WEBVTT 00:00:00.080 --> 00:00:02.430 The following content is provided under a Creative 00:00:02.430 --> 00:00:03.810 Commons license. 00:00:03.810 --> 00:00:06.060 Your support will help MIT OpenCourseWare 00:00:06.060 --> 00:00:10.150 continue to offer high quality educational resources for free. 00:00:10.150 --> 00:00:12.700 To make a donation or to view additional materials 00:00:12.700 --> 00:00:16.600 from hundreds of MIT courses, visit MIT OpenCourseWare 00:00:16.600 --> 00:00:17.263 at ocw.mit.edu. 00:00:26.760 --> 00:00:28.260 PROFESSOR: Today we have a lecturer, 00:00:28.260 --> 00:00:32.140 guest lecture two of two, Costis Daskalakis. 00:00:32.140 --> 00:00:33.960 COSTIS DASKALAKIS: Glad to be back. 00:00:33.960 --> 00:00:37.480 So let's continue on the path we followed last time. 00:00:37.480 --> 00:00:40.110 Let me remind you what we did last time, first of all. 00:00:40.110 --> 00:00:43.010 So I talked about interesting theorems 00:00:43.010 --> 00:00:47.010 in topology-- Nash, Sperner, and Brouwer. 00:00:47.010 --> 00:00:52.160 And I defined the corresponding-- 00:00:52.160 --> 00:00:53.970 so these were theorems in topology. 00:00:53.970 --> 00:00:57.820 Define the corresponding problems. 00:00:57.820 --> 00:01:00.990 And because of these existence theorems, 00:01:00.990 --> 00:01:04.300 the corresponding search problems were total. 00:01:04.300 --> 00:01:10.260 And then I looked into the problems in NP that are total, 00:01:10.260 --> 00:01:17.010 and I tried to identify what in these problems make them total 00:01:17.010 --> 00:01:20.040 and tried to identify combinatorial argument that 00:01:20.040 --> 00:01:24.150 guarantees the existence of solutions in these problems. 00:01:24.150 --> 00:01:26.850 Motivated by the argument, which turned out 00:01:26.850 --> 00:01:29.130 to be a parity argument on directed graphs, 00:01:29.130 --> 00:01:34.460 I defined the class PPAD, and I introduced the problem 00:01:34.460 --> 00:01:44.660 of ArithmCircuitSAT, which is PPAD complete, and from which 00:01:44.660 --> 00:01:48.330 I promised to show a bunch of PPAD hardness deductions 00:01:48.330 --> 00:01:49.530 this time. 00:01:49.530 --> 00:01:57.280 So let me remind you the salient points from this list 00:01:57.280 --> 00:01:58.130 before I keep going. 00:01:58.130 --> 00:02:02.020 So first of all, the PPAD class has a combinatorial flavor. 00:02:02.020 --> 00:02:04.250 In the definition of the class, what I'm doing 00:02:04.250 --> 00:02:09.610 is I'm defining a graph on all possible n-bit 00:02:09.610 --> 00:02:13.790 strings, so an exponentially large set 00:02:13.790 --> 00:02:16.370 by providing two circuits, P and N. 00:02:16.370 --> 00:02:22.260 P is a circuit of possible father, 00:02:22.260 --> 00:02:25.920 and N is the circuit of possible child. 00:02:25.920 --> 00:02:29.780 And given these two circuits, I establish a directed edge 00:02:29.780 --> 00:02:34.110 between string v1 and string v2, if they agree on their parent 00:02:34.110 --> 00:02:37.730 relationship, meaning v2 believes V1 is his father, 00:02:37.730 --> 00:02:41.280 and also v1 believes v2 is its child. 00:02:41.280 --> 00:02:43.745 In that case if this condition is true, 00:02:43.745 --> 00:02:47.020 I establish an edge between these two pairs of nodes. 00:02:47.020 --> 00:02:49.480 And I do the same for all pairs of strings. 00:02:49.480 --> 00:02:52.190 And in the end, I get a graph. 00:02:52.190 --> 00:02:55.650 And the problem end of the line is given these two circuits 00:02:55.650 --> 00:02:59.130 and the corresponding graph that they define on this set, 00:02:59.130 --> 00:03:04.030 if the all 0 string is unbalanced meaning different 00:03:04.030 --> 00:03:07.410 in and out degree, then I want you 00:03:07.410 --> 00:03:13.960 to find another unbalanced node, string, 00:03:13.960 --> 00:03:17.940 which is guaranteed to exist by the parity arguments. 00:03:17.940 --> 00:03:21.350 And PPAD is a class of all search problems in FNP 00:03:21.350 --> 00:03:25.580 they are reducable-- polynomial time reducible to this problem. 00:03:25.580 --> 00:03:29.480 I'll remind you also of structure of the graph defined 00:03:29.480 --> 00:03:31.130 by these two circuits. 00:03:31.130 --> 00:03:37.040 It's easy to verify that if my edge definition is this, 00:03:37.040 --> 00:03:40.340 then every vertex has in degree and out degree at most once. 00:03:40.340 --> 00:03:43.730 So the graph that's induced by these two circuits 00:03:43.730 --> 00:03:45.320 will have this form. 00:03:45.320 --> 00:03:49.410 And basically, I'm looking for all red points. 00:03:49.410 --> 00:03:53.280 For any of these points, any of these red points, strings, 00:03:53.280 --> 00:03:55.610 are solutions, any unbalanced string, 00:03:55.610 --> 00:03:58.180 except for the all 0 string. 00:03:58.180 --> 00:04:00.160 If the all 0 string is not unbalanced, 00:04:00.160 --> 00:04:02.219 then I don't want you to do anything. 00:04:02.219 --> 00:04:03.760 If it is unbalanced, then I'm looking 00:04:03.760 --> 00:04:05.170 for any of these red vertices. 00:04:05.170 --> 00:04:08.590 So that's the class PPAD, and it has a combinatorial flavor. 00:04:08.590 --> 00:04:12.050 So I'm defining a huge graph, we have these two circuits. 00:04:12.050 --> 00:04:15.390 And I'm asking you to find any of these red points. 00:04:15.390 --> 00:04:19.529 So on the other hand, sort of like the problems 00:04:19.529 --> 00:04:22.880 we were targeting, Nash equilibrium, Brouwer's theorem, 00:04:22.880 --> 00:04:25.090 had a more continues flavor. 00:04:25.090 --> 00:04:32.790 So instead of working directly with this problem, 00:04:32.790 --> 00:04:34.370 with the end of the line, and trying 00:04:34.370 --> 00:04:37.740 to reduce this problem too Nash and to Brouwer 00:04:37.740 --> 00:04:40.590 to establish PPAD hardness deductions, 00:04:40.590 --> 00:04:43.010 I actually introduced a problem that actually 00:04:43.010 --> 00:04:44.700 closer to this problem. 00:04:44.700 --> 00:04:46.430 It has a continuous flavor. 00:04:46.430 --> 00:04:49.940 And it was the problem ArithmCircuitSAT. 00:04:49.940 --> 00:04:51.760 There were a bunch of problems last time 00:04:51.760 --> 00:04:53.720 about the definition of the problem. 00:04:53.720 --> 00:04:57.630 So I decided to be more explicit about what it is. 00:04:57.630 --> 00:05:00.590 So basically I'm giving you a circuit 00:05:00.590 --> 00:05:04.020 that has two types of nodes, viable nodes, v1 through vn, 00:05:04.020 --> 00:05:06.740 and gate nodes, g1 through gn. 00:05:06.740 --> 00:05:12.470 Now gate node has one of six possible flavors. 00:05:12.470 --> 00:05:17.220 It could be an assignment gate, and addition gate, 00:05:17.220 --> 00:05:20.730 a subtraction gate, set equal to constant gate, 00:05:20.730 --> 00:05:25.346 multiply by a constant gate, and comparison gate. 00:05:25.346 --> 00:05:26.720 Depending on the type, it's going 00:05:26.720 --> 00:05:31.740 to have from 0 to the 3 inputs, to 2 inputs. 00:05:31.740 --> 00:05:33.900 And it always has 1 output. 00:05:33.900 --> 00:05:38.610 Now what I wanted to emphasize is that this graph doesn't 00:05:38.610 --> 00:05:40.840 have inputs-- input variables. 00:05:43.760 --> 00:05:46.440 Loops are allowed. 00:05:46.440 --> 00:05:49.280 And what I want to emphasize that I didn't emphasise 00:05:49.280 --> 00:05:55.430 last time is that variable nodes have n degree 1, 00:05:55.430 --> 00:05:57.470 and gates have 0, 1, or 2 inputs, 00:05:57.470 --> 00:05:59.370 depending on their type. 00:05:59.370 --> 00:06:03.000 Otherwise for instance, the out degree of a node 00:06:03.000 --> 00:06:04.360 could be arbitrary. 00:06:04.360 --> 00:06:08.790 The fan out could be arbitrary, it doesn't matter. 00:06:08.790 --> 00:06:10.050 But you have to respect this. 00:06:10.050 --> 00:06:14.060 Every variable node has n degree 1. 00:06:14.060 --> 00:06:18.920 Every gate node has 1, 2, or 2 inputs depending on the type. 00:06:18.920 --> 00:06:22.404 There are no edges between gates and gates, 00:06:22.404 --> 00:06:23.570 and variables and variables. 00:06:23.570 --> 00:06:26.270 There are only edges between variables and gates, 00:06:26.270 --> 00:06:28.060 and gates to variables. 00:06:28.060 --> 00:06:29.710 That's the input. 00:06:29.710 --> 00:06:33.080 The input is a circuit that has this form. 00:06:33.080 --> 00:06:34.840 And what I want you to do is I want 00:06:34.840 --> 00:06:39.165 you to find an assignment of real 0, 00:06:39.165 --> 00:06:43.200 1 values to the variables of this circuit 00:06:43.200 --> 00:06:48.870 such that the constraints of the gates are satisfied. 00:06:48.870 --> 00:06:51.510 And here are the constrains of the gates. 00:06:51.510 --> 00:06:53.530 So if they gate is an assignment gate, 00:06:53.530 --> 00:06:57.100 I want the output node, the node who's 00:06:57.100 --> 00:07:00.270 connected to the output of the gate 00:07:00.270 --> 00:07:05.240 has equal value to the node that's feeding into that gate. 00:07:05.240 --> 00:07:06.950 If they gate is an addition gate, 00:07:06.950 --> 00:07:12.750 I want that variable node who's connected to the output 00:07:12.750 --> 00:07:18.430 of the gate to be basically the sum of the values of the inputs 00:07:18.430 --> 00:07:22.439 to the addition gate, except I'm also going to threshold it. 00:07:22.439 --> 00:07:24.980 I'm not going to allow it going above 1, and so and so forth. 00:07:24.980 --> 00:07:27.810 So these are the gate conditions. 00:07:27.810 --> 00:07:33.270 So now what I said last time is that a satisfying assignment 00:07:33.270 --> 00:07:36.820 always exists for this problem. 00:07:36.820 --> 00:07:39.320 It's not a priori, so it a requires work 00:07:39.320 --> 00:07:41.750 and actually it is going through a fixed point 00:07:41.750 --> 00:07:46.960 to argue that there is always a solution to this problem. 00:07:46.960 --> 00:07:49.790 What I also claimed last time is that it's 00:07:49.790 --> 00:07:52.680 PPAD complete to find a satisfying assignment. 00:07:52.680 --> 00:07:56.390 So it's a natural starting point for deductions. 00:07:56.390 --> 00:07:59.290 What I also said last time is that in fact, 00:07:59.290 --> 00:08:05.660 I can allow some noise in the error, in the gate constraints. 00:08:05.660 --> 00:08:10.080 I can allow plus minus epsilon deviation 00:08:10.080 --> 00:08:13.200 from the gate constraints. 00:08:13.200 --> 00:08:16.140 And so this should be a minus epsilon. 00:08:16.140 --> 00:08:18.580 Sorry about that. 00:08:18.580 --> 00:08:21.860 And this epsilon is part of the input. 00:08:21.860 --> 00:08:25.950 I can give as input both a circuit and an epsilon. 00:08:25.950 --> 00:08:28.420 And I will ask you to satisfy the gate constraints 00:08:28.420 --> 00:08:30.530 with an epsilon. 00:08:30.530 --> 00:08:33.980 Now last time I also showed you the structure 00:08:33.980 --> 00:08:36.470 the hardness proof for Nash, which 00:08:36.470 --> 00:08:40.860 basically took generic PPAD end of the line problem. 00:08:40.860 --> 00:08:43.049 Today PPAD complete problem, end of the line, 00:08:43.049 --> 00:08:49.070 is embedded into geometry, into the 3-D cube. 00:08:49.070 --> 00:08:55.160 Then we define a version of Sperner's Lemma, which then 00:08:55.160 --> 00:08:58.110 introduced ArithmCircuitSAT. 00:08:58.110 --> 00:09:00.090 And I didn't show this part of the deduction, 00:09:00.090 --> 00:09:01.881 and I'm not going to show it because that's 00:09:01.881 --> 00:09:03.540 the complicated part. 00:09:03.540 --> 00:09:05.650 But I am going to show is how to go 00:09:05.650 --> 00:09:08.080 from this problem, the ArithmCircuitSAT 00:09:08.080 --> 00:09:11.810 to Nash equilibrium, just to show how easy it 00:09:11.810 --> 00:09:17.570 is to work with this problem and reduce to other problems. 00:09:17.570 --> 00:09:20.150 So that's where I want to focus on. 00:09:20.150 --> 00:09:22.350 So that's the review from last time. 00:09:22.350 --> 00:09:24.510 And this time, I want to talk about-- I 00:09:24.510 --> 00:09:26.260 want to show the PPAD completeness of Nash 00:09:26.260 --> 00:09:27.870 equilibrium. 00:09:27.870 --> 00:09:34.460 I'm going to briefly give two other examples that 00:09:34.460 --> 00:09:38.500 have come from combinatorics. 00:09:38.500 --> 00:09:41.690 And then lastly, I'm going to talk about other existence 00:09:41.690 --> 00:09:46.570 arguments and the complexity classes that they define. 00:09:46.570 --> 00:09:50.922 So I'm going to introduce these classes, PPA, PPP, and PLS. 00:09:50.922 --> 00:09:52.380 Before I do that, I also thought it 00:09:52.380 --> 00:09:53.629 was a question from last time. 00:09:53.629 --> 00:09:55.720 PPAD stands for polynomial parity argument 00:09:55.720 --> 00:09:59.170 in directed graphs, corresponding to the fact 00:09:59.170 --> 00:10:01.780 that this class is defined with this parity argument 00:10:01.780 --> 00:10:05.130 in directed graphs in mind. 00:10:05.130 --> 00:10:07.410 So let's focus on this reduction, from Circuit 00:10:07.410 --> 00:10:08.740 to Nash. 00:10:08.740 --> 00:10:13.370 I want to introduce a concept before I show the reduction. 00:10:13.370 --> 00:10:16.770 That concept is graphical games and polymatrix games, 00:10:16.770 --> 00:10:19.090 a special case of graphical games. 00:10:19.090 --> 00:10:21.650 So graphical games were introduced in 2001 00:10:21.650 --> 00:10:25.990 by Kearns, Littman, and Singh as something very natural. 00:10:25.990 --> 00:10:28.560 Basically they tried to capture situations 00:10:28.560 --> 00:10:30.140 where the payoff of a player only 00:10:30.140 --> 00:10:34.130 depends on the actions of a few other players, 00:10:34.130 --> 00:10:36.590 because of geographical, communication, 00:10:36.590 --> 00:10:38.020 or other constraints. 00:10:38.020 --> 00:10:42.675 So in the graphical game, the players are nodes in the graph, 00:10:42.675 --> 00:10:44.940 in a directive graph. 00:10:44.940 --> 00:10:47.990 And a player's payoff only depends 00:10:47.990 --> 00:10:51.780 on her own strategy, as well as the strategy of the players 00:10:51.780 --> 00:10:53.530 that point to him. 00:10:56.710 --> 00:11:00.310 For example, this guy's payoff depends 00:11:00.310 --> 00:11:03.390 on this guy's, this guy's, and this guy's action. 00:11:03.390 --> 00:11:05.670 Because all of these guys point to him, as well as 00:11:05.670 --> 00:11:06.380 his own action. 00:11:10.950 --> 00:11:15.680 A special case of these games was actually 00:11:15.680 --> 00:11:18.940 introduced much earlier. 00:11:18.940 --> 00:11:21.360 So polymatrix games are graphical games 00:11:21.360 --> 00:11:23.510 where the payoff functions of the nodes 00:11:23.510 --> 00:11:26.250 are actually edge-wise separable. 00:11:26.250 --> 00:11:30.310 So for instance, the payoff of this guy 00:11:30.310 --> 00:11:34.000 as a function of everybody's mixed strategy 00:11:34.000 --> 00:11:37.150 is separable overall edges that point 00:11:37.150 --> 00:11:42.530 to him of some pairwise player function that 00:11:42.530 --> 00:11:47.460 has to do with his action and his neighbor's action. 00:11:47.460 --> 00:11:49.700 That's what's written here. 00:11:49.700 --> 00:11:59.230 And it's not very hard to see that any-- so this is a utility 00:11:59.230 --> 00:12:06.750 function that depends on the two mixed strategies. 00:12:06.750 --> 00:12:10.730 And by assumption, these players randomize independently 00:12:10.730 --> 00:12:12.050 of each other. 00:12:12.050 --> 00:12:18.680 So any such expectation of a pair of players' strategies 00:12:18.680 --> 00:12:20.850 that's are a product can actually 00:12:20.850 --> 00:12:24.594 be written as a quadratic form. 00:12:24.594 --> 00:12:25.260 Do you see that? 00:12:29.100 --> 00:12:31.130 Let me write on the board. 00:12:31.130 --> 00:12:44.940 So again, Xv is the mixed strategy of player v. Xw 00:12:44.940 --> 00:12:50.870 is the mixed strategy of w. 00:12:50.870 --> 00:12:59.540 What I mean by Uwv Xu, comma Xw is basically 00:12:59.540 --> 00:13:06.180 an expectation over an action Su drawn from Xu, 00:13:06.180 --> 00:13:09.480 and action is Sw. 00:13:09.480 --> 00:13:13.960 Sorry-- Sv drawn from Xv, Sw drawn from Xw, 00:13:13.960 --> 00:13:31.590 independently of-- and that's just the sum of all the Us 00:13:31.590 --> 00:13:45.950 and all the Sws of U, W, sv, sw, and then the probabilities. 00:13:53.910 --> 00:13:56.340 And that's a quadratic form. 00:13:56.340 --> 00:14:00.260 So that's what I mean by this line. 00:14:00.260 --> 00:14:03.780 Because players play independently from each other, 00:14:03.780 --> 00:14:09.070 any expectation with respect to that product distribution 00:14:09.070 --> 00:14:11.249 is a quadratic form. 00:14:11.249 --> 00:14:12.790 I'm not saying something interesting. 00:14:15.480 --> 00:14:17.630 Good? 00:14:17.630 --> 00:14:20.210 So then a polymatrix game is really, 00:14:20.210 --> 00:14:24.460 because of this, defined by a directive graph. 00:14:24.460 --> 00:14:30.630 And then for every directed edge, and every-- there is 00:14:30.630 --> 00:14:38.730 a matrix that defines the quadratic form for that edge. 00:14:38.730 --> 00:14:41.964 So polymatrix game is easily described. 00:14:41.964 --> 00:14:43.630 You can describe by specifying the graph 00:14:43.630 --> 00:14:46.170 and then giving a matrix what every directed edge. 00:14:49.010 --> 00:14:52.170 So bimatrix game are two player games. 00:14:52.170 --> 00:14:54.200 So that's why these are called polymatrix, 00:14:54.200 --> 00:14:56.996 because you have many matrices. 00:14:56.996 --> 00:14:59.370 In a two player game, you only need to give two matrices, 00:14:59.370 --> 00:15:06.799 one for a two player game is just player 1, player 2. 00:15:06.799 --> 00:15:08.590 And you give one matrix for this direction, 00:15:08.590 --> 00:15:10.110 and one matrix for that direction. 00:15:10.110 --> 00:15:12.530 So that's a bimatrix game. 00:15:12.530 --> 00:15:13.813 This is a polymatrix game. 00:15:17.760 --> 00:15:23.360 Now what I want to do is I want to-- in order 00:15:23.360 --> 00:15:26.577 to reduce ArithmCircuitSAT to Nash, instead what 00:15:26.577 --> 00:15:28.410 I'm going to do first is I'm going to reduce 00:15:28.410 --> 00:15:33.320 ArithmCircuitSAT to finding a Nash 00:15:33.320 --> 00:15:36.450 equilibrium in a polymatrix game. 00:15:36.450 --> 00:15:38.250 That's what I want to do first. 00:15:38.250 --> 00:15:41.870 After I do that, then I'm going to reduce it to-- so here 00:15:41.870 --> 00:15:43.600 I have many players. 00:15:43.600 --> 00:15:46.340 I want to go down to two players. 00:15:46.340 --> 00:15:48.780 But the first step is to just go to multiplayer 00:15:48.780 --> 00:15:50.000 Nash equilibrium. 00:15:50.000 --> 00:15:53.790 Then that deduction is the easy part. 00:15:53.790 --> 00:15:55.670 Also this is an easy part. 00:15:55.670 --> 00:15:56.920 The hard part happened before. 00:16:00.200 --> 00:16:05.800 So now how can we reduce an ArithmCircuitSAT problem 00:16:05.800 --> 00:16:10.760 into a polymatrix game Nash equilibrium problem? 00:16:10.760 --> 00:16:18.180 Like in LP completeness, like in reductions for NP, RP hardness 00:16:18.180 --> 00:16:21.820 proofs, you have to give gadgets. 00:16:21.820 --> 00:16:28.890 You have to identify objects in polymatrix games that 00:16:28.890 --> 00:16:31.920 simulate the operations that happen 00:16:31.920 --> 00:16:34.030 in your circuit over here. 00:16:34.030 --> 00:16:38.870 So what I want to introduce is what is called a game gadget. 00:16:38.870 --> 00:16:41.030 These are small polymatrix games that 00:16:41.030 --> 00:16:44.350 do various arithmetic operations, which 00:16:44.350 --> 00:16:49.560 I can then put together to simulate an ArithmCircuitSAT 00:16:49.560 --> 00:16:50.060 problem. 00:16:53.586 --> 00:16:55.460 So what I want to do is I'm going to give you 00:16:55.460 --> 00:16:56.720 a flavor of these gadgets. 00:16:56.720 --> 00:16:58.670 So I'm going to give you the addition gadgets. 00:16:58.670 --> 00:17:03.010 I want to give you a polymatrix game that does addition. 00:17:06.740 --> 00:17:08.150 So it's going to have-- this game 00:17:08.150 --> 00:17:11.540 is going to have four players. 00:17:11.540 --> 00:17:14.760 Everything player will have just two strategies-- 0 and 1, 00:17:14.760 --> 00:17:18.300 two pure strategies, hence the mixed strategy of that player 00:17:18.300 --> 00:17:19.849 is going to be a real number in 0, 00:17:19.849 --> 00:17:21.780 1, which corresponds to the probability 00:17:21.780 --> 00:17:24.210 by which this player plays 1. 00:17:28.930 --> 00:17:31.030 So here's the structure of the gadget. 00:17:31.030 --> 00:17:33.230 So this gadget, I'm showing it here embedded 00:17:33.230 --> 00:17:36.630 into a potentially bigger polymatrix game. 00:17:36.630 --> 00:17:42.075 But the gadget itself is going to have four players, X, Y, W, 00:17:42.075 --> 00:17:47.580 Z. Now X, Y are what is called the input to the gadget. 00:17:47.580 --> 00:17:50.970 So these are players who point to W, 00:17:50.970 --> 00:17:53.930 but don't depend on the actions of W and X 00:17:53.930 --> 00:17:56.500 because there are no directions going the other way. 00:17:56.500 --> 00:17:59.730 So these guys don't care about what these players are doing, 00:17:59.730 --> 00:18:03.300 and they serve as an input to the gadget. 00:18:03.300 --> 00:18:08.030 Now this player, W, gets his input of players strategies 00:18:08.030 --> 00:18:09.840 in this gadget. 00:18:09.840 --> 00:18:13.810 While X only cares about the W is doing. 00:18:13.810 --> 00:18:16.050 So X and Y are going to be called the input 00:18:16.050 --> 00:18:17.470 players to the gadget. 00:18:17.470 --> 00:18:20.130 Z is going to be called the output player to the gadget. 00:18:20.130 --> 00:18:24.930 And W is going to be called the auxiliary player. 00:18:24.930 --> 00:18:28.630 Now what I want to do is I want to define-- the only thing I'm 00:18:28.630 --> 00:18:30.720 going to define is the payoff function of W 00:18:30.720 --> 00:18:32.020 and the payoff function of Z. 00:18:32.020 --> 00:18:34.320 And I'm going to define this payoff functions in a way 00:18:34.320 --> 00:18:37.930 that addition somehow happens at a Nash equilibrium 00:18:37.930 --> 00:18:39.250 of this little game. 00:18:42.820 --> 00:18:44.974 So I'm going to define the payoff succinctly. 00:18:44.974 --> 00:18:47.140 But then I'm going to convince you that these really 00:18:47.140 --> 00:18:49.630 correspond to tables. 00:18:49.630 --> 00:18:56.370 So I'm going to say that W is paid an expected-- depending 00:18:56.370 --> 00:18:57.240 on what he plays. 00:18:57.240 --> 00:19:02.020 So if he plays 0, he gets paid an expected probability 00:19:02.020 --> 00:19:04.870 X equal $1, plus probably X equals 00:19:04.870 --> 00:19:09.080 Y equals $1 if he plays 0. 00:19:09.080 --> 00:19:13.610 But if he plays 1, he only gets paid probabilities equals $1. 00:19:13.610 --> 00:19:16.720 So in some sense if he plays 0, he looks to the left. 00:19:16.720 --> 00:19:19.020 If he plays 1, he looks to the right. 00:19:19.020 --> 00:19:23.810 His payoffs are the sum of these two guys probabilities 00:19:23.810 --> 00:19:30.380 of playing 1 in one case, and this guy's probability 00:19:30.380 --> 00:19:31.990 of playing 1 in the other case. 00:19:31.990 --> 00:19:38.880 So that's the payoff function to W. Now you would ask me maybe 00:19:38.880 --> 00:19:42.170 how can you implement this payoff functions. 00:19:42.170 --> 00:19:45.100 And that's actually very easy. 00:19:45.100 --> 00:19:52.290 So here's the table for player W. So when W plays 0, 00:19:52.290 --> 00:20:01.950 his payoff is depending on the strategies of X and Y. 00:20:01.950 --> 00:20:06.310 I'm going to define it to be 0, 1, 1, and 2. 00:20:06.310 --> 00:20:12.370 Now notice that in expectation over X's and Y's strategies, 00:20:12.370 --> 00:20:17.360 he's payoff-- so if he plays 0, his expected payoff over X 00:20:17.360 --> 00:20:21.820 and Y's strategies is just the probability that X plays 00:20:21.820 --> 00:20:26.280 1 plus the probability that Y plays 1. 00:20:26.280 --> 00:20:29.580 Do you see this from this matrix? 00:20:29.580 --> 00:20:37.060 So what's the expected payoff to W when he plays 0? 00:20:37.060 --> 00:20:45.110 So this is 1 if Y plays 1, but X plays 0. 00:20:48.400 --> 00:20:56.300 And it's 1 if X plays 1, but Y plays 0, 00:20:56.300 --> 00:20:59.040 and 2 if they both play 1. 00:21:04.970 --> 00:21:08.680 And I claim that if you properly collect the terms, 00:21:08.680 --> 00:21:13.730 this is just equal to probability Y plays 1, 00:21:13.730 --> 00:21:18.982 and X plays 1. 00:21:18.982 --> 00:21:19.940 So that's what I claim. 00:21:22.700 --> 00:21:24.740 So I covered this line. 00:21:24.740 --> 00:21:28.160 This line is also easy to cover by saying that when W plays 0, 00:21:28.160 --> 00:21:34.720 his payoff just depends on what Z is doing. 00:21:34.720 --> 00:21:36.390 And it's going to be like this. 00:21:39.950 --> 00:21:42.560 So when W plays 1, his expected payoff 00:21:42.560 --> 00:21:44.670 is exactly the probability Z plays 1. 00:21:48.690 --> 00:21:50.380 So what I've written here is actually 00:21:50.380 --> 00:21:54.355 consistent with some tables that I'm hiding from this slide. 00:21:57.440 --> 00:22:02.210 So that's, I guess, that's what I want to write. 00:22:02.210 --> 00:22:06.230 And then similarly, Z is paid to play the opposite of W. 00:22:06.230 --> 00:22:08.270 What do I mean by that? 00:22:08.270 --> 00:22:13.220 I mean that Z's payoff, when he plays 0, 00:22:13.220 --> 00:22:17.010 is exactly 1/2, no matter what W does. 00:22:17.010 --> 00:22:24.070 But if he plays 1, his payoff is 1 minus W plays 1. 00:22:24.070 --> 00:22:26.420 Which again, you should be able to see 00:22:26.420 --> 00:22:29.400 that there's a table, a little table implementing these payoff 00:22:29.400 --> 00:22:31.220 functions. 00:22:31.220 --> 00:22:32.290 And that sounds weird. 00:22:32.290 --> 00:22:33.830 Why did I define it this way? 00:22:33.830 --> 00:22:36.270 Here's my claim. 00:22:36.270 --> 00:22:38.680 In any Nash equilibrium of a game that 00:22:38.680 --> 00:22:44.470 contains this little gadget, the probability 00:22:44.470 --> 00:22:46.900 that the output player plays 1 is basically 00:22:46.900 --> 00:22:49.990 the sum of the probability that the input players play 00:22:49.990 --> 00:22:54.530 1 thresholded at 1 of course. 00:22:54.530 --> 00:22:57.970 So if this little gadget that I define here 00:22:57.970 --> 00:23:03.960 is part of a bigger game-- now what can the bigger game do? 00:23:03.960 --> 00:23:10.090 So it can fit something into X. It can take the value of Z 00:23:10.090 --> 00:23:13.070 and use it in some other way, potentially looping around, 00:23:13.070 --> 00:23:18.550 doing anything and once, except the only inputs to W's payoff 00:23:18.550 --> 00:23:19.860 are X, Y, and Z. 00:23:19.860 --> 00:23:22.890 And the only input to Z's payoff is W, 00:23:22.890 --> 00:23:26.520 but otherwise the game can be arbitrary. 00:23:26.520 --> 00:23:28.690 So if this game that I define here 00:23:28.690 --> 00:23:30.930 is embedded within a bigger game, 00:23:30.930 --> 00:23:32.930 then in anything Nash equilibrium of that bigger 00:23:32.930 --> 00:23:36.670 game, the probability that this guy plays 1 00:23:36.670 --> 00:23:38.800 is exactly the sum of the probabilities 00:23:38.800 --> 00:23:43.280 that these two guys play 1, thresholded at 1. 00:23:43.280 --> 00:23:44.550 Now how can we see this? 00:23:44.550 --> 00:23:47.740 Why is that true? 00:23:47.740 --> 00:23:50.400 It's a little case analysis. 00:23:50.400 --> 00:23:51.330 It's very simple. 00:23:51.330 --> 00:23:52.290 Let's try to do it. 00:23:58.790 --> 00:24:03.820 So suppose that the probability that Z plays 1 00:24:03.820 --> 00:24:10.500 is smaller than the probability X plays 1 plus the probability 00:24:10.500 --> 00:24:13.050 Y plays 1. 00:24:13.050 --> 00:24:17.580 Actually, let's do something else-- 00:24:17.580 --> 00:24:23.560 it's smaller than the min between this and 1. 00:24:26.200 --> 00:24:27.420 What happens in this case? 00:24:27.420 --> 00:24:31.570 So what happens if Z is smaller than the minimum of these two 00:24:31.570 --> 00:24:33.880 values? 00:24:33.880 --> 00:24:37.954 What is W going to do in that case? 00:24:37.954 --> 00:24:39.900 AUDIENCE: Is he going to play 0? 00:24:39.900 --> 00:24:42.940 COSTIS DASKALAKIS: Yeah, because 0 gives him a high payoff, 00:24:42.940 --> 00:24:44.700 gives his a higher payoff than this guy, 00:24:44.700 --> 00:24:47.550 but because of that condition. 00:24:47.550 --> 00:24:59.159 So this implies that W is going to play 0 with probability 1. 00:24:59.159 --> 00:25:00.200 Now what does this imply? 00:25:03.230 --> 00:25:07.165 If W plays 0 with probability 1, what does Z do? 00:25:07.165 --> 00:25:08.290 If he plays 0, he gets 0.5. 00:25:08.290 --> 00:25:10.850 What if he plays 1? 00:25:10.850 --> 00:25:13.880 How much does he get? 00:25:13.880 --> 00:25:16.800 He gets 1 because of that condition. 00:25:19.540 --> 00:25:22.040 So he's going to play what? 00:25:22.040 --> 00:25:22.830 1. 00:25:22.830 --> 00:25:33.380 And so this implies that-- but how 00:25:33.380 --> 00:25:36.560 can 1 be smaller than the minimal of 1 and something 00:25:36.560 --> 00:25:37.290 else? 00:25:37.290 --> 00:25:38.550 That's can't be. 00:25:38.550 --> 00:25:42.280 So this is impossible. 00:25:42.280 --> 00:25:46.040 So do the other side, or maybe you already 00:25:46.040 --> 00:25:47.530 trust me that it's OK. 00:25:55.680 --> 00:25:58.752 Let's try to argue that this cannot be the case. 00:25:58.752 --> 00:26:00.210 I don't need to do the mean anymore 00:26:00.210 --> 00:26:03.730 because it can't possibly be that this guy plays 00:26:03.730 --> 00:26:06.220 more than 1 probability. 00:26:06.220 --> 00:26:09.200 So this is the only case I want to consider now. 00:26:15.480 --> 00:26:16.720 So what happens in this case? 00:26:19.560 --> 00:26:22.690 The same logic. 00:26:22.690 --> 00:26:24.944 What does W do? 00:26:24.944 --> 00:26:25.890 AUDIENCE: Play 1? 00:26:25.890 --> 00:26:27.848 COSTIS DASKALAKIS: Has to play 1 because that's 00:26:27.848 --> 00:26:30.690 the better strategy. 00:26:30.690 --> 00:26:33.057 And what does he do? 00:26:33.057 --> 00:26:33.765 AUDIENCE: Play 0. 00:26:36.874 --> 00:26:38.540 COSTIS DASKALAKIS: But 0 can't be bigger 00:26:38.540 --> 00:26:40.850 than the sum of two r probabilities, 00:26:40.850 --> 00:26:43.320 so this and that can happen. 00:26:43.320 --> 00:26:47.910 Hence, the only case that's possible is this one. 00:26:51.460 --> 00:26:53.180 So it's very simple. 00:26:53.180 --> 00:26:55.610 So the only realization that one had to do 00:26:55.610 --> 00:27:00.330 is that-- where the arithmetic happens? 00:27:00.330 --> 00:27:02.000 It happens in the mixed strategies 00:27:02.000 --> 00:27:03.010 chosen by the players. 00:27:03.010 --> 00:27:04.500 That's where it happens. 00:27:11.610 --> 00:27:13.000 So this is the addition gadget. 00:27:13.000 --> 00:27:15.302 Similarly, one can define other gadgets. 00:27:15.302 --> 00:27:16.760 For example, the subtraction gadget 00:27:16.760 --> 00:27:21.270 is really, really similar to this except with one change. 00:27:21.270 --> 00:27:24.540 I replace this plus with a minus. 00:27:24.540 --> 00:27:26.130 That's the only change I need to do. 00:27:29.500 --> 00:27:35.430 I'm not sure I want to do it, but in other words, 00:27:35.430 --> 00:27:41.140 this table-- I only change this table 00:27:41.140 --> 00:27:47.270 to have 0 here and minus 1 here. 00:27:47.270 --> 00:27:51.210 And I've exactly implemented that, the subtraction. 00:27:51.210 --> 00:27:53.880 And the proof is exactly the same. 00:27:53.880 --> 00:27:56.600 You can argue that in any Nash equilibrium of a game that 00:27:56.600 --> 00:28:00.120 contains this gadget, the probability that this guy plays 00:28:00.120 --> 00:28:04.330 1 is actually the difference of the probabilities 00:28:04.330 --> 00:28:09.370 that X plays 1 and Y plays 1, a probability truncated at 0. 00:28:13.020 --> 00:28:17.040 Any questions about what taking place here? 00:28:17.040 --> 00:28:20.820 What we're trying to do is we want take a ArithmCircuitSAT 00:28:20.820 --> 00:28:24.730 instance, and we want to create a polymatrix game that 00:28:24.730 --> 00:28:27.530 simulates that circuit SAT instance. 00:28:27.530 --> 00:28:31.660 And what we need for that is gadgets, so little polymatrix 00:28:31.660 --> 00:28:34.565 games that implement various operations. 00:28:37.530 --> 00:28:39.800 And I showed you addition and subtraction. 00:28:39.800 --> 00:28:43.740 But I claim that you can implement 00:28:43.740 --> 00:28:44.795 a bunch of other gates. 00:28:49.990 --> 00:28:51.790 To have a more succinct notation, 00:28:51.790 --> 00:28:57.220 I'm going to use probability that some node plays 1. 00:28:57.220 --> 00:28:59.480 I'm not going to differentiate that with an X, 00:28:59.480 --> 00:29:02.090 and the probability that X plays 1. 00:29:02.090 --> 00:29:03.820 With this notation, I claim that we 00:29:03.820 --> 00:29:07.090 can implement all the gadgets that we need to simulate 00:29:07.090 --> 00:29:10.220 a circuit SAT instance. 00:29:10.220 --> 00:29:11.765 So let's look at this table. 00:29:11.765 --> 00:29:14.870 So in all these lines of this table, 00:29:14.870 --> 00:29:18.000 Z is the output player of the gadget, 00:29:18.000 --> 00:29:20.880 X and Y are the input players of the gadgets. 00:29:20.880 --> 00:29:22.870 And potentially, the gadget's implementing 00:29:22.870 --> 00:29:26.110 each of these gates use auxiliary players. 00:29:26.110 --> 00:29:29.590 Like in the addition case, I was using one intermediate player. 00:29:29.590 --> 00:29:32.050 He was the auxiliary player. 00:29:32.050 --> 00:29:37.390 And the claim is that you can implement gadgets such 00:29:37.390 --> 00:29:41.230 that if any of these gadgets is contained in a bigger 00:29:41.230 --> 00:29:45.820 polymatrix game, then any Nash equilibrium of these bigger 00:29:45.820 --> 00:29:51.527 polymatrix game these conditions are 00:29:51.527 --> 00:29:58.700 satisfied by the output and input players of the gadget. 00:29:58.700 --> 00:30:02.010 And this is as long as you don't mess up with the gadgets, 00:30:02.010 --> 00:30:06.920 meaning that the bigger the game can have edges 00:30:06.920 --> 00:30:09.080 into the input players of the gadgets, 00:30:09.080 --> 00:30:11.490 and edges out of output players of the gadgets. 00:30:16.220 --> 00:30:18.700 But I claim that you can implement all of the gadgets 00:30:18.700 --> 00:30:20.160 that you need. 00:30:20.160 --> 00:30:23.710 So in particular, if you have an instance for ArithmCircuitSAT 00:30:23.710 --> 00:30:26.780 you can create the polymatrix game 00:30:26.780 --> 00:30:34.350 by composing gadgets for each of these gates, 00:30:34.350 --> 00:30:38.280 so that at any Nash equilibrium of polymatrix game, 00:30:38.280 --> 00:30:42.060 all gates conditions are satisfied. 00:30:42.060 --> 00:30:47.430 So in particular, you are-- by finding a Nash equilibrium 00:30:47.430 --> 00:30:50.140 of this game, you are solving the instance 00:30:50.140 --> 00:30:54.140 of ArithmCircuitSAT you started with. 00:30:54.140 --> 00:30:56.300 So that's the idea. 00:30:56.300 --> 00:31:02.590 Any questions about what happened so far in the lecture? 00:31:02.590 --> 00:31:08.150 Questions about this deduction, or how the gadgets work? 00:31:08.150 --> 00:31:08.650 Yeah? 00:31:08.650 --> 00:31:10.233 AUDIENCE: Does this give a valid proof 00:31:10.233 --> 00:31:13.130 that every instance of ArithmCircuitSAT 00:31:13.130 --> 00:31:14.370 has a satisfying ascendant? 00:31:14.370 --> 00:31:16.810 COSTIS DASKALAKIS: Yeah, that's correct. 00:31:16.810 --> 00:31:18.550 Because this is a game. 00:31:18.550 --> 00:31:21.230 Some by Nash's theorem, there is a Nash equilibrium. 00:31:21.230 --> 00:31:24.240 Hence, there is a solution to that. 00:31:24.240 --> 00:31:27.676 Because a Nash equilibrium is a solution to that. 00:31:27.676 --> 00:31:28.550 That's exactly right. 00:31:28.550 --> 00:31:30.200 Actually that's how we get it. 00:31:33.830 --> 00:31:36.000 One thing that's not exactly obvious, 00:31:36.000 --> 00:31:39.490 but you can argue that it's true is that actually not only 00:31:39.490 --> 00:31:40.990 there's always a solution, but there 00:31:40.990 --> 00:31:44.480 is a solution in irrational numbers 00:31:44.480 --> 00:31:47.860 with polynomial description complexity in the input size. 00:31:47.860 --> 00:31:53.430 So that requires some linear programming techniques. 00:31:53.430 --> 00:31:55.840 But that's also use it to show. 00:31:55.840 --> 00:31:58.250 A priori, Nash's theorem will give you 00:31:58.250 --> 00:31:59.470 that any solution exists. 00:31:59.470 --> 00:32:01.910 It wouldn't give guarantees about the description 00:32:01.910 --> 00:32:03.620 complexity of that solution. 00:32:03.620 --> 00:32:05.060 But using linear programming, you 00:32:05.060 --> 00:32:07.350 can argue that there's always a rational solution 00:32:07.350 --> 00:32:10.332 with polynomial description complexity. 00:32:10.332 --> 00:32:12.415 Are you interested in seeing any of these gadgets? 00:32:16.230 --> 00:32:17.590 I've only shown addition. 00:32:17.590 --> 00:32:19.310 Do you want to see anything else? 00:32:19.310 --> 00:32:21.450 Do you believe me that I can actually do that? 00:32:21.450 --> 00:32:23.260 AUDIENCE: The comparison one that 00:32:23.260 --> 00:32:27.290 has arbitrary output for a certain conditions 00:32:27.290 --> 00:32:28.680 seems interesting. 00:32:28.680 --> 00:32:31.229 COSTIS DASKALAKIS: OK, I can show that. 00:32:31.229 --> 00:32:33.020 So let me try to do the comparison gadgets. 00:32:33.020 --> 00:32:35.103 They're actually simpler than the addition gadget. 00:32:39.560 --> 00:32:41.560 So I want to implement a comparison. 00:32:41.560 --> 00:32:46.540 So I'm only going to have the input players and the output 00:32:46.540 --> 00:32:48.880 player. 00:32:48.880 --> 00:32:55.090 And the payoffs of the output player are-- let's see. 00:32:57.900 --> 00:33:00.010 So if you play 0, then his payoff 00:33:00.010 --> 00:33:07.070 is only depends on the X. And it looks like is. 00:33:07.070 --> 00:33:13.090 If he plays 1, let's see if it works. 00:33:13.090 --> 00:33:17.520 I haven't prepared this but it's easy to figure this out. 00:33:17.520 --> 00:33:21.140 Worse case, I'll back track. 00:33:21.140 --> 00:33:26.490 So now what I want to argue is that-- let's bring up 00:33:26.490 --> 00:33:28.090 the conditions that I need to satisfy. 00:33:32.150 --> 00:33:34.470 So I want to satisfy those conditions. 00:33:34.470 --> 00:33:36.460 So let's see if I do satisfy these conditions. 00:33:36.460 --> 00:33:39.640 So I messed it up a little bit, I guess. 00:33:39.640 --> 00:33:45.985 So I have to replace X here and Y here. 00:33:48.560 --> 00:33:50.080 Now let's see if that's true. 00:33:50.080 --> 00:33:59.760 So I claim that if probability x is bigger than probability y, 00:33:59.760 --> 00:34:04.720 then what do I prefer to play? 00:34:04.720 --> 00:34:05.810 I'm prefer to play 0. 00:34:12.440 --> 00:34:15.219 Similarly for the other case, when they're equal, 00:34:15.219 --> 00:34:18.290 anything is possible. 00:34:18.290 --> 00:34:20.460 It's that's simple. 00:34:20.460 --> 00:34:22.070 There's not [INAUDIBLE] going on here. 00:34:26.111 --> 00:34:26.610 Cool. 00:34:26.610 --> 00:34:30.920 So I've established that this direction, 00:34:30.920 --> 00:34:32.620 this part of the direction. 00:34:32.620 --> 00:34:36.190 Now I want to go down from a polymatrix game 00:34:36.190 --> 00:34:37.760 to a two-player game. 00:34:42.820 --> 00:34:45.000 That's the next part of the lecture. 00:34:45.000 --> 00:34:48.139 How do you go down to two players? 00:34:48.139 --> 00:34:58.170 Well the first thing to note is that all the gates 00:34:58.170 --> 00:35:02.600 can be implemented with bipartite graphs. 00:35:02.600 --> 00:35:10.390 So if you remember my addition gadget, 00:35:10.390 --> 00:35:13.710 it had the input and the output players 00:35:13.710 --> 00:35:17.430 on one side and the auxiliary player on one side. 00:35:17.430 --> 00:35:20.000 My comparison gadget does not satisfy this probability 00:35:20.000 --> 00:35:22.000 because the input players and the output players 00:35:22.000 --> 00:35:23.467 are on different sides. 00:35:23.467 --> 00:35:25.800 But I claim that you can implement actually a comparison 00:35:25.800 --> 00:35:30.150 by adding an X and an additional step with a bipartite graph. 00:35:30.150 --> 00:35:32.900 So I claim that all gates can be implemented 00:35:32.900 --> 00:35:36.400 with polymatrix games that have input and output 00:35:36.400 --> 00:35:41.680 players on one side, and auxiliary vertices 00:35:41.680 --> 00:35:43.280 on the other side. 00:35:43.280 --> 00:35:49.871 So I claim that all my gadgets are actually bipartite. 00:35:49.871 --> 00:35:51.620 So in particularly, I can color this graph 00:35:51.620 --> 00:35:54.730 with two colors, blue and red. 00:35:54.730 --> 00:35:59.850 And what I want to do is I want to create these super players. 00:35:59.850 --> 00:36:01.880 I'm going to call the lawyers. 00:36:01.880 --> 00:36:08.100 I'm going to have a red lawyer and a blue lawyer. 00:36:08.100 --> 00:36:11.150 And the red player is going to represent all the nodes that 00:36:11.150 --> 00:36:14.940 are colored red in here, while the blue lawyer is going 00:36:14.940 --> 00:36:16.810 to represent all blue nodes. 00:36:19.360 --> 00:36:23.330 And what I want to do is I want to define the lawyer 00:36:23.330 --> 00:36:24.310 game in the next slide. 00:36:27.120 --> 00:36:29.460 So every lawyer, so the red lawyer, 00:36:29.460 --> 00:36:35.240 his strategy set is the union, not the product, the union. 00:36:35.240 --> 00:36:37.040 That's important, otherwise my deduction 00:36:37.040 --> 00:36:38.990 would be exponentially large. 00:36:38.990 --> 00:36:41.380 My induction wouldn't be a polynomial. 00:36:41.380 --> 00:36:44.400 So as a lawyer, and that poses actually 00:36:44.400 --> 00:36:45.967 technicalities in the construction 00:36:45.967 --> 00:36:47.050 that I'm about to present. 00:36:47.050 --> 00:36:51.310 But every lawyer will have a strategy set, a pure strategy 00:36:51.310 --> 00:36:55.990 set, the union of the pure strategy sets of the clients 00:36:55.990 --> 00:36:57.010 that he represents. 00:36:57.010 --> 00:36:59.700 So the red lawyer has one strategy 00:36:59.700 --> 00:37:04.830 for every strategy of every red nodes. 00:37:04.830 --> 00:37:07.820 The blue lawyer has one strategy for every strategy 00:37:07.820 --> 00:37:09.150 of every blue node. 00:37:11.912 --> 00:37:13.620 Now what I want to do is I want to define 00:37:13.620 --> 00:37:16.190 the payoffs in the lower game. 00:37:16.190 --> 00:37:18.400 What are the payoffs in the lower game? 00:37:18.400 --> 00:37:21.200 Well, what happens if the red lawyer 00:37:21.200 --> 00:37:27.680 decides to play strategy i or red client U? 00:37:27.680 --> 00:37:31.860 And what if the blue lawyer plays 00:37:31.860 --> 00:37:34.840 strategy j of blue node v? 00:37:34.840 --> 00:37:36.990 What are the payoffs of the two lawyers 00:37:36.990 --> 00:37:40.450 in that choice of strategies? 00:37:40.450 --> 00:37:43.500 Well, it's going to be exactly the corresponding payoffs 00:37:43.500 --> 00:37:48.920 of the nodes of the polymatrix game. 00:37:48.920 --> 00:37:51.750 So the red lawyer will the payoff 00:37:51.750 --> 00:37:56.990 that U would have gotten in the polymatrix game 00:37:56.990 --> 00:38:00.410 if these two players played i and j. 00:38:00.410 --> 00:38:03.010 And the blue lawyer's going to get 00:38:03.010 --> 00:38:05.980 his client's value, so v's value, 00:38:05.980 --> 00:38:09.070 under the same choice of strategies. 00:38:09.070 --> 00:38:12.780 So in particularly, if there's no edge between those two nodes 00:38:12.780 --> 00:38:16.450 that blue and red lawyers chose to play, 00:38:16.450 --> 00:38:18.800 then nobody gets anything. 00:38:18.800 --> 00:38:22.390 The payoff here is going to be 0. 00:38:22.390 --> 00:38:27.730 And if there is a directed edge from V to U, 00:38:27.730 --> 00:38:31.120 then this is going to be 0, and that's 00:38:31.120 --> 00:38:33.190 whatever it is, and vice versa. 00:38:33.190 --> 00:38:35.686 But you get the idea. 00:38:35.686 --> 00:38:38.880 If the blue lawyer decides to represent a client, 00:38:38.880 --> 00:38:42.950 and the red lawyer decides to represent another client, 00:38:42.950 --> 00:38:46.240 then the payoff of the lawyers are the corresponding payoffs 00:38:46.240 --> 00:38:47.960 that the clients would have gotten 00:38:47.960 --> 00:38:51.120 had they played the strategies that the lawyers decided 00:38:51.120 --> 00:38:52.510 to play for them. 00:38:52.510 --> 00:38:55.280 That's the definition of the lawyer game. 00:38:55.280 --> 00:38:55.890 Is this clear? 00:38:58.810 --> 00:39:04.440 Now the wishful thinking is that if X and Y 00:39:04.440 --> 00:39:08.290 is a Nash equilibrium of the lawyer game, 00:39:08.290 --> 00:39:16.100 then if I look at the marginal probability distributions 00:39:16.100 --> 00:39:20.650 on the different nodes, on the different clients 00:39:20.650 --> 00:39:23.570 by their lawyer, then this marginal probability 00:39:23.570 --> 00:39:25.775 distribution are a Nash equilibrium. 00:39:28.830 --> 00:39:32.380 So the wishful thinking is that if I 00:39:32.380 --> 00:39:34.920 start with a Nash equilibrium of the lawyer game, 00:39:34.920 --> 00:39:38.340 then if I look at the distribution 00:39:38.340 --> 00:39:42.580 that the red lawyer, the marginal probability 00:39:42.580 --> 00:39:47.980 distribution that the red lawyer places on every red node 00:39:47.980 --> 00:39:50.770 separately, and the marginal probability 00:39:50.770 --> 00:39:53.590 distributions that the blue lawyer places on 00:39:53.590 --> 00:39:56.200 all blue node separately, so all this collection 00:39:56.200 --> 00:39:59.170 of marginal probability distributions, 00:39:59.170 --> 00:40:02.290 that this collection is a Nash equilibrium of the polymatrix 00:40:02.290 --> 00:40:03.344 game. 00:40:03.344 --> 00:40:04.510 That's the wishful thinking. 00:40:10.380 --> 00:40:12.340 Questions about what the wishful thinking is? 00:40:17.450 --> 00:40:18.950 Of course, this is wishful thinking 00:40:18.950 --> 00:40:21.800 because we know how lawyers behave. 00:40:21.800 --> 00:40:24.185 They only represent the lucrative clients. 00:40:26.920 --> 00:40:33.440 There's no reason that the red lawyer would have incentive 00:40:33.440 --> 00:40:41.010 to choose at least some strategy to place positive probability 00:40:41.010 --> 00:40:43.350 mass on every node he represents. 00:40:43.350 --> 00:40:47.210 Maybe in the Nash equilibrium of this lawyer game, 00:40:47.210 --> 00:40:48.910 some of these marginal distributions 00:40:48.910 --> 00:40:52.800 are actually undefined because the lawyers-- are ill 00:40:52.800 --> 00:40:55.330 defined because the lawyers place 0 probability 00:40:55.330 --> 00:40:59.440 mass on the strategies of those nodes. 00:40:59.440 --> 00:41:01.990 So that wishful thinking isn't going through. 00:41:01.990 --> 00:41:03.590 But there is some truth to that. 00:41:03.590 --> 00:41:05.290 So there is a way to fix it. 00:41:05.290 --> 00:41:09.150 So here's how we're going to fix the lawyer game. 00:41:09.150 --> 00:41:11.220 So we know what lawyers like. 00:41:11.220 --> 00:41:13.250 They like money. 00:41:13.250 --> 00:41:15.600 So we're going to define a high stakes game. 00:41:17.926 --> 00:41:19.300 But the lawyers are going to play 00:41:19.300 --> 00:41:23.230 on the side at the same time as the actual game 00:41:23.230 --> 00:41:25.160 that we're interested in. 00:41:25.160 --> 00:41:28.570 So for a lot of cash-- so those are the stakes of that game. 00:41:30.857 --> 00:41:32.190 We're going to do the following. 00:41:32.190 --> 00:41:35.200 So some terminology first. 00:41:35.200 --> 00:41:37.310 So suppose that we had lots of generalities. 00:41:37.310 --> 00:41:40.550 Suppose that every lawyer has n clients he represents. 00:41:43.370 --> 00:41:46.000 And let's label the red lawyer's clients 00:41:46.000 --> 00:41:49.690 1 through n, and the blue lawyer's clients 1 through n. 00:41:49.690 --> 00:41:52.500 If you know one of the two lawyers have fewer clients, 00:41:52.500 --> 00:41:55.960 then we can pad this with dummy players 00:41:55.960 --> 00:42:02.000 and that doesn't change the polymatrix game's equilibria. 00:42:02.000 --> 00:42:05.970 So suppose that both lawyers represent the same number 00:42:05.970 --> 00:42:07.840 of players, clients. 00:42:07.840 --> 00:42:11.970 And let's label both lawyers' clients one through n, 00:42:11.970 --> 00:42:13.080 in an arbitrary way. 00:42:17.600 --> 00:42:20.770 The strategies of the high stakes game 00:42:20.770 --> 00:42:23.050 are exactly the same as the strategies of the game 00:42:23.050 --> 00:42:24.320 that I showed in the previous slide. 00:42:24.320 --> 00:42:25.695 In particular, the red lawyer has 00:42:25.695 --> 00:42:28.660 the union of the strategies of blue nodes. 00:42:28.660 --> 00:42:31.530 And red lawyer has the union of the strategies 00:42:31.530 --> 00:42:33.400 of the red nodes. 00:42:33.400 --> 00:42:36.910 Now what's the high stakes game? 00:42:36.910 --> 00:42:40.840 Suppose that the red player plays any strategy of client j, 00:42:40.840 --> 00:42:44.310 and the blue lawyer play any strategy of client k. 00:42:44.310 --> 00:42:50.570 Then if they choose different clients, 00:42:50.570 --> 00:42:53.100 they both get 0 dollars. 00:42:53.100 --> 00:42:57.060 But if they choose the same client-- I mean, 00:42:57.060 --> 00:42:59.730 it's not the same client, it's the same label of a client 00:42:59.730 --> 00:43:02.090 because they each represent different clients. 00:43:02.090 --> 00:43:06.690 But if they choose different labels, then they both get 0. 00:43:06.690 --> 00:43:11.830 If the choose the same label, whoever it was, 00:43:11.830 --> 00:43:15.950 the red lawyer gets a lot of cash, 00:43:15.950 --> 00:43:19.070 and that blue lawyer loses a lot of cash. 00:43:23.810 --> 00:43:29.520 Again, so this game has the same strategies as the lawyer game 00:43:29.520 --> 00:43:31.960 that I showed in the previous slide. 00:43:31.960 --> 00:43:34.810 Except now in this game, all the lawyers care 00:43:34.810 --> 00:43:40.160 about is the labels of the clients 00:43:40.160 --> 00:43:42.680 whose strategies they choose. 00:43:42.680 --> 00:43:48.640 So if they choose a strategy of a client that 00:43:48.640 --> 00:43:53.060 has the same label, then the red lawyer gains a lot of money, 00:43:53.060 --> 00:43:54.820 and the blue lawyer lose a lot of money. 00:43:54.820 --> 00:43:58.440 But if they choose strategies of clients with different labels, 00:43:58.440 --> 00:44:01.700 then they both get 0. 00:44:01.700 --> 00:44:04.740 In other words in some sense, the blue lawyer 00:44:04.740 --> 00:44:08.300 is trying to avoid the red lawyer, 00:44:08.300 --> 00:44:13.810 and the red lawyer is trying to catch the blue lawyer. 00:44:13.810 --> 00:44:16.330 So now this is a simple 0 sum game. 00:44:16.330 --> 00:44:19.270 And it's not hard to see that in any Nash 00:44:19.270 --> 00:44:28.620 equilibrium of this game, the lawyers are 00:44:28.620 --> 00:44:34.190 going to represent every client with the same probability. 00:44:34.190 --> 00:44:37.830 So each lawyer assigns a probability 00:44:37.830 --> 00:44:42.500 exactly 1/n to the set of his strategies 00:44:42.500 --> 00:44:46.360 corresponding to each of his clients. 00:44:46.360 --> 00:44:48.240 So the high stakes game has the property 00:44:48.240 --> 00:44:53.160 that the lawyers represent all their clients 00:44:53.160 --> 00:44:55.430 with the same probability distribution, 00:44:55.430 --> 00:44:57.750 and you can divide the probability distribution 00:44:57.750 --> 00:45:01.590 in an arbitrary way within the strategies of each 00:45:01.590 --> 00:45:03.770 of their clients in this game. 00:45:06.500 --> 00:45:10.840 That's easy to see just by the symmetry of this games, that 00:45:10.840 --> 00:45:12.790 has to be true. 00:45:12.790 --> 00:45:18.310 And with these two definitions, the game 00:45:18.310 --> 00:45:21.310 that I'm going to do for my reduction from polymatrix games 00:45:21.310 --> 00:45:26.610 to two player games is going to be the sum of these two games. 00:45:26.610 --> 00:45:30.200 So this is the game that I defined earlier. 00:45:30.200 --> 00:45:33.230 And this matrix is a block matrix. 00:45:33.230 --> 00:45:38.070 So this is a block of strategies corresponding to-- this matrix 00:45:38.070 --> 00:45:45.410 is a matrix that is a constant matrix where everything is big. 00:45:45.410 --> 00:45:48.940 And this block corresponds to-- these 00:45:48.940 --> 00:45:50.830 are the strategies of client with label 00:45:50.830 --> 00:45:52.520 1 for the red lawyer. 00:45:52.520 --> 00:45:58.520 And this is the strategies of the client 00:45:58.520 --> 00:46:01.290 of the blue lawyer with label 1, and so on so forth. 00:46:01.290 --> 00:46:03.380 So this is a block matrix. 00:46:03.380 --> 00:46:09.730 And there are ms and minus ms in this diagonal blocks, 00:46:09.730 --> 00:46:11.960 and everything else is 0. 00:46:11.960 --> 00:46:13.560 So that's the high stakes game. 00:46:13.560 --> 00:46:17.940 It's played along blocks of strategies 00:46:17.940 --> 00:46:20.830 because all I care of is the label the client 00:46:20.830 --> 00:46:24.370 I'm choosing for each of these lawyers. 00:46:24.370 --> 00:46:26.480 Well, this is a more fine-grained game, 00:46:26.480 --> 00:46:29.490 where I not only care about which clients and choosing, 00:46:29.490 --> 00:46:32.440 but also which strategies of these clients I'm choosing. 00:46:32.440 --> 00:46:37.290 And I'm going to choose an m that overwhelms 00:46:37.290 --> 00:46:38.510 the payoffs in this games. 00:46:38.510 --> 00:46:41.470 So this condition is OK for what I'm about to say. 00:46:41.470 --> 00:46:45.650 But think of m as huge compared to the maximum utility 00:46:45.650 --> 00:46:49.440 in this game times the number of clients in that game. 00:46:52.902 --> 00:46:55.360 AUDIENCE: Just to make sure I'm understanding this so far-- 00:46:55.360 --> 00:46:59.365 so in the naive game, if the red guy chooses 00:46:59.365 --> 00:47:01.240 strategy and the blue guys chooses a strategy 00:47:01.240 --> 00:47:02.781 and the two vertices they choose from 00:47:02.781 --> 00:47:05.230 are not connected by an edge, then they both get 0. 00:47:05.230 --> 00:47:07.680 COSTIS DASKALAKIS: Yeah. 00:47:07.680 --> 00:47:10.110 And if they're connected, they get payoffs 00:47:10.110 --> 00:47:12.161 that the corresponding nodes would have gotten-- 00:47:12.161 --> 00:47:13.660 AUDIENCE: [INAUDIBLE] vertical edge, 00:47:13.660 --> 00:47:16.801 one is the one on the outgoing vertex will still get 0. 00:47:16.801 --> 00:47:18.176 COSTIS DASKALAKIS: Exactly, yeah. 00:47:23.290 --> 00:47:24.800 Any other questions about-- yeah? 00:47:24.800 --> 00:47:26.955 AUDIENCE: Even with a really large choice of m, 00:47:26.955 --> 00:47:29.771 can't it still mess up the values a little bit? 00:47:29.771 --> 00:47:30.770 COSTIS DASKALAKIS: Yeah. 00:47:30.770 --> 00:47:32.430 It will mess it up a little bit. 00:47:32.430 --> 00:47:34.280 So the Nash equilibrium of this game, 00:47:34.280 --> 00:47:36.620 the addition of these two games, is not 00:47:36.620 --> 00:47:39.227 going to have the property that this game by itself had. 00:47:39.227 --> 00:47:40.560 But it's going to be very close. 00:47:40.560 --> 00:47:42.892 And I'm going to be specific about it. 00:47:42.892 --> 00:47:46.910 AUDIENCE: So the choice of me is so that the isn't that big. 00:47:46.910 --> 00:47:48.910 COSTIS DASKALAKIS: Even for arbitrarily large m, 00:47:48.910 --> 00:47:53.230 I can't actually quite show that the lawyers represent 00:47:53.230 --> 00:47:54.690 their clients exactly equally. 00:47:54.690 --> 00:47:56.060 But I can bring it close. 00:47:56.060 --> 00:47:58.990 And the larger m is the closer the come. 00:47:58.990 --> 00:48:01.610 In particular, I can show the following statement 00:48:01.610 --> 00:48:08.540 that in any Nash equilibrium of the combined game, if X used 00:48:08.540 --> 00:48:12.700 the total mass that the red lawyer places 00:48:12.700 --> 00:48:19.630 on the union of strategies of node U, then that's about 1/m. 00:48:19.630 --> 00:48:23.760 But there is the matter that's the case with m, and similarly 00:48:23.760 --> 00:48:26.810 for the blue lawyer. 00:48:26.810 --> 00:48:32.160 So approximately they're representing all their clients. 00:48:32.160 --> 00:48:36.270 And if I choose m huge then at least my marginals 00:48:36.270 --> 00:48:37.970 are well-defined now. 00:48:37.970 --> 00:48:41.960 So there is probability mass on every client 00:48:41.960 --> 00:48:44.640 and I can define these marginal distributions. 00:48:44.640 --> 00:48:47.130 Now whether these distributions are useful or not, 00:48:47.130 --> 00:48:49.690 we are about to see. 00:48:49.690 --> 00:48:54.820 But what I'm saying here is that as far as deciding 00:48:54.820 --> 00:48:59.760 how to split the pie into my clients, 00:48:59.760 --> 00:49:01.950 really the large game is what matters. 00:49:01.950 --> 00:49:03.560 The high stakes game is what matters. 00:49:03.560 --> 00:49:09.230 Because if I make a mistake, m is huge, some huge [INAUDIBLE]. 00:49:09.230 --> 00:49:11.850 So for splitting, for deciding how 00:49:11.850 --> 00:49:15.490 to split my total unit of probability mass 00:49:15.490 --> 00:49:19.290 into my clients, only the high stakes game-- essentially, 00:49:19.290 --> 00:49:22.610 only the high stakes games matter. 00:49:22.610 --> 00:49:30.350 On the other hand, when it comes to having decided how much mass 00:49:30.350 --> 00:49:33.800 to put on the unit of a particular node strategies, 00:49:33.800 --> 00:49:38.900 deciding about the fine-grained decision of how to allocate 00:49:38.900 --> 00:49:47.570 that Xu into the different strategies on that node U, then 00:49:47.570 --> 00:49:50.790 actually only the small game matters. 00:49:50.790 --> 00:49:53.500 And the reason is that the payoff 00:49:53.500 --> 00:50:01.030 different of the red lawyer from strategies Ui and Uj-- 00:50:01.030 --> 00:50:04.790 so to distinguishes between strategies 00:50:04.790 --> 00:50:09.950 i and j of node U, the payoff difference between these two 00:50:09.950 --> 00:50:13.450 choices doesn't have m in it. 00:50:13.450 --> 00:50:17.250 If you actually look at it, m goes away. 00:50:17.250 --> 00:50:22.100 There's no m in the payoff difference between these two 00:50:22.100 --> 00:50:24.000 options. 00:50:24.000 --> 00:50:29.126 So which one is better doesn't have m in it, 00:50:29.126 --> 00:50:30.500 essentially doesn't have m in it. 00:50:30.500 --> 00:50:35.940 There is some m in here because this I'm 00:50:35.940 --> 00:50:39.330 summing over all nodes, and different nodes 00:50:39.330 --> 00:50:46.790 have different sum of probabilities. 00:50:46.790 --> 00:50:50.940 But I'm going to get to that. 00:50:50.940 --> 00:50:56.460 But trust me that when the red lawyer is trying 00:50:56.460 --> 00:51:03.820 to decide how to allocate this Xu probability that he has 00:51:03.820 --> 00:51:07.600 decided to allocate on the difference 00:51:07.600 --> 00:51:11.454 strategies of node U, he looks at this difference, that's 00:51:11.454 --> 00:51:12.870 the difference in the two payoffs. 00:51:12.870 --> 00:51:15.360 And essentially there's no m in this equation. 00:51:15.360 --> 00:51:20.239 So we're going to see when m gets into the picture and why. 00:51:20.239 --> 00:51:22.280 But this is just from the definition of the game. 00:51:22.280 --> 00:51:23.920 Again, this is exactly true. 00:51:23.920 --> 00:51:26.080 The difference of the two payoff is exactly this. 00:51:28.850 --> 00:51:34.530 Now from that equation, it follows 00:51:34.530 --> 00:51:37.840 that if they lawyer decides to put positive probability 00:51:37.840 --> 00:51:43.930 mass on a particular strategy i of that node, then for all js, 00:51:43.930 --> 00:51:47.070 for all alternative strategies of that node 00:51:47.070 --> 00:51:49.960 that he could choose, it better be that this payoff 00:51:49.960 --> 00:51:52.690 difference is positive. 00:51:52.690 --> 00:51:57.060 And that really, really looks like the condition of the Nash 00:51:57.060 --> 00:52:01.690 equilibrium conditions for the client, if you think about it. 00:52:01.690 --> 00:52:08.820 Because if I define the marginal probabilities 00:52:08.820 --> 00:52:11.190 of node U on these two strategies, 00:52:11.190 --> 00:52:17.610 on these strategies, then this is really the Nash equilibrium 00:52:17.610 --> 00:52:23.980 condition for that node U. The only problem 00:52:23.980 --> 00:52:29.540 is to go from this equation for the unmarginalized 00:52:29.540 --> 00:52:34.690 probabilities Vy, Vl's, to the marginals. 00:52:34.690 --> 00:52:37.150 I'm dividing with something, and that something 00:52:37.150 --> 00:52:39.880 is an equal for all these. 00:52:39.880 --> 00:52:42.810 So if it was equal, then these marginals 00:52:42.810 --> 00:52:45.440 would actually be directly Nash equilibrium. 00:52:45.440 --> 00:52:49.510 Because this condition I could just divide by 1/n, 00:52:49.510 --> 00:52:52.180 and this would be made marginal probability. 00:52:52.180 --> 00:52:54.480 And that's exactly the equilibrium condition 00:52:54.480 --> 00:52:56.590 for the client U. 00:52:56.590 --> 00:53:00.930 The problem is that these Xvs are not all equal. 00:53:00.930 --> 00:53:03.180 Yvs are not equal. 00:53:03.180 --> 00:53:06.780 So I cannot just divide by 1/n and claim that the equation is 00:53:06.780 --> 00:53:09.120 still true. 00:53:09.120 --> 00:53:13.010 But the point is that this error doesn't 00:53:13.010 --> 00:53:17.180 create too much problems in the equilibrium conditions. 00:53:17.180 --> 00:53:19.260 And OK, it's not going to be an exact Nash 00:53:19.260 --> 00:53:21.480 equilibrium for the polymatrix game, 00:53:21.480 --> 00:53:24.050 but it's going to be an approximate one. 00:53:24.050 --> 00:53:28.500 And because I can take n to be as large as I want, 00:53:28.500 --> 00:53:33.830 I can make this approximation go to 0 as fast as I want. 00:53:33.830 --> 00:53:37.350 And remember this ArithmCircuitSAT problem 00:53:37.350 --> 00:53:39.617 allowed some error in it. 00:53:39.617 --> 00:53:41.200 So what effectively is going to happen 00:53:41.200 --> 00:53:44.830 here is that I'm going to get an approximate equilibrium 00:53:44.830 --> 00:53:45.810 of the polymatrix game. 00:53:45.810 --> 00:53:47.980 So that would correspond to an approximate equilibrium 00:53:47.980 --> 00:53:49.854 of the arithmetic SAT problem I started with. 00:53:49.854 --> 00:53:52.070 But this approximation can back accommodated, 00:53:52.070 --> 00:53:55.140 and the problem is to PPAD hard. 00:53:55.140 --> 00:53:57.470 So that's approximately how the argument works. 00:54:02.420 --> 00:54:05.310 The fact that the disagreements aren't uniform 00:54:05.310 --> 00:54:09.230 doesn't really matter as long as n is chosen large enough. 00:54:11.770 --> 00:54:14.280 I went a bit too fast. 00:54:14.280 --> 00:54:18.240 I didn't mean for this to be very detailed. 00:54:18.240 --> 00:54:22.210 But I meant to convey the bigger picture. 00:54:22.210 --> 00:54:23.910 And the bigger picture basically says 00:54:23.910 --> 00:54:26.160 that you can analyze what happens in this lower 00:54:26.160 --> 00:54:28.460 game in two steps. 00:54:28.460 --> 00:54:31.930 In one step, you argue that the lawyers approximately 00:54:31.930 --> 00:54:34.680 represent all their clients. 00:54:34.680 --> 00:54:36.810 In the other step, you have to decide 00:54:36.810 --> 00:54:44.060 how these lawyers allocate their probabilities 00:54:44.060 --> 00:54:47.840 to the different strategies of a particular node. 00:54:47.840 --> 00:54:51.910 That leads you to write down this difference of payoffs 00:54:51.910 --> 00:54:56.720 that the lawyer is experiencing when he switches. 00:54:56.720 --> 00:54:59.500 This is tracking how much better the expected payoff 00:54:59.500 --> 00:55:04.150 from this strategies is compared to that strategy. 00:55:04.150 --> 00:55:07.210 And by the equilibrium conditions of the lawyer game, 00:55:07.210 --> 00:55:11.040 you get the inequality I showed in the next slide 00:55:11.040 --> 00:55:15.660 that if the lawyer decides to place positive probability 00:55:15.660 --> 00:55:18.980 mass to a strategy i of node U, then it 00:55:18.980 --> 00:55:24.236 must be that there is no alternative strategy, Uj, 00:55:24.236 --> 00:55:25.860 that would give them the better payoff, 00:55:25.860 --> 00:55:28.250 so you get this condition. 00:55:28.250 --> 00:55:32.730 And that's essentially the equilibrium condition 00:55:32.730 --> 00:55:35.550 for the polymatrix game, except that you 00:55:35.550 --> 00:55:37.720 need to normalize these guy. 00:55:37.720 --> 00:55:40.080 And when you try to normalize these guy, 00:55:40.080 --> 00:55:43.850 you run into the problem that the lawyers 00:55:43.850 --> 00:55:48.330 don't play uniform strategies for their clients. 00:55:48.330 --> 00:55:51.540 But they play approximate uniform distribution 00:55:51.540 --> 00:55:52.810 of their clients. 00:55:52.810 --> 00:55:56.190 So this inequality gets messed up a little bit. 00:55:56.190 --> 00:55:58.320 So effectively that means that the players 00:55:58.320 --> 00:56:01.490 are almost best responding. 00:56:01.490 --> 00:56:04.052 The marginal distributions on an approximate Nash equilibrium 00:56:04.052 --> 00:56:05.010 of the polymatrix game. 00:56:07.680 --> 00:56:12.090 Also because the polymatrix game came from my ArithmCircuitSAT 00:56:12.090 --> 00:56:14.640 problem, an approximate evaluation of that problem. 00:56:14.640 --> 00:56:18.670 But theses are all PPAD hard, as I pointed out earlier. 00:56:18.670 --> 00:56:21.340 That's the high level idea. 00:56:21.340 --> 00:56:24.270 And trust me, the details are not hard at all. 00:56:24.270 --> 00:56:24.770 OK 00:56:24.770 --> 00:56:32.850 So you just have to trust me that the approximate uniform 00:56:32.850 --> 00:56:34.990 distribution of the claim are true 00:56:34.990 --> 00:56:41.200 and that dividing by approximately 1/n here 00:56:41.200 --> 00:56:43.458 doesn't mess up this condition too much. 00:56:46.390 --> 00:56:48.940 So this was the end of the proof basically. 00:56:48.940 --> 00:56:51.520 That's basically how the proof goes. 00:56:51.520 --> 00:56:56.780 So if have this, it's easy to do this, 00:56:56.780 --> 00:56:58.220 and then it's easy to go here. 00:57:00.942 --> 00:57:03.377 Any questions? 00:57:03.377 --> 00:57:04.960 Now the reduction I showed you was not 00:57:04.960 --> 00:57:06.150 from the original paper. 00:57:06.150 --> 00:57:11.690 It was established by a follow up paper by Chen and Deng. 00:57:11.690 --> 00:57:15.490 In that original paper, we actually took this problem 00:57:15.490 --> 00:57:16.900 in reduced to 4-player Nash. 00:57:20.530 --> 00:57:23.152 The proof is identical to the one I showed you. 00:57:23.152 --> 00:57:25.610 The only reason that we went to four players instead of two 00:57:25.610 --> 00:57:31.330 players is that our gadgets were four-partite instead 00:57:31.330 --> 00:57:32.260 of bipartite. 00:57:32.260 --> 00:57:35.520 So we had four colors and four lawyers. 00:57:35.520 --> 00:57:38.900 Now why did we have four-partite gadgets? 00:57:38.900 --> 00:57:42.150 Well, we were being a bit silly. 00:57:42.150 --> 00:57:44.500 We had in our arithmetic circuit, 00:57:44.500 --> 00:57:47.540 we had an extra gate, which plus multiplication, 00:57:47.540 --> 00:57:50.750 not by a constant, but the multiplication of two numbers. 00:57:50.750 --> 00:57:54.276 And you can show that this-- but we didn't use it. 00:57:54.276 --> 00:57:56.710 We were being a bit silly. 00:57:56.710 --> 00:58:00.520 We had this gadget here that we weren't using. 00:58:00.520 --> 00:58:05.300 And it was hard to-- actually it's impossible to make it 00:58:05.300 --> 00:58:07.980 into a bipartite gadget. 00:58:07.980 --> 00:58:11.890 So in our recent paper, we made it into a four-partite gadget. 00:58:11.890 --> 00:58:13.570 And then, in the followup paper we 00:58:13.570 --> 00:58:16.980 actual managed to reduce it to a three-partite gadget. 00:58:16.980 --> 00:58:19.510 But then this multiplication gadget actually you can show 00:58:19.510 --> 00:58:22.430 cannot be reduced to bipartite. 00:58:22.430 --> 00:58:24.190 So these guys observed that actually we 00:58:24.190 --> 00:58:26.660 are not using multiplication. 00:58:26.660 --> 00:58:30.410 So you can implement all gadgets that we actually 00:58:30.410 --> 00:58:31.980 used with partite graphs. 00:58:31.980 --> 00:58:33.730 So you can go down to two lawyers 00:58:33.730 --> 00:58:37.190 instead of three lawyers. 00:58:37.190 --> 00:58:41.930 So it felt like leaving money on the table, but that's OK. 00:58:45.780 --> 00:58:50.451 That's my discussion of PPAD-completeness of Nash. 00:58:50.451 --> 00:58:50.950 Yeah? 00:58:50.950 --> 00:58:53.150 AUDIENCE: Obviously, in zero-sum games, 00:58:53.150 --> 00:58:55.890 it's easy to find an equilibrium. 00:58:55.890 --> 00:58:59.120 Is there something in between zero-sum games 00:58:59.120 --> 00:59:02.450 and general two-player games for which any hardness 00:59:02.450 --> 00:59:03.452 result is known? 00:59:06.410 --> 00:59:08.500 COSTIS DASKALAKIS: Yeah. 00:59:08.500 --> 00:59:13.780 A natural way to interpolate between zero-sum games 00:59:13.780 --> 00:59:17.450 and general two-player games is the following. 00:59:17.450 --> 00:59:21.950 So in a zero-sum game let's call R and C the payoff matrix 00:59:21.950 --> 00:59:23.460 of the two players. 00:59:23.460 --> 00:59:27.820 In the game of zero-sum, if R plus C, the sum of these two 00:59:27.820 --> 00:59:35.110 matrices is identically 0, as I claimed in the very first slide 00:59:35.110 --> 00:59:36.600 I gave last time. 00:59:36.600 --> 00:59:45.610 Now you could call a game rank-r if R plus C has rank-r. 00:59:51.190 --> 01:00:04.582 Now what is known is that rank-1 games can be solved in P. 01:00:04.582 --> 01:00:14.135 I believe that it's known that rank-3 games are PPAD hard. 01:00:20.970 --> 01:00:28.780 And you should look at the paper by Mehta, Ruta Mehta 01:00:28.780 --> 01:00:34.860 in '14, I believe, where she shows that. 01:00:34.860 --> 01:00:38.910 She might even be showing rank-2, but I'm not sure. 01:00:38.910 --> 01:00:43.760 Maybe there is some gap there. 01:00:43.760 --> 01:00:50.630 Maybe it's even-- so there is a constant where it's already 01:00:50.630 --> 01:00:51.430 PPAD hard. 01:00:55.660 --> 01:01:00.680 And while I'm here, an interesting open question 01:01:00.680 --> 01:01:06.950 is approximate equilibrium. 01:01:06.950 --> 01:01:15.970 So interesting open problem-- approximate Nash equilibria. 01:01:20.930 --> 01:01:22.640 Even into two-player games, we've 01:01:22.640 --> 01:01:24.190 don't know how to find them. 01:01:24.190 --> 01:01:27.980 So let me remind you what this. 01:01:27.980 --> 01:01:31.550 So a Nash equilibrium is a pair of strategies such 01:01:31.550 --> 01:01:35.990 that no one has an incentive to change his randomization given 01:01:35.990 --> 01:01:38.410 what the other player is doing. 01:01:38.410 --> 01:01:46.070 And epsilon Nash equilibrium is when this condition are true to 01:01:46.070 --> 01:01:47.630 within an order of epsilon. 01:01:47.630 --> 01:01:52.129 So no player has incentive, more than epsilon, additive epsilon 01:01:52.129 --> 01:01:53.670 incentive, to changed his strategies. 01:01:58.600 --> 01:02:06.055 So at most, additive epsilon incentive to change. 01:02:09.540 --> 01:02:12.170 Now what do we know about these problems? 01:02:12.170 --> 01:02:15.880 Well, if the input to your problem 01:02:15.880 --> 01:02:20.370 is a game and an epsilon, then it follows from those results 01:02:20.370 --> 01:02:21.600 that that's PPAD complete. 01:02:24.340 --> 01:02:28.220 Now it's even true that if your input is a game, 01:02:28.220 --> 01:02:31.480 and you have a prespecified epsilon that's 01:02:31.480 --> 01:02:35.040 inverse polynomial in the size of the game, 01:02:35.040 --> 01:02:38.010 so that the input is a game, and there 01:02:38.010 --> 01:02:42.190 is a inverse polynomial function, 01:02:42.190 --> 01:02:43.190 and that's your epsilon. 01:02:43.190 --> 01:02:45.290 And you want to find an inverse polynomial Nash 01:02:45.290 --> 01:02:49.420 equilibrium of this given game that's still PPAD complete. 01:02:49.420 --> 01:02:51.465 But what is not known is epsilon constant. 01:02:54.750 --> 01:02:56.620 How hard is it to find equilibria 01:02:56.620 --> 01:03:00.340 when epsilon is a constant? 01:03:00.340 --> 01:03:10.110 And if your matrices have entries in 0,1, 01:03:10.110 --> 01:03:12.830 then we know how to do this in time n 01:03:12.830 --> 01:03:16.760 to the order log n over epsilon squared. 01:03:19.570 --> 01:03:23.430 We know that there is no FPTAS for the problem because 01:03:23.430 --> 01:03:28.890 of actually these results, follow up results 01:03:28.890 --> 01:03:29.860 to these results. 01:03:29.860 --> 01:03:33.370 But what I claimed earlier about inverse polynomial accuracy 01:03:33.370 --> 01:03:36.940 prohibits an FPTAS for the problem. 01:03:36.940 --> 01:03:38.780 But a PTAS is possible. 01:03:38.780 --> 01:03:40.490 What we have is a quasi-PTAS. 01:03:40.490 --> 01:03:44.080 So if you have an n-th strategy game, 01:03:44.080 --> 01:03:46.740 and you're interested in some constant epsilon, 01:03:46.740 --> 01:03:50.720 then we can get you an epsilon Nash equilibrium, n to the log 01:03:50.720 --> 01:03:52.900 and over epsilon squared time. 01:03:52.900 --> 01:03:54.930 And because I'm looking at additive of epsilons, 01:03:54.930 --> 01:03:57.740 I'm also normalizing my payoffs to be 0, 1, 01:03:57.740 --> 01:04:03.950 so that epsilon is related to the maximum payoff of the game. 01:04:03.950 --> 01:04:05.270 So that's what we have. 01:04:05.270 --> 01:04:13.380 And it's a great open problem if you can improve this, or show 01:04:13.380 --> 01:04:15.280 a lawyer bond. 01:04:15.280 --> 01:04:17.990 That's a great open problem. 01:04:17.990 --> 01:04:24.530 Speaking of which, if the rank of this game is up to, 01:04:24.530 --> 01:04:27.415 I believe, logarithmic, we know how to get a PTAS. 01:04:33.590 --> 01:04:35.900 Now I don't have too much time left, 01:04:35.900 --> 01:04:40.460 so I guess I had two options for today. 01:04:40.460 --> 01:04:44.780 One was to talk about different problems 01:04:44.780 --> 01:04:52.270 that you can show PPAD hard, or to talk about other existence 01:04:52.270 --> 01:04:54.360 arguments. 01:04:54.360 --> 01:04:57.990 And I think I only have time for one of the two. 01:05:04.470 --> 01:05:07.940 The other option is to show other arguments of existence 01:05:07.940 --> 01:05:08.790 in TFNP. 01:05:08.790 --> 01:05:12.229 So two examples, or other argument? 01:05:12.229 --> 01:05:13.145 What do you guys vote? 01:05:15.990 --> 01:05:17.820 So who wants two other examples? 01:05:20.350 --> 01:05:21.620 OK, who's giving the count? 01:05:24.370 --> 01:05:26.135 Who wants other existence arguments? 01:05:29.960 --> 01:05:33.034 So I think there's a slight majority for other existence 01:05:33.034 --> 01:05:34.575 arguments, so I'm going to show that. 01:05:40.670 --> 01:05:46.010 PPAD is founded on the directed parity argument. 01:05:46.010 --> 01:05:49.870 Other natural arguments of existence in combinatorics 01:05:49.870 --> 01:05:51.880 are the following, and each of them 01:05:51.880 --> 01:05:55.950 is going to correspond to a different complexity class. 01:05:55.950 --> 01:05:59.230 This is a parity argument on an undirected graph 01:05:59.230 --> 01:06:02.160 if an undirected graph has a node of odd degree then 01:06:02.160 --> 01:06:05.260 it must have another node of odd degree, 01:06:05.260 --> 01:06:08.120 also known as the handshaking lemma, I believe. 01:06:11.730 --> 01:06:13.360 Well this is another simple one. 01:06:13.360 --> 01:06:18.300 If a DAG, any DAG has a sink, that's 01:06:18.300 --> 01:06:19.910 going to give rise to the class PLS. 01:06:23.210 --> 01:06:24.910 And that's the pigeonhole principle. 01:06:24.910 --> 01:06:26.970 If you have a function mapping n elements, 01:06:26.970 --> 01:06:29.840 n minus 1 elements, than there is a collision. 01:06:29.840 --> 01:06:32.460 That's going to give rise to the PPP. 01:06:32.460 --> 01:06:34.990 So PPA stands for polynomial parity 01:06:34.990 --> 01:06:40.310 on undirected graphs, PLS, polynomial local search, PPP, 01:06:40.310 --> 01:06:42.890 polynomial pigeonhole principle. 01:06:42.890 --> 01:06:44.820 And I'm going to define them formally 01:06:44.820 --> 01:06:47.180 in the next few slides. 01:06:47.180 --> 01:06:49.050 So I'm going to start with PPA because it's 01:06:49.050 --> 01:06:51.765 very similar to be PPAD, except there's no direction. 01:06:54.470 --> 01:07:00.920 And the input to the problem is a circuit 01:07:00.920 --> 01:07:05.200 that induces a graph of [INAUDIBLE] strings. 01:07:05.200 --> 01:07:13.760 So the circuit gets its input at nodes, and outputs two strings. 01:07:13.760 --> 01:07:16.570 Gets one string, outputs two strings. 01:07:16.570 --> 01:07:20.090 Now this circuit induces a undirected graph 01:07:20.090 --> 01:07:22.960 over all possible strings in the following way. 01:07:22.960 --> 01:07:25.620 There is a node between string 1 and string 2 01:07:25.620 --> 01:07:29.990 if v1 is in the output list of v2, 01:07:29.990 --> 01:07:32.120 and v2 is in the output list of v1. 01:07:34.940 --> 01:07:38.180 So this circuit gets n-bits as input, 01:07:38.180 --> 01:07:40.970 and has two n-bits as output. 01:07:40.970 --> 01:07:44.500 And no matter what it is, it doesn't use a graph 01:07:44.500 --> 01:07:51.370 over strings that places an edge between these two strings, 01:07:51.370 --> 01:07:56.310 if v1 is in the output list of v2, and vice versa. 01:07:56.310 --> 01:08:00.440 Now in the same reasoning as last time, the same spirit 01:08:00.440 --> 01:08:03.520 as last time, any input C is going 01:08:03.520 --> 01:08:05.660 to reduce a graph with a particular structure. 01:08:05.660 --> 01:08:07.970 What's that structure? 01:08:07.970 --> 01:08:10.250 The odd degree of every node is what? 01:08:10.250 --> 01:08:12.180 At most 2. 01:08:12.180 --> 01:08:16.910 Hence the graph that is induced by any given circuit 01:08:16.910 --> 01:08:19.825 is going to be a collection of isolated vertices, cycles, 01:08:19.825 --> 01:08:20.325 and paths. 01:08:23.410 --> 01:08:27.340 The odd degree node problem is the following. 01:08:27.340 --> 01:08:31.740 If the 0 string has odd degree, then I 01:08:31.740 --> 01:08:34.830 want you to find me another node with odd degree, 01:08:34.830 --> 01:08:38.140 which we know exists by the handshaking lemma. 01:08:38.140 --> 01:08:42.640 Otherwise, if 0 to the n has even degree, just say, yes, 01:08:42.640 --> 01:08:45.260 and call it a day. 01:08:45.260 --> 01:08:51.779 Now PPA is the class of search problems in NP 01:08:51.779 --> 01:08:54.740 that are reducible, polytime reducible, to this problem, 01:08:54.740 --> 01:08:58.605 to the odd degree node problem, whose graph structure 01:08:58.605 --> 01:09:03.979 is very similar to PPAD, except there are no directions. 01:09:03.979 --> 01:09:08.270 So any circuit C defines a graph of this form, 01:09:08.270 --> 01:09:11.069 except the graph is over exponentially many vertices, 01:09:11.069 --> 01:09:15.779 so we cannot just go and check every node, 01:09:15.779 --> 01:09:17.330 every node's degree. 01:09:17.330 --> 01:09:22.160 So the question is if 0 to the n has odd degree, 01:09:22.160 --> 01:09:24.960 then there must be another node of odd degree. 01:09:24.960 --> 01:09:27.490 Any of these nodes is a valid solution to the problem. 01:09:30.100 --> 01:09:34.240 Something that I didn't mention before for PPAD, 01:09:34.240 --> 01:09:36.080 but it's useful to mention, and I'm also 01:09:36.080 --> 01:09:40.080 going to mention it for PPA, if I insist 01:09:40.080 --> 01:09:43.140 on finding the other end of this path, 01:09:43.140 --> 01:09:45.870 of the specific path that started at 0, 01:09:45.870 --> 01:09:46.960 the problem is not in FNP. 01:09:51.420 --> 01:09:56.060 So it's crucial that given the unbalanceness, 01:09:56.060 --> 01:09:57.955 or the odd degree of this node, I'll 01:09:57.955 --> 01:10:02.020 allow you to return any other odd degree node. 01:10:02.020 --> 01:10:03.540 That's very crucial. 01:10:03.540 --> 01:10:06.630 If I insisted on you returning me 01:10:06.630 --> 01:10:11.290 the other endpoint on the path that starts at 0, 01:10:11.290 --> 01:10:14.080 that's above NP basically because there's no source 01:10:14.080 --> 01:10:18.850 certificate that this node in the other end of the path 01:10:18.850 --> 01:10:19.880 starting at 0. 01:10:19.880 --> 01:10:21.560 So it's very crucial that I allow you 01:10:21.560 --> 01:10:23.320 to return any odd degree node. 01:10:26.550 --> 01:10:30.490 Now an interesting problem that is in this class, 01:10:30.490 --> 01:10:33.950 and as far as we know, not in PPAD, 01:10:33.950 --> 01:10:39.520 is the problem Smith, which is given a Hamiltonian 01:10:39.520 --> 01:10:43.360 circuit in a 3-regular graph, find 01:10:43.360 --> 01:10:47.394 me another Hamiltonian circuit in this graph. 01:10:47.394 --> 01:10:49.060 Now this graph is not exponentially big. 01:10:49.060 --> 01:10:50.976 It's actually a graph that you can write down. 01:10:50.976 --> 01:10:53.490 So I give you a graph explicitly, 01:10:53.490 --> 01:10:56.200 and I also give you a Hamiltonian circuit 01:10:56.200 --> 01:10:59.310 in this graph. 01:10:59.310 --> 01:11:01.485 I claim that there is another Hamiltonian circuit. 01:11:04.640 --> 01:11:06.430 The question is find it. 01:11:09.184 --> 01:11:11.100 Now why is there always a Hamiltonian circuit? 01:11:11.100 --> 01:11:14.710 Well, that follows from a theorem by Smith 01:11:14.710 --> 01:11:18.640 saying that-- a theorem by Smith implies that there's always 01:11:18.640 --> 01:11:20.622 another Hamiltonian path. 01:11:20.622 --> 01:11:22.080 And let me actually show it to you. 01:11:22.080 --> 01:11:23.880 It's very simple. 01:11:23.880 --> 01:11:27.090 So here's is actually a copy from Papadimitriou's '94 paper. 01:11:29.660 --> 01:11:32.570 I guess there's a missing apostrophe here. 01:11:32.570 --> 01:11:35.980 So here's the input graph. 01:11:35.980 --> 01:11:44.800 And obviously, this is a Hamiltonian circuit. 01:11:44.800 --> 01:11:49.630 So I'm following the outer boundary here, going inside, 01:11:49.630 --> 01:11:51.070 and then coming back here. 01:11:51.070 --> 01:11:52.945 That's the Hamiltonian circuit. 01:11:52.945 --> 01:11:54.820 And here's the same circuit, except I removed 01:11:54.820 --> 01:11:58.250 one edge, x and y. 01:11:58.250 --> 01:12:00.770 And here I'm showing a bunch of operations 01:12:00.770 --> 01:12:04.480 you can do this Hamiltonian path-- this 01:12:04.480 --> 01:12:10.950 is a Hamiltonian path-- to create another Hamilton path 01:12:10.950 --> 01:12:13.480 with the same edge missing. 01:12:13.480 --> 01:12:17.410 Hence, out of that edge, you get another Hamiltonian circuit. 01:12:17.410 --> 01:12:20.230 And the [INAUDIBLE] is very simple. 01:12:20.230 --> 01:12:23.160 So what you do is the following. 01:12:23.160 --> 01:12:25.380 So you start with-- it doesn't matter. 01:12:25.380 --> 01:12:28.230 You can start with an arbitrary node. 01:12:28.230 --> 01:12:36.830 All the circuits I'm going to be defining, 01:12:36.830 --> 01:12:40.190 will have x as one of the two endpoints. 01:12:40.190 --> 01:12:44.160 But the other endpoint is going to be moving around 01:12:44.160 --> 01:12:47.480 and in the end, is going to get back to this vertex. 01:12:47.480 --> 01:12:50.770 Hence, I can knock that edge back and get the circuit. 01:12:50.770 --> 01:12:52.880 So I started with the circuit. 01:12:52.880 --> 01:12:56.630 I remove one edge, x and y. 01:12:56.630 --> 01:12:58.130 And I'm about to show you that there 01:12:58.130 --> 01:13:02.930 is another Hamiltonian path that misses the same edge, 01:13:02.930 --> 01:13:05.020 but is different than this one. 01:13:05.020 --> 01:13:06.370 So it's going to be that one. 01:13:06.370 --> 01:13:08.000 Now let me try to argue that that 01:13:08.000 --> 01:13:11.947 is another Hamiltonian path that is missing the same edge. 01:13:11.947 --> 01:13:13.030 How am I going to do that? 01:13:13.030 --> 01:13:18.370 Well, in this sequence of Hamiltonian paths 01:13:18.370 --> 01:13:20.710 that I'm going to define, x is going to stay fixed, 01:13:20.710 --> 01:13:22.720 y is going to be moving around. 01:13:22.720 --> 01:13:26.360 Now let's land somewhere in the middle, here. 01:13:26.360 --> 01:13:28.970 So now y is this guy. 01:13:28.970 --> 01:13:33.320 y, because the graph is 3-regular, 01:13:33.320 --> 01:13:36.580 and y is an endpoint, there's exactly 01:13:36.580 --> 01:13:40.420 one edge that's used in the Hamiltonian path. 01:13:40.420 --> 01:13:43.590 And there are two edges that are missing. 01:13:43.590 --> 01:13:50.605 I'm going to try to add both of them and get a circuit-- 01:13:50.605 --> 01:13:53.720 and get a path, a Hamiltonian path. 01:13:53.720 --> 01:13:58.850 What would happen if I tried to add this edge into this path? 01:13:58.850 --> 01:14:01.615 Well, that wouldn't be a path anymore. 01:14:04.380 --> 01:14:08.460 So because this guy would have degree 3 if I added that edge. 01:14:08.460 --> 01:14:11.480 Now have two kill one of the two edges adjacent to this node. 01:14:11.480 --> 01:14:14.620 And I'm going to kill the one that the maintains the fact 01:14:14.620 --> 01:14:17.000 that the graph is connected. 01:14:17.000 --> 01:14:21.960 If I add this edge, and I kill this edge, I'm screwed. 01:14:21.960 --> 01:14:24.820 The graph is not connected anymore. 01:14:24.820 --> 01:14:26.250 So I'm going to kill that edge. 01:14:26.250 --> 01:14:27.840 So what happens here? 01:14:27.840 --> 01:14:30.160 I added that edge, and I killed that edge. 01:14:30.160 --> 01:14:32.760 And this what I got. 01:14:32.760 --> 01:14:37.899 If I tried to add this edge, then this node 01:14:37.899 --> 01:14:38.690 will have degree 3. 01:14:38.690 --> 01:14:42.130 I have to kill either that edge or that edge. 01:14:42.130 --> 01:14:46.439 I can kill exactly one to avoid disconnecting my graph. 01:14:46.439 --> 01:14:47.980 And I'm going to be killing that one, 01:14:47.980 --> 01:14:50.290 and that's going to bring me here. 01:14:50.290 --> 01:14:58.610 So every Hamiltonian path has exactly two neighbors 01:14:58.610 --> 01:15:07.450 corresponding to adding one of the two missing 01:15:07.450 --> 01:15:12.310 edges from the y endpoint of that path. 01:15:12.310 --> 01:15:16.400 And there's only one way this thing can stop, 01:15:16.400 --> 01:15:17.330 by arriving at y. 01:15:21.280 --> 01:15:24.010 If I don't get endpoint y to come back 01:15:24.010 --> 01:15:26.070 to its original position I will keep going. 01:15:31.680 --> 01:15:35.000 In other words, like my proof [INAUDIBLE], what I did here 01:15:35.000 --> 01:15:37.510 is I defined a neighborhood relationship 01:15:37.510 --> 01:15:40.320 between Hamiltonian paths, which was 01:15:40.320 --> 01:15:43.360 adding an edge, one of the two missing 01:15:43.360 --> 01:15:45.530 edges from the endpoints. 01:15:45.530 --> 01:15:49.025 And another, the only stopping condition is if I get here. 01:15:49.025 --> 01:15:52.940 And that's why Smith is in PPA. 01:15:52.940 --> 01:15:54.890 Now it's a very interesting problem 01:15:54.890 --> 01:15:56.875 to show that this is PPA complete. 01:16:01.780 --> 01:16:04.660 Let me define the class PLS. 01:16:04.660 --> 01:16:07.950 That was define earlier by Johnson, Papadimitriou 01:16:07.950 --> 01:16:10.730 and Yanakakis. 01:16:10.730 --> 01:16:15.220 Now I want to implement a class that exploits this argument, 01:16:15.220 --> 01:16:16.402 that every DAG has a sink. 01:16:16.402 --> 01:16:17.860 The way I'm going to do that is I'm 01:16:17.860 --> 01:16:20.260 going to give you two functions. 01:16:20.260 --> 01:16:23.390 Function C is going to take-- both of them 01:16:23.390 --> 01:16:26.730 will have n-bits as inputs. 01:16:26.730 --> 01:16:31.760 This guy, C, will output a list of strings, 01:16:31.760 --> 01:16:34.010 a list of k strings. 01:16:34.010 --> 01:16:38.340 So it has n-bits as input, and k n-bits as output. 01:16:38.340 --> 01:16:42.790 While this guy is outputting some real number, 01:16:42.790 --> 01:16:44.800 I mean, some rational number, whatever. 01:16:44.800 --> 01:16:48.007 But I interpret the output of the circuit as a number. 01:16:48.007 --> 01:16:49.590 I interpret the output of this circuit 01:16:49.590 --> 01:16:50.965 as a list of other strings. 01:16:54.681 --> 01:16:58.010 I add that's between-- given these two circuits, that 01:16:58.010 --> 01:17:01.930 induces a DAG, I claim, in the following way. 01:17:01.930 --> 01:17:05.440 I add an edge from v1 to v2 if v2 01:17:05.440 --> 01:17:12.890 is in the adjacency list of v1, and the score of v2 two 01:17:12.890 --> 01:17:17.240 is better than the score of v1. 01:17:17.240 --> 01:17:21.290 Then and only then, I add an edge from v1 to v2. 01:17:21.290 --> 01:17:27.280 And obviously, this is a DAG now because I'm increasing my score 01:17:27.280 --> 01:17:28.480 as I go along. 01:17:28.480 --> 01:17:32.150 So I cannot come back to where I started. 01:17:32.150 --> 01:17:34.980 So the problem that I want to define 01:17:34.980 --> 01:17:41.020 is the FindSink problem, which is given two circuits as above, 01:17:41.020 --> 01:17:48.800 find an x that has a better score than any 01:17:48.800 --> 01:17:52.470 of the adjacent vertices. 01:17:52.470 --> 01:17:55.650 And such a thing has to exist because if I define this graph 01:17:55.650 --> 01:17:57.350 and find any sink of this graph, this 01:17:57.350 --> 01:17:59.930 will satisfy this property. 01:17:59.930 --> 01:18:04.240 The class PLS are all search problems in the P, 01:18:04.240 --> 01:18:05.830 but are polynomial time reducible 01:18:05.830 --> 01:18:08.850 to this problem, FindSink. 01:18:08.850 --> 01:18:12.970 The picture for this problem is this, 01:18:12.970 --> 01:18:15.060 exponentially large graph. 01:18:15.060 --> 01:18:16.910 But there is a DAG that's implicitly 01:18:16.910 --> 01:18:20.440 defined by these two vertices, by these two circuits. 01:18:20.440 --> 01:18:21.915 And all of these are solutions. 01:18:26.920 --> 01:18:28.640 An interesting problem in this class 01:18:28.640 --> 01:18:31.390 is the LocalMaxCut problem, a relative 01:18:31.390 --> 01:18:35.510 of the well-known MaxCut problem, which is NP complete. 01:18:35.510 --> 01:18:39.850 In the LocalMaxCut problem, I'm giving you a weighted graph. 01:18:39.850 --> 01:18:43.310 And I want a partition that's not globally 01:18:43.310 --> 01:18:45.920 optimal, but locally optimal, meaning 01:18:45.920 --> 01:18:47.780 there is no single node I could move 01:18:47.780 --> 01:18:50.060 from one to the other side of the cut 01:18:50.060 --> 01:18:51.231 to improve the cut value. 01:18:55.950 --> 01:19:00.860 Now if the weights were bounded and integral, 01:19:00.860 --> 01:19:04.390 this wouldn't be a hard problem. 01:19:04.390 --> 01:19:07.100 But for arbitrary weights, it's actually PLS 01:19:07.100 --> 01:19:12.732 complete to find a local maximum cut. 01:19:17.340 --> 01:19:20.110 Final problem, and I'm concluding, 01:19:20.110 --> 01:19:23.480 is final class is the class PPP that's 01:19:23.480 --> 01:19:27.120 trying to implement the pigeonhole principle. 01:19:27.120 --> 01:19:29.950 Here I'm giving you a circuit that has 01:19:29.950 --> 01:19:31.795 n-bit input and n-bit outputs. 01:19:38.610 --> 01:19:40.690 The collision problem that I want to define 01:19:40.690 --> 01:19:43.630 is given sets of circuits, either find 01:19:43.630 --> 01:19:47.660 me an x that maps to the 0 string, 01:19:47.660 --> 01:19:52.580 or find me a pair x and y that map to the same string. 01:19:52.580 --> 01:19:56.350 Clearly, by the pigeonhole principle if nobody goes to 0, 01:19:56.350 --> 01:19:59.470 there must be two guys that collide. 01:19:59.470 --> 01:20:03.830 So this problem is also total by the pigeonhole principal. 01:20:03.830 --> 01:20:07.590 So it always has a solution, no matter what the circuit is. 01:20:07.590 --> 01:20:10.150 And the class PPP is all problems in NP that 01:20:10.150 --> 01:20:11.520 are reducible to this problem. 01:20:14.730 --> 01:20:19.830 Finally, the hierarchy of problems I defined is this. 01:20:19.830 --> 01:20:23.600 P, FNP, there is total FNP somewhere here, 01:20:23.600 --> 01:20:25.130 which I don't show. 01:20:25.130 --> 01:20:28.210 And these are the relationships of these problems. 01:20:28.210 --> 01:20:32.790 I haven't shown that these arrows true, 01:20:32.790 --> 01:20:35.530 that this is a subclass, PPD's a subclass 01:20:35.530 --> 01:20:36.530 of these two subclasses. 01:20:36.530 --> 01:20:37.350 This is easy. 01:20:37.350 --> 01:20:41.800 This is a simple exercise we can think about. 01:20:41.800 --> 01:20:43.880 This is basically my introduction 01:20:43.880 --> 01:20:48.250 to PPA, PPAD and related classes. 01:20:48.250 --> 01:20:50.040 The final thing that I want to point out 01:20:50.040 --> 01:20:52.615 is answering a question that was asked 01:20:52.615 --> 01:20:53.990 after the previous lecture, which 01:20:53.990 --> 01:20:57.430 was why did you define these classes, 01:20:57.430 --> 01:21:02.760 and not just a TFNP complete problem. 01:21:02.760 --> 01:21:09.170 Why did you have to pay special attention to precise existence 01:21:09.170 --> 01:21:12.700 argument that gives rise to the guarantee 01:21:12.700 --> 01:21:14.960 that your problems are total? 01:21:14.960 --> 01:21:18.750 And the reason for that is that actually TFNP 01:21:18.750 --> 01:21:21.890 is not what's called a syntactic class. 01:21:21.890 --> 01:21:29.870 In other words, if I give you a problem with TFNP-- 01:21:29.870 --> 01:21:34.900 if I give you a Turing machine, you cannot decide whether that 01:21:34.900 --> 01:21:39.250 is computing a total problem, that no matter what the input 01:21:39.250 --> 01:21:43.940 to that machine is, there's always an output. 01:21:43.940 --> 01:21:50.210 So I had to pay attention to the specialized existence arguments 01:21:50.210 --> 01:21:53.030 because for specialized existence arguments, 01:21:53.030 --> 01:21:57.930 I know a priori that the problem is total. 01:21:57.930 --> 01:22:02.600 So in particular, no matter what circuit I give you here, 01:22:02.600 --> 01:22:04.120 I don't even have to check anything. 01:22:04.120 --> 01:22:07.720 No matter what input you give me, I know there is a solution. 01:22:07.720 --> 01:22:12.110 No matter what pairs of circuits you give me here, 01:22:12.110 --> 01:22:13.920 I don't need to check anything. 01:22:13.920 --> 01:22:15.930 I know the answer to this problem-- 01:22:15.930 --> 01:22:20.100 there's always an answer to this problem, and so on, so forth. 01:22:20.100 --> 01:22:23.490 No matter what you give to me as input, 01:22:23.490 --> 01:22:27.000 it is important to define complexity classes 01:22:27.000 --> 01:22:30.686 for which you can show hardness results to find 01:22:30.686 --> 01:22:31.435 complete problems. 01:22:34.220 --> 01:22:36.150 Otherwise, you would have what is 01:22:36.150 --> 01:22:39.790 called promise classes, which are not amenable to showing 01:22:39.790 --> 01:22:42.740 the completeness results. 01:22:42.740 --> 01:22:44.970 On that brief not, I want to stop. 01:22:44.970 --> 01:22:46.820 Thanks a lot.