Slowdown and stuff. Okay. So the work I'm going to talk about a joint work with the three people who used to be a grad student at CMU, but it now at the University of Michigan. Jason, who was also a grad student and just graduated. He's doing a post-doc at Berkeley. And David Harris who that who has no connection to the simulator is at the University of Maryland. It's good. So I'm going to talk about the Kcat problem. And really you should think of this is answering two questions. I want to look at objects called minimum K cuts and graphs. And I'm going to ask how many of these are there and how do you find them? Okay? And so what the Kcat? Kcat is just you're given a graph and you want to delete the minimum number, minimum weight of edges so that the graph falls into k pieces. I don't want any balance, I don't want any other criteria. I just want k pieces at the end. There. If k were two, this is the min-cut problem. You just want to break the graph into two pieces. And for this, we know how to find a poly-time algorithm that you can use esteem and guts and you can now output the best one. But when k is large and think of big, large constant for the stock. How do you find these two examples? Suppose I start off with the cycle and I want to get KPC, the old thing. If you think about it, if you delete some k edges of the graph. In fact, if the recycle evacuated, you should delete, any cages would do. And so obvious thing to notice is the number of minimum cuts, minimum Kcats, event tooth sockets. The number of minimum cuts. In the second graph that I'll keep in mind that the complete graph. And hand, if you again think about it, what happens is that the right thing to do? And maybe I'll try to alternate between the two slides so that you guys can see on both sides. The right thing to do is to just carve out k vertices from the graph on k minus one vertices from the graph and remain, the remaining piece gives you the character piece. And sort of the optimum in some, some quantity like that. But more or less, I squint really hard. And I, I'll think of this is pretty much like k times n, right? I'm cutting out k, k minus one, but k vertices from the graph. And each of them is removing n edges. Make sense? Okay, good. Also the number of minimum cuts Israeli like n choose k minus 1. I've written n choose k out there, but it's really n choose k minus 1. K minus 1 vertices is the minimum Kcat. And so if I asked you to conjecture, what's the maximum number of k cards? Minimum kick out the graph can have. A reasonable conjecture, would be anxious and really deserve moral of today's talk is going to be, we got to ask these two questions. How fast you compute Kcat and how many minimum cut Kcats data? It'll give you the punchline. The right answer is n choose k. I don't know how to show it. But I'll show you is a bound which is close. Though. We'll, we'll, let's, let's get to that. And okay. So let me tell you a little thing about prior results. I want I don't want to bore you too much. Actually, if you think about it, it's not clear so many cuts in the graph because you had esteem and cuts. You could find all pairs of esteem and guts and use that to find the minimum to cut off the graph. But if I'm asking you to find three cuts, this approach is not a good idea. I can tell you right off the bat because it runs into NP hardness. But it turns out in this remarkable paper of gold metal Hoffman, who showed that the, there was a deterministic algorithm which runs in n to the k squared time. And then this was improved by a randomized algorithm which runs in a board and to the 2 k time. And in fact, the thing to remember is really the look at the exponent and ignore constants for the time, but not the multiplicative constant value to this, like n to the two k time. This was made deterministic. And then for 10 years we didn't know how to do anything better to recently. And again, I'm going to just flash these. Don't don't spend too much time on these. Recently we managed to break through the barrier of entry to get at those things for unweighted graphs and stuff like this. But And the lower bound, which is about n to the k. So if you catch me later on, I can tell you that finding cake leaks and graphs is related to the problem minimum Kcat. To be the thing, if you ignore the numbers, the upper bound. The lower bound is n to the k. And we can ask the question, where did it come off? A result show is Duke is not the right answer. Maybe 1.98 K and come on if it's not 1.9 or not. Okay. Where can it be? Though? I'll, I'll tell you the maintainer, the randomized algorithm to find the minimum ticket in time n to the k times some nuisance terms that I don't know how to get rid of. And in fact, this algorithm can be used to enumerate all minimum K cuts in about the same running time. The bosonic. Yeah, so you can actually find, you can show, You can use this in a probabilistic argument to say that the number of minimum Kcat is only n choose n to the k. Not quite n choose k, which I believe is true, but I don't know, please. Oh yeah, exactly. So we don't expect something which is f of k times polynomial. In fact, we don't, yeah, so basically n to the k seems like a given almost two. This might be optimal under the right hardness conjectures. And we have written, I say, they vary with an algorithm to find this thing. But I'll actually tell you the algorithm and the algorithm is one line. And it's an algorithm many of you have seen before. You're going to see one of my favorite algorithm, like a beautiful island. The back in the nineties, David Karger gave an algorithm to find min cuts. And he did not use esteem and cuts. Use this algorithm which you might have seen in your undergrad or grad algorithm classes. They could graph, oh, by the way. So for the rest of the talk, I'm going to focus on unweighted graphs. But everything I say can be extended weighted. But now you are given an unweighted graph and you want to find a minimum Kcat. The suppose you want to find a minimum to CAC, how would you find it? You will run this algorithm or you could run this algorithm. And it just says whatever your graph is, choose a random edge in this graph and frontline it. And repeat top when you have two edges with me. And let me just do a little little cartoon out here. You can see on this graph on the left, I pick this bold edge and I contracted it. If I pick, I will remove all self-loops and irrelevant. But I'll keep parallel edges. And I do this again and again and again and again and again. And in this case, it turns out I'm left with these two vertices. Each vertex represents a shrunk set of original vertices. And this gives us a cut in the graph of the natural cut in the graph than one output. But this algorithm makes sense. Pick an edge at random, contracted. So this algorithm is toggles the min-cut algorithm. And galileo showed that it finds the minimum cut. You fix a gut midgut and it'll find it with probability one over n choose 2. Which by the probabilistic method means that there are at most n choose two minimum cuts in the graph. And you'll see a proof in a, in a second. Oh, actually you will see a proof right now. And the proof is, I want to show you the proof because the proof though, what's the, what's the idea? So it just says inductively, if the minimum cut is still there, the graph, I want to miss it with the reasonable probability. To look at the current stage of the graph. I'm shrinking down the graph that are remaining and support that haven't hit the optimal. Keita haven't shrunk any edge of the optimal cut so far. Because the cart is still there, the minimum cut is still there. Or make sure I don't pick an edge on this cochlea. Pick a nice either this side or on this. So the probability of screwing up is exactly the size of OPT, divided by the number of edges. I want to miss this cut. I'll hit Discard would probably opt divided by number of cutting edges. The optimum Min cut in a graph is at least the smallest degree. At most the smallest degree. So you could have taken the smallest degree vertex and that's the cut. The OPT can't be larger than the OPT is smaller than the smallest degree. On the other hand, the number of edges is at least the smallest degree times the number of vertices divided by two, because you're over-counting edges. So that's all I've written out there. And now the smallest degree cancels out. And you say that at the odd step, the probability that you hit this edge, you hit this cut is 2 over at most. So the probability that you don't hit this, it will take one minus du over our product, over our love. And I'm going to use my favorite inequality of all time. 1 minus x is approximately e to the x. And I knew that in the wrong direction actually. So you gotta be a little careful, but it works out. So it's really like e to the minus two are summed over all are, which are like the harmonic clapper. I'll give you a minute to just look at that. It's not no rocket science, yay telescopes, but it's much easier. I don't have to do the calculations. I'm cheating. But actually you can, pointing out is you can actually do the calculations carefully. And it's no more complicated than this. But you're going to see a lot of such, such tricks him a little bit. You might as well get used to this good. And you've seen one of my favorite algorithm, that's good. And then how do you do Kcat? Well, the algorithm is you stop and that k vertices remaining. And my claim is that this is going to find the minimum KYC or minimum kicked you fix your minimum ticket. You got to find that minimum kick out with probability at least 1 over n to the 2k. And if you plug in k equals two, you'll get back n squared. And the analysis is very similar. Or the probability that you'll hit the U, the optimal step is October, the number of edges. But now I have to actually use something bold. The one candidate for AAC is the sum of the k minus one smallest degree vertices. And one fancy way of writing that if K minus 1 times the average of the smallest k minus 1, the number of edges is r times the average of all the degrees divided by doing that, the average of the smallest k minus 1 degrees is no larger than the average of everybody. This is two times k minus 1 divided by table calculations. But now the calculation is a little messier. And here I'm cheating. Cheating, but not too much. It's everybody happy. This is the state-of-the-art circa 1960. And making this has been surprisingly difficult. Let me try to give you a sense of why what happens at the two examples. The other thing is that this argument, if we're going to break, oh, actually I haven't told you the important thing. I want to analyze this algorithm. And I want to show that it actually find the minimum cake. I could probably about 10 to the minus k, not n to the minus two. Yeah, Good. So if when r is smaller than about two k minus one, then those probability that bogus. And so that's what I meant by I'm cheating. So all these things are truly now they're approximations everywhere. And I'm actually, yeah, you have to work a little bit. You have to use the battle unbounded. That may be a bounded one would be enough, but I think there's something to be done. But don't look too carefully. So the 0.1 to make is that in this analysis, if I want the answer to be n to the minus k, then this, this kind of calculation that I'm doing must be two week. But let's look at where it is to pick on the cycle. D, the inequality that I wrote down. Let me just look at the left side. Opposite k number of edges is on the right-hand side. It's due times k divided by r. Being do lose. When I'm comparing October number of edges to this average over a 100, I'm already giving up too much. I can't afford. So you might think, Oh, maybe I should replace that middle term by something more sophisticated. Now let's look at the complete graph. Actually, if you, I won't bore you with all the details, but in the complete graph, at the beginning, the answers that are both correct. And these are approximate numbers. They are lower-order terms and things like this. But actually turns out that at the beginning, the approximation that we used is correct. But what happens is once you start contracting edges, the graph becomes denser. And the average degree actually does not really drop. To as the process proceeds, this approximation gets worse and worse to whatever I use in the middle. But whatever I used to approximate the left-hand side is going to have to be careful in how I have to somehow look at the dynamics of the process. That's the point I want to make for you. So what I'm going to do in the rest of the talk that I'm going to show to you that this random contraction algorithm, which is due to collagen and then Calgary and Stein is actually going to succeed with probability about 10 to the minus kx. And I want to give you the main idea of the proof. But actually the idea that I feel in this proof. And what's the first idea? The first idea is, well, I want to classify, okay? Actually, this is a good time to take a deep breath because we're starting to get into the proof and point out one thing that I'm going to do, which can confuse you, but I don't I want you not to be confused. We were talking about Kcat, so fun. Things which break the graph into KP's. I want to relate the flux of the graph. We just two cards, just, just a set of edges that break the graph into two pieces. Because we understand to cut this process will be useful to you, will see how I want to take all the ducats of the graph and study them. And I want to classify them as small, medium, and large. I want to show that there are very few small and medium cards. You'd like. Why are they doing this? Cuts are large. I will argue that the graph is a dance. To that the approximation we were using was due week. I'll give you a sense of what's happening. The graph will start off with some number of small, medium, and large cuts. As the process goes on, the small and medium cards will start disappearing. You'll be left only with large cuts. They're only large cuts. The graph is denser than we earlier than anticipated. The other approximation we can improve. And that's how we're going to get him. I'm sorry. Not the graph is actually. So earlier we said the main cart is like k times the degree. The degree. And the diagraph is n times the mean degree or our tanks, the mean degree. I want to say it's actually much more than the denominator had, much more Thompson. To the probability of screwing up is the property of being good is increasing. That's decimated. Because then they said, in the complete graph, initially it seems like the approximation for correct. But then later on the approximations might, I claimed was poor. And that's because the number of small and medium cuts will disappear. And that's the third piece of the puzzle. I'll give you a refined analysis of this random contraction algorithm, RCA, which will basically just say that the small and medium cuts will disappear after a little while. The graph is denser. And so we've been pretty good. So what does large me? Let me quantify that a little bit. And introduce two pieces of notation that I'm going to use for the rest of the talk. I want to look at this quantity Z, which is the optimal solution, the optimal Kcat divided by King and I in some average sense you can think of it. This is where a minimum to cut should live. Aortic or whatever. I don't know. This quantity. Medium is, if you ignore the sort of niggling term in the middle is between z and to see anything smaller is small. Z is OPT divided by k. This is what the gut should have on average some medium in between v and z model. This model is bigger than some notation to keep good, I'll try to remind you of this notation again. And when I say Phew, I just really mean a linear number lineage. The main idea hopefully makes sense. Yeah. So OPT is opt of the Kcat. I should have written OPT sub k. Maybe it's OPT. Opt divided by k is a natural normalization. Could talk about what the values of VR, but I don't know if that's really relevant right now. Let me just okay. So far so good. Let me just make sure how I'm doing on time. That's what the rest of the talk, maybe about 15 minutes. I'll talk about how to view the process using this sort of continuous generalization, which you don't have to do. But it really helped us come up with the result it. So I'll tell you this idea which I found useful. When frankly, what should you take away from tonight? Hopefully you'll take away a technique or an idea or two. Maybe. One idea that we had was that this process of taking an edge and contracting eight and then taking another agent contracting it was two sequential for us to reason about. So we tried to do this in some independent sort of way. But I will just explain what I did. But I said what we did because most of the smart ideas came from the students. I was generally just simply wrong. And then there's the second piece where I want to show that there are very few medium and small cuts. And for this, we'll use a couple of results from extremal set theory, which you might have seen in your discrete math classes. And they're very simple results. And we'll just stop. So here's the continuous. But imagine that every edge has this exponential clock. Basically it has a little exponential random variable sitting on it, which is called a file at some point. Exponential random variable. Also you can think of a little coin that you're tossing at every point in time. And when it comes up heads, you stop. You've probably seen this course is that maybe David Tokyo, an exponential random variable with rate 1 over thing that fires independently in any little Delta P is it fires independently with probability delta over z. And this is happening independently. And when an edge fires, you contracted process clearly and stop, and you have two vertices remaining the graph and that's it. Then the first observation is of course, had been contracted in random order. And I'm taking every edge have the same distribution. All the vegetal can be contracted. But now there's independence between them, which really makes thinking about the material. The things that are independent, and the probability of two events happening in some tiny, tiny delta is there. And this is of the many places where I'm cheating, but morally things are correct. Okay? So let's do the standard analysis. And everything's in expectation. When our motif is remain the same kind of analysis as before. By the way, the optimal k, Remember the same analysis before. It shows that you have about our z over two edges, at least. And if you're contracting every edge with probability delta over z, about our delta over two edges are contracted. And this is just happening independently to initially a lot of that being quickly contracted because everything's happening at the same rate. And then the process slows down as the few remaining. And if delta 0, the edges that I'm contracting, contracting vertices, there's really no cycles because they'll contribute Delta squared type tags. So really the number of edges went down or the number of vertices and don't move from edges to vertices. So from I went down by R times delta over two. So I contracted about one minus delta over two. So at every delta time, the number of edges went down by one minus delta over two. To after time t, the number of vertices remaining is n times one minus delta over two to the T Delta, whatever. Because remember, the scaling is little face down. In picture. I have plenty of time though that what's happening is the number of vertices starts off at them. And then it falls off exponent. In this way with the number of edges, the remaining after time t and e to the minus t over Tau. When we'll finish when t becomes to login. Because then I'll have less than one vertex or two vertices, whatever. Their time to login, this process will finance. And then the only thing that I need to do is, well, any fixed min-cut had been fixed Kcat. Kcat is going to survive for this time with probability e to the minus KT. There's another some calculations, the same kind of calculation. Each little delta Ps. This guy is going to die with probability delta over z divided by k, z, etcetera. But going back to this picture, I'm adding in a second fact, another calculation that the minimum kcat derives at sometime with probability e to the minus KT. So if I take my escape, the, if I say k, If I say t equals to login, I'll get e to the minus 2, Lorna. Thanks. K, which is N to the minus 2k. Do the same analysis as before. I don't want to lose your place. Good though. So there'll be a lower-order term out. Absolutely. And if you know the previous algorithm where I'm contracting one edge at a time, this is just, you know, it's, it's compressing time at the beginning by a lot and then compressing. Maybe if you guys don't mind, I'm going to stand out here. I feel most comfortable close to the slides. Otherwise I feel lost. Something is unclear. Holler. If you see something like this, of course, you shouldn't. You should look at it on a log scale. So now you're the log of the number of vertices and it's non-linear. So here's what we're going to do. But here's what I want to do in this proof. I want to say at this point in time, I want to give a better analysis that from this point onwards, the lines not going to go like that. It's going to go much faster. That's my two, I'll finish much earlier. And hence the probability of surviving going to be much better. Suppose there were only very few minimum medium Kcat, medium to medium. Remember it between z and ignore that annoying little darwinism. Okay. They should have come a little slower. Let's look at the first line. One of these two cuts. The probability that it survives at some time t is e to the minus t. Another one of these simple calculations say that Li Zhi each time a die, the probability delta v over delta period. Model of the story, all the medium two cuts are going to die out my time learning halfway here. In this picture. At this point, there are no more medium cartoony. And then my claim is that very few vertices have degree smaller than v. If a lot of them had degree smaller than V, just cut them out. That will give you a cake on the model. Surely. Ignore. You have too much words on this slide. Sorry guys, I apologize. Moral of the story. What's the model at that time? And in fact, let me go to this. What's the bonus? You've seen that too much analysis. The idea is at this point in time, there are about square root and federal money just from this line. And that it's a log scale. Most vertices have degree doozy. There are no more medium guards. With high probability. There are very few, have small degree. Most vertices have very high degree. And the point is that this means that we'll shrink down at twice the rate. You shouldn't have asked this question. So yeah, so right now I've even said the expected number of medium cards is less than one. I haven't shown you high probability statements by this kind of thing as a medium cut is going to survive. Absolutely. Absolutely. So I'll come to that adjustment totally. So what this is is Dana of idea. We'll hold off on that for just 1 second. I'm going to go linearly until here. At this point, there are no medium cuts in expectation. What's expectation between friends? Did very few small degree vertices to all vertices must have high degree. To our graph is contracting at a much more rapidly. Our calculation was too weak. And that's the idea. That's the only idea. So we'll finish at time 1.5 on it. And then I'll just do the same calculation. This is k to the minus 5, that's 10 to the minus 14, 15. I'm already when it does this make sense? To really, you know, if, if I think back on how we came up with this analysis to really then those were the two examples. The cycle, the cycle outgrew its usefulness very early. But really the complete graph, the spent lot of time. And really we understood that the problem was the medium and small guts. We said, suppose there were a few medium and small cuts. What could we do? Well, we could do this now. And in fact, we could do something better. We should explore the curve. The Dana talked about, you know, why should I do this piecewise-linear? At every point in time, I should try to see where it, where this is going. And so what we'll do is we'll write down a differential equation. And then of course, you don't want to do this in public, but how do you do this? So you can do one of two things. You can address more heads question, I just talked about expectations. Maybe I should do high probability argument. So that was our first argument. We showed that whatever the expectation was, that the expectation will be given by this differential equation. And we showed that we stay close to this expectation. But that's a lossy process. Another thing you could do is you could actually guess once you, once you work with this long enough, you say, well, the expectation should go like this. And you write down some horrendous equations. Which is where we were working with David Harris. And it was if I'm working with him on this, where we really managed to nail down the details. And then there's some details. But don't worry too much about, actually worry about that later. So that's the first piece is this continuous view. And frankly, it is this independence. I just wanted to bring out this exponential clocks idea. It just makes your life much easier. You say everything's happening independent. There's no correlation. Please. Though. It is okay because we had all we're conditioning on the minimum Kcat not die and the medium ducats disappearing. I think it's yeah, it should be totally fine. Perfect. Okay. So if you guys were a little lost toward the end of that, whether you're not too much calculation. Fine. Let's do the second piece. The second piece has pictures. More pictures. I can draw more pictures for this then for the previous one. I want to show that a graph has few medium, two goods also have unbounded number of small cuts. But my claim is that there are at most a linear number times this sum nuisance factor, medium two cuts in particular are two guts of size at most of the medium and small. And for sanity checks and offer the cycle, this quantity z is 1. Z remember is OPT divided by k. K, k by k is 1. How many cards are there? How many cards are there in the cycle of 51? Oh sorry, of size less than two. This should be strictly less than 0. The minimum to get his face due. To recycle. This quantity is 0. And for the clique, this quantity is just the singletons or medium. For this part. I want you to think of cut sets. You already do that, but I just want to emphasize this. In a graph. This is actually the edges of the card. I just want you to think about one of the sides. Pick your favorite side, smaller side. The pancreas. A cut is just the left side. I always put the small side on the left. And I'm going to use this work, really fashionable theorem from extremal graph theory called a sunflower lemma. Okay, what's the sand fly? I'm given a Sec system. So all these little ovals are sets. A sunflower is a collection of sets here, five of them, so that their intersection is 0, such that they intersect in a compensate for any two of them intersected the same as the intersection of all of them. And so they intersect them this way. The petals, the petals are disjoint from other petals. And then there's the core, which is common to all of this is the sunflower. And this old theorem of erudition rhabdo. Those that if they ignore the actual numbers, if there are many sets, then there must be a sunflower. Petals at the sets are of size d. That this is the quantitative thing. The question that people have been interested in trying to improve it is this quantitative bound tight? I don't care. For me, those bounds will just be all in terms of k to the k kind of thing. Notice that here there's nothing about the underlying set system. The underlying number of elements. It's only so many sets of size d, the sunflower with our petals. I want to slightly stronger, but it's a triviality to prove the flightless raw material. If there are n times so many pets, then the sampler with a nonempty caught the sunflowers could have EMT course. If you just have disjoint set of tests with an empty car. And you should really think of all of this will become k to the k, and this will be my claim. And this is the last thing I want to proof, the proof by picture, but it's still a proof. Let's assume no small cards. Remember, medium, between z and z, slightly more than z to be the medium cut. Mouthparts are smaller than this and big cuts are bigger than the pose that no small guts. And the data considered medium guts to be the only medium could be that the set corresponding to meet it. I want to claim we cannot have a sunflower with gay petals and a nonempty amongst the medium. Do you think of each of these as the left side of some medium? Cut? It like this. It cannot have too many. You might say What the heck is happening. Where did this come from an amine when actually, it's very natural. Let's, let's do this. Suppose we had Saturday. Okay? But let's look at the for a sec. Let's think this was a medium cut. The number of edges leaving this whole big glob. If less than 2000. Definition of medium. The right hand side is nonempty. But they had no small cards in this set system. In fact, there's no small cuts in the graph entirely to the number of edges leaving this side is bigger than this. That's the lower bound of the small ducts. There are no small cut to the number of edges leaving this side is also bigger than this bond. There's blood leaving the left hand side, blighted, leaving the right-hand side, but not too many edges, leaving the whole thing. Whether the remaining edges going, they must be going between the two sides. So at least, you know, if you just add these two things up, the data divided by 2 for double counting, the number of edges leaving is something like v over k. Now we're done. For any petal like this. There's many edges leaving the court going to that baton to do a lot of edges there. That a lot of edges, they're disjoint because the petals are all disjoint. Let's look at the last guy. How many edges are leaving this little powder blue factor? It's a medium. With each of these is 0. The number of these mosque. They can be even careful liver numbers. That's the main thing. Though. Cannot have a large sunflower, cannot have too many medium cut. Ooh, actually, paired ocean Grado are kind of annoying because they not only want the states to be a few, they also want this to be 124 that is due later in this talk. You look at, you look at other very trivial bounds. Okay? Let's look at this. If you had a whole bunch of small, thank you. Intersected them together. Suppose, let me suppose you had a whole bunch of each of which was smaller, medium. But there were a few cards. You put them all down. You got more than k regions out here, which are nonempty. That would contradict. The fact that the Kcat of five, at least whatever KV. If I want to be fancy out here, I can say that the dual VC dimension of the Sec system has to be on whatever. Basically these cuts can point is these cuts the way we've defined them when they intersect, the system cannot be too rich. That's the point. How rich? You can define that in many different ways. Some help you in one case, some help in the other case. You put this all together. You get the result. I'm going to stop prioritizing. But the main idea that just the fact that, oh, I took this discrete math class, I should remember those, those, those theorems to the dark me. It turns out to be useful ones. If you put these pieces together. Do small medium, got, I didn't really talk about small cards at all. And now here small and medium cards die out quickly. The graph is dense, your performance is bad. To get quantitative bounce like this requires a little more care. Actually the proof get more streamlined, but they'll more difficult to come, come at from the first time. So I told you how we came at these and I've left out the streamlining part. There are open questions. Will, this is a really beautiful open question I'd love to solve it. Can you show that the right answer is n? Choose to. Forget the algorithm. Hopefully the algorithm will also come, oh sorry, n choose k. I don't know or disprove it. Matrilineal used to say, we are to make with making conjectures. I don't know if anybody cares, but I'm willing to support. Given deterministic algorithm, we don't know good deterministic algorithms that beat. And in fact, we don't know of any deterministic algorithm which pizza for the unweighted case for small weights, yes, we can do something but for large weight and natural extension to hypergrowth k parts. But that's a little more esoteric. That's it. Thank you. Yes. Please. Absolutely. Absolutely. The loss of suffering out that while the fake loss of two, once the graph, the condenser, and you'll get a smooth tradeoff. Thank you. I'm sorry. Yeah, absolutely. So in fact, the piecewise linear thing only get the 1.5 K. And the way to get rid of that is to write this differential equation. We just tracks exactly how the improvement is going. And that will give you the angle. Just instead of doing this thing, once you start doing it in smaller and smaller pieces, you'll see that the answer will be n to the. But I don't know. I don't have a good answer to that. I do remember looking at the company hopefully at some point. But since it's trying to preserve like SD cards and stuff like that, it is now only, only the global cuts it. So it's entirely possible. So our worry was that the Gomory who tree was so preserving has set of structures which will richer than what we wanted. But you're right, I don't have a concrete response. Yeah. Absolutely. So we'll have a look. I'll take more and hopefully we'll have something interesting to respond to your question. So for k equals 2, this answer is correct. For n choose two is the right answer. I think for k equals three also the answer is, it's not, not, not, not. I should check with William. He was working with the student trying to get into 3D Panther. I don't know. The edges too. So it's David Karger have the proof. But it turns out there was an older proof of cargo van somebody and somebody Lomonosov Carson oven, somebody which gave that. I can dig up a reference. It's the stuff I didn't know because I know there are at most so many that actually does that. So it is indeed the paper that was mentioned in the garden of paper was giving the cactus decomposition. So deterministic, I didn't really spend any time on this. But deterministic algorithm for this, there is a paper of which proceeds in the following way. Very quickly. He's going to find a spanning tree of the graph such that the optimal Kcat is obtained by deleting about k edges of this spanning tree. This gives you connected components. And then combining them together in the right way. The heel find the spanning tree snip 2k edges and then recombine. How many ways in which you can snap two cages and to the Duke. That but of course, all the smart side ensuring that there exists such a tree and how you find it. And he uses some nice tree packing ideas which somehow is used in the midgut, develops very similar to cargoes nearly mirrored time. The dossier, the question, that's the only example I know. But that's also true for k equals 2, the only example, right? Yay. Okay, That's why I was getting a little confused there. David Carter recently had the sequence of papers talking about network reliability. Well, he, he gave new proofs, cleaner proofs of all the results. This is a nice side note. Thank you.