 So, this is what I will talk about. So, before anything let us fix some notation, fq is the field of q elements and the point is that there is some ordering of arbitrary ordering of elements does not really matter, pd is the set of polynomials of degree at most d, univariate polynomials. So, this is some basic notation that let us agree on this. The main problem here first I will tell you a picture of the problem. So, this is a grid let us pretend that this is fq cross fq right and we ordering the field elements a 1 to a q right. The point is that for every a i, I give you a subset a i of the field. So, in some like I give you a set of points for every element in the field right. So, for a 1, a 2 and so on a q. So, they need not be like contiguous anything they just some sets for every element and you are supposed to find out whether there is some polynomial of degree at most d that passes through these sets. So, this is a function right it is one if there is some polynomial that passes through all these sets and 0 otherwise and the point is of course, the degree must be restricted because this as long as a i's are empty there is some polynomial always passes through them. So, this is really the problem I will define it formally. This is the source as far as I know I think this is the only place was ever mentioned anyway. So, you are given an integer d which is a degree and your subsets a 1 to a q and output 1 if there is a polynomial such that on a 1 it takes value in capital A 1 and so on such that the polynomial is completely contained inside this in these sets the graph is completely contained in these sets. It is a function on q cross q square bits to 0 1 this q is the degree never needs to be more than q it is and the problem and this problem is that show that this is NP hard and it is I do not do that here. So, but as a combinatorial function this is still very interesting too. So, here is a picture again like so the given sets a 1 to a q find out whether there is a polynomial curve whose graph passes through these things. So, this is the thing right and so I should mention this it is not really relevant to the rest of the talk, but I should mention that this is a close related to what people do in coding theory it is like it is called list recovery. The point is you fix any q re code and you fix the you parameterize it by an integer l which is an integer not an alphabet or the point is that I send you a code word and for every coordinate you give me a subset of alphabets saying the code word is probably will is in one of these alphabets and the point is decoded in this fashion so I guess you can guess that read Solomon codes is what anti problem is related to and it is kind of an unparameterized in the sense that I am not telling you the degree before I am not telling you the code I am not telling you the l beforehand and it is a Boolean version like coding theories are usually interested in when this list size is small, but I just want to see if it is non empty at all. So, that is some kind of background good. So, here is the main result which the title might have suggested by the way we parameterize it to like let us agree on a degree beforehand this only makes whatever lower bound stronger because so let us any question so let us agree on a degree beforehand now it is just a Boolean function on q square bits right and let us denote this by d and the point is right any high touch circuit with and or not parity gates that computes this function must have size at least some exponential and sub exponential in the number of inputs right and yeah right this omega depends on this r. So, I am thinking of r as a constant. So, this is the result here I am sorry no there this is an easy 0 parity thing so there is a parameter by high touch and I did not use depth d because this right of course right. So, after this like barely a month ago Bogdanov told me that you know there was a paper in 2007 like by Bogdanov and Safra where the techniques also can be used to kind of do this, but we will do something related, but something I guess stronger and interesting. So, let us so far so good so let us continue. So, here is some relevant background for this. So, this is right so I am just defining a function it is a Boolean function on n bits and I am telling you that it can solve the p delta coin problem if it can distinguish between coins of bias slightly less than p and slightly more than p. So, that that is when it solves the coin problem formally like f if I give it Bernoulli independent random variables of p of times 1 plus delta it mostly outputs 1 and if I give it p time I might say that it mostly outputs 0. So, this is when it is like f solves the coin problem on the face of this it does not look too hard you count and you can probably that will probably tell you the correct answer, but you know we are in ac 0 and you cannot really count these things are not really possible here. Just just to make that formal majority does the majority function does solve the half order 1 over root n coin problem you need at least 1 over root n, but the point is as this delta gets larger as the gap between the biases gets larger the problem becomes easier. So, here is another problem like a tribes function solves the thing with a slightly more bias gap the slight detour, but what is a tribes function it is a it is a Boolean function on some k t bits where t is log k plus some constant whatever constant will make it solve the half 1 over log n coin problem and this is how the function look it is a DNF it is an or of ands and all the variables of every monomial are disjoint and they are all of the same size. So, this is a tribes function I drew it like this because it is a height 2 function circuit of linear size that is and so, there has been some like recent work this year by Lemma Srinivasa Srinivasan Tripati and Venkatesh Venkatesh show that essentially show this one like if I increase the height to h I can solve the half comma log n to the h minus 1 coin problem. So, ignore this tilde over here. So, tribes is an instantiation where h is 2 and big O of 2 is 1. So, you can you can solve the 1 over log n gap in the coin problem with this and I guess for this talk the extremely interesting second result is that this is pretty much if you ignore this tilde it is it is it this is the truth the any circuit that solves the half delta coin problem has its circuit size at least so much. So, you can plug in delta is 1 over log n to the h minus 1 and you see that barring this everything matches. And right this was also independently shown by Chattopadhyatamayi Lovettanthal. So, in pretty much at the same time right. So, far so good. So, the point is that half is not really special if I tell you this statement like you can probably prove this without much effort. It is like saying if I want to create a coin of bias one-eighth I just take the and of three coins of bias half that is pretty much what is going on here and half is nothing special. The whole point we need to go to a lower thing is that first anyway. So, let us see we have that any circuit that solves the coin problem has at least a certain size and we want to show that some function has at least a certain size. So, you can guess where this is going that this function does solve the coin problem and right and we need this because this function only works for R q for like you know we are working over R q for constant R there the bias at which it distinguishes lies at q to the minus r. So, more formally rather more elaborately when you plug in inputs of independent Bernoulli's of bias slightly more than q to the minus r you get slightly less than q to the minus r you get 0 mostly and if you plug in inputs of bias slightly more than q to the minus r you mostly get 1 and you know probably guess that the 0 statement is the easier part. So, I guess the main sub point is that is the one statement and this is one of the reasons at least I was very motivated by this even before the technology of the previous slide that how do you how do you chose such a statement that if you pick a bunch of points at random you will contain the you contain the graph of some polynomial at that time I could not do it and that was my main motivation for this. So, like I said why do you restrict? So, why restrict ourselves to this case? So, it is not known to be very easy in other cases. So, for example, why not something like this where r is something 1 over log q here you would have to show that it would solve the half comma epsilon bias you know half is a nicer number than 1 over some polyn. So, why do you restrict yourself to this point and the answer is I do not know how I will get to this at the end, but at the moment I do not know how like I can only do if r is a constant. So, we are here we are at the main point. So, the 0 statement is if you plug in independent Bernoulli's of 2 small bias you get 0 mostly and if you plug in independent Bernoulli's of slightly more than q to the minus r bias you get 1 mostly. So, I will start with the proof sketch proof the first part and like I said this is the easy part. So, what do we really want to show is that you pick a 1 to a q each containing independently by including each element with probability q to the minus r times 1 minus epsilon. So, yeah log. So, the point is epsilon is will be 1 over polyquial stop kind of showing that as a parameter just yeah, but here it is just very easy the point is right let x be the number of polynomials that you the number of polynomials that you catch in this set and the expected number of polynomials is of course, you sum over all polynomials this is a probability of containing each point in a polynomial and the q points in the graph this is easy and this is at most 1 this is right and yeah this is q e d right and for the second yeah. So, look. So, how would you want to do the second part just the ideas one guess would be just try and compute the second moment right, but that is no good the second moment is actually too large it does not right and I guess I can go into this a bit later, but the point is again you can quote and quote made to work, but it does not really solve any it does not really give any lower bound question so far yeah epsilon is something smaller than 1 this is small no no right right right. So, the point is if epsilon is smaller it is a harder statement to prove. So, here it is 1 which is more than here it is q to the minus 1 which would be a harder statement to prove than minus 1 third which is what minus 1 minus r over 3 was. So, I mean it is a point is like yeah this implies it for q to the minus 1 minus r over 3 that is the point yeah it is right the second moment is too large, but you can quote and quote make it work, but does not anyway does not solve the coin problem question so far. So, here is here is how the second part works is that you try and count the number of polynomials in the Fourier basis and what do I mean like define a function f define a function f from f q to the q define a function f on f q to the q to 0 1 right. So, this looks weird, but what is this function indicating the take any polynomial of degree at most d write down the set of all its evaluations that is a vector right. So, f is an indicator of all such vectors and the point is that you add up two polynomials of degree at most d you get a polynomial of degree at most d. So, this is the linear space. So, this is an indicator of some linear space and which we will call c from now on right and for every a i let g i be the indicator that x i is an a i. So, this is also so the way I am stating it this looks like this is a function from f q to the q to 0 1, but it really only depends on the i th coordinate. So, where there is no confusion I will just I am not abuse notation, but the point right. So, given these two x is exactly the inner product of f and product of g i's because you want f to be 1 in the sense that your vector should be a polynomial and every a i should also x i should also be in a i that is that is counting the number of polynomials right and what do you want to show is that if a 1 to a q are chosen by including every point with probability at least something then this quantity is more than 0. So, the point is that this is a normalizing factor I do not care about this is really where it is you know where it is happening. So, this will just agree that this is what we want to show is more than 0 and continue right. So, I guess one it is one of these things where the first if I tell you the first one and half steps it is the risk is pretty straight forward to work out and. So, I guess at least at that point the weird thing for me was that I was always used to using Fourier analysis to convert a convolution into a product, but here you kind of unfold this product into a convolution, but luckily things work out happily and this is what you get you sum over all the dual space and it is like as if you are splitting up this g i remember like c is the space of all the polynomial evaluations it is pretty straight forward things actually times some normalizing, but you ignore that it is only the non the zeroness or non zeroness is what matters here right. So, you remove the zero part the one where alpha 0 which is also lies in the dual space minus everything else it is a typo it is actually at least this much it is not equal to right because I do not know the directions of these numbers anyway. So, what can the hope be this is that this term really dominates and the rest of it is junk I do not know that is junk yet. So, I will just call it r. So, the point is that every g i hat has expected value 0 and you want to understand how this term looks like. So, very natural thing this is a square root and hope the cross terms go away right and you want to show this of course you want to show this is at most this much smaller than the main term what will be the main term right and luckily that is exactly what happens. So, you look at r question no it is an indicator function of a i like a i is a set right and whether this coordinate belongs to a i right and right. So, if you look at this. So, because of that when you square this you pretty much take the expectation all the way inside right. So, this is what happens here and I guess the relevant facts are you know the 0 Fourier coefficients of sets how they look like and you know the non-zero Fourier coefficients of set how they look like and really the main sub sub point is that this is much that is much bigger than this and. So, you want to hope that in most of these summands most of the product is dominated by these things and not these things that that is what that that would do it and yeah. So, that is a good thing and we can do that. So, we can say weight distribution of is one of these things where if I the dual space of. So, c dual c is the space of evaluation of all degree d polynomials the dual of c is the evaluations of all degree q minus d minus 1 polynomials. So, these things are well understood and get exact weight distribution as per we want. So, let us wrapping up this is small because you know we most of the terms are coming from here and. So, with high probability that is small r itself is small and on the other hand product of g i had 0 is large because g i had 0 the size of a i and if you pick points independently the set should not be too small and that is really what this thing and this is pretty much just a standard large deviation bound and therefore, it is large therefore, this is non-zero and that is it I am putting a box it is a q e d and sure it is also that x is what you expected to be high probability. So, that is that is all for this one. So, I guess sure I have a bit of time I can tell the other results. So, the point is that like I said only this only work when r was a constant when the degree was some constant times q, but it works whenever it you know you can prove it for any r, but you kind of you give up on solving the coin problem you just want qualitative sharp threshold and the whole you know I wrote like something should work in the paper, but I have had time to check since and it does work I will just tell you the proof pretty quickly. So, I said second moment does not work, but when you put quotes it works since I have time I will. So, point is why the second moment not work. So, you are picking a subset s square right you are picking a random subset here and for polynomials p 1 and p 2 the second moment essentially involves estimating terms of like this. If what is the probability that I contain p 1 condition that I contain p 2 right, but the point is the size of s is something like q to the 2 minus r the size of p 2 is q like a polynomial contains q points and this is much bigger than the standard deviation that you get when you pick such a set and. So, you have changed the distribution. So, what do you do you just condition on and then you most of the bigness goes away, but it is it is not all done. So, you have to use the you can show that the probability of it being non-zero is at least some half right and then you say that because it is a symmetric function all the influences are the same. So, by k k l that all the influence must be at least something by Rousseau-Margoulis if the influence is large the function must be growing pretty quickly at that point and so it must be close to once that that is a proof and that gives delta some log in, but the point is it works throughout without any hassle that is the first result other result. The second other result is so far we have been looking at containing full polynomials. So, what if I want to contain only a suppose I fix a degree and I give you subset and I pick a bunch of random subset like what is the largest agreement of this polynomial that this subset has and there is a theorem like that too. If you pick up if you want to contain a polynomial of degree D in a subset S then you contain at least so much I will explain what this means, but here is a picture like you are not containing you are not you are not completely contained in the you are not completely passing through these sets, but and the intuition is that this is what you expect not in a probabilistic sense in the sense that you cannot expect more than this by you that is easy to show that you will not contain more than this and the point is that this is just junk that is the truth. Questions. You need to cut interstate all the sets. No, what no like what is the largest agreement with the set in any polynomial have. It is not two sets this is the same set I mean some part of it is here some part of it is here. Yeah, so yeah this is an example this was to illustrate that they are not necessarily intervals. Yeah, just want to pass through them. So, anyway so this is the this is the second theorem and right I have like this have like the point is that this goes back to when I was saying I could not like does it solve the coin problem and like I have asked this question in many ways, but I guess one clean way to do nothing with the coin problem is to ask what is the influence of this function this is.