So now we can start. So we were talking about the symmetric group. Oops. We had introduced the cycle cycles in the symmetric group. So this was just to have an element, say, A1 to AR, where this is just a set of elements of set of numbers from 1 to n, such that sigma of ai will be equal to ai plus 1. And if i is smaller than r, and it's equal to A1 for i equal to r. So that really means the permutation permutes these elements cyclically. And obviously sigma of n is equal to n of k is equal to k if k is not an element in the cycle. We had then seen that every element in Sn has a cycle decomposition. So we had this theorem. Every sigma in Sn is the product. So sigma is equal to sigma 1 times sigma s of disjoint cycles. So disjoint cycles just means that the supports are disjoint. So the sets of elements which are actually permuted by them. And we also had seen that this product is essentially unique. And it's called the cycle decomposition of sigma. And we want to see that if we want to now use this to describe the conjugacy classes of Sn. So the claim is that what we want to show is that there's a one-one correspondence between the conjugacy classes for elements in Sn. So a conjugacy class is just an equivalence class under the operation of being conjugated. So where a is equivalent to g, a, g to the minus 1, or this case in Sn. So we want to have one-one correspondence between the conjugacy class in Sn and the partitions of the number n. So I have to remind you what the partition is. So the partition of the number n is a tuple, say n1 to say nr of positive integers. So that means that ni is bigger than 0, which are somehow ordered. So with n1 is bigger equal to n2, bigger equal and so on, bigger equal to nr, and the sum of the ni is equal to n. So for instance, the partitions of 4 are you have 4, 3, 1, 2, 2, 1, 1, and 1, 1, 1, 1. You can easily see that these are all. So in this case, there are five different partitions of 4. And now we want to associate to every conjugacy class of elements in Sn the partition and want to see that this is the one-one correspondence. So let sigma in Sn be a permutation. And let sigma equal to sigma 1 times sigma r be its cycle decomposition. So the cycle decomposition really means that these are. That sigma is a product of these disjoint cycles and that the union of the supports is the whole of the set 1n. So every integer from 1 to n occurs in one of these cycles. Some of these cycles might be one cycle, which then means they are the n-cells identity, but that's OK. So we can look at the so let l of sigma i be the length of the cycle sigma i. So that means sigma i is a l of sigma i cycle. So for instance, this thing would be an r cycle. The length of this thing is r. So obviously, this composition is well defined up to reordering them because it's the sigma i commute. So by reordering, we can assume that the length of sigma 1 is bigger equal to that of sigma 2 and so on. We can just decide in which order we want to enumerate them. And then the tuple, say, l of sigma 1, l of sigma 2, and so on, l of sigma r is called the cycle type of sigma. Now I claim that this thing is a partition of the number n. So clearly, these numbers, l of sigma i, are positive integers because they are the length of the cycle of the cycle. So it's some positive number. So the l of sigma i are positive integers. And we have ordered them in such a way that they are descending in this way. And we have also clearly that l of sigma 1 is bigger equal to l of sigma r by our definition. So the only thing we have to see is that the sum of them is n. But we have that sigma is equal to sigma 1 times sigma r is a product of disjoint cycles. So every integer from 1 to n occurs only in one of them. And the support, the union of the supports of the sigma i is the set 1n. This is what we wanted for a cycle decomposition. So that means that if we take all the numbers, so every number from 1 to n occurs precisely once among all the numbers occurring in the cycles. So that means the length of the cycle of a cycle is how many numbers occur in it. So every number from 1 to n occurs precisely in one cycle. So the sum of the length of the cycle is the total number of numbers that occur in all the cycles. So this number is n. So the sum i equals 1 to r, l of sigma i is equal to n. So this is a partition of n. OK. And now we want to see. So we can look at maybe some example to make sure we understand what we're doing here. So if we, for instance, take, again, the element sigma to be the element in S6. So 1, 2, 3, 4, 5, 6. And then what is it here? 6, 2, 5, 3, 4, 1. So then we can make the cycle decomposition. So 1 goes to 6, and 6 goes to 1, and 2 goes to 2, and 3 goes to 5, 5 goes to 4, and 4 goes to 3. So this is the cycle decomposition. So the lengths of the cycle are 3, 2, and 1. So the cycle type is this partition 3, 2, 1 of the number 6. So now we want to see what the cycle type has to do with the conjugacy class. So the claim is that two partitions are conjugated to each other if and only if they have the same cycle type. Lemma, two permutations, sigma and tau in Sn, are conjugated if and only if sigma and tau have the same cycle type. OK. So we want to see that. So we have to see what does mean to be. So we take our element sigma, the element is Bs. So this is the cycle decomposition of sigma. So now we want to take an element conjugated to it and see that it has the same cycle type. So let, maybe I call this in pi. Let tau be an element in Sn. And so we want to conjugate sigma with tau to see what happens to the cycle type. So if we form tau, sigma, tau to the minus 1, still conjugated element, what does it do? So we want to find out what this is. So we have this cycle decomposition. So we write sigma i equal tau, so ai, a1i, until a, say, m of i, i. OK. So I think it has a double index, no? So the i-th cycle goes from a1 to ami, but it depends on this. And then we put, so for all i from 1 to r and j from 1 to m of i, we put bi, b, that was maybe, bjii to be tau of agi. And we put, say, bi equal to b1i until a, b, m of i, i. So we make a new cycle by replacing the ai's, the aij's, by the bij's. And now what do we have? So if we take tau, sigma, tau, so tau, say, sigma, tau to the minus 1, and we apply this to some element, bij, what is it? First we apply tau to the minus 1 to this. The tau to the minus 1 of this is aij, no? Because tau of ai of r, here it is, yeah. So it was the other way around, but anyway, ji. So the tau of aij of aji is bji. So the tau to the minus 1 of bji is aji. So this is tau of sigma of aij aji. Now we know what sigma does to this, because sigma is the product of this disjoint cycle, sigma i. So applied to this element, it's just sigma i applied to this, which sends this one to the next one in the same cycle. So this is tau of, so there are basically two possibilities. This is tau of aji plus 1i. If i is smaller than mi, and it is the first one, so a1i, if j is equal to mi, no? That's how it is.  This one to the next, and so on, the last one comes back. And this tau applied to this. But now tau of this is the corresponding thing with b. So this is bj plus 1, i, j is smaller than mi, and it's b1i if j is equal to mi. So that means this thing permutes just the bijs, the bjis in the same way as here the ajis are promoted. So it follows that tau, sigma tau to the minus 1 is equal to, say, pi 1 times pi r, where pi i is equal to precisely this thing. We're going to write it on the next line, where, as I already wrote here, pi i is equal to b1i until bm i. So we see that what happens is actually very simple. If you have a cycle and you conjugate it with a certain element, then what happens is just that you get, again, a cycle of the same length where the elements in the cycle are replaced by the images under the element by which you conjugate. So in particular, we see that these have the same lengths as before. And so clearly, we have that the cycle type pi equal to pi 1 to pi r is equal to that of sigma. So we see that if two permutations are conjugated, they have the same cycle type. And now for the converse, it's actually not very much. And basically, we trace the steps. So assume, we basically just have to see that all the steps can be reversed. So assume that sigma and pi have the same type of type. So we can write sigma equal to sigma 1 times sigma r pi equal to pi 1 times pi r, where the length of the sigma i is equal to the length of the pi i for all i. So that means I can write sigma i equal to, say, a 1 i until a, say, mi i. And pi i, I can write as b 1 i until b mi i for just some elements a 1 i. So for this, just some elements in one n, and the same here. So as these have the same number of elements, and the union of all these, I mean, the union of the supports of them is up to the set 1n. We can find the bijection from 1n to itself by just sending each aji to the corresponding bji. So let tau from 1n to 1n. So I send aji, where in this decomposition, to the corresponding bji. We know that these are disjoint subsets of 1n, and the union is the whole of 1n. So this is a bijection of the set 1n to itself. And by what we have proven here, we know precisely, so by the above, we know if we take tau sigma tau to the minus 1, then this is equal to pi. Because we have precisely seen that this is, we are precisely back in the situation, and we know that if we conjugate in this way, we get that the cycle with the aji is replaced by the corresponding cycle with the bji. So it's anyway good to know that this conjugation has this very simple effect on the cycle. So as a corollary, we now can count the conjugacy classes of Sn. So the number of conjugacy classes of Sn is equal to the number of partitions of n. And basically, we have shown it proof. So we have seen that the map, which associates to a partition, so the map from the conjugacy classes in Sn to the partitions of the number n, which associates to a conjugacy class of an element sigma, the cycle type of sigma. So we have seen that, so what have we seen? We have seen that the cycle type of a partition depends only on the conjugacy class. So that means this map is well defined as a map from conjugacy classes to partitions. And we have seen that, in fact, two elements are conjugated to each other, so in the same conjugacy class, if and only if the partition is the same. So that means this map is well defined and injective. So we have to see it's surjective. And that is very simple. We just have, for every partition, we have to write down an element in the symmetric group, which has that cycle type, but that's kind of trivial. No, we just have to. So for instance, so let's say if p equal to n1 to nR is a partition of n, then we can take sigma. So we just have to have something where the lengths of the cycle are given by this. So it means the sum of d and i is equal to n. And we just take, so say, 1 until n1 as the first cycle. Then we take n1 plus 1 until n1 plus n2. So the length of this will be n2. And you can kind of see how it goes on. In the end, we have n1 plus nR minus 1 plus 1 until the sum of all d and i, which is equal to n. So this is a cycle decomposition of some permutation. Because these are just disjoint cycles. And we see that the length of the corresponding cycles is here n1, here n2, and so on. Finally, here it's nR. So this has cycle type. So we have this projection. So this was as much as I wanted to say about this cycle type and the conjugacy classes in SN. Now finally, to finish with the symmetric group, I want to talk about the sign of a permutation. So this is something that you might be even familiar with from linear algebra. Sometimes one of the definitions of the determinant is that you take the product over all ways how you can take one element out of every row and column. And you multiply by the sign of the permutation and you sum them all up. So it already occurs there. Now I want to introduce it here. So there's one n invariant. So a number you can associate to a partition of a partition is its sign. And in fact, the way how one usually understands it is it is minus 1 to the m, where m is equal to the number of transpositions, which one uses in, so the partition is called sigma, used to write sigma. So if I write sigma as a product of transpositions, there will be many different ways how to do it. But the claim is that whenever you write it, the number of transpositions you need is either even or odd. And this does not depend on how you write it. And minus 1 to the number of transpositions you need is the sign of the partition. Now, so this is what it is for some reason. It's not now the easiest to work with for me at this moment. So I give another definition, and then I show it's equal to this one. Because obviously here the problem is that whether this definition is well-defined or not. You could expect that sometimes you need an even number, sometimes a not number. But this is not the case.  So let n be equal to 2 be some positive integer. And so for sigma in Sn, the sign of sigma is, so it's called, I denote by epsilon of sigma, and I write down some crazy formula. So this is the sum over all i, j, sigma of i, minus sigma of j divided by i minus j. So these are all integers. So you can certainly do this. And the sum means it's the sum is over all pairs of integers i comma j with i and j, just integers from 1 to n, and i bigger than j. So I take this sum. So if you look at it in the first moment, you look at it, you might even wonder whether this is a sign. This is supposed to be plus or minus 1. So in the first moment, you might even not be sure whether this is plus or minus 1. It looks like a rational number. But we'll see the moment is not the case. So now let's study it. So the first claim is that epsilon of sigma is equal to minus 1 to the m, where m is not precisely this, but something similar, where with m equal to the number of pairs. Let me write it equal to the number of pairs j with, so again, ij pairs in 1n with i bigger than j and sigma of i smaller than sigma of j. So this is the number of elements where kind of the order of the elements is reversed by the permutation. This what? Well, you're right. This would not make any, I think I certainly want the product there. What? What? OK, let me see. No. Yeah, if I sum, I get 0. So if the product is OK, this term will never be 0, because i is different from j. And this is the permutation, so they always are different. So this was just a misprint, which, however, I did not. OK, this is certainly the product. I think now in the proof, it also is everywhere the product. And then the second statement is that this epsilon, so the map which associates to permutation its sign, is actually a group homomorphism. So secondly, the map sigma epsilon from SN to the set consisting of the elements 1 and minus 1 together with the product. So this is a subgroup of the, whatever, the rational numbers with multiplication.  So this map is a group homomorphism, so where these are really the integers 1 and minus 1 with the multiplication. Well, it's obviously subjective. But I mean, that's if n is at least 2. But I mean, that's kind of trivial. We'll see that in a, but it just means if you have a, for instance, if you just have a trans, according to this story, if you just replace two things, if you place 1 and minus, if you have two elements, say, you always have the transposition 1, 2, and this transposition will have, according to the part 1, the sine minus 1. And obviously, the identity, according to the same story, has always the sine plus 1. So it's a trivially subjective. So this is a group homomorphism. And third is that the statement that I used as a definition. So if sigma equal to tau 1 times tau k is a decomposition as a product of transpositions with tau i transpositions, then it follows that epsilon of sigma is equal to minus 1 to the k. So which was what I first said would be the definition. So let's prove it.  So let's see. Let's just write down, as before, product i bigger than 0 sigma of i minus sigma of j. So we are afterwards supposed to divide it by this. So we want to compare it with this. So let's look at what this is. Well, clearly, I can write this as a product of all i bigger than j, where sigma i is also bigger than sigma of j of the same thing times the product where now sigma i is smaller than sigma of j, sigma i minus sigma j. That's not very difficult. But now, if you exchange the role, if you here take the product the other way around, you get each time a sine minus 1. So this is equal, if I want product i bigger than j, sigma i bigger than j. OK. So anyway, so I can replace this by sigma j minus sigma i. And I get the sine minus. So if I do this here, I get minus 1 to the number of elements here. So I should maybe say that m was the number of pairs. So I claim this is minus 1 to the m times the product over all sigma i bigger than sigma j of sigma i minus sigma j. Because what do we have? So this is the product over all. So what I mean is that if we get this, I think I'm confused. So if i is bigger, so here we have this. In the second product, we can exchange the role of i and j. If we exchange the role of i and j, then we get now that j is bigger than i is smaller than j. Sigma i becomes bigger than sigma of j. And now we have exchanged i and j, so we get the sine minus. And so we get minus 1, if we do this precisely, m times, where m is the number of elements which occur here. m is the number of pairs ij, where i is bigger than j, and sigma of i is smaller than sigma of j. So these are precisely those which occur in the second product. They are precisely m factors. Is it clear, or should I start again saying it?  Should I explain it again? Or is it clear? What? It's fine. I hope it's fine for everybody. Anyway, so we have this. So this will be this. And now, so now what do we have here? What? What? Yeah. OK, maybe I write a little bit more. I write back what I had before. Sigma of i minus sigma of j. So first, I should notice that these are precisely m factors. Because m was precisely the number of pairs ij, such that i is bigger than j, and sigma of i is smaller than sigma of j. These are precisely those which are here. And now, in this last product, we want to exchange i and j, so i with j. So what does it mean? It means now that first, we get that now i is smaller than j, and sigma of i is bigger than sigma of j. So just the last factor looks like this. And here, we have sigma of i minus sigma of j, which precisely changes sign. So we still should write sigma of i minus sigma of j, which is the same as minus m. Because now this i, so we have exchanged the role of i and j. So now what was i before is now j and vice versa. So i is smaller than j, and we have this. So we have precisely m factors. So this part is minus 1 to the m times product i smaller j sigma of i bigger sigma of j sigma of i minus sigma of j. But now, if we put this together, here we have the product, so I've only dealt with the second factor. So this is the product of the i smaller than j sigma of i bigger than j, and here we have the same product with i smaller than j, but the sigma i's are the same. So the condition is now only on the sigma i. So if I now, this was a computation in between, but if we go on here, this is equal to product over all ij such that sigma of i is bigger than sigma of j sigma of i minus sigma of j. So this product means it's the product over all pairs ij such that sigma of i is bigger than sigma of j, because there's no condition on whether i is bigger than j or smaller, because we have used both. But now, sigma is a permutation of the set 1n. So here, if we apply i by sigma of i, we have just reordered the factors. So this is the same. So reordering the same as the product over all i bigger than j i minus j, because I just replace sigma of i by i. If I do this for all i, I get the same factors only in a different order. And so this means if I take the quotient, except that obviously I made one mistake, because I had this factor of minus 1 to the m here, which I dropped. And so the definition was that I'm supposed to take this product and divide it by the product of the i minus j. And what remains is just minus 1 to the m. Maybe you can go through it again also, but OK. So this is 1, then 2. So I want to see it's a group homomorphism. So it's somehow always the same trick, so which takes some time getting used to it, so maybe you can recognize it the second time. So we want to show it's a group homomorphism. So if I take a product of two permutations, then I should get also the product of these signs. So if I take sigma and tau elements in a symmetric group, then we have the product. So we want to compare epsilon of which one we want to tau, sigma tau. We want to compare this to epsilon of sigma times epsilon of tau. In fact, we want to show they're equal. So let's try to compute this. So according to this definition we have given, this is the product over all i bigger than j sigma of tau of i minus sigma of tau of j divided by i minus j. There's nothing wrong with that. So this is just a definition. Now we can certainly multiply this as one often do in analysis. Multiply this with one in a slightly complicated way. So I hope you remember the statement. So we just multiply this with one. So this is product i bigger than j sigma of tau of i minus sigma of tau of j divided by i minus j. So it was by i minus j instead. We multiply it by tau of i minus tau of j. And we multiply by tau of i minus tau of j divided by i minus j. If I want, I can also write product i bigger than j. So we have this. Now this is just we've just multiplied. We have multiplied by this and we have divided by the same. So there's nothing happening. And now let's look at this factor. So I claim, first I write it down and then I give it the reason this is the same as a product over all tau of i bigger tau of j. So it means again, the product over all pairs ij of different elements in 1n such that tau of i is bigger than tau of j of sigma of tau of i minus sigma of tau of j divided by tau of i minus tau of j. Why is that? So if i is smaller than j and tau of i, if i is bigger than j and tau of i is bigger than tau of j, then nothing changes. If i is bigger than j and tau of i is smaller than tau of j, we can replace exchange the order of i and j. So we replace i by j. So then i is smaller than j and tau of i. So that means, so in this, to compare these two. So if i is bigger than j, tau of i and tau of i is bigger than tau of j, we just take the factor here. You'll get the same factor here. Don't change anything. Don't change the factor. And if i is bigger than j and tau of i is smaller than tau of j, we replace i by j and j by i. Then we have now that i is smaller than j. So then the new i is smaller than the new j. And the tau of i will now again be bigger than tau of j. But what is the effect of this? Replace i by j. So now the denominator changes sign because we exchange these two. And the numerator exchanges sign because we flip these two. So then both numerator and denominator change sign. So we get this whole product, we get this product from this product by exchanging i and j in all the factors where this holds. And when we do it, each time we exchange them, both the numerator and the denominator change sign. So the whole thing doesn't change sign. And so they're equal. And then we have the same argument as we had to finish the argument here. If we take the product over all pairs with tau of i bigger than tau of j, tau is a bijection from the set 1n to itself. So we can apply this bijection. So it means that if we replace tau of i by i, we have just permuted the factors. So tau is a bijection. So for each value tau of i, for each tau of i, tau of j, we get the factor. And if we replace tau of i by i for all i, we get also the fact that we get the same factors because tau is just a permutation of 1n. So there's a bijection between the factors in the same way as we did it here. So it means this is equal to the set of all i bigger than j sigma of i minus sigma of j divided by i minus j. So let me say it once more again. I just say that these factors here are just the permutation of these factors. They are permuted by the permutation tau. And then obviously these two things together say that this product, so this together says that epsilon of sigma tau is equal to epsilon of sigma times epsilon of tau. Because here, this is epsilon of tau. And this one we find this epsilon of sigma. So this is part 2. And then the third statement was this one. So if we write sigma as product of transpositions, then epsilon of sigma is minus 1 to the number of transpositions we use. And for this, we only have to see that for transposition that this number is 1. So it is minus 1. So for 3 to prove 3, it is enough to show that epsilon of tau is equal to minus 1 for any transposition. Because as this is a s by 2, this epsilon is a homomorphism. So if I take epsilon of sigma, this is epsilon of sigma 1 times epsilon of tau 1 times epsilon of tau k, each time we get minus 1. So we get minus 1 to the k. OK, so we just see it. So for instance, if tau is equal to 1, 2, so the transposition which exchanges 1 and 2, then we just see, then obviously, epsilon of tau is equal to minus 1. Because there's precisely, if you look at this thing, then all stay the same. So if i bigger than j, then also tau of i is bigger to tau of j for all i, except for 1 and 2. And 1 and 2 are exchanged. So you get 1 factor minus 1, and all the other factors are 1. So 2, 1 is the only pair of i, j with i bigger than j and tau of i smaller than tau of j. And so this gives us minus 1. And so now, on the other hand, we know by the above that if you have a transposition, so now if tau is equal to i, j is a transposition, we know it's conjugated to this one. In fact, if sigma is a permutation in 1n with sigma of 1 is equal to i and sigma of 2 is equal to j, then we know, we have seen this, that if I take tau 1, 2, tau to the minus 1, then this is equal to i, j. So if we apply, again, our epsilon to it, this is a group homomorphism. So it follows that epsilon of, say, this permutation i, j. So epsilon tau, epsilon of 1, 2, epsilon of tau to the minus 1 is equal to epsilon of i, j. So this is some number, plus or minus 1. This is minus 1. This minus 1 times epsilon of tau times epsilon of tau to the minus 1, which is epsilon of tau to the minus 1. So now we are just in the group of 1 minus 1 with multiplication. So this commutative, so we can cancel these two and we get just minus 1. So we see that for any transposition, we get minus 1, and this proves the result. So finally, one can use this to introduce the alternating group. So the permutation sigma in Sn is called even if its sign is 1. So it means it can be written by an even number of transpositions. So the set of even permutations is called the alternating group in n-dettas and denoted En. And note that this is a normal subgroup of Sn. So An is a normal subgroup because it's the kernel of a group homomorphism with An is equal to the kernel of the map epsilon from Sn to 1, set 1 minus 1 with the multiplication. The kernel are those which map tau, the neutral n. OK. So one of the things which, so this group occurs sometimes, one of the things it occurs in is also in Gallo theory, I don't know whether we will mention that later. But the fact whether you can solve the polynomial equation by taking roots and so on, it depends on whether a certain group is solvable, which has something to do with having enough normal subgroups. And it turns out that the fact that you cannot solve all polynomials of degree 5 in terms of taking roots and roots of something is connected to the fact that this group An is simple. That means it has no normal subgroups, but we will see. That's actually not so, is 4n equal 5. So it's just a statement that, for instance, A5 is a simple group. I was thinking of making it an exercise, but it's somewhat difficult. And so maybe I don't. What? I still don't understand you. So what is not commutative? Well, I will have to think about the statement. So you say that, yeah, you could say like that. Yes, if you want. OK, yeah, there are certainly, in principle, well, I don't know precisely what that means now. So because if you, for instance, take the fact is, if you take S3, for instance, as a simplest example. So the, obviously, the even permutations are actually a cyclic group. So they're also commutative. The fact is only that the S3 is actually to isomorphic to the dihedral group in three letters. So it is somehow, it comes from two cyclic groups, but altogether it's not commutative. So it's, I don't know how this answers your question, but you can see it somehow. The way how the two come together is somehow, I mean, sometimes, so in this case, you see that the even permutations are actually commutative. And the quotient is also commutative, but the whole thing isn't. OK, so I took a bit longer than I expect. So now we will want to start to prepare for, in some sense, for the statement and proof of the Silov theorems, and we do this by studying operations of groups on subsets. So if we see operations on subsets. So if we have a, so see if G acts on a set X, on a set S, it will also act on the set of all subsets of S. We want to a little bit investigate this, because we will use it when we try to prove the Silov theorems. And anyway, it's interesting to know. So let's state this. So let G act on a set X, and let U be a subset of S. Then for an element G in G, we can form GU to be the set of all G. Maybe I write time because in operation G times U. So it's not U doesn't have to be G. It's not automatically coset G times U, the set of all G times U, where U is an element in U. So we can associate, we can make an element G in G act on a subset of S, but just taking this set we obtain by applying G to every element of U. So it's easy to see that the map G comma U, maps to G times U, defines an operation of G on the set of subsets of S, of G, on the set of subsets, because you know what you have to see. If you take the identity, it maps every U to itself. If you take the identity element in G, it maps every one to itself. And if you take the product of two elements, it's the same as if you apply first one element and then the other. It's kind of trivial. We can also see we can restrict this action to subsets of S with a fixed number of elements. So I claim if we just take the set of subsets of S, which have five elements, then this operation defines an operation on that set. And this is because as we have learned, the set, the multiplication, the operation, is always by bijection. G times from S to S, which sends an element S to G times S, is a bijection. It follows from the fact that you have an operation that each element operates by bijection. And so if you have a subset with a given number of elements applying G to it, the image will have the same number of elements. And now we want to define the usual things that we have for operations. So we can look at the stabilizer, GU. So if U is a subset of S, no. The stabilizer, GU, is the usual thing. It's the set GU, which is the set of all G and G, with respect to this operation, act as the identity. But what does it mean? It means such that G times U is equal to U. That's the stabilizer. So it means that the set U is mapped to itself. As usual, this is a subgroup of U. And maybe there might be some good. Yeah, well, it couldn't be one of U. So just as a warning, the stabilizer GU of U in this thing is not the set of all elements G and G, which fix all elements of U. So it is not. So GU is not the set of all G in G, such that GU is equal to U for all U in U, which would be the stabilizer of every element. It just stabilizes the set. So each element of the set is sent to some element of the set. But rather, the set of all G in G, such that GU times U, is an element of U for all U in U. Is that a problem? No. OK. Anyway, it's maybe obvious, but I just wanted to indicate. Well, I make one trivial proposition. It's a small remark. So let G act on a set S and let U in S be a subset. Then I'm asked myself, when the stabilizer of U is the whole of G, so then G is equal to GU, if and only if, U is a union of G orbits. OK. So we can see that's quite obviously, and that's basically obvious if you think of what it means. So G, by definition, G is equal to GU, if and only if G times U is an element of U for all U in U. That's essentially the definition. And for all G in G. So the set of all GU's, G times U's, for all G in G is the orbit of U. So that means it's the same as the set of all that GU is contained in U for all U in U. And the statement that, for each element, its orbit is contained in U is the same as saying that U, so for every element, its orbit is contained in U, is the same as saying that U is the union of orbits, namely precisely, the G of U for elements U and U. OK, maybe it's not a good moment to start something new now, so maybe I will stop now.


What's going on everyone? Sam here with late in sports cards, we've been 22. I'm going to need Prism Football Asia Team all 10 box break number one, random team style. So look again at your random results. As always, you can find those on the website under the results tab as well as the link in chat. Good luck guys. Here we go. One, a couple of silvers, potentially autos in here too. Very rare but they do come in every now and then. Rod Jeremy Rucker, James Cook. First one's going to be Laramie Tunsel on the choice exclusive to 44. Nice one there for the Texans and Silver Dax Hill. You get one Silver Rookie on the back each time as well. Rookies I believe are numbered to eight I want to say in this as well unless they switched it this year. Forty, Louisine, Romeo Dobbs, Tony Pollard. Nice one there for the Cowboys, two of 44. And a Silver, Caban Tibido. Yes, number to 44 or eight. It is confirmed on the box that it is eight. Check out. Nice one there Rookie for the Chiefs. Give me AJ Tyrell. Nice one there, nine of 44 for the Falcons. And Silver Malik Willis, some Titans. Sam Howell Rookie. There's Daniel Bellinger. We've got Derek Henry. Nice, 13 of 44. Grats Titans on that one. King Henry. Silver or David Bell. Still dealing dude. Kenneth Walker, nice canine. Got Jermaine Johnson. Going to be Trayvon Diggs. Five of 44, two off Jersey number for the Cowboys again. Cowboys doing well here. Very nice. Silver or Wandaille Robinson. Drop you guys a link again to Cosmic. Again just nine to go. 140 a spot. Get your random team in a half case. Let's get that filled up everyone. Again, last group break of the night right there. Only break with spots. Cosmic Chrome. Half a case. Got two half cases select coming up after. That's going to be fun. Nice. Silver Rookie Auto. Congrats on that Bucks. There you go. Tampa, Paul. There you go, buddy. Nice one there Paul B. Devin Lloyd, Jalen Tolbert. Justin Herbert. Very nice. 15 of 44. And how about a Silver Purdy on the pack? Hell of a box right there. Grats everybody. Got something there. Hell of a box. There in Mathis, Ty Chandler. And it's Micah Parsons again for the Cowboys. 7 of 44. There you go. Dallas doing well. Silver, Jamo. Jamo Draftkings. Silver Rookie. Jamison Williams for the Lions. Very nice. Once he's back, I'm excited for him.  Jake Ferguson. It's Mere White. There's a Jalen Naylor. That is going to be one of eight. Very nice. Congrats, Minnesota. Picking up the first Rookie Choice Exclusive. Minnesota is going to be Jonathan B. There you go, Jonathan. Silver Traylon Burks. See one more Rookie out of these last few boxes. Come on, baby. Let's see, let's see it. And some good ol' T-Mall. Tariq Wollin Rookie. Carson Strong. That's going to be a Nico Collins second year. 21 of 44. Going out to the Texans. And a Silver Clial Shakir for the Bills. David and Jamo. Rookie Auto for the Ravens. Two autos coming out of this. Very nice. Congrats, Baltimore. Going out to Nick H. There you go, Nick. Again, going to be a stud on the edge for the Ravens. For my Ravens, they share, hopefully. Three solid Rookie. Cole Comet, 244. 9 of 44 for the Bears. And Troy Anderson, Silver Rookie for the Falcons. Still a quick recap of our colors and our autos. We had Cole Comet, Nico Collins. It's a Jalen Nailer to eight. Again, that's going to be one of eight on the Rookie. All the rest should be to 44. Micah Parsons, Justin Herbert, Traymon Diggs, Derek Henry, AJ Turrell, Tony Pollard, Laramie Tunsel. Autos, we had David O'Jabbo, and K.Dotton. It's going to do it for the break, guys. Thanks again for the fill. Next up, we got Select. Let's go.


The last time I did Himalayas in a motorcycle was 81, I think, or 81 or 80, I'm not so sure. So, forty or forty-one years later, I'm back on the motorcycle in Himalayas. And well, at that time motorcycle was weak, a 250cc Czechoslovakian bike. It was wonderful for those days, but it's not the same as what we're riding today. Today we're riding 1100cc super-competent motorcycle. But at that time motorcycle was weak, but I was strong, young and strong. Today motorcycle is strong and I'm old, at least people think so.


what's going on everybody welcome back to another breaking news story and Naomi Osaka has pulled out of Wimbledon not going to England not playing on the grass courts in 2021 her team have come out with a statement let's go read it so this is the statement that Naomi Osaka's team has written on her behalf Naomi won't be playing Wimbledon this year she's taking some personal time with friends and family she will be ready for the Olympics and excited to play in front of her home fans so there it is Osaka pulling out of Wimbledon again having to get some more time to herself of course she said that she wasn't going to be coming back anytime soon in the statement she wrote after withdrawing from the franchise and she wasn't going to be seeing us in a couple weeks at Wimbledon and looking at the Olympics I mean that's a big big thing for her she's going to be in the spotlight she's going to have to prepare not only physically but mentally as well because she's going to get a lot of attention on her not only from the tennis public but also from the general sports world as well because the Olympics of course is not just the tennis thing it's an Olympic thing it's got a lot of different sports and being the home nation she is going to be one of the faces of Japan during the Olympics so a lot to deal with and we all know that she doesn't really like the attention she's going to get a hell of a lot of it in Japan but let me know down in the comments below do you think it's a smart decision from Osaka to pull out of Wimbledon she hasn't actually got that many points to defend at Wimbledon because in 2019 she lost really early on so it's not a big loss to her rankings why she won't drop down in the rankings very far if anything depends on the players around her so let me know down in the comments below do you think focusing on the Olympics is definitely the way to go for Naomi so Osaka she's pulled out of Wimbledon just like Rafa but she is going to be playing the Olympics


Hi everybody, let me begin this video with an apology. So I was supposed to give a talk I was scheduled to give a talk at the fall meeting of the division of nuclear physics for the American Physical Society I got an email on August 31st saying hey, congratulations. You're abstracts been accepted do these things to confirm your talk I didn't see the email. So the ultimate fault in all of this lays in me As a result of not seeing that email I didn't know that I'd been scheduled to speak on a Saturday on Halloween and I found out 30 minutes before my talk From a colleague of mine in the session. Hey, are you good to give your talk and of course I panicked So this is the talk that I would have given to my colleagues at that meeting Please don't take the fact that I didn't show up for my talk as a sign of disrespect for the nuclear physics community I actually because I'm so interested in particular in the electron-ion collider I obviously have a huge amount of respect for my colleagues in the nuclear physics community You can just chalk this up to me being really dumb at handling email and we can just move on from there, hopefully All right, so in that spirit what I'm gonna do now is I'm going to switch over This is the talk that I would have given at the DNP fall meeting I've timed it out for 10 minutes and I'm gonna start the clock in a moment And we'll go ahead and get started. So again, my apologies Let's take a look at some physics and see what things we might be able to do with the electron-ion collider in the future So in this talk, I will talk a little bit about charm jets as a probe for strangeness at the future electron-ion collider and As I like to begin all my talks I'd like to thank the organizers for this opportunity which I've squandered To talk a little bit more about some recent work that my colleagues and I have been doing looking at this very subject So with that in mind, let's go ahead and move to the outline I'm going to talk a little bit about our baseline electron-ion collider detector implementation We're using the Delfts fast simulation framework to do this work I'll talk a little bit about the physics process We're interested in studying and then the Experimental approach we're taking to isolating that process and then talk a little bit about how that can help us to probe The degree of strange-court contribution to the proton at a future electron-ion collider program And I'll offer some conclusions and outlook at the end So first let me talk a little bit about the baseline EIC detector implementation that we have made in the Delfts framework Now as you would know from the rest of this mini symposium here at the DNP fall meeting The electron-ion collider will be able to collide electrons and positrons on protons And of course heavy ions as well. We've selected a beam configuration 10 jev for the electrons and 275 jev for the protons which is believed now based on the hard work that's going on by the accelerator community in developing this Accelerator program to yield one of the highest instantaneous luminosities And so therefore we can rapidly integrate data and in one year we would expect to take about a hundred inverse femtobarns of data Okay, so for the students here. I've got a little explanation about what 100 inverse femtobarne means But basically if we have a cross-section process with a cross-section of 15 picobarne and we have a hundred inverse femtobarns of data We would expect about one and a half million events to be provided by the collider Now we have to detect them and reconstruct them and use them in analysis And what we're going to do is study intrinsic strange and charm in the proton in the nucleus I'm going to focus on strange in this talk to try to improve our understanding in particular of the strange part on distribution function And maybe ultimately we can get at things like the strangeness holicity contribution to the proton And this work is available in a paper that has been submitted to physical review D. So this is a baseline EIC detector It's got all the standard configurations, but because this is not a symmetric collider There's one side where the beam electron will tend to scatter and that's the electron end cap And then there's one side where the hadron remnants will tend to scatter from the ion or proton being interacted with That's the hadron end cap and then in between there's a barrel region that has to cap Capture the overlay of those sort of two halves of the events. So of course we expect to have a high Quality particle charge particle tracking covering a wide range of angles with sort of you know 85% to 98% efficiency depending on the last sort of 20 to 25 years of tracking detector development with about a 20 micron resolution In things like the impact parameters of those tracks An electromagnetic calorimeter covering a wide range of angle as well as important with good energy resolution and in this Work we would of course you can take a look at our paper But we basically advocate for a hadronic calorimeter as well with good granularity that covers the same region as the electromagnetic calorimeter and And gives us a good energy resolution so that we for can for instance do a reliable calibration of jets Now let's talk a little bit about the physics process of interest The physics process of interest is shown here in the left-hand side diagram This is the leading contribution of intrinsic strangeness in the proton where you get a strange anti-strange Quark that pops out of it into existence just as it's interacted With by a W boson emitted by the beam electron This then converts the flavor of the anti-strange quark to an anti charm quark and it with high enough energy We would expect then a jet of hadrons to be produced and for that jet to be consistent with having come from a heavy Long-lived hadron like those created by charm quarks now We've simulated this using pithia 8 and we find the cross-section predicted for a q squared or greater than 100 jev Squared to be about 15 picobarn. So that's the kind of hint process I hinted at earlier now the middle and right diagrams which involve intrinsic charm and also Gluon splitting to charm and these are entangled together inside the proton. They're very important These will also lead to the production of for instance charm jets But the the trick here is that the pithia 8 framework doesn't simulate these and we're ignoring them for now in this study They would lead to increased production of charm jets So by ignoring them we're effectively conservatively Underestimating our charm jet yield at the end of this analysis That said these diagrams would need to be disentangled from one another in any future EIC based analysis All right, so what do we find at the generator level? Well, this is a distribution of the As a function of transverse momentum the differential cross-section for producing the charm jets and all jets in red And as you can see here the charm jets are preferentially produced at fairly low transverse momentum That's where the largest relative rate of charm to all jets is actually located coming in at about three to four percent The charm jets are produced at relatively low angle to the the proton beam in this case They peak around 30 degrees, but they have a wide range when they make a jet We have jets of r equals one that we're using to do this study So we we use this to motivate not only good calorimeter coverage, but good tracking angular coverage We need to catch all the stuff inside of those large radius jets so that we can flavor tag Now speaking of flavor tagging identifying these charm jets is obviously a crucial ingredient in this work And you can see that illustrated here Delfts has a very nice event display built into it And this is in fact a charged current interaction producing a charm quark Through this flavor changing process We've got missing transverse energy of about 25 GeV from the conversion of the electron into a neutrino the jet PT is about 25 GeV so this is a rather high-energy charm jet and This is at pseudo rapidity of 1.6 and you can see here within the yellow jet Region that's been reconstructed. There are a couple of displaced vertices that would be consistent with coming from a long-lived charm Had drawn and that's what we want to get and we capture those kinds of jets with those displaced decays By using a very simplistic approach This is basically looking for displaced tracks from the interaction point and counting them All right, so we use the sign 3d impact parameter and we simply count the number of such displaced tracks that we find we Optimize the hyperparameters in this approach by minimizing the uncertainty on the final expected light jet subtracted charm Yield in this case So we're aiming for minimum uncertainty on the charm yield at the end after background subtraction and you can see here Why this is so effective? Charm jets tend to have a very high positive side SIP 3d value Whereas light jets these distributions are relatively symmetric and fall off quickly around zero, okay? So track counting efficiency on charm or light jets ranges from averages of about 20 percent Okay, that's for the charm jets and point four percent. That's for the light jets So that's the background that we have to subtract and we've optimized this by then effectively subtracting that background in the analysis With the kinds of conditions in this analysis We would expect this to yield about six thousand fully reconstructed charm jet bearing events Something in that ballpark at the end of a hundred and inverse of data taking about one year All right, so how does this probe strangeness in the proton? Well, the constraints on the strange court contributions to the proton do range over a wide area Based on PDF fits that have been done over the last few decades. So for instance in the CT 18 framework There are some fits to data sets that prefer a rather high enhanced Strangeness in the proton maybe an RS of about point eight six three Or others that allow a really suppressed amount of strangeness in the proton Maybe around point three two five and we use these two bookends as the sort of range of variation Of strangeness in the proton to see whether or not we could actually see the difference between these two levels of Intrinsic strangeness in the proton just by counting charm tag jets and in fact we can so in gray here as a function of reconstructed jet PT you have the relative uncertainty on the reconstructed charm jet yield after background subtraction at the end of the analysis and then in blue you have the corresponding amount of Variation relative to the suppressed strangeness scenario that you get in the enhanced strangeness scenario And there is clear daylight between those two distributions in this analysis It's extremely clear that that with an EIC detector with all the conservative assumptions We've made so far There's a lot of promise in being able to tell the difference between a very suppressed scenario for strangeness in the proton and a very Enhanced scenario for strangeness in the proton and something that lies in between should be readily accessible at the EIC with Even a very baseline detector design Again, and you know, we emphasize the need to be able to reconstruct jets at low angle with what large radius parameters And so to be able to calibrate those jets we would we would emphasize calorimetry and tracking in the design of a future EIC detector All right, so let me offer some conclusions and outlook So we did a lot of things in the paper I didn't talk about here like varying detector resolution to see what it does to flavor tagging and employing particle ID Algorithms or some future expected particle ID approach in order to also isolate single Particles within jets as a sign of heavy flavor. I didn't cover that here But you can take a look at the paper or you can ask me questions about it afterward There's lots of directions we can go in with full secondary vertex reconstruction jet substructure and flavor tagging and more realistic Particle ID implementations and of course I urge you to take a look at this We're very excited about the potential in the future as we continue to develop these studies for understanding more about intrinsic Stringes in the proton. Thank you very much


Live from San Jose in the heart of Silicon Valley. It's theCUBE, covering DataWorks Summit 2018. Brought to you by Hortonworks. Welcome back to theCUBE's live coverage of DataWorks here in San Jose, California. I'm your host, Rebecca Knight, along with my co-host, Jim McCobiellis. We're joined by Aaron Murthy. Arun Murthy, sorry, he is the co-founder and C Chief Product Officer of Hortonworks. Thank you so much for returning to theCUBE. It's great to have you on. It's been a fun time getting back, yeah. So you were on the main stage this morning in the keynote and you were describing the journey, the data journey that so many customers are on right now and you were talking about the cloud, saying that the cloud is part of the strategy but it really needs to fit into the overall business strategy. Can you describe a little bit about how you approached that? Absolutely, and the way we look at this is we help customers leverage data to actually deliver better capabilities, better services. That's the better experiences to their customers. And that's the business way, right? Now, with that, obviously, we look at cloud as a really key part of it, of the overall strategy in terms of how you want to manage data on-prem and on-the-cloud. You know, we kind of joked that we're sort of living a world of real-time data, we're just living it. And data is everywhere. You might have trucks on the road, you might have drones, you might have sensors and you'll have it all over the world, right? At that point, you know, we've kind of got to a point where enterprises understand that they'll manage some of the infrastructure but in a lot of cases, it'll make a lot more sense to actually lease some of it and that's the cloud, right?  That's kind of what the cloud is. So that is why we really fundamentally believe that we have to help customers leverage infrastructure whenever it makes sense, pragmatically, both from an architecture standpoint and from a financial standpoint. And that's kind of why we kind of talked about how, you know, your cloud strategy is part of your data strategy, which is actually fundamentally part of your business strategy. So how are you helping customers to leverage this? What is on their minds and what are, what's your response? Yeah, it's really interesting. Like I said, you know, cloud is, you know, cloud and infrastructure management is certainly something that's at the foremost, at the top of the mind for a VCIO today, right? And what we've consistently heard is they need a way to manage all of this data and all of this infrastructure in a hybrid, multi-cloud fashion, right? Because you know, in some geos, you might not have your favorite cloud vendor, you know, go to parts of Asia as a great example.  And that is why we fundamentally believe with it's really important to have, you know, give an enterprise a fabric with which they can manage all of this, right? And hide the details of all of the underlying infrastructure from them as much as possible. And that's data plane services, exactly. The hardwoods data plane services which we launched in October of last year, actually I was on Qube talking about it back then too. It's a, we see a lot of interest and a lot of excitement around it because now they understand that this, again, this doesn't mean we actually drive it down to the least common denominator. It is about helping enterprises leverage the key differentiators that each of the cloud vendors will provide. So for example, Google, which we announced a partnership with, they're really strong on AI and ML, right?  And for example, you can now go to Google Cloud and get TPUs which work great for TensorFlow. Similarly, a lot of customers run on Amazon for a bunch of the operational stuff, Redshift is an example. So this is a, the world we live in is that we want to help the CIO leverage the best piece of the cloud but then give them a consistent way to manage and count that data, right? We were joking on stage that IT is just about learn how to deal with Kerberos and Hadoop, right? And now we're telling them, oh, go figure out IAM on Google, which also is IAM on, it's called IAM on Amazon, but they're completely different. The only thing that's consistent is the name, right? So I think with the, we have a unique opportunity with, especially the open source technologies like Atlas, Stranger, Nox and so on, to be able to draw a consistent fabric over this in security and governance and help the enterprise leverage the best parts of the cloud to put a best fit architecture together, right? But which also happens to be a best of brain architecture. So the fabric is, everything you're describing, all of the Apache open source projects in which Horton works as a primary committer and contributor, are able to manage schemas and policies and metadata and so forth across this distributed heterogeneous fabric of public and private cloud segments within a distributed environment. Exactly. It's increasingly being containerized in terms of the applications for deployment to edge nodes. Containerization is a big theme in HTTP 3.0, which you announced at this show. So Arun, if you can give us a quick sense for how that containerization capability plays into more of an edge focus for what your customers are doing? Exactly, great point. And again, the fabric is obviously the core parts of the fabric of the open source projects, but you've also done a lot of net new innovation with data planning, which by the way is also open source. It's a new product and a new platform that you can actually leverage. You layer it over the open source ones you're familiar with. And again, like you said, containerization is, what is actually driving the fundamentals of this, the details matter, the scale at which we operate. We're talking about thousands of nodes, petabytes of data, the details really matter because a 5% improvement at that scale leads to millions of dollars in optimization for CapEx and OpEx. So that's why all of that, the details are being kind of fueled and driven by the community, which is kind of what we deliver with HTTP3. And some of the key ones, like you said, are containerization because now we can actually get complete agility in terms of how you deploy the applications. This is, you get isolation not only at the resource management level with containers, but you also get it at the software level, right? Which means if two data scientists want to use a different version of Python or Scala or Spark or whatever it is, they get that consistently and holistically that now they can actually go from the test-dev cycle into production in a completely consistent manner, right? So that's why containers are so big because now we can actually leverage it across the stack and the things like Minify showing up, right? We can actually- To find Minify before you go further, what is Minify for our listeners? Yeah, so we've always had NiFi, right? Which we, you know, real-time data flow management. NiFi was still sort of within your data center. What Minify does is actually now a really small layer, a small, thin, you know, library, if you will, that you can throw on a phone, a doorbell, a sensor, and that gives you full of all the capabilities of NiFi, but at the edge, right? And it's actually not just data flow, but what is really cool about NiFi is actually, it's actually command and control, right? So you can actually do bidirectional command and control. So you can actually change in real-time the flows you want, the processing you do, and so on. So what we're trying to do with Minify is actually not just collect data from the edge, but also push the processing as much as possible to the edge. Because we really do believe a lot more processing is going to happen at the edge, right? Especially with the A6 and so on coming out, there'll be custom hardware that you can throw and it can actually leverage that hardware at the edge to actually do this processing. And we believe, you know, we want to do that even at the cost of data not actually landing up at rest. Because at the end of the day, we're in the insights business, not in the data storage business. Well, I want to get back to that. You were talking about innovation and how so much of it is driven by the open source community.  How do we maintain that? How does that continue to be the fuel? Yeah, and a lot of it starts with just being consistent, right? From day one, James was around back then in 2011 we started. We've always said, we're going to be open source, right? And because we fundamentally believe that the community is going to out innovate any one vendor regardless of how much money they have in the bank, right? So we do believe that's really the best way to innovate. Mostly because there's a sense of shared ownership of that product. It's not just one vendor throwing some code out there trying to show it down the customer's throat. And we've seen this over and over again, right? You know, three years ago, we talked about a lot of the data plane stuff comes from Atlas and Ranger and so on. None of these existed. Now these actually came from the fruits of the collaboration with the community, with actually some very large enterprises being part of it, right? So it's a great example of how we continue to drive it because we fundamentally believe that that's the best way to innovate and continue to believe so. Right, and the community, like the Apache community as a whole has so many different projects that for example in streaming, there's Kafka and there's others that address a core set of common requirements but in different ways supporting different approaches for example to doing streaming with stateless transactions and so forth or stateless semantics and so forth. It seems to me that Hortonworks is shifting towards being more of a streaming oriented vendor away from data at rest though. I should say, HDP 3.0's got great scalability and storage efficiency capabilities baked in. I wonder if you could just break it down a little bit what the innovations or enhancements are in HDP 3.0. For those of your core customers, which is most of them, who are managing massive multi terabyte, multi petabyte, distributed federated big data lakes. I mean, what's in HDP 3.0 for them? Oh, for lots. I mean, again like I said, we obviously spend a lot of time on the streaming side because that's where we see. We live in a real-time world but again, we don't do it at the cost of our core business which is continuous VHDP. And as you can see, the community turned to its drive. We talked about containerization, massive step up for the HDP community. We've also added support for GPUs. Again, if you think about true ad scale machine learning. Graphical processing units, AI, deep learning. Yeah, it's huge deep learning, TensorFlow and so on. We really, really need a custom sort of GPU if you will. So that's coming. That's in HDP 3.0. We've added a whole bunch of scalability improvements with HDFS, right? We've added federation because now you can go from, you can go over a billion files, billion objects in HDFS. We also added capabilities for- But you indicated yesterday when we were talking that very few of your customers need that capacity yet, but you think they will. So. Oh, for sure. Again, part of this is as we enable more source of data in real time, that's going to, it's the fuel which drives. And that was always the strategy behind the HDF product, right? It was about can we leverage the synergies between the real-time world, feed that into what you do today in your classic enterprise with data at rest. And that's what is driving the necessity for scale, right? We've done that. We spend a lot of work getting, again, loading the total cost of ownership, the TCO, so we added a ratio coding. What is that exactly? Yeah, so a ratio coding is, it's a classic sort of storage concept, which allows you to actually, instead of, you know, HDFS, it's always been three replicas, right? So for redundancy, fault tolerance and recovery. Now, it sounds okay having three replicas because it's a cheap disc, right? But when you start to think about our customers are running 70, 80, 100 petabytes of data, those three replicas add up. Because you've now gone from 80 petabytes of effective data, we're actually to a quarter of an exabyte in terms of raw storage, right? So now what we can do with the ratio coding is actually, instead of storing the three blocks, we actually store parity. We store the encoding of it, which means we can actually go down from three to like two, one and a half, whatever you want to do, right? So, if you can get from three blocks to one and a half, especially for your cold data, right? The ones you're not accessing every day, right? It results in a massive savings, right? In terms of your infrastructure cost, right? And that's kind of what we're in the business of doing, helping customers do better with the data they have without having to spend more. Whether it's on-prem or on the cloud, that sort of, we want to help customers be comfortable in getting more data under management, you know, along with security and governance and the lower tort TCO. The other sort of big piece I'm really excited about at HTTP 3 is all the work that's happened in the Hive community for what we call the real-time database, right? As you guys know, you follow the whole SQL wars and the Duke space. Hive has changed a lot in the last several years. It's very different from what it was five years ago. The only thing that's same from five years ago is the name, right?  From there to drive it to with 3.0, it's now like with Hive 3, which is part of HTTP 3.0, it's a full-fledged database. It's got full asset support. In fact, the asset support is so good, it's that, you know, writing asset tables is at least as fast as writing non-asset tables now, right? And you can do that not only on- It's transactional database. Exactly. Now, not only can you do it on prem, you can do it on S3, right? So you can actually drive the transactions through Hive on S3. We've done a lot of work to actually, you know, you were there yesterday when we were talking about some of the performance work we've done with LAP and so on to actually give consistent performance both on prem and the cloud. And this is a lot of effort, simply because the performance characteristics you get from the storage layer with HDFS versus S3 are significantly different, right? So we have been able to bridge those with, you know, things like LAP. We've done a lot of work in sort of enhance the security model around it, right? Governance and security. So now you get, you know, things like, you know, column level masking, roll level filtering, all the standard stuff that you'd expect and more from an end-priced warehouse. Like, you know, we talk to a lot of our customers, they're doing, you know, literally tens of thousands of views because they can't actually, they don't have the capabilities that, you know, exist in Hive now. And I'm sitting here kind of being amazed that, you know, for an open source set of tools to have the best security and governance at this point, it's pretty amazing coming from where we started off. And it's absolutely essential for GDPR compliance and compliance with HIPAA and every other thing, every other mandate and sensitivity that requires you to protect personally, identify all the information. So very important. So really in many ways, Hortonworks has one of the premier big data catalogs for all manner of compliance or requirements that your customers are facing. Yeah, and James, you know what about it in the context of, you know, Datastore Studio, which we introduced. Yes. You know, things like consent management, right? Having... A consent portal. Consent portal. Indicates the degree to which they require controls over their management of their PII possibly to be forgotten and so forth. Yeah, it's right to be forgotten. It's consent even for analytics. Like within the context of GDPR, you have to allow the customer to opt out of analytics, them being part of an analytic itself, right? So things like those are now something we enable through the enhanced security models we've done in Ranger, right? So now, instead of actually the really cool part about what we've done now with GDPR is that you can get all of these capabilities on existing data, right, with Hive and existing applications by just adding a security policy, not rewriting your SQL query, right? It's a massive, massive deal, which I cannot tell you how much customers are excited about, because they now understand they were sort of freaking out that they have to go to, you know, 30, 40, 50,000 enterprise apps and change them to take advantage of the, you know, to actually provide consent and right to be forgotten. The fact that we can do that now by changing a security policy with Ranger is huge for them. Yeah. Arun, thank you so much for coming on theCUBE. It's always a fun talking to you. Likewise. Thank you so much. I learned of something every time I listened to you. Indeed, indeed. I'm Rebecca Knight for James Kobielus. We will have more from theCUBE's live coverage of DataWorks just after this.


In the midst of the war, on June 8, 1967, Israel attacked the USS Liberty, an unarmed American naval ship in international waters flying the American flag. It was a coordinated attack from the air and sea. The Liberty was identified by Israeli pilots as being American. There is no doubt about this. This was not mistaken identity. There are recordings. Despite clearly identifying the ship as American, by reading out the ship's whole number and seeing the American flag, Israeli pilots fired at the ship from above, and Israeli missile boats fired torpedoes from below. 34 American sailors were killed, another 174 injured. The American government tried to cover this up. They tried to hide this event from the press and the American public. Why are their greatest ally in the Middle East intentionally and callously attacked them? The most likely reason is that the USS Liberty was a spy ship. And as a spy ship, it was listening in on Israel's war plans. Israel didn't like that. So they tried to massacre everyone on the ship.


Hello everybody, Andrea Mayesca here with dental well tutoring. I'm sorry. I decided to clear my throat really really quick But I wanted to talk to you guys about something that happened to me yesterday actually So what do you do when your piezo or cavatron isn't working? So for those of you who are a dental hygienist you might either always use a piezo or always use the cavatron I have used both depending on the offices where I'm in I'm fine using either or but I do prefer the piezo But what do you do when you show up and it was working yesterday? Or it was working before your last appointment, but for this patient or just today It's not working and it's not like you can get down on your hands and knees and try to fix it If it's an internal unit you can't just move things around to try to fix it If it's just one you just simply plug in those are easier because usually it means you just have to Turn it off turn it on again. Maybe plug it back in again Maybe just keep on trying different tips, you know just weird things happen with these machines But all in all if it just stops working what do you do? So this happened to me the other day and it happened to me for my last patient when I was doing quad Scaling okay, so she was in for quad three for the full the full hour I'm just scaling quad three.  I'm not amazing at being able to get those black rings off around the teeth So I was annoyed. I was stressed out. I know the patient can't tell Even when I went to go use the water instrument I like to call it for my patients and she's had it every single time So she knows exactly what it is because I have explained it to her I went to go pick it up and then it just wasn't working. I kind of said to her. Oh our water instruments not working I'm just gonna change the tip I'll be right back and I kept on trying to get it to work for a good couple minutes, but it just wasn't working So instead of me saying to her, okay I'm not gonna be able to do your cleaning today because there's no way but I can hand scale everything I said to her, you know what? That's okay. I'll just use my handed hand instruments. You might feel a little bit more pressure but um Actually, she had full LA so full local anesthetic quad three so that actually helped me out But I said you might feel some more pressure today my water instrument for some reason isn't working So you might kind of see me dig more but things will look amazing Meanwhile inside. I'm thinking oh I have to hand scale for an hour. I won't be able to get it off I won't not all of it anyway, and that's what happens. So considering you know For a lot of patients if they were just in for their cleaning today Opposed to quad scaling. I would have said to them. Oh, there's more plaque than usual I can't clean everything today.  It was quad three for the full hour There's no way I was going to tell her that we can't clean everything did today We're going to have to bring you back for quad three. I just wasn't going to do it Okay, I wasn't going to do it. So I scaled as much as I possibly could thank God Thank goodness. It was my last patient because my hand was hurting. I was sore I was not happy. So thank goodness. It was the last appointment But I just simply said to her okay. Well, we did a great cleaning today The gums will probably bleed more than usual because there was more on this side But the next time we will continue on into quad four But I left charter behind you guys. I know I did because I wasn't able to get my piezo in there I know I did I left charter behind and I didn't tell the patient I left charter behind but what I'm going to do is at her next appointment. She's actually in on Thursday She's in for quad four. I'm going to obviously clean quad four I probably won't have time to go back to quad three and kind of finish it up But since it's quad four what I'm going to do is I'm going to obviously clean entirely quad four She will have local anesthetic again, and then I'm going to at the end of the appointment say to her I Am going to go back with my water instrument and just kind of clean a little bit everywhere And since I didn't have the water instrument last time I'm just going to spend an extra, you know two minutes making sure that it is as You know healthy as possible. So that's it so what I'm what I'm going to do is I'm going to go back with the Piezo because now it's working by the way and I'm just going to spend longer and quad three and make sure to get everything off So that's what I'm going to do. So you see how I did that You see how I did that so it just makes me feel so much better because I know I didn't get everything clean I know I didn't But that's just kind of how I handle the situation So if that happens to you guys just just kind of look and see like if it's some And I did actually have another patient where my piezo was not working once So not the same day, but another day and they had a lot of charter their lower anteriors I was able to use my hand instruments to get it off But it was really hard like it was really tenacious. So I simply said to them You know what I can't do your whole cleaning today because there's just simply not enough time If you could come back for like 20 minutes, that's it next time I will only obviously charge you for the 20 minutes of a cleaning and just do the rest of it So that's what I did because I couldn't get it all off with my hand instruments plain and simple Or if let's just say that that patient had a lot of charter everywhere I did the best I could with my hand instruments But I couldn't get it all off without Using the piezo that I will bring them back for a full hour the next time to do the rest of it You will probably be done early unless they did obviously have a lot But make sure that that the piezo or the cabotron is working when you bring them back Or you're just gonna have to work really hard those hand instruments. I love my piezo I like to use the piezo first and you should always hand scale afterwards anyway because you might miss something So yeah guys, that's what I do if it hasn't happened to you yet It will probably happen to you knock on wood though But that it doesn't happen soon and I hope that these little tips can help you guys any comments Please let me know I might have left something out Please let me know make sure to click like and make sure to subscribe Because I do always appreciate it and it's nice to know that people are actually watching me do these videos So thank you guys so much. Have a good day today. It looks nice outside. Enjoy the weather and I'll see you next time


Hello and welcome to the final AFC BTV preview show of the season. Neil Perritt, formerly of the Bournemouth Daily Echo, and now a member of our media team here at AFC Bournemouth joins me and we'll be looking ahead to another big week in the Premier League. Here's what's coming up. We'll be looking back at that fantastic one-nil victory over Spurs here at Vitality Stadium. We'll also be discussing the season in general and looking at all of the highs that have come with it. And finally, we'll be looking ahead to Sunday's game against Crystal Palace at Selhurst Park. But first, there's just one place to start, and that is with last weekend's one-nil victory over Spurs here at Vitality Stadium. Let's remind ourselves of the highlights. Son still has it. Danny Rose in support from the left-hand side. Danger once again. Rose is crossed. Morris Header travels again! Brilliant goalkeeping! Morris Header travels heading for the top corner, threw up his hand and climbed it away. Here's Son from the short corner now. Son tried to go past Lerla, but no way past the Colombian wall and Cherries will try and bring it away. Fraser can't quite manage to smuggle it clear. And then he goes down under pressure from Son. And Son's pushed Lerla over. That's going to be a red card for Son. You don't get away with that. Do an arm out at Jefferson Lerla. Son who didn't win. It's a moment of madness in the Korean. He's just back from suspension. He's flawed Jefferson Lerla, and he's off. I want to, I just don't understand what he was thinking there, Son. It was some sort of multiplication, but he just, two arms pushes him over, and the referee's got no option. Fraser was pushed by Son. Jefferson Lerla tried to take a quick free kick, and Son, for some reason, Son wasn't... That's what he's doing. 4-2-2-1. That's a bad challenge. Bad challenge on Simpson from one point. He's gone through the young man. That's another red card! This could get interesting now, Chris. Well, Jack Simpson has been left in the heat by that challenge from one point, and Spurs are literally falling like flies. We're getting a replay here, as he lost control of the ball point when he's come in with his studs up, sliding. He's made an attempt to get the ball, but it's a high one. He's caught him high. It's a horror tackle, lad. Four minutes remaining, and out of time. The ball says four. That's not at this moment, but... I'm going to wait to finish it. Game it, if he could get a goal. That's one of those. Does he go up with £2 million to place in the Premier League? A point could be very valuable at the moment. Crazy's ball in. The heads go up again, easy! Well, a last-minute header from Nathan Ackay there. Neil, it was a crazy game, wasn't it? And we didn't expect that at all. Absolutely mad. It was a mad game. It was the real proverbial game of two halves. Zoe, in the first half, Tottenham obviously had their full complement with players. Played very well. Mark Travers came in and made a fantastic debut. He made a string of fine saves, keep the score level at halftime. And then a couple of sendings off either side of halftime. Certainly turned the tide in our favour. Didn't look like the goal was going to come, but eventually Nathan came up trumps right at the death. And you mentioned the two red cards there. What was your take on them? Well, I have to say, I think the referee got both of them right. I mean, it was unfortunate because there was possibly a third red card in the game, an earlier red card with Eric Dyer. Made another bad foul and then looked like he'd fouled Callum for a penalty as well. But you can't go around pushing people around on the pitch like Sun did. And then it was an overzealous. Perhaps not too much intent with a tackle from fourth, but by a letter of the law, they both had to go. There's no question. And you mentioned that one on Callum Wilson that Eric Dyer did in the box. It looked a certain penalty, didn't it, at the time? Well, I think at the time I was looking sort of at the linesman and at the referee and you could argue that neither of them could actually see what happened. The first time we saw it, we weren't too sure, but we obviously had the benefit of seeing it on the monitors afterwards. And he's clearly just chopped Callum. OK, he wasn't looking. You could argue it might have been a little bit accidental, but it looked like a clear penalty, certainly second time round when we saw it. Yeah. And we mentioned just a minute ago, but Mark Travers, what an unbelievable debut that was? Well, unbelievable debut in more ways than one. Not only did he come in and deservedly get man of the matchmaking, three or four fantastic saves in the first half, but the fact that he was only 19 years old is only previous competitive experiences in the Southern League with Weymuth to come in and be so assured and so calm. And when we were watching him, he was virtually standing on the halfway line.   And we were there earlier in the season in the Premier League Cup against West Brom. I think after that game we both said what a fantastic performance he'd had, but it's completely different, isn't it, when you're live on the sky and you've got spurs in the Premier League? Well, exactly. We were at Hensford United that night. I think it was in about January or February in front of one man and his dog. And Mark Travers had a fantastic game that night, but you've got to put these things into perspective.  You know, this is the biggest league in Europe now, as has been proved this week by the four English teams getting into the European finals. You know, and we're going to finish 12th or 13th in this league and this young teenage lad has come in, like I said, with very little match experience behind him and performed like that. It was just, it was a debut to remember and you know, fantastic for the lad. And up the other end of the pitch, as you said a minute ago, it didn't quite look like that goal was going to come and they left it late, didn't they? Well, we've seen it so many times following Bournemouth over the years, these last, gas, dramatic goals, and you know, it's typical Bournemouth, if you like. You see, it looked like had a nil nil draw written all over it. You know, we were knocking on the door throughout the second half and Nathan, you know, not for the first time, like the Liverpool game a couple of seasons ago, comes up, Trump's powers through the box, fantastic header and sends everyone home happy. And he pushed up into midfield last week. What did he make of his general performance, the goal aside? Well, it's difficult to, it's difficult to ever pick fault with Nathan Ackie wherever he plays. I think he's played left back this season. He's obviously played centre back and midfield as well now. You know, he had that season at Watford where he was on loan, where he played a lot of left back there.  So, you never really know whether he's unhappy or anything. So, but he certainly, you know, he played superbly, superbly. And you wouldn't have thought, you know, losses to Burnley and Fullham at home. The last home win came against Chelsea and then Spurs come here. Final home game of the season and what a way to send the fans home happy. Well, exactly. And it's another one to chalk off because, okay, we've had a point against Spurs in a nil nil draw but we've never beaten them. So, I think it's just Manchester sitting now we want to try and get a point off them at some time when we can there. The only team in the Premier League I'm going to point off, but yeah, exactly. Sends everybody home happy for the summer. A win against Spurs and a win that's, you know, emphasised even more by the fact about what Spurs did during the week. You know, when was the last time that AFC Bournemouth beat a current Champions League finalist? Okay, you can look back and, you know, we've beaten Liverpool and the likes of Nottingham Forest to won the Champions League in the European Cup as it was. You know, but what a feather in the cap that is to beat someone like Spurs, it was just memorable. Absolutely. Well, there's been some brilliant moments throughout the season and Eddie Howe reflected on that earlier in this morning's press conference. Let's take a look at what he had to say. Has it been a season of progress, would you say? I think overall it has to be considered that way. I think there's been challenges for us. I think there's seen quite a big transition in the team this year due to loads of different factors, but injuries have been a key thing of the second half of the campaign. It's opened up opportunities for younger players. I think the average age of our squad has come down, or our team has come down, which I think is a really good thing for the future. So maybe that's led to some inconsistencies in our performances and certainly something that we're focused on in the summer for next year. Well, it's been another solid season in the Premier League, Neil. There's been so many amazing moments, haven't there. What stood out for you? Well, I think you've got to go back right to the start of the season. It was a record-breaking start, 20 points from the first 10 games unprecedented from Bournemouth before, and it really laid the foundations for preserving the status for a fifth consecutive season. Let's not lose sight of what an achievement that is, particularly, again, I refer back to what's happened this week, four English teams in the finals of the European Champions League and the Europa League. There we are. Little Bournemouth, if you like. Well, certainly that's how we used to be known, and we're going to be finishing in mid-table in a league that includes teams like that, clubs like that. You can't take away from what a fantastic achievement it is, and the other one to throw in there is the fact that there's only two teams in the Premier League who've been promoted and not been relegated. That's Bournemouth and Brighton. So that just shows you how difficult it is to stay in there. And we've pulled off some fantastic results this season. Obviously, we've just mentioned spurs. But that four and eleganths, Chelsea, that was incredible, wasn't it? That was a memorable night, but you've got to remember the game before at West Ham, the 2-0 one at home to West Ham. We were under a little bit of pressure going into that game. We were on not a great run. The fixtures, they're all tough fixtures, obviously, but West Ham at home and Chelsea at home, can you maybe get three points out of those two games and get six and all of a sudden a little bit of the pressure's off. So it was a fantastic January and two really important results. But every season you get important results. So I remember a few years ago, we went to Chelsea away in one and we beat Man United and West Brom. Three games on the trot. I think it was in a December time and all of a sudden it just eases that little bit of pressure. And of course, the second half in the season in particular, we've had a lot of injuries and you think if that hadn't been the case, that first half of the season, the four might have been replicated. Well, I think a lot of people lose sight of the fact that you've had half of the team ripped out, if you like, and not just like injuries for a few days or a few weeks, but these are season-ending injuries that people like Lewis Cook and Andrew Sermon and Simon Francis and Charlie Daniel. I mean, the list is going on and on. I don't think I've known a season like it in the Premier League where we've had such bad injuries. And we're going to finish mid-tables over here. You know, you've got to take your half to everybody. And individually, there's been some brilliant seasons out there. Callumawson, he's got in the England squad. Ryan Fraser, he's right at the top of the assist chart with Eden Hazard for most of the season. It's been a really good campaign individually as well, hasn't it? The list goes on again. You know, David Brooks has come in his first taste of the Premier League. Jefferson Lermer, let's not lose sight of the fact that first season in English football as well. You know, that's really difficult. Struggles a little bit with the language, obviously, coming from Colombia and having played in Spain.  You know, there's a lot of teams that would love to have someone like Jefferson Lermer as the heartbeat of their team.  It's just been... We could go on and on and on, couldn't we? We could. And then you mentioned David Brooks and Jefferson Lermer, it's their first season, both of them in the Premier League, obviously last year, David Brooks playing in the championship, but not week in, week out for Sheffield United. So for him to come into this team and, you know, play as regularly as he has, he's done so well, hasn't he? He's done tremendously well, like you said, not a bit part player at Sheffield United, but I know he had a couple of injuries as well, which restricted him. But, you know, taking it like a proverbial duck to water, if you like, and, you know, what Jefferson and David Brooks can achieve next season, and then, you know, I can't wait to see, really, because they've certainly put the platform in there for some great performances. Absolutely, and you alluded to it earlier, but a fifth season in the Premier League, that's, you know, something that would never have even been dreamed of 10 years ago. Well, I think the club's had five seasons in the second tier, two in the championship, and three in the late 1980s. You know, now we're going to be having five seasons in the top flight. I know that people keep saying you shouldn't keep harping back to, you know, the past, but, you know, let's not lose sight. You know, this club spent a long, long time in the bottom tier and the third tier. You know, and now we're sort of looking at the second tier and the first tier. It's just, you know, it's pinching yourself stuff, really. Absolutely. Well, it's been quite the season. We round off our campaign this weekend at Selhurst Park, but it was quite the evening here at Vitality Stadium when Crystal Palace visited earlier in the season. Let's remind ourselves of what happened that night. Mill Mill Bournemouth Palace, Smith's ball into the penalty area, picked up by Wilson, square to Brooks, left but in, and through his left foot. In off the underslaves, making waves in the top flight is David Brooks. One Mill Terries. Well, I'm quite an amazing move, and it was having Smith. He's got the cross, he went his left foot, drilled it across the face of goal. I think it was it Callum, or Callum held the ball off. Zahar gets a fortunate break of the ball, and now Van Artholk ghosting in left-hand side. Right, but it's shot! That is some equaliser from Patrick Van Artholk, who has lashed it into the same top corner that David Brooks found earlier on, and Palace have been threatening at the start of their second half, and they get the goal that their possession deserves. 55th minute, 1-1. Well, you have to say, it seemed likely the way they were going when it was the pass through. Just repeated the beat, of course, but he didn't have a lot of angle, decided to go with power and smashes it into top corner. Just a little suspicion on both sides. I think he was just the odd. The sumpterships they brought on have got extra defensive artillery here. There's a penalty, is it? It is a penalty! It's Lerma who's gone down! Bourbon's got a penalty! Yellow card in there as well for Sacco, but the referee had just literally worn them. And Lerma has gone down, face down on the edge of the penalty area, and he has won the cherry's a spot kick. Stannislas has got the ball. Let's have a look at the replay here. Well, as it comes in, Lerma's got it. It's just got a clout, a flailing arm from Sacco. Can Stannislas possibly win it? Right footed in the net! For Bournemouth against Crystal Palace. Bournemouth 2, Crystal Palace won. Well, I did have a mind, but I thought we'd give someone like Stannislas all the way. Well, a fine evening at Vitality Stadium there. Neil, it's going to be quite the occasion on Sunday, isn't it? 12th to be 13th, last game of the season. Anything can happen. Well, yeah, like you said, I mean, you know, you could say that it's a bit of a dead rubber, but far from it. You know, there's a substantial purse at stake for the team finishing higher in the Premier League. Crystal Palace and us have finished on the same points, Tally twice in the past three seasons. They're just ahead of us at the moment on goal difference. Their home form is not great. They are, I think, their second from bottom in the home league table in the Premier League. But where they've really turned it on is away from home, where they're fifth. And you think that's some achievement when you see you've got the likes of Spurs and Man United and Man City who are all at the top of that league. And then Roy Hodgson's done a fantastic job away from home. I think they've got something like 29 points away from home. Home, Cellhurst Park hasn't been a happy hunting ground for them this season, but they're going to be formidable opposition there as we've always found when we've been there. Where we're unbeaten, let's not forget. And it's also worth mentioning that 47 points, it's still very much doable. That would be the highest that the cherries have ever recorded in a single Premier League season. And another reason not to lose sight of how far the club has come this season. The manager was asked this morning about whether he feels the team has progressed this season and he thinks they have. Well, there you go. If they were to win on Sunday, it would be a record points tally. It's still a fantastic points tally anyway. Let's not take away from that. But all to play for, certainly all to play for. And in terms of the palace players, Zahar is the obvious one to look out for. But Aaron Wambasack has enjoyed a good season as well, hasn't he? Really a breakthrough season for him. He's really caught the eye with some fantastic performances. Very dangerous player to play against. Sort of player you'd probably want in your team. Unsettled defenders, very quick and stuff like that. So certainly want to keep an eye on it. More than want to keep an eye on it, if you like. And in terms of our injuries, unfortunately we're in exactly the same position as we were last week. The players that were out last week, Eddie Hall confirmed that they're not going to be back for this week.  Give these guys as long as they need and make sure that they're ready to hit the ground running at the start of pre-season. But then you can say, well, you know, the guys who played last week, so you know, they've got a 1-0 win against Spurs. So, you know, it's the same team. Then we've got a great chance. And we've talked a lot about a waveform this season and it's perhaps not been where Eddie Hall would have liked it to have been. But four points from the last two away games is nothing to be ashamed of, is it, Southampton and Brighton? Yeah, I mean, there was a run halfway through the season where, you know, we didn't win an away game for quite a long time.  We played Spurs. We played Manchester City. You know, if you look at the league table, not many of the teams in the bottom half have got points against the top six away from home. You know, and we're in that bracket. But where we've done really well, Zoe, is we've won away games that we've needed to win, at other places like Cuddersfield, like Brighton. And, you know, but for a great save at the end against Southampton, we would have won there for the first time in our history as well. So, you know, we've won where we've needed to. And that Spurs win last weekend. They really, really want to back it up this weekend, won't they, and end the season with back-to-back wins? You always want to go into the summer on the back of a win if you can. I know it keeps, you know, confidence high and everybody's buzzing going into pre-season or the break or international duty. So, yeah, everybody wants to get a win behind them and back-to-back wins. What would that be? The Spurs at home in Crystal Palace away? What a great way to finish the season. Absolutely. Well, it's going to be a really exciting game at Selhurst Park on Sunday. That's all we've got time for today. But if you've been joining us throughout the season, thank you very much. And we look forward to seeing you up at Crystal Palace on the weekend. Bye for now.