 All right, so for the next talk, this talk on the complexity of finding S factors, unfortunately the authors were not able to make it, but they recruited the local arrangements chair to give this talk on behalf of them. I didn't realize this is also an option in general, but here's what they might present to you. Okay, so I'll talk about the complexity of finding S factors in regular graphs. The authors are Sanjana Koli-Shetty, Lin Lei, Elia Valkovich and Mihail Sianakakis. Okay, so let me just tell you what the outline of this talk is going to be. We'll start with the motivation for the problem. We'll talk about their model and the known results. Then we'll talk about their results. I'll just give an overview and there were also some proof details and I've used some MacBook hacks to remove the parts of the slides which I didn't want to present, so they won't be presented and I'll conclude. Okay, so in complexity theory we are interested in understanding easy problems versus hard problems and constraint satisfaction problem is a very general problem which can actually model many interesting problems that we study in complexity theory, easy as well as hard. So what is a constraint satisfaction problem? It's a type of problem where one needs to satisfy multiple constraints on variables simultaneously. So what does this mean? For example, I can give you a set of linear constraints and I can ask you whether they are satisfiable, can these set of linear constraints be satisfied. As we all know, this is an easy problem which is the instance of constraint satisfaction problem. On the other hand, you could have graph coloring like problem where you're given a graph and you're asking does there exist a proper coloring for the vertices and as we all know, this is a very hard problem. By hard, we mean NP hard. So constraint satisfaction problem could scale from problems in P all the way till problems that are in NP and there are versions where maybe it's even harder than NP and so on. The constraint satisfaction problems that will be considered in this talk are on Boolean domain and they are stated in a slightly different language. So let me introduce some notation to just talk about the problem a little more carefully. A relation R is a constraint on the truth values of propositional variables xi. So any relation is a valid relation. You could say xi or yi or xi or negation yj. All of these are relations. Let gamma be any fixed set of relations. So now I develop a vocabulary using these relations. So I allow myself a finite set of relations using which I can talk about instances generated from those. So let gamma be a fixed set of relations. A gamma instance is just a conjunction of constraints coming from the set of relations gamma. I mean if you keep satisfiability at the back of your mind, you know what I'm talking about. Relations are just ors of variables and you're taking ands of those. So a constraint satisfaction problem over the alphabet gamma or over the set of relations gamma is a form of a decision problem where one needs to determine if a given gamma instance phi is satisfiable or not. So it's exactly what we understand just spoken in a slightly different language. And now the name of the game is to classify the complexity of CSP of gamma. So let's go through some quick examples. 3SAT is just a formula over 3CNF formula and those which are satisfiable from this set. So if I were to talk about them as constraint satisfaction problem over gamma, then the modeling would be that gamma will be this set of relations. You take relations where all three variables are non-negated or one variable is negated or two variables are negated or all the three variables are negated. And now an instance 3SAT is maybe a specific instance like this and and of these things. So there is a beautiful dichotomy known from some classical work of Schaefer which says that there are properties that these gammas must satisfy and if they do the constraint satisfaction problem for those gammas is trivial that is in P and in all other cases it is NP hard. So what it says is that it gives six cases. In those six cases if gamma satisfy those six properties any of one of these six properties then CSP gamma is in P in all other situations it is NP hard. So it is a dichotomy result. So the six cases are very easy to state either when all things are set to 0 it evaluates to 1 when all the variables are set to 1 it evaluates to 1. So basically constant RGS are binary the each clause is just a binary clause or it is a conjunction of Haun clauses. So a Haun clause is a clause where all but one variable is non-negated. So it basically is all but one variable is negated. All but one variable is negated. So X implies Y or Z or W or something like this. So when you write it as the other way round. At most one positive literal. At most one positive literal or an implication like this which essentially means you negate all the variables or implication. Am I correct? Okay. Thanks. And dual Haun clauses are just when exactly at most one variable is negated or the formula the relation is a fine form which means that you take X i parity X j parity X k equals C some 0 1 order. In all these cases it is in P constant satisfaction problem of gamma is in P and in all other cases it is NP hard. So this is a classical result and now many works have tried to generalize this result for some more notation the CSP gamma is called trivial if every instance in CSP gamma is satisfiable. So examples of CSP gamma in P are the trivial cases two SAT which is basically the third case that I mentioned NP hard are three SAT not all equal three SAT one and three SAT etc. So you go to any gamma other than the one that I listed it becomes NP hard just for the sake of familiarity this slide has been introduced. In this particular paper they consider something called as CSP k gamma where k stands for the number of times a certain variable appears in this formula so read k formulas. A variable is read k if it appears at most k times in a formula a read k formula is a gamma instance where all variables are read k CSP gamma k is a specialization of CSP gamma to the case where the gamma instance is a read k formula. And more specifically this talk will concern will be about CSP 2 gamma so it is not very difficult to see that read ones formulas are trivial because I mean are easy to solve because you just fix a variable to the appropriate value and move on it never appears again. Read three or more is equivalent to unbounded I think the reduction is not that difficult but I don't know it off hand read twice is where apparently the whole discussion is for a while and so this paper talks about read twice. So Schaefer's theorem does not immediately generalize for read k settings and that again is not very difficult to see we can see it quite easily maybe. So again these are using some older results so CSP 2 1 and 3 sat okay so 1 and 3 sat actually I should have spent a little more time on it on the previous slide 1 and 3 sat means 1 out of the 3 gets set to 1 you know any clause. They are literals so CSP 2 1 and 3 sat now this instance actually can be modeled as solving the following question does there exist a perfect matching in a 3 regular graph yeah it's not very difficult to see because you'll put clauses on one side and variables on the other and sort of a graph there I don't know how to do it I tried I don't know how to do this but anyway if it reduces if you believe that it reduces to this problem then it's solvable in P is known from a classical work CSP 2 not all equal 3 sat so again it's a 3 sat instance where a valid satisfying assignment does not assign the same value to all three controls in any clause not all equal 3 sat is trivial since every 3 regular graph has a so okay it reduces to finding a slightly general graph structure than a matching it's called a 1 2 factor and computing a 1 2 factor in a 3 regular graph is what the problem reduces to and that again is known to be easy. So all that we are saying with this is so the shifters categorization we put problems in P and NP right once you restrict to CSP 2 some more NP hard bit of problems come to this exactly yeah so even if gamma does not have those six properties even then the problem has become trivial yes. Good question. Okay great I thought somebody in the audience will know. A focus taken by an earlier paper of straight was to consider the set gamma that consists of symmetric relations and what is a symmetric relation it just depends on the hamming weight of the input so the relation evaluates to 1 as far as the hamming weight falls into some class of numbers so for example majority relation is a symmetric relation because it just depends on the number of ones in it etc so not equal is a binary relation and the spectrum of that is just one that means either one of the variable is set to one and another to zero or vice versa spectrum of a relation is the valid values that it can take where it evaluates to one so in the case of not equal for not equal to evaluate to one the bits better be set not equally in which case the hamming weight of the input has to be one similarly equal to K on as a K-ary relation the spectrum is zero or K or either all bits are set to zero or all bits are set to one not all equal 3 sat is a ternary relation where so okay let me not do these examples I think the definitions are clear on this slide so at this point we I could have really stated the entire result but I have to connect this talk to the other part of the title which is S factors so let me do that and then I'll be ready to state the results so every read twice formula can be converted into the one where it is exactly twice instance so every variable not just appears at most twice but exactly twice and when you do that it can also be converted into a graph instance and this graph instance is nothing very different from the one I already spoke about where variables are on one side and clauses on the other an S factor so the definition that was being asked has come up now I can define it more carefully here so we say a graph G has an S factor if there exists a spanning subgraph inside that graph where for every vertex V the degree of V comes from this set so a one two factor now maybe the definition is slightly clear the notation was different it probably should have been curly braces one two factor yeah and now here is the reduction from CSP gamma k problem to the S factor problem the two main characters of the story so far let R be any carry relation the complexity of solving CSP to R is the same as the complexity of finding an S factor in a K regular graph where S equals the spectrum of R so carry relation converts to K re graph so maybe what I said about getting a graph instance I may not be absolutely correct there so the graph instance that they are talking about in the first lemma is not the one I mentioned it's something else but I don't know what it is unfortunately I don't know I don't want to say more than this but let's believe this observation that the CSP problem reduces to finding an S factor problem in a specific graph so if you are able to solve the problem efficiently in certain cases it will give you some information about how hard the CSP two instances in those cases so that's the main important theme for this yes yes yes so that will come up single symmetric yes so when I state the result I will mention all the assumptions that have been made so far and as Shrikanth was pointing out the notation will change slightly and will not keep on putting these curly braces if it's understood that gamma only consists of one relation so actually we are talking about quite a restricted case but that itself is surprising given the known results in the area okay so now stating the results this one more slide before I can state the results the high level goal is to classify CSP to gamma as per the choice of gamma so this would be the highest possible goal that we want to achieve however so ideally we would like the read twice dichotomy version of the Schaefer's theorem so this would be a great achievement in this line of work however we are still very far away from any such classification and when we are stuck over the years at a point it's a good idea to look at some restrictions I guess so to narrow the scope gamma consists only of symmetric relations and as Meena was saying just a single symmetric relation okay so that is the setup we are in and here also the result is not a complete classification but some sort of a conditional classification which I can now state okay so I'll state the result in two slides the first slide we'll talk about when the relations has even parity so let R be any symmetric relation and let it have even parity set to L CSP to R isn't polynomial so this is a complete characterization in the even case they obtain a complete characterization when gamma is symmetric and a singleton either there is some even number in the spectrum of the relation in this case it is easy or L which is the halfway point of the parity of the relation that appears in the spectrum in this case also they have a polynomial time algorithm if not the midpoint but L minus one and L plus one are in the set they if it's a subset of the spectrum even in that case it's a polynomial they have a polynomial time algorithm if it is a zebra where zebra is basically take any interval AB and only pick out every alternate entry from there so that's called a zebra and if the spectrum itself is a zebra in that case also they have a polynomial time running time in all other cases it's NP hard so for even parity they have a full characterization as I said zebra is basically just a set S is said to be a zebra if it consists of the interval AB while only alternate element is in the set okay so I've told you one part of the result is this statement of the result clear yes yeah oh no no the total number of variables in your formula could be very large right every every clause if you think about satisfiability say correct correct for it for example it's the relation is x or y or z bar or something because x yeah x y then it's a singleton for instance but I think satisfiability if there is no yeah yeah that is still a valid on the other hand if it is odd arity relation they don't have a complete characterization and there is a slightly technical caveat but let's let me state the result as if there is no caveat first so if that caveat wasn't there we would have a complete answer so let me state it in that form first R is a symmetric 12 plus 1 a relation then CSP 2 R is in P in the following two cases either is just trivial or it's a zebra the spectrum of the relation is a zebra okay and in all other cases it is NP hard unfortunately this is not quite the full picture but it they also require that the relation is not always false so this is an added assumption they have to make and if one could get rid of this assumption maybe the question would be solved for CSP 2 gamma with singleton relation and symmetric I'll just give you one example where it is constantly false or not is unclear and therefore that case is not settled by this for instance so let me just give you the second bullet on the same page and then we'll see whether I understood what they are saying it is currently an open conjecture whether all five regular graphs have a 1 4 factor or not okay and this finding so suppose I asked you to solve this problem this is like asking you can you give a CSP 2 can corresponds to finding whether all CSP 2 are are satisfiable or not so for a relation if all instances are satisfiable that is what I mean by the red part I had written on the previous slide trivial means I think that a satisfying assignment whereas this means that the problem is just always true there is a slight distinction in fact this red part is added by me using my Mac and I was not clear looking at the slides why don't they have a full characterization and I was not able to make sense of this bottom comment and I didn't find the time to write to them in the meanwhile so that's a problem but from looking at the paper it was clear that that there is a caveat in the given statement and that relates to this example at the bottom of the slide correct correct correct so yeah so the statement is this and then basically whether a certain instance is trivial or not is not always easy to see and if we can say that for all instances that okay triviality is easy to check then this will be a full characterization and the statement triviality will not have to check then it is trivial or it is not yeah so it's not something that you check on the instance this is like the dichotomy there is a dichotomy but we don't know what it is I don't know whether it goes on this side or this side so looking at an R whether it is trivial or not is not easy to tell that is the same thing right what I said okay okay yeah correct correct you are absolutely right yeah that's true so this is where the status is now and I have described the result I think I'm just going to wrap up now so let me summarize they have given a complete classification for CSP to gamma when gamma consists of a single even arity relation unfortunately it is not a classification if gamma has odd arity it would be a classification as Meena was saying if one could tell if the instance was trivial or not and actually most of their paper actually talks only about not so much about CSP to but about actually these s factors and finding them all the proofs go via from what I could understand very superficially via solving problems about s factors so the questions they ask are also kind of related to the s factors is it possible to identify for which set s and s factor is always guaranteed to exist this will resolve the question that was raised on the previous slide is there a set gamma that does not so in one of the proofs they use this equal equal three instance to push a certain case and therefore the second question comes from there and of course the most neat question to ask at this point would be what if gamma is not a singleton but symmetric but consisting of multiple relations can we say anything about it so yeah with that