 We're happy to have young ho yo from Georgia Tech finish off this semester's district math seminar by talking about approximating TSP walks and subcubic graphs. So let me hand over the laptop to you. Thank you, Stephen. I apologize for the technical difficulties. Yeah, so I'll be talking about TSP walks and this this is joint work with Michael Michael who's a graduate student at Georgia Tech and distinction. So the main motivating problem for our talk is the is a famous problem known as a traveling salesman problem. So the problem is given a complete graph where you, you have distances or these costs on the edges. You can find a tour or a spanning cycle of minimum length. So there's a very well known problem of intensely studied problem in computer science and operation research. And this is, there's a special case that is a particularly important. The metric TSP where the distance is form a metric in other words, going from you to be can cannot cost any more than going from you to W and then to be so taking a detour can only cost you more. And this is a natural assumption in many real world applications. And so this is a, again, this is a very widely studied problem and there are even further special cases of this problem. So now, instead of just assuming an arbitrary metric, what if all the edges have, what if the distances come from some underlying graph where the minimum, the shortest path between these two vertices is the distance between these two vertices. So again, so the distance between you and me is the shortest path, given some underlying graph. This is called a graphic TSP. And even further there are special cases where for the graphic TSP we restrict the cubic and subcubic graphs for the underlying graph. Okay, and it's just a sequence of specializations and it should in principle be easier. But these are of course all NP hard. So they're all NP hard to solve exactly. And in fact they're all NP hard to even approximate with a certain factor. So let's first look at the traveling sales, the general traveling sales and problem. So this is the general problem is NP hard to approximate within any fixed constant factor. It's not difficult to see that if you're allowed these arbitrary weights. Then within even any constant factor approximation would let you solve the Hamilton cycle problem in under the graphs. So you cannot have hope for a fixed constant factor approximation for the general TSP. But for the metric TSP. It is possible to have a constant factor approximation. And it's not that you cannot do better than 123 over 122. So it is known that you cannot approximate the shortest length of a TSP walk or a close tour within the within this factor of the optimum, unless P is equal to NP. And even in the special case of the graphic TSP. It's still in approximately with the slightly better constant. And even when you get to subtubic and cubic TSP is still in approximately within some constant factor. All right, but, but there's this famous algorithm of crystal fetus that I'm sure most of most of people here will know that you can do a three half approximation for a general metric TSP. So this is what a beautiful algorithm from the 70s is it's a very old algorithm, but it's, it's a very elegant and a nice combinatorial algorithm that that will give you a tour that is guaranteed to be at most 1.5 times the optimal tour, given any metric instance, right. And this was despite many years of research, this three halves ratio was not improved upon for almost 40 years for any non trivial case of the TSP of the metric TSP. And it was only in 2005 that this three halves ratio was improved just slightly to three has minus fiber over 389. And this was for the very, very special case of the cubic graphic TSP, and further only on three connected cubic graphs. So this was in 2005 and this was considered a big breakthrough at the time that you could, you could ever beat this three halves ratio. And part of the reason that people thought maybe it wouldn't be possible to beat this three halves, which may maybe is part of the reason why it took so long to beat this bound is that like this crystal videos out and was so nice and beautiful that people that maybe this was like the right algorithm that it might not be possible to do better. But it turns out you could you could do better. And this was the first breakthrough result. And following this there was a sequence there was sort of a flurry of activity on this problem. So following this result for the general graphic TSP. They also improved on this three halves ratio by a tiny bit. So even a smaller amount. But this was also considered a breakthrough for the general graphic TSP, not just cubic graphs. And afterwards shortly afterwards this was improved to a more significant constant to 1.461 by Monty and Svensson. And much he actually showed that this same algorithm of Monty and Svensson yields a 13 over nine approximation just by improving the analysis. And this was further improved to seven over five by Sable and Bygen. And currently this is the state of the art so for a general graphic TSP the best currently known bound is seven over five which is 1.4. And so it is a significant improvement over this three halves approximation that stood for a very long time. In metric TSP. The three halves ratio stood for even even longer and it was only about a year ago that this three halves ratio was improved to three halves minus 10 to the minus 36. So this was just a year about a year ago that this was published. And this won the best paper award at the stock, which is one of the pop CS conferences so this this is also considered a big breakthrough result that you can improve this three half ratio for the general metric TSP. Okay, but so for this talk will be focusing on cubic graphs and sub cubic graphs. So let's first discuss. Okay, why do we even care about cubic and sub cubic graphs. So one reason is that they're sort of they're the simplest classes of graphs, where the, all of the hardness of the TSP still persist. So it is still MPR and it's still MPR to approximate within on a small constant factor. So the important reason for caring about cubic and sub cubic graphs, which comes from the standard linear programming relaxation of the traveling filter problem so this is called a sub tour, sub tour elimination, linear program, where you put weights on the edges. We're trying to solve this linear program where you put, we should think of it as an integer program where for each x e, I put it one to to indicate that I include this edge in my tour and zero to indicate that I do not include my edge in this tour. So I want to minimize the sum of the weights on the edges or the distance on the edges that I'm using. And I want to sort of give linear constraints to ensure that the edges I pick satisfy this tour constraint. So first of all, I want to make sure that for every vertex. I have exactly two edges coming out of that each every vertex. So for each every vertex. I look at the set of edges incident with that vertex and x. So I need to use exactly two of those edges. But just having this first condition alone will only yield you these district district unions of cycles. So I need to make sure I only have one cycle. And one way to get around that is to say, okay, if I look at any subset of the vertices, there shouldn't there's, there should always be at least two edges coming out of that subset. So there can't be any two disjoint components. And that that's encoded in the second constraint here. And the third constraint is just say X is it is non negative. So this is sort of the one of the most standard linear programming formulations of the TSP. And of course, so if we impose an integer constraint on the XC is and this solves it the TSP exactly. But of course we relax it to a linear program and see what what we can do with the linear program that we know how to solve. And it's there's a very famous conjecture that this linear program will always give you something that is within four thirds. The actual optimal integral tour is always within four thirds of the optimal value of this linear program. So this is another long standing conjecture in this in this in this area that this linear program relaxation will only make will only improve the optimal value by this third or third amount. And again, why do we care about subcubic graphs that it happens to be the case that this fourth third approximation or his fourth there's an integrality gap is achieved by subcubic graphs. In other words, subcubic graphs they in fact exhibit the worst case behavior in the in this context of this sub elimination. In your program and here is the example that that shows that you can get this fourth third ratio. So I take three long paths of the same length and I put two triangles at the two ends. And I put all ones on the pads and I put halves on the triangles. This graph with this edge waiting satisfies all the constraints in this linear program. But, but the optimal tour has to use each of these paths, well it has to use has to cross from left to right four times, essentially, because if you want to. If basically you're looking for a standing closed walk in this graph, you have to cross this, these these three paths four times an even number of times. And because there's three of them you have to cross at least four times which means. If the path if the length of the path is roughly K, then the optimal value of the TSP walk in the interval TSP walk will be roughly four thirds, or four times K. But of course, if you sum the X is here, you'll get roughly three K. So the subcubic graphs, they really exhibit the worst case behavior. So this is, this is really the primary reason for focusing on these subject graphs. So what progress hasn't made for cubic and subcubic graphs. So then the first result, first breakthrough on this three half ratio was on three connected cubic graphs. This was improved. So partially motivated by this fourth there's integrality gap conjecture. They showed that for three connected cubic graphs, you can actually do a fourth year's approximation. And again, this was considered for a while to be a natural barrier. But it turns out, oh, sorry, I choose. This was actually extended to connect the cubic graphs. And this was also extended to connect subcubic graphs. So for subcubic graph is actually, it's actually considerably more difficult than just looking at cubic graphs. But this, this was done. But in 2015, they also went through this four thirds ratio. So they improved, they broke through this barrier this natural barrier of four thirds by a tiny amount. So it's independently done. Zulin also broke through this barrier. And once you break through this barrier, of course, other people work on it to improve the ratio even further so this was further improved to 1.3 and to nine over seven. Most recently by the project crawl and mohar. Our top today here is to improve this bound even further for cubic graphs from nine over seven to five over four. So it's not a big improvement. So it's nine over seven is about 1.28 and five over four is what about 1.25 so it's about a 0.03 improvement. So let me remark that so for cubic graphs we have all these improvements, but for subcubic graphs, these were still stuck at four thirds we still have not broken through this four thirds bound for subcubic graphs. Okay, so let for the rest of the talk let me try to give you an idea of how we prove this five words approximation results so it's really just a purely mostly a graph theoretical result. We don't apply any sort of fancy algorithmic approximation techniques, we just prove a general bound on the length of TSP walks in cubic graphs, for instance, subcubic graphs. The way we do that. So let me state our main theorem so given a graph G. We're going to denote by TSP of G as the minimum length of a TSP walk, or in other words, the minimum length of a spanning closed walk, so a closed walk which is every vertex. And our main theorem is that given a simple to connect a subcubic graph with n vertices and and two vertices of degree two. The the minimum length of a TSP walk is bounded by five and plus and two over four minus one. Okay, and moreover such a TSP walk, you can find in polynomial time justify essentially following the proof. So we can construct such a TSP walk this that's guaranteed to be at most five and over five and plus and two or four. And in particular, if you give me a cubic graph that has no the vertices of degree two, that I'm going to my algorithm is going to return a tour of length five and over four, because n two would be zero if you gave me a cubic graph. So naturally we this this this theorem just yields a five fourths approximation for cubic graphs, even though our our actual main result is more general for subcubic graphs. But also let me note here that if you give me a general subcubic graph, we do not get four thirds approximation for this result. What we get because and two could be very large if and two is very close to n, then this is actually closer to three halves which is not an improvement. Although, it is possible to modify our arguments to essentially recover a four thirds result but we, we still don't know how to break through the four thirds barrier for subcubic graphs. So this was actually conjectured by the Bojack problem or her and in their previous paper where they put the 97th approximation ratio. And in this bound is actually best possible. So in particular there are infinitely many subcubic graphs that achieve this bound exactly. And even if you restrict the cubic graphs, there are infinitely many that achieve this bound exactly minus one. So I can't get a minus one, but there are infinitely many that achieve minus two. So the five fourths constant is best possible in terms of a general bound for the length of the TSB walks. Hey, and moreover, we actually characterize the extremal cases. So we characterize all the graphs where the TSP the minimum length of a TSP walk has length exactly this amount of five and plus and two or four minus one. Okay, and just one quick note here is that this requirement that the graph has to be simple is necessary. So if you take a distant internally distant union of three very long paths that are that have the same endpoints. But now if you double sort of the at the every other edge in the path so that every vertex in the path is incident with two parallel edges, it turns out that violates this bound. So the simple list condition is necessary. So, our main idea is to consider even covers of G, where an even cover we define as a vertex disjoint union of cycles and isolated vertices. So let me remark that in previous in most previous works on the TSP, the way we obtain an approximation is sort of to to first find a spanning. E a olerian graph, a spanning connected olerian graph, and then try to remove vertices to obtain an approximation. Here we're doing something. So we're doing it from the other side where we start with something. So with a spanning but not connected we can have many isolated vertices. We just start with some even spanning graph that has many different components and we just try to connect these even components. Even cover so you just for an even cover we think of disunions of cycles and isolated vertices. Given such an even cover how do you construct a TSP walk or a spanning close walk out of an even cover. The way you do it is you just make a spanning tree out of these cycles and isolated vertices where you treat each cycle as a single vertex. So, when you when you add so you're making a spanning tree. So the TSP walk that you get out of this even cover. Well, when you add one isolated vertex, you're forced to repeat the vertex that you left from to reach that isolated vertex. And if you add a cycle, you're forced to repeat the vertex that you left from in the original graph, but you also repeat the vertex in the new cycle. So we define the excess of an even cover to be two times and two times the number of cycles plus the number of isolated vertices. And it turns out that given any simple subject graph, the minimum length of a TSP walk is exactly equal to the minimum excess of the graph, plus and minus two. So really, this is an exact relation. So TSP walk, I can give you some even cover that has length at most that I mean under this relation has length at most the length of the TSP walk minus and plus two, and vice versa. And again, the picture should have in mind is that given uneven cover or distant union of cycles and isolated vertices, you make a spanning tree out of it and each time you add. So you end up on this spanning tree on the leaf of the spanning tree. And every time you add a vertex, you add one, you're forced to repeat one vertex and every time you add a cycle, you're forced to repeat two vertices. So we're really only going to work with these even covers. And it is also known that you can, given an even cover you can you can convert it to a TSP walk in linear time, actually, because you're really just finding a spanning spanning tree. So, again, what we want to prove is that the minimum length of a TSP walk is at most five and plus and two over four minus one. But we just showed that this, the length of a TSP walk is exactly the minimum excess of the graph, plus and minus two, where the excess of an even cover is defined to be to be two times the number of cycles plus a number of isolated vertices. So, so just just to rephrase as this result, what we're actually going to show is that given a simple to connect us up to the graph, I can find you an even cover of access this amount. I'm just rephrasing plugging in this formula for for our main deal. So let me try to. So what is our sort of main key idea. We're actually going to prove a stronger statement than this by asking our even cover to go through a specified edge or avoid a specified edge. So let's. So suppose we're given such a graph G and and and some specified edge just any edge in the graph. What if I asked for an even cover to contain this edge in a cycle. It is still possible. So this is bound for even covers I contain this cycle. And it turns out the answer is no. So the same example is to look at K for minus or the graph obtained from K for by deleting one edge. And if the edge, if my specified edge is the one edge that has degree three vertices on both ends. So. This is really so I have an edge and I have to internally just run paths with one vertex, one internal vertex in each of those paths. And I'm forced to use the middle edge, which means I have to use one cycle and one isolated vertex for the other path. So every even cover using this specified edge has access three. And of course, if you calculate this value and plus and two or four you get three halves. So the minimum excess is this and plus and two or four plus three half. So it violates our desire bound by a half. So that's maybe not good, but it only violates by one half that's manageable. Now what if we ask for an even cover to not contain this edge. We still don't get a bound and in this case actually the situation situation is much worse because if you just take a long cycle and just pick an edge and ask you to not contain this edge. Then you're forced to use every single vertex as an isolated vertex. There's as the only you can check contain any cycle because your graph is just a cycle. So that's sort of a, but that's, that's a bit trivial. So now what if you ask, but what if you're, you delete the edge and you're still too connected, what if G minus is still too connected, can you still, can you then prove this bound. Can you find an even cover not containing a that satisfies as the answer is still no. So now if you take K 23 so you should think of it as two vertices and you have three internally this train paths, each with one internal vertex. But now I add an additional edge between to the degree to vertices. But of course I'm asking my even couple to not go through this edge. So again by if my even cover is not allowed to contain my this new edge. Then I again have to contain two to these pads in the cycle and one isolated vertex so the excess is still equal to three and n plus and two or four is still three halves. So this again violates the bound by by a half, but the idea here is that, well, so K 23 plus an edge is really bad if you force it me to not contain the edge. But in this case, if I ask it, if I allow it to contain the edge, and I can actually do a lot better I can, it becomes a Hamilton graph I can cover the entire graph with one cycle. Similarly, in the previous case work with a K for minus an edge. If I, if I'm forced to go through this middle edge, then I have to have access three. But if I then allow it to not go through that edge, then I can again cover the whole graph with one cycle. So what we prove is that so if you force it to go through an edge or if you force it to avoid an edge, you might violate this bound by a half. But then you can, in that case you can save a little bit on the other case. So these two quantities should balance out if you force it to go through an edge or not do an edge these two quantities should balance out. They can't both be very bad. Right, so again, in each of these cases you can save a bit on the other side. What she showed is that we prove separate bounds. So we prove separate bounds for each of these two cases, and we again characterize the extremal cases, extremal examples in both of the both of these with both of these constraints. Right, so this is our actual main theorem. So let's assume just for simplicity that G minus E is to connect it. So we don't want these examples where G is just a cycle, and E is just one of these deleting E gives you a lot of these cut edges. So let's just assume that G minus E is to connect it. Then what we can prove is that, if you force me to go through an edge, then I can find you an even cover with a size and with excess at most this much with at most plus three halves, instead of plus one I allowed three halves. And similarly, if I, if I'm forced to avoid an edge, then I can, I can again prove a bound of at most plus three halves. And over the really important thing here that these two parameters balance out. So how do I formalize that one way to write it is like this. So I really want to focus on these plus three halves or plus constant. So if I look at the plus constant in the first case. And I also look at the plot the constant in the second case, those two constants sum up to at most two. And three halves here. In the other case, what I will get a one half. And if I get a three has by not containing the edge, then I'll get a one half by containing the edge at most one. And really for for for most of our work we really just focus on these additive constants after and plus and two over two. So, I'm not going to define a lot of notation. I try to avoid of dividing a lot of notation but this is a very important notation for his talk so we'll define these deltas and hat deltas as these additive constants after and plus and two or four. But there's this little weird thing here where I subtract minus two when so delta is when I'm forcing the even cover to go through an edge and had delta is when I forced to not go through the edge. And I'm going to subtract to in the definition of a delta. Because. Okay, so what what to rephrase our main theorem in terms of these deltas what we're showing is a delta of G and E is always at most minus one half. The hat deltas are always at most three halves. And the two deltas for a given pair G and E they always sum to at most zero. So okay why are we doing this minus to the definition of delta. So it's sort of, well first of all it leads to much simpler computations when we do the actual calculations. And the reason is, we're going to be piecing together these sort of chains of sort of blocks or chains of graphs, and we're going to combine all the cycles of each block that contain. We're going to be combining the cycles of each block. So we have these, let's imagine you have like a chain of blocks, and I want a cycle that goes through this chain. And then, I sort of I can take back a closure of each block I force an even cover through this. I joined the end vertices of this block I force and even cover through this new edge. And I get a sequence of even covers of each block. And I think I combined each of the cycles in each block. So the excess of the even cover in each block contains the cycle but but the cycle is not actually going to be used in the overall graph so I'm going to take away the cycle in the smaller block that is just used as a path in the overall chain. So that's sort of the intuition, and it actually has a natural interpretation as sort of the savings of going through the block, and the cost of not going through the block. So let me try to formalize this intuition a little bit. So again, this is just a definition of the deltas. So suppose I give you a, I'm given a specified edgy, and let's suppose that this edge is in the two edge cuts. Okay, so now let's, there's two edge cut to find two kind of components if I delete the two edges. And now let's close up each of the two components. I don't think I have a picture for this. So imagine you, I think the idea, hopefully the idea is clear, I have two edges, like dividing the graph into two components for each component I just close up the two ends of the two edge cut. All right, and sort of the things that really makes this work, or hopefully intuitive is that the delta of the overall graph going through this edgy is just a sum of the deltas of each of the closures. And similarly the hat delta of the overall graph, not containing E is some of the hat deltas. So the really the idea is that you should think of one side of the cut as being the main part of the graph, and one the other side as being some small part that I'm trying to decide whether or not I should go through this part, or not for this part. And the deltas tell me, if I go through this part, and I save what amount that I save by going through the part is exactly delta of that component. And if I don't go through the part and the amount that it costs me is exactly the hat delta. So remember, our main theorem is that the deltas are always at most minus one half, the hat deltas are always at most three halves. But if I go through a chain, or if I go through a component with a two edge cut, I can always save at least one half. But of course I have to balance this out, there might be other chains, other block, other components I can go through. So it turns out that this is right intuition. And I won't go through the proof, but the idea hopefully is clear. A cycle contains this my edge e if and only if it contains both edges in the two cut. Right, so this gives you a bijective correspondence between uneven cover containing the edge in the overall graph and pairs of even covers in each of the components. And if you just run through the calculations with the definitions, because we define the deltas to be minus two, because of this minus two in the definition of delta, you get exactly get this result. Right, so this is a restatement of our main result. So, the third condition is really saying that the two costs are always balanced. So the savings of going through a chain, or going through a component is, and if you, if you only save a half by going through a component then you also don't have to cost that much by not going through the component and vice versa. These tight components on this. So important to let's skip this. So, I kept saying the word chain because these are, this is what really happened if you give me a subject graph. I could have a lot. It's a too clear to subject graph. I can have a lot of these, if I have a two cut, then one component of the two cut is really you can think of it as a as a chain. You can, you could have you could have just one block in the chain. And we define the closure of the component or the closure of the chain as just closing up the end edges of the chain. And for each block, I can also think of it as closing the ends of the of the block in the chain. And similarly, if you run the similar calculation as before, what we get is that is that the delta of the chain is just the sum of the deltas of the closure of the blocks. And the hat delta of the closure of the chain is again the hat delta of some of the hat deltas of the closure of the blocks. So in other words, the savings that I get by going through a chain is the sum of the savings of each block in the chain and the cost of not going through the chain is just the sum of the cost of each block in the chain. Okay, and the chain is tight if only every block is good. Let's skip that. Now, now let's talk about the extremely example. So remember, the examples that violated our desire bound of plus one. The first example was K4 minus. Right. Okay, so K4 and you delete an edge. So it's sort of a similar construction. The rooted theta chain is you take an edge and you take two internally distributed paths with endpoints that are the same endpoints as the pictures here. And if this chain was just a path of length to and this then this is exactly just K4 minus much. And so this is a rooted theta chain. So the unrooted just a normal theta chain is just an internally disunion of three sub qubit chains. So I don't have a picture here but just imagine that instead of this a single edge or you also had another chain. Right. So these are technical definitions. So really the important example is this K4 minus. And more generally, if you had a, if both of these chains were just some paths, let's say both chains were paths of length K with K internal varities. Then you, if you do the calculations you will you get that you again violate this bound by one half. So what's really bad if I'm forced to go through this edge, then in the worst case it doesn't really matter which chain are you choose to go through and not do it they have the same sort of they have the same cost and the same savings. So this balance miscondition just says that they have sort of the same amount of savings and the same amount of costs. So it doesn't, it, no matter which chain you choose it's they're equally bad. It turns out that these are exactly the, the, the extremal graphs that these are the only grass at that violate our bound of plus one. So these are essentially the only extremely examples. So, so again here is our main theorem. So it turns out, we have this condition that G minus is too connected. It's actually only for this case here, if I'm forced to not go through an edge then I want, I don't want to have like this in country the situation where I have a long cycle. But in the other cases, if my edge is already in a two edge cut and I that's only better for me because it, and I'm going through an even longer chain essentially, and I can always save even more. And the condition is only really needed for the second statement. Okay, but we actually, so we characterize all the extremal cases so we, I can tell you exactly when equality holds. So, Delta of GNE so going through some graph, I can always save at least a half. And the only time I save only a half. So I would like to save more, but sometimes I can only save a half. In that case, I know that this pair GNE forms a tight balance rooted beta chain. So it's basically a picture like this where just imagine two long paths of the same way. Those are really essentially the only cases where I'm forced to, I can only save a half by going through this chain. Similarly, if G minus is too connected, and the cost of not going through this block is three halves. If equality holds, then G minus E is a minimal theta chain, which means I have three internally this joint chains with the same endpoints. And of course the simplest case is when each of the three chains are just paths of length two. In other words, this is K23. And Delta plus hat Delta is always a 0. So again, we're characterizing when these inequalities hold with equality. And we can also characterize the tight cases. Let's skip that. But what we get from this is we get a structural characterization of these data chains. Let me skip this right now. So the two cases we've been looking at where when Delta is minus a half, or when hat Delta is equal to three halves. But there are like there's one other case where Delta is equal to minus one and hat Delta is equal to one. And that's really what I would we're interested in the general like the TSP walk problem. For the TSP walk, we don't really care about forcing and or avoiding an edge, we just want the best even cover overall. So if Delta is equal to minus one half, then I know that hat Delta is equal to one half or at most one half, which means the additive constant that I get for an even cover by avoiding E is at most one half. So if if hat Delta is three halves, and I can delta is at most minus three half, which means, remember that Delta had a minus two in the definition which means the additive constant that I get for with the even cover is again at most one half. So there's a sort of the middle case where Delta is equal to minus one and hat Delta is equal to one with, and I actually get a plus one in the additive constant. But we can all we can also characterize those cases. So what we get is that if you don't care about going through or not through an edge, just generally, given a simple to connect the subject graph, the excess of G, the minimum even cover has access at most and plus and two or four plus one with equality if G is a K four or G is a minimal data chain. So, minimal data chain. I don't think I defined it clearly here but we can prove a structural characterization of it that what we can get is essentially K two three right, you have these three internally this joint packs of like two. You can blow up a vertex of degree two into a four cycle like this, it turns out that this operation preserved all the Delta's and have Delta's. And here I blew I took a K four minus an edge, and I blew up to two various degree for degree two, but you can you can keep blowing up. You get new versions of degree three and you can keep blowing them up. But so this is a K four minus a blow up of a K four minus. And if you blow up a K two three that satisfies on this bound with a plus one, and it turns out that those are the only graphs that have this extremal property. So, the minimum access of a graph is equal to this and plus and two or four plus one. So this is a quality if and only if either K four, or it can be obtained from K to three by blowing up degree two various sequentially to a diamond. All right. So, let me try to give you an idea of some of the proof here. So, let's let's let's try to put the first thing is the first thing is really the most interesting part. So if I if I'm forced to go through an edge. I have delta and most minus one half with equality if and only if I get a tight and balance with data chain. How do I prove such a thing well let's let's look at the case where G minus UV is disconnected. So if I delete the endpoints of e, then I get to this disjoint components. And of course, I'm forced to go through my specified edge e so of course I could go through, I could choose one of these chains to go through. And the cost of and the other chain I have to not go through it. Obviously, so I have to look at the cost of not going through the other chain, and I have to look at the savings that I get from going through one chain. And if I look at the if a, and if I look at the two choices that I have, and if I calculate out the cost, and it turns out you get, you get the, you get what it's going to do this. So, so we, so we can now assume that G minus UV is connected. And if my specified edge is in a two edge cuts. Then again, I can sort of, I can look at each component separately and I can sort of apply induction. So, let's assume that he is not in a twitch cut. He does not disconnect the graph by doing the end vertices. In the two cases that we get is is essentially look like this. So just up to symmetry if you just look at the structure of this the subcubic graph. These are the two cases that you get. And again so let's look at this left example here. I'm forced to go through my edge UV. I essentially have two choices I can go through you one. Come go across and go to V2, or I can go through you to and go across and to V1 to complete my cycle. Of course I can use you one and V1 but then I'm not using using this middle chain here. And it turns out that's just too costly. If I just go through the middle chain I have two choices and in turn, and if you do this sort of similar calculation of balancing out the cost and the savings of the chains, you get the right balance. And it's a similar idea here. So when if this chain if there is no like bridge between these two parts that what we get is we just get to one to the component. And now I'm forced to go through this edge. I essentially have four choices so I can choose either you one or you two and I can connect it to either V1 or V2. And now you look at all these four choices you look at the sort of the cost of the other change the chains that I'm not going through and the savings of the things that I'm going through you do this balancing out calculations. And you get the right now. And the second statement is much simpler because I can essentially just reduce it to the first thing because like what am I really asking for if I ask you to not go through the edge. This is the same thing as saying I just delete the edge and find an even color. Right because it's a much weaker condition in some sense so I just delete the edge and I just look at the area neighborhood around the edge. Okay, I, if I delete the edge and then I'm going to create a vertex of degree to, I'm going to suppress that new vertex of degree to and maybe maybe forced to go through the edge in the smaller graph and apply induction, or I don't force to go through the edge. And I can apply induction. And you get the right. So this is really just a reduction to the first statement. Let me just skip that. So let me just say a few words about the algorithm. So the outfit is essentially just follow the proof. So what we're going to do is I'm going to define essentially two algorithms were given a subject graph G and an edge E. I'm going to return and even cover containing the edge, or another algorithm that returns and even for not containing the edge. Right and because we have this structural characterization of all the extremely examples we can determine exactly when the deltas are exactly one minus one half and when the hat deltas are exactly exactly three halves. And in all other cases. So I probably should have said this earlier but all the deltas and had deltas are always half integral. Because it's, it's n plus and two over four right but n plus and two is always even because n plus and two you can think of it as n three plus two times and two. So the vertices of degree two are counted twice. And the number of vertices of degree three is even because it's just a number of odd degree. So the deltas are always have integral. If you're not in these extremal cases, then we know that the deltas and the hat deltas are at most minus one and one respectively. And of course we can't actually compute these deltas exactly because you. That's NP hard to do. But we can always give an upper bound for it, like what is the worst possible scenario. The worst possible scenario is that the delta is equal equal to minus one minus one and hat delta is equal to one so we just for the algorithm we just say we given up we give an approximate upper bound on the delta we just say, if it's in this extremal case of minus one half, you just say we just pretend it's the worst minus one half and plus one half case. If the hat delta looks like it's going to be three halves and we just we just assign three halves to it. And all other cases we just pretend is the worst possible case and just assign minus one or one. So we basically just compute this worst case estimate for the deltas. And essentially, and now we're basically just going to run through the proof in the entire proof with these worst case estimates. And because our proof works in the worst case. The worst possible case, like, our algorithm also gives us this guaranteed bound that. So our algorithm will eventually yield on even cover that has this bound where and plus and two or four plus the big delta. Plus two, or uneven cover not containing my edge with a with a constant of hat delta, but these had deltas are always at most. So again, the algorithm is essentially there's only one thing we need to do for the algorithm except follow the proof is just to say, because we can't compute the deltas exactly. We just give an worst case sort of upper bound for it. And we, it actually is, it does take some work to to determine when we're in these two extremal cases. But other than that, we just say it's just assume is the worst possible case. And because our proof proof actually still works in the worst possible case the algorithm should also give us even cover that's at most at worst this bound. In particular, if you give me a cubic graph, then my algorithm will return a graph with will return an even cover with excess at most and over four plus small constant. So to summarize, our main theorem was that was to look at even covers and prove this bound of n plus and two or four plus one. And we had equal we actually characterize the extremal examples as K for or minimal theta chains or in other words graph that can be obtained from K to three by blowing up degree two for a season to four to to four cycles or diamonds. And of course we prove this by actually looking at a more refined statement or stronger statement of even covers going through or not there. So, and because of this observation earlier that the minimum access is exactly related to the minimum length of a TSP walk. The immediate corollary is that given a simple to sub cubic graph, the minimum length of a TSP walk is five and plus and two over four. So, with equality if and only if either G is K for or this K to three blow up more than more of a we can find this TSP walk in polymer and immediate corollary of that is we get a 54 approximation for the cubic TSP. Some remaining open questions. So, our bound is this five fourth bound is sort is best possible, only in the sense that there are graphs that achieve this that have there are graphs whose minimum length TSP walks have have length five and over four. But of course like we do the way we achieve this approximation ratio is a okay we find a TSP walk of length five and over four. And we know every TSP walk has length at least end. So we know it's at at worst within five fourths of the optimum. So we're really not saying anything about any any sort of lower bounds or how close it is to the actual optimum. So it might be possible to actually like if you if you actually go down to look at the graph and say what what what is the minimum length of a TSP walk look like. And then it might be possible to improve the approximation ratio. So the are our upper bound is best possible in general but the approximation ratio could still be improved for cubic graphs. And it's possible that our sort of techniques can be applied to more general graph classes. So this is something we're still working on. It's, it's possible that we could sort of maybe prove this for bounded degree graphs or something similar. So I'll stop here. Thanks for your attention. Perfect. If we could all thank our speaker in some way chat here. So are there any questions for young hope on either audience. It doesn't appear to be so let's take our speaker one more time. And let's say, you know, for all of us here at USC, you know, this is our last day of classes so happy last day of classes and it's not quite the end of the semester, but this will be the last seminar that we'll have of the semester so don't give us a great sort of final talk of the semester I really appreciate it. But as far as as far as other weeks go, I'll have to keep you updated and you know probably a month or so and we'll have some some, you know, new new speakers then but thanks everyone for for a great semester.