 who let me introduce the speaker. Today's speaker is Alex Brands and he will talk about from his side on non-burien domains. Sorry, yours, Alex. Yes, thank you, Jacob. Thank you for organizing this seminar. It's been nice to be able to talk with the CSP community. I will say that my dorm Wi-Fi is not super good. So if something bad starts happening, please try to alert me by whatever means you can. I realize it may be difficult if the Wi-Fi is acting up. Let me just share my screen now, I'll start the presentation. So, yes, I'm gonna be talking about essentially a dichotomy for argument of PCSPs of the non-burien domain. This is joint work with Martin, who is here, and maybe joining later, also both at Oxford. Where did this idea start from? Well, we know that two-set is possible in real-time and three-set is an NP-hard problem. And there've been all sorts of variants and classifications and variants of these problems. And in this particular case, we'd like to know what happens in between two and three-set and whether there's a nice way to formalize this. And in a work by Austrian, they did find one natural way to generalize this into a promise problem, which is to say, which was to say, suppose now you're no longer distinguishing between satisfiable and unsatisfiable formulas, but very highly satisfiable formulas and unsatisfiable formulas. And in particular, the two cases we're gonna distinguish between are those where there exists a true assignment satisfying at least g-literals in every clause, rather than just one as a problem. And the formulas that have no satisfies, is that there are no formulas in our instances of these two cases. This problem here, where I should explain the parameters. So one g-case that means, is there an assignment satisfying at least g-literals per clause? Or is there no assignment that satisfies even one literal per clause? And they were able to show that this is NP-hard as soon as the fraction of satisfied literals that you're guaranteed is less than a half. And in a sense, this puts the hardness threshold between two set and three set, right next to two set and not near three set. So now our question is, what's the natural way to accept this problem to a larger domain? And we devised the one g-k set-set problem. So what does this look like? Don't mean size D. So this can be thought of just the integer one to D. And we have literals, which are indicator functions of some set. And in our particular case, we're gonna take the indicator functions of all elements of the domain, except a single element. And there are a bunch of simple reductions that show that this is the most interesting case. This is where the core of the results come in. And now, as before, we construct these KCNF clauses. So we have our literals here, F sub A1, which forbids X1 taking the value A1. As sub A2, sorry, the subscript should be a two. This forbids X2 taking the value A2 and so on. And our formulas are just gonna be conjunctions of these clauses. And similarly to before, with the Boolean problem, we wanna distinguish the two cases where your formula is highly satisfiable. So you can satisfy at least G literals in each clause. The formula is unsatisfiable, so you can't satisfy even single literal. And our main result is that it's similar to the Boolean result in flavor, that if the guarantee on the number of satisfied literals, you can have it in a clause that's high enough, then the problem is tractable, and otherwise it's NP hard. So yes, here, which again shows that our literals, are indicator functions of sets who forbid only a single domain value. And we can see that if we take S equals one, we recover the Boolean result. So in that case, this fraction here becomes one half. And then what does it mean to have a literal that forbids only one value? Well, it means it forbids either true or false. And you have domain size too, so that's exactly. The easier part of our result is to show tractability. And we have this randomized algorithm, which is actually an adaptation from an algorithm for two sets by Papa Dimitriou. So what does it do? It sort of does the obvious thing. You just choose any arbitrary assignment to your formula. And as long as the formula is not satisfied, you just choose an unsatisfied clause and you flip one of the literals. Yeah, we arbitrarily pick a clause C, choose one of its literals. And now there are many values that we could flip XI to, to satisfy the literal, because literal forbids only a single value. And we just randomly choose any of those. It doesn't matter. In the Boolean case, you've only got one force, but here there are many. And so what is the analysis of this algorithm look like? Well, suppose that we have a G satisfying assignment X star. So that means an assignment satisfying at least G literals in every clause. And then let XT be the assignment obtained at iteration T of the algorithm after repeating while loop for T steps. And now we claim that as T increases, the distance from XT to X star and expectation decreases. And this follows from a simple calculation, which depends on this inequality here. If you reverse the inequality, then this is no longer true. And that's why the algorithm doesn't work in the hard regime. So essentially we have a bias random walk, which reaches our assignment X star with constant probability in order of n squared steps. And you can make this probability fairly small and just change the constant here. One thing that's also interesting not only does it find a one satisfying assignment for the formula, but it actually finds something much stronger. It finds the G satisfying assignment X star. And this is something which actually appears often in the problem of CSBs. Is that when you wanna show tractability, you show it for the search problem, which is at least as hard as the decision problem. And when you wanna show partners, you show it for the decision problem. And it's not known in general, I believe one of the two are equivalent. So yeah, that's about like 30 years old now. Well, there are in fact, so we can look at what polymorphisms we might have in the easy regime. And in the Boolean case in this regime, there were many majority polymorphisms. Now majority doesn't immediately make sense over domain size more than two, but there's quite an easy generalization. Let me come back to this point. So plurality functions turn out to be polymorphisms. And what do those do? Essentially, they return the value that appears most frequently in the input. If there are ties, in this case, we just break that function becomes so much rougher that all polymorphisms set is quite important to pardon this proof. This just means that a polymorphism must return one of the values that it's input, which functions the best. So when G over K is strictly greater than S over S plus one, we have plurality polymorphisms of all arities, which means that the BLP, like your programming relaxation applies, you can solve it. And when we're right on the threshold where G over K is equal to S over S plus one, we actually don't have symmetric poles of all arities. Because you get some difficulty breaking ties, there's no good way to resolve it. There's other problems that have polymorphisms of infinitely many arities, symmetric polymorphisms of infinitely many arities, and that's the problem. In general, the question we wanna ask is, how are the polymorphisms limited in the hard regime? And if we look at our class of plurality polymorphisms, which interface nicely with the trackability, it turns out that these now have only bounded essential arity, but the problem still has other polymorphisms on bounded essential arity. We can't use them to derive hardness. That's what's done in the Boolean case with two plus S. And it turns out we can also show the polymorphisms of this problem don't satisfy other sufficient hardness conditions that we use elsewhere such as fixing sets or avoiding sets or even the quite strong condition of Epsilon robustness or sort of special conditions like lack of Olshek polymorphisms. So really all the known sufficient hardness conditions didn't hold for this problem. So we need either come up with a new hardness source or a new kind of property of the polymorphism. So what worked is this notion of smug sets and a smug set is a set of coordinates of your function whose values are equal to the output value of the function. So the set of I such that VI equals the output F of V. The reason they're called smug is that I guess they're proud or they're happy that the function has chosen their particular variable. This means I'll walk you through all the variables in the setup. So here in the case with K equals five. So we have five CNF formulas and the columns correspond to assignments that satisfy a clause. In this case, our clause is a disjunction of these literals. So the literal forbidding X1 equal to three, X2 equal to two and so on. And we can see that if our function F returns the following values. So on this row, it returns three. That means that these coordinates are smug sets. This, sorry, this group here is a smug set. And the smug set is exactly the set of coordinates equal to the output. We'll take in a subset of the first two coordinates would not happen with a smug set. And similarly, in the second row, this B2 unit of these two cells. And now we can see maybe that there will be a connection between the smug sets of a function polymorphism. And in fact, it's polymorphism if and only if some condition in the smug sets that we don't have any multi-set of smug sets of K-smug sets for F's coordinate is contained in that multi-set. Now I'll just sort of illustrate in this diagram what proposition holds. So you can see the output of the function here. Three, two, one, three, three. If our clause is the clause on the right hand side then it would not be satisfied. It would not even be one satisfied by inputting these values for X1 through X5 because you've set X1 to three but now you forbid X1 equals three and so on. So you're forbidding all of these. So this is exactly the case where F fails to be a polymorphism. And now what remains to check is that these were all legal inputs to the function which means that these are all G sets, fine, fine. In this case here, let's take, I don't think it's indicating, let's take G equals two. So we can check that each of these columns satisfies at least two literals in the clause. I mean, you can check just by checking all possible combinations or you can realize that the literals that won't be satisfied are the ones contained. So for example, these two here won't be contained for this assignment if third column if it's input to our clause. The first two literals will not be satisfied because they're contained in smoke sets but the last three literals will be similarly for this column. So this three satisfies the clause, three satisfies the clause and so on. This column here has four literals satisfied over the clause. So you can see that in each column here we put that note to smoke sets. Which means that all these columns form a really satisfying assignment. And that's G equals two, where G equals three is formed, legal inputs to the polymorphism is not even one set, so that explanation was clear. Please do stop me if you have a question. Yes, smoke sets provide new property polymorphisms to study and then we also required a new hardness source which is variant on the classical label cover problem. Achieve by adding layers of variables. So what does this problem look like? We take layers of variables, all ranging over here is one through M and our constraints are functions from a variable in one layer to a variable in a higher layer. And we say a constraint is satisfied by some assignment, sigma, from the variables to M. If the value you assign to variable Y is equal to the value assigned to the variable X after applying the constraint function. And now, suppose we have a sequence of variables of length L plus one that all have constraints between them. This is something that we're gonna call a chain of variables. Final definition, I believe, say a chain is weakly satisfied if at least one of the constraints in the chain is satisfied. And this is the hardness condition variant of label cover, which is similar to the other label color statements, cover statements, saying that it's not be hard to distinguish strongly satisfiable instances from those where not even a small fraction of the chains are satisfied very weak way. Excuse me, can I interrupt? Yeah. So for a chain, I guess you also want the condition that the composition of phi from X is zero to X two. Well, this phi from X zero to X two is the composition of phi from X zero to X one and phi from X one to X two, right? So these fies cannot be arbitrary. Yeah, I believe so because otherwise you could have a totally contradictory setup. I mean, that might be easy to detect in terms of complexity, though. I'm not entirely sure about it. You are saying that it's not needed to put the assumption here. Are you okay? I need to think about it. Okay. Thanks. Yeah, thanks. I need to think about it too. Prove that every polymorphism and the smug set of that most k-coordinate f has no more than l pairwise disjoint smug sets. Smug sets are preserved under pre-image liners. So this here means if G is a smug set of G, then you can instead pull back S and you'll again have a smug set. Interesting to note here is that the actual definition of smug is irrelevant in terms of getting hardness as long as it satisfies these three conditions. So when tackling other problems, one could just come up with an appropriate definition of smug. And then show that it's like these three and hardness will completely follow. So now three properties. So before I go on, I'll say that in our problem, the first property is very involved to show and the others are quite easy. So I'll give an explanation for the second two and the first one. I'll just mention essentially what this says is that for our problem, the number of disjoint smug sets is bounded by constant, which depends on the parameters k and G. Okay, let's suppose you have more than one. So we want to build up a multi-set containing each of these smug sets, bring back the earlier proposition since the proposition slide with the diagram is larger than k over k minus G. This allows us to take multiple copies to the stacks of our disjoint smug sets and then achieve this property here of having k smug sets of f and then that contradicts the fact that f is a polymorphism. Now, this property is also not too difficult to see in our case. So what do we say that G is the minor of f? Well, to compute the function G, you take its values and you plug them into function f according to the minor pi. So here's a picture. Pi takes values from G. It always looks like the direction of application is slightly backwards. Now, why do smug sets of G end up being preserved by pi inverse or pre-images under pi? Well, let's suppose that the last two coordinates of G form a smug set. So this just means there's some input where on the last two coordinates, the value is equal to the output of the function, which in this case is A. So these two values will be A and here we're forced to take B and see if it's different from A because the smug set has to contain exactly the same values equally. Okay, and now we take the pre-image of the set under pi and we cheat. And now we ask, is this also a smug set to minor identity? So we know that the two functions are equal in need of the case. And in principle, it could happen that a property of a set of coordinates has a property that's not preserved under images or pre-images of minors. For example, smug sets aren't preserved by taking the forward image of a minor. So now suppose that we have a smug set in F consisting of these last three coordinates, which means that there's some assignment BCAAA that has the value A. And now we ask, is the image of this set also a smug set of G? Well, now we run into a problem because when we wanna construct a value, a vector of inputs for G, there's no value that we can put on this coordinate. That'll be this vector here. You map it to a single one and then you're not guaranteed to have a smug set in G. And there are other properties of sets of coordinates that are satisfied by only pi or only pi inverse. So it can be kind of a game of playing around with them to find out which ones work and which ones you can combine with which hardness condition last set. The most complicated one. I'll only state the result. It says that in the hard regime, so where G over K is less than S over S plus one, probably more fissioned have a smug set whose size is better than G. And of course G is a constant for this problem. What does the proof look like? We construct a small smug set using properties of minimal smug sets and using conservativity. And quite a complicated induct argument is quite useful when changing the value of a mental smug set. You can then enforce the output of the value of the polymorphism because it's forced to take one of the values of it's in. So these three properties together conclude the hardness proof. And then there are some of these problems. Now, instead of taking literals that are indicators of a set with size just one less than the domain. So that you forbid a single element. Suppose that you allow arbitrary set families, a family of sets L and a power set of D. If you have these cases will reduce to set set. I think if all the sets have the same size then this reduces to set set. But now you can take set set of different size. And again, there are some trivial cases and reductions. So here this says, if you have an element that's contained in all your literals then the problem is tractable. Well, that's because you can just assign that one of the values to all your variables and all the literals will be satisfied. So it's sort of too trivial to be interesting. And again, you can make a conjecture about what the behavior of such problems should be. So if L is our set family and F max is the size of the largest set in L. And now suppose that we don't satisfy this condition here. So actually this should say the intersection is empty rather than the intersection is non-empty. So suppose we don't fall in this trivial case then we conjecture that the problem is tractable if and only if G over K is at least S max over S max plus one. Which is saying if I only decide the maximum size of a set in your collection that determines the complexity. And there's a bit of evidence for this conjecture so far. And my thought for that is that it actually performs better if your literals have smaller size because then there are fewer choices it can make when it's having a variable to satisfy the literal. But from the hardness side it's quite difficult to see what happens because we no longer have observativity and smug sets don't make sense when you don't have conservativity because smug sets correspond to having the same values in the input function, the output. Another class of related problems are hypograph color. In the Boolean case there was also a result on discrepancy colorings saying that it's NP-hard to distinguish a coloring, a two coloring as that's possible discrepancy from one. So that means G, even if you're gonna be a very strong property, it's still hard to find a two coloring that doesn't leave monochromatic hyper edges. And now we'd like to see what this problem means in our context for a larger domain. So what should the set up be? So the two G plus one essentially becomes the domain size S plus one on the parameter R plus A. So you can see that the two G plus one so you can think of this as a bunch of rows or columns of length S plus one and then a remainder so that the total arity isn't a multiple of S plus one. Now, what are our two cases that we wish? So again, strong cases that we have a coloring that's basically as good as possible coloring on this graph with discrepancy at most one. And you can see that'll mean you have our vertices of each color and then maybe one extra of some color depending on what they will be. So you could you couldn't get a stronger coloring property notice for this problem. And now for the weak property, again, we put the weakest possible condition every S plus one coloring of the graph is the least monochromatic hyper edge. So this problem we can check for to be NP hard. And now it's actually very close to a problem that's trackable. So if you get rid of this A here and consider instead an S plus one times R uniform hyper graph, then these two cases can be distinguished in polynomial time. So again, the promise is as strong as possible that we have discrepancy zero, which can now be achieved because the arity is exactly a multiple of the number of colors you have. So these promises are analogous in their strength. And the other side is that, again, you want to avoid a monochromatic hyper edge. So the largest possible. So tractability of this problem follows easily from tractability and this conjecture here is this conjecture difficult and not achieve exactly the same method as that. Well, mainly because again the polymorphisms are not conservative. So we can't use smug sets. And there's much more symmetry as well in a coloring problem than in a problem with a clause with literals. Because a clause with literals, the order sort of matters. You can't just permute the values arbitrarily as you can when you have a coloring. And then that gives a weaker guarantee on how the rows of inputs to polymorphisms interact. So this problem could end up being quite a lot more difficult. But I think it's an interesting generalization to consider. And that's all. Thank you. Yeah, thanks. Thanks, Alex. I have some questions. Libert. So there was this, you mentioned that in this more general situation where you have other unary constraints than just all but one. Let me see. So here. Here, this slide. So you say that randomized algorithm still works. Does it still have a block symmetric polymorphisms? So we can use the tractability result of Venkat and... I'm not sure. I haven't checked that for the tractability case. Uh-huh. But yes, I understand that that would be only plus a fine algorithm to be used. Okay. And one more question. So there is this general hypergraph coloring conjecture. This here. Yeah, I didn't see it before. So is it in any relationship with the hypergraph coloring result or not really? Like that it's, you know... You mean that have any relation to known results? Yeah, yeah, yeah. So if, I mean, hardness of this would imply hardness of say this result that is hard to find K coloring of two colorable hypergraph or such things. Yeah, I'm not sure which other problems are miss. It might imply other than set set which is obviously already solved now. But I think there are quite a few related problems with similar kinds of colorings. For example, with rainbow colorings or a mix of rainbow and discrepancy promise. Yeah, so it might be really good. Thanks. Some more questions? So maybe I'll ask a question about the hypergraph coloring thing. So here, if you make the soundness, not S plus one coloring, but two coloring, is that known? I mean, by known results on rainbow coloring or something. You mean in the conjecture here? Yeah, I mean, suppose you promise that there is a S plus one coloring of discrepancy at most one, that in particular implies that there is a two coloring. But then finding like such a two coloring might be harder, right? Because it'll make the hardness easier if you replace the second bullet instead of every S plus one coloring as every two coloring. So is that known? Yeah, because it shrinks the gap between the two. I don't know if that's known. I haven't thought about it yet actually. So this is the context some of the rainbow coloring results get. So but here you have a slightly stronger promise in the completeness case that the discrepancy is at most one. So that would be a good thing to look at. Is that actually, so rainbow color rank is also a, like for some cases, it's also a color rank of discrepancy at most one, right? Right, so if you take R equals one here and A equals one here, then it will imply. But that case of rainbow coloring is pretty hard. I mean, that's not known to be hard at least under unless you make some other additional assumptions. So this is even more general? Yeah, so if you take R equals one then A equals one that includes the rainbow coloring thing seems good. Okay, thanks. Strongest. I was hopeful that something would come out of this conjecture given that it's quite similar to this blue in case and that in the blue in case the two proofs are quite similar. I think they might turn out to be very different here. This conjecture may be a lot more difficult. I'm not entirely sure yet. Thanks. Some more questions? Okay, so let's thank Alex again. Okay. Thanks.