 Yeah. So, this is about the relationship between these two problems, regular separability and intersection emptiness. This is work with Georg Zetscher at Max Planck Institute for Software Systems. So, let us look at intersection emptiness first. So, you are given languages L1, L2 from some language class C and you want to know whether they overlap or not and we are in particular we are interested in when this problem is decidable say for instance this language class could be context-free languages and this is something that has been of significance in many areas in particular in verification of software and hardware and it relates to safety conditions in concurrent programs which can be thought of as in particular for instance intersection emptiness of context-free languages. So, what about a certificate showing that in fact, these two things do not intersect. This leads to the idea of separators and separability and we have the related problem of regular separability where we ask for a separator S which is regular which contains all of L1 and it is disjoint from L2 and if S is regular then checking this is usually easy for a lot of different language classes and regular languages provide a good trade-off between separation power and decidability. We could consider separators from other language classes, but in this talk we are going to talk about regular separability. So, more generally we can have not one, but two different language classes in the sense that L1 can come from C1 and L2 can come from C2 similarly for regular separability, but in this talk we will mostly focus on the situation where C1 and C2 are the same language class. So, our goal is to understand whether there is some connection between the decidability of these two problems. So, in general we know for instance that for visibly pushed down languages you have undecidable regular separability. So, result by recent result by Kobzinski 18 and decidable intersection emptiness. It is a older result by Alul Marisuzan, but for instance these languages are not closed under certain closure properties and there is coincidence of decidability in many cases. For instance, both of these problems are undecidable for context free languages and they are both decidable for one continent, they are both undecidable for one counter automata. So, the citations in brackets are for regular separability. Intersection emptiness typically something that is much older whereas regular separability has been of more interest recently and for integer was we know that both are decidable. So, one thing that was observed was that all of these language classes where there is this coincidence are full trios. In other words they are closed under rational transductions. So, a rational transduction can be defined as follows. So, you have a machine an asynchronous transducer which can be thought of as an automaton which produces output. So, here for instance you have some state P and then the machine is making a transition to Q and it reads something. So, here it reads nothing it is epsilon and it outputs A. So, you can also think of there is an abuse of notation I am just using the language of this machine as T to denote this language of this machine. It is some subset of sigma star cross gamma star where sigma is the input alphabet and gamma is the output alphabet, but here we are more interested in the notion where it is input output machine. So, we can generalize this notion in this sense given some language L which is some subset of sigma star you can look at the transduction T applied to L and collect all the words V such that there is some U in L such that U V belongs to T. So, this is the notion of a transduction applied to some language you get some other language and closure under the transduction means that when you apply different kinds of transducers to different languages in the language class you still get languages in the same language class. So, language classes which are defined by finite state control with data structure. So, lot of many different things like some kind of finite state control with say a Q or a stack these are these are all easily seen to be full trios and therefore, this is a natural and mild assumption on language classes. So, we are we want to know whether in this particular case for full trios is it the case that these two problems are the same having seen that they do coincide in many different cases and we show that in fact, these problems are independent. So, in the sense that there is inter reduction either way and we exhibit full trios C and D which are closed under rational transduction such that the intersection problem is decidable, but the regular separability problem is undecidable and vice versa. So, these counter examples are defined using special kinds of counter machines which we will come to in a bit and the focus is on undecidable regular separability and decidable intersection emptiness in this talk. So, let me just go through some key ideas used in the proof. So, for any injective function f which maps some space sigma star to gamma star you let us look at two sets A and B which are in the domain then they do not intersect if and only if their corresponding images f of A and f of B do not intersect. Whereas, this kind of behavior where intersection is preserved under these kinds of transformations injective maps this is not maintained for regular separability. So, this is the idea that we use. So, in particular here is a key lemma we will be using. So, for any two subsets S naught S 1 of sorry this should be natural numbers here if S 1 contains all non powers of 2. So, it contains yeah it contains all non powers of 2 then the unary language A to the S naught. So, A to the n is in this language if n is in S naught is regularly separable from A to the S 1 if and only if S naught is finite and disjoint from S 1. So, this is not difficult to see in the sense that if you take any unary language any unary regular language then it has its ultimately periodic. So, it has some period P and we are including these increasingly long lengths of numbers or words unary words in S 1. So, any infinite regular language must intersect S 1. So, this means that any regular separator cannot be infinite it has to be finite which implies that S naught must be finite. So, and in particular the distortion this f that we use is some kind of binary to unary conversion via what we call incrementing automata and then we apply this unary separability lemma. So, starting with a new with the language class which has undecidable infinity which means the infinity problem is given some language from some language class decide if this language is finite or not. So, this the undecidability of the infinity of the original language when you apply this distortion it results in the undecidable regular separability of the resulting language class. So, we will see how we do this. So, what are incrementing automata over a predicate class? So, we start with some P which is a class of numerical predicates. So, each P belonging to Cal P is a subset of N and we have an incrementing automaton over Cal P which is essentially a finite state automaton which has the usual finite set of states and input alphabet. The edge relation is slightly modified in the sense that in addition to reading an input letter from sigma union epsilon you could also either increment there is a single counter which you either choose to increment or you do not do anything. So, you are not allowed to decrement this counter and q naught is some initial state and for the final for the acceptance we have some final states, but now we also have some predicates P which belonging which are from the Cal P class. So, that at the end the counter value must belong to this particular P. So, you have these pairs q comma P such that after reading the word w you end up in state q and the counter value some number n which belongs to P and you have some finitely many such pairs. So, this is how you accept a word in this automaton model and so, we denote by IP the class of languages accepted by these incrementing automata over P. How is the counter? The counter is updated by this on the edges you can always increase by 1 or keep it 0. So, IP is always closed under rational transductions and it follows from the following lemma which says that any language such language is a finite union of languages of the form T A to the P where P sum predicate from your predicate class and T sum rational transduction and it sort of easily follows from the way it is constructed. You can see for every pair q comma P of accepting the final pair you have one such T A to the P and it is just a finite union of these kinds of things and so, we just note that this P is some arbitrary thing and this closed under rational transductions is just coming from the fact that this model is some kind of finite state machine augmented by some data structure. So, we hope that it would be of independent interest in other constructions. So, in particular we consider incrementing automata over pseudo R predicates. So, let me describe what these are. So, we start with some number W and you look at W as the binary representation of the some number and that is the number which is denoted U of W. So, for instance U of 110 is 6 and for any language class we define pseudo C to be all the predicates such that P is all of these binary values all these numbers that you get when you look at the strings in L as in binary. So, in particular we consider pseudo R where R is the set of coverability languages of reset vector addition systems with states. So, I will not go into the definition of what this particular model of computation is. So, what we really require are some properties of R for our proof to go through. So, R in particular is a full trio and is closed under intersection. So, these things can be shown using standard product constructions and it has decidable emptiness because it is a well structured transition system and we are looking at coverability languages and it has undecidable infinity and this is an old result by Duford-Fink-Ellen from 1998. So, our first theorem is that regular separability in this particular language class which is incrementing automata over pseudo R predicates is undecidable and recall that this was a reduction from the infinity problem for R to the regular separability problem for this thing. So, starting with some input L belonging to R we look at the language K which is basically 1 0 to the length of W for all words W which belong in R and this can be effectively constructed because R is effectively closed under rational transactions and we note that K is infinite if L is infinite. So, in particular because the shape of strings is 1 0 to the mod W these are all if you further look at new of such strings then these are all things which are powers of 2 because it just starts with a 1 and it has a bunch of zeros and K 1 is this unitary language which is a to the nu of K and clearly this belongs to I should do of R and we define K 2 to be all the non powers of 2 which you can further see as the binary encodings of strings which have 2 1s in it right. So, this is 1 0 1 star 1 and 0 1 star and in particular because this is a regular expression. So, this is a regular language 1 0 1 star 1 0 1 star. So, it is also a reset was language and we taking the binary encodings of these things. So, you see that this is again something which belongs to incrementing automata over pseudo R and now we apply the unitary separability lemma. So, we have K 2 which contains all non powers of 2 and by construction K 1 and K 2 are disjoint because K 1 just includes certain powers of 2. So, the only way that K 1 and K 2 are regularly separable is if K 1 is finite and thus we have reduced it I mean reduced from the problem of infinity for R. So, therefore, this problem is also undecidable. So, secondly we have decidable intersection emptiness and the proof strategy is to reduce to the truth problem of the following logic. We let us look at the positive existential fragment of Pressburger arithmetic extended by pseudo R relations. So, these are relations which are so, so, so, so, so unitary relations are the ones that we considered in the automaton model but here we actually have you could have binary relations which you think of in the way hopefully if you have seen automatic structures you can think of it that way as encoded into a single unarbitrary arity relation can be encoded in a way which can be recognized by a machine. And the proof of this lemma which shows that the truth problem is decidable is essentially an induction on the formula structure where we show that every relation which is definable in this particular logic can be recognized by a reset vase and it is very similar to an automatic structures proof if you have seen this. So, let us see how we reduce this problem of intersection emptiness of incrementing automata over pseudo R languages to this particular logic. So, firstly recall that for any L 1, L 2 these are finite unions of languages of the form some rational transduction applied to A to the p where p belongs to pseudo R. So, in particular we can consider just a single T A to the p. So, you are looking at T 1 A to the S 1 intersection T 2 A to the S 2 checking whether this is non-empty. And further you can apply the inverse transduction of T 2 to conclude that it is sufficient to look at T A to the S 1 intersection A to the S 2. So, this particular transformation is important because now this allows us to think of T as just a transduction on unary alphabets. So, it is a subset of A star cross A star whereas previously it could have been something else. So, in particular now when we look at x y which belongs to n cross n such that A to the x and A to the y belong to T. Then this is semilinear by Parich's theorem which says that any the Parich image of any regular language is semilinear. In other words it is there is a formula in Presburger arithmetic which is just n plus existential Presburger arithmetic which defines this relation T with two free variables. And further once you have we have already assumed in the base logic that we have these pseudo relations pseudo R relations S 1 and S 2. So, we can write a formula for T A to the S 1 as well as A to the S 2 such that the solution set is precisely what is required. And now we can write the fact that the intersection of these two languages is non-empty simply by saying there is some common point x which belongs to both of these. So, the second counter example is in the other direction namely to do with decidable regular separability and undecidable intersection emptiness. So, this uses higher order push down languages which are things defined for instance if you look at order two push down languages these are defined by using stacks of stacks and they have four different operations where you have the normal push and pop operations on a push down automaton, but also an order two operation where you can copy the top most stack which is basically a second order push operation or pop off the top most stack which is the second order pop operation. And now you can consider order three order four and the union of this entire chain for all k's is what we call higher order push down languages. And so one of the languages is higher order push down languages, but the other language is again something constructed using incrementing automata which is slightly different which we call power h and not pseudo h some details. So, what is important is to see that this is an asymmetric counter example. So, we have h versus power h unlike the previous counter example where we had i pseudo r in both places. So, it is not as nice of a counter example, but nevertheless for these two classes we have decidable regular separability, but undecidable intersection emptiness. And to conclude we have shown that regular separability and intersection emptiness are independent problems and we have shown a counter example in both directions. So, importantly the undecidability proof of regular separability for sorry this should be pseudo r ink of pseudo r is different from the previous proof. So, because a lot of the previous proofs which showed that regular separability is undecidable usually take the undecidability of intersection emptiness and then modify that to show that you also have undecidability of regular separability, but since here we are building a counter example where we need decidable intersection emptiness, but undecidable regular separability we need to use new techniques. So, one could say this is one of the key technical contributions of this paper to show that you can take the infinity problem for a class and then somehow make a new class where the regular separability reflects the undecidability of the infinity problem. And of course, we would like to see if there is a symmetric counter example for undecidable intersection emptiness and decidable regular separability. There are lots of open problems which remain in regular separability which has been an active area of research over the last 5 to 10 years. An important one being that for vector addition system with states many in the literature survey I think I showed that we saw that many of the sub cases have been solved for VAS, but the most general case remains open. Yeah, thank you for listening. Questions? When you mentioned about higher order pushdown. Multi-stack meaning you mean like two stacks. Yeah. No, but that is already touring complete, right? If you have two stacks for a pushdown automaton. So, there is no hope of getting the solubility results for things like that. So, your results are interesting but quite negative. Okay. Do you look for a class, another class of trio, for which you will have the equivalence between separability and intersection emptiness? So, I think the closure condition of rational transactions is sort of natural. So, one of the other suggestions that I have heard is that, so, maybe this infinity is the problem. So, what if you try to include that in the characterization? So, along with intersection some kind of finiteness property if you throw in, maybe that captures regular separability. So, I think that is an interesting direction to consider to see if for full trios you can get an alternate characterization but which adds to the intersection problem in some way but this we have not thought about. So, for that we already have a nice combinatorial characterization which is sort of why this question came about. I did not mention it here but we have for instance for piecewise testable separability we have this thing called the simultaneous unboundedness problem which is equivalent indecisibility to piecewise testable separability and it is an entirely combinatorial question which is I would say a very nice characterization. So, sir, few questions.