 Okay, great. Thanks everybody for making it out and thanks to Ava for agreeing to give a talk on such a short notice. So today we have our own Ava Sparka who will be talking about a counter example to a conjecture of Airdish. But take this away. Okay, so as you can see this is a joint work with Laszlo and Ina who graduated last year. And let's just start with this. So, a bunch of people. I'm pretty sure the list that I put up is not complete. Have shown that the connected graphs on n vertices and minimum degree delta have diameter at most 3n over delta plus one plus bigger one as n goes to infinity. And that bigger one term actually if I remember correctly can be changed to maybe minus one. So, and this is actually sharp. Here is an example. Basically what happens is that you have a bunch of complete graphs. If you look at the number it's roughly equally distributed delta plus one vertices are roughly equally distributed between three consecutive clumps of complete graphs here and between two consecutive clumps all vertices are present, except that the end you may want to blow out the first and the last clumps so the minimum degree condition is satisfied and if you compute everything out. This is an example that this can be achieved this diameter can be achieved. And finally, I'm not going to show the example because it's slightly more complicated catch it and smiths approaches if I missed from us these names. This bound above is short even for Delta regular graphs. So, at this park polar can to saw came up with the idea of, well, let's add something extra. Besides the minimum degree and the number of vertices in the graph. And they said, let's for for, let's forbid complete graphs of the certain size of slot graph. So there are two parts of the conjecture. If you have a graph that is K to R plus three and three R minus one divides delta, then as n goes to infinity the diameter is at most three R minus one over R times and over delta plus big O one. And I am rewriting it for reasons that will become clear later as three minus two over two R and over delta so three minus two over the maximum possible click size within the graph. They have came up an example that shows that this is actually sharp. Now the clumps that I have are sets of independent vertices. There are our clumps in a layer or column. And each clump has size that over three R minus one since three R minus one divides that this is an integer. This is why we need that condition and within a layer and between every two consecutive layer every edges are present except within the column. And once again at the end you might want to blow up some of the clumps to make sure that at the end the minimum degree condition are satisfied. So first of all, because you can only find a complete graph within two consecutive layers because edges are present between two consecutive layers but they are not present between layers that are further away. And the edges are only there between clumps and there are two R clumps within two consecutive layers. The maximum click size in this graph is two R so it's a K two R plus one free graph. If you take a vertex it is adjacent to every vertex in different clumps in its own layer and to all clumps in the other two layers so there are three R minus one clumps into which it's adjacent to vertices to each containing delta over free R minus one vertices so the vertex has degree delta indeed. And the number of vertices is D times R delta over free R minus one plus constant times delta. The diameter is D there are R clumps in every layer each containing delta over free R minus one vertices and you know the little magic we do at the end and the fact that there is one more layer gives us some contacts time delta vertices and that actually gives us that this graph achieves the upper bond that we need. So the conjecture that I showed you if true is sharp. G is K two R free and my condition on the divisibility conductive condition change to R minus one times three R minus one divides delta. Then as N goes to infinity, the diameter of G is at most this quantity. Let me not even read it out you can read it and here is the example that shows this as sharp. You have R and R plus one clumps R and R minus one clumps in the consecutive layers or columns. Plum size in the R clumps is R delta over R minus one free R plus two and in the R minus one clumps is R plus one delta R minus one free R plus two. Two consecutive layers contains two R minus one clumps so it's a K two R free graph and you know with a bit more work that I'm not going to show because it's just algebra. You can show that this is indeed a graph of minimum degree delta and the diameter is achieves what it needs to achieve. And just for comparison that little constant can be written as three minus two over two R minus one so three minus two over maximum allowed click size minus a teeny bit more and over delta plus big over. Now I would like to note here that we have the condition R delta bigger than equal to two. So we are talking about graphs that are K four or K six three so we are not talking about triangle three graphs here. But what would correspond to triangle three graphs in the part a actually that inequality has been shown and proven to be true by the same collection of people. In summary, at the spark polar can choose a conjectured that if you forbid a complete graph of size K plus one or sorry order K plus one. The diameter can be at most three minus two over K times and over delta plus big one as n goes to infinity in fact a bit less than K is odd. And we have a lower band on delta in this conjecture that comes from the divisibility condition. This is the corresponding version of the conjecture for triangle three graphs. Now we know that the diameter of any connected graph is at most free and times the free and over delta plus one. So if you want any improvement over this with or additional condition that is of the form C times and over delta the constant C has to be strictly less than three. So as this is even true for regular connected graphs and delta regular graphs are K that was from free for any improvement. We need the delta to be bigger than K minus two. And for this particular upper bound to be an improvement we need the delta should be at least three K over two. So we need to have some lower band on delta just not necessarily the one that that we have up there. So not much in not much progress has been made on this conjecture and Pete and on command that we to be working South Africa suggested to try to work on this but change the condition that says that oh you can have a click of size at most K here to the weekend condition that or graph is K colorable. And if we change the condition in change the size of the maximum click to the colorability condition the conjecture remain sharp if it's true, because the examples are K colorable. And we have actually managed to prove the conjecture for four colorable graph. So if you have a four colorable graphs of order n and minimum degree delta, the diameter is at most five falls and over delta minus one which is exactly of the form that the conjecture suggest. And or proof uses inclusion exclusion and the partitioning of the graph according to correlation patterns in the distance classes. And let me show you the structure that we can impose on the graph which helps with the proof. I'm not going to show you this proof I'm going to show you something else, but I still need these clump graphs here. So first of all, for the problem that we are considering it's enough to consider K color graphs with minimum degree at least that are edge maximal in the sense that if you if I insert any edge. The diameter decreases or the chromatic number increases above K. We start from a vertex V of maximum eccentricity, fix the K coloration and we use Li for the set of vertices that are a distance I from V. Because we have edge maximality two vertices are adjacent in this structure precisely when they differ in color, and they are either in the same distance class or in distance classes that are one apart. We're going to contract vertices in the same distance class and the same color into a single weighted vertex and the weight is going to be the number of vertices in the original graph. And the resulting weighted graph structure is going to be the clump graph. And if we are given a weighted clump graph, we can recover the graph that it comes from. In particular, if we have a clump graph where every two consecutive colors have at most K work to see the graph is K colorable that's not the, not the necessary, not the sufficient but the necessary condition and sometimes it is helpful. So, let's use clump graphs to describe graphs now. So first I want to concentrate on the free colorable case of sort K for three graphs. When two device delta delta is even, I'm going to consider this little gadget here so it has two, three, four, five, seven layers. I have the number of clumps I have is one, two, one, one, one, two, one. It's kind of a symmetric structure. The first, the third and the last clump has a single vertex. The third and the fifth clump has delta over two and the second and the panel to that clump has delta over two and delta over two minus one vertices per clump. I, and what we are going to do we are going to repeat these blocks, some P times. And in the first and the last block I'm going to make a little change, namely, then in the first clump the second in the second column this delta over two minus one changes to delta over two and in the last clump, the penultimate column in the penultimate column in the delta over two minus one the clump is going to change to delta over two. I claim that this graph is free corollable and has minimum degree delta. The free corollable is obvious because if you look at the structure, the number of clumps in any two consecutive layers is at most free. So that's fine. What about the degree condition, it's enough to check in the first half because the graph is symmetric. The middle vertex, if you look at the middle vertex it has two clumps of size delta over two as neighbors so the total number of neighbors is delta two times delta over two which is delta. If you look at the third layer the delta over two clump it the number of neighbors is one plus delta over two plus delta over two minus one which is delta. If you look at the say the size delta over two clump in the second column, the number of neighbors is one plus delta over two plus delta over two minus one which is delta and the other guy has one more neighbors so that's fine. If you look at the first layer, there is this little one, the number of neighbors is delta over two plus delta over two minus one plus one because it's adjacent to the previous guy which is delta. Or if it's the first clump, then both of those neighbors are size delta over two, so the minimum degree is delta. So this is indeed a graph that is free comparable and has minimum degree delta over two. The number of vertices well I repeat the block P times so P times the number of vertices in the block. And that's n minus two because the first and the last column has first and the last block has an extra vertex added and the diameter is P times the number of columns in this block minus one. Well, so the diameter is n minus two times the number of columns divided by the number of vertices in the block minus one. And if you compute everything out that's seven n over three delta plus one plus big over. And if you want a constant there that works for every delta that's actually seven third and over delta. So the diameter is seven third and over delta which is three minus two third and over delta as big over. And that increases this is the best constant we can put here so this is sharp. And this is bigger. This constant is bigger than the one in the air dish popular to the conjecture. And in fact, if we drop the condition that to devise delta, I can modify my formulas it gives the same thing for even delta. So it works with odd delta and the same numbers work out I'm not going to go through in five to stop I hope you believe me that this works out. So the problem here is that a then that is big enough this is going to be a larger diameter than the upper band allows us for in the air dish popular to the conjecture. In fact, if we just look at the seven n over three delta plus one number it's going to be bigger than the n over delta times the air dish popular to the conjecture. Then delta is bigger than 16. So for moderately large delta, the air dish popular to the conjecture fails in the free colorable case. So what happens if we try to look at larger numbers because you know it just might be a fluke because we have a small number so odd colorable case. Okay, two are free. If the same thing would work out we would have some blocks, repeated P times make little magic at the two ends to make sure that the additional edge that we have lost there doesn't drop or minimum degree below what we needed. And we would have something like this and minus two assuming that only two vertices are enough to make the fudging equals to P times the number of vertices in the block, the diameter is P times the number of colors columns minus one. And the diameter is number of columns divided by number of vertices times n plus because one, and it's going to be number of columns divided by how many times delta fits into the number of vertices that's all constant. So what can we do with this well. The next case that we can consider is the five colorable case. I'm going to show you the structure we found just when four divides delta because that's easy so five colorable case corresponds to are equals to three. We have 123456. 13 columns. And this is the pattern and the first every third column has a single vertex. So single color single vertex between two of these. I have some clumps of size that over four and the two columns in total has five clumps. The difference is that, oh, the total is five clumps in the first column, I have four clumps second two, three, then two, then one so four, three, two, one in the first column of these two consecutive columns and one, two, three, four in the second column. And if you look around you can see that the minimum degree is delta. And all because in two consecutive columns the sum of the number of clumps is five it's a five colorable graph and because I have 13 columns and five delta plus three vertices. The diameter of this graph is 13 and over five delta plus three plus big O one, which is 13 over five times and over delta plus big O one if you want the constant that works with every delta in front of then over delta. And that is three minus two over five. There is no little extra there so even in the five corridor order which is the K, six three case. If delta is big enough, this beats the air dish back to the conjecture. And in the general case the two are minus one colorable decay two are free I'm going to show you a block that has six are minus five columns and two are minus one delta plus two are minus three vertices, which will give you the free two is two over two are minus one times and over delta upper bound. And you can't put any constant that's better than this in front of the end over delta because the construction has diameter six are minus five and over two are minus one delta plus two are minus three vertices. This is going to be a counter example to the original conjecture for this case, then delta is big enough big enough means that it's bigger than two times are minus one times three are past two past two are minus three. So how does this counter example looks like let's start with the case when two are minus two divides delta. I will have six are minus five columns and the number of clumps in the columns is going to be one and then every third one is going to be one, the number of columns between two consecutive of these black ones there are going to be two columns. And the total number of clumps is two are minus two, and the size of the first compass to our minus two to our minus the two are minus four to our minus five and so until it goes down to one. And the other one goes up because the sum of those two numbers is always to our minus one. This is a two are minus one callable graphs because the sum of any two consecutive columns the number of clumps is to at most two are minus one. There are six are minus five colors columns. I'm giving you just the numbers that are not necessarily one but this is the same thing as I have up there. The edges within a layer and between consecutive layers are there because the number of clumps work out just fine. So what is the size of the clumps in the black single tone columns you have a single tone vertex. So that's just one vertex in the red and the blue columns every clump has size delta over two are minus two which is an integer according to our divisibility condition. And this is what we have in a picture this is an example of our equals to six. Now these little rectangles are there because I'm subtracting one there to make everything tidy. But let's think about what's happening here. Let's pick a vertex that has. Excuse me, you do see my arrow right. You see it. Okay good because otherwise I'm talking about stuff and you don't know what I'm talking about. Let's pick one of these single tone vertices in a layer. It has exactly to our minus one clumps as neighbors in the two neighboring to our minus two sorry in the two neighboring layers. So the size that over two are minus two so the degree is delta here, except here when I have this little minus one if I'm not at the end but then I have a one extra neighbor here so the single tone columns have the vertices in the single degrees delta. If I pick any vertex in these two consecutive columns well in the two consecutive columns I have two are minus one clumps. So that guy has two are minus two neighbors. Among the clumps each cell size delta where two are minus two that's degree delta and except you know if I managed to pick somebody here where I took off one but even in that case they haven't won extra neighbor. I'm on the left on the order right. So the minimum degree is delta everything works out here. What's the difference when two are minus two doesn't divide delta everything is the same. Except when I'm talking about the clump sizes that delta where two are minus two doesn't work out to be an integer I have to make it an integer so I have to put in some floors and ceilings. To tell you in every column how many of the ceilings I put in the rest of them are going to be floors. I pretty much described you everything that's going on. And in order to do that I'm going to say that D is going to be the remainder. Then I divide delta with two are minus two. That's what I need to figure out everything worked out in the divisibility case. So I just need to make sure that when I put in these ceilings. I'm going to put in both neighbors that get that gets a little extra plus one from the ceilings for every vertex is going to be at least D. And the ceiling numbers are symmetric so I'm going to describe what happens in the first half and then you just have to flip it around for the second half. So in the blue layers in the first half. I'm going to put in D minus one ceilings, as long as they fit. As long as they fit because the number of vertices that I have in the column is less than D minus one. I make all of them ceilings in the red columns. I'm going to look at not the previous blue, but the next blue guy. And I look at how many ceilings I put there. And I just put enough many ceilings in this red column so that the total number of ceilings surrounding this single black ones is exactly the so it's D minus the whatever many I put in the next blue guy. Except in the very middle and I'm going to show you what's happens there in the very middle. In the two columns I put the over to floor and the over to ceiling, many ceilings on those the over to that over to our minus two numbers. So why does this work out well, you might have noticed that I put D minus one in the first three here. That's simply because D is the remainder so it just works out the D minus one fits into the first three no matter what you are doing. I've selected my numbers in such a way that if you select one of these single tone vertices, the number of ceilings in the two neighboring columns is exactly D. And therefore, everything works out for those guys. That may be for the first one but then I have D minus one ceilings here and I have an extra neighbor on the side. If it's not at the end and I have a little plus one somewhere on one of these if it's at the end so the single tone vertices have degree at least delta. I picked a vertex in these two consecutive columns of bigger clump size. Well, the ceiling numbers have been chosen in such a way that in two consecutive columns I have at least the sometimes the plus one ceilings. So, if I'm unlucky and I pick a vertex that comes from a column that has ceiling or that comes from that has ceiling on it, even that vertex is still has at least D minus one many ceilings with clumps with ceilings in their size in their neighborhood. So, D minus one is not enough but to the left or to the right depending upon from which column I pick this vertex I still have an extra vertex. So everything works out fine. This is a graph with minimum degree at least delta, which is to our minus one colorable. And as the numbers work out the way I said, this is a counter example to the other spot black to the conjecture. So, this gives us the following update to the conjecture for every K bigger than equal to zero. If delta is at least three K over two and G is a case of K plus one free graph or a big compression K colorable graph or order of order and a minimum degree delta. And then the diameter of G is that most free minus two over K times N over delta plus bigger one. And, or examples give the disk conjecture is best possible. Sing this over a little bit more carefully. If delta is smaller than three K over two. Then the known and sharp free and over delta plus one upper band on the diameter is better than this number so we can't conjecture this for small delta. If K is even and three R minus one divides delta, then this conjecture is the same as the adage pop or lactosa and their construction shows that this is best possible. If the divisibility condition doesn't hold, then well we try the same sort of conjecture for the even K. That gives the exact same upper bound but not sure because I have a constant times delta plus another constant as before so this constant is the best that works for the in the form C times N over delta and I suspect that you could do the same thing with their conjecture. So this constant conjecture is the best possible for K even if you want the upper bound in the form C times N over delta. If the conjecture is odd then all construction shows that this is best this is best possible that that works for all delta simultaneously in this form, but because what we have is not exactly in this form or example is not exactly this form in the diameter it's sometimes delta plus something times delta plus another constant there, the adage pop or lactosa conjecture may still be true in a certain range of delta. It's me because it gives us a better construction and we don't have something that says that the adage pop or lactosa conjecture cannot be true we don't have an example like that. So, I promise you that I can at least give you some upper bound when I change the condition from or maximum click sizes not bigger than K to K colorability so I'm going to use clump graphs for that. We consider a maximum graph we fix a K coloration, a vertex of maximum eccentricity, Li is the set of vertices at distance I from V so I have L0 up till LD where D is the diameter. The clump graph is defined by replacing vertices of the same color with a single vertex and V the vertex with the number of vertices that I put together. I'm going to denote by CI the number of colors used in Li in the K coloration of the original graph, which is the same as the number of vertices in the clump graph. By little Li the number of vertices in Li in the original graph, which is the sum of the weights in the clump graph. So, we impose some extra conditions on our clump graph because we have we can so for any K colorable connected graph of order and diameter D and minimum degree at least delta. There is a connected K color graph on the same parameters, diameter D and layers L0, LD for which the following hold. Well the first and the last layer not only has just a single core but has only a single vertex. So if you are not talking about the very last layer, if I in the highest layer we have a single color then in the iPod's first layer we have at most K minus one colors. This is kind of obvious because in the iPod's first layer. It's the iPod's first distance class. The vertex is adjacent to a vertex in the previous class. Therefore, every vertex is of different color from the vertex it is adjacent to so when there is only a single color that color is missing from the next layer. The next one is the number of colors used in any two consecutive colors is as large as possible. Well how large can it be it cannot be bigger than K because the graph is K colorable. It cannot be bigger than the total number the number of the sum of the number of colors used in the two layers, because even if they are all different, you get the sum, so the minimum these two numbers. That means that when the number of colors used in the IS layer plus the number of colors using iPod's first layer is at most K, then Li and Li plus one doesn't share any colors. For the coloring argument you probably can think it's easier than I can describe it and the last one is if we have a layer with exactly K colors. Then it is not the panel. It's not the last two layers and the next layer contains at least two colors. This is something that we need in some of the proofs that we have. Prove a couple of other things too but we haven't find use for them so I'm not even going to bother with that. So we will say that a comb graph that satisfies these properties is a canonical comb graph. And for the problem that we are considering it is enough to restrict ourselves to these graphs. And I'm going to use a linear programming approach that uses the dual. So in order to do that I'm going to slightly rephrase my question. You can view the original question is that I fix a K, the number of colors used a delta lower band on the minimum degree and the number of vertices and how big can the diameter DB. I'm going to turn this and then the around a little bit so I fix K the number of colors delta, the lower band of the minimum degree and the diameter. How small and the number of vertices can be such that connected K color graphs of order and minimum degree at least delta and diameter D exist. Well, we can rephrase it as an internal integer in our programming problem. Consider a family, the family of unweighted canonical congrats so all the clump graphs that come from this problem that arise from connected K carbo grass with diameter exactly D and minimum degree at this delta but forget about the weighting. How can I assign weighting to this that results in the smallest and smallest smallest number of vertices and gives me a graph that I need. So we fix any of the graphs in this family and assign integer weights to all vertices. The goal is to minimize the total number of vertices to minimize the sum of the weights subject to the condition that I assigned weights in such a way that result in minimum degree delta, which means that if for any vertex of the clump graph if you add up the weights of its neighbors. It's going to be at least delta. So this corresponds to exactly to the problem I talked about this is an integer in our programming problem. And I promise that I'm going to use the dual so the dual is going to be flipping every single round so let's just think about what's happening here. The constant column in this problem has all deltas. The coefficients of the weights are all once. The matrix with which I'm multiplying the weights well that's going to be the adjacent symmetrics of my Canonical com graph. So when I turn this around the constant column becomes the column becomes the coefficients of the weights in the function that I want to maximize so that's going to be all deltas. The constant, the new constant column is going to be the old coefficient so that's going to be all once. And the new matrix is going to be the transpose of my matrix a transpose of any decent adjacent matrix itself so I'm going to come back to the same Jason semantics of the same graph so the dual problem. It's going to be fix a Canonical com graphs, maximize delta times the sum of eight subject to the condition that for any vertex, the sum of the weights of its neighbors that most fun. This is the dual of the previous problem. I wonder what we can say that if you fix any K bigger than equal to three and find constants you and see such that for all D and delta and all Canonical graph in the set of Canonical graph that arise from cake horrible graphs of diameter D and minimum degree delta. The optimum of these waiting problems of the dual problem is at least you times delta times D plus C times delta. Then for any such a graph again, the diameter is at most one over you times and over delta plus some other constant C prime. The fact that this is true is actually actually comes from the city trivial direction of the duality problem, the any feasible solution for the dual problem is at most the value is at most the value of any feasible solution of the original dual problem. And since the feasible solution of the dual problem according to our assumption is at least you times delta D plus C times delta. The number of vertices in any such graph is at least you times delta D plus C times delta, which if you do a little algebra gives us the So how can we use this using this we can show that if cheese a cake horrible graph of minimum degree at least delta, then it's diameter is at least three minus one over K minus one times and over delta. Plus we go one. So, this is not exactly what we want because what you want is three minus two over K, but at least it's an upper bound in terms of K. And this is actually a new upper bound for pretty much okay. We have proved the original condition three minus two over K which for K equals to four. And therefore as a consequence it's too for so for four callable graphs and therefore as a consequence it's true for three callable gaps as well so what do we get for free from this. And then for connected free callable graphs of minimum degree, at least delta, the diameter is at is at most three minus two over four and over delta. So we don't get anything new we get something that's a consequence of the fact that we already know the conjecture for four callable graphs. But as one over K minus one is at least slightly bigger even the half of two over K that we are shooting for the improvement that we have here in the upper bound is of the right order. So by the previous, if I want to show this by the previous theorem, it is enough to aside dual way that every layer has total weight K minus one over three K minus four, and the condition that the total weight of the neighborhood set of any vertex is at most one holds So how do we do that. So we can omit it canonical crump graph with layers as zero up to LD. I need to assign my weight. If in the is layer I have less than K clumps. So we can compute the total weight K minus one over three K minus four, evenly amongst the vertices so the total weight of the layer is K minus one over three K minus four, and because the number of clumsy eyes at most K minus one. Every vertex gets weight at least one word three K minus four, which will help us. So what if we have exactly K clumps we can't have more because graph is K colorable and every clump has different colors. Well, I'm going to divide the vertices in the IS clump into two parts excise the set of vertices that are adjacent to every vertex in the two neighboring layers and why I is the rest of the vertices so why I contains those guys who have at least one neighbor missing from the adjacent clumps for the vertices of excise I assign dual weight one over three K minus four and I'm assigning this other number to vertices of why I I have chosen this number in such a way that vertices of why I get the same weight and the total weight in the layer is still K minus one over three K minus four. So the total weight is K minus one over three K minus four in the layer and vertices of excise I get weight one over three K minus four. You should still have the question how do I even know that this is. This is a weight. In order for this to be a weight excise should not have all the vertices in line. How do I know that that happens. I have K clumps in this layer. If you remember from earlier that means that the previous layer has to have at least two vertices. This K clump. This is a K color graph has all the K colors that I can assign so some of the vertices share color with those two vertices from before so at least two of the vertices of this layer are not in excise so excise the size of excise between zero and K minus two, which means that this non this way that I assigned to vertices of why I is actually a weight. And this also means that because K minus excise at least two vertices of why I get weight at least half of one over three K minus four. So in summary, what I have managed to do with this waiting, the weight of a vertex is at least one over three K minus four, unless the vertex happens to be in one of those by eyes, meaning it comes from a column that has exactly K vertices, and it has at least one non neighbor in one of the adjacent layers. So every vertex regardless of where it's coming from has weight at least half of one over three K minus four. So the total weight of any two vertices at least one over three K minus four, and the total weight of every layer is K minus one over three K minus four. And if you remember that's enough to prove the theorem, as long as I can show that the condition holds, meaning that for any vertex, the total weight of its neighbors is at most one. So let's check that for convenience I'm going to set. So I don't have to worry about how I describe the boundary conditions, I'll mind the non existance L minus one and LD plus one to be empty set with this if I take any vertex X, it's coming from a layer ally. And it has only neighbors in the I minus for the IS and the I plus first neighbor, and you can leave out X from these layers because it's an ally. And it's not its own neighbor. Every layer has weight at most K minus one over three K minus four because that's what I said. So if the dual of eight of X is at least one over three K minus four then the total weight of its neighbors is that at most three times K minus one over three K minus four the weight of these three layers together minus the weight of X which is at least one over three K minus four so it's at most one. So far so good. What if the dual weight of X is less than one over three K minus four. Well then X comes from one of these Y eyes, meaning it has a non neighbor in an adjacent layer, meaning the neighborhood set is the three layers, leaving out X and Y. The total weight of X and Y together is at least one over three K minus four because the total weight of any two vertices at least this much. And the same logic gives that therefore the total weight of access neighbor is at most one. And this finishes the proof of the tower and that I showed you. So the only thing for which we could show something new. And we haven't managed to show it because for the fourth global case we can show anything new we know the conjecture is true. Are the three colorable graphs with a very different linear programming approach we have managed to show that for every three colorable graph of order and a minimum degree at least delta. The diameter is at most 57 over 23 and over delta plus we go one. And we need to show this because if you have been here last year at in a stalk you have seen the proof and you. And this is also long and, you know, tedious. In addition, we show that if the commonly called graph doesn't have in the commonly called com graph don't don't have singleton layers except for the first and last one which are forced to be that way. And actually, the diameter is free mind at most three minus two third and over delta plus because one which is the conjecture thing. This is kind of an interesting thing because if you remember the example that we gave five sevenths of the columns are singleton columns. So, the example that actually achieved this bond as close as we can tell so far has plenty of singleton layers and I'm not aware of any example that don't have singleton layers that gets close to this bond though there might be some. And just for comparison, you know, five, which is 2.5 is the known bond that was proven seven sir, which is 2.3333 is the conjectured bond conjecture constant bond. And this is somewhere in between the punch closer to 2.5. So what's left. What's left is find the maximum diameter of cake or gap or order and and minimum degree at least delta because we haven't found that we have a conjecture. Well, what can you do in this. You can create counter examples to the new conjecture if exist though at this point I believe they don't but you know I might be wrong. You can improve the existing upper bonds we have now three minus one over K minus one times and over the other roughly. You can make better lower bound conjectures then K is odd because if they exist because right now they approach as delta increases. The conjecture bound but for a fixed delta they never reached. It cannot be done so you can prove better upper bonds in the form eight times then divided by be times delta plus C. Also, if you remember, this wasn't the original problem the original problem wasn't about cake or graphs. It was about graphs that have no larger click size than K so do the same thing for the original problem. So, the examples that may be sharp, or we hope to be sharp are all very periodic. Is it true that the examples are all nearly periodic so one of these periodic things or a different one if you've managed to come up with one slightly maybe modified so it's not necessarily periodic. And that's all I finished is there any question. Thanks Eva. Let's all thank our speaker in some way and then we'll go ahead and open it up for some questions. We have any questions for Eva. Oh, I put everyone asleep. Okay, there's no questions. Thanks again, Eva and thanks everybody for making it out. Have a good weekend everybody. You do. Bye bye.