 So this week, we are excited to have Xiaonan Liu from Georgia Tech. She will be telling us about counting Hamiltonian cycles and cleaner triangulations. So take us away. Okay, thanks for the invitation. Okay. So, and thanks for the introduction. So today, I'm going to talk about counting Hamiltonian cycles in planner triangulations. Okay, so. So first, let's start with the very famous theorem, the full color theorem is every planner map is full face colorable. Yeah, this is. It was first proved by a panel and taking in 1976, but then in 1997 Robison Sanders Seymour and Thomas, Thomas gave a complete proof, but it is still computer assisted. Yeah, so here, what I want to talk is about the Hamiltonian cycles. So, if a planner map has a Hamiltonian cycle, then it can be easily full face colored. So why do we have this one, because if a planner map has a Hamiltonian cycle. So actually, we can just like draw. Oh, sorry, like, we can fix a drawing such that we have the Hamiltonian cycle here. And we have some, yeah, some edges between between them. So then we can know that actually the graph inside of this inside of the Hamiltonian cycle if we consider the deal graph, so we don't have a cycle. Yeah, because we don't have a vertex inside of this inside of this cycle. So the new graph doesn't contain a cycle. So the new graph is a tree just inside of the Hamiltonian cycle. So same for the outside. It's also a tree. So we know that we can color the tree with just two colors. So two colors inside and two colors outside so we can just use at most four colors to color this planner map. So we just want to, why do we want to talk or learn the Hamiltonian cycle in the planner graphs. So here. So we just want to find the condition to guarantee that the to get in the existence of Hamiltonian cycles in planner graphs. So here are some early results. In 1931 showed that the every four connected planner triangulation contents a Hamiltonian cycle. So the triangulation mean that every face of this planner graph is funded by a triangle. Okay, it's just a cycle of less three. And also touch extended this theorem to all four connected planner graphs. So he showed that every four connected planner graph contents a Hamiltonian cycle. So this is a very classical theorem. So, actually, this is like the best condition, because we can find a three connected planner triangulation that doesn't admit a Hamiltonian cycle. So yeah, we can show. I can give an example like this. So, for example, if we have, if we have a triangulation, a triangulation and vertex and vertices. So, then we know, by all our formulas, we know that it has three and minus six edges and two and minus four faces. So if we insert a vertex in each face. So like here, this is just this graph, this triangulation G. And if we like for each face we insert a vertex here. So then are now for this one for this new graph to it. We know that's the number of vertices of G should just be and plus two and minus four, because we have like this vertices. Yeah. So here, just because the vertices we insert actually form an independent set in age. So if we want to find a cycle in age, actually the length of this cycle can be at most two times. Yeah, to let me say it just because this vertices are independent. So if, if they want to be a cycle, it can be at most. Yeah, actually just to it. That mixes. Yeah, it's like, so the, like the length of the cycle in age for the length of every cycle in age can have can be at most two times. Yeah, because, because these are independent. Yeah, if they want to be a cycle, they must like work like this. So this is the red vortex. Yeah, the red vortex means that insert for that vertices. So yeah, we need to have like a black vertex here. So then we can have at most two and vertices in a cycle. Yeah, because the black like the black vertices means original vertices in G. So the red vertices means a new vertices we add. Yeah, so. So, in the graph, age, the cycle, yeah, we cannot have a Hamiltonian cycle in age, but age is a translation. So it is three connected. Yeah, so yeah, this is the best possible. Okay, so now our questions like, what can we say about the number of Hamiltonian cycles in for connected planet translations or in for connected planet graphs. Okay. Yeah, because now we know that they must exist. The Hamiltonian cycle must exist in such graphs. Okay. So here, actually, in 1979, how can we make Michael and Thompson conjecture that if D is a full connected planner triangulation and vertices, then D has at least two times M minus two times M minus four Hamiltonian cycles with equality if and only if G is a double wheel. Yeah, so this means that this conjecture showed like states a lower bound, a lower bound on the Hamiltonian cycle on the number of Hamiltonian cycles in for connected planner triangulations. And this lower bound is attained by the double wheel. So first let me just show what is a double wheel. So mean that we have an vertices in a graph. So here, we take out what two vertices and the remaining and minus two vertices from a cycle, and we connect this vertex to all cycles. All vertices in the cycle. Yeah. So here, let's just count the, the number of the Hamiltonian cycles in the double double wheel and vertices to show this lower bound. Okay, so here, the number of Hamiltonian cycles. So, actually, we have two ways, like, for the, like for this vertex, we can just choose. We know that in each Hamiltonian cycle, we have two edges incident with one vertex. So here, if for this one we choose to add just this two vertices are adjacent in this cycle. So then actually we have a minus two choices for this one. We choose to adjacent vertices here for the verdict verdict inside we must choose the choose like this two vertices must be adjacent in this cycle, then we can just form a Hamiltonian cycle. So, for this, for this kind, we have m minus two times m minus three, because like for, for this one, we, we have a minus three and minus two choices, and for the one inside we have a minus three, because just we cannot choose this one. Yeah, so then. Okay, so also we can just choose to like the two edges here, these two vertices are not adjacent in this cycle. So actually, then we will only have two directions. So one directions like we go from this vertex. Whatever it's just from here, and we get here and then like this. So from like the clockwise direction we have this one and the further. And also we can go from here. The music. Yeah. Like this. So we can only have like this two directions if this to this two vertices are not adjacent in this cycle. Okay, so then here it's like for this we will just for this protects we will have like minus two choose to just all pairs that we need to subtract. Yeah, subtract is like the pairs are adjacent here so we have like a minus two pairs here. We have so this mean that the pairs of vertices that are not adjacent on that cycle, and we use this one times two, because we have two directions. Yeah, so just. Yeah, if we simplify this one, we will get it should just be this is in minus two times in one or three. Yeah, and this one minus in minus two. Yeah, and then we get a minus two. This is minus two times two and minus six minus two. So yeah we take out to we get two times and minus. And minus four. Okay. Yeah, so this is just. Yes, the exact number of Hamiltonian cycles in the in a double wheel on vertices. So, any questions so far. So, yeah, so then in the same paper. Actually, they showed that every four connected and vertex planner triangulations has at least an overall log two and Hamiltonian cycles. Yeah, so the conjecture bond is a quadratic bond but they show an overall log it like this. And also, in 1999. So, by her and tag show that every five connected and vertex planner triangulation D. So, yeah, so for five connect is just if we improve the connect connectivity of the graph. The lower bound for the Hamiltonian cycle should just be the exponential and it's to to the omega. And to the one over four. So this is a lot. Yeah. So, but for recently, we have a lot of work about a lot of related work about this one. So, alarm D. Eldridge and Thompson showed that every five connected and vertex planner triangulation D. Actually, yeah, they improved the previous bound to to the omega and to the one over four to two to the omega. Yeah, so the explanation about and also breakman van clean clean clean put showed that for connected planner graphs have at least a linear bound a linear number of Hamiltonian cycles. So, we know. So in the conjecture, we can get the conjecture the four connected planner triangulations have like the omega n squared Hamiltonian cycles. Yeah, so here they showed that for connected planner graphs. So this is not triangulation have at least a linear number just something say times and Hamiltonian cycles. Yeah, and they use the use idea about the counting base. Yeah. So, now let's consider like, if let's consider the results on the four connected planner triangulations and vertices. Yeah. So here. Okay. Yeah, this one is just. So, yeah, previous. So they use it like the same method is a counting base they show that the D has at least 12. Yeah, this is also a linear bound 12 times and minus two over five Hamiltonian cycles. Yeah. So, low show that. So, here's like, low show that D has has omega n over log n squared Hamiltonian cycles. If D has a big old log and separating for for triangles. So this means the separating for triangles mean that so in the planner graph. We know that if we have a in planner triangulations, if we have a cut actually this card will just form a cycle means that here's like this is a full cycle. Yeah, so this for vertex we're just like bound something. Yeah, so something inside and something outside. Yeah, so from here is like this we have some restrictions on the on this triangulation is like if it has fewer separating for cycles. So it's kind of like a relaxed a relaxing of the five connectivity like that. Yeah. Okay. And then So, yeah, and I am my advisor show that the D has. So I just we just improve the loss result, and we showed. Oh, I forget. Sorry, I forgot to mention that this is a joint work with. I'm sorry about that. Okay. So, yeah, so improve those work. We just like, we show that D has omega n squared Hamiltonian cycles. So, if D has big old and over log to Why, why. So I have. Oh, okay. Okay. Oh, okay, okay. Yeah. Yeah, so here's like. Yeah, so it's just this bounce the omega n squared is the conjecture, but but here we have some restrictions on G. And also we show this this result is G has two to the omega n to the one over for Hamiltonian cycles, if G has minimum degree five. Yeah, so this is just like, because here we assume G is for connected, and we know five connected planner triangulations have, have, have a lot of Hamiltonian cycles. Yeah, actually, there's just some. Yeah, relaxing of the five connectivity. Yeah. Yeah, and also low and change recently showed that he has to actually he has exponential lay Hamiltonian cycles FG has big old and separating for cycles. Yeah, actually this results improved our previous result. Yeah, okay. So, No. Yeah, so here we just improve. So it's just, we just confirmed the conjecture we we mentioned asymptotically we show that if G is for connected planner triangulation and vertices, then D has at least say times and squared Hamiltonian cycles. Where I'll say is just this constant is less than one. Yeah, yeah, it looks ugly but yeah it's just. Yeah, actually this, this doesn't matter just because we just want this and squared. Yeah. Okay, and also in our paper we showed another result but maybe today I don't have time to talk about, talk more about this one. So we also show that it's like there exists a constant say greater than the real such that for any for connected planner triangulation and vertices in which the distance between any two vertices of a degree fall is at least three. D has at least two to the same times and to the one over four Hamiltonian cycles. Yeah, so this one is just so previously we showed that if he has minimal degree five, then he has at least two to the omega and to the one over four Hamiltonian so here we just like put like make the condition be the like any vertices of degree fall. The distance between any two vertices of degree fall is at least three. So this is better than the previous previous one. Okay. Yeah. So then, next, I will just talk about how we prove the first theorem. Yeah. So, actually, our idea is just, we just apply induction on the number of vertices of G. So, so how do we show this like, first, we just assume that any two vertices of degree fall G. Indeed, are not adjacent. If they are adjacent, we just want to apply induction. Okay, because here, this is good things like if we have two vertices of degree fall and they are adjacent in the graph G. So, if we contract this edge UV and let this new vertex be used to start. We know that this vertex is still a full connected. Yeah, it's just because if we assume that the number of vertices is large enough. Then we know that the new graph is still a full connected planner translation. So then, by induction we know that for this new graph we have say like say times and minus one squared Hamiltonian cycles. Yeah, and if we have this one. So it's just. If we have like, so for any Hamiltonian cycle in this new graph, we can modify this cycle to be a Hamiltonian cycle in the original graph. For example, if we do that, if the two edges is incident with your star. So this two edges. So in the original graph we can like to do this. Okay, if, if it's just like this, we can actually we can have two ways to do this. Yeah, we can also like go this way. So, yeah, then, if we just apply induction then we know that we have say times and minus one square squared, Hamiltonian cycles in the original wall. Okay, but to get the quadratic bond, we also need like a linear bound, like, like, say one times and to get to. So if we find this like say one times and like it's a new Hamiltonian cycles, then we can show this. So how can we get this one, because we just found that if we just do if we modify the cycles in the new graph, we just find all cycles in the original graph past the UV edge, this edge. So now if we want the if we want the say one times and new cycles. So, so we just consider maybe we can consider the Hamiltonian cycles in the in this graph that doesn't use the edge UV. So then we just consider. Oh, now, just here. Maybe we can just use these two edges, because if a Hamiltonian cycle use these two edges. Then we know it cannot use you the edge UV because otherwise we have a cycle a triangle on this Hamiltonian cycle. So, oh, so we can use this one or like your this two edges. Like this, because if the Hamiltonian cycle go through go through the two red edges, all the two grand edges, it won't use the edge UV. So then we just think maybe we can show that if if the if the Hamiltonian cycles through this to edge, the number of the Hamiltonian cycles through this to edge. It's linear in and then we are good. Oh, it's at least linear in and then we are good. So here, we just want to show this two things. So the, the two things like this is just what we want to show is just just invert this is yes a linear bound is just through any two edges e one e two in a given triangle of G. Yeah, so here, because here's like a want like the, the, the number of the Hamiltonian cycles passing through this two edges is at least linear in and so this is what we want to show first. Yeah. And the second one is just, we want to show that, because now, if we can assume that G doesn't have to adjacent degree for vertices, the remaining graph, it just in which any two degree for vertices are not adjacent. So we need to show such what such planner triangulations should, should they have at least a quadratic Hamiltonian cycles. Yeah, so, because this one is better because here, if assume that we have a full connected planner triangulation, and any two degree for vertices are not adjacent in this graph. So actually it's kind of like, like, to make, to make it to make this for connect for connect graph, like, a five connected graph, yeah, like that. Okay. So, here, first, we just consider the case that the two vertices of degree four are not adjacent. Yeah. So actually, this is just this, this result comes from, comes from two results in our previous paper. So maybe let first just let me just show you this one. It's like, here's like, because we know. So we consider this one. It's like, we know that if G has fewer separating four triangles, it, it will have a lot of Hamiltonian cycles. So we will consider like the separating triangle, separating four cycle in the four connected planner triangulation. And we want to analyze the structure inside of this separating four triangle. So here, the first result is like, this is led to be a near triangulation with out cycle. Say UV, yeah, it's like this one. So the outer is a near triangulation mean that so all faces except, except the possibly the outer face bounded by a triangle. Yeah, so here you can, we can see that all faces here. are bounded by a triangle, just the outer face. The infinite face is bounded by a cycle less fall. So here, we just consider like the Hamiltonian pass between any two vertices here. This lemma we consider that this two vertex are not adjacent in this full cycle. So we get like, if G minus VX. So here, we want to say the Hamiltonian, the number of the Hamiltonian pass between you and they in G minus V and X. So we, we, we get this conclusion, that the G minus VX, either it has at least two Hamiltonian cycles between you and they, like, we can have five to two Hamiltonian pass between you and they, all G minus VX is a pass, just like a pass you between you and they. Yeah. So here, then we know that all vertex vertices on this pass, just inside should have a degree full. Yeah, so this is just for two vertex vertices on this cycle which are not adjacent. So here, and we also consider like, if we have, if we consider like two adjacent vertices on this cycle we get you and X. So here we want to say the Hamiltonian cycle between you and X in G minus V and W. So here, it's like we also get like a similar conclusions like G minus VW has at least two Hamiltonian pass between you and X. G minus VW is an outer planar near triangulation. Yeah, so what does this mean is like if we delete V and W, the remaining graph is outer planar triangulation is just all vertices on the outer cycle here. Yeah, and if we observe that we find, we actually have two vertices of degree four that are adjacent. Yeah, so these two vertices have degree four and they are adjacent. If we only have one unique Hamiltonian cycle between you and X in G minus VW. Yeah. So, why do we want to do this is like, we also want to apply induction, because if we now if we consider the case that in which any two degree four vertices are not adjacent. And if we have a separating four triangle, such that the interior is not, is not that large, it just the interior has like, has some vertices, which is not, yeah, the number of the vertices inside of this separating four cycle is not that many. Then we can just do the contract the interior, just contract the interior to one vertex, and then apply induction. Because we know that if we contract the interior to one vertex, the remaining graph, the new graph is still for connected planar triangulation. And it meets Hamiltonian cycle. So then we know that, oh, sorry, we should like, we just use the induction maybe let me first like, if I assume that the interior has at most over four vertices. Okay, so if we come, and then we apply induction, then we know that it has say. Like say times. And like over three over four square Hamiltonian cycles. Yeah, because it's like the interior, the number of the vertices inside of this cycle is at most an over four, and then we have to. Yeah, after just contract this, the interior, and we get this is the number, the lower is like, it has at least say times three and over four squared Hamiltonian cycles in this new graph. And we can just like modify, because any just any to any Hamiltonian cycle we have like two edges incident with this contract tip vertex. So this means that. If you the WX. So this mean like in the original graph we need to find a Hamiltonian cycle Hamiltonian pass between you and they in this graph, minus the WX, just in the original graph, because we contract the interior. So, if to, we want to modify this cycle to be a Hamiltonian cycle in the original graph. So this is just, we want to just to find, yeah, like this find a Hamiltonian cycle cycle in your way. In between you and they in G minus WX. Okay, so here, because here we are in the case that the G doesn't has to adjacent degree for vertices so we know that for each such separating for cycle. So actually, we will have four. Oh, sorry, have to, at least two distinct Hamiltonian cycles Hamiltonian pass between you and they in this sub graph. Okay, so now then we just multiply this one by two. So this is the number of the Hamiltonian cycles in the new new graph. Yeah, so here we just get to times. Yeah, so here's just two times nine over 16 so which is greater than one. So, and then we are done. So here, now we just want if we want to do this one so it's like we need to find like a separating for cycle in this graph and the number of the vertices inside of the cycle should be less than an over four. Yeah, like that. Okay, so to get this one. So we, we just analyze this. So this is about. So here, we led to be a full connected planar triangulation on vertices, let I be an independent vertex. So this is just mean that actually here's like, either we can have like in this graph, either we can have like they and X, such that they have a lot of common neighbors in in the all there is a subset as a. So this is an independent set such that it has a large size, and it's a set your is no for cycle five cycle or diamond six, six like yeah we haven't defined this, but we just need to remember, these are some good properties for this for this independent set. Either we can get a large double wheel structure. So if we have this one. Yeah, all a large good independent set. So if we have this one mean that we have a lot of separating for triangles. And here we, we assume that no degree for vertex, vertex, no degree for vertices are adjacent, then we know they are not just a trivial for degree for vertex, we have something here. And yeah if we have a lot of such separating for cycles. Then by pigeonhole principle we know that we have a separating for cycle, which in which the inside the number of the inside vertex vertices is not that many. Yeah, so it's just kind of. So here, then we just need to know like here we have an independent set, which has a lot of good properties. Yeah, maybe I don't think I have time to Yeah, but maybe so here let's just. So here's just so I independent set. So here is a full cycle or five cycle in G, if this even as intersecting they say the size of this one is to an asset asset here is a diamond six cycle D. So this is the diamond six cycle. So this has come contents exactly three crucial vertices update. Yeah, so this is the definition me if asset who is like, if asset who is a full cycle or five cycle. So mean that we have to what have vertices on this cycle. Yeah, we can have actually we can have at most two vertices. So now, just remember that we get. So here's like this, the independent set we get here. So it's no full cycle no five cycle or diamond six cycle. So yeah, so this means actually that means very good because it's like in each full cycle we can have at most one vertex in that independent set. Okay, so then here. So let's define this one because we want to use this good independent set to find many, many distinct Hamiltonian cycles. So here. So we just consider because the independent set consists of vertices of a degree at most six. So here we deal with this vertex, according to their degree. It's like, if you has degree for we denote this for edges as a link of this vertex. If you has degree five or six, we denote this. Oh, sorry, we denote the edge such that he is incident with you and D minus E is for connected. If, yeah, like this. So here, for example, here for this edge. We delete this blue edge. So this three vertex we are just, we have something here, this three vertex we are just to be a three card. So it is not for connected. So this blue edge is not in a you. What we want to show is like, we want to delete some edges from for each, for each vertex from a you and then as the remaining graph is still for connected, because we know for connected has for connected for connected planar graphs have a have haven't have a Hamiltonian cycle. Yeah, so here. Yeah, we just show. Yeah, this is just show that if it is a for connected planar triangulation. And so here it's like, yeah, this is just no degree for vertex of D. Actually, yeah, this is stronger, but in the, it's just under the condition, mean that we can fall. If we find I, I independent set, set, set your rating no for cycle or five cycle, or diamond six cycle in G. If we delete edge from a you for each you, the remaining graph is still for connected. Yep. And we want to use this, this property to find many distinct Hamiltonian cycles. Yeah. So any questions so far. Okay, so then. Yeah, this is, this is something about path path and cycles. So we actually we use this to help help us to find the distinct. Hamiltonian cycles in the for connected planar triangulations. So here. It's like, yeah, we can just talk about this. This is just. So G is a graph and H is a sub graph of G. So H bridge of G is. So here, we can consider G as the whole graph and we have a sub graph H. So H bridge is a sub graph graph of G in view in deals to buy either edge in like this. So this edge is in is in G by nine H, but the both incident vertices are in the edge. So this is, I'm great. It's a break or each bridge. And also, like, if something outside of G, and we have something like the incident vertex here. Yeah. So this one. The fourth green thing is the is also an H bridge of D. So here, and the fall on each bridge be of these attachments of B on H, other vertices in they be intersecting H. So here's like, for the red bridge, we have like two for this bridge we have two attachments. The intersection has two vertices for the green bridge, we have like three vertices. Yeah, so the attachment, we have three vertices in the attachment. Yeah. So here, this is so the. This is something about the type of sand type cycle. So a pass or cycle in D is called a type of cycle or type cycle if every P bridge of D has at most three attachments on P. So here's like, if D is a panel graph, and we consider a P bridge. Yeah, so P is a pass or cycle. So here, if it is a tough pause mean that all other other break other P bridge has as most three attachments on P. What do we want to see this one because we want to like to show that in the full connected planner triangulation, the touch pass or touch cycle is a Hamiltonian side pass or Hamiltonian cycle in D, because if it is not a Hamiltonian cycle we have something of this pass or cycle is just something non trivial, and it has at most three attachments. So this is just because the graph is full connected. So this is not cracked. Yeah, so then we show. Yeah, we just want to show that this pass or cycle should be Hamiltonian. Yeah, just like this one. And also, this is just. So it's like, if P is a tough pause or cycle in D, such that every P bridge. So here's like we have a new sub graph is like say. If this P bridge contends an edge of this sub word, say, and this bridge we are just has at most two attachments on P, then P is called a say type pass or say type cycle in G. Yeah, so this is just for type pass or type cycle, we can have at most three attachments. But so here we can just understand it like if this bridge contains edge of some some graph say it can have at most two attachments on P, not at most three. Okay, so yeah, so here. Yeah, so here actually we have a theorem about the. We can show that if G is a two connected plan graph and say B is auto cycle and X in they say why in the G set matters X and E in they say, then G has a say type pass P between X and why such that he is in this. So actually here this the theorem, it just. So we can from the actually this the theorem show that for every four connected planner translation, or maybe for every four connected graph, we can find a pause. We can find a Hamiltonian cycle passing through any two edges in this. Yeah, let me show this one. So it's like, if we have like D for connected planner. And we have outer cycle here. Yeah, maybe see, and this is just we find. So if we want to just a fix two edges, E, and another, another one, maybe this is a one. So, okay. And another edge F. So here's like we can just. So for this F. Maybe we can just delete this. Like, yeah, so it's just. Yes, like we can. Oh, no, sorry. Let me just. So yeah, it's like we fixed two edges, E and F. So here. Yeah, so here just want to show there is this Hamiltonian cycle in this graph passing through the edges F and E. So here if I choose E is in E say, and let me see. Oh, okay, so yeah, so for this as we can just find this is X. Yeah, one is X and another is a Y. So we can find a hamlet, a tadpast between X and Y passing through. It's just passing through E. And why such that X and Y. Yeah, it's just we have. Yeah, so here we have the condition that E and F are coefficient. Yeah, so we choose. Yeah, this is the same. So, X is in they say, and why is a distinct the vertex from X. So here we have a part, a tadpast from X to why passing through E. And we know that since she is for connected so this pass should be Hamiltonian pass. So, and if we add this edge. So we know that it should be a Hamiltonian cycle and it passing through E and F, like this. So, yeah, then. So what we want to use a lemma from Jackson is about the past cycle. So what do we want to use it just we want to count that we want to find the Hamiltonian cycles passing through specific edges here. Yes, that my idea is just this one. Yeah, so. Okay, so the proof sketch. I've talked a lot about the proof sketch so here's just I want to tell you like. Yeah, so we apply induction of the number of vertices are first. This one is about the way showed that the number of Hamiltonian site so every for connected planner triangulation and vertices has omega and Hamiltonian cycles through any edges. IE to in a given triangle of G. So, yeah, yeah, this is just, it's kind of yeah just it's same is similar to the proof we did for the quadratic bond. So we just apply induction of the number of vertices. Yeah, and then we assume that no degree for vertices are adjacent. And for that part, and we. And then we assume that we, we can, we can find a good independent set of large size, and we use this good independent set to find a linear bound. Yeah, a linear bound Hamiltonian cycle through this to edge it. Yeah, so same this is. Same for the quadratic bond. Yeah, it's just a similar proof sketch. It's like, yeah, we apply induction as a number assume no two vertices of degree for not adjacent. Either it contains a separating triangle whose interior has at least two but at most and over four vertices, all G has a large independent set with good properties and we use this as to find quadratic bomb for Hamiltonian cycles. Okay, so last let's just give some open problems. Yeah, so now the conjecture the exact bond is still open. So, yeah, also we confirm this one. So confirm this one asymptotially but this exact bond is still open. Okay, and then next one is like maybe we can extend this results to graphs on surface surfaces of higher genius. Maybe we can have some questions on the upper bounds on the number of Hamiltonian cycles in and vertex planar triangulation. Yeah, so because before we just talk, talk about the liberal bound on the number of Hamiltonian cycles in four connected planar triangulation. Yeah, so here. So what about the upper bound. Okay, so, yeah, thank you. That's, that's all. Yeah, sorry that. Yeah. Everybody let's thank Sean on. And are there any questions for speaker. So do we do you have any conjecture, which four connected planar triangulations may have the largest number of Hamiltonian cycles. Sorry, let's let me get to you in a second. I had to mute it for a second. Go ahead. Sorry. So, so do you have any idea, which for now in the regulation may produce the largest number of Hamiltonian cycles. No, no, now we just know the lower bound for five connected planar triangulation is exponential two to the same times. Yeah, but we don't know the upper bound. Yeah, and we don't. Yeah, maybe we can make it bigger than the exponential. I guess maybe we could have. So, but maybe that's not correct. Actually, I guess it could be into the end into a safe house and like that. Yeah, but I, I, I don't find the extreme. I'm asking about the graph, not a number. That was that was Lincoln's question that she was answering. Can you phrase your question again. So, which four connected planar triangulation has the largest number of Hamiltonian cycles. I think that was the same. Was that the same question. I did not hear the other question. But for the other question, the answer was a number. My question is a graph. With or have Let me ask you. Oh, there we go. To the chat. Do you have any idea about the structures about the graphs that have the largest number of Hamiltonian cycles. No, no, no, I have no idea. So we had a recent paper on the minimum and the maximum not minimum and the maximum of the index in say for connected planar triangulations we did it in several graphs. And the double wheel was the minimizer, and we have a conjectured maximizer, and our conjectured maximizer maybe your maximizer as well. But this is kind of hard to count Hamiltonian cycles, even in small examples. Okay. Okay. So can you just help us your construction. No, same paint. The paper. Okay. I will just read it. Okay. So, sorry, that was a little bit. Are there any other questions for shout out. Okay. Thank you. Okay, so it's there are there are other questions immediately. Let's make our speaker one more time. Thank you everyone for coming out and have a good weekend.