 Okay. Well, thanks everybody for making it out. Today we've got Alex Rice and Osky from Iowa State University, and he will be talking to us about the edit distance function of random graphs. Go ahead and take us away Alex. Thanks for coming. Yeah, thank you for the invitation to your virtual seminar. Today I'll be talking about the edit distance function and related to random graphs. This is ongoing and kind of dissertation work. So the motivating question we have is, given a fixed graph F, how hard is it to remove all induced copies of F from a graph of density p. The cases that we'll take at first are F being a complete graph on three vertices also a cycle graph on three vertices. And the next graph we'll consider is K22 also known as C4. And the main problem is to consider what happens when F is a random graph. And I also claim that this problem is is connected to other sort of famous extremal questions one one being the Turan, the Turan number of a graph will we'll see we'll see that come up in this case. So what do we mean by hard work. And for this problem we're going to measure it in terms of what's known as the edit distance. So given two graphs on the same vertex set. We're going to record their edit distance as the cardinality of the symmetric difference of their edge set normalized by the number of possible edges. In this case we can kind of intuitively think of an edit as a unit of work. So how much work does it take to transform this graph to this graph. If the rule is that you have to add and delete edges with weight one over entries to on on each of those operations. So let's see what edges they don't have in common. So here are the edges in the first graph that do not lie in the second graph so we have to add those we have to delete those. And here are the edges that are in the second graph that are not in the first graph so blue edges we have to add red edges we have to delete. And if my arithmetic right we've deleted for added six. So that should be 10. And the right number to normalize by is 36. So their edit distance between these two graphs, at least as far as label graphs goes is point to seven repeating. Are there any questions up to this point. So let's let's make our first attempt. How hard is it to remove a k three from a graph. So one natural thing you could do to remove k threes from a graph is to make the graph by part that would be you would delete some number of edges in order to make a new graph we could call it G prime, where G prime is part eight. If you do this efficiently. Then in general the best you could do is by deleting about half of the edges. So if if your graph had density. P, then you have to have deleted something like P over two proportion of the edges in the worst case. That worst case would be attained by several things a complete graph with a bunch of isolated vertices that could also be deleted by a random graph of density P. And we could ask, like, in what sense is this type. If, for example, P equals one. What are we asking well we're taking some graph, we're taking a complete graph. Oops. And we want to know how many edges we need to delete from it to make it to make it triangle free. And this is in some sense the same as saying, how far is the graph from being triangle free. So you could think of the graph. Okay and my, okay and minus the edges in the deleted graph is triangle free. And we could call this graph maybe G prime. And how many edges could G prime and possibly have well this is this is Tehran's theorem that the number of edges in G prime is at most floor and squared over four. And so at the very least for people's one this should be tight and there are arguments to make to ensure that this is tight for other values of P. So we're going to graph what we've shown so far. So on the y axis will say distance. And on the x axis will say density. And what we've sort of argued here is that our, our function should take on the value, this being one, that a complete graph takes about half of its edges to be deleted. So this should be one over two, one, one over two. And at any other value of P. We've argued that the best you could do is P over two. Okay, so that that's sort of a functional way of displaying what information we've communicated so far. Are there questions about this. Feel free to turn your mic on and interrupt me there it's there's always a chance I miss chats in the comment. I really don't mind interruptions for these kinds of talks it's it's very easy to go fast when you don't want to. Okay, yes and I've already I've already explained this point Tehran theorem says that we're tight at that value. And we've already graphed what what this looks like for different values of P. Let's try a harder example, because this one, this one you could do in a first course in graph theory. Let's try something a little harder. So let's suppose that G is a graph of density P, and ask how hard is it to remove cycles of length four from C from G. Every time I say remove a graph. This necessarily has to be induced. So I'm always removing an induced graph. So if we translate immediately into the following. What if we want, we're, we can instead ask, what is the distance of your given graph G to the set of all graphs without induced C for that's just a direct translation. And this maximum being taken overall and vertex graphs of density P in both cases. And we can tighten this notation a little more describing the graphs that have no induced C for us just for the C for some of you in the room may notice that this is an example of a hereditary graph property. And we'll talk more about that on future science. So how can we remove all C fours efficiently but also indirectly. And then, okay, this is where I will pause and ask questions. So does anyone have a really basic way to remove C for and to do this we're using additions and deletions of edges. I think we have a mic on for something else. Leave all the edges, you can add all the edges. Okay, delete all the edges at all the edges. This is this is a softball. So the two constructions we're going to consider first is remove all the edges from G and add all the non edges. And the graph has density P. We know already about how many edits we've made. In the first case we've taken out all the edges. So the distance to this constructed graph is P, because we're always normalizing by entries to. And in the other case, our constructed graph is the complete graph and since we're adding all the missing edges. Well we're just adding in one minus P proportion of the edges in the complete graph. And we want to know how optimal this is. One thing I will say, before we move on is that if you use these two constructions, again, distance being on the y axis density on the x axis, then our distance to see for that we've measured is at most this red line. And most this blue line, where at the middle we have one house. And so when we ask is this efficient. What we're essentially asking is, is there another way is there another algorithm we could have used to do this in general where the curve is below the curve that we have at any point. Okay, that's what I've already said that the distance is at most this minimum of these two functions. And it turns out yes, there are there are ways to do this in a more efficient manner. So consider the following sort of structure on the leftover here we have something that looks like a complete graph, something that looks like an independent set. And in between I put gray gray here means it could be anything. The claim is that if your new graph G prime can be split in a fashion that is an independent or a clique on one side and independent set on the other with anything in between. Then I claim that that new graph has new C for. So why would this be true. Well, this is true sense. Where would you put a C for if you could embed it in here. So C for has four vertices. It's clique numbers to its independence number is to. So you would have to put two of its, one of its cliques here, and one of its independent sets here. But that's impossible just by just by inspection. This is a clique. And this is an independent set of cliques of size to an independent sets of size to intersect. And this is non trivially in this case so this kind of embedding is impossible. So if the graph that you produce G prime looks like this in some sense, then it naturally has no C for. So our new editing procedure is just going to mimic is just going to make a graph G prime that looks as much like our original graph is possible. The new graph has this decomposition. What we'll do is we'll randomly map about X one proportion of the vertices to the black region. Map the remaining vertices to the white ring region. And we'll do this ID for every single vertex. We're going to make no distinction about the structure of our graph G. The reason why we're we're almost entirely forgetting the structure of G and just remembering its density. Because when we do this. Okay, again, I should say that the vertices that we map into the black set. We're going to complete that graph that induce sub graph. And on the vertices that map into the white set. We will empty that graph and then that means that the new graph we created does in fact resemble the structure. So now we have to ask, how many edits did we make. Well, the only edits we made involved edges where the two vertices fell into this pile and edges or none. Edges that fell into this pile and non edges that fell into this pile. So those would be the two edits that we made. And that would just be represented by this kind of quadratic form that the expected number of edits that we made under this random mapping is about one minus P. X one squared plus P X two squared. And it's pretty easy to see how to minimize this. You just set X one and X two to be one minus P and P. And that's what we did in expectation something like P times one minus P. And if you compare the function. If you compare the two functions that we have. Yeah. So overall the distance that we have here from G to the form of C4 is at most this function. So we went from something like, I think I did read first, and we did much better. This has a peak of one half with height, a quarter. So again with distance on the Y axis density on the X axis. This construction is a sizable improvement, especially when he is equal to a half. Sometimes the structure can be quite complicated. So this you could kind of look at it and say it looks like a ball of rubber bands. But it's actually a template that's made from a strongly regular graph. It has the name q44 in some in some contexts. And this was actually entirely necessary for understanding distance to form of K to four. Complete bipartite graphs have a lot are end up being quite complicated for this problem. So let's take what we have so far and move it into a little more general setting and try to make things a little more formal as well. So a hereditary property of graphs is a family closed under relabeling vertices as well as deleting vertices. So far we've only considered hereditary properties generated by forbidding some small graph. And let's think about why that's actually a hereditary property. So if you take a graph that forbids some induced F, and you delete a vertex will clearly it still forbids F. And likewise, if you relabel the graph that forbids F, then you haven't done anything to the structure of the graph and so it should also forbid F. And in fact, any hereditary property can be written as forb of a family of graphs, which is to say, any graph that forbids all elements of the set F. Why do we study hereditary properties. Some would say that most interesting graph properties are hereditary. And that would be a bit bold for me a grad student say. So I will quote someone else. This is from a ball of Washington wine rights paper on basically quantifying a little bit more of Sherman and Zito's argument about the speed of hereditary properties. They go into a little more structural detail in this paper. And let's also do proof by authority. Olga alone and his student Yuri stuff also said that most interesting graph properties are hereditary. So we have, we have, we have authority figures who are making this argument. And maybe we should also try to believe it in our own right. You've seen lots of hereditary properties. So in an intro graph theory course you would probably see all of these properties being bipartite is hereditary interval graphs planar graphs are all hereditary. Not all of them are interesting for this problem planar graphs are sparse. So you don't really have to think about removing how to edit in order to make a graph planar you just delete all of the edges except a few. There's also money in hereditary properties, the focus in prize plus a $10,000 bounty was awarded to these four authors for their 178 page proof, which characterizes a perfect graphs. And there's a whole program of studying similar kinds of characterizations of graphs such as graphs that embed on certain surfaces. Does anyone have any questions up to this point I've been kind of speeding through. I'd like to think of this problem involving metric spaces. I suppose these lines could be a little bolder here. So if we take all the graphs on four vertices. So the same. If we take all graphs on four vertices. And we circle those up to isomorphism. In this case it's just circling them up to edge count. We get a smaller metric space. And the edit distance between different graphs is really just the number of edits needed the number of edges needed to take to get between the graphs. If you reduce it up to up to isomorphism. More interesting picture would be all the graphs on four vertices already reduced up to isomorphism. And we have to remember what is our problem. So we're considering the distance from a hereditary property. In this case we're we've circled all of the bipartite graphs. And we want to know what is the graph that is furthest from this of given densities. If we just wanted to know the one that's furthest we would look no further than over here on the left a complete graph is this from being bipartite, at least of density one, maybe this is the graph that's furthest of density, five, six and so on. Yeah, so that would be our optimum. And another picture that I have quite fondness for, because I like graph limits is the following picture. So in blue, I'm going to draw this as. HN. By that I mean this is the each of these is a each of these pancakes is sort of a slice of all the graphs of the given order. The whole picture is by the way GN if you wanted to give a notation for the graphs of order and in this sense what we're doing is, we want to know the graph, or the graph sort of in the middle of this, the graphs of density P that have the furthest distance to this outside disk. And for sort of convexity reasons that make more sense with graph limits this is this is sort of the right picture for this. So what is our original problem. The remaining question, originally asked by Alan and stuff was the following. Let H be hereditary. And I should mention every time I say hereditary I mean hereditary and non trivial in two senses. The one sense being that the hereditary property actually is infinite that it contains graphs of every order. And the other trivial case would be that it contains the set of all graphs or all but finitely many graphs. What they showed is what they asked us as follows. What if you want to know just the graph that is furthest from that hereditary property. It's actually a small lemma to actually prove that this limit is well defined that it's not just a limb soup. And for each and we are asking among the graphs of order and what is the furthest. This is a more restricted version of the problem we were originally asking, but it's in fact the motivating problem. What they showed is that for each hereditary property. There's a parameter P, governing the randomness of a random graph. This is the air to shiny random graph of density P star. And it really attains this maximum distance EDH star. Another thing that's worth mentioning is that anytime one of these results is shown it usually involves a couple applications of the regularity lemma. But there are also fairly slick proofs of these results using graph limits. And some of those are due to Bologna, Lovasa and Segati. The next sort of step that gets us back to our original general question is if you have a hereditary property, pick any value of P between zero and one and define EDH P as the limit of the maximum distance from the hereditary property that you could get if you're looking only among graphs that have density P. So let's put some sort of epsilon error term in here and it would give you the same quantity. We just chose to round to floor of P and choose to, but you could you could round within a small window of that to and still be considered density P. So what Bologna and Martin showed is that for all values of P. amount, this maximum is in fact detained by a random graph of density P, which is a bit strange because we're called that we're going to be studying sort of a stochastic version of the question, which was the distance from four of a random graph. And these theorems are all saying that the furthest graph from forbidding a random graph is another random graph, which is a little annoying to have to come to terms with philosophically. The other main tool that Bologna and Martin found was that this function, if you take as your argument P on the set 01 to 01 is continuous and concave down. So why is that useful for Alan Stovs question. Well, what it means is that if you were to graph and edit distance function. That's maybe not starting and ending at one, but concave down functions. Graphically speaking have a fairly easy to identify maximum. There are cases where it's flat. So it could have multiple maxima. But that's the whole benefit of noting this convexity or this concavity. So our question is, what are the maximum edit distance, the value P star and the edit distance function in general, if H is defined as forbidding a random graph. Okay, so this is a good moment to pause and and ask for any questions. Because there are a lot of things going on and we're going to go in a couple different directions. So the big question is how do we actually compute the edit distance function, presumably once we get the edit distance we can compute the maximum value as well as the place where that maximum is attained just by looking at the graph. So what we're going to define here is a notion called a colored regularity graph. We've already encountered one in fact we we took a template like this. In the CRG language, this is a black vertex. This is a white vertex and this is a gray edge. CRGs are a more diverse generalization of this. All vertices in a CRG are white or black. And there are no non edges, because we're coloring a complete graph or decorating a complete graph, and the edges of that complete graph are decorated as either white black or great. In this decoration, anytime we edit a graph to make it resemble one of these CRGs, what we're doing is completing everything that's black, emptying everything that's white and doing nothing to things that are great. And there's a good reason why we don't use gray vertices, more or less because of counting lemon if you wanted that. With these CRGs, you're pairing it with an optimization problem, much like the one we had before. We measure the distance from a generic graph G to a CRG like this along the path that we took when we completed and emptied accordingly. In this case, we recall that we had the formula one minus px one squared plus px two squared. We also talked about how we maximize that we're generally, well, you can unpack this expression as in terms of the matrix governing this quadratic form. And it's important to stress that we're optimizing over the set of x, where x is a non negative vector something to one, which you could also say as a vector in the standard simplex with two coordinates. Let's see what would happen here. So if if in fact this edge was black, you would have a one minus p here. If this edge were white, you would have a p here and here as well, because it's naturally a symmetric matrix. So for every CRG you have some sort of matrix here in later slides I might call this mk of p. So let's go to this example. And what I'm going to emphasize on this slide is that not all vertices of a CRG are actually useful. So by that I mean, it might be easier to edit to just a black vertex or just a white vertex that the area in between was too annoying to deal with, and that you might as well just complete your graph or empty your graph. It's a fairly depressing case, but it does happen. So in this case, the matrix corresponding to the CRG at the value p does have a one minus p here, one minus p here because of this black vertex and a p here because of the white vertex. And again these numbers all correspond to the amount of work that you did, deleting edges and adding edges. So here the quadratic form has a different kind of optimization. In this case, it's possible the optimum in in any case is obtained by setting one of the variables equal to zero, which that tells you that this is not a good structure that it was actually these two structures doing all of the work. So we need a notation, we need a good notion. We need a good definition to capture this. It seems to be a couple slides ahead. So we call a CRG p core. If the minimum value attained by any of these different weightings this x is attained by some vector where all of the weights are strictly positive. And another another way to say this is that it's the value you get this kind of quantity that's governing the distance. You should say here, the quantity that's covering the distance from your generic graph G to your new graph G prime edited according to K actually uses all of the vertices of K, which is to say you shouldn't have used any kind of sub CRG of K that you strictly speaking got more had to do less work if you included an extra vertex. So why are we talking about all this technical stuff involving CRGs. We need to remember that this problem is related to hereditary properties. So, in this, in these two examples, recall that C four doesn't embed into this CRG, which means it's a good. This is a good kind of CRG to associate to a property that forbids C four. And likewise, we had briefly discussed that K to four doesn't embed into this wild CRG. And for that reason, this is a good CRG to associate to the hereditary property that forbids K to four. For this reason, we find for any hereditary property the set of all CRGs which embed only graphs that satisfy H. If you only forbidden one graph, then KH is just all see all CRGs that don't embed your forbidden graph. Does that make sense. By the way, how much time do I have left about 24 minutes. Okay. Thank you. So, all of the CRG business is actually useful because of this theorem, which is a kind of joint theorem of two authors so in the same paper that Bella and Martin had their previous results. They also showed that CRGs are useful to get the edit distance function. But there could be an infinite number of CRGs and they didn't quite get that this in FEMA is always attained. This in did in fact show that for all P, this, this infimum isn't is attained. Why is it a minimum. Well, it's a minimum because we're always wanting to find the best CRG that is useful for editing a generic graph of density P. The minimum is always attained by a PCOR CRG because well if a CRG is associated to a hereditary property so is any sub CRG. And well if you had a CRG that had some vertices that weren't doing any work you might as well just delete them. So this minimum is always attained by a PCOR CRG. I have to define that here, but we will come back to this notion. Now we're going to talk a little bit more about our original problem. What are the PCOR CRGs or just generally what are the CRGs that don't embed our random graph. I did not define the random graph. So this is this is this can be thought of as a as a strange kind of coloring question. Let's go back to a little bit of basic graph coloring. So if, if you take K to be the CRG with W white vertices and all other edges gray, something oops, something like this. You use that CRG with all edges gray. Now the picture goes away. So we're mapping a graph into this CRG is the same thing as a W coloring of the graph, because what are we doing we have a we're mapping the vertex set of your small graph F into the vertex set of some CRG K but all of the vertices are white, and there's no edges in between. The gray area means that we don't have to worry about whether there are edges in between or not. And since all of these are white vertices, this is just a proper coloring of your small graph F. If you're using one fewer vertex than the chromatic number, then your small graph does not embed into that associated CRG. Moreover, we can compute the G function of this CRG. If there are, if there are W white vertices and no, no edges other than gray edges, then your quadratic form is pretty boring. It's only considering the edge deletions that occur on your your W many white vertices. And on those vertices, you're editing by deleting so so you have to be so this is proportional to P. And that's our G function there. And similarly, you could use the chromatic number of the complement and a CRG with only black vertices and gray edges. And you get this sort of general upper bound of the edit distance function concerning one forbidden graph F. And so based on this, Martin in 2013 conjectured that if you are forbidding a random graph. Then the edit distance function is no more than what you get from taking a minimum of these two chromatic bounds. So why are these the two quantities you see here. Well, because this celebrated theorem of Bologbash says that chromatic number of a random graph with P fixed is concentrated around its mean, and its mean is and not over to log one over one minus P event not. And again, this is the quantity that you're plugging into here. And when you work it out, you get this quantity. And similarly, you get a another quantity over here that you would plug in for B for the co chromatic number just by replacing one over one minus P with one over P should have a P not here, I suppose. So basically this conjecture of Martin is that you can't really improve this bound for a random graph. But yes, there should be a tilde here. Are there any questions about that. One thing I should also mention is what are the optimal values that you get out of this. So we have, we have this edit distance function graphed with density on x axis distance on the y axis. So we have two lines, red and blue. And the goal from Ellen and stuff was to compute this peak point, which you could call P star. P H star. If you wanted all that. And if you had this, if you had that this formula was asymptotically tight, then you would get these values for the, for the value of P star and the value of EDH star. So we have a more general setup that we could use instead. And this is a good benchmark to test whether there are ways to improve this bound. In the ordinary coloring case, we use a bunch of white vertices with gray in between, or a bunch of black vertices, also with gray in between. But there's something else we could do. We could take any CRGK and use a bunch of a multiple number of copies of it. So we could, for example, take CRG like this and clone it a bunch of times. So we would take that, make another clone over here, make another clone over here, and exclusively put gray in between. That would be this quantity we have called M times K, M times K. And so for any CRG, we could define that as a kind of chromatic number. So we let K sub K of G be the minimum M such that G actually does embed into M copies of the CRGK. And Bolo-Bosch and Thomason, this is a special case of their theorem, but it's the most useful version for what we're talking about. Bolo-Bosch and Thomason showed that they showed that for fixed K, that you can compute this K chromatic number of a random graph of any density that's fixed. Asymptotically, using quantities we already know, such as GK and, yeah, such as just GK. And you'll note that this is proportional to what it should be for the chromatic number, for the ordinary chromatic number. So this shouldn't come as too much of a surprise since this is an example of it. If you just plugged in K being a single white cortex, then this would work out exactly to be the quantity that it is for the chromatic number. So this is in some sense a generalization of that theorem. One important obstacle to this is this only explains constructions that use multiples of some fixed CRG. There's also a sort of pathological possibility that the CRG that you have has a tremendous number of vertices, in which case it wouldn't be fixed relative to N. It could possibly grow huge. And so it would be very hard to characterize when a CRG embeds into, or whether a random graph embeds into a CRG. If that CRG, if you have no control over any of its structure or even its size. So we introduce a sort of trimming. For this, we have to introduce a definition, which is reminiscent of things we've done already. If you had a CRG like this, and you put gray in between, then it's natural to kind of consider these two pieces as components. Let's say that the components of a CRG are the subgraphs that are held together by non-gray edges. So if you were to make black and white edges as strings, then these are the pieces that would come up if you grabbed a vertex and lifted. Lemma of myself and Martin is that this is useful for finding peak or sub-CRGs, as long as P lies in between one third and two thirds. So what the lemma says is that if you specify a tolerance and you start with any CRG, then you could shrink to a peak or sub-CRG so that two useful properties holds. One, the G function didn't change all that much, which is to say they're almost just as useful for embedding a random graph in terms of that chromatic lemma. The components are actually bounded order. And that's useful because we could have possibly been dealing with a CRG with an unbounded number of components in terms of N. So overall our argument for proving the result for the edit distance function associated to a random graph, first we trim any CRG that could possibly break the bounds that we already have. Again, the bounds that we're trying to break are this, so our pathological CRG might somehow have something below the peak point. If it has something below the peak point, we're not happy. Right, so if it looks like this, then obviously our original construction, one of these two constructions was not optimal. And by concavity, it's equivalent to ask if this function lies below the curve anywhere, then it also lies below the curve here. That's just a sort of concavity argument. So our trimming lemma, what it does is it shrinks the CRG into something more manageable. So now each of the components lie in a finite set. Since it's a finite set, all of them can be considered fixed in terms of our tolerance and our value of P between one third and two thirds. And we're just going to apply Bologbash and Thomason everywhere. We're going to cut our big graph, our big forbidden graph F into a huge number of pieces. And use Bologbash, Thomason to basically color each of those pieces using the elements of this graph, of this set of CRG's B, in the most efficient way possible. And by that I mean basically up to a tolerance the chromatic number of that little piece, the B chromatic number of each of those pieces. And we just assemble them all together. I mean, this is, this seems like a pretty straightforward argument once you get your hand around Bologbash's theorem, Bologbash and Thomason's theorem. The tricky part is actually this lemma, this piece. The trimming lemma took a lot more work than it might seem on the surface. The general algorithm for proving this trimming lemma is to take to shift immediately to a PCOR sub CRG by deleting vertices of weight zero, and then delete a vertex of lowest weight repeatedly until you reach a certain threshold for the value of the weight vector, the optimal weight vector. And then stop once you're at that point. I argue that, as far as the underlying graph is concerned, the degree is bounded. That takes an argument. But how do you show that a bounded degree graph has bounded components. Well, you would need to show that the diameter of each of those components is small. And that's what we end up doing. At this point we do have to go back and investigate some questions about PCOR CRG's. In Marshall and Thomason's paper, what they found is that PCOR CRG's need to have a very restricted color pattern. If K is PCOR and P is between zero and a half, then no edges are black, and no white vertex is incident to a white edge. So what that basically means is that the white vertices are isolated islands, and the other interesting components have only black vertices and white and gray edges. So this CRG would have five components. That's under this first case. So we're kind of assuming that this P is between zero and a half. And so it's possible for this thing to be PCOR. We're not saying it's PCOR for every P. It's just that this is a necessary condition, necessarily would have to look like something like this. And for P between zero and a half, you just swap the colors in this sentence. So actually it can't be PCOR and P equals one half because of this color restriction. The result classifies all one half core CRG's in general, and it also classifies all two vertex CRG's on any interval. What we introduce now is a new notion of a type of CRG that we'll call a Dalmatian CRG. We recall that we have components defined in terms of non-gray edges. The result states that on a certain interval, the only PCOR CRG's have these specific, have a very specific definition. We're calling this a Dalmatian CRG. Each of these is a Dalmatian CRG. So a Dalmatian CRG is a CRG that, the white edge relation is essentially transitive. So the only possible components are isolated white and black vertices and cliques of black vertices with white edges. The interval where this holds that this is the entire classification of PCOR CRG's is the interval one minus phi in verse one half where phi is the golden ratio. We don't really have a clear reason why the golden ratio is coming up here other than it solves the quadratic equation. And if you were to negate sort of take the photo negative of the same question, we're swapping black and white. And so the compliment of Dalmatian CRG's are the unique PCOR CRG's on the complimentary interval one half to phi inverse. What it does is it enables computation of edit distance function on this interval of positive length previous results only had characterization on the interval, or at the point at the point P equals one half. So now we can actually compute an edit distance function on a positive length interval without having to do any CRG magic and optimization. So the finalization conjecture, I think in a 2013 paper that asks whether this is possible, and at least on this interval it is possible. The way that the trimming lemma was actually proved was with what we'll call spectral prohibition. So we formed the underlying graph of a CRG by deleting all gray edges and forgetting all of the colors. So the CRG that we've seen a lot is should be 3k one plus all this. That's already there. Yeah, so the underlying graph of the CRG has these five components, which are just found by forgetting the colors, except for gray gray is a non non edge. And we say that a graph is P prohibited. If it cannot be an induced subgraph of any PCOR CRG. So I guess I should say an induced subgraph of the underlying graph of a PCOR CRG. In our paper we define this only in terms of CRG so we said a CRG is P prohibited. If it can't appear as a sub CRG of a PCOR CRG. So what we showed is that if your graph G has minimum eigenvalue lambda, then then the graph G is P prohibited on the interval that this is defined in terms of that minimum eigenvalue lambda. And the proof of this is kind of fun. You take that optimization problem and you infer that necessarily the matrix in the middle has to be positive definite, because you can cheat, you can decrease by exploiting a minimum eigenvalue. If it's negative or zero system future work. This is involved in the Iowa State discrete math RTG. The class involving undergrads, no no undergrads graduate students, a postdoc, and my advisor, and one of our mini goals is to widen the classification. It's tight if you only use Dalmatian CRG to classify, but we want to widen that interval a little more to see whether we could understand when wider intervals where we know all of the PCOR CRGs. The other result would be to widen the interval for which for which the approximation. The CRG trimming lemma also holds, and I am personally interested in understanding this question of hereditary property defined by forbidding a inhomogeneous random graph. So if you've seen other random graph models, you may even, you may have encountered those that are associated to to graph limits. There's a way that you can define this using a graph limit, and I am interested in what happens to this hereditary property if you're forbidding one of those. In particular, do you still get for some values that the graph looks like this. And this is a sort of Wild West territory because we just barely know things about inhomogeneous random graphs, we barely know that their clique number is concentrated for a generic inhomogeneous random graphs we don't even know where it's concentrated. So we have no right to also know where the chromatic number is concentrated. So we have no idea what this line and this line should even look like in the inhomogeneous setting. And those are sort of precursors that are necessary to start thinking about this problem properly. So this is my last slide. So I'll just say thank you for your attention. Thank you for having me at your seminar. Thank you, Alex. If we could all get some thank yous. Thank yous for Alex in the chat or turn your mic on and say thanks that work as well. Yeah, maybe some some claps in the camera. Give us those to put in. Okay, and oh, there you go. Like a little little emoji there. Okay. If anybody has any questions go ahead and ask away. Sorry. Can we do anything if the target graph is sparse instead of instead of dense. That's a great question. So the question is, whether we know what the edit distance function is if you forbid in a sparse random graph is that that's your question. Yeah, also, also you start with also a sparse graph. So you say start with, are you talking about a forbidden graph, or are you talking about if you want to know the distance of a sparse random graph to a hereditary property. So I guess there's a you have a starting family of graphs right then you want to reach another another family of graphs. So here I guess, both of the starting and and the target family are both dense. So those are stars. Are there any theory on that. So, I should say, the setting is sort of, you take any graph. So these are all the graphs of order n. So you can take some graph and then you highlight in blue are for hereditary property. So what you're proposing is that what if instead you had a some other subset and you wanted to know the distance from I don't know all these I. Yeah, so so we're talking about generically what is the furthest graph or the furthest graph of density P to H. But yeah, that's that's a totally natural question. What about for this triangle free graph to see for free graph. That's that's an interesting question and I don't have an answer for you. Yeah, I guess if we don't restrict the starting family, then when if we go from a dense to a sparse one then the edit function is always is it one or something like that. If you went from a. So again, your hereditary property is forbidding a random graph. Or no. Just like whatever we forbid just the family we want to go into is stars is all stars. Oh, you mean this family sparse or. Yeah, after we forbid that all these forbidden graphs and then whether this is sparse. Yeah, let's see this. No, no, I was saying like the other all the elements in HR sparse. Let's say the set of planar graph or something like that. Yeah, yeah, yeah, where H is a sparse or small. Yeah. So my understanding from this, I'm going to misspell this but China men. I'll blur the leather letters together so I don't make a typo and Zito. So they have a theorem that essentially classifies the growth rate of a hereditary property into different classes. And so there's also a later result of see Balog. I think I talked about this Balog Bolog and wine right, which gives a little bit more structural information in the various growth types. My understanding I might be wrong about this is that in the sparse setting, or when your hereditary property is small, then these graphs are either nearly empty, or nearly complete. So my understanding of that case and I think some of it some of this can be inferred from colored regularity graphs. So in that case, yeah, your intuition is right that it would basically be either a zero or one one that the distance would either be P or one minus P all the time. Because you would be moving to essentially a nearly complete or nearly empty set. You could have a different normalization to account for the sparse case or something. Josh, it's pretty hard to hear you. Oh, you could have a different normal is it can you hear me okay. Yeah, that's better. You could have a different normalization than juice to sparse case right. Yeah, so we have distance G prime edits, divided by interest to so you're saying like if you. If it was a graph of linear density, what would, what would that look like. Yeah, so you would only be considering the edit distance function on a short snippet, or maybe the density P is something like one over N. And yeah, I don't know. I don't know I think all of this is in the language of dense graphs, which is why the regularity lemma, and it's most useful forms actually kick in, but I don't know. I think these are good questions. I'm also interested very much in this other question involving what if you're forbidden graph is a sparse random graph, because we know it's chromatic number we know it's co chromatic number. It's just that the arguments haven't been made in this case and some of them might go wrong. If you've, if you've actually had something sparse. Do we have any other questions. Just a quick one. The, well, maybe quick. What is this finalization conjecture. Yeah, that's a good question. Do I get to make a new page. Lots of blank space there. Yeah, maybe I grab it and move it. No, I can't. So the finalization question would be. It comes in a couple of different forms. So basically the most general one would be if a territory. Then, does there exist. That's backwards even. There exists a subset of the forbidden CRGs. And I guess I should also say, can you read this and a is between zero and half. Does there exist an M such that the edit distance function is equal to the minimum overall elements of M. Or GKP. On the sense, let's say P belonging to a one minus a. So our proof works for a equal to one over golden ratio. Yeah, that's something we're working on right now. Do you have any other questions. Okay. Well, thanks again, Alex. And yeah, I guess Josh is there anything that you you want to say about the YouTube channel. To those of us that are here, or can subscribe. Like comment and subscribe. All right. Thank you, Alex. Yeah, thank you. I think this was not disastrous. So go ahead and upload it. All right. Thank you.