal strategy q for x, 00:46:37.850 --> 00:46:40.040 then y will take advantage. 00:46:40.040 --> 00:46:45.460 So, really, the game settles down to once-- 00:46:45.460 --> 00:46:47.630 so these should be equal. 00:46:47.630 --> 00:46:51.770 So when those are equal, what do I get? 00:46:51.770 --> 00:46:57.110 I have p-- it looks like I have 8 minus 7p 00:46:57.110 --> 00:46:59.000 there for the first one. 00:46:59.000 --> 00:47:01.670 And here, I have 4p minus 2p. 00:47:01.670 --> 00:47:04.640 That's 2p plus 1. 00:47:04.640 --> 00:47:07.074 Did I get that possibly right? 00:47:07.074 --> 00:47:08.038 AUDIENCE: Plus 2. 00:47:11.412 --> 00:47:15.550 GILBERT STRANG: 2, 4p minus 2p is the 2p. 00:47:15.550 --> 00:47:16.690 Oh, it's 2. 00:47:16.690 --> 00:47:17.640 Yes. 00:47:17.640 --> 00:47:19.190 OK. 00:47:19.190 --> 00:47:20.770 So those are equal. 00:47:20.770 --> 00:47:25.635 And that tells me that 9p is 6, and p is 2/3. 00:47:29.300 --> 00:47:33.800 So it turned out that the best strategy for y 00:47:33.800 --> 00:47:39.770 is 2/3 on the first column which didn't look so promising, 00:47:39.770 --> 00:47:42.950 and 1/3 on the second column. 00:47:42.950 --> 00:47:48.210 And by creating that strategy, what do these numbers come out 00:47:48.210 --> 00:47:48.710 to be? 00:47:48.710 --> 00:47:50.730 And they're supposed to come out equal. 00:47:50.730 --> 00:47:57.040 So 4p plus-- so 4 times 2/3 is 8/3, 00:47:57.040 --> 00:48:00.320 plus 1 minus p is 1/3 times-- 00:48:00.320 --> 00:48:03.140 plus 2/3 is 10/3. 00:48:03.140 --> 00:48:06.260 So I think that this is 10/3. 00:48:06.260 --> 00:48:10.250 And if that one isn't 10/3, I'm very surprised. 00:48:10.250 --> 00:48:14.220 So that's 2/3-- help. 00:48:14.220 --> 00:48:16.172 Is it? 00:48:16.172 --> 00:48:23.140 4p, 2p, 4/3. 00:48:23.140 --> 00:48:27.860 Yes, it's telling me they're the same. 00:48:27.860 --> 00:48:29.070 Did I set them equal? 00:48:29.070 --> 00:48:29.570 Yes. 00:48:29.570 --> 00:48:30.920 I set them equal to find p. 00:48:30.920 --> 00:48:33.330 So they have to be the same. 00:48:33.330 --> 00:48:38.070 Except for instructor mistakes, which never happen, I suppose. 00:48:38.070 --> 00:48:38.910 OK. 00:48:38.910 --> 00:48:41.580 So those should be equal. 00:48:41.580 --> 00:48:48.330 Then x has no way to do better, choose these columns. 00:48:48.330 --> 00:48:50.220 But he can't choose those columns freely-- 00:48:50.220 --> 00:48:55.590 those rows freely because y could take advantage, unless-- 00:48:55.590 --> 00:49:00.990 so this would give the best strategy for q-- 00:49:00.990 --> 00:49:04.030 sorry-- the best q for x. 00:49:04.030 --> 00:49:04.530 Yes. 00:49:07.533 --> 00:49:08.700 Do you see that picture now? 00:49:08.700 --> 00:49:10.860 There could be other-- 00:49:10.860 --> 00:49:13.470 it could be a much bigger matrix, of course. 00:49:13.470 --> 00:49:15.150 There could be other columns. 00:49:15.150 --> 00:49:19.830 Suppose there was a 0, 0 column. 00:49:19.830 --> 00:49:21.630 What difference-- what would be the effect 00:49:21.630 --> 00:49:25.620 on the optimal strategy of having that additional column, 00:49:25.620 --> 00:49:29.820 that additional option for y3? 00:49:29.820 --> 00:49:32.130 Well, he wouldn't take it. 00:49:32.130 --> 00:49:34.850 So let's make it more tempting. 00:49:34.850 --> 00:49:38.730 10, 10, 10. 00:49:38.730 --> 00:49:41.360 Oh, he would take that, right? 00:49:41.360 --> 00:49:41.860 Yes. 00:49:41.860 --> 00:49:45.760 That was not-- so I'm not sure what I am trying to say here. 00:49:49.360 --> 00:49:51.150 Yes. 00:49:51.150 --> 00:49:54.900 There certainly could be-- there could be rows that-- 00:49:54.900 --> 00:49:59.770 or columns, or rows for x, that don't enter the mixed, 00:49:59.770 --> 00:50:02.040 the optimal mixed strategy. 00:50:02.040 --> 00:50:04.060 A mixed strategy is some combination 00:50:04.060 --> 00:50:08.350 of strategies, some combination of pure strategies, 00:50:08.350 --> 00:50:11.500 like choose this column, choose this row. 00:50:11.500 --> 00:50:15.820 But some columns may not-- or rows-- may not 00:50:15.820 --> 00:50:17.800 show up in the best mixture. 00:50:17.800 --> 00:50:22.342 So I won't complete that partial thought there. 00:50:26.700 --> 00:50:31.260 So do you see that we have a duality theorem? 00:50:31.260 --> 00:50:35.470 First of all, you could write that as a linear program. 00:50:35.470 --> 00:50:39.030 The unknown is p for x-- 00:50:39.030 --> 00:50:40.190 for y. 00:50:40.190 --> 00:50:42.630 The unknown is q for x. 00:50:42.630 --> 00:50:50.370 So, actually, you just have one unknown in this simple problem, 00:50:50.370 --> 00:50:52.200 for this small problem. 00:50:52.200 --> 00:50:55.690 But you have a minimization, a maximization. 00:50:55.690 --> 00:50:58.260 They meet at the optimum. 00:50:58.260 --> 00:51:05.460 And the duality theorem becomes the theorem 00:51:05.460 --> 00:51:08.040 for two-person games. 00:51:08.040 --> 00:51:12.330 Linear programming matches two-person games. 00:51:12.330 --> 00:51:16.430 The point about two persons is very important. 00:51:16.430 --> 00:51:20.400 Three-person games are incredibly complicated. 00:51:20.400 --> 00:51:27.600 No theory at this simple level would solve three-person games. 00:51:27.600 --> 00:51:32.760 So that's where John Nash's Nobel Prize 00:51:32.760 --> 00:51:35.830 comes into the picture. 00:51:35.830 --> 00:51:38.340 So his Nobel Prize was in economics 00:51:38.340 --> 00:51:40.710 because of the wide applications, 00:51:40.710 --> 00:51:45.060 but he was able to analyze the problem of-- 00:51:45.060 --> 00:51:47.895 and for functions more than for matrices-- 00:51:47.895 --> 00:51:53.760 of a three-person or n-person game. 00:51:53.760 --> 00:51:57.070 So you know the story of John Nash? 00:51:57.070 --> 00:52:01.070 The book A Beautiful Mind and the movie A Beautiful Mind. 00:52:01.070 --> 00:52:03.560 So it's one of those movies that involves MIT 00:52:03.560 --> 00:52:05.750 because he was here. 00:52:05.750 --> 00:52:09.760 But you can't recognize anything about MIT in the movie. 00:52:09.760 --> 00:52:12.762 It's like Good Will Hunting. 00:52:12.762 --> 00:52:17.360 You're maybe in some basement of some remote building. 00:52:17.360 --> 00:52:20.360 But anyway. 00:52:20.360 --> 00:52:22.550 It was of course a tragic-- 00:52:22.550 --> 00:52:26.750 tragic, and then cheer, then wonderful, and then tragic 00:52:26.750 --> 00:52:29.060 again for John Nash. 00:52:29.060 --> 00:52:33.650 He had-- so when I met him, he was 00:52:33.650 --> 00:52:36.470 going to teach linear algebra, one section, 00:52:36.470 --> 00:52:39.530 but that never developed. 00:52:39.530 --> 00:52:41.310 That was when he was-- 00:52:41.310 --> 00:52:45.710 his mental state was going downhill. 00:52:45.710 --> 00:52:53.010 And he moved to Princeton and just stayed. 00:52:53.010 --> 00:52:55.640 And then the wonderful thing was that he improved, 00:52:55.640 --> 00:52:58.820 and then the very sad thing was his death 00:52:58.820 --> 00:53:04.640 in a car accident on the way home from the Nobel Prize. 00:53:04.640 --> 00:53:09.090 So quite a story, an amazing story. 00:53:09.090 --> 00:53:10.430 Yes. 00:53:10.430 --> 00:53:17.180 So that's specific optimization problems, linear programming 00:53:17.180 --> 00:53:19.250 and two-person games. 00:53:19.250 --> 00:53:27.240 And I hope that Friday it will be Professor Sra. Anyway, 00:53:27.240 --> 00:53:30.540 the lecture will certainly be about stochastic gradient 00:53:30.540 --> 00:53:31.200 descent. 00:53:31.200 --> 00:53:31.800 Good. 00:53:31.800 --> 00:53:33.350 Thanks.