 to present my results here. Today I will tell you about no rainbow problem and I will tell you about the complexity of the surjective consensus section problem. And I will start with the definition of the consensus section problem, which you probably know because this is a CSP seminar, but I will repeat. So let A be a finite set and then let gamma be a set of relations on A, which we call the constant language. Then for every constant language gamma, I can see the decision problem CSP over gamma, given a conjunction relations from gamma, and we need to decide whether it has a solution. And my usual example, if we are on three-element domain, we have two predicates. For this instance, we don't have a solution. For this instance, we have a solution, for example, 000. So what we know about this decision problem? The complexity of this problem is known from 2017, from our result with Andre. So we know that if gamma has a weakening polymer field, then this problem can be solved in polynomial time. And in all other cases, the problem isn't p-hard, so in p-complete. So for CSP, everything is known. We know which problem are easy, which problem are difficult. And today, I want to talk about surjective consensus, consensus section problem. So everything is the same, but this time, we are looking for a solution which is surjective. So again, for every constant language gamma, we have a decision problem, CSP over gamma, given a conjunction relations from gamma, and we need to decide whether the formula has a surjective solution. That is a solution such that any element of our domain A appears in this solution. So we have additional global constraint that the solution should be surjective. Example, if you are on three element domain, we have just one predicate, x is less than or equal to y. Then, for example, this instance has a surjective solution, for example, 0, 1, 2, 2. But if you consider this instance, then in any solution, a variable should be equal. So it has a solution, but it doesn't have a surjective solution. Okay. And the main question I'm addressing today is what is the complexity of the surjective consensus section problem for different constant languages gamma? Okay. One trivial example. It seems that we have three elements in our domain, three colors, and we have just one predicate equality. Then any instance of this problem can be viewed as a graph like this one. And for example, this graph has just two connected components. And it's clear that it cannot be colored into three colors. So we can color it into two colors. And this means that this instance, this concrete instance doesn't have a surjective solution. And also it's easy to see that this problem is solvable in polynomial time. So to solve this problem, we just need to count connected components. And if we have more components than the elements in our domain, then we have a surjective solution. Otherwise, we don't have a surjective solution. Okay. So for this constant language, everything is simple. And then I wanted to show an example when surjective CSP is harder than CSP. So it seems that we have three element domain, zero, one, two, and we have just one predicate in our constant language. X is not equal to Y plus one. So this predicate is not symmetric. That's why when we draw a graph, we want this graph to be directed. Okay. What is easy to see? It's easy to see that this instance always has a trivial solution, for example, zero, zero, zero. Okay. So CSP over this language is trivial, because we can always say yes. But if we try to find surjective solution, for example, if we start with red color, then we may see that all these vertices are either blue or red. And if we try to find the solution, and if we put blue color to some vertex, then we should get this. Anyway, my point is that this instance doesn't have a surjective solution. And actually, we can prove that this problem isn't complete. I don't have a trivial proof of this fact, but this is just an example when surjective CSP is harder. So it's easy to check whether we have some solution, but it's pretty hard to check whether we have a surjective solution. Okay. So let's compare CSP and surjective CSP. What we know. As I already mentioned, the complexity of CSP is known for all gamma. The complexity of surjective CSP is widely open. It's easy to see that CSP over gamma can be predominantly reduced to surjective CSP. In fact, we can always add dummy variables we never use. And using these variables, we can make any solution surjective. So we can take any instance of CSP and look at this instance as an instance of surjective CSP. And as we saw from the previous example, sometimes surjective CSP is harder than CSP. Okay. Another trivial observation is that surjective CSP can be reduced to CSP over gamma with constant relations. And here we are talking about Turing reduction. How we do this? Assume that we want to find a surjective solution. Assume that we have S elements in our domain. So to get a surjective solution, we need to guess S variables, taking all different elements from our domain. So we just guess this S variables. And then we add such constraints. So we put these values to the corresponding variables. And as a result, we get an instance of CSP over gamma with all these constant relations. Okay. And it's easy to check that we have polynomial and many guesses. And that's why this is polynomial reduction. So we can review surjective CSP to CSP. Okay. And Hubechen conjectured that it's also true in another direction. So he conjectured that surjective CSP over gamma is equivalent to CSP over gamma with all constant relations. And actually, this conjecture holds for two element domain. So we know that for any constant equation two element domain, the complexity depends only on this on the complexity of CSP over gamma with all constant relations. Okay. Another reason why this problem is interesting and should be studied is the graph homomorphism problem. So what is graph homomorphism problem? It seems that if you have a graph H, then H coloring problem or graph homomorphism problem is the following problem given a graph G. And we need to decide whether there exists a homomorphism from G to H. And standard example is when graph H is just inequality on three element domain or just triangle, then finding a homomorphism is equivalent to graph three coloring. So actually graph homomorphism problem here is equivalent to CSP over inequality. And we know that CSP in some can be formulated as not a graph homomorphism problem, but structures homomorphism problem. Okay. But yes, today I want to talk about surjective graph homomorphism problem. Again, we have a graph H and for a given graph G, we want to find a surjective homomorphism from G to H. And in my opinion, this is very natural question. You just given a graph and you want to find the surjective homomorphism. Okay. What we know about this problem? The complexity of graph homomorphism problem for undirected graphs was described in 1990. And in 1998 it was proved that graph homomorphism problem is as general as CSP. So for any constant language grammar, we can find a graph H, undirected graph such that CSP over gamma and homomorphism over H are equivalent. So we know that these two problems, CSP and graph homomorphism problem are in some sense equivalent. Yeah. Okay. And from our result with Ante, we know the complexity of this problem for any graph H. What we know about surjective graph homomorphism problem? We know that the complexity of this problem was described for all partially reflexive forests in 2011. Then in 2005, the complexity of this problem was described for all graphs of size four other than C4F. C4F is just four cycle, reflexive four cycle. And then in 2011, this problem was, it was proved that this problem, surjective graph homomorphism problem for surjective four cycle is NP-complete. And in 2017, it was proved that this problem is NP-complete for six cycle. And so what is great about this surjective problems? The great thing is that even for very simple graphs, the complexity of the problem is not trivial. Because if we look at, I don't know, 2015, when CSP dichotomy conjecture was not proved yet, we didn't have any concrete example of a constant language with unknown complexity, any simple concrete example. And here it's really interesting because you can take any graph like six cycle. And for this six cycle, the complexity of this problem is highly non trivial. So it's a little difficult question and it was open for many years. Okay, this is another motivation why we should study this surjective problem, surjective graph homomorphism problem. Okay. And now my favorite surjective CSP. Consider two element domain. So we have just zero and one. And we have just one predicate x1 equals x2 or x3 equals x4, just disjunction of qualities. What is the complexity of this problem? Let's consider a concrete instance of this problem like this. Again, we can look at this instance as a graph. But this time, instead of edges, we have pairs of edges. So if you look at the first constraint, it's just this pair of edges. This is this pair of edges, this pair of edges, and so on. And so what it means to find a surjective solution, given pairs of edges, and from every pair I need to, we need to choose one edge, so that the obtained graph is not connected. And in my opinion, it's really a natural question and it's really beautiful. You just given pairs of edges and you need to choose one edge from every pair. For example, for this concrete example, this instance has a surjective solution like this. So we can choose one pair from every pair. Okay. Okay, so what is the complexity of this problem? Let's start over. Consider this instance. So we have one variable on the top, one variable on the bottom, and all the remaining variables are connected with both of them. So we add this constraint that every variable is either equal to y or equal to that. So if we consider this instance then what? It's easy to see that this graph is connected if and only if y and z are connected. Or in other words, this instance has a surjective solution if and only if y and z are different. So in any surjective solution of this instance, y and z should be different, because any other variable is either equal to y or equal to that. Okay. And that's why we may assume that y is zero, that is one. Because in any surjective solution we either have this, y is zero, that is one. Or another way around, y is one, that is zero. Okay, so we may assume that y is zero, that is one. And then it's easy to see what we can do. Then we can add this pair of edges. And adding this pair of edges means adding this constraint x1 equals x2 or x4 equals one. Okay. And if we add this pair of edges, this constraint, this equivalent to adding this condition x3 equals x4 or x1 equals zero. Okay. And so what I'm talking about, I'm claiming that csp over this constraint to English can be reduced to our surjective csp. Yes, because we can, the first two constraints, they already showed how to get them. And the last two constraints are just constant relations x1 equals zero or x1 equals one. And to get this constraint, we just identify xi and y and or xi and z. So what I'm saying, I'm saying that surjective csp over gamma is NP complete. And this follows from this reduction. So we defined reduction from csp over this constant language to surjective csp over gamma. And this reduction is really easy. Okay, let me try to explain the main idea again. So we start with this instance. And what we know, we know that in any surjective solution, y and z are different. And they take all elements from your domain. And we still have a freedom to choose xi. And that is the main idea. So we start with this instance, and then we add all the constraints we need. And that's how we prove that this problem is NP complete. And I'm not the first one who proved this. As I said before, the complexity is known for two element domain. But for me, this is a cool example. And this is the first interesting surjective csp solved. Okay. And now no rainbow problem. This time you have three elements in our domain, zero, one and two, or you may think about three colors, red, blue and green. And then I consider a relation, ternary relation, which consists of all tuples such that not all elements there are different. And that's why it's clear why it's called no rainbow problem. So if all colors are different, it's rainbow. And that's why this relation is no rainbow. Okay. Then no rainbow problem is just surjective csp over r. So given a conjunction of relations r, and we need to decide whether there exists a solution. So we need a solution containing all three colors. And yeah, we are interested in the complexity of this problem. What we know about this problem? You know that this problem appeared in graph theory many years ago. I don't know when exactly. We know that Manuel Badirsky offered 50 euros for the solution for the complexity of this problem. And it was also many years ago, I believe 10 years ago or so. I know that Hubechen formulated this problem as a concrete conference inspection problem with unknown complexity. And he presented this problem many times at the conference. Probably even I had two talks by Hubei about this problem. What is also trivial to see and easy to see. Csp over r is trivial because we always have a trivial solution, zero, zero, zero, zero. Okay. So Csp is easy. And what if we consider surjective csp over r with all constant relations? It's clear that Csp over r with all constant relations. Surjective csp is equivalent to Csp over r with all constant relations. And that's why this problem isn't the complete and NP hard. But then in 2012, it was proved that even if we have just two constant relations, this problem isn't the complete. So surjective csp over r and just two constant relation isn't the complete. And one of the main result of my talk is the fact that surjective csp over r is NP hard. And that's why it's NP complete, of course. Okay. And now about the proof. How do I prove this? What is the idea? Actually, what I tried to do, I just tried to generalize my proof for the junction of equalities. So what I tried to do? This time I have three elements in our domain. And so I need three variables, v, u and w. And my plan was to find and to build an instance such that in any surjective solution, v, u and w should be different, should take all elements from our domain, zero, one and two. So that's what we want. That's what we want to achieve, but we don't know how to achieve this. Okay. But anyway, I just tried to copy what I had for the equality. So in the equality case, I had just one piece like this. And here we have three pieces like this. So what else I want? Yeah. And assume that we achieve this. Assume that somehow we know that v, u and w are just zero, one and two. Okay. And what we also want? We want all x's to be from the set zero, one. And to achieve this, we just add this constraint r of v, u, x, i. Okay. So r says that not all variables are different, which means that x, i should be from the set v, u and v, u is just zero, one. Okay. We want all y's to be from the set one, two. And to achieve this, we add this constraint. We want all z to be from the set zero term. And we add this constraint. Okay. What else we want? We want y's and z to be uniquely determined by x's. So the only freedom I want to have in x's. So that's what I want. I want y, i to be equal to x, i plus one, not the three. And I want z, i to be equal to x, i. And to achieve this, I just add these three constraints. Okay. And for example, you may check that if x, i is equal to zero, then u is one, then y, i can be equal to two, then y, y, y, i is one. And then if you look at the third constraint here, if y, i is one, v is zero, then z, i can be equal to two, so that i is equal to one. No, to zero, yeah. So anyway, you may check that these three constraints guarantee that we have these two conditions. But we still have our assumptions that v, u and w are exactly zero, one and two. Okay. Then again, we want to encode constraint satisfaction problem, which isn't the hard. And again, we want to encode the same constraint satisfaction problem. So I want to express this constraint. And to do this, I just write this r of x, i, x, j, y, k. Because if x, i, and x, j are equal, then it's true. If x, i, and x, j are different, then one of them is zero and one of them is one, then y, k can be equal to two. So y, k is one, which means that x, k is zero. Okay. And if I want to write this constraint, I just write this. Okay. So I almost have everything because I can encode these two constraints. And that's why I can encode csp that is NP hard. But now I want to be sure that v, u and w are actually different. And to achieve this, I just add all possible constraints that are always true. So I just checked what constraints are correct for any choice of axes. And so I found this constraint. And I just add the two of them. So for example, for every i and every j, I add the constraints that x, i, z, i, and y, j cannot be all different. And why this is true? x, i, and z, i, what we know about them. If x, i is zero, then z, i is also zero. So then it's true. If x, i is one, then z, i is two. Then this is true because every y is from the set one, two. So the problem is that now I'm not explaining why the proof works. And I'm explaining how I get the proof. So I started with the first part, then I added all these constraints. And it turned out that this is enough. It turned out that after I added all these constraints in any subjective solution, v, u, and w are all different. So somehow I was just lacking. And so I proved that csp over this constraint language is reduced to subjective csp over r. But the general idea is exactly the same as for disjunction of equalities. I started with an instance. Okay. Yeah, we'll keep it. I started with an instance having this property that in any subjective solution, three concrete variables take all the values from your domain. And this is the main idea. Okay. And that's how I prove that this problem isn't incomplete. Okay. If you have any questions, please ask. Okay. If you don't like this proof, I have another one. Let's prove that csp over not equal on two-element domain can be polynomial reduced to subjective csp over r. How we do this? Consider an instance of this csp. And now I want to assign a binary operation fi on the element domain to each variable xi. And what I want to achieve, I want fi to be the first projection if xi is zero and I want fi to be the second projection if xi is one. So what I'm trying to do is something similar to what is popular for promised csp when we consider minor conditions and so on. So what I do, I'm saying that I'm claiming that the instance i is equivalent to the following conditions. These conditions are not actually minor conditions, but when I invented them, I really wanted to write minor condition, but I didn't succeed. Anyway, what conditions should we write? First of all, I want all these operations fi to preserve our relation r. And we know what operations preserve r. We have two types of operations preserving r, either the separations are essentially unary or the image of the separation has just one or two elements. So we have just two types of operations preserving r. That's what we know from clone theory. Okay, second condition. I want to get the same function after identification. Then I don't want to explain now the source and the first condition. And then for each constraint not equal over xi xj xk, I add the constraint like this. And what is the meaning of this constraint? If fi, fj and fk are the same projections, like projections on the first variable, then I get a contradiction because I get zero one two, which violates the relation r. But if at least one projection is different, like fk is the second projection, but fi and fj are the first projections, then it's okay because what I get is zero one zero. Okay. And what I'm claiming, I'm claiming that all these conditions can be written as an instance of csp over r. Because what I actually do, I encode every function fi with nine variables. So for me, fi of zero zero is one variable fi of zero one is another variable and so on. So I encode every binary operation fi with nine variables on the domain zero one two. And then I can write all these conditions as an instance of csp over r. And what I want, I want csp over not equal to be equivalent to surjective csp over r. Okay. And in one direction it's really easy because if you have an instance of csp over not equal with a solution, then to get a solution of our surjective csp, we just use this assumption that fi is the first projection if xi is zero and fi is the second projection if xi is one. And you can easily check that the first condition holds because all our operations are projections. The second condition holds. You can check the third condition. For example, if we read the first part, if fi is the first projection, then we get x, y and fj returns as x or y. I'm not sure whether it's possible to for anyway, you can easily check that all the conditions hold. And if you look at the fifth condition, this condition holds because for fi being projections, this is just equivalent. Okay. So in one direction, it's really easy to check this. How to check in another direction? So assume that these conditions are satisfied. And if you write this instance of csp over r, this instance has a surjective solution. How to prove that we can find a solution of csp over not equal? That we can find a solution to the instance e. What we do? Consider this function g and consider three cases. If this g is a bijection, then what we know? If this g is a bijection, then from the first condition, we know that every fi is essentially unary. And then what we do, we just put xi equals zero if fi depends on the first variable. And we put xi equals one if fi depends on the second variable. And that's how we get a solution. And again, using condition five, we can prove that this is actually a solution because it doesn't matter whether fi is just projection or bijection of one variable. This fifth condition works in the same way. Okay. And then we need to consider two more cases when image of g has just two elements and when image of g has just one element. So g is a constant. So if image of g has just two elements, then we can easily derive from the first two conditions that image of every fi has just two elements. And these are the same elements. And that's why we cannot have a subjective solution because the third element of our domain can never appear. And the only remaining case when g is a constant. And then we can find fi and fj such that in the image of fi, fj, we can find all three elements of our domain. And that's why we need condition three and condition four. So we can actually prove that we can violate one of these four constraints, which you can see in the condition three and condition four. Okay. So that is an idea. And if you look at this proof, you will see that this proof is not very different from what you saw on the previous slide. So again, what we know, we know that in any subjective solution, this g of x should be a bijection, which means that three different values should appear in a three concrete places and three concrete places. So the idea is very similar, almost the same. And the only difference is that here it's easier to explain why it works. It's easier to look at the separation than just to check many cases proving that in any subjective solution, v, u and w are different. Okay. Is it clear? Yeah, probably I should ask you which proof is better. Okay. So the first main result of my talk is a theorem saying that no rainbow problem isn't become late. And here I want to repeat Kubitschian conjection describing the complexity of subjective csp over gamma for all constant languages gamma. So it says that the complexity of subjective csp and complexity of csp over gamma with all constant relations are the same. And this would be really amazing because this would characterize everything. And this is really simple and everything again depends on polymer physics. So that's what we actually want. And if you look at my results, my result confirmed the Kubitschian conjecture. And so after I proved that no rainbow problem isn't be hard, I spent couple months trying to prove Kubitschian conjecture at least for three element domain. And I know something for three element domain. So I checked many cases and in each of them, I actually proved this conjecture. And this was to the moment I found the contra example. So I actually found constant language gamma such that suggestions csp over gamma can be sort in parallel time. But csp over gamma with constants isn't be hard. So unfortunately, this conjecture doesn't hold. Okay. And now I want to show you this context example because it's really simple and I hope you can understand why it works. So we have just one five RE relation. So here columns are tuples. Okay. And if you look at this relation, what you can see, you can see that the first variable should always be zero. The next three variables should always be from the set one to and the last variable can be can be any element. Okay. And what I claim, I claim that suggestions csp over R can be sort in parallel time. Okay. We know everything about the complexity of csp. So we know that csp over R with all constant relations isn't be hard. And the worst thing is that if you define relation R prime like this, so R prime is just projection of R onto the middle three variables. R prime is just projection. You may check that surjective csp over R prime isn't be hard. So we have five RE relation and for this five RE relation, the problem is easy, but we can see that projection and for the projection, the problem isn't be hard. And this is really terrible because obviously every operation preserving R preserves R prime. And this means that the complexity cannot be described in terms of polymorphism at all. It's not only about Hubei conjecture. It's just in general. You cannot try to condition that you need some kind of polymorphisms for the problem to be talked about. It's not possible. Okay. So if you look at the three lemmas, lemma two is known, lemma three can be easily explained. Let me try to do this. So if we look at R prime, this is a constraint on two element domain. And for two element domain, we know everything. So the complexity is the same as if we just add all constant relations. And if we add constant relations to this R prime, then we get csp which isn't be hard. Okay. So lemma two is known, lemma three is known. So the only open, not open question, but the only lemma without proof here is lemma one. And now I will try to explain why lemma one works, why surjective csp over R can be sorted in well enough time. Okay. So that's what I want to prove. I will need another relation sigma. This sigma is just equality on two element domain one two. Okay. And what is the idea? The main idea is that if you look at this relation R and you restrict the last variable of R to the set one two, then after this restriction, relation R comes very easy. So it's just a conjunction of sigma and unary relations. So if you restrict the last variable to one two, then relation becomes very easy like this. If you restrict the last variable to the set zero, then again, your relation is very simple. Okay. These are just two simple observations. And how we use this? How to solve surjective csp over R? So if somewhere we restrict the last variable of R to a smaller domain, we just simplify as we showed before. And we achieve one consistency, simplify and so on and so on. Keep doing this. And after we did all of this, we have two options. Either we still have some R and in this R, the projection onto the last variable is the set zero one two. So if we still have some R, then what we do? Then I claim that we always have a surjective solution. And to get the surjective solution, I just send variables with domain zero one two two. I send variables with the domain one two to one. And I send variables with the domain just one element set D to D. So that is the idea. So if I have an instance, first I simplify. If after all the simplifications, I still have this relation R, then I can just try the surjective solution. Okay. And it's really easy to check why it works. Because if you look at the relation, then the only difficulty is in the middle, is in the R prime part. Yeah. But in this middle part, I can always put one to other variables. That's how I can use the complexity. So, okay, so what I do again, I simplify R, if it's possible, if I restrict the last variable of R to a smaller domain. And if after all the simplifications, I don't have any R, then I have an instance where there's only sigma relation and unary relations. And this instance is really simple. So I just simplified something. And I ended with sigma and unary relations. And of course, I can check whether I have a surjective solution or not. Okay. So that is an idea. And that is, I don't really like this trick. I believe this trick somehow can be avoided, but I don't know how. But right now, we have this contract example, that the complexity cannot be described in terms of polymorphism because we just take a projection and as a result, our problem is more complicated. Okay. This is the proof of lemma one. And I'm almost done. But I was too fast. Let's summarize what we have. For two element domain, we know the complexity. And this complexity agrees with Hubei-Chen conjecture. For many graphs H, the complexity of surjective graph home office problem is known. And you can look at the survey of Padirsky-Paramarty from 2012 to see a lot of examples of graphs H and their complexity. And all of these results were consistent with the Chen conjecture. In 2014, Hubei-Chen proved that if gamma has only essentially unary polymorphism, then surjective CSP over gamma isn't be complete. And actually, this result is sufficient to prove classification for two-element domain. Because on two-element domain, it's always true that, yeah, the problem isn't be hard if and only if we have only essentially unary polymorphism. Okay. This year, in March, I proved that no range of problem isn't be hard. Then I found a contract example to Hubei-Chen conjecture. And then a bit later, Hubei-Chen published a unification of the hardness results. So Hubei-Chen generalized my idea for the no rainbow problem and published a paper explaining this approach. So actually, something good is happening. And this is happening this year. So this is a good problem to think about. So I really recommend. And we still have a question. We still know nothing. We don't have any conjecture on how to classify the complexity of surjective CSP. Even for three-element domain, we don't have a conjecture. And I'm not sure about Hubei, but I really don't have any conjecture on how to formulate this. Okay. The final slide of my talk. I want to show Hubei-Chen conjecture again. And I want to ask, what conditions do we need for this conjecture to hold? For example, does this conjecture hold for surjective graph homomorphism problem? So if we have just one binary relation in our constant language, is it still correct in this case? Because in my contract example, I essentially use the arity of relation. So my relation should be of high arity. That's what I use. What if we forbid this? What if we consider just one binary relation or probably several binary relations? So is it true that the conjecture holds for binary constant languages? Or another question, what if our constant language is closed under projections? So I'm just trying to give my contract example. So if it's closed under projections, is it true now or not? But of course, the main question for me is I really want another simple concrete constant language with unknown complexity. And that's where I hope Hubei-Chen can show me some. Because right now I don't have any concrete example. Okay, that's actually everything I wanted to tell you. Thank you for your attention. And yes, special thanks to Hubei-Chen, who is trying to use me to the problem. And actually, I was so much attracted to this problem that I printed and posted this problem at the Moscow State University. And I even offered 500 Hubei, not 50, but 500 for the solution. And I was lucky because I don't need to pay now. Anyway, thank you. So thanks for the talk. Very, very inspiring, very disconcerting, at least a little bit. Are there some questions, remarks? If we still have some time, I can repeat something. So if I may, I have a few questions. Sorry, then I'll go ahead. Yeah, so you said you don't know any concrete relation for which you don't know the complexity. What about like you mentioned cycles? Like, is for every cycle it is known if you want homomorphism? Because you said like even C6. Yes, yes, yes. It's a good question. Actually, yeah, I don't know. Probably you're right. This is another good example, but I'm not sure whether eight cycle is simple enough. Yes, probably it's simple enough. Yeah, it's a good example. I should ask Hubei whether it's known or not. Probably it's not known because if for six cycle, it was both in 2017. So it's really recent. Yeah, probably right. Yeah, so I wasn't sure about these graphs and the graphs. So is it open even for like proper graphs or symmetric relations? Symmetric binary relations or you mean digraphs in general? So is it undirected or directed? Okay, C4F is, yeah, these are undirected. Yeah. These are undirected. So it's open for undirected as well. Yeah, I, yes, I think it's open for undirected. But I'm pretty sure that you, it's better to ask Manuel Badirsky. I'm pretty sure he knows more about graphs and certain problems. I think it's open. And so one more question or maybe two more. So in your example, the subject is yes, which is tractable, but your country example. So you claim that one consistency solves the problem. Is the correct algorithm? Okay, is that, well, okay, you reduce to this equality. Yes, yes, yes. I use this reduction first. I'm not sure it can be. Okay, okay. Yeah, but yes, yes, what I'm saying is that if we do this reduction, every time we reduce the domain of the last variable, we do this reduction. And also we achieve one consistency. And then I claim that, yeah, everyone consistent has a solution. Everyone consistent having R after all these reductions. Because using using R we can, like here, the problem is that, okay, if you look at the relation R prime, the problem is that we need to use both one and two. It's easy to use one because one, one, one is correct, but it's not easy to use two. And here I use the hips variable of R to put two there. So I put two there and everything else became trivial. So I put one in other places. So actually I tried to think about this just to find a new conjecture, but I didn't succeed. And if I may, one more question. So you prove that this tractability is not preserved under existential quantification, right? Yes, yes. That's what you were unhappy about. So is it closed under conjunction? Or do you have country example showing it's not closed under conjunction? So even say partial polymorphisms? No, I believe for partial it should be correct because we can always let me think. Yeah, for partial it should be correct because if you replace one relation by a conjunction, you still use the same elements in both of them. Yeah. Okay, thanks. Yes, so for conjunction, yeah, it definitely works. And that's another direction I wanted to think about. I wanted to invent some other types of polymorphisms which would work here. Or, I don't know, some kind of positive primitive definition, but not really positive primitive. I don't know. If I may make a comment, Hubei Chen has in his 2014 papers results in these directions. So some safe that you can use. No? Could you repeat, Mother? I think Hubei in his 2014 paper that you also quote here. Yeah, he proposes some safe gadgets that you can use to make complexity reductions. Yeah, yeah, yeah, yeah. And so it would be some partial clone theory, but there should be more theory than just partial clones, hopefully. Yeah, but as far as I remember, his idea from this paper is that it seems that we can see that all, I don't know. So you have a constant language. And we can see that all binary operations preserving this constant language. And as in Galois correspondence, we define this relation of arity, size of your domain to the power to your arity. Yeah? So this relation can be defined using just conjunctions. And that's what you can use. So to define this relation, you don't need essential quantifier, and that's why you can use this idea, this structure for subjective CCP. That's what I remember. But I believe now Hubei Chen understands much more. For me, it was difficult to read his paper because it's too abstract for me. But I believe the idea is very similar to what I did in my proof for the disjunction of equalities and for no rainbow. I just start with some instance, with some freedom. So I can choose freely some variables, but I have a general property that if I have a subjective solution, then in this subjective solution, these several variables give me everything. So anyway, I'm pretty sure that Hubei Chen knows much more about how to generalize all these hardness results. And I really want to ask him whether he has another conjecture now. Probably he's not here today. I think he's not here. I have some more questions, conjecture? I would like to thank Dima that he solved this and for the nice talk. You really deserve the 50 euros now. But let me ask you, did you submit it already for publication? Yes, I submitted this. I have a scientific advisor, Barnaby Martin, who sent me advice where to send my papers. So I submitted my paper to the Soda conference. I got some replies. The guy didn't understand my second proof, but he agreed with the first one, so I hope it's enough. So has the price been already paid off or but now are waiting until it gets reviewed? Wait for a physical act to be transferred. I can also make bank transfers, but get it published. That's important. Inflation. Yeah, it's inflation, so I'm going to get much more. After the Russian events, I already have 10% more. Okay, some more questions? If not, let's thank Dima again.