 Good afternoon everyone. I want to start exactly as I started yesterday by thanking Taylor and Ben and Mike and Lauren for all the work they've done to put together this conference. It's yesterday I expressed the hope that it would be great. Now I can express the fact that it's been great or so far has been great. I want to, yesterday I talked about a problem in geometry. Today I want to talk about a problem in differential topology. Somehow this seems strange again in a graph theory conference, but actually we will see that underlying this problem I'm talking about, there is a lot of very interesting combinatorics. And in fact how I got interested in this problem was a colleague of mine taught me what was a purely combinatorial problem. And after I thought about it for some time he said, well actually this is a real problem because it has a motivation in differential topology. So I'm going to spend a chunk of the talk giving you some background and telling you about Morse functions and telling you what this combinatorial problem is that we're looking at. And then towards the end I will mention some results. It's all joint work with Tina Carroll at Emory and Henry College in Virginia. But before I start I want to advertise another conference that's going on. On the week before the next mighty, the fifth Lake Michigan workshop on combinatorics and graph theory is going to happen at Notre Dame the weekend of April 21st and 22nd. And we're very pleased to have Ryan Martin give three tutorial lectures on edit distance in graphs and Bridget Tenner give three tutorial lectures on combinatorics of Coxeter systems. So it promises to be a good weekend too and we have sponsorship from IMA and Notre Dame so there will be some support for travel. So advertisements over, talk begins. Morse theory is basically the business of understanding the shape of a manifold by studying smooth functions on the manifold. And it's definitely best to lead this with an example. So here's a nice manifold sitting inside the plane. There are some very natural smooth functions that you might put on this manifold to try and say something about the shape. And the most natural set of smooth functions are you specify some line and you ask about perpendicular distance of points on the manifold from the line. Well you can really think of this as the process of taking that line and moving it across the manifold and looking at the various cross sections you get. The cross sections you get will exactly be the level sets of this function perpendicular distance from the line. So what do the cross sections look like as I move this line across the manifold? Well I start with the empty set because the line isn't hitting the manifold and then at some moment it will be tangent so that the cross section will be a single point. But that's fairly degenerate. Immediately the cross section will become two points and it will stay being two points for a long long time until again I become tangent. It will degenerately become a point and then it will become the empty set for a long long time. Different lines might lead to different behaviors of the evolution of the level set. So for example if I took this line here then again I'm going to start with the empty set. I will soon get to a point where the level sets are pairs of points. Then I will get to just a short stretch where the level set consists of a set of four points. And then I will get back to the level set consisting of two points and then again it will be the empty set. So two different lines here are giving us two different sets of level sets as the line moves through the manifold. And so for example you can already say that the perpendicular distance from a line is enough to distinguish this manifold from for example the circle. For which perpendicular distance from the line will always give you the same evolution. It doesn't matter which line, you will always go empty set, two points empty set. Perpendicular distance from a line is just one example of a smooth function you might have on a manifold. So now let me give you a more robust definition of a Morse function. So for my purposes it's so excellent Morse function is the terminology I should use but I will just say Morse function. A Morse function is going to be for any manifold, some particular manifold and small print it's nice manifold, smooth, compact, oriented, without boundary. All of that doesn't matter because for the most part I will be thinking about Morse functions on the circle or on the sphere. A Morse function is a generic smooth function. So smooth function means that it doesn't have any discontinuities, doesn't have any problems, doesn't have any points of non-differentiability. What do I mean by it's generic? Well it shouldn't do anything that a typical function wouldn't do and one thing that means is that the critical points should all lie on distinct level sets. If I happen to have the value 5 be two different critical points I can just perturb my function ever so slightly and these two critical points will become on different level sets. And the other notion of degeneracy that I need is that it shouldn't be the case that I have any points where the second derivative vanishes at the same time where the derivative vanishes. So for example if I'm looking at the real line I don't really want to consider a function like x cubed because x cubed is looking like it's reaching a maximum at zero but then it fools you, it fakes you out and it continues to increase and that's because the second derivative vanishes at the same moment where the first derivative vanishes. I guess more generally what I'm saying is that if you form the matrix of second partial derivatives that should be an invertible matrix at a point where the derivative vanishes. So these are properties of a random function that a random function would almost surely have. So when I say generic function this is always what I'm meaning. So this does not sound like it's fodder for a combinatorics or graph theory talk. I'm talking about smooth functions on a smooth manifold this is an inherently continuous topic. But in fact there's a lot of combinatorics that goes on when one thinks about Morse functions and we'll start to see why over the next few minutes. And in particular back in 2006 Vladimir Arnold raised a essentially combinatorial question about Morse functions. He asked okay you give me a manifold how many Morse functions does it have? Well of course the answer is infinitely many so you need to put some kind of control on how you're counting to get a genuine enumeration question. So first of all this really should be up to some notion of equivalence of Morse functions. And secondly you probably want to parametrize the Morse functions by something to make even the collection of equivalence classes be finite. And so what we're going to do is parametrize by the number of critical points. So the question here is if you're given a manifold can you say something up to some notion of equivalence and we'll have to say what the notion of equivalence is. How many Morse functions are there and this is now beginning to sound a little bit like a combinatorial question. So in his original discussion of this problem Arnold introduced one notion of equivalence which he called a geometric notion of equivalence. I'll give the definition here but really what the definition means is what you want to carry. So two functions are supposed to be geometrically equivalent if there are these orientation preserving diffeomorphisms one from the manifold to the manifold the other from the reels to the reels with the property that you can get from one function to the other by doing some suitable compositions. What this really means is you have two functions on the manifold. You're allowed to shift and stretch the manifold so that the critical points line up exactly. And then you're allowed to shift and stretch the real line so that the values taken at the critical points match up exactly. So this is probably best seen by a picture. Let's say my manifold is the circle my one Morse function is perpendicular distance from this line. My other is perpendicular distance from this line. These two different Morse functions really are geometrically equivalent because I just need to rotate the circle slightly to make the two critical points match up perfectly. And then what are the values of the two critical points? Little a is the global maximum. Capital a is the sorry little a is the global minimum. Capital a is the global maximum. Little b is the global minimum. Capital b is the global maximum. That means this number here is smaller than this number. This number here is smaller than this number. That means there's going to be an increasing function of the non-decreasing function of the reals that maps little a to little b and maps capital a to capital b. This is fancy language orientation preserving diffeomorphism. I need that language when I want to talk about general manifolds but when I'm talking about the real line all I mean is non-decreasing function. Actually a monotone function. A monotone increasing function. It's strictly monotone increasing function. Okay so we can start playing now the game on some manifolds and actually I won't talk about very many complicated manifolds. I'll talk about the sphere S1 and the sphere S2 mostly. A Morse function on S1 which is just the circle it's pretty easy to see that it's going to have an even number of critical points because you locate the global minimum and then on either side of that there will be local maxima and then you'll next have a local minimum and a local maximum and local minimum and you go back around and you have to end up with an even number of points half of them local minima half of them local maxima. One of those local minima is the global one of the local maxima is the global maximum. So let's ask the question how many geometric equivalence classes of Morse functions are there that have a certain number of critical points and since I know that it's an even number I can parameterize it by a generic even number. It's convenient to do 2n plus 2 so that I'm starting at n equals 0 and what's the answer at n equals 0? n equals 0 means I have two critical points a global minimum and a global maximum there is just going to be one. There can be only one, there's nothing to do. What about when n is 1? Now I'm talking about functions which have four critical points. There are two possible functions I think and here are the schematics of the two possible functions there's a global minimum at this point on the manifold and then I have a maximum, a minimum, maximum, maximum, minimum, maximum but the question is is this maximum here on the right a local maximum or is it a global maximum and I claim that these two schematics lead to inequivalent Morse functions so let's put in some specific numbers let's say the global minima are 1 and then the global maxima are 4 so I can certainly shift around one of these circles to make and stretch and scrunch it if necessary to make all of the critical points match up with each other but then I'll have a 1 matching up with a 1 and a 2 matching up with a 2 I'll have a 3 matching up with a 4 and a 4 matching up with a 3 and what increasing function of the reals maps 1 to 1, 2 to 2, 3 to 4 and 4 to 3 there aren't very many of them, there aren't any of them so these two guys are inequivalent and this idea leads very quickly to a characterization of what the geometric equivalence classes look like for a general number of critical points so the equivalence classes are basically going to be permutations of 1 up to 2n plus 2 where the permutation starts with 1 that's going to be my local minimum and then what I do is I read off all of the other critical points let's say going to me I'm going counterclockwise to you I'm probably going clockwise so I read them off going clockwise and I just see where do they stack in the list of critical points ordered by the usual order and if two functions lead to two different permutations then this problem of an increasing function of the reals mapping a higher number to a lower number and the lower number to a higher number is going to come up so they will be inequivalent and the property of these permutations is that if I'm at an even point in the permutation which is going to be a local maximum it has to be bigger than both of its neighbors and if I'm at an odd point which is a local minimum it has to be smaller than both of its neighbors let's have a picture here are two Morse functions at least I've mapped out their critical points and one of them is corresponding to the permutation one five three six two four the other is corresponding to one four two five three six those are different permutations they're different alternating permutations notice that the five is above the three and the one the one is below the five and the four the four is above the two and the one and so on and it's easy to convince yourself that these lead you to different Morse functions and that you can get an equivalence class of Morse functions out of any such permutation so now the question of how many equivalence classes there are is the purely combinatorial question of how many alternating permutations there are and of course this is very very well studied you can just go to online encyclopedia and very quickly you'll find that the number of alternating permutations if you plug it into a generating function an exponential generating function it will give you the tangent function so these numbers are extremely well understood you can do a little bit of asymptotic analysis and discover that the growth rate of the number of equivalence classes once you take the logarithm is around 2n log n so everything is lovely on the sphere for this geometric notion of equivalence I want to now mention a second notion of equivalence this is the one that the results that I'll talk about will be related to this was a notion of topological equivalence that Livu Nikolescu introduced at the same time as he began studying Arnold's notion of geometric equivalence so for topological equivalence what I'm not going to do is look at the level sets of these functions I'm going to look at the sub-level sets so you give me a real number 10 I'm going to look at the set of points on the manifold whose value at the function is less than or equal to 10 and as 10 moves through the reels this collection of sub-level sets is going to evolve it's going to start from the empty set and it's going to work its way up eventually to the entire manifold and at various points that collection of sub-level sets is going to change topologically the points are exactly where I cross critical points and that means that there is really a discrete set of sub-level sets so for example if I have n critical points and I'm going to have a sequence of m plus 1 sub-level sets and so for example for this Morse function here my first sub-level set is empty when I'm looking at the sub-level sets of input anything less than 1 there's nothing so I get the empty set then I go past the first critical point 1 and what I start getting is an interval around the global minimum then I get to adding an interval around the local minimum 2 so the sub-level sets evolve to 2 intervals and then I add in the number 3 and what happens is it merges the 2 intervals around 2 and 1 so now again I'm looking at a single interval and then finally I add I cross the critical point 4 and I've gone up to the entire circle for this guy the picture goes exactly the same I start with an interval around 1 I put an interval around 2 then I merge the 2 intervals with the interval around 3 and then finally I get the entire circle adding intervals subtracting intervals what's the word I haven't said multiplying intervals yes so let me define now what is the topological equivalence on Morse functions I'm going to say the 2 Morse functions are topologically equivalent if there are orientation preserving maps mapping the sub-level set sequence of 1 onto the sub-level set sequence of the other term by term so if you topologically can tell the difference between the evolution of the sub-level sets so let's ask the question how many topological equivalence classes are there so I said that combinatorial objects crop up in almost every instance of this problem so you know there's going to be a combinatorial object in the next few minutes question is which combinatorial object is it going to be so there is only one sub-level Morse function that has two critical points so n equals 0 when n equals 1 these are the two geometrically unequivalent but they are topologically equivalent and you can do nothing else so there is still only one when you go to n equals 1 what about in general well there is definitely going to be more than one now because the function on the left starts with an interval and then you add two it goes to two intervals then you add three it goes to three intervals then you add four you merge two of the intervals then you add five you merge the intervals again so you are finally down to a single interval then you add six you get the entire sphere but the evolution of the sub-level sets goes differently for this guy you start with an interval you add two you get to two intervals then you add three and now you have one super interval again you step up and now you step down again you add four that steps you up again you add another interval you add five that merges you you step down again and then finally you add six and you are the entire manifold ok so I will ask the question again Sean what word is missing you are nodding you are thinking step up step down step up step up this problem is clearly going to be a Catalan path or a Dick path problem so the non-extreme sub-level sets are non-empty unions of intervals and each time you cross a critical point one of two things can happen you either add an interval or you merge two existing intervals in other words you subtract an interval and so that means that the equivalence classes correspond exactly to Catalan paths or to Dick paths you start at zero zero you end at two n plus two zero you either take steps up or down and you never touch the x-axis except right at the very end so these are the two paths corresponding to n equals two and the number of equivalence classes is the most well-known sequence in combinatorics maybe after the sequence of binomial coefficients the end Catalan number tn one approximately four to the n ok so on the circle everything is lovely we know how many geometric equivalence classes there are we know how many topological equivalence classes there are both problems have led to nice combinatorial objects let's up the ante a little bit and let's think about the sphere s2 so more functions on the sphere smooth functions mapping the sphere to the real line are a little bit more complicated than functions from the circle they also have the property that they have always an even number of critical points but now it's not n local maxima n local minima it turns out that if you have two n plus two critical points on a function from s2 to the reals exactly n of those points and the remaining points will be local maxima or local minima but they won't necessarily be equidistributed between local maxima and local minima so just as yesterday there was a point in the talk where I said if you don't want to listen to anything else I have to say here's a nice thing to think about between now and the end of the talk and this is a nice exercise to think about why is it that smooth functions from the from s2 to the to the reals have some critical points so now the parameterization taking two n plus two is the number of critical points should make a little sense I'm going to be mostly thinking about s2 and now the number n here is exactly the number of saddle points of the functions that I'm thinking of so now we can ask the same two questions again how many geometric equivalence classes how many topological equivalence classes when n is zero there's a global minimum and a global maximum and there is only one up to any notion of equivalence here is probably how you should think about these functions this is probably the easiest way to visualize them think about them as surfaces as elevations on the globe and so what is happening when g is zero is for example you're looking at the latitude function so at the south pole you take the value minus 90 at the north pole you take the value plus 90 and then as you move from the south pole to the north pole you're slowly increasing all the way around okay so when you go to one saddle point interesting things start to happen and there are basically two types of behaviors that can happen and Arnold separated those two types of behaviors by distinguishing between Mount Elbrus and Mount Vesuvius what's the difference between these two mountains well one of them has two peaks and the other has a single peak except that it's been cut off it's got a bowl where it should have a peak and these two guys are going to be different in terms of Morse functions why am I thinking of these as Morse functions well here's how I'm imagining my surface of the earth I'm imagining that the global minimum of the earth is located at the south pole and I'm imagining that as you move up from the south pole towards the north pole you are just gradually rising no critical points and all the action happens at the mountain range that is sitting close to the north pole in one case the mountain range is Elbrus in the other case the mountain range is Vesuvius so these are Morse functions even though they just look like pictures of mountains maybe they would look more like Morse functions if I drew them a little bit more schematically here's Elbrus with its two peaks here's Vesuvius with its with its crater and its bowl these guys are geometrically inequivalent and that's essentially trivial because one of them has two local max and one local min that's this guy here one of them has two local min and one local max that's this guy here the other local min by the way is D D is the south pole or the point of infinity which is actually the global min so they're geometrically inequivalent so let me try to understand the evolution of the sub-level sets of course they start with the empty set then you get to a single disk that's going to be the disk around the south pole once you cross the value of the local minimum then as you increase what's going to happen next well what's going to happen next with Mount Elbrus is that you take in the number C you cross the critical point at C and now your sub-level set is the entire surface of the globe except for a little disk around A and a little disk around B so it's S2 minus 2 disks and then when you cross the critical point B you've gobbled up the whole of this smaller peak and so now your sub-level set is everything except a little disk around A and then finally you cross A and it's everything what happens for Vesuvius you start with the empty set but now when you cross the second lowest critical point B what you are now seeing is two disks you're seeing a little disk around B and then you're seeing the huge disk around the south pole then finally when you or not finally but when you cross the critical point C then you're back to the situation that all that is missing is a little disk around A and then finally you go to S2 so this is the two different sequences of sub-level sets and I will say in a moment that these are why these are topologically equivalent but of course you can convince yourself that S2 minus 2 disks is not topologically equivalent to 2 disks I think I have a little picture showing that in a few minutes so I won't say any more now okay so I've introduced the two notions of equivalence and the object I'm going to be working with which is the sphere what about geometric equivalence well I'm not going to tell you any facts that I can justify in this it's difficult Arnold had speculated that the growth rate of geometric equivalence classes was exactly the same as it was for S1 and Livu Nikolescu about four years later he verified that speculation so what he did exactly was he said let me form the generating function the exponential generating function whose coefficients are these numbers I want to find out and he did some nice asymptotic analysis and some nice generating function analysis and he discovered that the inverse of this generating function is given by some explicit integral this doesn't come out of left field because on S1 the exact same thing happened plug the numbers into a generating function you get the tangent function the inverse of the tangent function is indeed an explicit integral but it's just the function is a little bit more complicated so that left a question Livu asked the question for the topological equivalence classes what is the growth rate what is the number of topological equivalence classes and that is a question that remains open although we now can say a lot more about it than we were able to say a little while ago so let's think about the evolution of sub-level sets on the sphere the sub-level set sequence always starts empty and always ends with the entire sphere and actually I'm going to conflate the entire sphere and empty because I want to think about starting empty and ending empty otherwise the sub-level sets are going to be unions of disks except some of the disks might have holes punched in them that's what we've seen in the two examples for Mount Elbrus this is what we saw we see that in the examples because we saw that one of them was s2-2 disks but here's my little picture here is s2-2 disks and if I just take one of these two disks and stretch so it becomes the outer boundary then what I get is a disk with a hole taken out of it so in both of these cases yes the sub-level sets are always unions of disks some of the disks might have holes in them so what happens when you cross a critical point well I won't go into a great deal of detail on this I'll say all the things that can happen but I won't say a great deal about the topology that's underlying this so a couple of different things can happen either you have your bunch of disks with holes from them and a new disk arrives you just pick up a new area around a certain critical point another thing that can happen is that a new hole gets punched in one of your disks another thing that can happen is a disk that doesn't have a hole disappears the last thing that can happen not the last thing that can happen the last thing that can happen is that you had a hole and the hole gets filled in those are all fairly obvious things that can happen and you can imagine how they would happen as you can cross a critical point but there's one other thing that can happen that is quite complex and that adds life to an interesting problem and that is that it's possible especially when you cross a saddle point that you can take two disks that have punched in them and you can merge the two of them and you keep the holes but now you've just got one super disk that has as many holes as the sum of the number of holes here and here these are the only things that can happen as you cross a critical point as you're watching the evolution I would not have been interested in anything to do with this problem if that's how it had been presented to me but no one day Nikolasko came to my office and he said David I want to tell you about the Greek wedding game I said oh that sounds fun so he said David here's how the Greek wedding game goes you start with an empty table and then at each step either you add an empty plate to the table or you add an olive to an existing plate or you remove an empty plate and you smash it to the floor this is why it's the Greek wedding game or you remove an olive from an existing plate and presumably you eat it or you merge the olives on two of the plates and you take the now empty plate and you smash it to the floor these are the things that can happen at a Greek wedding said Nikolasko to me notice that these are exactly the operations that happen when you cross a critical point so he framed the problem in terms of this appealing game of plates and olives and the game is going to end the first time that you return to an empty place because that corresponds to when you finally crossed the whole manifold and his observation was that the number of topological equivalence classes is exactly the number of games plates and olives of length 2n plus 1 now you can do a little bit of computation you can figure out what this value is for some small values this turns out to be how the sequence starts you can go to OEIS you won't find the sequence on OEIS there's an unfortunate issue here and if you search for topological equivalence classes of Morse functions on OEIS you will find the sequence it won't be the same as this because there was a minor computational error in the original paper from 10 years ago that Tina and I were able to clean up so this is the correct sequence we just haven't entered it into OEIS yet but even still you don't look at the sequence and say oh it's obviously some translation of the catalan numbers or something it's not anything obvious so the question is what is this sequence how is it growing, how is it changing there's a very nice lower bound on the number of games and the lower bound connects to another very beautiful combinatorial object Ferrer's diagrams and Young's lattice on Ferrer's diagrams there's a very simple encoding of a state of the game via Ferrer's diagram if you happen to have a plate with two olives on it that will correspond to three boxes in a row of your Ferrer's diagram if you have a plate with one olive that would correspond to a row with two boxes if a plate corresponds to a row with one box any configuration leads to a Ferrer's diagram and the directed graph on the states of the game of plates of olives turns out to be exactly the hassa graph of Young's lattice on the set of Ferrer's diagrams so for example what's happening as you go from here to here you had a plate with one olive and you added a second olive and when you go from here to here you had a plate with one olive and you added an empty plate so the moves in Young's lattice on Ferrer's diagram Young's lattice is simply one thing is below another if it can be contained inside so this guy sits inside this so it's below it in the partial order this guy sits inside this so it's below it it also sits inside this so it's below it the edges of the hassa graph of this poset correspond exactly to all of the moves in the game of plates and olives not exactly it corresponds to all the moves except this interesting complex move where you merge two plates so if you can count how many ways there are to start here and end here after two n plus two steps then you will know how many Morse functions there are topologically well you'll have a lower bound on it you'll know how many don't use any of these interesting complex moves Stanley has lovely work on differential posets that was inspired by this particular lattice and you're able to read off of that that the number of zero zero walks is exactly two n minus one factorial and so a lower bound on the number of topological equivalence classes is two n minus one double factorial which turns out to be two over e to the power n so that's our lower bound and when this lower bound was first published in 2008 the question was asked is there anything like a matching upper bound but none was known so in 2006 Nikolesko asked the question what is the growth rate and in 2017 on math overflow he made a specific speculation and his specific speculation was that the growth rate of this quantity when you take the logarithm is n log n in other words this lower bound of two over e to the power n, n to the power of n is really the truth at least at the logarithmic level I have to say when I saw this speculation on math overflow I was absolutely horrified and the reason is that a couple of years earlier Tina and I had thought about this problem and we worked on it and we got an upper bound and the upper bound shows that this logarithm is asymptotic to n log n and we thought that's nice but who cares we really want a more precise asymptotic result and I never even actually mentioned it to Nikolesko because I thought he isn't going to care about this and then one day I logged on to math overflow and I saw that he posted this question where he said I would really really really love to know what the logarithm of this number is what it is asymptotically and I was thinking oh my god we've known this for four years and here is why I was really horrified because he posted it at 5 o'clock in the afternoon and I logged on at 11 o'clock in the evening and there had been something like 250 views of the problem and so I was thinking oh my gosh 250 people are now thinking about this so I had to hurriedly write a sketch of a proof and posted on math overflow so the lower bound is 2 over en times n to the n so the lower bound on the logarithm is n log n plus a constant and so his speculation was that this lower bound of the logarithm was essentially true and notice that for geometrical equivalence classes the logarithm of the number of equivalence classes was like 2 n log n so this is saying that there really is a significant difference between the topological notion of equivalence and the geometric notion there are very many few topological equivalence classes then there are geometric equivalence classes so with Tina I wanted to show that the logarithm of this number is indeed asymptotic to n log n and we were able to get some upper bound that we were happy enough with and that upper bound is 4 over e to the power n times n to the n so of course that raises the obvious more refined question does this nth root limit exist and if it does exist where inside in 2 over e 4 over e does it lie I have fairly good reason to believe that it's neither 2 over e nor 4 over e it's something in the middle and probably towards the in the last minute or two I'll I'll say something that might lead you to at least think that I have some reason to think this so let me spend a little while now every talk has to have a proof let me spend a little while giving you a sketch of how you might get an upper bound of n log n so there are going to be 3 key observations but only the first key observation is necessary for an upper bound of n log n the remaining key observations are necessary to try and drive that constant down all the way to 4 over e so if you only want to listen to one thing for the next 10 minutes this is the one thing to listen to and then you can definitely switch off the key observation is that if there happen to be t olives on the table at any moment then there's a bound on the number of distinguishable plates on the table plates are only distinguishable by how many olives they have on them and I claim that if you only have t olives then you can only create around square root t distinct plates and that is nothing more than the fact that the sum of all the numbers up to square root 2t is equal to t so the best you can possibly do with t olives is put none of them on one plate one on one plate two on one plate three on one plate and so on and you get square root 2t different numbers and you've exhausted your t olives so here's how I'm going to count the number of games of plates and olives the first thing I'm going to do is I'm going to specify the type of move I make at each step maybe I add a plate maybe I add an olive maybe I remove a plate maybe I remove an olive and then I'm going to specify the specific move that I make given the type of move that I made let's see how that goes adding plates is easy there's only one way to add a plate adding olives there's only ever going to be square root 2n ways to add an olive because there's only going to be at most n olives ever at play if I have to only take 2n plus 2 moves if I added n plus one olives then I'd have to remove them and that would be 2n plus 2 moves already before I'd even done the add an initial empty plate so there can only be at most n olives ever so there can be at most root 2n options for adding an olive what about removing a plate well I have to deal with the complex move here and the complex move says I choose two plates from among the at most root 2n and that number is about n and what about removing an olive well again I have to choose a plate and so they're at most around root 2n options so now I just plug in everything and see what happens and what happens is the number of games is at most there are 2n roughly moves and at each move I have to say what type of a move is it plate up olive up plate down olive down so that's a 4 to the power of 2n then I have to pay a price of n to the power of number of plate minus moves then I have to pay a price of square root 2n to the power of olive plus moves and to the power of olive minus moves now what am I going to do with this so the number of plate plus moves is the same as the number of plate minus moves because every time I add a plate eventually I'm going to have to remove it so I could really think of this as square root n to the power of p minus p plus this is square root n O plus plus O minus the number of olive plus moves is the same as the number of olive minus moves so this is 2 times the number of olive minus moves and the 2 gets rid of this square root n and what I have is n to the p minus times n to the O minus well that's just n to the power n because n times I remove something corresponding to n times I add something so all of these guys together just give me n to the power n and then I have a bunch of constants like this root 2 raised to the power of junk this 4 squared raised to the power of n and I can gather it all together it's certainly going to give me no more than 32 and so on this one slide there's the complete proof that the upper bound is really a constant to the power of n times n to the power of n I have to say that when I wrote it on math overflow it was a lot longer than this and Livu said that's great but he sort of said it half heartily and then I give a talk on this a couple of weeks ago at Notre Dame and he said to me afterwards thank you now finally I understand so it's gotten much simpler so obviously I'm giving away things in this proof so how can I improve this count well what I want to do in the last few minutes is maybe suggest two ideas that allow me to improve this count significantly and then that will show where is the one missing idea I want to argue to you that really very few games involve many plate moves and in order to do this I'm going to distinguish between two types of plate moves simple plate moves where I remove an empty plate and complex plate moves where I merge two plates now before in my count I was giving myself a factor of n for either of these guys but really I could give myself a factor of 1 for these guys so for each one of these guys that happens I can save a factor of n and so that means that if there are more than around n log n of these guys I save a factor of n to the power n log n which is 2 to the power n so if by hitting this n log n with a large constant I can hit this 2 with what turns out to be a small constant so well no sorry the c will turn out to be a large constant because what I can do is if I know that there are enough of these plates simple moves I can depress the 32 to the power n by an arbitrarily small base raised to the power n so I can dispense very quickly with all of those games in which there are a lot of simple plate moves but I also wanted to dispense with all of the games in which there are a lot of complex plate moves and this seems much more difficult what I have to do is past thinking about the olives for a moment and again I'm going to distinguish between two types of olive moves there is the first time I add an olive to a plate and then there's a later time that I add an olive to a plate the time I add an olive to a dirty plate the observation on this slide is that there are at least as many first olive moves as there are complex plate remove moves because if I remove plates in a complex way 20 times then 20 times I'm taking away a plate that at one point had an olive on it so there has to be at least 20 times when I added an olive for the first time and I was paying a price of square root of n for both of these guys but for the first ad moves I only pay a price of one if I tell you I'm adding an olive to a plate for the first time I'm adding it to an empty plate and there's only one way to do that so if I have a lot of first olive ad moves then I also get a depression but a lot of first olive ad moves means that there are a lot of complex plate moves and so other way around I do beg your pardon completely the other way around if there are a lot of complex plate remove moves then there also happen to be a lot of olive remove moves and therefore again the 32 to the power n drops by a huge factor so what I've argued in this slide is that you get to assume that the game that you're looking at basically involves very very very few plate moves it's all about adding and subtracting olives because the games which involve lots of plate moves contribute negligibly to the total count they get below the lower bound how about an upper bound on games that have few plate moves well underlying each game there's a catalan path of how many olives there are at any given moment you might as well assume that that's the entire game because I'm now assuming there are few plate moves the number of options at a step of high t is about square root 2t now I'm not saying square root 2n I'm actually taking into account how many olives I have equivalently the number of options at a step of high t is 2t because every time I have an up step I have a corresponding down step at the same height I can pair up the up steps and down steps in a catalan path so basically what I'm doing is I'm enumerating catalan paths with a weighting the weight that each step gets is two times how high up it is on the path and so the key observation number three is that an awful lot is known about dickpats an awful lot is known about catalan numbers so let's leverage that and here's what has to be leveraged there is a lovely result that the weighted sum of catalan paths of length 2n weighted by the products of the heights of the up steps this is not an asymptotic result it is exactly 2n-1 double factorial this is an example here this one contributes six this contributes four this contributes two this contributes two this contributes one they all add up to fifteen which is a double factorial of an odd number so I don't get to say that my total count is 2e n to the power little of n times n to the n because I also had this niggling factor of two and it's that factor of two that is the gap between four over e and two over e so let me end with a summary the main result that I've discussed is that this limit if it exists lies somewhere between two over e and four over e so the open question is what is this limit if it exists I feel like the lower bound is really giving away a lot because it's ignoring these complex merge moves so the lower bound is probably not right but the upper bound is also giving away a lot because it's using a a far from typical upper bound on how many distinct parts there are to a Ferrer's diagram I'm saying that if you have t boxes you have at most root two t or root t distinct parts but actually typically you have fewer than that there's a result of wealth that says you have square root six over pi square root t distinct parts with high probability so that strongly suggests that this four over e is also giving away something where to go next well my first inclination was to think about the s3 and beyond although I was cautioned recently that I might be getting into a quagmire if I do that because Milner has shown that to try and solve this problem for s4 is equivalent to classifying knots in space so I don't think that I'm going to be able to contribute greatly to that problem however there is one very interesting direction to take this work that Arnold was very interested in just before he died and that other people who think about this problem are very interested in which is not the sphere but the torus or other manifolds that are embeddable in r3 of higher genus and there should be some other very interesting combinatorial structures that emerge from these problems so I think this is a fruitful opportunity for interaction between topology and combinatorics I want to end with two thanks I want to thank Livu Nikolascu who ten years or about five years ago tempted me with this problem and I want to thank you all for your attention yes Doug the merge moves from what you were saying later there's some restriction of how many olives are on the plate when you do a merge move no there is no restriction the merge move is simply that you take two existing plates and you merge them together so even if they're both empty it's okay if they're both empty then that would be equivalent to remove an empty plate and so topologically it would make no difference or if one of them was empty and the other had olives on it that would be equivalent to remove an empty plate so it won't change the count so when you were doing your count I think you were saying when you were merging things they were not empty and that would be the reason why yes I guess I've lost my pointer but I guess in retroactively I can do my little asterisk these differences end up in the asymptotics making no difference but yes one has to be quite careful it's dangerous to sweep anything under the rug when smart people are listening there are no other questions let's thank David one more time