 So welcome to the workshop. We're really happy with such great attendance on expansion complexity and lifting theorems Really happy to have our professor Arquitif Chattopadhyaya here for the first talk I've known Arquitif for a long time He goes PhD in McGill and I heard of him because he solved the problem that I was really interested in which is To prove a lower bound for set disjointness in the multi-party number on forehead model Which was a really great result So I got to know him a long time ago from that and then he did a postdoc in Toronto, and he's been at TFIR for Seven years. He's a number of remarkable results in communication complexity and circuit complexity, and he's also great speakers. So Without further ado Thank You Tony for the kind introduction so My idea for asking the organizers to Organize a workshop on extension complexity and lifting theorem was to learn about extension complexity and lifting theorem Because I really don't know anything about it about extension complexity and their gift back to me was that I am the first speaker So you'll have to bear with me with whatever little I know. I only know a little bit about lifting theorems Thanks to a bunch of very smart students that I had But yeah, I mean most of this is classical stuff a lot of stuff has happened and I mean, I think several speakers will follow me up and Tell you more about them and more about those those very interesting work So I will assume some familiarity with communication complexity, okay, so I Will quickly recall so Alice and Bob. Oh By the way, the thing that I'm going to since there is no space for Writing on the whiteboard. This is based on joint work with Mihal Kautsky Bruno Lough and Shagnik Mukhapadhyay also a very recent work with You are filmers Sajin Korot, Urma and Tony So we will see two styles of lifting theorems but first let me Talk about the basic setup. So there's Alice and she gets x Bob gets y as you all know and Then they start communicating according to some Predetermined protocol they have to compute some function f Which could be a partial function or a or even a search problem? okay, but I will assume it to be function for now and And I will assume that you know for simplicity. Let's just assume that they talk one a bit at a time So there are rounds in each round the protocol specifies Who's to talk and if Alice is to talk Alice sends one bit based on what she sees on her side So it's some arbitrary function Okay, and this is something that you all of you know. Okay, so at the end of the earth's round They output the answer Okay, so the protocol which I will call pi pi of x y should be exactly equal to f of x y This is the goal Okay, and now there's another model which is even more elementary in which case you have some input which I will call Z So this is the first bit of z and this is the nth bit of z and you have some other function. Let's say h of 01 to the and in this model which is the query model You have a decision tree. So it's a pro it's it's an some algorithm Which queries a particular bit of z at the root maybe it queries the z1 then it queries the second If this is zero this it goes this way and then maybe it queries the 17th bit and so on and then at the leaves they are labeled either zero or one or Whatever the range of the function is Because this may be also again a partial function or a search problem and so for every possible input Alice this this query algorithm There is a fixed path because I'm assuming deterministic. There's a fixed path that gets traveled Quaring each bit at a time and you reach a leaf and then you output whatever the leaf has labeled, right? So this looks like a really really simple model where you know, there is no Alice no Bob and The algorithm doesn't get to see Any input bit for every bit that it wants to see it has to pay a cost of one unit, okay? So so that's the setup and What we are going to do today and what lifting theorems the kind of lift if lifting theorems that we will talk about is that sort Of a very interesting connection between you know hardness if a function is hard in this model It means meaning that you have to in the worst case query a lot of bits before you can give the answer then Associated or related function becomes hard in this very different looking model called communication model Okay, this is the goal of lifting theorem. So here's an example Which all of you know now here, of course you you're given two sets x and y and what you have to output is Is one if and only if you know x Intersection y is the empty set Okay, so you can think of them as the characteristic vector of two sets And you want to find out if the two sets are disjoint only then you want to output one Here on the other hand and the query model. Let's look at this function or Okay, very simple function. Yeah, so this or of z is equal to one if and only if there exists an I So that z i is equal to one Okay, now this is something of course you have you know seen many times But if you think about it Somehow the hardness of set disjoint is a very hard function in the communication complexity model The hardness is associated with this fact that what can Alice and Bob do because you know, they've got this Alice has got x and Bob has got y now what they want to find out is Basically if there is an x i and y i says that x i y i if there is an i says that x i and y i are both One right so in some sense the function can be rewritten as You sort of do first bitwise and of x and y you get an n-bit string and then you want to find out whether there is a One in that n-bit string So although Alice has full access to x and Bob has full access to y the relevant information seems to be in the String if I may call it z which is equal to the bitwise. This is the bitwise and of x and y Right and just looking at x and y they have very little information about this z Okay, if x i is zero then of course at that position x i and y i cannot be one But then Bob doesn't know that right so the so the inform a they seem to be almost Information wise as much blinded as a query algorithm is when the query algorithm doesn't get any access to z Right, so they have to find out for each bit and it seems that to find out for each bit Alice has to tell Bob Look, I've got here a one. What have you got there? Okay, so this is of course a very very vague idea because We will see a counter example very soon about this that this of course doesn't happen always Okay, this happens only Only in some special situations turns out for said disjointed this can be formalized and that's the formal proof That indeed because the query complexity of or so we know that and this is the notation I will use the query complexity of or this is deterministic decision tree query complexity of or is equal to n This is of course informal Yeah, this implies that the Is also equal to omega n okay, and lifting theorems would try to make this form Okay, so And in let me also remind you that in communication complexity Many functions that we deal with like said disjoin this inner product all of these can be thought of as First Alice and Bob have to do a bit wise certain operation, and then they have to compute some other function. So for example You know Said disjoin us is can be written as or of and or nor of and right inner product can be written as x or of and and So I'm just reminding you If you think about it most commonly used functions have this kind of composition property, and it's something to do with the Incompatibility of the outer function and the inner function that makes the problem almost as hard as the decision tree version for the outer function okay but So what I'm trying to say is that one may expect that the decision tree complexity Reflects in the communication complexity and another reason why you should think of that? This is very natural is suppose in general. I were to give you some function capital F Which is F composed with so this is a general setup F composed with G okay, and One natural communication strategy for computing F would be to look at the decision tree of F Okay, so F is some function from Z to 0 1 and G is some gadget Okay, so each bit of Z is being hidden. Yeah, it's being distributed or imported by G Okay, so a natural communication strategy for for for capital F is to look at the decision tree Protocol for small F and whenever and this was something Tony was saying yesterday Whenever you know say here I want to find out what is Z1 Well, I can't find out what is Z1 for that. I need to do the first instance of G X1 Y1 If I evaluate that then I get Z1 Okay, and I will employ the best communication protocol that I have for G Okay, and use that to evaluate Z1 and then I know Z1 and Alice and Bob would now at this point Both knows that one and maybe Z1 maybe this evaluation gives me one So I go to the right side and now this was maybe Z5 now. I evaluate G X5 Y5 Yeah, so this is the fifth block and the of X and Y and then this will give me Z5 And then again knowing that Alice and Bob can continue with Simulating this decision tree. Okay, and if you do that then this strategy Shows that the deterministic communication complexity of F is at most the decision tree complexity of small F times the communication complexity of G Okay, the question is is this optimal the set this join as example or inner product suggests Well, yeah, this probably is optimal, but this is not always the case here are two Examples that I would like to come back sort of counter examples that I would like to come back Later also at the end of the talk because this will lead us to a very interesting subject. I think so Trivial pathological looking example. Suppose I take XOR of XOR Okay, now XOR which is the parity function of course has very large deterministic communication Query complexity until and unless you know all the bits. You have no idea about XOR so you have to query everything and get and only then would you be able to know the answer right and the come so You know if I look at XOR of XOR I Would expect that the that the communication complex if such a theorem was true always then I would have expected That the deterministic communication complexity of this function would have been very large But of course, this is not true because XOR of XOR Squishes into XOR and so Alice can just Take the XOR of her bits and send it over in one bit to Bob and then Bob can just evaluate the answer So this looks like a pathological example. So some some care has to go into this Okay, you can't of course say that this is always true Okay, this looks like okay. This can be easily avoided. So now one important thing is we would also like to assume as I said my Function the outer function f I want to assume as little about it as possible For instance, it may not be a Total function. It could be a partial function In fact the lifting theorems are often used in the context when the outer function is a partial function or a search problem So here's another example suppose you know, I say that f is a function which is The input to f sorry input of f. So my promise Is the following Either Every zi is equal to zero or there exists a unique I such that zi equals one so I Have an n-bit string and I'm giving you the promise that either all of them is zero What there's exactly one the hamming weight is one and you have to distinguish these two cases Now if you go to the query model, this is the hard instance of the or function, right? This is this is the one on which you will spend most of your time because you will keep querying for a one You will never get it until the adversary and when he has exhausted you he will show you the one Or he may not show you the one, right? So this is very hard. This takes n bits Okay, so this is going to be the communication complexity. So this is let's say promise star The decision tree complexity of promise star is still omega n right now, let's look at F which is promise star of equality now equality I equality is just another important function equality is And of Exor So equality is a classical example of a hard function in deterministic communication complexity and What you would like to Think that this function Because you have composed it with a hard function here this function should have very large communication complexity in particular suppose I had Instances so let me write it here. This is an instance of equality and this is m each equality string is an m bit string This maybe I should write it larger right So you would you you you think that this should take n times m bits of communication Because this by the naive protocol whenever you would like to evaluate over here You have to spend m bits and I showed you a convinced you that this the decision tree complexity of the promised or is Is very large right, but it turns out It's a cute exercise to show that you know the The communication complexity of this f no matter how large is n is actually order m Okay, because of this promise Okay, so When you Yeah, exactly. Yeah, I mean it's promised in the sense that you can assume that they are free to answer whenever the promise is broken They can answer anything. Yeah, just as in this case. I gave you the promise if the z is were different You could answer anything right despite that despite knowing that it was the hardest possible Thing so you would have you could have hoped this now this requires a bit more thought it's not it's not hard I mean, you know, okay, I'll leave it with a curious interesting exercise So the point is you know what I'm trying to say is that not every gadget Is is is useful in rendering the communication model to the query model? Yeah, meaning or or you have to be very careful about this Okay, so you can't you can't hope that you know in every situation this would have Okay, and it's it's This particular one I would like you to work it out because you will see that the that the fact that the communication Sort of model has this side information. Yeah in the sense that okay, there's some core. I mean there's correlation things are correlated Okay, I mean it is not independent and Alice and Bob are able to see all of x and all of y that side information can be used in situations to bypass the following the knife following the knife algorithm So now let me try to say the main theorems So in some sense these theorems are rather delicate about I Mean, it's amazing that they work when they work that they work so very generally Yeah, yeah, so one of the theorems will show. Yeah, okay delicate in the sense Okay, one of the theorems will show that if you picked a large enough random gadget this will work Okay, so Regardless of what the other functions Okay, so I want to give you two theorems as I said From these two works one this will generalize a Series of works first by Raz and Mackenzie back in the early 90s and then Tony with Mika Gus and Tom Watson's yes So for this promised or composed with this equality. Yes, if M is equal to one Yes, then in that case this just becomes this said disjointness problem Equality is an XOR and that makes a huge difference. Oh Said disjointness was an and Right. So what I was saying was that this is generalization of Raz and Mackenzie And I think Raz and Mackenzie were the first ones to prove a lifting theorem of the kind That we are that we'll talk about but I mean as a Disclaimer, I want to say that there are many instances of lifting theorems depends on what you are lifting to what? Okay, so there are lifting theorems which which are more algebraic in nature very important example of this is Sherstoff's lifting theorem independently also discovered by she and zoo where he showed that you know a different measure of One-party functions like approximate degree Can be lifted into randomized communication complexity? Okay, we are not talking about such lifting theorems the lifting theorems that we will talk about are when you want to lift lift Specifically the decision tree complexity measure So in order to state our lifting theorem I want to tell you about two other notions that will be useful one of which you might have already seen so How do you prove lower bounds against communication? Protocols the basic building blocks. I'm sure all of you know of a communication protocol is that R-R-R rectangles So in other words if I have a communication protocol of cost C then the communication matrix Which is? This matrix One associated with all possible inputs of Alice the other associated with all possible inputs of Bob and this is a Boolean matrix You know so each entry here is the value of f If this is x, this is why this is x y Right So that's the communication matrix and this communication matrix is broken into two to the C Monochromatic sub matrices by a communication protocol a communication protocol just you know takes a certain cut here as Alice speaks she divides her input onto two parts then Bob speaks Bob divides his input onto two parts And if you do this you will see that it gets broken up into chunks and each of this chunk At the end at the leaf each leaf represents a Submatrix and these submatrix have to be monochromatic Because if I assume a deterministic protocol then you know that there's no no no no possibility of error So it basically breaks down this communication matrix into two to the C at most to the C monochromatic sub matrices okay, and these sub matrices are what I will call combinatorial rectangles and One way to prove lower bound against communication complexity Communication protocols is to say well you're going to break it down into two to the C Monochromatic Submatrices then at least one of them the average size of that such a sub matrix is two to the minus C the fractional size right and You if you show that there cannot be a monochromatic rectangle of size fractional size to the minus C Then you are done right then then then you cannot have C bits so now I'm going to talk about a property of of this so We are in this situation F is F composed with G And I'm going to talk a property of this gadget G because I want to prove a lifting theorem meaning, you know Some property of G will ensure that no matter what F is the decision sees a goal is to say What property of G? ensures That the knife protocol is a simple protocol right so DCC of F is equal to Yeah, we saw that this is all not always true right I gave you two examples What we want to find out is what is that a sufficient property of G that? ensures that this holds Okay, and more generally Since I don't have time let me also say the same thing should hold for randomized communication So there's a difference between epsilon and epsilon prime. This is technical stuff So yeah, so you want these things to be true. Okay, so it's all about G There is because it's G's somehow encoding this in such a way that as I said even though you have access to Massive amounts of side side information even under promises of weird correlations. Yeah, it doesn't matter G is so hard that Alice and Bob are basically forced to follow this knife strategy of querying bit by bit Okay, so we want to find out an interesting Properties of G that that is sufficient for for such purposes okay, so In order to talk about those properties of G. I will have to talk about this properties about rectangles So I want to talk about two things one is as the title of my talk Said one is hitting distributions They seem quite different and the other is discrepancy so Now what hitting distribution says is that you know so I look at G I look at all the monochromatic rectangles of G now there are two kinds of monochromatic rectangles of G Yeah So my question is in the composition G like The examples you gave G is a very low arity function, right? Maybe just on one bit The low the lower the better for applications. Yes, that's true And we also assume that so each copy of G would act on like disjoint set of areas. Yes. Yes This is block composition. Yeah This notation somehow traditionally this was you so this block. There's no overlap. Yeah, that's very important Okay, so there are there. I mean I will talk about a C monochromatic Rectangle of G this should be clear C is some C is in zero one, right? So it's either zero monochromatic rectangles of G or one monochromatic rectangles of G now I Want to say that not only G is hard in the sense that it has no small monochromatic rectangles, but The so That's true of a random function a random function will not have large monochromatic rectangles Okay, but now I want to say something which is which is not true of a random function It's okay at one point in one since it has no large monochromatic rectangles That's like a random function, but you know this at the same time it has It has not too large monochromatic rectangles. It has somewhat large monochromatic rectangles and this somewhat large monochromatic rectangles are sort of nicely distributed And what do I mean by that? formally I have a distribution for each C hitting distribution that I will say sigma C which samples Okay, which samples a C monochromatic rectangle at random. So this is some distribution, okay, and Under this sampling if you give me in this matrix any larger rectangle This is some large rectangle not necessarily monochromatic, of course It is so some large rectangle the probability when you sample according to sigma C a C monochromatic rectangle This rectangle is R R intersects with your sampled rectangle Let's say u cross v Yeah, this is very small Sorry intersects. Oh, this is large. Yeah, okay So what do I mean by large? So there are two parameters here the probability must be the probability of not intersecting is very small like delta and the largeness of R the size of R so R is Greater than equal to 2 to the minus h large. So I'm always talking about fractional fractional largeness, okay So I want to have these distributions So no matter which rectangle you give me if I have two distributions Sigma C that is Sigma 0 and Sigma 1 you want to hit with a zero monochromatic rectangle I'll sample from Sigma 0 and then you know, I will If I sample such a rectangle u cross v according to my distribution Sigma 0 then very likely I'm going to hit This you're given rectangle Okay, and I can do that also with Sigma 1 in particular therefore that means that your large rectangle is not monochromatic because I've been able to hit it with non-zero probability with Very large probability both by a zero monochromatic rectangle and a one monochromatic rectangle So in particular your your rectangle is not it cannot be monochromatic The the the simple condition that we talked about before but you see you need to have these things These are not satisfied by a random gadget. It's very delicate distribution. Okay, so this I will call If G has this property, so then yes No R has to be large enough Well u cross v some distribution that I mean I mean u cross v lies in the support of my distribution And I'm sampling according to that. I mean who knows what it what it for all R for all R Sorry for all R. Yeah, u cross v is being sampled u cross v is being sampled according to Sigma 6 u v is sampled according to Sigma 6 so when this happens I say that G So this is said G has the Delta H Property seems like barely I will have the time to state the two theorems. Okay, so the first theorem is the following that Okay, I'm again stating this informally because yeah, you can check out the paper and look at it more carefully so assume that So you're looking at f n composed with G B Yeah, so the so the so the block size of G is b bits and you have n instances of G and you're creating this function and Let's assume that one is less than it. Okay. This is an important Epsilon is some some fixed Number between 0 and 1 and G has the heating Delta H heating property then I can show I'm intentionally writing it in this way Because this will suggest how the proof goes so Yeah, so what this means is if you give me a communication protocol of This much cost for this protocol then I will be able to extract a decision tree So that's how the proof goes you've given me an arbitrary protocol, which is working in weird ways We have no idea But using the fact that G has this heating distributions. I will be able to extract a Decision tree for f with cost this much and the important thing is, you know, this cost is being is being Slashed by this this age Yesterday you had a question about why should this saving take place? Well, the intuition is very clear. Why should the saving happen? Yeah, because G is H hard. Okay, so until and unless you have communicated sort of age bits Okay, so so, you know, you have here the inputs of Alice and Bob Yeah, not this. I mean in order to so you you are evaluating G on each of these blocks, right? And this the the individual instance of G is you should assume it's H hard You require this shows clearly that this heating distribution a trivial corollary of the heating distribution Is that the communication complexity of G is H? Okay, so until and unless you spend H bits on a focused coordinate You know nothing about it. This is something you have to prove, but I'm saying this is the intuition, right? So every what H bits worth of communication you are revealing perhaps too much about one bit at that point Aha, you know the one of the one of the coordinates has become very clear, right and so on Average You know this is happening. Okay, so this is this is the intuition to expect. This is very clear. Okay, but you have to formalize this Okay, so that's that's theorem one and Yes, yes for all epsilon Yeah, and delta we assume is less than one over hundred whenever delta is one over hundred this will work for all epsilon in Epsilon is any any parameter No epsilon does not depend on ends. Oh epsilon in principle could So epsilon is between let's say now for now epsilon is a fixed number between zero and one and Whenever delta is less than one over hundred this inequality is satisfied Okay, so that's the first theorem The second theorem is Yes So if g was chosen to be a random function, right the thing that's unclear is if whether a random function satisfies the heating distribution No, it's the random function will not satisfy. Yeah, you can prove that a random function with high probability will not satisfy This theorem does not imply that no for a random function No And now I'm going to state the theorem with Tony and others where where we will be able to show that a random function works The only the okay, let me state the other theorem and then I'll say the the the way these things compare For which this works, uh, there are many Examples, uh, although a random function doesn't satisfy it But the examples that but many of the cardinal problems and communication complexity in this regard do not behave like random function for example the index indexing function the inner product function the gap hamming problem All of these would satisfy this, okay And so, you know Okay, let me first state the theorem for the other thing. So what is discrepancy? Okay I mean if you have taken a course in communication complexity of surely seen this but let me still state it again Okay discrepancy is so here so far Uh discrepancy in some sense will seem like an overkill. Okay, but it's very powerful So if you have a communication protocol it Basically, uh, we all know it partitions the communication matrix into 2 to the c More completely monochromatic rectangles, right now it turns out now suppose I say that I have forget that not only My function is hard in the sense that not only does it not have non large monochromatic rectangles If your rectangle is slightly large It has very small discrepancy meaning, you know, if you think of it as being points are being colored by 2 colors 0 and 1 The coloring has very small discrepancy Okay, in the sense that they're roughly balanced either your your your rectangle is extremely small Or they're roughly balanced. Okay, so, uh, formally you would want to say that the discrepancy Of a rectangle are For a function f is exactly equal to the probability Is this clear so so you're looking at Yeah, so x y is chosen under the uniform distribution So for now in this talk, I will only deal with the uniform distribution. Okay, so you sample So basically you want you you've given some rectangle Okay, and you randomly choose an x y and what's the probability? So it's basically, you know, the the probability mass of The ones inside this Minus the probability mass of the zeros inside is the difference of these two probability masses It's a very natural way to measure, you know, if you had a monochromatic rectangle Then this discrepancy would just be the probability mass of this rectangle Right because yeah everything is one colored the same way And now you're looking at the difference of these two masses Okay, and if they had perfect, uh, balanced coloring then this would become zero Right, so the smaller this difference is The less biased this rectangle is and therefore this is Being even further from being monochromatic Okay, so it turns out That this is a very powerful quantity If you worked on extractors, this is this may this may This may remind you of an of an extractor like property Because I'm using that word because I talked about hitting distributions there. This is somewhat similar to an extractor But when you have this very powerful property, then you can in fact show that not only so maybe we should say g here because we are talking about g Yeah, so so not only is g hard for deterministic protocols. It's even hard for randomized protocols Okay, and our theorem 2 is now saying And and oh the important remark is if you choose a random function g on b bits So it's it's a fact that you know For a random g The discrepancy sorry, this is discrepancy of a rectangle and the discrepancy is The maximization of this quantity over all possible rectangles Okay, so discrepancy of discrepancy of g Is maximize this same quantity r Over all possible rectangles now for a random g discrepancy of g Is extremely small is is like about 2 to the minus I'm assuming g is over b bits Okay, so for a random g is extremely small exponentially small okay That's the nice thing. So most g's will have this property that will have a small discrepancy and the theorem 2 with With uh, you are Sajin or and Tony Is the following result again Informally because I think I'm running out of time is that same setup now f is f n of Gb, okay, and Let's say The discrepancy of g Is Is is is at most 2 to the minus eta b for some constant eta Okay And b Is at least c times log n Where c is some constant that depends on eta Okay, then we can say two things Okay, so we make we made a stronger assumption and we are able to prove much stronger things, which is that not only now The the the deterministic query hardness lifts to the hardness for deterministic protocols But also as you can expect the randomized hardness Lifts the randomized decision tree hardness lifts to the randomized communication hardness Okay So the advantage of this theorem is that as I said Is that it applies to yeah random gadgets g and you also get a randomized lower bound, right? Such a randomized lower bound. We don't know how to obtain using Hitting distribution property, okay, or any any any any let's say, uh, natural generalization of hitting distribution property Okay Sorry in place of b. Yes. Yes. Yes. Yes Because the communication complexity of of g would be about log 1 over discrepancy of g Okay, so let me now just try to say a few very Brief things about how the proofs go, okay So The point is in each of this In both these theorems what you would start with is some arbitrary communication protocol I will stick for now with the deterministic protocol Okay, so you're given some deterministic protocol And you want to extract a decision tree out of this deterministic protocol With absolutely no idea how the internals of the protocol is working, right? But if you think about a protocol a protocol itself is also a tree, right? You have alice Who talks then you have bob And then again alice And so on Right and here the leaves represent as I said Rectangles, okay, I'm talking about deterministic protocols Okay now The point is alice of course is no longer in this protocol Quaring one bit. She's computing an arbitrary function on his on her side And so is bob, right and we want to extract from here Another decision tree. This is some sort of a tree and we want to extract another decision tree where you will be for We'll be just querying One bit at a time So the idea is So I want to construct this decision tree and the decision tree. What is the input the input is Some z in 0 1 to the n right so the point is the general Strategy would be to delay querying a bit Okay, there will be some events that will trigger me to In the query in the in the query protocol to to to make a query And when does that happen at the beginning things are nice because you know the protocol has not said anything So, you know all possible inputs for now if you were just watching The transcript of this protocol how which path is this executing in the protocol tree at the very beginning It could be you know nothing you it could be the whole set of inputs, right? Then you know alice compute some function and maybe you know, this is one So you go down this path when you go down this path Some information about alice's side is leaked, right? And if alices Revealed too much information about a particular bit Then that's a cause of alarm because you know alice's input might start looking like fixed getting fixed in that particular coordinate right, so the idea is that you want to sort of Keep a measure about each coordinate, okay And as long as it's some sort of a measure complexity measure and as long as that measure is large You want to say well that bit that particular coordinate is far from being revealed Okay, in the two theorems we use two different measures and when that measure is low on a particular coordinate Then you will make a query and if you make a query Again the intuition is is is clear. Why did the measure go? Why did the measure becomes low on that particular coordinate because alice and bob chose to communicate Something specific about that coordinate if they have done so Then you then then the natural expectation is Well, they've only communicated c bits and most of these c bits were talking about this coordinate Then the other coordinates if once I query about this and project myself onto the other coordinates Very little must be known about these other coordinates each time I make a query I must While alice and bob chose to make the this particular coordinate. I'm saying it in a very vague way Alice and bob chose to Reveal a lot about this coordinate You know, so this coordinate became very very Fixed looking like the other coordinates Remained unfixed. So when I project I should gain back something. This is the this is the rough idea okay, and in the case of The second theorem this measure is about this measure That that captures this this intuition is the following. It's it's a mini entropy based measure So what you want to say is if you have two random variables x and y right In fact, just let's first talk about One random variable you have a random variable x and so x is Formed of x 1 to x n right each xi Let's say is in 0 1 to the b yeah, so We want to talk about the density Of x Now what is the density of x the density of x if it is It is equal to let's say gamma then what I want is that For all subsets of n Right, so this is the some subset of coordinates that I chose to pick Yeah the min entropy Of x projected down to that sub Those those those set of sub those set of coordinates right. This is just a projection of x right that well if things were perfect this min entropy would be equal to i times the cardinality of i times b that's the maximum maximum possible entropy that you could have Well, you wouldn't have that but you would have that at least gamma times the size of i times b so i times b is the maximum possible min entropy because you're you know, you're yeah You're over b bits that i coordinates But you want to have at least a certain fraction of that and this is true for every possible subset i So then I say that the density of the random variable x is x is gamma yes, why isn't x I just uniformly distributed because you are just No x is not uniformly distributed x is a random variable I have not assumed that x is uniformly distributed or anything x is not uniformly distributed Okay, and similarly I could define y Arbitrally distributed, but the point is x and y are independent because I'm in a rectangle Okay, so so I could define that and at each Roughly very very roughly at each node of the protocol tree The invariant that I I want to keep an invariant that will preserve this this uncertainty about the coordinates is that the min entropy of both x and y Sorry, the density of both x and y measured in terms of min entropy is large Okay, these two properties discrepancy and the deltas property. Mm-hmm discrepancy is strictly stronger than the deltas property. They're they're not comparable They're incomparable one doesn't imply the other As country because you said like we have a stronger assumption here They're free able to prove a stronger result, but there's not strictly a stronger assumption. No, no, no Yeah, no, it's not a stronger result in the sense that okay. We can prove the same result here, but we can prove additionally Uh result about randomized lifting. I said it's stronger in that sense Yeah, but the properties seem incomparable Why is it uh the case like like you said a random function has high discrepancy, right? Has small discrepancy. Yeah, just No, but then why does that kind of not seem to imply like maybe with like a for a weak Like parameters delta h that it it it should have delta h property Like there it depends on what I said delta to be right You want the the rectangles Uh, you know to be to be well spread out the monochromatic rectangles, right? And if you're given a sort of a large rectangle, yeah, uh, then If you think about it if you if your rectangle the monochromatic rectangles are not too large In size not sufficiently large in size then you have no chance of catching a largeish rectangle Okay for in order to have this Distributional thing You you you if you think you will see that you need to have Rectangles of a sufficiently large size and a random and a random g does not permit that So that is why you can't have these things nicely spread out