Okay so let us continue with our discussion of Montel's theorem so you know so this is essentially Montel's theorem so what you do is you take so it applies so you know it is a version of Arzela Ascoli theorem adapted to the case of analytic functions okay and so as I told you Arzela Ascoli theorem everything happens on a compact set alright you need compactness on the set on which your functions are defined alright but here you know of course analytic functions are defined only on open sets on domains okay and we consider them on open connected sets namely domains. So what you have to do is you have to put all the requirements only on compact subsets okay so let me write this down so let script F be a family of analytic functions defined on a domain D okay inside the complex plane so D is an open connected set and all the functions in the family script F they are defined on this domain and they are analytic functions okay. Suppose script F is uniformly bounded on D so okay so here is again so I impulsively wrote down something that is too much to expect so you know when you want to when you want when you do the Arzela Ascoli theorem you will say you have family of functions defined on a compact set and it is uniformly bounded on the compact set okay then the Arzela Ascoli theorem is in gives you an equivalence between 2 statements once one statement is equic continuity of the family at each point and the second statement is that every sequence of functions in the family has a uniformly convergent subsequence okay. Now so you know so you know to expect a family of analytic functions to be uniformly bounded on a whole domain is too much okay so you should modify this and say that it is uniformly bounded on compact subsets of the domain okay so I will change this because this is too much on compact subsets subsets of D okay. So see a property that holds on compact subsets is called a normal property okay a normal so if you have convergence on compact subsets is called normal convergence if you have uniform boundedness on compact subsets it is called normally boundedness okay so I can state it as you know f is normally bounded on D okay. Then so you know what is so what is your Arzela Ascoli theorem the usual sense will say that you know if you have this uniform boundedness then equic continuity is equivalent to the fact that every is equivalent to the statement that every sequence has a uniformly convergent subsequence. Then the following are equivalent number one script f is equicontinuous equicontinuous at each point of D okay and the second condition will be every sequence in this family script f has a uniformly convergent subsequence but now again you should not expect uniform convergence on a domain always you should only expect normal convergence that is you should expect uniform convergence only on compact subsets. So the second statement should be written carefully you should say that every sequence fn in f has a subsequence fnk which converges normally that is uniformly on compact subsets of D okay. So you know the statement so if you compare this with the Arzela Ascoli theorem the statement is in the Arzela Ascoli theorem your family of functions is not analytic in the general ascoli theorem the family of functions is not analytic it is only continuous family of functions okay but they are complex valued of course. So the analytic condition is not there but you have a weaker condition which is just continuity okay then and in Arzela Ascoli theorem the functions are not defined on a domain they are defined on a compact subset of the complex plane and the condition on f is that it is uniformly bounded on that compact set now that is replaced now the compact set is replaced by a domain okay therefore the condition of uniform boundedness is restricted only to compact subsets of the domain alright and then you have then when you have uniform boundedness then the Arzela Ascoli theorem the philosophy is that equicontinuity is the same as the existence of subsequence that converges uniformly. So equicontinuity condition is going to anyway it is a continuity condition so it remains as it is but the existence of a subsequence which converges uniformly that also you should expect only on compact subsets so that is why we put this we say that for every sequence you get a subsequence that converges normally alright. So this is Montel's theorem and actually again you know the implication 1 implies 2 is what we are going to prove the implication 2 implies 1 is you can prove the same thing I mean if you know 2 implies 1 can be proved just in the way that we have done that one could do for the Arzela Ascoli theorem okay it is a proof by contradiction okay. So what I want to tell you is that condition 1 will always be true okay condition 1 will always be true because of Cauchy's integral formula for the derivative of an analytic function which will give a bound therefore what will happen is that 1 will always be true if the family is uniformly bounded and therefore this is always true therefore this is always true okay. So what I want to state is that this first condition is superfluous the first condition always holds okay simply because you are not working with just continuous functions you are working with analytic functions and analytic functions you know for analytic functions you have a good bound for the derivative and in fact you have a bound for all orders of derivative derivatives of every order at a point because of the Cauchy integral formula okay so 1 is always true and therefore 2 is always true so 2 is always okay. So let us look at the proof of this so what I will do is I will just I will give the proof of 1 implies 2 which is the which is a slightly technical thing okay take a point z0 in the domain and choose rho greater than 0 such that the disk mod z-z0 less than or equal to rho is contained in the domain okay. So certainly you can do this you can since the domain is an open set z0 is an interior point so there is a disk surrounding z0 which is contained in the domain and you take a slightly smaller disk its closure will also be contained in the domain and call that radius as rho okay. Then note that by Cauchy's integral formula what you will have is you see so you know my diagram is like this here is my domain D and here is my point z0 that is this I am taking this circle centered at z0 and I am going to take radius to be equal to rho-epsilon where epsilon is a very small quantity alright and what is Cauchy's integral formula Cauchy's integral formula for the derivative will tell you that for any analytic function on this closed disk alright the derivative of the function at the center of the disk is given by 1 by 2pi i integral over this boundary circle mod z-z0 is equal to rho-epsilon you will get g of zeta t zeta by zeta-z0 z0 square okay this is the Cauchy's integral formula for the derivative right this is just Cauchy's integral formula right for g analytic in this disk okay epsilon very small so this is I am just writing Cauchy's integral formula I am doing anything else right now in particular so this is you get a bound you get a bound by you know putting by parameterizing this so you will get g dash of z0 is equal to 1 by 2pi i integral from 0 to 2pi so you know the points in the z on this disk can be parameterized as zeta equal to z0 plus rho-epsilon into e power i theta this is how I can parameterize that circle okay so where theta varies from 0 to 2pi so if I transform this integral to a real integral to an integral based on a real parameter so what I will get is I will get mod I will get I will get g of so I plug in that d zeta is going to be rho-epsilon i e power i theta d theta divided by this is going to be zeta-z0 is rho-epsilon the whole square e power 2i theta okay this is what I am going to get right and now what I am going to do is I am going to take modulus and note that the modulus of the integral is less than or equal to the integral of the modulus okay I am going to use that inequality which is always used whenever you are estimating integrals it is otherwise known as the ml formula so or it leads the ml formula so mod g dash of z0 is going to be mod of this thing on the right but that will be less than or equal to you know if I take mod you must outside I will get 1 by 2pi alright and then the modulus of the integral is less than or equal to integral of the modulus so I will get 1 by 2pi integral 0 to 2pi and I will get l so let me put m into rho-epsilon this mod of i e power i theta is going to be 1 and mod d theta is just d theta because theta is increasing along this interval and here I am going to get rho-epsilon square and this is again going to be 1 and so this is what I will get where what is this m, m is bound for the modulus of g on this boundary circle okay where mod g is less than or equal to m on mod z-z0 equal to rho-epsilon okay so here this is just an inequality and you know if I calculate this if I integrate 0 to 2pi d theta I will get 2pi this 2pi will cancel this 2pi I will simply get m by rho-epsilon this is the boundary okay this is for an analytic function on this which is analytic on this closed disc now what I am going to do is I am going to apply this to all the functions in my family script F to all the functions in this family see if you take all the functions in this family alright there of course analytic therefore they are analytic on such discs alright and the point is they are all uniformly bounded so I can find a single m which will work for all the functions okay and therefore I will get this uniform bound for all the derivatives okay and that is good enough to tell me that the family is equicontinuous okay so let me make this statement for any since f is uniformly bounded on mod z-z0 less than or equal to rho you see you have the uniform boundedness on compact subsets of d okay therefore I am applying this uniform boundedness I have this uniform boundedness on this disc closed disc and this closed disc is compact because it is closed and bounded alright so this is a compact subset of d therefore this family is uniformly bounded on this okay so which is compact there exists an m such that mod f is a system of m on mod z-z0 you have this so if you so you have mod f dash of z0 is a system or equal to m by rho-epsilon as explained above so this inequality that this estimate I have got for the model of the derivative I apply it to f alright so I get this right and now what I want to say so what this tells you is that this is for what this is for every f in the family. So what you have got is you have got that m by rho-epsilon is a uniform bound for all the derivatives okay so what you have got is that all the derivatives are uniformly bounded at that point now you see now it is a fact that if derivatives you know if you have family of functions whose derivatives are uniformly bounded then that family is equicontinuous okay so I will state this as a lemma lemma 1 if so let me write g is a family of analytic functions functions on u which is domain in c and g and t and g dash which is equal to the set of all a small g dash where g belongs to g the derivatives is uniformly bounded in a neighbourhood of z0 in u then g is equicontinuous at z0 so I am setting this fact that the uniform boundedness of the derivative imply uniform boundedness of the derivatives in a family implies equicontinuity of the family uniform boundedness of the derivatives at a neighbourhood of a point in a family implies equicontinuity at that point okay and the proof of this lemma is pretty easy it is just given by simple I mean if you want to prove it for real functions it will follow I mean if you are working with real valued functions on a close bounded interval or on an interval okay then the proof will come by applying the mean value theorem okay but if you are working with complex valued functions you will have to use integration so what you do is see you know the situation is that so you know I have this u and I have this point z0 and you know I have this disc centred at z0 radius some r k mod z-z0 less than r r is greater than 0 okay this is inside u I can find such an r because after all z0 is an interior point of u and u is a open set and what I am given is that all the derivatives is uniformly bounded in the neighbourhood therefore there exists an m such that mod less of g dash is started at m in mod z-z0 less than r so this is given to me all the derivatives are bounded alright and now how do you show that the family is equicontinuous at z0 so what you do is you know you calculate g of z-g of z0 modulus okay so of course this is for all g in g okay so small g is in script g so small g dash is in script g dash and it is given that script g dash is bounded uniformly bounded in a neighbourhood of the point okay and now what is mod gz-g z0 you see what you can do see this is integral it is integral along the straight line path from z0 to z of g dash of z dz e dash zeta d zeta of course you know if you integrate g dash you will get g because after all derivative of g is g dash you can integrate g dash to get g mind you g is analytic g dash is also analytic okay and therefore this integral is actually independent of the path chosen in a simply connected neighbourhood of z0 and of course we are always considering this disc surrounding z0 which is simply connected okay. So for example you can take z to be any point here and you can take the you can simply take the straight line segment from z0 to z and you can integrate alright but then again you use the fact that modulus of the integral is less than or equal to the integral from z0 to z mod g dash of z mod d zeta okay and but what is mod g dash is you know uniformly bounded by this m so I will get this is equal to this is less than or equal to m times and integral from z0 to z mod d zeta will give you the length of the arc from z0 to z which I am considering and that is I am considering that to be a line segment so I will simply get mod z-z0 this is this is the value of integral from z0 to z mod d zeta okay normally when you integrate mod d zeta along the path you will get the arc length but now I am integrating along the straight line path so I will simply get the straight length of that straight straight line segment which is the mod z-z0 but you see so this is true for all g in script g okay so what this tells you is so given epsilon greater than 0 there exists delta which is equal to epsilon by m such that mod z-z0 less than delta which is epsilon by m implies mod g z-gz0 is less than epsilon okay for all g in g this is what you get given an epsilon I am able to find a delta okay the delta only depends on this of course this delta I fix the z0 delta depends only on epsilon okay and this z0 but it is independent of the small g in script g because this m is uniform for all g dash okay but what does this tell you this actually means that the family script g is equicontinuous at the point z0 that is the definition of equicontinuity the definition of equicontinuity is that this epsilon delta definition for continuity should hold at a point for an epsilon you should get a delta which works for all the which works simultaneously for all the functions in the family. So I have got a delta which depends only on epsilon this delta does not depend on g okay that means that the family is equicontinuous at z0 okay so this implies g is equicontinuous at z0 so that gives the lemma so what the lemma tells you is that whenever you have uniform boundedness in a neighbourhood of a point okay then you will have equicontinuity at that point okay. Now if you now look at what we have written here the uniform you have uniform boundedness of the derivative derivatives at that point okay and therefore it will work also in a small neighbourhood of this point alright and therefore you will have equi by this lemma you will have equicontinuity at every point in the domain. So you have this bound at z0 okay and it is uniform for all f alright so what will happen is it the same you can get a bound for all points in a small neighbourhood of z0 okay and therefore the derivatives are all uniformly bounded in a small neighbourhood of z0 and if you apply the lemma you will get that the family is equicontinuous at z0 so this is a statement that one is always true okay that the family will always be equicontinuous okay. So for analytic functions the derivatives will be bounded just because of Cauchy's formula okay and since derivatives are bounded equicontinuity will come automatically because that is what the lemma says that whenever derivatives are uniformly bounded you get equicontinuity alright. So by the lemma f is equicontinuous at z0 at each z0 in the in your domain okay so for the point of the story is that you know equicontinuity is automatic just because the derivatives are all automatically bounded because of Cauchy's integral formula right. So everything comes from just uniform boundedness uniform boundedness of your family of analytic functions on compact subsets will automatically give you uniform boundedness on compact subsets of derivatives you will get uniform boundedness of derivatives in compact neighbourhoods of each point and that will give you equicontinuity at that point and in this way you can cover all the points so you get equicontinuity everywhere right. So this part one this is always true okay and so I will have to now do this I will have to show that so this is always true so therefore this is always true so the only thing I will have to do is I will have to show that give me a sequence here I will have to show that I can cook up a subsequence which converges uniformly on compact subsets okay. So what I will do is okay so let me retain this so start with a sequence enough okay I already know that the whole family script f is equicontinuous at each point of the domain alright start with a sequence now what you do is so we do a clever construction okay so the construction is you know see I want to basically apply Arzela-Ascoli theorem but you know Arzela-Ascoli theorem the usual Arzela-Ascoli theorem that I want to apply will only work on compact subsets therefore you know what I have to do is I have to find I have to chop down this domain into a union an increasing union of compact subsets and apply repeatedly Arzela-Ascoli theorem on each member of the union and then apply a diagonalization argument okay. So what you do is you do the following thing so this is the trick this is the trick of chopping up a non-compact set okay into a union of compact subsets which cover it so what you do is for every n greater than or equal to 1 let En be so you know you look at mod z less than or equal to n this is the disk this is the close this centered at the origin radius n okay. So you know if so if I so let me draw a diagram so that you can think a little bit just for motivation so you know you have suppose my domain is like this okay of course the way I have drawn it the domain is already the closure of the domain is already compact okay but let me do the following thing let me just remove this so that you know you can think of the domain as being probably unbounded this is part of the boundary of the domain alright. So this is my domain D and this is the boundary of D okay now what you do is you look at mod z less than or equal to n alright so that is going to be so you know if I take n I am going to get a disk like this okay and if I take n plus 1 I will get a bigger disk so this is n and this is n plus 1 okay as n increases these close disks they cover the whole complex plane okay the union of all these disks as n goes from 1 to infinity is the whole complex plane alright and now what I will do is you see I intersect it with the set of all points in the boundary the set of all points in the domain D such that the distance of that point to the boundary is greater than or equal to 1 by n look at this rather funny condition so the condition is I am looking at all the points in the domain okay which lie inside this disk and whose distance from the boundary by distance of course I mean perpendicular shortest distance okay the shortest distance from the boundary is at least 1 by n okay that means I am avoiding points whose I am avoiding points in the domain whose distance from the boundary is less than 1 by n okay. So if you if you think of it like this if you think of it like that then what will happen is you know so here is this is the portion of the boundary this is the portion of the boundary alright and what I am doing is I am avoiding all the points whose distance is so if I take this smaller disk okay if I intersect the smaller disk with which is mod z less than or equal to n with the domain alright what I will get is I will get this okay this is what I will get alright this is the intersection of the smaller disk mod z less than or equal to n with the domain alright and of course this is the portion of the boundary this is the portion of the boundary that intersects the disk mod z less than or equal to n right. Now what you do in this boundary you throw out all those points in the domain okay you take do not take all the do not take all of the shaded region but throw out all the points in the domain whose distance is less than 1 by n okay. So it means that you know I am throwing out all points here now I am just I am just avoiding all points close enough whose distance is so you know this is what I am throwing out okay I am throwing out this because all points here the distance with the boundary is less than 1 by n I am throwing that out I am just throwing out a piece of the domain which is close to the boundary okay and it is this shaded set which is e1 I mean this is en this shaded set this is en okay. If you take en plus 1 what will happen is that I will get this whole intersection minus I throw out all points whose distance from the boundary is less than 1 by n plus 1 which is smaller distance than this okay so you see as n becomes larger see you can see something that is happening as n becomes larger I am covering more and more of the domain because after all as n goes to infinity these discs will cover the whole complex plane therefore as n becomes larger I am covering more and more of the domain and I am what I am throwing out is lesser and lesser I am throwing out points very very close to the boundary of the domain okay therefore in this way I will cover the whole domain so what you must understand is that and of course you know this is a compact set okay this is a compact set and this is a close set alright this is a compact set and this is a close set and therefore the intersection is continues to be compact so the moral of the story is that en is compact for every n and union n equal to 1 to infinity en is your domain your domain has been chopped up into compact sets so this is the I mean this is the clever trick that one needs to make use of to be able to apply Arzela Ascoli theorem okay so en is compact for everyone for every n union of all the en's is d and of course you know en you can see that you know en plus 1 will contain en plus 1 will contain en alright en plus 1 will contain en so it is increasing it is an increasing sequence alright and another beautiful thing is you take any compact subset of d any compact subset of d will be contained in a sufficiently large en okay any compact subset of any compact subset subset of d is contained in a sufficiently large in en for en sufficiently large so you see these are the these are the properties of this of the en's the en's are all compact their union is d they are increasing and any compact subset of d is contained in en for en sufficiently large okay and this is essentially you can think of this as chopping the domain out into I think instead of even saying chopping I should say you know your it is more of a filling out the domain in terms of an increasing sequence of compact subsets the en's fill out the domain the union is the whole domain okay now since each en is compact okay I can apply Arzela Ascoli theorem because you know each en is compact and each en is a compact subset of d but on compact subset sub d I have uniform boundedness because it is normally uniformly bounded okay and of course Iq continuity is already there okay so I can apply Arzela Ascoli theorem on each compact subset okay so what I do is I do the following thing I again I cleverly use again a diagonalization argument and if you recall already in proving the Arzela Ascoli theorem we use a diagonalization argument okay so we again use a diagonalization argument apply Arzela Ascoli theorem to e1 to the sequence on e1 okay e1 is a compact subset of d and I have sequence of functions on e1 that sequence is uniformly bounded why because the sequence is part of the family script f which is uniformly bounded on compact subsets therefore it is also uniformly bounded on e1 alright now Arzela Ascoli theorem will tell you that there is a subsequence which will converge uniformly on e1 okay so to get a subsequence fn1 of fn that converges uniformly on e1 okay now what I do is I take the subsequence and apply Arzela Ascoli theorem to it on e2 okay so I go to the next bigger set okay now apply Arzela Ascoli to this fn1 on e2 okay to get fn2 a further subsequence which converges on e2 uniformly okay now what I will do is I will take this fn2 and apply Arzela Ascoli theorem to it on e3 and I will proceed like this okay by induction we get a subsequence fnk of fnk-1 which converges uniformly on e for all k greater than okay alright and now comes the big deal now what you do is you from this seek from all this you take the diagonal subsequence that will give you a subsequence of the original sequence that will converge uniformly on compact subsets of t and that is the end of the proof okay so again you know I will again draw that the same kind of diagram that we drew for the diagonalization argument in the Arzela Ascoli theorem the proof of Arzela Ascoli theorem see so you know you have the situation like this you have you have you know you have fn so you have fn1 and then I have the subsequence fn2 then I have the subsequence fn3 and so on and so this is you know this is some fi1 fi2 fi3 and this is this is the sequence it converges uniformly on e1 okay then here I have fj1 fj2 fj3 that converges uniformly on e2 then I have fk1 fk2 fk3 that converges uniformly on e3 okay and it goes on like this and now what I am going to do is I am going to take this diagonal subsequence okay you define fm where fm is equal to mth member of fnm okay then fm is contained in the intersection of all these subsequences which is in of course everything is a subsequence of fn converges uniformly on compact subsets of t and that finishes the proof and why does it converge uniformly on compact subset of d because you take any compact subset of d any compact subset of d is in some en and okay and so you take any compact subset of d it will be in some ek but on ek capital fk capital fk plus 1 capital fk plus 2 etc they all converge uniformly on ek okay so this sequence of functions eventually converges uniformly on every compact subset of d therefore it converges after all convergence itself is an eventual thing it has to happen only beyond a certain finite stage alright therefore you get the last statement that this sequence converges uniformly on compact subset of d okay and that gives the proof of 1 implies 2 okay where 1 is already true we have seen so we have proved 2 is true okay and that gives the proof of Montel's theorem okay so you see it is a so the clever thing is to chop the domain up into these increasing sequence of compact subsets and then to repeatedly apply Arzela's theorem and again get a apply a diagonalization argument okay so I will stop here and we will continue with the proof of Riemann mapping theorem in the next lecture.


I want to start from 30. I'll give you one minute. I'll be prepared for it. You have to start. OK, I started, go ahead. Recording in progress. OK, we have David Hedler, who will tell us about prime-size independent anglers for global-driven QAOA. Please start. OK, thanks to the organizers for accepting my talk. I don't think they're yet in the room. So today, I'm going to be presenting some work of mine which is about finding optimal angles for QAOA in a way which is independent of the problem sizes which that we're solving. And to do this, I'm using a somewhat unconventional driver known as the Grover driver. To summarize this presentation today, there's a version of QAOA called Grover QAOA using a Grover driver rather than a regular driver. And when you use this driver, you take into account a lot less structure than you would if you used a regular driver. And as a result of this, we can calculate the performance of this version of QAOA independently of the size of a problem that we are solving. And therefore, we can optimize this performance to obtain optimized angles which are for the average case of certain problems or in the infinite problem size limit. And this would facilitate a few-shot implementation of QAOA where you've already found your optimal angles and therefore, we can just sample from a quantum computer the good solutions to the problems we want to solve. So a brief recap of QAOA. We are preparing a state on a quantum computer which consists of P layers of operators. And each of these layers consists of a problem and driver Hamiltonian, as we see in adiabatic quantum computing. Apart from these, these layers are independently applied. They're never applied simultaneously. And so these P layers are controlled by 2P parameters. And the idea here is that if we can find, if we have a strategy for finding good parameters, then sampling from this, the state produced by this circuit, will hopefully produce good solutions to combinatorial optimization problems. And of course, AUC provides us with a set of angles such that if we have an infinitely deep QAOA circuit, we should always find the ground states of the problem we're solving. Conventionally, we would use single-cuber X drivers, and as in adiabatic quantum computing. And however, in this talk, I'm talking about the Grover driver in which you're kind of driving transitions between all possible pairs of states. And so if you are using the single-cuber X driver, you are biasing structure in your problem with respect to flipping bits. And of course, this means the arrangement of your objective function or the energy landscape of your spin glass or combinatorial optimization problem matters. And so you've got this hamming distance metric for how different states of your system are. A Grover driver, however, it's not biasing any structure in your problem. There's no kind of distance metric for how far away two states are in this space of possible states. And therefore, only the overall distribution of the values of your cost function affect the expectation value of your problem Hamiltonian, which is what we want to maximize in QAOA. Of course, to caveat this, this Grover driver is harder to compile on advice than single-cuber X drivers, though in the QAOA context, we're in a gate-based context. So it's not quite as fatal as in adiabatic quantum computing. However, it's still not quite a NISC-friendly thing to be doing. So because this Grover driver only sees the overall distribution of the values of our cost function, we can model our problems as random variables. And so if you have a cost function C of z, we can model this as sampling from a continuous variable. So if I picked a random string, this could be modeled by sampling from a random variable. And therefore, this cost function would be described by a probability density function. And if we find some continuous probability density function, which describes the likelihood of a random string of our optimization problem taking a certain cost function value, if we have a continuous PDF, then this is equivalent to taking a limit of infinite-sized problems, assuming we get some scaling right. So this overall distribution stays static with increasing problem sizes. And so if we know a PDF for describing the distribution of our cost function, we can derive an analytical formula for the expected value of our problem Hamiltonian under the Grover QA way state. And so to illustrate this, in the plot on the left-hand side, we see this is a 15 qubit SK model. And the dots we see here are individual configurations of that model and the associated energy. And if you imagine sourcing this from low to high, then you obtain the figure in the middle, which is a quantile distribution function, which you could then invert to get a cumulative distribution function. And from a cumulative distribution function, you could differentiate this and get a PDF describing the likelihood of sampling a certain solution quality. And so for p equals 1, so layer 1 QA away, with pen and paper, you can derive the solution quality for the mean solution quality you would sample from a problem given a PDF. And we obtain this formula here. It depends on C bar, which is the mean of the problem, which is very simple to calculate, B, which is just a shorthand for this e to the i, b to minus 1, and finally gamma, which is the characteristic function of a PDF describing our problem. And a characteristic function is simply that a Fourier transform of the PDF describing your problem. And so if we know the PDF, this characteristic function is fairly easy to obtain. For p equals 2, we can obtain a similar expression, but we need to use computer algebra because it's fairly large. But the dependencies are the same. It depends on the parameters of the QA away circuit and this characteristic function. And so it's easy to evaluate this any point in our parameter space. For arbitrary depth, we can derive an expression and there are 2 to the power of 2p terms in this expression where p is the depth of your QA away circuit, the number of layers you're applying, and therefore the number of parameters, 2 to the power of the number of parameters that you have. And this formula, importantly, there is no dependence on the size of the problem we're solving. We just need to know the characteristic function. And there isn't enough time to explain all the notation here, but we have 2 to the p terms to evaluate here and you can write a Python script to evaluate this function. And it simply depends on the characteristic function of our problem. So what does this look like? So we are looking at number partitioning because this is a problem where we can analytically derive the overall PDF describing the problem. And so in number partitioning, you start with some values and the objective of this problem is to split these values into two sets, such that two sets are equally, the sum of the elements in each of these sets is equally large. And so we start with some, say, uniformly distributed values. And from this, you can obtain a cost function of our problem. From this cost function, we can derive that this has a chi-square distributed PDF. Before you transform this PDF, you would obtain a characteristic function with this form, which can simply be substituted into one of the formulae on the previous slides. And we get a fairly complicated expression for the mean solution quality of the states that we sample from our QAOA state. And we can numerically minimize this to obtain some optimal parameters that are independent of the problem size. So on a surface plot, so here we see on the left-hand side, this is a finite size instance of number partitioning. And the expected value of our QAOA state for different points in the parameter space. And so we can see that as the number of qubits increases from left to right, we have this convergence to a terminal landscape, which is the landscape of derived analytically from this PDF of the problem. And the optimal parameters I derived on the previous slide, this corresponds to this, the darkest region here on the right-hand plot. And so for number partitioning, we can minimize the large formula on the bottom, which is the expected value of our QAOA states, the solutions we might sample from the QAOA state. And we obtain these curves of the parameters at different depth. So the dot on the far left of this plot is the depth 1 optimal parameters for this number partitioning problem at infinite size. And the lines represent the higher depth and therefore greater parameter count versions. And likewise, as we increase the depth of our QAOA circuit, we are increasing our solution quality. So in this context, a lower expectation of our problem Hamiltonian is better, because we're looking at the residue when we subtract the sums of the values in two partitions. So the main caveats to this work we probably don't know the PDFs at infinite size for most interesting problems. But I imagine this is a solvable issue and could at least approximate them for most problems. We need sufficiently large problem sizes for this approach to work well. But from some of the plots we've seen, and I have an extra slide which also shows this, 12 qubits is quite a large number of qubits in this context, or 12 bits in our problem. It's hard to compile a Grover driver, but this isn't totally fatal, but probably not a very NISC friendly thing to be doing. The main problem I think with this work is that if you went through the effort to compile a Grover driver on a device, you would probably sooner compile your objective function to a threshold function, which evaluates to one if you satisfy a condition or zero otherwise. And then you're essentially running Grover's algorithm. And this would probably perform better than QA away in all these problems. And so if you use an approach like this where you use statistical properties of a problem to find optimal parameters, then you don't need to use your quantum computer very much. You only need to sample from your final states. And the main value I see in this is I think it's probably possible to extend this method to more complicated drivers using more complicated statistical information of your problem. And this work is ongoing. And when I initially started this work, my objective really was to determine if Grover drivers were performing better or worse than single-cube X drivers as in the work of Eddie Farhi, who published a paper where they look at the performance of QA way on infinite-sized SK models. And I haven't considered SK models with this work yet, but this is something that I can use this method to compare the performance of these Grover drivers, which is the no-structure to usual drivers, which hopefully should exploit structure on infinite-sized problems. And I have a quick extra slide. So this slide I'm just showing that the optimal parameters as we increase finite-instance sizes for p equals one and p equals two, we are getting closer and closer to the terminal values, the infinite-sized values as we increase our problem size. Great, and I think that's my time. So does anyone have any questions? I have a question from the internet. So, yeah, I guess it is, if you have... What would be the purpose of Grover-driven QAOA which gives you an approximate solution if p is one or if it's a small number of iterations? Why would you use that instead of just implementing Grover's algorithm, which gives you an exact solution? Well, I think the answer is I'm interested in this from a theoretical perspective. I don't think it would be a useful algorithm to run on a quantum computer, but I hope this leads on to more interesting work, which could be more useful because we can find optimal angles for more relevant drivers. Okay, can I ask my second question then? You're probably aware of the paper by Betel and Kleisch that says that finding optimal beta and gamma is an NP-hard problem, and so it makes me wonder what's the catch, is it because they're not using Grover drivers? Is it because you need, eventually, you'll need an exponential number of samples to properly estimate the mean of this distribution? And I guess I'm wondering about the relationship of estimating the quality of estimate of the mean of this distribution and how that converges to good quality solutions. Is that an approximate convergence itself? Okay, so, I mean, I would expect, so in these plots here, I am optimizing some formula which has an exponentially sized parameter space in your number of parameters, so I would expect that as you increase the depth here, it's a hard problem to optimize these parameters. I wonder if you're talking about it being an NP-hard problem if you're actually dealing with samples from the quantum computer, or? I don't remember, but it's NP-hard in both N and P in this paper, but this, I guess, is probably assuming using the conventional method for finding the conventional drivers for beta and gamma, and I wondered if that was the difference because this doesn't look to be an exponential time. It's certainly not exponential in the number of qubits. It's probably at least, yeah, it's exponential in the number of, in the depth. Yeah. So there's a mismatch there, and it would be interesting unless you've got a polynomial time algorithm for an NP-hard problem, in which case congratulations. Let me be the first to congratulate you. All right, thank you. How sensitive is this to your parameters? It doesn't look like the solution quality is that sensitive to how you've set your beta and gammas. I think when you have the kind of energy landscape, when you use a Grover driver, it does tend to be very smooth. So I think the optimization task for a Grover driver is significantly easier than it would be for a more complicated driver. So if we don't have enough time, we have to move to the next talk. We can thank the speaker.


Hello, I'm Rachel Zenberry, Chair of the Arlington Redevelopment Board, also known as the ARB, and I will be taking you through Warrant Article 20 for the 2020 Special Town Meeting. Article 20 is a zoning bylaw amendment submitted by the ARB to address alternatives to existing parking requirements in the B3 and B5 districts, which cover building blocks in East Arlington, Arlington Center and Arlington Heights, Arlington's primary business districts. The intent of the article is to encourage business development by removing unnecessary barriers that exist within the zoning bylaws. The zoning bylaw currently allows the ARB and the zoning Board of Appeals, also known as the ZBA, to evaluate and make a determination through their respective special permit processes to allow for parking reductions in business, industrial and multifamily zoning districts. This amendment would provide the ARB or ZBA with a new option to reduce parking requirements to as low as zero, only in the B3 and B5 districts where the businesses do not have the ability to create new parking. The encouragement of business development in our main business districts aligns with the Towns Master Plan, which was adopted in 2015, the Arlington Heights Neighborhood Action Plan, and policies being discussed in the development of Connect Arlington, the Towns Sustainable Transportation Plan. The main points of this article, as indicated on these slides, include to address the challenge that many businesses face in creating new parking on existing lots when parking is either unnecessary, infeasible, or available through other options by providing special permit granting authorities with the ability to reduce the requirement to provide parking to as low as zero. Multimodal transportation options are available in these districts. To allow for the ability to reduce parking requirements for business districts as low as zero in these two districts, this would be provided as an additional option available to the ARB and ZBA where specifically appropriate, not as a default. And to proactively address an issue that can hinder business growth and expansion, the amendment is limited to business uses. As background, the B3 and B5 zoning districts make up three major business districts, Capital Square in East Arlington, Arlington Heights, and Arlington Center. There are 109 properties located in these zoning districts. The maps shown here identify the location of the parcels in the B3 and B5 zoning districts. There are no properties owned B3 or B5 outside of these areas. By definition in the zoning bylaws, the B3 and B5 districts activities are oriented towards pedestrian traffic, and in the case of B5 centralized parking. It's important to note the relatively small scale of these parcels with coverage of the lot at or near 100% in many cases. Additionally, most of these properties are listed as architecturally significant structures on the local historic property inventory, and in Arlington Center are also part of an historic district. In these two zoning districts, it is often impractical, infeasible, and unnecessary to create parking in a business. In the current zoning bylaws section 6.1.5 allows parking reductions with consideration of shared parking, offsite parking, and transportation demand management or TDN measures. These can help address employee parking as well as offsite customer and visitor parking. A parking analysis or impact assessment can also be part of the board's review process. However, currently the zoning bylaws only allow the special permit granting authority to reduce to 25% of what is required in the required parking tables. Existing parking requirements in Arlington zoning bylaw can create a situation where the space required for offsite parking for a development or change of use can be almost as large as the use itself. The town bylaw places a priority on preservation of buildings. Lastly, the current supply of on-street and centralized parking is often sufficient to meet or even exceed demand. This bylaw does not change on-street parking or ADA accessibility requirements. In summary, this amendment supports a number of the town's strategic plans aligned with the definition of pedestrian oriented activities in the B3 and B5 zoning district and eliminates a significant barrier to the development of new businesses in Arlington's three main business districts. We hope that you will vote favorable action as recommended by the redevelopment board. Thank you.


No, that was that. You ever seen a fish eatin' one like that? 6 in the morning. Here we go. Alright guys. Welcome to another vlog. We got like an hour to sleep here guys. Watched the last vlog we got here like at 5 in the morning. 6 right now. A little, a little over six. Gotta go to the boat. Load all the rods, hit the ramp, hit the water all day. Today is gonna be a long... Yeah. About to throw this bad girl in the water. Alright guys, in the water right now. We're about to get this day one of practice. I'm proud of her. Only day of practice started right now. I'm absolutely dead. When you get 4 hours of sleep in 70... Out of 72 hours you get 4 hours of sleep.  Especially if that drive last night, that was about 12 hours.  Oakley's about to hop on the boat. We're about to go grind today. Figure something out. Figure out a few patterns. First spot. Let's get this started. Let's go to Cheddar Bay. I'm gonna put this away now. I'm telling you man. I got a catch on this juice while I'm here. Jesus Christ. That didn't work. I just slide it over the top of the trees. Okay. Second cast. Already stuck. You just see a 10 pounder come on. Oh, huh. That was nice. What? I hooked some braid. I'll hold the water braid. That tells me everything about why I got stuck. Wow. That's a way to start right there, right? That is a way to start right there. Second cast. Hooked a big water braid. Where did you see Shad popping right here on the bank? Shad popping everywhere right here. Off the bank. On the bank. Dude, I'm telling you. I can't throw another color, bro. I believe in it so much that I just can't. I'm sure I got a few in my box, yeah. He's a little baby squeaker. I'll take that for the fifth cast. Little something. That's like the first time I haven't seen one choke that. Look at that. And I'm not going to lie, this one is the smallest fish I've ever caught on this thing. Really? Probably the smallest fish I've ever caught on this Chudderbait. First little guy. On a lay down on it with a chatter bait. Alright, I'm going to keep fishing because you're too small to fit in my line. To fit? He'll fit. He'll fit now.  I'm not going to lie, that 10 pounder caught, he didn't fit. He was bent. I'm going to be real with you, man. I'm going to be 100% real with you.  It's a little lucky tackle box bait. Look at that. They might for that bait. That's literally identical, dude. I'm going to try it real quick. Oh yeah, you should get one there. Dude, he hit it so hard. Oh my God. He hit it so hard. I hooked him in the booty. Dude, I was like, I freaked out. Think about like a bass hit times 10. I was going to catch a fish, right?  That's my question. He choked it. Alright. Oh yeah, look at that. There you go. We're getting somewhere. I think what's his son gets up. There he is. Yeah. He's a good one. Yeah.  He's a big one, bro.  I'm not even kidding, bro. Oh, it's the freaking turtle. I thought that felt so weird, bro. It was hard for me to catch up to him. How do I catch a turtle on a Carolina rig? I got him in the arm.  I got you. Alright, buddy. Come on. That's jerk bait fish right there. There's a fish on there. Yeah, there is. It's a big one, too, bro. Oh my God, bro. Looks like he's about four. Maybe more than three. He's about three, three and a quarter. That's a little, that's a chunker, bro. That is a healthy fish. See that? Really thick and heavy fish. Let's get a release on him. There she goes. That's crappy.  That's a good one, too. How many species? Oh my God. You want some dinner tonight? Good night, dude. See, y'all old crappy. Slipping him on a sandwich. Slipping him on a sandwich. Oh, there's a school. Don't throw in there. Don't throw in there. Do not throw in there. I know, I know. A smallmouth. The school's a smallmouth.  Yeah, but he's right there. He's about, he's about four or five pounds. Smallmouth. Smallmouthy. Smallmouthy. God, we're just catching every species today. Yeah. Call it. Cropy. A bluegill. A smallmouth. A largemouth. A turtle. A turtle. Jesus. Turner's to the left. It's getting a little... Oh my God. What species is this, man? That's a bass. A little baby. A little baby. A little baby. That's a bass in the middle. That's him. That's a good one, too. Let me look at him. Oh, there's two. Dude, look. That's a good one, bro. That's a good one. That's about four pounders. See him? See his tail? Yeah. He swam off. A baby, he is.  I ain't going to take my foot off it. I'm going to work our way to where we can hit the mouth. If we catch one in the mouth right there, we're not going to hit any more mouths like that. Jesus Christ. Nah, I got him. Pretty fish, guys.  On this mud line. Why does it have a freaking brim? Now I caught a perch. I ain't going to lie. Yeah. Have you ever seen a fish eat one like that? Literally, the bill is down its throat. I put her up. Look at that shad. Look at that shad. Look at the bait jump. That's kind of how you need to find it. That bank looked too good. She's a very heavy fish. You could be a little bit over that, but let's get a release on her. See you, baby. Woo. How you going, pitcher? Yeah. That's a three. Hope you guys enjoyed this video. If you did, be sure to drop a big like, comment below, and subscribe, and hit the little bell. There is a ton of fish catches. I did not put it in this video, guys. We caught a ton of fish this day. Also, a lot of fish that we did not get on camera, sadly. But I did pack it full with a lot of fish catch action, but I hope you guys enjoyed this video. Be sure to drop a comment below on what you guys want to see next. Also, I want to be posting some tip videos this week. So comment below what tip videos you guys would like to see. Thank you guys so much for the support, and I'll catch you all in the next episode on Thursday.


let's add some basic to the light to a very simple game that we have this game was created only using visual script and you can move a character around the screen but there's nothing else happening actually the trees are not even there you can go over them because it's only a background but it's gonna be enough for us to demonstrate this simple example to start adding lights and shadows I will first add a light to the kinematic body that we have here this is our character let's add a shadow and it's going to be a light 2d now that we have the light 2d we need to add a texture to it without a texture the light is not gonna be happening to make a texture for a light you can use any image where the white is going to be the light and transparent is going to be the darkness so in this example you have like this kind of effect I will provide a link to download this sprite that I'm using right now in the description but I already imported it into the project and I have it here so let's open the light 2d node and let's drag and drop the texture you can see the size of it here let's add it here on the light and as soon as I do that you see that the image already gets brighter around where it is so we have that working right now the next thing you want to do is enabling shadows so whenever you open this section you can click here on enable and that's pretty much it you can now add a new node which is going to be a light occluder 2d this is everything that will block the light and create a shadow so let's build one here so you can see there are new options here at the top of the editor this means that we need to create a vector we have a warning here which is like you need a polygon first you can draw it which is going to be the occluder let's do one for this tree let's say I'm gonna start clicking it will ask me no polygon resource do you want to create one yes let's create one so I start clicking here now here and here okay and I click it again to finish it so now I have this polygon that I can modify anytime and as you can see on the editor I already see like a shadow so this is going to be the part that will create a shadow from the tree so if you see here I can move the character in the editor and I already see the shadow being drawn on this scene I can move it away and you see that the spread is going further and further and the the scene doesn't have any shadow or anything but now that I'm moving it closer you can see it again and I can move it so I'm going to apply the same to all the trees and I can do so by just clicking here and duplicating with ctrl d and I can set the shadow to the rest of the trees like so okay and now let's create an eye one movie for this entire house part let's create a new one like the cd clicking create one and now I will do something like this let's see just a block okay so that's gonna be for the house okay so now if I run the game you can see that all the trees emit a shadow and if I move the character around it's going to be casting the shadow on the obtutors also for the house like if I'm behind the house like I cannot see anything in front of it like any light and if I move in front of it you will so that's a very simple way of adding 2d lights and shadows you can play around with it there's a lot of extra options that you can touch here on the light you can change the color of it maybe you want some red you know and or maybe like a green doesn't matter you can also play with the occluders and maybe the last things that you want to know is that this light here you have the item cool mask this is a layer where the light is going to be displayed so if you want some lights to be blocked by some elements and some other is not here if you see in the light occluder we have the light mask this is the same layer as the light so if I remove it I don't see it because it's not on the same layer this will work if you want to add multiple lights and you want those lights to be casting different shadows in different characters but I will let you play with that if you want that's it for today and I hope you like this small tutorial I would really like to thank my patreons and if you want to help me making more of these videos you can visit my patreon here and yeah see you next time


If we're wanting to win back women, win back our families, win back our country, we've got to stop giving our sex away. We're slaves to it. And it's not just with women, it's also with pornography. Stop watching porn. According to libido domanandi, E. Michael Jones states that when the Israelis wanted to take over Palestine, one of the very first things that they did when they came in was they littered the entire cities with pornographic material. Papers, magazines, pornography. Then the next thing they did is they went in and they took over three of the four television stations and broadcast pornography 24 hours a day. So these people were on lockdown, right, and this was back in the, what, 1980s? And they were forced to stay home and stare at a screen that they would be hopefully getting news from, but instead porn. Think about 2020 lockdowns in Pornhub. There's a reason why, because you make the men very weak. You make them very passive.   They don't need to use bombs and bullets.


It looks like a city, and in every way except name, it is one, a very sizable city. But it is more than a city, much more. Again, except in name, it is a university, really a multi-versity. And it is an infantry training center. America's probably the world's largest. This is Fort Benning, Georgia, the United States Army Infantry Center and home of the infantry school. Fort Benning is vast, over 182,000 acres, 620 square miles of forest, underbrush and open country. The Chattahoochee River gives it an amphibious dimension for water crossing problems. Plus 23 firing ranges and 1,222 problem sites. On any working day, its population hovers at the 60,000 miles, 50,000 military and 10,000 civilians. Another 50,000 people linked to Fort Benning live in its environs, Phoenix City, Alabama and Columbus, Georgia, making the total population serve over 110,000 people. Many are the families of active and retired military personnel. The fort itself is named after a distinguished Confederate general, Henry L. Benning of Columbus. So there is good reason why local ties with Fort Benning are strong. This closeness is actively expressed by a monthly meeting of representatives of Fort Benning, Phoenix City and Columbus. This is the civilian and military council. The commanding general of Fort Benning sits with the civilian counterparts, the mayors of Columbus and Phoenix City, the provost marshal with the chiefs of police, the fort Benning information officer with newspaper publishers, radio and television executives and educational leaders, the post chaplain with local representatives of the church, and there is a cross section of leaders in the business, industrial and civic communities as well. Fort Benning moves to a military beat, but there is a strong family heartbeat as well. It shows in its recreational life and in its social activities. A community feeling makes itself helpful in many ways. Faith is a powerful ingredient in the spirit of Fort Benning regardless of creed or denomination. It is particularly meaningful and sustaining to many families that live in the vicinity of Fort Benning. Families of husbands and fathers who serve in a foreign land. Fort Benning trained infantrymen are skilled and resourceful soldiers. They rightly claim a lion's share of this nation's combat successes. The Benning mission is simple but lofty. To produce the world's finest infantrymen and combat infantry leaders, a mission symbolized by its towering statue of a rifleman and the traditional cry of the infantry soldier, follow me. The heart of the training effort is lodged in the new six-story infantry hall. The Fort Benning complex and its huge educational plant are under the direct control of a major general, who is both commanding general of a post and commandant of the school. He is aided by an assistant commandant for school activities and a chief of staff who coordinates the activities of a large directorate and a special staff. On any given school day, approximately 12,000 officers and men, including students from 57 allied countries, throng corridors and classrooms with another 2,000 engaged in outside studies. They study a wide-ranging spectrum of subjects, 240 grouped in 27 courses, whose effectiveness is constantly being evaluated by the director of instruction and his staff. There is every type of visual aid, including closed circuit television. Two TV studios are kept busy turning out video training tapes. All grades are computerized to record and evaluate the progress of the students and the effectiveness of teaching methods. But more important than any computer or teaching device is the human relationship between students and teachers, most of whom are veterans of Vietnam. They are aware that the students, often of senior commissioned rank, have come to learn and are eager to be challenged. This huge complex teaching program produces over 56,000 highly trained infantry specialists and leaders each year. The largest group of graduates is turned out by the school's Airborne Department, over 25,000 annually. Despite its rigorous training and hazardous mission, there never is a shortage of volunteers. In three crammed weeks, they learn the fundamentals of military parachuting, how to exit a plane, the correct positions for body and feet, and the proper landing technique. A high point, literally, is the jump from the 250-foot tower where the student has the experience of controlling a parachute during an actual descent. He is taught how to follow through on the ground by controlling the parachute despite adverse winds. In the final week, the student must make five jumps to qualify. On the last jump, he carries a full combat load. There is tension in the jump plane, but training, conditioning, and the presence of an experienced jump master calms the students. Whatever hesitation they feel, there is none when they jump. It is a memorable moment for each newborn paratrooper. He has proved himself to be a superb embodiment of the spirit of Fort Benning. A member of a fighting elite emerges from the Ranger Training Command of the Infantry School. It not only graduates some of the world's toughest, most resourceful fighting men, it also develops combat leaders. They are capable of operating for an extended period over any type of terrain in either a conventional or counter-guerrilla environment. The training is deliberately severe and challenging in order to toughen the Ranger's body and mind. A series of confidence tests develop control of normal human fears, such as fear of water, height, and bodily injury. Engineers estimate the trainee hits the water with an impact of 50 miles an hour. Hand-to-hand combat teaches him aggressiveness and confidence in his ability to defend himself with his hands and feet.  He moves to the mountains of Georgia for training and mountaineering skills and conventional warfare. He becomes adept at handling rope, using it to scamper up and down sheer walls with ease, employing it as a lifeline for litter evacuation, building rope bridges so that he can cross difficult terrain with a minimum of time and effort. The emphasis is on patrols, day and night, of exhausting length and intensity. They continue in the swamps of Florida, where training shifts to counter-guerrilla warfare. This part of the course teaches the Ranger to operate in thick underbrush and in water. The hunger, fatigue, sleeplessness, and other hardships he must endure often exceed those he will have in combat. The course concludes with an offensive operation against jungle-trained experienced troops who simulate a guerrilla force. Superbly trained, daring, and resourceful, it is no wonder the Ranger was given the accolade of the Army's ultimate soldier. The Army's need for highly trained small unit leaders has long been answered by Fort Benning's superb officer candidate school. The candidates move through a rigorous 23-week course, striving for the gold bars of a second lieutenant. Each man becomes proficient in the operation and maintenance of all infantry weapons. Most important, he learns how to handle troops in combat. He is confronted with a variety of realistic battlefield situations which tax his skill and command ability to the utmost. The grueling months of training are climaxed by the most strenuous exercise of all, a company assault on an enemy position supported by combined arms. During the exercise, each candidate rotates through different command positions in the squad, platoon, and company. Only when he proves to the satisfaction of watchful officers that he has the quality of leadership can the candidate hope to win a commission. This is only a part of the immense and complex training effort carried on by the infantry school. The large numbers of men, weapons, and vehicles necessary to stage over 320 instructional problems performed approximately 12,000 times annually are provided by the 197th Infantry Brigade. This brigade is the world's largest, over 7,000 men strong. To stage these demonstrations, the 197th draws foot soldiers from three infantry battalions. Tank and artillery from armored and artillery battalions. Combat engineer companies and a chemical platoon from Brigade Headquarters Company. And a scout dog detachment. Every skilled infantryman and infantry leader was once a beginner, and these recruits are raw beginners indeed. Every week, nearly 900 inductees enter Fort Benning's Infantry Training Center to receive eight weeks of basic combat training. The transition from civilian life is abrupt for these youngsters, but they are helped to adjust by their drill sergeant. These experienced sergeants are hand-picked non-coms, all volunteers. There are three per platoon, and they stay with the same platoon all the way through basic training. The fledgling soldier learns how to use and care for his weapon. He learns the ways of the army, and is toughened mentally and physically for rough bodily contact by training exercises like the use of the Pugelsteck. In eight fast-moving weeks, the awkward rookie has taken on considerable polish and self-confidence. He has become impugned with the spirit of the infantry. He is now ready for advanced individual training as an infantryman or schooling as a technician. Fort Benning is the birthplace of air mobility. The large numbers of helicopters and crews required for training in this type of warfare are provided by an aviation group, one of the largest tactical army aviation organizations in the United States. It teaches gunship pilots and door gunners the art of suppressive fire. It also conducts a variety of transitional training programs for helicopter crews and gives aviation mechanics the maintenance skills for direct and general support. Entirely separate from the aviation group is the Lawson Army Aviation Command. It is primarily a service and support operation for the infantry senate. Lawson easily ranks among the top 20 of the most active airports in the United States, civilian as well as military, with over 200,000 takeoffs and landings yearly. Thus, in many ways, Fort Benning operates as a major and very active metropolitan area, self-sustaining to an extraordinary degree. Like any large city, it must have a large number of municipal services to keep it running smoothly. These are provided by personnel of the center troop command, which includes a women's army corps detachment. There are police and fire protection services, transportation facilities, fuel, finance, communications, engineering services of an unlimited variety, and an army field printing plant. This giant plant operates around the clock to meet the instructional needs of Fort Benning and the 360,000 students and instructors in the ROTC, National Guard, Army Reserve, and Army Infantry non-resident programs. One of Fort Benning's most vital installations is Martin Army Hospital, a modern nine-story structure. It offers a complete range of medical, surgical, diagnostic, and pathological services. Martin has a 930-bed capacity and can be expanded to meet future needs. One of the hospital's happiest functions is the output of brand-new Americans, as many as 2,000 babies are born here each year. Hardly less impressive in size and specialization is the Army Dental Clinic. Its dispensaries and laboratories provide a complete range of dental services. Scattered throughout Fort Benning are a host of tenant units, all concerned with improving the combat capabilities of today's and tomorrow's infantrymen. Among them are the infantry board, which tests and evaluates newly developed weapons and equipment. The Infantry Agency, whose concern is the development of future combat material and doctrine. The Human Resources Research Unit, whose personnel study training needs and human factors that determine combat effectiveness. A marksmanship training unit, which promotes individual effectiveness with weapons and prepares Army shooters for Olympic and international competition. Fort Benning is huge and marvelously complex in its varied activities, but its spirit is not to be found in its massive educational plant or in its overwhelming statistics. Its spirit is alive, vibrant, warming the blood of the fighting men who train there. It is in the young recruit who attains manhood in the rigors of combat training. The officer candidate who discovers in himself the magic power of leadership. The ranger who makes the wilderness his home. The airborne men of steel who learn to assault the enemy from the sky. Their spirit is found in the families of these men who share their trials and triumphs and give them tireless devotion. There are very special words for these people at Fort Benning, words expressed by the President of the United States.  You are a community of courage. The spirit of Fort Benning finds its most exalted expression in the epic poem, I am the infantry, which is a stirring climax to graduation exercises held in infantry hall. I was there from the beginning, meeting the enemy face to face. For two centuries I have kept our nation safe, purchasing freedom with my blood. I am the infantry, follow me. From bull run to apomanics, I fought and bled. Both blue and white were my colors then. Two masters I served and united them strong, proved that this nation could right a wrong and long endure. I am the infantry, follow me. At Chateau Thierry, first over the top, it was I who cracked the Hindenburg line. In the arcane I broke the Kaiser spine and did not come back till it was over, over there. I am the infantry, follow me. A generation older at Batan, I took a beating, then bowed to return. I invaded Tunisia on the African shore, dug into the sand at Anzio. The English Channel, stout beach defenses, and the hedgerows could not hold me. I broke out at Saint-Laus, unbent the bulge, bolted the Rhine and stormed into the Nazi heartland. Hitler's dream and the Third Reich were dead. In the Pacific, I hopped from island to island. I did return to hit the beaches and to set the rising sun. I am the infantry, follow me. In Korea, I gathered my strength around Busan, crossed the frozen Han, outflanked the Reds at Incheon, and marched to the Yalo. Follow me. A bayonet and the wings of power keeps the peace worldwide. My allies in the mountains and forests, in the jungles and rice paddies, I teach, I aid, I lead. Follow me. Where brave men fight, there fight I in freedom's cause. From conquered bridge to heartbreak ridge, from the Arctic to the Mekong, the Queen of battle always ready, then, now, and forever. I am the infantry, follow me. Follow me. Follow me. Follow me.


This is the place where Camffit's story begins. The village of Mola in Zimbabwe. I went there to do a study into the reasons why girls were not at school. This was the place where the Tonga people were shifted when the Kareeba Dam was built in 1956. I was expecting to find resistance to the idea of girls' education from parents because everything I had read told me that that was the reality behind girls' excuse from education. But what I found was poverty, deep poverty and a rational decision making which meant that boys who had the best chance of paid work were favoured. So it wasn't the poverty of the culture of the families. It was rather the pernicious culture of poverty and it was all around me. I could see it in the thin frames of young mothers. I could see it in the distended bellies of the children. And so we began with 32 girls. And in the intervening years we have grown into five countries to support 1,260,000 girls. And over that period not one single family has turned down the offer of support for their girls. Not one, not Christians, not Muslims. How have we done it? We'll be put girls at the centre. They are our primary clients. And we recognise that around her there is poverty but there are other forms of capital. We dignify and extend those forms of capital. So parents and families are not reminded of what they do not have but they are reminded what they can bring to the table. The social capital and the family. The care and the love of families. Their encouragement to girls. Their preparedness to make sure that they have time to study at home. The institutional capital of the communities. 5,065 partner schools that we work with. The churches, the mosques, the chieftancies. Chiefs are among the strongest advocates. We work with them to support the education of girls and also to speak out against child marriage. Knowledge capital. I saw a sign on a headmasters wall in Zimbabwe and it said somewhere in the world there is a committee discussing your life and you are not invited. Now in every community we have a committee. A broad based committee that comes to the table to design and implement with us. One of the greatest achievements I think of Camford is the emergence of Cammer. An alumni of extraordinary young educated women who are now driving change. Cammer members are lawyers, doctors, teachers and they are set to grow to 190,000 over the next five years. They understand poverty because they've lived it and they have escaped from it through education. And they're philanthropists. They're bringing new financial capital to the table and they have supported themselves from their own incomes. 63,274 children in 2013 alone change makers from within. Here's one of the Cammer members, Ruka. Ruka decided that what her community needed was high quality protein available in her community of Northern Ghana. And so she established a chicken farm which has grown substantially. Last year she won the Mandela Washington Scholarship and was awarded by President Obama. And here Dr Raniraro Machungaidzi. She is our first doctor, one of the first 32. And here she is working with patients and she is changing her hospital from within. She says to nurses perhaps who are unkind don't speak to that woman that way. She could be my mother, Abigail. Abigail, often very young, brought up by her very poor grandmother. Now she is one of the 15 youth advisors to the UN Secretary General, Ban Ki-moon, on education. And Angeline McWenderry. Angeline whom I've known since she was a small child now married, Angeline Mormirwa. She is here speaking at Brookings. She is our director of Southern and East Africa. And here she was speaking alongside Michelle Obama. Extraordinary young women, all of whom from a background of deep rural poverty. We cannot afford to neglect the potential of girls and young women. Every problem that we face can be solved through the education of girls, child mortality, maternal mortality, food insecurity, child marriage. There is no problem that is easier to solve than through girls' education. So join us because when you educate a girl everything changes. Thank you.