 Hello, everyone. Welcome to the next seminar. It's my greatest pleasure to introduce Aditya Apatukuchi. We'll talk about some inappropriate of hypergraph coloring problems. Thanks a lot for having me. Is this audio? So everything I'll present is this joint work with, based on joint work with Pair Austin and Amai Bengali. Pair is currently in KTH Stockholm and Amai is in U.C. Riverside. I plan to present proofs in more or less the almost all details. So please feel free to stop me at any point and I'll take care of this. I just want to make any point if anything is not clear. Okay. So I guess some basic notations. So for a set, for a collection of K sets of 1 to N is called a K uniform hypergraph or a K graph for short. The set N are called the vertices and the elements of H are called the hyper edges. So I'll forget to say hyper once in a while. And the hyper edges are denoted by E, F, etc. In fact, I'll only use E to denote a hyper edge. So that's some basic notation. For the most part in the stock, K is something for the most parts. K is large but fixed constant and N is growing. So K is, okay. So that's the definition of hypergraph terminology and the colorings of hypergraphs. So a C coloring is just, I'll use K to denote a coloring and the elements of the set are called colors. K is proper if every edge has at least two colors. So in words, no edges monochromatic. K is rainbow if the edges in every, sorry, the set of colors in every edge is the set of total colors that you have to work with. So that means every edge sees every color. So that's the, so this is the definition. Last bit of definition, I guess. It is C colorable if there is a proper C coloring. And it's a C rainbow colorable if there's a rainbow C coloring. So again, so these are just definitions. So here are how problems look like. So here's a problem template. The input is a C colorable K graph. So you have the promise that it is C colorable. And you want to output a proper decoloring of this hypergraph. So for such a coloring to always exist, you at least need to ensure that D is at least C. So then you can ensure that such a coloring is always there. In fact, so the case when C is equal to D is essentially the question is at C colorable. So if you can do this question and you can do you can, if you can do this, you can do this in pretty much the same time. Okay. So, okay. So this is a famously NP hard problem. So it's NP hard case to that means it's a graph, and you're talking about at least three colors or it's NP hard if it's a, if it's a hypergraph of higher uniformity and you're talking about at least two colors. Okay. So we're interested in like, yeah, D is more than C. For now, for example, let's say K equal to two. Even here, recently, like the best known results, which are fairly recent, as you can see, there's NP hard to color a graph, a C colorable graph with the colors of these at most to see minus one. The better bound is it's NP hard to color for these at most C choose C over two minus one. So this looks like the better bound, but really this is worse for small C. So the best thing that we know for graph is that it's NP hard to five color three graph. So this is kind of state of the art for graphs and upper bound you can color it using some end to the point, I think one nine colors. So, right, so we, so we care about higher values of K also. Okay. It's NP hard if so I'm going to show you a bunch of results. Pick them out with I thought they were somewhat relevant and they're kind of local maximum of this type of results so I'll just present them. So we know that for K equal to four, four uniform to colorable hypergraphs, you can't carry with log into the omega one colors. It's NP hard to color, then this is by Bengali for K equal to eight and C equal to two or K equal to four and C equal to four. You can't color it with fewer than two to the poly log in colors. For K equal to three and C equal to three D needs to be. Yeah, for K equal to three C equal to three, you can't color it with some essentially log in many colors. I put a star because it's not really NP hard to quasi NP hard but it's for for this talk I don't I wanted to distinguish between them. And I guess one, one range that's of interest is the K equal to three and C equal to two and then D equal to. I'm sorry. I need some some epsilon over here. Okay. So essentially yeah so this this result by the new reggae and smite this is what will be interested in fact it's not the result itself but it's kind of the technique that they use which will be the basis of a lot of the proofs that I will talk about today. Okay so so so that's that's coloring. So here's another problem template. So now the input is a C rainbow colorable k graph so everything is same except for this you have a stronger guarantee on the coloring and you want to put a proper decoloring of it. So, this was studied in the con like this was studied as a weaker notion, a much stronger notion of colorability first in this paper by Austin Wilson and also. And just as a sanity check. We can check that he has an observation. So if it is C rainbow colorable right sees more than two, then it is to rainbow colorable. And that's easy because for example, say, suppose this was an edge and this is how you rainbow color it. So you can identify the first four colors with just the color red and the last three colors just a color blue and that's a rainbow colorable graph to rainbow color was exactly the same as to colorable. And if it's too colorable then it's also decolorable for the more than two. So this is definitely something that exists. The point is to find it. Right, so so so that is the problem template that I'll talk about. And this problem, they'll call RC for rainbow color K as in uniformity C as in rainbow colorability and D as in the output number of colors. These are the problems parameterized. Here's one range of parameters where C equal to K like it's K uniform K rainbow colorable hypergraph, and you're looking to color to two colors. And this was known to be that you can do this efficiently. This is by McDermott like this was randomized algorithm, which was later made deterministic. So so this is like the extreme you cannot possibly have K plus one rainbow color K uniform graph. So in this extreme case, it's a it's a it's efficiently doable. And in fact pictorially this is how it looks. So what is a KK what is K uniform K rainbow colorable graph. So you have a bunch of color classes. So yeah, so so these are the color classes, and every edge contains exactly one vertex from each color class so they just the transverses over here. Okay, and what the previous result said was that if it is carrying book color that is if it's k part time. Then you can efficiently to color it. Okay, so okay, I guess I'm, of course a caveat is that you don't know the colors, you don't know this partition yet so that's kind of the difficulty here. Okay, but in this particular case, you can efficiently to color it. So so this is what we know. And this is a, it's a conjecture by break and say can go to me just that this is pretty much the limit, which is just that if you take a K uniform K minus one rainbow colorable graph to coloring is empty hard for K at least three. So I say K at least three I really mean K at least four because three is easy. It's easy and since it's known. Okay, and towards this like, I guess some kind of partial progress is if you have a K uniform graph, which is K over two rainbow colorable coloring with any constant number of colors is empty hard. Okay, but but this is a much weaker condition than this. And so, like I said, K equal to three is easy and since that this is just the two colorability of three uniform hypergraphs that there is a dimension in the very first part. So really, for this was the smallest unknown case. And that that that was where like, so that was the first theorem this joint work with the pair and so the previously smallest case is now taken care of and to try and extend it, it works for K uniform K minus order square root K rainbow colorable graphs to coloring is empty hard. So this is supposed to be K minus one if the conjecture is true. So it's it's K minus order square root K. So, so. Right. Okay, and the proof. Kind of follows the new regards my framework and I'll explain the details and explain the proof in more or less full details so just stop me. So here are some again some basic preliminaries so the reduction is from a label cover instance. So label cover means that yeah you have a bunch of variables, which so these are the variables. And these are the constraints between variables and so they have this a bipartite graph structure for now. And right so the constraint is called five. So this is a special type of constraint but the set of constraints is called capital five. And this is the. These are the. So this is how the label cover instance is defined. Every x in Xi, which is Xi means X one or X two, every X and Xi takes values from some range. This kind of a specified with the problem. And in this case, these are also projection constraints in the sense that for every assignment to X, there is at most one assignment to why that satisfies this constraint. So these are called projection constraints. And I guess, if I say X gets a, then I just define its projection to be B. So that next time, I don't have definitive reading. So, in fact, the instance is slightly more complicated. You have a bunch of, you don't just have two layers, you have layers of vertices, and you have some extra conditions on the structure of this graph also. So, so the relevant theorem is the following. So this is a label cover instance. So this is a theorem. So this is again, then the new reggae's Mike and then we're going to show me what they gave, which is for L greater than two and R, which is, I mean, for fix a number of layers, and R is kind of the range of every alphabet the range of the variable that every variable can take. And given a label cover instance, it's empty hard to distinguish between their existing a satisfying assignment or their every assignment satisfies at most a tiny fraction of constraints between every two layers. Okay. So, so this is the theorem. This is the hard problem that we reduce from. In fact, so there's a bit more detail than I would. So there's a bit more detail than is really relevant for this group. So I just give a kind of a stripped down version. And then you can believe that that is a hard problem. So, so like, for now we're just sticking with two labels is just two layers of variables. Given a label cover instance with ranges in R1 and R2, which are both reasonably large, it's NP hard to distinguish between their existing a satisfying assignment. And every assignment satisfies that most some say one over our poly R fraction of constraints. So this graph looks, think of this graph as some pseudo random or complete graph so you have, you take subsets of vertices and subsets of variables and they have as many constraints between them as you'd expect so roughly speaking. But but for the details that I'm giving that this is really good enough. So I guess the main thing I would like to showcase are the gadgets. So, here's a gadget for for the rainbow for this special 432 case. It's a following the vertex set is just 012 to the end. It's all 012 to the end. And this is the hyper graph. This is the gadget R3M, which is, you know, it's a three uniform hyper graph on the vertex set 012 to the end. Such that the hyper edges are defined if the UVW form a hyper edge, if the following happens for every INM, you see at least two of these three alphabets in every coordinate. And for at most one coordinate you see exactly two alphabets. So just to make it concrete. Here's an example. So here's UVW. You see all three alphabets again like all three, all three, all three, all three in this exact one coordinate usually two alphabets and everywhere else you see all three. Okay, so this in this case this is kind of called the noisy coordinate. You have at most one noisy coordinate that is a hyper edge relation. So this is easy is that the independence number is at most two, two, two thirds, two thirds fraction of the vertices. Because if you take more than two third you see a coordinate with all three repeating. Yeah. So, so this is easy. What is was not easy for us. Again, it's, once you know how to do it, it's pretty simple. It's that this is not to colorable. And something that we don't know how to do but I think is really interesting that something to do with in approximation is that the independence number is actually less than half of the vertices. It's the hamming ball that goes just the maximum independence. It is a hamming ball. So whatever, however big you can make that is the independence. And this has nothing to do with. I mean, the, again, the lower, I'm sorry, the lower bound is easy. And the conjecture implies whatever this theorem is. So that's not how we prove this theorem finally. So, I guess that's a hint. It's a nice problem that I like regardless. So here's a reduction. So, I guess I mentioned at least one more reduction and they kind of follow similar template. Again, so yeah, so these are close to a DRS. So, right, here are your two layers. So what you do is the following. The constraint x, y, like to pick two variables that are joined by constraint. So firstly, replace every variable with a replace every variable with a copy of this type of the cabinet. Okay, so this is how I represented like so when I talk about x, the orange hue is the copy of the hypergraph associated with that vertex. So consider three vertices in this hypergraph and one uvw and one vertex z. Right. So you put a hyper edge if the following happens. Firstly, these three form a hyper edge themselves. Okay. And secondly, in every coordinate ui, it's a uk, vk, wk and z projection of k, you see all three alphabets. So, in the hypergraph in all but one coordinate you see all three alphabets, but whatever coordinate that noisy thing is it's made up for by its projection on the z vertex. So, so that's how you place the hyper edge. Okay. And the completeness is pretty easy. It's a following. Suppose you have an assignment that satisfies all of them have to show rainbow three colors. So, suppose you have an assignment that assigns a to x. Look at the coordinate a you just assign whatever is here as a color of you. You color it zero one or two, and you just assign whatever is here to the color of that vertex. So, firstly, let's just verify that this works. Consider a hyper edge. Firstly, this must form a hyper edge in a three itself. So, in all the non noisy coordinates. If a is non noisy, like that that means you see all three colors right here. But if a is noisy only see two colors, but then because it's a projection constraint, the remaining color is made up for by this vertex here. So, so, so you see all four. That's strange. I'm sorry, this seems to be a technical difficulty. So, for the construction, you a w and z the projection of a gives you all the three colors, which is what you need for the, which is what you need for the rain bonus. Okay. So, so that's a completeness. And there's a soundness. So you have to show if it is too colorable, then there is an assignment satisfying at least some say one over 64 fraction. So, in the proof sketch it'll be more like half but some constant fraction of constraints, more than what the within the range of the table cover theorem that we use. So, suppose, okay, so here's your constant. So, this is how you consider the two colors are red and blue, which is this color and this color, these are the two colors. Now, we know that this hypergraph is not too colorable. So it must have a monochromatic edge. So, so this is a monochromatic edge you want to be one and consider three such I mean, like, for every vertex you have a you want to be one of you to you have a monochromatic edge, and even for why. So I guess one thing is that I've drawn all of them in the red color. And again, so that is a detail that's really by standard averaging and pigeonhole architecture can get that like you can assume that all of the monochromatic edges have more over all of these vertices have more than one third red. So all of these clouds have more than one third red vertices. So the point is it's not just to color you have a monochromatic edge, because the independence number is two third any color class more than two third has a monochromatic edge. So you have, if you might as well assume that you have a good and this is really what will be a problem, something you can't assume when you go beyond two colors, as a kind of a spoiler. But let's assume that this is how your situation looks, you have X1 X2 X3 all of them are connected to why, and they all have red monochromatic edges and all these clouds have a good fraction of red vertices. Right. And so, and consider, sorry, yeah, and consider these such that it forms an edge with all of these monochromatic edges. It forms a hyper edge in the construction with all these monochromatic edges. So, let a1 a2 a3 be the noisy coordinates in these hypergraphs. So this, again, believe, suppose they, they don't have it. It's the easier part. So suppose they have a noisy coordinate. Let a1 a2 a3 be the noisy coordinates in these hyper edges. The point really is that when you take the projections. Suppose they project onto distinct points. So suppose this is a1 and suppose a1 projects here, a2 projects here, a3 projects here, like they have distinct projections. The point is every projection places a lot of restrictions on Z. For example, any red, so for, yeah, let's suppose that these are the actual coordinates here. A1 has 001. That means if a red vertex, any red vertex cannot have two here because then that would form a hyper edge. So all the red vertices must have either zero or one in this point. And similarly, so which is what this is. If for any z prime, which is red, the projection of the first coordinate cannot be two because two is missing here. The projection of the second coordinate cannot be zero. The projection of the third coordinate cannot be two. So this means a fraction of red edges is at most a 2 third cube because you placed already three restrictions on it. But when you assume it's at least one third, that's a contradiction. You can't have that. So you only project on to at most two values, but two is not very important. You present some onto some bounded number of values. And the point is you pick right and here's how you assign, here's how you assign, find an assignment to the vertex. For every excite, just assign it to its noisy coordinate that was part of the analysis. And for why you just pick randomly among these you satisfy at least half the constraints on in expectation. So that that's what this is. So you get excite the what what is dictated what what you analyze as a noisy coordinate and why is just one of the two things are at random. So the probability that a constraint is satisfied is half on expectation is at least half. And that is the end of the proof. So I guess it's a question so far. So the only detail that did not give are how you get one third red edges and how you just managed to remove the extra the other but but those are just standard averaging arguments. Okay, so like I was so since I was trying to kind of advertise the gadgets anyway. So this was the that was a gadget for 432 right so the gadget for higher uniformity looks as follows. So I think I confused D and K. But but anyway so for higher uniformity this is a gadget that works. The vertex set is again some one to D to the end so the vectors which are which are over alphabet of size D, and you have a D uniform hypergraph. And the hyperage relation is given by as follows if this so D minus. So D minus all the coordinates you see at state J. All the all of them must be less than R. So in other words like pictorially how does it look. So when you write down x1 to xd, you want to see all D alphabets in every coordinate. So the total number of missing alphabets summed over all the coordinates is at most R. So suppose here you only see D minus one here you only see D minus one and so on for R. So that's it. Every variance you see D. That's hyperage. And I guess the point is that this is also not too colorable and it's used in a somewhat similar way to the hardness that I was talking about. Can I have a question about a budget. So R equals one and the right parameters for the for the rest is the same budget as before right. Yes, R is R1 and D3. It's the same thing as before because you only see one missing coordinate in total. Okay, so, so really so so the hope was to show the hardness of two colorable right so the hope was to want maybe try and increase this two to three like that that was the hope at this point and yeah. Unfortunately we have only partial progress towards that and I'll describe the partial progress. So you don't really have rainbow colorable as a as a promise you have almost rainbow colorable. So this is what CS colorable means, which means you can color the vertices using C colors so that every edge sees at least as colors. So this is somewhat of a generalization of rainbow coloring. So, most every edge gets most colors so that that's how we should think of it and essentially pretty close. And by the way, so the problem template is as follows, given a CS colorable hypergraph find a property coloring. So here you have to, again, think of C as close to D so that the existence of this. Sorry, C close to S so that the existence of this guarantees this because as student that's that's not obvious either. So let's name this RC as in rainbow coloring K uniformity CS colorable and output T colors. So that that's kind of thing. Yeah, by the way, if S is equal to C, this is just C rainbow colorable. So, a theorem we proved here was, again, RC, K, K minus square root CK K minus three square root CK C is NP hard. So you can increase this at the rate of dropping C more than dropping S. So, so that that that was a second that was a second result. And so, again, this is also I'll describe the gadget and I think the difference, the reason it was the reason it's not R's KCC is that probably the gadget should have been better. But, but anyway, I'll describe the gadget. A slightly simplistic one. We again so the vertex it is again a in length strings over alphabet of size D, and it's a D uniform hypergraph if and the edge relation is is if the number of coordinates where you don't see all D is at most D. So the number of coordinates. Yeah, so Yeah, so yeah, so I'm not counting what alphabets you see and what you don't see but just the number of coordinates. So you don't see if you're at most because this technique cannot distinguish between these things. And more specifically the bound on the chromatic number cannot use this extra information. So this is this is a this is a theorem, which is that this hypergraph is not. So, as you increase this noise, the chromatic number increases linearly. It's not decolourable for the data than to prime for any odd prime. So this. Yeah, so this uses kind of this uses like some similar ideas as low versus proof of the Knesser graph chromatic number. It's pretty interesting. I don't know. I don't know. So, okay, forget the dependence on D, the dependence on P, P square. So unlike in the previous case, I don't really have a strong conjecture here, but peace. Yeah, this seems to So there's a gap. It's not this. It is this and so in fact, even D equals two is interesting here. So, okay, anyway. Yeah, and I think this was actually a known conjecture. I just I couldn't find any source for this, but I think this is a known problem and same kind of thing manifested to here. Okay, so that's all I have to say about rainbow coloring. So here are some open problems. So the first is this main conjecture that was on that that has been unsolved as of yet. And this is that now the smallest open cases K equal to seven because in a in a somewhat recent work by Griswamy and something like they take care of five and six. So whatever it was presented. So whatever we did with with a pair and a man, it doesn't work for five. But yeah, it can be made to. In fact, they do something much more general. So you don't want to just color the graph you want to output the rainbow coloring of the graph. And the second thing which I think should be more easy is this one just just instead of CS just just remove the almost rainbow colorability. Okay, so so that's that's that's what I wanted to say about rainbow coloring. Yeah, can you go back a couple slides to the square root secret even more that came in a square root secret. So it's a dependence on C required there on the on the completeness part. But does it come from the fact that you just need the decoding to be guaranteed. Right, right, right. So I mean like you're asking is something like is something like a K minus one C. No, no, I was talking about the general problem where you mentioned that you don't need to use every color in the edge. So you had K minus square root CK and K minus three square root secret. Yeah. So there is the dependence on C needed in the, like can I get something like K minus root K and K minus three root K for every secret. Or does it come from the guarantee that's needed. Because usually the analytical results give that any C coloring is not possible, right. And that she doesn't appear in the completeness usually. Yeah, because they come with an independent set guarantee to write usually. Yeah, yeah. So these things can't these things don't get it. Yeah, so somehow you do think that he has to appear. That I have. So there are no low bonds for these low bonds. So I think this is also something that I know at least some people believe. Okay. So, yeah. Yeah, okay. Thanks. So that's about rainbow coloring. But, okay, but I'll show you uses the similar kind of proof technique with other kinds of gadgets to get other reductions. So the results won't be state of the art they'll be kind of subpar but they're new and hopefully they're Yeah, the proofs are new and there's some hopeful direction for improvement like again like as long as it's just a matter of finding a gadget. So let's get back to problem template one and I guess yeah I should show I'll just show one example of this. And let's stick to four uniform, three colorable graphs and you want to output it's not rainbow anymore it's just four uniform and three colorable graphs, and you want to output some poly log and coloring of this. Okay. So, in fact, the NP hardness of this is known for C equal to two but like in different ways. Okay, actually, yeah. Okay, if I replace it by two even there the proof has like similar ideas here, but I'll just show it for three to show different types of gadgets. So this is a gadget. A set just a set subset of 012 to the end. Okay, and that's a gadget what is the property I want from this is that for every subset of T of size at least delta s delta some parameter that will be tuned. There are two points that agree on fewer than T coordinates. This is kind of, this is a gadget here. So now you'll get, okay, as a, I guess, spoiler here you'll get independence and guarantees because of things like this. Right, the gadget is the fault. Yeah, so this is the gadget you have a subset of one to three to the end. So it's that for every large enough subset of this, there are two points that don't agree on that agree on fewer than P places. Yes, the ages agreement right which is n minus the having distance of between ages and a number of points where they match. So what's an example. S is three to the end is an example of this. And it's not at all obvious this one, the fact that this is an example this is a theorem by Franklin Tokashika and I'll just read a question. This type of, so this type of conditions are seem to be pretty hard to prove for three to the end, but the hope is that they use really different techniques so okay. Again, I'll describe the reduction again briefly here. It will be more or less the same flavor. So, yeah, so we start off with a label cover instance it has some kind of soundness guarantees and all the expanded like all of the previous things and only two layers. Oh, by the way, it's also like a caveat so all of this is going to be it's quasi and be hard. But like I said like for the sake of this. I won't distinguish between. I won't make the distinction some kind of empty harnesses. Okay, so here's a deduction. Okay, so, so, consider two variables or in projection constraints in the same kind of soundness etc. And now consider two variables X1 X2 and with job with a common neighbor why, and put a copy of S again like similar to the previous case put a copy of the gadget for every vertex. And look at you one you two and V1 V2 for every. So this is that the opposite way. So, right for every and for every and be whose projections match pick you one you two V1 V2 where you see at least two entries in these four places. And then you put, and I guess I did not draw that thing and then you put a hybrid between these. So for every A and B that have the common projection, find two vertices here to what is there with that show you at least two different points and you put the common and put in itself. Again, so when I tell the problem the completeness this kind of clear. Okay, so, again, suppose you have an assignment. I'll tell you how to color. So, so let A be an assignment to what this is over of you and me, and the coloring is exactly as you'd expect it's just color every vertex you with whatever you see here. Because like, you see at least two different values in both of these things so you can't have monochromatic edges. So that's the completeness. So the soundness we kind of use this agreeing property. In fact, so so this is the you get, like I said you get an independent set guarantee you get that you can't even distinguish between being three colorable and having independent set of size or something. Okay, so you have to show that the independence fraction is at least delta, then there exists an assignment satisfying at least some fraction of constraints. The way to do it is the following. You have. This is how the, this is how the hypergraph looks. So, I'll use this pink color to denote the independent set. So if you have an independent fraction of at least delta by similar averaging arguments, just assume that it takes a delta fraction of every cloud. Now, now use the fact that if you take at least a delta fraction of the cloud, you have two points which agree on fewer than T places. Okay, and in every two of them. So in every one you see two points that agree in very few places. And moreover, the projections of these agreements must pairwise intersect because otherwise they form a hyper edge. I guess the punchline is that if you have an intersecting family, some vertex must have a high degree. So that means some assignment here is present in a lot of projections. Just, again, if you pick randomly or at least set that fraction of satisfy that fraction of increase and it's a very similar argument. So that's, I guess, without, so it's just the skeleton of the argument without any details. Okay, and this is pretty, yeah, it's pretty simple, I guess. So, here's the main open problem. So what, so what stopped us from getting better bounds on the coloring is the size of the gadget actually. So if you can construct gadgets with smaller size that have a similar property, then you actually get a much better balance. In particular, if you can find such a gadget of previously the gadget was something like three to the right. If you can find this kind of gadget on two to the polylog and many points, then you can take D. So, so K equals four C equals three, and D equals something like two to the polylog in. So, yeah, so I guess at this point the main hope comes from the fact that these techniques are pretty different and so this stuff to be done. Right. And that's, that's all I have to say thank you. Yeah, thank you for the talk. That's really nice. Are there some questions. Andre. Yeah, well, I guess, lots of your your gadgets, they have a very similar structure. The vertices are on tuples. Yeah, your domain and your condition is that in every coordinate something is satisfied. That is very close to the definition of a polymorphism. Right. Which is like this right you exactly this you have on tuples. And then you have a collection of tuples and every coordinate some conditions is satisfied and then the the output values are supposed to satisfy some some condition. It's it's it's not exactly that right because of noise. Yeah, it's it's very close. Would that also capture this kind of thing for example, because I'm only talking about a subset of the zero one to the n tuples now. Yeah, not sure about this. What do you think. Yeah, I can see it. I don't know. But I'm like confused about the uniformity here like if you want to make it into a polymorphism. No, yeah, I'm just like thinking how does it connect to this polymorphisms I don't I don't really see it. Maybe just two values. I don't know. It's it's kind of that your enemy polymorphisms. They have the range that depends on and somehow right. So, yeah, binary polymorphism might output, you know, four possible values churnary polymorphism eight possible values and so on. Because that's not necessarily a problem. No, well, this because this is the depends on on and which is the idea. Yeah, yeah, yeah, we don't know how how this business works with polymorphism you know when one is not a constant but you can grow them. Some other questions. Yeah, so maybe I will have a question. So this, this kind of topological argument. In your proof it's used in this. Nezah Nezah like result right so you're you're analyzing the gadgets. How does, how does the proof compare to the Buddhist. It's everything is a generalization so can you look for anti portal pairs of points right. So instead of that you have a sphere and you define some free group action, and you want a point orbit whose orbit should have be the same, have the same value. And so that kind of forms a hybrid that that means you have a monochromatic hybrid. So, with that kind of generalization it's the exact same. Okay, cool. Thanks. Another questions. If not, let's thank you again, virtually. And like if you want to socialize again, the meeting is open just admit yourself. I should probably stop the recording.