 So let me introduce today's speaker, which is Simon Knoijer from the Udresden. He will talk about network satisfaction for symmetric relation algebras. It's yours. Thank you, Jakob. So this is John Drog with Manuel Budijewski. And I would like to start with an overview of what we'll talk about. So I will introduce a class of computational decision problems to you. They are called network satisfaction problems. And the result of our paper is a complexity dichotomy for this class. So the way we obtain this result is by a connection to infinite domain CSPs. And I will show you this connection and also will give you some details of the proof. From the CSP side, what we used is, of course, the universal algebraic approach to finite as well as infinite domain CSPs. So this is precisely dichotomy theorem by Andre Bulato for conservative CSPs from 2003. And more on the infinite domain side, we use a Ramsey type theorem by Hubek Knoijer. Okay, let's start with an example of a problem in our class. And therefore consider a finite set of integers and think of them as distances. So one to five as distances in the plane. And now there's a computational problem that is given a graph and undirected graph that is labeled by subsets of the set D. So for example, one for five at this edge here. And now the question is, can you choose for each edge and distance that is given in the label set such that all triangle inequalities hold on all triangles of the graph. So what do I mean with triangle inequality? Of course, the label of an edge from A to B should A to C should be less than the label from A to B plus the label from B to C. And we think of them as distances. Okay, how could this look like? We choose this assignment here. And now we have to check, are all triangles, is there a sign for all triangles compatible with the triangle inequality? For example, here we have three, four, five. So all possible triangle inequalities hold, of course. Another triangle in our graph is this one. Also here you can check this is fine. There's the last one, and this is this. Here we have, of course, that four is not less or equal than one plus two. So this was not a good choice of distances. And yeah, you can see that there's also no better choice. So this is instance we would recheck now. Okay. So this was an example in more general. Can I, what are you with a simple question? Do you also care about like longer cycles? So if you have a graph that has no triangles then. Only the triangles, only the edges of the induced triangles of edges. Okay. So in general, we have this finite set D and inequalities we formalize now more general in a set B that consists of at most three elements subsets of D. And now the problem, which is generalization of the example from before, network satisfaction problem of DB. We have given an undirected graph G with edge labeled by subsets of D as before. And we ask, can we choose from each label and element such that now on each triangle, the three, the set of the three elements is in B. Yeah. So how now let's try to match this with the example D is the set of distance one to five and B are all subsets XYZ of D. We're all trying all possible triangle inequalities. And for the instance that we saw before, you would now check, okay, for the first triangle, three, four, five is in the set B. As well as two, four, five is in the set B, but one, two, four is not in the set B. So in this sense, we reject this instance or we would reject this assignment. Okay. And now our result is for classification of complexities for problems of that form. And the classification is that they are polynomial time tractable or empty complete. And additionally, we can give them DB. So this, this table DB, these two data types, it's decidable which of the two cases holds. And of course, this means that there are no NP intermediate problems. And one could see this result as a partial solution to a question by Robin Hirsch. And he asked, can we point out all the P time tractable network satisfaction problems of relational algebras. And he called this problem the really big complexity problem. So, of course, there's now the question what have these network satisfaction problems of relation algebras to do with the class of problems that I introduced. And therefore, let me give you a very short intro to relation algebras. And they are basically sets of binary relations. So think of them as a finite set of possibly infinite binary relations that are closed under union intersection and relational product, the binary relational product. They are used to model temporal and spatial reasoning problems and AI and some examples that you might know are Ellen's interval algebra point algebra and read you connecting Calculi like RCC five and RCC eight. So the history of study of relation algebras started with Tarski and some other names Hodgkins and Maddox. What what's important for us is that every relation algebra has well defined network satisfaction problem a computational problem that is called network satisfaction problem and decision problem. And the complexities for such problems in this. Yeah, for for for relation algebras are only partially known so only for some examples like basically the names that I mentioned before so ends interval algebra point algebra. There are 18 small relation algebras there the complexities known and maybe interesting there are also examples known for finite relation algebras with an undecidable network satisfaction problem so it's not the case that we are always in NP for in general for network satisfaction problems. And in this sense, our result can be stated as classification of network satisfaction problems of symmetric relation algebras with a flexible at so only for the context of the title and some motivation of this work. But this is all about relation algebras because we would like to see something about CSPs and therefore start with some basics for for infinite domain CSPs. And here we start with set F of finite finite set F of finite relational tau structure so tau is a finite relational signature and B is a possibly infinite tau structure. And then there's a class of finite tau structures for of F and these are all the finite tau structures that do not embed any of the structures from the set F. And for example, you can think that F could be the K3 the complete graph on three nodes. And then for both F would basically all finite graphs that do not embed and trying that do not have an induced triangle inside. As some example. On the other hand, H of beam are all the finite tau structures that embed into the possibly infinite structure B. And we call the structure B finitely bounded if there exists some finite set F such that H of B is equal to four both F so we can describe the finite sub structures of B by means of a finite set of forbidden structures. Another important notion for infinite domain CSPs is homogeneity and this is that every homomorphism of two finite sub structures of B can be extend to an automorphism so whenever you see two pieces that are isomorphic in your infinite structure you can shift them by a global automorphism of the whole structure. And, yeah, there's an important property for classes of finite structures amalgamation property that is in connection to homogeneity of the infinite structure. So what about what's amalgamation. For any three structures a B1 B2 such that a embeds into B1 and B2 so you can think of, and we have two structures B1 and B2 and they have in common overlap in the structure a. Then there exists another structure in your class, in this case forb of F in forb of F such that you can embed B1 and B2 together in the structure C and this behaves well with the structure a so in formally we have this commuting or this equality here. And now there's a connection between dimension connection between homogeneity and amalgamation property by fascist theorem. And we get that if a class forb of F has amalgamation property then there exists a unique countable structure such that the age of this structure is equal to forb of F and the structure is homogeneous. And we will make use of this theorem. So, therefore, CSP of an infinite structure are all finite structures that have an homomorphism to be. And the observation that we use is that for every MSP there exists a finitely bounded homogeneous structure such that the network satisfaction problem defined on, yeah, is equal to the CSP of this infinite structure. So this is basically the core of this motivation or this connection between network satisfaction problems and CSPs. So why is this? How do we get this? We think of D, so D is the set of distances you can have in mind as a relational signature with binary symmetric symbols. So distance is symmetric and they are binary. And now we define the finite set of finite D structures as follows. All two element structures with more than one non-empty relation. So all things like this where you have two different distances between two nodes. So distance one and distance two and all, yeah, also if you have three different distances and so on. Everything that is not exactly one distance. And moreover, we have all the three element structures that induce a triangle that we don't like, yeah, in terms of the set B. So the labels are not in the set B. Now, of course, whenever I write such a picture and talk about now the D structure, so the signature is D, then with the picture I mean that these two elements here are in the relation four. And now the class four of F has the amalgamation property. And this is relatively easy to see because whenever you have such a diagram here, so two structures B1 and B2 and something in common. And if you don't have a bad triangle, then it's fine to basically draw, take this structure that you see here as the amalgam. So you don't have to put new relations and this is fine to have some free amalgamation. We call this. And by Francis theorem, of course, we have an homogeneous structure B prime with a B prime is equal to four both F so by applying the theorem. And as a last step we extend B prime by all binary first order definable relations so we take all the unions of the relations more or less of our signature and extend our signature such that we have all of them in our signature. And now what's why is the network satisfaction problem of this table DB basically the same as the CSP of the infinite structure be now up to some rewriting of signature so start with an instance of the network satisfaction problem looks like this. And now we add new edges and new relations with a new symbol F and this new symbol F means that two elements in the structure B are in no distance of the set D. So, of course, it could be or it's yeah, there are elements in the infinite structure be that are not in any of the relations named by the set D, and this new relations so no distance is denoted by F. So free or flexible or whatever you want. If you now change the this point of view that the label on this edge means these two nodes are in relation to or for then this is precisely a structure in the same signature SP. And it holds that this is also in the CSP. So it's also an accepted instance if the former was accepted. Okay, and for the rest of the talk I would like to call such structures be such infinite structures be in the class FA for flexible atom or pre amalgamation whatever. Yeah, yeah, that arise from such a data type DB SS I out you in this observation. Okay, so now I can give you a second version of our result and always have in mind so the infinite structure be has some some guys connected to this data DB we will make use of this and then. Now we define a finite structure called the type structure of be or maybe also type structure of DB one could say, and the domain of this finite structure is the set D together with two new symbols and one of the symbols you already know this is the symbol for the no distance integration in the structure be and a symbol it for identity and we put in this finite structure or finite all subsets as you know re relations and internally a relation where XYZ is in age if one of the three options here hold so first of all. The set XYZ is a good set in our in our data set be the second option would be the it element is not in the set but f is there and then we also have a have some some good table and XYZ is in age and the last option is that there's. One identity element and only one other element at most one other element so this somehow. Models now a triangle where one one edge has an identity label then the two other edges should be the in the same relation but we'll make this clear in the following and. Okay, of course, figures polymorphism for the finite structure D is a homomorphism from the six power into the structure itself with this identity here and the second version of our theorem would be for every infinite structure in this class FA so. Given by such tablets DB then the CSP is polynomial time tractable if this tight structure D of the structure be has a figures polymorphism and otherwise it's anti complete. And okay, of course, the structure D is finite by definition and there are only finitely many homomorphisms for on the six power so you can decide whether D has a figures polymorphism and this proves this more over part of our theorem that it's decidable which of the two cases. Given that this theorem is true. Okay, how do we get how can we prove this so be is in the class of a and the first observation is that the CSP of the structure is an empty so this is basically because be is finitely bounded for finitely bounded structures you can. Guess relations that you have to add and after guessing you can check polynomial many situations if the bounds are there or not and recall that we have only a finite set of bounds so this is clearly checkable in polynomial time if you get. And the first non trivial step is that for this finite type structure D there's a polynomial time reduction from the CSP of P to the CSP of D. And of course if this is true then we can use the CSP dichotomy for for finite domain CSP is and if D has and signals polymorphism then. The CSP of D is in P by the dichotomy and in fact we don't use the full strength of the dichotomy but only for conservative finite domain CSP is. And of okay if we haven't P time reduction then this immediately implies that we have that the CSP of B is also in P. So. The second big step is that we assume that D does not have figures polymorphism and then also by the dichotomy we get that the CSP of D is NP complete. And main part of our work is to more or less prove that there's a reduction from the finite domain CSP to the infinite domain CSP and this is. With the help of Ramsey type theorems as I mentioned by who which kind of children and others. And then we would have that infinite domain CSP is also NP complete. So let's have a look at the first and the easier case about tractable cases and as before we are in the same setting and recall the type structure we have this domain. Which is the set D together with these fresh symbols and we claim that for B in the class FA then there exists a polynomial time reduction from the CSP of B to the CSP of its type structure. And this is done by the following reduction. So this instance of the CSP of B recall we added these these new edges and labeled them by F and say these two do not have a distance. And now we reduce this and so first of all we add for each edge in this graph here new variable and index correspond to which edge. Then I put these red bubbles here for each triangle. So whenever the three edges now build a triangle then they are in the in the relation H. And of course we have four of them. And the last step is that we add for each the relation of each edge here so X2 and X3 are in the relation two or four. And this is now a unary relation at this for this variable two three and also for all the others. And this reduction works and is straightforward to do this like this. Now the second step for this tractable cases is of course that we use the dichotomy and I will call a finite structure now conservative if all subsets of A are unary relations. So this may be not totally common to have this notion for structure this bus of course compatible with the notion of conservativity for clones or. And then of by definition the type structure D is conservative. We put it all unary relations in this structure. And I would like to remark here that there already existed a notion of type structure introduced by police can one day. And this finite type structure was used in exactly the same way we used our type structure for reducing the infinite domain CSP to the final one. And yeah in a in a very nice way but this final type structure of police can one day would not be conservative and our type structure is conservative and this is what we heavily will use in the hardness proof later. Okay and of course for conservative structures we have the dichotomy now yeah in this version here so which is maybe not the original one. But if we combine this with our reduction from before then we have that if the type structure has a secret spot in office and the CSP of B is in P and this is everything for the algorithmic side of our result. Okay so let's have a look at the second the hard cases. So also here I would like to start with some definitions and therefore okay polymorphisms of B are all homomorphisms from the and power of the structure into the structure itself for any M. And we call and polymorphism canonical if for all tablets x1 to xn from the structure B and all automorphisms alpha one to alpha n. There exists another automorphism beta such that whenever you first shift the tablets x1 to xn by the automorphisms and then apply f it's the same as if you apply f directly on the tablets and then shift it by the automorphism. Okay, in other words this means that canonical function in induces a function on an n-ary function on the set of orbits for all m-orbit so for all orbits of the action of the automorphism group on n-tables of the structure B. And there are some observations so our structure B is homogeneous and only has so all important relations are of arity 2 and this means that we don't have to care about m orbits but only care about two orbits so orbits of tablets of length 2. And another thing is that the two orbits can be identified with the set D star so this is basically because this is the why we choose to set D star in this way. So D star was the domain of our type structure and we have that all the elements there correspond to the relations to the basic relations in the structure B and in this way we get a one to one correspondence of the orbits of the infinite structure and the elements of the finance structure D star. And most important we have the following if polymorphism of B is canonical then this induced function on the orbits which is an induced function on the set D star or an n-ary function on the set D star is in fact a polymorphism of the type structure D. And this is also true in some sense in the other way whenever you have a polymorphism of the type structure D you find an infinite and polymorphism of the infinite structure B that is canonical and that induces exactly the finite operation on the finite domain with which you started. Okay, so last definition here, I would like to read in this notion of canonicity a little bit and this is for two elements A and B of D star so basically two orbits of our structure. And then I call an polymorphism AB canonical. If the same holds, if we restrict ourselves to tablets x1 to xn from the set A or orbit A or B. So, this is not all orbits now, but only two, but also the same holds. So, if we shift, so we stay in the orbit by shifting of the automorphism and then apply F then this is the same if otherwise you would start with applying directly on the tablets x1 to xn the operation F and then shifting with the automorphism better. Now, such an canonical AB canonical polymorphism does of course not induce an operation on all the orbits, but on a subset of orbits on the subset AB of two orbits in this case. And, okay, one thing that's important here that it really induces an operation on the set on this subset of orbits is because we are the structure B is closed and all binary first order definable relations so the polymorphism that that preserves all these relations has to put in, if all its inputs come from orbits A or B, then its output also is in the orbit A or in the orbit B, because the relation orbit A or orbit B is preserved by all polymorphisms. This is why this is really a function on a subset of two orbits. Okay, and now for another time the conservative CSP dichotomy but now more technical and we have that for finite conservative structure then either there exist two elements in the domain such that all polymorphisms, the restriction of all polymorphisms on AB to the end is a projection and then and the completeness or hardness follows, or on the other side, Paul of A has in figures operation as before. But what does mean, we are in the case that the polymorphisms of our type structure D do not contain in figures operation. However, we have that there exists by one two elements a star B star in D star, which is the domain of the type structure such that one is true. And this is now the starting point for our hardness proof so all things that we collected now so DB where our data is in the infinite structure B, the finite structure D and two elements from the domain of the of the structure that have this special property from from before. And now the first step is that we prove that if B does not have a binary injective polymorphism, then the CSP is hard. Okay, this is basically what we would like to prove in general here in this in this part in this hardness proof, but it's a little bit easier under this assumption. This is basically because the element it in our in the domain of the structure D is somehow independent of the data from DB, because we all in all cases. Whenever you have another another another DB table, you add the new element in the same way. I mean the definition stays the same only. And the second step is that if B does not have does have a binary injective polymorphism then it has a canonical binary injective polymorphism. And for this step we use their Ramsey theory. Next step if this implies that it the element it is not part of a star B star, because otherwise on the induced function. And if the polymorphism F is now canonical then it uses a function on the orbits. And if it is on the other hand injective this means that the induced function is symmetric for all all sub algebras it X in D star. Because injectivity means whenever your input is it or X the output is not it because this would contradict injectivity. Okay, so it is not here. And next step is that we show that if that B does not have an A star B star symmetric polymorphism and by this I mean an A star B star canonical. So it's not canonical in general but canonical on on the orbits A star B star and the induced operation on A star B star. Think of it as a partial function in on the type structure is symmetric is binary symmetric. And again here there is heavy use of the Ramsey theorem. Step five. This implies that all polymorphisms of B are a star B star canonical. And if this holds for all polymorphisms it's relatively easy to see that the CSP of B is then hard because there are not so many cases that can apply. And you can again use the Ramsey theorem and get that this would contradict your assumption about the elements A star B star in the type structure. So six is not not not very hard. But the most interesting step is so this number five. And this is what I would like to show you here. The rest of the talk and in step five we would like to prove this proposition. So you see step five here four implies that all polymorphisms are A star B star canonical. So in other words assume that B has an injected polymorphism and does not have an A star B star symmetric polymorphism. Then we get that all polymorphisms of B are A star B star canonical. Okay, and we start with the negation of two and we use this lemma which is inspired by a work of Woody's can Pinsker. And it says that when for all elements a B in our in the domain of our type structure or you can also think of a B as two orbits, then B has an a B symmetric polymorphism and the induced function on a B. A maps tablet a B to a and B a to a, of course. This is equivalent to that every primitive positive formula find that satisfies that has four free variables and it's satisfiable with relation a on the first two variables and relation B on the second two, as well as relation B on the first two and a on the second two. Then we get that phi also satisfy a on the first two and a on the second two. Okay, and okay, of course we do our assumption is that we do not have an A star B star symmetric polymorphism. The two options is not a star B star means not not mapping both to A star and not making mapping both to B star so we get two formula. Phi A star and Phi B star. And they have the properties that are mentioned here in two of the lemma. Okay, and now we use these two formulas, Phi A star and Phi B star to define a new formula file. And by the following so in my picture I denote. Yeah, variables are the nodes here and the small edge is the relation after the free relation the flexible relation dotted line is a star and the full line is B star. So we have a formula phi A star that isn't for every formula or for three variables. And we know that it's possible to have this configuration so a star in on the first two and B star on the second two. It's also possible to have B star on the first two and a star on the second two but it's not possible by definition to have a star and a star on the first and second two. And that holds for the formula Phi B star. This is possible, this is possible, but this is not possible. And now we build a new formula file. And this is done by the following the three variables are the top and bottom x one x two and y one y two and all the other points here are existentially quantified. Okay, the relations between so the three variables are always relation F. But if we define this in that way then we have a nice property that if you start with a dotted with a full line here in top, then by the green bubble here we get that here. And we cannot have the full line so then in some sense there needs to be the dotted line. And now we apply the red bubble and we get that there cannot be the dotted line so there needs to be the full line. So we have full line here. And again the green one and we have that whenever you start here with the full line you have the dot line here. And this works also in the other direction if you start here with the dotted line and you get by red, the full line dotted and full. Okay, so we have full line if and only if dotted line here and also in the other way around. And the key thing here is that the relation F and the existence of an injective polymorphism ensure that this formula fire is really set up. And this implications always mean if if there is some type of then it has this property, but the main things are the power of our setting here that we have this relation F and the injective polymorphism and ensures that such topics exist. This is important. And with this formula five we can now define a new formula q equivalence, which is also for every. And so this is straightforward we get that we have on the first two coordinates a star if and only if there's a star on the second two coordinates, as well as B star if and only if B star on the first and on the same. So this is now very, if you have to find and this is easy to get. And what's also easy is that if such a relation is pp definable and therefore preserved by all polymorphisms. This means that all polymorphisms are a star B star canonical so you can take the the preservation of this of this relation also as an definition for for a star B star Okay, and so this was all basically what I want to talk about. Let me finish with last sentence so our work also confirms the so called infant domain tractability conjecture about reducts of finitely bounded homogeneous structures. The class FA is a small subclass of the class of all reducts of finitely bounded homogeneous structures. And, yeah, in this sense, and one thing that would nice in the future of course classify more general classes of network satisfaction problems. And from the relation algebra on what Asian site is one possible option could be to drop the symmetry assumption so all our relations that we did with were symmetric and would be also interesting to have directed edges directed colored edges in some sense. Okay. Thank you for your attention. Yeah, thanks. Thanks for a very nice talk. The questions comments. Now you cannot see my screen. No questions. Well, maybe I can ask a question. Yeah. So Simone, you were mentioned in the beginning, some really big complexity problem. As it was called by Hirsch, but it's not maybe immediate to everybody. Well, not immediate to me why it's so really big. Could you say something about this. So, yeah, I mean, so this this question has different levels and there's one level that I would call not not not very easy for for our infinite domain perspective. And this is only a subclass of relation algebras where we already already don't know a classification. So this is not the full problem of Hirsch seems already relatively hard to solve with with our, I would say advanced tools for infinite domain CSPs. But in general, yeah, there's, I don't know that in general you cannot apply the infinite domain CSP stuff or theory. And there, I think there are no, no ideas how to do this in general, or if there is even a possible classification in some sense, I don't know. Maybe. Okay, thanks. It's just a hard problem. Maybe I will have a question as well. So you at some point you compare to your type structure to to that of Anton. So what is exactly the difference between the two. So basically the airity of the tuples that you would take so so in the type structure of manual and on one you would in our setting since we have bounds of size three so we forbid structures of size three. And we would need to take tuples of size three and look at the orbit structure of these three tuples. And of course, the conservatility only applies to the two tuples, because they are the part of our signature and if you would have your your type structure on three tuples, you are not long on conservative and Yeah. Okay, so, so it's like, basically just you are restricting the RAT to two and that makes And then you also have to be careful because if you have the higher airity then you there's no need to have this relation age you you can choose other relations that encode the structure. So the good finite structures in your infinite structure be. But if you are not on the three tuples, then you have to add something different to to to to, yeah, to ensure that this correspondence of polymorphisms hold so that the behavior of the canonical polymorphism is a polymorphism of the type structure. And this is why there's no relation age in the type structure of on 209. Okay, thanks. Some more questions. So, sorry, so is it that the two types of type structures are essentially the same like up to BP into the finality or something or what's what's the exact situation. Up to the interdependent. I just might be. Yeah, I don't know could you but I mean they they should have the same same polymorphisms right to the polymorphisms are just the like, yeah, yeah, yeah. Yeah, yeah. Yeah. Yeah, you're right if you go back to the canonical clone then this might. Another question. One last to ask one last opportunity to ask a question. So, yeah, I wanted to ask so does the ground theory of approximation simplifies things here. Yeah, I think this is true. I think. At least the first step. I mean, yes, the general feeling is that this should simplify most of this. But this is not not checked yet. But I think the maybe this first step in the hardness proof so going to infinite going to the injective. And so, okay, this is not really the theory of smooth approximations but you can find some nice lamata in the paper and there are definitely some things to do better and easier. And the general feeling is that, yeah, should be simplified. Okay, cool. Thanks. I mean, in your case you have this unique interpolation right. So like when you want to extend this clone homomorphism say that you can just take some function which is interpolated by some canonical function and consider the image. But in your case it's it's kind of happening in the sort of obvious way right because you have the same behavior on that two elements that you already have the some behavior realize that so. Yeah. Okay. Okay, so let's thanks again.