WEBVTT 00:00:00.050 --> 00:00:01.770 The following content is provided 00:00:01.770 --> 00:00:04.010 under a Creative Commons license. 00:00:04.010 --> 00:00:06.860 Your support will help MIT OpenCourseWare continue 00:00:06.860 --> 00:00:10.720 to offer high quality educational resources for free. 00:00:10.720 --> 00:00:13.330 To make a donation or view additional materials 00:00:13.330 --> 00:00:17.207 from hundreds of MIT courses, visit MIT OpenCourseWare 00:00:17.207 --> 00:00:17.832 at ocw.mit.edu. 00:00:21.544 --> 00:00:23.460 PROFESSOR: Ok then, let's start with problems. 00:00:23.460 --> 00:00:29.414 And as we find concept issues, we'll get to concept issues. 00:00:29.414 --> 00:00:30.830 So, we talked about shortest-path, 00:00:30.830 --> 00:00:32.390 but we talked about shortest-path 00:00:32.390 --> 00:00:34.730 in a very odd way, right? 00:00:34.730 --> 00:00:35.360 I'm a coder. 00:00:35.360 --> 00:00:39.260 So for me I'm used to, all right I go to one of these lectures, 00:00:39.260 --> 00:00:41.600 I hear a problem, then I get out of the lecture 00:00:41.600 --> 00:00:44.430 with an algorithm and the running time, right? 00:00:44.430 --> 00:00:45.930 This time we got out of the lecture 00:00:45.930 --> 00:00:47.660 with no algorithm and no running time. 00:00:47.660 --> 00:00:49.650 So, what's the point? 00:00:49.650 --> 00:00:52.090 The point is that we learned some analysis tricks 00:00:52.090 --> 00:00:55.620 that we can use for any shortest-path algorithm. 00:00:55.620 --> 00:00:58.470 And the advantage of knowing that-- by the way too easy 00:00:58.470 --> 00:01:00.720 to forget, at least until you see the first real 00:01:00.720 --> 00:01:02.960 algorithms-- if you know that, if you 00:01:02.960 --> 00:01:06.589 have to modify an algorithm, you can still use the same analysis 00:01:06.589 --> 00:01:08.880 to prove that your new algorithm is going to be correct 00:01:08.880 --> 00:01:11.164 and that it's going to be fast. 00:01:11.164 --> 00:01:12.580 They're really nice tools to have, 00:01:12.580 --> 00:01:15.500 especially if you have to do that on a problem set 00:01:15.500 --> 00:01:20.020 or on an interview, or who knows, maybe even on a quiz. 00:01:20.020 --> 00:01:21.950 We'll see if that happens or not. 00:01:21.950 --> 00:01:24.530 We didn't talk about any shortest-path algorithms. 00:01:24.530 --> 00:01:27.130 So, we're not going to assume shortest-path algorithms. 00:01:27.130 --> 00:01:32.530 Instead we're going to use them as if we had them. 00:01:32.530 --> 00:01:33.270 No, wait. 00:01:33.270 --> 00:01:34.181 I don't like that. 00:01:34.181 --> 00:01:35.180 That's kind of annoying. 00:01:35.180 --> 00:01:36.090 Let's not do that. 00:01:36.090 --> 00:01:37.320 Let's do something else. 00:01:37.320 --> 00:01:39.320 We didn't talk about the shortest-path algorithm 00:01:39.320 --> 00:01:39.820 in lecture. 00:01:39.820 --> 00:01:42.940 Let's build one right now, how about that? 00:01:42.940 --> 00:01:45.230 Let's turn to the graph. 00:01:45.230 --> 00:01:46.340 Vertices and edges, right? 00:01:52.820 --> 00:01:54.430 Let's see if I can make this work. 00:02:17.040 --> 00:02:18.200 Suppose we have a graph. 00:02:18.200 --> 00:02:24.690 We have V vertices, E edges and all the edges 00:02:24.690 --> 00:02:27.200 have a weight that's non-negative. 00:02:27.200 --> 00:02:31.290 Sorry, let's make it positive just to make things easy. 00:02:31.290 --> 00:02:35.430 It's smaller than W, so it's not too big. 00:02:35.430 --> 00:02:38.220 And they're all integers. 00:02:38.220 --> 00:02:38.720 Go. 00:02:44.490 --> 00:02:46.157 There's brute-force, OK. 00:02:46.157 --> 00:02:47.490 So, brute-force would mean what? 00:02:47.490 --> 00:02:49.510 Enumerate all the paths? 00:02:49.510 --> 00:02:50.607 OK. 00:02:50.607 --> 00:02:52.440 Let's go over something a little bit better. 00:02:56.200 --> 00:02:58.820 Does anyone remember the algorithm 00:02:58.820 --> 00:03:00.850 that was taught in class? 00:03:00.850 --> 00:03:02.020 The structure? 00:03:02.020 --> 00:03:03.670 No. 00:03:03.670 --> 00:03:08.340 Klaus mentioned Dijkstra and mentioned Bellman-Ford. 00:03:08.340 --> 00:03:09.490 Let's write them up here. 00:03:15.110 --> 00:03:17.100 And they gave us the running times, 00:03:17.100 --> 00:03:19.250 which we'll try to remember in a bit. 00:03:19.250 --> 00:03:24.160 But, we don't know algorithms, so no Dijkstra, no Bellman-Ford 00:03:24.160 --> 00:03:27.000 for us. 00:03:27.000 --> 00:03:29.780 What do we know? 00:03:29.780 --> 00:03:31.890 Let's start with a simpler case. 00:03:31.890 --> 00:03:33.430 What if there are no weights here. 00:03:33.430 --> 00:03:35.050 What if the graph looked like this. 00:03:42.460 --> 00:03:43.600 You're making my life hard. 00:03:49.050 --> 00:03:51.090 Like this. 00:03:51.090 --> 00:03:51.870 No costs. 00:03:51.870 --> 00:03:54.607 All the edges have the same costs. 00:03:54.607 --> 00:03:56.190 What would we do to solve the problem? 00:03:58.740 --> 00:04:06.790 Say we want the shortest path from A to E. 00:04:06.790 --> 00:04:08.330 AUDIENCE: BFS. 00:04:08.330 --> 00:04:10.400 PROFESSOR: BFS. 00:04:10.400 --> 00:04:11.800 How's everyone feeling about BFS? 00:04:15.150 --> 00:04:16.899 What does BFS do, really quickly? 00:04:19.579 --> 00:04:21.080 How do you run BFS? 00:04:21.080 --> 00:04:23.103 AUDIENCE: Take a starting node and then you 00:04:23.103 --> 00:04:24.580 can search all it's neighbors. 00:04:24.580 --> 00:04:25.730 PROFESSOR: OK. 00:04:25.730 --> 00:04:27.590 Pick a starting node. 00:04:27.590 --> 00:04:30.410 Then, look at its neighbors? 00:04:30.410 --> 00:04:33.290 AUDIENCE: And then you enter it through each of its neighbors 00:04:33.290 --> 00:04:36.940 to find its neighbors and so forth. 00:04:36.940 --> 00:04:39.222 PROFESSOR: Sounds an awful lot like BFS. 00:04:39.222 --> 00:04:40.680 What's the difference between them? 00:04:40.680 --> 00:04:43.720 AUDIENCE: Well, you go through all of these neighbors 00:04:43.720 --> 00:04:45.650 before you start going through any of it. 00:04:45.650 --> 00:04:47.280 I mean, you go through all As neighbors 00:04:47.280 --> 00:04:48.402 before you take a neighbor and go 00:04:48.402 --> 00:04:49.700 through all of its neighbors. 00:04:49.700 --> 00:04:51.560 PROFESSOR: Ok, that's the difference. 00:04:51.560 --> 00:04:53.250 And it makes a huge difference as we've 00:04:53.250 --> 00:04:55.490 seen in topological sorting last time. 00:04:55.490 --> 00:04:57.990 So we have a list and we're going 00:04:57.990 --> 00:05:01.070 to use it to keep track of the nodes that we need to visit. 00:05:01.070 --> 00:05:05.460 We start with A. We numerate all of As neighbors. 00:05:05.460 --> 00:05:08.510 And the ones that haven't been seen 00:05:08.510 --> 00:05:10.580 get stuck at the end of the list. 00:05:10.580 --> 00:05:13.690 A, B, C. A is out of the list. 00:05:13.690 --> 00:05:19.230 Then we take out the first node from the list, B in this case. 00:05:19.230 --> 00:05:24.260 We numerate all its neighbors, A, C, D, E. Out of all these, 00:05:24.260 --> 00:05:26.075 we take the neighbors that we haven't seen 00:05:26.075 --> 00:05:27.700 and we put them at the end of the list. 00:05:32.460 --> 00:05:34.330 take out C, we look at its neighbors. 00:05:34.330 --> 00:05:35.040 We've seen them. 00:05:35.040 --> 00:05:36.790 We take out D, we look at his neighbors. 00:05:36.790 --> 00:05:37.350 We've seen them. 00:05:37.350 --> 00:05:39.183 We take out E, we look at all its neighbors. 00:05:39.183 --> 00:05:40.040 We've seen them. 00:05:40.040 --> 00:05:43.060 So this is BFS. 00:05:43.060 --> 00:05:44.830 BFS has this concept of levels. 00:05:44.830 --> 00:05:48.890 And another way to look at levels is, if you start at A, 00:05:48.890 --> 00:05:52.670 and you draw a circus of radius one, 00:05:52.670 --> 00:05:55.561 it's going to have all the nodes that are distance 1 from 8. 00:05:55.561 --> 00:05:56.060 Right? 00:05:56.060 --> 00:05:58.080 Exactly one edge. 00:05:58.080 --> 00:06:01.790 Then you draw a circus of radius 2 00:06:01.790 --> 00:06:06.550 and you get all the nodes that you can reach in two edges. 00:06:06.550 --> 00:06:10.630 And BFS calls these circles levels. 00:06:10.630 --> 00:06:11.940 A is at level 0. 00:06:11.940 --> 00:06:13.590 This is level 1. 00:06:13.590 --> 00:06:15.860 This is level 2. 00:06:15.860 --> 00:06:18.710 If you look at this list and you write down the levels, 00:06:18.710 --> 00:06:21.860 you have 0, 1, 1, 2, 2. 00:06:21.860 --> 00:06:26.610 If you had another node out here, say F, 00:06:26.610 --> 00:06:28.630 this would be discovered when we get 00:06:28.630 --> 00:06:33.200 to D. It would go all the way here at the end, 00:06:33.200 --> 00:06:34.900 and this is level 3. 00:06:34.900 --> 00:06:38.480 In BFS the levels are always increasing. 00:06:38.480 --> 00:06:40.950 If you keep track of parent pointers, 00:06:40.950 --> 00:06:43.900 the parent pointer of a node, at some level, 00:06:43.900 --> 00:06:47.490 will always point to a node at the previous level. 00:06:47.490 --> 00:06:51.770 So, in BFS we get the shortest paths in terms of edge count. 00:06:55.310 --> 00:06:56.240 Make sense? 00:06:56.240 --> 00:06:57.240 This is a recall, right? 00:06:57.240 --> 00:06:58.580 It is not new stuff. 00:06:58.580 --> 00:07:01.350 Everyone happy with it? 00:07:01.350 --> 00:07:02.850 If it wouldn't have costs, we could 00:07:02.850 --> 00:07:06.750 use BFS to compute shortest paths. 00:07:06.750 --> 00:07:09.006 But, we have costs. 00:07:09.006 --> 00:07:12.160 How do we deal with that? 00:07:12.160 --> 00:07:15.950 Intuitively, I'd like to make BFS go across this edge 00:07:15.950 --> 00:07:19.260 faster than it would go across the edge. 00:07:19.260 --> 00:07:21.930 If I could do that, if this guy could get stuck in the queue 00:07:21.930 --> 00:07:24.750 before this guy, I would be happy. 00:07:24.750 --> 00:07:27.090 Intuitively, that's what I'd like to do. 00:07:27.090 --> 00:07:29.648 How do we do that? 00:07:29.648 --> 00:07:32.812 AUDIENCE: Sort before adding? 00:07:32.812 --> 00:07:33.520 PROFESSOR: Sort-- 00:07:33.520 --> 00:07:35.800 AUDIENCE: Sort your neighbors sorts. 00:07:35.800 --> 00:07:36.850 PROFESSOR: Sort? 00:07:36.850 --> 00:07:39.110 OK. 00:07:39.110 --> 00:07:41.960 AUDIENCE: What if you kept track of it in a menu? 00:07:41.960 --> 00:07:45.830 Kept track of your queue as a priority queue. 00:07:45.830 --> 00:07:46.955 PROFESSOR: Congratulations. 00:07:46.955 --> 00:07:48.300 You deserve a Turing Award. 00:07:48.300 --> 00:07:49.510 You have discovered Dijkstra. 00:07:53.240 --> 00:07:55.410 OK, let's go for something simpler. 00:07:55.410 --> 00:07:57.160 I just say this is Dijkstra. 00:07:57.160 --> 00:07:59.350 We'll talk about it in lecture on Tuesday. 00:07:59.350 --> 00:08:02.270 Very good. 00:08:02.270 --> 00:08:03.932 AUDIENCE: Like, have an adjacency list 00:08:03.932 --> 00:08:06.824 and, I think using what he said. 00:08:06.824 --> 00:08:10.500 Just sort it by cost-- 00:08:10.500 --> 00:08:12.490 PROFESSOR: OK, let's see how this would work. 00:08:12.490 --> 00:08:18.510 We have A, and we're pulling A out of the list. 00:08:18.510 --> 00:08:22.250 And we see its neighbors are C and B. Right? 00:08:22.250 --> 00:08:23.610 The distance to C is 2. 00:08:23.610 --> 00:08:25.660 The distance to B is 4. 00:08:25.660 --> 00:08:29.580 So we start with A. Does this mean 00:08:29.580 --> 00:08:32.419 that I'm going to put in B before C? 00:08:32.419 --> 00:08:34.537 I'm going to put in C and then B? 00:08:34.537 --> 00:08:36.872 AUDIENCE: No, it doesn't really matter 00:08:36.872 --> 00:08:38.740 because at the end you just sort the part. 00:08:38.740 --> 00:08:41.954 I guess you could add in the costs. 00:08:41.954 --> 00:08:44.049 Like associate the costs of B with 4 00:08:44.049 --> 00:08:48.009 and C with 2 within your adjacency list? 00:08:48.009 --> 00:08:50.300 PROFESSOR: So, that's how we would keep track of costs, 00:08:50.300 --> 00:08:50.690 right? 00:08:50.690 --> 00:08:51.280 One way. 00:08:51.280 --> 00:08:52.274 AUDIENCE: --for the queue would be A 00:08:52.274 --> 00:08:54.857 and then the value would be like a [INAUDIBLE] B and its cost, 00:08:54.857 --> 00:08:57.620 and the [INAUDIBLE] C and its cost. 00:08:57.620 --> 00:09:00.530 PROFESSOR: So you're saying that what I would have in this list 00:09:00.530 --> 00:09:05.020 is B with a cost of 4? 00:09:05.020 --> 00:09:09.150 And C with a cost of 2? 00:09:09.150 --> 00:09:12.610 Presumably, the other way around, right? 00:09:12.610 --> 00:09:13.265 C goes first? 00:09:20.890 --> 00:09:22.440 Like This? 00:09:22.440 --> 00:09:26.220 And then I take out C, I explore the neighbors so and so forth? 00:09:26.220 --> 00:09:28.770 The moment I did this I already lost the battle. 00:09:28.770 --> 00:09:36.760 Because, see this path A, C, B, length 3? 00:09:36.760 --> 00:09:38.660 If I put both of them in the queue, 00:09:38.660 --> 00:09:41.700 then the parent pointers are going to look like this. 00:09:41.700 --> 00:09:44.271 So, I already have a bad path from B to A. 00:09:44.271 --> 00:09:46.920 AUDIENCE: OK, well don't add that node then. 00:09:46.920 --> 00:09:48.755 PROFESSOR: OK, then how do we do it? 00:09:48.755 --> 00:09:52.635 AUDIENCE: I guess you can talk to each of them 00:09:52.635 --> 00:09:54.490 and see which one's lower and then, only 00:09:54.490 --> 00:09:56.850 add that one if it's the lowest. 00:09:56.850 --> 00:09:58.500 So it's more like depth search then-- 00:09:58.500 --> 00:10:00.416 PROFESSOR: Well, you two need to talk together 00:10:00.416 --> 00:10:02.530 because you're getting to Dijkstra, too. 00:10:02.530 --> 00:10:06.120 You're both going to rediscover Dijkstra. 00:10:06.120 --> 00:10:09.370 So that's great, but I'm looking for something simple. 00:10:09.370 --> 00:10:10.940 This is too complicated. 00:10:10.940 --> 00:10:12.740 If you're going to do Dijkstra now, then 00:10:12.740 --> 00:10:14.697 what are going to do on lecture on Tuesday? 00:10:14.697 --> 00:10:17.557 AUDIENCE: More Dijkstra problems? 00:10:17.557 --> 00:10:20.140 PROFESSOR: Well, I think Shreeny wants to talk about Dijkstra, 00:10:20.140 --> 00:10:21.280 so let's not get there. 00:10:21.280 --> 00:10:24.514 AUDIENCE: How do you do this without doing Dijkstra? 00:10:24.514 --> 00:10:25.180 PROFESSOR: Good. 00:10:25.180 --> 00:10:27.150 That's exactly what I want to know. 00:10:27.150 --> 00:10:29.970 [LAUGHTER] 00:10:29.970 --> 00:10:31.791 So, let me give you another hint. 00:10:31.791 --> 00:10:34.040 The first hint was that I want to go through this edge 00:10:34.040 --> 00:10:36.380 faster than I would go through this edge. 00:10:36.380 --> 00:10:40.744 The other hint is that we're allowed to change the graph. 00:10:40.744 --> 00:10:43.380 AUDIENCE: Is this what we talked about in lecture, 00:10:43.380 --> 00:10:46.491 or are we thinking of something else? 00:10:46.491 --> 00:10:48.240 PROFESSOR: He might have talked about that 00:10:48.240 --> 00:10:51.610 at the end of lecture. 00:10:51.610 --> 00:10:54.960 AUDIENCE: Were you putting the current weight 00:10:54.960 --> 00:10:58.390 for each node inside the circles? 00:10:58.390 --> 00:11:01.330 PROFESSOR: No. 00:11:01.330 --> 00:11:02.800 AUDIENCE: You could add it and then 00:11:02.800 --> 00:11:05.740 if you could find a better path to B? 00:11:05.740 --> 00:11:07.620 With rows relative to A, then you 00:11:07.620 --> 00:11:09.500 could just delete that path? 00:11:09.500 --> 00:11:10.626 And B to 4? 00:11:10.626 --> 00:11:12.626 So at the beginning you add them all, but then-- 00:11:12.626 --> 00:11:15.312 PROFESSOR: You're risking to have exponential running time. 00:11:15.312 --> 00:11:18.357 AUDIENCE: (LAUGHS) 00:11:18.357 --> 00:11:20.190 PROFESSOR: Those are all valid optimizations 00:11:20.190 --> 00:11:21.570 under some constraints. 00:11:21.570 --> 00:11:24.250 Like, this one's good if you don't have too much memory, 00:11:24.250 --> 00:11:27.730 but the time goes up. 00:11:27.730 --> 00:11:30.810 I want my graph to have edges with no costs, 00:11:30.810 --> 00:11:34.250 because if I have costs, this is not going to work. 00:11:34.250 --> 00:11:36.950 BFS is not going to work. 00:11:36.950 --> 00:11:37.700 We can't tweak it. 00:11:37.700 --> 00:11:41.110 If you could tweak it then that would be another Turing Award. 00:11:41.110 --> 00:11:42.561 AUDIENCE: So, create dummy nodes? 00:11:42.561 --> 00:11:43.810 PROFESSOR: Create dummy nodes. 00:11:43.810 --> 00:11:46.930 How do I do that? 00:11:46.930 --> 00:11:49.780 AUDIENCE:Put a band of node in between A and C? 00:11:49.780 --> 00:11:50.480 PROFESSOR: OK. 00:11:50.480 --> 00:11:51.390 Why am I doing this? 00:11:56.510 --> 00:11:59.710 So now I have two edges and they have costs. 00:11:59.710 --> 00:12:00.600 AUDIENCE: One. 00:12:00.600 --> 00:12:03.092 Then you put three nodes in between A and B. 00:12:03.092 --> 00:12:03.675 PROFESSOR: OK. 00:12:07.120 --> 00:12:10.980 And now here I have four edges, right? 00:12:10.980 --> 00:12:12.990 1, 1, 1, 1. 00:12:12.990 --> 00:12:17.990 There was a 4 here before and there was a 2 here before. 00:12:17.990 --> 00:12:20.040 Do you guys see what's happening? 00:12:20.040 --> 00:12:24.520 So, instead of one edge of cost C, I have C edges. 00:12:24.520 --> 00:12:31.010 So, now BFS is going to take C steps to go through that edge. 00:12:31.010 --> 00:12:33.230 And all the edges now have the same cost. 00:12:33.230 --> 00:12:35.275 This looks like a problem that BFS can solve. 00:12:35.275 --> 00:12:36.150 We know it can solve. 00:12:36.150 --> 00:12:37.930 We proved that it can solve it. 00:12:37.930 --> 00:12:39.260 Well, CLRS did. 00:12:39.260 --> 00:12:41.140 We take their word for granted, sort of. 00:12:45.140 --> 00:12:46.977 This works right? 00:12:46.977 --> 00:12:48.060 Is everyone happy with it? 00:12:48.060 --> 00:12:49.770 Does it make sense? 00:12:49.770 --> 00:12:55.160 AUDIENCE: So basically, we get through C 00:12:55.160 --> 00:12:57.250 before you'll get to B on the other path. 00:12:57.250 --> 00:12:58.000 PROFESSOR: Yep. 00:12:58.000 --> 00:13:00.230 This is the path of length 3, so this node 00:13:00.230 --> 00:13:02.730 is going to get in the queue first. 00:13:02.730 --> 00:13:03.610 Then this node. 00:13:03.610 --> 00:13:04.980 Then these two nodes. 00:13:04.980 --> 00:13:09.970 And they're going to get in the queue from C. 00:13:09.970 --> 00:13:12.620 Let's see what happened here. 00:13:12.620 --> 00:13:14.440 This is really important because this 00:13:14.440 --> 00:13:17.170 is what you're going to be doing in real life. 00:13:17.170 --> 00:13:20.000 Chances of having to build a new algorithm? 00:13:20.000 --> 00:13:20.700 Kind of slim. 00:13:20.700 --> 00:13:22.250 Why don't you want to do that? 00:13:22.250 --> 00:13:25.240 Say you take Dijkstra and you tweak it a little bit. 00:13:25.240 --> 00:13:27.910 If you did that, you have to go through the analysis 00:13:27.910 --> 00:13:29.320 and prove the running time again, 00:13:29.320 --> 00:13:30.800 prove the correctness again. 00:13:30.800 --> 00:13:35.090 Time-consuming, error-prone, it's really hard. 00:13:35.090 --> 00:13:37.380 However, what happened here is that we 00:13:37.380 --> 00:13:39.880 have some problem that we want to solve. 00:13:39.880 --> 00:13:41.630 Say we have the raw input for that problem 00:13:41.630 --> 00:13:44.910 that we're trying to solve. 00:13:44.910 --> 00:13:48.365 That is this graph that I draw the board. 00:13:51.740 --> 00:13:54.890 And we apply some transform and we 00:13:54.890 --> 00:13:57.120 get an input that's suitable for an algorithm 00:13:57.120 --> 00:13:58.080 that we already know. 00:14:05.670 --> 00:14:07.460 I'm going to take a shortcut here. 00:14:13.120 --> 00:14:14.870 We apply the transform, right? 00:14:14.870 --> 00:14:17.550 We took this graph and we built a new graph in. 00:14:17.550 --> 00:14:19.740 And this is hopefully easy. 00:14:19.740 --> 00:14:20.720 Right? 00:14:20.720 --> 00:14:22.320 Take an edge and split it up. 00:14:22.320 --> 00:14:23.070 Put in fake nodes. 00:14:23.070 --> 00:14:23.840 This is easy. 00:14:23.840 --> 00:14:26.410 We know how to code that. 00:14:26.410 --> 00:14:28.676 And now we have this algorithm that you already know, 00:14:28.676 --> 00:14:30.300 and we're treating it like a black box. 00:14:33.180 --> 00:14:34.187 We know it's correct. 00:14:34.187 --> 00:14:35.270 We know it's running time. 00:14:35.270 --> 00:14:36.290 We're not questioning them. 00:14:36.290 --> 00:14:37.910 We're taking them straight from the textbook. 00:14:37.910 --> 00:14:38.780 They have to be right. 00:14:38.780 --> 00:14:40.571 By the way, what's the running time of BFS? 00:14:48.620 --> 00:14:49.426 AUDIENCE: V plus E? 00:14:49.426 --> 00:14:50.300 PROFESSOR: Very good. 00:14:55.050 --> 00:14:57.110 Then BFS is going to give us a path, right? 00:14:57.110 --> 00:14:58.830 And the path is going to look like this. 00:14:58.830 --> 00:15:07.700 It's going to be A, fake node, C, D, 00:15:07.700 --> 00:15:13.112 fake node, E. If you're giving this back to the guy who 00:15:13.112 --> 00:15:15.070 gave you the problem, they're going to be like, 00:15:15.070 --> 00:15:16.210 what are these fake nodes? 00:15:16.210 --> 00:15:17.770 I don't know anything about them. 00:15:17.770 --> 00:15:19.320 This doesn't make sense. 00:15:19.320 --> 00:15:23.940 You need to take this raw output and you 00:15:23.940 --> 00:15:25.280 need to transform it again. 00:15:25.280 --> 00:15:28.110 Based on what you did this transformation, 00:15:28.110 --> 00:15:29.510 you have to read out the output. 00:15:29.510 --> 00:15:31.110 You have to interpret it. 00:15:31.110 --> 00:15:33.970 So, this is either interpret or readout. 00:15:33.970 --> 00:15:38.230 And then you get a nice output that's 00:15:38.230 --> 00:15:40.700 the output to your original problem. 00:15:43.280 --> 00:15:47.490 This process here is the most likely way 00:15:47.490 --> 00:15:50.110 in which you'll use your 6006 knowledge. 00:15:50.110 --> 00:15:53.460 And this is replaced by any algorithm that we taught. 00:15:53.460 --> 00:15:55.319 This is replaced by any real life problem 00:15:55.319 --> 00:15:56.235 that you'll encounter. 00:16:01.580 --> 00:16:02.844 There's one missing step here. 00:16:02.844 --> 00:16:04.260 We know the running time for this. 00:16:04.260 --> 00:16:08.650 Let's find out the running time for the whole thing. 00:16:08.650 --> 00:16:11.220 In order to do that, we have to figure out 00:16:11.220 --> 00:16:13.390 given this original graph. 00:16:13.390 --> 00:16:15.180 Let's say that now, in the original graph, 00:16:15.180 --> 00:16:18.220 we have V prime vertices, E prime edges, and W 00:16:18.220 --> 00:16:20.760 prime weights. 00:16:20.760 --> 00:16:24.100 We have to convert this into V plus E. 00:16:24.100 --> 00:16:26.800 We have to see when we make this transform, how many vertices 00:16:26.800 --> 00:16:27.300 do we get? 00:16:27.300 --> 00:16:30.262 How many edges do we get in terms of the original graph 00:16:30.262 --> 00:16:31.928 so that we can compute the running time? 00:16:34.530 --> 00:16:38.670 In the original graph, V prime vertices, E prime edges, 00:16:38.670 --> 00:16:40.180 W prime weights. 00:16:40.180 --> 00:16:43.540 In this new graph, how many edges do I have, at most? 00:16:51.014 --> 00:16:52.246 AUDIENCE: Something times E? 00:16:52.246 --> 00:16:52.870 PROFESSOR: Yep. 00:16:52.870 --> 00:16:53.369 Perfect. 00:16:57.410 --> 00:17:00.100 Each edge has cost at most W. That 00:17:00.100 --> 00:17:04.361 means we're going to split it up into most W edges. 00:17:04.361 --> 00:17:06.166 How many vertices are we going to have? 00:17:14.462 --> 00:17:16.920 AUDIENCE: W prime? 00:17:16.920 --> 00:17:19.510 PROFESSOR: W prime, E prime. 00:17:19.510 --> 00:17:20.776 AUDIENCE: Plus V prime? 00:17:20.776 --> 00:17:21.776 PROFESSOR: Plus V prime. 00:17:21.776 --> 00:17:23.349 Don't forget about the original ones. 00:17:23.349 --> 00:17:25.640 They're still there. 00:17:25.640 --> 00:17:29.875 So, total running time for BFS on this new graph is? 00:17:37.778 --> 00:17:40.200 AUDIENCE: --E prime. 00:17:40.200 --> 00:17:44.610 PROFESSOR: Put it the other way around, plus W prime, E prime. 00:17:48.240 --> 00:17:50.340 This is the running time that we have. 00:17:50.340 --> 00:17:53.300 Now, does anyone remember Dijkstra's running time 00:17:53.300 --> 00:17:55.060 and Bellam-Ford's running time? 00:17:58.870 --> 00:18:00.623 So what is the running time for Dijkstra? 00:18:00.623 --> 00:18:04.970 AUDIENCE: V log V plus E? 00:18:04.970 --> 00:18:06.593 PROFESSOR: Almost. 00:18:06.593 --> 00:18:07.980 AUDIENCE: V log V? 00:18:07.980 --> 00:18:09.002 PROFESSOR: Oh, wait. 00:18:09.002 --> 00:18:10.850 I think you're right. 00:18:10.850 --> 00:18:14.451 V log V plus-- 00:18:14.451 --> 00:18:16.840 AUDIENCE: E? 00:18:16.840 --> 00:18:19.680 PROFESSOR: I already named this from CRS? 00:18:19.680 --> 00:18:20.670 Or is this what we got? 00:18:23.630 --> 00:18:25.646 Entry is a fancy key to give this running time. 00:18:25.646 --> 00:18:26.845 AUDIENCE: [INAUDIBLE]. 00:18:26.845 --> 00:18:29.470 PROFESSOR: You can get this, but you need the really fancy data 00:18:29.470 --> 00:18:30.720 structure for it. 00:18:30.720 --> 00:18:36.980 In real life, the running time looks more like E log V. 00:18:36.980 --> 00:18:38.160 And how about Bellam-Ford? 00:18:49.970 --> 00:18:52.226 So let's take this running time for Dijkstra 00:18:52.226 --> 00:18:53.600 because this is what you're going 00:18:53.600 --> 00:18:54.929 to implement in real life. 00:18:54.929 --> 00:18:57.220 And we can argue about that after next lecture for now. 00:18:57.220 --> 00:18:58.760 Take my word for it. 00:18:58.760 --> 00:19:02.480 And let's compare it to what we got here. 00:19:02.480 --> 00:19:05.990 V is smaller than E in most same cases. 00:19:05.990 --> 00:19:09.450 Let's say that this is actually W prime, E prime. 00:19:09.450 --> 00:19:12.630 If you compare this with this, you'll 00:19:12.630 --> 00:19:18.820 see that, if W is smaller than log V, well, smaller equal, 00:19:18.820 --> 00:19:22.420 then this algorithm is not worse than Dijkstra. 00:19:22.420 --> 00:19:25.440 It's on par with Dijkstra. 00:19:25.440 --> 00:19:28.760 For graphs with small costs, we discovered an algorithm 00:19:28.760 --> 00:19:30.720 that solves shortest paths. 00:19:30.720 --> 00:19:33.380 You walked into this recitation knowing no algorithm 00:19:33.380 --> 00:19:35.060 to solve shortest path. 00:19:35.060 --> 00:19:36.440 Now we have an algorithm. 00:19:36.440 --> 00:19:38.480 Already in a much better position. 00:19:38.480 --> 00:19:39.030 Right? 00:19:39.030 --> 00:19:40.620 And we didn't invent anything new. 00:19:40.620 --> 00:19:42.420 We don't have to prove correctness. 00:19:42.420 --> 00:19:46.490 All we did was one of these transformations. 00:19:46.490 --> 00:19:49.589 How's everyone feeling? 00:19:49.589 --> 00:19:50.630 Those things makes sense? 00:19:50.630 --> 00:19:51.370 Everyone happy? 00:19:54.770 --> 00:20:00.025 Let's talk about another problem that also uses this structure. 00:20:10.037 --> 00:20:11.620 Suppose we have the same graph that we 00:20:11.620 --> 00:20:17.040 had before, or actually, any graph with V vertices, E edges. 00:20:17.040 --> 00:20:18.910 Now we know how to compute shortest paths. 00:20:18.910 --> 00:20:22.250 So we can assume these as black boxes. 00:20:26.620 --> 00:20:29.150 We're going to use these as they are. 00:20:29.150 --> 00:20:31.540 We have a graph with V vertices and E edges. 00:20:44.640 --> 00:20:46.435 Let me copy the costs really quickly. 00:20:55.660 --> 00:20:57.720 Suppose this is the highway system, 00:20:57.720 --> 00:20:59.700 right, like in Google Maps. 00:20:59.700 --> 00:21:01.320 And we have two brothers. 00:21:01.320 --> 00:21:03.957 They start off from a place and they 00:21:03.957 --> 00:21:05.290 have to end up in another place. 00:21:05.290 --> 00:21:07.520 They're going to drive together. 00:21:07.520 --> 00:21:10.330 They want this to be sort of fair. 00:21:10.330 --> 00:21:12.240 Every time they go between two cities 00:21:12.240 --> 00:21:14.950 they're going to switch seats so that none of them 00:21:14.950 --> 00:21:16.330 drives too much. 00:21:16.330 --> 00:21:21.357 So say the brothers are Tim and Jim. 00:21:21.357 --> 00:21:22.190 The names are wrong. 00:21:22.190 --> 00:21:25.420 This is actually a problem from last year's quiz. 00:21:25.420 --> 00:21:26.700 So, by the way, real problem. 00:21:26.700 --> 00:21:30.100 You can pay attention now. 00:21:30.100 --> 00:21:32.700 These are two brothers. 00:21:32.700 --> 00:21:34.010 They're going to alternate. 00:21:34.010 --> 00:21:34.510 Right? 00:21:34.510 --> 00:21:36.691 One of them is going to drive across one road, 00:21:36.691 --> 00:21:39.190 then the other one, then the first one, then the second one. 00:21:39.190 --> 00:21:40.970 So on and so forth. 00:21:40.970 --> 00:21:45.950 Now, two Tim has a much better sense of direction than Jim. 00:21:45.950 --> 00:21:47.450 That's just how things are. 00:21:47.450 --> 00:21:50.880 And if you ever driven long distances, 00:21:50.880 --> 00:21:53.310 the hardest thing to do is to get from the city 00:21:53.310 --> 00:21:54.379 to the highway. 00:21:54.379 --> 00:21:56.420 Like, if you have to drive from here to New York. 00:21:56.420 --> 00:21:57.410 Hardest thing to do? 00:21:57.410 --> 00:21:59.800 Get out of Boston and onto the highway. 00:21:59.800 --> 00:22:02.160 Then, at the end, the hardest thing to do? 00:22:02.160 --> 00:22:03.780 Get from the highway into where you 00:22:03.780 --> 00:22:05.630 want to go through New York traffic. 00:22:05.630 --> 00:22:06.420 Everything else? 00:22:06.420 --> 00:22:08.890 Piece of cake. 00:22:08.890 --> 00:22:13.055 We want Tim to handle both of these situations. 00:22:13.055 --> 00:22:17.220 The driving path must start with Tim and must end with Tim. 00:22:17.220 --> 00:22:19.260 Otherwise, God knows, they're going to crash. 00:22:21.850 --> 00:22:24.730 First off let's convert this into graph terms. 00:22:24.730 --> 00:22:26.060 And then let's solve it. 00:22:29.619 --> 00:22:31.952 AUDIENCE: You might convert what's already in the graph. 00:22:34.620 --> 00:22:36.150 PROFESSOR: I have this constraint 00:22:36.150 --> 00:22:39.989 that they're going to switch seats and that Tim has to start 00:22:39.989 --> 00:22:40.780 and Tim has to end. 00:22:40.780 --> 00:22:43.540 How do I phrase this in mathy terms? 00:22:43.540 --> 00:22:44.300 In graph terms? 00:22:52.430 --> 00:22:54.621 AUDIENCE: We're starting it where and ending where? 00:22:54.621 --> 00:22:56.370 PROFESSOR: We're given where in the graph. 00:22:56.370 --> 00:23:00.420 We're starting at S, ending at T. Some points in the graph. 00:23:00.420 --> 00:23:05.500 AUDIENCE: And people have to drive with it an entire edge? 00:23:05.500 --> 00:23:07.658 PROFESSOR: Ya. 00:23:07.658 --> 00:23:13.488 AUDIENCE: So basically, Jim has to-- whoever 00:23:13.488 --> 00:23:17.170 the good driver has to-- to drive on the edge that's 00:23:17.170 --> 00:23:19.345 adjacent to E and S. 00:23:19.345 --> 00:23:20.220 PROFESSOR: Yep. 00:23:20.220 --> 00:23:21.552 OK. 00:23:21.552 --> 00:23:24.710 AUDIENCE: You physically need, like an odd number of edges. 00:23:24.710 --> 00:23:25.890 PROFESSOR: Yep. 00:23:25.890 --> 00:23:29.060 So I want the shortest path with an odd number of edges. 00:23:29.060 --> 00:23:30.490 Why odd number of edges? 00:23:30.490 --> 00:23:33.620 If we look at the edges, Tim's going to take the first one. 00:23:33.620 --> 00:23:37.210 Then Jim, then Tim, then Jim, then Tim, then Jim. 00:23:37.210 --> 00:23:40.410 No matter how many I have here, it has to end with Tim. 00:23:40.410 --> 00:23:42.710 So, I'm going to have an odd number of letters here, 00:23:42.710 --> 00:23:44.460 therefore, odd number of edges. 00:23:44.460 --> 00:23:44.960 Right? 00:23:44.960 --> 00:23:51.730 So I want the shortest path that has an odd number of edges. 00:23:58.570 --> 00:24:02.140 And as the hint we're going to use that trick over there. 00:24:02.140 --> 00:24:02.760 Yes? 00:24:02.760 --> 00:24:05.596 AUDIENCE: But, the shortest path isn't necessarily 00:24:05.596 --> 00:24:07.093 like the fairest, right? 00:24:07.093 --> 00:24:11.090 Like, in terms of distributing driving? 00:24:11.090 --> 00:24:13.680 PROFESSOR: One of them is going to drive one more road, 00:24:13.680 --> 00:24:14.840 such is life. 00:24:14.840 --> 00:24:16.345 Sorry, one more edge segment. 00:24:18.915 --> 00:24:20.290 Tim's going to drive three times, 00:24:20.290 --> 00:24:21.992 Jim is going to drive two times. 00:24:21.992 --> 00:24:24.450 AUDIENCE: Right, but like, in terms of the weights? 00:24:24.450 --> 00:24:25.283 PROFESSOR: Oh, yeah. 00:24:25.283 --> 00:24:27.680 We don't care about that. 00:24:27.680 --> 00:24:29.522 Jim doesn't know where he's going anyways. 00:24:29.522 --> 00:24:31.980 AUDIENCE: Just a minimum, we have to have like three paths, 00:24:31.980 --> 00:24:33.117 right? 00:24:33.117 --> 00:24:35.117 We don't just want Tim to get out to the highway 00:24:35.117 --> 00:24:37.158 and be like, oh look our journey's not that far-- 00:24:39.650 --> 00:24:42.940 PROFESSOR: So I went to the path with the smallest cost. 00:24:42.940 --> 00:24:45.787 So, smallest total weight. 00:24:45.787 --> 00:24:47.620 AUDIENCE: In theory it could be length, too. 00:24:47.620 --> 00:24:48.530 Right? 00:24:48.530 --> 00:24:50.071 PROFESSOR: Well, if it's length, too, 00:24:50.071 --> 00:24:53.850 then it's not going to be good because they're going to crash. 00:24:53.850 --> 00:24:54.690 AUDIENCE: Oh, I see. 00:24:54.690 --> 00:24:56.592 They have to switch every time. 00:24:56.592 --> 00:24:58.800 PROFESSOR: Tim can't drive for two consecutive edges. 00:25:08.117 --> 00:25:09.950 AUDIENCE: You could use breadth-first search 00:25:09.950 --> 00:25:14.410 and, once you get to the end point, 00:25:14.410 --> 00:25:20.267 or made all paths that are [INAUDIBLE] and then, 00:25:20.267 --> 00:25:23.105 of those paths, go through to see 00:25:23.105 --> 00:25:24.880 which one's the lowest cost? 00:25:24.880 --> 00:25:26.380 PROFESSOR: You're thinking enumerate 00:25:26.380 --> 00:25:28.910 all paths of even length? 00:25:28.910 --> 00:25:30.830 Just to make sure I got it. 00:25:30.830 --> 00:25:32.570 So this is cool because it let's me 00:25:32.570 --> 00:25:34.290 show something else that's cool. 00:25:34.290 --> 00:25:38.320 So, enumerate all paths of even length. 00:25:38.320 --> 00:25:40.160 Something else I heard, enumerate 00:25:40.160 --> 00:25:42.070 all the shortest paths. 00:25:42.070 --> 00:25:44.110 For both of these, let's look at this graph. 00:25:53.830 --> 00:25:57.964 S, T, all the paths have length 1. 00:26:00.910 --> 00:26:02.740 How many ways are there to get from S to T? 00:26:06.010 --> 00:26:06.970 AUDIENCE: Two to three. 00:26:06.970 --> 00:26:07.970 PROFESSOR: Two to three. 00:26:07.970 --> 00:26:10.740 Two ways to go across this diamond. 00:26:10.740 --> 00:26:13.080 Two ways to go across this one. 00:26:13.080 --> 00:26:15.040 Two ways to go across this one. 00:26:15.040 --> 00:26:15.990 Right? 00:26:15.990 --> 00:26:17.520 2 times 2 times 2? 00:26:17.520 --> 00:26:19.270 8. 00:26:19.270 --> 00:26:25.370 If I add another diamond, how many paths? 00:26:25.370 --> 00:26:25.950 One more? 00:26:29.220 --> 00:26:32.600 So the number of even paths, or the number of shortest paths, 00:26:32.600 --> 00:26:38.380 the number of whatever paths I can think about is, order of? 00:26:41.446 --> 00:26:42.200 Come on, guys. 00:26:42.200 --> 00:26:44.421 Math. 00:26:44.421 --> 00:26:44.920 Rrrr! 00:26:44.920 --> 00:26:45.420 Go! 00:26:45.420 --> 00:26:47.294 [LAUGHTER] 00:26:47.294 --> 00:26:49.907 AUDIENCE: 2 to a number of diamonds? 00:26:49.907 --> 00:26:50.740 PROFESSOR: Which is? 00:26:54.700 --> 00:26:58.520 Roughly to the vertices. 00:26:58.520 --> 00:27:01.040 Exponential in the number of vertices. 00:27:01.040 --> 00:27:02.670 Not very good, not very good. 00:27:02.670 --> 00:27:05.660 We can't do this. 00:27:05.660 --> 00:27:07.687 So, it's better that we talked about it here, 00:27:07.687 --> 00:27:09.270 then that you try to do it on the quiz 00:27:09.270 --> 00:27:11.709 and you have to discover this on your own, right? 00:27:11.709 --> 00:27:12.250 Good comment. 00:27:12.250 --> 00:27:14.708 AUDIENCE: Didn't understand what you meant by that lecture. 00:27:14.708 --> 00:27:15.600 Now it makes sense. 00:27:15.600 --> 00:27:17.380 PROFESSOR: That's completely different. 00:27:17.380 --> 00:27:18.610 AUDIENCE: That wasn't the thing he was talking about? 00:27:18.610 --> 00:27:19.920 PROFESSOR: That's different. 00:27:19.920 --> 00:27:22.800 AUDIENCE: OK. 00:27:22.800 --> 00:27:25.070 PROFESSOR: That's where relaxation can go wrong. 00:27:25.070 --> 00:27:27.660 So if you show that if you relax the edges in a random order, 00:27:27.660 --> 00:27:29.840 you can have an exponential number of steps. 00:27:33.077 --> 00:27:35.410 This shows that there are an exponential number of paths 00:27:35.410 --> 00:27:40.161 for any graph, no matter how good your algorithm is. 00:27:40.161 --> 00:27:42.160 So we're not going to be able to enumerate them. 00:27:42.160 --> 00:27:43.201 Let's try something else. 00:27:47.981 --> 00:27:49.945 AUDIENCE: You have another suggestion? 00:27:49.945 --> 00:27:51.890 Just look at all the-- 00:27:51.890 --> 00:27:53.700 PROFESSOR: So, you're doing this. 00:27:53.700 --> 00:27:56.035 You're looking here. 00:27:56.035 --> 00:27:57.800 Everyone's looking here. 00:27:57.800 --> 00:27:58.880 How about this? 00:28:03.236 --> 00:28:04.690 AUDIENCE: We should do that? 00:28:04.690 --> 00:28:05.510 PROFESSOR: Ther's that, right? 00:28:05.510 --> 00:28:06.360 There's a transform. 00:28:06.360 --> 00:28:08.430 You need to transform your graph in some way, 00:28:08.430 --> 00:28:10.040 then run the algorithm. 00:28:10.040 --> 00:28:14.110 And then reading off the outputs should be easy. 00:28:14.110 --> 00:28:16.382 I claim that the hard step is the transform step, not 00:28:16.382 --> 00:28:17.398 the read off step. 00:28:24.982 --> 00:28:26.815 Let me give you one more piece of intuition. 00:28:30.860 --> 00:28:34.130 We're going to keep track of states, somehow. 00:28:34.130 --> 00:28:36.120 Remember Rubik's cube? 00:28:36.120 --> 00:28:38.980 We had one vertex for each state. 00:28:38.980 --> 00:28:43.280 A state representing the configuration of the cube. 00:28:43.280 --> 00:28:48.360 Here, suppose I'm looking at the path from S 00:28:48.360 --> 00:28:51.530 to D. The problem that I have with this 00:28:51.530 --> 00:28:55.480 is that, if I have an algorithm that tells me the shortest 00:28:55.480 --> 00:28:59.980 path from S to D is something, I don't know if it's even 00:28:59.980 --> 00:29:00.820 or if it's odd. 00:29:00.820 --> 00:29:02.990 It's not keeping track of that. 00:29:02.990 --> 00:29:04.090 It's missing some state. 00:29:04.090 --> 00:29:06.450 The state that it's missing is whether the path 00:29:06.450 --> 00:29:09.470 that I used to get here is even or odd length. 00:29:09.470 --> 00:29:10.990 I should do something to the graph 00:29:10.990 --> 00:29:12.396 to keep track of this state. 00:29:16.590 --> 00:29:17.990 AUDIENCE: Can you BFS it once? 00:29:17.990 --> 00:29:23.744 To plan-- You BFS once. 00:29:23.744 --> 00:29:27.076 Then you that B and C are all like 108, 00:29:27.076 --> 00:29:28.980 and then you know that E and D are 208? 00:29:32.210 --> 00:29:36.650 PROFESSOR: B 208 if I take this path. 00:29:36.650 --> 00:29:41.510 It might be 108, might be 208. 00:29:41.510 --> 00:29:42.260 But, this is good. 00:29:42.260 --> 00:29:43.926 This is what you'll do on a quiz, right? 00:29:43.926 --> 00:29:46.680 You come up with an idea and then 00:29:46.680 --> 00:29:48.470 you're going to think about it for a bit. 00:29:48.470 --> 00:29:51.000 If you get stuck, try to convince yourself 00:29:51.000 --> 00:29:52.255 that it's wrong. 00:29:52.255 --> 00:29:56.480 If things get too hard, discard and look somewhere else. 00:29:56.480 --> 00:29:57.780 This is thought process, right? 00:29:57.780 --> 00:30:00.197 You think of something, backtrack, backtrack, 00:30:00.197 --> 00:30:02.530 backtrack, and eventually you get to the right solution. 00:30:13.824 --> 00:30:15.490 Let's go over the solution for this one, 00:30:15.490 --> 00:30:19.870 and then I'll give you a hard one that is sort of like this. 00:30:19.870 --> 00:30:22.970 Then I'll want the solution from you for that one. 00:30:22.970 --> 00:30:25.140 I'm going to draw this graph in a bigger version, 00:30:25.140 --> 00:30:26.660 so I'll have to erase this. 00:30:29.390 --> 00:30:32.650 So this is a cool concept. 00:30:32.650 --> 00:30:36.366 You will be able to do a lot of things with it. 00:30:36.366 --> 00:30:38.302 AUDIENCE: Is this one solvable by Dijkstra? 00:30:38.302 --> 00:30:39.754 PROFESSOR: Ya. 00:30:39.754 --> 00:30:40.713 AUDIENCE: Just curious. 00:30:40.713 --> 00:30:42.420 PROFESSOR: Yeah, we'll run Dijkstra on it 00:30:42.420 --> 00:30:43.353 after we transform it. 00:30:51.570 --> 00:30:52.770 These are big nodes! 00:30:52.770 --> 00:30:53.900 Why am I doing it this way? 00:30:53.900 --> 00:30:57.580 AUDIENCE: (LAUGHS) Is that like, Seattle, Portland, San 00:30:57.580 --> 00:30:58.930 Francisco? 00:30:58.930 --> 00:31:02.000 PROFESSOR: It's because I'm going to need room for copies. 00:31:02.000 --> 00:31:05.750 I'm going to make a copy of each node to keep track of state. 00:31:05.750 --> 00:31:06.594 AUDIENCE: Gosh. 00:31:06.594 --> 00:31:08.010 PROFESSOR: So, instead of one node 00:31:08.010 --> 00:31:10.850 I'll have two nodes for each node. 00:31:10.850 --> 00:31:12.580 And I have two nodes because this 00:31:12.580 --> 00:31:15.700 is what lets me keep track of state, odd path or even path. 00:31:19.000 --> 00:31:21.150 Let me erase the edges because that's not 00:31:21.150 --> 00:31:22.525 how it's going to work, actually. 00:31:27.210 --> 00:31:32.900 Instead of having a node A, I'm going to have a node A even. 00:31:32.900 --> 00:31:38.816 And I'm going to have a copy, A odd. 00:31:38.816 --> 00:31:45.196 E even, I'm going to have to copy, V odd. 00:31:45.196 --> 00:31:50.480 C even, copies to C odd. 00:31:50.480 --> 00:31:57.300 D even, make a copy to D odd. 00:31:57.300 --> 00:31:59.120 Wait, this is E, sorry. 00:31:59.120 --> 00:32:01.808 Sorry, my bad. 00:32:01.808 --> 00:32:09.110 [QUIET TALKING] 00:32:09.110 --> 00:32:15.390 Suppose I got to A using an even length path. 00:32:15.390 --> 00:32:18.900 If I take the edge from A to B, I 00:32:18.900 --> 00:32:21.670 will get to B using an even length path, or an odd length 00:32:21.670 --> 00:32:22.992 path? 00:32:22.992 --> 00:32:24.450 AUDIENCE: Odd. 00:32:24.450 --> 00:32:26.950 PROFESSOR: So, if I get to A using an even length path, then 00:32:26.950 --> 00:32:31.110 from there I'll get to B using an odd length path. 00:32:31.110 --> 00:32:31.890 Sorry, four. 00:32:31.890 --> 00:32:35.204 Using this exact road. 00:32:35.204 --> 00:32:37.414 AUDIENCE: A really messed-up graph. 00:32:37.414 --> 00:32:39.080 PROFESSOR: That's a nice way to name it. 00:32:39.080 --> 00:32:40.410 [WOMAN LAUGHING] 00:32:40.410 --> 00:32:42.510 I've heard worse. 00:32:42.510 --> 00:32:46.240 If I get to A using an odd number of edges, 00:32:46.240 --> 00:32:49.170 how many edges will I have and I get to B? 00:32:49.170 --> 00:32:50.160 AUDIENCE: [INAUDIBLE]. 00:32:53.160 --> 00:32:56.120 PROFESSOR: So this edge became these edges here. 00:32:58.670 --> 00:33:01.240 Now let's do the edge between B and E. 00:33:01.240 --> 00:33:04.730 I get to be using an even number of edges. 00:33:04.730 --> 00:33:08.681 I take B. Do I get to the even or to the odd? 00:33:08.681 --> 00:33:09.264 AUDIENCE: Odd. 00:33:12.010 --> 00:33:14.030 PROFESSOR: I get to B using an odd path. 00:33:17.190 --> 00:33:20.364 What are the lengths of these paths? 00:33:20.364 --> 00:33:20.864 AUDIENCE: 5. 00:33:23.324 --> 00:33:23.990 PROFESSOR: Cool. 00:33:26.870 --> 00:33:28.940 God, this is starting to look ugly, you're right. 00:33:28.940 --> 00:33:30.190 [LAUGHTER] 00:33:30.190 --> 00:33:33.570 Let's do A, C and stop there. 00:33:33.570 --> 00:33:35.180 A even is connected to? 00:33:35.180 --> 00:33:36.870 AUDIENCE: Odd. 00:33:36.870 --> 00:33:40.890 PROFESSOR: C odd, by a path the length. 00:33:40.890 --> 00:33:43.550 And A odd is connected to C even by a path of length. 00:33:47.389 --> 00:33:48.680 Is this starting to make sense? 00:33:54.120 --> 00:33:57.866 So all the edges are going to look like this. 00:33:57.866 --> 00:34:02.090 AUDIENCE: So are those like two independent graphs? 00:34:02.090 --> 00:34:04.120 PROFESSOR: Well, I have edges connecting them. 00:34:04.120 --> 00:34:06.080 But, I like your question. 00:34:06.080 --> 00:34:08.730 The reason this is ugly is because I'm 00:34:08.730 --> 00:34:12.320 looking at a 2D projection of a 3D graph. 00:34:12.320 --> 00:34:15.040 If I had a holographic display, which would be a lot nicer, 00:34:15.040 --> 00:34:17.065 then I would do something along these lines. 00:34:21.750 --> 00:34:22.699 Two pieces of paper. 00:34:25.260 --> 00:34:28.100 I would draw the original graph, and call it the even graph. 00:34:28.100 --> 00:34:29.710 Put it here. 00:34:29.710 --> 00:34:32.409 I would draw the graph again, call it the odd graph, 00:34:32.409 --> 00:34:34.679 and put it here. 00:34:34.679 --> 00:34:36.050 So this is my 2D graph. 00:34:36.050 --> 00:34:38.400 This is my map, the original map. 00:34:38.400 --> 00:34:39.330 And this is state. 00:34:39.330 --> 00:34:40.663 The third dimension is state. 00:34:43.889 --> 00:34:46.909 When I'm at a node, and I used an even number of edges 00:34:46.909 --> 00:34:51.830 to get here, I'm going to go take one edge 00:34:51.830 --> 00:34:55.489 and go to another node using an odd number of edges. 00:34:55.489 --> 00:34:57.390 So all the edges are going to go from here 00:34:57.390 --> 00:34:58.840 to here, and from here to here. 00:35:02.050 --> 00:35:04.130 Does it makes more sense now? 00:35:04.130 --> 00:35:06.550 So the reason that looks ugly is because you're 00:35:06.550 --> 00:35:08.580 seeing the graph like this. 00:35:08.580 --> 00:35:11.240 If you could see like this, and maybe play with it 00:35:11.240 --> 00:35:13.529 a little bit, it would make more sense. 00:35:13.529 --> 00:35:14.570 It wouldn't look as ugly. 00:35:18.150 --> 00:35:20.650 There are two graphs, the even graph and the odd graph. 00:35:20.650 --> 00:35:26.290 And the edges between them represent you doing something. 00:35:26.290 --> 00:35:29.550 When you take the highway, you transition from an even state 00:35:29.550 --> 00:35:32.180 to an odd state, and from an odd state to an even state. 00:35:32.180 --> 00:35:35.840 So that's why the edges are the way they are. 00:35:35.840 --> 00:35:37.720 Now let me see if you guys get it. 00:35:37.720 --> 00:35:39.250 What node do I start from? 00:35:39.250 --> 00:35:41.788 What node do I want to end up in? 00:35:41.788 --> 00:35:45.080 AUDIENCE: A, and you're ending up at B? 00:35:45.080 --> 00:35:45.580 [INAUDIBLE] 00:35:49.860 --> 00:35:51.630 PROFESSOR: OK. 00:35:51.630 --> 00:35:54.526 AUDIENCE: Oh, there's two-- 00:35:54.526 --> 00:35:55.900 PROFESSOR: I heard three answers. 00:35:55.900 --> 00:35:57.700 One of them is correct. 00:35:57.700 --> 00:36:00.940 Do you want to vote or do you guys want to figure it out? 00:36:00.940 --> 00:36:02.025 Talk it out? 00:36:02.025 --> 00:36:03.100 Fight? 00:36:03.100 --> 00:36:04.830 AUDIENCE: A even, E odd. 00:36:07.490 --> 00:36:10.860 PROFESSOR: We start from A even and we end up at E odd. 00:36:10.860 --> 00:36:14.224 Why do I started at A even? 00:36:14.224 --> 00:36:15.890 AUDIENCE: We've seen exactly zero cities 00:36:15.890 --> 00:36:16.370 when we started out. 00:36:16.370 --> 00:36:17.869 PROFESSOR: Yep, so the original path 00:36:17.869 --> 00:36:20.490 has length the 0, which happens to be even. 00:36:20.490 --> 00:36:24.400 And we want to end up with an odd length path. 00:36:24.400 --> 00:36:26.360 So if we go from here to here-- see how 00:36:26.360 --> 00:36:29.800 because we have two nodes out of each original node, 00:36:29.800 --> 00:36:32.386 the path is going to alternate between even and odd. 00:36:32.386 --> 00:36:34.010 And is going to keep track of the state 00:36:34.010 --> 00:36:35.593 that they didn't have before, and it's 00:36:35.593 --> 00:36:38.540 going to just do the right thing. 00:36:38.540 --> 00:36:39.240 It's magic! 00:36:39.240 --> 00:36:41.464 It works! 00:36:41.464 --> 00:36:43.505 Let's see what is the running this new algorithm. 00:36:46.560 --> 00:36:48.570 Say my original graph had V and E. 00:36:48.570 --> 00:36:52.016 How many new vertices do I have? 00:36:52.016 --> 00:36:53.910 AUDIENCE: Two kinds. 00:36:53.910 --> 00:36:56.106 PROFESSOR: How many edges? 00:36:56.106 --> 00:36:56.772 AUDIENCE: Twice? 00:37:01.080 --> 00:37:03.080 PROFESSOR: Each edge is copied exactly twice. 00:37:03.080 --> 00:37:04.580 Which algorithm am I going to use? 00:37:04.580 --> 00:37:06.919 All my weights are positive because they're 00:37:06.919 --> 00:37:08.460 the time it takes to drive somewhere, 00:37:08.460 --> 00:37:09.840 the distance or something. 00:37:09.840 --> 00:37:12.659 So which algorithm do I use? 00:37:12.659 --> 00:37:13.950 AUDIENCE: Breadth-first search? 00:37:19.367 --> 00:37:21.200 PROFESSOR: We can't use breadth-first search 00:37:21.200 --> 00:37:23.587 because you have weight. 00:37:23.587 --> 00:37:26.170 So breadth-first search is not going to find the right answer. 00:37:26.170 --> 00:37:27.220 AUDIENCE: [INAUDIBLE]. 00:37:27.220 --> 00:37:28.970 PROFESSOR: Oh, you want to use this thing. 00:37:28.970 --> 00:37:29.510 AUDIENCE: Yes. 00:37:29.510 --> 00:37:30.250 PROFESSOR: Well, so we don't have 00:37:30.250 --> 00:37:32.780 to worry about that because now we put the shortest path 00:37:32.780 --> 00:37:34.999 algorithm in the black box and we have them. 00:37:34.999 --> 00:37:36.540 AUDIENCE: Oh, so we can use them now. 00:37:36.540 --> 00:37:38.070 PROFESSOR: We use these black boxes. 00:37:38.070 --> 00:37:40.360 But what we need to know is when to use this black box 00:37:40.360 --> 00:37:42.000 and when is this black box. 00:37:42.000 --> 00:37:45.420 Which boxes faster, by the way? 00:37:45.420 --> 00:37:47.904 Please, please give me the right answer. 00:37:47.904 --> 00:37:48.795 AUDIENCE: Dijkstra? 00:37:48.795 --> 00:37:49.670 PROFESSOR: All right! 00:37:49.670 --> 00:37:50.540 Dijkstra is faster. 00:37:50.540 --> 00:37:51.080 [LAUGHTER] 00:37:51.080 --> 00:37:54.000 So whenever we can use Dijkstra we will use Dijkstra. 00:37:54.000 --> 00:37:55.651 When can't we use Dijkstra? 00:37:55.651 --> 00:37:57.400 AUDIENCE: When it's negative, [INAUDIBLE]. 00:38:00.250 --> 00:38:01.440 PROFESSOR: Negative-- 00:38:01.440 --> 00:38:02.190 AUDIENCE: Weights. 00:38:02.190 --> 00:38:05.280 PROFESSOR: Negative weights. 00:38:05.280 --> 00:38:09.100 If I have weights greater or equal to 0, 00:38:09.100 --> 00:38:10.630 I'm happy I can use Dijkstra. 00:38:10.630 --> 00:38:15.030 If I have weights that are smaller than 0, well, whatever. 00:38:15.030 --> 00:38:17.260 It's going to be slower, but I can still solve it. 00:38:17.260 --> 00:38:19.280 AUDIENCE: It's like arbitrary negative weights. 00:38:19.280 --> 00:38:21.070 PROFESSOR: Yep. 00:38:21.070 --> 00:38:22.267 Arbitrary negative weights. 00:38:25.250 --> 00:38:30.214 So all the edges here are positive, so I'm going to use-- 00:38:30.214 --> 00:38:31.105 AUDIENCE: Dijkstra. 00:38:31.105 --> 00:38:31.480 PROFESSOR: Dijkstra! 00:38:31.480 --> 00:38:32.150 Very good. 00:38:32.150 --> 00:38:37.625 So the running time will be, order of-- 00:38:37.625 --> 00:38:39.315 AUDIENCE: V log V? 00:38:39.315 --> 00:38:43.190 PROFESSOR: V log V plus E. If you're a theory person, 00:38:43.190 --> 00:38:47.870 if you're in real life it's E log V. 00:38:47.870 --> 00:38:52.070 This is exactly the running time for the original graph. 00:38:52.070 --> 00:38:54.710 The transformation only increased the graph size 00:38:54.710 --> 00:38:55.850 by a constant factor. 00:38:55.850 --> 00:38:57.619 So, same running time. 00:38:57.619 --> 00:38:58.410 Pretty good, right? 00:39:01.650 --> 00:39:02.900 OK, let's do one more problem. 00:39:02.900 --> 00:39:04.149 Let's do the hard problem now. 00:39:11.110 --> 00:39:13.110 Are you guys getting ready for the hard problem? 00:39:20.477 --> 00:39:22.185 AUDIENCE: If we hadn't seen this probably 00:39:22.185 --> 00:39:23.351 wouldn't have thought of it. 00:39:23.351 --> 00:39:25.390 I wouldn't even think about this. 00:39:25.390 --> 00:39:28.490 PROFESSOR: Well, that's why we taught you this. 00:39:28.490 --> 00:39:29.770 So I wanted to do two things. 00:39:29.770 --> 00:39:32.061 I wanted to show you that trick, which is a cool trick, 00:39:32.061 --> 00:39:35.740 keep track of state by multiplying 00:39:35.740 --> 00:39:37.660 the vertices that you have. 00:39:37.660 --> 00:39:39.620 And I wanted to go through some wrong solutions 00:39:39.620 --> 00:39:41.245 and figure out why they're wrong so you 00:39:41.245 --> 00:39:42.851 can develop an intuition. 00:39:42.851 --> 00:39:44.850 Because, when you're going to get a new problem, 00:39:44.850 --> 00:39:46.790 you're going to think of something. 00:39:46.790 --> 00:39:48.820 Unless you're really good, unless you 00:39:48.820 --> 00:39:51.310 are destined to win a Turing Award of some sort, 00:39:51.310 --> 00:39:53.834 the first solution will probably be wrong. 00:39:53.834 --> 00:39:55.750 Even if you're destined to win a Turing Award, 00:39:55.750 --> 00:39:57.670 the first thing that comes to mind will probably be wrong. 00:39:57.670 --> 00:40:00.135 I'm willing to bet Dijkstra didn't think of Dijkstra 00:40:00.135 --> 00:40:05.010 off the top of his head when he heard about the problem. 00:40:05.010 --> 00:40:07.630 What you want to do is get an intuition. 00:40:07.630 --> 00:40:10.620 So you think of a solution and it starts getting complicated, 00:40:10.620 --> 00:40:12.680 or things start looking wrong, you 00:40:12.680 --> 00:40:15.122 want to back out and think of something else. 00:40:15.122 --> 00:40:17.580 The more things we can consider, higher chances that you're 00:40:17.580 --> 00:40:19.430 going to stumble upon the correct solution. 00:40:19.430 --> 00:40:22.070 So that's why we are building an intuition for bad solutions. 00:40:22.070 --> 00:40:23.800 [LAUGHTER] 00:40:23.800 --> 00:40:25.700 I'm not just saying, hey, give me a solution 00:40:25.700 --> 00:40:26.660 and now I'll tell you it's wrong. 00:40:26.660 --> 00:40:28.710 It's not just to embarrass you or something. 00:40:28.710 --> 00:40:30.251 There's that intuition building step. 00:40:30.251 --> 00:40:32.770 That's really important. 00:40:32.770 --> 00:40:36.224 There's this network that we keep talking about. 00:40:36.224 --> 00:40:38.140 There's this highway network, except this time 00:40:38.140 --> 00:40:39.306 it's a bit more complicated. 00:40:42.480 --> 00:40:46.950 So for each edge I have two things. 00:40:46.950 --> 00:40:50.080 I have a fuel cost, which is constant. 00:40:50.080 --> 00:40:54.130 A fuel cost is a function of the length of the road. 00:40:54.130 --> 00:40:58.240 But now, I have these realistic highways where I have traffic. 00:40:58.240 --> 00:41:01.700 If you try to go from Boston to New York, 2 AM? 00:41:01.700 --> 00:41:03.140 Three hours. 00:41:03.140 --> 00:41:03.850 Rush hour? 00:41:03.850 --> 00:41:05.100 Six hours. 00:41:05.100 --> 00:41:07.910 So we have to keep track of this in some way. 00:41:07.910 --> 00:41:10.910 Well we're going to split up the day into minutes. 00:41:10.910 --> 00:41:12.750 Say you have 10 minutes in a day. 00:41:12.750 --> 00:41:15.340 Can anyone tell me what M is, really quickly? 00:41:15.340 --> 00:41:18.810 Not a number, a formula, something. 00:41:18.810 --> 00:41:23.625 AUDIENCE: What's 3,600 times 24? 00:41:23.625 --> 00:41:27.090 And that's seconds. 00:41:27.090 --> 00:41:28.580 PROFESSOR: All right. 00:41:28.580 --> 00:41:30.526 So this is how many minutes in a day. 00:41:36.330 --> 00:41:39.720 This is how many minutes we have in a day, right? 00:41:39.720 --> 00:41:42.610 For each edge, we're going to have a function that's 00:41:42.610 --> 00:41:44.906 the time cost of the edge. 00:41:44.906 --> 00:41:48.800 So it's the time cost of this edge, and it's going to say, 00:41:48.800 --> 00:41:50.820 if I start at a certain time, it's 00:41:50.820 --> 00:41:54.510 going to take this many minutes to go across the edge. 00:41:54.510 --> 00:41:57.800 For each highway I know how much time 00:41:57.800 --> 00:42:01.610 it's going to take to go across it, given when I start. 00:42:01.610 --> 00:42:04.990 And I know much fuel I'm going to consume. 00:42:04.990 --> 00:42:06.720 By the way, edges are directed now, just 00:42:06.720 --> 00:42:08.660 to make our life easy. 00:42:08.660 --> 00:42:13.500 This graph is going to be like this. 00:42:13.500 --> 00:42:15.464 Let's see if I can get this right. 00:42:15.464 --> 00:42:16.380 AUDIENCE: [INAUDIBLE]. 00:42:19.495 --> 00:42:20.870 PROFESSOR: So the real reason I'm 00:42:20.870 --> 00:42:26.530 having this is the time to go across the highway 00:42:26.530 --> 00:42:29.090 might be different, depending which way you go. 00:42:29.090 --> 00:42:29.670 AUDIENCE:Oh-- 00:42:29.670 --> 00:42:31.320 PROFESSOR: Getting into Boston in the morning 00:42:31.320 --> 00:42:33.240 versus getting into Boston in the evening. 00:42:33.240 --> 00:42:34.550 Which one's easier? 00:42:34.550 --> 00:42:37.392 Sorry, versus getting out of Boston in the morning. 00:42:37.392 --> 00:42:39.100 Intuitively, I'd say it's probably harder 00:42:39.100 --> 00:42:41.770 to get in in the morning because people are going to work. 00:42:41.770 --> 00:42:42.960 And it's easier to get out. 00:42:42.960 --> 00:42:44.540 But ya, I don't know either. 00:42:44.540 --> 00:42:46.916 AUDIENCE: The 95 one's a hurdle. 00:42:46.916 --> 00:42:47.720 [LAUGHTER] 00:42:47.720 --> 00:42:48.886 PROFESSOR:We have an answer. 00:42:48.886 --> 00:42:51.470 [LAUGHTER] 00:42:51.470 --> 00:42:53.160 Our edges are going to be oriented. 00:42:56.990 --> 00:43:00.255 I want to find an itinerary that satisfies two constraints. 00:43:00.255 --> 00:43:03.530 AUDIENCE: [INAUDIBLE]. 00:43:03.530 --> 00:43:05.810 PROFESSOR: Ya, that's not good, right? 00:43:05.810 --> 00:43:07.442 New York and New York. 00:43:07.442 --> 00:43:10.181 [WOMAN LAUGHING] 00:43:10.181 --> 00:43:12.430 So I want to see how fast I can get to my destination. 00:43:12.430 --> 00:43:14.740 That's my top priority. 00:43:14.740 --> 00:43:20.580 If I get to New York at 5:00 PM, I want to get there at 5:00 PM. 00:43:20.580 --> 00:43:23.120 I don't want to get there at 5:01, for example. 00:43:23.120 --> 00:43:25.952 But if I have three ways to get there at 5:00 PM, 00:43:25.952 --> 00:43:28.900 I want to choose the way that's the most eco-friendly. 00:43:28.900 --> 00:43:32.370 So, the least amount of fuel. 00:43:32.370 --> 00:43:35.350 So out of all possible ways, the fastest way? 00:43:35.350 --> 00:43:38.170 Out of all possible fastest ways, 00:43:38.170 --> 00:43:39.880 the way that consumes the least fuel? 00:43:42.690 --> 00:43:44.740 Go. 00:43:44.740 --> 00:43:48.470 You have two minutes. 00:43:48.470 --> 00:43:49.020 Maybe five. 00:43:49.020 --> 00:43:52.130 AUDIENCE: And we have the same costs for the-- 00:43:52.130 --> 00:43:55.440 PROFESSOR: Each edge always has the same fuel cost. 00:43:55.440 --> 00:43:58.010 But, depending on when you start, going across it 00:43:58.010 --> 00:44:00.440 is going to take a different time, too. 00:44:00.440 --> 00:44:02.720 AUDIENCE: And we don't know what those times are? 00:44:02.720 --> 00:44:05.076 PROFESSOR: They're going to be different for each edge. 00:44:05.076 --> 00:44:07.086 AUDIENCE: One guess, take Dijkstra. 00:44:07.086 --> 00:44:09.134 Find the shortest path length and we 00:44:09.134 --> 00:44:11.872 look for all path lengths that are that length. 00:44:11.872 --> 00:44:13.150 I don't know. 00:44:13.150 --> 00:44:15.760 PROFESSOR: So we have the issue that-- I might still have it. 00:44:15.760 --> 00:44:16.830 Oh no, never mind. 00:44:16.830 --> 00:44:19.251 Sorry, I erased it so it's not on your mind anymore. 00:44:19.251 --> 00:44:20.750 We have those diamonds that show you 00:44:20.750 --> 00:44:23.240 that there might be an exponential number of paths 00:44:23.240 --> 00:44:25.274 with the same length, or the same distance. 00:44:25.274 --> 00:44:26.815 AUDIENCE: And it doesn't help that we 00:44:26.815 --> 00:44:28.995 know we should cut off the-- 00:44:28.995 --> 00:44:30.150 PROFESSOR: Nope. 00:44:30.150 --> 00:44:32.750 So we need to do the process that I just erased. 00:44:32.750 --> 00:44:35.384 Transform the graph in. 00:44:35.384 --> 00:44:37.300 The thing is, now you're missing state, right? 00:44:37.300 --> 00:44:38.220 This misses state. 00:44:38.220 --> 00:44:40.180 This tells you how many edges you 00:44:40.180 --> 00:44:42.360 have to get from source to destination. 00:44:42.360 --> 00:44:46.290 But we're not keeping track of some state that these vital. 00:44:46.290 --> 00:44:48.980 Transform the graph to keep track of state, 00:44:48.980 --> 00:44:52.590 run, say Dijkstra, and then interpret the output. 00:44:55.800 --> 00:44:56.777 What state do we need? 00:44:56.777 --> 00:44:57.610 Let's think of that. 00:44:57.610 --> 00:45:00.182 AUDIENCE: Fuel cost, I think. 00:45:00.182 --> 00:45:00.890 PROFESSOR: Sorry? 00:45:00.890 --> 00:45:04.137 AUDIENCE: The two costs is the same for each edge. 00:45:04.137 --> 00:45:06.482 It's just the time is different. 00:45:06.482 --> 00:45:07.889 To be in traffic. 00:45:07.889 --> 00:45:12.370 AUDIENCE: But once we have found the fastest spot 00:45:12.370 --> 00:45:17.202 that we wanted to see cut into our fuel cost range 00:45:17.202 --> 00:45:20.424 and which one do we want to take. 00:45:24.710 --> 00:45:27.900 Thinking about the priority, our priority is time. 00:45:27.900 --> 00:45:34.983 So we want to save fuels as a state, to check it? 00:45:38.447 --> 00:45:40.030 PROFESSOR: The problem is I don't even 00:45:40.030 --> 00:45:41.363 know if fuel costs are integers. 00:45:41.363 --> 00:45:44.550 I can't keep fuel as a state. 00:45:44.550 --> 00:45:45.769 How many copies would I have? 00:45:45.769 --> 00:45:47.310 For each vertex how many copies would 00:45:47.310 --> 00:45:48.490 I have if fuel is a state? 00:45:48.490 --> 00:45:51.160 Who knows? 00:45:51.160 --> 00:45:53.160 So let's try something else. 00:45:53.160 --> 00:45:55.070 But, you're on the right track. 00:45:55.070 --> 00:45:56.975 Let's take a variable and make it state. 00:46:01.830 --> 00:46:05.923 AUDIENCE: So for every vertex can you keep 00:46:05.923 --> 00:46:15.480 track of what's the shortest time it took to get there? 00:46:15.480 --> 00:46:19.105 PROFESSOR: What's the shortest time it took to get there? 00:46:19.105 --> 00:46:20.480 How would you keep track of that? 00:46:20.480 --> 00:46:22.398 Would you make that state, or how 00:46:22.398 --> 00:46:23.606 would you keep track of that? 00:46:27.377 --> 00:46:28.210 I'm not disagreeing. 00:46:28.210 --> 00:46:29.520 I'm trying to understand your-- 00:46:29.520 --> 00:46:36.292 AUDIENCE: So for example, let's say if you have B, right? 00:46:36.292 --> 00:46:41.380 Let's say it takes like one hour to go from A to B. 00:46:41.380 --> 00:46:44.590 PROFESSOR: It depends when you start, by the way. 00:46:44.590 --> 00:46:47.020 How much time it takes depends on when you start. 00:46:47.020 --> 00:46:48.940 If you start at 8:00 AM, it might be an hour. 00:46:48.940 --> 00:46:53.980 If you start at 9:00 AM it might be two hours. 00:46:53.980 --> 00:46:57.600 AUDIENCE: Do you need to have a finite number of states? 00:46:57.600 --> 00:46:59.380 PROFESSOR: It's nice if it's not infinite, 00:46:59.380 --> 00:47:02.110 but it doesn't have to be constant, like we had before. 00:47:02.110 --> 00:47:05.690 It can be more than constant, for sure. 00:47:05.690 --> 00:47:07.506 So what are you thinking? 00:47:07.506 --> 00:47:09.740 AUDIENCE: I want to use time as a state. 00:47:09.740 --> 00:47:11.380 PROFESSOR: I like that idea. 00:47:11.380 --> 00:47:13.880 Let's try to do it. 00:47:13.880 --> 00:47:16.180 We're going to use time as a state. 00:47:16.180 --> 00:47:20.070 How many vertices am I going to have for each vertex? 00:47:20.070 --> 00:47:22.060 So for each original vertex, how many vertices 00:47:22.060 --> 00:47:23.518 am I going to have in my new graph? 00:47:26.046 --> 00:47:27.545 AUDIENCE: Three or four [INAUDIBLE]. 00:47:31.150 --> 00:47:33.950 AUDIENCE: The sum of all times? 00:47:33.950 --> 00:47:38.010 PROFESSOR: So if I say, hey, I'm at this vertex at this time, 00:47:38.010 --> 00:47:39.840 then I'm going to have M vertices 00:47:39.840 --> 00:47:42.230 for each original vertex, right? 00:47:42.230 --> 00:47:43.190 AUDIENCE: Right. 00:47:43.190 --> 00:47:46.540 PROFESSOR: I promise that the resolution of time is minutes. 00:47:46.540 --> 00:47:48.780 I promise that this thing gives you 00:47:48.780 --> 00:47:51.055 an integer number of minutes. 00:47:51.055 --> 00:47:53.920 AUDIENCE: Can it be put into days, though? 00:47:56.597 --> 00:47:58.930 PROFESSOR: Suppose we can get from source to destination 00:47:58.930 --> 00:48:01.257 in one day. 00:48:01.257 --> 00:48:03.090 So let's see how it builds this graph first, 00:48:03.090 --> 00:48:06.440 and then we can figure out the rest of the things. 00:48:06.440 --> 00:48:10.810 So for each node I'm going to make M copies of it. 00:48:10.810 --> 00:48:13.670 Suppose in my original graph I had a node. 00:48:13.670 --> 00:48:16.150 I'm going to have M copies of that node. 00:48:16.150 --> 00:48:18.820 And each copy has the original vertex, 00:48:18.820 --> 00:48:20.900 and the time when I'm there. 00:48:20.900 --> 00:48:21.400 Right? 00:48:21.400 --> 00:48:23.210 There are V of these. 00:48:23.210 --> 00:48:27.570 And there are M of these. 00:48:27.570 --> 00:48:28.408 Yes? 00:48:28.408 --> 00:48:29.783 AUDIENCE: It totally makes sense. 00:48:29.783 --> 00:48:31.400 It's a great idea. 00:48:31.400 --> 00:48:32.810 PROFESSOR: Yeah, you had it! 00:48:32.810 --> 00:48:33.880 Great idea. 00:48:33.880 --> 00:48:34.660 So far so good. 00:48:34.660 --> 00:48:35.160 [LAUGHTER] 00:48:35.160 --> 00:48:36.785 AUDIENCE: Now they need to be connected 00:48:36.785 --> 00:48:38.180 to the appropriate next one. 00:48:38.180 --> 00:48:41.490 PROFESSOR: Let's hear how we do that. 00:48:41.490 --> 00:48:48.370 Suppose we have an edge from U to V. 00:48:48.370 --> 00:48:49.990 How are going to connect this? 00:48:49.990 --> 00:48:52.810 How are we going to transform this? 00:48:52.810 --> 00:48:54.850 How many edges are going to make from that edge? 00:49:01.244 --> 00:49:03.452 AUDIENCE: Do you have to do it from every start time? 00:49:03.452 --> 00:49:05.290 I guess. 00:49:05.290 --> 00:49:08.490 So you'd write M edges. 00:49:08.490 --> 00:49:09.360 PROFESSOR: M edges. 00:49:09.360 --> 00:49:16.216 From U and a start time to V and what? 00:49:16.216 --> 00:49:19.138 AUDIENCE:T plus T C of T. 00:49:19.138 --> 00:49:22.880 PROFESSOR: T plus T C of going through that edge. 00:49:22.880 --> 00:49:23.529 Right? 00:49:23.529 --> 00:49:24.195 AUDIENCE: Right. 00:49:24.195 --> 00:49:27.864 PROFESSOR: T C are going from U to V at the stop. 00:49:32.640 --> 00:49:35.070 Ya, like this. 00:49:35.070 --> 00:49:36.260 So you start from a time. 00:49:36.260 --> 00:49:39.000 So the edge points from the time that you start up 00:49:39.000 --> 00:49:42.610 until the time when you'd be able to finish. 00:49:42.610 --> 00:49:45.800 So, up until the time where you'd be off the highway 00:49:45.800 --> 00:49:47.480 and die in the next city. 00:49:47.480 --> 00:49:49.250 And we need one more type of edges. 00:49:49.250 --> 00:49:51.890 There's one more tiny trick. 00:49:51.890 --> 00:49:55.170 If I'm somewhere in Massachusetts, and I know that 00:49:55.170 --> 00:49:56.470 it's really bad now. 00:49:56.470 --> 00:49:59.110 I prefer to wait out for a few minutes, go to a bar, 00:49:59.110 --> 00:50:02.640 and then come and drive-- OK not to a bar-- go to a food place, 00:50:02.640 --> 00:50:03.380 then drive later. 00:50:03.380 --> 00:50:04.850 [LAUGHTER] 00:50:04.850 --> 00:50:06.570 AUDIENCE: --vertical edges that-- 00:50:06.570 --> 00:50:08.404 PROFESSOR: --represent waiting. 00:50:08.404 --> 00:50:10.990 AUDIENCE: You have to add one minute between them. 00:50:10.990 --> 00:50:14.220 PROFESSOR: How do I do that? 00:50:14.220 --> 00:50:18.000 AUDIENCE: From U to U with a cost of 1. 00:50:21.360 --> 00:50:26.126 PROFESSOR: So from U to U at 20 plus 1. 00:50:26.126 --> 00:50:29.170 I like this better. 00:50:29.170 --> 00:50:31.730 What's the fuel cost of this? 00:50:31.730 --> 00:50:33.810 AUDIENCE: Zero. 00:50:33.810 --> 00:50:36.010 PROFESSOR: So here this edge had fuel cost at C, 00:50:36.010 --> 00:50:37.510 Then the new edge is going have fuel 00:50:37.510 --> 00:50:41.514 cost for the ones on the top. 00:50:41.514 --> 00:50:43.180 AUDIENCE: Is it proportional to minutes? 00:50:43.180 --> 00:50:44.600 Speed? 00:50:44.600 --> 00:50:46.970 PROFESSOR: It's the same. 00:50:46.970 --> 00:50:49.860 F C stays the same, no matter when we go through the highway 00:50:49.860 --> 00:50:54.320 if C is a function of the distance of the road. 00:50:54.320 --> 00:50:58.380 This edge becomes M edges with the same cost. 00:50:58.380 --> 00:51:00.690 And then I have to have vertical waiting edges. 00:51:03.840 --> 00:51:05.360 If we had the holographic display 00:51:05.360 --> 00:51:09.140 that I talked about earlier you'd have M sheets of paper 00:51:09.140 --> 00:51:10.430 this time. 00:51:10.430 --> 00:51:12.300 You start at time 0 at your source 00:51:12.300 --> 00:51:17.350 and then your edges go represent the moves that you can make. 00:51:17.350 --> 00:51:20.360 So, you could start in Boston at 8:00 AM. 00:51:20.360 --> 00:51:22.570 And you could take I-90 and end up 00:51:22.570 --> 00:51:26.850 in-- I don't know Massachusetts city names-- Albuquerque 00:51:26.850 --> 00:51:29.420 at 9:00 AM? 00:51:29.420 --> 00:51:31.080 [LAUGHTER] 00:51:31.080 --> 00:51:32.000 AUDIENCE: Amherst. 00:51:32.000 --> 00:51:32.980 PROFESSOR: Amherst, OK. 00:51:32.980 --> 00:51:34.354 So you started in Boston 8:00 AM, 00:51:34.354 --> 00:51:37.240 you end up at Amherst at 9, 10:00 AM. 00:51:37.240 --> 00:51:37.740 Right? 00:51:37.740 --> 00:51:39.450 So this is one edge. 00:51:39.450 --> 00:51:41.490 The edges represent the moves that you 00:51:41.490 --> 00:51:42.810 could make in this graph. 00:51:46.110 --> 00:51:47.230 Does this makes? 00:51:47.230 --> 00:51:51.475 AUDIENCE: I don't understand the two last lines there. 00:51:51.475 --> 00:51:54.427 I don't really get what we were doing there [INAUDIBLE]. 00:51:54.427 --> 00:51:56.510 PROFESSOR: Let's see what we're trying to do here. 00:51:56.510 --> 00:52:01.270 We're saying that, if I'm a node U, and I'm starting at time T. 00:52:01.270 --> 00:52:03.770 AUDIENCE: Ya, so that's like one piece of paper. 00:52:03.770 --> 00:52:05.340 PROFESSOR: Yep. 00:52:05.340 --> 00:52:07.250 I'm going to go on the road, right? 00:52:07.250 --> 00:52:09.950 I'm going to go on the road from U to V. 00:52:09.950 --> 00:52:12.064 So where am I going to arrive? 00:52:12.064 --> 00:52:14.630 AUDIENCE: There at V at some time plus that. 00:52:14.630 --> 00:52:16.546 PROFESSOR: Now we have to figure out what time 00:52:16.546 --> 00:52:17.620 we're going to arrive at. 00:52:17.620 --> 00:52:20.610 The time that we arrive at is the original time, 00:52:20.610 --> 00:52:23.520 plus whatever time I'm going to spend on the road. 00:52:23.520 --> 00:52:25.040 What's the time I spend on the road? 00:52:25.040 --> 00:52:27.940 It's not constant because I have that timetable that 00:52:27.940 --> 00:52:29.820 includes traffic. 00:52:29.820 --> 00:52:31.680 So this is what this tells me. 00:52:31.680 --> 00:52:34.680 This is what this big, ugly formula is all about. 00:52:34.680 --> 00:52:37.450 It says the timetable for this edge assuming 00:52:37.450 --> 00:52:39.950 you start at time T. That's all there is. 00:52:39.950 --> 00:52:40.635 Nothing else. 00:52:40.635 --> 00:52:42.260 AUDIENCE: What's the second line, then? 00:52:42.260 --> 00:52:44.490 PROFESSOR: So the second line is waiting. 00:52:44.490 --> 00:52:47.870 So if I don't have this then I'm constrained 00:52:47.870 --> 00:52:50.760 in that I have to drive all the time. 00:52:50.760 --> 00:52:52.804 I go from here to Amherst, then I 00:52:52.804 --> 00:52:55.220 have to go from Amherst to somewhere else, and keep going, 00:52:55.220 --> 00:52:56.178 keep going, keep going. 00:52:56.178 --> 00:52:59.270 AUDIENCE: [INAUDIBLE]. 00:52:59.270 --> 00:53:01.850 PROFESSOR: Yeah, this is where I wait for a minute. 00:53:01.850 --> 00:53:03.940 If I wait for a minute, I don't consume any fuel 00:53:03.940 --> 00:53:07.049 and I go from time T to time T plus 1. 00:53:07.049 --> 00:53:08.320 AUDIENCE: OK. 00:53:08.320 --> 00:53:10.860 PROFESSOR: Does this make sense? 00:53:10.860 --> 00:53:13.110 Do we want to analyze the running time for this really 00:53:13.110 --> 00:53:14.060 quickly? 00:53:14.060 --> 00:53:14.917 AUDIENCE: Sure. 00:53:14.917 --> 00:53:16.000 PROFESSOR: How many edges? 00:53:19.325 --> 00:53:21.230 AUDIENCE: E times M? 00:53:21.230 --> 00:53:24.401 PROFESSOR: OK, let's do vertices because vertices is quick. 00:53:24.401 --> 00:53:26.190 AUDIENCE: V times M? 00:53:26.190 --> 00:53:31.430 PROFESSOR: So, edges is, you're saying, V times M? 00:53:31.430 --> 00:53:31.970 Almost. 00:53:31.970 --> 00:53:33.039 AUDIENCE: Pause there. 00:53:33.039 --> 00:53:34.580 PROFESSOR: So there's a pause, right? 00:53:34.580 --> 00:53:38.440 How many pause edges do I have? 00:53:38.440 --> 00:53:45.154 AUDIENCE: V-- M. 00:53:45.154 --> 00:53:46.695 PROFESSOR: This is how many vertices? 00:53:46.695 --> 00:53:48.380 This is how many edges? 00:53:48.380 --> 00:53:52.255 Plug this into which algorithm, Dijkstra or Bellam-Ford? 00:53:52.255 --> 00:53:54.170 AUDIENCE: Dijkstra. 00:53:54.170 --> 00:53:56.460 PROFESSOR: Done! 00:53:56.460 --> 00:53:58.501 Yay, we solved the hard problem. 00:53:58.501 --> 00:54:00.810 AUDIENCE: How do you ensure that the time's right? 00:54:00.810 --> 00:54:03.305 Do you have to go through and see if there's a path? 00:54:03.305 --> 00:54:04.930 PROFESSOR: How do we read the solution? 00:54:04.930 --> 00:54:05.430 That's good. 00:54:05.430 --> 00:54:06.388 That's a good question. 00:54:06.388 --> 00:54:07.290 I like that. 00:54:07.290 --> 00:54:11.100 So, we're going to have M vertices of the destination. 00:54:11.100 --> 00:54:12.450 Alright? 00:54:12.450 --> 00:54:14.250 Let's see how they're going to look like. 00:54:14.250 --> 00:54:15.800 First off, if you start at 8:00 AM, 00:54:15.800 --> 00:54:19.810 maybe you're not going to make it to New York at 8:01 AM. 00:54:19.810 --> 00:54:22.010 The vertex that says New York at 8:01 AM 00:54:22.010 --> 00:54:26.180 is probably going to have a cost of plus infinity. 00:54:26.180 --> 00:54:27.887 First off the really early times are 00:54:27.887 --> 00:54:29.470 going to have a cost of plus infinity. 00:54:29.470 --> 00:54:30.770 Not going to happen. 00:54:30.770 --> 00:54:34.310 Then, at some point, the cost is going to become finite. 00:54:34.310 --> 00:54:36.910 That's the fastest way you can get from Boston to New York. 00:54:39.740 --> 00:54:41.800 using the least amount of fuel. 00:54:41.800 --> 00:54:45.490 So the cost that you have there is the answer to our problem. 00:54:45.490 --> 00:54:47.590 AUDIENCE: So basically, when you get there you 00:54:47.590 --> 00:54:49.550 have to enter it through all M. 00:54:49.550 --> 00:54:52.210 PROFESSOR: All M vertices that correspond to the destination. 00:54:56.917 --> 00:54:58.500 The costs are going to look like this. 00:54:58.500 --> 00:55:01.732 They're going to be infinity, infinity, infinity all the way 00:55:01.732 --> 00:55:03.190 up, until some point in here you're 00:55:03.190 --> 00:55:06.110 going to have your final answer. 00:55:06.110 --> 00:55:08.100 The cost to get there the fastest-- 00:55:08.100 --> 00:55:10.940 see, you can get there at 3:00 PM. 00:55:10.940 --> 00:55:13.850 This is how much fuel you have to spend to get there at 3;00 00:55:13.850 --> 00:55:14.690 PM. 00:55:14.690 --> 00:55:18.410 But, if you're willing to wait until 3:01 PM, 00:55:18.410 --> 00:55:21.500 you're going to have the fastest cost you can have for that. 00:55:21.500 --> 00:55:24.500 If you're willing to wait until 3:02, 00:55:24.500 --> 00:55:26.650 you're going to have the answer for that, too. 00:55:26.650 --> 00:55:29.550 So here you're going to get the whole trade-off curve of, 00:55:29.550 --> 00:55:31.384 if you're willing to wait for a few minutes, 00:55:31.384 --> 00:55:33.508 or if you're willing to wait for an extra hour, how 00:55:33.508 --> 00:55:34.470 much fuel you can save. 00:55:38.810 --> 00:55:43.270 So, I think that's cool about your questions. 00:55:43.270 --> 00:55:44.600 You had a question, too. 00:55:44.600 --> 00:55:47.064 AUDIENCE: Why did we have V times M 00:55:47.064 --> 00:55:48.910 in the E prime expression? 00:55:50.972 --> 00:55:52.930 PROFESSOR: First off, we have E times M, right? 00:55:52.930 --> 00:55:54.660 We're good with these. 00:55:54.660 --> 00:55:56.710 These are the waiting edges. 00:55:56.710 --> 00:56:01.400 AUDIENCE: So we are waiting after every minute? 00:56:01.400 --> 00:56:04.370 PROFESSOR: The waiting edges are vertex time 00:56:04.370 --> 00:56:05.890 to vertex time plus 1. 00:56:05.890 --> 00:56:08.270 How many vertices? 00:56:08.270 --> 00:56:09.913 V. How many times? 00:56:09.913 --> 00:56:11.120 AUDIENCE: M. 00:56:11.120 --> 00:56:15.866 PROFESSOR: So it's V times M. Because at each minute, 00:56:15.866 --> 00:56:17.490 you can be at a certain city and decide 00:56:17.490 --> 00:56:19.860 to wait one more minute to stay the same city. 00:56:23.130 --> 00:56:25.935 OK Does this make sense? 00:56:25.935 --> 00:56:26.560 Thank you guys! 00:56:26.560 --> 00:56:28.800 I'm really happy we solved the hard problem! 00:56:28.800 --> 00:56:30.520 That's good.