 Okay, thanks for the invitation. Thanks to you, to our... To our organizers, so we'll see him. Obviously everybody's liking this idea. I think quite know what to do with this request. So I thought it would be fun to talk about something, a problem that everybody's probably heard of back in the day, perhaps. And some recent work that I've been doing with George Shakin at Oxford University. So the Frobedius postage stamp problem is, well, a simple example is the post office has an unlimited supply of three and five cents stamps, but due to the current crisis, they're unable to get any other denomination stamps. So what exact postages can you pay with some combination of those two stamps? So the smallest amount you can pay, zero. But after that, well, you obviously can't pay one or two cents. You can pay three cents. Four cents, you can't pay five, you can. Six, but seven, there's no obvious combination. And if you just keep on going, you find, well, here's the list. And now, hold on, I wanna see if I can get a cursor I can use. Cursor I can use. There is a way, but I'm not seeing it. Anyway, let me, oh, can you see my cursor? Yes. Oh, you can, okay, great. So I didn't need to even ask it. Okay, so what we see is that at eight cents, we start finding many combinations of three and five cents stamps, and it seems to go on. And then if you had actually worked it out yourself, which you may have done so, you'll notice that there's a pattern here that going from eight cents to 11 cents, the difference seems to be just add another three-cent stamp. So in fact, we can go from eight, nine, 10 and add another three-cent stamp each time and get 11, 12, 13. And then we can proceed by induction and obtain any integer greater than or equal to eight. I don't know if that was too fast, but it's trivial mathematics. You just add three cents, add it again and add it again. So what it means is that we can get any integer number greater than or equal to eight as an eight combination of three and five-cent stamps just by starting with eight, nine, 10 and adding on three-cent stamps. So what's the proof concept? You cover each congruence class mod three and then you keep on adding multiples of three. So what we saw on the previous slide was that you can cover every non-negative integer except one, two and four and seven. So this is an exceptional set that the numbers you can't represent as linear, positive linear combination or non-negative linear combination of three and five. So try and keep these numbers one, two, four and seven in mind, we'll keep on coming back to this example. Okay, so let's try and generalize this idea. So let's give an A and B-cent stamps. What is the set A and plus Bn? Obviously, it's really a number three problem where we're just taking positive integer or non-negative integer linear combinations of the positive integers A and B. So I'm gonna call PAB my set of numbers that are represented as positive linear combinations of A and B. And one, it's not hard to see that if G divides A and B we can just divide it out and we'll get G times what happens if you add A over G and B over G. So for that reason, we'll just assume without any loss of generality that A and B have G, C, D, one. And so what was the idea with three and five? We wanted an element of each congruence class mod three and well, I'm not gonna do it exactly the same way as before, I'm just now, I'm just gonna say what's the minimum number we can represent in each congruence class mod three? Now, maybe what I should have said is if I've got up here, if I've got three times n plus five times n and I wanna think about that mod three, the three times n is irrelevant, it's just a five times something that's relevant. So to get the minimum numbers, you just do zero times five, one times five and two times five and that'll get you the minimum representative in each congruence class mod three. So what integers can we represent as non-negative linear combinations of three and five? Well, we take that zero times five and we can add on any multiple of three, the two times five then add on any multiple of three and the one times five and that's the set of integers represented. So what integers are missing in each of those congruence classes? Well, here we started at zero, so obviously nothing's missing. Here we started, let me start here, here we started five. So what numbers are missing that are two mod three? Well, just two. And here we're looking at the numbers that are one mod five and we started at 10. So the number's missing a one, four and seven. So you see the exceptional set appearing one, two, four, seven and this is kind of a perhaps more logical way to go about constructing it. So we could also do it the other way around. We could have started with mod five and look at the smallest number in each congruence class mod five, which are zero through four times three and the numbers missing in each congruence class and you see again it's one, two, four and seven. So here's two different ways of coming to the exceptional set in a way that generalizes. So here's the general idea. We've got A and B, which are code prime and we're gonna look at the smallest representatives in each congruence class mod B. These are just the multiples of A and the multiples from zero and B minus one and then the numbers that are represented are those multiples plus any multiple non-negative multiple of B. So what's missing? What's the exceptional set? I'm gonna write, so I'm gonna write an occasion where you get all the non-negative inches minus some finite exceptional set and what is the finite exceptional set? You take each of these arithmetic progressions mod B and you look at all the numbers that are less than M times A. So this is what's represented and then what isn't represented just like in the example up here is the numbers that are smaller than M times A in that arithmetic regression. So what's nice is we can see it's a finite set. If you want to count the number of elements of a set and do various things, so there's all sorts of history of this problem. But I wanna bring in a modern twist on this whole problem. Perhaps I should have said one of my motivations for working this problem is that it's an old problem, but there's a new subject of additive combinatorics which is very interesting in adding things together and looking at the structure, but with some control. And it's all about structure. So let me give a modern twist on this whole problem. So in the list of representations I gave up to 13, there was a unique way of representing each number. But once you get to 15, you could place five 3 cents stamps or three 5 cents stamps on your envelope and both would get you to 15 cents. So the question is, which is the better choice? So given that we're all trying, well, there's been so much less emissions thanks to nobody moving around and we want a green world. The hint was I guess in the color there. To save a planet, we wanna use less stamps. So the choice is that 15 cents is three 5 cents stamps. Another perhaps reason you might want less stamps is your envelopes only a certain size. And if all your stamps are the same size, then you want fewer stamps. So what we're gonna be interested in is not just what can be represented with three and five cents stamps, but if you've only allowed to put 20 stamps on the face of your envelope, what numbers can you represent by a combination of three and five cents stamps? So I wanna do a slightly strange thing. So I'm going to look at n times my set. This is a sort of notation for n times your set. Well, what I'm going to do is allow m and n, m times three and n times five, m and n going up to capital N. And whatever's left over, I'm going to say it's a multiple of zero. This is very convenient because I can just write everything as l times zero, three times n, five times n, where the sum of these coefficients is capital N. So here, the set A is all combinations of zero, three and five on a multiplier. I'm adding a set to itself n times. So what we've seen is that for this set, the exceptional set is one, two, four and seven. And what I wanna do is understand what n times A looks like for, n is one, two, three, four, et cetera. So let's just do a calculation. So two times A, it looks, there's not much structure to be seen in here. Three times the sets, I'm just taking some linear combination of three of these guys, maybe repetitions. This is what I get, which, well, we start to see in the middle there, the eight, nine, 10, 11 that we saw before, right? We're just missing down this end, we're missing the one, two, four and seven. We get the stuff in the middle and then we're missing a couple of numbers at the end. So that's what three times A looks like. It's everything you'd expect. Well, you couldn't, with three stamps, you can't go to bigger than three times five. So the range for three times A is zero to 15. And what we're interested in here is what's missing. So if I look at 40 today, I see in the middle here, eight, nine, 10, 11, 12, 13. Oh, that's a typo, the 14. Sorry, I should have put that in. And we're just missing 17 and 19 at the end and the one, two, four and seven here. So already we're starting to see a structure that three times A and four times A, they're missing an exceptional set for A and they're missing a couple of numbers at the end and just two. So let me just make the observation that since we can add zero, if I take three stamps, that's gonna be a subset of four stamps because I could have just added zero. And so these are nested sets and they're all the end set is always between the integers between zero and five and we're missing some. Okay, so what we're gonna be interested in is take this set, subtract n times A. And what are we missing? We know we're missing exceptional set at the beginning, but what we've seen in these, at least so far when we've done the calculations, this is two integers missing at the end. So let's just continue looking. Four A is in five A and what I'm gonna do is take a bunch of elements and add five, which will be the new elements I can get in five A. And then when I do that, I mean you can verify it if you'd like, you can see I get every number between zero and 25, except 20 times 24, which you'll notice are just five more than these two numbers. So a pattern is emerging. And in fact, just this proof of picking these numbers for the next one, I just did 17, 19, 21, 23 and 25 and add five. By induction, one's able to prove that you get everything between zero and five and you're missing the exceptional set at the beginning that we know can never be represented as many combinations of zero, three and five, but somehow we're always missing two numbers at the end. So what are the, oh, the other thing I should say is that if you think about zero, three and five or zero, let's look at this one, it's missing one, two, four and seven. And let's think five N minus one would be nine, five N minus three would be seven. It's also missing seven and nine. So it's not just for N greater than or equal to three. These are also set minus these, it's just that there's some overlap between these two sets. So for all N greater than or equal to one, this is true, easily proved. And the next question is really, can we understand what on earth why minus one and minus three? So we have to do a little calculation here and that's simply what we're gonna do is, and this is partly why we've formulated things the way we did, is we wanna look from the other end from the five N ends of what's going on. So if I have a number between zero and five N and I write it as L times zero and M times three and five times N with L plus M plus N equals capital N, then what I'm gonna do is look at five N minus K and then, well, straightforward enough calculation is a linear algebra, but here's the beautiful thing. It ends up five N minus K is just like N times zero plus N times two plus L times five. So it's a linear combination of, well, N numbers, L plus N plus N of N, but now with the set zero, two and five. And what do you think the exceptional set is for zero, two and five? What numbers aren't represented? Well, it's actually easy to see because we just have to look mod two, right? We just have to put the congruence classes. Now zeros are represented, so zero, two, four, et cetera, represented. What's the smallest odd number? It's five. So the only odd numbers missing are one and three. So there's a proof that that's exceptional set. And, well, in the notation we're gonna use, five minus our original set is zero, two, five. That's just the calculation we did up here. So what we've proved is that this five N minus three, five N minus one is, well, five N minus the set one, three, which is five N minus this exceptional set, which is a modest sectional set, it's just five minus A. So what we see is that for this set A zero, three, five, a number of the set of integers represented by N, a linear combination of N numbers from the set A is all the N numbers from zero and five N minus an exceptional set for A and an exceptional set that appears from the other end. So, yeah, so that's cute. But here's the thing, this is kind of where we're sort of in the modern formulation of private across that that's one of the beautiful things bad. The set of integers to itself enough times structure appears. And here the structure is you have a whole interval except some very understandable endpoints. So the general theorem is for any set of three elements, so two stamps and zero, which are co-prime, you get, well, here I'm gonna have my numbers be zero, A and B, we get all the numbers between zero and B times capital N. And the only things that are missing are the exceptional set, which we proved earlier on was finite, at this end and then the exceptional set at the other end. That's all that's missing. Now, the funny thing is actually that although this has been reproved many times in literature, it seems to us as the first time to note that actually this actually works for all N greater than or equal to one. So it's always true that you get N times A is exactly this. So question, can we generalize this for a larger set of stamps? Well, okay, what am I saying generalize? Can we generalize the structure theorem? And well, here I'm just being very specific about is it true for all N greater than or equal to one? Let's leave theorem. So if I just take the set of stamps one, nine and 10, then we can see those exceptions, right? Because I can just add one to itself as many times and get every integer. And when I look at 10 minus A, well, it's the same set. So there's no exceptions for 10 minus A. So given that's the case, we note that one, if this above theorem was true, A would represent every number between zero and 10, but it doesn't. We're gonna have to wait a while before multiple of A represents eight, for instance. So let's just, we're gonna have to search for the correct generalization. In this example, we can see eight will be represented by eight times a set. It is like eight times one, but it doesn't appear in seven times a set. And more generally, if A was zero one, B minus one, B, then B minus two, so the equivalent of eight in this example appears here, but it doesn't appear in the set before. So the best we could hope for here in if we're gonna generalize arbitrarily large sets is n bigger than B minus two. So that's going to be our first goal in this book. And well, okay, the goal is just proof for sufficient large start off with. So we saw if A has three elements, then we get, it starts at one. And there are A's for which we have to go from B minus two. And here's the correct generalization. A set of K plus two elements, between them they're co-prime, everything can be represented. So we can proceed by four. We could just simply, so it turns out the most convenient thing to do is to split into the integers modulo the largest number here, B. And to take the linear combinations in here, we're gonna do exactly the same thing. We're gonna look at the minimal number represented each arithmetic progression and then add multiples of B to that. That's kind of the idea. So let me just formalize some notation. Unfortunately, we're gonna have to have a slightly complicated notation. So for my set A, I'm going to look at the numbers that are represented as linear combinations of elements like non-negative linear combinations. And remember B was the largest element for my set A. And I'm gonna look at the smallest number in the arithmetic progression A mod B that is represented. And what we know is if you represent this number, then you can add B to it as many times as you like. So every number in that arithmetic progression that's at least as big as this is represented. So the exceptional set is just gonna be the union of the numbers in this arithmetic progression that is smaller than this minimal element. So I'm just going to try and prove a theorem that things work when you're sufficiently large. So what I want to do is look at ends in the arithmetic progression that are bigger than the minimum represented here. And smaller than, well, if I look at NB minus N, if I go to the other end, this is the corresponding number, right? The smallest number represented in the arithmetic progression minus A mod B by the set B minus capital A. And actually, what's gonna make that very easy is this an induction argument to work. So if this interval is represented for N, then the interval, the corresponding interval is represented with N plus one because you just add B to the largest element here and you get the largest element here. Couldn't be easier. So induction works. So we just have to find a first capital N where this works. And then all the subsequent capital N we've got the theorem for. Okay, so here I'm just gonna give you, it's a little bit notation promised proof, but it's not hard. So I'm gonna take capital N where I'm gonna start my induction to be the sum of these two numbers. What are these numbers? Well, I defined the smallest number represented in each arithmetic progression mod B and I need to know what's the first time it appears? What's the smallest capital N in which it appears? And so there are kind of predictions. So that's this number, capital N, A comma A. And so I'm going to prove for this number that everything that I're into will live representative. And the trick is I'm going to look at whether I can represent N or the number at the, I'm gonna try and represent it from the other end if you like. And this sums obviously the B times N and then because capital N is this thing I can say that's actually equal sign. So what that means is that N is either less than this or this is less than that, less than or equal to, right? Otherwise we'd have a problem. So I'll just work with this one. This is the same argument, just a little bit more complicated. So remember I said N was the starting point in arithmetic progression plus the multiple of B. The multiple of B here is nicely bounded just by this inequality. Therefore, when I try and get my number N it's the starting point plus K times B. The starting point we know is in that multiple of A. We just proved that that many B's are in here. And so N is an N times A. That was it. So it's just the start of the induction on the previous slide I showed. There is an induction but once you got it for capital N you got it for capital N plus one. And so what if we proved, we've proved that we do get our structure. So if we've got a set of integers that are co-prime so that we can represent every integer by some linear combination, positive or negative that if we look at N times the set where we only do positive linear combination but non-negative linear combination no more than what's exactly at the end of them then we all the number of N zero and B times N minus the two obvious exceptional sets once N is sufficiently large and this is what's sufficiently large. Okay, so this is not, I've got a compute also secondary thing so let's try and understand that little better. Can we be a little more explicit about this bound on capital N, slow bound? So what I want to do is I wanna show that I wanna get some bound on these terms. And let's just think about definition. It said that I'll take the minimum number represented in arithmetic progression A mod B and I'll write it as a sum of elements from my set oh these subscripts are not the same as those forgive me for that. So these are, this is the representation by elements from the set A and this capital N is the fewest summons. So here's my claim that the fewest summons is no more than B. In fact, no more than B minus one in fact. So why is that? What would happen if if the fewest summons was greater than B? The division whole principle. So does that work? A document is kind of ready as A1 to AN where N is greater than B. Then I can just look at these numbers zero, A1, A1 plus A2, A1 plus A2 plus A3, et cetera. Telescoping all the way. And I've got B plus one different objects here. So two of them are congruent mod B. So if I just take the difference of the two that are congruent mod B, say it's the jth one and the I minus first one, then I get some sum like this which is congruent to zero mod B. And what I'm gonna do is I'm just gonna subtract it from this total sum. So there is some sum, which is zero mod B. I take it out. If I take it out, then this sum without delete it, it's the same as the original sum which is congruent to A mod B. But this is a smaller sum. I've used less elements and that contradicted the minimality. So if we have a sum that's too long, we can simply erase any subsum that's zero mod B. That's, I mean, that's sort of a general idea that if you've got a minimal sum that's A mod B, you can't have a subsum that's zero mod B. So with this, we're able to say that each of these numbers are less than B, less than or equal to B. And so this bound here is two B, worse two B. But can we do matter? We knew that in the set zero one B minus one, that was the sort of worst case we had was B minus two not two B. And in this particular case, where we know that one of these numbers can be large, it's feasible that the other one isn't. It's feasible that this being large forces this to be small. That if we can understand when one of these numbers is large, maybe that structure will allow us to say something about this one coming from the other direction. So again, here's the idea. We're going, so now I'm gonna just suppose instead of saying up to B, I'm going to suppose that one of these sums has got more than B over two elements. That's not true. The two things add together to less than B and I've got what I want. So let's suppose that we've got some minimal subsum that's a mod B and that total is that there's more than B over two elements and there's no subsum zero mod B. So I'm going to add one element to make life a little bit more pretty, which is just B minus A. So now I'll get a sum of N plus one elements of zero mod B with no subsum being zero mod B. Again, if a subsum here is zero mod B, then it's complemented zero mod B and one of those will be a subset of those. So this is sort of an interesting little combinatorics problem, mod B. You have a large number of numbers, mod B, more than B over two of them. There's sum of zero mod B and no sum is it, subsum is zero mod B. So you can sort of ask yourself, how do we do that? Well, one idea would just be to take positive integers, all of which are less than or equal to, which sum up to B. So by taking positive integers that sum to B, all of them must be less than B. So no subsum could be B. So that's a pretty easy construction of a set of integers that sum to zero, a big set of integers that sum to zero mod B with no subsum being zero mod B. So the question is, are there any other possibilities? So let me just talk, I thought I had another thing there. Let me just talk about, how about just taking a sum being two times B? So I've got a lot of integers summed together. So I say two thirds of B and the sum is two times B. Well, it could be a whole mixture of integers, but if I have, if it's two thirds B, there's probably gonna be a bunch of one, two, some threes amongst them. And if I've got a bunch of one, two, some threes, then we can take subsums of those and sort of cover little intervals. So as we sort of move up towards B, it will be hard not to cover the number B itself by subsum. So one way to avoid it would be, perhaps to take each number AI to be two or something like that, except maybe A zero. Now you've got to be a little bit careful because if B was even, you would hit B. But for instance, if I took the sum A zero plus A one up to A and to be two B and all of them were two and B was odd, that would work. And in fact, that's the only other possible but this is wonderful theorem by South Chevin Chen in a theorem in the journal numbers, in a paper in journal number three. It's a combination of several papers and they prove that if you do have a sum that's zero mod B with no proper sum being zero and there's more than B over two elements in the sum, B of two plus one elements, then you actually, what really is underlying it is you have a bunch of integers that sum up to B which are positive integers and then the A is just some multiple of those mod B. That's the trick. So obviously if a subsum of the A j's was zero mod B the corresponding subsum here in the C j's will be zero mod B. So you can see this construction works but what's amazing is this is the only possible construction. And in fact, South Chevin Chen went down to B over three but it's very complicated and it's a subject that's crying out for more people to work in it. I think there's a lot to be done here that would be very beautiful to understand such sums. So yeah, so using this, it wasn't hard. I mean, just one of these sort of annoying chasing around arguments but it wasn't particularly hard to prove that we actually do get that n times A is what you guess. There's always what you guess. All you need to do in B and N minus the obvious exceptional sets when N is bigger than two times integer part of B over two. And let's just see how that compares. This means that we can take, this number is gonna be B minus one will be as B is odd or even and our conjecture was B minus two. So we just failed to prove the conjecture that B minus two would be best possible if it's doable, if it's correct and we just failed to get it. And okay, we certainly believe that B minus two is correct. The problem is that as soon as you go to B the subsum, a sum of size B over two the sum is zero and has no proper subsum just being zero, mod B. Then you've got in Safchev and Chen they point out there are lots of new cases. So one has to go through all those cases very carefully and it's a nice starter problem for a PhD maybe is to actually prove that this is the right answer by chasing through all those cases. What might be true in general is that you could do slightly better depending on the size of a set E that perhaps this number N A should be B plus two minus the size of the set. And here's an example where you get equality. So in our example, zero one B minus one B A was four that's why we've got B minus two and this might be the best you can do in general. Okay, so that's what happens in if you've got an arbitrary set of postage stamps and you wanna do N of them then we have an exact structure theorem from some point onwards. And again, what I should say is in literature something like this was proof of N from some point onwards for us the main interest is to getting as good a bound as possible here. So what I want to do now is say well let's try and think about this in higher dimensions. So here kind of a very elementary problem you just get a set of positive integers you're looking at all the linear combination as positive or non-negative integers. Now I wanna move to Z to the end but let's just focus on Z squared because we can kind of formulate the problem in two dimensions and the formulation works in a higher dimension. So why did we start the set at zero? Well, translation is not a big deal when you're doing additive things, right? So it makes sense to start with zero and similarly in higher dimension we're gonna start with the point zero zero. In the next thing that we did in one dimension is to have the GCD of all the integers being one. What was the reason for that? Well, the reason was that if you look at all the linear combinations positive or negative of those integers you would get a whole of Z. It wouldn't be, you know you wouldn't be barred by some congruent condition. So if we're gonna go to high dimension or two dimensions, what we'll insist on is when you take all of the linear combinations of a set positive or negative you get a whole of Z squared. That's the natural generalization. And then another thing that happened was we said if A is a subset of or goes from zero to B and it's some subset of the integers and then including zero and B then N times A is a subset of the integers between zero and N times B. And the reason was that zero to B is actually the convex hull of the points in A. So if we're gonna generalize the higher dimension we wanna do the same thing. So it's fairly standard. The convex hull is you take the elements of A you take all the linear combinations non-negative linear combinations where the coefficient sum for one and that's your convex hull. And then it's clear that N times the set these set of points in A this finite subset of Z squared belongs to N times the convex hull. And it's gonna be convenient for us to take that hull to infinity. So what is sort of the lattice points that could be represented by non-negative linear combinations in because like when we worked in Z we just had the integers greater than or equal to zero. We cut out inches less than zero we had a ray in one direction. So in two dimensions we're gonna have some sort of cone we might have a lattice point here and lattice point here you can see my picture and we might have some in between but if we take the convex hull we just go out from the convex hull. Okay, so that's gonna be a convenient location. So in two dimensions we're asking to find that subset of Z squared if you look at it what it generates over Z it generates all of Z squared it includes zero zero and we're gonna ask for representation inside the convex hull. So this is just saying what we had on the last slide and again what we're going to, well firstly we're going to say if we think n times a what is the set of everything that's represented and by some linear combination of elements of a and we're gonna have an exceptional set which is take everything in the cone. So that's everything that could feasibly be represented and remove the ones that are represented and that's the exceptional set. So just the analogy to what happens in one dimension. So first question is the exceptional set from that we saw it was in one dimension that was one, two, four, seven if you remember in our original case. And if we're gonna formulate the theorem we probably need to understand what happens when you go to the far end and take off the exceptional set. Okay, let me start with the exceptional sets. So B minus A what was that? Well, I didn't sort of say it when we worked with it but what we were sort of doing was recalibrate instead of sort of starting from zero and looking at the set going towards B where you're starting with B making that kind of a zero and looking in the other direction. So B minus A is like a recalibration it's moving where your perspective to that vertex on the convex hole. So what makes sense in a generalization is to look at for each A and A the set little a minus capital A and that's the perspective from that element of a convex hole. So we'll be looking in that direction and that gives us the guess that N times the set will be the Z squared lattice points in N times the convex hole minus these exceptional sets if N sufficiently large that would be the thing you'd naturally conjecture to be the right theorem now. So let's try an example. So, okay, so I had two options. One was to try and get latex to print more as 400 dots. And I figured the amount of time myself dotting my graph paper is going to be considerably less. So here it is. I'm sure some of you will write to me tell me how I could have done the second, but anyway. So anyway, here's the set I'm going to use. So we've got zero zero we've got two zero we've got three zero so the convex hole will be just the positive quadrant, right? If we've got things sticking out in that direction and I've just thrown in one more element which is one one. So we're going to take everything in the positive quadrant and one can, you just run the computer or you just get your handsaw and you quickly find out this is what's represented. And what can we see? Well, it's pretty much filled in here, right? And if we look along here just at the start there's some exceptions down here and there's some exceptions up there. So again, we're taking all non-negative linear combinations of these vectors and we're looking at what's representing what isn't and what we see is there seems to be things not represented close into the boundary here and close into the boundary here but it looks like there's a pattern so this will probably go on to infinity. So it doesn't look it looks like the exceptional set is no longer finite. So what we're going to have to do is understand the exceptional set if we're going to make any progress. And we also want to prove that basically the exceptional set's only near the boundary that something's here in the middle life's easy. So how did we manage that in one dimension? We said, well, when we were working mod three we looked at the smallest numbers represented mod three and then we just added multiples of three. So what we're going to want to do is look at all of z squared and if we're going to use the same argument do some modular arithmetic. Well, modular arithmetic and z squared you've got a mod by a sub lattice and so we'll look at the sub lattice generated by two zero and zero three. So if we think about that, look, we've got if you can follow my cursor we're representing four three five three four four five four these things here, right? This little six tuple here and I can add to them two zero to get to this six tuple here. I can add again two zero to get to this six tuple here and alternatively I could start again here and add zero three and get to this six tuple here and this six tuple here and then zigzag my way up to infinity. So here's the plan. I start with that and I add two zeros and add zero three and what I've really done is I've added non-negative linear combinations of two zero zero three with set of representatives of z squared modulated sub lattice and that gives me everything out there. So what I've proved is that everything that's not represented in this case is close to one of the boundaries. It's this chunk here we're done. So let's be a little more explicit about the exceptional set. If we just instead, so this is I took the graph I had before and I put across where the number was missing and let's just have a look if I go down here it's pretty clear you can just see that everything's spaced by a two zero, right? So there are three arithmetic progressions here spaced by two zero and if I look up here it's a little more tricky to see out here you can see zero three spacing the next line down you can see a zero three spacing. Here we've got two missing guys but then you add zero three and there are two missing guys, two missing guys, et cetera. So the exceptional set is nine arithmetic progressions where we've got a common difference which corresponds to the vector that defines the boundary. So this is it as written out and well our surprise is that the exception is that it's in finite but it is a finite union of one dimensional objects. So this has been a little more explicit and so our exceptional set is in fact the finite union of sets of form, a vector, a starting point plus P of B where B has one element other than zero zero. This is in dimension two. So if we go to dimension M we have the same kind of general theorem. So we have the same hypothesis, A represents everything plus zero then when you take N times A you get everything inside the convex tilt minus these exceptional sets so the sets that go to infinity there's the only thing missing from some point onwards. So unfortunately our proof isn't explicit. I'll explain very briefly why it's an explicit in a minute but so that's a bad part of the proof. A good part is we can determine the general structure of a set A. I'll explain also we can be explicit in certain cases so this work could be done here. So the structure of exceptional set in general is that it's a starting point plus the numbers represented by some subset of A where the subset is well the elements that subset in fact I should have said they're all linearly independent of one another. It contains N minus one element so it's less than the dimension here and B was a subset of A. So just like we saw in the example that's what the exceptional set always looks like. In general you can prove that the exceptional set always stays a bounded distance away from the boundary of N times the hull. This was proved in some cases in very different contexts by Simpson-Tiedemann in 2003. And a corollary of what we've done is a surprising theorem of Kavansky is that once you get to a sufficient large point you can always write the number of points inside N times A as a polynomial in this capital N of degree the number of elements in A. Sorry, the degree, the dimension that you're working in. And this in 1992 was a big, it was in real algebraic geometry. So this sort of idea of, well they're talking about counting points and polytopes comes up in different fields and Kavansky proved this as I said in 1992. It's quite a fuss in his area and his proof was a little bit different from us. So what he did is construct a finitely generated graded module over C extended by K variables where K is the size of set A. Each, so M1, M2 is the finitely generated graded module. Each of these MNs was a vector space of the complex numbers of dimension N times A. And there's a theorem of Hilbert that allowed him to prove that if N is sufficiently large these elements of a graded modules, sorry, N times A was the dimension of this space. And the theorem of Hilbert allowed him to show that the number of points inside that module would be or the dimension would be a polynomial in it. So I have no idea really what I just said. I'm just reading it out. But yeah, there are very different proofs in this world. The problem, I mean, there are many problems with this proof in terms of how much you can get your hands on things. Subsequently, in about 2000, Nathanson generalized it to a more combinatorial look. And then together with Ruzia, they used essentially elementary ideas to prove the same theorem. They, so a little different because they didn't need to access all the structure in order to prove this theorem. So there's more in what we've done because we have all the structure. So what is this polynomial? It's rather mysterious. I don't know if you've tried to use air-heart polynomials. So you've got some shape, it's corners on lattice points. And you try and count the number of points inside. There is a polynomial that will tell you. And this polynomial, and the first few are obvious, like the top coefficient is the volume, right? Number of lattice points inside the shape. The next is to do with the volume of the sides, one dimension down. The third coefficient, you not only have the volume of the n minus two dimensional sides, but you've got to take account of the angle between those sides. And they're just mysterious. I mean, how to compute them, how to really understand them is very hard. So in terms of actually writing down these coefficients of the polynomial, it's very tricky. So, strangely, we can have a fairly explicit description of what this thing looks like, but we struggle to count because of the difficulty of constructing coefficients in such polynomial lists. So I'll just say a couple of the tools we use. So there's a beautiful theorem by Hera Theodoron, better known for complex analysis. And what we break this down to is if we have this set A, then what Hera Theodoron's theorem says is if you have a vector inside this convex hull, then you have a linearly independent subset of size n plus one, which V is inside that convex hull, I should have written this as convex hulls. So we can, instead of looking at some arbitrary set, we can break it down into sets with which are linearly independent. So a bit like when we're working mod B, before we only need to think about the elements zero and B. Once we've filled in the ends side, we're able to do that in higher dimension, but there you have, it's a little more tricky to include everything. So we needed this theorem that Hera Theodoron had needed. Another interesting theorem that we use is by, well, I first learned a bit in a famous combinatorial number 30 paper man, but the algebraic parameters know of the theorem that this was proved by Dixon some time earlier. It's funny how the same ideas proliferate in different directions. So here we're going to take numbers in our N and we have a partial ordering on them, which is that A is less than B if all the coefficients of A are less than or equal to all the coefficients of B. And if we have some subset of all of the numbers with coefficients all greater than zero, what the finite basis lemma says, are you a finite such that everything in our original set is greater than or equal to something in the finite subset. So rather beautiful theorem. You just have some arbitrary set of points with non-negative coefficients and there is some subset of that set which is less than or equal to every element of the set. So it's a beautiful theorem and really key to what we're doing. And the problem is you can see it's not explicit. It's just there exists a finite subset and the proof's a little hard to do anything terribly explicit. So yeah, maybe that is one get to the description of what's missing. And so this tells us something, but what we want is what's not included. And then you can kind of revisit this lever and look at it from the other direction and you can get an explicit description of what vectors aren't in there. So let me just finish by saying we have this problem of inexplicitness that we'd like to do better than. What isn't hard to, so but we had these theorems in one dimension where we just had a three element set and we're able to just say for n greater than or equal to one you get exactly this structure from the start. So can one generalize that same argument to higher dimension? And the trick actually is that you can do it if in n dimensions you take a set of size n plus two including zero zero. And it has to also have this property that there is a convex hull and the one of the points is inside the convex hull. So there are cases in which we can express it. This is actually new and is not on the archive pre-print but will be at some point when we decide we're done. So let me leave it at that. Okay. Well, thanks. Thanks very much, Andrew. Thank you. We could all unmute and clap, for example, to thank- Oh my God. It doesn't sound like two hundred and three people. I'm sorry. Okay. So I'm going to stop recording now and then we can ask the questions after. Okay. Three, two, one, I'm stopped.