 We are very happy to have the GU back and we'll be talking today about polynomial Chi binding functions for graph classes. So, take us away. Right. Thank you very much for everybody for coming. It's really nice to be back. You know, hopefully I can we can see that the new, the new renovated the console. Yeah, so today I want to talk about something slightly different from what we do here. It's about structure graph theory. So in particular, we want to talk about some kind of relation between coloring and the structure of graphs. So this is about polynomial Chi binding function for graph classes. So what is the background. So we say a graph. So given a family of graph H we say a graph is H free. If this graph G doesn't contain any H in the current age as induced sub graph. Okay, so here we are we are mainly talking talking about forbidden induced sub graph. And then the commit commit a number is the least number of colors to property color graph. The click number is the largest size of the largest number vertices of the complete graph. So we know clearly that the committed numbers at least the click number because if we have a click of omega vertices then we need every vertex need to be colored differently so we need at least omega colors. So the problem is that can we bound the committed number by by click number by some function of the click number. So if there is a function as such that the committed number of G is upper bounded by F of Omega. Then we say that this graph class is kind bounded. So color color color bounded, and then the function that's here F is called the binding function. Okay, so, in general, the family of all graphs are not kind bounded. So there are multiple evidence for that first of all, cut and Michelle ski have some constructions that for any end, there exists triangle free graph. So triangle free then we know the click number is two. And then but the chromatic number can go actually high. So there's no function that such that the committed numbers upper bounded by a function of the click number. And then another construction famous construction this odish probabilistic proof that there exists a graph with arbitrarily high growth and arbitrarily high chromatic number. Which means that if we have a finite number of forbidden graph. Okay, and then if all the graph in the courage are not trees, then this family will not be kind bounded because if if all of them contain some cycle. If the list is finite, then we can make get a good that is larger than all the, all the cycles in the graph and then, therefore it will forbid all those graphs in in the in creation but then we can also get the committed number arbitrarily high. So, so there's no such a function that bond, bond to the committed number. Okay, so in general it's not true so we. So, obviously we would care about for which kind of graph classes can we can we have. So, so we just said that if we have a final list of forbidden graph and then one of them and they have to be the order for the graph class to be kind bounded. They have to contain a tree. So this famous conjecture by Yaffas and some of which is our seminar. Yes. Yeah, so this is Yaffas summer conjecture that for every tree. The class of T free graph are kind of bounded. So this was made a while ago but we we actually know very little about this conjecture so. So, so here are some partial progress. So, one of the breakthrough breakthrough is Yaffas, Samaritan user. They show that it's true for all trees of readers to in triangle three graph, and then care stand and paradise kind of extended this idea of using this template idea, and then they prove that it's true for all trees of readers to. So this particular P5 is kind bounded. Okay, and then later on there are there are many other kind of extensions of these results to larger larger graph but in general this is still a wide open conjecture. All right, so after we know a graph classes kind bounded. The next question is, you know, how good is the function is. So here we say this graph class is polynomially kind bounded if we can pick the F subset the binding function is actually polynomial. And then there's a very bold conjecture made by as far as that in his or her thesis dissertation that every kind bounded graph class also polynomially kind bounded. Obviously because because the contract is still open so we haven't found any example that is kind bounded and then not polynomially kind bounded. So, but I think this is a very bold conjecture, because we know so few polynomially kind binding graph classes. Okay. There are. So this problem has some important connection to the other China contractor. So what is what is the early China contractor, basically says that if we have an inverted graph, but also H free for any H, if we forbid H in the graph, then it will contain a clique or independent set of size, a polynomial with respect to what we know by these probabilistic method we can construct a graph that without clique and independent set of the larger clique independent set is log log in. But here we are saying that if we if we break kind of their pseudo randomness in the graph, and then we say by saying H free, then we actually can find a very large clique or independent set in the magnitude of end to the some constant H. Okay, so what is the connection. And so if we know other China contractors, very hard it's still open for p five free graph so if we cannot do that, let's prove something you know, even harder. So here. The idea is that the polynomial bond is important because if we can prove age free family is pulling on polynomially high bounded, then it also implied that the this order China contractor true for that particular family. So here, we have been working on this p five free, whether, whether graph, whether they are kind of polynomial bounded. Okay. This is very hard. So let's. Now after we say okay some graph class is probably normally bounded next question is, you know, can we get the best binding function. What is the best finding function well, the best one is committed number equals to the click number. Okay, we cannot go lower than that. So this is called the perfect graph. This is the graph family with binding function as of X equals to X. This is a very good characterization of such graph class. Yeah, this is characterized by forbidden, what we call holes and anti holes. So a whole is just in the cycle of length at least four just in the cycle. And the anti whole is just a component of the whole. Okay, so there is this very famous. This is a very strong perfect graph theorem proven in 2006 by, by these people that a graph is perfect if and only if they are odd holes and odd anti holes free. Okay, so this is a very, very famous breakthrough at that time. And then we also know that, in particular, before three graphs are perfect. Okay. So, you know, the next best one is the what we call the, when the binding function is chromatic number of g equals to the click number of g plus one. Then we call it satisfy the vising bound. Why, why, why is it called a satisfied advising bond, because the advising theorem has that has that has that's bound. So, the chromatic index, I think we all know that is the, you said the color the vertices we color the edges is the minimum number of colors required to color the edges, so that no two incidence edge share a color, and then advising prove that for the chromatic index is upper bounded by maximum degree plus one. In particular, this, if we look at the line graphs of some graph, then the chromatic number of the line graph is the chromatic index, and then equivalently so we know the for line graphs for this line graph the chromatic number of G is at most the click number of G plus one. So that's why we call if some graph class satisfy the this bound and then we say satisfy satisfy the vising bound. Right, so let's let's zoom out. Okay, so here, the main areas in this kind of the main questions in this area are, first of all, for which trees, first of all, are some kind of key free graph kind bounded. That's the first question to answer. And then if they are kind bounded. So for which graph class class can we know they are polynomially kind bounded. And if we know they are polynomially kind bounded. Then you know what is the best kind binding function for such a free graph. And so these are the kind of the order of the questions. So if you know if we leave now for some reason then here are the two main things that I want to show is that. We have this family called tea broom free grass and then so together with little shooter and we show that this family of graphs are polynomially combined and that we'll talk about what is the definition. And then also, and then for the for more exact fun with with shooter and you will show that for triangle and then for free for some folk graph, it says satisfy the vising bound. Okay, so although I say satisfy the vising bound here, it sounds cool but what is it really saying is that if it's triangle free then the click number is to right then so satisfy the vising bound that just means that it's smaller equals to two class one. So it's just saying that triangle free triangle for free graph, three colorable. So here that is what we, what we approve. Okay, it's okay so far. All right, then. So let's, let's start for the tea broom free graph. Okay, I want first of all want to kind of talk about why do we talk about people free and what is people. So first of all, so before free graph we know they're probably we know they're probably normally bounded because it's the perfect graph. Okay. And then k1 t free graph is also probably normally bounded. Because so, if we talk about k1 t free graph, the binding function is very easy is this upper bounded by the Ramsey number of t Omega, because the maximum degree of a k1 t graph is upper bounded by Ramsey number. Because if we fix the vertex, and then we look at this neighborhood. If it has more than it has at least Ramsey number of t Omega many vertices, then either it has a large click has a click of Omega, which violates and then together with that vertex it violates the number of clicks, number of the size of the click, and then all it has a key independent set, which together with that vertex is a star. So, this is a very easy to show that they are polynomially chi bounded. So, if we look at what is an immediate superclass of both graph classes. So let's have some structure that is slightly larger than people, but also slightly larger than k1 t. So, this is what we so if we even easier if you look at k12. So just like a star of three vertices. So this is the next immediate superclass of these two it contain the people. And then you also contain a k have like a three star in the graph. So this is what we call a chair. So if you're, if you're, if you're not interested you can spend some time trying to redraw this picture, they can, it can be rejoined to look like the shape of a chair. Or sometimes it's also called a fork. Okay, so even for the for this graph. Well, we don't know. So we didn't know whether we know these two are probably k1 t and people are polynomially bounded but the next immediate superclass we don't know whether they're probably more polynomially chi bounded. And so that is the motivation for studying this graph class is the next natural thing to to study. So more generally, if we generalize this, what we call the chair by attaching attaching more leaves on the right side. This is called a tea broom. And so, so officially is graph obtained from this key plus one star by 70 by 70 by the one edge to make it one side slightly longer. Okay, it is known. Yeah, so we know because it's really is to and then just now we say, so it is kind bounded but we don't know we didn't know whether it is polynomially kind bounded. Okay, so, so. This question was made official make public in a survey by Schremer and render off into the 19. They asked for chair free graph. Let's start from simple. There exists a polynomial chi binding function for the family of chair free. So that is why we were kind of interested in in this problem. So there are some partial progress there were some partial progress on this problem by forbidden some additional thing instead of chair, if we'll be if we forbid one more for a bit K to two. So, so then we have more constraints, then we know is linear, linearly bounded by the committee by the by in with respect to the click number. And then, if we forbid, I mean, obviously, obviously for the smaller graph is make your graph class more restrictive. And so, also, if we forbid some F, which is some five vertex graph. You know, it then is the proof is quadratic. So there are some partial progress on forbidden some subclass of the chair free class. And then in the in the same article, several men actually conjectured that if we just forbid chair, then the class of chair graph is perfectly debatable, but we don't need to care about the definition. The consequence that he conjectured that it amidst a quadratic high binding function. Okay. Okay, so here. Our main result is that we first of all want to make answer here. I can render out the question by showing that for keep room free graph is polynomially high bounded. In particular, the binding function is little old of the click number to the power of t plus one is, is little old some Ramsey number. And then for the, for the, for T equals to which we want to address the first conjecture is that for all chair free graph, then the committed numbers upper bounded by some constant times the is quadratically tie bounded. So this answers the conjecture of several men. So this is all joint work with a little shooter and you. And then, in addition, if we forbid some kind of some additional structure for a bit complete bipartite graph, then we can lower the exponent a little bit by improving it from t plus one to t minus one plus two over t plus one. But here these two the first two are kind of the main results. Okay, so let's talk about the, what is the kind of the proof idea. So usually, so it is kind of so what we want to do is basically do some structural analysis on on the T broom free graph. And then, once we get some structure, then we color them in a nice way. This definition are pretty standard, then the end of the neighborhood and I have some set is the ice neighborhood of us, and then and at least I is the ice neighborhood and beyond, and for some set as. Then we also say that a vertex is complete to a set if this vertex V is adjacent to everything in the set. If it's anti complete means V is not adjacent to anything. And then these neutral means it's not complete, not anti complete. So it's adjacent to something and not adjacent to something else. So the main idea is that we want to, and they usually also this is also some idea that already existed is that we want to take some maximal global structure takes and what they call the template, but really it's just some global structure. And then we want to look at the neighborhood of that structure and look at a second neighborhood. And then, if we pick this structure nicely, then we can kind of get a good structure of the whole graph. So the picture, the template, the structure we pick is just a q part q part type graph. So it's just a maximal complete q part q part type graph, where one of the part has t vertices, and then the other part have only one vertices. So here, it just some maximal complete q part type graph and so we have this vq is one vertex and then we want to be q minus one. So this is a t vertex. So take a maximal such graph. And then we want to consider the what's the structure of the neighborhood of q, and then a second neighborhood of q, second neighborhood and beyond of the q. And then the idea is that we want to apply. It's impossible to entirely know the structure of the graph so we want to apply induction when appropriate. And then, so the idea is that we want to color, first of all, for some take away some part that is easy to color, color them polynomially, then color the rest of the graph using induction. And if we have some, if we have a function f of omega minus one, plus some polynomial, and then we can make the f degree higher, and then to to to absorb those polynomial terms. So, that is the idea. So let's first of all get some structure that is easy to color. So that if we look at the structure of this. So the idea is that if you look at the structure of this template. This is called bound. So these are very easy to show so I will skip the proof that the for the second neighborhood beyond of this to the maximum degree is the most some three times the Ramsey number of T omega. So we can we can just color them using new colors. So then we don't need to worry about second neighborhood beyond just look at the first neighborhood and of the of the of that structure. And then, so, and then what we do is we can partition again partition the first neighbor put into something that is easy to color. Okay, so in particular we partition them into, let me check the time. We partition them into AB, Z union w. So later I will draw a picture. Okay, what is the, how do what do we partition them by so we have this queue which is maximal template. Vq, and then Vq minus one Vq minus two, by the way to be one. So this, this be which with which is the neighborhood vertices in the neighborhood of Q that is complete to to all the other complete to be one to be Q. So this vertices again by using this structure condition that it is T boom free is easy to show that the maximum degree of those vertices are bounded by the Ramsey number of two omega. And then we also, there's also a part, what we call a so these are the vertices. So this is the vertices are the vertices that is neutral to Vq, neutral to Vq means some connection to Vq and some non-edge to Vq. These vertices we all they're also very easy to to bond. We proved that the number of vertices in A are not much. It's also some polynomial for omega. Okay, so those two we can we can again color them differently. And then what is left with are these two to group which will try to color them by induction. So the remaining vertices are either what we call Z, which are the vertices that is complete to Vq. So we already removed those that are neutral to Vq. So, so we have a bunch of Z that is complete to Vq. And then we also have a W that is anti-complete to Vq. That is the others are all polynomially colorable so we don't care. So now we want to cut this. So here we are in here, we want to color Z union W using induction. But the problem is why can we color them by induction? In order to apply induction, the quick number has to drop. So we again need to get some need to analyze the structure a little bit in order to apply induction on the union of these two. We can apply induction on Z, but why when we include W we can also apply induction. So again we need to get some structure. All right, so now we'll focus on what is the structure within Z and W. The idea is actually quite simple. So maybe let's forget about this text. I will just draw a picture and then it should be clear, you know, what is going on. So we have this Vq, we have this again we have this template Vq and then we have a bunch of vertices Vq minus one to V1. The idea is that so this we can partition. So we have W and then which is anti-complete to Vq and we have Z which is complete to Vq. The idea is not complicated. So first of all, this W, we can partition them by looking at which vertices among V1 to Vq that they are adjacent to. So here also we can look at the definition. For any subset, let's say I, which is a subset of the vertices in V1 Vq, we just write, we let WI to be the set of vertices in W that is adjacent to exactly I. So we can partition all the W by which vertices in V1 to Vq minus one, they are adjacent to. Okay, so by doing by doing the structural analysis. First thing we show is that if I, I prime are distinct then WI and then WI prime, they are so suppose they are adjacent to different thing, then they are they are anti-complete. So for different I and I prime, these two blocks are anti-complete to each other. So that's nice because we can reuse the color so that they don't interfere with each other. Okay, what is the next thing is let's look at the edges between Z and WI because we want to color them together using induction. So the idea is that each vertex, suppose let's take a component of WI, assume they are connected, then each vertex in Z, each vertex Z is either complete to WI or anti-complete to WI. So there's no neutral thing going on between the vertices in Z and WI. So that is the next thing we show. And then what we have, so for each of the WI, basically I can look at, because each vertex is either complete or anti-complete to WI, I can look at for the Z, I can partition them by looking at what is the neighbors of WI, what is the non-neighbors of WI. So here the nice property is that for each of the WI, if I look at what is the neighbors of WI, and then look at the non-neighbors of WI, it is complete to each other. So this is the third, the neighbors and non-neighbors of components of WI in Z are complete to each other. Okay, so here then this will give us all the nice information that we need for doing the induction. So basically, what we have now is that basically for each of the WI, I can make a cut of the Z and then they are complete to each other. And then for some other WI prime, I can make another cut, okay, and which also complete to each other. So this, so then what we can do is now let's cut it for each of the WI. And then so in the end what we have is that, so let me just throw it here. We can partition Z into kind of, so this Z, we can just cut it step by step. Then we have a bunch of blocks that are complete to each other. And then for each of the WI is adjacent to some subset of the blocks, they are complete to some subset of the blocks and anti-complete to some subset of the blocks. So this is exactly the structure we need for doing the induction. So when we apply induction, now we use the fact that, you know, if when we have the function, the polynomial function we pick is convex. If the, if the click number add up to omega, then f, the function of the function of the click number added up will be smaller equals to f of omega minus one. Because there has more than two blocks, okay. And then, so you see that we can apply induction. So let's say each of the set have omega one, omega two, omega three, all the way to omega t. Then we can, so this sum of the omega is less or equals to omega minus one, because we have some vertex in here too. Yeah, so then by the convexity of the function, f of those sum of the f of omega i is also a most f of omega minus one. Okay, so then, so we can color them, color those in the action. Now how to color the remaining W. Well, it's natural, what do we do. For W i, we are going to reuse the colors used in Z in particular for the part for the colors used in the part that W i is not adjacent to, then we can reuse those colors into W i. And then because W i and W i prime they're all destroyed. So we can just reuse them for each of the W i. So that is, once we get this structure, then the induction step is kind of natural. Okay, so here, all of these stuff are just saying that that thing. Okay, it's okay. And then so what we have now is that basically, if we commit a number of W union Z for technical purpose I need to remove some, some set W zero, but those W zero are also polynomially colorable. Okay, so what we have is that this W union Z, the committed number can be handled in one setting by induction and then it's upper bounded by f of omega minus one, plus some constant. Okay, so that is the main idea of the proof. Then what we can do is just add up all these things. So we have then f omega minus one plus some polynomial term. In order for the induction work, you know, we can just pick some polynomial, such that the polynomial is one degree higher than the polynomial error term that we have to absorb. So that is the that is the main idea of the proof. So if I free graph, then this is much harder. We don't have a very good structure to analyze. So this is this method doesn't quite work. Okay, but for people free graphic works. Okay, so there are some reasons result. So the people free graph is this structure. And then we allow tea tea leaves. And then so this recent result by Alex Scott Simone as a sparkle. They just, they generalize this, and then to show that it's also true for all to even make a bigger double star free graph. So this contain all graphs with diameter three. Now P five is a graph with them to fall. So the next question is, is the class of P five brief P five free graph polynomially kind of on it. Okay, so here. Yeah, here are some open questions. So actually, in the survey, she and me are a renderer actually can informally conjecture. So we prove that the chair free graph are quadratically co bounded. So they kind of informally conjecture that it is the Ramsey number of three omega. So chair chair remember is this, this graph. The informally conjecture is Ramsey number of three omega. So which is like sub quadratic. So it will be interesting to know which one is the truth. Okay. Yeah, another question is that the Steve Raman conjecture that kind of conjecture that this care for free graph is perfectly divisible. Okay, let's skip the definition. And then also another question is for that for if we additionally forbid. What we showed is that is big O of omega of t plus one t minus one plus two over t plus one. Okay. I think it could be improved to maybe all of omega to the t minus one remove this fraction. So can this be improved. Maybe refine the argument a little bit to to make it a little bit better. Okay, then the most significant problem is for p five three. That's it to meet a polynomial blinding function, if it does, it will imply the older China conjecture for p five three graph, which is the remaining piece. So all the other five vertex graph for all the China has been soft. We showed for C five that this year. So the only thing remaining is P five, and it's compliment for five vertex graph is P five, and it's compliment and then there are there are equivalents so showing one implied the other. So P five is the last piece, if we can prove this, it will be, it will be nice. So is that any binding function for people. We know, yeah, we know there exists a binding function is this type on it, but we don't know whether it's polynomial. But because it's true for all trees of radius two. Yeah. Okay, the next one is okay, instead of. Okay, now, now, the next goal is that let's talk about some specific finding function, the rising bound. So recall that graph class satisfied advising bound, if committed number of trees upper bounded by click number class one. Okay, so here we are forbidden one graph is hard. Okay, so so now. More, we are considering forbidden a pair. So a pair AB actually forbidden one one graph I think most of them doesn't, I think they don't satisfy the visor bound so a pair AB is called a good advising pair. If, if we forbid AB, then they satisfy the visor bound, but also why is it called good is that both of these conditions are needed. So none of neither a free or be free is redundant. Okay, so then then we call it good. And this is actually pretty important because here's that in 1984 show that K5 minus the edge K13 is a good advising pair. So it seems like a harmless pair to forbid you know why do we forbid those. It turned out, this is actually strengthening of the of the rising theorem. So we just say that for line graph line graph satisfy the rising theorem. And it turned out the line graph can be characterized by forbidden nine forbidden set of nine graphs, and then these nine graphs imply include this K5 minus E and K13. So here is saying that if we forbid the nine if we forbid two, and then it's also satisfied advising bound. Okay. And then we say a good advising pair is AB is called saturated. If basically the idea is that you know if we have to pair, which is a rising pair. Now if I take a sub graph, take a smaller graph, a prime. And we're still advising pair because we have we have forbidden a smaller graph, which kind of make our graph class more restrictive. So we want to know what is the largest AB that we can we can we can take. So that's the AB is is a good advising pair. So here, a rising pair is called saturated. If for every other saturated, so every other good advising pair a prime B prime that is contained in AB. Yeah, there's no other AB pair that contain both a and B. So it's the is the kind of the maximal under sub graph containment. Okay. So, the question now is, okay, for which pair of graphs AB is maybe a good advising pair. So this is this research is initiated by a render off in 1998. Kind of give some fundamental work that if AB is a good advising pair, then we have we know something about AB. So it has to be a tree, not contained in before. And also B has to be contained in this five structures. So K five minus E. This is the HVN is K four K three and then diamond. And the first that is true is true for some subset of them. So if B is a subset of HVN of K five minus E then be chair is a good advising pair. Okay, so this question is still hard. Okay, so let's start from, let's try to prove as much as we can. So the first step is to consider the problem where determine all the advising pairs AB, such that any AB free graph is three colorable. So let's look at all the AB pairs. So that's give us three colorable graph. Okay, so what is the progress on this. So render off again in in show that if AB. So again is enough is sufficient to only consider the saturated vision pairs. And then if we take subgraph, then they're also advising pairs. So if AB are saturated vision parents that imply three colorable, then we know something about AB. A has to be either K three or K four B has to be belongs to people. H this graph H, and then also a fork. So this is the fork. They also call it a trident. It looks like a trident. And then so if a equals to K four, then we know exactly what B is B is people. And then if a equals to K three, which is the triangle, and then B is either H or a subset of the fault. So here this is the room that that this is the open thing that's left B has to be a subset of the fault. Well what B is. So to show that, okay, B is, is two for for B is equals to E and then be equals to B and cross, both of both of them are the subset of folk. Okay. Now the question is, how about folk. These two are the kind of the largest subset of the folk. Now, he couldn't resolve the focus. And if we can show that be equals to folk then. So then the saturated pairs are exactly be equals to H and be equals to fault. So this is the part that is left open. So he conjectured that K three folk are also saturated saturated vision pair. He conjectured in 1998, that if G is a K three folk free graph, then the committed number is satisfied advising by Zimbabwe. So there is a breakthrough by fans, yeah, and you in 2014. They show that if the graph is C five three, the odd girth is at least seven. And then it's true. And then what we do what's with shooter, and you we prove that we can remove the C five three condition and show that indeed K three for three are three colorable. So that is the, that is the result. And then. So what what is one corollary is that in the in the set of triangle three graph actually K three folk graph are these results will automatically imply for Paul Paul Paul three graphs. Okay, so what is this Paul. Paul is this K three plus some some additional edge. And so there's an observation by render out that K three B free graph is a saturated vision pair if and only if Paul be free graph is also I can we can replace the triangle with the pole. Okay, so, so that implies also the Paul B is a saturated vision pair if and only B is the two graph H and folk. What is the, well, what is the idea. The idea is that we want to suppose is not true. Suppose it's not true then we want to take a minimal example. So let's take a vertex, minimal example. Right. It's not true there's committed number at least four, and then, and then it's K three for three, and then but committed number of at least four. Then we know something first of all, for any UV in the graph, the neighborhood of the V. So there's no what we call no neighborhood containment. This is because if we have a view is very easy if there's a view where the, the, the neighborhood of these contain in a neighborhood of you. Then we can delete one of the vertex and then by the minimality the other, the remaining graph is three colorable but then I can use same color of you to color V. And then we are, we are good. And then by the fancy you result, we also contains a five cycle. Five cycle is actually the hard case, because if we have a. So, let me introduce the default. Okay, so if we look at stare at this folk graph. This is the, we have the degree for vertex is the center. And then we say there are the two, the two, the two length to path. We call the long arm, and then the two length one path to the leaf are called a short arm. So we have to long arm and to short arm. If we have a C seven, then it's easy to find two, but and also again we are forbidden as induced subgraph. Okay, if we have a C seven is easy to find the long arm to long arm, because we can just go along the cycle. And then we don't have to long arm, because the this is a connection. So that's why C five is part if we forbid C five. If it's a C five three then you know is, we can get the two long arm easily. Okay. So, and then also we get showed by some kind of also another observation is a trick that we very often use is that suppose I have a vertex of degree three. If I know the vertex of degrees three. Then, if I remove it. I have the neighborhood and then the by the vertex mean, the neighborhood is three colorable. This vertex of the, the neighborhood have to contain all the three colors, otherwise I can extend it to this V. And also, we can get kind of get more structure out of this local structure because this vertex one also need to have some neighbor of color two and three. Otherwise, if this vertex one doesn't have a color to then I can recolor. So this recolor and stretch the strategy. If this w vertex doesn't have a neighbor color two then I can recolor w with two, and then color V with one. So using this kind of so having this degree three vertex is nice. Okay, so here, using this recolor strategy we also show that she has no to cut. Okay, then the, let me give a proof sketch of the rest of the proof. So here's basically what we have what what what I have in mind when we kind of do the, do the, do the proof that it, this is some structure that doesn't have pay three. Doesn't have a triangle doesn't have a fork. So whenever we have to long arm and to short arm, we all we always have some edge between the long arm to arm to to break the fork. So this is the structure that kind of where we where we can't use the the the folk free pay three free to do any further structural analysis, but then this graph is like three color. And I'm think having this picture in mind as the as the extremal example in my in my mind to kind of go towards this structure. So the rest of proof kind of try to say that the other cases are not possible. So step one is that okay we take C5 we know there exists a C5 and then what we want to show is that. Again, look at the first neighborhood second neighborhood. And then the thing we show is that there's no two vertex in the second neighborhood that has two independent path to the to the to the cycle. Okay, again, here they don't they don't have it so I'm trying trying to exclude that structure. So step one, take like seven to eight pages to show that there's no two vertices that go to the same from the second neighborhood to the to the cycle. In either case, there's no, there's no, no sub thing. And then another kind of 10 pages. Now we some some technical thing we take a cycle that minimized actually here is the minimized the second neighborhood and beyond the vertices, and then using that condition. We show the same re coloring strategy and then also kind of using the K3 folk free condition. We show there's another another two things that we want to forbid. There's no partial diamond. So in particular if there's a, if there's a you want a one B one then there's a B one then you want has to connect to be one so there's no partial diamond here. So another, some structural analysis to show that this cannot happen. So there's one vertex in the first neighborhood cannot have two neighbors in a cycle. So all of these kind of combination of the structural analysis and the re coloring strategy. And then so once those are handled. We have a lot more structure to deal with. So then the, the final step is to finish off the analysis by looking at the number of degree three vertices in the in a cycle. If every vertex in C has degree three. Remember that the trick we say that just by re coloring strategy, we can get a lot of structure from degree three vertices. And then we can show that is three colorable, which give us a contradiction. And then, similarly, we're using different kind of strategy and structure analysis. The other case, if there are two consecutive two consecutive degree three vertices, and at least one degree. Yeah, if there are two consecutive degree three vertices is also can be handled. If there are exactly two vertices of degree three. So when there are degree three vertices, we mostly use re re coloring strategy. Okay, now. And then we get to get down to the, to the, to the kind of the extreme example that I have in mind, we get a most one one vertex of degree three, which means most of vertices on the cycle have degree four degree at least four so we have something like that. In that in that kind of context we have very nice. And then together with all those conditions we have is easy to three color the graph. So that is the kind of the general flow of this of this result. Okay, for this part. So what what is left, the left is that for the, for the saturated vising pair, render out conjecture that these are the older saturated vising saturated vising pairs. So the k five minus each error. Yeah, all of this. But so what's we, what's we take off is that it's remember. So all the three here already handled the top three, the three in the first row are handled. Paul folk. Just now we mentioned is the same as triangle for three. So these are crossed. So I think the only cases left are this. If we replace the triangle with k four minus the edge k four minus each and k four minus e folk. Then we will have all the saturated vising pairs. So it will be interesting to know whether they are indeed saturated vising pairs. And then, yeah, the most problem, the most interesting problem here is still does p five free and makes a polynomial binding function. Yeah, I think. Yeah, that's all I have. Thank you. So any questions for you. Still, it's near polynomial. They have a near polynomial bound, but it's still not known whether it's polynomial. Yeah. This p five together with some other five vertex graph. So there are a lot of studies on p five and then forbid another five vertex graph, all the other graphs are quoted polynomial. P five is this as this is the obstacle. Any other questions. So for the Yeah, find some good template and then look at the neighborhood and then all some recently the cut cut idea and then remove the cuts that the rest of them are colorable and then if if the cut size is small, then we are good if the cut size is large than the other part can be colored easily. Some, some idea like that. Let's thank our speaker again. And thanks everyone for coming out. And I guess, well, I'll see you next week.