 So of all the accolades we've heard, wonderful things we've heard about Barry, applied mathematics I don't think has been mentioned in the first bunch of talks. And the theme of this talk is the following. Anybody who does mathematics, no matter how focused, no matter how abstract, runs into outsiders. And you're at a faculty meeting, you're at a dinner, and some chemists, a social scientist, or you're a mathematician, I have a question, you know, can I ask you, you know, and every once in a while you get interested in do a little something and some applied math comes out of it. Now, it's hard, of course, to, you know, how do you keep track of somebody's output in that dimension, and I'm going to take advantage of the fact that the outsider that asks Barry the things that I'm going to talk about is me. So over close to 50 year period, I've talked to Barry, we're friends and talk about all kinds of things, but I tricked him, whatever, into thinking about various things that I'm going to tell you three of them. These are three stories. The first is about shuffling cards, and that's going to be my favorite story about stories, and the second is about the books of Plano, and the third is about walking on the streets of Paris, and, okay, so most of my mathematical adventures start with some real world problem, and let me set the scene for the first instance, so I was working at Bell Labs and I was trying to figure out the optimal strategy in a certain card game, and it was much too complicated to solve or get to the end of it, so we were doing it by simulation, try to make certain choices in how many cards each point is worth and then have it play against itself and optimize, that sort of thing. And so in order to do that, we needed lots and lots of random permutations of 52 cards, let me remind you how you generate a random ordering of a deck of cards, suppose you had 52 cards as memory registers or in a stack, then you get the computer to generate a random number between 1 and 52, say 17, and you transpose cards 1 and 17, and then between 52 and 52, say 5, and you transpose 2 and 5, and then between 3 and 52, maybe 3, so you do nothing to 3, etc. So you do that, and assuming that the random number generator works, or that you're happy with that, that generates a random permutation, right, because any card can be on top and then any card next, etc. Okay, so we ran the simulations and they took, well, I don't know, 30 hours of CPU time on a big machine, so it was a lot of effort, and after it was all done and we're looking at something was wrong, I don't know what does it mean something is wrong when you're looking at pages and pages of simulations, but something was wrong, something wasn't monotone, and so we start looking at the code, I mean there's no more painful thing to do than that, looking at somebody else's code, and after two days, I sent to the lady, Mary and ghetto, I'm desperate, how did you generate your random permutations, and she said, oh yeah, you said that fussy thing, transpose random with one random, she said I made them more random, I transpose random with random, and I said, ha, and I said, you have to redo the simulations, you have to rerun the code, and she actually cried, I remember, and said, you know, but it can't make any difference, she went to her boss and he went to his boss and he came down and yelled at me, your mathematicians are crazy, you know, she did 100 random transpositions, that has to be enough to mix up 52 cards, and so I really wanted to know the answer to the question, how many random transpositions does it take to mix up 52 cards, I mean that's the background for this story, and so just to put it in context, one, so you're with me, so suppose a deck of cards starts out on the table, one up to 52 in order, card one, card two up to 52, you pick a card at random with your left hand, you pick a card at random with your right hand, and you switch them, and you do that a lot, okay, and you allow your left hand to equal your right hand, because otherwise there'd be a parity problem after an even number of switches, you'd be in an even position, and I want to know how many transpositions does it take to mix up cards, in my instance the lady did 100, the first time that the bridge league went from hand shuffle cards to computer shuffle cards, they did 60 random transpositions, not knowing the usual algorithm, how many does it take, so let me make math of that for a second, so I'm working on all permutations of N is 52, and the probability thing that I just waved my hands about, you could code it up this way, it's a probability distribution, the probability of a permutation sigma is well 1 over N of sigma is the identity, and it's 2 over N squared if sigma is a transposition, and 0 otherwise, that's my basic step in my switches, and then repeated transpositions we model as convolution, so I write it down just for a second p star p of sigma, the chance of being at a given arrangement of a deck of cards after shuffling twice is equal to the sum of p of eta, p of I guess it's sigma a dangerous, here summed over eta, so that's just ordinary convolution, and p star k of sigma is the chance of being at the arrangement sigma after k switches, so there I've written it down, and then is the uniform distribution u of sigma, which is 1 over N factorial, all permutations equally likely, and I want to know how large this, well, Pankeray actually was the first one to show, the 1890s that p star k converges to u, when k gets large, which just says if you shuffle a deck of cards a lot, it gets all mixed up, and you already knew that, and I want to know how fast, and so in order to tell you that, I have to say one more definition, stop for a second too, and so the distance between shuffling k times and uniform, this is one good way of defining it, p star k of a minus u of a, and so here's what this says, a is some set of arrangements of the deck of cards, maybe all arrangements of the deck of cards with the ace of spades is in the top half, and p star k of a is what's the chance that the deck is in that set of arrangements after k shuffles, u of a is if the cards are perfectly mixed, what's the chance? You take the difference between those two numbers, and then you take the worst case of that, that's the distance that we often use, total variation, and with all of this specified, there's now a math problem, finally, given epsilon bigger than zero, how large k, so that the distance of shuffling to uniform is less than epsilon, that's now a math problem, and there are a lot of algorithms for generating random, I don't want to answer that right now, I'd be happy to discuss it offline, not the right time for that question, but I'd be happy to discuss, there are lots of algorithms, I just try to say the standard one, well I don't know either, but I told you the one that they used, so that's yet another question, so with Miradaj Shah-Shahani, another longtime friend of Barry's, we figured this out, and the way theorem comes out is this way, if k is one half n log n plus cn, c is bigger than zero, I'll explain this, then the distance of shuffling to random is less than or equal to 2 e to the minus c, and this c is the same as this c, and what that means, well that means if you want to know how many times to shuffle to make this less than 1 in 100, you write equals 1 in 100, you solve for c, and they put that into here, and that gives you the answer, and there's a lower bound that matches this, so that was, I want to do that, I want to learn how to do this, and then learn how to do that, and so that's what a theorem looks like in the subject, I think this was the first such theorem in the sense of a short quantitative bound, and I, in 1985, I gave this same talk in this same room, Barry and David Mumford were interested in having somebody in the math department who knew about combinatorics and probability, and asked me to give the job talk, although I didn't know it was a job talk, but anyway I did talk about this, because I just written it all down, and I should say by the answer, by the way, the answer is around 400, I mean, or something like that in order to, and we know why it's not random when it's not random, and that did explain the anomalies in our results, and after my talk, Barry, as usual, I saw him talk after talk, came up and said, that was great, and thank you, meant a lot to me, still does, and he said, but the reason your argument works is because the set of transpositions is a conjugacy class in the group, and what you're really doing is taking the element in the group algebra, which is the sum of all transpositions, plus the identity, and then raising it to a high power, well you could do that for any group, you know, any lead type group, and he was going on and kind of sketching out a program of research, and we're friends, and after about five minutes or something, I said, yeah, but what's the story? Why would anybody care about doing whatever I was just doing on a lead type group? And Barry didn't know what I was talking about, and after another five minutes of my trying to explain, he said, ah, you see, he said, what you don't understand is that in mathematics, somebody else makes up the story. There's some truth to that, and it's a wonderful history if you, you know, the character theory that I did in order to prove this theorem was done by Fermanius for just the craziest reason, you know, is that the determinant of a, of a circulant is a product of linear forms in the entries, and I mean, just who would care about that? The character theory came out of that, so stranger things have happened now. So, but Barry's thought stayed with me, even though I do, I, I told you why I was doing this problem, but I just came, it sometimes takes a while for the ideas to mature. I just came from, from six months at MSRI where a group of us are doing exactly what Barry laid out, that is, take a lead type group, say GLN over a finite field, and take a conjugacy class like something like transfections or any conjugacy class, in fact, these are all almost simple groups and almost anything will generate, and then you can do the analog, make the random walk, and with a group of very wonderful group theorists who I've corrupted, it's his fault, but I've corrupted them, there's a kind of unifying result which is more or less, but really more or less, that is essentially always, for any lead type group and any, any conjugacy class, not the identity, etc., the rank of the group is necessary and sufficient in order to, in order to have things be random, and so took a while, but I did finally follow, and I have no idea why we're doing it, it's just the math is nice, and now I want to say, let's see, what else do I want to say about this particular thing, really the reason I told it to you is it is my favorite story about, about stories, but one thing that came out of this result, just a lot of smart people here, this particular thing, if I graph it, so this is k, the number of shuffles, and this is the distance of p star k minus u, how close we are, and this distance that I'm working in is the difference between two numbers which are between zero and one, so it's between zero and one, and it's zero if the measures are singular, and it's, sorry, zero if the measures are the same, sorry, and it's one if the measures are completely historically supported, and what this theorem says is something a little surprising, it says that this distance behaves with a sharp cutoff, and this cutoff happens at a half n log n, that's where this cut, that is this, going to zero happens in the second term of the asymptotics, okay, so if you were to make a plot, and it's an outstanding conjecture in the field, let me say it twice, take any set of permutations that generate a symmetric group, any wave shuffling that you like, and start shuffling with them, that there is a sharp cutoff, that is there is a, this cutoff phenomena happens for any generating set, and a marvelous development that came from this, we made a theory following what I told you about, which says if you know all about one way of shuffling, if you have very, very sharp answers for one way of mixing things, then you can get good answers for essentially any way of shuffling, it's called comparison theory, and that's now a machine on its own, and using comparison theory, Rijuk, Chirach, and Helfgat, I think an inverse order, proved that you know if you pick two permutations at random, they generate either s n or a n with probability, well, one, and they generate s n with probability three quarters, so pick two permutations at random, say a transposition in an n cycle, but any two permutations, pick two permutations at random, the mixing time is of order n q log n, and that's the worst, and we conjecture that there's a sharp cutoff of this sort, and in fact we conjecture it for any set of generators for any finite simple group, but there's just a handful of examples and this is one of them where we've been able to prove that there are cutoffs, in particular for a transposition in an n cycle, we don't know, I believe that there's an explicit constant such that n q log n plus c is the right answer, but we don't know, so those are math problems that come from this development, but more than that, it is a typical example of how I get involved in a problem and what happens from it. Okay, so that's enough about shuffling cards and you should be happy because I just finished teaching a graduate course on the mathematics of gambling and they had to sit through four weeks of shuffling, so this is all you're going to hear about today. Okay, so in case, you know, you're tired, it's the last part. Wake up, new taught history. So my second story, I want to tell, those of us who've been participating, I hope most of us were lucky enough to have heard Elaine Scarry talk about Plato's Poetics on Wednesday, it was just made these wonderful texts come to life. Well, I'm going to talk about Plato, it says it right there for a few minutes, and so one of the things that we don't know, so Plato wrote a lot of famous books, the ones that have come down to us, and we don't know what order he wrote them in. Okay, so Plato's Republic was written early, Laws was written late, and the others, and it matters if you're a Plato scholar, that is, because he changed his opinion, it matters, you know, if you're trying to interpret, how did he come to? So that's a question, how to seriate the books of Plato, and there's been endless articles written about that, you know, well, he was whilst here and while there, you just can imagine the conversations going on and on, and around the late 1950s, a Plato scholar named Brandwood decided to go get some data, and Plato said that he systematically changed his rhyming pattern over time, and so what Brandwood did was recorded for every sentence the last five syllables and the sentence, whether the stress is whether they were longer or short, and so, for example, Plato's Republic has around a thousand sentences in it, and for each of those thousand sentences, Brandwood recorded a binary vector, it was in, you know, this kind of language, of length five, you know, long, short, short, short, long, if that's what those symbols mean, and then the notion was, the thing that he did and that I did was let's look at, well, so there are 32 possible binary vectors, all possible patterns from all short to all long and everything in between, and so I would have a function f of x, if x is a possible binary pattern of length five, which is how many times did that pattern appear in Plato's Republic and in the other books, f of laws of x, and the idea was to look at, for example, Plato's Republic, which we know was written early, and see if these numbers show some pattern, and then if that pattern is different in laws, we could use it to serigate. Okay, that was the notion, and using, I'm not going to say in detail how that came out, but if you're interested, I think if you type in Percy and Plato, it will come, that's a paper I wrote with Julia Saltzman, and not quite what I want to talk about, but that's an example of analysis of binary data. Well, there's all kinds of binary data. It could be that I'm looking at a string of zeroes and ones, x1 up to xk, which maybe of length 12, which was one or zero as that worker was employed or not in that month, and the current population survey gives me 5,000 workers and vectors of length 12, and then it tells me what f of x is for those workers, and again, you could try to make sense out of those numbers, and there's lots of, a class of students, and you gave them a true-false test, and you have how many students had this pattern. So there's lots of examples of binary data that one wants to analyze, and what I want to get to, so that you know this is how it started, I want to get to my own personal elliptic curve, so I've got a ways to go. So when we analyze data of this sort, we often replace the data by various averages. For example, it might be that I looked at all, all, I sum, how many people have, how many people, or how many sentences began with one, how many sentences have a one in the I place, how many sentences have, you know, one zero in places, I and J, etc. So we often analyze such data by taking averages of the different functions, of the functions that come up, and then that helps clean up noise and allow us to make sense, and from this analysis I was led to find a quite simple striking pattern in the republican laws that were quite different, and when we use that to seriate the books, the majority opinion was vindicated, I'll put it that way, and it has had some impact on this tiny little area of scholarship, the fact that some math also pointed in the same way as other arguments. Okay, so if you have data, so I'll now suppose we're given one function on binary k-tuples, and you are going to take certain averages of it, and so I'm going to specify a class of averages, let s be a subset of 2 to the k, a thick subset of k-tuples, and then you could define the rate on transform f bar of y, so I'm given f, and y is a binary vector, and this is the sum of f of x for x in s plus y, so s is something like a ball, and what you're doing is taking your function and averaging it over a ball around y, that's a kind of common thing one will be doing for a second at any rate, and so you replace your data or you buy some smoothing of it, and now a natural question if you're doing stuff like that, if you're massaging data in that form, is well, can I have I captured all the juice in the data? Well, one pale version of that is given a bunch of averages that is fixed in s, for example, and suppose I tenu all of these averages, can you reconstruct f from it, or is there something hidden in the averaging process that is now invisible to you? So, kind of math question is given s and f bar of y, for all y can we reconstruct, and this is the ray-don transform that is invertible in this situation and there's no shortage of math and developments about properties of replacing a function by its averages over a collection of sets, and this is one that came up for me that I was interested in many other things, but I want to get to my favorite with the curve, and I've got to go pretty directly to it. And so let me tell you some news, there's bad news and good news. So the first piece of bad news is that well, first of all the good news is the start of Barry's conference, the start of the good news if the size of s is odd, you can always reconstruct okay, that's not hard to see but sometimes you can reconstruct it it's important to reconstruct when the size of s is even and okay, well the first piece of bad news is that for most sets s of size to k you can't reconstruct so the radon transform isn't invertible and that's uniformly in k actually, okay, and here's the second piece I'll keep down here, maybe you won't be able to see it a second piece of bad news is given a single s you can ask him to be given it can you reconstruct that problem it's a yes or no question, that problem is NP complete so there's not going to be some simple algorithmic way of doing it for general s so that's maybe bad news for certain sort but here's some good news for many natural s's because typical s worst case s, of course, the trouble could happen but for many natural s it's okay that is, you're not losing any information and the class I'm going to tell the class that led to an elliptic curve was bowls in the hamming distance so let s be equal to the set of all x such that the number of ones in x is less than or equal to r, say okay, that's a ball fixed ball where this is the number of ones in x the hamming metric so suppose I replace a function by smoothing it can you reconstruct it this is a math question that came up in the middle of doing the analysis and so now it starts to depend on r so okay, so it starts to depend on r so I have two parameters now the diameter of my ball and k the length of my vectors so when r is 1 so everything within 1 of me it's okay if and only if k is odd so that's a theorem when r is 2 so everything within 2 of me you can reconstruct if and only if k is not a square so mostly you can reconstruct r is 3 it's okay if and only if k is odd and not a square it's getting more and more okay and when r is bigger than or equal to 4 just slightly bigger balls and it's okay that is you boys have any reconstruction except for finitely many k okay so that's already uses some classical piece of number theory but it's not okay so I started in working on the question of well alright what happens with 4 I know that there are only finitely many bad values of k can I find them what's it about and I want to just write down what this has to do with anything so let me write it down so when r is 4 it's bad that is you can't reconstruct if and only if this polynomial z to the fourth minus twice 3k minus minus 4z squared plus 3k times k minus 2 has an integer 0 so there's a diafantene type condition the k's which are bad the k's such that this is the fourth for people who like orthogonal polynomials and that's when when you can reconstruct well okay this is work done with Ron Graham who then had a team of programmers and we went to the computer and so the bad values I'll write here so what are the bad values of k well it was bad for k equals 1 2, 3, 8 17 66 1521 15,043 okay so sometimes it was bad and we looked pretty hard and we thought well is it possible that's all now for those many people in this audience this is essentially an elliptic curve let me write it down that way by an elementary transformation so I have two parameters z and k here so this is down so just to try to make it look more like something this audience might relate to this is equivalently this is the curve u squared is equal to 6x squared times x squared minus 1 plus 9 and if you prefer bias dross form u squared is equal to 4v cubed minus 57v plus 53 so you needed to find integer solutions of these equations and we know that there are only finitely many and it's a question about whether we found the largest one or not and it's just a little corner of work but I was very pleased to have an elliptic curve that I met on my own terms without somebody saying well this is a famous thing there it came I mean I just came from seriating the books at Play-Doh so I went to Barry and I said hey I've got my own elliptic curve and he looked at me but we're friends so he listened up and he got very very interested in it and he put the word out whatever that means but Barry calls you and says and I asked him why are you so interested in that curve thank you of course thank you but why are you so interested in that curve and he said you know curves that come up in nature are a cause that are often have a way of reappearing and you know curves that actually come up in a real problem deserve respect and I thank you and through a chain of events but it really did start with Barry and I wrote their names down and I'll mispronounce them but let me at least try to say it Stoker and Devager we found all the integer solutions except for this list of k's you can always invert the radon transform when you are averaging over balls of size 4 anyway it's a little kind of corner thread of applied math but it was one in which Barry really did make a difference there are papers it's actually not easy to show that you found all the solutions maybe it's easier since they did it but ok so again I would wake you up when you were awake anyway so here comes the third talk and this talk this is about wandering around on the streets of Paris and here's how it goes so I'm interested in many things some talents language is not one of them so there I am I was teaching it because you walk and you walk and you walk and you walk but my French is just poor and so I noticed I was using an algorithm to maneuver around not a map of course but that way but that way before I kind of recognized it I'll go that way I was wandering around the streets of Paris using the algorithm that streets that I walked more in the past I was more likely to choose in the future and let those of us who have not done such a thing cast the first stones and at some stage I decided to try to make a math problem out of that that sounds like something I should be able to mathematically so let me tell you what I did let's take the simplest case let's take a triangle here it is and I'm going to put weights one on each of the edges and imagine a random walker who starts at A and then picks B or C at random maybe he goes to C and every time he crosses over an edge he adds one to the weight of the edge and when you're in a new vertex you pick where to go with probability proportional to the weight of the edges leading out of you so that's what I was doing and I just said it as math so now the question is what happens just a triangle so here two thirds of the time I go back and now I'm here again and now three quarters of the time I go back so one natural thought is you wind up dying on an edge you go back and forth and that's what you do and the second thought is what can't matter what happens at the beginning you wind up spending your third of your time in each vertex there's mostly only two natural guesses and both of them are wrong yes that I could think of but it's a simple enough problem that it must yield to thought it's just a triangle after all and I kept thinking I'd see it and eventually I blasted it out and it was hard work and I want to tell you a sense about it first let me tell you the sort of bottom line of it so again if NAB is the number of times I cross AB after N steps and I look at NBC N over N and NAC so this is the proportion of times I've crossed each edge so this is a vector on the simplex the numbers it's 7 to 1 this converges to a limit theta 1, theta 2, theta 3 on the unit simplex and N goes to infinity almost surely but this limit is random so if you started with a fresh triangle and you did it again you would converge almost surely but the limit where you converge to and each time it will be different and so then it's a natural question well there's some limiting probability distribution on the simplex and I know lots of them, my community knows lots of them and I did figure it out I'm going to write it down in a second but it's nobody's friend not one of the standard ones I just write it down not far away, I almost remember it but let me so this is okay the limiting density this has a continuous density it was xy plus xz plus yz to the one half divided by x plus y this is x plus z to the three halves times y plus z to the three halves and pi is a normalizing constant and here I guess this is x this is y and this is z those are so it's a funny density I once was at CCR and I said I'll give 100 bucks if anybody can show that this pi is the right constant it's an integral over the simplex and some kind of yen did it and I paid 100 that was the hard part is to get them to take them anyway so this is the limiting proportion of time that you cross each edge and some funny limiting probability distribution now so let me just say this the same story I just told you is true on any finite graph so if you take a graph it could be a square it could be any kind of finite say simple finite graph connected you start random walk with reinforcement on all the edges and you move and every time you add you add one to the edge the same story is true that is the proportion of times you cross each edge tends to a limit that limit has a density which is absolutely continuous and there's an expression for it which is a little bit worse than this but it's worse than this in an interesting way the description of the limit uses the first homology group of the graph and it's the only theorem I know about in probability that where anything about homology comes in and I'd be happy to be corrected about that but now what does that have to do with bearing so as part of our long conversation I have often tried to teach bearing probability how to think a little bit like a problem was and it hasn't been easy so one of the ways that I tried is I said well you know and he wants to learn I mean speak all languages including mine and so I said well next time I'm in the middle of some problem we can do some calculations together so you get a feel for what it's like so let me say a sentence about what is behind this this random walk with reinforcement that's what it's called generates a stochastic process so x naught where you are say that's a and then there's x1, x2, etc so if you wander around where you are is random and I'm giving them names x naught, x1, x2, etc and I guessed from examples or from having done things like that for years that this process for any graph this process was what's called partially exchangeable and let me try to explain what that means so if you take two possible realizations so all the strings will start with A so for example suppose you start A, B, A, B, C, B, A that's one way your process could start and another way it could start is if I switch these two blocks B so I just took this block and this single block and switched them, that's all I did these two strings have the same transition count matrix that is if I make a counting matrix here's A, B, C A, B, C, how many times in a string did I go from A to B, etc well it's 0 along the diagonal how many times in this string did I go from A to B so from A to B I went twice from A to C no times, ok from B to A once twice, ok, etc well we can just do it from B to C I went once from C to B, C to A C to A, 0 so that's the transition count that has the same transition count matrix you can get, that's if and only if you can get one string to another by switching blocks in this way and the claim is that for random walk with reinforcement if two strings have the same transition count matrix they're assigned the same probability and I verified that in small examples and thought it was probably true but I hadn't actually written out a proof and we were walking back from the science center to his house or something, I told him the story and he said well let me try that, I'm going to write it down and the next day he wrote out a two page proof and honest, you know, combinatorics from ballistic proof that for any graph a graph that random walk with reinforcement is called partially exchangeable well I built a machine which is called Diffinating's Theorem for Markov chains and so this condition is called partial exchangeability so partial exchangeability is equivalent so a process is partially exchangeable is equivalent to it being a mixture of Markov chains that is if and only if for every n and all x1 up to xn the probability assigned to x1 up to xn the probability that your process goes through this string I suppose it starts at a fixed point is equal to the integral over the simplex of the product of theta ij to the ni ij mu d theta so here this is the simplex it would be the three simplex if I was working with the triangle but it's the simplex of all indexed by the set of all edges and so the claim is that partial exchangeability is equivalent to their existing a measure on the simplex fixed measures such that for all possible strings the chance of seeing that string can equivalently be said as pick a transition matrix pick a Markov chain from this probability on Markov chains and then run that Markov chain for Markov chains so his calculations allowed me to use this theorem and then knowing that there was a limiting measure it's the only first way we knew that there was a limiting measure I could then use the existence of a limiting measure to calculate by combinatorics what are the chances and this is what came out so that's a third example and so one part of this talk of course is as much as I am an applied mathematician statistician what do we do for a living so I've given you three examples they're mostly problems that come from some real world connection and shuffling cards and analyzing binary data and well walking around the building in many cases even though you've done a specific problem that came from an application if you take the math seriously something happens in particular I'll tell you one sentence of what happened from this triangle problem of course you can do this on any graph so take an infinite graph like take the integers and so put one on every edge of the usual graph of the integers a random marker starts at zero and every time she crosses an edge she adds one to the weight of the edge so there's a restoring force back and when you're at a new vertex move to one of the vertices adjacent to you with probability proportional to the edge weights and ordinary random walk on Z is recurrent you come back to zero infinitely often so it must be if you have a restoring force it must also be recurrent and that's true it wasn't so easy to prove it was Robin P. Manel's thesis at MIT but in two dimensions it was open for 30 years it was my most famous open problem that is, is random walk in two dimensions with recurrence is it with reinforcement is random walk recurrent and finally Pierre Therres proved that it is recurrent in fact it's recurrent on any lattice just to give you one more sentence about that if you take an infinite tree our favorite object with the infinite binary tree new random walk with reinforcement if you add one each time it's not recurrent if you add five each time it is recurrent and five is the square root of some solution of the transindulic region so there's a phase transition that's another result of Robbins and so it's an interesting problem and it came from this well the breakthrough came because the measure my crazy measure this mu you can't see it but there it is I didn't write it down but this mu which involved the homology group of the graph in interesting ways they've been studied by physicists Tom Spencer and Martin Zernbauer for doing super symmetric string theory and in an unbelievably long hard piece of analysis they figured out enough about this measure so that we could do the asymptotics that we needed and then they got interested they've never heard of different eddies theorem or this way but it's the same measure that came out and there's a very healthy interaction between the group of physicists and the group of probabilists and if you're interested it'll be an aimed workshop in about nine months from now on the interaction these are three instances of how Barry helped one person in applications I bet that everybody in this audience at least those of us who know Barry have a similar story it would be interesting to try to bring them together I'm going to end with a warning I don't think I've successfully ever finished teaching Barry probability but we're both still around