 I'm delighted to have my dear friend and a member of our advisory council to kick off tonight's program. Julia Kamp also happens to be the director of data science at NYU and her specialty is quantum computing. So she's here with us tonight to introduce tonight's event. Please welcome Julia Kamp. Thank you, Cindy. It's always a pleasure to be here and I'm every single time blown away by the caliber of speakers. And that will be that I have the pleasure to introduce this time that Cindy manages to get here and also by the enthusiastic crowd that I see every single time. So it's a big pleasure for me every single time to be here. Maria Chudnovsky is a brilliant and very shining mathematician. Before I speak to her career, she works on graph theory, which happens to be one of these rare topics in mathematics that at the same time gives mathematicians a chance to do very subtle sophisticated math of various kinds. You can do statistics, you can do combinatorics, you can do various kinds of math in a very sophisticated way. And at the same time these are objects that everybody understands to some extent, graphs. I mean we encounter them in our everyday life. So I work in data science where graphs are basically a tool that without which we couldn't live. We use graphs to represent data in various ways and probably you all have heard of for instance the six degrees of separation. So these are social network graphs where we connect to people when they either they're connected on Facebook or they know each other or some such thing. And then we can make inference about the shortest path between two people and the conclusion is that the size is six or less. And that's the sixth degree of separations in graphs that you probably all heard of. There's graphs in biology, protein networks you probably also heard, communication graphs, people study random walks on graph, percolation on graphs. It's really a fascinating object that goes back in mathematics of course several centuries. But it's particularly pervasive and relevant nowadays in particular with the advent of a lot of data and large data which relates to my work. So Maria is as I said a brilliant mathematician and we are very fortunate to have her here. It's a special pleasure to introduce her. She completed her undergraduate degree in Israel at the Technion and then did her PhD in Princeton. And I think either towards the end I think of her PhD or very early after did a really major proof together with coworkers really major theorem in graph theory at the very beginning of her career. And that's the strong perfect graphs theorem which earned her the Falkerson Prize with her co-authors Robert and Seymour and Thomas. The prize was given in 2009. And then she remained in Princeton and her career basically interleaved Columbia and Princeton at various stages. So she became an assistant professor and then associate professor and then she got a named chair in Columbia and is now back since for the last five years has been back to Princeton as a professor. Maria has received various numerous I would say awards and honors. In particular she received in 2013 the MacArthur Award which is also known as the Genius Award. So that's a big you know a big award and a big honor. And it's not so often given to mathematicians actually you know it's given all across. The biologists get it. Physicists get it. Various you know. So it's highly unusual and shows you know the quality of Maria's work that was honored. And she in 2014 was an invited speaker and the International Congress of Mathematics which is an equally big honor in mathematics. So without further ado let us welcome Maria for this talk on parties, donuts and coloring. Well thank you very much Julia for the lovely introduction. It's a pleasure to hear you speak about me. And thank you thank you for inviting me to speak here and thank you everybody for coming. It is not often that people come to listen to me talk about my work for an hour. It's a rare opportunity that I'm very excited to have. So this talk is called parties, donuts and coloring. And the way it's structured I'll talk about three problems each of which is very simple. It's a kind of problem that if you're a mathematically oriented teenager you learn about it in your early teenage years. But then it turns out that these problems are related to research that's still current and still cutting edge. And I'll show you some theorems that people are working on right now. So this is a structure that's three completely independent parts to the talk which is good. And each of them starts simple and then I'll connect it to more modern math. And I should also tell you that this talk begins with a prologue. So you know how in a book you have something, you have a prologue and then you have the main story and the pages are numbered differently in the prologue. They're the same in this talk. The pages of the prologue are numbered with Roman numbers and then we switch to regular Arabic numbers. And the history of this prologue is that I gave this talk at NASA about a year and a half ago and they had a lot of suggestions about how I should make it better. And the first day I got a little bit upset and I said, no, this is my talk. This is my talk, this is the talk I give. And I thought, well, I don't know. They're NASA. Maybe they know what they're talking about. Maybe I should at least think about what they suggested. And so I made all their suggestions and the talk became much better. So they're NASA. They know what they're talking about. And I'm happy to tell you that the talk I'm presenting to you is NASA approved. So the first contribution from NASA is the little section on the history of the subject. So the Wikipedia tells us that graph theory was started when Leonard Euler published a paper in 1736 about the problem of the seven bridges of Königsberg. Königsberg is a city which back then was in Prussia. Those of you who are interested in history know that it went back and forth between many different countries. It must be a wonderful city, but that's not the topic of this talk. So the city of Königsberg is situated on the river Pregel. And so it spans two banks of the river and there are also two islands in the middle. And the different parts of the city are connected by bridges, by seven bridges. I made the drawing, so please be accepting, be kind. So these are the banks and these are the two islands and the blue is the water. And then this island is connected to this island by bridge and this island is connected to that bank by this bridge. This island is connected to this bank by those two bridges. So you understand the notation in this picture. And the question was design a walk that crosses every bridge exactly once. And so you have a handout and I'm going to pause for a couple of minutes and I invite you to think about this problem very briefly. Can you design a walk in the city of Königsberg that crosses every bridge exactly once? All right, so I say that people are still thinking, but I'll continue talking. So the point of this exercise was not necessarily to get you to solve a problem that Euler, the famous mathematician, solved, but just to give you the experience of thinking about this problem so that the next thing I'm going to say is interesting for you. At least somewhat. So the answer is it's impossible. There is no such walk. That's probably one of the reasons why it's taking people some time to find it. And the reason it's impossible is let's think about what a walk like that would look like. So you would start somewhere. And then so you'd start at some landmass and then you would cross a bridge, get to another landmass, leave that other landmass, get to another landmass, enter it on a bridge, leave it on a bridge. So what happens except for the starting point and the ending point and every other landmass, you exit exactly as many times as you enter. And that means except for possibly two landmasses, which are your start and end, every other landmass in this picture should have an even number of bridges touching it. And exit and enter and exit and enter and that's true everywhere except start and end. So now let's look at this picture. So it has four landmasses and this one touches three bridges. This one touches five bridges. This one touches three bridges. This one touches three bridges. So nobody has an even number of bridges, let alone all but two. And that's the reason why you can't do it. So that was a problem Euler solved but a much more important thing that Euler did was the model, what we now call the mathematical model that he built in order to solve the problem. And obviously it doesn't matter that it's bridges and it doesn't matter that it's a river and it doesn't matter that it's the city of Kirigsberg. The only thing that matters in this problem is that we have some points and there are some allowed connections between them that you can walk along. And the question is can you do something in terms of these points and these connections? And that's what Euler did. He understood what the correct abstraction is. He understood what of this problem remains if you want to think about the essence of it. So he used different terminology but today we call this object graphs. A graph has vertices so the landmasses will turn them to vertices. Here they are. This one represents this. This one represents that. You see their position on the plane is similar to the top picture. And then the connections between the points represent the bridges. So for example, this point is that island, this point is a top bank. There is one bridge between this island and the top bank. There is one edge, we call it edge, joining this vertex to that vertex. Between this island and let's say the bottom bank there are two bridges. Between this vertex and this vertex there are two edges. That's how you understand the model. I think it's better that I describe it like this than if I give you a precise definition. All right, so Euler built this model. And then he understood that the only thing that matters in order to answer this problem, in order to answer this question, is how many edges are incident with each vertex? How many bridges touch each landmass? So what's the question now? The question is, can you design a walk in this graph where you walk between vertices and all you can do is walk along edges? And can you design a walk in this graph that traverses every edge exactly once? And Euler said no, because at every vertex except the start and the end you need to enter and exit the same number of times. So every vertex except possibly two needs to have what we call even degree, needs to have an even number of edges touching it. But here that's not true. All vertices have all degree. All vertices touch an odd number of edges. This is three, this is three, this is five, this is three. All right, so that was Euler's victory. And it started graph theory and today to this day we call this an Euler walk. Now that I talked about this, let's turn to page two in your handout and let's see if you can find an Euler walk there. So you can notice that in the picture of page two, it is true that only two vertices have odd degree. And the question is, can you find an Euler walk? All right, so maybe we've thought about this enough. So I'll start the solution and then maybe somebody wants to join in. So the first thing we notice is that in this picture, there are two vertices with odd degree. This one sees three edges, this one sees three edges, and every other vertex has even degree. And we already said they start and end at the odd degree, or what we said is everybody who is not start and end is even degree. So they start and end better be the odd degree vertices, so I have to start here and finish there, or maybe the other way around. So does anybody have a walk that starts here? Does anybody have a walk that starts here and finishes there, or maybe the other way around, and would like to draw it? Okay. Oh, please, come on up. So you draw it here. It's a document camera. You draw it here, but it doesn't ruin anything. It's amazing. Amazing. Thank you very much. All right. Thank you. Yeah, unfortunately I don't have prizes, but... All right, let me show you mine. Mine was different. Mine went like this. All right. And there are all kinds of things you can do. Once you've figured out the start and end, there are all kinds of things you can do. All right. So moving on. So now here's a related problem, which is called an Euler tour. So what's an Euler tour? It's like an Euler walk, except you came with a car and you parked it somewhere. And so now it's not enough to just cross all the bridges once. You also have to come back to your car and go home. So in other words, it's like an Euler walk, except you have to start and end at the same point. Right? And now we can already say it in terms of graphs. So you have a graph and you want to walk, moving between vertices, only walking along edges. And the rule is you have to cross every edge exactly once, or you have to traverse every edge exactly once, and you have to start and end at the same point. And if you think about it for a minute, you realize that if such a tour exists, then the degree of every vertex needs to be even. Every vertex needs to touch an even number of edges because in every vertex you enter and exit the same number of times. There's no start and end anymore. You enter and exit everywhere. All right, so let's do another exercise. On page three of the handout, there is a picture of a graph with all degrees even. So there's a chance it has an Euler tour. And the question is, does it? Can you find one? I guess there's two separate questions. Does it and can you find one? Okay, would anybody else like to share their solution? This is a pen, this is a paper. Thank you very much. All right, well, thank you. All right, let me do mine. This is my prerogative as a speaker. I get to do mine. Uh-oh, uh-oh. Okay, let me not do mine. Okay, so I didn't go here. You didn't see that. Now I go here. All right, so I suspect our walks are possibly the same walk traversed in opposite directions. I'm not sure. Okay, so in fact, it's a theorem that in order to exist if and only every vertex has even degree. If I give you a graph and you can see that every vertex touches an even number of edges, then there's an Euler walk. And if you think about it for a minute, it doesn't matter where you start. It's a closed walk. So whatever you do, you can start wherever you want. And then another point is this is actually a better problem than the first problem I described to you. Because why? Because if you understand this problem, you understand the previous problem. What does it mean to look for an Euler walk in this graph? But remember, we decided we had to start here and finish there. So let me put an edge here. And now let's look for an Euler 2. And it's the same problem. Starting here and finishing there is the same as starting here and coming back here. So somehow the closed walk problem is better. All right, so this completes the NASA portion of the talk. And now we continue with my talk. And as I said, my talk consists of three problems that I want to describe to you. And then I'll explain their connections to modern mathematics, more modern mathematics. So the first problem is, true or false, of every six people at a party, either there are three who know each other or there are three who don't know each other. At the other party, you get your favorite six people or your least favorite six people. And among them, there is either three who know each other or three who don't know each other. So it is true. This fact is true. And let's try to prove it. And maybe let's try to prove it by drawing a picture. Here is the first person of the six. Here are the other five. And you can see, I'm a graph theorist, so you can make them a little more human. So this guy, if I look at those five, either he knows three of them or he doesn't know three of them. If you have five people of them, either he knows three or he doesn't know three. So case one, he knows three, then I can assume it's these three. Now, if no two of these three know each other, that's three people who don't know each other, if some pair of those knows each other, let's say this one, then this and this and that are three people, every pair of which knows each other. That was case one. That was when there are three people that he knows. That he doesn't know. I can assume it's these three. If they all pairwise know each other, that's three people who know each other. If, let's say, this pair doesn't know each other, then this and this and that are three people who pairwise don't know each other. Now, I'm sure you didn't quite follow this proof, but I'm sure you did follow the fact that it's the same thing over and over again. So what that means is that we need to build a model that would make the symmetry clear. If you kept making the same argument, that means there's some symmetry hidden in there and you just haven't found it yet. So let's do that. Let's build a better model. Here it is. So I'm going to build a graph with six vertices. The vertices represent the six people at the party, and I'm going to draw an edge. I'm going to color the edge red. If two people know each other, I'm going to color the edge blue if two people don't know each other. So here is the model. Now, what am I looking for? I'm looking either for three people who know each other. That's a red triangle, a triangle where all three edges are red, or three people who don't know each other. That's a blue triangle. That's a triangle where all three edges are blue. So in other words, I'm looking for a monochromatic triangle. I'm looking for a triangle where all edges are the same color. Well, now the symmetry is clear. There's no difference between red and blue. If what I'm looking for is a monochromatic triangle, then there's no difference between red and blue. So now let's do this proof again. But in this language, you can see how much better it becomes. Okay, so here is vertex number one. Here are the other five. He is joined to... Either he is joined to three of these red edges or by blue edges, and there's symmetry between red and blue. So I can assume he is joined to three red edges, and this is going to represent red. By the way, this is going to represent blue, because these are the two things I know how to draw. So if let's say this edge is red, then fantastic. This is a red triangle. This is red. If all the edges are blue, if all the edges among these three are blue, then this is a blue triangle, right? Exactly the same proof. I said exactly the same words, except much clearer because we got rid of all the things that didn't matter. We only kept the essence. Okay, so... And by the way, what if five people came to the party? Is it still true that there is a blue triangle or a red triangle? The answer is no, because of this party, right? These are five people. The outer cycle is pairs that don't know each other. The inner cycle is pairs that do know each other. There is no monochromatic triangle in this picture. So we just finished proving Ethereum, which in graph theory we would describe as R of 336. So what that means is that if I'm looking for... If I take a graph with some number of vertices and all the edges present and I color the edges red and blue, then in order to guarantee that there is either a red triangle or a blue triangle, I need six vertices. Six vertices is enough and I cannot do it with fewer than six. So here we proved that six vertices is enough. Here we proved that I cannot do it with fewer than six. So together we proved an equation. And graph from the theory is this. For every R and S, there exists an N, so that out of every young people at the party either know each other or as don't know each other. So you give me your favorite R and S, a million and a trillion. And there's a big enough party so that either a million people know each other or a trillion people don't know each other. And a fancy way to say it is R of S and T is finite. No matter what S and T are, there's some number of very big but finite that would give you a monochromatic S configuration or a monochromatic T configuration. All right, so that's very nice. It's finite. Now what if I actually wanted to know what size it is? That's a much harder problem to get the right answer. So for R33, it worked very well. We immediately got it to be six. For R44, it's known to be 18. It's much harder. For R55, we don't know the answer. Can you believe it? For R55, we don't know the answer. It's between 43 and 48. And in 2017, people proved that this is the gap. Until 2017, the gap was bigger. So it's amazing. For R66, the situation is even worse. And Paul Erdos, you might know who he is. He's a legendary mathematician who worked in combinatorics and in number theory and all kinds of things and probability. And he used to say that, to describe this problem, he used to say if aliens invade the earth and say we're going to kill everybody, unless you calculate R55, then we should collect all the mathematicians and all the physicists and all the scientists and all the engineers and get them to calculate R55. However, if aliens invade the earth and say we're going to kill you all off unless you calculate R66, what we should do is collect all the mathematicians and physicists and scientists and engineers and send them to fight the aliens. So this is round it there. Okay, so this is very nice. This was all known in the early 20th century. But now let's generalize it. And if you generalize it, you get some cutting-edge math. So let's say... So far, what did we do? We said I take the complete graph. I take a very big graph where every pair of words is adjacent and I... two colors the edges. I color the edges with two colors. And I want to know, can I get a monochromatic copy of this or a monochromatic copy of that? That was the question so far. Now, what if I don't start with a complete graph? What if I start with a graph where some pairs are adjacent and some pairs are not adjacent? And now I'm going to color all the edges. So some edges are red, some edges are blue, but some pairs, there's not an edge there. And then I want to know what can I say... And so now I have some little target graph H. It used to be a monochromatic triangle before, but now it's some other H. And I want to know, for what graphs G, no matter how I color it, I can get a monochromatic H. So in that case, we say that G is Ramsey for H. Okay, so that's the concept you can think about. And then an interesting question to ask about this is, what about two different targets? Are they the same or are they equivalent or are they not equivalent? What does that mean? That means, can I say a big graph contains H1 if and only if it contains H2, right? In the sense every two coloring contains a monochromatic H1 if and only if it contains a monochromatic H2. And then you can ask which pairs of graphs are Ramsey equivalent, right? Which targets are the same? For which pairs H1, H2, is it the case that if some big graph contains one, then it contains the other? And that's an interesting question. That's like people approving theories about it right now. For example, it was recently shown that if you take a graph on n vertices and then next to it you draw a graph on n minus one vertices, that's not the same. That's not equivalent to, I'm sorry, a clique on n vertices. So n vertices all pairwise adjacent. Next to a draw a clique on n minus one vertices. n minus one vertices all pairwise adjacent. So that's graph one. Graph two, just a clique on n vertices. These two are not Ramsey equivalent. And they're very small. And that's a theorem from like three years ago. Just to give you a sense of how we suddenly go to cutting edge math. Another question you can ask is, why do I have to two color? I don't have to color red and blue. I can color as many colors as I want. What can you say then? So then it turns out, right? So then you can talk about R, S1, S2 up to SK. So then it turns out that's still Ramsey, right? So if I give you a big graph and a big complete graph, all pairs are adjacent and I color every edge with some color, but the total number of colors is fixed. And I ask, how big does the host graph need to be in order to contain a small complete graph, monochromatic small complete graph. Turns out that number is still finite. And actually it's the same proof basically as for two colors. But still it's a nice generalization. All right, problem two. Everybody who went to sleep can wake up. Everybody who feels like they've done their part can go to sleep. Moving on. Problem two. So problem two is about map coloring. What's map coloring? You're given a map, right? On a piece of paper or on a globe, which actually is the same. And you want to color the countries in such a way that if you look at this map from space, you can tell where one country is and another country starts. So what does that mean? That means that if two countries share a border, you want to color them with different colors. And here, share a border means they share a little interval of the border. They just share a vertex that doesn't count. If you look at that from space, you can see it even if they're the same color. Okay, so the question is, this is a setup. I give you a map. How many colors do you need in order to color it in this way? And the theorem is four colors is enough for any map on Earth, which is kind of amazing. This is probably the most famous theorem in graph theory, except I haven't told you the whole truth yet. You need another assumption. Otherwise, this is completely false. So you shouldn't allow Alaska. Countries should be contiguous or as we say in mathematics, connected. I think contiguous was the last word I learned in English because in math, you don't need it to say connected. But actually, this morning, I was practicing. So really, it's the last word. Okay, so let me show you why you need to assume that the countries are connected. So otherwise, here is an example. So here's a bit of country one, and it touches country two and country three and so on up to country 100. Here is a bit of country two. Remember, each country consists of many islands. So here's a bit of country two. It touches country three, country four, country five, up to 100. Here is a bit of country three, and it touches a bit of country four, a bit of country five, up to 100. And you can keep going like that. The last thing I'm going to do is a little bit of country 99 touching a little bit of country 100. So what did I do? Oops, thank you. There's a reason. It doesn't fit. So if you allow countries to have islands, then the theorem doesn't work. But if you insist that the countries are connected, then the theorem is true. And so it's a lovely fact. It's a surprising fact that four colors is enough for any map no matter how complicated you make it. So let's first see that you actually need four, that you can't hope to strengthen it tomorrow in proof three. So here are some maps that need four colors. In the top map, so you see four countries are colored with four colors, purple, green, red, and blue. And how can I convince you that you really need four colors? Well, it's because every pair of countries touches. Every pair of countries shares the border, so you have to color each country in a different color. Now, okay, so maybe that's the only reason. Maybe if you don't have four countries all sharing an interval of the border, then three is enough. But no, because look at this picture. So the picture here is this is one country, and then the triangles are additional countries. And I claim, and so here there are countries that don't touch it, but our definition, this country doesn't touch this country. But I claim you still need four colors here because why? So suppose I managed to color it with three colors. So I can assume this outside ring is colored color one, and now I have colors two and three left to do the triangle. So I can assume my color is two, and then here's a three, then he must be a two because he touches a three, and he must be a three because he touches a two, but now what about him? He touches a two here and a three there contradiction. So that's an example of a more complicated map where you need four colors. So don't try to improve it, this is the honest truth if you assume the countries are connected. And it's a nice fact, it's a beautiful fact. It was conjectured by a cartographer in 1860 or something like that, and he noticed that every map he tried to color four colors were enough, and then mathematicians decided they were gonna take on the task and prove it's true. But what's even more important about this theorem is how much research is generated. So it was finally proved in 1972, right? So from mid-1800s to 1972, that's more than 100 years. And it really produced a lot of important things in mathematics. So for example, in early in the 20th century, there was a proof by Kempe that turned out to be full, to be wrong, he didn't prove this theorem. But he came up with this tool that are called Kempe traits. And it's one of the most important tools in graph theory to this day. I can't tell you how many theorems are proved using Kempe traits. So who cares if he didn't prove the four color theorem, he invented Kempe traits. And then in 1972, and finally it was proved, it was the first theorem that was ever proved using a computer. Apple and Hacken did it, so in the 1950s they built the first computer in 1972, less than 20 years later. Apple and Hacken said, not only can you use a computer to do five plus six, you can prove theorems with it. And they did it. They had a proof done by computer, or mostly done by computer. So that's just such a good theorem, right? It brought so much good to the world. Okay, so that's the four color theorem. So now let's generalize, let's get closer to more modern mathematics. So, okay, so far we're talking about countries on Earth, which maps on Earth, which means a map that I draw on the globe, right? On the sphere. Now what if I take some other surface? What if I take the donut, which in French language you call the torus? Or what if I take the pretzel, which in French language I call the double torus? Or general surface was g-holes, which in French language you call surface of genus g. What can you say about that? What if I draw a map there? How many colors do you need? So, you know, maybe if the surface becomes sufficiently complicated, then for any number of colors, you can draw a sufficiently complicated map so that you couldn't color it with that number of colors. But the answer is no. For every surface there's a constant so that c-colors is enough. And then just like with Ramsey theorem, here you can also ask what's the best constant, right? So you can prove there's some constant, what's the best constant? And it turns out... So first of all, this theorem is not very hard. If you know what Euler's formula is, and it's not very hard, it all goes back to Euler. So if you know what Euler's formula is, this theorem is not very hard. And in fact, for all surfaces, except the sphere, so you can get the best thesis way. You can prove that whatever the easy proof gives you is actually best possible. For the sphere, it's not true. For the sphere, the easy proof gives you the six-color theorem. It tells you that you can color with six colors. And I guess I'm not sure when that was proved, but I would say at least 50 years before four was proved, right? So six is easy. Five you can do using camping chains. So the four colors, so to get a finite C is not very hard. To get the best C for more surfaces is not very hard. To get the best C for the sphere is very hard, but it was done. And this gave us the whole part in graph theory called topological graph theory. And what that means is they want to know how to color maps on surfaces, but if I tell you something additional about the map. So for example, I showed you that... I showed you that you need four colors for maps on the sphere, but what if I tell you something additional about the map? And it turns out the good thing to say is something about what rings of countries, rings of adjacent countries look like. And so there's a recent theorem that three colors is enough if you don't have a ring of lengths less than five. Whatever that means. But it's something people think about. When can you improve four? What kind of information can you give me so that I can improve four? Or more generally, what kind of information can you give me about the map on the general surface so that I can improve this best possible C? All right, so moving on. Problem number three. Now, problem number three is actually something I work on, so I can talk about it for a really long time. But I won't, because it's five to eight. But I'll talk a little bit. And then possibly I'll get to slightly more technical stuff. Okay, so interaction to problem number three. Every graph theorist hears this story that I'm about to tell you, and I have to tell you right away. I have no idea if there's any truth to this, if there's any problem like this, or if just a graph theory professor somewhere in the world invented it, and then everybody else repeated it. So I apologize in advance to anyone who knows anything about cell phone towers or frequencies, but that's the problem I grew up on, so I'm going to tell you. So the problem is you have cell phone towers and they transmit at various frequencies, and you want to make sure that their transmissions don't interfere. So like if two towers are close to each other, then they might interfere, so you have to give them different frequencies to transmit on. But maybe if there's a mountain between them, then even though they're close, they don't interfere, so you don't have to give them different frequencies to transmit on. So some story like that. And the point is you need to buy the frequencies from the government, so you have to solve this problem, assign a frequency to every tower, but do it with as few frequencies as you can. So that's an example of what we call a graph coloring problem, because how do you model this as a graph? You build a graph whose vertices are cell phone towers, and this is a model with seven cell phone towers, or seven vertices, and then you put an edge between two vertices if there's a possibility of interference. And now you need to assign frequencies, which for abstraction we call colors, to each vertex, and the rule is if two vertices are adjacent, then they get different colors. And this is called graph coloring. Here's the same thing said again, given a graph color its vertices, so that if two vertices are adjacent, they give different colors. Now, one thing I can try to do is give every vertex a new color, that would certainly work, but the goal is to do it as few colors as possible. Another way to think about it is you want to partition the vertices of the graph into sets, and the rule is that within a set, you're not allowed edges. In a set, you're not allowed interferences. And you want to partition into as few sets as possible. Now, I don't know if the cell phone tower story is correct, but I can tell the story from my life, which is correct. So, when I got married about eight years ago, like everybody else, we had to make a seating chart for our wedding. And everybody who's ever involved in organizing a reception knows that this is kind of a pain, but I decided I was going to impress my future family by saying, I'm going to do it, and I'm going to do it in five minutes, there. And indeed, I did, and the reason I could do it is because you can think of it as a graph coloring problem. So, the vertices of the graph are the people you're inviting to your reception, your friends and family, and then you put an edge between two vertices. If they really don't like each other, you can't put them at the same table. So, this is edges, and now you need to partition your graph into sets, into tables. The rule is two people who can't sit at the same table don't put them at the same table, don't put them in the same set. Okay, so, and that's it, and now we want to do it with as few colors as possible because you need centerpieces and they cost money. But... But... So, okay, but, I mean, those of you who know something about graph theory know that it's a very difficult problem to find a coloring of a graph with even a good approximation of an optimal coloring, it's hard. But the thing is if your friends and family are reasonable people, then this graph doesn't have too many edges. It sometimes happens that your graph doesn't have too many edges, meaning only a few people here and there really wouldn't sit at the same table, and otherwise it doesn't matter, otherwise it's fine. And so then... I think this talk will go in a very different direction. And what happens is if your graph doesn't have too many edges, then it is an easy problem. Then you can use something called the greedy algorithm, which is a very easy algorithm, and find an optimal coloring, and that's exactly what I did. I didn't tell anyone what I used, the greedy algorithm. And now my family likes me. Okay. So, this was my story about graph coloring. And so now, okay, so now let's get back to math. Why don't we go back to math? Why don't we go back to math? Let's get back to math. Why is it some... So, let me just say this number, chi, is called the chromatic number, and it's the smallest number of colors you need to color the graph, right? The smallest number of tables you can use for this particular part. So... Why would chi be big? Why if I give me a graph, why would I need many colors in order to color it? One reason is that maybe there's something that's called the big click in the graph. This is all pairwise adjacent, and you have to color each of them with a different color because they're all pairwise adjacent. And so we immediately prove a theorem that says chi is at least omega. Omega is the size of the biggest click in the graph. The number of colors you need is at least the size of omega. Is at least omega, at least the size of the maximum click. And then one might ask, well, maybe that's what we call in math the only obstruction. Maybe that's the only reason why you can't color a graph. If a graph doesn't have a big click, then you can color it with few colors, with not too many colors. And the answer to this question is also no. Because of a couple of examples, so here's one. These are what we call odd cycles. Odd cycles. So a cycle means a bunch of vertices. You draw a bunch of vertices in a circle and you make consecutive pairs adjacent and non-consecutive pairs non-adjacent. And that's called the cycle. And the length of the cycle is how many vertices you draw in the circle. So for example, this is called C5 for cycle of length 5. This is called C7 for cycle of length 7. And you can do it for general K, right? C2K plus 1. By the way, my 5-year-old came to my talk at 4. And he, at this point, he got very excited. And apparently what he wanted to know is who is general K? So you can do it for general K. And I claim in all these cases the chromatic number, the clique number, and all the same. Because why? So how big a clique can you get in a cycle? The only adjacent sets of vertices are consecutive pairs, right? You can't get three pairwise adjacent. So the clique number, for each of these graphs, the clique number is two. And then what's the chromatic number? Well, let's see if we can color this graph with two colors. So I can assume he's colored one and then he must be colored two because he has a neighbor who is a one. And then he must be one and he must be a two and he must be one and he must be a two. But then he's stuck because he is adjacent to a one and he's adjacent to a two. So you need three colors. So omega is two, K is three. So fantastic. We constructed an infinite family where omega is two, where omega and K are not the same. Maybe this is it. Either you can color with omega colors or it's an odd cycle. So then it's still not quite because here's another family. So these are called infinite anticycles. So what's an anticycle? You put an odd number of vertices as an infinite anticycles. I meant to say odd anticycles. It's an infinite family of odd anticycles. So you put an odd number of vertices on a circle and you make consecutive pairs non-adjacent and non-consecutive pairs adjacent. So like here. Consecutive pairs and non-adjacent, non-consecutive pairs are adjacent. Same deal here. And you can prove that if you have two K plus one vertices, then omega is K. The clique number is K and you need K plus one colors. So that's another family where K and omega are not the same. So maybe this is it. This is basically yes. And I'm going to explain to you in what sense it's basically yes and I'm going to stop. So, okay. Let's ask for something a little more interesting than just whether K and omega are the same. Let's make a new condition that says however I delete vertices from the graph, K and omega are the same. The clique number and the chromatic number are the same. So the graph is called perfect if however you delete vertices from it, the clique number and the chromatic number are the same. So for example, this graph is perfect. It has four vertices. The clique of size four is four colors. However you delete a vertex, it's still a complete graph. This graph is not perfect and you can see it right away because the clique number here is three. The best thing you can do is a triangle but the chromatic number is four. And the reason for that is if you could color this with, I mean it really did this argument so if you could color this with three colors, this would be one and now this five cycle would be colored with two colors and we already proved you can do that. So I mean this is an example of this. Certainly a graph that's not perfect but it doesn't demonstrate my point. My point was that we want a slightly different condition than just K equals omega. So let's have an example where the graph on the surface looks like, maybe on the surface is not a good expression to use in the graph theory top. On the face of it looks like it's perfect but actually it isn't. So here's this graph. In this graph the clique number is three because of the triangle on top. The chromatic number is also three because you can see I have three colors with colors one, two and three. However it's not perfect because if you delete the top vertex then you're left with this guy on the wrong way around but here the clique number is two and the chromatic number is three. So that graph is not perfect even though it itself has a K equals omega. So now let's try to think about what perfect graphs are. So Ken, is it possible that in a perfect graph I can delete vertices and make an odd cycle? And the answer is no because for odd cycles K and omega are different but for perfect graphs however you delete vertices it's the same. In a perfect graph however I delete vertices I get a graph with K and omega the second. So certainly if a graph is perfect then it doesn't contain odd cycles and it doesn't contain odd anticycles in this sense of deleting vertices and the thing is it's even on leaf. So the strong prefer graph theorem as Julia mentioned in her introduction tells you that the graph is perfect if and on leaf no odd cycle and no component of an odd cycle is perfect if and on leaf no odd cycle and no component of an odd cycle can be obtained from it by deleting vertices. So the only if part is easy the only if part is already proved. Proved. The if part is hard. This is a conjecture that was opened for 40 years and then a group has proved it. So the proof was 150 pages long which is long for a proof. I realize maybe not everybody knows how long a normal proof is a lot. And then a few years after the original proof Paul Simmer and I found another proof which was only 110 pages long but there was something else good about it. So when you do proofs in math you have some intuition, you have some idea and then you write it down and you formalize it and it comes out and now you have a proof. So there is sort of this gap between an intuition and a proof but first you have to have an intuition. And then every now and then you don't have an intuition. You just do a calculation and somehow miraculously it comes out. And when you get a proof like that you feel very uncomfortable because you don't actually understand why it's true. So for all you know there might be a mistake in it. So we don't like proofs which came from a calculation we didn't understand and not from intuition. And the 150 pages the last 50 pages of it we didn't understand why they worked. We kept checking and other people kept checking and it seemed it was probably true but we really couldn't understand why it all aligned like that. And in the 110 page proof we got rid of the set. We replaced it by another argument which we understood. So the proof got somewhat shorter and much truer or much more comfortable. And then the last thing I want to say is that another open question in the field was whether you can recognize perfigraphs efficiently and efficiently means in polynomial time and the answer is yes this theorem was proved a few months after the strong perfigraph theorem was proved even though it has no connection to the strong perfigraph theorem it doesn't use the strong perfigraph theorem barely even uses the same methods but somehow by the time we could prove this we had enough understanding about the field that we could also knock off that. And let me stop here. Thank you. All right we have time for some questions. You raise your hand and I will bring the microphone. Hi with a problem like the bridges of Konigsberg it's kind of intuitive for someone who's not a graph theorist how drawing the pictures could lead to a solution and trying different ways. But when you get to your 150 page proof type problem I have a hard time believing that you sit at your desk drawing graph like the bridges of Konigsberg. I spent a lot of time with my desk drawing graphs. So some of these have got to be solved by computer can you make a connection for me going between doodling so to speak and maybe I don't know whether it's matrix math I'm guessing. So sometimes you can reduce the problem to a finite number and use to check. And then after you've done that if you're good at it you can feed them to a computer and the computer will check them. I'm not sure this answered your questions. This proof didn't use a computer so I cannot make this connection for this because it's not there. But for example for the four color theory what you prove is that there is 300 graphs that if you can four color each of the four colors then you can color any map with four colors. There are some reduction rules to those 300 and then if you color that you can expand it back to where you started. So then you feed the 300 to the computer. You're at Princeton which if memory serves was the mother load of operations research work in the last century. Has your work in graph theory contributed any extensions to linear programming and dynamic programming in those fields that you can speak of or are you strictly on the theoretical end of things. Yeah I think it's more the other way around. I think every now and then we used tools from combinatorial optimization like the ellipse method and the whole idea of dynamic programming. I'm not sure specifically what I did helped optimizers but who knows. Have there been any interesting problems relating to social networks that graph theory has I don't know had to input on. So certainly I mean the study of social networks as far as I'm concerned obviously I'm biased is a branch of graph theory. Right I mean you have this networks which are graphs and you want to know things about them. I was actually talking to somebody in my office just a few days ago and he was explaining to me that they were looking at pairs of friends on Facebook and they wanted to see if you look so let's see so you look at person 1 and you look at his friend 2 and now you want to see among the friends of person 1 how many friends that person 2 have. And they did right and sort of not just how many but what does it look like if you just look at that graph what it looks like. And they had a heuristic that predicted that person 1 and person 2 as falses and what they did was look at this graph and the heuristic is if person 2 knows many friends of person 1 who are not know each other among themselves that means person 2 has connections with person 1 in many different areas and that probably means there's falses or at least some how related. The second best hit was sibling as far as I understood. I was very interested when you brought up the cell phone problem because it is a real problem. New York City should have had an auction X amount of years ago for its wavelength and it's been delayed and one of the reasons is what you mentioned that it's people will are supposedly going to pay more for the tower. I'm simplifying it. But the problem is that many of these towers are adjacent and so not only do they have to figure how much is coming through but if one has more wavelength just to say than the other and it's near that can also cause problems. So maybe you could offer your services to the city of New York. If the city of New York will take them I'll offer them. Alright, so one more hand for our speaker. Thank you for coming.