It's my pleasure to introduce our coverage of ribulose. The e-commerce is a factor B. So yes, these are all proximation algorithms for us. Thanks for the invitation. So the topic of today's seminar is group fitness and combinatorial optimization. The joint work with several current and former students. Though it's so here are the people who have been involved over the years in this project. So she knew change was my undergrad. He's now at Facebook. Chang who was a former postdoc, was faculty at UIC, Brandon who was my grad student and now he's teaching professor at Duke. Xiao, who was an undergrad with me and is now at Stanford. Charles knew who was my undergrad and now he's a grad student at MIT. And the last one, my current graduate student, Yang Shen and coming home. So I'm going to present a bunch of papers deserves from a bunch of papers, but it's very cohesive and there are a bunch of open questions. So I like talking about this work through agenda problem that I'm going to talk about is what is called a committee selection in social choice literature. So you have a set of N voters and you have a set of M candidates and committees. A subset is any subset of candidates. And each committee has a certain weight or size. For the purpose of this talk, we can assume that the weight of a committee is just the sum of the weights of its candidates. Think of each candidate having a size. The committee has size, which is the sum of the sizes of the candidates. Some of our results will extend much beyond this. And every water has a preference ordering over possible committees. I mean, if I give you one subset of candidates and another subset of candidates, a voter, say I prefer this over that, and the preference could be weak. At any point you have questions just ask me and I want the seminar to be somewhat informal. Order to be peak means that it could be in different. Okay. A strong preference would be that I strictly prefer this to that week would mean that I'm I could be indifferent. So I give you a weight limit or a budget k. So the size of my weight of my community should not be more than k. And I want to find a fair committee whose total weight is at most k. Haven't defined what fairness is. I'll come to that in a bit. But this is the overall goal of what we want to do. So just let me first talk a bit about what does it mean for a vote or to prefer one committee or what the other? So each one has a weak ordering of a possible committees. You can take a utilitarian view where you can see that each voter has a utility function that maps committees are subsets of candidates to real numbers. And the only property that we will enforce for most of the talk is that the preference that this utility is monotone, meaning that if I have S and S prime, where S is a superset of S prime, then the water gets more utility from S than from S prime, right? I strictly, weakly prefer superset to a subset. For sure, I mean, but, but we need this assumption for our results. I agree that there could be preferences which don't satisfy this property. But for most of what we'll see that will satisfy this problem. We're optimizing for the water as you prefer more stuff to less stuff. Okay? So one classic setting that is captured by this is what is called multivalent elections. So in this water, so you're given a bunch of candidates and the water comes in every box? Yes, for a subset of the candidates. He doesn't really know for any candidate. I mean, typically you Mark Yes, What a bunch of candidates to others who are indifferent to us. And you need to choose k winning candidates. And those form your parliament or committee or whatever. A voter prefers S to S prime if they have more yes candidates in S compared to S prime, right? I mean, so it's like a voter simply counts how many yes candidates were chosen in the committee. And they prefer committees which have more or less candidates, is a classic setting and core social choice. It has been studied for more than 100 years. There's another problem called participatory budgeting, or what I, what you can think of as the knapsack problem, which is these candidates are public projects. And every project has some cost to provision it. Let's call it SI. And it gives a utility to each to have the water. And you need to choose a subset of projects whose total cost escapes. So the city has some budget K for doing these projects. And voters come in and say, Okay, this project will give me so much utility and so on. The cost of the projects are fixed. And you either, the city can either choose a project completely or not use it at all. So it's an integer allocation. The utility of water gets the status, some of the utilities of the projects that are chosen. And there's not like a toy problem. Pb elections are actually widely run around the world. In fact, the total amount of money that has been allocated through participatory budgeting is order of 100 million. And we have done some applied welfare. We collaborated with the towns of the cities of Durham and Greensboro and not Carolina to like do some kind of advertising to make sure people vote in elections, in the participatory budgeting elections better. So this is like this one. I mean, I'm sure the regions in Georgia that run these kind of elections. Similarly, if we look at the clustering problem, you can think of it as committee selection. Clustering. Given a bunch of data points to set of endpoints, wanted to choose a set X of K centers. And every point, data point has a cost, which is the distance to the closest center, right? So you, you, you open centers every points gets assigned to the closest center. And the disutility of a point is the distance to as closest center. So the way to map it to committee selection is that the voters are the data points. Think of people and they're located somewhere. The other data points. A potential center as a candidate. And a committee is a set of K centers that you choose, That's your clustering solution. And data points only derive utility from the closest center. And a data point prefers a committee or solution x to y. But the closest center, the center to which it is assigned, an x, is closer to it than the center to which it is assigned. And by that, I mean, natural. So far so good, right? Or no question so far. So maybe you can think of network design problems. Suppose you have a single sink and you are given a graph where every edge has some length G, right? Think of it as some kind of road network. And the city can maybe spend some costs, see on the edge and speed up the road segments so you can view it as reducing the length from GE to H0. Okay? Again, you have a quarter, the city has some total budget on how much money you can spend in improving these rows. And for every n naught V, the disutility will be the length of the shortest path to t. Okay? And so if you don't spend any money on this edge, you get length G. If you spend money as the length reduces to HCI. Okay? And the mapping, the committee selection is the candidates at the edges, quarters at the end nodes. A committee is a subset of edges that you purchase. Then n naught prefers S to S prime or a subset S to S prime of the shortest path to its shortest path to t is smaller than S competitors. But here are a bunch of, I mean, so it's, the committee selection problem is a fairly general problem. You can view a lot of combinatorial optimization problems as falling in the springboard. The washer fairness mean something like participatory budgeting. Individuals can have different preferences on what to fund, okay? By fairness, what we mean is that the outcome that we find or the city finds should fairly represent different subgroups of individuals but present in the population. And similarly, if you're doing some kind of clustering and medical applications, you need to ensure that the casting solution is spherical, different subgroups, and you don't incur a large error on some subgroup of agents. This is still vague. Formalize this a bit more. So. So you want some notion of group fatness. I want to be fair to different subgroups. But we want to do it in a way that you don't know the identities of these groups. There could be all possible groups in the population. So we borrow a notion from economics called stability to define our fairness notion. So every demographic slides should feel that they have gotten fair representation. And this demographic group should not deviate and choose their own committee of proportionately smaller size, such that all individuals in that demographic get higher utility from this new committing compared to the old Committee. So if this property holds, then we say the solution is fair or stable. The more formally, given some committee excavate a coalition S of waters. So here SS for the waters called blocking. If they can choose a committee of weight, which is k times their fraction in the population. But if they are, say, 10% of the population, they get 10% of the budget cake, and they can choose a community within that budget. And all voters in this coalition S, prefer this new committee y to the original committee x. If this happens, then this coalition S of voters is a blocking coalition for committee X. And a solution x is set to be stable if there are no blocking coalitions. This definition makes sense. So this would be strongly prefer. If the weakly prefer, then yeah, they don't count. So strongly prefer, at least our theoretical results hold for strongly prefer. So for an individual, the amount of budget they get is really small. So you can regulate down, it becomes standard proportionality. But, or these kind of problems, they won't give very meaningful results. As an example. So here's just a picture. So you have waters, they find some committee. There are a subset of waters and they all prefer this other committee of smaller size. All right? Then this is unstable. Okay? So this solution concept is called the code in economics. It has been around for a 100 years. It has a fair taxation interpretation. And this noodle from 1919, it was probably there even before that. For each voter pays. So there's your total budget that the city has. Each put the N voters, each water-based 1 over n paths towards banging the committee. So there's some thought experiment, right? Thanks for displaying some equal amount of tax. So every committee member CE, costs $1 per unit, weight or size. The coalition, as voters can buy a committee, they can pull their tax money together and buy a committee of that size. Okay? So basically fairness says that no coalition of voters can spend their own tax money in a way that benefits all of them better than what society provides them. Okay? So such a solution will be Pareto efficient, that it can't simultaneously be better for all the waters. There's also scale invariant, meaning that every voter is only using its own utility function to decide where to move. So you can scale up the utility function for one water. It won't change the definition. This is called scaling. But though, for example, I mean just to show that dislike standard things like K-Medians will not give such a solution. So suppose I give you like a clustering instance, there are n over three points which are very spread out. And there are two groups of n over 3 points which are very tightly clustered than k-means, might build two centers here and one center there. Okay? But these two and over three points only got one center. But they're entitled to two centers because they are two-thirds of the population. So they can, they're entitled to two-thirds of the resources, which, in which case the resources are three centers, right? But decoding everyone could improve if you put two centers here. Okay? So such a solution that k-means solution which puts two centers here, may not be stable. So in other words, the users and B and C pay more tax, but they only get one park or school or whatever. What happened. Okay. So previous work in this space. So there's a discrete version of what is called core stability in economics. The notion, of course, stabilities from the sixties. And the general idea of proportional voting and stuff dates back to something called cumulative voting from the 18 sixties. And the earliest paper I could find on it is due to group in 1881, right? But groupings, what I'm coming to that Good question. So what is openness does, does for the multi, when I re-election setting where you have the candidates have yes snowboards. And the utility of a voter is simply the number of yes candidates in the committee. That is output. The existence of a stable solution is an open question. It's known to exist if not all groups of voters can deviate, but they have to have raphe cohesive preferences for them to deviate, exit. They are all democrats and then they can deviate. And that's fairly recent work. And this achieved the algorithm that achieves this stability if the coalitions or restrictive is an algorithm that has been around in this voting literature for more than a 100 years. Again, it's called proportionate approval voting. It's very similar to what is called Nashville fat. That might be something you might be familiar with that. Okay? If the preferences are, if the space of allocations is continuous and your preferences are convex over that space. But think about it as you can allocate any fraction of candidates and your utility function of the voters are concave. Then stable committees always exist. And that's the result from fully in economics than 970. And this is shown by a fixed point argument. This fractional thing. But we want to study the discrete version. And in fact, for the discrete version, we know that for problems like clustering on participatory budgeting, stable committees need not exist. Exactly. Stable committees need NOT exists, is reasonably simple to construct examples where it doesn't exist. That's where we are looking to be a bit more formal. So I'll present what result we show, but before that I'll introduce some notation. So given two committees, S1 and S2 at v S1, S2 be the number of voters who strictly prefer S2 to S1. The committee I've given a committee S of weight k, S prime blocks F there enough voters that can deviate and pay for S prime. But that's what we define. So, so for a committee S prime to be blocking the number of voters who prefer S prime to S times the tax money per water is enough to pay for S prime. Okay? So if this property holds, then S-prime blocks S to S is not steep. As the committee SSTable, if there is no S prime that satisfies this condition. Good. So we know that, okay, this exact definition we can't satisfy. So we'll do approximation. What is approximate stability? The committee S prime c approximately blocks F box S, but think of C as a constant bigger than one, like two or three or something like that. So the amount of the number of voters will deviate to S-prime is not only enough to pay for S prime, but it's also enough to pay for a constant factor more than S prime. Okay? So these two, you need a lot more water to deviate from S to S prime, right? To satisfy this definition. So this will be a harder condition to satisfy the Committee of Eight case c, approximately stable if for every committee, but all other committees S prime, the number of voters who deviate to S prime is less than t times the scale down weight of the weight of S prime, right? The N over K is just like the scaling factor. Doesn't make sense. So basically the condition should not hold. So just write that down. Okay, So far so good. The main result is that, uh, if C is 32, 32 approximately stable community will exist for any monotone preference over the of the waters. And they can have any monotone preferences under the weight of the committee. As a function of the candidates could be any sub-additive function. There is the more general setting for which we can show that a 32 approximately stable community always exist. And this captures all the clustering, might even our elections and so on, everything that represented so forth. And this is the result and stopped 20, I'm going to present a more result is our towards the end of the paper. But I felt this is the most interesting desert and there are bunch of open questions on puppet slot. And we know that for clustering, we can't get exactly stable. In fact, we can't get 2 minus epsilon approximately stable as well. So they are bound to an upper bound of 32. I'll give you a proof sketch of the upper bound because I feel a proof technique is interesting. So what we do is we define a randomized relaxation of stability. And there'll be different from the kind of relaxations that economists have used will be a new randomized relaxation. And we'll show that. So we call this relaxation stable lotteries. And then we'll show that a two approximately stable luckily always exists. I'll present a nearly complete proof of this statement. And then what we do is once we have this randomized notion, can treat it as a fractional solution. What we do is I iterated rounding on top of that issue or 32 approximately stable committee. I won't present much details for this because I think the interesting thing is the existence of this table lotteries. Okay? Any questions so far? Okay, so what is a largely a lot revealed for us will be a distribution. So you are given this weight K on the committee size. A lottery will be a distribution of our committees whose weight such that each committee in this distribution has weight at most k. For example, if K is $40 and I give you three projects or three candidates that cost 10.20.30, then it could be a distribution over the Committee on the left, which costs $40, and the Committee on the right, which costs $40. But for this is allowed this is not allowed because the Committee on the right OS $50. Okay. So a distribution over this committee and this committee is allowed. What this committee, and so every committee in their distribution should have eight at most stable lottery to define that. So let's define, let's call a lottery by delta. The committee S prime c approximately blocks a lottery. If you look at, so you draw a committee S from data and look at the number of voters will deviate to S prime. Okay, So for different members of this, in the support of Delta, look at the committees and the support of Delta, but different committees. The number of voters who deviate will be different, take the expected value, okay? So the expected number of voters times the tax money by water enough to pay for the size of S, t times the size of it. Okay? So it's just taking the previous definition and putting an expectation on top of it. So lotteries, Delta I C approximately stable. Before every committee S prime. The expected number of voters who deviate is not enough to pay for the, the committee S plane. But good to my claim is that for C equals two, there is a lottery delta that satisfies this definition. Regardless of what their preferences are, what the voters and regardless of what this weight function is, roughly regardless of what the wavefunction. So, so just to be more formal. So there's your lottery. We just say a distribution over four committees. And suppose the probability delta is P1, P2, P3, P4. Okay? So take some S prime, but the amount of money that you get from S1 is P1 times the number of voters who deviate to S prime times K over N. Then for S2, S3, S4, that total money can't pay for constant factor times W weight of ESP. So our goal is to construct this lottery such that for c equals to this condition holds for every aspect to the algorithm. So you can think of it as the algorithm chooses. So you can think of S prime as being some kind of attacker riders kind of kill your stability notion. The algorithm is trying to defend against that attack. Then I can think that the algorithm chooses some defending lottery data. Let's call it delta d, d for defender. And we need to show that for c equals to that. Regardless of what the attacker chooses, attacker chooses committee SA k, the expectation over delta ib of sort of the amount of deviation minus the cost of the attacking committee is less than 0. Okay? So this has two. So there has to exist a delta D such that for all essay this condition holds is the same as saying, if I want, if I min over SD and max over essay of this quantity, it has to be less than 0. Okay, good. Now you stare at this and say, okay, the zero-sum game that switch the max and the mint. So if we switch the max and the min, that's saying that the adversary chooses and attacking lottery Delta a. And I want to show that a max over del t times the Min over the defending committee of the same quantity is less than 0. Okay? So, so now the adversary goes fast and chooses an attacking lottery. The algorithm goes next and says, Okay, this is my defending committee that I choose, right? And that will protect against deviation. We will construct a defending largely. Tesco is equivalent, but the other way, I'll show it, this will be delta ID. Yeah, you're right. Okay. So, so there's some attacking latte del dy, right? And these are the probabilities in the attacking lottery. And I want to choose one defending committee. The virtues are defending committee. The number of voters who deviate here to S1 will be this VSD S1 number of voters who deviate here as VSD S2 and so on. So if you look at the expected amount of this, it's not enough to pay for the expected size of the slot. Okay? Times the expected C will be two. Okay? So the question is, I give you this as input, how should I choose SD? So just to write it a bit more formally, um, in what does it, what does it say? If I look at the number of voters who deviate, I can write it as sum over all the waters. Okay? So if I fix a Delta a and defending committee SD, right? What do I want to show? I want to show some over the waters of the probability that water will deviate. So this will be the expected number of voters who deviate. K is less than the expected weight of the Attacking lottery. Okay? So for any Delta a, I want to find an SD such that this condition holds. Now what I'll do is I'll try, I'll show a stronger property that water by voter. This probability will be small. But they're just writing out the expectation nothing. You can have a stronger condition. So given an arbitrary adversaries lottery, they'll die. I suppose the expected weight of the committee is in the lottery, is some constant factor B timescale like B could be less than one. We will be less than one. I mean, think of BSA 1 tenth. Okay. What does have arbitrary preference ordering over the committee? And we find a defending committee SD of 8 K such that for every water V, the probability over essay that the water will deviate is less than twice B. Okay? So basically water by water, if I show this constraint, then what I can do is this quantity less than to be. I can glad to be here. And, and just multiply it out, I'll get that this condition holds. Okay? So basically I am showing it voted by water. Okay? And the algorithm for solving the dual is now really simple. What you do is you sample the bees, think about B has been 10 percent, right? So what I'll do is I'll sample 1 over 2 be Committees from the lottery delta t and take the union of these to construct my defending lottery defending committee SD. Sd will not be a distribution because you're doing sampling. The expected weight of SD is really simple, right? Because I've chosen maneuver to be committees at random. There's one over two b times the expected rate of the attacking committee source, attacking committee, I'd be times k will get k over two. Now you can use miserable. Yeah. Yeah. So you said 10 percent writes a 1 tenth. This would be 1 fifth. So you sample five committees or five committees from the distribution folder. And I'll just add them up. Just take it really. Just take the union. Yeah, if b is bigger than 1.5 you anyway, I wanted to show like some poor approximations are bigger than 1.5. Now, we need to use some dependent, we need to use dependent rounding to make sure that that SD deterministically doesn't have weight more than k. But the details of that are straightforward, not a big deal. Okay? And now the basic idea is that look, take we take some voter V and suppose the, if you look at the attacking lottery, suppose the probability, suppose the committees in these ranking, hey there S1, S2, S3, S4. So V prefers S1 to S2, S2, S3, S2 to S4. And they have some probability in the lottery, which is P1, P2, P3, P4. And what am I doing, right? I'm picking one over two b committees at random from here. Right? Now it's like simple math to show that the probability over the choice of SDI, Okay, That, that the water, I mean, so what should happen, right? That the attacker should be tackled, pick something at random. Suppose attacker pigs this. Now I'm taking, as a defender, I'm picking one over two b committees. All my choices how to write lie on that side, right? Then the water will deviate. But attacking committee for the probability of that happening will be less than two elementary math and that will work on that. So that's, that's how you solve the dual, the given and attacking lottery. There is how you'll find a committee defending committees such that no, What I mean, the probability any water deviates is less than 2 B. So the expected number of voters who deviate is not enough to pay for the Lawton. Do we solve the dual? We need to go back and solve the primal. For that we can, I've used a standard like zero-sum game algorithms to say that, okay, you have an oracle for solving the dual. We just do multiplicative weights or something and that'll solve the primal 40. Only problem with this kind of argument is that our proof is existential and this part of our algorithm doesn't run in polynomial time. Because you have an exponential number of committees. What we can show using this duality is that there's a lottery delta ID or defending lot identity such that the expected number of voters who deviate to any S prime cannot pay for the weight of his due within a factor of. What, what this means is that given this delta ID, any S prime is kind of loading the ranking of many voters because there are not enough voters will deviate to play for the weight of S-prime. Okay? So, so if S prime, so suppose I give you two waters, those rankings are like this. Then S prime kind of is like lowering the ranking. Right? Now. We still need to find the core solution. We haven't found it yet. What we have thought is like, you have found a stable lottery. I had to go from lottery to a deterministic committee. E, it used monotonicity. You do. The union is better and the subadditivity will get used to with somewhat like in this part, I guess we all get the exact because this, these candidates have different sizes. So I don't know how to, if I sampled these many, I don't know how to construct it. So that deterministically, this committee has size less than k. So I want, in the end I want a distribution over committees. Like I don't know how to find a feasible commit. That's an open question in the sense that I feel that for most reasonable preference structure, you don't need a factor of two, but we don't know how to show it. You could show it for if your committee has size 3. There are three candidates in the committee. We could show it, but it's like a hairy calculation even then. But there should be some nice rounding type of argument to show that you don't need this factor of two, but we don't know. No, no. If the well, I'll come to that in a bit, but you're right. Okay. So now what you do for this, so you have this primal solution, which is a lottery. We want to construct a deterministic committing to what you do is you sample one committee from this lottery. Okay? Now, sort of the lemma that we show is that this committee has to be high up in the ranking of many voters. And here I'm being a bit informal. And so, so basically as primates like that. And if you sample one committee from your lottery will be high up. It'll dominate S prime for many of the waters. Right? Now, what those, what does you don't have to worry too much about them. They won't deviate. And what you do is recurse. So you consider all water such that the committee that your sample is low in the ranking and this low, I mean, we formalize it in our paper and it's a formal definition. Armando put it here, it's a bit technical. And what we show is that the number of such voters will be significantly smaller than, and it will be n times some factor which is much smaller than one. Now you just need to focus on those waters because they are kind of your bad voters for that community that you chose. What do you do is you scale down the weight of the committee k. And you now take just these voters who were kind of low down in the ranking high and resolve the problem. Okay? So just because on the smaller problem, we just these bad voters and you want to find a stable committee of smaller size. And finally, what we'll do is just add to take all the solutions that you find and take the union of those. Okay? So you'll get a bunch of committees are geometrically decreasing weight. Take the union of all of those. And what you can show that the final committee has weight at most k, it will be 32, approximately stable, with a whole bunch of technical details, but that is a main idea. We can improve the results for some special cases. They might even our elections where the candidates, whether voters have yes, no votes and they get utility one from a yes candidate and 0 from a no candidate. Then you can, for that setting, you can, and each candidate has size 1 and you want to choose k candidates. For that setting, we can show that exactly stable lottery exist. And the open o, the open question is doesn't exactly stable, luckily always exist, right? I mean, it may not have any, okay, maybe we're considering two general a problem. But I feel that this existence, that stable lattes always exist should be independent of like sort of the utilities. I mean, as long as an additive utilities and your function is additive, I think it should hold. We couldn't find a counterexample, but we don't know how to prove it. But one nice thing about clustering is that if you have a committee to which a bunch of waters deviate too, then you can easily show that. This committee has the five center side. There'll be one center which is also a deviating pick a subset of those water is going to be greater to one central to the you only need to consider deviating committees of size one. Now if you look at the primal-dual algorithm for solving the zero-sum game, you can show it runs in polynomial time. It's easy to see that runs in polynomial time, the rest of our rounding and so on, both the runs in polynomial time. So our clustering, this entire algorithm runs in polynomial. But even for multi-unit elections, you're deviating committees could look big and we don't know how to make it run in polynomial. Yes, the open question is, is there an efficient algorithm for the general case? And also a factor of 32 is like really lose, right? I mean, our lower bound is two. We tried hard to find something worse than two. We can't. I feel that there is an entirely different algorithm that can get the 32 down quite a lot. That is also efficient. Efficiency is too much to ask if you're looking at general utilities and subadditivity and so on. But for reasonable special cases like multivalent elections, we're doing approximation. There is no reason for our algorithm to run an exponential time on top of them. Access. So if you have, say, a participatory budgeting where the utilities are additive, then I mean, sort of like essentially solving knapsack, right? That different voters. So there, there is no reason, there is nothing underlined in the problem which that should make it exponential time. You are additive, say, and if you have something like more gender like submodular, suppose I give you like a value or a lot of demand article. Can you make a polynomial? I don't know. Much time do I have? There's a different notion of approximation that one can think about and there's approximation on lotteries. Even then we don't know polynomial-time. In fact, that is the bottleneck. The rest of our rounding runs in polynomial time. But getting the fractional solution is what we don't know how to do it. But it could be that there is an algorithm for the integer solution that doesn't use rounding and that runs in polynomial time. We don't know for different notion of approximation is under Utilities. And here we can show somewhat stronger results. But for special cases, pension of reducing the weight of a deviating committee by a factor of 32, which is what we're doing, right? We can say that a voter can only deviate if a utility increases by some factor of c. That, I mean, the previous slides, I was saying that the voter deviates if they get more utility. But it could get like an epsilon more utility, they're still deviate. But suppose we say that they can only deviate if the utility goes up by some factor, right? But we don't reduce the weight of the deviating committee. We don't we don't scale down the tax money. So formally for a committee of weight S, which is a committee of weight k, committee S prime is blocking. If a voltage VD VA stress prime, if the utility increases by a constant factor and the number of such deviating voters is enough to exactly pay for this committee, but at least pay for the committee S prime. So the approximation factor is on the utility and not on this Taxman. Okay? That makes sense. You shifted the approximation to the other term. But for this, we, we need something more than, I mean, we need to have some utility function actually specified. Target committee see approximately stable if there's no blocking committee according to this definition. And for this part I'll assume that the weight is additive across the candidates. Before please consider clustering. That committee is a set of K centers, water. It has utility from the closest center of water, deviate store blocking committee, the distance enclosed by a factor of c. The natural definition. But what we could show is that there is a simple greedy algorithm that actually is a 2.42.411 plus square root two approximation. And the algorithm is very simple. We call it greedy capture. And it turns out that it was also, we might have reinvented, it was known in a different context. All points start out uncaptured and your solution set is initially empty. And what you do is you continuously grow balls are found every point. Now. If there are N over K uncaptured points in a ball around a center J. Then J captures those points and you add j to your solution set. And you just keep growing this ball even after this. So j has been added to the solution set. If it encounters a point that is uncaptured you just captured. As an example. These are the points. You start with like really small balls around them. Keep growing the boss. So suppose N over K is three. Okay, so what you do is the center has captured three points. Though you open this, this middle point and you say that the, all these three points are captured. Keep growing. This big ball hasn't captured the points yet. This big ball hasn't captured yet. But this ball in the center has captured the fourth. At this point you can stop because this point can never capture three points. And this point can never capture three, right? And your final solution is just this one. Does you output a solution of size one? That's it. Now, proof is really simple. So suppose there was some deviating center I that captures n over k points, right? But suppose your solution was not stable. And then there's some deviating Center captures a bunch of points and our k points within some radius r. And you didn't open the center in your actual solution. But what must have happened? But there must have been some point inside this ball which was already captured by something that you opened. But it must be some point x0 DeVos captured by a center G that you actually open. That's the only reason why I wasn't opened as a center. Which means that take any point w and this ball, it will travel at most distance three times r to reach G. Which means that this point won't deviate. If you say C equals 3. The only deviates if it's distance goes up by a factor of three. But there is a center which is within a factor of three of the distance to i. Okay? So this point won't deviate one. That's a contradiction. So this shows that this algorithm is 3 approximately stable. If you are a bit more careful in the analysis, you can filter down to one per square root. And there's a lower bound. I forgot what the lower bound this, but there is a lower bound for the last five minutes. I'll just present a recent result I hadn't you have in this upcoming soda. This is for the participatory budgeting problem. So for every hotel, from the utility function is submodular function of other candidates. And the weight of a committee is a sum of the weights of the candidates. And then what you can show is a polynomial time algorithm that gives a constant approximate code for your approximation is on the, you did it. For this, we use a different fractional relaxation. Suppose every candidate has chosen to a fraction XA in our fractional relaxation, plus x I can lie between 01. Now your size constraint is simply that the quarter fractions to get the weight fractions to which you choose the candidates has to be at most k. Now, going to define a fractional notion of utility. For that, we use just classic multilinear extension of the submodular function. And what this says in English is that suppose every candidate I is included independently in your community with probability equal to x psi. And what is the expected utility that you will get? Okay? That would be a continuous function that is called a multilinear extension of a submodular function. This Magdalena extinction is not, it's concave only in positive directions. Otherwise it's kind of not concave. And what you do is you consider the Nashville fair objective, which maximizes the sum of the logs of the multilinear extensions, is sum over all the waters of the log of this utility function. Or you can think of it as maximizing the product of the, of these fractional values. The issue is that this function is, this UV is not really a concave function, is only concave in positive directions. But nevertheless, what we can show is that you can find a local optimum to Nashville. For a local optimum, I mean, that you continuously increase the fraction to which you choose one candidate and decrease the fraction to which another, you choose another candidate as long as your national fair objective is increased. So at the local optimum, there is no pair of candidates such that increasing the fraction of one and decreasing the fraction of another improves your objective. This you can compute in polynomial time. It needs some work to show it, but it's not hard. And our main theorem is that this local optimum is approximately stable fractional committee. Once we have this, then you can again do this iodinated rounding on unsatisfied motors and so on. And it gives like a 68 approximate integer committee. I'm not happy with the factor of 68. That's what we could show. I don't have time, right. Amount of time. Okay. Yeah. And yeah. So let me skip to the next. Let me present this. So since I have two minutes, there's another notion of this, another way to compute these fractional solutions in the setting, that is what is called an equilibrium. Now I'm presenting this because it's like an interesting concept that we've dug up from the old economics literature that I don't think many people aren't familiar with. So let me just present it. There's a market clearing solution for these public goods kind of problem and actually view it as a fraction or exact fractional core. Utility function is additive for the candidates. And what you do here is what this market clearing solution does is that for each candidate I, and for each voter be it computes a price, let's call it P IV. And such that the total price of a candidate is its weight to the total price is the sum of the prices of the voters. Now, give each water it's tax money, and ask it according to these prices by your favorite allocation. The bullet buys an allocation X V. Okay? Now, at this equilibrium, What are this market getting solution? What happens is all the waters by exactly the same solution. Because it's not, I mean, the solution has to be common to everyone, right? So there exist prices such that if every voter behave selfishly according to those prices and buys an allocation All by exactly the same allocation. In fact, this allocation will lie in the fractional. Cool. Okay? The difference with the classic Fisher market for private goods is that an official market? That in this equilibrium, the pore water prices, it has an official market. The prices are common like the prices for goods. They're common. But here are the prices I pour water. And he hadn't done in equilibrium, the prices yield a common allocation for all the agents. All these XPS end up being identical for the water. But as an official markets, since it's private goods, prices are common, everybody gets a different allocation. First flips the what is common and what is not combed. The theorem from falling in the 1970s that the Lindell equilibrium always exists. And it lies and the exact code as long as that utilities are continuous and concave. And the proof of this is a fixed point argument. It's not computationally efficient. And what you can show again using I iterated rounding is this gives a tough, you can approximate integer coordinate. Your utilities are additive. They can improve the approximation factor using this playground runs in exponential time. The open questions are, the nice thing about the equilibrium is that it gives a pricing rule to just price the goods and ask everybody to make their own decisions. So in practical settings, these rules are kind of nice. All right, so the question is, is there a polynomial time rule that actually gives a constant approximation, but in nonequilibrium gives you an exact answer, but takes exponential time. There's some recent work by others that gets a lot of approximation, but, but it's computationally efficient. The question is, can you merge these two somehow? And can we improve the approximation factor of a simpler algorithm, right time you have a 67 approximation and it's terrible. And our utility approximation works for a submodular valuations. What about XOS are subjective utilities? Dad, you can't define multilinear extension, so you need other ideas. But we don't know any lower bounds. The best lower bound is only a constant. And so this is bad. Yeah, thanks. That's no, I don't know any hardness. So in, in some simple settings, it actually can be computed in polynomial time. But for even, but for the participatory budgeting setting, I don't know if it is hard to compute it. I spent some time thinking about it, but I don't have a hardness. So basically because you said you can, somebody give us a very special case of non-linear compliment answer is not clear at this non-linear complementarity problem is actually hard to solve. It almost looks like efficient market like the kind of the way you write a fixed point. Market is a fixed point. I have complementarity problem. So, but inefficient market, right? So it's not clear that you would be possibly converts liquid, but in some cases we could rewrite it as a convex program. But for the other cases, we don't know if it is hot. You can, I mean, it's like a zero-sum game, right? So you can view it as like this primal-dual process, but we don't, We didn't think too much about whether you can biochar because like some kind of separation oracle kind of problem and what you can do it. When we talked some about it. Can you think of it as a tool? Works? Just like you just thought of it as like a solving a zero-sum game. But, but the problem is that in the primal, the number of committees you have is exponentially large and we don't know if there is like some small support for it, that if you only increase weights there it would solve the problem. I mean, that that's where maybe a separation oracle kind of thinking might help. But it's the dual has the single step of the dual has a polynomial time solution. So the dual article is polynomial times solvable. But once you apply multiplicative weights on top of that, anything on top of it, or even ellipsoid on top of that, I don't know because the primal has exponentially many of those committees with your loan meant that it wasn't. Obviously we want to find a primal solution. So at stake, unclear how it would work. But Ugandan it a rule. No, no, no, no, no, no. I don't think we can solve it or they're just one step of the dual the console. But I don't think it gives then it wasn't obvious how to like, I'm stupid or the other we should we should discuss offline. There there's other people. It's not comfortable.