WEBVTT 00:00:00.080 --> 00:00:02.500 The following content is provided under a Creative 00:00:02.500 --> 00:00:04.019 Commons license. 00:00:04.019 --> 00:00:06.360 Your support will help MIT OpenCourseWare 00:00:06.360 --> 00:00:10.730 continue to offer high-quality educational resources for free. 00:00:10.730 --> 00:00:13.350 To make a donation or view additional materials 00:00:13.350 --> 00:00:17.217 from hundreds of MIT courses, visit MIT OpenCourseWare 00:00:17.217 --> 00:00:17.842 at ocw.mit.edu. 00:00:21.257 --> 00:00:22.090 LING REN: All right. 00:00:22.090 --> 00:00:24.030 Let's get started. 00:00:24.030 --> 00:00:28.250 Today's topic would be distributed algorithms. 00:00:28.250 --> 00:00:31.900 We will look at two new algorithms. 00:00:31.900 --> 00:00:35.210 But they are similar to what you have seen in the lectures. 00:00:35.210 --> 00:00:38.930 So it will also be a review of some concepts in lectures. 00:00:52.250 --> 00:00:56.130 So our first example would be, again, 00:00:56.130 --> 00:00:59.390 [INAUDIBLE], the simplest example. 00:01:05.670 --> 00:01:11.730 But this time, the network and topology would be a ring. 00:01:21.810 --> 00:01:25.700 So in the lectures, the example we see 00:01:25.700 --> 00:01:28.570 is a click, meaning they are fully connected. 00:01:28.570 --> 00:01:29.070 Right. 00:01:29.070 --> 00:01:32.340 Every node can talk to every other node. 00:01:32.340 --> 00:01:36.240 There, if you remember, our solution for everyone 00:01:36.240 --> 00:01:39.100 to generate a UID or a random number. 00:01:39.100 --> 00:01:45.260 And if you are the maximum, then you output, you're the leader 00:01:45.260 --> 00:01:48.610 You can do that because, yeah, you are connected to everybody. 00:01:48.610 --> 00:01:52.000 So you immediately know what random number everybody 00:01:52.000 --> 00:01:52.640 generates. 00:01:52.640 --> 00:01:55.870 And you can compare whether you are the largest. 00:01:55.870 --> 00:02:00.830 Now, the idea is the same if we have a ring. 00:02:00.830 --> 00:02:04.940 So you want everyone to generate a ID or random number. 00:02:04.940 --> 00:02:07.550 I'll just say ID from now on. 00:02:07.550 --> 00:02:11.780 And you want to collect everyone else ID 00:02:11.780 --> 00:02:13.840 so that you know whether your ID is 00:02:13.840 --> 00:02:18.326 the largest among all of them. 00:02:18.326 --> 00:02:18.826 OK. 00:02:24.754 --> 00:02:28.212 AUDIENCE: [INAUDIBLE] 00:02:29.480 --> 00:02:30.188 LING REN: Pardon? 00:02:30.188 --> 00:02:33.152 AUDIENCE: [INAUDIBLE] 00:02:34.289 --> 00:02:34.830 LING REN: OK. 00:02:34.830 --> 00:02:39.270 The question is where did the comparison happen? 00:02:39.270 --> 00:02:42.200 So we first need some way to pass 00:02:42.200 --> 00:02:44.810 the numbers around such that everyone 00:02:44.810 --> 00:02:47.370 has everyone else's number. 00:02:47.370 --> 00:02:47.870 Right. 00:02:47.870 --> 00:02:49.460 You have the number of everyone, then 00:02:49.460 --> 00:02:52.791 you can compare whether the largest equals yours. 00:02:52.791 --> 00:02:53.290 OK. 00:02:53.290 --> 00:02:55.000 If the largest equals yours, then you 00:02:55.000 --> 00:02:56.041 know you are the largest. 00:02:56.041 --> 00:03:01.130 And you're going to output, I'm the leader. 00:03:01.130 --> 00:03:01.630 OK. 00:03:01.630 --> 00:03:05.300 So the difficulty is just how to pass the numbers 00:03:05.300 --> 00:03:08.220 around so that everyone sees everyone else's number. 00:03:11.120 --> 00:03:12.790 Any ideas? 00:03:12.790 --> 00:03:13.520 Simple solution. 00:03:39.755 --> 00:03:41.590 Well, there's not much you can do. 00:03:41.590 --> 00:03:43.990 You are just connect to your two neighbors, the left one 00:03:43.990 --> 00:03:44.740 and the right one. 00:03:54.800 --> 00:03:56.835 So how do you propagate the information? 00:04:06.134 --> 00:04:07.130 Go ahead. 00:04:07.130 --> 00:04:10.616 AUDIENCE: You can take the maximum of your ID 00:04:10.616 --> 00:04:14.579 and neighbor's IDs and kind of broadcast that. 00:04:14.579 --> 00:04:15.120 LING REN: OK. 00:04:15.120 --> 00:04:16.519 So what does broadcast mean here? 00:04:16.519 --> 00:04:17.935 AUDIENCE: Like tell your neighbors 00:04:17.935 --> 00:04:21.720 what the largest of your value [INAUDIBLE] 00:04:21.720 --> 00:04:23.230 LING REN: OK. 00:04:23.230 --> 00:04:25.170 And yes. 00:04:25.170 --> 00:04:32.650 Let me call them A, B, C, E, F. Then, yes, C can tell D 00:04:32.650 --> 00:04:35.167 or can tell B, but how does C tell E? 00:04:41.498 --> 00:04:47.267 AUDIENCE: It waited around to I guess propagate the maximum. 00:04:47.267 --> 00:04:47.850 LING REN: Yep. 00:04:47.850 --> 00:04:53.970 So it would be just, say, everyone talks 00:04:53.970 --> 00:04:55.660 to its right neighbor. 00:04:55.660 --> 00:04:58.930 And then B has A's IDs, C these B's ID, D has C's ID, 00:04:58.930 --> 00:05:02.360 and then continue and pass around next time. 00:05:05.390 --> 00:05:08.260 So just to make it perfect clear. 00:05:08.260 --> 00:05:15.280 Say they generate some random IDs that's 5, 10, 20. 00:05:15.280 --> 00:05:19.035 And then in the next round, A would send its ID to B, B 00:05:19.035 --> 00:05:22.470 would send its ID to C, C would send its ID to D. 00:05:22.470 --> 00:05:25.520 And in the next round, B would pass this information along 00:05:25.520 --> 00:05:28.930 to C. And C would pass the information to D. 00:05:28.930 --> 00:05:31.320 And just continue. 00:05:31.320 --> 00:05:36.240 And eventually everyone will have everyone else's ID. 00:05:36.240 --> 00:05:38.720 So how many rounds do we need? 00:05:51.790 --> 00:05:58.001 If they are in nodes in the system. 00:05:58.001 --> 00:06:00.292 AUDIENCE: I think probably in just one direction or two 00:06:00.292 --> 00:06:02.110 neighbors. 00:06:02.110 --> 00:06:05.190 LING REN: I think you can do it either way. 00:06:05.190 --> 00:06:09.914 If you propagate both ways, it probably 2x faster, yeah. 00:06:09.914 --> 00:06:11.390 AUDIENCE: [INAUDIBLE] 00:06:11.390 --> 00:06:12.870 LING REN: Yeah, correct. 00:06:12.870 --> 00:06:14.700 It's just O(n). 00:06:14.700 --> 00:06:16.810 But to keep it simple, let's say we just 00:06:16.810 --> 00:06:18.240 propagate in one direction. 00:06:18.240 --> 00:06:19.030 That's also fine. 00:06:19.030 --> 00:06:21.340 Still O(n). 00:06:21.340 --> 00:06:25.983 So how many messages are sent in total? 00:06:25.983 --> 00:06:28.388 AUDIENCE: n squared. 00:06:28.388 --> 00:06:29.487 LING REN: Pardon? 00:06:29.487 --> 00:06:30.320 AUDIENCE: n squared. 00:06:30.320 --> 00:06:31.399 LING REN: n squared. 00:06:31.399 --> 00:06:31.898 Yep. 00:06:34.590 --> 00:06:37.660 Is that obvious to everyone? 00:06:37.660 --> 00:06:38.160 OK. 00:06:38.160 --> 00:06:40.132 Give an explanation. 00:06:40.132 --> 00:06:41.715 AUDIENCE: Yeah, at each round everyone 00:06:41.715 --> 00:06:44.900 sends a message to its neighbors. 00:06:44.900 --> 00:06:45.900 LING REN: Yeah. 00:06:45.900 --> 00:06:48.030 Yeah. 00:06:48.030 --> 00:06:50.350 Every round, everyone sends a message. 00:06:50.350 --> 00:06:54.550 Or you can think of every message as propagated n times. 00:06:54.550 --> 00:06:58.700 And there are n messages in total. 00:06:58.700 --> 00:07:00.171 OK. 00:07:00.171 --> 00:07:00.670 Yeah. 00:07:00.670 --> 00:07:04.180 That's definitely a solution. 00:07:04.180 --> 00:07:07.000 Well, you can imagine, it's probably the naive solution. 00:07:07.000 --> 00:07:09.220 And can we do better than that? 00:07:14.566 --> 00:07:16.996 AUDIENCE: In this case, are we assuming they 00:07:16.996 --> 00:07:18.454 know that there are n of them? 00:07:21.380 --> 00:07:22.280 LING REN: OK. 00:07:22.280 --> 00:07:23.370 Good question. 00:07:23.370 --> 00:07:25.989 So if they know there are n of them, 00:07:25.989 --> 00:07:27.530 yep, then they know how to terminate. 00:07:27.530 --> 00:07:30.230 So your question is how to terminate, right? 00:07:30.230 --> 00:07:35.070 If they don't, eventually you will receive your own ID 00:07:35.070 --> 00:07:36.230 on the other side. 00:07:36.230 --> 00:07:38.850 If we keep sending left, eventually you 00:07:38.850 --> 00:07:40.600 will receive it on the left port. 00:07:40.600 --> 00:07:44.265 And, yeah, that's an indication of termination. 00:07:47.498 --> 00:07:47.998 Question? 00:07:47.998 --> 00:07:49.664 AUDIENCE: Are we assuming the trajectory 00:07:49.664 --> 00:07:51.165 that these are unique? 00:07:51.165 --> 00:07:51.790 LING REN: Yeah. 00:07:51.790 --> 00:07:55.480 We usually assume that, yeah. 00:07:55.480 --> 00:08:00.420 Either it's a unique user-- sorry, UID. 00:08:00.420 --> 00:08:02.260 What does U stand for by the way? 00:08:02.260 --> 00:08:06.630 Or if you generate random numbers in a large range, 00:08:06.630 --> 00:08:08.180 it's very unlikely that they collide. 00:08:20.560 --> 00:08:21.230 OK. 00:08:21.230 --> 00:08:22.480 So can we do better than that? 00:08:37.990 --> 00:08:40.700 You have some idea? 00:08:40.700 --> 00:08:47.560 AUDIENCE: Once something [INAUDIBLE] 00:08:47.560 --> 00:08:52.204 where you only pass it to the one that's bigger. 00:08:52.204 --> 00:08:52.870 LING REN: Hm-mm. 00:08:52.870 --> 00:08:53.200 OK. 00:08:53.200 --> 00:08:54.490 I think you have two ideas. 00:08:54.490 --> 00:08:58.080 And binary search, we'll see how that works later. 00:08:58.080 --> 00:09:01.250 So you said only forward something that's larger? 00:09:01.250 --> 00:09:02.580 0 00:09:02.580 --> 00:09:04.746 AUDIENCE: Yeah, like only forward things are larger. 00:09:08.014 --> 00:09:09.930 LING REN: OK, yeah, that's on the right track. 00:09:09.930 --> 00:09:13.730 So one obvious thing we can do is 00:09:13.730 --> 00:09:18.920 that so we actually don't care what all the IDs are, right. 00:09:18.920 --> 00:09:23.100 We only care whether a certain ID's the largest. 00:09:23.100 --> 00:09:29.060 So for example, in this case, when A send its ID 5 to B, 00:09:29.060 --> 00:09:32.210 B knows this 5 won't be the leader. 00:09:32.210 --> 00:09:32.710 All right. 00:09:32.710 --> 00:09:33.970 Because 5 is too small. 00:09:33.970 --> 00:09:36.220 It's even smaller than his ID. 00:09:36.220 --> 00:09:40.190 So we can choose to drop this message. 00:09:40.190 --> 00:09:44.720 There's no point in passing that message further. 00:09:44.720 --> 00:09:46.810 Same thing for this message. 00:09:46.810 --> 00:09:49.420 C knows that 10 is too small. 00:09:49.420 --> 00:09:53.980 And it doesn't have to pass it along. 00:09:53.980 --> 00:09:57.452 AUDIENCE: [INAUDIBLE] 00:09:58.940 --> 00:09:59.940 LING REN: Oh, the ID? 00:09:59.940 --> 00:10:03.460 It's just the integer each node chooses at random. 00:10:03.460 --> 00:10:06.210 See. 00:10:06.210 --> 00:10:10.280 Yeah, the only purpose of the ID is to break symmetry. 00:10:10.280 --> 00:10:11.720 So, yeah. 00:10:11.720 --> 00:10:14.230 Like we have seen the lecture, if they don't do this, 00:10:14.230 --> 00:10:17.284 if they don't have any unique identifier, 00:10:17.284 --> 00:10:18.950 they won't be able to select the leader. 00:10:21.570 --> 00:10:25.120 And when they have a unique number, 00:10:25.120 --> 00:10:28.100 then they can select, yeah, the largest one or smallest 00:10:28.100 --> 00:10:30.210 one in some way. 00:10:34.490 --> 00:10:34.990 OK. 00:10:34.990 --> 00:10:37.190 Does this optimization make sense? 00:10:37.190 --> 00:10:42.650 We can cut [INAUDIBLE] messages that 00:10:42.650 --> 00:10:47.365 have no chance of becoming the leader. 00:10:47.365 --> 00:10:49.031 AUDIENCE: What is the upper bound of it? 00:10:49.031 --> 00:10:51.310 Is it still n squared? 00:10:51.310 --> 00:10:52.060 LING REN: Correct. 00:10:52.060 --> 00:10:52.559 Yeah. 00:10:52.559 --> 00:10:57.300 That's a very good question and very good answer, by the way. 00:10:57.300 --> 00:11:00.660 But how effective is this? 00:11:00.660 --> 00:11:04.510 Well in average case, we may be able to drop some messages. 00:11:04.510 --> 00:11:08.190 But there pathological case where it actually 00:11:08.190 --> 00:11:09.270 doesn't help at all. 00:11:09.270 --> 00:11:12.530 Say this is the ID we choose. 00:11:12.530 --> 00:11:15.450 Then when there's 20, it send around. 00:11:15.450 --> 00:11:17.180 B cannot drop it. 00:11:17.180 --> 00:11:20.320 Right, because it may be the largest. 00:11:20.320 --> 00:11:23.240 And when this 10 is sent to C, C cannot drop it. 00:11:23.240 --> 00:11:26.130 And when 20 comes along, it also cannot drop it. 00:11:26.130 --> 00:11:30.420 So no node can drop any message. 00:11:30.420 --> 00:11:32.585 Its worst case is still n square. 00:11:51.770 --> 00:11:52.270 OK. 00:11:52.270 --> 00:11:58.320 So let's think about the binary search equivalent, yeah, 00:11:58.320 --> 00:12:00.240 or how to binary search in this case. 00:12:07.968 --> 00:12:11.130 AUDIENCE: So binary is [INAUDIBLE] what? 00:12:11.130 --> 00:12:12.380 LING REN: Yeah, good question. 00:12:12.380 --> 00:12:15.920 It's not very obvious binary searching to what. 00:12:18.800 --> 00:12:23.530 I Yeah, just some binary idea that 00:12:23.530 --> 00:12:26.510 will give us a n log end bound. 00:12:26.510 --> 00:12:29.640 So actually this is the better logarithm. 00:12:29.640 --> 00:12:31.982 But once you have a log n, you can probably 00:12:31.982 --> 00:12:32.773 get-- OK, go ahead. 00:12:32.773 --> 00:12:35.360 AUDIENCE: How about kind of merging, 00:12:35.360 --> 00:12:39.019 like instead of finding the [INAUDIBLE], 00:12:39.019 --> 00:12:43.355 finding like log maximum, then [INAUDIBLE]. 00:12:43.355 --> 00:12:43.980 LING REN: Yeah. 00:12:43.980 --> 00:12:46.448 That's on the right track, yeah. 00:12:46.448 --> 00:12:48.281 AUDIENCE: Like clusters of some [INAUDIBLE]. 00:12:50.615 --> 00:12:51.240 LING REN: Yeah. 00:12:51.240 --> 00:12:55.440 That's definitely the right idea. 00:12:55.440 --> 00:12:58.411 Let's detail it a little bit. 00:12:58.411 --> 00:12:59.410 How do you carry it out? 00:13:16.274 --> 00:13:17.266 Hm-mm. 00:13:17.266 --> 00:13:19.746 AUDIENCE: Divide the two parts. 00:13:24.210 --> 00:13:27.170 Like kind of the [INAUDIBLE]. 00:13:27.170 --> 00:13:29.440 LING REN: And, OK, so then it's let 00:13:29.440 --> 00:13:33.940 them select their leader respectively and compare 00:13:33.940 --> 00:13:35.710 which one is larger. 00:13:35.710 --> 00:13:37.590 It's interesting thought. 00:13:48.975 --> 00:13:50.955 AUDIENCE: First, for example, A, B 00:13:50.955 --> 00:13:53.430 finds the maximum between these two, 00:13:53.430 --> 00:13:57.885 the C, D finds the maximum E [INAUDIBLE] maximum. 00:13:57.885 --> 00:14:00.855 Kind of merging, considering them as one node. 00:14:02.877 --> 00:14:04.460 LING REN: I think the first difficulty 00:14:04.460 --> 00:14:06.870 I see is that if you cut it by half, 00:14:06.870 --> 00:14:09.157 it's no longer a ring, right. 00:14:13.149 --> 00:14:17.150 AUDIENCE: Also they can't cut themselves in half. 00:14:17.150 --> 00:14:17.830 LING REN: Yeah. 00:14:17.830 --> 00:14:18.190 Yeah. 00:14:18.190 --> 00:14:18.700 Correct. 00:14:18.700 --> 00:14:20.000 Yeah, yeah. 00:14:43.952 --> 00:14:48.532 Yeah, it's definitely not an easy problem, not 00:14:48.532 --> 00:14:49.240 a easy algorithm. 00:14:54.370 --> 00:15:01.330 And the idea is to, well, you had the right idea. 00:15:01.330 --> 00:15:07.380 That we want to-- what's the word? 00:15:07.380 --> 00:15:10.030 Let the weak candidates shut up early. 00:15:12.630 --> 00:15:17.790 So what I mean by that is say we have several round, so this B 00:15:17.790 --> 00:15:25.690 and A and C. We will that B only propagate to A and C first. 00:15:25.690 --> 00:15:30.600 If B is the local maximum among A, B, C, then 00:15:30.600 --> 00:15:32.390 B will try to talk further. 00:15:35.000 --> 00:15:36.430 Increase its range. 00:15:36.430 --> 00:15:40.140 If B is not the local maximum of A, B, C, then, yeah, 00:15:40.140 --> 00:15:41.600 it can be quiet. 00:15:41.600 --> 00:15:44.460 Doesn't need to send messages anymore. 00:15:44.460 --> 00:15:46.550 If B succeeds in the next round, then you 00:15:46.550 --> 00:15:51.320 go further increase its range to try to talk to more people. 00:15:55.670 --> 00:15:56.310 OK. 00:15:56.310 --> 00:15:58.010 So how does it work in detail? 00:16:00.910 --> 00:16:07.950 Well so in round I, we will let a node 00:16:07.950 --> 00:16:12.950 send this message up to 2 raised to I hops. 00:16:16.190 --> 00:16:20.485 In this case, is 1 and in next round is 2 and then 4. 00:16:24.090 --> 00:16:29.460 If at any point some mode like along this range 00:16:29.460 --> 00:16:32.940 decides that you are not the local maximum, 00:16:32.940 --> 00:16:37.790 then they will reply that, yeah, you no longer 00:16:37.790 --> 00:16:40.600 need to send messages anymore. 00:16:40.600 --> 00:16:45.360 If this message successfully reaches this endpoint 00:16:45.360 --> 00:16:48.980 that 2 raised to I hops, then this guy will respond. 00:16:48.980 --> 00:16:52.890 And this guy, if it still thinks you are the local maximum, 00:16:52.890 --> 00:16:57.520 then it will respond the message while saying, yeah, continue. 00:17:03.250 --> 00:17:03.750 OK. 00:17:03.750 --> 00:17:11.190 And if the sender receives the continue message on both sides, 00:17:11.190 --> 00:17:14.190 then they it continue into the next round. 00:17:14.190 --> 00:17:16.063 Otherwise, it will go inactive. 00:17:20.272 --> 00:17:21.230 Is the algorithm clear? 00:17:35.581 --> 00:17:37.664 AUDIENCE: If the problem reaches the next maximum, 00:17:37.664 --> 00:17:41.729 like next the node each has to send a message. 00:17:41.729 --> 00:17:42.770 LING REN: Say that again. 00:17:42.770 --> 00:17:47.180 AUDIENCE: Like after receiving like don't sending it, 00:17:47.180 --> 00:17:49.262 like do you have to choose one node, which 00:17:49.262 --> 00:17:50.542 has to send the message? 00:17:50.542 --> 00:17:51.250 LING REN: Oh, OK. 00:17:51.250 --> 00:17:54.110 In the first round, everyone send their messages. 00:17:54.110 --> 00:17:54.610 OK. 00:17:54.610 --> 00:17:57.820 Then some of them will go inactive 00:17:57.820 --> 00:17:59.930 because they learn they are not maximum. 00:17:59.930 --> 00:18:02.630 And the remaining, the surviving ones, 00:18:02.630 --> 00:18:05.350 will, yeah, continue sending messages. 00:18:05.350 --> 00:18:07.410 And then, yeah, half of them probably 00:18:07.410 --> 00:18:08.911 will die in the next round. 00:18:08.911 --> 00:18:10.785 And the surviving ones keep sending messages. 00:18:14.315 --> 00:18:14.815 Make sense? 00:18:21.038 --> 00:18:23.483 AUDIENCE: How do you return messages through I? 00:18:23.483 --> 00:18:26.906 Like if you [INAUDIBLE] its neighbor and not the node that 00:18:26.906 --> 00:18:28.276 do either. 00:18:28.276 --> 00:18:28.900 LING REN: Yeah. 00:18:28.900 --> 00:18:32.340 So we will send the message of this form, 00:18:32.340 --> 00:18:35.970 say, well, some message. 00:18:39.200 --> 00:18:43.970 And then we will send the hop and the direction, 00:18:43.970 --> 00:18:47.260 either left or right. 00:18:47.260 --> 00:18:50.160 And this hop will initially be set 00:18:50.160 --> 00:18:56.150 to 2 raised to I, the number of round. 00:18:56.150 --> 00:18:58.505 Then when this guy receives the message, 00:18:58.505 --> 00:19:05.420 it will increment the hop count and pass it along. 00:19:05.420 --> 00:19:08.170 And every node when forwarding the message 00:19:08.170 --> 00:19:12.220 will decrease the that h by 1. 00:19:12.220 --> 00:19:16.490 And finally when it reaches here, that number becomes 0. 00:19:16.490 --> 00:19:23.020 And when a node sees a message with 0 hop count, 00:19:23.020 --> 00:19:26.220 it's going to reverse it, send it in the other direction. 00:19:26.220 --> 00:19:36.220 And, again, set it back to 2 raised to I. 00:19:36.220 --> 00:19:41.040 This message I should say ID. 00:19:44.770 --> 00:19:48.780 And at certain point, a certain node 00:19:48.780 --> 00:19:50.920 may decide this ID's too small. 00:19:50.920 --> 00:19:52.300 It doesn't have a chance. 00:19:52.300 --> 00:19:56.710 Then the I can directly send it in the opposite direction, 00:19:56.710 --> 00:19:58.974 replying a message saying, yeah, you are too small. 00:20:02.610 --> 00:20:03.290 OK. 00:20:03.290 --> 00:20:09.200 So any more questions on the algorithm itself? 00:20:09.200 --> 00:20:11.090 If not, what's the next step? 00:20:16.425 --> 00:20:17.395 AUDIENCE: Time. 00:20:17.395 --> 00:20:19.030 LING REN: Yeah, time complexity. 00:20:19.030 --> 00:20:21.210 I already claimed it's n log n. 00:20:21.210 --> 00:20:21.800 Is it? 00:20:27.800 --> 00:20:30.194 This is round. 00:20:30.194 --> 00:20:30.860 This is message. 00:20:33.900 --> 00:20:35.270 This is also message complexity. 00:20:38.820 --> 00:20:39.320 OK. 00:20:39.320 --> 00:20:40.345 So why is it n log n? 00:21:15.190 --> 00:21:18.034 AUDIENCE: The log n [INAUDIBLE]. 00:21:18.034 --> 00:21:18.700 LING REN: Hm-mm. 00:21:18.700 --> 00:21:22.055 So we have a certain number of rounds. 00:21:22.055 --> 00:21:23.305 So how many rounds do we have? 00:21:28.530 --> 00:21:31.010 AUDIENCE: Two. 00:21:31.010 --> 00:21:32.000 LING REN: Yeah, maybe. 00:21:32.000 --> 00:21:33.936 But, yeah, I'll just say log n. 00:21:33.936 --> 00:21:34.436 OK. 00:21:37.280 --> 00:21:42.810 Because you are increasing your hop length. 00:21:42.810 --> 00:21:48.250 And we're going to compute like how many nodes are still 00:21:48.250 --> 00:21:55.490 active each around and how many messages are sent in the round. 00:21:55.490 --> 00:22:02.630 So, well, the number of nodes active 00:22:02.630 --> 00:22:07.651 will just be this number if we start from 0. 00:22:07.651 --> 00:22:08.150 Why? 00:22:08.150 --> 00:22:11.210 Because the first time everyone is active. 00:22:11.210 --> 00:22:16.210 In the next time, only 1/3 of them will be active. 00:22:16.210 --> 00:22:20.100 But we said we are conservative here at we put 1/2. 00:22:20.100 --> 00:22:21.080 Right. 00:22:21.080 --> 00:22:23.490 Next round is actually 1/5. 00:22:27.180 --> 00:22:29.350 If it's local maximum, it means like these two 00:22:29.350 --> 00:22:32.625 and those two will go inactive. 00:22:32.625 --> 00:22:39.200 But we put this as a upper bound. 00:22:39.200 --> 00:22:44.660 So this is the number of nodes that are still active. 00:22:44.660 --> 00:22:52.890 And they will send the message up to this many hops. 00:22:56.285 --> 00:22:58.710 OK. 00:22:58.710 --> 00:23:00.955 And there are two directions. 00:23:06.460 --> 00:23:08.400 And you send a message back. 00:23:08.400 --> 00:23:11.760 And then, yeah-- sorry, send the message forward 00:23:11.760 --> 00:23:15.175 and someone will reply. 00:23:15.175 --> 00:23:18.615 But in the end, this is n log n. 00:23:24.117 --> 00:23:28.500 Yeah, and I think I got 8n log n. 00:23:28.500 --> 00:23:32.664 So my recitation note says it's 4 log n, not entirely sure 00:23:32.664 --> 00:23:33.330 what's going on. 00:23:33.330 --> 00:23:33.830 Yeah. 00:23:33.830 --> 00:23:36.330 But you can double check whether this is correct 00:23:36.330 --> 00:23:38.170 or the recitation note is correct. 00:23:58.728 --> 00:24:01.190 AUDIENCE: What is I again? 00:24:01.190 --> 00:24:04.790 LING REN: I's the number of round in the I-th round. 00:24:04.790 --> 00:24:05.290 Yeah. 00:24:05.290 --> 00:24:07.230 This many nodes are still active. 00:24:07.230 --> 00:24:11.020 And each of them will send a message of this many hops. 00:24:28.090 --> 00:24:30.970 And I didn't mention whether the network 00:24:30.970 --> 00:24:32.680 is synchronous or asynchronous. 00:24:32.680 --> 00:24:35.180 And it turns out it doesn't care. 00:24:35.180 --> 00:24:38.680 Some of them can work for both synchronous and asynchronous. 00:24:38.680 --> 00:24:40.960 Apparently, it works for synchronous networks. 00:24:40.960 --> 00:24:45.710 If it's asynchronous, then what changes 00:24:45.710 --> 00:24:49.620 is that different nodes are in different rounds. 00:24:49.620 --> 00:24:52.650 A certain node may be far ahead than the others. 00:24:52.650 --> 00:24:53.590 But it's fine. 00:24:53.590 --> 00:24:59.660 Eventually, they will converge to the correct result. OK. 00:24:59.660 --> 00:25:01.505 So let's look at the second problem. 00:25:07.330 --> 00:25:10.070 Well, problem's defined even simpler. 00:25:10.070 --> 00:25:15.140 We just want to count how many nodes are out there. 00:25:15.140 --> 00:25:19.235 We want the algorithm to work both synchronously and 00:25:19.235 --> 00:25:19.860 asynchronously. 00:25:23.560 --> 00:25:26.260 By that I just mean we have a network. 00:25:33.130 --> 00:25:36.220 Well, say you have a lot of nodes after that. 00:25:36.220 --> 00:25:39.440 Just want to count how many nodes are there 00:25:39.440 --> 00:25:40.170 in this network. 00:25:45.440 --> 00:25:49.312 So I'll give you, say, one minutes to let's first come up 00:25:49.312 --> 00:25:50.270 with a high-level plan. 00:25:56.350 --> 00:25:58.543 Is the problem clear? 00:25:58.543 --> 00:26:02.471 AUDIENCE: In the worst case, [INAUDIBLE]. 00:26:02.471 --> 00:26:05.389 The worst case. 00:26:05.389 --> 00:26:05.930 LING REN: OK. 00:26:05.930 --> 00:26:06.888 I haven't defined that. 00:26:06.888 --> 00:26:11.460 Let's not worry about complexity now for now. 00:26:11.460 --> 00:26:14.380 The complexity will depend on number of nodes 00:26:14.380 --> 00:26:22.330 and the number of edges, E. Let's just 00:26:22.330 --> 00:26:23.610 get it functionally correct. 00:27:53.070 --> 00:27:53.630 OK. 00:27:53.630 --> 00:27:56.090 Anyone share a high-level strategy? 00:28:01.130 --> 00:28:01.820 Go ahead. 00:28:01.820 --> 00:28:05.257 AUDIENCE: So each node will start 00:28:05.257 --> 00:28:11.149 like IDs of the other devices [INAUDIBLE]. 00:28:11.149 --> 00:28:15.077 So basically like it's going to send propagates [INAUDIBLE] 00:28:15.077 --> 00:28:18.023 sets of each node. 00:28:23.424 --> 00:28:25.310 LING REN: On what edge? 00:28:25.310 --> 00:28:26.495 AUDIENCE: Furthest one. 00:28:26.495 --> 00:28:27.120 LING REN: Yeah. 00:28:27.120 --> 00:28:30.274 It wasn't on both edges? 00:28:30.274 --> 00:28:32.409 AUDIENCE: Itself first. 00:28:32.409 --> 00:28:32.950 LING REN: OK. 00:28:32.950 --> 00:28:38.030 So then maybe send the 1 here and send the 1 here there. 00:28:38.030 --> 00:28:41.870 Then this node will think it has a children, it as a child, 00:28:41.870 --> 00:28:43.060 right, which is this one. 00:28:43.060 --> 00:28:46.430 Let's just say this is the entire network. 00:28:46.430 --> 00:28:50.772 And what message does it send to this guy? 00:28:50.772 --> 00:28:51.772 AUDIENCE: Reinforced it. 00:28:57.450 --> 00:28:59.440 LING REN: Well, you probably need send two here 00:28:59.440 --> 00:29:04.480 because you have one and this is possibly its child, right. 00:29:04.480 --> 00:29:06.484 But then we're double counting this node. 00:29:06.484 --> 00:29:07.150 See the problem? 00:29:13.442 --> 00:29:16.589 AUDIENCE: Do nodes have their IDs, like unique IDs? 00:29:16.589 --> 00:29:17.380 LING REN: Oh, yeah. 00:29:17.380 --> 00:29:18.460 They have their IDs. 00:29:18.460 --> 00:29:20.680 AUDIENCE: We can't send the IDs instead of the IDs 00:29:20.680 --> 00:29:22.589 through the end node. 00:29:22.589 --> 00:29:23.130 LING REN: OK. 00:29:26.300 --> 00:29:29.103 We are going to send all the IDs. 00:29:29.103 --> 00:29:30.186 AUDIENCE: Yeah, neighbors. 00:29:43.497 --> 00:29:49.269 And after n steps, it's going to have the [INAUDIBLE]. 00:29:49.269 --> 00:29:49.810 LING REN: OK. 00:29:49.810 --> 00:29:51.550 Yeah, that's an interesting thought. 00:29:51.550 --> 00:29:56.260 And then you may still need the root. 00:29:56.260 --> 00:30:01.750 And that will have everyone's ID. 00:30:01.750 --> 00:30:04.710 And then see how many unique I's are there. 00:30:07.600 --> 00:30:08.100 OK. 00:30:08.100 --> 00:30:09.060 Interesting. 00:30:09.060 --> 00:30:12.455 Does this algorithm work? 00:30:12.455 --> 00:30:13.790 Yeah, I don't see any problem. 00:30:22.376 --> 00:30:23.810 OK. 00:30:23.810 --> 00:30:28.650 But let me still repeat what our algorithm 00:30:28.650 --> 00:30:34.900 is because it's closer to what's in the lecture. 00:30:34.900 --> 00:30:42.305 So we're going to find a spanning tree of this network. 00:30:44.960 --> 00:30:47.720 A spanning tree means, well, like I 00:30:47.720 --> 00:30:49.135 have to cut one of these edges. 00:30:52.840 --> 00:30:56.860 If I have a tree, then I can have every child report 00:30:56.860 --> 00:30:58.610 to its parent. 00:30:58.610 --> 00:31:05.130 Like how many offsprings, including myself do I have. 00:31:05.130 --> 00:31:11.510 And this node will sum up all its children 00:31:11.510 --> 00:31:12.670 and report to its parent. 00:31:18.370 --> 00:31:20.440 Does everyone get that? 00:31:20.440 --> 00:31:23.550 So first, we'll find a spanning tree. 00:31:29.466 --> 00:31:37.154 And second, we'll have child reports to parent. 00:31:50.790 --> 00:31:54.470 AUDIENCE: How can we find the spanning tree? 00:31:54.470 --> 00:31:55.470 LING REN: Good question. 00:31:55.470 --> 00:31:59.770 So in the lecture, we have seen algorithm 00:31:59.770 --> 00:32:12.105 that find BFS spanning tree for synchronous networks. 00:32:15.330 --> 00:32:15.830 OK. 00:32:15.830 --> 00:32:17.380 This is review of the lecture. 00:32:20.732 --> 00:32:21.440 How does it work? 00:32:35.290 --> 00:32:39.280 Each node will, say, we need to first choose a root. 00:32:39.280 --> 00:32:43.350 And our root will just send a message to its neighbor 00:32:43.350 --> 00:32:46.950 and, yeah, saying, you are my child, you are my child, 00:32:46.950 --> 00:32:49.000 you are my child. 00:32:49.000 --> 00:32:53.300 And then every node upon receiving this message 00:32:53.300 --> 00:33:00.724 from the parent will search among its neighbors. 00:33:00.724 --> 00:33:01.690 All right. 00:33:01.690 --> 00:33:04.260 So the neighbors that haven't got a parent 00:33:04.260 --> 00:33:08.515 will acknowledge this sender as the parent. 00:33:12.400 --> 00:33:12.900 OK. 00:33:12.900 --> 00:33:13.910 That's a little messy. 00:33:27.650 --> 00:33:33.380 So this node will search to the leaf node. 00:33:33.380 --> 00:33:36.250 But then it will also try to search for this guy. 00:33:36.250 --> 00:33:38.630 But this guy already had a parent, then, yeah, 00:33:38.630 --> 00:33:44.210 it will say I already got a parent and blah, blah, blah. 00:33:44.210 --> 00:33:47.330 This will give us a BFS spanning tree. 00:33:47.330 --> 00:33:48.080 What does it mean? 00:33:48.080 --> 00:33:50.075 It's a spanning tree found by BFS. 00:33:52.620 --> 00:33:54.680 Does it work for asynchronous network? 00:33:59.680 --> 00:34:00.395 Go ahead. 00:34:00.395 --> 00:34:01.172 AUDIENCE: This version doesn't. 00:34:01.172 --> 00:34:02.672 But there's a different version that 00:34:02.672 --> 00:34:05.110 contains an edge for relaxation technique. 00:34:05.110 --> 00:34:05.840 LING REN: Yeah. 00:34:05.840 --> 00:34:06.470 Correct. 00:34:06.470 --> 00:34:10.931 And so why does this version not work for a synchronous network? 00:34:10.931 --> 00:34:12.306 AUDIENCE: Because different nodes 00:34:12.306 --> 00:34:14.290 can be on a different round number 00:34:14.290 --> 00:34:18.270 so that a longer path could end up going to the node even. 00:34:18.270 --> 00:34:19.290 LING REN: Yeah, exactly. 00:34:19.290 --> 00:34:23.030 So let me give a concrete example. 00:34:23.030 --> 00:34:25.080 What does asynchronous network mean? 00:34:25.080 --> 00:34:28.580 Is that, well, messages travel at different speed. 00:34:28.580 --> 00:34:32.760 Say this link for some reason is temporarily down. 00:34:32.760 --> 00:34:37.030 And this node doesn't receive the message from the root. 00:34:37.030 --> 00:34:41.210 And then this message travels very fast. 00:34:41.210 --> 00:34:43.000 And this message also travels very fast. 00:34:43.000 --> 00:34:47.480 So this message may reach this node earlier. 00:34:47.480 --> 00:34:52.300 And then this node will think of it as a child of this node. 00:34:52.300 --> 00:34:52.850 OK. 00:34:52.850 --> 00:34:55.889 And then this message comes along, finally. 00:34:55.889 --> 00:34:58.490 But this node says, yeah, I already have a parent. 00:34:58.490 --> 00:35:00.870 And I'm going to reject you. 00:35:00.870 --> 00:35:03.070 AUDIENCE: So it still makes a spanning tree? 00:35:03.070 --> 00:35:04.220 LING REN: Yes. 00:35:04.220 --> 00:35:05.330 Good point. 00:35:05.330 --> 00:35:09.143 This is not a BFS spanning tree but it's still a spanning tree. 00:35:09.143 --> 00:35:09.870 All right. 00:35:09.870 --> 00:35:12.600 So in our problem, we're totally fine with it. 00:35:16.340 --> 00:35:16.840 Yeah. 00:35:16.840 --> 00:35:20.210 But you do have to know if you really want a BFS spanning 00:35:20.210 --> 00:35:22.580 tree in an asynchronous network, then 00:35:22.580 --> 00:35:25.002 you have to like record the distance and do the relaxation 00:35:25.002 --> 00:35:25.960 and so on and so forth. 00:35:29.229 --> 00:35:30.170 OK. 00:35:30.170 --> 00:35:33.740 So we will just use this algorithm, the BFS spanning 00:35:33.740 --> 00:35:37.139 tree algorithm, just run it asynchronously. 00:35:37.139 --> 00:35:38.680 It doesn't find the BFS spanning tree 00:35:38.680 --> 00:35:40.610 but it finds some spanning tree. 00:35:46.990 --> 00:35:47.490 OK. 00:35:47.490 --> 00:35:48.850 How does this algorithm work? 00:35:56.290 --> 00:35:58.030 We will have several variables. 00:35:58.030 --> 00:36:01.510 The first one is parent. 00:36:01.510 --> 00:36:04.540 We will initialize to this undefined single. 00:36:13.900 --> 00:36:14.900 OK. 00:36:14.900 --> 00:36:20.350 Then every node will pass around this search message. 00:36:20.350 --> 00:36:23.170 I'll use a slightly more shorthand notation 00:36:23.170 --> 00:36:24.185 than from the lecture. 00:36:32.746 --> 00:36:33.250 OK. 00:36:33.250 --> 00:36:38.920 Let's say the code I wrote is for a process u. 00:36:42.392 --> 00:36:44.880 OK. 00:36:44.880 --> 00:36:51.410 If we receive a message, a search message, from v 00:36:51.410 --> 00:37:01.135 to u, that means u-- sorry, v is trying to become a parent. 00:37:01.135 --> 00:37:01.635 OK. 00:37:05.130 --> 00:37:11.530 And if I do not have a parent yet, 00:37:11.530 --> 00:37:25.310 I should set the parent to v. OK. 00:37:25.310 --> 00:37:26.432 So what's the next step? 00:37:38.990 --> 00:37:41.120 Now, we got this search message from our parent. 00:37:41.120 --> 00:37:43.200 And we have to pass it along. 00:37:49.687 --> 00:37:52.431 AUDIENCE: Should we send it to the like 00:37:52.431 --> 00:37:53.690 add your child [INAUDIBLE]. 00:37:53.690 --> 00:37:54.830 LING REN: Oh, yeah. 00:37:54.830 --> 00:37:55.330 Great. 00:37:58.746 --> 00:38:03.310 Yeah, we first need to respond by saying, OK, 00:38:03.310 --> 00:38:07.300 I'll use a shorter notation. 00:38:07.300 --> 00:38:12.860 Send these a cue that that's the message that will 00:38:12.860 --> 00:38:14.640 be sent to v at some point. 00:38:17.350 --> 00:38:21.330 The message we send is parent 1. 00:38:21.330 --> 00:38:23.640 Parent 2. 00:38:23.640 --> 00:38:30.107 This is a response to v saying, yeah, you are my parent. 00:38:34.400 --> 00:38:34.990 OK. 00:38:34.990 --> 00:38:50.720 Then else, we also have to respond by saying parent 0. 00:38:50.720 --> 00:38:54.820 You are not my parent because I already got some other parent. 00:38:54.820 --> 00:38:55.320 OK. 00:38:55.320 --> 00:38:56.474 Missing a step here. 00:39:03.590 --> 00:39:05.255 If we receive a search message. 00:39:05.255 --> 00:39:06.500 Go ahead. 00:39:06.500 --> 00:39:08.935 AUDIENCE: We send messages to all the other nodes. 00:39:08.935 --> 00:39:09.560 LING REN: Yeah. 00:39:09.560 --> 00:39:17.170 We need to pass it to all my potential children. 00:39:17.170 --> 00:39:26.050 So I use this comma u, meaning the neighbors of u. 00:39:26.050 --> 00:39:27.870 And then I'll send them a message. 00:39:31.959 --> 00:39:32.500 What message? 00:39:35.410 --> 00:39:35.910 Search. 00:39:47.041 --> 00:39:47.540 OK. 00:39:59.465 --> 00:39:59.965 OK. 00:39:59.965 --> 00:40:06.000 So, well, naturally since we have a search message 00:40:06.000 --> 00:40:08.130 and we know how to deal with it, now we 00:40:08.130 --> 00:40:10.620 are sending this parent message with better deal with it. 00:40:10.620 --> 00:40:11.940 Right. 00:40:11.940 --> 00:40:15.940 So then the next chunk of code should 00:40:15.940 --> 00:40:25.030 be if we receive this parent message, I'll say parent b, 00:40:25.030 --> 00:40:28.540 meaning this message comes with either true or false. 00:40:32.590 --> 00:40:37.854 I received this message from some node w. 00:40:37.854 --> 00:40:38.700 OK. 00:40:38.700 --> 00:40:40.780 I'm still u here. 00:40:40.780 --> 00:40:43.910 Because I just send all the message to all the w's, 00:40:43.910 --> 00:40:46.765 to all my neighbors, and they should give me a response. 00:40:55.430 --> 00:40:55.930 OK. 00:40:55.930 --> 00:41:03.530 If b is 1 that means this particular w 00:41:03.530 --> 00:41:06.810 take me as its parent. 00:41:06.810 --> 00:41:09.360 Make sense? 00:41:09.360 --> 00:41:10.210 OK. 00:41:10.210 --> 00:41:14.280 So I'd better have a list of my children. 00:41:14.280 --> 00:41:16.500 I want to keep track of that. 00:41:16.500 --> 00:41:17.000 How? 00:41:20.320 --> 00:41:25.480 We will create a new variable for children. 00:41:25.480 --> 00:41:28.890 It's going to be initialized to what? 00:41:32.100 --> 00:41:33.190 Yeah, empty set. 00:41:37.730 --> 00:41:42.040 And now if this b is 1, then I'm going 00:41:42.040 --> 00:41:45.182 to put w in this children list. 00:41:49.910 --> 00:41:51.170 OK. 00:41:51.170 --> 00:42:01.700 I'm leaving some space here because our root 00:42:01.700 --> 00:42:03.020 should be slightly different. 00:42:03.020 --> 00:42:06.050 So every other node will send search messages 00:42:06.050 --> 00:42:07.954 when it receive a search message. 00:42:07.954 --> 00:42:09.120 We need someone to initiate. 00:42:12.260 --> 00:42:14.130 Make sense? 00:42:14.130 --> 00:42:20.800 So if u equals root, we say the root is v0. 00:42:26.460 --> 00:42:28.300 So two things should happen. 00:42:28.300 --> 00:42:34.260 First, we should set its parent to some special value. 00:42:34.260 --> 00:42:35.150 Just say root. 00:42:40.050 --> 00:42:44.527 And then I should copy this blob of code to here. 00:42:58.163 --> 00:42:59.160 OK. 00:42:59.160 --> 00:43:04.720 It's a little crowded but I hope it's still legible. 00:43:14.930 --> 00:43:15.430 OK. 00:43:15.430 --> 00:43:19.050 This is already almost the correct algorithm, 00:43:19.050 --> 00:43:22.020 except that we don't know how to terminate. 00:43:22.020 --> 00:43:24.720 If we wait long enough, then everyone 00:43:24.720 --> 00:43:29.110 will receive all the responses and everyone will know 00:43:29.110 --> 00:43:32.721 its parent and this child list. 00:43:32.721 --> 00:43:33.720 But how do we terminate? 00:43:38.322 --> 00:43:39.472 AUDIENCE: After n rounds. 00:43:39.472 --> 00:43:41.680 LING REN: Yeah, that's one way to do, after n rounds. 00:43:41.680 --> 00:43:44.710 But the whole point is we are trying to find what n is. 00:43:44.710 --> 00:43:46.420 We don't know how many nodes are there. 00:43:46.420 --> 00:43:50.920 AUDIENCE: If we receive reject for all, [INAUDIBLE]. 00:43:53.920 --> 00:43:55.551 LING REN: Say again. 00:43:55.551 --> 00:44:00.999 AUDIENCE: For all the search, if we receive a reject. 00:44:00.999 --> 00:44:01.790 LING REN: Oh, yeah. 00:44:01.790 --> 00:44:06.850 If you receive a response, either parent 0 or parent 1 00:44:06.850 --> 00:44:11.370 from all your neighbors, then your job is pretty much done. 00:44:11.370 --> 00:44:14.813 But the others have not. 00:44:14.813 --> 00:44:15.760 All right. 00:44:15.760 --> 00:44:17.700 So you send a message very fast. 00:44:17.700 --> 00:44:18.920 They responded to you. 00:44:18.920 --> 00:44:22.930 And they are still working very hard. 00:44:22.930 --> 00:44:23.630 OK. 00:44:23.630 --> 00:44:27.490 So then we need to use the technique from the lecture 00:44:27.490 --> 00:44:31.050 that's called converge cast. 00:44:31.050 --> 00:44:36.430 So everyone will send a done signal when it is done 00:44:36.430 --> 00:44:38.016 and all its children are also done. 00:44:41.404 --> 00:44:41.904 OK. 00:44:44.830 --> 00:44:50.150 To do that, I'm going to define a new variable called searched. 00:44:50.150 --> 00:44:53.816 Searched means someone has responded. 00:44:53.816 --> 00:44:54.750 OK. 00:44:54.750 --> 00:44:58.670 In this case, w has responded. 00:44:58.670 --> 00:45:01.070 I'll put it into the search the list, 00:45:01.070 --> 00:45:05.001 no matter whether it accepts me as parent or reject. 00:45:05.001 --> 00:45:05.500 OK. 00:45:05.500 --> 00:45:09.230 Then naturally I need to define this variable here. 00:45:24.699 --> 00:45:25.200 OK. 00:45:25.200 --> 00:45:38.310 As a last step, If we search the list equals my neighbor list, 00:45:38.310 --> 00:45:43.660 that means everyone has responded 00:45:43.660 --> 00:45:51.630 and all my children are done, then 00:45:51.630 --> 00:45:55.410 I need a new variable called done. 00:45:55.410 --> 00:46:03.388 That's another list tracking who has finished and who has not. 00:46:07.628 --> 00:46:08.128 OK. 00:46:08.128 --> 00:46:09.600 AUDIENCE: So what's the first one? 00:46:09.600 --> 00:46:10.530 LING REN: Say again. 00:46:10.530 --> 00:46:11.930 AUDIENCE: What was the first one? 00:46:11.930 --> 00:46:13.140 LING REN: This one? 00:46:13.140 --> 00:46:16.975 Searched means someone has responded 00:46:16.975 --> 00:46:18.961 to the search message. 00:46:18.961 --> 00:46:19.460 OK. 00:46:19.460 --> 00:46:25.170 Done means all its children are done. 00:46:25.170 --> 00:46:27.975 I haven't write how done is defined. 00:46:27.975 --> 00:46:29.130 Give me a minute. 00:46:29.130 --> 00:46:34.780 In this case, I'm going to send my parent a message. 00:46:34.780 --> 00:46:38.460 This is the step of converge cast. 00:46:38.460 --> 00:46:39.257 What do I send? 00:46:39.257 --> 00:46:40.590 I'm going to send them I'm done. 00:46:48.318 --> 00:46:49.780 OK. 00:46:49.780 --> 00:46:52.620 Then whenever we create a message, 00:46:52.620 --> 00:46:56.440 we should deal with that message. 00:46:56.440 --> 00:47:02.821 If we receive a message I'm done, what do we do? 00:47:02.821 --> 00:47:03.320 OK. 00:47:03.320 --> 00:47:11.480 From w, then we're going to mark that node as done. 00:47:26.030 --> 00:47:26.830 OK. 00:47:26.830 --> 00:47:28.470 There's this other point. 00:47:28.470 --> 00:47:32.450 So someone has to initiate the done signal. 00:47:32.450 --> 00:47:34.550 That's going to be our leaves, right. 00:47:34.550 --> 00:47:37.160 Because when this condition check, 00:47:37.160 --> 00:47:38.510 they don't have any children. 00:47:38.510 --> 00:47:40.130 Their children list is empty set. 00:47:40.130 --> 00:47:42.270 And their done list is also empty set. 00:47:42.270 --> 00:47:45.580 But they're going to send the I'm done signal first. 00:47:45.580 --> 00:47:47.740 And then in the median nodes, we'll 00:47:47.740 --> 00:47:52.281 send a done signal when all its children are done. 00:47:52.281 --> 00:47:52.780 All right. 00:47:52.780 --> 00:47:57.150 So this is the converged cast version. 00:47:57.150 --> 00:48:00.930 Only gives us a termination point 00:48:00.930 --> 00:48:04.380 of our spanning tree search. 00:48:04.380 --> 00:48:08.090 We haven't count the number of nodes in the network. 00:48:11.072 --> 00:48:15.330 But that's a small modification. 00:48:15.330 --> 00:48:19.700 We're just going to include that number in the I'm done signal. 00:48:19.700 --> 00:48:24.190 So then I need to define another variable called 00:48:24.190 --> 00:48:27.460 total that's initialized to 0. 00:48:27.460 --> 00:48:31.040 This variable will attract how many 00:48:31.040 --> 00:48:35.560 nodes do I have in my subtree, including 00:48:35.560 --> 00:48:36.720 me and all my children. 00:48:39.740 --> 00:48:43.040 Then when I send the I'm done signal, 00:48:43.040 --> 00:48:47.370 I'm going to send this-- sorry. 00:48:47.370 --> 00:48:50.700 That's not right way. 00:48:50.700 --> 00:48:52.790 When I'm sending the I'm done signal, 00:48:52.790 --> 00:48:57.980 I'm going to send my total number of offspring with it. 00:48:57.980 --> 00:49:03.330 And when I receive, one of my children 00:49:03.330 --> 00:49:05.110 reports that I do have t children. 00:49:08.310 --> 00:49:12.070 I need to increment my total by that amount. 00:49:21.601 --> 00:49:22.100 OK. 00:49:22.100 --> 00:49:24.240 I made a mistake again. 00:49:24.240 --> 00:49:28.060 Should be total plus 1. 00:49:28.060 --> 00:49:29.010 Right. 00:49:29.010 --> 00:49:30.930 Because I'm counting all my children and then 00:49:30.930 --> 00:49:32.030 I should include myself. 00:49:43.426 --> 00:49:44.800 So that's the complete algorithm. 00:49:49.370 --> 00:49:49.870 Yeah. 00:49:49.870 --> 00:49:53.970 One purpose is just to create a different angle to look 00:49:53.970 --> 00:49:55.070 at distributed algorithm. 00:49:55.070 --> 00:49:58.310 Usually just draw that network graphs. 00:49:58.310 --> 00:50:00.980 But sometimes it's helpful to think about how 00:50:00.980 --> 00:50:02.227 the code actually works. 00:50:05.703 --> 00:50:06.203 OK. 00:50:10.180 --> 00:50:12.170 That's all for today.