 Are there any difficulties, questions about the previous lectures, because you have had this working. Are there any difficulties about the previous lectures, because you may have had time to look at them over the weekend, no difficulties, ok. So, let us continue, so let me recapitulate what we did in the last lecture. So, we were discussing the directed sand pile models. So, we said that you know that you can think of them as wooden blocks on a staircase and they are much more plausible picture of a realistic sand pile than the BTW model, because you know there is some surface on which things happen which looks more like a real sand pile, ok. So, then we said that in this case the delta matrix is upper triangular. So, then the determinant is easy to find you know it is just a product over all the diagonals all the for a 2 D problem you can think of it as a sand pile like this. Its height when it topples it throws out 2 grains 1 to each downward neighbor and so the allowed heights are 0, 1 at each side total number of configurations is 2 to the power n and all these configurations occur with equal probability in the steady state, ok. So, then for this model because the recurrent space of configurations is somewhat simpler. It is easy to calculate properties of the system in the steady state like what happens if you add 1 grain. Well if you add 1 grain then with probability half nothing happens because it goes to a site with height 0. So, you can calculate Avalanche properties like properties that number of topplings s equal to 0 is just half probability that number of topplings s equal to 1. So, it should this side should topple, but when it falls 2 things down these should not topple. So, the probability is that this is height 1 and these are height 0 and so this is 1 by 8, ok. And then you know you can extend this further is equal to 2 will be that you know this side should topple then it should come down to something and here the height should be 1. So, that when 1 particle comes it topples further it falls down, but this should not topple and this should not topple or you can have a configuration which is like this and this side topples and this side topples and then they do not topple and this one also does not topple and both of these have equal weight which is 1 upon 2 to the power 5 into 2 and so on. But we also show that for largest the Avalanche clusters form wholeness clusters. So, there are no holes the only the boundaries do random walks these are unbiased directed random walks downward that each boundary takes left or right step with equal probability and so you can calculate the probability that s is bigger than equal to x goes as 1 upon x to the power 1 third probability that duration greater than t goes like 1 upon t to the power half and of course, you can actually get the exact formula for this probability of s for arbitrary s, but that involves some generating functions and you know hyper geometric functions or some such thing. So, I do not want to get into that just now, but it is the reference is given and then we said that model is more interesting because it is the equivalent to other models. So, one we discussed was shiedegger river networks and the shiedegger networks were defined on a big graph. So, if from each side there is a downward direction there are two possible downward directions, but geographically one of them is chosen and we say that we imagine all kinds of landscapes and in the ensemble of such networks each downward direction occurs with equal probability or you pick a random sample you do not know which one it would have been it would be equally likely to be left or right with equal probability. So, you generate a configuration then in that configuration you ask what that the large scale properties of this network what does it look like. And the answer is that if you look at any node then with some probability nothing will coming into it. So, the river outward will have net flow one in our units, but you know if you ask what is the probability that the river out was a net flow two then there are some finite number of configurations which you can calculate and all these can be done and the key point was that if you go to some site in the river basin. So, there are all kinds of rivers. So, I look at one particular site and ask what is all the water which is flowing down from this site that is called the catchment basin of the river at that point. So, that catchment basin has the statistics of the Avalanche river and the Avalanche clusters in the directed sample model. So, these things have other names. So, you know these things are sometimes called. So, a typical picture of this cluster they called staircase polygons because you can draw them like this and each and now the axis may be this is the vertical and this is the horizontal then these are staircases then two possible staircases join up to form a polygon and this is called a staircase polygon and we are looking at the statistics of staircase polygons and that is the way this thing can be sold exactly without too much hassle. So, let me just say that. So, we define this Horton's law which is that the number of rivers of rank are divided by number of rivers of rank r plus 1 is a number which is something like 4 which is the same for all geographical river networks. So, you can ask what is the value for this 2D model and interestingly that number has not yet been computed exactly you can of course, determine it in simulations, but since the model is exactly soluble you may ask that can I actually calculate it from first principles and should be possible, but it has not yet been done. There might be other related questions which you can think about yourself which have not been answered or which have not even been asked, but we will not go too far into that one. Then we discussed Takayasu model aggregation model and this is actually the same as the previous one. So, not much need be said even though it is when it was originally proposed it does not look exactly like a river network model it is a model of aggregating particles on a line or some such thing, but if you imagine that the time history of the aggregation model as a picture then that picture looks like a river network geographical picture. So, you have to actually realize that the space time picture of this model is the same as that picture. So, there was one thing which we did not do which is actually an important model. So, I will do it today. So, this model so this is actually this is the end of the recap. So, this is equivalence to water model. So, again this model is much older than the sand pile model and it was studied a lot and people found it interesting to study and then it is very nice and that this new model turns out to be related to the older well studied model. So, let me define the water model again you have a lattice water model is very interesting because again you know the physical systems being discussed are somewhat different. In water model you have a population where people make they vote. So, each of these is supposed to be a person he sits at a node he can talk to his neighbors and is influenced by them and there is some issue like there are two political parties A and B and he can either vote to A or vote to B, but if all the friends are voting for B then he tends to vote for B that is the model. Then what happens as time evolves then. So, the statement is that each side has a voter opinion of the voter time t is equal to A this thing the opinion of the voter is not a very fixed event it is a stochastic event he looks at its neighbors and with some probability he chooses one view and with some probability chooses the other view. So, we will write down the probability this is just equal to the number of neighbors divided by total number of neighbors. So, if all the neighbors are voting for party A then he will also vote for party A with probability 1 if three of them are voting for A one of them is voting for B then he will vote for A with probability 0.75 if half of them are voting for A half for B he also votes with probability half for A and B and then he has a vote and now at the next time step his neighbors may change their opinion based on his opinion. So, the system evolves in time you start with some opinion assigned to various people, but there is a tendency in the model for people to have same opinion. So, what happens is that as the time evolves you will find that there are clusters of regions where people have similar opinion and these clusters tend to grow in time. A particular feature of this model is that once there is no hole in a region if everybody is voting A then this part will not change their opinion the change of opinion only occurs if there is some region voting A some region voting B then at the boundary there will be people who will have some people voting A some people voting A they can change their opinion with some probability. So, all the dynamics occurs at the boundary of this different regions voting A and B and then what was found is that there is a something called coarsening which is that as time goes by the region which votes one way keeps on becoming bigger and bigger in time. So, the mean size of cluster of one opinion keeps on growing and for large at very large times everybody will just pick up one opinion and then nothing will occur further. So, that was the model, but it was very nice to understand this one because A it was one of the simplest stochastic dynamics models and it is an opinion exchange model or whatever you want to call it and ok. So, what makes it so interesting is that see this is like an Ising model because the two opinions are up and down states and the probability of up state is more if the neighbors are up and so on. So, it is like a ferromagnetic Ising model except that the transition rates here do not actually satisfy detailed balance condition. So, it is a non equilibrium model and so it becomes interesting to see what happens here and the reason people got interested in this model was that usually when you have all these time dependent models then you try to write evolution equations for average quantities and they satisfy this b b g k y hierarchy. So, how many of you are familiar with the word b b g k y hierarchy 1, 3 ok. So, no it is not a problem I did not expect all of you to be familiar with the word already. So, what it means is the following that you know so, there are variables which we will call s i of t this is the site index and this is the time the site I took to be two dimensional, but you can work in any other dimension also if that is not really the issue ok. So, s i of t is a variable which evolves in time. So, you can start with some initial state and you can ask what is this average value of a how does the average value of the opinion at time at state site I evolve in time suppose I give you the initial state. So, this one it turns out that because it is a Markov process you can write some evolution equation for this d by d t, but that typically involves two point correlation functions of the state. Then you try to write an evolution equation for the two point correlation function that involves a three point correlation function and three points will involve higher ones. So, there is a hierarchy of equations n point correlation functions involve n plus 1 point correlation functions and so on and then of course, the problem is hard to solve because how do you close this hierarchy, but it turns out that for this simple model the equations close and the one spin expectation value only depend on the one site correlation functions and then you can solve them in closed form. So, of course, the reason you can close this hierarchy is because the rates in the model have been chosen in a particular way and that makes the problem soluble and the corresponding problem for the Ising model will not be soluble. So, let us not get into this one we will take this to be our problem. So, in our case I want to consider the following case suppose I have a system in which everybody has opinion A, but there is only one percent who has opinion B sorry ok. So, origin has opinion B everybody else has opinion A how will this evolve? So, what I expect is that at some time or the other the opinion B will die and everybody will have opinion A and then after that the system does not evolve. So, I might want to know how long does it take for the system to for this opinion B to die out or it may be possible that the opinion B grows bigger and bigger and from 1 you get to 2 or 3 or 4 or 5 and it just grows without bound or some such thing and even then. So, I would like to know how what is the time evolution of this ok. So, in this case what we will imagine let because of ease of drawing pictures let me work in 1 plus 1 dimension. So, at one time there are all these people and this one has opinion B and everybody else has opinion A. Now, at the next time so, what we will do is we will do odd even evolution odd even evolution says that let me draw it and then we will describe. So, these are the even sites and these are the odd sites. So, odd sites at time t equal to 0 these are the even sites and they these have opinion A these have opinion B. Now, this guy at odd site 1 this is 2 0 this is 2 x equal to 2 this is x equal to 4 this is x minus 4 minus 6 x equal to plus 2 x equal to plus 4 x equal to plus 6 and so on. And this site is x equal to plus 1 it has these 2 neighbors and I want to find his opinion at time t equal to 1. So, what we will do he will look at his 2 neighbors and count how many of them have opinion 1 or 2 you know opinion B divide that by 2 and that gives you his probability of picking that yes. So, it had some opinion earlier and that does not matter because it just looks at the neighbors and picks one of their opinions. In fact, the way we say it he picks a neighbor at random and picks his opinion as his own pick the neighbors opinion as one's own opinion with equal probability just picks a neighbor that is the same result as this. You can say that he looks at all of them and he just chooses a random number or picks a random neighbor and picks his opinion these are the same result. So, this guy I am trying to find out what is this person's opinion at time 1. So, what will it be it has one neighbor which is A one neighbor which is B probability half it is A with probability half it is B what about this guy this one is also probability half it is A probability half it is B this one with probability 1 it will be continue to be A you know because both neighbors are A. So, this will be A A A, but this one will be B with probability half this will be B with probability half and these are independent events. So, here this is I can use colors these guys are let us say these are opinion A is the usual opinion this is the new special opinion. Now, here this guy can become green or it may stay you cannot see from here it will be redone. So, I am saying that suppose there is a site this one which has an opinion B and the two neighbors have opinion A bad drawing this is constant time line these have opinion A this site has two neighbors down two downward neighbors he picks one of their opinion at random. So, this is likely to be B with probability half this one is also likely to be B with probability half all these other sites all these other sites will continue to be A and their upward neighbors will continue to be A. So, this is my model. So, the claim is that the evolution of the A and B as in this model is the same as the toppling process in my original model because if some site topples then you know it sends in whether the next site will topple or not if one of the things topples below then it topples with probability half if both of them topple then it topples with probability one. So, yes yeah so, this one is now the time direction this is time and this is the lattice and I am looking at the opinion yeah this is the one dimensional. Voter model in 1 D is equivalent to the 2 D directed sand pile model and the similar thing for other cases. So, now, I do not have to go for that all the yes time evolution of a 1 D E case that was the x coordinate actually the site in a this 2 D lattice is called odd if it is x plus t is odd. So, here what happens is that let us imagine that there is this there are these sites also the odd sites at time 1 will look at that neighbors at the previous time. They have their opinion, but they do not remember it at the next time they only look at the neighbors opinion and pick one of them. So, you imagine that there is this x equal to 0 1 2 3 4 5 6 and then this is t equal to 1 t equal to 1 x equal to 1 t equal to ok like that sorry this goes like this. Then in this problem the lattice breaks into two parts which do not talk to each other one is the sites with x plus t odd one is the sites with x plus t even. So, I only draw one of them I do not draw the other because they do not talk to each other we can discuss one of them at a time. So, I only look at this site here and it affects these two and then they affect these and you know like that. So, I do not look I only look at odd sites at odd times and even sites at even times yeah, but they convey it to the even particle and then they forget. So, if I am one of those sites let us say even site and I want to form my opinion. Now, I just look at what was the opinion of my two neighbors at the previous time step both of them are odd ok. So, that is what t term is my opinion and then this opinion I give my vote I tell everybody can see what I have voted next time somebody else will see oh my neighbor will say oh he voted this way that the other neighbor voted this way. So, I will vote this way and so on ok. So, the claim is that if you work with x t lattice then odd and even sub lattice is decoupled they do not talk to each other because of my strange dynamics and then the evolution of this particular model is exactly the same as in the directed sand pile model ok. So, in particular if I calculate the probability that x and t plus y. So, this is the s takes values s equal to plus minus 1 let us say. So, this will be equal to half s of x minus 1 t plus half s of x plus 1 t ok. So, it is related to the mean opinion of the previous of his two neighbors x minus 1 and x plus 1 is the previous time, but now this is a linear equation of evolution it is a s of x t is a function of two variables and it satisfies this equation. So, can I solve this as a function of x and t given the initial condition and the answer is yes we have seen this equation before because I have studied the random work problem it involves the same equation is the probability of a random worker being at x at time t plus 1 is the sum of the probabilities at the two neighboring sites at the previous time step. So, same equation is encountered in the random work problem and the solution is given in failure or you can work it out by yourself right. So, this problem is easy to solve then I can look for some other questions which also can be solved. So, in particular I can ask you know. So, what happens is that this b what happens is that the b evolution look like this it had some set of b sites, but eventually they died and everybody became a afterwards. So, what is the size of this b cluster in this space time plot that statistics with the statistics of Avalanche is my San Pine model. So, then you know there is a large amount of literature on the water model which we will not get into just now, but one learns a lot from other previously studied problems. So, I am just making the contact that oh well this looks like a previously studied problem. So, I lot of stuff is known. So, maybe I can read up on this and many answers are known, but anyway we figured out many answers, but the correspondence is interesting because it also works in 2 plus 1 dimensions or 3 plus 1 dimensions identical no change everything goes through. So, next yes. So, this can be called a probabilistic cellular automaton evolution rule. So, there is a lattice each site evolves by looking at what the neighbors do and so, this is a particular probabilistic cellular automaton can also think means it has a preference over for one particular thing he has a preference of his own. So, those kinds of models have been studied in the water model context, but they will not become equivalent to a san pile model. So, what happens is that usually if there is a preference for one candidate in an election of preference for one state then in these models those people kind of prefer that state and the state is after sometime that state is preferred. So, quite often you find everybody goes into that state. However, if it turns out that here is a system in which some 40 percent of the people prefer state A and 60 percent of the people prefer state B and you evolve it in time then this fraction on the average does not change even though in the end everybody becomes the same state. So, what happens is the probability 40 percent everybody will pick up the state with probability 60 percent everybody will pick up the state and the mean does not change mean remains 0, but the state becomes unanimous unanimity on the average. So, we will not discuss this further let me take next topic. So, this was equivalence to water model just once again there was some question. So, next is there is a very useful notion called waves of topplings. So, what we do is the following we have this square plate is cent pile I add a particle and I topple, but what I do is that I topple this particle once, but after that I hold it I do not topple at this side I topple everywhere else and then I look at the final answer. So, what can happen is that after sometime all the toppling has died and there is some this sides have changed and this sides outside will of course, this sides have toppled and they will throw out some grains, but now everything is quiet and stopped in which case this is the final configuration and I am done ok, but there is another possibility is that this side toppled, but it threw out 4 grains, but all these 4 grains toppled and they threw out one grain back each and the new height is still 4 ok. In which case now I topple it once more and topple it again and so the new things will topple and they will usually spread out, but they do not spread out as much as the last time and then I look as everybody is become quiet no everybody has become quiet, but is my initial height at now stable or not if it is unstable I topple it once more. So, avalanche using the abelian property you can break it into sub avalanches where you hold the origin topple it once let everybody else relax turns out everybody can relax at most once at every other side will topple at most once ok, proof by induction and you know it is a trivial proof you know the rules of the modulus trivial all the proofs are trivial just have to think straight ok. So, then what happens is that avalanche breaks set of waves of toppling and this set of waves the regions of these waves form nested structure clusters. What is meant by nested clusters? Nested clusters means say something as simple as this that I take this origin it topples all these sites then you can prove that there will be no holes in this cluster ok. The proof is the same if all the four neighbors topple that site also has two topples so it cannot have holes ok. So, now this all these sites have toppled once and then there will be set of sites which have toppled twice and there will be set of sites which have toppled three times and so on. And finally, there will be a set of sites which have toppled four times let us say where one of the neighbors has only toppled three times then now the avalanche will stop ok. So, any avalanche can be broken into a subset of these avalanches which are waves of toppling 1, 2, 3, 4 some number which varies from avalanche to avalanche, but then it stops and then I can say what is the size of the first wave? What is the size of the second wave which is the total area covered by the first wave of toppling, second wave of toppling, third wave of toppling and so on. And these are interesting statistics in other models this does not happen it is not as simple as you can have structures like this. In other models you may have sites which topple once and there are sites which topple twice and there are sites which topple three times ok. And they do not have to be connected to each other in any simple way. So, this nested structure is a particular property of this particular model and the undirected 2D model or may be it has also happens in higher other models, but not every other model no no no it can be 15 or it can be arbitrarily large for very big lattices ok. In fact, so you can take very simple case like you can take an L by L lattice and every site is initially height 3 and then you add one particular particle at the origin and ask how many topplings will occur at the origin and the answer is order L ok. Yes, but it is not enough to be a billion there are some additional properties which are required. So, we will get to it today itself in I will explain later another model where this does not happen it is called the stochastic sand pile models where the toppling directions are arbitrary ok. So, it is not just a billion property it is a billion property plus the fact that all the sites are the symmetric transfer rules ok. So, it is interesting to study the statistics of these avalanches or the statistics of sub avalanches which are the waves of toppling. So, statistics of waves. So, the original sand pile model the way we have defined it still remains not fully solved in the sense that people cannot determine the critical exponents corresponding to how many topplings will occur if you add one grain. So, the probability of s topplings goes as 1 upon s to the power tau, but I do not know the value of tau exactly analytically yet ok. So, people have tried to study the variations of this question how you can understand them and it turns out that one of the variations is you say what is the probability of first wave has exactly s topplings s 1 ok. So, I can ask this one I can ask for the probability that the second wave has s 2 toppling you know s equal to s 1 plus s 2 plus s 3 total number of topplings is equal to topplings in the first wave plus second wave plus third wave and so on ok. So, if we were not able to answer the original question there is not much hope of asking a more detailed question which is the newer one. However, it turns out that if you modify the question a little bit then the answer becomes possible and the modification is the following. So, I start with my senpile model and I make a time series s s 1 s 2 s 3 s 4 s 5 s 6. So, I added a grain and I found s 1 topplings then it is relaxed and then I added a grain I found s 2 topplings and then stopped and then s 3 s 4 s 5. So, I have a long time series of how many topplings occurred in the next avalanche that is called s i here s i is a time series of events of toppling numbers and my computer has generated these and I want to calculate the average of s which we actually did, but I want to get the probability distribution of s and so on which we did not do ok. So, this one is a hard problem, but suppose I go to a different variation of this and I break this s into s 1 1 s 1 2 s 1 3 s 2 1 s 3 1 s 3 2 s 4 1 s 3 2 s 4 1 s 3 s 4 2 s 4 3 s 4 4. So, what I have done is that this entry goes into this 3 entries, this entry goes into this entry, this entry goes into this entry. So, each avalanche is broken into sub avalanches and I write each of them separately. So, now, I have a different time series whose length is different and now I can ask in this time series if I pick entry at random what is the probability that that entry will be 7 or any other number ok. So, it turns out that this answer is calculable the difference being that this ensemble is different from this ensemble. If you have one event which is very big then it will give rise to 6 or 7 entries. While the original in the original model the probability that s equal to 10 to the power 6 is very small, but now that 10 to the power 6 event will give rise to 15 10 to the power 4 size events or some such thing. And so, the net probability is bigger and this one one can actually compute exactly ok, how can we compute it exactly you can show that in the new ensemble each configuration like this. So, what happens if I look at this configuration there is some site which is here and all you know. So, this is after 3 stroplings this is my configuration. So, I ask is this configuration stable or not recurrent is this configuration recurrent or not suppose. So, if I change this from 4 to 3 will this become recurrent the answer is if it is yes then you are done if it is no then the original configuration will not burn fully. So, there will be a cluster which will remain unburned ok. So, let me say this again yeah this ensemble the ensemble is. So, I started with the time series then I convert the time series into a longer time series. And now I pick an entry at random in the new time series that is my ensemble all the entries are whatever they are I pick one point at random and say what is the probability that this entry will be this number. How is the conversion? How is the conversion the answer is that. So, there is a number s number s gives rise to some possible number of children which is the number of entries in the lower list. This number is a random variable ok, but then I look at I give equal weight to each child. And then I calculate the average value attached to each child that problem can be solved exactly because the new ensemble is more tractable for some reason. No, because I do not know how to break one of these entries into smaller set this one I broke because I broke the avalanche into sub waves of toppling. It is some wave of toppling I do not know how to break further in a natural way I can break it at random, but I cannot break it in a natural way ok. So, what happens is that here some big avalanches have some weight here the big avalanches have more weight because they have more you give them preference because you count how many children they give rise to right. It is like in a population if you give voting right to each individual or voting right to each individual depending on how many cars they have then of course, the ensembles are different in the second one the richer people get a preferred you know status. So, in this new ensemble the average value of s is different than in the previous one ok and the distribution of s is also different and. So, it turns out which we will not show, but I will quote just now. So, this is called s waves and this is 1 upon h w and the proof goes by showing that each of these elements of this ensemble is actually equivalent to a spanning tree that is what we our original ensemble was, but with one break. So, if you take a system covered by two spanning trees instead of covered by one spanning tree ok. Then all these spanning trees will occur with equal probability and that will correspond to the configurations corresponding to each of these sub ensembles and. So, that is the statement and I the proof is given in the pip references and I do not want to give it here it is a little bit involved ok. So, next thing is there any question so far yes please ok. The correspondence with the spanning trees is this to each of. So, I can say the following that take one entry from this lower ensemble. Can I attach a configuration to it? The answer is the configuration to it is the one which is obtained when you know. So, this was like this was first wave, second wave, third wave, fourth wave. So, the corresponding configuration of the sand pile is the configuration after you have toppled twice or after you have relaxed two waves, but not the third one and so on. So, you look at these configurations they have a different statistics ok. So, what is the one to one corresponding? So, here one site had height 4 that was the key point about you know the difference between this site is not in the recurrent configuration because one site has height 4 the origin. So, now if I convert the origin to site 3 then it becomes part of my old configurations, but then will it be recurrent? The answer is no it would not be recurrent because if it were recurrent then you would not go further ok. So, it turns out that the natural correspondence of the configurations here is to spanning trees with one link missing. You take a spanning tree break one link and see what happens it breaks into two spanning trees. What is the size of the smaller spanning tree? That is a question you can ask and the corresponding answer is that it is 1 by s and that answer corresponds to the case that if you pick an ensemble at if entry at random in this longer list what is the plurality that will be s ok. So, that is the statement ok yes ok. So, I said that you take these avalanches and break them into waves of topplings. So, s will be suppose this particular avalanche broke into 7 waves of toppling will be s 1 plus s 2 plus s 3 plus s 7 where s 1 is the size of the first wave s 2 is the size of the second wave s 7 is the size of the seventh wave then it stops then I add another grain and that is the you know it will have its own entries and so on. Now, I look at the sequence of the waves of toppling corresponding to my evolution of the sand pile and. So, this suppose I had generated the original series with 10 to the power 7 grains added one after another the number of entries in my list will be 10 to the power 7. Now, in because one entry has been converted into several entries the new list will be bigger than 10 to the power 7 it may be 5 into 10 to the power 7. I treat all of these entries as equally likely and pick one of them at random was the plurality that that new entry will be s that the distribution is called this thing probability of s waves in the waves ensemble. This ensemble is the size of waves sequence in time series of waves not time series of avalanches. Yes, but the sum of the numbers are equal in this of course, the total sum will be but since the number of entries has changed the average value of s will change in the new ensemble it is still proportional to l square, but it is a smaller number. So, now, we want to go to different question which so this is called the a billion distributed processors model it is also called a billion networks. So, the idea is this we have found that in studying all these different kinds of models it was nice to have this a billion property because it had to solve some problems which were kind of difficult to solve in most other similar models of self organized criticality. So, can we extend the idea of this a billion property somewhat and see a wider class of models where you can use this result can we generalize this problem result to a bigger class of models and the answer is yes and the general idea is computer networks is where you can use this. Nowadays you know computer networks are very important and you want to model them also and I guess the details of the computers are not very important relevant, but one way to think about computers is like this that there are local computers and they are connected to each other by some links and so we will take this this is the details of the model. So, there are set of computers they are connected to each other by some links a computer has an input stack of messages and output stack. So, what it does is that if the input stack is not empty it takes in one entry and reads it and based on that entry it does some calculation and goes into some internal state. So, there is an already computer has some internal memory there is some internal state reads in that input based on that input it goes into some other internal state and it can send out some messages to other neighbors. So, based on the input which is called an instruction based on the instruction and its internal state it goes into another internal state and it sends some specific messages to some other computers given. Now, those computers and this message goes into that computers input stack and sits there and if you know when the time comes it will be processed and that computer will also send out some messages to other people and so on. So, if all the input stacks are empty then thus network is idling there is no work to be done. Then you take one of those computers and put in one instruction then it will do some work it will send some messages to other people those people then will do some work and they will send some messages to other people may be and eventually the computer will either stop or it will keep on working forever. So, we will of course, want to study cases where the computer eventually comes to stop no point in discussing cases where it goes on forever we just want to avoid the cases where it goes on forever. So, the what the computer does it goes from one quiescent state where every computer is idle and all the input stacks are empty to another quiescent state where all the computers are idle and all the input stacks are again empty. In between there is this period of burst of activity where some messages were sent from one computer to another from the second to third one and so on and so forth. And we can measure the size of this activity that will be called the avalanche in the computer science is a burst of activity of the computers. You can think of the computers as neurons then this will be burst of activity in the neurons. So, that is my basic model and I want to understand the behavior of this model. So, what happens in computer networks is that there are all these networks they are working, but at the time when you send some instruction you do not know how much time the computer will take to execute that instruction. So, after some time the answer will come back to you let us say, but you cannot be sure of how much time it will take to execute the instruction ok. So, of course, people design protocols for parallel processing in which there is this guy he has some problem he wants to solve. So, he breaks the problem into 15 parts and sends it to 15 other computers to solve. They send some answers and they come back and you have to collate the answers and do something with them ok. So, the basic idea is that the final answer you want should not depend on the order and the speed of different processors involved in the calculation ok and your protocol should be able to ensure that this is what happens ok. So, for example, let us consider income tax processing. So, you know in India it is at the end of June, but in other country there is some date by which you have to file in your papers and the computer processes them and either tells you that ok you have paid extra or you have paid less and so much money is returned to you or they ask you for extra payment ok. Now, all kinds of people are filing their input from all kinds of places, but hopefully the answer does not depend on the order in which different people have filed their papers. Your process you have filed in something the answer comes to you the answer should not depend on whether Mr. A who somebody else filed the paper ten days ago or three days ago or some such thing that is a reasonable requirement to put it. So, the way we write computer programs to process it will try to ensure that this is what happens the outcome does not depend on the order in which messages arrive at a computer and the it is the same independent of the order. Obviously, this does not work in every possible case because I may have an airline reservation system in which you know people is put in request and you are assigned tickets, but of course, the order matters you cannot the answer cannot be independent of order for airline reservation if you put in earlier you get a reservation you put in later you say sorry nothing doing right. So, it is not true for all possible problems, but some subset of problems you would like the answer to be independent of the order in which the processing occur. So, for example, for example, I can imagine I do not know you have to calculate pi two million places of decimal that is my job. So, I do it like this that I break the problem into sub parts and ask fifteen other computers to do something and they bring me back the answer and then I add up all the answers. So, for example, let us not give that example in great detail it is easy to imagine cases where you break the problem into parts give the part which sub processor and then collate the answers right. So, it is best done in sorting problems like if I have a big list I want it sorted then I make break the list into parts get one person to sort one list another and then put the sorted list three sorted list together to form a single sorted list that is how the sorting goes. So, anyways so what we require is that the answer require in many cases not in all that answer is independent or speed of processors. So, how do we ensure it? So, this is the result the result is that if you can ensure that each computer in your network each node has the i billion property each node is a single computer it has some inputs it has some outputs. So, for each possible sets of inputs the answer does not depend the order in which the inputs arrive then the whole network will have the same property proof is by induction you know problems which involve only one transaction which involve two transactions three transactions and so on. So, it is rather clear and the range of applicability of these kinds of stuff is also reasonable. So, we want to construct examples of examples distributed processors. So, the first example is of course, the BTW model. So, here the computer is very simple it has input stack and it takes if there is entry one it says my height is one if one more message comes it says my height is two that is the internal state of the computer. Once the internal state becomes three and some entry comes it says make your internal state zero and send four messages to four neighbors right. So, that is the model and that is clearly a billion and so it is an example of this a billion distributed processors model is rather a simple example ok, but now we can make much more complicated rules for processing the input. So, in particular the rules can be periodic in time. So, periodic toppling rules. So, for example, the rules can be that the height is if the height is more than three then send out three to these neighbors then the next rule is that if the height is more than four then they send out four then it is three and four and three and four and the critical height keeps on jumping between three and four three and four three and four three and four and the a billion property remains as you can easily check and all the analysis we did we will go through ok. The period does not matter the period can be very long ok. So, we were trying to make up some nice models of periodic toppling rules and a very sort of interesting variation is that you do not keep any particles at the site. As soon as a particle comes to a site it is sent away, but you send it to. So, it is called a Eulerian Wokker model. So, the model is defined like this there is a lattice at each side there is an arrow which is marked towards one of the neighbors ok. So, that is my configuration at each side there is an arrow which goes towards one of the neighbors this one goes this way this one goes that way this one goes that way and so on and these are just static. Now, a particle is dropped at one site then it sees oh I am here there is an arrow mark this way. So, it tilts the arrow to the right by 90 degrees rotates the arrow and then takes a step in the direction of the new arrow direction ok. Now, it goes there there is also an arrow mark there. So, there he will look at the arrow direction rotate it by 90 degrees and take the step in the new direction and it keeps on taking steps like this till it finally, leaves the lattice you know may be at some site here the arrow may be marked this way. So, then it goes out. So, you go from a static configuration to another static configuration then you drop another particle like this ok. So, that is my dynamics seems fairly straight forward you add one particle at a time and that particle does a deterministic and walk because once I am in drop a particle it has a fixed rule it says go there the guy says go there this guy says go there keeps on going until it leaves and then you drop a new particle and it does. So, what is interesting about this problem the interesting point about this problem is that as the worker moves in this medium it modifies the medium and that affects the next worker which will come whose motion will be affected by the medium in turn affects the worker. So, if the worker comes back again then that worker will be affected or if somebody else comes later then that person will be affected. And so, it has this long term cumulative memory at some level and it gives rise to long range correlations that is the yes please. Because it modifies it tells the arrows no. So, as the worker before the worker went the arrows were in some direction after the worker has gone the arrows have moved and they stay put there. So, the medium is modified by the motion of the worker no the. So, the rule was you can make other rules, but the rule we chose was that when he comes to a site he sees an arrow then he rotates the arrow and then goes into the direction of the new arrow. So, he modifies before he moves at the site you can do other names you know these are names, but basically he goes in some direction based on local rules. And we have chosen the local rule that first he rotates the arrow and then moves you can make the other rule of course, which is equivalent, but that is what it is yes. So, the update rules are not stochastic. However, where you add the particle may be stochastic you add the particle. So, what happens in this model that there is this stuff you add the particle here it does some work and leaves. And then you add the particle here it does some work and leaves and then you keep on adding and then you keep on seeing. And the point is that the motion of the worker is different from a random work now, because it is affected by the medium and we would like to know how much it is affected by the medium. It generates long time correlations it is a very simple example of some generating some memory in the system. No, so the arrows of course, have four directions. So, they keep on it is a good question may be in general it is possible, but in this particular set of rules it is not possible we will show, but first the proof of the Abelian property. Suppose you had any you can actually drop one person and so, a i is the act of adding a grain at site i and relaxing the configuration of arrows. The configuration of arrows is called c and this is c prime and then of course, I can do a j is equal to a j a i of c this is what I want to prove a i a j is equal to a j a i no wait until he leaves the system ok. So, I can imagine that suppose at sometime there are two people who are two workers present then this guy has his own local rules you know he has to tilt this one and move that way and this guy has to I can go to this guy and move make it step move one step and go to this guy make it move one step then make it move two steps and then come back here and do it in any order I like until they leave the system and in the end there will be a final configuration of arrows and will it be independent of the order in which I have done the update yes please yes very good no they do not move together. If they come to the same place then we will just imagine that one of them has come earlier and then he moves according to the first rule and then the second one moves according to the second when they come together they are treated as one just after the other yes yes that is what we are trying to argue that it is that if you start with A what I was previously I said topplings commute here if you take these two configurations and you move this one one step and move this one one step this will you can do in any order and you will get the same result and then we said that this implies that you can have any number of workers and you can topple them or move them in any order and you will get the same final result they are all identical so they do not have any names so I cannot tell okay well these are independent statements you can prove each of them yourself okay so so this model is actually very interesting for various reasons so let me just do one more so I can take a configuration I can work on a torus where there is no boundaries okay so it is a periodic graph with no ends and I just put in one worker and the configuration of arrows and this worker will move and he will change the arrows and then it will come back and change the arrows again it keeps on reconfiguring the arrow configurations and eventually what happens eventually it has to go into a cyclic state because the arrow configurations are finite in number the position of the workers are finite in number so whatever configuration you have we will eventually repeat okay so but this number of configurations is very big it is four to the power lm if lm are the sizes of this torus but it turns out that this worker is very smart he organizes all the arrows in a way such that eventually he gets into a cycle and the length of the cycle is just four lm eventually you start with any configuration it will go into a cycle of length four lm in which each cycle each bond is a traversed exactly once in each direction so in one cycle it will organize the arrows in a way such that eventually it will come back to the origin after traversing each bond exactly once in each direction so it takes four lm steps and it comes back to the origin and this happens for each initial configuration no four lm steps but there is only one worker and so the number of bonds is two lm and each bond is traversed exactly twice in one cycle one this way one that way okay so the worker somehow organizes the medium into so that it gives you a periodic cycle okay and so this property is actually true for all possible graphs you can make doesn't have to be square lattice doesn't have to be you know triangles either can be this graph so at each node you have some output directions and we just say that the there is a rule which says from this direction you go to this direction to go to this direction and periodically repeat which order you repeat also is arbitrary choose any order you like at any vertex and start with any configuration of arrows and let the worker walk around eventually he will get into a cycle in which each bond is traversed exactly twice in one cycle is valid for all arbitrary graphs so it's a very nice and interesting result and such tours of workers were first studied by Euler so in honor to Euler we call this the Eulerian worker model is the reference to Euler known to everybody no because I think it's a good historical reference so it is nice to mention in case you don't remember know or remember how many of you know the original Euler's cloning spread bridge problem at least ten people know okay so let me mention because it's a historical remark it's not really relevant for my work but I think it is good for you to be aware of the history of the subject and so this is a recreational math problem so they said that there is some city I might get some details wrong it has some bridges and there is a person who starts from one city one place and he has to find a circuit such that he can visit each bridge exactly once and come back to the origin or can you do this this was the question and Euler proved that in the graph which was the traditional problem which had been studied for hundreds of years by kindergarten children this was not possible I think the final result was not very profound but the method he used to prove this gave rise to the subject called graph theory because he was the first to convert this problem from recreational mathematics into a graph theory problem he said there are these vertices and there are links between them and you make a walk between them he said that each site will have even number of visits and then if you want each of them to be stayed then there are odd number of coordination number or some such thing and so there was a proof but the subject his proof which was sort of recreational mathematics problem for some you know it was known for a long time but the analysis of the problem was sufficiently novel and interesting and it gave rise to a whole subject graph theory for which lot of people have been studying for so long afterwards so the fact that the original problem was kind of simple is not a derogatory negative comment it is that you know it depends on you if you can study it well then you can get profound results from elementary problems ok. So, that is the context but since this problem gave rise to the whole subject of graph theory I think it is an interesting problem and this our stuff is just a additional ingredient into the same general problem a walker which visits each bridge now it says it has to visit each bridge in each direction exactly once and how will it do it and you know so there are some the circuit is the Euler circuit and how many circuits there are different possible and all these kinds of questions you can ask and they may and may not have good answers ok. So, that was the historical comment so what is known about this problem Yes, no no no the movement of each walker is fully determined by the background on which the arrows the walkers movement is fully determined by the arrows ok once he come here you have to look where is the arrow then rotate it and then go in that direction ok. So, the only randomness in this problem comes from the initial distribution of arrow you can start with the configuration with some configuration of arrow and put save the walker is here what happens and without going into the details of the problem I will tell you oh eventually you go into a periodic cycle with this period ok that result is sufficiently non trivial yes please sorry in this version with periodic case there is no sink, but then the walk goes on forever it goes into a periodic cycle then you look at the properties of the cycle there is another version in which you add a sink such that it can leave the lattice then I can ask questions like ok how many steps does it take on the average to leave how many what is the distribution of step sizes and so on they are analogues of the corresponding problem for the sand pile after all the walker is just a grain of sand which moves into the neighbor and does some kind of a walk and eventually leaves ok and the way it works depends on what the rest of the grains are doing ok ok. So, what is known about this problem? So, we will look at the version with the sink slew drive slew drive means you add a particle and let wait until it goes out of the system then you add another particle and so on ok. So, you can define operators a i I will work with this lattice square lattice ok and then you can show that these operators at different sites will commute and they satisfy operator algebra also and we cooked up the rule. So, that the operator algebra is exactly this because if you add 4 walkers at the one site any site then one of them wherever the arrow is one of them will rotate it once and go there means when 4 of them will rotate eventually the arrow will be back in the same place and there will be 4 arrow 4 particles at the 4 neighbors. So, that is like adding one particle at each neighbor. So, this equation is the same as the equation for the sand pile operators. I gave the proof just now know I said that if you add 4 particles at one site then one of them will go in this direction one of them will go like that one of them which ever way you do ok. So, this is obvious from definition and abelian property I what I do is I add 4 particles I first move this one and then this one and then 4 ones and then let them move later and so this result is immediately clear. So, this operator algebra of these operators AI is the same as in the sand pile model. So, then you know in some sense we were hoping to factorize the original problem there were too many grains in the original problem that is hard to study. Now, you can study one grain at a time because you can study one operate one walker at a time. However, that does not work out because the statistics of abalances of this model is not the same as in the other model the. So, forget all the other things if I take this model like this. So, I have this lattice size L by L let us say I add a grain at random then how many steps do I have to take in order to leave the lattice. Well as we argued earlier or as we can argue now this number is of order at least L ok, but in the original sand pile with the finite probability you could did not add anything and in the balance stopped. So, there was a finite probability that s equal to 0 s equal to 1 s equal to 2 s equal to 3. Now, all the if L is big then most of the time you will have to take many steps before the particle leaves. So, there is no distribution at small L there is only a distribution at large L ok. So, in this case from probability of s as a function of s there is something like this where this value is like L squared just order of magnitude L ok. So, it has a different statistics. So, that is very interesting or surprising you have the same operator algebra, but somehow it does not give you enough information about the avalanche statistics ok. So, then we said that let us try to modify this model further. So, suppose you make the model in which the particles come, but they do not leave a single particle cannot leave, but if there are two part two workers who come to a particular site then both of them leave one after another one of them goes in that direction set by arrow and the next one in the next direction ok. So, now you again have the same thing like the sand pile model except now the critical height has become 2 instead of 4. So, the particles they will be in the recurrent state they will be 0 and once and then when you add something it will go to if it you add at 1 then it will topple and then you know that avalanche will propagate in some way and can we understand how it goes. The critical height can be reached even on a square lattice you can reset the critical height now to 2 or 3 or even 5 if you like at 5 5 grains will you know ok. So, the model can be generalized the statistics of ever. So, this is called height arrows model because it has both variable called height at the site which is the number of grains at sitting at a site and a direction and if just a generalization of the sand pile where there were only heights and the Euler model where there were only arrows in the recurrent configuration and now you have both heights and arrows and you can study the statistics of this and see what happens ok. I think this model has been studied in some cases, but the full analysis in all cases is still not done. So, that is what I am going to do for you to try to do I just convince you that it is an interesting model beyond that it is not easy to study fully yet ok. So, next yes 3 no here 2 stochastic toppling rules stochastic or arbitrary. So, now what we will say we do not want to keep the toppling rules periodic even let us forget rid of the periodic condition. So, what happens is that at each site there is a stack of colored cards ok. So, when a walker comes here he sees a colored card or which says go to north. So, he goes north and the colored card is removed because it is work is over then the next colored card comes up which says go to wherever it wants to say ok. So, at each toppling there is a new rule, but all these rules are already listed in a stack and you come to the site it says if the height is less than 4 then stay put, but if the height is more than 4 then topple like this. So, that is what it will do, but once it topples then the that entry is removed and the new thing pops up that you know this colored card goes and then the next one comes up and the next one will have its own rule whatever rule it has the rule is if the height is more than this then go to this is this neighbor ok. Now, it is much more general model, but the abelian property still continues to hold proof is the same as before there is no difference ok. So, now what is the interest in this more general model. So, the model is right now very general. So, we want to make it less general to make it tractable. So, we make a particular choice that all the entries in the stack are go to n e s s e n I will define. So, it says if bigger than 3. So, it you know that this entry says that if the number of grains is already bigger than 3 then 1 goes to n 1 goes to e 1 goes to s next one says that if z is bigger you know z has to be big enough. So, that this rule follows otherwise you cannot go then both of them go to n then the next one you make any set of rules like this, but let us say that there are only finite number of colors 10 different cards and you just put them all 10 of them in some random order and the frequency of this rule is p 1 and the frequency of this rule is p 2 and the frequency of this rule is p 3 and p 4. So, finite number of rules with finite number of probabilities. So, then yes please no no no each site has independent stack and they do not know it is other stack they only are autonomous systems. Yes please no. So, the idea is that the stack is there the stack has been put in place what are the entries of the stack you got free to choose, but you put in the stack okay how you make the stack is entirely up to you we are saying that it is made up randomly it is a random stack with this distribution okay, but now given the stack it is a deterministic evolution because you know if I add a if there is a fixed set of grains then they will evolve in some way, but I can still choose the order in which I topple and the order will not matter and you will get the same result independent of ordering okay. So, now I have my rule which is the following that if the height is more than 2 then 2 particles go like this if the height is more than 2 2 particles go like that and there is a randomly they have been put in there are 2 colors of instructions and in each stack they have been randomly put in what happens a you can show there is an abelian property, but b now there is an independent description of the system the system is defined like this it is a stochastic toppling rule the stochastic toppling rule says if you come to a site then with some probability you go like this with some probability you go like that. So, in each time you topple you independently choose which way you want to go okay, so that is a stochastic toppling rule, but now the result for the abelian property is still true what it involves the key point here in the proof is that you have to have this notion of pseudo random number generators the point is all the randomness in the random number generators is only apparent randomness they are actually deterministic because there is this random number generator inside which is a fixed program. So, even though the actual pattern is deterministic it may appear to be random, so that is what we are using we are we have built a stack which evolves in a deterministic way. So, the stack is there now the toppling of the particles is deterministic according to the stack, but the stack itself may appear to be random. I may for example, say that inside the computer it says that just take the next random number and see if it is odd then do this if it is even then do that and then it will appear like a random stochastic evolution rule okay. So, these are called stochastic toppling rules that is possible, but now it is a correlated work because the rules can be that two particles live together. So, it is not a random work it is a much more general setup and it may say that with some probability you throw out two particles with some probability you throw out three particles where you throw out will also be specified with some with various probabilities. So, these stochastic sand pile models have been studied actually for a long time soon after Peierbach's paper there was a paper by Mana where he defined a stochastic sand pile model where he said that oh well you know 4 heights is too much let us work with sand pile with only 2 heights 0 and 1 if the height becomes 2 then 2 or more that is what he said then all the particles live in different directions okay. So, that model was studied numerically easily, but they were not very easy to analyze it theoretically, but if you make the rule different you say that at each toppling it is decided independently but at each toppling the number of grains transferred is prefixed it can be different at different toppling but it is prefixed to be 2 or 3. So, this top card says throw out two particles in these directions two particles then I throw out two particles the next entry says throw out three particles then I will throw out three particles and so on, but then the problem becomes a billion sorry what does a billion mean now. So, you start with a configuration stable configuration and you add a grain then with some probability you do this with some probability you do that then I can evolve it and there is a set of final configurations with different probabilities. Now, if I add another grain I can do it and then you will get another you know in each of these, but if I have this configuration then it evolves in one way if I have that configuration evolves in some other way. So, what does a billion property mean now it means the following a i of c when I add a grain I at configuration c there will be different possible final configuration c prime each of them occurs with some given branching ratio probability. So, I multiply that and that is the action of this operator on this state and so a i is a matrix, but now it does not give rise to a unique configuration for each configuration you add it gives rise to a linear sum of configurations which is taken to be the linear sum of probabilities ok. And so then my time is just up, but let me add 3 minutes and then we will stop. So, let us take a model in which you add a grain then if the height becomes more than 2 then 2 grains leave each in a random direction independently. So, each particle decides where it goes and both particles may go in the same way ok. So, then the operators a i satisfy this algebra a i to the power 2 is equal to a i 1 plus a i 2 plus a i 3 plus a i 4 it is the same it says that adding two particles will always topple, but you know with some probability 1 by 16 both of them will go north with some probability they will go east with some probability one of them will go north one of them will go east and so on and so forth. So, now you have a similar operator algebra, but it involves a linear combination of terms ok and then can what can one say about these kinds of equations. Again you know the operator algebra is still the same and thus they will still be simultaneously diagonalizable then each a i can be still replaced by a complex number to solve for the eigenvalues. And then I have couple set of algebraic equations solving for eigenvalues and then you know then can we understand what is the steady state of this system that is the state which corresponds to all a i equal to 1 or some such thing ok. So, that let me stop here now and we will continue in the next lecture.