 Okay, so this week, next story, this week we are excited to have Anton Bernstein from Georgia Tech, talking about county and colorings of triangle free graphs, and this talk will be on the board so hopefully this is legible. That was our main goal. And I will be trying to monitor chat so if you have questions and you would like to type them in, I can try to relate it for you, but other than that, let's take a slide, Anton. Thank you, Stephen. Thank you everybody for coming and for joining. Thank you. Thank you everybody for doing such a tremendous work on setting all of this fancy new technology out. I hope that this goes well and you cannot imagine how much of a pleasure it is to give a real live board talk to a real life audience, but you know I might be a bit rusty so you know, don't judge me too harshly. Yeah, so I'm going to talk about coloring counting cowlings of triangle free graphs and all of this is joint work with Tyler Breselton, read your cow and a can come from Georgia Tech, the students at Georgia Tech. So, what's the general setup that we work with we have a graph G which is going to be triangle freeze throughout the talk. Now, and I might forget to mention this again, but it's always triangle free delta is going to stay for the stand for the maximum degree of G. Also, and is going to be the number of vertices and M, the number of edges in G, and what we're interested in is coloring G with as few colors as possible. Which is, I guess the question of what is the best upper bound and the chromatic number of such a graph you can get is this is a classical classical question there is sort of a, what we know is that there is a lower bound that's impossible to beat. It could be at least, let me say this way. One half, maybe plus and lower the term times delta over log delta. This is need to freeze and would check from 92. They actually got the constant right there are early results from late 70s early 80s by Bolabash and costage can mother independently who showed that at least there was a constant you can put here. So the order of magnitude is at least delta over log delta, but actually we get one half here how do you approve this by looking at the random delta regular graphs that's that's the band for those. On the other hand, it turns out that this order of magnitude actually is is best possible. The question famously showed in 96 that actually for any such triangle free graph the chromatic number is bigger delta over log delta. So this is most the constant times delta over log delta for some value of the constant which is not one half but some something independent of Delta. It's actually this is this, this is this kind of a curious papers never has been published and it's impossible to find and different sources side different constant factors here so we don't actually know what the constant was. But in any case it was some, some value maybe around 10 or something like this, she's been slowly improved or time until very recently Mike Malloy. In 2019 in a breakthrough paper actually argued that they can reduce the constant all the way down to one. And this is this is the best possible best known currently known bound. So this was a remarkable result for several reasons why is that the methods used in Johansson's original proof could not really be, you know they have had been pushed to their limit before and you cannot reduce the constant all the way to one there. Another thing is that there's still a gap of a factor of two between these constants but we don't even know that these graphs triangle free graphs have large enough independent sets to be colored with fewer than this many colors. And even already proving that the independence number is large enough would be a tremendous breakthrough it would improve bands on off diagonal Ramsey numbers know all other stuff. So somehow this really as much as the best that we can kind of hopeful with our current understanding of the universe. Okay, so this this this result is the starting point of our investigation. So we want to study is the number of calories now if you know that that the chromatic numbers less than this much so there isn't at least one coloring with this many colors, but can we argue that there are many, rather than just just one. So let me introduce this notation. This is going to be the number of G comma q. This is going to be the number of colorings of G using q colors. Number of proper key colorings of G. And so the main the main result I want to talk about today is a lower bound of this quantity for triangle free graph as G in the regime where Q is at least this much so we know that at least one calories is by my lawyer's theorem. So this is our result. Let me give you some intuition for where the band comes from the band turns out to be sort of what you would kind of naively guess the answer should be. So what is the naive calculation. Well, imagine just coloring every vertex randomly. So there's a set of colors you give us what a random color to each vertex, maybe F is a random q coloring. Any proper color is the rise the same as asking what's the probability that the random coloring is is proper. Now if I take two vertices say you and V that are adjacent in my graph. What's the probability that they are colored differently in F. Well, each of them is colored randomly is one of the key colors so this probability is equal to one minus one over two, there's only one of a key probability that they get the same color. So if these events for different edges happen to be independent from each other we could just take a product and see what the probability is right. So if these events are independent for all the edges, or religious, then the probability that F is proper would be exactly equal to one minus one over q raised to the power of however many edges. And this is the same as to say that the number is equal and the same that the number of two colorings is equal to one minus one over q to the power M times q to the end, because this is how many. Right. Okay, well this relies on this weird assumption that these events are independent, and they're in general not independent fact they would only be independent of the graph has no cycles because any cycle has it has a dependency here. So, true only if run no cycles. Yeah, so so in that case the calculation is exact but that's not what we want to worry about. On the other hand, if you think about this a bit. Even if they're not independent if you go if you take an even cycle then these events are going to be correlated positively, the probability that they all happen at the same time would be actually higher than the product of the problem, just rather than lower. So it will seem to only help us, and this intuition is justified by this theorem of, I want to say Chevara Chekwari but I'm not 100% sure pronouncing this correctly. Oh, thanks. Yeah, who argued that this is this is a lower bound to the graph is bipartite if there are no odd cycles or cycles or even this is a lower bound. And what we argue is that, well, if you don't forbid all odd cycles just the shortest ones the triangles, then this isn't quite the lower bound but you can get a bond that's pretty close. So here's our is the main theorem of today's talk. I guess how I want to say this. Let's say for all epsilon greater than zero. If delta is large enough, and q is at least one plus epsilon delta over log delta, then the number of proper q colorings of G is at least one minus one over q to the M times one minus a little bit times q to the N, where this little bit delta is technically which comes out of a proof of something like for the right this kind of weird expression like this but the main point is that in the regime where q is at least one plus epsilon delta over log delta this is this goes to zero as delta goes to 20. So this is some error term, thank you this is an error term that you can plug in there in the, except to this error term band is exactly. Okay. Any questions. The main, I should put it in the box. This is the main rule. Yeah, I should say that really we can prove something stronger, which I will not talk too much about today. The same is true for list colorings, for example, if you give every vertex of its own list of two colors and you color and you count how many colorings that with the same parameters for q, you get at least this many colorings from those lists. So with correspondence coloring or DP coloring is the same holds from that setting as well, if you don't know whether it's not very much. Okay, so, let me mention a few corollaries of this. Yeah, so this is stated for all q that's at least this much right, but in particular if you take, just use this bound for q and flag that in, then you get some some explicit formula. For example, if you're using the bounded q is at least one plus one. Delta over delta and also here there's an explicit dependence on M. So somehow the more the fewer edges you have the better the boundaries. But since maximum degrees delta M is certainly at most delta over two. And if you plug these two numbers into the formula, you can get something like this is at least e to write it this way. After their term in the exponent. Okay, whatever this is. Maybe it's not so important to know what this formula is, but let me derive a corollary of this statement now. Suppose I don't care about the number of colorings I just want to count independent sets in my graph well a q coloring. This is just a sequel you can think of as just a sequence of q independent sets as all just joined and cover the whole verdict set. This means that the total number of q colorings is at most the number of independent sets, raised to the power q, certainly, you know, this is most what you can get by just taking q independent sets of the Charlie, right. So now if I raise, you know, if I take you through to both sides and plug this in. I guess, yeah, these two statements together will imply that the number of independent sets is at least, again, something. Okay, something like this, again some formula, but I want to by type that this actually this results independent sets was already known, and we obtain it as, you know, sort of the simple corollary of our calculations. So this is the theorem of Avis Jensen Perkins and Roberts from 2018, and they obtained this bond using rather, rather different, different techniques. And one other thing I should mention that this bond actually is optimal in general there is no up to the value of the slower the term this is possible. That's what they showed. So we recover this optimal bound on the number of independent sets by just saying well you know look at the number of colorings and use this trivial inequality. Actually, well this bunch of independent sets is optimal this applies that at least in some regimes are bound on the number of colleagues also has to be optimal. Actually it's optimal for all possible values of q. And if you say that as well is optimal up to the value of the error term for all, for all q. So, so for any q and delta. There exists a triangle free free delta regular graph G such that the number of q colorings of this graph is at most. In a slight slight error term here, this most this much. So, so really this name bound actually cannot be improved and triangle free glasses. Yeah, maybe, maybe some of you can guess how you get the graph G. This is this is this is true with high probability, hold this and the quality holds for random delta regular graphs, and they're triangle free with positive probabilities this implies that actually most triangle free graphs have this one. Okay. Any questions. Alright, so then in the remainder most of the remainder of the talk. I'd like to schedule the proof of this for you. This. So this, the proof is simple enough that they can explain pretty much most of it in this lecture which is kind of nice. And let me tell you how we came up with this. And as I mentioned there is this, there are these results. These bonds on chronic number 23 graphs, the earliest one is duty Johansson, and that proof. This is something called the nibble method. So, don't worry too much about what this is but there's some kind of standard probabilistic technique reconstruct the coloring by you know bit by bit that's what's called the nibble method because the color a few verses at the time. So you use the local lowest local number repeatedly to argue that the coloring has the properties that you want. And this is sufficient to prove a band of the form be go delta over log delta on the chromatic number but this constant can not really be reduced below something like four is really the limit for this for this method as it turns out. So, key contribution of Molloy was to use instead of the nibble technique use something called the entropy compression method. And this is a technique that's, that's been invented that that's been invented originally to provide an algorithmic proof of the lowest local Emma, but then it turns out that in many situations you can apply directly to combinatorial problems to get better results. And this is this is one of the instances of this. Actually, later turns out turns out that you know if you applied the lowest local Emma, a lopsided law so clever and some kind of clever way that you can avoid the use of entropy compression and still get Molloy's result that this is again a different argument. So, in a recent development in the field of understanding things like entropy compression a lot dilemma. Nancy Rosenfeld came up, just in 1920 with a new counting inductive counting technique, which can serve as an alternative to entropy compression to entropy. So, entropy compression itself was kind of a slightly technical tool to your bulky technique to use this is a very elementary method that recovers. Pretty much all of the same results and sometimes gives better results. And one byproduct of this is that it not only proves that something exists but actually gives you a low amount on how many of these things there are kind of it comes out of the method directly. So what we're going to do we're going to apply this. This, this idea to the question of coloring triangle free grass and somehow out of that the lower bound of the number of such colorings will come out. But it's a bit more complicated than just using this so let me now start to explain how the perfection goes. Okay, let me raise this. Yeah, so so let me write this. See here so we remember that Q's is just one plus epsilon delta, delta, and I think that delta is very large. Yeah, okay, so how does the proof go. Well all of these arguments Molloy's and Johansson's proof, and a lot of other arguments in this area you don't build a coloring at once you first find a partial coloring with some kind of nice properties and then you argue that these are going to extend it to color the rest of the first, and we're going to do the same thing so first going to focus on partial colorings, and then try to extend them to color in the photograph of colorings. So since we're trying to extend this coloring let me introduce some notation so imagine F is a partial Q coloring of G and maybe x is an uncolored vertex. x is the vertex. And when it's uncolored I'm going to write that F of x is blank, the color assigned to this blank. And then we will want to color this verdict somehow but you want to keep the coloring proper and since x already has some colored neighbors we cannot use the same color as you wanted them. So let's write down this notation else of F of x this is going to be all the colors that x can use still. So I'm going to write all colors alpha, and I'm going to say that my set of all colors it just wants for Q for convenience, such that x has no neighbor, why, with F of y, be equal to. So this means that if I color x by alpha it's not in conflict with any vertices that the color already. And I want this to be large. Right the more colors there are available for example freedom I have when I color it that's good for me. Let me introduce one more piece of notation. I take one of these available colors. I'm going to define this thing with degree sub F of alpha, x. This is something that's called the color degree. Let me write this and explain why this is relevant. This is the number of all neighbors of x, why, such that first why is blank. And alpha is available to why. And you can how many of these things there are. Okay, so what is this, what does this say. Well, alpha is an available color for X, so you might consider coloring X with alpha. The problem is that why is also in color we'll need to color that and alpha is available for why so we also consider coloring why with alpha. So we want X and Y if they both are colored alpha is going to be a conflict between these. So this counts how many of these conflicts that can be involved in the color alpha. You know what possible problems we can create by coloring X with alpha. And so this we want to be small. So one day. We want the size of the set to be as large as we can make it and we want these color degrees to be as small as we can make. Right, that's that's the approach that we take. So in fact, let me say that F has a flaw at X, if first F of X is uncolored, sorry X is uncolored, so F of X is blank. And either the size of this list is small, let me say it right this way it's less than L where L is some parameter I'm going to write in the second one this is but this is some fixed numerical parameters because it's like our desired size of this set. Or, or there is some alpha in L F of X with the color degree greater than some other parameter D. So either the color degree of this is too large or there are too few available colors. Let me say that this is bad for us where let me write these parameters right here. I'm going to take it to be something like this. Okay, whatever this is some number doesn't really matter and D is going to be just L over say 25. The parameters numbers 25 and one half are more or less arbitrary the point is that D is kind of significantly smaller than L. Okay, so that's my definition and we don't want any of these flaws we say F is good. If there are no flaws. Here's a fact. That's kind of a standard fact in this whole theory is that once you have a good coloring you're done in the sense that well if you just want to get one color and if you have a good coloring partial color you can extend it to coloring of the whole graph. Every good partial coloring can be extended to a coloring of all. You can color all the uncolored versus really creating conflicts, the way you prove it just it's just a local Emma an application of local and your color and colored verdicts randomly by taking random day color from the available list available colors. And you can check that the conditions of local Emma sets. So it's kind of a well known thing and that's how all of these proofs. Johansson's proof and Malloy's proof they both construct a good partial coloring somehow, and then extended to color in the whole graph. And this is going to be to count how many of these good partial colors there are first, and then for each of them to count how many ways there are to extend it to a coloring of the whole graph and this is going to give us some lower bound and the number of coloring so we Okay, so to achieve this we're going to use a tool that was introduced in more ways proof. We need a slightly sharper version of this but it's essentially essentially the same. Key tool is some kind of coupon collector like lemma, due to, I want to say, yeah, it's due to Malloy came came up and his proof. Let me try to explain what the situation here is. So, we want to get rid of all of these flaws so we look at a color at a verdict sex has some neighbors. We're going to call them why one way to this is why I like a, where case is it most delta maybe it has fewer than delta neighbors. And then there's the rest of the graph. And of course these neighbors can now have neighbors there. But there are no edges between them because the best triangle free that's going to be important. So how can we generate a partial coloring of this whole graph we're going to do this in two steps. We imagine already having some partial coloring outside of the neighborhood of x, let's call it G. This is some kind of partial coloring here. Now what happens to these vertices. Well they have some neighbors that may be colored now. So they have their own list of available colors. And if I look at the verdicts say why I, which color can this verdicts have, well it can have either from its list, this L sub G of why I, I guess, right, these are the colors are able to why I know, or it can also be blank because we're doing partial coloring. We're blank. And the same is true for each of these vertices and moreover because there are no edges between them these choices are completely independent from each other. They cannot come, you know, they cannot conflict we can assign to each of these any of the colors from available list or blank and this is going to be a perfectly fine partial coloring of this. So this random choice for each of them independently, this generates some kind of new partial coloring of, I call it F which is G here and then these random choices in the neighborhood. And now we can ask well what is now the available colors for X and what are the color degrees there. And, and the lemma says that if you look at the expected size, how many colors X expects to have available to me this is pretty large. This thing. As it turns out with case the degree of X in this picture. So why am I calling this a collector coupon collector lemma is the standard coupon collector lemma that you learn in the basic probability class theory class is a special case of this for all of these sets L sub G of why I just equal to once for K. Right. If every verdict says that all of the same colors available to it, then what's happening we're just picking elements from once for K randomly, we pick, sorry, once for Q randomly we pick care of them, and we're asking how many elements are not fixed that's literally what this is. And we're saying that the expected sizes is this which is easy to calculate what Molloy observed is that, no matter what these sets are actually whichever subsets you pick here, they still get the floor back. So by construction, this is, this is at least to L, where else this parameter there. So in other words, it's, you know, the expected sizes like twice larger than what we actually wanted to be this gives us enough room to prove that the probability that there is a flaw. This is very small is is the most say P, which is some something exponentially small, let me say something like this. Yeah, so the point is this is tiny. Again, what does it say that says that was the probability first that this size is actually less than half of the expected value and that happens very rarely. There's the second condition about the college degrees that also you could show happen. Okay, so this is the limit of the piece. Any questions so far. Yeah, so I would like to point out this this lemma really relies on the fact that coloring is a partial for the proof of the lemma is crucial that these versus why I are allowed to be blank. In this stream case for all of these sets of all colors are just one color per each vertex. If you were not allowing them to be blank, they would have to pick exactly those colors and you would have nothing left for X. But, but now was probably to one half the same color in that situation. That's good. That's how it's sufficient to get. All right. So, all of this was preliminaries. Now let's actually start the, the argument. Well, since we want to understand how many good partial calories that first we're going to just get a lower bound on the total number of partial calories, good or otherwise. Count. All partial problems. And for this we're going to use this lemma specifically this, this part. This, this is going to make an appearance. So let's say, let's introduce some notation C sub P of G. This is the set of all partial colorings with Q colors or Q is fixed. No, of G, and also for a subset of the vertices. Let's say C sub P of you sub default partial C sub view this is the set of all partial count on just you. So everything outside of use on colors at this point or like if you like it's a set of. This way it's partial colorings of the sub graph induced on you. Okay. So, in other words, it's a partial colorings whose domain is a subset of you so not everything in you has to be colored because the coloring is partial. And the lemma. So the key lemma is, we want to understand how this the size of the set changes that we as we have vertices to you one by one, so that you can then use in the document. So we have a subset of the vertices and we have some verdicts x that's not in here yet. And let's say k be the. This is the number of neighbors x has in you. Then the size of the CP of you, plus x this is my notation for you are the verdicts x 2. This is at least this quantity times. That's right. Yeah. The usual formula times this. Okay. Hopefully we'll be able to recover. Yeah, okay. So this, now, before I say how this is proved right once we have this, just apply this repeatedly started with the you being the empty set there is one partial coloring that nothing is colored right then you start adding versus one by one each time you multiply the quantity by this factor. So this implies that they. And if you do the calculation you'll see that the total number of partial counts of G is at least all of these one minus one over cues race to these parts exactly adapt to one with one with you to the power of total number of edges. In terms of these cues, you know, which for short x comes in to the end so you get exactly a bond of the sort of this we want for the partial calories that the stuff can still be in college. Okay. Does make sense. Yeah. So, proof or proof sketch at least. How do we prove this well we just applied apply this lemon so let's see. We want to count calories on you plus x right. So first color you then color X right, let me write it this way the sum of our all colorings f of you only how many ways are there to color x once we fix the coloring on you. Well access to have one of the available colors all blank. So it can be the number of colors available to currently plus one, because of the blank. Actually, I don't care about this plus one. So let me just do this. It's at least this much. So how are we going to deal with this. Well, let's let's do this in two steps again. Let's say that this is equal to some overall let's first color the vertices that are not neighbors of X. So let's remove the neighborhood of X. So the G. I'm going to try to recreate the picture there right I have a G which is a coloring of everything except for the neighbors of X. Then I need to extend it to the neighborhood let's go f. Let's let's say ext of G is the extensions of G to the neighborhood. These are the extensions of G to the neighborhood of X so we already called everything except the neighbors now we color the neighbors, and now it was the same thing. Okay, but now. So this says that for every particular choice of G, the average of this over a random extension is at least this much. Right. Therefore the sum, which is what we're looking at here is at least. So this is the sum overall of these G's. So if the average is at least that then the sum is at least one minus one over Q to the K times q times how many of these coloring several. Okay. And I promise I'll fit this in this box is so this is just a constant doesn't depend on q let's take it out. And what will be left is the sum overall G's of the number of their extensions. Right so G is a coloring of everything except the neighborhood and then we count how many ways they're to extend to the neighborhood. This just tells me the sum was just how many coloring so you know everything outside the neighborhood was the neighborhood there are, which exactly gives us what we want times the number of all the colorings of you this is exactly the band that I played. Okay. Everybody, everybody happy. Every second. Yeah, so see that wasn't hard right. Okay, so this is a case I kind of want to keep this on the board says the case I raise this broadly. Yeah, we have a band. Step to count the good company. Yeah, so we know this. Step to instead of all partial calories we're not counting the good ones. This is the key. This is the key part of the program. And remember good means it's a partial coloring without flaws so. The floor means that a vertex is uncolored and it has either too few available colors orders and one of the available colors was too high over quality. So we want to avoid both of these, and we're going to try to do a similar thing as we did here right it was no some kind of induction we're adding vertices one by one. We're going to do the same thing but in slightly unexpected way. Somehow the obvious idea would be to say, well look at I can do sub graphs and count how many good calories there are and advertises one by one that's not what we're going to do. And this is the subtle part of the proof. So let's say C sub G of G this is the set of all good calories, good partial calories. So this is what we want to count. And now I'm going to say that for a subset of the vertices C sub G of the subsets, these are going to be calories that are good on you. So this is one of all partial calories of the whole graph G. So these are not only on you there also works outside of you that maybe call it, such that F has no flaws at the vertices. Maybe I'll. And at the vertices. We have X with the second neighborhood of X contained in you. Okay, this is kind of a convoluted definition. So first the set you here doesn't tell us which vertices I call it it tells us where flaws are allowed to be the bigger the set you the fewer flaws we're allowing. Secondly, when forbidding flaws not only inside of not for all of the verdicts that's in you, but only for those vertices that the verdicts itself is in you its neighbors are in you and their neighbors are in you. So we can't forbid the floor to be there. What is this defined in this way. The idea is that, so the reason why we need to look at the colors of the whole graph, essentially what's happening is that this adjust the probability distribution, and some kind of way in the calculation that actually makes the proof go through. But this condition here ensures that. Okay, so we have this coloring it's something inside of you scouting something outside to check whether or not this property holds I don't need to know what happens outside of you. So whether or not there is a floor to verdicts is determined by the coloring of this verdicts its neighbors and their neighbors. We need to know the vertices on college, we need to know what its list of all the colors is this is determined by the colors of the neighbors. And we need to know the college degree which somehow depends on the available list of colors for the neighbors, which is determined by their neighbors. So, since if the second neighborhood of access containing you, I know whether there is a fluid X or not by only looking at you doesn't matter what happens outside. That's why the definition is given in this way. Okay, now we have the lemon. And the lemon says, so so now the ways define the big year is the smaller the says, because we're putting more and more constraints. But we want to say that doesn't shrink too much as we are vertices to it. So if you have you a subset of vertices and we have X. So if we are designing you, then the size of CG of you plus X is at least one minus a little bit. And the size of CG of you where this data is something very small, something exponential, something like this, some kind of exponential small term. So the size shrinks when you adverse but only by this tiny factor, or factor I guess it's close to. Okay. So we are going to use this Rosenthal ticking this in this step. But first, let's see what happens when you combine the lemon was the statement already know this implies that. Yeah, how do we do this. Well again we need to start with you being the empty set. Well, so so what is, you know when you use the empty set was said well calories are the good on the empty set. These are just all the calories the empty set means there are no restrictions, we don't care about where the flaws are. This is the same as this as this for for G, all the partial columns of G, we have a lower bound on this. Now we do this inductive argument so this shows that the number of good colors and G is at least one minus eight into the end. So we got this little bit of an error term from this, but this is, this is not okay. Yeah, so let me start the proof here. Yeah, so this is the Rosenthal technique. And it's, I mean, it's, you know, I'm trying to advertise this technique but it's so simple that you might have not noticed where the technique is being used. It's you only realize how simple this is after you compare this to other things that proved you much more arduous methods. Other results is much more complicated methods to, to, then you can get them kind of easily from this. So first key idea is that we're going to prove this climate itself by induction. That's the first thing. The induction on the size of you. So it's so we're assuming that we already know this for all smaller sets. Here's what we do. We want to understand this quantity. Number of columns good on you plus x want to compare this to the number of kinds of the good on you. Well, let's write it this way, the number of colors that the good on you minus size of some set F where what is F is the colorings that we need to subtract. This is the columns that are good on you, but not on you close. This is kind of a triviality, as you can write this way. So now to get the lower bound on this I need to get an upper bound on this thing. What can I raise. I'm going to keep. We haven't used this yet. So this Chuck have gone is going to, is going to shoot eventually. Yeah, but this I won't be anyone. Okay. So how are we going to bound F. Well, first observation is that the size of F is as moist, let me write this. And then I'll say why this is true. Okay, what is F sub u here. These are all the colorings in F such that F has a flow to you. So I'm giving vertex you and I'm go over all the versus in the second neighborhood of us. Why is this the case. Because a coloring in F has to be good, not good on you plus X. So there's going to be something some flowing it at some vertex whose second neighborhood is entirely contained in you plus X, but the second neighborhood is not contained in you, because the coloring was good on you. So it needs the verdicts has to be a distance at most two from X. So it has to have a flow at some verdicts like this. Now we're just going to bound each of these individually. In fact, let's me start writing this right away. How many of these we have at most something like the square plus one. I'm going to tell the square plus one times, you know, something which is abundant the size of the set. Okay, this is in a very awkward place. Maybe we'll remember this. Okay, in the script one collector setting the probability of having a Floyd X is less than P, which is something tiny. Okay. Let me maybe at least write the value of p here. Yes. There's also this data now. Okay. Yeah, so claim was the claim. The size of a few is at most this thing. So we can bound the size of a few as in terms of these numbers p and data times to piece tiny. And this is okay something that's actually going to be close to one. This really says that it's a tiny tiny little fraction of this. So, once we have the claim little box. If you can prove this then we plug this in here. P one minus eight delta. And this you can you can show is at most somehow because he's so small as much for smaller than eight or whatever it was. So this is just a calculation that for a large enough delta this inequality fold. Okay, so. Okay, and then we have this and we plug it in here, we get exactly the result that we want, because we want to get this one minus eight. So, so the only thing that's left to do is to prove the claim. Well, let's do it. So, I kind of want to write this. Yeah, pretty easy. So, so we're going to do a similar thing to the induction that you did before we're going to color everything except the neighbors of you, and then we're going to color the neighbors of you. So, let's say s is the set of old G's, which are partial colorings off. V of G minus the neighbors of you size that G is good. On B of G minus. I guess close I want you to be in college. Something. Yeah, so then the size of F of you is. And so for each of these G let's say, we have this ext G, which are all extensions of G to the neighborhood of you. And let me call it F ext. This is the flawed extensions. So these are extensions that have a fly to you. Right. So this is again the coupon collector setup we have everything I'll accept neighbors then accepted neighbors and ask what are the colorings where they are connected to you. Well, where the coupon collector lemma, this ratio, the probability that we get a flaw, which is exactly going to this ratio. This is the most P as the coupon collector. And now, now, now I can write this. So size of F of you is equal to the sum overall G's in s. Well, I guess it's this is most some world is an S of this. Because because there are more conditions after coloring here has to be good on all of you. We're not even saying this we're just saying is good and you on on surface review. You minus neighbors. Okay, but in any case this is the most P times the sound. All of these G's of the number of all extensions. This is by this. And this, while you color everything then accept the neighbors then you call the neighbors this gives you P times the number of and the result in color will still be good on you minus neighborhood. Okay, we get this, but we want to compare it to the color gallons of the good on you. This is where the inductive hypothesis comes into play. We start, you know, we know it's already good on you minus the neighborhood. But all the vertices in the neighborhood who lose a factor of one way this data each time by the inductive hypothesis. And that's where this one was added to the delta. So, by the inductive hypothesis. This is at most P over one minute data, delta. Which is exactly what you want. So now let me just to finish off. Let me briefly say what what happens next. So far. So this was the kind of the heart of the argument. We've counted these good colorings but we're not quite done yet. So first, we need to know how many ways there are to extend a good coloring to a proper. We already know there is at least one that's a classical result but we need to know how many. So this is step three. So we're going to just use the usual lowest local Emma here. To show that a good coloring G with say K on colored vertices vertices can be extended in at least something like this. So what's happening is, well the lowest local limit is if you color each vertex using using an available color randomly, then this positive probability, you're going to succeed you're going to get a couple counting the whole thing actually the lowest like a local even tells you lower bound on the probability, and you can translate it into counting result and you get something of the sort. This number is greater the more and color vertices you have this somehow is important. And then there is the last step, just I guess step four, you know, put everything together. So what's what's happening is, we know how many we know lower bound a number of good coloring. We know for each of them and lower bound a number of extensions. The problem is that the same coloring of the whole graph can be an extension of different good coloring. But so what we need to do is do some kind of double counting to account for that. And double counting to obtain the lower bound on C. And double counting relies on the fact that this grows with K. Some of the more and color vertices you have the more extensions, which helps. Okay, so this is kind of reasonable standard double counting arguments are going to go in there. Let me just say that this step is when we get most of the error term. So far, our error is like this a tech quantity which is very tiny. And then the distribution to the error term house comes from this step, and it would be interesting to understand if that can be somehow removed or improved. But I will talk more about it. So, thank you. We think our speaker. Are there any questions for you, Tom. It's hard to answer the sort of question it's, it seems that it is like, it seems that every problem which we know how to approach using the compression. So far, we can do, you know, we can we can frame it as a Rosenthal counting argument as well, although in this case somehow, you know, the argument has to be somewhat as they explain some would settle like this definition of these, you know, these inductive sets is a bit is a bit unusual and there was there's a recent paper by by the peers appeared on the archive a couple days after hours did by these two authors where they also use Rosenthal method to reprove model is theorem. So to get an upper bound on the chromatic number, but they do it differently again so it seems that somehow there's no way to sit down and like translate the proof in the Rosenthal stuff so we had to do this, this weird thing where we define the coloring being good on you in this complicated way, and their analysis is like they don't even talk about good colorings at all they somehow, you know, do a different sort of thing, but it seems that even in this situation still we can get the same result if we, if you think of it. There are other situations when translation is more straightforward but yeah it's not 100% clear how you would form a main problem is the study was a question is this unclear how you would formalize what it means to have an empty compression style argument. Yeah, depending how you interpret it the answer is probably yes. Yeah maybe. Yeah, yeah. I didn't do these two steps but already I essentially prove to you that at least there is one coloring because because it's kind of a standard local application and these two steps are pretty pretty straightforward. I'm just trying to sell the Rosenthal method, it's so easy. Any other questions for Anton. And thank you for those of you who attended on zoom. I hope that was sort of legible. So, I think Jerry says thanks for sharing the excellent talk via zoom so I hope that it was, you know, mostly legible.