 All right. Well, thanks everyone for making it out to the first street math seminar of this academic year. We're very excited to have our own Drew Meyer kicking off this seminar this year and he's going to be speaking about anti Ramsey number of edges just joined rainbow standing trees in all graphs. Take it away Drew. Yeah, thanks Steven for the introduction and thanks everybody for making it out. I'm excited to finally have something of my own to present in the seminar. So this is a result that Lincoln's you and I worked on in the early part of this summer. And I really like this result because I think that it is a it's very reasonable to present almost the entire thing in 50 minutes. So I think that, you know, if there's any point that maybe definitions have been lost, or you know, maybe I'm going to fast or something please tell me to slow down or feel free to any points interrupt and ask me what's going on. So to sort of get our feet wet with the type of problem that we're going to be talking about. I want to first do this kind of toy example, which asks the type of question that we're going to be dealing with, but in a small enough setting to where we can really see what's going on. And the question that I want to ask is, is how many colors can I pack on to the edges of a K for so as to avoid a rainbow spanning tree as a sub graph here by rainbow I just mean that every edge gets a different color. And so, as with any question like this, you know, you should ask yourself, is the question interesting does it sort of make sense. I can think well if if I color all of the edges of K for with the same color, then certainly there's no rainbow spanning tree. Why well because spanning tree would have to have three edges and necessarily all those edges are read. So okay, I can do it with one color. What if I color every edge with a different color. Well in this case, certainly there's a rainbow spanning tree and in fact any spanning tree is rainbow necessarily because every edge got a different color. And so the problem at least is interesting and you know where does the switch happen between impossible impossible, or at least maybe not impossible but at least there exists a color. And so let's let's start with K for and let's do two colors. So our goal is to find a coloring of the edges of our K for with two colors that avoids a rainbow spanning tree so I can give you such a coloring. I'll color the following edges red. And this last loan edge blue. And my claim is that there's no rainbow spanning tree in this graph. And the reason really has to deal with this partition of the vertex set here. So why can there be no rainbow spanning tree. Well, any spanning tree of our K for must connect you to be in some way. And you can pretty quickly justify yourself that well hey there's no way to do that without using to read edges. For example, maybe I use the edge UV in my spanning tree, in which case another one of these red crossing edges on the inside must be used to get back to the remaining part of the graph. If I don't use that edge UV then clearly I use two of the crossing edges so certainly in this case, no rainbow. Another argument is that the spending they must have say edges. So by the region whole place you must have a repetition. Sure. Yes, that is that is a is another approach. But this is a little bit leading to what we'll see later this idea. So, so yes that's another approach and probably a simpler approach. But this idea will be present later on as well. So no rainbow spanning tree here. Okay, what if I jump up to three colors. So let's draw our K for here. And maybe I'll color these two edges red. Maybe these two edges blue. And maybe these two edges green. And so do we have a rainbow spanning tree here. Yes, I think I see one. I could take this red edge. This green edge. And this blue edge. And there certainly is a rainbow spanning tree. And I'll just use our ST from now on to to talk about these things. And so of course this doesn't show that for any three coloring. I have a rainbow spanning tree, but it turns out in the case for K for three is the six excuse me two is the maximum number of colors that I can use and avoid a rainbow spanning tree. And so this example motivates the following definition. An edge colored graph H is called rainbow. If every edge of H receives a different color, exactly what we talked about before. And the anti Ramsey number of T edge disjoint rainbow spanning trees, which we'll denote by RK and T is the maximum number of colors and an edge coloring, which has no T edge disjoint rainbow spanning trees. Now an important thing to to mention here is this last sentence that in this case in the edge edge disjoint case, repeat colors are allowed in distinct trees. So what do I mean. Well, if maybe I have two trees T one and T two, the color red is allowed to appear within the edges of both T one and T two, but only once within each tree. So the color red is allowed across trees but not with it. And at this point, when I gave this talk on Tuesday in the graduate colloquium. There's a very good question that was asked and the question was, suppose I knew for some K in that if I color it with say our colors. I always have, let's just say a single rainbow spanning tree in this case, the arguments going to apply to T edge disjoint rainbow spanning trees as well. But let's suppose I know that anytime I color K in with our colors. I have a rainbow spanning tree. Is it obvious that when I then go to our plus one colors. I still am guaranteed a rainbow spanning tree. And the answer to that question is yes. All I need to do is just pick any color. Let's say maybe blue that appears somewhere on the edges and recolor all of the blue edges a color that existed somewhere else let's say red. I'll color all the blue edges red. And now I've colored with our colors. So I have a rainbow spanning tree. And either that tree has a red edge or it doesn't. If it doesn't have a red edge well then when I go back to my old coloring it's the same tree so we're still good. If it does contain a red edge, then either it remains red when I go back to my other coloring, or it flips blue. And if it flips blue, well blue didn't appear anywhere in my first coloring. So I still have a rainbow spanning tree. So that's the argument there. So I want to take a second to talk about the history of this problem. So in 2001 by Lestaki and boxman were able to answer the question for a single spanning tree. So they were able to show that the anti Ramsey number for a single spanning tree is equal to n minus two choose to plus one. And I think that really the best way to get an idea of where these things these these numbers are coming from is to talk about the lower bound. So I'll show you the lower bound construction. The lower bound construction is going to sort of persist throughout the the rest of this talk. So what is the idea for a single spanning tree. Well what we'll do is we'll break off a K and minus two, and we'll color every edge within here distinct colors. Okay, so how many colors does this add well of course the number of edges in K and minus two, which is a minus two choose to. And then just like we did with our toy example at the start will isolate two vertices and color every single edge back in to our K and minus two red. And so again by the same argument and this adds our plus one color. And now by the same argument that we made before any spanning tree must connect you and V to the K and minus two, and there's no way to do it without using two red edges. And so there's no rainbow spanning tree with this coloring, which tells us that our K and one is greater than or equal to the above number and minus two choose to plus one. And of course, finding the lower bound construction is not too difficult. The difficulty lies in now saying, okay, I've got any coloring with one more color show that I can always find a rainbow spanning tree. And that is the topic of their paper. Let's talk about for more trees. Akbar and Alapour in 2007 were able to finish the case for two edge disjoint rainbow spanning trees, and they have this number, which is one more than the previous numbers. Now we get to add an extra color to see where this color is coming from. We can do a similar type of argument for the lower bound. So we're looking for a coloring within minus two choose to plus two colors with no rain to edge disjoint rainbow spanning trees. So what's the construction. Again, we'll break off a K and minus two and color this with distinct colors. So this adds in minus two choose two colors. And now our vertices, which are isolated you and V will now add two colors, a single edge of one color, and everything else will color red. So why does and so of course this adds my plus two. And so why does this have no two edge disjoint rainbow spanning trees. Well, I think it's fairly easy to see for the same argument before that the first rainbow spanning tree which we're claiming to exist must use the blue edge. Now that blue edge is gone. Now we run into the same problem that we ran in the last argument that there's no way to connect you and V back to the can minus two without using two red edges. And so this lower bound construction gives us that our K and to greater than or equal to and minus two choose to plus two. Okay, great. So what about more trees. So young Beckham in West in 2016 almost settled the whole problem. So they showed that for any positive integer T anti Ramsey number of T edge disjoint rainbow spanning trees is equal to the pattern continues in my to choose to plus T for in larger than this quantity here. And then something slightly different for when and is exactly equal to two T. So first of all, where does the n equal to two T come from. Well, if our graph is even to contain T edge disjoint spanning trees let alone rainbow spanning trees, it must be able to contain the T edge disjoint spanning trees. So what are the different edges with that add to our graph. Well each tree has n minus one edges, and there's T of them. So we have T times n minus one edge is necessarily, and therefore and needs to be large enough so that the complete graph has enough edges so and choose to needs to be larger than this value. And of course this is true if and only if T times and minus one is less than or equal to n times and minus one or two, which gives us that to T is less than or equal to. And so the question is really only meaningful for in larger than at least two T. Now of course if in a smaller, since the graph can't even contain T edge disjoint rainbow spanning trees were allowed to just color every single edge distinct colors. And so the anti Ramsey number for anything smaller would just be the number of edges. Okay, so I mentioned that this almost solved the, the entirety of the problem, except for this pesky. square root term here. And so, then in 2020 Lincoln and you were able to completely solve the this problem by eliminating that gap that was left by on Beckham and West. And so I should mention, where does the lower bound for this come from. Well the argument is exactly the same. We'll just break off a K and minus two. These are all distinct colors. So this gives us in minus two choose two colors. And now what we'll do is very similarly to before will color in minus, sorry T minus one unique colors here. Sorry I shouldn't have used yellow will do T minus one distinct colors there and then we'll color everything else read. This runs into exactly the same problem that we had before. The first T minus one trees must use at least one of those distinct colors in the crossing. But then when we get down to the last tree, there's no way to connect you and V again without using two red edges. So that's the lower bound there. And so this completely solved the problem for the case of the complete graph can. And so recently, there's been some interest in finding anti Ramsey numbers when we instead play the game on a different host graph. So this there was this paper from Jalu and Jang very early on this year I think February or some some something around there this year where they instead decided to play the game on the complete bipartite graph. And so they have this result here, and you'll notice that they almost tell the whole story again, except for this square root factor. And this is because they followed very closely the paper of young back in the West, but instead playing the game on the complete bipartite graph. And so G Lincoln and I thought well hey maybe we can apply the same techniques that Lincoln and you did in 2020 to finish the case for KN to finish the the problem for the complete bipartite graph. And it turns out that actually we can do a little bit better. And so to state our results. First, a couple of definitions. If G is a graph and P is a partition of its vertex set CR PG is going to be the set of all crossing edges of G with respect to the partition. So these are the edges whose endpoints are in different parts of the partition. And then EPG is the set of all non crossing edges. So these are the edges with both endpoints in the same part. So just maybe to draw a picture. These blue edges are the crossing edges. Maybe something like this. And these red edges are the non crossing edges. And so our results as the following for any multi graph G. I'm not going to talk too much about the fact that everything applies to multi graphs. You know if there's any loops, we can color all of those unique colors since they don't add anything to spanning trees and then none of the results that we need differ when we talk about multi graphs so for simplicity just keep thinking of everything as a simple graph but nothing breaks in the case for multi graphs. So here is a partition P not of the vertices of G, which satisfies the number of crossing edges is less than T times size of P minus one, then the anti Ramsey number is equal to the number of edges of G. I'll talk about in a second where this is coming from I'm sure, probably some of us already know why this is the case. This is in some sense like the degenerate cases like the not very interesting one so this is the reason that I've sort of started this first, sorry, paused the slide here. Just to take a second and say this is in some sense the degenerate cases and really the result is here that otherwise the anti Ramsey number of any host graph for T edge disjoint rainbow spanning trees so remember we set out trying to finish the case for complete bipartite graphs but instead we were actually able to determine the anti Ramsey number for any host graph of T edge disjoint rainbow spanning trees is given by this maximum, where we take the maximum overall partitions with at least three parts. The number of edges contained within that partition plus T times the size of the partition minus two. And in a couple of slides I'll talk about a little bit more where this number is coming from and hopefully make it a little less mystifying but right now I'll say it relates back to those lower bound constructions that we were working with a second ago. Okay. So I want to take a second to show that the results that were known about the complete graph and the complete bipartite graph can be redirected quickly I'm not going to go through all of the details here but I just want to give you an idea of how you can use our results to derive the known theorems. And it turns out this should really say f of s. If f of s is defined as this maximum where we only look at the partitions with exactly as parts. It's not too hard to show that that function is actually concave up. And then remember we also have that T times size of P minus two term, which is of course concave. So their sum is concave. And so we only need to check the maximums at the boundaries when the size of P is equal to three. And then it's just arithmetic you just check and it turns out that what our function spits out is the same as the known results. And then in a similar fashion, because that function is concave up for the complete bipartite graph. You can also close the gap from Jalu and James results and get this this result here, and another class of graphs that it's fairly easy to check. It could just be the multi-partite graphs. It's essentially the same thing as the bipartite graphs just slightly more notation. Okay, so let's continue. I would be remiss to give a talk involving the edge disjoint spanning trees and not mention Nash Williams. So, you know, one of the most famous theorems in graph theory and certainly for me, I think it might have been the first the first time where I saw sort of something where the necessity was so clearly obvious where but the sufficiency was just mystical to me. You have to be very clever to prove the reverse direction but I can show you the proof of the forward direction and you know a few minutes, which I'll do right now. So let's take some, let's first suppose G contains T edge disjoint spanning trees. And let's take any partition of the vertex set. Maybe it looks something like this. And so there will be edges crossing this partition, maybe something like this. So this is part one part two part three part four. And what we'll do is we'll build an auxiliary graph where the vertex set or the partitions, the sorry the parts in the partition so this is P one this is P two this is P three, and this is P four. And we'll add an edge exactly when there's at least one crossing edge with endpoints in either partition part of the partition. And now this auxiliary graph has a nice property and that property is that it must be connected. If it's not connected. Well then when we blow back up to the original graph there must be some disconnect between these parts. Is that an issue. Well we claim that G contains at least one spanning tree, since we have one spanning tree, G has to be connected. So this clearly can't happen. So this means that this guy, our auxiliary graph has at least the number of parts minus one edges. And this argument that I just made, we can do T times over, we can remove P minus one edges and still contain a spanning tree. And therefore, the argument holds T different times. And so we must have at least T times size of P minus one edges crossing edges in the original graph. So, again, you know it only takes two or three minutes to convince yourself that the necessity is true. You know, as with so many theorems and in graph theory the sufficiency is very difficult. And you have to be really clever to prove. Okay. So then there's this theorem of Shriver from 2003, which has a very, very similar flavor and is slightly more applicable applicable to our problem. So an edge colored graph G contains T color disjoint. Okay, color disjoint rainbow spanning trees in the color disjoint case. Now we're no longer allowing repeat colors across the different trees. So every single color that appears within the T edge disjoint spanning trees must be unique. So we contain T color disjoint rainbow spanning trees, if and only if for every partition of the vertices G, the number of colors which cross the partition is at least this Nash Williams type T time size of P minus one. And so I'll mention this a couple times in the future. It's not directly applicable to what we're doing because we have a little bit more freedom right we're allowed to distribute colors to different trees, but nonetheless it is still very useful. Okay, so I want to go back to our theorem and show the lower bound to you to get a little bit of information on where this is coming from. In the next line for any graph G if there is a partition of vertices with not enough crossing edges, then the anti Ramsey number is equal to the number of edges of G. This is directly Nash Williams. Why, because if I have a partition, which doesn't have enough edges in the crossing, I can't even contain T edge disjoint spanning trees, let alone T rainbow edge disjoint spanning trees. So I'm going to color every single edge a new color, and there's not going to be T edge disjoint rainbow spanning trees. So this is why I mentioned this is somehow the degenerate case. Okay, but for the more interesting part. We made the claim that the anti Ramsey number and any graph of T edge disjoint rainbow spanning trees is equal to this maximum here. And so I want to take a second to show you the construction for the lower bound. This is what we want to show, we want to show that this anti Ramsey number is at least this large. So what we need to do is exhibit a coloring with at least this many colors with no T edge disjoint rainbow spanning trees. And so how do we do it. Well what we'll do is we'll take a partition with at least three parts and we'll select T times size of P minus two minus one crossing edges and color them in distinct colors. So this here is the reason that we need to have at least three parts. If we only chose partitions with two parts, well then this number would be equal to minus one, which of course makes no sense. So we need to have at least three parts. And then what we'll do is we'll color the remaining crossing edges in a new color, which will call see not, and the remaining non crossing edges all in distinct colors. Okay, so let's draw a picture and see what's going on. So maybe here is our partition with at least three parts. And so everything in here is going to be distinct colors. So how many colors does this add. Well this is definitely the number of edges. Non crossing edges. Then what we'll do is we will select T times the size of P minus two minus one colors. So here we have T times the size of P minus two minus one crossing edges, which will all be unique colors, and then everything else is going to get the color green in the crossing. Okay, so we have the prescribed number of colors. Why does this contain no T edge disjoint rainbow spanning trees. Well every spanning tree. Must have size of P minus one crossing edges. And of those crossing edges at most one can be green. Okay, so at most, at most. Maybe let me phrase this, we'll say, because at most one can be green. At least size of P minus two are not green. And since we're claiming we have T edge disjoint rainbow spanning trees. This implies T times size of P minus two, not green. Crossing edges, which is a contradiction, we only colored T times size of P minus two minus one things in the crossing not green. Okay, so that gets us the lower bound. And now the remaining difficulty is to prove the upper bound so we want to color with one more color show that no matter what we always get T edge disjoint rainbow spanning trees. And so what is the difficulty. Well, unlike in the color disjoint case of Shriver, we're allowed to have deficient by partitions or sorry just deficient by partitions. What do I mean by deficient. Well, I mean partitions which have many duplicate colors right in Shriver's case we needed to have many many distinct colors in the crossing. But for us, we can take those deficiencies the duplicate colors and distribute them across the trees and somehow fix the deficiencies, according to Shriver. And so using lemma formalizes what I've just said and gives us a technical tool to use. So G is an edge colored graph. If we have a family F1 through FT of edge disjoint rainbow spanning forests of G now so the forests are allowed to have duplicate colors across them. We say that F1 through FT has a color disjoint extension in G. Exactly what you expect to happen, we can extend each of the F1 through FT in a color disjoint way meaning all of the edges that we use to extend the FIs do not appear anywhere else. And we can extend them each to to a spanning tree in an edge disjoint way as well. So that's the idea there. And then in 2020 Lincoln and you were able to give us the following result which has a similar flavor to Shriver and Nash Williams. If G is an edge colored graph and F1 through FT or a family of edge disjoint rainbow spanning trees. If for every partition of the vertices of G. So we have a similar Nash Williams T times size of P minus one, but then we have this quantity here which we'll talk about in a second, then you can extend the fi. And so what is this doing. So G prime here in the inequality is the graph that we get when we remove all of the edges, which have colors that appear somewhere in the fi. Okay, so you remember the, the color disjoint extension means we're extending color just disjointly so we don't want to include any of the colors that we used in the fi. And so how does this help us. Well you can imagine if I've got a partition. That is deficient in the sense that maybe it's got many repeated red edges. Well what I can do is I can now distribute. I have three and maybe you know maybe we've got more. I could distribute those three red edges to the first three of these forests say, and then what happens to this inequality. Well the left hand side, the colors in the crossing decreases by most one. So this decreases by at most one, I could have only lost one color in the crossing by do this by doing this, but the number of crossing edges with respect to P of the fi has increased by three. And so in some sense I'm fixing these deficiencies right before just the number of colors crossing was not enough. And so I removed some of the duplicate colors added them to the fi and bump this left hand side up to make it less deficient. Okay, so that's the kind of game that we're playing. So, from now on, let's assume that G is a graph which satisfies Nash William so it does have T edge disjoint spanning trees and is colored with this many colors. It's some way to get an idea on just how deficient partitions can be if it were the case that some partitions could have, you know, terribly few colors crossing we might be in trouble because it maybe would be likely that we can't fix them with this extension And so it turns out actually that the partitions can't be too deficient. So I before I state that I'll give you two definitions so a to PG is going to be the number of edges within the partition minus the number of colors which appear within the partition. And so this this number is a little bit mysterious but to me it kind of is like how far away from our lower bound example, we are. So what do I mean by this will remember that in our lower bound construction. Every edge within the partition got different colors. And so this number would be zero for our lower bound. And then let's the second quantity is C of PG, and this is going to be the number of colors which appear within the edges, and in the crossing, sorry within the non crossing and crossing edges. And again, this is somehow how far away we are from the lower bound construction, you can imagine that, or sorry, you remember that the lower bound construction had no colors shared within the parts, and in the crossing. Okay, and then we can do sort of a slick application of the inclusion exclusion principle on these two sets to get this equality here. So where is this coming from. And certainly I can write the number of colors on the edges of G as the number of colors in the crossing union with the number of colors in the edges. Sorry within the parts. And we have this minus EPG. And then applying the inclusion exclusion principle here. We get this is the number of colors in the crossing of PG. And then we have less the number of colors on the edges within the parts. And then what do we have, well we need to subtract away those colors which belong to the crossing, and within the parts, and that's our C. So we have minus CPG. And then we have minus of edges. Yes. And so now the plus C colors within minus the number of edges, that is of course our negative Aida. So these two guys combine these two terms, combine here to give us our minus Aida PG. Okay so that's where this equality is coming from. And so why is this equality useful. Well since the number of colors is larger than this maximum overall partitions. We get the following corollary. So you can you can imagine that for any partition with at least three parts. I'll just slap a greater than or equal to on here and place this maximum over here. And then since this maximum is taken overall partitions with at least three parts, whatever partition with three parts I chose, I can replace this maximum with that partition, and then the minus EPG will cancel. So you get the following corollary that if P is a partition with at least three parts. Excuse me, if P is a partition with at least three parts, then the number of colors which cross P is at least, and this is exactly what we had before, we add across the Aida PG, CPG, and the plus one, and then the number of edges of PG cancel. And two so this doesn't apply for partitions with size to which I'll call bipartitions in the future. Why, well because we need to use that fact that the maximum is taken over only partitions of size, at least three, but it does work out if it so happens that the number of colors on the edges of G is strictly larger than the number of edges within the parts of our partition. And then we get a similar thing. So why is this result this corollary nice. Well, for lack of a better term, it means that we can be kind of lazy with how we construct the fi. This is probably the most technical thing in the in the paper. But really what I mean is this thing in parentheses that when we're building the fi to use the extension lemma, we only need to ensure that when we're adding edges into the fi. We are not increasing the number of colors, this quantity on the right too much. And so, again, for lack of a better word it means we can be kind of lazy with how how we construct the fi. Because our deficiencies are not too large across many partitions. So with that corollary, we are ready to start the proof. Okay, so let's, for the sake of contradiction suppose the G is an edge minimal counter example that does not satisfy our claim. Okay. And so what we'll do. First, we'll suppose that there exists a partition with at least three parts with the number of crossing edges equal to the minimum, according to Nash Williams. So in this case, the number of colors on the edges of G, we can calculate that exactly. This maximum, you know you can rewrite the number of edges within the parts as the number of edges of G minus the number of edges in the crossing. Oops, sorry, not number of colors number of edges in the crossing. And so finding this maximum is equivalent to finding the minimum of the crossing number of crossing edges. And here we've achieved the minimum possible number of crossing edges with this partition. And so we can calculate exactly the number of colors on the edges. And that number of colors is equal to the number of edges minus t plus one. Because remember we've colored with this many colors plus one colors. Okay, so how is this useful. Well, it turns out that for any bipartisan p prime, the number of edges and p prime should be equal to the number of edges of G minus their crossing. Well that number, the smallest the crossing can possibly be is T for bipartisanship. So this is less than or equal to the number of edges of G minus T, which is strictly less than the number of colors on the edges of G. So why is this useful. Well now every single partition satisfies our corollary from before, and it means that we can be again for lack of a better term lazy with the way that we construct our forests. Then all you need to do, since the number of edges or sorry the number of colors is very large, then we can sort of very easily just greedily construct the fi. And it turns out that they will satisfy the lazy condition, and this fixes all of the partitions with respect to the extension lemma. And we're good to go. So that handles the case when there exists a partition, which meets the minimum. In my opinion, this is sort of the less interesting case, and the more interesting case is is case two, where for any partition with at least three parts, the number of crossing edges is at least T times size of P minus one, plus one. Now, as with any, you know, assume our graph is edge minimal, what we'd really like to do is somehow remove an edge, show that it still satisfies the assumptions, and then get to apply our hypothesis. So how would we do that. Well, we'll do that with the following claim. We claim that there exists an edge with color multiplicity at least two, whose removal maintains that for every partition, the number of crossing edges is at least the minimum, according to Nash points. So, there's a couple of things here. First, it doesn't matter what edge I remove. All partitions with size at least three are still happy. They still have at least T times size of P minus one crossing edges. Why, because I've only removed one edge. And since we're in case to every partition with size at least three has one more than the minimum necessary. Okay, so we certainly do not need to worry about any of the, the partitions of size three. So then the question is what about the partitions of size two. Well, if any sort of bad bipartisan exists in the sense that removing an edge makes it have not enough crossing edges. Then there must be exactly T crossing edges in that partition. And so now my claim is that there can only be at most two such bipartitions. Okay, and to see that. First, what we'll do is we'll take their intersection. Okay, so suppose P1 and P2 are two bipartitions for which there are exactly T crossing edges in their their crossing. And then if we look at their intersection, the number of edges in the crossing of their intersection is certainly at most the sum of the crossing edges with respect to P1 and P2. If they're disjoint then there's equality, or they could share some crossing edges in which case, you're still smaller. And we know that number is equal to two T by assumption. That's an issue. Well, no matter how I intersect two bipartitions, unless they're the same bipartisan, there has to be at least three parts in that new partition. And so we have that the number of crossing edges in the intersection of the two bipartitions is at most two T from the previous line. That's three equal to T times the size of P1 intersect P2 minus one, but this partition P1 intersect P2 definitely has at least three parts. And so that contradicts the fact that we're in case two. So at most one bad bipartisan can exist. So let P be that bad bipartisan. Since we're in case two, this maximum is less than or equal to the number of edges of G minus T minus one. And so the number of colors on the edges is at most number of edges of G minus T. Certainly, there must be at least one color that has multiplicity to if every color has multiplicity one meaning that they only appear on one edge, then G contain T edge disjoint rainbow spanning trees to begin with. So if C one is the number of colors with multiplicity one and G, it's definitely true that C one is less than or equal to the number of colors minus one, which with the above inequality gives us less than or equal to the number of edges of G minus T minus one. So if we graph G two, which we get by deleting all of the edges that have colors appearing elsewhere on the graph at least once. Then the number of edges of G two is definitely equal to the number of edges of G minus C one right after I remove every edge with a unique color everything else contributes an edge to G two. And then with the above inequality, we get that the number of edges of G two is equal to T plus one. So does this help us. Well, since P was a bipartisan, which is bad in the sense that if I remove any, it has T edges in the crossing. I know that there's T plus one edges with multiplicity at least two. And so at least one of them must be outside of the crossing of P. And so that edge, I'm allowed to remove in our previous claim is proven. So what have I shown shown that I can take an edge whose removal maintains Nash Williams for the existence of T edge disjoint spanning trees in the graph after I've removed the edge, and also that edge doesn't lower the number of colors on the graph. Okay, so how does this help us. If we're chosen to be edge minimal, we must have that the anti Ramsey number of G without a of T edge disjoint rainbow spanning trees is less than or equal to the maximum that we've seen many times now. And now how does this maximum relate to the maximum for G. Well, this quantity here for any partition the number of edges within the parts can only decrease. I say that I said could ever add an edge by removing an edge. And therefore this is certainly less than or equal to the maximum if we replace that with G. But now, how many colors. Did we color the edges of G without you with, well the same number of colors that we colored G with, because he was chosen to have color multiplicity at least two. T edge, didn't remove any colors. And so our G of T is colored with this many colors, plus one. And therefore, G without a must contain T edge disjoint rainbow spanning trees, and therefore, so does G. And that finishes the proof. So that's all I've got for you guys. Thanks everybody for listening. And here are a bunch of references. And the paper is on archive if you'd like to take a deeper look. Thanks, Drew. Everyone could thank you in some way. And let me open up for questions. I would like to ask a question which is a conceptually simpler problem. Since it would have fewer quantifier. So you have a graph, which color the edges. It's a polynomial time algorithm to find the maximum number of these joint rainbow spanning trees. I don't immediately see a clever way. It's definitely an interesting question. So you're saying that I was this long literature so so you're saying you're, you're inputting a graph with a car. Yes, and you want to output the maximum number of edges disjoint rainbow spanning trees. Yeah, that's an interesting question I don't I don't know much. I don't know much about us. Are there any other questions for Drew. I have a quick question about your main results. So you applied that to different families of of graphs or use you had some application slide I think you called it or or something. And get there. So you had these three families you have the complete graph, the complete heart type graph and then multi part multi part type graphs. So this might be obvious I haven't really thought about how this works or not but when you apply your main result to these families of graphs does something special happen to where it becomes concave up or would it be concave up in a lot of cases or for that. Part of the reason is that it's, it's easy to calculate the these maximums at the boundary. I'm not sure what qualifications you would need to ensure that this function is concave up for an arbitrary graph. Maybe Lincoln can maybe Lincoln can say something more than than me there but I don't I don't immediately see for for an arbitrary graph how to determine whether this function is concave up. This commission can justify I think you can construct a graph which is not concave up, but you can take some graph and that brought up. But in general is it's we don't know how to test the graph is a function of up or not. The reason we included here because some people solved for the company by photograph, and they have a gap, we had to address for this special case what is the solution. And more generally, you know it's the third one is registered easier generation for the second case. I'm just curious if there's any intuition was to other families of graphs that this might be like nicely applied to or something like that, but I understand that I guess the original question was to solve the bipartite case. Yes. Very. Thank you. Any other questions for drew. All right, if not let's thank drew one more time. And I appreciate everybody coming out today. I don't know if this is bad to say quite so early in the semester but I am ready for a long weekend and that being said I hope everyone enjoys their long weekend. So, I think that's all for for today so I'll see everyone next week.