 It's really great to be here as I for many, many reasons and not just temperature. So, this is joint work and just to explain where it comes from. So the first group of authors were all center around the second author urban Yuri, and I think they are all former current or former students of urban. But the bottom is current and I think Chuan Chi is also current. But this was during my Fulbright visit to rain union was a great time it was in fall of 2019. And then I followed this up and I started doing some of this work with Chris Cox and actually I think most of the talk will cover the stuff with Chris but it's the same basic topic so. So, come together. Okay, so I think you all are familiar with this but let me just start with the slide and with the motivation for why one even considers the planar graphs and. So the question that was an old question was can every map be colored with four colors and the drawing of the graph and the plane without crossing edges is called a plane graph and if you can move this even further. I cannot. Okay, that's as far down as it goes. So it's regions form a planar graph. So, or sorry, form a plain map. So what you're seeing here is a plain map and the graph with the vertices of course is a plane graph. We use the term plane graph to distinguish from planar plane graph is actually a drawing in the plane. Sorry. No, it's okay. So, because this is boring and everybody's seeing this. Sorry, I don't know how to make that. Okay, let's. I'll try to remember what I'm actually. We use the hours. We don't use the hours. Because because we're not there. Okay. Okay. So that's just some notation of run us it's very familiar if you have done. V is a set of vertices is a set of edges at this is the set of faces. And I guess this theorem is Oilers theorem and certainly Oilers formula or at least one of many oilers. And it's V minus e plus equals two for any plan graph. And a graph for which you're playing graph drawing exists is called a planar graph and this is the most interesting planar graph because it's the most dense. It's a well if you judge density in a certain way. And it's K five minus an edge and the edge that's missing is. It should be online. Alright. So the degree of a vertex is the number of edges incident to be you're familiar with that the degree of a face is the number of edges incident to the face, except for bridges which count twice. And so if you look at this drawing you can see there's a couple of three faces. The outer face has many has a very large degree where you see some of the leaves level 12345 degree face. All right, and there is what I what I was just calling a handshake. I'm not for playing graphs. And so for every plane graph and I'm sort of emphasizing it's a plane graph here so I'm picking some particular drawing in the plane. So twice the number of edges is equal to the sum of the degrees of vertices, which is equal to the sum of the degrees of the faces. So if you've done planar graph stuff you know that they're every planar graph there's a dual graph, the dual graph. The vertices become faces and the faces become vertices. Okay. And there is there's an interesting corollaries here. The interesting thing is that we have bounds on the number of faces in terms of the number of edges for trivial reason that the degree of every faces at least three, except in the very trivial case where your graph has exactly one edge, and that edge counts twice, but there's not a number of edges there for that face, but otherwise is at least three so number of faces is at most two thirds number of edges, and if your graph happens to be bipartite, you can't have any three things. And so it's at least four. And again, the exceptional cases that if there's exactly one edge. And so the number of faces is at most one half the number of edges, but if you plug that back into Oilers formulae gets a couple of very famous formulas. The fact is that for a plane graph or plane graph, the number of edges is at most three v minus six. And if it's by apartheid it's at most to be minus four, except for the case where you have a K two. In fact, these formulas work, even with isolated vertices in fact isolated vertices only help. So the actual only exception is K two. And I guess K one, but all right. And so this immediately allows us to conclude two different graphs are not plain K five cannot be planar, because it violates the first condition. It violates first mission. The second graph K three three. The first condition is okay because it has nine edges and three times six minus six is 12 is perfectly fine, but it is a bipartite graph and therefore it violates the second one. And of course if you do something silly like but degree to vertices and amongst the edges, and of course that's not planar also because topologically it's the same thing. And of course that the famous result of Kurtowski and Wagner. Kurtowski, she has no subdivision and K five K three three. That's true if the graph is planar, and it has no minor of a K five K three three. And usually think about subdivision so I think that's for me that's the easiest thing to think about your mileage may vary if you actually know I need to apologize, which I do not. All right, so Hakeem and Michael. Define the maximum number of copies of the graph and invertex planar graph. Well maybe they didn't define it but we're retroactively assigning that quantity to them. And it's a typical sort of turn counting thing. So the, the number, the maximum number of copies of an edge in a planar graph is three and minus six. Now we established an upper bound you do have to establish a lower bound, but type established below about, and actually established that they're an exponential number of planar triangulations and shockingly got pretty much the exact value. I think there's a one plus one there but that's, that's the number of planar triangulations. So maybe do you have the three and minus six is achieved, but for large enough and it's achieved by a huge huge number. And they also went to counting try and they established that if I was at least three, then their maximum number of triangles was three and minus eight minus six minus eight. So you don't quite get a triangulation. So the example that they got was exactly this thing here. I was calling it the purse, you know, it looks like a fancy purse you might say it's a mask that ties away all the way back. But they got three and minus eight triangles. Okay, so reasonable results. They also got C fours. The same thing the same thing maximizes it. And it's roughly and squared over two. Now they, they put a lot of work into the lower order terms and if you know where my education comes from, I'm really focused on the first order terms, and second order terms. And of course, you are C four is a K two two, that will come up later so you can if you have a C four, then you can generalize it in two ways. You generalize it as a cycle, or you generalize it as a complete bipartite graph. And we'll do both. Then, after lying dormant for a while, along and Carol picked it up. I think this is like notice of paper or something like that. So it's very early. So they got a nice perfect formula for counting stars. So, this is K one K. So it's a star leaves. This is the formula. The main term is of course, you're going to fix K and go to infinity. And that's roughly, and to the cat, two times, okay, okay, factorial is the first order term. They have this term of N minus four times four choose K, which is of course zero of days bigger than four. And then this last term, okay, it's the same extremal branch. It's this. It decided, well, we got K one K, what about K two K. And this theorem looks really complicated. But, you know, for my selfish purposes, if I say, I only care about large enough and then the first line, and the third line, and the last line are the ones that I care about. Okay, and let me go to infinity. But there's, there's several interesting configurations which I really won't discuss that happen if n is small. So some things are allowed to happen when it is small. That's that doesn't happen when it gets larger. Okay, so again, if you want to rewrite this the way I would like to it's roughly and to the K of a K factorial is the first order term, and then they're smaller terms that are order and to the K minus one. And I'll just remind you that the extremal examples this one is. In particular, this is just a special case I'm pulling this out. If you pick the path on three vertices, then we get n squared plus three minus 12. And a group consisting of four of my co authors and Casey Tompkins established the exact value for C five and they got a couple of annoying configurations, I'll consider them annoying but they're also beautiful. And because I've done enough Tixie for this talk, and it would involve a lot more to do the seven configuration five is not so bad. But event is sufficiently large it's roughly two and squared. And let's see they got P for again some really nasty configurations and equals eight and seven. It's, I would say that it's sort of coincidence that it works for five and six that we've got the same formula for and at least nine, though, you get seven and square. And they proved that if you had any path, then they got the right excellent they figured out the floor of K minus one over two plus one. And there are these. There are these constants that show up that upper world constant. All right. And then this is a very recent result by Chatham Lou from 2021. It was just posted a lot of months ago. And it says for all each there exists a case such that this number is theta of N to the case of you pick pick a graph that you're trying to count. And you want to count the number of maximum number of copies of it in a planar graph established that based on that graph there's some parameter K, which he actually is able to describe. And then you get upper and lower bounds of the order of it. All right. I think it's useful in analyzing this result to see why the extremo example gives you the value of dots. So you get seven and squared plus jump. So what what we first look at is we sort of count these according to where they're appearing in the mask. So this first example has one edge sitting in the middle of the mask you barely see it but it's red. And then, and then you choose this one and this one. So, the way you might look at it I guess is maybe you could choose the edge first and believe there are n minus three edges to choose. And then I have two choices for the vertex to pick two choices for the direction to go up the right. And then I have n minus four choices for the last. So that's how you get four of this type, but there are other types. There's type like this. And so for this one, you have two choices for which direction to go. Is does it have this degree one vertex here, or is it safe. Does it sit on the left or does it sit on the right. I'm trying to help the people who are online so I like to point so. And so there are two n squared of this type. And there's n squared of this type. Of course, it's two times that lines to choose to so it's a square. And there, this is where you take the handle or the strap of the mask. And then that's your center and she and then there's a smaller order of some others. For instance, the strap of the mask could be could be a pendant edge. Because those are asymptotically. Okay, so, so that's P five. Yeah, that's before that's P five so notice that the order of magnitude has changed. So the order of magnitude was n squared for P two and P three. So n squared and seven and squared and now we haven't cubed. And basically they all look the same, at least for the higher order term, if you want to cubic term, then what you have to have is that the second and the fourth vertices have to be these huge degree verses. If you think about that, if you restrict yourself to this configuration, it's sort of obvious that that's going to be exactly how you want to count this. The second and fourth, the vertices do not have high degree, you can guarantee that you'll have one lower order. And we could strip this, and it doesn't change the asymptotic cell. Notice that we couldn't strip this for the path P for for P for we needed those that the extra edges in the mask, and the extra edges edge of the handle for. So what do we do when we got the P five and you will notice we do not have an exact answer here. So, and I can give you about three different points in the proof where that will be help us. So, so we just let and be sufficiently large and we didn't even worry about what the second one was when that was the case we just got the coefficient and notice that we knew that it had to be constant times ended three, they had established that before. So, we were just working for the constant. And the big one right here, which is what I really haven't heard of. And it's very similar to the handshaking llama for planar graphs, but it's, it's quite a bit stronger. You let any greater than or equal to K greater than or equal to three and let G be a planar graph on vertices, such that you have a subset of K versus case at least. And then you have this front loaded degree sequence. So, not only is the number of edges bounded, but if you take a small subset of vertices, then, and small subset say 20. Even though the sum of the degrees is going to actually overall could be as high as six. But if you just say 20 vertices is going to be two. Now, eventually the six K term catches up with. If K is large enough than that actually contributes. If you think about a small number of vertices and basically some of those degrees is two. So this is exceedingly trivial. And it has to do with looking at us, and then the bipartite graph between us and everything that's not as the sum of the degrees of the graph induced by us. Well that's a planar graph. So that's B by six K minus 12. And then, if you take the rest of the edges, then that's a bipartite planar graph. And so that's bounded by two and minus four. This is why we need a K to be at least three. But if K equals to I think it's pretty clear that basically this formula works as well because no vertex can have degree more than minus one. So you, you don't quite get this in quality, but you get. Okay. So then we took a look at this and we said, well, if I want to count the number of p fives in the graph, what can I do. I can order the degrees. And I can just say, let's focus on the second and fourth vertices of my path. I can just fix, fix some v i and v j. I compute the code degree, that's the number of third vertices. And then I look at the degrees of the second and fourth, that will give me the first. So this formula is pretty easy to see I think. But of course the code degree is greater than is less than or equal to the minimum of the two degrees. Well, we know what that is we put them in order, and v j was just designated to be that. So, with this degree sequence, the number of p fives is bounded by the summation of the degree of v i which is the higher degree vertex times the degree of v j squared. Now point out that this might be a wild over count. I know not concerning myself with a lot of them I've thrown a lot away here. In the example that I showed you we didn't really count the vertices that were say, next to each other on the in the mask as being second and fourth. So, I mean this, this could be a crazy over count. But we didn't care, because we only care about asymptotics so why not. And so we said, well, this is this is a problem that we're trying to solve. Can we actually get the answer, because we know the lower bound event cubed because we know it from that construction. So we just have to match it. And so we just have to maximize that thing subject to that thing subject to a linear number of constraints and notice that the second constraint there. I mean, all I need to do is take the first K vertices because I ordered them by degree. And so this is just a degree camera not even focusing on that graph theoretic structure, let alone the topological structure. And then we, we had that one last thing, six out of minus 12 for some value at some value K, the second constraint becomes trivial. And so we have to fill in this last. Turns out the solution is n cubed plus big O fence squared so we got what we wanted. The proof is sort of a case analysis of how bad the beginning of the sequences, the beginning of the sequence is not too bad it turns out this is small anyway. The beginning of the sequence is very, very good, we're very close to our mask. And if it's somewhere in between it's annoying needs some end to be larger than 11,664 to actually make the proof go through. I don't know why it was so easy to pluck out but somehow 11,664 work. So we reduced it to this to this optimization problem and we're perfectly happy with that. And then we were done and I came back and Chris was back at Iowa State and we were all masked up and we, it had been a year since I worked on this problem and we started discussing it. And so what about p seven for a variety of reasons p six is a lot harder p seven is easier. And I'll tell you why it's easier to do p seven, because I have second, fourth, and sixth vertices. But with p six, they don't alternate nicely. So that creates a problem. And it turns out that we got four over 27 and our stream will graph is different. It's not the same. Irvin had conjectured that this was the kind of structurally larger paths, and he was right for p seven so probably right. So, this, the construction here, I, you might think of this as some sort of hexagon but I think of it as a trying that you have three vertices of huge degree. And there's a K to T in between the T is roughly and I could put extra edges in there but I'm not going to bother because it doesn't affect the products, but those extra edges would affect it if it were p six. Well, the same sort of argument passes through for c six. So p seven is very similar to see six in that sense, and you get n cubed over 27. You may notice that our error terms are not as nice as others, although we don't have a little out we actually get a step. There's no, there's no particular magic there it's just that method that we used to prove it couldn't get a second order term better than four minus one fifth but we're pretty short. It should be n cubed in the first case and squared. It's the eight and over four to the fourth, and the configuration is what we might expect it is, and it's big O of N to the four minus one fifth. One note I would make is. This is what I would call, I'm going to call it later a planar blob of a four second. Which I'll get into hopefully we have some time. And so we got this general result which worked for all ends, but was not quite perfect. The exponent on and had already been established. It was established for paths and I think I appreciate was published for cycles already by my co authors. And I think that's also in newspapers so. So that that was established but here we get upper and lower bounds on constants. The other point out is that I think these are actually really good because maybe it's deceiving that the upper bound has a one of rent minus one factorial, the first line. The other one has basically an M to the M in the denominator but that's if you take the ratio that's exponential rather than factorial and so. So it's not so bad. But that was that was our general result for paths and cycles. And then we started beating on our general result to sort of get other values. We were able to get 10 and 12. And we got the conjectured results for 10 12. We didn't bother much with the lower order term now now we have a little love. We didn't, we have a sort of general approach to things and it doesn't, it gets the first order term but it doesn't, it doesn't really worry too much about the second. And in fact, it generalizes a lot more than this. So, what we could do is let each the thing that we're counting be some planar graph, and then we want to blow it up. And so we put these strange curly braces around the K because we were inventing notation. And so the planar law of H by K is to replace each edge with a K to K. So what I've drawn here is an example of C six blown up by three. And so you've seen that that I want to emphasize that this is an example of something that we are counting. But remember that in the examples of the things that were extremal, they were also blocks planar blocks. Little bit. And what we were able to show was that if you had a graph like this, we were able to get upper and lower bounds again that differ by. We should differ by. Yeah, something is exponential, but not factorial. But in the case where each the original graph that you're blowing up, either has minimum, either K times delta minus one is at least two. So either Delta has to be at least three, or K has to be at least two and Delta has to be at least two. So those two cases. And K is at least nine and also works but for some reason we couldn't get anything for Delta equals one and K between eight. And that's just an artifact of the proof that's not, there's not I don't think there's much reason why that's. And in fact, if K was really large we actually got the answer we were able to improve the upper bound. So if K is a planar graph, where K is roughly this thing, then, then we were able to get the coefficient there. And the lower bound is not anything mysterious that's, that's the blow up of H by the right amount and over K something like that. So we will have all our results get the upper bound. All right. And then there's sort of a general approach to this with which Chris likes a lot and I do too. And I think it's, it's sort of a neat way to look at graphs and away from the planar setting or anything like that. So let V be a vertex set and let mu be a probability mass on the pairs of that vertex set so you might as well consider your underlying graph to be complete. But it's a probability mass, well, you know, it's everything's between zero and one. So I'm waiting the edges between zero one and some of the weights is one. And for any graph G, I wanted to find the, the new value of the graph itself to be the product of the news on the edges. So, think of G as a cycle. So this, this V vertex that I'm not even giving the graph itself a name because I might as well assume that it's complete. And G might be a cycle or path. So we define mu of some graph G to be that. And, and any vertex we sum over all the weights sitting at that vertex, and that's mu bar. And so what we do is, we basically show that if we're trying to count what we're trying to count. What the graph should look like is one of these planar blocks sort of. So, what's going on in the first line of this odd path reduction here. So I assign a probability mass to the edges. Then I have new bar of x one which is coming basically the weighted degree of x one the weight sitting that x one. And then the product of the weights of the edges of Xi Xi one, and then the weights of the other vertex. I would point out that this is odd path so what's happening is in P five. Then am here is equal to two. P five has two sort of hinge vertices that matter. So the am is not the number of vertices but it's the number of vertices minus one divided by two. And then, among all, among all probability masses, there's some finite being, we find the supremum of this expression. And we prove that the number of pads of the point two m plus one is at most row them over two times and the m plus one plus the value of this small order. And that's the general approach. And Chris and I both like this, just this idea of let's put probability masses on a graph. So this is done for various, for various things it's not terribly new but it's, it's interesting that it counts this especially for these, these types of things. There's an even cycle reduction theorem and not much has changed here, except that we're taking the product of a new for all edges in every cycle of the complete graph. And we should have had it in the previous graph. So, here this is here I'm selling over all the paths. And here I'm selling over all the cycles in the complete graph induced by the vertex. That's it. The, the thing that that gave us the results here was the blow up reduction theorem. And then here, the first line is very complicated to take some overall instances of this graph h and then take the probability of the edges along that, that graph and then we find the supremum overall, overall probability masses, and then that gives us the upper value and why, again with this one fifth. Okay, so just a couple of notes about about this thing, the reduction theorems are not structural theorems. And that means that the probability distribution attempts to simulate a planar blow up, but we can't guarantee that the maximum mu that we get is not say a blow up of a k five. So, all we're doing is, it's comes from the very first paper, we take this question, and then we create an optimization for that has nothing to do with graphs underneath. It's entirely possible that the new is has five choose to positive values on a thought, you can't guarantee that this shouldn't happen because I can't think of a graph that we can construct that will produce such a thing. But we weren't able to prove it and we got frustrated sort of saying of course it's not true it has to be planar you can't have, you can't have a non planar configuration that and we just, we couldn't prove it. Would have, would have made some of our later results like the 12 cycle simpler, and we use Lagrange multipliers to compute these, these coefficients. I wish to talk here to be specific, precise, but if you don't remember never learned those. That's just a gradual. Alright, so let's consider the cycle to do this example. If we use an equal distribution on the edge set of KM, then this this coefficient that we, we compute this data is exactly this. So you want to find all the cycles in a complete graph on m vertices m minus one factorial over two cycles to choose by doing m cycles on vertices and minus one. Yeah, yeah, you have to divide by two because you can go the other way. Yes. Right. And of course each one would get equal weight one over m choose to. But if you look at the equal distribution on the cycle itself. It's a different expression it's one over m to m, but in fact that's large enough to trust me but for all strictly greater than three I think it's of course it's equal for m equals three because k three and see three are the same thing. But if I was at least four, we get a straight in quality here, which means that we do not want to put our weights on the on the M clique we would have it on the rather have it on the cycle. Unfortunately for us those are only two examples of probability masses that we could choose either, you know, you want to do a six cycle. Well, you could six cycle divide by two. So, now I'm going to write on the board that we can see, but, but if you, let's say we want to do C 14, which is the next case. So we know that put equal probability mass. C seven is better. But there's no a priori reason why we even need seven verses, our probability mass could be on 20 verses and there could be the some nice configuration that gives you more 14 cycles, then, then you would possibly get from that so. So, and, you know, we can we wouldn't even be able to go through all the cases. I suppose you could but all the different probability masses on a graph on seven vertices. So anyway, do what this says an equal distribution on C is better than a equal distribution. Okay, so for C 10, we did actually go through some of this so so maybe I'll just give you a flavor of what's going on. So what we proved was that the optimal distribution for C 10 is given by a five cycle where the new value of every action is one fifth. And if new is optimal the current push talk of conditions end up giving us two inequalities. And, and let's see so. So we have these two conditions for every vertex, if we assume new is optimal. So these come from if you if you have an optimal one, then we have different ways of sending an epsilon from one place to another, and that provides a contradiction to optimality. So they these two conditions for the edges, and then for the pieces. Now, as I said, compared new to the prime, which it's the same sort of thing, which sets new prime equal to zero rescales the other edges so if you have some configuration, find one of them is positive anyone of them, delete it. And then we scale, compare it to what you have before, and it turns out that that's not better. So, the corollary of that ends up being that the support, the total number of vertices on which you have support is five which is great. Now I only have five vertices to worry about. But then it also gives that the sum is two over m which in this case is five, which it's not too hard to figure out that that implies that we're going to have one fifth one inch. So the limit that from before it gives that the new bar is greater than 0.355 for every V in the support. But since the sum of the support is to that means the support has to be five. Because if you have at least 0.355 and have six things. That's the six times 0.355 is greater than two. So you can't have six things to support. But since we're concerned with counting the 10 cycle which reduces to the five cycle, then we need at least five vertices to actually count this thing all. So the support must be equal to five, including Kush Tucker gives you bar is less than because of this new bar is less than equal to two fifths so quality must hold, and that gives you the first and second phase. And then what we did was, we did a compare mutated prime, which sets me for this, you know, does it differently scaling. So for this third lemon, we had to assume the first couple of limits we had to assume, in particular to. But we're going to do now we have five vertices so I can get a handle on five vertices. So we actually do something a little bit more clever to show this. And again, the idea of the proof is just to do a rescaling. And then corollary is if Z, Z satisfies that, then you will be is greater than zero five. And so the idea of that is set Z will be two thirds satisfies the lemon. And so new is greater than two fifths. And since the sum of all the edges. And the checks is equal to two fifths, the degree of the must be less. And that's it. So, so in fact, the support of new must be a maximum degree to graph on five vertices with at least one five cycle and that's the five cycle and that's all you have to support it. It's not terribly long. It's sort of hard to describe and talk like this, but, but it's not so bad. It's the C 12 was more complicated and more cases. What I will point out is that for and I don't know if I have this in slides the support for C 14. This is true. I think it's just eight. But it's not seven seven is easy. You told me it was seven I could, I could run through the green fished up conditions. Here it's eight and I'm not sure if there's some really nice algebraically defined graph on which I can count a lot of C sevens, and then that will give the count C 14. I'm not sure. So, there are open questions and all of these are the first, the coefficient of the first order term that we don't know. So what is the count for C M for even cycle starting 14 or the path for odd path starting nine. The odd cycles are harder, I think significantly harder, our meta theorem that we use. Proves again the right order of magnitude the right exponent for for cycles and paths, but it's not good at computing the exponent and that's because are the coefficient. The reason why it's not good at computing coefficient is that these extra edges that I could delete in the mask and center of the mask and the strap of the mask. It's absolutely necessary for odd cycles and for even paths. So, and that makes it hard to deal with, which I think you might be able to deal with but we just couldn't. We couldn't do anything general enough to make it work. Yeah, so okay so I did have it in slide so for C 14. If we were to compute beta of seven, we could get the support those eight vertices, not seven, which is what we would like. In general, actually we didn't have to be at a time in computing this beta of M, you require less than 1.26 times m vertices. So the number of vertices, at least in the cycle case is not crazy it's barely more than the number of vertices we're looking for. We have no reason, no ability to prove that. So I think, you know, I, if you really want to look at this as a sort of different problem. We say for a given graph H, what is the new on some can for all values and the maximizes the probability of finding an H by choosing E of H edges independently at random for me. This is this generalizes what we were doing, but this may or may not actually solve the problem with counting graphs and planar graphs. One of the reasons is the answer might be a configuration it's not planar. Then we don't know what to do, we have an upper bound but we don't. So this is sort of the general question that that I think is an independent interest and just choose edges at random for some distribution, but what you're interested in is maximizing the result maximizing probability and not the expectation. And, you know, can you, can you take our results and sort of easily get exact results. It seems likely. Okay, we have our results we know that a blow up of a C five is is the right answer, at least numerically for finding copies of C 10. Can we get the fact that can we get some sort of stability that says, if we are not close to that structure, then we are not close to the number, and therefore you use stability, and then say, if we're close to that structure, we can move a few edges here and there and put in straps and put in extra edges in the mask and get something more precise. And I don't know that seems that seems like it might be a lot of work. That's all I have. Do we want to stop sharing or not, I don't know. Oh, control well, I guess you just so. Everything works because we don't even worry about the top logic questions. What works for us, what works for us is that the graph has to be a linear number of edges so the class of grass that we're considering instead of planar class can have a linear number of edges, actually, it can be slightly more than linear but let's say we have no copy of a K3T for some value. And of course a planar graph satisfies that more, but that's all we need for a reduction first. And so it works for any genus because for every genus there exists some T so that K3T cannot be drawn on that size. So it works for us for more complicated topology and we looked that up recently and after somebody asked similar question and but I don't think it actually shows up. Yeah, are there any other questions for Ryan? I don't know. The largest argument. I don't know. I don't know. I mean, I think we got lucky in that what we got was a blow up. And so, so that was sort of the impetus for everything is everything can be treated as a blow up and if it can't be treated as a blow up, I don't know. But yeah, I mean my inspiration, it always is, is back to some of these lemma. So, you know, it's, that is senses of blow up. I mean, it takes some sort of, it takes things that reduces it to a finite amount, which is what we read. But it was really that five path thing and noticing the second and fourth vertices were the ones that mattered and everything else was dependent on those. And when she saw that, then the generalization was ready to go. I apologize if I covered Irvin's talk. Oh, okay. Any other last questions for Ryan? Doesn't look like it. So let's say Brian one more time. And thank you everyone for making enough. And I guess we'll, we'll, we'll convene there. Okay, or. Yeah. Okay.