Okay, so yeah, one one's like I'll turn it over to you here but I just wanted to say just a couple things before we get started as one welcome to the seminar, just a quick note on just the fact that attendance is open to everyone so despite graduate student being in the title there is sort of a wide variety of backgrounds here so I just want to draw you know just to be respectful with your friends and colleagues. So if you have any questions, if you have a question for us the easiest way is to just go ahead and politely interrupt at some point. If you have a question you can also type it in chat. I can try to keep an eye on that and really it's speaker. If you have sort of a longer question we'll leave a little bit of time at the end. If anybody wants to ask anything. Right, so I guess with that let me let me go ahead and turn it over to Sarah Griffith from Brown University and she'll be telling us about the tropical study room. Yeah, so I'll, I'll be talking about this algebraic object that is really important to a subset field of algebraic geometry called tropical geometry. It's equivalent to another gadget called the Max plus semi ring, which I'll mention later and sometimes it is also called the tropical semi ring. So a note on the name so the subject was sort of founded by a Brazilian mathematician. He named it. Apparently in, in honor of him. I don't know what his thoughts are on the name I think there's a, there's a general feeling that the name tropical geometry is a little distaste. Yeah, but I don't have a better name for it. Min plus semi ring is therefore both like, let's me avoid the term and it's it's more accurate because there are two possible rings that may use. So tropical geometry really leans on this on this min plus semi ring, and it's not really obvious how it comes about.  Personally, I never found that very convincing. So what I'm going to do is like not so much teach you like tropical geometry proper, which is kind of like theory laden and a little and hawk. But I'll just try and convince you that the min plus semi ring naturally encodes interesting things. So I'll give you for applications of increasing complexity. So the first three have to be like pretty broadly accessible the fourth one will be more ag ish. Okay, so let's go. So the first problem that we can use the min plus semi ring to address is the problem of finding minimum weight paths on graphs. So by a graph here. Well you can work with more generality than this but what I mean is you've got a set of vertices you've got a set of edges you have at most one edge between the two vertexes. And you have non zero weights on each of the paths. So if you are interested in given two vertices specified ahead of time, finding the shortest possible path from one to the other. So linear algebra, as it is useful everywhere is useful in graph theory. So you may be familiar with something called the adjacency matrix for the graph. So for this graph right here. The vertices are labeled v1 v2 v3 v4 and each of those corresponds to column of this matrix called the adjacency matrix of the graph. So the v i corresponds to the ice column. And we put a one in if the vertex, this column corresponds to is adjacent to the vertex corresponding to the corresponding to the vj or j is the number of rows you go down. So for example this corresponds to v1, and then we have a zero here because we don't consider v1 to be adjacent to itself we have a one here because we consider v1 to be adjacent to v2, and so on. So this you can prove my induction has a really nice property, which is if you take the adjacency matrix, and you raise it to the nth power. And the way they entry will be exactly the number of paths from v i to vj. Now, there's kind of some some data associated with our graph that isn't represented in the adjacency matrix, which is, we insisted that our graph edges would have non zero weights. What we can do is we can look at the weight matrix. So this isn't completely filled in yet. But this corresponds in a very similar way to pairs of vertices. So, for example, this still corresponds to v1. We still have a one here, because the path from v1 to be sorry the edge between v1 and v2 has weight one. On the other hand the edge between v1 and v3 has weight two, and that ends up getting represented as a two here. Now what are these little question marks doing here. Well v1 is not adjacent to v4. So we want to use some kind of like linear algebra construction to use the weight matrix to calculate possible possible weights of paths. It's not obvious what we should put in these entries. So one of the first questions is, can we fill in these question mark spots and then use some kind of linear algebra or variation thereof to find these minimum weight paths. Sorry, I just had a quick question. So when you say powers compute the path between vm vj, like if I raise it to say the third power and I look at the ij entry. Like as in if I raise it to different powers won't the ij entry be different. And so, Yeah, yeah, sorry, I may have omitted a word. The the nth power computes the number of paths of length and specifically. Okay, thank you. Yeah, my apologies. Thank you for clearing that up. Let's see. So let's try just like naively trying to compute some paths from v1 to v4 and see if it suggests some kind of like algebraic operation that we can relate to the matrix. Okay, so one thing is, the strategy we're going to pursue is going to involve taking powers of this weight matrix and reading off some kind of some kind of entry that we've generated by that method. Now, what we might expect is that, say, the nth power gives you the minimum weight path of length n between two vertices. That's what you can read off the entries, because as was just clarified, that's what you get with the adjacency matrix. We've already managed to find a path that lands at the endpoint we'd like. It would be convenient if we could just consider a path of length and to be say a path from the vertex we're trying to go from to the vertex we're trying to go to, and then a bunch of like stand still moves of weight zero. So we're going to include not moving not moving as a valid step that increases the path length by one. Okay, so let's look at different path lengths and see how we can potentially get from view one to view four. So if we have paths of length one. Well, we could stand still so I didn't bother drawing that in. Or we can take what the weight one path to v2, or the way to path to v3. As paths to v4. These are like infinitely bad, because they don't end at v4. We'll assign these right here, the value infinity. If you look at the paths of length two, we can go to v2 and stand still we can go to v3 and stand still, we can go to v2, and then to v4. And in that case the total weight of the path is one plus four, these first two were still infinitely bad. So we can go to v3 and then before, and the total weight of that path is two plus one seems to be a stray infinity here I'm not sure what that's doing. Probably an example I deleted. On the other hand, if we look at like paths of length three. We've got all the all the prior ones, essentially interspersed with do nothing moves. For example, this one can be regarded as the path of length to append to do a do nothing move. So that's one plus four plus zero. And so on, is that clear to everyone. I'm going to take those yes please feel free to interrupt me or like ask for a backtrack or something I don't mind at all. So, once we've got all the possible weights of paths to v4 of a particular length, which since we allowed stand still moves will include paths of shorter length, essentially. What we need to do is we need to look at the possible path lengths that we determined, and we want to take the minimum among all of them. Okay, so, in order to formalize and make rigorous this kind of naive calculation. We define the so called men plus semi ring. So, this, as a set consists of the real numbers and also infinity. And it has two operations. So I'll call this, I guess, oh plus and I'll call this oh times. Times is great because it's not a tensor product, but whatever. Okay, so what are the operations. So, the plus operation, the O plus operation is going to take the minimum of two numbers. So that's the comparison step we did at the end where we got a bunch of paths of different lengths and we found the least one. The times operation is going to add two numbers together. So we're going to add in like the usual sense of real numbers. Now, why is, why is, why are these defined these way this way well they they enjoy the usual distributive and associative properties that you associate with a ring. So the big difficulty there is that there's no, there's no subtraction.  So there's a canonically isomorphic object called the max plus semi ring, where instead of O plus taking the men of two numbers. It takes the max. So it's the same thing. But you know some sometimes one convention is used sometimes the other. So, our earlier sort of naive calculations together with this definition suggests that we let the weight matrix have remaining entries, infinity, in order to indicate that we think of the distance between non adjacent vertices as essentially infinite. Okay. Now, if you take w to the end I wrote three here for some reason specifically. That's why. So it's pretty clear that the minimum possible path length from view one to view for. Let's see. Will be of length at most two. Now if you take w three, which is certainly long enough, which is certainly a high enough power to capture the paths of length to then the IJ entry will in fact to be the minimum path length from via to VJ. Because if you look at the site. There's a third power with respect to these operations. So to multiply two matrices together, you do completely ordinary matrix multiplication, but wherever you would multiply. You do times and wherever you would add, you do O plus. So if you get out will encode minimum path lengths between vertices of length at most whatever power of the matrix you're taking. I have a question, is that the weekend. Yeah, so when I say, let's see, the minimum. Yeah, that's just a typo, the minimum weight. Wait, sorry. I may have been misspeaking this whole time. Okay, so is, is the meaning of that clear. Wonderful. Thank you. Okay, so here's sort of the heuristic this suggests, whenever we have an optimization problem. We can encode it and max or equivalently men plus. When we can do two things we can run processes together with linear benefits or costs. So that's encoded in the operation of times. And almost extreme of those combinations of processes will matter. So that's O plus. So to relate that to what we just did. So we can combine the two processes of moving from V1 to V2 and from, I guess this was V2 to V4. And we can combine these two processes of moving these paths. And then we can compare which of the two produces the most extreme outcome and we end up with the minimum weight path. So less abstract example. You can put water pipes or transit methods end to end. And in that case, the one with the least capacity will end up controlling the total capacity of the combined ones. Or you can put them side by side. And in that case, the two capacities just like add to each other. So that's the kind of like optimization problem that we might want to look at. Okay. So problem two is completely different. So this is a question about finding approximate solutions to numerical questions. So everything on this page will become clear when I do an example.  And we have an equation that consists of taking a product of a bunch of these FIJs and adding them together and seeing when that's zero. So suppose we want to understand the solutions to that equation. They are potentially horribly complicated. But it could be that their magnitudes are not so complicated. So, in that case, what we do is we write each FIJ as a product. So the product will consist of two parts. So there will be a part that we don't really care about that much, which is a part of roughly constant magnitude. And then there will be a kind of magnitude control term, consisting of 10 to roughly the logarithm of the FIJ. So once we have this, we can rewrite the equation using these.  And then this is multiplied by 10 times while you add the exponents that you came up with together. So now your equation looks like this. Okay. Well, if we're going to have a solution to this. So that two of these terms must be canceling each other out. So what that means is that these magnitude control parts of each of these have to be about the same. So, in other words, we have to have that the max of these sums that are occurring in the exponents occurs twice. Does that make sense to everyone? Okay, I'm seeing some notes. So that's encouraged. Okay. So let's take a let's take a deliberately horrible example. So let W be a virus trust function. So these you may recall are the. They're the continuous but nowhere differentiable ones, I think. Anyway, the point is that they're discussed in. And they're bounded. And we can like adjust the bounds a little bit something like that, whatever. So we'll say that zero is less than or equal to the magnitude of the virus trust function, which is less than or equal to two. And now we consider possible solutions to zero equals 50 plus virus trust of X times 10 to the X squared minus 100 plus 10 times virus trust of X times 10 to the X plus three. Okay, so I mean this is, this is, this is disgusting. Completely impractical, except that surprise this equation is engineered so that each of these terms has a really simple magnitude part. So, we can look at that through the manipulations down here. Okay, how do they end up with oh yeah yeah yeah. So we're sort of factoring as much magnitude as possible out of this part right here we want to put it in the prior form it's almost there but not quite. And when we do that. We have hours of 10 that are living inside come out and live with this 10 right here. And our equation is now five plus wire stress of X over 10. So very consistent magnitude times 10 to the X square plus one minus one plus virus trust of X over 10 again very consistent magnitude times 10 to the X plus. Okay, 32, but whatever. Let's see. I think, I think there was supposed to be a zero here and the rest of the slide is correct. Okay, there we go. Okay. So the solutions have to be near the sets where these two exponents match up. So in other words, near where x squared plus one is equal to x plus 32. So the possible solutions can only only occur near x equals one half times one minus five root five, or x equals one half times one plus five root five. Guarantee of existence here. This is just a necessary condition. Okay. So a quick question from the chat is how do you. Sorry, I guess there are two questions. Magnitudes here today mean absolute values. Let's see. I don't think I quite caught that was the question like does the does the magnitude consider absolute values or something. Yeah, or by the magnitude is that does that mean the absolute taking the absolute value. Yeah, yeah, absolutely. So we're going to apply like, essentially this logic with non ordered fields. And yeah, so that'll battle that won't arise until part four as far as as far as this application part. If you if you actually wanted to show existence I think you would need to like do a more careful analysis involving something like the intermediate value theorem. And then you would probably want to care about sign. But just for producing this like necessary condition that tells you that sort of bounds the region. Well, okay, I guess it doesn't boundary, whatever it kind of tells you where to look. You don't really need conditions on the signs in order to narrow it down that far. And there, there's a second question about how you know, how many factors of 10 to factor out here. Yeah, good question. So, the actual answer is that we just want these, these parts in here to have like roughly similar magnitude. Yeah, so there's no there's no like canonical answer for that. See anything else. Some folks are answering the questions in chat. Okay, great. Okay, so the third problem. So this is proof. I warned you guys on Twitter, this is going to be contrived. So, you know, hang in there. So the third problem is finding redundancy while optimizing over time. So, in this scenario, you're designing a protocol for powering down a nuclear reactor in very bad circumstances. By the way, this is hypothetical. So the power, the overall power on the plant P is going to go to zero over time. I've drawn this, I've drawn this nice diagram. And I'll go through and sort of explain, like, what's up with it and roughly what it's trying to encode. So, there are six systems by which I'm referring to these like gray boxes right here. I'm sorry, I didn't. I feel like I didn't respect color blindness in this palette very well so my apologies for that. Let's see. So, these are the six systems right here. So, each of them receives power from three sources. So we've got the main power. And then we've got something called the fizzy modulator here on the left, and we've got something called the spinning modulator here on the right. So, these are these are real nuclear reactor technical terms although it may not seem like it. No, they're not that was a lie. So kind of the idea here is that the power coming out from the fizzy modulator to these various systems, and the power coming up from the spinning modulator to these various systems is a function of two things which is first off, power as a whole, and second, how much of the main power you choose to shunt to those modulators. Okay, this is this is let's let's put this slightly more formally. So the thing you control is how hard the modulators work that is how much power they're outputting to various systems as a function of the main power. And then each of these receives power based on both the main power and the functions you determined that tell you how hard the modulators are working. So each system's level of activity is a function of the inputs. I've drawn up above here so these are the equations so for example. This, this system right here has the equation one over p times s, which means as the power goes to zero. So this, this term will say that the system works harder, and this term will say that depending on how hard the spinning modulator is working, the system will the work of the system will vary according to that, according to a factor. Okay, so each system's level of activity as a function of the inputs is shown above. And the goal is to have as many systems with little activity as p goes to zero as possible. So, you know this is, this is just like basic and clear reactor science I assume you all knew this coming in. Okay, actually anecdote. So I used to work at, I used to work at Java juice, and I was sitting in the back, reading one of the newspapers results and looking through the classified ads and there was an ad that said, nuclear power operator wanted no experience necessary. And I was really curious about that so I called the number and it turned out you were signing up to work on a nuclear sub. So, there you go. And the equations for the various systems up above here. So these color coded bars are supposed to indicate sort of like how much power it's getting from each of these and that's reflected in the equations but you don't worry about it. So, here's how, here's how you optimize having as few systems active as possible. The power goes to zero.  And what you're trying to do is you want as many of these to be minimized. Yeah, you want as many of these, which in some cases I've combined the terms to be minimized as p goes to zero as possible. So how did we get these, how did we get these functions. Well, we know that f is equal to p to the x, where x is like some some variable we can choose that determines how f will behave in terms of p, that's the control you have. You set a number x, and then that controls f in terms of p. Well, if we substitute that in up here, then we get the equations down here. So you want as many of these to be minimized as p goes to zero as possible. Okay, so if you look at the exponents of each of these. What we're asking is that as many of 2x plus 2x plus y 2y plus 2x y minus one and three need to be minimal as possible. So, if we look at the men plus arithmetic encoding of the question that I just indicated we were trying to solve. We're trying to get multiple monomials to matter in the following equation right here, when I say multiple monomials to matter these pluses are taking the men of each of these. And so you want multiple men's to be the same. Is that. Is everyone sort of on board there.  Yeah, so this is this is all real valued stuff. Yeah, so these x's and y's are variables that you will input actual numbers into.  Let's see. Okay. Any further questions. I'm particularly curious as to whether I got the role of x and y across. Okay, we can come back if it was unclear, please, please interrupt me. Okay, so let's draw the points in the XY plan where multiple of these monomials end up being the minimum. Well, if we do that. We get this. So these are the coordinate axes. And this is exactly the set of values of x and y for which multiple of these monomials end up mattering to the tropical outcome. If you've done some tropical geometry before you will recognize this as the tropical curves associated with the with tropical polynomial ring up here. Okay, well, if we look at this, we see that we need to examine the following possibilities. So this is, I'm not quite sure why I wrote this this way but this is what x and y are sort of by definition. And the ones that Mac that minimize the, the most number of things have to be located at three for negative one zero, negative two zero, or negative one three. So these line segments right here are what you get when two of the monomials end up having the same men, and these vertices right here are what you get when three of them end up having the same men. Okay, so we can also read off from this. There's a request would you mind revisiting what how x and y are defined. Yes, absolutely. Let's see. Yeah, so the control that you have over the system is what the output of the fizzy modulator and the spinning modulator is in terms of the overall power of the system. And what that means is that you have the ability to choose a number x and a number y. The contribution of the fizzy modulator to the behavior of each of these systems will be determined by p to the x. And likewise for the spinning modulator. You'll set s equals p to the y, and each of these equations up here. And we'll determine how hard the system is working. So x and y are real numbers that are that are under your control. Let's see. Okay. Another thing we can read off from this is that you can't minimize activity for more than three systems at once. Which I think is is like very non obvious from the presentation of problem or even from the tropical polynomial. Okay, so this is more AG right here. So this kind of relates AG stuff to the to the previous stuff I've talked about. And I'll try to be like explicit enough that the AG terminology doesn't just like overwhelm you if you're not like familiar with it. Problem four is this is this is not an actual technical term I just made it up for this is dilation stability in very affine varieties. This is an actual term, I have no idea why it's very strange. So if you have a close sub variety why of an algebraic tourists TN, you call that a very affine variety. Let me indicate what an algebraic tourists is. It's related to but definitely different from the product of s ones that you think might be used to if you are not like AG brain poisoned. It's the product of affine space minus the origin with itself and times. So it's, I guess, you know in the cases where you where you can say this it's homotopy equivalent to the tourists that you used to. But yeah, in AG that's what we call the course. Okay, so suppose we have a very affine variety. I should mention, if you, if you don't know what a variety is this, this might be a little rough, but you can think of it as a complex manifold, possibly with singularities, and you'll be fine. So if that's if that's more approachable. Just think of that. Okay, so one question is, what does why look like near the origin. So this is maybe not exactly the question will end up answering but it at least leads us to like think about the question will end up answering. So the usual AG answer is that there's a procedure called blowing up that you can do at the origin and AN. And then you can take the closure of your very affine variety within the blow up. And if you look at the exceptional, the exceptional fiber, you get a sub right of PN that tells you what directions, your why is coming in at. So, you don't quite need to understand the details there, but it's a procedure that exists.  So, you know, algebra is considered that geometry but like, you know, normal people may not be as happy. So you might ask, can we understand this question numerically. Well, let's say for the sake of simplicity that why is actually like a hyper surface inside the tourists. And by that I mean that why is defined by the vanishing set of one equation. One function f inside of the coordinate ring of the tourists which looks like this. So, that means you've got a field and you would join a bunch of variables and their inverses and you take one polynomial in that ring. Okay, so we're going to experiment with this suggested concept by seeing what the limit as the approaches zero of f of z looks like as we approach zero from different directions. Okay, so this is not a well defined notion yet. So I'll figure out what that means, we would like to do something similar to the reactor problem. So in the reactor problem, we had one parameter in that case it was the power, and we set each of these variables to be defined in terms of that parameter. So the parameter is sort of like was, it was a regulator that we could compare the other variables to. Okay, so technical technical term. And so if you have a valued field, which means a field equipped with something called evaluation that sort of axiomatizes properties of taking a logarithm, which goes from the unit group of your field to the additive group of our the image being referred to as gamma. Then we say that that valuation that generalization of the logarithm is called split. If there's a choice of splitting for this shortage X sequence right here. There's a theorem that if K is algebraically closed, then a splitting always exists. It is also a theorem that didn't write down that your choice of splitting actually like doesn't matter for anything in the theory. Two choices of splitting will always give you like a canonically isomorphic collection of results. So why, why would we want a splitting. So the point is to mimic this property of power series where you can calculate evaluation by factoring out a bunch of powers of the relevant variable. So, let me actually write down what that, what that indicates. Let's see. So we write, if we have a function F that we want to take the logarithm of, we write it as F equals two to the W, F tilde with the valuation of F is equal to W. So splitting lets you always, always do that. Okay, so assume that K is valued and has a splitting. So, and for the most part, it's okay and tropical geometry and the valuation is trivial. You kind of recover like more, more torque geometry stuff when you do that. So let's assume otherwise for a moment. So geometrically, if you have this ring right here. K join a bunch of variables in their inverses and K is equipped with evaluation. The picture you want to have is this. You have a n dimensional tourists. So that's each of those variables gives you one factor of affine space minus the origin.  So I don't really have. I don't know if I can like explain if you're not sufficiently, you know, unhealthy in an AG way. Why that's like the correct way to think of it. But that's kind of like what the algebra of such a ring is including. Okay, so, in particular, I would suggest that you think of this variable T that arises in the splitting as being actually a lot like the zi. It's just that it's encoding movement along sort of this infinitesimal access. So you can kind of think of your variety as TDN times. Negative epsilon epsilon cut out zero. Sarah, can we go back to where he was introduced. Yes, absolutely. So T is yeah so T is T is the splitting right here. So that means if you have a particular value of your valuation in mind. Then you can write it as the image of some variable, and this will be like compatible with the relevant group operations. So the way to think about that is if for example you have like a discrete valuation so this is like the case of looking at power series at a point. Okay, for complex for holomorphic functions. Well the, the image of the valuation map associated with that will be the integers. And for each of those integers. There is a function Z to that integer. So that the valuation will output it. So that's like a sufficient explanation. I don't know if I can think of this is like a uniformizer. Yeah, I don't know. Let's see. So I mean yeah certainly certainly like in the in the discrete case. Yeah, take a uniformizer. Okay, I guess what I want to ask is so we have this variety. And so you have this wise in the torus, and it is in which is in a and K affine case piece is required to be this valued field. Right. And for example it could be like this, this run series. Yeah, absolutely. Okay, let's see. So, here we're going to remember that that F is the is the function that our wise define defined as the vanishing set of. So we're going to learn how to take what's called the tropicalization of F. So the way you do this is that polynomials in this this ring right here. Have a canonical. This ring is canonically a vector space of this coefficient fields with basis the monomials. So you write it in terms of that basis. So you've got this, the sum right here these are the coefficients associated with particular exponents. And you define the tropicalization of F to be essentially, you take exactly what you wrote down and you replace the operations with O plus and no times. You know it's important that you do this like canonical decomposition first, and then you get something in the tropical semi ring that is your usual tropical semi ring. Hang on a sec union infinity. There we go.  Okay, so for an example of this procedure. So if F is this thing right here. So we're going to rewrite this in terms of no meals and the C sub I. And let's see. Oh yeah that's right. Yeah I forgot to mention, you do something to the coefficients. What you do is you take the valuation of the coefficients. So replace operations and take the valuation of the coefficients. Sorry. Okay, so you write these things in terms of the no meals and then you factor out as many powers of tea as possible. And then that will tell you what the valuation of the coefficient is. So for example, here. T has valuation zero. Here T minus one times T cubed has valuation three.  And here you get a coefficient of three and so on. Okay. Now, if we substitute. We do the w I for Z sub I, and w I here is some choice of weights that are in the image of the valuation. Then the tropicalization of F evaluated at the W sub I is calculating the exponent of the term which will dominate as T goes to zero. So like filter through. We're almost done by the way I know we're almost out of time. Things are winding up. Okay. So, for example, if we have F of so here are weights are we choose w one equals one choose w two equals negative two. So, if we evaluate F of T to the one comma T to the negative two, which is this substitution right here. Then we get out one minus T times T to the negative two plus T minus one times T squared plus T minus one. And as T goes to zero the term that will dominate here as well is the one with the most negative exponent. Which is that's right here to the negative two. This will kind of fade away. Okay. So, as it happens if you take the tropicalization of F and you apply it to the weights one negative two. Then you're computing exactly the same thing you're taking the minimum of negative two, which is this exponent here to which is this exponent here and zero which is the exponent associated with this. And you get negative two corresponding to that. Okay. Okay, so suppose that's only one of the terms, only one of the monomials for a particular choice of weight w contributes to top F of w. Then, as T goes to zero F will approach that monomial. So, there's a couple possible cases here. So if alpha this exponent here is is zero. Then F is approaching the coefficients. And so the vanishing set of F will approach the empty set. And F will approach a non zero constant. If alpha is not equal to zero. Then F is approaching this monomial right here. And so the vanishing set of F will degenerate to a bunch of axes and planes and infinity of an. To be clear, we might have like, say, Z one over Z two is what we end up with with a particular choice of weights. Let's see. No, that's not a good example. Okay, I'll come back to it after I after I go over time. So from, from TN's point of view what this is saying is that again the vanishing sets of F is just like disappearing from our field of view as T goes to zero. So, this suggests the definition call the points are and we're multiple monomials contribute the vanishing set of top of F. So if we restate what we just said, the vanishing set of top of F encodes the weightings of variables for which simultaneous dilation according to the weights can yield the generations of why the actually live inside of TN. Okay, that's all I have to, that's all I have to present. So I should have stepped an example for this in here, but it doesn't matter. I went one minute over time anyway so it is actually a brilliant decision on my part. Let's go ahead and thank Sarah for a wonderful talk. Anybody, anybody have any questions. Sarah, can you go back like a page or two. No, this one, this one, this page. So, in your example, there was one plus T. This had no power of T. So shouldn't be when you take the valuations that should this be a negative infinity or infinity. I'm sorry for the X negative to the negative two. Sorry, I believe you're muted. Sorry. Yeah, so this says this has a valuation zero. So you can, because it's because it's a unit essentially in the, you know what I mean in the in the valuation ring associated with it. Yeah, you know, like one every T has valuation negative one T has valuation one so if you multiply the two together you get like something that's pretty similar to one plus T so it has valuation zero. It's weird to see there's zero in the coefficient. Yeah, yeah. Yeah, I think it's best to just try not to think about it. You know, which is arguably true of all of us but So there is there's a question chat. We get off here. The question is that when we pick off the trivial valuation this recovers some work stuff. Any idea or hints on what that might be. Yeah, let's see. So for one thing. If you take a polynomial and you have a If you take a tropical polynomial and anything about this for a sec. Okay, no, no, the correct statement is, if you have, if you have a polynomial and the valuation on your field is trivial. Then, when you take the associated tropical hyper surface. So, for example, like the curves we saw in the case of two variables that is like the one dimensional skeleton things. What you will get out will actually be the skeleton of a fan. You'll get something fan like like with non trivial valuations you can get sort of like polyhedral complexes. Like this. And then with with trivial valuations, you will only get things that that look like if you take a vertex.  You'll get things that look more like that. So in general the connections between like, between torque varieties and tropical geometry are pretty close. So my suspicion is that if torque stuff were developed today would probably be written in terms of tropical geometry. I want you to sort of like a canonical source in the connections there which is, there's a textbook Diane McLean and chapter six is all about connecting torque varieties and and tropical stuff. Any other questions. Thank you. Thanks for once more.


Hey, everyone. So I guess Mark mostly went over the introductory material on this with somewhat better slides. So as you know, we use the random oracle model a lot to model protocols where we still don't know how to prove things in the standard model under standard assumptions. And in the quantum world, we want to be able to port this over, so for which purpose we have the quantum random oracle model. But the proofs in that model are much more difficult and often give looser bounds. And we would like to, of course, be able to distinguish between some cases the proofs give looser bounds, and some cases the attacks are actually better. So the most well-known example is Grover's algorithm. And so in that direction, we want to sort of develop techniques that work in the classic ROM that still work in the Q-ROM. So you heard in the last talk how to do this for recording the adversaries queries and responding adaptively. But there's still a large number of caveats on that. And so it requires quite a lot of expertise to wield. So a simpler technique, but a weaker one, is this one way to hiding technique that sort of still works in the Q-ROM. So here we're trying to make that easier to use, apply to more cases, and have at least in some cases a lower security loss. So the original work comes from Unruh, who's one of the authors on this paper in 2015. But his one way to hiding only works in very restrictive settings and has a quadratic security loss in the number of queries Q that the adversary makes to the oracle. And in here, we sort of generalized it, removed a bunch of the restrictions, and tightened it so that it can be as low as linear loss and linear in the depth of the queries that the adversary makes to the oracle, which practically speaking could be much less than the total number. So an outline of the talk is a brief, brief discussion of the differences between the classical versus the Q-ROM. And these new one way to hiding results. And then some examples of how you can use them to prove things, sort of time permitting in a relatively sane way.  And so you suppose that the attack would still work if the hash function really were a black box or a magic eight ball or something. And so this formalizes a lot of intuition that you might have when designing a protocol, such as the attacker can't know anything about the hash of a message without actually knowing what that message is. And it allows you to furthermore extract the message because it has to be presented as an oracle query before the adversary knows anything about it. And so you have all these proof techniques, right? Where the simulator can control the random oracle, can see all the queries, can choose answers adaptively, can rewind the adversary and so on. And furthermore you get some very simple information theoretic limits about what adversaries can do on certain problems. So for example, if you're given h of x, then the adversary can't find x with probability more than q plus one over the domain size because while he gets q shots to query it on random oracle queries and then maybe one guess with his output. So there's a sort of trivial proof of that almost in the classical random oracle model. But in the quantum random oracle model things get a little bit more complicated. So the adversary can now query the hash function in quantum superposition. So as Mark said, right, you could imagine like he puts in the sum of amplitude times x and gets out the sum of amplitude times hash of x but that doesn't actually work because it's gotta be reversible. So it's actually this, you know, x, y and then hash of x gets x ordered into y but same basic idea. And so it's been shown that quantum adversaries can do more in this model with fewer queries on some problems and on some they can't and so it's kind of hard to tell intuitively which ones those are going to be. And it's also very difficult to record queries to oracles. So if the oracle is a uniformly random quantum oracle then the previous talk says you can sort of record the queries approximately and there's a little bit of loss in recovering them. The simulator can in some cases respond adaptively but again there's a lot of caveats on that. So we're gonna show a technique that can recover some semblance of tightness and proofs without necessarily getting into the analysis required to figure out quantum recording of queries or to analyze too much into the quantum states. But before we do that, one note on depth restrictions.  So you wanna make two to the 64 queries to an oracle for your attack. If you're Bitcoin then you're doing this four times a second every second. And that works because the Bitcoin network has a huge number of computers that are operating in parallel or rather ASICs operating in parallel but if you were to do this sequentially even if you could do it every clock cycle at five gigahertz it would take more than 100 years. So practically you can say essentially no adversary on computers that we know about is going to be able to do two to the 64 sequential work but they might very well easily be able to do two to the 64 parallel work. So in addition to the number of queries we're going to use a depth of queries a sort of a circuit depth where the things that add to the depth are oracle queries which isn't necessarily less than or equal to Q. And all this really does is in the analysis it replaces query with round of queries. And the reason we're doing this is that while in the classical random oracle model it doesn't necessarily help you.  In the Q ROM it will help. So for example Grover's algorithm is going to depend on the depth of the quantum queries and not just the number of them. So onward to one way to hiding. So the classic version of this idea is very simple. So if you're doing a proof by a series of games then you're sort of cheating the adversary a little bit more in each game step. And one way that you might do that is by replacing the previous random oracle with a slightly different possibly less random oracle. So if you have two oracles actually they don't have to be random. They can just be any oracles classically that agree everywhere except on some like possibly small set S. Then an adversary can't tell them apart without querying an element of that set classically. It's pretty obvious, right? Because they behave the same everywhere else. And so in particular if you have a simulator that's observing the oracle queries then they can extract some X in that set if they can recognize it. And the adversary's probability of telling the two oracles apart is necessarily less than or equal to the probability that the simulator extracts an S. And so this is how you'll do, the simulator can't tell apart the encryption of two messages unless the adversary can't tell apart the encryption of two messages unless he can tell the simulator the RSA problem answer or whatever or the discrete log of something CDH problem. This is how you're gonna construct your proofs in general. Of course it's also possible that the simulator is not able to recognize elements of S like if it's a CDH problem for example. And in that case at worst it can randomly choose one of those queries and gets that that was the element of S. And then if the adversary is telling things apart with some probability that's less than the number of queries times the probability that the simulator is extracting things. So again, while this talk is sort of nominally in the random oracle model this is gonna work with things that are not necessarily random oracles. They can be oracles drawn from any sort of arbitrary distribution. They can be fixed oracles. The adversary still can't tell them apart without querying the place where they differ. So the quantum case is a little bit similar. So this appeared in Unruh's work in 2015 but with a lot of restrictions. So again if two oracles are the same except on some set S the adversary can't tell them apart without querying that set. But it might query a superposition of many inputs some of which are in the set and some of which are not. And they might have certain amplitudes possibly small amplitudes on the set and that would give some chance of telling them apart and so on. So the one way to hiding theorem says that again this is assuming to start with that the simulator cannot recognize the elements of this set. So it will randomly choose some query or since we're doing this depth restricted model randomly choose a round of queries. Measure all the queries in that round and then output them. And it's possible that one of these is in fact in this set and call that event guess. So whereas before the probability was the distinguishing probability was at most Q times the guessing probability. Here it's at most 2D times the square root of the guessing probability. And this will end up actually sort of being more like two times rad QD guessing probability because this here we're outputting like multiple possible guesses. And in the previous one we're only outputting one. Another fact that comes from the same theorem is that the difference of square roots of probabilities of two events occurring not just the distinguishing probability this is also bounded by the same value. And that's gonna be true for the other theorems in this work as well because of the structure of the proof. And the main point of this is that usually in these oracle proofs at some point you cheat the adversary so hard that he can't win. So if for example the adversary can't win here and this probability is zero then you can clear the square roots and you won't have this obnoxious square root. Whereas up here if you clear this then you still have a square root in the bound. So this raises the question of like well what happens if the simulator can recognize the set being queried. And so you could imagine well you just measure whether the input is in that set or not and that's what we do but you can't do it exactly like that. So suppose you have some quantum algorithm that recognizes an element of the set. The idea is that you get this quantum input x and then you run the recognizer on it and you measure only the output of the recognition function and not the element x. And then that tells you whether it's in the set or not. And probably once you recognize that it is in the set then you're going to measure the x and output it but if it's not in the set, if it's not the thing that the simulator is looking for then you just return the output of the oracle and continue on. And so this is what we call a punctured oracle. It's the hash punctured by this set that you're recognizing. And the reason it's called punctured is that as long as the adversary makes queries that are measured not to be in the set then his behavior doesn't depend on the value of the oracle there. So it sort of effectively removes those inputs from the domain of the random oracle. And it also follows interestingly that here we had a guess given b talking to h in this bound and since it's symmetric it's obviously also true with b talking to g but those bounds are not actually the same. It could be that in one case the adversary finds a thing and then he keeps querying it over and over and over again. So if you guess a random query to look at then you're always gonna get yes on the one side and on the other side he doesn't query it over and over again. But here the finding probability is actually the same on both sides. So the main lemma that we have on this that slide was just the description. So the main lemma is that if you have these two sets then the bound now has d or rather d plus one inside the square root. So before it was 2d square root of p guess now it's 2 square root of d plus one p find. And you can actually, this is not in the paper but it's not difficult to tighten this d plus one to a d in here but not here. This one is still d plus one actually. Also the difference between a h and a h punctured is bounded by this smaller term as our a h punctured and not find, a g punctured, a g punctured and not find. Some fairly large number of things that really only differ in the case where you found an element of s. And so what this is saying, so up leveling a minute from the equations, what this is saying is that measuring whether the adversaries queries are in this set will disturb the adversaries state. That's unavoidable because that's the observer effect in quantum mechanics. But it only will disturb the adversaries state by an amount that's proportional to the chance that he actually was querying an element of this set. And therefore proportional to, or the square root of that actually, but proportional to the square root of the chance that he was querying an element of the set. And therefore proportional to the square root of the chance that you got the answer that you were looking for. If the adversary is not very good at finding elements of this set and querying them then measuring the queries or measuring whether they're in the set doesn't change it very much. So I don't wanna go into the proofs of these theorems. They're not terribly difficult, but the basic ideas that they are based on the sort of geometry of quantum states that quantum states are like unit vectors in this high dimensional complex vector space. But basically you can think of them as just being like vectors, right? And so the idea is that you can bound the probabilities of differences in probabilities of final measurements having some distinguishing probability or whatever based on the distance between those quantum states in a Euclidean measurement. So it's well known if you do quantum computation that this thing is bounded by something called the trace distance. In fact, that's basically the definition of the trace distance. And both that and this difference of square roots are bounded by another quantity called the Bohr's distance which is more complicated but is roughly the expected Euclidean distance between the two states or at least is bounded by that. So that means that if you can prove something in geometry about how far apart these states are or how far apart they can be for a given value of probability of find or probability of guess, then you will get these bounds on the distinguishing probabilities. So a final question that is actually pretty interesting to me but is an important detail in these proofs on using semi-classical oracles is what is the finding probability, right? So if you have a classical oracle, takes classical input, gives classical output and you have a criterion that holds with probability epsilon everywhere, then and the adversary has no other information about where to find it, then the finding probability is necessarily less than q epsilon, less than the number of queries he makes times the probability that any one of them succeeds. But if the oracle gives quantum output, then it could be up to q squared epsilon or I guess dq epsilon as we'll see later. But that's because of Grover's algorithm, right? Grover's algorithm lets you get a quadratic or almost quadratic speed up on searching for some rare property in an unstructured function. But in this case, you have a situation where the adversary makes a quantum query and he gets a classical output, right? Is this in the set or not? And so it's not clear intuitively whether this is more like the classical case or more like the quantum case. It turns out to be more like the classical case. So it turns out that this can speed up over the classical case by a factor of asymptotically exactly four. So there's some algorithm that will search at least in some cases of unstructured function for a value satisfying a predicate with a semi-classical oracle for it that's four times faster than the classical case but not more than that. So a result of this is that actually the fully quantum case of one way to hiding can be factored into this search theorem plus the semi-classical case with a loss of about two in the bound but it's interesting to note that asymptotically at least it follows from the other two theorems in the paper. So as a summary of these various one way to hiding methods, like imagine you're trying to distinguish, the adversary is trying to distinguish two oracles, he does it with probability delta and the simulator is trying to extract a value of the set where they differ and does it with probability epsilon then we have these various bounds. So in the classic case, if you can recognize it then you get deltas less than epsilon and if you can't, you get deltas less than q epsilon. In the quantum case, we've improved Unruh's bound from two q-rad epsilon to something like two-rad dq epsilon but I guess we'll get to that in a minute. But the difference is that, the bigger difference is that instead of having to be uniformly random oracles that differ in one uniformly random place, now they can be arbitrary oracles, possibly jointly distributed in some complicated way that differ in some arbitrary set of places. So we've removed a lot of restrictions on that. Additionally, if the simulator can recognize the set that you're looking for, then it improves to d plus one inside the square root. And then, as an interesting corollary, this was proved after this work but it's based on Gendry's talk. You can use this oracle recording technique to, if the simulator can simulate both of the oracles, like for example, if they differ only at the preimage of some function and the simulator knows the function but they don't know, and they know the difference but they don't know the preimage, then you can get a bound that doesn't depend on the number of queries the adversary makes. So that's kind of nice but they have to, there's still a restriction on that. They have to differ in only one place. So finally, time check. Okay, let's get to a couple of applications. I guess I probably won't get to all of them with four minutes of these techniques. So the simple one is let's reprove bounds on Grover's algorithm. So suppose that you have a function that you're searching for, it's got some property with probability epsilon at every point and let's compare this to a function that doesn't have that property anywhere. So h is zero everywhere. So these differ at the set S, which is exactly where, you know, g of x is one and the adversary wants to output an element of that set or just to simplify the analysis, you might say that you want that if you take the adversary and then you run g afterwards, which you could call a sub one of g, then you get one. And so the one way to hiding theorem tells us that the square root of the probability that that happens minus the probability that happens for h is bounded by the guessing probability. But in fact, it can't happen for h because h is zero everywhere. It doesn't have that property anywhere. And also h has no information about this set because it's zero everywhere. So the guessing probability is at most like the number of queries, although we added a query here and a depth. The number of queries over the depth, because that's the average number of queries the adversary is making every round. And so when you plug that into this equation above and remove the square roots, then you get the probability that you find an element with this property in the sparse random function is at most four d plus one q plus one epsilon. So in other words, and I think this was well known before, but it's bounded by the depth of the adversary's computation times the number of queries he can make, not just by the number of queries squared, for example. So it doesn't parallelize very well. Another simple application, maybe I'll skim over this, is that if you have a function fk of x, which is h of k comma x, where h is a random oracle, then this is a pseudo random function. So this is fairly obvious in the classical world. And in the quantum world, indeed you can get the PRF advantage is sort of less than this bound. And this bound is also very similar to the cost of a Grover attack on the seed of the pseudo random function or the key of the pseudo random function. So it's about what you would expect. And this comes just sort of straight from the one way to hiding lemma. So then I wanna mention also briefly the Fujisaki Okamoto proof, which is slightly more involved. But the basic idea is, so suppose that you have an in-CPA secure public key encryption algorithm that's randomized. So it takes an input of coins and you wanna de-randomize it because you need to use it in Fujisaki Okamoto or whatever. And so you'll set the coins equal to the hash of the message. And since it's in-CPA secure, it's hard if you have two messages and a challenge that's the encryption of one of them with some unknown coins to figure out which one of these messages it is. And so the proof technique is that you have your hash H and you can also define a hash G that is a variant of H that actually returns those coins, but you don't know what they are when the adversary queries the right message. So you can't simulate this oracle. But you do know where it differs from the oracle H. It differs at like at most M0 and M1. Actually only one of those, but you don't know which one. And so if you plug this into the semi-classical one-way to hiding lemma, you get that the one-way advantage or the square root of it minus the chance that it returns M and doesn't find in a punctured version, which you can arrange to be zero, is bounded by this finding probability, which expands out to the one-way advantage is less than D plus two times the finding probability. And this is the probability that the sort of find thing either measures M0 or it measures M1. And so this is a distinguisher on the classical, I mean, there's a quantum computer, but it returns a classical result, a distinguisher on this NCPA secure encryption system because it returns either M0 or M1, but the adversary doesn't actually have any information about the one that wasn't the encryption that you got. And so that means that by the search theorem, there's a small chance that he finds it. And the chance that he finds the other one is bounded by that plus the end advantage because it's a distinguisher against the randomized encryption scheme. And so in sum, he has this limited advantage that's like D plus two times the end advantage, so that's not good, that's hopefully not tight, but we don't know. Plus this term, which is actually just a grover attack on the message, which is what you would expect times two or something. So anyway, so this is a useful technique. It's been used already in a few papers doing mostly Fujisaki Okamoto kind of stuff. And I hope that we can sort of simplify it further and like synthesize this with the other techniques that we have, like recording oracle queries so that they can be sort of more accessible and easier to use. So that's all. We don't have time for questions, but Mike will be around. Bye, guys.


Well, welcome everybody to the wrap-up of day one of the boff and hacking sessions of Academy It's Monday September 9th, 2019 And we've got a whole bunch of boffs to get through and the first one I would like to call up is the Maui boff Is Camilo here? One of the night trucks people Maui has mysteriously vanished darn Yeah, Marco's gonna do a little bit on Maui come on up Speak into the mic. You'll be recorded, but you no one else can hear you Okay, so I I arrived not at the beginning so I I don't know all of it But Yeah, there was a bit of a discussion on where to Where to put Maui how to to position it Okay, yeah, I don't really remember Yeah, but but Maui The knee trucks part was was more on their plans on plasma mobile if they may want to have Some slightly different user experience with maybe a different home screen We have infrastructure for for that we also discussed if If there was some some things that can be improved in the in the In the collaboration so That it feels a bit less of a of a different thing so Depending how how things will evolve they may they may start to work on the one plasma UI upstream plasma UI directly or or maybe wanting just some Some Some some little customizations another and other things for Maui was since they They value a lot having the application on Android But they are still not on our Android CI so Exactly They they reach to the conclusion that they should They should get to into our Android CI so Exactly what what they need to do on their part to have them just worked there. So they They will be at the Android both Yeah, that's it. Thank you Marco and in the afternoon they moved on to the get Lab get lab off. So I'd like to hand it off to Ray. Thank you So I think we had a pretty good Session I counted about 35 40 people in the room talk briefly about our community product And Alessia who's I mean both Alessia and Yacopatou go back and catch their trains I think we appreciate a lot of questions especially when Alessia was going through the workflow of Committing code doing reviews relating them to issues Appreciate your questions. I mean there are I mean a couple of pain points that you that we go through as well like Selecting range or a range of lines like when you're commenting on something that's something that yeah, we're familiar with Good question about federated user profile. We'll definitely have to check on that But if I miss anything big let me know and then Yacopo kind of finish it off with His experience contributing to get lab as a community member and hopefully we'll see some of that as you guys continue to kick tires But that's about it Thank you Ray and Rounding off the day in that particular room was the Higg and Kirigami documentation Fabian Liepmaier And So we did a short presentation of the human interface guidelines and talked about Problems with using them and then we had quite a longer discussion about the Kirigami documentation Yeah, so We we do have right now the API docs of Kirigami, which is which is awesome even though The presentation of those dogs will have to improve but this is another set of problems What we want to also have is a different website, which is the developer documentation more high-level compared to API docs so with many tutorials and explaining many concepts Which are more about the interaction of the various parts rather than the documentation of one class and We will try to do that and see if if it can be used as an example for other KD projects or if it's a Good idea all together or not we have Some volunteers to work on some parts of this website We will We have a mailing list that we never really used so we will take the mailing list as kind of repurpose it as the central point to where to coordinate the work on documentation writing and see how This will go and yeah, hopefully to make it way more useful for new people Thank You Hig folks Moving on to the next room. We've got in you one oh two We had promo stuff all morning by Ivana and Paul Paul's here Okay, so we started off by looking at the state of promo where we had advanced and We found that Nearly all our figures was up set for in something like the dot and we were wondering why that was like that So we are thinking of maybe giving it a new look and try and see what sort of topics we can make We can publish there to attract more traffic to that. We're finding that the headlines of the dot do a Raise interest on social media, but people are not clicking on them to go to the site so we thought of several strategies we could try to Increase traffic to the dot Then we talked about announcements and we know that for example the last one for apps 19.8 1908 there were some people that were not we're not happy with it We had some our reasons to do that and we we think that we probably didn't communicate them So we're gonna think we're gonna think of ways of We're gonna think of ways no we thought of ways of communicating with the developers better And so they could submit there we could have a two-way Feedback so that we could Use their input for to make the announcements better and they could use our input so that they know what sort of information We need to Ensure that they that they are fairly represented in the announcements then we talked about KD arc that the website which is Is we found several problems with the website with the landing page of the website and We talked about possible redesigns that could be used to make it more visitor friendly and to do some to get users to actually Get to what they're looking for when they go to KD arc Carl Schwann will be having a buff about this tomorrow and We promo forwarded some ideas and He's gonna talk with developers tomorrow From those two things we will come up with a design that will make that better. That's it. Thank you KD promo Moving on to the next room How many Davids have we got here right now? One oh Dave okay several Davids hosted a kf6 planning message meeting so one David Right so we had a session about framework 6 and the relation to Qt 6 and we discussed a number of topics one of them was When do we actually branch out kf6 and start using Qt 6? It's actually too early for that. It's gonna be you know, maybe another six months. We want to use Maybe when the first Qt 6 alpha is out then we can start doing that But before that we still have a lot of work. So we discussed a number of Things that out that should be investigated or done and actually so we have a new board on fabricator It's empty right now, but we will fill it with the It's not empty anymore. Oh, it was empty for a long time, but it's not cool and Then the work of actually figuring out all of the cleanups that have to be done that is Quite some work in itself and we thought maybe we need to do an actual sprint for that possibly early December So that because that's basically what we did at Rwanda in 11 that was quite some years ago for For frameworks 5 and that that has led to a number of very interesting results So we would like to re redo that another time and then Once once we start branching out 6 the idea is that When we start releasing from work 6 then we will still have a kind of a stable branch for frameworks 5 so that We can still do Less frequent but still some releases of frameworks 5 in parallel Maybe every three months or four or whatever there are security issues or whatever so We managed to avoid branches until until now, but at that point we will actually have to have branches Yeah, then we discussed a number of you know specific issues What do we want to get rid of how do we make a support library of these kind of things? But that needs more investigation There was only a single Dan involved in the next boss and it's delegated Yeah, there was an unfortunate K alarm accident for Dan so that ended up with me Yeah, we in the pinball we discussed various topics that Currently relevant in in that area The first one is finalizing the move of K contacts and K calendar core to frameworks That should hopefully be executed within the next week So the October release of frameworks will have those two included Then we discussed what to move next to frameworks mainly driven by requests from the plasma mobile team and it looks like the the next one to look into is the Duff protocol implementation necessary for Kaldar and Carter and Then we looked into Removing some of the pain due to external Users depending on internal API And that seems to be in a large extent to due to address book access And the way forward there seems to be that report those two K people which is the abstraction for that independent of Akonali And then we discussed creating something similar like K people for calendar Again driven by my plasma mobile and push on specifically and I just heard that Nico is already implementing this That could for example replace the very ad hoc integration that we have for the plasma calendar applet and have that on a on a cleaner and better way to access actual calendar and data and Yeah, in general we discussed on how to get the plasma mobile PIM team and the dinosaur PIM team closer together and then Collaborate better We discussed on how to get a sunshine into the application release That is waiting for Kevin to finish the current release and then we'll hopefully be able to integrate that Oh, yeah, and finally Following the kf6 discussion one part that we need to look into is The heavy use of cross in the account wizard Which led to in discussion on how to generally Redesign the whole account handling in K mail And making more use of the work that is ongoing that has already been done in K accounts And how to bring all of that together Yeah, I think that's it Thank you dinosaurs Moving on to the next room you two oh seven had a release team Meeting I'd like to call on Albert's to talk about the releases Yep, so we talked a bit about the pain points. We are having with the KD applications releases David mentioned that he has already fixed some of those for the frameworks releases. So we're gonna steal that from him There's also some pain points related to how we update upload applications to flat up Alish has some ideas we might want to implement. So there's work on that Nobody from the plasma team show up, so I don't know if they have problems or not There might be a plasma both at some point where that will be discussed Yeah, that's basically it we forgot to make to took notes So don't do that when you start above start decide that we who will take notes You will have you won't have the problem. I'm having right now of not remembering what we talked about And that's it. Thank you Or maybe you could delegate that works, too Moving on to the next room snap craft and snaps was run by Hattles And maybe John wants to talk about it John is in hiding But it was actually run by Igor who is I guess not here, so he's really in hiding and he gave an introduction to packaging stuff with snaps and how they work and how you make your programs with them and how they install and then sat down with him for a couple of hours to go with some problems and Find many problems and and maybe fix them Let's see if I can get how old to show up for this one open QA for KDE Kai is not going to talk about it So once again we talked about open QA just like last year the result is exactly the same as last year We want to use it more and we are not quite sure how We look at the bits about at what we already have and it's not good enough So we wanted to make it to so we want to make it better But the road there is a bit unclear what we certainly want to have is more tests and more comprehensive tests But yeah, there's some foundational things that are still not in place to actually enable Making tests or creating tests easily. So that's a bit of a blocker Thanks, Harold mama rock Well, you wrote it you're on the wiki yourself Next time delegates writing the wiki entry on users on user support Yes, okay, we did actually gather people from all kind of Corners of KDE who are closely or largely relates to user support be this documentation be this wiki be this Sis admin etc We did a short state of the user support Being aware that there are channels where we get user support questions in which are simply not Handable because they don't even provide a chronological order like Facebook or Twitter Who tend to mix up everything there is a similar problem on reddit web user can just vote up or down valuable interesting stuff and Replace it with junk. So those are not platforms really suitable for we have the forum We have IRC we have the mailing lists all of those have their advantages, but of course also their downsides and ideally we would want to have better suited medium that can Eventually replace the forums The problem is for example for all the multimedia applications who are the biggest applications within in KDE like Cree dark KD and life DJ cam who use the forum as a user support very extensively But are limited by its capabilities of using Images and and media files. So the other thing is that if we are working on new Channels of user support sis admin must be on board because they will have to implement it and we don't want to overload their work So the thing came up with the suggestion to use this course which looks very promising Especially when it comes to using screenshots We will Have to talk to this course people because there are some features we would need like for example the categorization of entries Else we will end up with endless lists of Applications and if we can't class them in categories as we do currently in the forums This will be not really easy access for users and Ben also mentioned that currently this course needs Docker and This might eventually not be the best solution But this might eventually also something for the sis admin buff on Wednesday to talk about I'm not not my specialty We want to avoid that people go ask questions in mailing lists Which are not meant for we get regular German questions asking in KDE docs eng Which is meant for English documentation writing? No idea how they come up with that mailing list we get User support questions on the wiki. I Mean if we miss these questions It's not really a surprise because nobody looks at that the discussion feature in the wiki is not meant for user support So ideally we would like to have a better suited User support for the users easier to find that there is also a necessity from the Projects to really take care of user support. We want to make it easier for them The problem currently we have on the forum is when you don't immediately answer a question you will not get the updates because somehow this doesn't work and You get updates, but only there has been a question But no mention of the content of it or the content of the discussion, which is not necessarily something You want to look at when you are Developing but we came to the conclusion that we need to make it better But we need to also have all the projects Contributing because a project which has users need to take them seriously because the day you don't We stop taking care of users We will lose them and the project with no users can close shop Thank you a project with no users can close up shop But it's a good thing that we have goals to guide our future and Lydia is going to talk about the office you ran Yes, we talked about the goals And we did three things the first thing was looking back at the previous goals and what we learned from them what We can improve for example one thing that we can improve is making it clearer what Kind of tasks can be picked up for each of the goals and then in the second part we looked at the new goals and what Kind of immediate next steps can be done there one of the things we identified is a landing page that Leads people to the places where More information about each of them can be found as well as for the previous goals And the third thing we talked about is that With the new one new goals being voted in that does not mean that the work on the old goals stops But quite the opposite. They're still very much At the center of what we do right? They just Integrated in everything we do now and we can shift focus a bit And rounding out the day in you to oh eight a was a rather sleepy boff on neon for everyone John do you want to talk about neon anymore? We They were not fireworks. There was just some sleepy people discussing submissions with this neon most of which can be resolved by updating to a new version of cute which is handy because there's a new version of cute out that we want to update to and Installing packages that for sort of external various proprietary things that people have problems with and looking at that And one of those all work arounds the user experience is not necessarily great and it's worth Worth us working ahead to make that great And for everyone who was awake during lunch hour there was the KDE EGM that's a traditional thing on the Monday and Frederick is going to tell us all about it Yeah, so some participate it. I don't want to talk much about it We actually saw that business in the last year was Conducted in a good fashion so the board was relieved which is awesome, and we had election of two new board members Where are you guys near fitters and eight? Congratulations, I think Thank you very much That wraps up the first day of is in does anyone feel slighted or missed did anyone do anything interesting? They want to talk about right now If not, then I wish you all a pleasant evening