 So hello, everyone. Welcome to the first seminar of 2021. Today's talk is by Akbar Akim of Simon Fraser University on the complexity of CSB-based ideal membership problem. Yeah, thank you. Right. So, as Jakub said, I will be talking about the complexity of CSB-based ideal membership problem. And this is based on joint work with Andre Floto. So I'm going to start off with some definitions to state the problem. So let me start off with what do we mean by ideals and some like introduction to this algebraic geometric stuff. So through this talk, let F be a field, for instance, a field of complex number or real numbers. And then by fx1 to xn, we denote the ring of polynomials or f or maybe often I use capital X. And then an ideal is the subset of this ring of polynomials, basically a set of polynomials that satisfies three properties. So the first one is zero belongs to the ideal. The second property is that the ideal is closed on the addition. So if you have F and G in the ideal and then you sum them up, the result is also a polynomial in the ideal. And then the third property is sort of the ideal is absorbing the multiplication. So if you have this polynomial F in the ideal and then another polynomial H from the ring of polynomials and then you multiply them, you get something in the ideal. So it has these three properties. Let's see an example of ideal. So let's say we have these polynomials f1 to fm in the ring of polynomials. So we define this subset, which is basically the summation of hi fi, where hi are from ring of polynomials. So this gives us a set of polynomials, like sort of generated by this f1 to fm. Right. So you can check that this set, the way it is constructed satisfies the three properties of an ideal. So this set is basically an ideal. And it is called the ideal generated by f1 to fm. Okay. For instance, let's say I have polynomial one, so basically polynomial degree zero. So the ideal generated by this polynomial is basically the entire ring of polynomial. So as long as you have one in the generator set, then you get the whole ring of polynomial. So there is this fundamental theorem by Hilbert called like Hilbert basis theorem, which says every ideal of the ring of polynomial has a finite generating. So this generating set is not unique, is not unique, but it's, we have a finite generating set. So this gives us a way of representing ideal. So let's, let's get to the definition of ideal membership problem. So the input of this problem is an ideal and another polynomial f. And the question is, does f belong to i? So I'm going to check if the input polynomial f belongs to the ideal. Right. And as I said, this ideal i is represented to us by its generators, f1 to fm. And just to remind you the definition, if we have these generators, then the idea is going to be set of these polynomials, right. Okay, so in a more like a stronger version of this problem. So basically the search version of this problem we have this input polynomial f and then this generating polynomial f1 to fm. And now the question is, does there exist h1 to hm such that the equality holds. So basically you can generate f using this multiplicative polynomials h1 to hm and the generating polynomials, right. And so we want to find such h1 to hm from the ring of polynomial such that the equality holds. And this h1 to hm are called nullish.proofs of the fact that f is inside the ideal. Okay. So I'll give you a couple of seconds to stir this and digest. Right. Okay. So how do people generally like attack this type of problem. So basically for the search version when you want to find those proof nullish groups h1 to hm. So basically the first thing that comes to mind is like start dividing two polynomial division. So we have this input polynomial f and then this generators f1 to fm. And then if you start dividing f by this f1 to fm, we get the remainder, right. And if the remainder is zero, then we can say that the polynomial, the input polynomial f, belongs to the ideal. Okay. But we may not be lucky and the remainder may not be zero. Right. In that case, we cannot say anything for sure. We cannot say f is not in the ideal. Why is that? Because the order of division matters. Right. The way that we divide this input polynomial by this generating polynomial, it matters. Right. So let me elaborate on that by an example. So for instance, let's say this input polynomial is f, x squared y minus x y squared plus y. And then my generating polynomials are f1 equals x squared and f2 is x y minus 1. Then the ideal I'm looking at is generated by f1 and f2. All right. So now let's start dividing f by f2. Right. So if we start dividing f by f2 first, then we get this result which says f equals 0 f1 plus something times 2 and the remainder is minus x. Right. But if we start dividing by f1 first, and then divide by f2 after that, we get remainder zero. Okay. And the point that I want to make here is that this generator set, this generator set of ideal, they may not be well behaved in terms of division. So if you are unlucky and you get the nonzero remainder, then you cannot say for sure that the polynomial is not in the ideal. Right. So this is like basically the beginning of the story of Grobner bases where people are interested in finding sort of good generators. Right. So, well, now the more robust approach to solve this search version of ideal membership problem is to construct Grobner bases of the ideals, whatever they are. So the good property of this Grobner bases, basically, they are generators of the ideal, and they are pretty well behaved in terms of polynomial division. Right. So, basically, the most important property of this Grobner bases is that the remainder is unique, no matter in what order we do division. Okay. And so in order to solve the IP problem, what we do, we have this ideal, which is generating polynomial, then from those we construct Grobner bases. And then as soon as we have the Grobner bases, we do division polynomial division as long as we can in any order we want. And if they're the remainder is zero if and only if the polynomial is in the ideal. All right. Yeah, but life is not as simple we there are some drawbacks here. So first of all, this grobner bases are very sensitive to monomial ordering. Whenever we talk about polynomials or polynomial division, we have to set fix the monomial order, for instance, lexicographic order, or graded lexicographic, right. And if you change the ordering that you are working with, you see like drastic changes in the complexity of construction constructing Grobner bases. And we will see that this monomial orders, we don't like them in terms of PC, later on in the talk. And another drawback is that, so there is this general algorithm called book burgers algorithm for constructing Grobner bases. So this algorithm is sort of non deterministic. It's quite tedious to analyze the complexity of Grobner book burgers algorithm, which construct Grobner bases. So, while there are some upper bound and lower bound for the upper bounds for the complexity of this book burgers algorithm, but quite serious to analyze this algorithm and see for your specific problem which you are interested in, is it going to give you a polynomial time algorithm to construct Grobner bases or not. Okay. So this was about search version. Now let's go to the decision version of problem which we are not interested in finding those proofs h1 and hm h1 to hm. And we are just interested to say yes or no, the polynomial is in the ideal or not. So here we are interested in the geometry of variety. All right, let me define what variety. So a variety of an ideal is basically the set of all zeros of polynomials inside the idea. So it's a subset of fn, right. And their f is your field and this variety is basically the set of all zeros of polynomial. Okay. So despite strong, we can say that we can say the following every F, every polynomial in the ring of polynomial that vanishes at every point of the variety, long to the idea. So if you have a representation of the points in the variety, and you can check evaluate the polynomial, the input polynomial F on those points, and for all of them you get zero, then you can say the polynomial f belongs to the variety. Okay. So, for instance, if the ideal is generated by one, basically the ideal is the entire set of them. And the entire ring of polynomials, then the variety is empty. So basically all the polynomials are inside the ring of polynomial, which is obvious, right. So I have put this radical condition in brackets. So in order to, there are some technicalities here. So in order to be able to apply this strong knowledge to us, we need the ideal to be radical, whatever that means. But we will see the way that we construct the ideals for CSP problem. They automatically be radical, so we can apply this strong. Okay. And yeah, so we have these two theorems, which are, we can strong knowledge to them, but and is a basically what I said is basically the intuition of these two. So let's see an example of some combinatorial ideal. For example, let's say I want to two color a triangle. So I have this triangle, X, Y, Z. And I want to two color the vertices of this triangle. Okay. And the colors that I have are zero one, let's say. So first of all, this vertex X, it can have only two colors. Right. So I have this polynomial equation, which says okay X times X minus one is zero. So forcing X to be zero or one. And for simplicity, let's say you are working in real numbers. Right. So we have the same polynomial equation for why which forces the image of why to be zero or one. And the same thing for Z. All right. And now we have set of constraints, which says basically, well, I want different color for for the edges screen. So adjacent vertices get different colors. All right. So I need to encode this using polynomial. So I have this polynomial which says X plus Y minus one equals zero. So because I have this edge X, Y, then I want them to have different colors. The same thing for Y VH. So I want them to have different colors. And the last edge is X and Z and X. So now I have this domain constraints, basically, domain polynomials and this edge constraints. Now we can say this graph is too colorable. If and only if the polynomial system that I explained here has a solution. Okay, if you find that's fine assignment for this polynomial, then you have a two colorable. You have a two color. But we know that the graph is not too color. All right, let me continue and okay, let me rephrase this problem in terms of ideal membership problem. So first I generate this ideal, which contains all the polynomials from the previous slide at his generators. Okay, then the graph from the previous slide has a has no two coloring. And only if one belongs to the ideal, basically the variety of the ideal is empty. Well, we can say that you can see that you can express one using this polynomial. And yeah, so one is in the ideal, which implies the variety is an empty set. So there you go. No two color. All right. So let's see some application. So this ideal membership problem is like has been a start for a long time like between mathematicians and what they have done quite amazing job. But recently we have. So this ideal membership problem has some applications in computer science for CS people right. So for instance, this ideal membership problem can be can be used as a proof. But as we saw in the case of two coloring. You can formulate your problem, such that it witnessed that an instant has no solution. Okay, as we did it before what we, what we do is we include our problems through some polynomial. And then check if one belongs to the ideal. If one is in the ideal then no solution. If otherwise, there is a solution. And in the more in more strong, then we want to find these proofs that are witnessing that one is in the ideal. Okay. So basically this nourishment of the proof and in the proof complexity, people are interested in the size of the smallest proof that you can find basically. The size constraint could be like a degree constraint or maybe number of number of monomials that you want to use. All right. So let's see another application of this ideal membership problem. Sort of in terms of optimization. So let's say we want to minimize this polynomial are of x1 to xn subject to some polynomial constant. These constraints are as follows. P1 of x1 to xn is called zero and up here. So an SOS, some of the square certificate of a lower bound for this polynomial. So let's say we want to prove that this polynomial certified that this polynomial are is at least data. So what do we do? We write this R minus data as a sum of a square polynomial. This is always positive because is the sum of a square polynomial. And then the idea membership part. So this part is zero for an assignment be satisfied all this system of polynomial. So this term is zero. And this term is always positive. So the whole thing witnesses that R is at least data. Okay. And while the degree of this sum of the square certificate is basically maximum degree of this guy that you have here and maximum degree of this guy you have here. All right. So this way we can prove like non negativity. We can define non negativity of a polynomial. So let's say we have the set of polynomial and before P1 to PL. And then we say this polynomial as the sum of a square proof of non negativity from P. If we can have this equality. So the same as before, this term is always positive because we have some of the square. And this term is always zero. So the whole thing is positive. So we have a proof of non negativity of that. So, while the main appeal of this as well as proof systems or basically the source certificate of non negativity is that we can with the existence of a degree D as well as certificate can be can be sort of formulated as a visibility of semi defended program. So this inside the semi defended program you have this monomials if you have a bounded degree. You have this monomials and then you are trying to find coefficients for this one. And if you have one degree the number of monomials that you have to consider is bounded. So then you have a like bounded size semi definite program and you try to find the coefficients. Right. So this common misconception that if you have some of the square, which is of low degree. If you know the existence of a low degree sum of square then you can find it basically as so as is degree automatically. Right. So, yeah. As I said it means like if you have a low degree so attribution, then you can find it. So this STP can be solved efficiently. Right. But it turns out this is not the case. And basically the sum of the sum of the squares are not degree automatically. And this is due to all down in 2017. It came up with a country example is that this is not the case and we need some extra condition for us to be automatically. And those extra conditions are that not only we need this low degree condition, we also need to have some bound on the coefficients of the polynomial that we are using. So that's where the ellipse of method is going to stop the city. Right. So in terms of these conditions that we are looking for can be can be satisfied provided that the IMP part of the problem is polynomial time solving. So if you can solve the IMP part. Then you can get the automatable SOS proofs. Right. And this is due to rubber band run by in 2017. All right. So we have seen some application we saw the definition and example and some applications. Now let's go to the IMP problems coming from CSP. All right. So in CSP, we have this instance P, and which is the instance of CSP of a where a is a relational structure with some basis. Let's say we denoted by T basically domain zero to R minus one. And then in the variables, let's say they are x12 xn. And we have some set of constraints. And each constraint has a set of positive set of variables and sort of the relation and coming from this relational structure. So now let's say we want to encode this problem in terms of ideal membership problem. So for every constraint that we have in the instance, we introduce a polynomial f sub C, which is on those variables inside the constraint I want to x, I want to xik and the zeros of this polynomial are exactly the samples from R. So sort of what we did in in the two in two coloring example, right? Furthermore, for every variable we introduce this domain polynomial, right? So I want every variable to get the value from this set. So we have a polynomial of degree R, which which forces this x i to get a value from this set. So as soon as we have these polynomials, we put these polynomials together and generate an ideal. So put all these polynomials together and it gives us an ideal which corresponds to this instance of CSPA. Okay. All right. So ideal membership with respect to a relational structure. Let me define the problem. So input is a CSP instance P of CSPA where a is a relational structure with some variables x1 to xn and we have this input polynomial f. So we construct the corresponding ideal to the instance IP. And the question is whether this f belongs to this ideal. Okay. And well, also find proof that it's possible. And we have a like sort of a easier version of this problem where we enforce the degree we bound the degree of input. So this polynomial f, we say, okay, let's let it have at most degree at most D. This problem is denoted by ideal membership of the. Okay. All right. Well, remember at the beginning of talk when I was talking about this new list of stuff, I was saying, okay, we need the idea to be radical to be able to apply this new knowledge that's right. So, since we we are adding all these domain polynomial, then the idea we are generating is always radical. So, you're good in that sense, we can apply this knowledge to them. So, since we have the equivalent formulation of ideal membership problem, right? Okay. So in this equivalent equivalent formulation, again, we have this instance P of CSPA is a relational structure with some variables, and this polynomial, which is coming from ring of polynomial. And now the question is, is every solution of this instance, the CSP is a zero of f. Okay. And if every solution of this CSP instance is a zero of f, so basically you take this solution, you plug it into f and evaluate f in on that solution. If you get a zero for all of them, then you can say f belongs to the idea. Right. So, here it's somehow motivating to work with this variety instead of like working with those group members, right? And in CSP we like this variety, right? We are more comfortable with this variety. So, as a corollary, if this ideal membership is co-emptive, is in co-emptive, meaning that you can verify the no instance. I give you a solution to my CSP, plug it in in the polynomial and it gives you non-zero, which is a no. So you can verify this no instance in polynomial type. So, imp is co-emptive, isn't co-emptive, right? All right. So some research questions. So the first question is that we have this relational structure A, and can we construct Grobner bases for any instance of this ideal membership problem? So let's say P is an instance of CSPA. So can you find the Grobner bases for IP, the ideal of corresponding to that CSP instance? The second question, so if we have this one, if we construct Grobner bases, we can solve the search version, obviously the decision version. The second question is that for which relational structure, we have a small knowledge that is approved of the fact that f belongs to the ideal. Okay. This is another question. For which relational structure, the problem IMPA or IMP sub-DA, a restricted version, is polynomial times all of them. Okay. So now we are slowly getting to the, studying the geometry and basically the operations of these relational structures. All sort of polymorphism. All right. So let's see what we know so far. So in the Boolean case, we have dichotomy, which classifies all the cases. And it is as follows. It's almost the same as the shepherds dichotomy. So let me get to the details. So in 2019, Monaldo, most importantly, he started to study this problem in a way that CSP people like. So I study basically the properties of this relational structure. And he proved the following. So if A has a majority polymorphism, then IMP of A is polytyped. Okay. So while in Boolean, when we have majority polymorphism, basically the instances are two settings. Right. And the proof was that a Gromner basis of this instance, if we have majority polymorphism, they have degree at most two, and they have good coefficients. Right. So what we do, we enumerate over all this degree to polynomial. Right. And we have, and we have the, we have our Gromner basis. Okay. And we have our majority, majority polymorphism. And the second result is as follows, if we have a semi-latid polymorphism, then IMP sub D of A is polynomial time for any D. So just note that in the majority case, we don't have any degree restriction on the input. But in the semi-latid case, we have this degree restriction. All right. So this is the proof idea. So when we have semi-latid polymorphism for this over Boolean, then the instances are anti or anti-homestead or homestead. Like depending if you have max or me. So then it turns out the Gromner basis includes only binomial polynomial. So polynomials, including only two monomials. And then they have good coefficient. All right. So if you have this bounded degree, if you have this bound on the degree of input polynomial, then you can enumerate over all these binomial polynomial and get your Gromner basis in time into the D. Okay. That's why we have this degree restriction for in the semi-latid case. Now the other remaining case is the affine case. So basically, the result is due to most Raleigh and Bharati. And they say if we have a fine polymorphism, if we have an affine polymorphism, then the IMP again we have this degree restriction on the input is polynomial time for any D. All right. So in this case, if you have a fine polymorphism, it means the instance is a linear equation, Mach 2. Oh, sorry. It's Mach 2 instance. And then, so this case in my opinion was like the most difficult case. And so basically we have these linear equations, Mach 2, then we need to lift them to have polynomials over real. The first thing is not always polynomial times is not always efficient to do it in polynomial time. So we need to come up with some tricks. Right. I'm going to skip the details. So it's like more tedious than the other cases. And on the hardness result, so unlike the shepherd's day, if we have this constant operations or basically this constant relation, the problem is going to be complete. So IMP 2 of A is going to be complete. And the reason that we have these two here, because we only have two constant zero and one in Boolean. If you want to have a polynomial that represents both of them, sort of speaking, then you need to stick with two here. Okay. All right. So this was what we know about these Boolean case. Now we are interested in to like have a higher level look at the problem and see if we want to extend results to higher domain. Or if we can, like, follow these tools that have been developed in CSP for the dichotomaterials and use them here for this ideal membership problem. Okay, well, let's start with something easy. We wanted to we want to check if we can expand the relational structure that we have by console. Okay, does does that change the complexity of problem or not. So let's AC denote the structure a where with the you know the structure a with some added constant relation. All right. All right. So we have this theorem which says, okay, if you add this constant relation doesn't change the complexity of the problem. Basically, I am P of AC is pull time reducible to I am P of a with a blow up in the degree. And this blow up in the degree is coming from the fact that we are trying to encode this constant relation with polynomial. So it increases the degree of the input polynomial. Yeah. Now, another simple observation is that. So, since we know that one is in the ideal if and only he has no solution, then we can say the following that whenever the CSP of CSP of AC is empty, complete, then we get a comfy, comfy, complete version of ideal membership problem. So if your CSP is difficult to solve then I am P is also difficult. So let's talk about something more advanced like pp interpretability. So this notion of pp interpretability is basically what tells us that if we if we can use this algebra instead of relational structures. So can we just talk about algebra and the operations into the relational structures or not. We say this structure a pp interpret structure be if the base set and the predicates of p can be p to define a don't bother about the definition is completely vague. Just so this pp is shorter for primitive positive. So basically you can express whatever you have in be using these relational structure that you have in a. So it turns out that if you have this pp interpretability, then the problems are reducible to each other right. So if a pp interpret p, then ideal membership of ideal membership of ship of be is for the time reduced to the ideal membership of a weather some blow up in the degree. Okay. So this blow up is because this a and B might might have like different domain. Right, so as so since we have this pp interpretability result and this means we can use algebra. So we can replace this relational structure put it aside and talk about algebra and its operations. So I am P of algebra a denotes the clock, the class of problem IPP, where B is the final language invariant under the operation of this answer. All right, let me just let me walk you through this walk you through an example right. So let's consider similarities. And we want to. We want to reduce the problem or similarities of over a domain of three elements to the bulletin case. Okay, so let's say a as base 012, and it's invariant on the semi lettuce operation of this algebra. Let's say the same lettuce operation is the following, which we have where we have a global maximum. And that's basically zero is less than two and one is less than two. And we don't have anything between zero and one. So it's not conservative. All right. The maximum zero zero and one is two. Okay, now consider this Boolean algebra P, and we have this instance. Carefully be, and we want to reduce this problem. And in the, in the algebra B we have gained the semi lettuce point, semi lettuce operation, which says, okay, zero is less than one. All right. So let's do the reduction. We use I am P of a to ampere. Okay. So, there is this result in like a universal algebra with that if you have semi lettuce polymorphism then your algebra is a sub algebra of this Boolean algebra to some power. Okay, so basically we have this mapping, which stands zero. The domain of a was zero one two, and the domain of the was zero one. So we send zero zero one, one to one zero and two, one natural way. So now, we need to first transfer and transform an instance of CSPA into an instance of CSP and we know how to do it. So we replace. So let's say we have this instance P of CSPA, and we want to construct this instance P star of the SPP. So we replace every variable X with a pair of variable X one and two. So this, this is because we have this. So here in this, in this algebra, we have zero one and one two and one more. So for every variable, we need two variables. Right. All right. So and in every country, we replace the value a for X with GA for X one and X two. Okay. Now, and for the domain constraint for every X, we add this constant, which says X one and two has to take one of these values. So the solutions of P are in one to one correspondence with solutions of P star. Okay. Now, since we have these instances P and P star, we can construct the corresponding ideals. And now we need to change the input ideal input polynomial. Okay. So the mapping G that I talked about, we can always interpolate it with the polynomial of a constant degree, right. And the same holds for G inverse, we can interpolate this G inverse mapping by polynomial. So in this case, polynomial is simple. So it takes X one and X two and gives you X one plus X one and X two. Okay. So now the reduction is as follows, we have this instance P, we convert it to instance P star from the Boolean algebra. And then we convert this polynomial input polynomial F to input polynomial F star. And how do we do that? Well, simply we replace every current of X with G inverse of X one and X two. Okay. Now, we have this demo which says, okay, if F is in the ideal of P, so F is in the ideal of P, if and only if F star is in the ideal of P star. Okay, there we go. We have our reduction, right. But so this reduction works quite well in terms of decision version. But in terms of search version, we don't know what's going on, right. So let's say I have defined a proof where F star is in the ideal of P star, so basically proof like this. But can we recover a proof for F being inside P ideal of P? This is something interesting that we don't know yet. Well, in terms of more polynomial cases. So as I thought, as I said, we have these similarities for every domain. But still we have this restriction on the degree. And our results follows like by these results from universal algebra and the one time response in Boolean. Then we have that the imp of dual discriminator. The imp of majority operation is poly time. So just remember for in the Boolean case we have the imp of majority is polynomial time but in Boolean case the only majority polymorphism or operation that we have is dual discriminators. All right. So now we consider this dual discriminator for any fixed domain. And it turns out the problem is poly time. So, well, some of the proof ideas, basically we have this geometric result about the dual discriminators, basically the variety of these dual discriminators. Then, while we have three types of constraints, if we have dual discriminator and it turns out this unigame constraint or this permutation constraint cause a lot of trouble to construct group. So what we do we preprocess the instance, get rid of this unigame constraints and then we get a group number basis after that. It's easier to get a group number basis after that. Right. And once again, we don't have this degree restriction for that dual discriminators. Or another case which is like the most difficult one is that so for Boolean we have this polynomial time. Now we want to consider that P where P is a prime. So I am PT that P is for the time solvable. And here again we construct this group number basis for these sort of reconstruct this truncated group number basis each where we only go up to degree T when we are constructing this group number. It's a long proof and yeah, so I think we can talk about it after the talk if someone is interested. Let me finish up with some open questions. Well, obviously, what are the more, what are more polynomial cases. Then the second interesting question is, what is the connection that this problem IP has to some standards to be taken. For instance, consistency. Well, we know that this local consistency stuff. Basically thing and our consistency solve all these all these bounded degree CSB straight, but why cannot we have something like this for the IMP. Or basically, it's consistency. What does it mean in terms of IP? That's the third interesting question would be like proof recovery. Just as I told you in terms of semi lattices, we can have this reduction, which works beautifully in terms of decision version. But in terms of proof recovery, you don't know what is going on there. And then we saw this low degree restriction, for instance, for semi lattices we have this degree restriction, but for majorities we don't have it. So why do they correspond to right. So by the way, if you want to have dual discriminators for on ternary domain, and if you want that like full grown up basis, then you don't have degree two, you have to go to degree three. So there's something interesting going on there. And the last thing that I didn't talk much about it. This is this like book burgers algorithm. This is just that this general algorithm to construct growth in a basis. So this algorithm uses this notion of as polynomial. Well, if you're interested, we can talk about later but it's really interesting to see what the polynomial corresponds to in terms of. So somehow in the dual discriminator case, if we do this preprocessing and get rid of this unique game concept, unique game constraint, then we don't need to compute this is polynomial anymore. So this book burgers algorithm becomes polynomial. Yeah, so thank you for listening and please ask questions if you have any. Thank you for a very nice talk. Yeah. There are questions comments. This is Peter. Question. Yeah. So I'm not sure I caught that correctly. Do you know for general imps if you have a majority polymorphism, whether that's tractable then. What do you mean, general. I think for if, if the bullying case. Oh, yeah. Okay. I think I understand your question for bigger domains. Right. So for general majority, we don't know anything yet. So, yeah, so this dual discriminator is sort of a special type of majority operations right. So we know everything for this dual discriminator, but for majority we don't know, like beyond bullying what is going on, we don't know. And it's like a super interesting because we have this boundary degree restriction for the semi lattices. We don't have this in bullying case. But we don't have this such a restriction for majority. Well, but no, never mind, because majority is in bullying case is just dual discriminator. Yeah, so to answer your question we don't know anything. Yeah. More questions. Yeah, so if I may just like, do you know anything on top of these like some concrete examples what I mean is this say this three element to some lattice or say product of two z twos or Z four. Right, so that's a really interesting question. Well, we can get the. Actually, this is what we are working on. So this, we can get the sort of multi sorted PSP. We can have like a result for multi sorted PSP. And in like in terms of. So basically the next step beyond this linear equation, but P is basically linear equations in a billion groups right when we have this fundamental theorem of a billion groups or we can write it as a direct sum of direct sum of these P groups. So, and over there we need this multiplication that's two times that's two right. So, for the decision version. It seems we can do it. Right, but for the search version, we don't know. So even like, not just Z two square but Z four is do it also all a billion groups. Yeah, but decision version. So there is a like really other pointer. Yeah, the decision version seems to be okay. Hopefully even a billion groups. But as far as I understand this multi sorted stuff should work well. Each, each variable has its own domain and then in the construction of these ideals, we shouldn't face any major problem, but some more questions. Okay, okay. It's maybe hard to answer but so this ideal membership is harder in general than the CSP problem. So somehow what does it tell us the tractability of I am. Does it tell us something more in terms of CSP, I mean, can you formulate, you know, if you can solve MP, then you can do this and this is your CSP instances not only decide but you know something. So, well, I guess I know what you mean so this basically this imp is a sort of generalization so this. So if you want to do refutation, refutation of CSP instances is basically membership of one in the ideal. Right. So but in the in the like general setting we don't so the input polynomial doesn't have to be one it can be of any degree. In that sense is more general than CSP. And did it answer your question or I know this. So, so it's more general but whether you if you can solve the general problem, does it give us anything about the CSP, you know, which you can formulate in terms of CSP that somehow not only you can decide instances but you can do this and this. Honestly, I don't know. So, well, honestly, like, I don't know anything about that because so. Well, we don't know even what is this consistency mean in terms of ideal membership problem right. So, well, I don't know. Probably it means something but we are far from knowing that you think. Yeah, so last chance for questions. Maybe I have another one then. So when you when you got the tractability result for IMP was it always by using criminal basis or did you have any alternative approach. Right. So, well, it depends if you want to work in for the search version or the decision. So, as I said for the decision version, you don't need to compute the scrub the basis. Not necessarily needed. So what you can do is just for the decision version you find the satisfying assignment to your instance, which is a non zero of the input. If you find such a satisfying assignment, then you're done. You can say the polynomial is not in the idea without using any growth basis. But for the search version. Unfortunately, we don't know any other approaches, other than growing the basis. Right. So, it's so that it comes to how do you do this polynomial division. If you don't have grobner basis, then it's not building a, maybe, I don't know, maybe there are some other notions of good basis, other than grobner basis, but I'm not aware of them. Yeah, so just in summary for the decision version, you can get around with the grobner basis. But for the version. For the search version, you mean you want to kind of nullstellensatz proof. Yeah, so in the search version, like you said, okay, F is in the ideal. So prove it, give me a bit more. Okay, I give you this time, which is some of this polynomial. There you go. This is your equality.