Welcome everyone to the last session of the workshop. This talk will be by Suhail, who is a graduate student at TFR, supervised by Arkadev. Yes, I was told that he did his undergraduate IT Guwahati. And then he reminds people of Arnold Schwarzenegger. I guess you will see some Arnold Schwarzenegger jokes during his talk. I guess he has done some recent groundbreaking work on separating communication and approximate rank. And today he's going to talk about something related. Thanks, Mokran. The Arnold Schwarzenegger thing, you can contact me offline for that. I will tell you later. OK, so my talk is titled Lifting with Exor. It's not yet clear what my talk is about. That will be more clear next slide. But as you said, I'm from TFR Mumbai. And I work under Arkadev Chatapati. I'm doing my PhD. And this talk is based on work that I did with my advisor and colleague Nikhil Mandi, who is now at Georgetown University. So in this talk, we're going to be talking about communication complexity for its own sake. It's been, the applications of it have been shown since the morning. And I think by now it's a somewhat honorable goal to just analyze communication complexity. And I'll be focusing at just getting a better understanding of communication complexity. And in the simple case where we have two parties, just Alice and Bob. And they're computing a total Boolean function. It's not a search problem. It's just a Boolean function. And it's total. And the way Exor functions come into the stock is that they are really structured. And some aspects of Exor functions lift very easily. Yet they are a very rich class of functions. And there's a lot of unknowns in the world of Exor functions. And there's very interesting open problems that I'll get to some of them in this talk. So I'll start off the talk by introducing some conjectures in communication complexity that will be relevant. After talking about these conjectures, I will switch over to Exor functions and what lifting theorems look like when you have Exor functions. And then eventually I'll tie them together. So let's start off with a communication protocol. This is what a deterministic communication protocol looks like. It's a tree. We've seen this before many times today. But let's just do the dance one more time. So when Alice gets an input x and Bob gets an input y, what happens is you look at the protocol. Here it says this is an Alice node. She has some function to compute here, which will tell her what to output. That function just depends on her input x. And so the x and y will follow the tree. It'll follow a path down to the leaf. And you will accept x comma y if it reaches a leaf that is labeled 1. And what kind of inputs reach a one leaf? So let's take this leaf here. It's those inputs on which the x will answer 0 at this node and 1 at this node. And those y's on which Bob will answer 1 on this node and 0 on this node. So it's a product set. Any x that satisfies that and y that satisfies that will reach this leaf. And let's look at the function that is computed by this communication protocol. We can look at its truth table as a matrix. And the rows are indexed by Alice's input, the columns by Bob's inputs. And the entry at the x comma y location is the value of the function f of x comma y. And we can build the truth table using the communication protocol. So if you take that leaf L that we had seen before, there's some product set which reaches it. So all those inputs actually give the output 1 according to this communication protocol. And summing up over all the 2 to the, at most 2 to the c leaves, where c is the communication cost, we see that this matrix actually has rank at most 2 to the c. And it's important that all these rectangles are disjoint, otherwise you couldn't have just added them up to say that the rank is small. And this gives us our first lower bond. So every rectangle is a rank one matrix. So our first lower bond says that if you have a small cost protocol, its rank must be small. So if you have a function whose matrix has large rank, it must have a large cost protocol. And that gives us the lower bound d of f is at least log rank of f. And this is probably the most fundamental conjecture that follows from this is, is this lower bound tight? Is it the case that for all Boolean functions f, this lower bound is polynomial, polynomially tight. And this is back from 1988. It connects a communication complexity measure with an algebraic measure. And similar log rank conjectures have been known to be true and have been used in lifting theorems quite often before. And they have been very useful in such cases. It would be very useful if this were actually true. The thing is we don't really know whether it's true at all. Our best upper bound is deterministic complexities at most square root of rank. So there's a huge gap between log rank and square root of rank. And our best lower bound to say that this is false is that we know that alpha is at least two a result by Gaspitasian Watson quite recently, actually. But what we do know is that if you restrict this rank decomposition to be non-negative. So we're actually summing up rectangles which are all non-negative rank one matrices. And if you look at the non-negative rank of our matrix, then this conjecture would actually be true. In fact, with alpha equal to two. d of f is always less than log square non-negative rank. So this is the most famous conjecture. There is an analogous conjecture for randomized communication. So let's just spend a few minutes looking at that. So in randomized communication, Alice gets a string of random bits, Bob also gets a string of random bits. They're independent. And again, you do the same thing. The probability that x, y is accepted is the probability that x, y reaches one leaf. Because now these functions depend on these random bits. So x, y is going to reach this leaf L with some probability. But we still have a nice product structure. The probability of reaching this leaf L is the probability that x answers red into the probability that y answers blue. So if you look at the communication matrix, well at least this probability of acceptance matrix has small rank. The probability that you reach a specific leaf L is a rank one matrix by this product property. And the probability that your input is accepted has rank at most two to the c. So although you're not exactly computing the truth table of your function anymore, you're computing something that's very close to it. The probability of accepting. If your protocol is actually a correct protocol, that's very close to the actual truth table. And we know this matrix on the right has small rank and this is pretty much the definition of saying that the matrix on the left has small approximate rank. So there is a matrix that is close by to the matrix on the left which has small rank. So therefore this matrix has small approximate rank and this gives us our randomized lower bound of the randomized communication complexity is at least log of the approximate rank. Please feel free to stop me if you want me to go slower if you have any questions or any clarifications. So this gives us again the next conjecture, the log approximate rank conjecture. This conjecture, it's exactly analogous to the deterministic conjecture, except it has an interesting history also. So it appears in the literature in 2007 in a book by Lee and Shraiman, but it actually appears in some place in 2005. This is the Wikipedia page for communication complexity. And two years before it appeared in the literature, this conjecture was there on the Wikipedia page. I don't know how that happens, but yeah. Yeah, so yeah, I should credit Forge God who is the user who made the edit. Don't know who that is. So what is the nature of this conjecture? It's interesting because it implies the log rank conjecture. So if you prove this, we've got that awesome result. And we know that beta is at least two because of set disjointness. And we actually know that it's at least four because of another recent work. And as far as upper bounds go, approximate rank itself is the best upper bound that we know of right now, order approximate rank. So we don't even know the square root like we knew in the deterministic case. And okay, so these are two conjectures. Let's just look a bit at non-negative rank. So as I mentioned earlier, the conjecture with non-negative rank is actually true for deterministic. This is a well-known result. And you can conjecture that the same thing would be true for approximate non-negative rank. If you want randomized, it's still a combination of non-negative rank one matrices. So maybe approximate non-negative rank is actually tight. There's actually a more reasonable conjecture here that because the lower bound that we have is actually the maximum of the approximate non-negative rank of F and of F complement. We don't even know if those are related yet. So it makes more sense to actually conjecture this variant of it. And in fact, these people did actually conjecture this variant of it. So we've got three conjectures. Let's just cache this knowledge for now. This is the communication complexity portion of the talk. We have the log approximate rank conjecture, which implies the log rank conjecture. And we have the non-negative variants, which are that R of F is order log to the constant approximate rank and this more reasonable variant. So now let's move to XOR compositions. So what is an XOR composition? If you have an N-bit string, let's refer to these strings as Z. When you're composing with XOR, you're coming up with two N-bit strings, X and Y, such that if you take their bitwise XOR, you get Z. And this has, well, intuitively, it feels really nice because it encodes this N-bit string into two N-bit strings. And if you look at any of the N-bit strings by themselves, X or Y, you don't get any information about Z, which is kind of what you want from a good composition in order to have a lifting theorem. And the fact that XOR is good for this purpose was actually known back from 1919 and was actually patented in some form. But the patent has expired, so we will continue to use it. And so this is the formal definition of an XOR composition, what it looks like. So, yeah, as I said, neither Alice nor Bob have any idea about any bit of Z. So if they want to actually compute F composed with XOR, one thing they can do is, again, look at a decision tree for F. They want to find out the ith bit. They can communicate XI and YI to find out the ith bit and continue. But XOR functions can actually do much more. So parity is really hard for decision trees. But you can easily compute parity if you're given it as an XOR function. So you want to find out the parity of Z. All you have to do is Alice computes the parity of X and sends it to Bob. Bob computes the parity of that along with Y. And that is the answer. So what we expect to lift is not from decision trees, but from this model called parity decision tree. So the allowed queries are all parities. So this is what a parity decision tree would look like. You have a hidden input. There's a set. Each node has a subset of N associated with it. And the way you go down this decision tree is you look at the subset at the root and you find out the parity of those bits of Z. If that parity is zero, you go to the left. If that parity is one, you go to the right. And so on, you reach a leaf and you output that. It's exactly the same as you would expect it to be. And which inputs reach the sleeve, it's exactly the set of inputs that satisfy the constraints. And each of these constraints is a linear constraint in F2. It's a parity constraint. So the set of inputs that actually reach the sleeve form a fine subspace of F2. And PDTs can have very interesting properties. This is one of them. For example, if you look at the function R, it's hard for deterministic PDTs because you have to reject just the input, which is the all zero input. Everything else you have to accept. And any leaf at depth D has a large fraction of inputs. It has a two to the minus D fraction of all inputs because each time you're halving, when you say that it satisfies some constraint. So if you want to accept just one leaf, one input, you have to have that leaf at depth N so that it captures a small fraction of inputs. But if you look at randomized PDTs, R is really easy to compute. In fact, in a constant number of bits. So you randomly sample a subset and you find out the parity on that subset. If your string is the all zero string, you're going to get the output as zero. If your string was not the all zero string, it's easy to see that you'll get the output one with probability half. So if you lift this with XOR, what you would expect is that the deterministic communication complexity is omega N and the randomized communication complexity is order one and that's exactly what happens. And R of XOR, or technically, R of XOR is the equality function and it separates randomized communication complexity from deterministic communication complexity. And if you have some other composed functions, as far as I know, there's no other function which will show such a separation. And there's a reason for this. It's that if you're lifting query complexity instead of parity decision tree complexity, query complexity doesn't show a separation between randomized and deterministic. So you wouldn't expect it to, it wouldn't actually lift to a communication complexity separation. So it's these kind of properties of XOR functions that actually are the backbone of this talk. It's what I find fascinating and it's why I like studying XOR functions is that they have these weird properties and if we can find more of these, we will have more of these properties translating into communication complexity. So for example, let's see what else we can get. So we saw that a randomized parity decision tree can compute OR, so similarly it can compute AND. And in general, it can actually compute affine subspaces. That means if you give me some affine subspace S and your question is does X lie in the affine subspace S, then a randomized parity decision tree can do it with very few queries. It's just an adaptation of the OR algorithm. So say you're given some PDT and you're given a specific node in this PDT and your question is given an input X, will it reach this node? So this is again an affine subspace query so an RPDT can do it very easily. And that actually gives it the power to balance PDTs. So if I had a PDT which is actually lopsided, so its depth is really large but it doesn't have that many leaves. Then I can use randomized queries to actually compute the same function with a balanced RPDT. So log of the leaf complexity of this PDT is an upper bound on the depth of this randomized parity decision tree. And it's an open question whether randomization can help any more in parity decision trees. That is, is there any function F for which RPDT of F is much smaller than log of the parity decision, log of the number of leaves of the parity decision tree. So this is one of the open questions that I wanted to highlight. What this will give us in communication complexity if it turns out that there's nothing else that RPDTs can do. This will actually, we would expect it to lift to this statement. Is there an XOR function which is easy for randomized communication but hard for deterministic even equipped with equality queries? So P to the EQ protocols and the real communication model that we saw earlier. So this is, I find this to be a quite interesting question. Recently it was closed for non-XOR functions Chattopadhyayalovit and Vinyals showed that there is a function which shows which is easy for randomized communication but hard for P to the EQ protocols. But it wasn't an XOR function. So it's still an interesting question to find out whether composed functions can have this kind of a separation or whether all composed functions are too structured and that all the randomized protocols are essentially deterministic protocols with oracle access to equality. So this was one of the open problems. This upper bound and RPDT, it's even, it's a non-adaptive upper bound, right? Is the upper bound true even for non-adaptive randomized parity decision trees? I don't see how that would follow. I don't think it should have. It's so to understand PDT is better and to find out. So I mentioned that there that you can lift with XOR. To understand what lifts with XOR, we have to understand how we will represent a function using its parity decision tree representation. So how many people here are familiar with Fourier analysis over Boolean Fourier analysis? All right, cool. Okay, so I'll just do a quick run through of this. So if let's take, for example, the AND function. Here I've taken it in the, where it's taking inputs in plus minus one instead of zero one. And the reason I'm doing this is because parities, we want to deal with parities and parities are very easy to deal with in the plus minus one world. The parity of Z1 and Z2 is just the product Z1 into Z2. The number of minus ones, whether it's even or odd, is going to tell us what that answer is. And Fourier analysis tells us that, given any such function, there's a unique representation as a multi-linear polynomial. And the measure of AND that we care about, well, there are these two measures. One measure is the sparsity of the function, which is the number of non-zero coefficients. So in this case, the sparsity of this three bit AND is eight. And the other property is the L1 norm, which is the sum of the absolute values. In the case of AND, they all have the same absolute values. So it's just eight into one by eight. So the L1 norm is one. And these, so in normal decision trees, you have degree playing a major role and approximate degree playing a major role. In this case, the best measures that we have are stuff like sparsity and L1 norm. So we knew that every leaf of the parity decision tree is in a fine subspace. So we can add up these leaves. So if you have a depth k, r, p, d, t, each leaf is an affine subspace with k constraints. That's like an AND of k variables. So it has sparsity two to the k. And you're adding up at most two to the k leaves. So your sparsity of this function is at most two to the two k. And for similar reasons, it's L1 norm is at most two to the k. And if it's computable by a depth k, r, p, d, t. So an r, p, d, t is just, you're first tossing some random coins and then choosing a parity decision tree. So it's a convex combination of p, d, t's. So an r, p, d, t is that you could look at it as you have a distribution over many p, d, t's. This one is chosen with probability p1, p2, and p3. And the probability that z is accepted is equal to p1 into the first p, d, t of z, p, d, t, one of z plus p2 into p, d, two, two of z, and so on. And each of these has L1 norm at most two to the k. So this whole thing also has L1 norm at most two to the k. And it approximates the function that we want to compute because the probability approximates the function. It's clear? Cool, so again we get an approximation lower bound. So it's L1 norm is at most two to the k. It's approximate sparsity is not yet clear because this format of writing it will actually have a lot of terms from, it depends on how many parity decision trees we have in our support and that could be huge. So we don't yet have an upper bound on approximate sparsity based on the complexity. By not yet, I mean not by this slide. It will come soon. So with these measures, let's see how it lifts. So lifting is really nice with XOR because for example, the sparsity of F is exactly the rank of F of XOR. There's no calculation that we need. It's very simple. There's no complex analysis that we need to do. The approximate sparsity, which is just the minimum sparsity among all nearby real functions, also lifts to approximate rank with a slight loss over here. And that's needed. Okay, I just didn't mention this. So what this implies is, if you actually have a result that your PDT is, this is the log rank conjecture, but in the PDT world, that it's at most polynomial in the sparsity, in the log of the sparsity. So if this is true, then because of this lifting, it implies that D of F is at most polynomial in log rank of F because sparsity of small F is rank of capital F. And similarly here, you would have, if the analog of the log approximate rank conjecture is true in the PDT world, then it'll also be true in the communication world. There's more nice lifting that happens here. So the L1 norm of F is exactly the same as the L1 norm of F of XOR. Now it might be a bit weird that we're talking about L1 norm of F of XOR, but that is not a mistake over there. A reviewer once corrected us saying, you probably don't mean that, but it's actually what we meant. The reason we meant it is that there actually is a conjecture here that says that randomized communication complexity is at most polylog in the L1 norm. So we, so the L1 norm also lifts. The approximate L1 norm also lifts exactly. So L1 norm is really nice. This was supposed to be justification for why the L1 norm of F had a good fashion. So this is, yeah, and it's... Right, so we don't have the other direction of this. So the correct measure to use here is probably something called the trace norm, which is pretty much equivalent to this. And it does have, the lower bound does actually hold over there, it's at least the trace norm. But back when this was written, so Grolmish actually had reason to look at this. In the multi-party setting, he actually showed that if you had enough players, this conjecture is true. And he didn't find any counter-example in the two-player case. And so it's a, yeah, it was a fair conjecture back then. And probably the best of all the lifting theorems we have, definitely the best of all the lifting theorems we have for XOR functions, is this result by Hatami, Hoseini, and Lovett, which actually says that your deterministic PDT complexity does lift to the deterministic complexity of the composed function. So the takeaway from this isn't that all these measures lift, it's just that there are a lot of measures that lift. You don't have to remember these exact statements. You just want it to be complete. There are actually other measures that also lift. There's the weakly unbounded randomized complexity. And that also lifts from PDT, our PDTs to randomized communication. Yeah, but that was, it's a bit esoteric. So what this last theorem means is that if we already know that sparsity is exactly the same as rank, and we know that PDT is polynomially related to D. So this, the conjecture that LRC is true for XOR functions is exactly the same as the conjecture that PDT of F is polynomially related to the log of the sparsity. Now, in this talk, I'll want to get to approximate rank. And I've gotten from our PDTs to approximate spectral norm. So all I have to do is go to approximate sparsity, and then I can go to approximate rank. So this next slide is just going to be connecting approximate L1 norm and approximate sparsity. And it turns out that they are the same thing up to a log n factor. So up to an additive log n, log of approximate sparsity and log of approximate spectral norm are the same. And the reason is quite straightforward. So if you look at a function and you write it as a polynomial, so z sub s is the monomial where we multiply all z i i n s and its coefficient is g hat of s. What I want to do now is I'll just sample from this polynomial g. So I sample a monomial with probability proportional to the value of the coefficient, to the absolute value of the coefficient. And so for example, here we have this, we've sampled one monomial and we're normalizing it. And this sampled polynomial on expectation is exactly our polynomial, because we're just sampling coefficients from the natural distribution. So there's some details here, but I think I'll just rush through this slide. So as you sample more and more monomials and take their average, we know that the expectation is g of z. So we expect that it's actually concentrated around g of z using something like a Chernoff bond. And that's exactly what happens. So if you take t to be large enough, which is order spectral norm squared into n, then the sampled polynomial approximates g on all its inputs. And this sampled polynomial, this sampled polynomial actually has small sparsity. It has sparsity exactly the number of samples or at most the number of samples. So we actually got this, we have this result that the approximate sparsity of g is related to the spectral norm of g. We have that statement. And therefore the approximate sparsity of f is related to the approximate spectral norm of f. It's a bit of a messy slide, so I don't want to keep it up for long. But what's important here is that the log approximate rank conjecture for XOR functions is equivalent to the log approximate L1 norm for XOR functions. And this implies Drollmisch conjecture, because if your randomized communication complexity is almost, and is always polynomially related to the log approximate rank, and therefore it's always polynomially related to the log approximate L1 norm, then it definitely is not going to be higher than the exact L1 norm, because approximate L1 norm can only be smaller. So this does imply Drollmisch's conjecture, even actually for non-XOR functions. So let's just get back to the familiar world of communication complexity. We had these conjectures already with the log rank conjecture for XOR functions. We're going to add Drollmisch's conjecture, which is implied by the log approximate rank conjecture. And it apparently also implies the log rank conjecture. And just for fun, we'll actually add one more conjecture, which I'm not going to talk about. But this slide is really to show you that this field is really rich. There are a lot of interconnections between various measures, and these are the statements. So we've covered communication complexity, the conjectures I wanted to talk about. We've covered a basic intro to parity decision trees. Now, let's just analyze the correspond, this conjecture over here, Drollmisch's conjecture. So this is actually the route that we took while we came up with the counterexample to the log approximate rank conjecture. It's never been presented like this before because I never actually had the time to do that. It was always shorter talks. But this time, I have enough time to go about it the way we did, yeah. So we were actually analyzing the Drollmisch's conjecture. Back then we didn't even know that it was implied by the log approximate rank conjecture. It seemed interesting in itself. Why would R of F be related to the spectral norm of F? And so its corresponding PDT question was what we analyzed. Why would a function with small Fourier L1 norm be easy to compute with randomized parity decision tree? So the easiest example of a function with small Fourier L1 norm is an and. It has L1 norm one. Or in generality, we have affine subspaces that have L1 norm one. And what is something, so these are actually easy for randomized PDTs. So these are not going to give us a counterexample. What we thought the next simplest example would be the sum of a few disjoint ands. Since they're disjoint, disjoint in the same way that rectangles affine subspaces were disjoint in the tree examples. So if they're all disjoint, then we can actually take their sum and we'll still get our rebound on the L1 norm of the function. So for example here, let's take S1, S2 and S3 to be three different ands. So S1 is saying Z1 is equal to zero and Z2 is equal to zero. It doesn't care about Z3. So these are subcubes. We've seen this before because we're in a lifting theorem workshop. And all of these are disjoint because if S1 is zero, then so if something is in S1 and S2, then that means Z1 is both zero and one, which in these are bits, not qubits. So let's just generalize this example to a larger case where we have M different subcubes. If I want to make them all disjoint, I can use the exact same trick from the last slide. For any two subcubes, say S1 and S2, I have a coordinate Z12, which should be set to zero in S1, but to a one in S2. So this means that S1 and S2 are disjoint. And in general for any SI and SJ, there will be a coordinate ZiJ that makes sure that SI and SJ are disjoint. And this example can be rephrased in a much prettier representation, which is the sync function. So the sync function has a nice description. It's on M2's two bits, which is exactly the same as the number of bits we needed in this example. To be thought of as you have a complete graph on M vertices and you put your variables on the edges of this graph. So Z12 is put on the edge between V1 and V2 and ZiJ is put between VI and VJ. And when you assign a value to these M2's two variables, we want to know what the output is. So let's just assign some values. What it does is it actually orients the edges. So if a particular edge, the variable there is set to zero, we interpret that as meaning the edge points from the larger vertex to the smaller vertex. And if it's one, it means it points from the smaller vertex to the larger vertex. So this gives us a complete orientation of all these edges. And the question we want to ask is, is there a vertex in this graph, which is a sync? And this is quite similar to the, and sync has been studied before, but never from this angle. It's query complexity was studied for, I'm forgetting the name of the conjecture right now. Yes, the under areas work up conjecture, yes. And some people have also studied this because it's like a converse winner. You're having elections between these people and you want to know if there's somebody who's won all their elections. And yeah, some people have also told me that sync is a very pessimistic name for this. You should be calling it the winner function or something like that. But as we'll see, sync is a pretty nice name to keep for this function. We'll see it later on. So the definition is this. If there's a sync in this graph, then we output one. If there's no sync, then we output zero. And we know, because this is the same function as the function we saw earlier, that this is basically a sum of M subcubes. The subcube where V1 is a sync, the subcube where V2 is a sync and so on. They're all disjoint because VI and VJ cannot both be syncs. There's an edge between them. So the spectral norm of sync is at most M and because of the equivalence between approximate spectral norm and approximate sparsity, we know that the approximate sparsity of sync is at most M to the four. So both of them in the log, when you take the logs, both of them are just log M. And you lift it with XOR and these measures just simply lift because XOR is really nice. So even for sync composed with XOR, the L1 norm is at most M and the approximate rank is at most M to the four. One further measure that this has, which doesn't that directly lift from the PDT function, is the approximate non-negative rank of F is actually small too. So why is this? This is quite simple. So sync of X comma Y is actually a sum of sync VI of X comma Y, which is asking is VI a sync, right? And finding out whether VI is a sync actually just, it's equivalent to just doing an equalities. For example, if you look at V1, V1 is a sync if and only if Alice's bits on these edges is equal to Bob's bits on those edges, then their bitwise XOR is all zeros and V1 is a sync. Similarly, every vertex being a sync is easy to compute via randomized communication protocol. So each of these matrices are actually matrices that come from a small randomized complexity communication complexity protocol. So each of these have small approximate non-negative rank. And since you're just summing them, the whole thing also has small approximate non-negative rank. So that M to the order one is actually just M to the five. All right, so what we proved in our recent result was that the randomized PDT complexity of sync is omega M. The randomized communication complexity of sync of XOR is omega M and that the particle number is also omega M, but I haven't introduced that, so it's useless. So what does that do to our conjectures? So these conjectures are all now false. So log approximate rank conjecture is false because log approximate rank is order log M, but the communication complexity is omega M. Similarly, Grollmische's conjecture, the parity kill number, and the strong log approximate non-negative rank conjecture are also false. Well, the next thing to see is the general form, the more reasonable form of the approximate non-negative rank conjecture. Well, there we weren't as lucky because the lower bound method that we used, it actually shows that if you take the complement of sync of XOR, that has large approximate non-negative rank. That's actually how we show that sync of XOR is hard to compute for randomized communication. So there, the bound that you would get there is omega M, which is tight. So that conjecture is still true, but sync actually syncs one more conjecture, and I couldn't have said winner is one more conjecture, so sync is a good name. And that's the quantum log rank conjecture. So log approximate rank is also lower bound for quantum complexity, and this conjecture asks whether it's polynomially redacted. And soon after our work, there were two works that went ahead and disproved this conjecture as well. Markrand is one of the people. And, yeah, so this was the sunken conjectures. So let's talk a bit more about this conjecture, which is something that I am very curious about because I feel that there might be an avenue to continue and disprove this, but yeah, I'll show you what I mean. So what we did was we showed that we came up with a function whose one inputs were actually a disjoint union of subcubes, a disjoint union of M subcubes, but the problem was that its zero side had large log approximate, not negative rank. So if we could come up with a function where F inverse of one is a disjoint union of few subcubes and F inverse zero is a disjoint union of few subcubes, while keeping its RPDT complexity large, then you would have good reason to believe that the general log approximate non-negative rank is also false. But it turns out that you can't do it with the union of subcubes. There's an elegant proof that follows from some work done by R.N. Furt and Hausler. And the corresponding conjecture, if you don't just limit yourself to subcubes, because remember, subcubes weren't the only thing that had smaller spectral norm. Even affine subspaces in general had small spectral norm. So if you try to extend this to disjoint unions of affine subspaces, you might actually be able to find an example like this. And the proof that you can't find an example with subcubes does not follow through to the affine subspaces case. So I think it would be very interesting if you could find out whether this is possible or whether it's not. And so to summarize what I wanted to say, we know that XOR functions behave well according to some measures that lift very easily. We know that PDTs are not well understood and there's a playground there that we can continue to search in. And there are many juicy questions. One is the question that I mentioned at the beginning. Are randomized PDTs the same as what I've written there as conjunction PDTs? So the measure of leaf complexity of a PDT. So if you have a PDT with small leaf complexity, you can actually balance it using affine subspace queries is what I said. So you don't actually need to go to our PDTs, this is exactly equivalent to this other model where you actually have affine subspace queries. And the question is, are these two models the same? Because these two models are. And if you think parities are really messy to work with, there's a conjecture here that doesn't even involve parities. The question of our conjunction PDTs the same as randomized conjunction PDTs. So the queries you can make here, you can query a single bit or you can query a conjunction of bits and does randomness add any power there? We know that without the conjunction, randomized doesn't actually add power. And as far as I know, this is open. Another open problem is can we close the avenue that I just mentioned on the previous slide? Can we better the closeness between randomized complexity and approximate rank? So we show that log approximate rank and randomized complexity can be far apart. And in doing so, we showed that syncs randomized complexity is up to fourth power close to its approximate rank. And we want to know whether you can actually make this closer because the best upper bound we have is order approximate rank itself. And another very interesting question is can we attack the log rank conjecture using XOR functions somehow? Because the way we used, so we said that sync was a summation of easy randomized complexity problems, right? And what this meant was that this has small approximate rank, so this summation also has small approximate, should. It has small approximate rank, but its randomized complexity increased a lot. And that's what actually gave the separation. If you replace these with stuff that have small deterministic complexity, so that means that these have small rank, exact rank. So that means that this would also have small exact rank. And now the question is, can its deterministic complexity be much larger? And in that case, that's actually not true. If this is a summation of stuff that has small deterministic communication complexity, then this function also has small deterministic communication complexity. And yeah, so the log rank conjecture is a Titanic conjecture, so it makes sense that it's not syncable. And yeah, there are other questions. And I haven't introduced the necessary measures, but there are many more questions that come up. If you look at PDT's and you just wonder, what are these strange properties of PDT's? And yeah, I think I'll open the floor to the public. So can you use the word Titanic in this context? Because I think that you actually think it is false. Oh, thank you, that's a good question. So no spoilers yet, I don't know what happened to the Titanic. So I actually think it could be true, but that's what I thought about the log approximate rank conjecture as well. After we had disproved it, we just went through our proofs wondering where we went wrong until we were convinced that no, it was actually not true. The same thing, we're still at that stage for the log rank conjecture. We don't know whether it's...


Let's say one of the things that it's very nice to see it's not It's because we have a new square. It's a very nice attraction that we have lately We have some places that they let's say wanted to renovate it and it's very nice the square and also It's called Pazari very nicely used to gather for some drinks and etc Raki is one of the alcohol that it's very well known in Albania But even wine I would say But you're getting the food We have a lot of specialities that people can try But mostly where a bit meat oriented Let's say people but even for other people that want to try something else we have Sweet or not sweet one, but I would say I don't want to See see the surprise that we will have during the conference. I cannot reveal them But regarding sightseeing I would say If people can see even a bit more I would highly recommend them to go even in other cities We will see during Libokon how much time we will have To go to another city that's difficult But in Tirana we can go at some at one of the museums that it's in the center It's the house of leaves that is a very nice place and I really like that the history that people can learn there September maybe it's knowing there it depends But even there it's very nice, you know with telephoric going there and enjoying a nice evening in Tirana


Good day everybody. This is Dr. Sanjay Sanyal, Professor of the Department Chair. This is going to be a demonstration of certain dissection specimens of the spinal cord. So we have removed the occipital bone from here. This is a prone cadaver. We have done a post chest for saccharinectomy. And then we continued down and we also removed the paraspinal muscles. We removed the laminate of the cervical vertebrae from the cervical thoracic, lumbar, sacral, and then after we have split the dura to show the spinal cord. In an earlier dissection, we had mentioned the upper part of the spinal cord and the spinal nerve roots and many other features here. Now we shall focus on and we shall go right up to the bottom of this dissection to show you the lower part of this dissection. We have, as I mentioned, we have split the dura and the arachnoid. And we have reached the lowest part of this dissection at this region that we can see here. This is referred to as the lumbar cistern. What's the meaning of the word lumbar cistern? Lumbar because it's in the lumbar region. It extends from L2 to S2, vertebral levels. Cistern means an enlarged subarachnoid space filled with CSF. What are the contents of this lumbar cistern? The contents of the lumbar cistern, as I mentioned, one is CSF. The second important content, of course, are these multiple branches that we can see here. These are the branches of the lumbar sacral and the coxigel nerves, which is referred to as collected with the corda equina. The third content, of course, will be a small structure which we cannot see, which is a non-neural structure called the phylum terminale. This is the T2-sidiacal MRI, the lumbar region to show the lumbar cistern, the CSF, the corda equina, and the conus medullaris. And this is a lumbar myelogram AP view to show the lumbar cistern, the conus medullaris, and the corda equina. Now let's see a few quick words about end of the spinal cord. End of the spinal cord is here. It is located in this region. This is referred to as the conus medullaris. The spinal cord ends approximately at the upper border of L2 vertebra and it is also referred to as the medullary cone. This conus medullaris, we can have an important syndrome here and that is referred to as the conus medullaris syndrome. That happens when there is any one of the intradural tumors like, for example, appendiboma, astrocytoma, or we can have HIV lymphoma, or we can also have arterial infarction of the region of the spinal cord. Incidentally, fracture of L1 vertebra can also produce the same syndrome L1 vertebra refers to this region of the spinal cord, which is the sacral of the coccygeal segments. When there is a compression of this region of the spinal cord, the segments which are involved are the S3 to CO segments, sacral third to the coccygeal segments. Usually the manifestation will be suddenly a non-sit and it will usually be bilateral. Manifestations will be more severe pertaining to the parasympathetic portion that is the bowel bladder and male sexual mechanism namely importance because the S2-3-4 segment is important for migratory center, defecation center, and the male erection in the ejaculation center. That is referred to as the conus medullaris syndrome. We can get another syndrome which refers to the portion above that and that is referred to as the epiconus syndrome which extends from segments L4 to S2 which can also be caused by pretty much the same causes and the patient will have very severe difficulty in extension and lateral rotation of the hip joint because of the paralysis of the gluteus maximus muscle. Additionally, the person will have weak anchor jerk which is root value with S1-S2. So that's about the conus medullaris syndrome and the epiconus syndrome. Now let's talk a few quick words about the cauda equina itself. This cauda equina looks like a tail of a horse that's why it is called cauda equina. The cauda equina can also be involved by medius tumors, intradural tumors, they can be nerve root tumors, lipoma of the cord, dermoid, neurofibroma, and it can also be caused by HIV lymphoma and rarely it can be caused by a central herniation of the nucleus pulposus of the lumbar vertebrae. This is a T2 MRI of the spinal cord showing an HIV lymphoma in the region of the cauda equina. The cauda equina syndrome usually occurs gradually and it is usually somatic symptoms will be more severe in the sacral saddle area. Saddle area refers to the area which a person sits on on a saddle and usually it is unilateral more common than it is bilateral. So there will be severe pain and there will be muscle weakness in this region. So that is heavy saddle anesthesia and heavy saddle muscle weakness. In contrast to the conus medullaris syndrome, the bald bladder symptoms, the autonomic manifestations will be very few or very limited. There is yet one more condition which in this is not visible and that is called the phylum terminal. The phylum terminal is a neuroglial tissue which extends from the conus all the way down through the lumbar cistern and it then pierces the dural sac and it goes and gets attached through the sacral canal to the coccyx. That phylum terminal has got the internal portion which is called the pial part and an external portion which is referred as the dural part. In a small percentage of people the phylum terminal can be very short thick and stout and it can pull on the lower part of the spinal cord and that is referred to as the phylum terminal syndrome. It can produce again symptoms pertaining to the lower part of the spinal cord that is the patient will have lower motor neuron bald bladder symptoms and will have also have weakness of the sacral region, the weakness of the ankle jerk or the phylum terminal syndrome and it may require transection of the phylum terminal. This same phylum terminal if it pulls hard then it can also produce traction on the upper part of the spinal cord and can produce downward displacement of the medulla through the ferrimid magnum and that is referred to as herniation syndrome. I had briefly mentioned that when I was talking about supra dintorial compression. This is a T1 sedatum MRI to show cerebellar vermus herniation through the ferrimid magnum in a case of anodicary malformation and this is a T1 sedatum MRI of an anodicary malformation to show cerebellar vermus herniation. So therefore this condition is referred to as a tethered chord syndrome. So these are the few quick points which I want to mention to you about the three regions that the clinic of significance, the lumbar system, the cauda equina, the conus medullaris and the phylum terminal. Thank you very much for watching Dr. Sanjeev, Sunny and Sunny out please like and subscribe if you have any questions or comments please put them in the comment section below have a nice day.


Hello everyone, this is Ahmed Dan from AhmedDan.com You are watching one of my New Zealand travel videos and today I am taking you to the iconic Auckland Sky Tower The Sky Tower is the tallest freestanding structure in the southern hemisphere 328 meters or 1075 feet high and the 25th tallest tower in the world Auckland Sky Tower defines Auckland's skyline because of its height and shape The breathtaking views offered unobstructed views of Auckland Having water, harbours and volcanoes surrounding Auckland made everything even more scenic There revolving restaurants, cafes, ice cream shops, gift shops, casinos and many other establishments in the tower Thank you for watching this video Forget about subscribing this channel, forget about liking this video There will be more videos coming up so I will see you shortly Thank you Do you want to try it?  Yes Do you want to try it? Yes No No I think you should try it Yes Yes


What I'm doing is I'm setting up for my final project. My final project, so I'll show that tomorrow morning and go over everything I learned from making art for 100 days in a row. Here's a little sneak peek.  Don't want to show you too much. Don't want to show you too much. Anyways, it's been a really fun project. I'm excited to finish it tomorrow. I hope you're having a wonderful Thanksgiving if you're in Canada, we just had ours, so. And I hope you're making stuff. See you tomorrow for day 100.


Hello everyone, this is Alice Gao. Welcome to another video on probabilities. In the previous video, I talked about two rules, the sum rule and the product rule, and we can use both of these rules to perform inference using the joint distribution. It's really nice when we have the joint distribution, but most of the time in practice, we don't have it. It's actually difficult to come up with the joint distribution for a particular scenario. So instead, we often have some prior or unconditional and conditional probabilities that we can use to describe the random variables we're interested in. In this video, I'm going to talk about two more rules, the chain rule and the base rule that we can use to perform inference using the prior and conditional probabilities. Here's an example of some prior and conditional probabilities we can use to describe our home scenario. Remember, we've simplified it so that we're only thinking about three variables, alarm, Watson and Gibbon. And these numbers are carefully constructed with an underlying model in mind, but you might not be able to see it yet. We will talk more about it when I introduce Bayesian networks. The kind of tasks we are going to do with these probabilities are the same. So we are still wondering how do we calculate a probability over a subset of variables, and then how do we calculate a conditional probability? The first one, the chain rule is going to be useful, and for the second one, the base rule is going to be useful. Let's look at the chain rule first. I have several examples here, one for two variables, one for three variables, and then the general case for any number of variables. Let's look at the two variable case first. This formula right here looks exactly like the product rule that we've seen earlier. So for the simple case, they are equivalent. Another version of this is you can order AB differently, and then you can get equivalent. Version where this is equal to probability of B given A multiplied by probability of A. So both of these are the same. Now for the three variable version, for this version, we can begin to see a general pattern for how the chain rule works. So if you look at this three variable version, it seems like we are taking these three variables and sort of order them in some way. And then given this order, we came up with this expression. You can think about coming up with the expression backwards, where we have the last variable in the order, just the prior or unconditional probability of C. And then the second to last variable, we have the conditional probability of B given C. And finally, the first variable, we have the conditional probability of A given B and C. So if you try to extract the general pattern, the pattern seems like every variable in the sequence is going to depend on all of the variables that come before in the sequence. Given some sort of ordering of all of the variables. So this is exactly the general pattern. For the most general case, we have n variables, and you can think about them ordering. Again, the order is from right to left, right? So x1 up to xn. And then given this ordering, we have a giant product so we can decompose this joint probability into a product of prior and conditional probabilities. And this product in general, you can write it in this form where this big notation in the front is about the product. And then you can see every variable in this product. So xi is the current variable that we're looking at. Every variable is going to condition on all of the variables that come before in the ordering. Right? And if you want to write out this entire expression, it looks like what I have on the next slide. Right? You can make an example for yourself for four variables or even more and then try to write out this expression. Right? And notice that the expression, there are many versions of this expression and that depends on how you've ordered the variables. Let's practice using the chain rule. I have two quicker questions on this slide and the next slide for you to practice using the chain rule. Same as before, I will only tell you the correct answers in this video and you can watch a separate short video to see how I derive the correct answers. For this first question, we want to calculate the probability that the alarm is going, Dr. Watson is calling and Mrs. Gibbon is not calling. You can go back to the previous slide and look at the prior conditional probabilities that I gave you, but to make all the information available right here, I picked out some of the probabilities that could be important for this question and put them on the slide. So for this question, the correct answer is 0.063. So we want to calculate the probability of A and W and not G and this 0.063. Question number two, what is the probability that the alarm is not going, Dr. Watson is not calling and Mrs. Gibbon is not calling. So for this question, we want to calculate the probability of not A and not W and not G and the correct answer is 0.486. Even if you got the correct answers, it might be worthwhile to watch the separate short video anyway because I will try to explain different ways of solving the same problem using the chain rule. That's it for our first task. So in order to calculate a joint probability over maybe a subset of the variables, we can use the chain rule. Next, let's think about how do we calculate a conditional probability? In particular, the task we have in mind is we want to flip a conditional probability. So when do we actually want to flip a conditional probability? This is actually quite a common scenario. Think about it in real life. Often we have some knowledge about that one thing could cause another. So we have some knowledge about what are the possible causes for a particular phenomenon. For example, we would know a particular disease is going to cause some particular symptom on a person or we know that if there's a fire, then the alarm is going to go off, right? So this is the kind of causal knowledge that we often know. It's great to know this kind of knowledge, but they are often not helpful in practice when we're asking questions. Because when we're asking questions, we often do not know whether a person has a disease or we do not directly observe whether there's a fire or not. Instead, we often observe the evidence or the symptom. So in the case of a medical diagnosis, we often can observe that a person has some symptoms and then we want to ask how likely it does this person have a particular disease based on the symptoms observed. And similarly, in the fire alarm setting, we often observe whether the alarm is going off or not and we want to infer whether there's a fire going on. So you can see that the kind of knowledge we have does not really match up with the kind of reasoning we want to do, which means, well, based on our knowledge, we might want to be able to flip the conditional probability and calculate the conditional probability the other way around. This is the famous base rule, so it allows us to take the probability of y given x and flip it to get the probability of x given y, and in order to do this, we need the probability of y given x plus the probability of x. You will see that we don't directly need to know the probability of y as the denominator because the numerator already gives us enough information to derive it. Now you should not memorize this rule at any time you should be able to derive it, for example, when you're stuck on a deserted island, so here is how you can derive it. Remember the two different ways of writing out the product rule for two variables. So if we have probability of x and y, depending on the ordering, we can write this out as the probability of x given y multiplied by the probability of y. This is ordering y first and then x next, or if we choose the other ordering, this is equal to the probability of y given x multiplied by the probability of x. Given these two ways of writing out the same expression, you can see on the right, we can simply how do we get to our version of the base rule? We can simply take the probability of y here and move it to the right. This is going to give us our version of the base rule. The second point I want to make is regarding the p probability of y in the denominator. In fact, in order to calculate the probability of x given y, it's not necessary to directly know the probability of y. But to understand this, first of all, let's do a little bit of mathematical derivation. I'm going to take the expression of the base rule and expand the denominator. The numerator stays the same and then take the denominator and apply the sum rule in reverse. So we expand it by adding x into the expression. p of y is equal to p of y and x plus p of y and not x. We don't care about x, so when we introduce it, we need to introduce all possible values for this variable. And then let's expand the two terms in the denominator further by using again the product rule. The numerator again stays the same and the first term in the denominator actually ends up being the same as the numerator. We are calculating the probability of y given x multiplied by the probability of x. And the second term will also order things so that not x comes first and then y comes next. Like so. This is one part of the story. Now, because we are using this base rule and in fact, using this formula, we end up calculating a probability distribution. So this equation is only showing you one probability in the distribution. The other probability in the distribution is probability of not x given y. So I can do a similar derivation, but let me just write down the end result. So to write down end result is actually quite easy because the denominator is going to be the same as the denominator that I just derived above. Where the numerator ends up being the second term in the denominator. Let's look at what's happening with these two expressions that we just derived. Well, we are trying to calculate the distribution. One probability in the distribution is probability of x given y. The other one is probability of not x given y. So this distribution says, let's assume y is true. So among all the worlds in which y is true, what's the probability that x is true versus what's the probability that x is false? And we derived two expressions and notice if you compare these two expressions, the denominator we have are exactly the same. Whereas the numerator, in one expression, we have the first term in the denominator and in the other expression, we have the second term in the denominator. So an alternative way of thinking about this is that how can we calculate this distribution? We can calculate the two numerators and after calculating the two numerators, we just need to take these two numbers and normalize them. So by normalizing, I mean take these two numbers and divide each of these two numbers by the sum of them. This is exactly what's happening when we divide both of them with the denominator. So if you think about this way of calculating this distribution, you will realize that p of y is simply a normalization constant that we're using here. A normalization constant basically means we take a bunch of numbers and these numbers might not make up a probability distribution because they might not sum to one. Well, what do we do with it if we want to convert it into a probability distribution? We'll take all of these numbers and divide each number by the sum of them. This is one way to convert this into a valid probability distribution. So this derivation gives us a new way of thinking about the base rule and also gives us a new way of calculating this distribution. This new procedure has two steps. First, we calculate the two corresponding values, which are these two. Probability of y given x multiplied by p of x and probability of y given not x multiplied by p of not x. And then the second step will normalize these two values so that they sum to one. So from this description, you can see what is the minimum set of values we need to calculate this distribution. We do not need to know p of y to start with. All the quantities that we need to know are p of x and we do not need to know p of not x. We can just say 1 minus p of x is p of not x. And then we need to know p of y given x and we need to know p of y given not x. Given these three quantities, we have enough information to calculate p of x given y and p of not x given y. I hope this gave you a new way of thinking about the base rule. Let's practice using the base rule. Again, I have two questions and I will only tell you the correct answers in this video and watch the separate video for the process. For this first question, we want to calculate the probability that the alarm is not going given that Dr. Watson is calling. And the correct answer here is 0.8, so option C. For the second question, we want to calculate the probability that the alarm is going given that Mrs. Gibbon is not calling. And the correct answer here is 0.08, which is option E. That's everything for this video. After watching this video, you should be able to do the following. Calculate a joint probability over a subset of the variables using the chain rule. Calculate a conditional probability using the base rule. Interpret the base rule in a different way where the denominator in the equation is you can think about it as a normalization constant. Thank you very much for watching. I will see you in the next video. Bye for now.


How's it going everyone Caleb here with late in sports cards? You're gonna be ripping 2021 national treasures football four box full case break Number 120 pick your team style Gonna read off the teams you do have in the break. Good luck everyone Thanks for joining Andres a Cardinals Neil P Rams Austin B Bears Bernie S Jaguars Braden K Texans Brian S Chargers carry W football team Got Corey S Colts Dave D Vikings Edgar B Titans Browns Eric R Jets Harper B Bills Bengals and 49ers John L Seahawks Justin R with The chiefs buccaneers Ravens Cowboys and Steelers Kevin H Lions Kyle S Patriots Matthew R Falcons and Raiders Max G Broncos Nicholas T. Dolphins Richard W Panthers Sam A Eagles Stewart Z Packers Saints and Giants. Good luck everyone Let's get this going Yeah, that's that is a pretty cool. That is a pretty cool one How do you pick to be a team? So just check out the brakes on the website here is a link to all of the different breaks that we do have You can check them all out right here And when you do find a pick-your-team style break you can select your team The teams do go fast sometimes, but with pick-your-team style breaks you pay the price You pay different prices for each team in the break So just find the brakes on the website that are pick-your-team style To give it a look You know Yep, all right here we go. Yeah, someone was asking about that good call Here we go. Good luck everyone. Thanks as always Yeah, like I said look for the pick-your-team style Yeah, I did Blake. He already showed me the Charizard Craziness Blake. I don't know how he does it man. I knew he was gonna pull that as well. I Knew it Should show great That's so lucky man.  Yeah, team all breaks are cool Mall breaks are cool. It's some exclusive parallels and stuff for sure the MOOC NFL gear trio materials Packers Aaron Jones Aaron Rock Aaron Rogers Aaron Jones and AJ Dillam Panthers this time 17 of 35 Shuba Hubbard dual patch They're really affordable prices to move. Is that the one you hopped in as well? Gotcha, yeah, I know we got a bunch of different team all in that are were breaking specifically at the national Yeah, I saw that yep Elijah Molden. That's to 10 there Titans Elijah Molden But yeah for sure the MOOC. Yep Safe force can break that one Oh, he'll crush it. He'll crush it to 10. That's the nice thing about the national is it's like so flexible now We'll make it work how we should Acetate here on the Thailand Wallace 10 of 25 the future. That's a cool card Ravens Football team all interesting. I saw Kenzie was getting in some basketball over here Jack Lambert here Two of 25 Steelers This is just unreal Bernie S 44 of 49 vertical RPA Trevor Lawrence boom Feel the burn again.  Oh my goodness another one for the stack man You go will nag that great three color. Holy cow next one up Wow, I Can't believe it man. I can't believe it truly. It's unreal. I Saw the tealot right when I opened the first box My hands were a little bit shaky right off the rip. Oh my goodness. No way No way so so sick All right empty Got the midnight black here number to 20 miles Gaskin Go Dolphins a Lot a lot That's a Zeezo Jollari to 99 there Giants Century materials Titans here Vince Young to 99 Titans Treasured patches Heinz Ward to 25 This one there to 25 Khalil Herbert to 99 bears This one there to 99. Oh my god What is going on? One of one Mac Jones laundry tag boom What is happening? Oh?  Oh man, that'll get megged Horizontal patch autograph 8 of 15 Cornell Powell very nice there eat a 15 Packers autograph 24 of 25 Jordy Nelson Packers Yes, yes We've got patch autograph Kyler Murray to 99 nice one there Cardinals to 99 I Can't believe these first two boxes man. I'm really hoping that this case can just continue at this rate this is so sick and 55 of 99 rookie jumbo prime SIGs Hedarius Tony That's awesome nice booklet there Giants To 99 nice patch You go Giants 55 of 99 What a start to go. Oh my goodness, man. I love this product It's just the product. It always treats me right Here we go. Let's continue on. Let's continue on. All right, face cards Pat Friar use to 25 Steelers AFC North or a Steelers these two divisions and to 99 Josh Allen bills Century materials is Mike Williams number to 49 with the triple or treat tricolor patch your chargers to 49 Colossal patch here gonna be Steelers chase Claypool to 25 There you go. This one there on the chase Claypool Beautiful It is Bernie. It is Nice one here printing plate patch to a one-of-one Very nice dolphins That's sick the one-of-one a patch Mag that for you nice one dolphins Very cool Step here falcons. It's gonna be 93 of 99 Kyle Pitts on the patch Yeah, correct. Oh, yep. He is number one. Yep Suppose that still counts even though it's a one-of-one You go on the pits Texans Nico Collins here rookie jumbo prime signatures 83 of 99 Nice one there Texans Very nice Yeah, that's awesome Lane. That's awesome.  Here we go Crossover RPA on the nausea. I knew that Lane. I'm serious action. Here we go Good luck AJ Brown to 11 number to the jersey number and Jarvis Landry to 99 Browns and sterile Henderson to 99 Jumbo patch To 99 Jumbo patch Rams treasured patches Tyler Habe two of 25 Rams again. We have a Joe Burrow 18 of 99 franchise treasures Joe Burrow Bengals Next up here. It's gonna be Steelers and Broncos NFL gear quad materials to 99 Najee Harris Jerome Venice Jamal Charles and Javante Williams That's to 99 Broncos. Yep. Yeah, that's gonna be randomed at the end Very interesting here 17 of 99. This is gonna be from 2020 NT rookie jumbo prime signatures James Morgan vertical patch autograph 24 of 49 Amir Smith Mar set Go Vikings Solid patch autograph is gonna be Marcus Allen 68 of 99 And one more boom hiding at the end 13 of 49 Davis Mills Horizontal patch autograph boom Great Mills, there you go to close Texans last card Very nice We'll mag the Mills Great ending there and yeah, what a case is right? There you go. That will get magged up Let me get this random going first guys, and then we'll do a recap for you. Just one sec This random first here and see one 20 All right, let's do that random So we've got the Steelers. We've got the Broncos. Let's roll the dice the team on top will land the duel Or excuse me the quad Five times here team on top wins it It's one two three four five Broncos congrats on the card you go Broncos right patch cards gods Joe burrow Tyler Higby Darryl Henderson Chase Claypool Mike Williams Khalil Herbert's Pines Ward Vince Young Shuba Hubbard NFL gear trios material Aaron Rodgers Aaron Jones AJ Dylan The Cowboys quad QB Melvin Gordon Kurt Warner Jordy Nelson auto Jack Lambert Thailand Wallace Elijah Molden Marcus Allen Kyler Murray and Amir Smith Marceps James Morgan rookie Jebo Prime SIG Nico Collins for the Texans and a Kaderis Tony book One of one to a printing plate patch Elijah Moore to 20 Najee Harris crossover That's gonna be also to 10 Cornell Powell to 15 To 49 Davis Mills Vertical RPA one of one Mac Jones Sick one there on the laundry tag and to 49 Trevor Lawrence Vertical RPA. Thanks again so much. Hope you did enjoy it flawless coming up next tonight


So I started to play Hatsune Miku Project Diva Mega Mix Plus and I am addicted.  I love it so much. Just not. I gotta end it. I gotta end it I don't wanna do it. I didn't want to stop playing. I didn't want to stop streaming I am in love with the game. The game is great, and I guess you guys all want to know How is the heme dog gonna kill? Well, if you don't put the hands up emoji in your comments I am gonna be forcing the heme dog to put a gun emoji into the comments.   No way No way we're having this problem with the game. It's not loading up. It's not loading up guys guys It's not loading It has a 23 gig Dude, it failed three files. It failed three files for the validation and it's got 23 gigs 23 gigs. What is my like? Great start innit? So during my 23 gig download I decided to listen to the playlist that Moors and Ghost and Pals made for me and man was it good Okay, this reminds me of traffic jams a lot. I gotta bring it home Holy wait coffee down wait. Oh my god. This character. I love this character designs so much Dude this sounds this is way better than traffic jam. Holy moly The video is a bit quiet. Is it? Is that better? I'm adding this to my playlist. I'm gonna add it to my playlist like do I do it now? Nani the fart Oh But the song is a bang Shit, I may have just shat myself. Okay, so now it's finally time for my first time experiencing the game There's a person This thing's happening. It's skit. There's stick action No fail mode game. I'm gonna be pro-professionale. Oh my god We got a music video. I'm not gonna be doing those things. I'm watching music video. Okay Is it just all bees come on change it up, please. Oh Hey, okay, this is pretty good pretty pretty good start. Oh, yes, let's go. Oh I'm professional. Oh Dude, this is so awesome. I love this game Oh god, it's so hard with the music video because I want to watch the music video. Oh my god Okay, not bad Everything they could have gone wrong perfectly wrong and I'm here for it This was the start of something great landing on something like that in a slipper Ouch, I was having a fun time. What song do I do? Oh? Oh my god, it's a fiber difficulty.  This song used to be on my playlist But I think it was on my Japanese my Japanese. Um, it was on my Japanese jam Playlist hey 56.36% not bad. Okay. It's pretty bad, but it's not bad I was clearing the levels. Oh, well, I was supposed to do two at once I've passed the border. I'm getting good at 50. Oh, there's an ad. It was great.  I kept failing. I'm dying. Oh more. Thank you for subscribing and fail.   Do you have a problem with me tree? Huh? Imagine if I kicked it in like 20 apples fell I Because all I wanted to do I just wanted to touch this branch I just wanted to touch it. Okay, I just wanted to touch it because I was like there's so many apples And we caught one of these apples fell in the foreground. No, but an apple fell on my head ground Life, I'm saying away from the street. Let's sit down for a bit because that tree terrifies me All the chair this chair is breaking and then it happened the disappearance of Hudson Amigo Alrighty, we're gonna do disappearance of Hudson Amigo. God bless yourself for doing this one Hey, one of my friends Recommended it to me, so I have to don't die.  I'm doing good The game just froze that was the game's fault not me. Oh I wasn't getting a score, but I was playing a lot better. I mean like it was like hitting them a lot Okay, we're gonna try that again Do I at least do better I did because I got zero percent last time.  Hey How because after that my goal was just to clear any map because I was just doing peep peep boop But all of a sudden came game of time. Maybe I need my glasses on man. Maybe I just need my glasses, you know You know, maybe maybe I just need my glasses on game of time now then how do game of time go? Time to be a gamer. Come on pass guys I Can see now there's something covering my eye. I think it was the wind. Yeah. Yeah. Yeah, I think the wind You know The wind the wind was a bit strong. It just just caught me off guard Yep Didn't work out. No, I'm bet me no gamer me failure I'm even ashamed because I failed 3d tomorrow, which is a song. I absolutely love I have fallen in love with that song and as well with lag train lag train has become a song that I love so much But it's a shame. It's not on project either So sad. There's actually depressing to be fair, but then something immaculate happened Spectacular, it was like I was Harry Potter. It was a wizard 101 Did I pass? I just passed. Oh my gosh. I just passed I like 50 points if my zero one For one 50 point for one. I'm all week. Oh I'm happy I pass Well, this game is great swiping right Works out.  It starts with a Sega Yeah, okay Sega I hear you I really hear you but what I really needed was Some confidence because I was not that confident on my first try last stream the audio like I had two things for audio So when I was watching the replay I was like, oh, the audio sounds weird. Hopefully doesn't sound weird this time because I fixed it Oh, this sounds fun. Okay, so this is okay. Just letting you guys know. This is the first one I should have probably done a warm-up. I should find like a good warm-up one, which is probably rolling girl But let's do this. Oh god. Oh, this is cool Oh, oh god Oh my god. No, no guys. I just woke up. I just woke up. Oh I'm struggling to see I'm struggling to keep up with it.  Oh no. I just passed. Oh My god, I need more coffee man. I need more coffee.  This instrumental Is so good Oh my god, I'm doing bad though. Yeah, I need to add this song to my playlist as well Am I missing? Am I missing Godzilla? Wait, what? Oh, I'm supposed to hold it. This song is so nice I'm just At least we put we we've cleared two in a row. This is noob Okay, so now I got some questions for you. What is your favorite song in this game? Do you think I'm all right as a noob and why haven't you followed my twitch yet? Really? Why haven't you come on tell me why? Yeah, for so because you can't be bothered Fair enough the link is in the description down below now. Something weird happened. Okay. I was on a roll A major role. I just kept winning It was actually insane. I want to do this. I want to do this one. I'm doing this one. I'm doing this one knife Dude, this is a banger as well I want more songs with these two with this duo. I'm doing good ish good ish not great Good ish. Oh, yes, go me wait. There's gonna be more isn't there? Yo, yo Pink light the pink light is a vibe, you know, it actually looks pretty cool Oh Boom boom boom boom Logan's on his rubbing game. I think soon we know who'll be playing osu. Don't dude. Don't don't get me into osu I've seen when dude. I've seen I've seen you play osu. I've seen how addicted it is projects. Okay. We'll be next I just need a device that can do it Oh, it's that I love the songs. I'm out Dude, the song's so good. Hello. How are you doing? I'm happy that I get to watch one of your streams I recently was watching one of your youtube videos and it's always a great vibe Hey, thank you. Thank you for joining the stream and thank you for liking the videos I got such a great. Okay. I got I got a video that I want to do on gigapay And the title that I thought of is gigapay's music makes my pee giga Okay, this just sounds so cool Oh Yes Yes, this vibe Wow I've never heard this voice before.  I love the character design. I love the sound of the voice I want to do more videos where um, I do it just like on the vocaloid or voice bank Because I did one on like, um I did one on meku. I'd love to do it more get to actually experience more of vocaloid Instead of just like the producers because I also really want to get into the producers and all their music, but goddamn Oh, how am I supposed to? Oh god. I was fat. Oh my god. Oh my god. Oh my god. Wait, wait, wait. Oh my god Why am I failing? Why am I failing? This is not that hard I could have done so much better, but I was I choked at the end. I choked I choked But yeah, I would like to do more stuff like on the the voice banks. Oh Kaito and meku we got to we got to Kaito and meku we got to Kaito and meku we got to Hmm Well, I got so many calls in a row. I found so cool. Oh, this is awesome Ah Why do I choke at the end? Yes, then show off them dance moves. Is it what what what is this? Is it ren and len ren and len ren and len? I want to do a video on them Ah, dude, we're doing the song. Uh, I'm sorry I wanted to do this song last stream because I listened to dance robot dance the other day And I want to listen to alien. Yes Oh man I love this song. Yes Oh my god. That's the best I've done. Yes. Let's go. Oh my god And then something really weird Wasp wasp wasp the game crashed. Do I have the dlc? Do I have the dlc? How do you know? Let me check on the actual steam store because that makes it so much easier. I don't think I do Oh, that's weird. That's really weird. All of a sudden I have it What Could have possibly happened. Oh, no Did my game just crash? I think my game just crashed.  There's no update or download happening right now. Absolutely not Uh, uh, uh Very suspicious right now. I mean no nothing suspicious happening right right now whatsoever. Oh the doubt. I mean the The the game has almost finished being broken Oh, what's that? My game is ready to play. Holy 244 song list. Where does it even say that? How did you even see that? Oh on the bottom left? I see so just down there a bit Ha ha. So after that weird moment of the game crashing I was having a blast and then someone challenged me to do a level six song I could currently not do that at my skill level, but I gave it a go You recommend 2d dream fever if I haven't played it yet. Alrighty, you've convinced me. Oh, yo, I'm happy That was the best one I've done so far Oh Oh Six difficulty you want me to die. Let's do it.  Oh my god Wow I thought I was actually gonna fail that was so difficult At like well at one point it got real difficult because I just couldn't see anything Like it was just too many flashing lights and I was like, oh god. I'm struggling to just see where all the Movement is okay. Wow. I have not failed one yet I did it. I actually did it. I have not failed yet I somehow did a six star level and it was actually really fun I am so happy with myself and so happy with how enjoyable this game is But the thing that made me super happy What's this? Is there a way that I could send you a fan art you just did for me? Oh, yeah Uh, uh, yeah, I have a uh, there we go Oh, this sounds cool This sounds really really cool No, I was so close to 100.  You posted it.  Wait, I need to share this on stream. How do I share this on stream? Can you guys see that? Isn't that awesome? I love that so much. Wow I want to I want to I want that That is actually so goddamn good. Thank you so much for that. Wow. I am. Wow, that's awesome I need to put the glasses on now There we go Wow How do how do people have turn you're happy that I like it? It's awesome. What do you mean? Wow, I I am amazed by that artwork. Thank you so much. Oh, and I poor god. I pause chat. Whoops for me PFP material right there. Yeah, could I actually make that like a Dude, dude, I don't know. Oh, dude. They'll be or oh, that's awesome. Cut. Wow You started doing it halfway through the stream. So it was a quick drawing What? What that was quick Dude, I can't even do a stick figure You draw fast dude. You draw like the speed of light I actually shown everyone in my life that bit of fan art. It was just so awesome I love it so much But then my winning spree continued.  We're doing the song Okay Oh, oh, oh god. They kind of confused me What has happened? There's so Oh my god. Oh my god No, no, no No, no, no, I'm not paying I failed Oh, there's two more. No, I failed. Oh my god. I failed because of Nyan cat No, my first fail today.  It literally ruined me because I keep failing afterwards Okay, we'll do Pinocchio P after we have to do Pinocchio P Oh Oh god, I'm oh man.  I can't believe Nyan cat is the downfall of me. I can't believe it Pinocchio P never fails to serve. You ain't wrong. Oh my god torturing me It was too much. It was too much. It was too much But I was a dummy. I somehow put the game into hard mode without realizing It was 6.5 Oh, I did hard mode. I was on hard mode I was on hard mode. That'll be why I was struggling Fuck, okay, let's do it on normal mode this time. Look, I'm just um, you have to deal with that And of course I have to as well, but after that silly move I became a gamer Oh, yeah, wait, does that mean I did Nyan cat in hard mode as well? I want to die. That wasn't intentional No, do you think I wanted to This is so much easier Yeah, okay. I need to get good with like the um, like the the sprint ones, you know Ah, I get to listen to the whole song Oh god, that was pretty bad. That was a bad ending. Oh, there we go Wow Okay, oh, yeah. Yeah. Okay. That's a vibe. Why are there any costumes? No 100 let's go 100 100 100 Oh, oh Dude this song is a vibe Oh, yeah, I need to end the stream after the song because I have work At least we're ending on a banger. How yeah Banger I did all those wrong. Oh my god. Oh, that was a pretty bad challenge Dude, that was a good song and I love the music video Oh, damn it. I don't want to end the stream again. I don't want to go to work And yeah, that is all for this video. I hope you guys enjoyed. I really love this game I'm gonna be streaming it a lot and I cannot wait to play project sekai in the future But until then I'm gonna be playing a lot of project TV over on my stream twitch.tv I'm ljl. There is a link in the description down below. It was so much fun I I I think that's a game that I'm gonna be playing a lot I wanted to play it during my own time, but I thought maybe Maybe I should just you know play the first few times on stream and then you know do some practicing So I'm I get better. Okay, it's actually running. So I need to go inside. Um, but thank you all for oh my god No, no, no each raining. Okay. Thank you. Goodbye


In this study, researchers developed a new method to improve the performance of quantum-dot light emitting diodes, cleds, which use quantum dots made up of multiple layers of materials. The researchers used a technique called bromide decoration to reduce the number of defects in the quantum dot surfaces, which improved their ability to emit light. This resulted in a significant increase in the quantum dots photoluminescence quantum yield, PLQI, the amount of light produced per unit of energy input, from 39.7% to 86.2%, as well as a substantial boost in the device's external quantum efficiency, EQE, from 0.74% to 5.46%. Additionally, the turn-on voltage of the device was reduced from 6.7V to 5.9V. These results demonstrate the effectiveness of the bromide decoration technique in improving the performance of cleds. This article was authored by Zhishuai Xing, Zhenwei Ren, Wen Linxia, and others. We are article.tv, links in the description below.


Elemental looked really bad when I saw the trailers. Looked terrible, atrocious, and Pixar lately hasn't been impressing me that much. I remember Buzz Lightyear, that movie was atrocious, could have been so, should have been so much better. This is Pixar for crying out loud. Anyway, that out of the way, I really liked Elemental. Let's talk about it. Here's the deal. The reason I didn't think Elemental looked good based on the trailer was because it presented itself as a very paint-by-numbers approach. A Romeo and Juliet love affair that we've heard a thousand times before. The girl and the boy can't be together. They come from different walks of life. She's fire. He's water. Those elements don't mix! But this movie mixes these elements so seamlessly. And the story is so cleverly done that it pulls you into the somewhat generic plot. Now, there is another plot going on here that isn't really prevalent in the trailer, and that's the storyline between the daughter and the father. And it's also one that we don't get a lot of in movies anymore, so it was kind of a breath of fresh air. Ember's the main protagonist here. She's voiced by Leah Lewis. Fantastic voice performance. Really dug this character. She works for her dad at his shop. Her dad an immigrant in this elemental world where fire, and water, and plant, and all the different clouds. That might be all of them. They all come together and live in perfect harmony. Kind of. They're also kind of segregated.  It's all gonna go up in flames. We can't have that. Ember doesn't want to take over her dad's job, but that's exactly what's gonna happen. That's the roadmap that's been laid out in front of her. Her dad is her best buddy. They get along perfectly. How does she disappoint her old man? Well, she'll learn, of course, as this film progresses, how to communicate better. Because she has a hot temper, because she's made of fire. It makes sense. It just makes sense. And a real way to keep down the flame is to introduce a little water, and that little water is Wade Ripple. He's the doofus, fumbly, whiny baby man that she's gonna fall in love with. Now, I will say, upon first meeting Wade, I wasn't impressed. I'm like, this guy's crying a lot. Like, he's cried more in this first five minutes of his introduction than I've cried in my life. Because I'm a man, and men don't cry! As the movie progresses with him, though, you just realize that water people are that way. They express themselves in a much more vivid, open matter, as opposed to fire, who lets things bottle up. Before they explode, you're understanding the puns, you're understanding all the metaphors and the symbolism. It's all there. It's all right in your face. Like I said, the story works very well. Ember has to balance her home life with this new love life that she's got going on. And of course, there's gonna be conflict.  But it looks amazing. Trailers, again, didn't really do much for me from a visual standpoint, either. It wasn't until I got into the theater and really took in all the magic going on that I thought, oh my god, there's so much detail in this film. And the way the elements are presented in different art styles, yet it all kind of seamlessly blends together, it's honestly wonderful. Now, I don't think this is top tier Pixar because we do have a pretty formulaic plot. I do think it is Pixar, though. This is one of the gems. This is one of those stories that you get from them where they take characters and individually they grow and they help you understand a thing or two as well. The movie has terrific pacing too. It's a rom-com. Lot of comedy, some good action. Of course, there's heart. You gotta have heart. It's Pixar. That's what they do, baby. There was one negative, a big negative for me at least, which was the opening 10 to 15 minutes short. Remember those? We haven't seen one in a while because a lot of the Pixar movies lately have been going straight to Disney+, which is a huge mistake, I think. Especially in the case of Luca, that movie was fantastic. Anyway, Up is the little short ahead of time. We follow the old dude and the dog-dog. I forgot the old man's name. Doesn't matter. This thing sucked. I couldn't wait for it to be over and it was a really bad introduction to Elemental. I hope that Elemental isn't sharing the same quality that this thing is because it's got fart jokes, it's got slapstick humor. It just doesn't work. And fun fact, it's been a while, I think, since parents have taken their kids to a film that has a little teaser short thing ahead of time. They didn't know what was going on.  Is this Elemental? They had no idea because Disney just throws it up there. They don't say, you know, the movie will show after this little short presentation. No, it just goes right into this thing. You get the castle logo and then boom, you're in Upworld. And the mom turns to her and says, I think it's just a commercial. After 10 minutes, the son turns to his mom and says, this is a long commercial. And I just sat there like, these people don't know anything about movies. Mmm, my own farts. Mmm. Disney has two animated films right now in the theaters. The Little Mermaid Elemental. If you're on the fence about which one to go to, which one to bring your bratty kids to, Elemental is the easy pick. Support new creative films. Even if that new creative film is kind of old hat as far as the story arcs go, it's still doing something different. Little Mermaid's a pathetic, crappy carbon copy of something much better that came before. Shove that junk out. Go to Elemental. It's worth your time for sure. It's an easy watch for kids, too. The kids in the theater were hooked. They were all over it, even the one that was running up and down the aisle dancing. I see you, Betsy. Sit down. No one cares. You're not that impressive.  If you don't have children, is it worth going out and seeing? No, probably not. Unless you go to animated movies all the time that are aimed more at families with kids, then yeah, it is.  I like going to this stuff. I'm a big Disney fan when it comes to their original content, especially Pixar. I would have saw it if I didn't have kids. If you're like that, you'll like that. And there you have it. That's Elemental, a movie that surprised me. It was better, way better than I thought. Again, not the greatest Pixar movie, not even in their top 10, but even a number 11 or a number 15 on the Pixar list is really damn good. Let me know in the comments if you saw Elemental or if you're planning on seeing it. Like this video if you kind of enjoyed yourself. You don't have to like the video even, but please hit the like because that helps me. Subscribe to the channel if you like what I'm doing here. I post tons of movie content each and every week. Would love to have you stick around. And after all that, I'll simply say, see you later because of the water. It's an element. Looking up. Thanks for watching.