Beautiful morning in the beach. Good morning. Hi, I'm Lavi. And I'm Oli. And this is our hero, Bumblebee. Together, we are attempting a Guinness World Record to become the youngest pair to circumnavigate the globe by motorcycle. Join us for season three here in South America. Good morning, world. Welcome back to the channel. It's day number 291 on our circumnavigation around the globe by motorcycle. Good morning, Bumblebee. Good morning, Bumblebee. We're here camping in front of the beach next to this little village called Pichacoy, just off of the Ruta 5 highway in Chile. Yesterday, we finally arrived here on the Pacific coast of South America. Yesterday was our first time seeing the Pacific ocean, so we decided why not camp just right in front of it. Yes, we had a really peaceful night and the night sky was just amazing here. And now that we've hit the west coast, it's time to make our way north. So let me show you guys where we're heading today. So here's Santiago and we are here on the coast. Over the next three days, we're trying to make it onto the next page to Antofagasta, but today we are gonna be heading up along the coast here past La Serena and trying to make it somewhere between La Serena and Valenar. Our aim is to get as far north as we can today so we don't know exactly where we're gonna end up or where we're gonna camp tonight, but it's already 7.15 and we have to pack up everything, so better hit the road, let's go. It's a beautiful morning at the beach. Sandy start. Oh, yeah, look at this. Traction by... Analyst, I guess. I liked it, traction by Dunlop. Oh, I miss it too. I can't fault it at the moment. I mean, it seems to be doing fine with this terrain. Yeah, absolutely. It's nice and smooth, eh? Yeah, good old Analyst cat-prex. Always a little bit tricky in the morning. Yeah, if you wanna get to the nice wild camps, you gotta go through a little bit of difficult roads. Get a little bit more. Yeah. And you know, I have been watching those Dakar rally videos, so... Skill level reached. All right. All right. Here we go. Coming back onto the Ruta 5 highway. From now on, it's easy. 725 miles to Anteuphogasta. Woo, let's go. Ha, ha, ha. Yeah, it's cool. And I was actually really scared that this part will be really, really hot, but we checked the temperature this morning and it's just about 25 degrees during the day, so it's really, really nice riding temperature. I'm so happy about that. In Santiago, it was like 35 degrees. Yes. So it's nearly 10 degrees colder here on the coast. And right now it's only 16 degrees, so... So it really struck me, a quite nice ride today. We just stopped here at the rest area at the service station and it's really cool because here in Chile, on some rest areas, they have tubs here where you can fill up water and it's drinkable water. So I thought I'd use the opportunity and fill up all our water bottles because we are using the water for cooking, brushing the teeth, drinking, of course. So we're having one and a half liter here. And then we have this camel bags here, which are two liters. So we have about five to six liters water with us and this will just last us for the day, actually. So we have filled up for the day. Perfect. Yummy. 1.5 kilos of granola, finished. That's a lot of granola. That was the biggest bag ever. We just saw a big sign saying queso, cheese. It looks like they are open. Yeah, it does look like they are open. Awesome. Queso! Hola, buenas. Oh, queso. Hola. ¿Cuánto cuesta una? Esto está bien. Oh, okay. ¿Es diferente, queso? Tienen mucho mas vía. I think it might be like a mature and a fresh. Cool. Muchas gracias. Is that? Es cabra. I think it's quite nice. Yes, is it so nice? Yeah. Perfect. Hello. Hello. Hello. We got ourselves a nice circle of cheese, a whole block. And so that was 8,000. So 8 pounds. Cool. When we stopped, we weren't exactly expecting goat's cheese, but you know, we'll roll with it. We'll see how it goes. I mean, it's like a little bit of variation, hey? There we go. Adios. He's just chilling. Look at this guy. We have some more of these cactus hedges that we saw yesterday coming up here. And some of them flowering. No way. Yes. Is that a flower? In between you can see some of them having like red flowers, I think. Oh, beautiful. Or something red. I don't know if these are flowers or not. I don't know. Yeah, there's some beautiful cactus covered hills around us. Very nice. Yeah, and the ride so far is pretty chilled. Yeah, we've managed to do 122 miles already this morning. And it's not even midday, so we're doing incredibly well. Yes. And the temperature is just 23 degrees. So it's just the perfect riding day, really. Ah yeah, the petrol station is coming up as well. Maybe it's time for a little lunch break as well for us, hey? Some breaded cheese. Test out this new goat's cheese. This is one of the only fuel stations for like another 100 miles or so. So it's always good as we're getting a bit more remote to just fill up whenever we can. Yeah. So did you tie up your hair then? Oh my god. Oh my god. The real purpose of their hair is revealed. We bought some butter in Santiago. And the butter is just everywhere. Like, it's so melted. Look at this mess. So we have to use it up now because it just gets hotter and hotter. And then we have some nice avocado on top. And of course, our new cheese. So it will be a really nice feast. Whoa, look at this. I really didn't expect it, but it's super hot. It's like parmesan cheese. Yeah, it's mature, hey? Wow. Have you ever had mature goat's cheese before? No. No, me neither. Bon appetit. Woo hoo hoo hoo. Just coming up to La Serena. This is one of the bigger towns or cities in this area. Yeah, it was actually really unexpected. We went just like over the hill and then there was this massive town here popping up. Whoa, and they've got some high-rise condos here. Look at this. Directly on the ocean. A lane just for Bumblebee. I mean, it's a really big town here, actually. We're doing miles down for today. So we're doing really well. I saw a sign over here at the petrol station for some papaya cheese. Hugo de Papayas Helados. Cold papaya juice. Dos, por favor. Two. Two. 2,000. 2,000. 2,000, yeah, it's perfect. And what do you have here? Is this also papaya? And where do they grow the papaya? Because here's desert. It's from all the La Serena cities. Yeah, crazy. La Menteca. Beautiful. Where are you from? England. England? England. I learned English in La Serena. I learned English. 10 years ago, I went to Germany. No way. Especially in Germany? No, in Israel. No. I was in England. We are from the Papayas. Yes. Nice to meet you. Thank you. Yes, yes. Papaya juice. Look how healthy we are. I'm not sure. It tastes like they add sugar to this. It tastes really nice. It's beautiful. But I think it might be some papaya and then sugar and water. I'm not sure. Very good. Thank you so much. Thank you. Amazing. Gracias. Ciao. Yeah, that was a really lovely papaya juice. I liked it. Yeah, and now we are rested. We're ready for the final leg of the day. So we want to get somewhere around another 100 miles for today. And then I'll put us up to 300 for the day and well within our two-day goal of reaching Antefagasta. Yes. Let's go. Let's do it. We're just coming up to our camping spot for the night. 294 miles so far today. Woo. It is time to come off the route of five. Oh, yes. We found the spot on Ai Overlander and it says that it's quite hidden and people had a good night. So let's see if that is true. Oh, yes. Well, it's certainly hidden from tarmac, that's for sure. Oh, no. It just said, actually, just somewhere along this road here, actually. OK. So I would say we were just right a little bit down and see. I mean, the road doesn't look that bad, actually. No, no, it looks fine. And with our new tires. Traction by Anlas. I think actually up and over this hill and then we'll be like death row hidden from the road. Yeah, I mean, I don't know where this road goes to. I'm not sure. I can see nothing. Oh, OK. I tell you what, just here, before this hill, this will be good. Somewhere here. Let's have a look what have we got. OK, wait a second. Yeah, maybe some of the sandy bits here. Wow, look at this place. Proper desert out here. OK, if you move that rock just down there, the big one, and then I'm just going to ride through those two little bushes. Yeah, just about all right. There's definitely a dry riverbed, though. Do you think it's likely that it could have like a rain coming down? No, it is. We're in the desert, I guess. Hey, OK, awesome. Let's set up camp then. Good evening, guys. We made it. Happy new life. Yes, we've set up our camp, and we're now hiding under the shade of Chris's Parasol. Thanks, Chris. Let me show you guys around. So we've got the tent set up in front of us here, and around us is just all these bushes. We're in this dry riverbed here, so it's nice and soft and sandy. And look at it out here. It's quiet. It's peaceful. It's beautiful. And we just saw on the sign a little while back that we've officially entered the Atacama Desert region. Yeah, so it's super exciting that we actually camped the first night here in the Atacama Desert. Yes, how awesome is that? Wow, I think we're going to have a really, really nice night here. I think so too. But that's it from us today. At the end, we did 295 miles, which is about 470 kilometers. And we hope you enjoyed the video. If so, please give us a thumbs up. Subscribe to the channel. Share the video with your friends and family. Comment below. And we will see you next time. Thank you.


Alrighty, I got a really quick construction update for you guys today because it was raiding and I had to fly my drone in the quickest little break in the clouds I possibly could and then it started raiding on my drone so I had to bring her back before she broke, but there is an awp booth finally installed on Tundra Twister and the platform for Tundra Twister is at its lowest point now too. So they are rushing, the stairs are in as well, they are going full force on this thing. They're already getting ready for offense on the Q-Line side of Tundra Twister to block off the forest so definitely moving a lot quicker whereas Snoopies is starting to slow down now, like I'm kind of shocked by the switch in pacing because if we go on over to Snoopies, not much else has occurred over here, they're going really slow on that second launch building in terms of installing the side paneling, they're about to start installing the decal on the side of that first launch barn, it's going to have I believe like crosses like you would on a bar, not like X's almost with those red beams. So super excited about that. Scaffolding has gone up at the very back of the station so I'm not sure what that's for nor is it a good sign in terms of timing because I really thought that'd be the ride to open and then here's Fastlane. Anyways, thanks for watching today's update, I know it was really short. Have a good one, see you at the park.


I am going to invite our distinguished visiting professor Peter Winkler to come and kick off tonight's talk. Please join me in welcoming Peter. Thank you, Cindy. Tonight we're very lucky to have, speaking to us, Penny Haxel, who is a research mathematician. She got her PhD at Cambridge University in England and is now a professor at the University of Waterloo in Canada. So Penny is one of the best in the world in her field. But what is her field? So you could get some clue by the fact that she is a member of the world's only department of combinatorics and optimization. Okay, so that sounds pretty impressive. It really is, actually. So combinatorics is the mathematics of individual separate things like whole numbers or people. Okay? And optimization is the mathematics of trying to find the best way of doing something. So what Penny is going to do for us tonight is she's going to take the cruise director's point of view, as you will see, and talk about some problems and get you involved in actually seeing how these things work. So please volunteer when it's a lot of fun. But in the end, you will have seen a very important algorithm and a very important theorem for mathematics. So there's really good, hard, and serious stuff behind it. But you're here to hear Penny and not me, so I'm going to turn over the show to Penny and welcome her aboard. Thank you very much. It's really wonderful to be here in the Museum of Mathematics. We're going to go on a cruise today, cruise on the math C. And here's our ship. You probably saw it on the poster. And every cruise comes with a cruise director. Cruise director is responsible for making sure everyone has a good time. So what I'm going to do is describe to you a couple of my favorite topics in combinatorics, actually in a field called graph theory. But I'm going to try and explain them to you as problems that the cruise director might need to solve. Okay, that's one of them. So here's our ship. Now the cruise director knows that many passengers, when they're first getting on the ship, they think, feel a little bit kind of anxious and ask themselves on this great big ship with so many people, how am I actually going to meet people and make friends? It seems kind of intimidating. So cruise director is an experienced guy. And what he's going to do is before people come to the ship, even when they register for the cruise, they're going to fill out a questionnaire. And the questionnaire is going to ask them about their interests. So here are some interests that the cruise director has a feeling that many of the cruise passengers might have. I hear he's got a list of about 20 here. You can think in general he has a list of N. And he's going to ask everybody to indicate a few of these, let's say five, that they're interested in. Now why does he want to do this? He has the following plan. And in the first few days of the cruise, he's going to partition all of the passengers into groups so that if two people are in the same group, they'll be put together in groups in there at dinner maybe or on excursion buses. With this property that if two people are in the same group, they will have at least one common interest. So there'll always be something to talk about with that person you're sitting next to. You might both find that you're interested in photography. And then you can have a conversation about that. So it's a kind of icebreaker idea to get people to meet each other and make friends. Okay, so that's the plan for the cruise director. So he's going to ask people to indicate five of these 20 interests. And this is his goal. He wants to partition the passengers into groups. And he'll do this in advance. And he'll do this maybe differently on different days, mix things up. So that every pair of passengers in the same group are going to share a common interest. Okay, so that's what he wants to do. He has some requirements. We're thinking about N as being about 20 and K as being about five. We want the number of interests to be big enough so that everyone has some of those interests. And then the number that they indicate to be reasonably small so that there are at least that many that they're interested in. Okay, and the general question here is how many groups does he need? So this question turns out you can describe as a very combinatorial notion. It's called coloring in graphs. Now I'm actually going to try and skip most of the more specific terminology to graphs and that sort of thing. So on my slides you might see the occasional title which maybe I haven't explained. But I'm going to try and suppress those details so that we don't get too bogged down in the terminology. And last time it did talk a couple of hours ago, it kind of took too long. So I'm going to skip some of that stuff. Okay, so here's our question. How many groups do we need? Well, here's an example. And this is in fact what the cruise director has in mind. I can easily do this with 20 groups. I've got 20 interests. I can do this easily and nicely with 20 groups because in the first day I can put into group one some of the people who like topic one and into group two. Some people who like two. And everyone likes some of these topics so I can definitely put everyone into a group. And maybe on the next day I can do it differently so people would be mixed up and enjoy meeting each other and talking about these different things. Okay, so he's quite happy about that solution. But then it happens that there's a complaint comes from the dining room staff. The dining room staff has to set up each day how many tables of each group and so on. And it's too many groups for them. 20 is too much work. Too much work to set up each day. Please minimize the number of groups that you can use. Okay. So the cruise director needs to think about this one. So he wants to ask, can I do better than 20? Okay, so here's his idea. Instead of 20 groups, starting with 20 groups, why don't I first of all start with about 15? And I'll do the same thing. Make sure into group five only into group five people who are in group five all like topic five and so on. But now instead of taking 20, I only take 15 and I make sure that every passenger whose interests list contains at least one of one up to 15 gets into one of those first 15 groups. So suppose I do that and so far I've used 15 groups and who's left? All the passengers that are left have exactly the same list of interests. It's exactly 16, 17, 18, 19 and 20. Those are all the ones who didn't make it into the first 15 groups. Okay, but these passengers, well, wow, they have tons of stuff to talk about. They all have exactly the same list of interests. So they'd be really happy all to be in one group. So here's then a big improvement. Instead of using 20 groups, I could use 15 for these and then just one more for this last one. Okay, so that's a big improvement. So we could do it with 16 instead of 20. Now, here's one of the titles that I'm not describing to you. There's a technical notion which captures this same idea, except instead of thinking about 20 and 5, you could think about any n and any k. That's what we're doing here. So if the cruise director goes a little bit further down this road, he realizes that maybe I can do even better than 16. Why don't I do the following? I'm going to try putting all passengers with an interest in 1 up to 11 in the first 11 groups. Why? Well, let's think about what's left over. The interest left over are, well, 12 up to 20. There were 9 of those. And remember that everyone has 5 interests. So if I ever set a size 9 and everyone is interested in at least 5 of them, many two people are going to share at least one interest inside that group if they didn't make it into these ones. So that's even better because we can use 11 groups for all of these passengers. And then for the remaining passengers, all of them are interested in at least 5 out of these 9 interests. And they can all be in the same group. So that's great because we get what we want and the number of groups we used is not 20, not 16, but 12. Even better. Okay, so the cruise director is very happy with this solution and the dining room staff after a bit of grumbling agree that 12 is okay. So that's great and we're going to set off on our cruise. Now, of course, the cruise director is still wondering to himself, could I have done it with fewer than 12? Okay, but I don't have time to think about this now. So he's going to come and think about it later. And I will tell you that he would have some thinking to do because in fact this question where this was N and K instead of our specific 20 and 5, this question was actually a very famous open problem. In 1955, it was conjectured by an mathematician called Knazer that this number, which corresponds to 12 when N is 20 and K is 5 and more generally it's N minus 2K plus 2 for N and K is that the right number, is it the smallest possible number of groups we could get for this problem? So we're going to come back to this question a little bit later but the cruise director now is very busy because an exciting event is coming up on the boat and that is Tango Night. Very exciting event, the ballroom dancers are very excited and many other people who have been learning dancing on the boat as well. I'm excited about Tango Night. Here are our Tango instructors, as you can see they're very serious and they're going to come on the boat and give an intensive full evening of Tango lessons and they require that couples pair up and then they dance with each other the whole night. They don't want shifting around, they want people to get used to the dancer that they're working with. Now the cruise director, however, is going to be responsible for pairing up people who are going to be participating in the Tango Night.  Well, he's a questionnaire guy, so he's going to ask the passengers to fill out another questionnaire. This time he's going to ask all the dancers, who would you like to dance with? More specifically, please rank in order of preference your favorite potential partner for Tango Night. So everybody who's participating in this and we're going to imagine we have the same number of gentlemen and ladies, it makes it easier. They're going to each rank the others in strict order who they like best and so on. Now what we want to do here then is assign a pairing. Here I've written instead of ladies and gentlemen, I've just written letters and numbers. And what does this mean? This means that person one likes D best of all and then J, then E and H and A is one's least favorite. Okay, and similarly B likes 10 best and then 4 and then 1 and so on. So here everybody has ranked the other side of this diagram called a bipartite graph in order of preference and the cruise director is responsible for pairing these people up in an appropriate way. So we need to wonder what is a good way to pair these people up taking into account their preferences. Okay, so here is what the cruise director would like to do. What he'd like to avoid is somebody messing up his assignment of partners. So why might such a thing happen? Well, suppose we have a gentleman N and a lady L who would both rather be dancing with each other than with the partners given to them by the cruise director. That would be unfortunate, right? Because then maybe they would just abandon their partners and want to dance with each other and then that totally messes up the assignment. Okay, so the cruise director would like to avoid this situation. Okay, so this notion of this type of matching of the numbers and matters, this is a well studied and important notion called stable matching. I'll mention to you it's also called stable marriage. So it suggests you a longer term commitment than just a night of tango dancing but we're just going to think about it this way. And we say a matching, so a pairing of those participants is stable, I'm going to call it M, that stable it means that for every pair like this, number and letter for us, if it's not in the matching, so if it's not our assigned pair, then either N prefers his partner in the matching to L or L prefers her partner in the matching to M. Okay, so this is a notion of stable matching. So cruise director wants to know, I've got this list of people and preferences, is there always a stable matching? Does such a thing exist? Okay, so what we're going to do is find out that there is and moreover we're going to find a way of finding a stable matching and to bring this algorithm to life. What we're going to do is implement this on the human computer, which is right here. Okay, so I'm going to tell you what we need for the human computer. The number one thing we need is 20 volunteers and I will tell you what volunteers are going to be doing. First of all, I can assure you there'll be no dancing involved. So don't worry about that. We're going to have two groups of people, both of size 10. They're going to be the standards and the sitters. So standards are going to stand. Each stool is going to have one standard behind it. And standards, you're going to be standing still during the algorithm. And we're also going to have 10 sitters. The sitters are going to be sitting in the stools facing out like this. And the sitters are going to be getting up and moving around to different stools. Okay, so I would like to invite 20 volunteers to take part in this activity. Kids especially are encouraged to do so. And I'll give you a hint, kids, it's more fun to be a sitter. Yeah, leave your balloon on your chair and please come. Yes, please come, lots of people. Very good. Yeah, if you just pick up the cards and just hang on to them for a second. Thank you. Very good. Yes, plenty more. Yes, I need. Okay. Yeah, we just have a seat and hang on to the cards for a moment. Thank you. You can face outwards. Sitters, face outwards. Yes, very good. Okay, excellent. Cool, sitters. And now I need 10 standards, please. 10 standards. We need quite a few more standards, please. Come, come, and be a standard. Yes, if I just stand behind this chair. Yeah, I need two more standards here. Thank you. Okay. Another standard. Another standard back here. You can't really see because it's way back here, but okay, good. I still need a standard back here. Essentially, we need two more standards. Come on. Come, come. Very good. Thank you. Thank you. Please come. Very good. Thank you very much. So we need a standard behind this chair. Yeah, and we have one more chair over here. I need a standard behind there. Excellent. Okay. Now, let me describe what you're going to do. Now, you're holding a card. Oh, now there are two cards being held currently by the sitters. Sitters, please give a card with the big letter to the standard behind you. Great. Okay. Now, the card that you're holding, the big letter or number is your name. Okay, so you are A. Hello, A. And you are nine. Hello, nine. Okay. Now, I'm going to tell you what the small numbers or letters across the top of your card mean. They're exactly the same as what you see there. And that's just a record again of who is your favorite. Okay. So if you look for A, so A for example, if you look at A, you can see the other side as well. And you like one best and then three and then nine and so on. Okay. And nine. Nine. What do you like? Who do you like best? You like E best and then B and so on. Okay. So everybody clear who they are and who they like in what order. Is it clear to everybody? Okay. Okay. Now, here is how the algorithm is going to work. When I say go, the sitters are going to jump up and proceed. And I would ask you, sitters always go counterclockwise just so we don't have bumping into each other. Always go counterclockwise. Please proceed to your favorite letter and try to sit in their stool. Okay. Not yet, not yet, not yet. Okay. Please, when I say go. Okay. Please proceed to your favorite letter and try to sit in their stool. Now, standards, what you're going to do is if there's nobody sitting in your stool, let the person sit there. But then, if someone else comes and you have a choice of who sits in your stool, choose the one you like better. Okay. Okay. So now, sitters, after you try to sit in your favorite, the stool of your favorite, maybe you'll be sent on, then you have to go down your list and move to your next favorite and try to sit in their stool. Okay. This has a lot of lessons in life. This exercise. Okay. So let's imagine, so it's now a bit hard to see the screen, but here, for example, we might notice that one, sitter one and sitter ten are both going to head for D's stool. So D, you are going to have to choose between one and ten. Okay. And that's dictated by your set of preferences. Okay.  Okay. So is it clear to all participants what you're doing? Okay. Wonderful. So is everybody ready? Okay. And I will just try to get myself out of the way here and I will say, go. I'll tell everybody that E is really hot property. He's got tons of people. So who do you have? Okay. Everyone sitting? Wonderful. Everyone sitting? Great. Excellent. So you've calculated a matching. So let's see if you have the right answer. Okay. One who you match to. Very good. Two. Three. Four. Five. Six. Seven. Eight. Nine. Ten. Excellent. Everybody. And I will tell you a secret in the first session. They didn't get the right answer. You are pros. Okay. Wonderful. So you've done very well. You've found a stable matching. I'm just going to say a word about why it's a stable matching. And without going into too many details. But remember what stable means. It means if we have a pair who's not in the matching, then one of you, at least one of you is happier in the matching than with that pair. So let's just take an arbitrary pair who's not in the matching. Let's look at, say, J and 10. Okay. So you're not in the matching. So this means that one of you has to be happier.  Five. I'd rather leave. You'd rather leave. How about you? You're happier. Very good. Yeah. So if we think about what happened during the matching, you would have received your five. And then if 10 came, you would have just said, aha, it did happen. Okay. Very good. So there's an example. And this is true in general. That every pair that's not in the matching, either the sitter didn't reach that, so the number didn't reach that letter because that number was already, when the algorithm terminated, was already off.  Very good. Okay. So that is excellent. And we have found that this is, this, we have found this stable matching. And that's a stable matching. That's great. So now what I want to do is, well, can we play again? You can. Now I want to think, you want you to think about one question before we play again. And that is, ask yourself, was it better to be a sitter or a standard? So if you think about what happened with the sitters, as the algorithm went on, things got worse and worse for you. Right? You had to keep lowering your standards and moving to another chair, right? Moving down your list. Whereas the sitters, the standards, things got steadily better for you, right? You got somebody in your chair, but then whoever came, you got to choose someone better. The things got better and better for you. So you might want to wonder, is it better to be a sitter or a standard? So you can think of that in the back of your mind. Now let's test it by switching sitters and standards. Okay. So now numbers are going to stand up. Keep your cards. So that's you. You are that person. Switch. Okay.  Now new sitters, please sit down just so we can see for a moment. And I'm just going to rearrange the numbers so that it's easy for the sitters to find you. Okay. So I'm just going to, you, okay. So where's number one? Okay. Great. So let's say number one is going to be arbitrarily here. So let's put number two here. Thank you. Number three. You're in the right spot. Right on. Number four. Perfect. Number five here. Number six. Good. Number seven. Yep. Number eight. And I need number nine there and you need to move one step that way. Perfect. Okay. That's great. And sitters, it doesn't matter where you start because you know, okay, everyone knows what to do. Now sitters, you're the letters and now you're going down your list one by one. Okay. And standards, you tell the sitters who come to you who should sit and who should go on. Okay. Ah, well that's the one that we found before. But we're also going to address another question while we're doing that and that is, is the staple matching unique? Is that the only one or is there a different one? Okay. Everyone ready? Okay. Go. Please go counterclockwise just to avoid confusion. Okay. Okay. Oh, you're my last name. Yeah. I'm not going to answer. You're my only take or so far. Oh, you're my last name. Yeah. I have a last name. Well thank you. I'm sorry. I'm sorry. I'm sorry. I'm sorry. I'm sorry. I'm sorry. I'm sorry. I don't know. I'm sorry. Oh. I'm sorry. I'm sorry. It's okay. It's okay. We want to get it before. Six. Six. Six. Six. Six. Six. Six. Seven. Seven. Seven. Seven. Seven. OK. Have we got sitters in every chair? Is everyone settled? Yes? No? Yeah? Is every chair got a sitter? Every chair got a sitter? Yes? OK. Very good. So you have found another you have found a stable matching. Let's find out whether it's different and compare it to the other one. OK. There's the other one. OK. Now let's make sure and let's check if it's right. OK. One. Who you match to? Two. Three. Four. Five. Six. Seven. Eight. Nine. Ten. Excellent. You guys are pros. Very good. OK. Now. Yeah. So now let's observe. Is it different or is it the same? Now G, for example, who you match to? And how is it before? Six. Yeah. So G and three were matched to each other in both of them. Now if you study the list, you might see that actually G and three were made for each other. They were both top of each other's list. And that's the fact about every stable matching, in fact, would contain that edge. However, let's see. How about two? Who were you matched to before? And now? And who do you like better out of D and C? So you like D better. Now, when you were a sitter, you got a better match. Notice? So for you, it was better to be a sitter actually. Let's see someone else who's different. Let's see. Who else is different? How about seven? You were? And? You're worse off now. Now at your standard. Uh-huh. OK. So we've learned a couple of things here. First of all, we learned that stable matching is not unique. As a matter of fact, there can be many stable matchings. And we have got some interesting information about whether it was better to be a sitter or a standard. Thank you very much, volunteers. Please take a seat. OK. That was excellent. So we learned some things there, as we mentioned. In fact, there can be many, many stable matchings.  And then if you switch them around, if there are several, you in fact get a different one. And that's the answer to the question, whether sitters or standards better off. In fact, and it's a very strong statement, if you think about all stable matchings, every sitter gets the best possible partner in this algorithm amongst all stable matchings, all simultaneously. The sitters don't even have to compete with each other. Each one of you sitters gets the best possible match in any stable match. And on the other side of the coin, every standard gets the worst possible. So that's one of the features, interesting features about this algorithm. And maybe a life lesson there. Because what were the sitters doing? They were going out there, going for their top all the time, and they had to suffer a lot of rejection. But in the end, in fact, they are better off. Okay. Well, now, the cruise director is pretty happy with the way things are going on the boat. People are enjoying themselves. They're going off on excursions. He's getting a little more leisure time and maybe gets to walk on the beach a little bit. And he's going to get back to thinking about that question. Those 12 groups, was that the best I could have done? So he's going to think about that again. This is, don't worry about it, it's some notation that goes with the fact that this general question, we could ask this general question, about not just 20, N being 20, and K being 5, but we could think about N in general. Turns out to be useful to think about it like this, and K being something general. So this is a general question. But this is what the cruise director is thinking about. Did we really need 12 groups? So let's just remind ourselves what we're talking about. If we look at all lists of size 5, so that's what I mean by a 5 subset, out of 1 up to 20. And if someone comes and puts them into 11 groups, I'm going to think of those as 11 colors, will there always be some pair of disjoint lists of size 5 that are the same color? That's the question we're thinking about. So I just write the general question here in terms of K and L, but it's the same question, and just how do these numbers work out? This is N, and this is K, and the number of colors turns out to be this number. But don't worry about that. So instead of thinking about 20 and 5, which are kind of huge numbers, we're going to think instead about a sort of test case when I just have N is 6 and K is 2. So once more, here's the question. Here I have written the list of all of the pairs of numbers from 1 up to 6. And somebody has colored them in three colors. I hope you can see this. It's green, purple, and orange. So every one of the pairs out of 1 up to 6, someone has colored in green or purple or orange. And we are wondering, because before, with the setup that the cruise director used before, he would have used four colors from 1 up to 6. And here we're just using three. And we're wondering, would this have worked? And here I'm just making a list of the parameters. Again, don't worry about that. This is our question. So everyone has colored all of these pairs with three colors. Is there always going to be two of them that are disjoint and the same color? Okay. So that's what we're going to think about. But for some reason, the cruise director gets distracted, perhaps by beach ball. And we're going to think for a while about spheres.  Okay. So what we want to say about spheres. Now, there is a notion of a sphere in actually any dimension. I'm going to use this notation SL. That's the L dimensional sphere.  Don't worry too much about this stuff that's written here. But think about, are your favorite sphere, which is the sphere that we think about when we say sphere, is the two sphere. This is called a two sphere. And what is it? It lives in three dimensional space. And it happens to be the set of points that are equidistant from the origin. Let's say the sphere of radius one is all the points that are distant exactly one from the origin. And the origin is right here, right in the very middle of my sphere. Okay. So that's the notion of sphere in two dimensions. This is a two dimensional sphere. But I will just tell you that there is the equivalent notion in any number of dimensions. You might wonder why it's called a two sphere when it lives in three dimensional space. And the reason is that locally it behaves like two dimensional space. What does that mean? It means that if we were teeny weeny little people living on this sphere, it would look flat to us.  And indeed, it looks flat to us. In fact, it was believed to be flat for a long time, erroneously. Okay. So that's the notion of the sphere. Okay.  Now, another notion on sphere, hemisphere. So what's the hemisphere? If I take a point A on the sphere, like say this one, the so-called open hemisphere centered at this point is the following thing. It's a set of points on the sphere. And it's this. If I turn the sphere so that that point is right in the very, very middle for you, then the open hemisphere centered at that point is basically the part of the sphere that you can see. Okay. So in particular, if I point directly at you the north pole, then the open hemisphere centered at the north pole is exactly all of the northern hemisphere, all of those points that have positive latitude, but the actual equator does not count. Okay. And that's the way to think about it. If you're kind of infinitely far away, it's the part of the sphere you can see when you look as stare directly at the north pole. Okay. That's the open hemisphere centered at a point. It has some a definition, but this is the best way to think about it. Okay. So the cruise director, for some reason, is going to think about special sets on the L sphere. We'll think about the two sphere. And he's thinking about the following question. If I have two numbers, k and l, positive integers, how many points do I need to place on the L sphere so that every open hemisphere like that contains at least k of them? So let's say it for our two sphere. What's the smallest possible size of a set of points on the two sphere so that every open hemisphere contains, let's say, at least one of them? So we're going to think about this question for some reason. And, but we're in this imperfect world and we all have balloons and stickers. So let's ask it in the following way. What's the smallest number of sticker dots, which is on that little strip of white, white strip that you have? You have some sticker dots on the strip. And we like to wonder how many do you have to place on your balloon so that no matter how you turn the balloon and look at it from some decently far away, oh no, a tragedy has occurred. Here comes another one. Okay. How many dots do you need to place on your balloon so that no matter how you turn it, you can always see at least one of them. Okay, so let's think about that for a few minutes. Obviously, if you just take your sticker dots and plaster your balloon with sticker dots, then no matter how you turn it, you'll see quite a few of them. But imagine that each one of those dots cost a dollar. Okay. And try to do it with a smaller number. Okay, so let's think about that. You have to use your imagination. Of course your balloon is not perfectly spherical, but imagine that it is. And the dots, you also have to think about them as having zero area. Of course they don't, because you wouldn't be able to see them. But if they had zero area, you would also need infinitely good eyesight. But trying to just set those little issues aside, let's think about this for a few minutes with our balloons. And if you think you can know the answer for this, that you can always see at least one of them, you might want to venture into variations like, how many do you need to put there so you can see at least two of them? Or maybe even at least three of them. Okay, so let's think about that for a few minutes. Okay, should we have a think about this now? Do anyone have a feeling about how many dots you need? Yes? I think this is great. Okay, any other suggestions? Yeah? I have five. Okay, any other suggestions? Yeah? I think I did it in four. Okay, any other suggestions? Yes? Three. Okay, another three. Okay, you did it with six? No, I didn't. You did it with six. Okay, very good. Okay, so turns out that the answer, you're hovering around the right range definitely, the answer is four. You need four, and you can do it in four. So here's a quick proof for why you need at least four. So take your balloon, and suppose you put only three dots on it. I'm going to do the following.  It's like an equator, but through the two dots that I chose out of yours. Okay, so I've got two dots on that equator. Now, if you've only got three dots all together on your balloon, then one of the hemispheres does not have a dot in it, right? So if I turn that hemisphere to you, then you can't see the two that are on the equator, and you can't see any there. So that's the reason why you need at least four, and in fact you can do it in four. So well done if you did it in four, but even if you did it in five, that's still pretty good. Okay, any feelings about the second question? How many do you think you need to guarantee two? That's definitely getting harder. Any guesses? Eight. Okay, eight is certainly enough. In fact, the answer is for if you need to guarantee two dots, the answer is six. So it's again the smallest possible for the same reason. If we think about that same proof, we would need at least six, and it turns out that six is possible. If you want three dots in every view, then eight is the right number. Very good. Okay, there's a theorem that tells us about this. I'm going to temporarily call it the balloon dot theorem, and this is what it is in general for every k and l. So we're talking about the l-sphere and k. There is a set w on the l-sphere. It has size 2k plus l, and every open hemisphere contains at least k of those dots. So on our two-sphere, there we have four. We're trying to see one in every view, two, six if we're trying to see at least two in every view, and eight if we're trying to see at least three in every view. Great. Okay, well, it's not clear why we're thinking about this, but this is what the cruise director was thinking about. Now, I want to tell you just a couple more things, and again, I want to go into here. I have the technical definition written here, but don't worry too much about it. Every open hemisphere is an example of a so-called open set in the sphere, and what does open set mean? Well, the technical definition is written there. Basically, it means it's a set of points on the sphere with the property that if you take any point in that set, and if you stay close enough to that point, you will still be in the set. So the open hemisphere centered at the north pole, there is again the northern hemisphere. I'm claiming is an open set according to this definition, because, well, if you're in New York, then actually for even a thousand miles around you, that area is still in the northern hemisphere, but even if you are standing one inch from the equator, if you drew a circle around yourself of radius half an inch, then that would still be in the northern hemisphere as well. So that's an example of an open set, and this is just a technical notion. Don't want to dwell on it too much, but I wanted to mention it to you because I want to tell you this very important theorem, a beautiful theorem with many implications. This is Borsuch's theorem. So Borsuch's theorem involves also this notion of antipodal point. The antipodal point to a point on the sphere is just minus that point. So the antipodal point to the north pole is the south pole. The antipodal point to New York is... No, it's just somewhere off the coast of Australia. It's kind of somewhere over there by the look of it. It's definitely in the water. Yep. It has the notion of an antipodal point. So Borsuch's theorem says that if you have a sphere, the sphere as L, so we think of L being 2, and if you have this many open sets, L plus 1, open sets on the L sphere that cover the L sphere, that means every point on the sphere is in at least one of them, then one of these open sets contains a pair of antipodal points. So let's just say it in our world with the two sphere. If we have three open sets that cover the two sphere, then at least one of them contains a pair of antipodal points. Okay. Well, where were we? We have wandered very, very far from the beach where the cruise director was wondering about those 12 groups. Here's the question once more. Here's our test case. So again, let's just think about this one. We have colored all of the pairs of the numbers 1 up to 6 with three colors, and we're wondering, does this mean that always has to be two disjoint pairs of the same color? Okay. Well, and now I'm going to do something rather amazing with these spheres and things, because instead of these numbers 1 up to 6 and all the pairs of those, I'm going to think about all the pairs of that set W, which is size 6, remember, that set W is size 6 on the sphere that has the property that however you turn the sphere, you see at least two of them. Okay. That's the set of size 6 I'm going to think about here. And so I'm going to transfer this coloring of the pairs of 1 up to 6 to the same coloring of the pairs of the points that I see here. Okay. Now there's a balloon dot theorem. We know this set exists. Now I'm going to now define some open sets, three of them, on the sphere. And here's how I'm going to do it. I'm going to take, for every point on the sphere, I'm going to put this point, I'm going to take this point and look at the hemisphere centered at it. We know that there are going to be two elements of W that you can see in the open hemisphere. Let's say these two. I'm going to look at those two and I see that they're numbered three and four. And so I'm going to put this point into the purple P set because this pair is purple according to that. Okay. Now for another point, I'm going to look at that point, look at the hemisphere centered there. And again, there are going to be at least two elements there they are of W that are in the open hemisphere centered there. And those two are numbered 1, 2. And so I'm going to say, okay, this point is in the G set. Now because of the balloon dot theorem, however, all the points on the sphere are going to get into at least one of these sets because of this property that every open hemisphere contains at least two points in W. So what I've done there is defined three open sets in the sphere. And this is the right thing about it. They cover the sphere and they're open. And to just get a little bit of a feeling of why they're open, I'll just say it. If this point is say in the purple set, it means that you see a purple pair. Now if I wibble the, so you're just a weenie bit around so that the, instead of just A, you see all the little points around A in a very small neighborhood. Well if I do that slightly enough, you're still going to be able to see that same pair all the time. So all of those points are also going to be in the P set. So that's the kind of feeling for why these sets are open. Well, what did Borsuch theorem say? It said, if I have three open sets that cover the two sphere, then one of them contains a pair of antipodal points. So one of these sets, the purple, the G, the green, or the orange, must contain a pair of antipodal points on the sphere. What does it mean? Let's say, let's suppose that there are two points, antipodal points, that are both orange. Okay, so what does it mean? It means, let's call them x and minus x. So if I show you x, the hemisphere around x, you see a pair of points that's colored orange. I am looking at minus x. So I see the hemisphere around minus x. And this point is also orange, so it means I see an orange pair as well. But look, your hemisphere and my hemisphere are completely disjoint, right? So it means that your orange pair and my orange pair are completely disjoint. So that says that in fact it's true. If we take this coloring, any coloring of the pairs of one up to six, which I kind of just transferred to the six points that are here on the sphere in a special set, any coloring of those with three colors by Borsuch's theorem and the balloon dot theorem, I put those things together and I find that I must get a pair of disjoint pairs. And in my example, apparently they're orange. And here I see an orange pair. One, five, and four, six are disjoint and colored orange. Okay, so this answers finally the cruise director's question. And he can be relieved now that he can tell the dining room staff that I could not have done any better than 12 groups. Now here I've been talking about six, two and he was thinking about 25. And in general we think about n and k. But all of these things, the same idea tells us that all of these things will work even for general n and general k. So cruise director is delighted he's finally answered that question and he can go home happy. Okay, so we're approaching the port now. Our cruise is almost over. But now we encounter actually something that looks dangerous, a great gale. Looks terrible for our boat. But in fact, this is not that kind of gale. This is a very good kind of gale. And someone I wanted to particularly show you, this is David Gale. Wonderful mathematician, sadly no longer with us. And why do I especially want to mention him? Well, the balloon dot theorem, it's not really the balloon dot theorem, it's Gale's theorem. So this is a wonderful theorem due to David Gale. It has a lovely proof as well, but it's kind of a little bit complicated, too complicated for this setting. So I'm not telling you about that. But it's very nice and elementary. That's the balloon dot theorem. And our algorithm for stable matching that we implemented on the human computer, that's the Gale-Shapley algorithm for stable matching. Right, very important algorithm, very nice and beautiful theorem when combined with Borsuch's theorem gives a proof of the Knazer conjecture, which I mentioned to you at the beginning. And I also wanted to mention David Gale for another reason, and that is because he was for many years a great advocate of the idea of a museum of mathematics. And he actually has an, had an online museum of mathematics, it's called Math Sight, and it's still maintained nowadays, and lots of fun stuff about math for just anybody who would like to play with it. It's wonderful. I shouldn't pass over Shapley either. Shapley got the Nobel Prize for Economics in 2012 in part for his work on this algorithm. Very important work. Okay. So there are many more nice things I can say about the Gale-Shapley algorithm. Maybe I will just mention a few more things because it is a very notable algorithm. It's very efficient. We implemented it here on ten vertices, on ten pairs. Our computer here was running not quite as fast as a modern computer, but it was more fun and more meaningful. But running on a modern computer is an extremely efficient algorithm. It can handle many thousands or whatever of vertices, and it's been used actually for some important problems. I will mention that we did it on a complete, on a so-called complete bipartite graph. Everybody ranked everybody here, but in fact, that's not necessary. You can have a kind of partial graph with only partial rankings, and it still works, and it still gives you a stable matching, but it might not necessarily match everybody. So that's just something that can't be avoided, but it still works just as well. Yeah, and it's been used in several practical situations for matching resident doctors to hospitals. Also, the New York and Boston public school system have used it for matching students to schools and other settings as well. So these big problems, it can solve very nicely and quickly. Okay, so we have come into port now. I hope you enjoyed the cruise. I'm just going to put here my references. They're the references to David Gale's papers. The proof that I told you of the Knazer conjecture, that was due to Balinese. This is using Gale's theorem, but that's based on a slightly earlier proof, and this was a real key to understanding this conjecture, is to realize that it, in fact, it does have something to do with spheres and open sets and antipodal points, even though the statement doesn't look like it at all. All these notions, the so-called topological notions, can be used to prove it, and in fact, really to this day, we don't genuinely know a different proof, a proof that doesn't somehow use these notions even if it's not explicit. Yes, so, right, we've docked. I hope you enjoyed the cruise, and I'd be happy to answer any questions you have.  So if we run it on the same graph, and with the same sitters and standards, okay, that's important. If the sitters are the sitters next time, then we will get exactly the same solution, but if we switch the sitters and standards around, as we saw, we got a different solution. There are many other stable matchings, there could be many other stable matchings, but that algorithm just finds those two, and those are actually special. In the whole set of stable matchings, those ones are special, and they're special in the way, actually, that I mentioned, the sitters are best off, amongst all stable matchings, and the standards are worst off, and the other way is the other way. If you mix it up, so that you get different results. Yeah, so that's an interesting fact, right, if you start from different positions, for example, or people go in different order, because notice that we didn't tell people to go one by one, which you kind of think of an algorithm. Because that's more like real life. I'm getting to the key question, though, for me, was if you mix it up, you don't go in the same order. Is there an optimization in that area, where if you just keep trying? The number of steps could be different, so you might converge to a solution quicker if you start from a different position, for example, but the solution will always be the same, if you run that algorithm. What happens if two of the people who are choosing have the same exact list of dancers have the same ranking? If, say, two sitters, two letters, for example, have exactly the same, which could happen, right? Because it might be that the dancers, there's a clear list about who's a better dancer than someone else, right? So maybe to have exactly the same list, then who's going to make the decision? The decisions are going to be made by the other side. So someone will get to choose who they prefer. Is there any way to get a solution that is equally good for the standards and sitters? That's a very interesting question. So the set of all stable matchings, it actually has a very nice structure. It's actually called a lattice, and that's just a technical word, but it's actually possible in a lattice to kind of move around from one stable matching to another. And there's a specific way of doing that. If we put those two stable matchings that we found on top of each other, some of the pairs would be the same. Now, G and 3, you might remember that you were made for each other, and you are going to be actually in every stable matching. But there are some others who got different stable matchings according to how we ran the algorithm. And there's a way of kind of switching some of those people to step from one stable matching towards the other one. And in this one, in fact, in this instance, there were only two stable matchings. But that's because I set it up in such a way that there would be a lot of action going on here. So if you just take a sort of arbitrary one, you'll actually find there are many more stable matchings, and you can move around in this lattice of stable matchings. And so there are many. And you can ask, what would it mean to be kind of more fair? And even that's a kind of bit open to interpretation, what it would mean. So you can try optimizing various things. If we introduce a cost function on the matching, or an algorithm that would minimize the cost of the matching? Yes, another very interesting question, and it's definitely studied. And yeah, I'm going to say there are results on that, but I don't have them at my fingertips. But it's a very practical question to ask. So maybe there's a weight also on each edge, and you might want to find the maximum weight stable matching, for example. So yes, it's certainly studied, but I'm afraid I can't tell you off and off of my head what the results are. Give another hand to Penny.


So, welcome to the second discussion that we are going to have on the contents of the first week. So, I had some very interesting questions from my teaching associates here in the previous discussion and they weren't able to exhaust all their questions and comments in the first discussion. So, here we are we have several more inputs from them in this discussion and let me hand over to them. So, they could present their inputs and questions one by one. Yes, over to you. Sir, in week one it was mentioned that the signals are reasonable functions. I don't get it. Can you explain a little bit what are reasonable functions? Well, you know reasonable was an informal term I used.  Right. So, I will write it down. Reasonable function or reasonable signal which I can or rather reasonable function which I can call a signal. Pick a simple exponential. Right. So, you have x of t is equal to e raised to the power minus say t by 2 for t greater than equal to 0 and 0 for t less than it is an exponential it looks like this. 0 all the way up to 0 as a function of t and then rises to 1 at t equal to 0 and then decays exponentially. Now, this is reasonable. I mean you would not understand why it is reasonable until I give you an example of an unreasonable function. But one thing I want to point out I have taken this example intentionally because here we have a discontinuity here. So, in spite of the discontinuity in these function I have still called it reasonable. So, reasonable does not mean that it cannot have discontinuity. In contrast, let me now take an example of what I call an unreasonable function. So, let me define x t in the following way. x t is equal to 1 whenever t is rational and 0 whenever it is irrational.   Now, in this tiny interval there are an infinite number of points where the function takes the value 1 and an infinite number of points where the function takes the value 0. In fact, it is an example of a function which is continuous nowhere in a certain there is there is no sense of there is no sense of you know the properties of the function are changing even over a small I mean I am speaking informally right now. I am not really being too rigorous mathematically, but in this tiny little interval of point 0.0002 you have so many changes taking place in the function and infinite number of changes in fact. We are not trying to talk about this kind this is an we do not really encounter these kind of signals in real life. And even if we wish to idealize real life situations is unlikely that we would like to take these kinds of signals as an idealization that is what I meant by unreasonable.  Broadly speaking reasonable functions are those where you are likely to be able to use them to understand real life situations even if by a little bit of idealization. And unreasonable functions are those which are unlikely to occur even by a large stretch of imagination when you try and model real life situations that is one way to understand the answer to the difference between unreasonable and reasonable. Of course, we will probably say a little more about this in subsequent videos. So, you could also wait that discussion little bit. Sir, so this is in reference to the RC circuit example you have taken the videos. In that context we discussed the properties of additivity homogeneity and hence linearity follows. So, my question is would an RL circuit also exhibit these properties and if so could you just demonstrate them or show an example of that? That is a very good question. In fact, this question helps me take two issues all at once. One is of course to show that the same principles that we develop for the RC circuit can also be carried to the RL circuit. And the second to illustrate how one very important principle in electrical circuits namely the principle of duality actually helps us make an abstraction or is ground to abstraction. Now, let me take the answer to this question first. So, let us take the RL circuit. Now, the first thing is you could have two kinds of RL circuits. You could have a parallel RL circuit and a series RL circuit. In both cases of course one can draw an analogy. Let us take a parallel RL circuit first. And let us take a series RC circuit for reference. I will first draw the parallel or the analogy here. Now, here I will write resistance capacitance. Here I shall intentionally write conductance which is the reciprocal of resistance and inductance. And here of course you remember that I had written an equation relating the input and the output voltage. So, I have the input voltage being VIT and the output voltage being VOT if you like. And I had related the input and the output voltage. In fact, I said that VOT that is the voltage across the capacitor plus the voltage across the resistor which is R times the current in the resistance which is R times C times VOTDT. This is the total voltage equal to the input voltage. Now, this is the equation that describes the series RC circuit. In contrast, let us look at the current input in the circuit. So, I have IIT if I would like to call it that input current to this circuit which branches into a current flowing in the conductance and a current flowing in the inductance. Let me call the output current as the current flowing in the inductance. And now I have two components to IIT. IIT is equal to of course that is simple essentially by using the Kirchhoff's current law at this node. Just like you use the Kirchhoff's voltage law here in this loop, you are using the Kirchhoff's current law here. So, IIT is equal to IOT and of course the voltage across the conductance.  It is going to be the current in the conductance multiplied by the resistance. You can similarly write down an expression for the remaining part of the current here. The voltage across the conductance is the same as the voltage across the inductance here. And how much is the voltage across the inductance? It is essentially LDIOTDT that is the voltage and you multiply it by the conductance and get the current. So, this is the current contributed by the conductance. Now, look at the similarity of these two and I am sure you understand that essentially we are trying to solve one problem. You know, if you solve one problem, you have kind of solved the other problem already. Now, this is actually behind what is called duality in electrical circuits. So, duality is just one instance of abstraction in electrical circuits. Duality says resistances and conductances are dual. Of course, resistance and conductance are really the same object, but inductances and capacitances are duals. So, the roles of voltage and current can be interchanged. Now, that is just one level of abstraction. So, you solve one circuit and you have already solved another circuit. That is so useful to have. What we are saying in signal and systems is much more. You understand one system and you have understood many other systems which are of a similar kind. Now, I put an exercise to the class. I have taken the dual. So, I have taken a parallel RL circuit and a series RC circuit and I have drawn an analogy between them. So, suppose I took a series RL circuit, how would I put down a linear system description for that? I am going to put this as a challenge for all of you who are participating in this course and let us see if some of you can answer this on the discussion forum. So, very good. So, I am very glad that there are some interesting questions that have come up. We will now continue this second discussion and I just wanted to mention. Another example of the reasonable and unreasonable function and we have discussed to some extent this kind of example even in the regular videos, but I am I repeated it here because I want to make it clear what a reasonable unreasonable meant at least at an informal level. So, now with that let me invite a few more questions. Yes. In the videos we actually discussed about systems. So, there were systems which have electrical input and electrical output. So, my question is whether we could use something like a transducer as a system because it can have a mechanical or a temperature input and it gives out an electrical output. Absolutely. That is a very good question. In fact, it is important to know that one could have different kinds of input and output. The nature of the input and the output could be different. In fact, right now of course, we are confined to a context where the independent variable is the same. So, the independent variable typically is time. So, for example, you know you could have a temperature input. You could be measuring the temperature in the room as a function of time and it could be they could be a essentially a temperature measuring device which generates an electrical output and electrical voltage or an electrical current by whatever mechanism. So, you could very well have a temperature input and electrical output or the same pressure input could then generate an electrical output, a pressure transducer that is possible or you could have transducers that take you from one non-electrical domain to another non-electrical domain. In fact, I would like all the people participating in this course to come out with several examples of these kinds of systems where you have different kinds of input and output. That is the domain from which the input comes is different from the domain in which the output is. But of course, we would like to take situations where the independent variable is the same. The independent variable is naturally time in many contexts. In some contexts, it could be space. Now of course, I must emphasize that the independent variable could also be different. For example, if you look many of you might be familiar with a cathode rheosiliscope. In a cathode rheosiliscope, we give a voltage waveform for example, as an input. Now the voltage waveform is a function of time and it is displayed as a function of space on the screen. Now in a certain sense, many of the dynamic displays that you have are also examples where you are making a translation from a time-based input to a space-based input. That happens, think about it. So, it is not necessary that the independent variable needs to be the same. But I would like you to first in the discussion forum to come out with examples where the independent variable is the same and then to come out with examples separately where the independent variable is different. So, the answer to your question is yes indeed and I am in fact putting further questions to you to encourage you to think further. Very good indeed. We will have some more discussion in subsequent discussion sessions. We need to conclude this one at the present moment. Thank you.


welcoming all to MSP lecture series on interpretive spectroscopy. So we are discussing about UV visible spectroscopy. In my last lecture, I was discussing about the DD transitions and not the conditions. Let's continue knowing more about that. Electronic spectra of transmittal complexes. Here, what would happen is, essentially electrons are exited from the ground state to one of the exited states. Such electronic transitions require high energy compared to other transitions. So during electronic transition, we also come across low energy vibrational rotational transitions because each electronic level is associated with several vibrations levels, which in turn are associated with our having several rotational levels. So in electronic spectra, such transitions are too close in energy to be resolved into separate absorption bands. The energy required for vibrational and exaltation as very minimal compared to the energy required for electronic transition from one level to another one. As a result of that one, what happens? They cause broadening of the absorption band in DD spectra. As a result, the bandwidth observed in DD spectra are in the range of 1000 to 3000 centimeter minus 1. So in a free gaseous metal ion, if you consider, DR builds are degenerate, and hence, no DD transitions are observed. But in a complex, degeneracy is lost because of ligand field splitting the orbitals into two or more levels. For example, if you consider octahedral level, we'll be having octahedral field, and we'll be having T2g, composite of dxy, dxz, and dyz orbitals which is lower in energy, and then higher energy, Eg level, composite of dx, minus y square and dz square levels. But in tetrahedral, opposite happens. So E will be the ground state in the homo level, and T2 will be the lumo level. So the magnitude of this delta O or delta T depends on the nature of the ligands, and affect the energy of electronic transitions, and hence, the frequency of absorption maxima. For example, if you are considering weak field ligands, of course, you can look into spectrochemical series. How we arrived at a position, a given position for a ligand in the spectrochemical series is considering the donor and acceptor properties of the ligands, and if you recall what I taught in advanced transmittal chemistry that all ligands, whatever you come across, can be simply classified into three categories. One is pure sigma donor ligands, such as water and ammonia, all neutral oxygen donor ligands, and also alkylamines and arylamines, on the other hand, the next one is sigma donor and pi donor ligands, such as halogens. Here, what we have is low energy field sigma orbitals, and low energy field pi orbitals are there. When such ligands interact with the metal, what would happen? Both of them will be donating to the metal, and hence, the CFSC, the homo-lumo gap, drops relatively compared to what we saw in case of pure sigma donor ligands, such as ammonia and water. The third category is sigma donor and pi acceptor ligands, example, carbon monoxide, phosphines, and nitrogen heterocycles, and other things. So in this case, what happens? They have low energy field sigma orbitals, and high energy, empty pi orbitals are there. And in this case, because of the sigma donation, and then pi donation back from the metal to the ligand, what happens? The energy gap between homo-lumo increases. As a result, this complexes are more stabilized, and also, in case if you want to have a DD transition, it requires very high energy. So that means the magnitude of delta O depends on the nature of the ligands, and affect the energy of electronic transition, and hence, the frequency of absorption maxima. Extent of splitting is related to the ligand position in the spectrochemical series. So the ligand position spectrochemical series very, very important in electronic spectroscopy. So here, just look into all the orbitals are classified into four categories here. There is some reason behind classifying like this. So first, D0 and D10. D0, D10 complexes are usually colorless, not always. Why they're colored? That I will come back soon. And D0 and D10, you cannot unspeak DD transition. But on the other hand, if you look into D1, D4, D6, D9, they have some similarities. In the same way, if you consider D2, D3, D7, D8, they also have some similarities. And D5 is unique. And what are the similarities between D1, D4, D6, and D9? So D1, we have one electron. D4, one less than half field electronic configuration. And D6, one more than half field electronic configuration. And D9, one less than completely field electronic configuration. So that means here, they have some similarities. And they show very similar spectra when you consider electronic or UV visible spectrometer, when we subject those into absorption. And the other hand, if you look into D2, D3, D7, D8, they have similarities. Because in D2, we have two electrodes. And D3, we have less than two electrodes for half-shell, half-field electronic configuration. And D7, we have two electrodes, more than half-field electronic configuration. And D8, we have two less than completely field electronic configuration. So that means again, they have similarities. And the spectra look almost identical, except for some anomalies here and there. And these compounds are colored. And D5 compounds are little bit pale colored ones, especially when they are tetrahedral compounds. And they're almost colorless when they are octahedral homo-liptic complexes having centrosymmetry. We will learn all those things in detail. But now let's look into HGI2. It is brick-red color. And here, electronic configuration, if you consider mercury, D10 electronic configuration is there. So no DD transition is there, but still it is intensely colored. When you consider potassium permanganate or potassium dichromate, metal exists in their highest possible oxygen states. That means Mn is in plus seven state. All electrons have been removed. And it has a D0 electronic configuration. But still, Km4 is intense purple in color. And similarly, if you look into potassium dichromate, chromium is in plus six state, and D0 system, and also intense orange-red color. And similarly, if you look into bismuth triiodide, we have D10 S2 electronic configuration. It's orange-red in color. And if you look into prussian blue, here we have iron 3 and iron 2 all there. This is also very intense in color. We will see one by one why. What is the origin of color in these compounds as you cannot anticipate any DD transition? So iodide has very high polarizability, which results in anionic charge getting easily transferred to the mercury 2 plus cation. So this process releases some energy which falls in the visible spectrum. Hence, we can say that compounds like MnO4, Hg2, et cetera are colored. That means what we come across is a term called charge transfer transition. When we talk about charge transfer transition, the moment we look into the electronic configuration, we should be able to tell in which direction charge is transferred. For example, if you look into D10, it is metal to ligand charge transfer transition. On the other hand, if you look into KMnO4, it is D0. It is ligand to metal charge transfer transition. And if you look into prussian blue, here we have iron in 3 state and iron in 2 state. So here it is metal to metal charge transfer transition. So these sum of these charge transfer transitions are responsible for the appearance of intense color, the respective colors in these compounds. So now we should come back to the selection rules of electronic transition. So when we were writing microstates, we saw several possible exercise states. And then if electrons are promoted to each state from the ground state, then electronic spectrum would have been very complicated. And elucidation or interpretation would have been very difficult. But however, all these transitions are governed by certain rules. That we call it as selection rules for electronic transition. What are those? So electronic transition may be classified as intense or weak according to the magnitude of epsilon maximum that corresponds to allowed or forbidden transitions. That means what are allowed transitions? These transitions have a maximum value tends to 4 or more. And property of their occurrence is very high. These are generally due to pi, pi star transitions, allowed transitions. And then example, pi, pi star transition in 1, 3 butadiene shows absorption at 217 nanometer. And epsilon has 2,900 represents an allowed transition. Then if allowed transition term comes, there should be forbidden transitions. And here if you look into it, n to pi star transition is forbidden. For this transition, e maximum is generally less than 10 or rise to 4. For example, if you consider n to pi star transition of saturated aldehydes and ketones, which exhibit a weak absorption of low intensity near 285 nanometer and having e maximum value less than 100 because they are forbidden transitions. So that means which transitions are forbidden and which transitions are allowed, we can look into it now. Not all theoretically possible electronic transitions are actually observed. That's what I told you because they are bound by a selection rules. Selection rules are there to distinguish between allowed and forbidden transitions. And allowed transitions occur. Forbidden do occur, but much less common and are of low intensity in nature. So now the first one is leopard selection rule. What is leopard selection rule? The transition should involve delta l equals plus r minus 1. If that is involved, they are called leopard allowed and have high absorbance. That means during electronic transition, azimuthal quantum number should change. It should not be, delta l should not be 0. This is about leopard selection rule. And then what is spin selection rule? During the electronic transition, electron does not change its spin. This electron is getting excited to the higher level. It should stay something like this. This is allowed. And then if this electron, if it goes like this, this is not allowed. So that means what it says is delta s equals 0. That means during electronic transition, electron does not change its spin. If the upward spin is there, it should go and sit in excited level, something like that. It's not allowed to go and sit like this. This is exactly opposite to NMR selection rule where in order to see nuclear spin transition, flipping of nucleus when you are applying radio frequency in a direction perpendicular to the applied magnetic field, the frequency of that one matches the present frequency of the nucleus, then what happens? It would start wabbling and coming out from more and more coming out of the axis of magnetic field and its flips. So there, if you consider in NMR, delta s equals plus or minus 1. One should remember, whereas in case of electronic transition, it is delta s equals 0. In nuclear transition, it's opposite. Upward spin will become downward spin. So here, the spin state doesn't change. That means d d transits are strictly speaking leopard fermedon. That means we should not anticipate any d d transition at all because it says delta l equals 0. And all d rs have the same delta l value. And of course, since they are leopard fermedon, that can be seen from the epsilon value. Molar absorptive coefficient will be in the range of 5 to 10 liters per mole per centimeter minus. When transition metal forms a complex, then why do we observe d d transition? When transition metal forms a complex, M is surrounded by ligands. Mixing of d and p-arbals may occur. And as a result, transition is no longer d d in nature. Of course, you can also recollect from Welles-Born theory about hybridization in case of octahedral complexes. When octahedral compound is formed, the hybridization we are giving is d2 sp3 or sp3 d2. So that means s and 3p and 2d-arbals, dz square and dx-my-square will combine to generate a set of six hybrid-arbals, which are having the properties of all the ligands which are involved in this hybridization. So as a result, what would happen? The d-arbals present in sp3 d2 are no longer d-arbals we observed in free gaseous metal ion. So that means mixing of d-p-arbals may occur as a result, transitions are no longer. When that is split into t2g and eg, they are no longer d-arbals of same property that we saw when it is not kept in the ligand field. So now, because of the mixing, they lose original, so d character. As a result, what happens? It becomes leopard-alloyed, and hence we see d-d transition. So because of slight relaxation in leopard rule, d-d transitions are observed. Mixing is common when complexes lack centrosymmetry. Mixing is even more pronounced when they lack centrosymmetry. So now, let's look into tetrabromomanganate 2 minus. This is tetrahedral. So manganese in plus 2 z is a d5 system. And then, if you look into pentamin chlorocobal 2 plus is octahedral with unsymmetrical substitution. And both the compounds are colored. This compound is also colored. This compound is also colored. So in case of MnBr4 2 minus, if you consider, and it's a tetrahedral compound, we have something like this. So here, if you see d5 system, all the electrons have a spin like this. And here, if you promote one of this electron, it should go here, or here, or here. If it goes here, both the electrons will be having the upward spin. And that is not allowed. That's the reason this is spin-permanent transition. But nevertheless, in tetrahedral compounds, that is allowed. And you can see, they show color because they lack centrosymmetry. And same thing is true in case of this one. We have d7 here. So here, any of these electrons with downward spin can go. And transition can be seen. And here also, it doesn't have centrosymmetry. It's unsymmetrical substitution is there. These compounds are colored. But on the other hand, if you look into homolypti compounds such as hexamin-cobalt 3 plus or hexamin-copal 2 plus, they have centrosymmetry. No mixing of P and D orbits are observed. And as a result, they are not colored or very pale colored in nature. So then, why this color comes, despite having centrosymmetry? That means when the metal to ligand bond vibrates, so that the ligand spends an appreciable amount of time out of their centrosymmetric equilibrium position, as a result, small amount of mixing to occur and low intensity transitions are observed. So that means when they are vibrating with respect to the mean position, it so happens that all the ligands will be at a different position from the central metal atom. As a result, they come out of centrosymmetric equilibrium. And hence, this is essentially due to the mixing of P and D orbitals. And that results in low intensity transitions. So now, you can see the type of transitions. Spin-fibrant transitions will be having very less epsilon value, typically less than 1. And the example is hexa aqua manganese 2 plus. It's a homolypti compound having centrosymmetry. And again, having D5 electronic configuration, the spin-burburden, and also Leporte-Furburden, that is. And if you consider Leporte-Furburden and spin-allowed, they will be having epsilon value anywhere between 1 to 10 and then R 10 to 1,000. For example, if you look into hexa aqua titanium 3 plus, a D1 system here. So here, centrosymmetry is there. Octahedral compounds homolyptic, that's the reason this value is low. And then if you look into tetrachloronucleate 2 minus, this tetrahedral compound, centrosymmetry is not there. And then it is a spin-allowed. And here, the value is about 10 to 1,000. So it's slightly better. And in case of charge transfer, we can see 1,000 to 50,000, for example, manganate ion, potassium per manganate. This is essentially ligand to metal charge transfer transition. So now, let's look into the system and type of spectra we will see. And also, when they are allowed, leopard and spin-allowed. So when a transition is leopard-allowed and spin-allowed, that has to be charge transfer transition. And they will be having very high epsilon values. Example, TiCl6 2 minus. And then, a partly-allowed and some PD mixing is there. We'll see DD transition. Here it is 500. And for example, if you take tetrahedral compounds of bromo and chloro. So tetra-bromo cobaltate or tetra-chloro cobaltate. And then, a leopard, a firmadon, and spin-allowed. We have quite a few. Here it is, again, termed as DD transition. Epsilon value is in the 10 to the 10, where we have centrosymmetry is there. For example, hexa-acquatatin M3-pes or hexa-covalidium 3-plus here. Then, partly leopard-allowed and spin-firmadon are those where we have typical D5 system. Epsilon value is very low, 4 here. And if it is spin-firmadon and leopard-firmadon, the value is very, very minimum, 0.02 epsilon value. In those DD transitions, very weak, and very weak, intense in nature. Intensity is very, very weak. And these are especially having D5 system having centrosymmetry. An example is hexa-acqua manganese 2-plus. Then, let us look into what are the factors that affect crystal-field stabilization energy and absorption wavelength. So, the absorption wavelength is directly proportional to the crystal-field stabilization energy we see. And that is governed by spectrochemical series. I have given here a very extended spectrochemical series here. Most of the ligands have included here. The magnitude of delta depends on the charge on metal. Metal measures in plus one state, plus two state, or plus three state. And also it depends on position in periodic table. That means whether it's a 3D series, 4D series, or 5D series. And also the ligand field strength. All these three factors have an influence on the magnitude of crystal-field stabilization energy. And here, increasing with increasing field strength so wavelength decreases. So that means as crystal-field stabilization increases and homo-lumo gap increases, as a result, what happens? High energy is required, wavelength decreases for those things. And you can see here, in case of octahedral, the energy gap between homo-lumo is very high. And that is slightly less in case of tetrahedral. And here, in case of sway planar, we are considering this transition here, dz squared to dxy level. So now let us consider metal to ligand charge transfer transition. If the metal is in a low oxygen state, that means it is electron rich. And the ligand possess low-laying, empty orbitals. They should have pi star in case of aromatic ligands, or carbon monoxide, or n-heterocyclic carbene, or sigma star in case of phosphine. Then a metal to ligand charge transfer transition occurs. We should remember here, metal is in low oxygen state. When the metal is in low oxygen state, it has more electrons. It is almost vulnerable for oxidation. That means it can readily give electrons. So those metals, when they are combined with ligands having low-laying, empty orbitals, like pi star or sigma star, then a metal to ligand charge transfer is smooth. And you can see MLCT here. MLCT transistors are common for coordination compounds having pi acceptor ligands. Whatever the back donation we call it, metal to ligand back donation, it's nothing but metal to ligand charge transfer. Upon the absorption of light, electrons in the metal orbitals are excited to the ligand pi star orbitals. So metal pi orbitals like DXY, DYZ, and DXZ combine with pi star to generate a set of molecular orbitals, bonding and anti-bonding. And this bonding will be occupying the electrons taken from the metal. This we call it as metal to ligand charge transfer, or metal to ligand back bonding. So metal to ligand charge transfer transitions result in intense bands. Example, if you consider TRIS 2,2-bipridel ruthenium 2-plus, this orange-colored complex is being studied as the exact state resulting from this charge transfer, which has a lifetime of microseconds. And the complex is a versatile photochemical redoxyre agent. So other examples are tetracharmonyl phenanthraline tungsten, and also 2,2-bipridel tricharmonyl iron compounds. If you, several examples you can find, which exhibit metal to ligand charge transfer transition. So let's learn more about charge transfer transitions in my next lecture. Until then, have an excellent time. Thank you.