 Thank you. So today I'm going to talk about coloring graphs containing no K5 subdivision. And this is joint work with Qiqin Xie Shijie Xie from Georgia Tech and a professor Xing Xinyu. So several like a long time ago, people find that for every map, it is possible to color it with only four colors, such that no adjacent areas are using the same color. And this is finally proved several decades ago by Appel and Haken. And below you can see as an example is a map of the United States colored with only four colors. And this is the well-known, the four color theorem. And if we state it more rigorously, it says that every planar graph is four colorable. And here planar graph is defined in a rigorous way that if we can draw a graph on a plane such that its edges intersects only at their endpoints. And by the way, for the four color theorem in 1997, Robertson Sanders Seymour and Thomas gave a cleaner proof, and which is more recognized by the mathematics society. And we can see here this description of a planar graph is more like a geometric way. And how can we describe it in a more combinatorics way. There's another definition here. So a subdivision of a graph is defined by resulting from inserting vertices into edges in the graph, zero or more times. So it's like we divide each edges into paths with this joint internal vertices. And with this definition, a planar graph could be defined or characterized in the way that a graph is planar, if and only if it contains no K5 subdivision or K33 subdivision. So in this way, the four color theorem can be further described as if a graph does not contain K5 subdivisions or K33 subdivisions, then it can be four colored. So there are two kinds of forbidden patterns. When people start to think of what if we only forbid part of those patterns, like what if we only forbid K5 subdivision or what if we only forbid K33 subdivision. In the K33 case, it is shown actually earlier by Wagner that if G is a graph with no K33 subdivision, then G has a cut of size at most two or G is planar or G is a K5. So after we have shown the, after we have known the four color theorem, we know that in the case when G is planar, it is four colorable. Of course, when G is a K5, it's not four colorable, but this is a very special case. And when G has a cut of size at most two, there is actually is kind of like a special structure for a coloring. So suppose this is a, suppose this is a two-cut of a graph. It's hard to draw on the laptop. Suppose this is a two-cut of a graph, then there are only two cases about the coloring on those two-cut vertices. They are either in the same color or in different colors. So when those vertices are in different colors, or in other words, we can divide that graph into smaller structures where we consider one part of the separation with an adding, oh, adding an edge to those two cut vertices. Sorry, this is ugly. So in this case, we know we can find sort of a coloring of the half part of the separation. Well, we know that the colors of those two cut vertices are different because we added an edge to them. And we can do it to both sides. And after that, we get colorings for each side such that the colors for the cut vertices are different. And then we can merge the coloring so that we can find such a specific coloring for the whole graph. And this also happened for the case when there's a one-cut. So whenever we can find some kind of coloring for both sides, we can combine them together if there's only one vertex in the cut set. So here we can see it's kind of clear with the case, with the coloring properties in the case, the graph does not have a k-3-3 subdivision from this Wagner's theorem. So what can we have when we only forbid a k-5 subdivision? Actually, there is a more general conjecture by Hayosh in the 50s that for any positive integer k, every graph containing no k-k plus one subdivision is k-colorable. So k-k plus one here means the complete graph on k plus one vertices. And so the question we were discussing is actually a special case here when k equals four. So this Hayosh conjecture is shown to be true for k less than or equal to three with a not very hard way, and it's open for k-k plus four and five. But after several decades, Catalan find a counter example for the cases when k is at least six. So the conjecture is disproved for a large case. And later Erdisch and Feitzlow, which show that this conjecture fails for almost all graphs in a random graph setting. So this conjecture now remains open only for the k-k plus four and k-k plus five cases. And the problem we discussed before is exactly the case when k-k plus four here. So that's another way of illustrating the background of that problem here. And so let me state again that the problem here is to consider the coloring properties of graphs for beating k-five subdivision. And there are some previous results on that. One of those, probably the most famous one of those is the Kelman Simmers conjecture, and it's proved several years ago by a group from Georgia Tech that if a graph containing no k-five subdivision, then that graph is planar, it omits a cut of size at most four. That is saying that if we have a five connected graph containing no k-five subdivision, then it must be planar. And of course, if it's planar we know it's four-colorable, so the only case we need to concern is the cases when g is not five connected, or it omits a cut of size at most four. And to consider that question we need to introduce like some special structure here. We call it a Hayosh graph. We consider a counter example to the Hayosh conjecture for k equals four. That is a graph with no k-five subdivision, and it's not four-colorable. So as long as we can show such a counter example does not exist, then that conjecture would be true. So in the sense of proof by contradiction we consider such an example, and moreover we consider minimum one. So we define the Hayosh graph by the graph is not four-colorable, it contains no k-five subdivision, and subject to those two items, the number of vertices is minimum. So Hayosh graph is a minimum counter example to that conjecture. And there are some previous results on the property of such a Hayosh graph. In 26, Yu and Zickfield showed that every Hayosh graph must be four-connected. That is as long as there exists a cut of size at most three, then it wouldn't be a minimum counter example. The sense is sort of similar to the example we just saw with a two-cut. If there is a three-cut we can sort of consider the different cases within the coloring of those three-cut and see, because that is defined to be the minimum counter example. So we know that each part wouldn't be a counter example, so in each part there would be a four-coloring. And then we sort of consider how to combine those four-colorings, and it could be shown that each Hayosh graph must be four-connected. And later Sun and Yu showed that if G is a Hayosh graph and T is a four-cut in G, then G minus T has exactly two components. That is saying that if there are more than two components hang at the same four-cut, then some similar argument could hold that either it could be four-colorable or they can sort of find a K5 subdivision in that cost structure. So the only case remaining is that the graph is four-connected but not five-connected. There exists a four-cut, or we'll say that for any four-cut in the graph, there are exactly two components at that four-cut. And with Professor Yu and other two more senior students graduated from Georgia Tech, we show a stronger property of such a Hayosh graph. So we can see the figure here that we know for each Hayosh graph there exists a four-cut, and there are two components of the four-cut. And we show that if one side of that four-cut is not trivial, that means it's not one for text with four edges to the four-cut, then that set cannot be planar. More rigorously, we show that if G is a Hayosh graph, then G does not omit four separation G1, G2 such that G1 is planar and those four-cut is in order on the outer cycle and the number of vertices containing those four-cut vertices is at most six. Or the other words, if it is planar and those cut vertices are in order, then that must be trivial. So in the remaining time, I will briefly illustrate our proof idea to this result. Before that, let's say why we would like to consider this kind of structure. This is because we know there's a characterization of the two-linked property by Seymour that as long as, so if you are given two vertices, as long as they are not in order, like of an outer cycle with a planar area or as long as it's not in the situation of the left part in this graph, then we can always find a two-link between those, any two-link between those four vertices, given that that part is two-connected. So we would like to show this because we want to argue that actually as long as we find a four-cut, we can find any kind of linkage between those four-cuts in further steps. So next, let me start to illustrate how we prove this property. Firstly, we would like to find some kind of a wheel in the planar part, and here a wheel is defined a bit differently with the common concept. So we define a wheel by a vertex and a cycle, well, a vertex and its neighbors and a cycle, like the closest cycle in the planar part surrounding that vertex. So we could have more vertices on the path connecting two neighbors of the center vertex. And further, in this case, we would like to consider sort of a happy wheel that we, like the case we like, is that if there is a vertex on the path connecting two neighbors, and that vertex is not a neighbor of the center vertex, we wouldn't like it to be one of the cut vertices. We're saying the cut vertices, if the cut vertices are on the wheel, then it could only be the neighbors of the center vertex. And with that, so if we can find such a wheel, then actually use the connectivity and some other coloring arguments, we can find four paths, four distinct paths connecting four neighbors of the center vertices to the cuts. And here, actually, there could be more neighbors to the center vertex but we only need four of those to connect to the cut vertices. And as long as we can find such a structure, so the next step we need is just to connect, find the two link of those four vertices on the other side on the non-planar side, and this is actually a K5 subdivision. And the branch vertices are the center vertex and the two neighbors of the center vertex. So we can see the wheel gives you eight edges, and then the other two edges is given by those paths linked to the cut vertex and connected on the other side and goes back to the cut and goes back to the other neighbor of the center vertex. So if we can find such a structure in each step like this, then we can find a K5 subdivision in the graph, which contradicting the assumption that it is a Hayosh graph which is a cut, which wouldn't have a K5 subdivision. So the last step is kind of straight is we can find directly find such a two link as long as the other part is non-planar using the two linkage statement. And we know that if one part is planar, the other part cannot be planar, giving that the whole graph is not four colorable. Because otherwise, if the whole thing is planar, it must be four colorable and it wouldn't be a counter example here. So we can find the two paths in step three for free. But in step two, it is not that straightforward anymore. So when we were looking for those four paths, there could be some obstructions that forbid us to find such four distinct paths. One common problem is that some certain type of five separations could cause problems. So if there exists some kind of five separations, we can, so the vertices in the middle would be less than what we need and then we cannot find such four paths. And to get rid of that problem, we actually we would consider, we would also consider five separations within that planar part. So in step one, we need to find good wheels inside a separation of order at most five. Can I ask a question? So when we say both sides cannot be planar, and then otherwise they are four colorable, even if the both sides are four colorable, how do we know we can patch the colors? Oh, actually, so by both sides are planar here. Actually, it's planar with the vertex, with the cut set. Yeah, with the cut set in order, in order of an order cycle or in order on a facial cycle. Let's see if I can draw it here. So it's kind of planar in the way that such four cut vertices are like this. And there could be other structures in the middle. So those would be the four cut vertices in counterclockwise order. Or this is back drawing. So here we say it's, so in the, so in the segment we actually say it is G1T planar. If you look at the second line from below, it's saying it's G1T planar, it means that all the cut vertices are on a facial cycle in order. And if we also have, it is G2T planar, that means the cut vertices are also on a facial cycle in order then we can naturally merge the two coloring together. Okay, thank you. Good point. Thank you for pointing that out. The only thing, so next we need to find such good wheels in four separations, as well as in five separations as we discussed, we need that for the process of finding the whole path. So next we discuss those situations with four separations and five separations. We have a four separation. So we first show that if we have a four separation and one part is planar, then there must be a good wheel. This, this is actually not very hard to argue. So one, one example is that in some cases, if we cannot find such a good wheel, we can actually find some kind of reducible configurations. That is, if such good wheel do not exist, then we can find some local structures that gives us the property that if the, if some smaller graph is four colorable, then the whole graph would be four colorable, which contradicting the probability of the way we take it. So one example of the or the most easy example is that as we as in the statement, we say that we only consider the cases when there's a full cut, and there are at least two vertices on the planner side. So in the case where when there are exactly two vertices, there is only one way to connect those vertices to the cut to maintain the full connectivity. That is the left, the left picture. So if there are only two vertices in the planner part, then they must be adjacent, and they must have two common neighbor in the cassette, and each of those need to connect to one other vertex. And if we have such a local configuration, then actually we can argue that by removing those two vertices UNV and add an edge between X and Y, then we get a smaller, we get a smaller graph. And because of the minute melody, we know that in the smaller graph is not anymore a counter example. So it's either contains K five subdivision, or it is four colorable. We know that the smaller graph wouldn't contain K five subdivision, because if it contains, then if it uses the new edge X, Y, then we can replace the new edge by the path X, UV, Y in the original graph. And if it didn't use the path that we can just find the exactly same K five subdivision in the original graph. So as long as the smaller graph obtained by deleting vertices UV and adding edge X, Y, if in the new graph there exists a K five subdivision, then in the original graph, there must also be a K five subdivision. So we know that in the new graph, there cannot be K five subdivisions. And since it, it is not a counter example so we know the new graph must be four colorable. So if we take a full coloring of the new graph. And since there's an edge between X, Y, we know X and Y are in different colors. So we, as long as we can find the extension of that full coloring to the original graph, such that it is proper that we have shown the original graph cannot be a counter example as well, and contradicting our very beginning assumption. So, next we just need to extend the coloring to the original part. So suppose X, Y are in, so in in this new graph X, Y and R in colors one and two. So if color color of X, say, say color one didn't show in Z and W or say if the color of C is not one and the color of W is not one that we can give vertex V the color one. And then we can find the color for you because X and V are in the same color. And thus X, Z, V, W could use at most three colors. And if one of C or W is in color one so we don't have that anymore. So suppose if Z is in color one, then we can firstly color vertex V because it now sees color of Z, color of W and color of Y, it sees three colors. So there must be at least one choice for V such that the whole thing is still proper. And after we colored V, we consider U since it has four neighbors and X and Z are in the same color. So it also it sees at most three colors. So we can readily find a color for you such that the whole thing is still proper. So with those, we can see that either in either case, as long as there is a proper coloring for the smaller graph, we can also find a proper coloring of the original graph. So that is actually that is the property of this local configuration. And this gives us that we would never have this look this kind of local configuration in Haill's graph. So with so we actually we found several different so called reducible configurations and using those we argue that because we cannot have such local structures and the since the whole graph is full connected, we can always find a wheel in within any full separation. It's probably a good point to ask, are there any questions. And we can do similar arguments for five separations, because there are more vertices in the cut set, it is a bit more complicated. So we, as in the previous argument, we actually can show that as long as it is not trivial. It is not one vertex connecting those four vertices, then there must be such a good wheel. But in five separations, there are several exceptions. So if we have a five separation as the white vertices on the outer cycle as you can see the, we could have such six obstructions that does not give us any good wheel, or it is not reducible configuration. So in these five separations, we show that the planner part either contains a good wheel or it, it is exactly one of the following structures. We can see that when there's only one vertex in the five separate separation, we don't, we don't actually have any control on it. So if there are two vertices, still we, we cannot do a lot with it. And if there are three vertices in the middle, then we can show that this is the only case where it is not a reducible configuration configuration and there is no good wheel. And then if there are four vertices in the middle, this is the only case. And if there are more than five vertices, then there must be a good wheel. So one example for that is that if we have four, sorry, if we have three vertices in the middle of the five, the planner side of the five separation, and it is not in the way of the one we have here. We can see here, we have three vertices. And those two are those three are actually in the past of lungs to. So what if, inside of that we have a cycle of lungs to the actually we can see this is also a reducible configuration. Similarly, deleting a vertices UVW and add edges PQPRPS. So again, we need to argue that as long as the original graph does not contain a K5 subdivision, then the new graph cannot contain a K5 subdivision. And if the new graph admits a proper five coloring, we can extend it to the original graph to find a proper five coloring. So here the argument for K5 subdivision is a bit more complicated than the previous one, because here we actually added three edges in the new, in the new graph, but we cannot find three disjoint paths to replace those three edges in the new graph. But since we know that we are looking for K5 subdivision, and if, so the only case we use three edges incident to one vertex is that that vertex is a branch vertex in the K5 subdivision. So we consider two cases, whether vertex P is a branch vertex in this new graph or well in the K5 subdivision in the new graph, or P is not a branch vertex. When P is not a branch vertex. So here is this K5 subdivision it uses at, at most two edges incident P. So, in the original graph, we can see that we can always find two disjoint paths to replace any two edges here. So for example for to replace edge PQPRPS. We can use paths PVQ and PUS in the original graph. And actually that holds for any pair of edges for PQPR we can use PVQPWR and then for PRPS we can use PVWR PUS. So if the K5 subdivision uses at most two edges, at most two new edges then we can find these trend paths to replace them. But if the K5 subdivision uses P as a branch vertex, then it only use one edge incident to P outside of the separation. So if P is a branch vertex and the K5 subdivision uses all those three new edges, then it uses at most one edge outside of this construction. So we can actually use a vertex V to replace the vertex P here. And the outer path is replaced by VP goes out and then PQ replaced by VQ PR replaced by VWR and PUS replaced by VUS. So briefly if so we could still add edges in the new graph such that we cannot find this trend path in the original in the original graph as a reducible configuration as long as it could be dealt with any cases in the K5 subdivision. So that was that argue that as long as there is a K5 subdivision in the new graph, then we can find a corresponding one in the original part. So again, we know that the new graph must be for colorable as the number of vertices is smaller than the original, the original graph or the highest graph. And lastly, we know that there's a full coloring in the new part. So we say a vertex P is in color one. And, and further we know that QRS are all not in color one because there are edges. In the original part we can extend this coloring. So P is in color one and we give W the same color as P give W the same color as P, and since QRS are not in P so it wouldn't create any problem. And, and after that we find the color for you, because you see the color of T the color of S and color one here so it sees at most three colors, so we can find the proper color for you. And lastly, we give a color to be because it sees only the color of you the color of Q and color one. And lastly, we can find a coloring for UNV after that. So that is a more complicated example for a reducible configuration. And with five separations we can have discussed a lot of that kind of local structures and show that if there's no good will then we must have one of the following construction and using that actually we can see that in the cases with more vertices in the middle, there are a lot of fixed edge in the, in the local structure here, and actually we can use this local structure to help us either find a five K five subdivision, or find a full coloring. With all of those argument, we can show that in the, in the health graph, if we have a full separation and one side is not trivial, then we can show it cannot be planner, such that the cut vertex are in order in a facial cycle. So, this is mainly what we proved in this in in this result. Okay, I think that's what that's all for today. Thank you. We could all find a way to speak up to thank our speaker. Let me open the floor up for questions. This might be a stupid question for the for the highest conjecture. If we cannot prove is K plus one subdivision free then it's K-colourable. That's one colourable, what is there an upper bound for the for the colour, chromatic number? For small, for small case, I'm not sure for large case. I'm not quite sure I think for, for, for had we go for had we go there's sort of such kind of results that I have seen that's years ago, but I don't quite remember to remember the paper for had we go that if, if you forbid, like, I think that's for like KK minor and then the chromatic number could be a colour some could be a factor of log, log, log D something. Not quite sure for maybe someone from the audience could answer it. So the, the, the, the counter example Catherine gave. Okay, so, okay. What, what is the, the colours. What is the chromatic number of the, of the counter example. Maybe I'll check that. Yeah, check it. Yeah. Any other questions for shop on. Okay, if not, thank you again. And I think I'll just say next week, we're on fall break so we're going to skip a week for our street seminar so we'll be convene again in two weeks. But I think, I think with that, we'll go ahead and break for the day so thanks everyone for coming out and thank you again for an excellent talk. Thank you, thank you for the invitation. So, right.