 and this is my first time kind of organizing this sort of thing so I realized that I messed up and forgot the treats. I won't have to do that. I apologize. Because today we will be math bags in bed. We have our colleague Darren Parker who's going to talk to us about multi-compositions taking a graph apart and putting it back together. Thank you very much. It's nice to see you all here and it's a joy to talk about one of my favorite mathematical topics. And as I was thinking about this talk, I was thinking about a good metaphor and this made me think of it because my kids and I love jigsaw puzzles. This is the jigsaw puzzles dot com and the good way to think about how a jigsaw puzzle works. So they're in the picture and they break up into pieces and then they're all spread apart. And so you have these different things that are happening in jigsaw puzzles. You have that whole picture which is kind of your target and you know that going in what your target is. And then you have all these pieces that you're supposed to put together. I don't have enough time to do this now. So I just wanted to show you how I was thinking about this and how it helps me. My father had a thing about how to think about some of the problems. So what are the expenses of jigsaw puzzles? Well, you need a collection of pieces that you're allowed to use. And in most cases, in the physical case, it's whatever's in the box. And that's all you have. Actually it's really restraining because you can only use the pieces once and you can't like bend them to make them fit in the right way no matter how hard you try. My children always give me a hard time when I try to do that. You need rules of putting the pieces together. In this case, the rules are pretty rigid in a real jigsaw puzzle. It'll be a little looser when we start talking about graphs. And the third thing is you need a criteria for being fit. And that's kind of what they started out in that picture. The picture is complete. Now you have your final picture here, your target picture. And we're going to apply this process to graphs. I think pretty much everybody in here knows what a graph is, but if you don't, just in case you don't, I'm going to say it anyway. So here's a picture of a graph, a visual representation of a graph. And in general, the abstract definition is you have a pair of sets. E is a set called a vertex set, and that's represented by the dots up there. And that could be any sets. We just represented them by dots traditionally. And then the elements of a vertex set are called vertices. E is a set called the edge set that contains unordered pairs. And up there in the picture, the unordered pairs are represented by the lines from between the vertices. 345, you might see this. So let's focus this graph a little closer and see how we can think of it as a puzzle. The other thing is this is my final picture, my puzzle. And you can actually think of it in terms of these three pieces. So it doesn't look exactly right. So I have to move around and I have to bend those 1, 2, 3 a little bit. And then this is the rule about putting them together. I'm going to think of this one here as having snaps on it. Power factor. I really wish I would have bring something with snaps on it and put it together. But you can snap this vertex on this one, this one, this one. So you know how that works. Yay, there it snaps. All I have is PowerPoint and then graph. And we're going to help you out. So those are the rules for putting them together. You can see that more or less they're the same pictures on the bottom of this little set. Now, we're going to say that I'm done with that. And again, I want you to notice that these pieces that we had, they're all graphs themselves. So the end picture, the end puzzle is a graph. And the little pieces that you use are also graphs. So, to talk about the mathematically, let G be any graph and let T be a set of graphs. And we're going to think of G as the finished puzzle in the final picture. We're going to think of T as the box, right? The set of pieces that we used to actually make the puzzle. Again, a little bit looser rules. We put the graphs of NT on G by sort of snapping the vertices onto the vertices of G as we did before. And finally, it's a little bit different than your normal puzzle, is we can use many copies of the graphs of T as we want. So if we have one piece that we really like, you can make copies of it and use it again. But the one little rule of that is you do have to use every piece in the box. At least once. At least one in the T. And then if you want to get a proper mathematical definition, these are called T-multi-D compositions. This is the partition of the edges of G into copies of graphs in T in which each graph T is used at least once. So in each of the partitions you'll have a copy of the graph. And when you put them all together, you get the whole graph. So let's look at this final picture. It's a lovely graph. I like this one. And here are our pieces. This one piece. We have this triangle that we're going to use over and over and over again. Not the most exciting puzzle you ever saw in your life. But once the best I can do, I'm powerful. Alright, so we're going to start here with these seven vertices. And we're just going to start filling in triangles. Notice I sort of stressed it out into a different shape. That's okay. That's part of the rules. Not like the regular pieces. They won't break over for anything like that. They're very stretchy. There's another one. At least I won't let that one cool down before I put down the next one. There's the third one. And we're going to keep on going. There's another one. We keep on putting in triangles until we get all the edges. Now we can't use an S twice. That'll be against the rules. And we're going to keep going until we get to the end. Won't we get to the end yet? Yes, we'll get to the end. Okay. And we get the whole one by putting in various copies of the triangles. Makes sense? Okay. And this graph actually is... This is a nice graph. This graph, this is a complete graph. Seven vertices. And it's very, very common in graph decomposition that your finished piece, your finished puzzle, is used to get a complete graph. So let's go ahead and talk about those. These are things that most people in the room probably know about. But we use complete graphs to define that. It's just a graph. We'll call it KM. It has N vertices where N is at least two. Right? It has N vertices. And every pair of vertices has precisely one N between them. So here are two pictures. That is the complete graph K7 that we saw before. Seven vertices and all the edges are there. But the one over on the right that's missing some edges. No. So here's a fun... Yeah, most of us in here, this is KNF. This is going to be important to us later. Well, let's count. So one way of counting them is you can say you have N vertices. Each vertex shares an edge with N minus one other vertices. Right? Because you have one for each one. And so all that means there are N times N minus one. This is right. Oh, you're right. Not right. Yeah, what happens is each edge gets counted twice. So we have two vertices, B and W. Each edge gets counted as an edge coming out of B. And the same edge gets counted as an edge coming out of W. So we each count twice. So yeah, I'm going to divide by two. Sorry about that. Great. So the reasons that I do it with my colleagues has to do with the question of what if you want to use more than one type of puzzle piece. Right? So we were using just triangles before. And that's actually a really rich history of really interesting problems of graph decomposition with just one puzzle piece. They can be really hard problems. But we were looking forward to doing more than one. And in particular, the big question comes up is how do you choose pieces? Now, one way you can do it is you can just say, I'll choose my favorite two or three graphs. Those will be my pieces. And then see what happens. You can do that. But Gueda and Devon found a more systematic way of finding puzzle pieces. So what you do is you take our fixed, complete graph, KT. And this KT isn't necessarily going to be the one we're going to use in our final puzzle. We're just going to use it to find the pieces. And the way we find the pieces is we take those edges and break them up into two sets. Partition them into two sets. Each one is going to be a graph. H1 and H2. We have some rules because we don't just want any two graphs. We want to make sure that H1, H2 use all the vertices of KT. And what I'm going to do by that is all T vertices in KT should have an edge. Each one should have an edge in H2 and each one should have an edge in H1. So each graph should not have any isolated vertices in KT. And then the second thing that should happen, they shouldn't be copies of the same graph. This would be a long, not very pleasant process of finding one graph that we already had to do anyway. So we want to make sure that there are really two different puzzle pieces. So the terminology that we use is if TE, H1 and H2 are graphs, and if T is greater than or equal to 2, we call H1, H2 partition KT if each vertex of KT is incident with some edge of each HI. And finally, if H1, H2 and non-isomorphic graphs in other words are different, then we call H1, H2 a graph pair of words T4. So we kind of have two questions to worry about here. First of all, if we're given a T, what are the graph pairs that we can get? How many different boxes can we get from a puzzle? And then the second question, once we have our pieces, what kind of puzzles can we make out of them? So we're going to do the first one. Look at what kind of boxes we can get. And just to make things not too crazy, we're going to look at graph pairs of order 4. So we're going to look at K4. We're going to break it up into two graphs and see how many are going to actually satisfy all the rules. Excuse me, a graph pair of order 4. H2B, some kind of graph pair. See, an order 4. And just to make things organized, let's suppose the H1 has the smallest number of edges between the two. Okay, so now the first question we may ask is how many edges can H1 have? And in particular, what's the smallest number of edges? Well, you can't have just one. Because if you have one edge that's two vertices between them, you have two isolated vertices, right? You get all four. So that's not really... Yeah, we need all four. And so if it has two edges, then you can do two edges, but they cannot share a vertex. Because if you have two edges sharing a vertex, you actually want to get three vertices. We need all four. Yes, so sharing is often three vertices. And that means your H1 would have to be this particular graph. And I do want to point out, this is really one graph. It has two isolated edges, but it's really one graph. They come in pairs. I saw the time. Okay? All right, so here's what H1 would look like with K4. Yeah, you can see the colors pretty well. See these two red edges over here? That's our H1. And then H2 is whatever's left over, right? It's loaded into two pieces. And there are these two... Here, let me interest that for you. There you are. It looks like a cycle of four vertices. So there's one more way. And obviously those are not isomorphic. And they both use all four vertices. So that's great. That's great. So we have ourselves a graph for our order four. And in fact, that's the only way you can do it if the smallest one has two edges. Okay? Well, can H1 have three edges? That's a good question. So we can answer that. Well, let's put it in two cases. The first cases are if our H1 looks like this. In other words, all three edges share a common vertex. That one has a good constant problem if we put the other edges in of K4 and then split them apart. Whoops. That red one only has three vertices. Okay? So that means at least two of them, at least two of the edges are not connected to one another. All right? So something like this. See the red ones, right? That's not the whole graph. I'm missing an edge. But at least two of them have to be separate. Okay? Now, the thing to notice is I have to add it every day somewhere. And it's got to be one of the remaining four. But no matter which one I add, it's always going to be a path graph. So if I add this one, for instance, if you have path here, to here, to here, to here. And similarly with any of these others. So I'm going to choose the one that looks like an N. But no Y. That was kind of fun. But there's our N. Right? Maybe you have a path of length one, two, three. And we go ahead and split that up. And there's your, it looks pretty good. All those stuff, we have two graphs. They use all the vertices. There are no isolated vertices on either of those graphs. Um... Yeah. Let me turn the left one sideways. They're the same graph. So we just have one fellow piece. And that's no fun. So we can't have any graph pairs where one of them has three edges. Each one can't have four edges either because there are only six edges all together. And each one is supposed to be the smallest one. Uh... That's two of the edges. Right? So for four, we have two times P, two P2 is what we call both the two isolated edges. And then C4 is what we call the four-cycle. So it seems like you have some game in mind here. How come you can't have the same... How come each one and each two can't have the same one? Well, because then you have the same puzzle piece in your... What's wrong with that? Well, then you can just say, let's have one puzzle piece and let's use it. So we like... The whole point of this was to have a puzzle with more than one piece. More than one type of piece. So if you do one, if you have one piece, and those are the gravity competition problems, those have already been looked at a lot in the last 60 years or so. And so we were... Actually without a way to end, what we're looking for is ways of finding graphs that have... that are different. Different from what we're looking at. Yes? I forget. Are you allowed in your each corner to have an isolated vertex? You cannot have an isolated vertex. Yeah, you can't. K4 is actually the smallest complete graph which you can have without care. Alright, I feel better now. And we've looked at graph number 4 and graph number 5 and graph number 4 and graph number 5 and we've looked at both of those type of pieces. So that's six boxes that they looked at in their... This is the back in the early 2000s. So that's just finding the pieces. Now we've got to figure out what puzzles we need to make. So now the question becomes, do complete graphs have multi-decompositions of graph pairs of order 4? So for which complete graphs can we actually use these puzzle pieces to make the whole picture? And it turns out not that unusual. So here are H1 and H2 and one thing to keep in mind about H1 and H2 is that H1 has two edges. H2 has four edges. So any way that you put them together you're always going to have an even number of edges. So that's going to lead a lot of complete graphs. That means that this number that we derived before, this n times n minus 1 over 2 has to be an even number. So what is that going to happen? Well, if n is even, we can write the number of edges as n over 2 times n minus 1. N being even means n minus 1 is going to be odd which means that n over 2 has to be even but that means that n has to be multiple 4. If n is odd, we can do the kind of the same thing. Now we're writing the number of edges as n times n minus 1 over 2 is odd now. Which means that n minus 1 over 2 has to be even n minus 1 over pasta then be a multiple 4 which means that n is one more than multiple 4. So we've narrowed it down considerably. There's a lot of complete graphs that aren't even candidates, simply because they don't have this right kind of of edges. Okay, so here's the sort of half theory we get. If you have a multi big competition for it then n is either 4K or multiple 4. So let's look at the multiples of 4. Let me see. So K4 is the easiest one. We got a 4K4, right? Yes, that would be easy. That was boring. So what about K8, is that boring? K8 does have 28 edges and I don't want to fill in the whole thing but it was hard enough generating these pictures as it is. I don't want to go through all those. So I want to use some of my previous cases. Being induction people, we kind of understand this. And I'm going to organize it like this. So imagine the top, right? The top, we look at all the edges that go along the ones on the top. That's going to form a copy of K4. But we just show that you can do a multi decomposition of that. So imagine it done, okay? Same with the top, imagine it done. So so far we have two 4 cycles and two of the 2P2s that we put in and that takes care of the ones on the top and the ones on the bottom. And now all that's left are those edges that go between. Right? So that graph is the complete biopartite graph K44. So I'm not even going to bother with the ones on the top and bottom, just the ones in between and I'm actually going to fill in and this is actually important with 4 cycles. So there's 4 cycles, I twisted it for you. So there's one 4 cycle, let it cool down and then I'm going to stress it to the right and then I'm going to do the exact same thing with the other 2, right? I'll stress that one to the left and then down. So that takes care of everything, right? The top got taken care of and bottom got the K4 case and in the middle got taken care of and we just did it. Great, let's look at K12. K12 we're going to do the same thing, right? We're going to have 4 vertices on the top, 8 vertices on the bottom, we just did K8 that's our induction hypothesis, right? So we're going to consider those done and then take a step further. We're going to ignore the 4 vertices on the right on the bottom and just look at what remains of K4-4 again, it's the exact same thing we just filled in, right? So we did it, it's done. And the nice thing is, the one that's remaining after that is 4-4, so again that's done as well. And so now you can see K16 is going to work the same way. You're going to have 12 on the bottom, they're done 4 on the top, it's done and then you have what, 1, 2, 3 3 copies of K4-4 which you filled in exactly the same. So they're all possible. We're going to be at the most decomposed forever. So let's look at when we have one more than the most 4-4. Let's start with K5, it's actually the hardest one. Now we need at least one each, at least one 4-cycle and at least one of the two P2's. So let me put it in the 4-cycle, it doesn't matter what I put it in. I think I'll complete that, we have lots of symmetry, so I can play whatever I want. So there it is. Now the first question is I certainly can't fit it in using those same 4-verses. So at least one of the edges of my other 4-cycle has to be that red one over there. And in fact I wanted to keep walking along my 4-cycle and talking back to where I started. So here's my first part of my walk. Now my second one, I can't go there, I haven't taken, can't go there. That's my only path to take on my 4-cycle walk. And now I learned the exact same problem. The vertex that I just got to, there's only one edge left that hasn't been taken up. And that's that one there. Unfortunately that's not a 4-cycle, but a 3-cycle. So even if I wanted to, I couldn't put in a second 4-cycle. So if there is a possibility that I have a multi-decomposition, it's going to be one 4-cycle, that's four, and then I have six more left over and that's going to be three So let's go ahead and do all the 2P2s. Okay that's a 4-cycle. Now I can't do that obviously, right? No sharing for the Cs. So what I have to do is something like that. Now that's just one of the edges of my 2P2. The other one is somewhere else, but for sure it does not include this vertex. Somewhere on the other four. So that's one of them. Now the second one also, just because every edge is coming out of that lower right, vertex has to be somewhere. So there's the second one. Again, the other edge, I don't know where it is, it's one of the other ones. But for sure it does not include that lower right vertex. And then there's my third one. And I'm done! Only I'm missing one, that bottom one, right? And so I'm really not done. So that's impossible. So K5 I can't do. So that's it. What about K9? Well, I'm going to do this. The reason I'm going to do this is because we've already done K8. We did that in the multiples of four case. So consider that done. And then that only leaves me, oh yeah, all these vertices, all these edges right from this one down here. Recycles in there. And I can't even fit in a 2P2. Well here's what I'm going to do. Now this we call, this means we have to recall something about the way we can start to K8. That's a multi decomposition of K8. One of the things we did is we took each of these sets of four and did what we did on K4 of it. So in particular, each of those sets of four have four sides of that. Let's get part of that to multi decomposition. So here's what I'm going to do. I'm going to keep everything the same, right? The same multi composition before except I'm taking out those two sides. Those edges. I'll have to put them in again at some point. But I'm taking them out, because I'm going to create some space for myself. And then what I'm going to do is I'm going to throw the rest of the edges with copies of 2P2. So here's one of them. There's one coming down. I don't know if you saw that one turn black over there on the left. That's the other one. So I've got to do these in pairs, right? That's how the graph works. And then every time I do this, I'm going to have one going up and down, and then one of the green ones turning black. There's four of them. There's the other four. Look, that was smart. I got those so I could create some room there. Smart me. And then I've got all the edges. It made me a little bit nervous about what I just did. And that is what I do. I took out two 4-cycles and replaced them with a bunch of 2P2s. I've got to make sure, there is a danger that all I have done is just take out all the 4-cycles and replace them with all. So now I have something with all 2P2s. I've just been using one type of puzzle piece. Let me see if I can remember what happened before. Did that happen in this case? Yeah, you have to remember what happened when we filmed it. So these came from decomposing K4s, right? But then we had the ones between the top and the bottom. Do you remember what we filled those in with? 4-cycles. Okay, so we're okay. So we have 1, 2, 3, 4 4-cycles in there from doing the K8. And so this turns out to be a really powerful technique when you're constructing these multi-decompositions. So that's sort of what happened here. So the first thing we did was we arranged the vertices of K9 into two sets. One was 8 vertices, one was 1. Okay? We added some multi-ade which we already did. So we say it's done. And that's it, right? Except that we take out copies of at least one more of the graphs in that set of graphs, in that graph here that we know to be in K8. And we actually, in this case, we need to know them more. We need to know where they work, right? We need to know when was that first set of 4 vertices and when was that second set of 4 vertices. Incidentally, this is sometimes not easy to do. It's sometimes impossible to do. This is something that gave us space when we're working on other problems. And then we used these edges in addition to the ones that we needed to fill in anyway to fill out the rest of the edges of K9. And then at the end, we made sure that we had at least one of each graph, right? That the ones we took out were either replaced later or there were more that we didn't take out. In this case, there were more that we didn't take out. All right. So this gives us, we did an endowments theorem. This is from, oh, I just put the year. 2002, 2003, somewhere around there. You always have a T-multi-decomposition of order 4 as long as, precisely when you have a multiple of 4 vertices in your complete graph, or one more than a multiple of 4 except for you can't have 5. And that's sort of, I showed you K9, but there's the same kind of recursive construction of these multi-decompositions that you use as well. Module 4. Well, what do we want these three different kinds of graphs? And in fact, this is the problem that I have worked on with my collaborators and that we actually continue to work on. This problem has a lot of links. Well, we construct them in the same way, right? We have, we take a complete graph, KT. We take that, now we break the edges up into three different sets. Each one forms a different graph, different in that they have to be mutually non-isomorphic. And not only that, but there can't be any isolated vertices. So they have to, each of those graphs has to touch each of the vertices in the complete graph. And then, what's the third thing? The third thing is that they partition the edges of KT. And then we call that a graph trip over T. So now we're dealing with a puzzle that has, where the box has three pieces that we can make up, right? And we're still trying to get a final picture of a complete graph and we don't want to know which complete graph will work for various reasons. Okay? So this doesn't seem like too much worse than what we had, right? So we had graph pairs were four, there was one problem. Graph pairs were five. I didn't derive this, I'm not going to. The five of those, all right? So we have triples, how bad do those be? Well, first of all, the smallest complete graph for which you have any graph triples is going to be K6. So that's what we saw, right? We'll do K6. And we got 131 graph triples of order six. So we're thinking, wow, this is going to be a big long paper with 131 zeros. It's not. Don't want to be like that. It's not, it's not. Yeah. So don't have to do something. And let me give you an example of the type of reasoning that we used. So here's K6, and you're on the color of the edges so that you can see the, you can't see the graph. Okay, so there's a triple. And you have these smoothies on the left, smoothies on the right, but one that I really like, and I like this one in the middle, I seem to like P2. So this is not two P2, it's like the last one, it's three P2. Okay, so this is one possibility you have for a graph triple. And let's see what's the proof of, you know, the proof of finding all the T-multi- or all the KNs, for which there's a T-multi-D composition. So let's talk about the K6. You already know we're going to have K6. So of course there's going to be a multi-D composition. Sorry for that. So we can bring that up. Okay, fine, fine. Okay, that's fine. All right, so now let's go from K6 to K9. So let's give you by three, this is not number four. All right, so we like to do the same thing as we did before. You can see I have the same kind of picture, but I have the six on the bottom, so there's our K6, and we just sort of do our multi-D composition. So that leaves us with three edges up top, and then with 18 edges going to the bottom. And I didn't like those swigly graphs on the side, so I want to stick with three P2, which is no problem, except that I start taking care of those edges on the top, and that is a problem, right, where I'm trying to do it properly, without sharing the vertex. Can't unless it's on the bottom. It's okay though, right, because we've been in this situation before where we can take stuff out and put other stuff back in. So I'm going to take out my three P2s from the bottom. Now I'm going to have a thing without a net here because that was my only three P2 that was in my multi-D composition. K6, right, there's only one. And now I have nothing left, except for those other two graphs. All right, those other two squiggles. That's okay, okay, so I feel a little nervous, but there, there's my three P2s. I feel a lot better now, right? So basically I took one out, okay, so that would cool down, and one of the ones coming up and down, and one of the ones on top going up and down. There's one. Up and down, up and down, up and down, it's finished yet. Right, so there we go. So the rest of you can just build the three P2s. And oh, one of the little more, sorry, sorry. I didn't mean to leave you out. So that's K9, but this is that we can do the same thing with K12, right? Between nine on the bottom, taking out three, and then we can play that same kind of game for K15, and in particular, anything that's a multiple of three. So that's zero, one, three. Okay, so that's good. So that takes care of infinitely many of the KNs. And that means we only have to figure out where you're one more than multiple of three, or two more than some, one more than three, or two more than three. And then the other thing you may notice is that we have this one graph triple that we were working in. It's one box, it's one puzzle box. But we really, as far as the proof is concerned, we only, only one of the pieces showed up in the three. That was the one with the three P2s, right? So really, I could have repeated that proof with any graph triple, as long as one of the graphs in there was the three P2s. And that's awesome, because there are 26 of those. So I added 131, just with the, I gave you a partial proof. I took care of, at least for multiples of three, 26 graph lengths out of 131, which is a tremendous relief, tremendous relief. All right, so of course, we have to still do one and two model three. I'll do one model three. We should start with K7. We'll start with K10. K7 is hard. K7 is hard. I'm not going to do K7 here. Other K10. And actually K10, even easier than K9. We've got an extra one on top. So filling that in with three P2s is no problem at all. And so, doing the same thing we did, we put in three P2s, three P2s, three P2s, and that is up and down. Up and down. Up and down. So one more. Yep. Same kind of thing happened. And then again, you have these three constructions. You go from 10 to 13, 13 to 16, and induction is here with me. All right, and so our final theorem on this particular one, where just one of the graphs was three P2s, went something like this. And this was with, I owe a great debt to my collaborators. I don't know the way down, and Stephanie Edwards, I owe many other debts besides that medical, who worked on this while I was at the University of Dayton. And so if you have a graph triple, more than six, where the H1, the first one, is the moment three isolated edges, that's our three P2. If it is an integer at least six, less than six, we have no prayer of having a multi-day competition, because it's too small. If it ends in multiple of three, then the KN has a T multi-day competition. We saw that. It ends one more than multiple of three, but not seven, then it always has a T multi-day competition. So 10 and up. Well, it's hard to give you the precise statement of what happened at K7, if I can't show you all the triple, because I'm not going to. But the upshot of it is sometimes it works and sometimes it doesn't. Okay. And then next, if you are two month three, and if H2 and H3 both have six edges, and what happens there is if H2 and H3 have six edges, H1 has three edges, you're always going to get a multiple of three. And unfortunately, three K plus two is never a multiple of three, so you have no chance of having a multi-day competition. But in all the other cases, as long as the arithmetic works out, right, in all the other cases, it does have a T multi-day competition. And then I shouldn't be composition to your puzzles, but we actually approve some results about what happens when you can't get the whole thing. And sort of ways of doing it where you do the best that you can. So best that you can is, you mean that you fill in as many of the edges as you can. That's called a multi-packing. Yes, multi-packing. And one where you use maybe one or two or sort of the edges more than once, and that's called a multi-covering. And all these three put together are called multi-designs. So if you see my paper and you see multi-designs, where's the multi-decompositions, multi-designs encompasses all those things, encompasses these multi-decompositions and encompasses the multi-packings and the multi-coverings. So we determine that for all of these that we did. So we don't come just through the ones where there are the three times P2. We also, whoops, okay, so here's our final results. We also did the ones where, and this is work with the, I mentioned my collaborators, this is back in 2006, we saw a lot for all triples whether we had one of the graphs had three edges, and that was the three P2, or all the graphs had five edges. That was one of the other ones we did, and that took care of 38 triples. So actually, most of those came from the three P2, either 26 of those, and only 12 of the five edges. But you don't need to see, the five edges are the hardest one. And in most TVs of the edges, you have yucky base cases to worry about. But in any case, one, let's see, let's see, let's see, let's see more impressive at the time. It's not in 2006, I don't know how it's been, hearing about it there, and this was actually this year, and last year that I was working on this with Stephanie Edwards and Charles Hughes, I haven't heard from folks about it, and he's in the computer science department over at Hope. Actually, he thought this tremendously, what happens in these proofs is, you know, an mathematician, you know, the recursive instructions. And we had all these set out, it took us a while, but it was hard. We had all these set out, and then the base cases were just awful. I mean, first of all, we have, what was it, 93, I remember in particular, I think it was 93, 93 cases, the triples that we had to worry about. But that wasn't the bad part about it. The bad part about it was how big they were. They were awkward number of edges that we had. They were like 7-inch graphs that we had to fit in there. This is yucky, right? And then that number took a million, not a million, but a whole bunch of base cases. And, you know, we don't have any highly systemized ways of going through these. It's all ad hoc, right? We have to specify the multi-d competition. We use the force, you know? We can't try to figure out how to do these things. But Charles Hughes helped us tremendously by giving us an algorithm that actually, okay, here's the graph I want to fit in. I want to take out the sticks from the bottom and leave the other edges in there. Can we do that, right? And so, you know, crank, crank, crank, crank, no, no, no. Yes, everybody really helped out our project a whole bunch. But we solved the problem for the remaining cases. And this, we submitted this one in June of this year. So, I mean, we're looking at other types of problems. We're looking at other types of puzzles, actually. And the complete graph is sort of nice and pretty. We like those in the final picture. But now, we're thinking, well, what if we can do some more interesting things? And in particular, a way to endow them with other things where they can make various types of products and graphs. And like Cartesian products and tensor products. I don't even know what to do as well. But we all have only neat products. And how can we do multi-d compositions of them? And other types of multi-designs. And that's probably what we're going to be looking at. Another possibility is actually, I mean, graph triple to order six, why not graph triple to order seven? Now that we have Chuck, he can actually help generate those. There goes 131 triples. We did those all by hand. Yeah, I know. It took a long time. And I missed like, when I was going through that, I missed like two or three that we had to come back. So I go back there. So Chuck is going to help us with that. Good for Chuck. I'm over early. Any questions? Go. I was just wondering, in terms of what the computer scientist from home did for you, was that more of a brute force thing that we were going to do with the graphs and we want to try and, we suspect that we can use these graphs to build a K whatever. Is that a brute force thing or is it some sophistication that uses other observations? I think it was a matter of checking all the cases. You said that kind of algorithm. What we really needed was, we had all the base cases. So for instance, I'm trying to remember, so like for two models, for K8, we're doing a mod of six or mod eight. I think we're doing mod eight for our prior induction. So we had eight as one of our possibilities. And so we have the recursive constructions for that. So we knew that we had all these possibilities that we had to go through. And so, you know, if we left us, there were going to be some cases where we have to do like one triple at a time. Because, I mean, we are lucky here, right? Because if you can get six on the bottom where you can say that's a triple and fill the other ones with just one type of graph, that takes care of a whole bunch at once. But there were some where we had to do repairs and then some where we had to do the whole triple. And that's just one at a time. And then there was a simply awful. And so what happened, two things happened when we used Chuck's software. First thing, some of the ones that we just couldn't do, we didn't know how to do, his program could do. And then some of the ones that we thought we could do, we had to use two of them to do, he got number one. So that went from like, what, multiple? Or four, right? Multiple four shortening of the proof. So it was pretty amazing. We had months of work ahead of us. And with this program, we got it done in a couple weeks. It was awesome. Can we back up a couple slides? Sure. It was the, that actually got right there. Okay, the last two cases, the second and the last one says if H2 and H3 both have six edges, the last one says if neither of H2 and H3 have six edges. What if one of them has six edges? You, let me see. Okay, so if your H1 has three edges, right? The other two have to add up to a multiple of three. So in your possibilities for three, you have 15 edges altogether in K6. Right, so you could have three, you have three six six, three five seven, three four eight. You do that right? Those are the three possible combinations of edges in your graphs. So if you had three six, well the length has to be 15 minus nine. So it has to be, yeah. Yeah, you're right. So logically, so if you think of your logic, you have to do all the cases. Those are the ones too. Yes. The other two are our empty. Yes. So your problem deals with the existence of these multi-compositions. Have you considered rating them a little bit? We talked about that for about a minute. Because that's a hard problem. That's a really hard problem. So one thing is, and so one difficulty is, I have, but I think there are two different multi-compositions. How do I know if I don't order the vertices in a different way, I don't get the same thing? So I mean, algorithmically, that's a tough one. I'd like to do that someday. But again, my colleague who used to be all my petitions were laughing. So maybe Chuck would laugh at me. He might not laugh at me. He might not laugh at me. Because let's do a global, or at least we can get something, you know, a polynomial time algorithm that will do it. But I don't know. Actually, another interesting question is, we're being pretty easy online what we demand of our rules for putting the puzzles together. You just have to use one of each one. I don't know if you know what's going on in that one, but you have one of the two swigglies and then everything else is 3p2. It's kind of a boring puzzle, if you think about it. Not very challenging. But what if you did something like, okay, I demand that you use this many of this graph, this many of this graph, and this many of this graph. Can you do it? That would be, that would actually be a really interesting problem. And for which, you know, is it going to be the case, no matter how, if all of the arithmetic works out, right, that you can make it, that would be, that would be an awesome theory. Right? Because essentially, that's what we did here. As long as the arithmetic worked out, except for K7, as long as the arithmetic worked out, you could always find a multiple combination. And the only time is K7, which is too close to K6, and then when you, when you're forced to have a multiple of the edges. So could you do something like, okay, I have a KN, I know how big my pieces are, if I do this number of pieces, this number of pieces, this number of pieces, so that the edges add up to the number of edges in KN. Can you fill that one out? Can I just fill that out? Of course. So you've been talking about decomposing complete graphs, but you also mentioned you'd like to do this for others. Sure. Sure. That seems to raise the possibility that you can, you know, consider some graph invariance, like productivity, or degree conditions, or something like that. Do you have any theorems yet? No, I mean, well, we haven't even, we just got into this together. So no. But you mentioned degree, though. I mean, degree is very important in the proofs. And in fact, this actually brings up K7. One thing that's really important in K7 is using, like, a different balance on degrees to show when things could not happen at all. Oftentimes, degrees tell you you can't do stuff more than they tell you you can't. They always say, maybe when they think you can, they say definitely not when you can't. So degree is very helpful. So there might be some more general theory about that. Oh, there are problems in graph theory that a multi-decomposition helps you to understand. But if you want to know something about the graph, G, can you brief it? That's a good question. That's a good question. Decomposition in general is important in the study of design theory. I'm not going to tell you what that is because I'm not sure what it is myself. I just know there's a connection between the way we decompose these things and block design theory. That's what they're called. All right. But there's some connection to design theory that has some importance somehow outside of graph theory. At least, as I understand it. So there are other places. I don't know about multi-decompositions. Actually, I suppose that our term multi-design came from that connection between decomposition and design theory. I don't understand what it is. Maybe if I can get out of it to explain it to you. I don't know anything else that hinges on our business. Thank you all for coming.