WEBVTT 00:00:00.000 --> 00:00:17.400 align:middle line:84% [SQUEAKING] [RUSTLING] [CLICKING] 00:00:17.400 --> 00:00:21.080 align:middle line:84% PETER SHOR: And this network could be all sorts of things. 00:00:21.080 --> 00:00:32.720 align:middle line:84% I mean, it could be pipes, roads, 00:00:32.720 --> 00:00:34.820 align:middle line:90% electronic communications. 00:00:34.820 --> 00:00:42.000 align:middle line:90% 00:00:42.000 --> 00:00:45.820 align:middle line:90% And each edge has a capacity. 00:00:45.820 --> 00:00:53.120 align:middle line:90% 00:00:53.120 --> 00:01:01.280 align:middle line:84% And the question is, how much whatever can you 00:01:01.280 --> 00:01:08.070 align:middle line:90% get from a source to a sink? 00:01:08.070 --> 00:01:15.960 align:middle line:90% 00:01:15.960 --> 00:01:19.520 align:middle line:90% So here, we have an example. 00:01:19.520 --> 00:01:21.840 align:middle line:90% This is a network. 00:01:21.840 --> 00:01:26.520 align:middle line:84% This 6 says that only six units of flow 00:01:26.520 --> 00:01:30.140 align:middle line:84% can pass over this edge at any given time. 00:01:30.140 --> 00:01:32.120 align:middle line:90% This says 4. 00:01:32.120 --> 00:01:33.520 align:middle line:90% This is 3. 00:01:33.520 --> 00:01:35.600 align:middle line:84% And the question is, how many units of flow 00:01:35.600 --> 00:01:39.360 align:middle line:90% can you get from s to t? 00:01:39.360 --> 00:01:45.160 align:middle line:84% And we will call the source s And the sink t. 00:01:45.160 --> 00:01:47.580 align:middle line:84% And these are the standard letters that are used for them. 00:01:47.580 --> 00:01:50.880 align:middle line:84% I don't know whether they stand for anything or not. 00:01:50.880 --> 00:01:54.820 align:middle line:84% I mean, clearly S presumably stands for source. 00:01:54.820 --> 00:01:58.880 align:middle line:84% But I don't know why t is the sink. 00:01:58.880 --> 00:02:01.570 align:middle line:84% So the question is, how much flow 00:02:01.570 --> 00:02:03.670 align:middle line:90% can you get from here to here? 00:02:03.670 --> 00:02:08.530 align:middle line:90% 00:02:08.530 --> 00:02:15.330 align:middle line:84% Well, let's try putting flow from here to here. 00:02:15.330 --> 00:02:17.950 align:middle line:84% Does anybody see a path you could run flow on? 00:02:17.950 --> 00:02:26.410 align:middle line:90% 00:02:26.410 --> 00:02:28.230 align:middle line:90% OK, this is an easy question. 00:02:28.230 --> 00:02:31.690 align:middle line:90% What is some path from s to t? 00:02:31.690 --> 00:02:35.570 align:middle line:90% Let's say s to-- 00:02:35.570 --> 00:02:41.110 align:middle line:84% oh my, I have left out the numbers on the nodes. 00:02:41.110 --> 00:02:43.490 align:middle line:90% So this is 4. 00:02:43.490 --> 00:02:46.250 align:middle line:90% And this is 5. 00:02:46.250 --> 00:02:52.690 align:middle line:84% So S to 2 to 4 to t, you could run three units of flow 00:02:52.690 --> 00:02:53.470 align:middle line:90% along this path. 00:02:53.470 --> 00:02:55.810 align:middle line:90% So let's put them there. 00:02:55.810 --> 00:02:59.130 align:middle line:84% Now, are there other paths you could take? 00:02:59.130 --> 00:03:00.730 align:middle line:90% Yeah, of course, there are. 00:03:00.730 --> 00:03:04.070 align:middle line:84% You could run two units of flow from here to here. 00:03:04.070 --> 00:03:10.770 align:middle line:90% So that's 2, 2, 2. 00:03:10.770 --> 00:03:14.530 align:middle line:90% And is this all? 00:03:14.530 --> 00:03:16.970 align:middle line:84% Well, no, because you could run two units of flow 00:03:16.970 --> 00:03:23.770 align:middle line:90% from here to here to here. 00:03:23.770 --> 00:03:27.510 align:middle line:84% And we have two units of flow already on here. 00:03:27.510 --> 00:03:29.530 align:middle line:90% So this gives you 4. 00:03:29.530 --> 00:03:32.390 align:middle line:84% You have 0 units of flow already. 00:03:32.390 --> 00:03:33.690 align:middle line:90% So this is 2. 00:03:33.690 --> 00:03:35.350 align:middle line:84% You have three units of flow here. 00:03:35.350 --> 00:03:38.730 align:middle line:90% So that gives you 5. 00:03:38.730 --> 00:03:42.690 align:middle line:90% Now, is that the maximum? 00:03:42.690 --> 00:03:44.380 align:middle line:90% Does anybody see any more paths? 00:03:44.380 --> 00:03:54.700 align:middle line:90% 00:03:54.700 --> 00:03:55.620 align:middle line:90% Yes? 00:03:55.620 --> 00:04:00.615 align:middle line:84% STUDENT: You could do one from s 2, 3, 5? 00:04:00.615 --> 00:04:04.580 align:middle line:84% PETER SHOR: Yeah, you could do one from here. 00:04:04.580 --> 00:04:08.160 align:middle line:90% Wait-- oh, you're right. 00:04:08.160 --> 00:04:09.580 align:middle line:90% This was 4. 00:04:09.580 --> 00:04:10.260 align:middle line:90% No, wait. 00:04:10.260 --> 00:04:11.420 align:middle line:90% Yeah, this was 4. 00:04:11.420 --> 00:04:14.340 align:middle line:84% But you could run one up here, 4. 00:04:14.340 --> 00:04:23.460 align:middle line:84% You could run one down here, 1 and 3, and one up here, 00:04:23.460 --> 00:04:25.220 align:middle line:90% and one up here. 00:04:25.220 --> 00:04:28.100 align:middle line:84% And there's still one more path you could do, 00:04:28.100 --> 00:04:34.820 align:middle line:84% which I guess is one here, one here, one here, one here, 00:04:34.820 --> 00:04:35.800 align:middle line:90% and one here. 00:04:35.800 --> 00:04:37.380 align:middle line:90% So that's pretty complicated. 00:04:37.380 --> 00:04:39.540 align:middle line:90% But this would be 5. 00:04:39.540 --> 00:04:41.900 align:middle line:90% And this would be 2. 00:04:41.900 --> 00:04:48.220 align:middle line:90% And this would be 3 and 2. 00:04:48.220 --> 00:04:49.660 align:middle line:90% No wait. 00:04:49.660 --> 00:04:53.160 align:middle line:84% And you could run one here from 4 to t. 00:04:53.160 --> 00:04:54.120 align:middle line:90% So that's there. 00:04:54.120 --> 00:04:56.020 align:middle line:90% And now, we have a flow. 00:04:56.020 --> 00:04:59.300 align:middle line:90% 00:04:59.300 --> 00:05:00.820 align:middle line:84% The flow through every edge is less 00:05:00.820 --> 00:05:02.320 align:middle line:90% than the capacity on that edge. 00:05:02.320 --> 00:05:03.740 align:middle line:90% So that's good. 00:05:03.740 --> 00:05:12.840 align:middle line:84% And we have gotten 9 units of flow from here to here. 00:05:12.840 --> 00:05:17.300 align:middle line:90% 00:05:17.300 --> 00:05:21.540 align:middle line:90% OK, I guess I copied down the-- 00:05:21.540 --> 00:05:23.538 align:middle line:84% no, I didn't copy anything wrong. 00:05:23.538 --> 00:05:25.580 align:middle line:84% I'm just looking at the green numbers rather than 00:05:25.580 --> 00:05:27.226 align:middle line:90% the blue numbers. 00:05:27.226 --> 00:05:31.670 align:middle line:90% So here is a flow of 9 units. 00:05:31.670 --> 00:05:39.100 align:middle line:90% 00:05:39.100 --> 00:05:42.860 align:middle line:84% So what we're going to do later is write down a linear program. 00:05:42.860 --> 00:05:44.540 align:middle line:84% And the solution to this linear program 00:05:44.540 --> 00:05:46.780 align:middle line:90% will be the maximum flow. 00:05:46.780 --> 00:05:49.380 align:middle line:84% But first, I want to talk about cuts. 00:05:49.380 --> 00:05:52.180 align:middle line:90% So how could you prove-- 00:05:52.180 --> 00:05:59.150 align:middle line:90% 00:05:59.150 --> 00:06:02.520 align:middle line:84% how can you prove that this flow is maximum? 00:06:02.520 --> 00:06:19.950 align:middle line:90% 00:06:19.950 --> 00:06:23.950 align:middle line:84% Well, let's look at these three edges-- 00:06:23.950 --> 00:06:27.350 align:middle line:84% this one, this one, and this one. 00:06:27.350 --> 00:06:33.930 align:middle line:84% There's four units of capacity on this edge, 00:06:33.930 --> 00:06:35.830 align:middle line:84% two units of capacity on this edge, 00:06:35.830 --> 00:06:39.590 align:middle line:84% and three units of capacity on that edge. 00:06:39.590 --> 00:06:43.230 align:middle line:90% And so this defines a cut. 00:06:43.230 --> 00:06:50.350 align:middle line:84% And at most, 9 units of flow can cross this cut. 00:06:50.350 --> 00:06:53.220 align:middle line:84% So that shows that 9 units of flow is maximum. 00:06:53.220 --> 00:06:57.350 align:middle line:90% 00:06:57.350 --> 00:07:09.150 align:middle line:84% Cuts-- a cut is defined as a subset 00:07:09.150 --> 00:07:20.210 align:middle line:84% of the vertices or subset containing s but not t. 00:07:20.210 --> 00:07:31.270 align:middle line:90% 00:07:31.270 --> 00:07:39.230 align:middle line:84% And so to make this a subset of the vertices, gosh, what we do 00:07:39.230 --> 00:07:44.950 align:middle line:84% is we make the subset this guy and this guy. 00:07:44.950 --> 00:08:05.440 align:middle line:84% And maybe we should call it C. Edges from C to C-bar 00:08:05.440 --> 00:08:06.700 align:middle line:90% are edges in the cut. 00:08:06.700 --> 00:08:13.680 align:middle line:90% 00:08:13.680 --> 00:08:19.140 align:middle line:84% And let's say the capacity of an edge is u of the edge. 00:08:19.140 --> 00:08:23.320 align:middle line:90% 00:08:23.320 --> 00:08:33.640 align:middle line:84% So summation over E equals i j inc j in C-bar. 00:08:33.640 --> 00:08:36.847 align:middle line:90% 00:08:36.847 --> 00:08:42.949 align:middle line:84% u of i j equals, I guess, the value of cut. 00:08:42.949 --> 00:08:48.320 align:middle line:90% 00:08:48.320 --> 00:08:55.960 align:middle line:84% And it should be fairly easy that the maximum flow 00:08:55.960 --> 00:08:59.520 align:middle line:84% is less than or equal to the minimum cut 00:08:59.520 --> 00:09:02.400 align:middle line:84% because if you have a flow from s to t, 00:09:02.400 --> 00:09:07.160 align:middle line:84% it must cross every edge of the cut. 00:09:07.160 --> 00:09:11.600 align:middle line:84% So if the flow is 9 units, then-- 00:09:11.600 --> 00:09:13.260 align:middle line:84% or rather, if the cut is 9 units, 00:09:13.260 --> 00:09:17.010 align:middle line:84% then you can't have a flow larger than 9 units. 00:09:17.010 --> 00:09:20.160 align:middle line:90% 00:09:20.160 --> 00:09:24.320 align:middle line:84% And the theorem we will prove later. 00:09:24.320 --> 00:09:34.830 align:middle line:84% The value of max flow equals the value of minimum cut. 00:09:34.830 --> 00:09:38.600 align:middle line:90% 00:09:38.600 --> 00:09:47.650 align:middle line:84% So you probably recognize this as being somewhat something 00:09:47.650 --> 00:09:50.250 align:middle line:90% like LP duality. 00:09:50.250 --> 00:09:59.650 align:middle line:84% And this is the primal solution of LP. 00:09:59.650 --> 00:10:06.560 align:middle line:84% And this will turn out to be the solution of the dual LP. 00:10:06.560 --> 00:10:09.290 align:middle line:90% 00:10:09.290 --> 00:10:18.530 align:middle line:84% And by the strong duality theorem of linear program, 00:10:18.530 --> 00:10:20.730 align:middle line:90% this guy equals this guy. 00:10:20.730 --> 00:10:25.750 align:middle line:84% So the maximum flow equals the minimum cut. 00:10:25.750 --> 00:10:29.050 align:middle line:84% So there's a bunch of things we have to do 00:10:29.050 --> 00:10:31.490 align:middle line:90% before we prove this theorem. 00:10:31.490 --> 00:10:33.070 align:middle line:90% So what do we do? 00:10:33.070 --> 00:10:38.490 align:middle line:90% 00:10:38.490 --> 00:10:40.740 align:middle line:84% First thing we do is write down the linear program. 00:10:40.740 --> 00:10:58.810 align:middle line:90% 00:10:58.810 --> 00:11:00.810 align:middle line:84% Does anybody have suggestions for what 00:11:00.810 --> 00:11:02.110 align:middle line:90% the variables should be? 00:11:02.110 --> 00:11:07.450 align:middle line:90% 00:11:07.450 --> 00:11:08.570 align:middle line:90% Yeah? 00:11:08.570 --> 00:11:12.596 align:middle line:84% STUDENT: How many units you load to each. 00:11:12.596 --> 00:11:14.970 align:middle line:84% PETER SHOR: Through each edge, exactly. 00:11:14.970 --> 00:11:20.770 align:middle line:84% So the variables are going to be x sub i j-- 00:11:20.770 --> 00:11:26.650 align:middle line:90% that's the flow over edge i j. 00:11:26.650 --> 00:11:30.890 align:middle line:90% 00:11:30.890 --> 00:11:35.350 align:middle line:84% And any suggestions for constraints? 00:11:35.350 --> 00:11:41.380 align:middle line:90% 00:11:41.380 --> 00:11:42.580 align:middle line:90% Yeah? 00:11:42.580 --> 00:11:44.080 align:middle line:84% STUDENT: For any you don't know, you 00:11:44.080 --> 00:11:47.010 align:middle line:84% put the srt, like maybe the flow inverse is the [INAUDIBLE]. 00:11:47.010 --> 00:11:47.760 align:middle line:90% PETER SHOR: Right. 00:11:47.760 --> 00:11:51.040 align:middle line:84% So flow conservation, I guess, you could call it. 00:11:51.040 --> 00:11:55.980 align:middle line:90% 00:11:55.980 --> 00:11:58.030 align:middle line:90% And let me write this down. 00:11:58.030 --> 00:12:15.600 align:middle line:90% 00:12:15.600 --> 00:12:18.810 align:middle line:84% Summation, hope I'm getting this-- 00:12:18.810 --> 00:12:21.340 align:middle line:90% 00:12:21.340 --> 00:12:29.460 align:middle line:84% j such that i j is in the set of edges x sub i j. 00:12:29.460 --> 00:12:36.300 align:middle line:84% So this is a set of things going into node i 00:12:36.300 --> 00:12:41.490 align:middle line:84% and a set of things going out of node i equals-- 00:12:41.490 --> 00:12:46.900 align:middle line:90% 00:12:46.900 --> 00:12:47.480 align:middle line:90% no, wait. 00:12:47.480 --> 00:12:48.938 align:middle line:84% This is the set of things going out 00:12:48.938 --> 00:12:53.140 align:middle line:84% of node i j and the set of things going into node i. 00:12:53.140 --> 00:13:03.140 align:middle line:84% And let's put a minus sign here so that k such that k i is in a. 00:13:03.140 --> 00:13:06.300 align:middle line:84% So this is a set of things going into node i, 00:13:06.300 --> 00:13:12.100 align:middle line:90% x sub k i equals 0 because-- 00:13:12.100 --> 00:13:15.060 align:middle line:84% linear programs, it's nice to write down, 00:13:15.060 --> 00:13:19.380 align:middle line:84% so you have the variables on the left side of the equations 00:13:19.380 --> 00:13:23.500 align:middle line:84% and our inequalities and the constants on the right side 00:13:23.500 --> 00:13:25.700 align:middle line:90% of the inequalities. 00:13:25.700 --> 00:13:29.260 align:middle line:84% And there's another set of constraints. 00:13:29.260 --> 00:13:32.260 align:middle line:84% Can anybody figure out what it is? 00:13:32.260 --> 00:13:33.740 align:middle line:90% Yep? 00:13:33.740 --> 00:13:37.198 align:middle line:90% STUDENT: [INAUDIBLE] 00:13:37.198 --> 00:13:37.865 align:middle line:90% PETER SHOR: Yes? 00:13:37.865 --> 00:13:38.970 align:middle line:90% STUDENT: The capacities? 00:13:38.970 --> 00:13:40.178 align:middle line:90% PETER SHOR: Yeah, capacities. 00:13:40.178 --> 00:13:42.310 align:middle line:84% We need to use capacities somewhere. 00:13:42.310 --> 00:13:48.150 align:middle line:84% And x i j is less than or equal to-- 00:13:48.150 --> 00:13:51.670 align:middle line:90% we call them u of i j. 00:13:51.670 --> 00:13:59.070 align:middle line:84% And also, xij had better be greater than or equal to 0. 00:13:59.070 --> 00:14:02.950 align:middle line:90% And what do we want to maximize? 00:14:02.950 --> 00:14:08.670 align:middle line:84% So what we want to do is we want to maximize some of the things 00:14:08.670 --> 00:14:16.547 align:middle line:90% into the sink, which is node t. 00:14:16.547 --> 00:14:19.130 align:middle line:84% But actually, we're going to do this a slightly different way. 00:14:19.130 --> 00:14:23.690 align:middle line:84% We're going to write out a new constraint summation. 00:14:23.690 --> 00:14:27.390 align:middle line:90% 00:14:27.390 --> 00:14:30.870 align:middle line:90% Let's write down for all i. 00:14:30.870 --> 00:14:49.170 align:middle line:84% And I want to write down k such that kt x sub kt is in a. 00:14:49.170 --> 00:14:56.210 align:middle line:84% So this is all the things going into the sink. 00:14:56.210 --> 00:15:01.510 align:middle line:90% x sub kt is equal to z. 00:15:01.510 --> 00:15:03.290 align:middle line:90% And we want to maximize z. 00:15:03.290 --> 00:15:08.385 align:middle line:90% 00:15:08.385 --> 00:15:13.650 align:middle line:84% I did not leave enough room, so maximize z. 00:15:13.650 --> 00:15:38.690 align:middle line:90% 00:15:38.690 --> 00:15:45.320 align:middle line:90% So that is the linear program. 00:15:45.320 --> 00:15:51.720 align:middle line:84% And I have a sneaking suspicion that I messed up 00:15:51.720 --> 00:15:53.520 align:middle line:90% a sign somewhere. 00:15:53.520 --> 00:15:56.360 align:middle line:90% But it doesn't really-- 00:15:56.360 --> 00:15:58.340 align:middle line:90% I mean, it doesn't matter. 00:15:58.340 --> 00:16:00.260 align:middle line:84% I mean, this is a linear program, 00:16:00.260 --> 00:16:01.770 align:middle line:90% which does give the solution. 00:16:01.770 --> 00:16:03.520 align:middle line:84% So it doesn't really matter if I messed up 00:16:03.520 --> 00:16:08.800 align:middle line:84% a sign, except then when we take the dual, it's going to be-- 00:16:08.800 --> 00:16:12.930 align:middle line:84% it's not going to be quite as nice. 00:16:12.930 --> 00:16:33.280 align:middle line:90% 00:16:33.280 --> 00:16:37.240 align:middle line:84% OK, let's just keep on going and write down the dual. 00:16:37.240 --> 00:16:43.320 align:middle line:84% And if I got the sign wrong, we'll fix it after that. 00:16:43.320 --> 00:16:45.420 align:middle line:84% I want to say the sign is not wrong. 00:16:45.420 --> 00:16:52.640 align:middle line:84% This is a perfectly good linear program for max flow. 00:16:52.640 --> 00:16:59.005 align:middle line:84% It's just I am a little worried that, when we take the dual, 00:16:59.005 --> 00:17:00.630 align:middle line:84% we're going to have to flip some signs. 00:17:00.630 --> 00:17:05.268 align:middle line:90% 00:17:05.268 --> 00:17:10.319 align:middle line:84% So why don't I take the dual now? 00:17:10.319 --> 00:17:22.160 align:middle line:84% To take dual, each equation gives a constraint, 00:17:22.160 --> 00:17:24.839 align:middle line:90% or each inequality-- 00:17:24.839 --> 00:17:27.079 align:middle line:84% no, each variable gives a constraint. 00:17:27.079 --> 00:17:33.330 align:middle line:84% Each constraint gives a variable. 00:17:33.330 --> 00:17:36.770 align:middle line:90% So we have two classes. 00:17:36.770 --> 00:17:38.930 align:middle line:84% Well, actually, I guess we have three classes 00:17:38.930 --> 00:17:45.130 align:middle line:90% of equations and inequalities. 00:17:45.130 --> 00:17:49.110 align:middle line:84% But I just want to use two of them. 00:17:49.110 --> 00:17:57.010 align:middle line:84% I'm going to call this one alpha sub ij, and this one beta sub i, 00:17:57.010 --> 00:18:05.050 align:middle line:84% and this one beta sub z, because this really is a lot like this, 00:18:05.050 --> 00:18:07.750 align:middle line:90% except slightly different. 00:18:07.750 --> 00:18:11.260 align:middle line:84% And I really should make this minus z equals 0. 00:18:11.260 --> 00:18:23.290 align:middle line:90% 00:18:23.290 --> 00:18:32.590 align:middle line:84% So how do we get the variables of the-- 00:18:32.590 --> 00:18:37.610 align:middle line:84% how do we get the constraints of the dual 00:18:37.610 --> 00:18:40.210 align:middle line:84% and the objective function of the dual? 00:18:40.210 --> 00:18:43.410 align:middle line:84% Well, the objective function of the dual is just the right-hand 00:18:43.410 --> 00:18:45.350 align:middle line:90% sides of the-- 00:18:45.350 --> 00:18:49.062 align:middle line:90% 00:18:49.062 --> 00:18:53.520 align:middle line:84% right-hand sides of the constraints of the primal. 00:18:53.520 --> 00:19:07.450 align:middle line:90% 00:19:07.450 --> 00:19:10.370 align:middle line:90% So we will have minimum-- 00:19:10.370 --> 00:19:14.430 align:middle line:84% because the original primal LP was a maximization, 00:19:14.430 --> 00:19:16.530 align:middle line:90% we want to have a minimization. 00:19:16.530 --> 00:19:27.230 align:middle line:84% And the minimization summation alpha ij u of ij subject to-- 00:19:27.230 --> 00:19:32.940 align:middle line:90% 00:19:32.940 --> 00:19:39.860 align:middle line:84% OK, we're going to have a equation for each variable. 00:19:39.860 --> 00:19:46.580 align:middle line:84% So let's say the equation corresponding to variable x sub 00:19:46.580 --> 00:19:51.660 align:middle line:84% ij is just-- you look at all the places 00:19:51.660 --> 00:19:54.300 align:middle line:90% where xij appears in equation. 00:19:54.300 --> 00:20:01.540 align:middle line:84% So x sub ij appears in beta sub i and in alpha sub ij. 00:20:01.540 --> 00:20:08.140 align:middle line:90% 00:20:08.140 --> 00:20:19.740 align:middle line:84% And we should have alpha ij plus beta sub i minus beta sub j 00:20:19.740 --> 00:20:23.220 align:middle line:90% is greater than or equal to 0. 00:20:23.220 --> 00:20:27.900 align:middle line:84% So it's greater than or equal to because these 00:20:27.900 --> 00:20:29.700 align:middle line:90% were positive variables. 00:20:29.700 --> 00:20:33.740 align:middle line:84% And when we go from a maximization to a minimization, 00:20:33.740 --> 00:20:35.300 align:middle line:90% you're minimizing this. 00:20:35.300 --> 00:20:38.480 align:middle line:84% You want a greater than or equal to sign in your constraints. 00:20:38.480 --> 00:20:46.579 align:middle line:90% 00:20:46.579 --> 00:20:48.510 align:middle line:90% Now, there's one more-- 00:20:48.510 --> 00:20:54.780 align:middle line:90% 00:20:54.780 --> 00:21:00.860 align:middle line:84% well, I guess z was also a variable in this. 00:21:00.860 --> 00:21:12.660 align:middle line:84% And so there's an equation, a constraint corresponding to z. 00:21:12.660 --> 00:21:15.980 align:middle line:84% And z appears in the objective function. 00:21:15.980 --> 00:21:18.580 align:middle line:84% So this should be greater than or equal to 1. 00:21:18.580 --> 00:21:22.030 align:middle line:90% And z also appears in-- 00:21:22.030 --> 00:21:26.150 align:middle line:90% 00:21:26.150 --> 00:21:29.950 align:middle line:84% well, I just want to say it only appears in one equation. 00:21:29.950 --> 00:21:31.750 align:middle line:90% That's beta sub z. 00:21:31.750 --> 00:21:36.190 align:middle line:84% So beta sub z is greater or equal to 1. 00:21:36.190 --> 00:21:37.740 align:middle line:90% No, actually it appears in-- 00:21:37.740 --> 00:21:43.570 align:middle line:90% 00:21:43.570 --> 00:21:44.070 align:middle line:90% yeah. 00:21:44.070 --> 00:22:01.150 align:middle line:90% 00:22:01.150 --> 00:22:07.210 align:middle line:90% Yeah, so this is the dual. 00:22:07.210 --> 00:22:10.210 align:middle line:90% STUDENT: [INAUDIBLE] 00:22:10.210 --> 00:22:12.830 align:middle line:90% 00:22:12.830 --> 00:22:14.078 align:middle line:90% PETER SHOR: What? 00:22:14.078 --> 00:22:15.950 align:middle line:90% STUDENT: Why do [INAUDIBLE]? 00:22:15.950 --> 00:22:17.250 align:middle line:90% PETER SHOR: Not negative 1? 00:22:17.250 --> 00:22:20.350 align:middle line:90% 00:22:20.350 --> 00:22:27.470 align:middle line:84% Because I should have written this equation minus and plus. 00:22:27.470 --> 00:22:34.590 align:middle line:84% And now, this left-hand side is things leading into-- 00:22:34.590 --> 00:22:38.710 align:middle line:84% oh, this is beta sub t not beta sub z. 00:22:38.710 --> 00:22:39.840 align:middle line:90% I'm sorry. 00:22:39.840 --> 00:22:41.090 align:middle line:90% That was a very good question. 00:22:41.090 --> 00:22:42.630 align:middle line:90% Beta sub t. 00:22:42.630 --> 00:22:53.150 align:middle line:90% And this thing is k-- 00:22:53.150 --> 00:23:01.390 align:middle line:84% this should be edges leading into t. 00:23:01.390 --> 00:23:05.590 align:middle line:84% And if we look down here, edges leading into i 00:23:05.590 --> 00:23:07.070 align:middle line:90% have a minus sign. 00:23:07.070 --> 00:23:10.430 align:middle line:84% So to make it consistent, this has to have a minus sign. 00:23:10.430 --> 00:23:12.750 align:middle line:90% And then this is a plus sign. 00:23:12.750 --> 00:23:20.370 align:middle line:84% And we have beta sub t greater than or equal to 1. 00:23:20.370 --> 00:23:22.000 align:middle line:90% That was a very good question. 00:23:22.000 --> 00:23:26.740 align:middle line:84% And that explains why I was worried about the signs. 00:23:26.740 --> 00:23:27.240 align:middle line:90% Good. 00:23:27.240 --> 00:23:39.720 align:middle line:90% 00:23:39.720 --> 00:23:43.050 align:middle line:84% So what do solutions of this equation look like? 00:23:43.050 --> 00:23:51.720 align:middle line:90% 00:23:51.720 --> 00:23:59.080 align:middle line:84% Well, how do you get cuts out of solutions to this equation? 00:23:59.080 --> 00:24:01.630 align:middle line:90% And this is a very-- 00:24:01.630 --> 00:24:20.920 align:middle line:90% 00:24:20.920 --> 00:24:25.236 align:middle line:90% this is a rather tricky thing. 00:24:25.236 --> 00:24:31.320 align:middle line:90% 00:24:31.320 --> 00:24:33.690 align:middle line:90% So for this, we have-- 00:24:33.690 --> 00:24:40.560 align:middle line:90% 00:24:40.560 --> 00:24:53.040 align:middle line:84% we put numbers on the nodes of this network. 00:24:53.040 --> 00:25:00.760 align:middle line:84% And we want to maximize alpha sub ij times the capacity 00:25:00.760 --> 00:25:07.760 align:middle line:84% and subject to the capacity beta i minus beta j 00:25:07.760 --> 00:25:10.120 align:middle line:90% is greater than or equal to 0. 00:25:10.120 --> 00:25:23.410 align:middle line:84% And first, I want to say, for optimal alpha sub ij 00:25:23.410 --> 00:25:29.170 align:middle line:84% is equal to, well, either alpha ij-- 00:25:29.170 --> 00:25:32.450 align:middle line:90% 00:25:32.450 --> 00:25:37.010 align:middle line:84% oh, did I write forget to write down alpha ij bigger than 00:25:37.010 --> 00:25:39.210 align:middle line:90% or equal to 0? 00:25:39.210 --> 00:25:50.130 align:middle line:84% Because alpha sub ij was an inequality in-- 00:25:50.130 --> 00:26:00.690 align:middle line:90% 00:26:00.690 --> 00:26:05.190 align:middle line:84% the constraint corresponding to alpha sub ij was an inequality. 00:26:05.190 --> 00:26:09.530 align:middle line:84% So alpha ij is greater than or equal to 0. 00:26:09.530 --> 00:26:13.350 align:middle line:84% And the beta sub i's are unconstrained. 00:26:13.350 --> 00:26:18.010 align:middle line:90% 00:26:18.010 --> 00:26:21.570 align:middle line:84% For the optimal, well, you want to make alpha sub ij 00:26:21.570 --> 00:26:25.810 align:middle line:84% as small as possible, subject to the constraints 00:26:25.810 --> 00:26:29.690 align:middle line:84% that alpha sub ij is bigger or equal to 0 and alpha sub 00:26:29.690 --> 00:26:34.370 align:middle line:84% ij is bigger than beta j minus beta i. 00:26:34.370 --> 00:26:39.090 align:middle line:84% So you want to make alpha ij equal to maximum of 0 00:26:39.090 --> 00:26:44.980 align:middle line:84% and beta j minus beta sub i and the optimal. 00:26:44.980 --> 00:26:51.910 align:middle line:90% 00:26:51.910 --> 00:26:57.610 align:middle line:84% So we're going to put values on the nodes. 00:26:57.610 --> 00:27:03.970 align:middle line:84% And we're going to make alpha sub ij for maximum 0 00:27:03.970 --> 00:27:06.410 align:middle line:90% and beta j minus beta i. 00:27:06.410 --> 00:27:18.580 align:middle line:84% Now, suppose each alpha ij equals 0 or 1. 00:27:18.580 --> 00:27:22.660 align:middle line:84% And now, what we have is we have-- 00:27:22.660 --> 00:27:30.645 align:middle line:84% and I'm just going to say here, because we 00:27:30.645 --> 00:27:32.520 align:middle line:84% want to make everything as small as possible, 00:27:32.520 --> 00:27:35.140 align:middle line:90% we'll take beta sub t equal 1. 00:27:35.140 --> 00:27:48.100 align:middle line:84% And we'll take beta sub s equals 0 because if there is no-- 00:27:48.100 --> 00:27:58.900 align:middle line:84% because if s-- if s is one of these-- 00:27:58.900 --> 00:28:03.020 align:middle line:84% if we're looking at ij or i equals s for this, 00:28:03.020 --> 00:28:06.460 align:middle line:84% then really we don't have a beta sub i. 00:28:06.460 --> 00:28:11.790 align:middle line:84% And the equation is alpha i sub j minus beta j 00:28:11.790 --> 00:28:13.040 align:middle line:90% is greater than or equal to 0. 00:28:13.040 --> 00:28:15.020 align:middle line:84% But you can just make everything uniform 00:28:15.020 --> 00:28:17.540 align:middle line:90% by setting beta s equal to 0. 00:28:17.540 --> 00:28:19.040 align:middle line:90% So we have beta t equals 1. 00:28:19.040 --> 00:28:19.990 align:middle line:90% Beta s equals 0. 00:28:19.990 --> 00:28:26.340 align:middle line:90% 00:28:26.340 --> 00:28:28.700 align:middle line:90% So we have 0, 0. 00:28:28.700 --> 00:28:30.110 align:middle line:90% OK, I should put some-- 00:28:30.110 --> 00:28:34.340 align:middle line:90% 00:28:34.340 --> 00:28:36.420 align:middle line:90% This is 1. 00:28:36.420 --> 00:28:41.500 align:middle line:84% Let's make this 1, this 0, et cetera. 00:28:41.500 --> 00:28:44.460 align:middle line:90% 00:28:44.460 --> 00:28:46.740 align:middle line:84% So alpha sub ij is the maximum of 0 00:28:46.740 --> 00:28:48.780 align:middle line:90% and beta j minus beta sub i. 00:28:48.780 --> 00:28:51.500 align:middle line:84% So that means that this is going to be 1. 00:28:51.500 --> 00:28:53.460 align:middle line:90% This is going to be 0. 00:28:53.460 --> 00:28:54.850 align:middle line:90% This is going to be 1. 00:28:54.850 --> 00:29:01.860 align:middle line:90% 00:29:01.860 --> 00:29:05.280 align:middle line:84% This would be minus 1, except that's less than 0. 00:29:05.280 --> 00:29:07.580 align:middle line:90% So this had better be 0. 00:29:07.580 --> 00:29:09.300 align:middle line:90% This is 0. 00:29:09.300 --> 00:29:11.660 align:middle line:90% This is 0 for the same reason. 00:29:11.660 --> 00:29:13.830 align:middle line:90% This is 1. 00:29:13.830 --> 00:29:20.110 align:middle line:90% This is 1. 00:29:20.110 --> 00:29:20.900 align:middle line:90% And this is 0. 00:29:20.900 --> 00:29:24.190 align:middle line:90% 00:29:24.190 --> 00:29:34.030 align:middle line:84% The 1 edges are a cut because the 1 edges are just 00:29:34.030 --> 00:29:38.150 align:middle line:84% the edges leading out of the things-- 00:29:38.150 --> 00:29:43.110 align:middle line:84% nodes with alpha-- nodes with beta i equals 0 00:29:43.110 --> 00:29:46.710 align:middle line:84% and to the nodes beta i equals 1, which 00:29:46.710 --> 00:29:50.670 align:middle line:90% was our definition of cut. 00:29:50.670 --> 00:29:54.410 align:middle line:84% So that means that this is a cut between-- 00:29:54.410 --> 00:30:01.093 align:middle line:90% 00:30:01.093 --> 00:30:02.010 align:middle line:90% which nodes are zeros? 00:30:02.010 --> 00:30:04.570 align:middle line:90% 00:30:04.570 --> 00:30:08.900 align:middle line:84% Between the green nodes and the other nodes. 00:30:08.900 --> 00:30:11.590 align:middle line:90% 00:30:11.590 --> 00:30:18.110 align:middle line:84% So there's one here, one here, and one here. 00:30:18.110 --> 00:30:21.830 align:middle line:90% So when you add up the-- 00:30:21.830 --> 00:30:25.790 align:middle line:84% you look at the objective function, 00:30:25.790 --> 00:30:34.830 align:middle line:84% it's just the sum over all of the edges in the cut 00:30:34.830 --> 00:30:36.830 align:middle line:90% times their capacity-- 00:30:36.830 --> 00:30:39.850 align:middle line:84% or 1 for all the edges in the cut times their capacity. 00:30:39.850 --> 00:30:42.140 align:middle line:84% So it's just the value of the cut. 00:30:42.140 --> 00:30:46.790 align:middle line:90% 00:30:46.790 --> 00:30:57.390 align:middle line:84% So this is dual objective function-- 00:30:57.390 --> 00:31:00.970 align:middle line:90% 00:31:00.970 --> 00:31:04.460 align:middle line:90% value of cut f. 00:31:04.460 --> 00:31:07.030 align:middle line:90% 00:31:07.030 --> 00:31:11.560 align:middle line:90% Beta sub i is in 0, 1. 00:31:11.560 --> 00:31:17.960 align:middle line:84% Now, here comes the trickiest part of this whole thing. 00:31:17.960 --> 00:31:25.040 align:middle line:84% Why can we assume it's the minimum cut rather than this? 00:31:25.040 --> 00:31:27.160 align:middle line:84% In other words, why can we assume 00:31:27.160 --> 00:31:31.600 align:middle line:84% that the optimal solution of the dual 00:31:31.600 --> 00:31:34.760 align:middle line:84% has integer values rather than fractional values 00:31:34.760 --> 00:31:36.300 align:middle line:90% for the variables? 00:31:36.300 --> 00:31:42.240 align:middle line:90% 00:31:42.240 --> 00:31:51.840 align:middle line:84% Look at-- OK, we're going to look at this graph in another 00:31:51.840 --> 00:31:52.880 align:middle line:90% way-- 00:31:52.880 --> 00:32:00.100 align:middle line:84% graph above, wrong in another way. 00:32:00.100 --> 00:32:03.200 align:middle line:90% 00:32:03.200 --> 00:32:13.440 align:middle line:84% So let's put s beta sub s equals 0 on the left side, beta sub t 00:32:13.440 --> 00:32:28.560 align:middle line:84% equals 1 on the right side, and beta sub 2, beta sub 3, beta sub 00:32:28.560 --> 00:32:30.330 align:middle line:90% 4, beta sub 5. 00:32:30.330 --> 00:32:40.800 align:middle line:90% 00:32:40.800 --> 00:32:44.400 align:middle line:84% I want to say that there's no reason these beta i's have 00:32:44.400 --> 00:32:46.240 align:middle line:90% to be in order. 00:32:46.240 --> 00:32:56.120 align:middle line:84% Put beta i's online at value of beta i 00:32:56.120 --> 00:32:57.740 align:middle line:84% in the optimal solution of the dual. 00:32:57.740 --> 00:33:02.360 align:middle line:90% 00:33:02.360 --> 00:33:06.930 align:middle line:84% So if beta was equal to 0.2, then the optimum solution 00:33:06.930 --> 00:33:11.090 align:middle line:84% of the pool at this point would be 0.2. 00:33:11.090 --> 00:33:14.610 align:middle line:84% And now, let's look at the edges in this graph. 00:33:14.610 --> 00:33:23.530 align:middle line:90% And this is 6. 00:33:23.530 --> 00:33:27.690 align:middle line:90% This is 4. 00:33:27.690 --> 00:33:28.880 align:middle line:90% This is 2. 00:33:28.880 --> 00:33:32.130 align:middle line:90% 00:33:32.130 --> 00:33:44.010 align:middle line:84% There is an edge 3, value 3 going from beta 2 to beta 4. 00:33:44.010 --> 00:33:46.210 align:middle line:90% So this is 3. 00:33:46.210 --> 00:33:51.530 align:middle line:84% There's an edge from beta 3 to beta 4. 00:33:51.530 --> 00:33:53.020 align:middle line:90% So that is 3. 00:33:53.020 --> 00:33:58.050 align:middle line:90% 00:33:58.050 --> 00:34:02.370 align:middle line:84% There's an edge from beta 4 to beta 5. 00:34:02.370 --> 00:34:07.410 align:middle line:84% Wait, there's an edge from beta 5 to beta 4, 00:34:07.410 --> 00:34:11.810 align:middle line:90% which is this way, 2. 00:34:11.810 --> 00:34:16.150 align:middle line:84% There's an edge from beta 4 to t So this is 8. 00:34:16.150 --> 00:34:18.929 align:middle line:90% 00:34:18.929 --> 00:34:20.310 align:middle line:90% And this is 2. 00:34:20.310 --> 00:35:06.620 align:middle line:90% 00:35:06.620 --> 00:35:31.400 align:middle line:84% So the claim-- edges crossing a vertical line form a cut. 00:35:31.400 --> 00:35:35.260 align:middle line:90% 00:35:35.260 --> 00:35:38.570 align:middle line:90% So let's look at this edge. 00:35:38.570 --> 00:35:44.900 align:middle line:90% 00:35:44.900 --> 00:35:50.440 align:middle line:84% This vertical line, this has edge 2, 3, and 4 crossing it. 00:35:50.440 --> 00:35:57.460 align:middle line:84% And we have beta s and beta 2 are on the left side. 00:35:57.460 --> 00:36:05.380 align:middle line:90% So that is just this cut. 00:36:05.380 --> 00:36:07.752 align:middle line:84% You have s and 2 on the left side of the cut 00:36:07.752 --> 00:36:09.960 align:middle line:84% and all the other edges on the right side of the cut. 00:36:09.960 --> 00:36:13.420 align:middle line:84% And we have 4, and 2, and 3 crossing them. 00:36:13.420 --> 00:36:15.980 align:middle line:90% Now, let's try another cut. 00:36:15.980 --> 00:36:19.030 align:middle line:84% And I probably should use a different color. 00:36:19.030 --> 00:36:23.780 align:middle line:90% 00:36:23.780 --> 00:36:31.390 align:middle line:90% So let's try this cut. 00:36:31.390 --> 00:36:35.420 align:middle line:90% 00:36:35.420 --> 00:36:38.660 align:middle line:84% Now, we have 5 and t on the right side and the other four 00:36:38.660 --> 00:36:40.180 align:middle line:90% edges on the left. 00:36:40.180 --> 00:36:44.690 align:middle line:84% So what cut does that correspond to here? 00:36:44.690 --> 00:36:47.860 align:middle line:90% 00:36:47.860 --> 00:36:51.430 align:middle line:84% That's not a different color I'm sorry. 00:36:51.430 --> 00:36:54.260 align:middle line:90% 00:36:54.260 --> 00:36:57.350 align:middle line:90% Let's try making it purple. 00:36:57.350 --> 00:37:04.510 align:middle line:90% 00:37:04.510 --> 00:37:10.980 align:middle line:84% And I see I have drawn things wrong. 00:37:10.980 --> 00:37:13.950 align:middle line:90% 00:37:13.950 --> 00:37:20.710 align:middle line:84% I left out an edge because there's an edge from b3 to b5. 00:37:20.710 --> 00:37:25.110 align:middle line:84% So that should look like this because we 00:37:25.110 --> 00:37:27.470 align:middle line:90% have three going to 5. 00:37:27.470 --> 00:37:29.330 align:middle line:90% And that edge had capacity 4. 00:37:29.330 --> 00:37:39.710 align:middle line:90% 00:37:39.710 --> 00:37:45.270 align:middle line:90% So the value of this cut is 2. 00:37:45.270 --> 00:37:50.070 align:middle line:90% No wait, it's 4 plus 8. 00:37:50.070 --> 00:37:51.670 align:middle line:90% That's 12. 00:37:51.670 --> 00:37:55.430 align:middle line:84% And we don't count this edge 2 because it 00:37:55.430 --> 00:37:57.510 align:middle line:90% crosses the cut backwards. 00:37:57.510 --> 00:38:01.430 align:middle line:84% And we only count edges that cross the cut forwards. 00:38:01.430 --> 00:38:09.140 align:middle line:90% And here, we have that cut is-- 00:38:09.140 --> 00:38:15.630 align:middle line:90% 00:38:15.630 --> 00:38:19.040 align:middle line:84% OK, I need a purple line here, purple line here. 00:38:19.040 --> 00:38:32.310 align:middle line:90% 00:38:32.310 --> 00:38:34.120 align:middle line:90% No, those are the wrong-- 00:38:34.120 --> 00:38:36.950 align:middle line:90% 00:38:36.950 --> 00:38:38.100 align:middle line:90% I'm sorry about that. 00:38:38.100 --> 00:38:40.610 align:middle line:90% 00:38:40.610 --> 00:38:43.610 align:middle line:84% So 5 and t are on one side of this cut. 00:38:43.610 --> 00:38:48.990 align:middle line:84% So this, this, and this edge is not across the cut 00:38:48.990 --> 00:38:51.190 align:middle line:84% because it crosses the cut backwards. 00:38:51.190 --> 00:38:55.550 align:middle line:90% So the value of that cut is-- 00:38:55.550 --> 00:39:01.100 align:middle line:90% 00:39:01.100 --> 00:39:01.600 align:middle line:90% no. 00:39:01.600 --> 00:39:05.080 align:middle line:90% 00:39:05.080 --> 00:39:06.380 align:middle line:90% These vertices are in the cut. 00:39:06.380 --> 00:39:09.420 align:middle line:84% So the edges are the ones that go from something not in the cut 00:39:09.420 --> 00:39:10.420 align:middle line:90% to something in the cut. 00:39:10.420 --> 00:39:18.880 align:middle line:84% So that's 4 to t, 3 to 5, and 5 to 4 is backwards edge. 00:39:18.880 --> 00:39:24.940 align:middle line:84% So the value of that cut is 8 plus 3, which is 11, 00:39:24.940 --> 00:39:26.100 align:middle line:90% which is bigger than 9. 00:39:26.100 --> 00:39:27.520 align:middle line:90% So it's not the minimum cut. 00:39:27.520 --> 00:39:40.360 align:middle line:90% 00:39:40.360 --> 00:39:42.680 align:middle line:84% STUDENT: I think the beta 3 to beta 5 00:39:42.680 --> 00:39:44.860 align:middle line:90% might have the wrong value. 00:39:44.860 --> 00:39:46.400 align:middle line:84% PETER SHOR: I think you're right. 00:39:46.400 --> 00:39:51.000 align:middle line:90% So beta 3 to beta 5 is 4. 00:39:51.000 --> 00:39:53.660 align:middle line:90% And I put down 3 on here. 00:39:53.660 --> 00:40:00.440 align:middle line:90% 00:40:00.440 --> 00:40:05.880 align:middle line:84% So let's see beta 3 to beta 5, 3. 00:40:05.880 --> 00:40:08.520 align:middle line:84% So this is supposed to be a three. 00:40:08.520 --> 00:40:09.220 align:middle line:90% Thank you. 00:40:09.220 --> 00:40:20.640 align:middle line:90% 00:40:20.640 --> 00:40:27.080 align:middle line:84% So we have a cut of value 11 for there. 00:40:27.080 --> 00:40:31.080 align:middle line:90% So these are cuts. 00:40:31.080 --> 00:40:40.770 align:middle line:84% Now, I want to claim the value of the dual optimum. 00:40:40.770 --> 00:40:56.610 align:middle line:90% 00:40:56.610 --> 00:40:58.810 align:middle line:84% Well, I want to label these cuts. 00:40:58.810 --> 00:41:19.470 align:middle line:84% So let's label this one as C1, C2, C3, C4, C5. 00:41:19.470 --> 00:41:26.010 align:middle line:90% 00:41:26.010 --> 00:41:33.250 align:middle line:84% And let's assume that the betas are put down in order because we 00:41:33.250 --> 00:41:38.650 align:middle line:84% can always renumber the variables so 00:41:38.650 --> 00:41:46.090 align:middle line:84% that the third variable is the one right after the second cut. 00:41:46.090 --> 00:41:51.750 align:middle line:84% So the value of the optimum and the dual is equal to summation. 00:41:51.750 --> 00:41:54.370 align:middle line:90% 00:41:54.370 --> 00:42:04.890 align:middle line:84% Well, summation alpha sub ij u of ij, 00:42:04.890 --> 00:42:24.970 align:middle line:84% which is equal to summation beta j minus beta j minus 1 times c 00:42:24.970 --> 00:42:28.110 align:middle line:90% sub j minus 1. 00:42:28.110 --> 00:42:32.250 align:middle line:90% 00:42:32.250 --> 00:42:34.450 align:middle line:90% Why is that? 00:42:34.450 --> 00:42:37.040 align:middle line:84% Well, let's look at this picture again. 00:42:37.040 --> 00:42:43.810 align:middle line:90% 00:42:43.810 --> 00:42:53.300 align:middle line:84% So we have alpha i sub j equals maximum of 0 00:42:53.300 --> 00:42:55.280 align:middle line:90% and beta j minus beta sub i. 00:42:55.280 --> 00:43:00.100 align:middle line:90% 00:43:00.100 --> 00:43:10.580 align:middle line:84% So this is equal to summation over forward edges beta j 00:43:10.580 --> 00:43:14.380 align:middle line:90% minus beta i u of ij. 00:43:14.380 --> 00:43:19.380 align:middle line:90% 00:43:19.380 --> 00:43:21.780 align:middle line:90% Now, let's look at this edge. 00:43:21.780 --> 00:43:25.140 align:middle line:90% This edge appears in two cuts. 00:43:25.140 --> 00:43:31.940 align:middle line:84% And its value in the first cut is beta t minus beta 5 00:43:31.940 --> 00:43:35.000 align:middle line:90% times the value times 8. 00:43:35.000 --> 00:43:37.900 align:middle line:84% The value of contributing the second cut 00:43:37.900 --> 00:43:43.180 align:middle line:90% is beta 5 minus beta 4 times 8. 00:43:43.180 --> 00:43:48.100 align:middle line:84% So adding them up, when you sum up all the cuts, 00:43:48.100 --> 00:43:54.800 align:middle line:84% you get beta t minus beta 4 times 8 for the value 00:43:54.800 --> 00:44:05.020 align:middle line:84% this edge contributes to the dual, which is exactly what it's 00:44:05.020 --> 00:44:05.840 align:middle line:90% supposed to be. 00:44:05.840 --> 00:44:12.020 align:middle line:90% 00:44:12.020 --> 00:44:13.880 align:middle line:84% OK, so this is the trickiest part. 00:44:13.880 --> 00:44:17.360 align:middle line:84% So should I explain it some more? 00:44:17.360 --> 00:44:31.120 align:middle line:90% 00:44:31.120 --> 00:44:40.220 align:middle line:84% So the cuts only come from the edges 00:44:40.220 --> 00:44:42.540 align:middle line:84% where beta 5 is bigger than beta 4. 00:44:42.540 --> 00:44:47.250 align:middle line:84% So this edge does not contribute anything to the cut. 00:44:47.250 --> 00:44:49.750 align:middle line:84% If there's an edge that goes forward, 00:44:49.750 --> 00:44:55.990 align:middle line:84% it contributes beta t minus beta 5 times 8 to this cut and beta 00:44:55.990 --> 00:44:59.830 align:middle line:84% 5 minus beta 4 times 8 to this cut. 00:44:59.830 --> 00:45:01.390 align:middle line:84% And adding those two things up, you 00:45:01.390 --> 00:45:09.790 align:middle line:84% get beta t minus beta 4 times 8 to the total objective 00:45:09.790 --> 00:45:16.210 align:middle line:84% function of the dual, which is just what it's supposed to be, 00:45:16.210 --> 00:45:20.090 align:middle line:84% because alpha ij is equal to beta j minus beta i. 00:45:20.090 --> 00:45:45.430 align:middle line:90% 00:45:45.430 --> 00:45:47.900 align:middle line:84% Oh, wait, I think I wanted to keep that. 00:45:47.900 --> 00:46:30.150 align:middle line:90% 00:46:30.150 --> 00:46:40.270 align:middle line:84% Objective function of dual is summation beta j 00:46:40.270 --> 00:46:45.940 align:middle line:84% minus beta j minus 1 c sub j minus 1. 00:46:45.940 --> 00:46:53.320 align:middle line:90% 00:46:53.320 --> 00:46:58.760 align:middle line:84% But summation beta j minus beta j minus 1 00:46:58.760 --> 00:47:04.160 align:middle line:84% is equal to beta t minus beta s is equal to 1 00:47:04.160 --> 00:47:06.470 align:middle line:84% because beta t was 1, beta s was 0. 00:47:06.470 --> 00:47:10.160 align:middle line:90% 00:47:10.160 --> 00:47:17.720 align:middle line:84% Let beta j minus beta j minus 1 equals lambda j. 00:47:17.720 --> 00:47:24.400 align:middle line:84% So this equals summation lambda j c j minus 1. 00:47:24.400 --> 00:47:30.440 align:middle line:84% OK, maybe lambda j minus 1, lambda J minus 1, c j minus 1, 00:47:30.440 --> 00:47:35.900 align:middle line:84% where summation lambda j minus 1 equals 1. 00:47:35.900 --> 00:47:42.280 align:middle line:90% 00:47:42.280 --> 00:47:56.932 align:middle line:84% This is an average of values of cuts 00:47:56.932 --> 00:47:59.480 align:middle line:84% just by the definition of average. 00:47:59.480 --> 00:48:02.440 align:middle line:84% But the average of a bunch of things 00:48:02.440 --> 00:48:05.360 align:middle line:84% is bigger than the minimum of the bunch of things. 00:48:05.360 --> 00:48:12.600 align:middle line:84% So this is greater than or equal to the minimum cut. 00:48:12.600 --> 00:48:16.520 align:middle line:84% And in the optimal solution of the dual, 00:48:16.520 --> 00:48:18.640 align:middle line:84% you take the minimum cut, you can 00:48:18.640 --> 00:48:23.460 align:middle line:84% put zeros on the beta sub j's that are to the left of the cut 00:48:23.460 --> 00:48:26.340 align:middle line:84% and 1 on the beta j's to the right of the cut. 00:48:26.340 --> 00:48:28.360 align:middle line:84% So you get the minimum value of the cut 00:48:28.360 --> 00:48:30.560 align:middle line:90% for the solution of the dual. 00:48:30.560 --> 00:48:38.880 align:middle line:84% So the value of the dual substitution 00:48:38.880 --> 00:48:43.010 align:middle line:90% is equal to the minimum cut. 00:48:43.010 --> 00:48:46.570 align:middle line:84% And we know that the value of the primal solution 00:48:46.570 --> 00:48:50.410 align:middle line:84% is equal to the maximum flow because that's 00:48:50.410 --> 00:48:52.210 align:middle line:90% the way we set up the LP. 00:48:52.210 --> 00:48:55.520 align:middle line:84% So the maximum flow is equal to the minimum cut. 00:48:55.520 --> 00:49:04.170 align:middle line:90% 00:49:04.170 --> 00:49:11.450 align:middle line:84% So the next thing I want to do, which I don't think I'm 00:49:11.450 --> 00:49:14.070 align:middle line:84% going to quite get through all the lecture notes today-- 00:49:14.070 --> 00:49:19.650 align:middle line:84% but what I've done today so far is the thing 00:49:19.650 --> 00:49:21.930 align:middle line:90% that I really want you to take-- 00:49:21.930 --> 00:49:26.370 align:middle line:84% the lesson I want you to take home, that looking at the duals 00:49:26.370 --> 00:49:30.250 align:middle line:84% and analyzing them can give you very interesting theorems 00:49:30.250 --> 00:49:33.200 align:middle line:90% about linear programs. 00:49:33.200 --> 00:49:37.620 align:middle line:90% 00:49:37.620 --> 00:49:44.090 align:middle line:84% I want to say this is a fundamental principle 00:49:44.090 --> 00:49:46.330 align:middle line:90% in combinatorial optimization. 00:49:46.330 --> 00:49:52.410 align:middle line:84% And if you actually get a linear program from somewhere, 00:49:52.410 --> 00:49:54.710 align:middle line:84% one of the first things you do is you look at the dual. 00:49:54.710 --> 00:49:55.610 align:middle line:90% You stare at it. 00:49:55.610 --> 00:50:01.930 align:middle line:84% And you say, what does this mean in terms of the actual problem 00:50:01.930 --> 00:50:03.490 align:middle line:90% you're studying? 00:50:03.490 --> 00:50:07.890 align:middle line:84% And often, it will give you some insight into the problem. 00:50:07.890 --> 00:50:09.030 align:middle line:90% Sometimes, it won't. 00:50:09.030 --> 00:50:18.210 align:middle line:84% Sometimes, it just looks like a bunch of variables 00:50:18.210 --> 00:50:19.870 align:middle line:84% that you can't understand at all. 00:50:19.870 --> 00:50:37.180 align:middle line:90% 00:50:37.180 --> 00:50:38.880 align:middle line:90% I think we can erase this now. 00:50:38.880 --> 00:50:59.340 align:middle line:90% 00:50:59.340 --> 00:51:18.840 align:middle line:84% Suppose you have a maximum flow problem with integer capacities. 00:51:18.840 --> 00:51:29.860 align:middle line:90% 00:51:29.860 --> 00:51:33.350 align:middle line:90% Then the solution is an integer. 00:51:33.350 --> 00:51:43.060 align:middle line:90% 00:51:43.060 --> 00:51:53.810 align:middle line:84% Proof-- the dual solution is a cut. 00:51:53.810 --> 00:51:56.380 align:middle line:90% 00:51:56.380 --> 00:52:09.980 align:middle line:84% So it's a sum of integers because the capacity of a cut 00:52:09.980 --> 00:52:11.760 align:middle line:84% is the sum of the capacities of the edges. 00:52:11.760 --> 00:52:14.540 align:middle line:84% And the capacities of the edges are integers. 00:52:14.540 --> 00:52:34.030 align:middle line:84% So we can ask the question, does the primal solution also 00:52:34.030 --> 00:52:37.040 align:middle line:90% have, I guess, integer flows? 00:52:37.040 --> 00:52:44.950 align:middle line:90% 00:52:44.950 --> 00:52:46.790 align:middle line:90% Because maybe it doesn't. 00:52:46.790 --> 00:52:51.830 align:middle line:84% I mean, there are certainly solutions where it doesn't. 00:52:51.830 --> 00:52:59.990 align:middle line:84% I mean, you can imagine coming up 00:52:59.990 --> 00:53:07.470 align:middle line:84% with a solution where you have, I guess, 6.5 units of flow 00:53:07.470 --> 00:53:12.710 align:middle line:84% flowing over this edge and 1.5 units flowing over that edge 00:53:12.710 --> 00:53:15.330 align:middle line:90% and adding up to 9. 00:53:15.330 --> 00:53:17.730 align:middle line:84% So I guess I can't do arithmetic. 00:53:17.730 --> 00:53:21.830 align:middle line:90% 00:53:21.830 --> 00:53:26.590 align:middle line:84% But you also know that all the edges, all the flows going 00:53:26.590 --> 00:53:30.030 align:middle line:84% across a cut have to be equal to the value of that cut-- 00:53:30.030 --> 00:53:32.040 align:middle line:84% the edges and the capacities in that cut. 00:53:32.040 --> 00:53:34.510 align:middle line:84% So you know some edges have to be integers. 00:53:34.510 --> 00:53:36.870 align:middle line:84% So the question is, do all the edges-- 00:53:36.870 --> 00:53:40.990 align:middle line:84% can you find a solution with all the edges integers? 00:53:40.990 --> 00:53:46.030 align:middle line:90% And the answer is yes. 00:53:46.030 --> 00:53:49.670 align:middle line:84% And I don't know how to do it easily, 00:53:49.670 --> 00:53:51.790 align:middle line:90% working from the linear program. 00:53:51.790 --> 00:53:57.470 align:middle line:84% So there's another algorithm for integer flows-- 00:53:57.470 --> 00:54:02.780 align:middle line:90% algorithm for max flow-- 00:54:02.780 --> 00:54:05.710 align:middle line:90% 00:54:05.710 --> 00:54:08.610 align:middle line:84% which I'm going to call augmenting paths algorithm. 00:54:08.610 --> 00:54:17.150 align:middle line:90% 00:54:17.150 --> 00:54:33.235 align:middle line:84% And the idea-- find a path, push as much flow 00:54:33.235 --> 00:54:34.810 align:middle line:90% over this path as possible. 00:54:34.810 --> 00:54:47.480 align:middle line:90% 00:54:47.480 --> 00:54:53.720 align:middle line:84% So as much flow over this path as possible, 00:54:53.720 --> 00:54:58.240 align:middle line:84% you will take the minimum residual capacity 00:54:58.240 --> 00:55:02.080 align:middle line:90% in every edge and the path. 00:55:02.080 --> 00:55:10.440 align:middle line:84% And that means that the residual capacity is always an integer 00:55:10.440 --> 00:55:14.360 align:middle line:84% because the flows are integers and the capacities are integers. 00:55:14.360 --> 00:55:18.180 align:middle line:84% So that means that the whole thing will be an integer. 00:55:18.180 --> 00:55:19.300 align:middle line:90% So let's try this. 00:55:19.300 --> 00:55:23.760 align:middle line:90% 00:55:23.760 --> 00:55:25.400 align:middle line:90% And let's see. 00:55:25.400 --> 00:55:27.480 align:middle line:90% We have a simple network-- 00:55:27.480 --> 00:55:39.960 align:middle line:90% 00:55:39.960 --> 00:55:49.400 align:middle line:90% 2, 2, 3, 3. 00:55:49.400 --> 00:55:50.600 align:middle line:90% These are the capacities-- 00:55:50.600 --> 00:55:52.040 align:middle line:90% 2, 2. 00:55:52.040 --> 00:55:57.760 align:middle line:84% And let's put another edge from here to here with a capacity 3. 00:55:57.760 --> 00:56:00.030 align:middle line:84% And we're going to direct all the edges to the right. 00:56:00.030 --> 00:56:05.870 align:middle line:90% 00:56:05.870 --> 00:56:06.370 align:middle line:90% Good. 00:56:06.370 --> 00:56:10.200 align:middle line:90% 00:56:10.200 --> 00:56:13.160 align:middle line:90% So the first step-- 00:56:13.160 --> 00:56:19.900 align:middle line:84% let's take the best path to push flow from here to here. 00:56:19.900 --> 00:56:25.820 align:middle line:84% So let's take 3 along here, 3 along here, and 3 along here. 00:56:25.820 --> 00:56:29.664 align:middle line:90% 00:56:29.664 --> 00:56:35.810 align:middle line:84% Now, how much capacity remains in the graph? 00:56:35.810 --> 00:56:42.390 align:middle line:84% Well, there's 0 left in this edge, 0 left along this edge, 00:56:42.390 --> 00:56:44.730 align:middle line:90% and 0 left along this edge. 00:56:44.730 --> 00:56:50.370 align:middle line:84% So it looks like we have a cut between the left side 00:56:50.370 --> 00:56:54.410 align:middle line:90% and the right side with-- 00:56:54.410 --> 00:56:59.690 align:middle line:84% and you can't put any more flow across. 00:56:59.690 --> 00:57:01.530 align:middle line:84% But of course, you should know-- you 00:57:01.530 --> 00:57:05.410 align:middle line:84% should realize that you could easily put a flow of 4 00:57:05.410 --> 00:57:07.770 align:middle line:90% on the original graph. 00:57:07.770 --> 00:57:16.590 align:middle line:84% So you have to be clever to use this algorithm. 00:57:16.590 --> 00:57:22.650 align:middle line:90% 00:57:22.650 --> 00:57:28.280 align:middle line:84% So how can we push more flow across this graph? 00:57:28.280 --> 00:57:31.490 align:middle line:90% 00:57:31.490 --> 00:57:33.960 align:middle line:84% I want to claim that we could use this path. 00:57:33.960 --> 00:57:41.610 align:middle line:90% 00:57:41.610 --> 00:57:47.210 align:middle line:84% So this path, we push a 2 across here, push 2 across here. 00:57:47.210 --> 00:57:49.890 align:middle line:84% And now, we're going the wrong way in the edge. 00:57:49.890 --> 00:57:51.970 align:middle line:84% But we can do that because there's already 00:57:51.970 --> 00:57:54.570 align:middle line:90% a flow of 3 across this edge. 00:57:54.570 --> 00:58:00.930 align:middle line:84% And so to get a flow of 2 this way, we just reduce this 3 to 1. 00:58:00.930 --> 00:58:06.730 align:middle line:84% And then we can push another flow of 2 across here 00:58:06.730 --> 00:58:10.130 align:middle line:84% and another flow of 2 across here. 00:58:10.130 --> 00:58:15.530 align:middle line:84% So you can push flow across edges backwards, as long 00:58:15.530 --> 00:58:21.870 align:middle line:90% as there is a flow left-- 00:58:21.870 --> 00:58:24.560 align:middle line:90% 00:58:24.560 --> 00:58:29.060 align:middle line:84% as long as there's a flow already along an edge. 00:58:29.060 --> 00:58:34.510 align:middle line:84% So I want to claim define residual network. 00:58:34.510 --> 00:58:42.220 align:middle line:90% 00:58:42.220 --> 00:59:01.020 align:middle line:84% If an edge has flow x sub ij less than u of ij, 00:59:01.020 --> 00:59:10.380 align:middle line:84% the new capacity is u of ij minus x sub ij. 00:59:10.380 --> 00:59:15.580 align:middle line:90% 00:59:15.580 --> 00:59:26.620 align:middle line:84% If an edge has flow x sub ij greater than 0, 00:59:26.620 --> 00:59:36.860 align:middle line:84% there's an edge in residual network with-- 00:59:36.860 --> 00:59:42.552 align:middle line:90% 00:59:42.552 --> 00:59:57.060 align:middle line:84% so let's call this edge eij e sub ji with capacity x sub ij. 00:59:57.060 --> 01:00:00.700 align:middle line:90% 01:00:00.700 --> 01:00:07.860 align:middle line:84% So what is the residual network for this graph when we 01:00:07.860 --> 01:00:12.240 align:middle line:90% put the orange flow across it? 01:00:12.240 --> 01:00:21.350 align:middle line:90% 01:00:21.350 --> 01:00:24.710 align:middle line:90% Well, this had capacity 3. 01:00:24.710 --> 01:00:31.230 align:middle line:84% So now, that edge has capacity 3 in the opposite direction. 01:00:31.230 --> 01:00:32.290 align:middle line:90% This edge has 0. 01:00:32.290 --> 01:00:36.750 align:middle line:90% So that has capacity 0, 2, 2. 01:00:36.750 --> 01:00:41.210 align:middle line:84% This edge also has capacity 3 and flow 3 in the forward. 01:00:41.210 --> 01:00:47.150 align:middle line:84% So now, we get capacity 3 going in the backwards direction. 01:00:47.150 --> 01:00:51.150 align:middle line:84% And this one, we have capacity 3 in the backwards direction. 01:00:51.150 --> 01:00:54.553 align:middle line:84% And this has capacity 3 in the forwards and the capacity 3 01:00:54.553 --> 01:00:55.220 align:middle line:90% in the forwards. 01:00:55.220 --> 01:00:59.640 align:middle line:90% 01:00:59.640 --> 01:01:01.500 align:middle line:90% So this is a residual network. 01:01:01.500 --> 01:01:12.198 align:middle line:90% 01:01:12.198 --> 01:01:14.670 align:middle line:84% And now, you can see that there's a flow of 2 that 01:01:14.670 --> 01:01:19.710 align:middle line:90% goes bum, bum, bum, bum, bum. 01:01:19.710 --> 01:01:23.270 align:middle line:90% So we push this flow of 2 here. 01:01:23.270 --> 01:01:26.770 align:middle line:84% And we get-- here, I'm going to redraw this. 01:01:26.770 --> 01:01:33.270 align:middle line:90% 01:01:33.270 --> 01:01:46.150 align:middle line:84% We get 2, 2, and 1 going this way, and 3 going this way, 01:01:46.150 --> 01:01:50.180 align:middle line:90% and 2, 2, and 3. 01:01:50.180 --> 01:01:55.550 align:middle line:90% 01:01:55.550 --> 01:02:01.670 align:middle line:84% So now, we have a flow from the source to the sink with value 5. 01:02:01.670 --> 01:02:03.790 align:middle line:84% And that's the best you can do because you can see 01:02:03.790 --> 01:02:06.630 align:middle line:90% that there's a cut of value 5. 01:02:06.630 --> 01:02:10.270 align:middle line:84% With just taking t and everything else, 01:02:10.270 --> 01:02:13.890 align:middle line:84% you can get at most value 5 across this cut. 01:02:13.890 --> 01:02:33.200 align:middle line:90% 01:02:33.200 --> 01:03:04.360 align:middle line:84% So claim-- if there is no flow in residual network, 01:03:04.360 --> 01:03:11.750 align:middle line:84% then there is a cut in original graph. 01:03:11.750 --> 01:03:18.200 align:middle line:90% 01:03:18.200 --> 01:03:23.160 align:middle line:90% So if you have a value-- 01:03:23.160 --> 01:03:26.400 align:middle line:84% so I guess I should have started this 01:03:26.400 --> 01:03:41.440 align:middle line:84% with suppose you have a flow of value z. 01:03:41.440 --> 01:03:45.840 align:middle line:84% If there is no augmenting path in the residual network, 01:03:45.840 --> 01:04:11.320 align:middle line:84% then there's a cut in the original graph of value v. 01:04:11.320 --> 01:04:35.210 align:middle line:84% If there's no flow in the residual network, let s be-- 01:04:35.210 --> 01:04:48.590 align:middle line:84% maybe c be set of nodes you can reach from s. 01:04:48.590 --> 01:04:53.970 align:middle line:90% 01:04:53.970 --> 01:04:58.530 align:middle line:90% So let's say we have this thing. 01:04:58.530 --> 01:05:03.350 align:middle line:84% So what nodes can you reach from s? 01:05:03.350 --> 01:05:05.610 align:middle line:84% Actually, you can't reach any nodes from s 01:05:05.610 --> 01:05:08.910 align:middle line:90% because this is at capacity. 01:05:08.910 --> 01:05:10.410 align:middle line:90% This is at capacity. 01:05:10.410 --> 01:05:24.970 align:middle line:84% So we could have made this capacity 4 and this capacity 4. 01:05:24.970 --> 01:05:29.290 align:middle line:84% And this is still a good flow in the residual network. 01:05:29.290 --> 01:05:32.520 align:middle line:84% And now, what nodes can we reach? 01:05:32.520 --> 01:05:38.290 align:middle line:90% 01:05:38.290 --> 01:05:44.560 align:middle line:90% Well, we can reach this node. 01:05:44.560 --> 01:05:47.850 align:middle line:90% 01:05:47.850 --> 01:05:52.490 align:middle line:84% We cannot reach this node because this is at capacity. 01:05:52.490 --> 01:05:54.400 align:middle line:90% We can reach this node. 01:05:54.400 --> 01:05:58.110 align:middle line:90% 01:05:58.110 --> 01:05:58.610 align:middle line:90% Oops. 01:05:58.610 --> 01:06:05.010 align:middle line:90% 01:06:05.010 --> 01:06:15.900 align:middle line:84% And we actually can reach this node because it's at capacity. 01:06:15.900 --> 01:06:17.300 align:middle line:90% But we cannot reach this node. 01:06:17.300 --> 01:06:19.020 align:middle line:90% And we cannot reach this node. 01:06:19.020 --> 01:06:21.580 align:middle line:90% So we have a cut here. 01:06:21.580 --> 01:06:26.580 align:middle line:84% And that corresponds to this cut here. 01:06:26.580 --> 01:06:42.740 align:middle line:84% So every edge from c to c-bar, the flow is equal to u of ij 01:06:42.740 --> 01:06:45.420 align:middle line:84% because, otherwise, you could push flow 01:06:45.420 --> 01:06:51.620 align:middle line:84% from c to c-bar and every edge from c-bar to 01:06:51.620 --> 01:06:55.900 align:middle line:90% c x sub ij equals 0. 01:06:55.900 --> 01:06:58.820 align:middle line:84% Because otherwise, you could push extra flow 01:06:58.820 --> 01:07:02.740 align:middle line:90% backwards along this edge. 01:07:02.740 --> 01:07:19.020 align:middle line:84% And that says the cut equals some c c-bar u of ij 01:07:19.020 --> 01:07:19.770 align:middle line:90% equals flow. 01:07:19.770 --> 01:07:22.700 align:middle line:90% 01:07:22.700 --> 01:07:28.300 align:middle line:84% So if you cannot find any augmenting paths, 01:07:28.300 --> 01:07:35.200 align:middle line:84% if you already have a partial flow, then you're done. 01:07:35.200 --> 01:07:37.290 align:middle line:90% You cannot get any further. 01:07:37.290 --> 01:07:42.700 align:middle line:90% 01:07:42.700 --> 01:07:45.220 align:middle line:90% Yeah, I mean, then you have-- 01:07:45.220 --> 01:07:48.640 align:middle line:84% that is the value of the maximum flow 01:07:48.640 --> 01:07:53.500 align:middle line:84% because you have found a minimum cut that is equal to your flow. 01:07:53.500 --> 01:07:57.980 align:middle line:84% So this says that you can use the augmenting paths 01:07:57.980 --> 01:08:03.740 align:middle line:90% algorithm to find max flows. 01:08:03.740 --> 01:08:08.310 align:middle line:84% And it will give you an integer flow 01:08:08.310 --> 01:08:11.430 align:middle line:84% if all the capacities are integer, which 01:08:11.430 --> 01:08:16.069 align:middle line:90% is what we wanted to show. 01:08:16.069 --> 01:08:20.590 align:middle line:84% And actually, there's an application 01:08:20.590 --> 01:08:24.950 align:middle line:84% of this, which maybe I'm going to get through. 01:08:24.950 --> 01:08:26.219 align:middle line:90% And this is König's theorem. 01:08:26.219 --> 01:08:29.649 align:middle line:90% 01:08:29.649 --> 01:08:30.149 align:middle line:90% Oops. 01:08:30.149 --> 01:08:48.990 align:middle line:90% 01:08:48.990 --> 01:08:49.830 align:middle line:90% All right. 01:08:49.830 --> 01:08:57.670 align:middle line:84% In a bipartite graph, the cardinality 01:08:57.670 --> 01:09:11.050 align:middle line:84% of the maximum matching equals cardinality. 01:09:11.050 --> 01:09:13.790 align:middle line:90% 01:09:13.790 --> 01:09:15.470 align:middle line:90% of minimum vertex cover. 01:09:15.470 --> 01:09:33.183 align:middle line:90% 01:09:33.183 --> 01:09:34.100 align:middle line:90% Let's look at a graph. 01:09:34.100 --> 01:09:44.430 align:middle line:90% 01:09:44.430 --> 01:09:46.830 align:middle line:90% So this is a bipartite graph. 01:09:46.830 --> 01:09:53.830 align:middle line:84% And what would be a maximum matching? 01:09:53.830 --> 01:10:02.030 align:middle line:84% Well, I mean, we clearly should put this vertex 01:10:02.030 --> 01:10:05.800 align:middle line:84% in the matching, this vertex in the matching. 01:10:05.800 --> 01:10:07.860 align:middle line:84% I guess we need another vertex here. 01:10:07.860 --> 01:10:12.920 align:middle line:90% 01:10:12.920 --> 01:10:16.243 align:middle line:84% And we can put this vertex in the matching and this vertex 01:10:16.243 --> 01:10:16.910 align:middle line:90% in the matching. 01:10:16.910 --> 01:10:20.200 align:middle line:90% 01:10:20.200 --> 01:10:28.120 align:middle line:84% So the maximum matching equals 4. 01:10:28.120 --> 01:10:31.480 align:middle line:84% So the maximum matching is just the maximum set 01:10:31.480 --> 01:10:34.030 align:middle line:84% of edges such that no two share an point. 01:10:34.030 --> 01:10:37.120 align:middle line:90% 01:10:37.120 --> 01:10:52.070 align:middle line:84% Minimum vertex cover-- set of vertices that touch every edge. 01:10:52.070 --> 01:10:57.360 align:middle line:90% 01:10:57.360 --> 01:11:07.240 align:middle line:84% And clearly, vertex cover is greater than 01:11:07.240 --> 01:11:14.040 align:middle line:84% or equal to the matching because there 01:11:14.040 --> 01:11:18.102 align:middle line:84% must be a vertex that touches one endpoint of each 01:11:18.102 --> 01:11:19.310 align:middle line:90% of the edges in the matching. 01:11:19.310 --> 01:11:24.580 align:middle line:90% 01:11:24.580 --> 01:11:28.560 align:middle line:84% I mean, you're not going to get less than that. 01:11:28.560 --> 01:11:31.560 align:middle line:84% But Konig's theorem is that these two quantities 01:11:31.560 --> 01:11:33.940 align:middle line:90% are actually equal. 01:11:33.940 --> 01:11:51.580 align:middle line:84% And proof-- maximum flow, mincut. 01:11:51.580 --> 01:12:02.210 align:middle line:90% 01:12:02.210 --> 01:12:06.890 align:middle line:84% We want a flow whose value is equal to the matching. 01:12:06.890 --> 01:12:08.070 align:middle line:90% So what are we going to do? 01:12:08.070 --> 01:12:16.770 align:middle line:84% Well, we're going to take s on the left 01:12:16.770 --> 01:12:20.170 align:middle line:84% and put an put an edge of weight 1 01:12:20.170 --> 01:12:23.690 align:middle line:90% to every vertex in the matching. 01:12:23.690 --> 01:12:28.450 align:middle line:90% And now, we're going to-- 01:12:28.450 --> 01:12:32.450 align:middle line:84% and let's do the same thing on the right. 01:12:32.450 --> 01:12:39.450 align:middle line:84% We'll put an edge to the sink from every point 01:12:39.450 --> 01:12:41.810 align:middle line:84% on the right side of the bipartite graph. 01:12:41.810 --> 01:12:46.690 align:middle line:84% And these capacities are going to be 1. 01:12:46.690 --> 01:12:52.150 align:middle line:90% And now, let's see. 01:12:52.150 --> 01:13:16.924 align:middle line:90% 01:13:16.924 --> 01:13:26.200 align:middle line:84% There is a max flow with integer values. 01:13:26.200 --> 01:13:29.860 align:middle line:90% 01:13:29.860 --> 01:13:34.060 align:middle line:90% And what that means is that-- 01:13:34.060 --> 01:13:39.100 align:middle line:84% oh, and we can let the capacities of these middle edges 01:13:39.100 --> 01:13:40.370 align:middle line:90% be infinity if we want. 01:13:40.370 --> 01:13:48.660 align:middle line:90% 01:13:48.660 --> 01:13:53.300 align:middle line:84% Well, what that means is that there 01:13:53.300 --> 01:13:58.780 align:middle line:84% are going to be some number of edges in the middle which 01:13:58.780 --> 01:14:02.480 align:middle line:90% have weight 1 here. 01:14:02.480 --> 01:14:05.380 align:middle line:84% So I want to claim that those are clearly a maximum matching 01:14:05.380 --> 01:14:16.460 align:middle line:84% because, on the right-hand side, if there's 01:14:16.460 --> 01:14:19.220 align:middle line:84% 1 unit of flow from here to here, 01:14:19.220 --> 01:14:24.900 align:middle line:84% then no other edge could use this vertex. 01:14:24.900 --> 01:14:28.020 align:middle line:84% So if you have a flow from here to here, 01:14:28.020 --> 01:14:33.000 align:middle line:84% then this edge is in the matching. 01:14:33.000 --> 01:14:34.920 align:middle line:84% And neither of these edges is in the matching 01:14:34.920 --> 01:14:36.620 align:middle line:90% and similarly on the left. 01:14:36.620 --> 01:14:42.860 align:middle line:84% So the maximum flow gives you a maximum matching. 01:14:42.860 --> 01:14:47.660 align:middle line:90% Dual is a cut. 01:14:47.660 --> 01:14:50.900 align:middle line:84% And that means that there's a cut whose value 01:14:50.900 --> 01:14:53.220 align:middle line:90% is equal to the maximum flow. 01:14:53.220 --> 01:14:57.940 align:middle line:84% And there are going to be some edges on the right, which 01:14:57.940 --> 01:15:03.940 align:middle line:84% are cut, and some edges on the left, which are in the cut. 01:15:03.940 --> 01:15:16.700 align:middle line:84% And I don't think I did that right. 01:15:16.700 --> 01:15:26.140 align:middle line:84% But if the edge is in the cut, then this vertex 01:15:26.140 --> 01:15:28.220 align:middle line:90% is in the vertex cover. 01:15:28.220 --> 01:15:33.340 align:middle line:84% And now, we know that there aren't any other paths from s 01:15:33.340 --> 01:15:34.220 align:middle line:90% to t. 01:15:34.220 --> 01:15:39.940 align:middle line:84% And that means that we know that every vertex, every edge 01:15:39.940 --> 01:15:41.560 align:middle line:84% is covered by at least one vertex 01:15:41.560 --> 01:15:44.500 align:middle line:84% because, otherwise, there would be a path from s to t 01:15:44.500 --> 01:15:45.680 align:middle line:90% going through that edge. 01:15:45.680 --> 01:15:59.830 align:middle line:90% 01:15:59.830 --> 01:16:00.730 align:middle line:90% Let me see. 01:16:00.730 --> 01:16:17.790 align:middle line:90% 01:16:17.790 --> 01:16:21.390 align:middle line:84% OK, I did not prepare this quite properly enough. 01:16:21.390 --> 01:16:24.790 align:middle line:90% So this is an edge in the cut. 01:16:24.790 --> 01:16:26.850 align:middle line:84% This, I believe, has to be an edge in the cut. 01:16:26.850 --> 01:16:39.635 align:middle line:90% 01:16:39.635 --> 01:16:42.130 align:middle line:84% And you can just say this is an edge in the cut. 01:16:42.130 --> 01:16:44.950 align:middle line:90% 01:16:44.950 --> 01:16:46.600 align:middle line:90% There should be one more thing. 01:16:46.600 --> 01:16:52.040 align:middle line:90% 01:16:52.040 --> 01:16:56.740 align:middle line:84% And that means so there are three edges in this cut. 01:16:56.740 --> 01:17:00.200 align:middle line:84% And I think that cut actually cuts all the paths 01:17:00.200 --> 01:17:02.920 align:middle line:90% from left to right. 01:17:02.920 --> 01:17:09.200 align:middle line:84% So we should have a maximum matching of only 3. 01:17:09.200 --> 01:17:13.560 align:middle line:90% And well, it's easy-- 01:17:13.560 --> 01:17:16.780 align:middle line:90% 1, 2, 3. 01:17:16.780 --> 01:17:20.480 align:middle line:90% 01:17:20.480 --> 01:17:23.170 align:middle line:84% And you can't get any more things in the maximum matching 01:17:23.170 --> 01:17:23.670 align:middle line:90% because-- 01:17:23.670 --> 01:17:27.680 align:middle line:90% 01:17:27.680 --> 01:17:30.200 align:middle line:90% because there's a cut. 01:17:30.200 --> 01:17:33.560 align:middle line:90% So this is Konig's theorem. 01:17:33.560 --> 01:17:36.060 align:middle line:84% And I seem to be completely out of time. 01:17:36.060 --> 01:17:40.720 align:middle line:84% So I will see you next time on Thursday 01:17:40.720 --> 01:17:44.910 align:middle line:84% when we start a new topic on information theory. 01:17:44.910 --> 01:17:51.000 align:middle line:90%