 So, welcome everyone. This week we have Stefan Wagner from Uppsala University and he will be talking about subtrees of grass and so please take us away. Thank you very much. It's a pleasure to be giving a talk in your seminar. It's my third time I think I was actually physically in South Carolina twice before doing talks then. This is my first time in the seminar, I assume, so let's see how it goes. Thank you for inviting me anyway. I hope you will find some interesting bits in my talk, which as you said is about subtrees of grass. I will be talking about this revisional properties, mainly probabilistic questions around subtrees in graphs. So what do I mean by subtree? Let's start with the subtree of trees. That's actually how historically the whole story started. The subtree of the trees, simply any non-empty subgraph, that's again the trees, like the bit that's indicated in right here. The six vertices, the fullness part, that's a subtree of the bigger tree. Any non-empty subgraph, including single vertices are counted as subtrees. Well, one thing combinatorialists like to do is to count things, including subtrees. And once you count things, it's also not entirely unnatural to associate a polynomial with your objects, similar to say the independence polynomial, the matching polynomial, and many other graph polynomials that are around. So I want to associate the subtree polynomial with a tree where the coefficient of x to the k is simply the number of subtrees that have k vertices in the tree team. So to give you an example here is a simple tree with five vertices. So what's the associated subtree polynomial here? It's got five subtrees of all the one. That's the five single vertices, of course. And there are four subtrees of all the two, one associated with each inch. Four subtrees also of order three. So it's this one, that one, and two here along the four vertex path. Another four, three of order four, you get each of them by removing one of the leaves. There is one, there is one, and the long path is also one. And the final one is, of course, the whole tree itself. So the leading coefficient of the polynomial, the coefficient of the highest power is always one. Now, what can we say about the distribution of the coefficients in this polynomial? What do we expect there to appear? Now to get some kind of idea, it might be interesting to look at a large random tree, like this one, this has 100 vertices, and see what the associated polynomial looks like. What seems to be the pattern with the coefficients. Now when you do that, it looks like this. And I mean, this looks very suspiciously like a normal distribution, which is only not the unreasonable to conjecture that if you have a large tree, then most of the time you might be close to a normal distribution. We will make this a little bit more rigorous later on. Right. So this was a large random tree, but what about some very special types of trees? I mean, the simplest types that come to mind would certainly be the path and the star. And for this, the coefficients and therefore also the distribution are fairly easy to determine. So if you take a star, then vertices, then the number of subtrees of order K is n minus one choose K minus one provided the case at least one, because all these subtrees are simply obtained by taking the center of the star and then any subset of the leaves. The only exception is the initial coefficient is one, which is n, because it also includes the single vertex subtrees, where you take one of the leaves. But I mean the large end limit, this single coefficient doesn't really matter so much. And we know of course that the binomial coefficients will converge to a nice Gaussian curve in the limit. So we have a normal distribution in the special case of the star. So if you look at the path, the situation is actually quite different. Because every subtree of a path is again a path, and it's uniquely characterized by the two endpoints when you once you fix these two, you know the entire thing. And the longer the subtrees become the fewer possibilities you've got for the endpoints. So it's difficult to convince yourself that the coefficients are simply as k equals n plus one minus k. So the coefficients are in minus one, in minus two, in minus three and so forth. If you were to think of this in terms of probability distributions, then in the limit, you would not get anything that's normal but rather triangular distribution. So there is at least one example of a sequence of trees where you're not getting that normal distribution. And if you play around with just paths and stars actually, then you can generate a whole lot of interesting distributions. That's by mixing and matching. One such way is to take a broom where you take a long path, length k, and then at one end you attach L leaves. It depends very much on how k and L relate to each other as they grow big. What happens to the distribution of the coefficients? So when L is fairly large, so if you have feeling many leaves, specifically if k squared divided by L still goes to zero, then the leaves will dominate everything. So the long path doesn't play much of a role anymore and the behavior of the coefficients is like the behavior for the star basically. So if you suitably normalize, you do the usual things, attract the mean, divide by the standard deviation, you would actually converge to normal distribution. So here's sort of the critical phase where the paths in the star are in a way equally powerful, in which case the limit distribution you will get is a convolution of a uniform distribution and a Gaussian distribution. And where does that come from? So almost all the trees, sub-trees, will contain the center here. And then they are composed of some subset of leaves and some sub-path here that can have any length from one up to k. So this bit, the path then basically contributes a uniform distribution because one, two, three up to k are all equally likely, whereas here you have another binomial distribution as before. So then it's exactly in the case where k squared and L are of the same order of magnitude. They're sort of equally powerful in the sense that they have the same variance order of magnitude wise. So you get the convolution of uniform and a Gaussian. And finally, if k is even larger, so if the path gets very long, then it starts to dominate and you just have convergence to uniform distribution where basically the choice comes from the length of the sub-path here. And that dominates the distribution. So you can get all sorts of things. And if you play around even more, you can even get weird situations where the distribution of the coefficients taken to the limit becomes discontinuous. So if you carefully balance two stars attached to a long path, like I did here specifically, you make the path length two to the n and then three n leaves on one end, n leaves on the other, then it turns out that this is just trying to obtain a distribution which has a continuous part and a point measure. And it's a mixture of the two. So that comes from there being two types of trees, essentially one containing the second center and one not containing it. In order of magnitude, it's sort of equinumerous. So equally many trees of the one type is of the other and the one type produces a point measure and the other produces a uniform part. So to summarize, in the random case, we seem to be getting something that's normally distributed. But if you take very specific sequences of trees, then the distribution of the coefficients is not necessarily always close to normal. So you can get funny things as limiting distribution. All right, but what is typical. So what do you know, what do you expect for most cases and that's what I'm trying to answer a few years ago. With nine of Amazon. So specifically we're looking at conditions for a normal distribution to occur. It turns out that the conditions you need are the following. So you want many leaves. That's one thing you should have at least linearly many leaves. It's constant C times the order and then someone. Yeah, more unusual condition on non-branching paths. So non-branching path is simply a path in your tree where each of the internal vertices has degree two, so it doesn't branch out. And you don't want these to get too long. Basically, you don't want any long paths in your tree that do not branch out at all. Specifically, the condition is that these should not get longer than a little bit below the square root of the size. And if these two are satisfied, then it turns out that the distribution of the subtree sizes does in fact always converge to Gaussian distribution, normal distribution. So suitably normalized here means that as you do in probability theory with your central limit theorem, you subtract the mean and then you divide by the standard deviation. So as to get a normalized random variable that has mean zero and then variance one. The sequence of distributions you get in that way will converge to a Gaussian. Right, so both of these, in a way, try to avoid the tree gets too path like. So path is of course the most possible number of leaves you can think of, and it has also the longest non-branching path. So, so both of these are somehow about not being too tree. It's important here to notice that the sequence of trees that I'm considering here can be a deterministic sequence of trees, even though I'm stating something about probability distributions with each of the trees and associating a probability distribution, but the trees themselves might form a deterministic sequence like the sequence of stars. Right, so evens can in principle be taken as deterministic here. It's not necessarily a sequence of random trees here. The conditions might look odd, but they're, as it turns out, essentially based possible. So we can't drop them. We can't drop the first one. We can't drop the second one. In each case, there are counter examples of sequences of trees we don't get normal distributions. It's not even possible to remove the epsilon. So even that bit is sharp. Fewer than linearly many, that's not, few than linearly many leaves. It's not good enough if you have non-branching paths of length about square root the size of the tree. That's also not good enough. So you can't even remove the epsilon. It might be possible, admittedly, to remove the epsilon and replace it by some log factors. So we didn't try to optimize it completely there. So there might be some room for improvement, but you can't do without it. The conditions are actually not so bad. So if you take a random tree, then the probability that it satisfies them will actually tend to zero quite rapidly as the size goes to infinity. So if you fix a C, you have to fix a sufficiently small C, of course. Because if you were to say to take C greater than one, you can possibly have more than C times the size of the tree many leaves, but take a small value of C and small value of epsilon. Then once you fix them, the probability that a random tree will satisfy both conditions for 10 to one as the size goes to infinity. So conditions are not too restrictive. In particular, the conditions do hold if there are no vertices of degree two. So trees without vertices of degree two have different names. They're sometimes called homomorphically reducible or series reduced. There are different names for that. Basically, that's trees that branch out of every vertex. And for such trees, there is actually a conjecture due to Jameson who started the investigation of sub trees of trees back in the 80s. The coefficients of the subtree polynomial will always be unimeraled, except at the very beginning. Unimeraled means that the increase first and then the decrease from some point onwards. Without having several peaks, the exception being at the beginning, because the first coefficient is always the number of vertices. The next one is always the number of inches. So that's always one less and then from that point on it goes up. The coefficients are connected to the unimodal for trees without vertices of degree two. We cannot confirm the conjecture, of course, without results. Because sort of asymptotically being normally distributed doesn't mean that the coefficients are necessarily in the model because it's an asymptotic statement. But there's some evidence in favor of this conjecture. Right. Now, we move a step further and look at arbitrary graphs now. So what can we say about arbitrary graphs that are no longer trees, you can still define sub trees and we can still define a subtree polynomials all these things makes sense. For example, the tree that the graph G that I have here is, of course, four one vertex sub trees, five two vertex sub trees associated with the edges. And eight three vertex sub trees. That's a little bit. Harder maybe to count but if you do it carefully you will find eight. And there are also eight four vertex sub trees that's the exactly the spanning trees of this graph. So the subtree polynomial associated with this graph with seven four x plus five experiments eight x cube plus eight x to the four. So all the definitions still made perfect sense. But so the study of the subtree polynomial is fairly recent compared to subtree polynomial for trees. While the subtree polynomial of tree was introduced way back in the eighties to the base of my knowledge the first occurrence of the subtree polynomial for graphs in the literature is only a few years ago. Let's look at the leading coefficient begin. So, unlike the situation for trees. These polynomials no longer have leading coefficient one. So, there might be more than one spanning tree. In this case there are eight. And if we pick one of the subtrees at random in total there are 25 year, then it has a probability of eight over 25 to be spanning eight of the 25 are spanning trees. So, we denote this probability by P of Jesus eight over 25 in this case the probability that a randomly chosen subtree is spanning. For trees. This wasn't such an interesting value because, well there was only one spanning subtree in each case. So this was simply one divided by the number of sub trees, which becomes very very small so you have a low probability that a randomly chosen subtree would be spanning. But as this example suggested it could be that there is an actual decent probability for randomly chosen subtree to the spanning. To get an idea how large that probability might be, we can look at the densest graph possible complete graph K and because they are counting sub trees is actually not so difficult we know that K and has entity and minus two spanning trees. And more generally for any fixed K, the number of sub trees with care vertices is simply this product because I have introduced K ways to choose the vertices and then K to the K minus two ways to pick the edges to form this subtree on these vertices. So, I can now work on the probability for random subtree of can to be spending. It's this questions, number of spanning trees divided by total number of sub trees. Slide change of variable in the sum, turns it into this. And now I look at the quotient of the health term here and numerator, some basic calculus reveals that this limit is actually one over ETL times health factorial. And as a consequence of that, you find that the probability that a randomly chosen subtree of the graph is spanning tends to one over the series which is e to the one minus one every. So for large aim, you have a probability that goes to a constant. So it's fairly large, perhaps surprisingly large. And we know even a little bit more because what we worked out here this one every to the L over L factorial. This is the ratio of sub trees that miss out somehow on L of the vertices, compared to the number of spanning trees. This tells us is that probability that a random subtree has in minus all vertices also converges. It converges to a to the minus one every the number we've seen before divided by ETL L factorial. So this is a personal distribution. If we make a random subtree, then the number of vertices it's missing that will converge to a personal distribution. So the subtree sizes are asymptotically also distributed to be more precisely in minus the subtree sizes are distributed like that. So one might wonder if this is not typical. So what we've seen before for the star that we get a normal distribution seem to be typical there. So we might not be tempted to conjecture that this kind of behavior is typical for dense tree graphs from instead of trees. So this dichotomy there, the dense graphs and the sparse graphs. The sparse, the denses is of course a complete graph and then we have seen that many of the sub trees are spanning were at least close to spanning, most of them are close to spanning. On the other hand, in the large tree. So the sparse is kind of connected graphs where where this whole business makes sense. All the sub trees are not spanning or not even close to span. So one could come up with a conjecture like this. If you have a sequence of graphs, then the probability that a randomly chosen subtree is spanning will go to zero. If and only if the density of the each density goes to zero. So it's number of edges divided by number of pairs of vertices, a quantity between zero and one. Maybe have a dense graph you might expect that probability for random substituting spanning goes to one or not necessarily to one but to some some positive constant perhaps if you have very sparse graph then you expect this probability to be small and to go to zero or sparse sequence of graphs. As you can see from this question mark here next to word conjecture. This is not actually anything that anyone ever conjectured. So very quickly that this fails. And in the in the first paper that studied this quantity and all duties for authors chain Gordon McPhee Vincent 2018. Everybody observed that the relation isn't quite that simple. If you put example take a complete graph, and just attach a relatively small path somewhere of length say routine. So this is negligible in the large scheme of things, the small part of the, of the vertices belong to this path, of course, so the density is still very very high. The probability that a random subtree spanning will actually go to zero because any subtree that is spending will have to contain that entire path. Which is rather unlikely because they're for each subtree that contains the entire path. There is also substituting contains the entire path minus one vertex also one that contains the entire path minus two vertices and so on so they're about routine sub trees associated with each spanning tree that are not spanning. So you have that probability to go to zero, even though the density is not only bounded away from zero but even goes to one. So it's, you come up with counter examples like this one fairly quickly. So the problem you realize is certainly that globally the trees, the graph is dense but there is, there is a small bit that is just very sparse and might want to avoid that so. What they did to rectify this is they made the additional assumption that your graph should be a transitive. So you want the dense graph order of magnitude in squared many ages, and you also want each of the gross to be a transitive. So as to avoid the situation where some parts are sparse and some are things. And then they conjecture that the limited theory of the probability is some is positive so you have a reason be large probability to have a spanning tree when taking a subtree of random. And then the counterpart to that would be a conjecture on sparse graphs or what they conjected here was that if you have into some smaller power many ages, the probability for a random subtree to spanning should go to zero. We can take to these two things which are. Well, in a way, the natural things perhaps to conjecture. Once you've started, once you've studied a few examples. Indeed, the first conjecture is true. So, one can if I prove something a little bit stronger. However, in the in the same in the same spirit, the flavors very much the same. Instead of assuming each transitivity which is fairly strong. I make the condition that the minimum degrees is large. So I don't want any parts where my graph is is sparse every every version should have a reasonably large degree. And this implies the conjecture on each transitive graphs because each transitive graphs are rather special they are either regular work by the type by regular. So, they would satisfy this if they are dense. Now, if we're assuming a minimum degree condition like this, then it turns out that we can explicitly bound the probability P of G science. It's great and equal to e to the minus one of alpha where alpha is the constant here and some error term. So for large in the bountiful is basically be to the minus one of rock. So, how does one prove this. How do we get such a low amount. The key idea is a double counting argument, where start with a graph minimum degree bound below. We count the number of pairs consisting of a spanning tree like this here, and some subtree of that spanning tree that is missing k vertices. Right sir. One possible peer industry consists of that ribs spanning tree and the blue subtree of that spanning tree. Maybe one of the pairs in this example counted by P one because the subtree s is missing exactly one of the vertices. Now we count this number in two different ways we can start counting with a so we can start counting with tea. So if we start counting with the spanning tree, then, well, we have SN of G number of spanning trees of G many choices. And for each of them, we have certainly at most in choose k choices for s, because that's the number of ways to choose any cave vertices to remove. Some of these choices might not give me a subtree, because if I remove something in the middle of the tree might fall apart. There are no more than inches came any choices for the spanning tree for the subtree s. On side chosen tea. So I can find this number of pairs by inches k times the number of spanning trees. And then that's my end to the care of a k factorial is the usual up above and on the provisions. Then we get a corresponding lower bound by starting with s. So, this is a little bit trickier, not all that much. For each of the cave vertices that are not in this. You notice that there are at least delta minus k neighbors needs. Right because k is the number of vertices that are missing is. And there are at least delta neighbors delta being the minimum degree. So at least delta minus k of the neighbors are in fact needs. So, what I can do now, given s is to, sorry, extended to a spanning tree by just attaching each of the came missing vertices, one of the vertices of s. And I can do this in at least delta minus came any ways because they're at least that many inches going directly to us. So I have at least delta minus k to the power came any ways to extend as is to a spanning tree. And so I have a simple lower bound as well for every k up to delta. So I have a nice chain of inequalities. Using also the delta is bounded from above delta is greater than or equal to alpha in alpha minus k to decay. So it's less than or equal to into decay okay if it really is in. So for small k, the minus k here doesn't actually matter so much. So I get that is in minus k, the number of sub trees within minus k vertices is at most. This factor times the number of spanning trees plus some irritant. It turns out the irritant doesn't matter so much large values of chaos negligible. So if I sum this overall k, then I get the total number of sub trees on the left, and the sum of these guys. Overall k times a sub d the sum is here simply to the one of alpha plus an irritant, which I'm not going to talk much about. So that's precisely the desired inequality of the total number of sub trees here, and I have bounded it from above by some constant times the number of spanning trees. So this is the key idea of the proof this double counting argument where I count pairs of spanning tree and the subtree contained within it into different ways. And what about sparse graphs. This was now for dense cross with a minimum degree was bounded below. So sparse graphs it turns out that even the revised conjecture is not quite correct. And you can still construct sequences of sparse graphs for which this probability has a positive limit. One way to do that is to take a long path and into complete graphs at the end. So it doesn't need to be huge can take into some small power there. Then it turns out that actually the graph is very, very sparse. It has only linearly many ages. However, if you do the calculations you find that most of the sub trees will have to contain the entire long path and vertices on both ends. At least heuristically probability that a subtree is spanning is probability that subtree is spanning in the one complete graph times probably spanning the other complete graph, which explains this into the minus two. It's just the probability for a complete graph squared. So if you take. So highly structured graphs where some parts are very dense and some parts are very sparse you can still construct nasty counter copies. What's the next best thing, if you can't even fix the conjecture like this. So well, you might want to try random graphs instead, because random graphs will never look like this. There are some very uniform, all the parts looking more or less the same. It turns out then that in this case both the conjectures do hold true in probability. If you consider the classical average any random graph, GNP where the probability P that an edge is included is allowed to depend on the vertices. If you consider such random graphs, then it turns out that as in ghost infinity. If P goes to a positive limit, you have convergence of P to a positive limit as well. And one can even calculate the limit. It is a deterministic limit no longer random so you have convergence in probability to a constant. Whereas if we have sparse graph so if P goes to zero within, then the probability P of G goes to zero. Maybe not surprising also looking at the first part, if here you let P infinity go to zero then that's expression will go to zero as well. So what goes into this proof main ingredients are similar, it's the same double counting ideas before mixed with some typical random graph arguments. So we know that random graphs are almost regular. In some sense, most of the vertex degrees are very close to PN. Then we know quite a bit about trees. Most trees have about N over E leaves which is useful. Because the number of sub trees that a spanning tree has can be bounded in terms of the number of leaves, namely like this. So the one inequalities is easy. The lower one here, you can take any subset of leaves and remove them and you get a subtree. The other bound is slightly more complicated, but also not too difficult to prove. And you see here that if K is comparatively small compared to L, then they're sort of the same asymptotically they're both about L to the K of a K factorial. So, you can estimate the number of sub trees on the spanning tree quite well in terms of the number of leaves. And in addition to that, you also know that most of the trees will have about this many leaves. Yeah, you can combine it quite nicely in the double counting argument. A couple of corollaries follow from this. So now knowing something about the distribution. You can also determine a bit more. We have a small behavior that we have observed for a complete graph. So for every fixed K, the number of sub trees that are missing K of the vertices divided by the number of spanning sub trees. So to well this constant here in probability, so that's exactly the probability associated with a constant distribution of parameter EP infinity up to the scaling factory to the minus one P infinity. One can determine for instance the average number of vertices in the random subtree. It's not to be very close to him. Not surprising based on what we know now we know that most of the trees are spanning or at least close to spanning. So, the average number of vertices in a random subtree will be close to in and the difference will converge to constant as well. All of that also. We can use what we've shown also to say something about the total number of sub trees, because we can now relate the total number of sub trees to the total number of spanning trees. And Johnson proved something about the distribution of the number of spanning trees in random drops. I think it was years ago I think it was. So you share it that logarithm of the number of spanning trees is asymptotically normal. And now we know that there is not such a big difference between number of spanning trees and number of sub trees. Their question is about constant so we can prove this statement as a corollary. Let me finally say something about the subtree polynomial and its roots. Sir Brown and more studied the roots of the subtree polynomial of the tree. This is also quite classical the roots of many different graph polynomials have been studied in the literature and there are a lot is nearing about them. So for the roots of the subtree polynomial specifically. We found that they form roughly a ring and annulus. This year, this figure shows all the possible roots of subtree polynomials of trees up to over the 14. And you see one dot here at zero, which is always there but the rest is in this annulus shape. In fact, they proved that all the roots of subtree polynomials are necessarily confined to some disk. So the disk is not like this. It's a little bit bigger than what we see here but it's not too much bigger. So it's injected further than generally the roots will line in a certain annulus. That looks like it's not in the circle out here and the other circle out there, both center that negative off. It couldn't quite prove the annulus bit but they did at least obtain that there is some disk to which the roots of the subtree polynomial are confined. This is in contrast to other graph polynomials. So for example, if you take the independence polynomial of graphs, the roots all over the place they like the instant plane. For matching polynomial, it's getting entirely different. They are the roots are real, but they can get arbitrarily large. So it's somewhat unusual behavior perhaps. Now what about the subject polynomial of dense graphs. So what we have just shown and could in fact be used also to get something out about the roots of the subject polynomial. Because we cannot asymptotically evaluate the subtree polynomial less as in goes large for random graphs. So for each fixed x, the value of the subtree polynomial x divided by the leading term will converge to some constant depending on x or every complex x other than zero. And as long as you keep it fixed. And the idea is simply to use this personal approximation that we have observed so the is in minus cake efficient can be well approximated by is in times this personal term, and then you just evaluate the series. This is slightly heuristic of course, because one has to take care of error terms for this is in a nutshell what one can do. Right. So we know something about the behavior of the values of the subtree polynomial in large random graphs that are also dense. And if we look at this further we see that well this here is so just a pre factor. And in this series never zero. We expect that this can be made rigorous with some uniformity arguments that the roots of the subtree polynomial of large teams cross will cluster around zero. So because this is basically the only bit from where we can get really zeros. So if you have a large random growth, GMP, then for some function are in the first year, the maximum modulus of all the zeros is bounded power of a message almost truly. I'm not exactly sure what the optimal choice of our vain is. I'm working on this kind of questions with PhD student. So a goal here is also to do randomize this. And it would seem that when can indeed replace the random graph it by a minimum degree condition as we have previously so if we just impose something on the minimum degrees. Then one can also bound the roots, showing that the roots of the subtree polynomial for Danes graphs are rather different from the roots of the subtree polynomial of trees, more generally sparse groups. So some directions for current and future work. One might want to get analogous results for other similar graph polynomials. Instead of sub trees one can of course consider similar quantities like the number of connected sub graphs this has been studied fairly recently and some other variants. What can we say about other random graph models. So, as I'm trying to pass on distribution roots of the subtree polynomial this is the thing I mentioned. Can we do this for deterministic things cross as well so I showed it for random cost for deterministic things cross this is still some work in progress. I'll finish with an open question that is not so much related to what I showed you now. I showed you some results about this quantity P of G the probability that a random subtree spanning. To the best of my knowledge, it is not even learn what the maximum of that probability is given the number of vertices. It seems reasonable to conjecture that the maximum is obtained for the complete graph. And as far as I know, nobody can prove that this point in time so that's my final challenge to the audience. Thank you very much for your attention. Thanks again for having me, and I'll welcome any questions you might have. Thank you. Let's all thank our speaker in some way. And are there any questions. Yes, I got a question for you. This result about the random graphs, having my probability of being spanning. I wonder is it. Is it enough to know that all the induced sub graphs have similar density. Is it, presumably not the randomness but somehow that, that all parts of the graph look the same, right that you can get it. It's likely that one can weaken the conditions even further. So basically you need to somehow make this double counting argument work. It's, it's not so crucial that it works for absolutely every single vertex you can probably allow yourself some leeway and say, occasionally this detail that I don't know what exactly an optimal condition would look like. Probably, as you say, you can replace it by something like the density of a reduced sub graph is bounded away. Yeah, I have a question that can counting subtree somehow be related to counting sub forest because sub forest seems to form like a metroid or a simplicial complex and I don't know, maybe there is more that each other, or it's a completely unrelatable issue. I certainly not completely unrelatable I see no immediate way to let's say, if you can count the number if you know the number of sub trees all sizes you would you would know anything about the number of sub forests of all sizes. But certainly the, the questions are very much related and instead of sub forests there's also a variant of that, which is a rooted sub forest so where each of the components of the sub forests, it's a root. The reason why these are interesting is that they occur as coefficients in the directories polynomial of the Laplacian sir. There are several other counting questions that are obviously similar flavor even though I'm not sure if one can sort of easily translate questions from one to the other. That's for our speaker. Right. If not, then thanks again for for a wonderful talk. Yeah, I think, I think we'll go ahead and wrap it up there and I'm glad to have seen you seen you I've read some of your work in my head and saw my some of my research so Thanks again and have a great weekend everyone. I hope to see many of you in person again at some point. Thank you, Stefan. Thanks. Thanks again for having me and have a good weekend.