 Okay, well then, thank you everyone for coming out to a first discrete seminar of the semester here. We're very happy to have urban Yuri from the many the Alfred ready Institute of mathematics. I apologize about if I mispronounce your name, by the way, but we're very happy to have you come out from the alphabet ready Institute of mathematics to pick up the seminar and talk about hexagon free painter grass, so please take us away. Thank you very much. First of all, I am very honored to give a talk in Columbia, let me see. Just one remark. Okay, I'm checking. Okay, it works. Actually, I give a talk if I remember what went to three years ago in Columbia, let's do you remember that I visited you in 99 was a talk but I don't remember when it was long ago you were at Wanderbilt, I think. Yes, I, as I drove there from, from there, it was 99 spring, more precisely, I do not know. I know the semester. And I would like to keep this habit. I hope you do that you invite me that after 23 years again and I will give a talk after 23 years again. What do you think, can we do that? Let's see. It would be nice. I will be 90, you will be just close to it. Okay, more seriously, I would like to speak of hexagon free planner graphs today. This is an extramar graph subject, but not the usual one. Of course, if I see hexagon free graphs, hexagon free graphs, then of course you do know that. I would like to, to speak of planner graphs, I mean, extramar planner gas programs a little bit. Okay, of course, if you are speaking of planner graphs then you have to say, you have to remember, I was so wrong. It is a very young serum. It says that the planner graph of n vertices as at most three and minus six edges. Actually, it was just the consequence of the eyes for you. And if you like you can call it extramar graph theory, but to tell the truth, if you want to make it really extramar graph theory, then you have to use Kurovsky serum from 1930. Well, this serum says that the graph is planner if and only if it doesn't contain any subdivision of K5 or K3t or if you like topological, there are some other versions, but no you just one version. And the point is that it means that if a graph doesn't contain something in the subdivision of K5 or K3t, then we can determine the maximum number of pages of an n vertex graph. And this to tell that it is quite normal that this is extramar graph theory, but somehow nothing happened for a long while in this direction somewhat later, which means 1979 hockey me and Schmeichel proved the following that the number of trying n vertices is at most three and minus eight, and we have infinitely many extramar cases. I do not want to speak of it anymore. Or another thing is that the number of four cycles in a planner graph of n vertices and is at least five, then the number is at most n square plus three and minus 22 over two. It's not a classical extramar result because here we are speaking of the number of trying this number of four cycles. But if I mentioned this then let me mention one thing which was open. I mean they left open together with my students, Paulos, Salia and Zamora Tomkins was not my student but he was a student of Katona. So they do not seem to be Hungarian, but these are five Hungarian on search in some sense. I mean they studied in Hungary. So anyhow, we proved it hasn't appeared yet that if any is greater than record yet, then the number of five cycles in a planner graph is at most 200 square minus 10 and plus 12, which, and in this way we saw the problem of hockey me and Schmeichel. Okay, here there are two very special graphs I do not want to speak of them. This one. And this one, but the main point is that here we have infinitely many graphs, which have this property. It can be checked, not so easy to come but I mean not extremely complicated but it is not 100% obvious that the number of five cycles is exactly this number. And are you referring to two graphs on your on your slides. Yes, you don't see it. I see your title slide still. No, we don't see anything. My goodness. Oh my goodness. I see your title slide that's all I see. I don't know that's Okay, then maybe I draw the graph. If I do, do you see it. I think you see it. Maybe running your slideshow that might, that might make it so that we see. I have no idea what we have to I see it of course, but I don't know why you do not see it. Do you know what should I do? Let's see. That's all you needed. Diabetic. Yeah, that's it. Yeah, but what do you have to do with it? Click on it. And I don't know if you're going to be able to do that. My goodness. You didn't see any slides until now? We just need your feed or your title page. The very first page. My goodness. You can check them to the hook because it doesn't come up. So try to reshare. Maybe it will fix it. So close PowerPoint and share again. What do you see now? Nothing changed. My goodness. The PowerPoint. Yes, I, of course, I click sometimes it happened. Somewhere on your screen is probably the zoom message that says it's like in red. It says stop sharing, maybe try that and then try resharing. If I click on the sharing that what will happen, it will be the same. Well, you're sharing something right now because we still see your original PowerPoint. So you need to stop sharing first. Okay, I start sharing it again. My goodness. I didn't try it. I cannot do anything. No, it is absolutely died because I cannot do anything either. So when you click share screen, what do you see? Actually, I think my zoom was lost. You're still here for on our ad. We see you. Yes, but. I see post on the post attendee zoom, which is not the normal zoom. I'm afraid of that. Why restart it. Okay, we see your, your PowerPoint now. Yes. But if I start with the other details. Do you see it now? Interesting. So it looks like so we still see the original one. So I think the slideshow that you opened is probably opening in a new window. Sure. That doesn't usually happen. Do you see this at least? No, it is. No, we don't. Anything. Yeah. Now we see click on the second slide. Yeah, I can see it now too. Yeah. Just it is not a slide show. Okay. So don't try. I'm going to the. I don't. I don't. I don't. I don't. I don't. No, I didn't do something. I don't. I don't have a. I don't. Yeah. Yeah. This is not the slideshow but actually this is what you can see also and this is enough, is it okay? Yeah, this is okay. If I do it in this way. That's fine. You don't have any animations or anything, right? No, I will not have any animation, but it is not so important. Okay. If this works for you, then I think that'll work for us. Okay, sorry for it. I do not know what was it. Thank you. Okay. Okay, so anyhow. So what we proved that the number of five cycles in a planar graph is at most once can minus 10 and plus 12 and then actually these graphs are the sharp examples. Of course, this is the interesting example. Because this is infinitely many, these two are just very special cases. Okay, and the next is a very similar result that the maximum number of passes of lengths three in a planar graph is at most it can be determined seven and spare minus 32 and plus 27. I think this is the same class of same collection of all cells. And with some other people with Debarongos, Ryan Martin, Paoloz, Adiso Paoloz, Nikas Alia, Zhuang Qixiao, and Zamora again most of them except Ryan Martin is my, but the others are my students. We proved that the maximum number of passes of lengths four in a planar graph, and this is interesting that we couldn't determine it. Just a simple to get is n cube, plus some error term. And actually this is the graph, which has that many passes of lengths four. And one more result of this type which is because it is very interesting to, together with Oliver answer and my students, we prove that if I have a planar graph of analysis, then the number of induce pentagon is at most n squared over three, plus little line. And now I would like, this is the problem that you will see some other things which are not so important here. So if I grab that, so now I would like to turn to classical extreme results. I'm sorry if you can see something which is not for you but because it is not slides so that's why it is here. So, everybody knows to run serum graph doesn't contain key RSS sub graph, then the number of edges at most the number of edges in the to run graph and that for the equity holds only if and only if this is a to run graph and a to run graph this is what everybody knows. R minus one classes of vertices, and they are as equal as possible. And they are joined to each other, if and only if they belong to the same classes. So one of my favorite results that the number of edges that took it cycle free graph is at most into the one plus one over K times a constant, but we do not know how sharp it is, I mean that we do know that it is sharp. So let's take it apart from the constant for cake was three cake was two three and five. Okay, and now I would like to continue with classical extreme problems in planet graphs. Classical means that not generalized to run number but really, you would like to determine the maximum number of edges in a graph in a planet graph which doesn't contain something. In this case, there is no such a problem like in the problem of even cycles of the order of magnitude is always linear so there is no problem like that. Of course, the constant will be important, and you could say that maybe that's why this planet extreme graph problems are not interesting in some sense, you're right that, but this is not a good question for planet graphs. Well, it doesn't mean that the constant to determine the constant coefficient of n is not interesting. And the other thing is that somehow the constructions will be pretty interesting. So this is what I would like to define the planner to run number of a graph age is XP, referring to the fact that this planner and age is the maximum number of edges in the planet graph on end vertices which doesn't contain it as a sub graph. If you like to do to run serum then it's, it will not be a two interesting thing as I said that the graph doesn't contain any triangle then it's easy to see just using ILS formula that the number of edges that most one minus four for already it is true for K2 and minus two and not nothing interesting. And the next case already that if you forbid K4, if the graph doesn't contain any a click of forward this is then the maximum number of edges is at most three and minus six, and this graph again is the external case, of course it is easy to check that it doesn't contain any K4 and of course it's a triangulated graph so the number of edges is exactly three and minus six. And if you would like to continue then then of course it's even more nonsense for K5 actually planner graph doesn't contain a K5 so there is no theory here. That's why nothing happened for a while. And to tell the truth, just in 2015, now then initiated the study of classical to run that problem. When the host graphs are planner, and we would like to study this extremely number. Okay. And I noticed that a planner graph is doesn't contain any C4 has at most 15 times and minus two over seven edges. If any is at least four. And for C5 we have a similar number 12 and minus 33 over five, we need some lower bound the first case it is trivial but the second is a little bit less natural that for small and it is not true. And it's not surprising because if any is very small if you have just few vertices then then this planet the condition is weak. It will be more interesting if you have more and more vertices. But anyhow, if any is large than most of the qualities are sharp, at least for infinity many values of n. That's the center of them. And now I would I arrived at the subject of this talk hexagon three planet graphs. It was 2019 when land she and song proved that if I have a n vertex planner graph, which doesn't contain any C6 any hexagon than the number of edges is at most 18 times n minus two over seven. And this is true for all. And at least six, and we have equal even then is equal to nine. Okay, but we have this case and then is equal to nine but to tell you that we do not have a general case. So now I would like to stay at the main cell and what I would like to speak of today. I would like to point out that if Jesus C63 plan plan graph or planner graph on at least 18 vertices. We have we need this condition. Then the number of edges is much less or some but less than the estimate above. So 18 over seven. And here we have five and five over two. So actually the number of edges that was five and over two minus seven. So, this is the certain but I would like to speak of to tell the truth what I would like to prove that someone different statement. This is too connected and see six free. And all the degrees are at least three, so the minimum degree is at least three, then the number of edges is at most it will depend on to the second state condition and also important. If you have a vertex of degree today you can delete it. Just somehow, this is nice to have this condition sometimes I can tell you. This is not an important thing to connect to this is important because we will see that we will have some trouble when the graph is not too connected. But before I speak of this problem before hexagons, I would like to go back to the pentagon free construction. And, okay, I will not explain this picture but they have this kind of greed, and then they make some modification, you can see what happens in the middle, or if you cannot see that it is not a big problem because I do not want to speak of it. I would like to speak of this. No, it is even more complicated. So again, the operation is that somewhere in the middle. These two vertices are replaced with some other graph, and then we did this. And then we get more and more complicated graphs. In the paper, I do not want to present it I do not want to understand it either, because I suppose that something is wrong with this construction. Okay, I mean, the construction is perfect just, I couldn't imagine that there is no simpler construction. Okay, I know I would like to show a simpler construction just for the pentagon free planet graphs. Here I have a pentagon free planet graph. This is the hexagonal grid. Wonderful. But then what we do is the following. On every edge. I didn't want to do this. Okay, so here. If I touch it, then it will move. I do not want to do that. Anyhow, you might remember how it was done. So here, every vertex was. So here this is better. So here every vertex remainder, but every edge is half. I take a middle vertex on each edge, and then around every vertex. I draw a key for, and then it is quite clear to see that. A five cycle. And then the question is what can we say about this graph. It has roughly 12 and over five edges. I'm sorry that it is moved. If you want to prioritize this construction and draw the boundary smartly, then you can get exactly 12 and minus 33 over five edges. So then it is a much nicer pentagon free construction. It is not that it's not a interactive construction or whatever. Actually, it is coming from the hexagonal grid. Now I would like to speak of the hexagon free construction, which will be somewhat similar, but here I definitely have to mention a story what I heard from Alakaz. Unfortunately, he died relatively young. I mean he was 61. And he worked together with Polaris. And then they arrived at some point that. Okay, this is a very nice serum. And then this. It is clear that this serum that the serum but the proof is sharp for the pentagonal grid. I mean that we have to use regular pentagons. And then you really like I said that Uncle Paul. We do not have hexagon. We do not have pentagonal grid. So, so what does it mean. And then I said that that's why we have to construct it. If it exists, then we don't have to construct it. And then they constructed a pentagonal grid, whatever it means, I do not want to mention it now. But sometimes this is really the point that we have to create something which is, which doesn't exist. What happened here. Back to that. So we have to construct a pentagonal grid. But I will show it later something this has changed. Now, anyhow, first I would like to. I'm repeating this serum again. And unfortunately being this technical condition, because otherwise it is not, it is not true for small and in this case, if we have this technical condition that it is true for and greater than equal to six. And why is it so we have this very simple graph. We have a number of vertices. And which is five in this case, and the number of edges is nine, which is five and over two minus 3.5, which is too many. We would like to have a minus seven here. And of course you can continue for example, if you put together three copies of them, then n is equal to 13 and ease the number of edges is 27 which is still five and over two minus 5.5 so it is still too So we, these examples are counter examples to the general upper bound. And that's why we would like to because we would like to avoid this. These examples that's why we have to assume that the graph is too connected. And now I would like to speak of the half time I agreed what I would like to construct. The trick is the following that we have to take the hexagonal grid, and we take. Oh, I'm sorry, I didn't want to do that. So here, we have this edges, and in each hexagon you can see that the same edge, the bottom right edge, and we put a halving vertex on this, and then in a smart way. You can consider just three rows you can consider if you like several rows, and then you can connect the upper and lower vertices in a smart way, and then we have 10 K plus seven vertices. In this graph if you have, if the length of this grid is roughly K. And then, or there is one more choice. If you have, if you would, you can delete five vertices which are in this red set, and you can add this red edge and then it is also a similar example. And then what we do is the following and this is more interesting. And that, what did we do in the case of the pentagons, if we had a vertex of degree three that it was replaced by K for, but in this case it is not dense enough. So, rather what you would like to do if I have a vertex of degree three, then here instead of a K for a K five minus is put here what is a K five minus one edges missing from the K five because K five is not planning. And then here this K five minus is put into the. This will replace this vertex, don't forget that we have this having vertices, or if I have a vertex of degree to this is the problem that in the hexagonal grid. There is no vertex of degree two, but in this. Let me call it have to go to the grid. We have a degree two, and we do the same. We use these two vertices, and then still we put this K five minus minus. Okay. And then we take this grid. And we take this graph and then embedding this K five minus pieces. And then we get n equals 36 care minus 30. So this is the number of vertices and the number of edges is exactly 90 k plus 63. I skip the computation but it is easy to check. Okay, so this is the construction. The second is basically the same just. We delete this five vertices and we do the same, and then we will have a construction when n is equal to 18 K. Plus 10. And now I would like to speak of the proof, not because it is simple. The proof is more than 20 pages. So if you have time I can tell you know, to tell that was I cannot tell you, I don't remember for the details. But at least I would like to explain what happens, because the basic idea is is, I think it is pretty interesting of course, in the applications, it is very very technical. So this is actually a concept, or maybe not so special anyhow triangular block. What is a triangular block triangular block is nothing as we take a plane graph, and we take an arbitrary edge. If he this edge is not contained in any three phase. We could not try and we'll be careful three phase, then we call it a trivial triangular block. Otherwise, because simply we construct a triangular block in the following way. Start with this edge, and if there is a triangle containing it is a bounded three phase more precisely counting containing it, then this other two edges are added to this age. I take another edge E prime in this graph and I search for another boundary three phase containing a prime. And if you have, then we add these edges, maybe sometimes it is just one edge because maybe we use two edges. Anyhow, these three phases added to that set of age, and so on we repeat it as long as we can. And finally, we obtain a maximal sub graph, and this is what we call a triangular block, and this is what we call be easy to check that it doesn't matter what edge is the starting point in this block, it will be the same. We will get the same triangular block. Maybe it was a little bit complicated so immediately I would like to explain what happens here. So what kind of triangular blocks do we have in the construction. What is that nothing else just key five minuses. So actually these are the only triangular blocks in the construction. Unfortunately, we have much, much more different types of triangular blocks, but it is easy to see that it cannot be too big. And the main idea of the proof is the following. We would like to prove that the number of edges is at most five and over two minus seven. But if you use I last for me, this is equivalent to this inequality that seven times F is a number of faces plus two and minus five E, this is less than or equal to zero. This is what we would like to prove. And this is what we would like to prove. Let me say triangular block wise. We compute more or more precisely estimate the contribution of each triangular block to this function seven F plus two and minus five E. Look at this triangular block what we have in the construction. Suppose it. It is put to replace that vertex of degree three. Then look at this, this triangular block with this tree. I call it Lex. This is the number of edges. This is the easiest part. We have nine edges here. No problem in this block. What is the number of vertices in this block. Not the number of vertices, the contribution to the number to the number of vertices in the graph. What is enough be here we have these two middle vertices. This is two. And here we have three vertices in this outer triangle. The contribution of this vertices is at least one half because here I have another block where I have this red edge. This is a block. Maybe we have more blocks than the contribution is not one half but once we have one force in this construction is exactly one. So it is at most or in this construction is precisely two plus three halves. And what is the contribution to the number of faces. Maybe if I have this triangular block then inside it contains five blocks, this is five faces. And then outside here we have these three edges, and under this is outside we have a face, but this face cannot be a cannot be bounded by a short cycle. No cycle, no five cycle, no four cycle, whatever, no triangle because it's a triangular block but even so it means that here we have at least a seven cycle and the contribution of this tree edge of this triangular block to the faces is one over seven one over seven one over seven and we take it three times. And then I got messages getting worse and worse. You see it. I cannot, I cannot open it. I lost it. Okay. Yeah, we just see the same, the gray screen. Yes, I know what you see. This is not what I would like to show you for a second it disappeared. Okay, here it is. The other version. If I have this five size this block, and it has just two legs you remember, we placed it to replace a vertex of the 22 then in this case, the competition will be different, but it is easy to see that again it is seven plus two and minus five is at most zero, but what happens in this case. Actually, that is a trick here. The number of edges is nine. This is clear. The contribution to the number of vertices this to middle vertex, middle vertices. This is to two vertices. This is one half one half. And here the upper vertex because there is no other block here. This contribution is one again. So the contribution to the number of vertices is three times one, this, this three vertices on the vertical line, plus two times one half. And what is the contribution to the faces here below we can have an at most a seven gone. So this is one over seven. On the other hand, here we have five faces, but to tell the truth we can have just one seven one on this side. So it is one over seven on this side to we cannot have a seven on a seven on here because here this is a vertex of degree one. Actually, the contribution to the number of faces is two over seven, one over seven below, and these two edges together contribute just one over seven because we have this joining edge so here you can have just at most a seven one, and then it is just one over seven of this seven one. It's a little bit tricky, but it can be done. And this is the trick that actually what we compute this is the contribution to the triangular blocks of to the, to this function, seven F plus two and minus five E. Here. There are some blocks here are the first two cases what I mentioned short in the picture here I have five other types of blocks. No, I have bad news, not just this five we have I don't know how many but a lot of other books for each of them, we have a similar competition, sometimes the proof is even more complicated, and even more it is not enough, because for example this kind of problems here I have this block in the middle, this two triangles sharing an edge is one block this is be, and we have four outer blocks be one bit to be three before they are trivial blocks they are just one edges. And in this case, the trick doesn't work. However, if I compute the contribution of this five blocks together, then this is correct. What is this block in the middle here, we have five edges. What is the contribution to the number of vertices here I have two vertices this is to, and because here I have three blocks be one bit to, and be quite similarly here, it is one set plus one set. So the contribution to the number of vertices at most two plus two over three. So the contribution of this block to the faces. Here I have two faces inside. And unfortunately here, we cannot see one over seven one over seven, just opposite we can see that here we have a quarter. We cannot have a triangle because then the block will be bigger. So it is one over four four times. This is it. So this is four over four. And if you add up then it is seven plus one minus five it is for sir, which is not less than or equal to zero. However, if I take this block and be one bit to be three before together so this five words that look at the contribution of be one. It is just one edge, it's a trivial block so the number of edges is one. What is the contribution to the vertices. At least it is one half. Here I have three blocks at least so it is one third. No, no, here, we cannot have that this is the, this is the whole here we use that the graph is has minimum degree three. So we would like to avoid this technique it is actually it is one said one said here we must have something. We must have something, otherwise the vertex of degree two. So the contribution of bi to the number of vertices is two over three. And quite similarly, the faces can be here inside we can have a foregone it is one over four. And on the other side we can have one over five. We can have a five gone we cannot have another for one because then we have a six cycle. So, this is how we come, and then in this case, we obtain again that seven f plus two and minus five is negative 31 over 60 which is better than what we expect. But anyhow, it can be estimated pretty well. There are a lot of tricks, really plenty of tricks like that, but this kind of tricks are just few, but to prove that we are at least of a triangle box is complete etc. This is a very technical thing. So this is what makes the proof pretty technical, but the main idea is once more that we consider the contribution of the triangular blocks to this function. And it is always less than or equal to zero or more precisely it is not true. Here you can see that sometimes it is not less than or equal to. Where is it. Here we have this for example, we have a similar case here that this is one half and then we have to find out something else. So, then we can see that the neighbor neighbor in blocks to and then together with these blocks. So we actually consider a partition of the blocks. And this is the proof, but once more than the good idea is that we can see that the contributions to this function, this function is nice because it is homogeneous. So we can add them up easily. And this is how we can prove it. So this was the sketch of the proof but the idea of the proof. And now I would like to mention just a couple of things. The general conjecture that if K is at least seven but at most then that we do hope that if n is large enough and the number of edges at most three times K minus one times N over K, minus some constant. So actually the construction will be similar. You remember that in the C 53 construction, we put K fives k force to replace the vertices in this construction, K five minus is if case bigger than then actually maybe six vertex maximum can be put there, or seven vertex it depends on K. And then in this way, this is what you can check that that it is true that this is the best construction. Why do we have this condition for that case less than equal to 10, because if case more than 10, then of course, maybe we can use some 10 vertex. So we have a maximum planar graph, but the question is do we have a Hamilton cycle so sometimes if we if bigger planar graphs, plain graphs are put there, but we do not create a short. A long cycle, there is no Hamilton cycle, for example the number of vertices is 12 and then, but there is no 12 cycle there, then actually this can be used. So probably this is true up to K question but then we have to modify even the conjecture and maybe the construction to. We have that rules. This is a, I do not want to say that it is an easy problem. It's an almost hopeless. There's a way to prove it because the number of blocks will be more and more, and this is extremely complicated, but at least we have a construction. We have a plain graph with all faces of lengths k plus one and with all vertices having degree three or two you remember this is what we can have and replace each vertex to the maximum planar graph of k vertices. So this was the construction and and then we are in big trouble because this proof might be used but it is extremely complicated. We started to prove cake with seven, but it was so long and so complicated that we stopped it. So that's all what I wanted to say thank you very much for your attention and I'm sorry for this technical problem so I do not know what happened. Thanks. Yeah, if everyone could think of it in some way. And are there any speaker, sorry, are there any questions for our speaker. I have a question there bound for four path three planar graphs. Is there a conjecture for K past three linear graphs. This is an interesting why it is so complicated. I don't remember if there is any conjecture probably there is but but I do not remember now I'm sorry. It's surprisingly more difficult. But the conjecture, what is the conjecture that's the second. I didn't think about the generalized to remember. I'm sorry, I don't remember for the conjecture. But I can. In that paper what I mentioned, this is available on air type, and I'm sure that we wrote a bit conjecture, but it wouldn't prove it. Or maybe I'm sorry I don't remember. I'm a little bit suspicious that maybe I don't remember for the construction because you do not have a nice construction. I'm sorry I don't simply I don't remember. I have a quick question. On your very last slide, I think it was the last one, you had said that, you know you have the bounds, or you have the conjectured bound up to K equals 10. And I'm not sure if I fully understood the reasoning but my question more so is, if you were to pass cables 10 do you think that there is a similar bound, or do you know anything about anything about that. So what happened here in the construction let me go back to this. The vertices are replaced by K five minus, or earlier a K four, if for example, you have K equals seven so there is no seven cycle. So here, you would like to put a six vertex maximum planet graph, which is perfect, and then if you put them together, a similar way that like in the hexagonal grid, but it's not agreed. It contains what is the degree to then it will be okay. And this is the conjecture for K equals seven for example, but this construction doesn't work because after a while. You can use not just maximum planet graph. Here, for example, for K five minus this is, I think it's a unique planner graph on the, however, if case bigger than this maximum planner graph with say K equals 12 vertices or 11 vertices is not unique. And then, if you use. A graph which doesn't contain a Hamilton cycle, then there is no 12 cycle here. So actually, in the 12 cycle free construction you can use 12 vertex maximum planet graphs here. This is a phenomenon which is changing that at the beginning there is no problem. So K four has a Hamilton cycle is K five minus has a Hamilton cycle. However, if and if key is large then there exists some key vertex graph, which is maximum planner and still it doesn't contain a Hamilton cycle. Of course, not just a Hamilton cycle is important after a while, the question will be what is the longer cycle what it contains, but this is a very difficult direction. It will be okay, but anyhow, it makes the problem different. So the general conjecture will not work. I mean the general conjecture is that we have this structure that replace each vertex to the maximum graph of k vertices. If the maximum planner graph of k vertices doesn't contain a case cycle. K plus one, but it doesn't contain a K plus one cycle, then you can use that one to be careful here we have links K plus one so which is forbidden this is a K plus one cycle. So there is a shift here. Sorry. So the main point is that what cycles do appear in this maximum planner grasp what we embed. And this will change the construction. Does it make any sense. Yeah. Thank you. Let me just. Let me just ask one last time are there any questions for our speaker, before we end. Thank you very much. Thanks for working through all the, you know, the technical difficulties there. I really appreciate it. So yeah, I'm sorry for that sometimes it happens. And the main problem is that I do not understand what happens in these cases. I mean, when I'm organizing a seminar, then this problem sometimes occurs and somehow finally, luckily, we can solve it but this was the worst case and I was the speaker that I couldn't. Yeah, it's okay I think we all we all got the whole talk still so that worked out well. Thank you again and thank you for coming out and I know it's it's later later where you are almost 930 now so.