 Oh, yeah, thank you. Well, thank you for organizing this. This is such like a gem and a hard time. This is a very, very awesome idea. So can you guys see my screen and everything? Okay, so the only problem I have is that sometimes I will scroll my screen and it won't start scrolling on the main one. So if I'm talking and then it doesn't make sense, just let me know and I'll try to. Yeah. Okay, so yeah, this is called vertex distortion of lattice knots. And I guess I should provide a little background because hopefully this doesn't make sense at all. So I went to a liberal arts school, which I loved, but it was low on resources for glad grad classes. So the topology I had, you know, didn't have any algebra or any analysis to help me along. So I tried to come up with an idea or something fun that you can do that didn't require much machinery. And this is that. So my name is Nicholas. And I worked with a professor at San Jose State, Mary and can PZ. Okay. And I assume most everyone here knows what a knot is and all that but I because this is recorded. I'd like to just start from the beginning and make sure that everyone. You know, if you haven't seen a symbol or something that you know we're all on the same page. So the technical definition is that I'm not as an embedding of S1 and S3 typically think of these as being smooth. Now I think being such a vast field it's kind of a it's kind of a quaint little definition so few words but really what you're looking at is a closed curve in in our three. It doesn't intersect itself and I get to play around and move things. And I liked the big question is, you know, when are these equals so these three are in equal if I have these in my hand and I get to push things around. If you're unfamiliar with S3, you know, I think when you Google knot, you probably will come up with something like a, you know, a sailor's. So clearly that's not a knot but it'll it'll just be like a line like this and it has two ends. Well, if I think of this as being closed in a ball. And then once I touch the edge of it I'm going to consider those to be the same point so really this S3 definition is probably the one that you think of. And if you haven't seen this before because once I hit the edge here and the edge here I close it up. And I get something that's a loop opposed to just something that like you tie your shoes and you have the S. Okay, so everyone knows what it is. So I guess just to save time if I were to give you this piece of string and in the real world, the laws of physics, you're allowed to push things around you can't cut it. And that's essentially what the ambient isotope is going to keep track of. You actually can't say if you do know some of these terms you can't say it's just an isotope. It's kind of weird but you can you can create a knot and you can shrink it down and technically shrinking a tangle down to a point is okay if it's isotope, but the ambience means that I'm, I'm actually adjusting the space around it. And that translates to not violating any laws of physics if I had these things in real space. And that kind of speaks to why, why study these things. Okay, if you haven't seen a righto meister move. If, if I'm allowed these three moves this is all I need to do if I give a planar diagram. Okay, I see a lot of shaking heads people know what this is. I'm mainly, if someone would like to come watch this video later it's for them. So take a look at these I only need to do these three moves to, to somehow describe this complicated ambient isotope which is a very fundamental idea. But the reason why to study these and giving you all these terms is this this the problem that I'm going to talk about actually has an application to has an application to biology, because certain DNA forms a loop. So if you've seen if you took, think I took AP bio and bacterial DNA is exactly this it forms a loop. And you like to know where does it cut itself and so forth so these these things are are studied there. So the problem I was told is that it's not nice like this diagram it's more like you get a plate of spaghetti, and it's just all full. And you say what not is this. And it's very hard to tell what's not it is so we have to come up with some type of an invariant that is going to be this so low, low crossing knots are not as bad so those rules I showed you above, that's what's happening here. So each one of these will call a diagram, or I guess I had been calling him confirmations but roll sin and some very famous. Not theorists call it a diagram. So if you decide to Google these this is a better term. So I diagram is a specific embedding. Okay, and then I have a an equivalence class which means anything that I can get to with an ambient isotope or those three rules. So each, each eight of these is a diagram, but they all belong to one class. And I just need to determine the class of one of these because they're all it's an equivalence relation so this one is called the not in the first slide. So therefore these all belong to the bracket of the sometimes you use a you. Okay, okay, so now here's some more technical things you may not have seen before so there's something called the distortion of a not and notice that this doesn't have a bracket right now so this is a diagram so it's a specific embedding so you're dealing with a closed curve like a picture on a piece of paper and I the distortion, it looks complicated I take the distance in the space. So this one we know. I'm in the numerator I have this new metric that I'm, I have the distance along it's an embedded type of metric that I have I to I choose two points on the not and then the length of the of the arc that is the shorter of the two. So if I do two points on a closed curve, there's going to be sort of two paths that I can take from one to the other, and I take the shorter of the two. And I take now the supremum so I'm trying to find what's the worst ratio of this, and that is called the distortion so I have a nice picture for that. So this actually you can just think, probably what the first thing you came to your mind, or at least for me was, you know distorted like you like a picture, like you try to have something in Photoshop and just store it. And that's actually what is trying to measure. So both of these embed a, an unnoted circle, and the one on the right seems to be more distorted. It looks like an ellipse. So the distortion of the one on the right is larger than the one on the left, is it for diagrams. Now this isn't a quick question if you don't mind. Yeah. So is the metric on the not the same. It's even. Sorry, are you basically like parameterizing it like a, like an, like just a certain like a curve. Yeah, yeah, yeah. The integral was headed to six hours a week so I can. Yeah, just you, you take the Euclidean length of that work. Okay, so you can turn this into invariant. Can I also jump in really, I don't know if you saw in the chat, but people are saying that if you do full screen mode and use the arrow keys, then you might not have to scroll. I don't know if that makes it easier or not, but. Oh, I actually that's a setting on my iPad that I can do here. I think this is going to be better now. Okay, cool. Is that good. Yeah. Okay, let me have, I'll also my okay yeah my chat has gone away. So I will not be able to see a chat. Yeah, what will monitor the chat for you. Okay, great. Okay, so you can turn this into an invariant for the class. And by, and how to do this is that you take the infimum over all diagrams that represent that class. So hopefully, sink in for a second you do a supremum and then you do an infimum and hopefully you think about it for a few seconds and you're like wow that's like impossible. I don't want to think about that, because any, any little adjustment counts as a new thing that you have to run this, this argument for. For example, here I select the P in the queue and then okay great I have to do it for every pair of points. So obviously infinitely many, but if I make any slightest adjustment to that curve k one or k two, I have to then do it all over again. And then that's what's called the distortion of not so because this is fairly nasty. I wasn't studied too much, other than it's when it was initiated so Gromov was setting this and say oh what a cool idea and I think that we're going to have to be greater than pi over to. So yeah, so he showed actually it's greater than pi over to and it's pi over to when it's exactly this geometric circle, so something that looks like the unit circle this k one. Okay, and then someone showed a pair of people showed that if it's not it you have to be greater than five pi over three. And in fact this is kind of the extent of the. So there's one more very famous one, but this was kind of the extent of what was shown for this for a long time. In fact, the only not class that has his distortion calculated is this is the and not was just Gromov result. So it turns it turned out to be a, you know, it still is but it's a very hard problem. So there's approximations that show the tray foil is probably five power three. Okay, and so he asked, is there any upper bound to this so as not to get more complicated let's say crossing as they get more crossings. Perhaps, do I bellow out like can I not distort a diagram so far, I think they said like can you get above 100 and pardon show that there, there is none so if I take. So if you do know not theory this is probably like a very obvious place to start these are called torus not but for now let's just say they're not and they depend on integers p and q. And as p and q get big the distortion gets big. So this was, this was a very well received sort of annals of mathematics results. And it doesn't seem to be that deep and it's, I would say it's deep for respect of how difficult this problem is. But there's just there's not a lot you can hold on to is my point so what what I decided to do was well I probably don't have a good chance at solving something like that so let's look at a simpler case so these polygonal knots are something that's made up of sticks. And it's, it's hard to, in this time to give a formal definition but hopefully you can just see this picture. So they're piecewise linear straight lines. And they form a knot. So the more specific type is a stick lattice knot. So you're allowed to go if I'd like to give these an orientation you're allowed to go up and down left and right. But that's it, you must be confined to this, this lattice. So the formal. It's R cross C cross C business but really go up down left right and you want to end on the, the lattice integers so she did the three questions on this and breaks for a question. Maybe just a quick question. So, yeah. I was just going to ask why, why do you need the factor of R in there as opposed to just having it on Z three or something. So I'd like to be able to be a not so like a continuous curve and that's what allows that so if, if I choose and let's say the origin and I'd like to stick that goes just straight up. So people at some point have you know 00 pi. So that's okay I like I need I need two of them as I'm moving two of them have to be fixed integers. And then I end on another triple of integers. And my screen is doing that thing I talked about where it's not moving the next one so let me try to fix this. So these are studied. The big question in this field is, is how few sticks do I need. So it's a pretty cool question to ask if I'm constrained to this piecewise linear I'm not allowed to curve. If I'm a lot if I want to do just sticks not stuck to the lattice that's something else but for us, how few sticks do I need in the lattice to create a certain type of die. Okay, and the step length is the length of this, this arc so that'll be an integer as well. And the vertex set is going to be just the, just the Z three points that are traversed. Okay, so here's a, here's the picture of the trade foil. The reason why I need that are is because while I'm traveling I'm going to travel over some like not so nice rational irrational point but that's okay, as long as my end points are. This Z three. So this is actually how few sticks you need. So the stick invariant, the lattice stick number variant of the triple is 12. You can't make it with fewer than 12. You can't make the half length with fewer than 16. So it's what you think, like I have a square here, and then a square year. So this is like 2222 horrible picture but hopefully you see that, you know, kind of like how you normally would do a hotline. But now there's great. Yeah, pretty close 12 figure eight is 14. You're going to need more than 15 sticks in general. Cool problems. It's a fun thing to think about. Okay, but what I want from this is now to adapt that distortion to this field. So I'm going to call the vertex distortion the lattice. It's a very similar thing except I'm not going to check all points. I'm going to just check the vertex points. So I'm going to just check these bold ones. So that goes from checking an infinite amount to a finite amount. So I can replace the supremum with the max and turns out still very hard, but it's at least like obtainable opposed to the previous one. So I can turn this into an invariant. If I just now take the infimum over all of them. So I'm going to try to find the worst case for a diagram and then the best case for all diagrams. That means I'm going to change the metric now to a, a little l one to the taxi cab or Manhattan metric. So this is the metric of, if I'm only allowed to go down the streets left and right. I think I have a picture of that coming up. So for example, if I have two points here, I'm not going to measure this length. I'm going to measure just the, I guess, to be more technical kind of like the, the length of segments. I'm going to sum up the coefficients of the basis, or if I have like one zero zero one. And you know to get to this point I need three of these and two of these so that the five is going to be my metric. Okay. This is what we showed. So for any confirmation of the not in the cubic lattice, if you find out that the diagram you have has a distortion of one then then you know you're the not sounds obvious but it's, it wasn't. But this is good. This means like the result and the idea that we're coming up with, you know, mirrors the, the smooth case. And then also, we have like the foundation to, to that to pardon's result that there's, there's no upper bound distortion. So we don't have it for a class of what we'll see as tourist nuts, but we have specific ones that are very special. They're like this, they're minimum, or they're reduced and they have the fewest amount of sticks or we're thinking that that probably means that the class itself should follow suit. So these are the main, main theorems if you guys have questions on these. These pictures came from a Cohen Adams paper. They put these in here and they said, Hey, these are, these are tourist knots, but they didn't show how or why. So we did that and we showed, you know, the actually when it comes up, it'll make more sense. Questions. Okay, so like I said before, so these are the, these are the, okay, I think technically they're called staircase blocks. If you're a common tourist, I called them, we called them taxi cab paths for a long time, but I think staircase walk is the thing. So this is, this is going to be a key idea. So I want to show that if I, if I say I run the distortion and I figure out it's one, then the thing I have is the, and I do want to go through the proof of this just because I think it shows kind of how fun this problem is. So from 00 to one, one, there are two paths. And then we can kind of iterate these and these are all the possible paths. And I will note the staircase walk staircase path is the most efficient way if I'm restrained to this metric. And I like to go from one point to the other. I have many options, unlike, you know, the, the geodesic of a, the Euclidean metric kind of like the straight line is going to give me the best way. There is more than one best way to get there. And I'll explain in the next one, but the, that's how you can check. So if your path has the same length as what a staircase walk has, then you are a staircase walk. And you are the most efficient way to get from one to the other. Okay, so this is going to be the slide to talk about. So let's see. So if I have a distortion of one, that means that the largest thing that I can get is, is one. And if I choose any other points, I'm going to be less than this because the distortion is going to be the supremum. But I can't, I can never be less than one, like I can't be closer along the knot that I am in the space. So this is also constrained below by one. So this tells me if I take any pair of points, the ratio I have is going to be one with the length, the shortest length along the curve and the distance in space. Hopefully that makes sense. So I'm going to pick a point here. I guess I'll just call this one P. The nice thing to, it's like a good little homework problem for discrete math classes that these are all, all, all of these stick lattice knots are going to be even length. That has to do that if I, if I want to be a closed loop and I, I go five meters out, and I have to end where I started I got to come five back at some point. It's going to be even. So what I can do is I can always choose its antipodal point. Okay, so this is the antipodal point of this. Okay, but kind of like I talked about before, so the shorter of the two path lengths, divided by the distance, the L one distance is going to have to be one. So that says that this, the green one is a, I'm going to not call it a taxi. This green one has to be a staircase. So it's like the most efficient way to do it. But this also says, now since this is an antipodal point, the other path is of equal length. So it actually says that this one must be a staircase as well. Okay, and then I want this idea is let's say I had this note. I've gotten pretty well that's a horrible drawing that I've gotten pretty good at remembering how to draw these. So I want this idea of a minimum bounding box. So I have my shrink a dink and I put in the oven and I so I have my shrink a team. I'm going to put it in the oven. It's going to shrink around my not. So it's going to, I'm going to end up getting, you know, a box that kind of just rest and it's going to, it's going to hit on the on some corners here. So this is another idea. It's an idea of the bounding box by a corner. So a corner is a corner if it's it's somehow a extrema point, meaning that for the X, the Y and the Z it's somehow either the smallest X value you can get the largest X value can get the smallest Y. So everything lives to one side of a corner. Everything lives to one side is not the right point, but one oxen. Okay, so, in fact, is so if I look at this P point. Everything lives a bug into the right. Everything lives above into the right because we verified that both of these are staircase paths. And that's kind of like the idea staircase path is that I never buckle over. So in fact, the point I picked, which was arbitrary is a corner. So therefore every point must be a corner. And then you just run through a few cases and you, you can see these are the only two you can get. So this was hard because the pictures we kept drawing were not these two. So it's just a bunch of counter examples not knowing it. But this we spent a while we so it does if someone finds this interesting, you know, we, and we were looking for a way to talk about this. So we originally tried to do write a Meister move so that like you can do it. And what you do is you can you get something that you can always unravel that's the big idea here. But to avoid so right a Meister moves are defined for the smooth case so we didn't really have it for the this stick case and they do some people just assume that you can do it but in just to avoid that we had this different this different method. But if I run this thing and I get a one out then I have to be one of these two so I am the end not any questions on this one. So I'm going to move to the next theorem. Okay, so the next idea, I think is a lot more beautiful but not as much generalization. So I'm going to show that there's something there is some collection of diagrams that get more and more distorted with our version of distortion. So I'm going to spend a lot of time just kind of talking about what those knots are and how you construct them. I actually, I would hope that someone who's interested in coming a torques takes this because you can kind of get the idea now distortion this vertex distortion tells how efficient your pads are. So if you have distortion one you're like the most efficient to get from one to the other. You're going to go as quickly as possible but if your distortion is 100. That means there's going to be a pair of points that you're going to be close in space so you know you live right next door your neighbor, but you have to go all the way around. So, I think for a graph theorist this could be an interesting idea. Anyway, so we do want to construct knots they're called torus knots. I don't have enough time to just say so I torus not means that so torus is a donor. And the knots here are drawn on the surface of this donut so they kind of spiral around go around here over. So how many times in the long run do I go around the longitude and how many times would go around a meridian. So those are both. So longitude is like the long way around the donut and then you go through the whole that would be meridian. And that's what the P and Q represent. And you can show that all torus knots look like this for relatively prime P and Q, we're going to do P and P plus one so those are obviously relatively prime so they describe torus knots. So we call an Adams show that the amount of sticks you need is going to be greater than equal to six times the P for P P plus one that actually has to do with the bridge index. So you have to be six more than the amount of bridges. If you do know the idea of bridges this kind of makes sense for sticks because you would expect you do at some point you go up at some point you go left. At some point you need to go right then and forward so you need to do all six up down left right forward back. Okay, so I had this idea of how to tabulate these if I was better at computer science I think this would be a great method to just compute these instantly so I'm going to say I'm going to orient these. And I assume you guys probably understand this but I'm going to so a way means that I have a close curve and I have some arrows on it. So there's a path I'm going to follow away I'm going to go around. So if I start somewhere I'm going to start the origin. If I go up, I'm going to say that this is z plus, and you know if I go x plus direction and so forth. And so to describe this, I just give links I say, I'm going to do a z plus an x plus a y plus the z plus is going to be 2p minus one length, and then I just I gave you all this data data and you can, you can create one. You can create a chart of this and you read this. So the first stick is a z plus the first z length is a 2p minus one that says I do z plus is a 2p minus one. Now I do an x plus. So I look for the first x that's a two, I go through here and so forth, six p sticks, and I get something that looks like all my screen. So I look at the type of stick I want so the first one says that I do a z plus in the sequence, and then I look at the first z, which is the 2p minus one and that tells me then the length, how far I do it. So you got to check is this a loop even do I get back to the point it started with, do I self intersect, and do I get the not that I want to actually get a torus not. So if I do this, these took a long time, but I get this very cool. Not here. I actually gave this talk at an LGBT conference. So the rainbow fit in perfectly. I actually had this thing 3d printed. Which turned out to be really useful to calculate a bunch of things. In fact, now that we're looking at this one I might say a few things about it. So if you look at any of these sticks. And if I tried to contract any of them, you. So you can't really tell because there's obviously space here but in terms of the cubic lattice I can't move any inward at all. There's something it's like as tight as it can be. And that's the idea. We didn't show the same result as part in but these look to be very special like these are as tight as you can be as few sticks as possible. Very cool to play around with. So to show that these are the torus not I made the torus that they live on. This took a long time learning how to do this in Mathematica. So you can count the P and the P plus one. So the long way around as I come around here. I pop out up here and here here are the sticks that I wind up hitting to get the. Actually, so I hit P of them right here. See if I can't do a finer. And then I come up at the top. And then so that corresponds to coming here here and then I get the P plus one by hitting this one, and then to count the meridian so the whole. This is the reduction I was talking about so same idea. Here's a different picture of it. I think it was easier to describe on that 3D print. This is kind of lame. You know it's it's going to increase your distortion without really changing the geometry of the tangle. It's kind of like a cheap trick. That's what I coined it at first. So these are as tightly packed reduced this way. I'm going to bring up point in. It's going to drag these in and create a plane I just don't want it to intersect as I go. I'm not going to talk about the proof of too much of this because I think it's it's just kind of boring unlike the other one, but this is the big idea I'm going to think of these as all being pointed line segments that have some type of orientation. I'm going to describe all of these so I'm going to do a lot of summing of arithmetic sequences and do a lot of arguments of this this coordinate never intersects with this coordinate. This is a picture from one of the proofs of them so this maybe this gives you the idea of how you wanted to, you know, talk about these things. This is the path that so this is the path that it has a very high distortion so these these two red points are very close but to get to them I have to travel far far around the knot. So these are the points that I selected each time. And they are exactly the distortion of these diagrams for small knots and then once we I got up to like 2020 plus one. It took like five hours to parameterize and right around 10 it flips like the points start being right here instead. So instead of, you know, trying to calculate this for all of them I say well here's going to infinity because so the distortion is the greatest of them. And so it's got to be greater than if I choose these each time, and these each time go to infinity. So the distortion which is going to be worse has to go to infinity as well. So I selected these two because it's right in the middle so if you count these. And the ones in the middle of course there's some times where there's an even amount and you choose. I think I chose the larger one. Yeah so here's the 2223 so hopefully you can appreciate that this may get a little cool picture though. 2728 I had to leave my computer on overnight. Cool. Oh is someone talking. I see lips moving but muted. Maybe. I was talking to someone else. Sorry. Okay, so these are the conjecture so this is the next step. So this paper is getting pretty hefty so we cut it short. This one is a little different than what we showed so these are the brackets. So we want to show the same grommet result kind of that if you are the if you get a one for the whole class then that whole class is the I'm not so we have one direction from the proof. Same thing for the other one so those the stronger result is show that the class of TPP plus one goes to infinity as well. Both. So this one we're very close in fact I think if we were just more clever it's right there this one. So this is the same method. I might need a little more machinery to do that. Both are fun. Both do not require a very hefty. Analytical topological to an undergrad to do it. That's I hope an undergrad decides to watch this one day. This is where I started with the smooth distortion. So I created these plots so this this maybe can show you why the the other way is a little more interesting. So I half of this is redundant, but I choose so I parameterize the I have my I start here and I parameterize so you know length one is right here and I don't need to be integer so maybe pies right here to parameterize the length of this and that that corresponds to choosing one of these so maybe I choose the point that's, let's do 60 along it, and then I choose the point that's 20 along it, and then I find where these intersect, and that gives me the distortion of the point at 20 along and 60 along. And this is where we started turned out to be a little too much that the computer can handle to calculate all the time. But very interesting, you always get something that looks like there's these peaks and valleys it kind of looks like an egg crate. You have these peaks and those valleys and these little dead zones, which means that you are very efficient in your path here, and you're very inefficient when you choose these pairs. So there's like a very natural question is like, can I use this easier stick distortion to then describe the smooth distortion. I think yes I think you know you could smooth the corners here and you could talk about it. You can, there's probably something to be said I wonder if you compare the points that I picked maybe they always line up in a certain area here and there's way to describe them. So this was, there was a master student that decided to do a project very similar to this. The original idea that I wanted to do is like, and one of those plateau problems so if you've seen that where if I have a closed curve like a knot. And it's fixed it's rigid it's not in topology anymore. What's the minimal area surface that that's the boundary of. So you try to find a, you know, perhaps you picture you have this disk, you know, clearly the best area you get is right here but you know you might be concerned if you have a bubble that comes up. Clearly this wouldn't be the minimal area. So I wanted something to do with knots and you know how can you describe a minimal surface so that's why I first said well why don't we make it rigid and make it a stick and then talk about it. And then we saw this distortion got distracted but this is the same. This is similar to that this is saying how instead of how few sticks. I need to create the knot, how few tiles do I need to create a surface that the not can live on. So how few squares do I need to make a tourist. You know, the natural thing is I remove the one here and I have the ones similar game to play. So she came up with some very nice things that do. And that's the last slide I have. But there's any questions I'd be more than happy to talk about something. I'm going to leave it on one of the nice pictures. Yeah, those are some beautiful pictures. But yeah, let's let's clap. So, so it's distortion related to the bridge number of the knot. I'm not directly I the distortion is the bridge index I mentioned here. It doesn't it more has to do with the stick number, which kind of makes sense because if you think of like a bridge position. It seems like an, I guess, when I way that I've, I'm trying to find a blank spot here the way that I've seen bridge. So some people define it kind of like a Morse function. So I have. So I'd like to, you know, as much as I can I stick to this plane. And then, you know, oh no right here I must come up in the plane to go here and then I need to get in the bridge. And then I get to stick on this plane. So that kind of makes sense that I have. Relates to the amount of sticks because there's no way I can avoid going up and then say left and then down. So that, you know, necessitates. Oh, I must have at least, you know, three sticks additional now. But nothing I know relates it to the distortion, there could be. Okay, thanks. I'm kind of curious what goes into the computer calculations and like you're saying at some point. You know, and I think that ceiling might be my fault. I think I, I don't, I don't know if I know how to use Mathematica efficiently enough to where it doesn't time out because I'm so clumsy with it all. What I did with these, once again may not be the best way is I parameterized these so I parameterized each segment by its arc length. So here, you know, so for example here like from here to here, you know, I have this big piecewise and I say I give it a function, and I, and I give it a parameter for that. And then, you know, I do, you know, potentially 200 of these, and then it makes them all, it gives me a whole set of points that I get to, you know, you can run through a calculator. So around 22, it starts taking a very long time. And the smooth way, I think I would hope I mean maybe someone is, I know people are better at Mathematica than I am but maybe someone is better enough to say, Oh, this is an efficient way to describe the distortion because you for small ones like for the I actually ran the smooth one through just a formula like oh take all pairs of points and divide by their distance and it actually it was okay I could do that. But anything bigger than that. I wasn't clever enough to program efficiently. So somehow in this computation you really are having to do like all of these pairwise comparisons. Yeah. So I did have Mathematica to, you know, make sure I did it correctly but in a sense it was all kind of by hand, checking all pairs to make sure. And then once I got big enough. I just said you know I'm going to just check the ones where I think are going to get bigger. So like this pair right here. And then I just checked those. And then what I did is is actually way easier by hand than by computer. But what you do is you pick this point. So you have this table, and you say, Oh, that point is always to P minus six, say, and the other point is always, you know, seven. And then you, you have to be general but you sum up these links. So you actually sum up a arithmetic sequence like you would in a pre cal class, and you get a quadratic function that depends on P. So you can actually find a closed form for those pairs of points that I picked. Not a groundbreaking result, but a very cool one. And I think it leads the way to say that hey this is a useful thing to study. It follows what your intuition says so it's worth your time.