 Crescent configurations in 3D. Reset everything, apologies. Okay, so A.V. was a leader, Malachi Alexander, Shan Zhang, and Yi Chen Wei. Today we are going to talk about the crescent configuration in three-dimensional. So motivated by the same question as mentioned in the 2D group, we wanna expand the work down before in the work to 3D. So first I'm gonna introduce the definition of a crescent configuration in three-dimensional. It's quite similar to the 2D one. So for a set of n points in R3, the crescent configuration should have exactly n minus one distinct distances where each distance labeled d i will appear i times. So for example, d one, one times and d two, two times, et cetera. But a slightly different condition is that to also be in the general position, a crescent configuration in 3D should have no more than three points in a plane and no more than four points on a sphere. It is actually an analogy of the 2D ones that we can expect to higher dimensions if we want. So just to begin with an example, this is a crescent configuration in three dimensions. D one is the black line. So there's only one of those. D two, there's two distances of square root of five and there's three distances of one. And so one of the questions that we're really interested in is how many of these can we find? And so what we did was we started off with a complete graph and then we asked, okay, how many ways can I place three edges with length D three? And what it ends up coming out to be is three different ones. You can perm you all the different possibilities but they're all gonna end up looking like this eventually. So you can just throw out all the other ones. And then you ask, okay, how can I put the other two for the D two distances? Well, there's three ways to do that. So we get nine possible situations. But if you notice, if you look at the top one, if I were to kind of like rotate or reflect across this diagonal, I get this one right here. And so there's some copies in the middle of it. So what we do is we kind of find mappings between the graphs and we show that there's actually four different isomorphism classes. So you don't need to check all nine. And so I'm just gonna label them one, two, three and four. And so you guys have seen the first one. The second one is this one. There's a third one and then there's the fourth one. So we can find all four of those. That's great, right? So we have all the four point cresting configurations are three. So naturally, what about five points, right? So here's an example of one on five points. We just changed, well, we didn't change anything. We added fourth color, which is D four in green. And so naturally I wanna do the same thing that I did before, right? So I'm gonna start with a complete graph on five points. I'm gonna find all of the different isomorphic graphs that are on four edges. And then I need to find all the ones that all the D three distances, right? But that's a little bit much, right? And so each one of these has three subcases as well. So there's like 60 for each one of these. And there's six of them. So 360 times and checking all the graph isomorphisms for that, that's a little bit crazy. So we need to find a better way. And that's where they're gonna talk about the algorithm. So this algorithm also come from the paper that Paulson student wrote. And basically what they do in 2D is they first get all the permutations of the distance and create a adjacency matrix. And they find all the four by four sub matrices and to catch all the cyclic case and linear error case. And then they just eliminate all the duplicate cases and check the geometric realizability using the Cayley-Mingere matrix. However, we cannot completely just implement this algorithm into the 3D cases. So we did a little modification on this. Since our algorithm is also based on the 2D one, there's some similarities between this one and the one mentioned before. So both of us beginning with finding all the isomorphism classes. So as Melika just talked about, we have many cases to discuss by hand, but by reducing them to isomorphism classes, we have a reasonably smaller number to go one by one. And the second similarity is that we also check the geometric realizability using the Cayley-Mingere matrix. This is one used in the paper written by Paulson's students and we expand this to 3D. However, there are differences between 2D and 3Ds. So first is that we no longer need to check if there are cyclic case or if there are linearity cases. Because the condition mentioned before that there are no more than three points lying in the same plane, automatically ensures no such cases will occur. However, more than what they've done before, we need to check if there are five points lying in the same sphere. However, the method for this is quite simple. We just can randomly choose four points and we can calculate the sphere defined by the four points and check if the fifth one is in it. Following by this is a fourth example. We can think just by stretching this middle point out to 3D, so seemingly like we have a question configuration in 3D, but notice the four points in the bottom lie in the same plane, so we cannot get examples in 3D just by modifying the 2D ones. Then I'm gonna briefly talk about how we use the Kalimantra matrix. The general form of the matrix is quite similar to that in 2D and we also use the same theorem mentioned in the paper, which states let M be the Kalimantra matrix of a set of N plus two points. Then these points lie in RN if and only if the determinant of the matrix equals zero. To apply it to the five points in 3D, we can easily see what we need is the determinant of the whole matrix to be zero. However, we need to ensure no four points lie in the same plane, so for each five by five sub matrix by four points in the configuration, we need the determinant to be not equal to zero. And then we just would like to say thank you to Park City Mathematics Institute and the Institute for Advanced Study for Hosting us, Dr. Paulson for the amazing project to work on and Samantha Fairchild for helping us present. Thank you for listening.