Mathematical Games Hosted by Ed Pegg Jr. [Episode 1: Collection of Points and Lines]

Hello, I am Ed Pegg Jr and we are starting a new series called Mathematical Games, which is the more fun side of mathematics. And I hope to show some of the things I've learned over the past, oh, 50 years or so. Mathematical Games was the name of a column by Martin Gardner, who I was heavily influenced by when I was very young. I, I read all of his columns and tried to improve his results when, when I got older and into programming. And I found with computers being a billion times more powerful than what the 1950s and 60s had access to, that it was very easy to improve various results. And as a result, quite a few advances were made within recreational mathematics just by about anybody who wanted to try something. So I'm going to touch on some of those things. And let's start off. I run the demonstrations project here and there are some images, here are some of the various recreational mathematics related items that I've done. This is 10 of the over 300 items I've put together on the demonstrations project. So an example of one of the points in line sort of items I've been doing over the past few months is we have this graph here and it doesn't look all that nice. And I was asked if I could make this graph look better. So in something called graph data, we have the same graph and various embeddings where basically you take these points and you move them around there. These are all basically the same graph, but we're trying to find one that looks better. Well, none of these look particularly good here. So I decided I would try throwing out various points where basically I'd take a point and disconnect the lines and see if any look better. And this one looked pretty good. So I wondered if I could take this graph, put this at the center and have all the edges be the same length. And with a series of complicated steps and solving this equation, I found this. This is that same graph, but here we have it in a nice symmetrical embedding where all the edges have the same length. And we can check that this is the same graph by using isomorphic graph. So with that we managed to get pretty picker pick pretty pictures of all graphs of a particular type for a column that Dr. Wolfram is working on. And here's the graph I made prettier here. And I think I came up with nicer pictures of some of these here. So we have various graphs which are points and lines, and we came up with nice pictures for all of them. So one fun project I tried looking at five years ago was the greatest number of unit lines you could get from from sets of vertices. And here's one where I just looked at roots of 3 and 11. And in the problem called the chromatic number of the plane, with something called a Moser spindle, you can toss the object and. And this item outlined in red here is a Moser spindle. If you toss this on a plane, you'll need at least four colors to color the vertices so that no two vertices are the same color. And this is part of what's called the chromatic number of the plane problem. And when I put this demo up that had lots of Moser spindles and Golem brass all packed together. A friend of mine like this a lot and he wondered if it might be a good idea to try to expand the roots. Everything here has a root of 3 or 11. He wondered if roots of 5 and 7 might help the problem. And I replied I didn't think that would help. Aubrey de Grey then found a big bigger graph with this many vertices. And he basically solved the chromatic number of the plane problem in 2000. What was it, 2018? It was the biggest result of the year. And I almost talked him out of solving the problem. So I was wrong with this. That happens quite quite frequently where you make a mistake in mathematics. So I'm, I'm one that's frequently on the wrong side. However, I still have a lot of fun doing this, but this is probably my biggest mistake so far in recreational mathematics. So we can go to the easiest cases here. Points and lines. One thing that happens with lines is sometimes they are vertical or horizontal. So you want a method that returns a line so that everything is, is predictable by a computer. So here we have a normalized version of the line so that x, Y and the intercept always have values. And this is called the norm diversion of a line. With things like this. This is also called the Hessian form. With things like this, it's sometimes easier to throw problems at a computer. And the computer never chokes on anything because X equals zero won't be something that causes problems. We can also do the same thing for planes. Three points define a plane. And here we have various planes that come out. These are also all hessian forms. And if you look at our Wolfram function repository, we've got a Hessian plane function that returns these these items. This is a way of working with, with planes that will work with computer programs more nicely because it can, it will always have values in, in predictable places. We can also look at, at circles Three points define a circle. So here's various circles that you can get from three points. And this is, you just need to solve this matrix and it will give you the equations of the circles involved. You can also get the center of a set of three points by using Circumsphere in Mathematica. And here are four points defining two parabolas. Pretty much any set of points, you can use those points to define something. And this is one of the problems that Isaac Newton worked on where he wondered what parabolas would go through a set of four points. And he came up with a method for finding these parabolas. And my function in this demonstration basically uses Newton's old code from the 1600s. I did basically copied his technique right into my code and it worked fine. This is one of the frequent nice things about recreational mathematics and computers and programming is you can take old code and if the person knew what they were doing, it just works. And Isaac Newton knew what he was doing. So I was able to just take what, what he wrote about parabolas and just use it. And with that we can take a, a heptagon or sorry, a 13 gon and use it to come up with 13 parabolas and 13 points. So that any pair of points is on exactly one parabola and any pair of parabolas goes through exactly one of the numbered points. And that's what's called a, a perfect configuration or a order for projective plane. And this is when I was making some projective plane pictures, I was trying to do nice ones. And recently when I tried to come up with new pictures for projective planes, I took a look on Google Images to see what was out there that might be nice. And pretty much all the nice pictures that I saw there for this particular case were my pictures. So I figured I would come up with another new picture since I'm the one who frequently makes these. I showed graphs earlier most of the or many of the graph embeddings for difficult graphs on either graph data or on Wikipedia or embeddings that I came. Here's Sylvester's four point problem. If you have four points or five points or more, what is the odds that all the points will be on the convex whole boundary? But turns out if you use just four points, the odds that you won't have, oops, there's a problem with that point there. So I'll need to go into this code and debug why that didn't change color there, but the chance that four random points will be in a On a convex hole, that one point won't be inside, the other is one. Oops, if it's, it's 2/3. If you pick all the points to be within this bottom triangle. And we can try this out here, it looks like 2/3 of these are indeed acting as convex holes. When we also go to a full square, and the chance here is 25 over 36. This is a problem Sylvester solved over a century ago. And if you go to five points, you get this series of odds. 25 over 36 or 4, 49 over 144 or 5 and so on. And it's a little bit simpler for triangles. This is kind of an odd thing about how points and lines will work together. This, this simple problem of a convex hole. Here's five points on a convex hole. Here's four points with one on the interior. This is Sylvester's four point problem. If we go up to five points, we can get a conic section where you just solve this matrix. And for example, with, with five points, you can either get a hyperbola or an ellipse, usually just those two. It's possible to get a circle or a parabola, but the, the chance of that just happening randomly is virtual. And with nine points you get what's called a cubic. And cubics have many more types of forms. So here's some examples of various cubics. One odd thing about cubics is if you have a rational point, these say these three points here, if these two points are rational, then they will intersect a third point which is also rational. And most of the advances in cryptography use this sort of elliptic curve method for coming up with difficult items to crack. Or there is a branch of cryptography involving elliptic curve methods. And this is, this is one way you can play around with them. There's a lot of deep math in here, but really all I'm doing at the moment here is moving points around. The computer is handling all the stuff in the background. And at the top here is the polynomial that has that cubic equation. We can use these cubics to come up with interesting recreational math type objects. For example, here are three numbers from negative 7 to 7 that sum to 0 mod 15. For example, 7 plus negative 4 plus negative 3 is 0 mod 15 or negative 4, negative 6, negative 5 is negative 15, which is 0 mod 15. And there are a whole bunch of lines here. And this is actually a record for the, this is the maximum number of lines of three you can put with 15 points. And if you add a 16th point, you'll notice there's faint green lines here. There's six faint green lines. If you put a point at infinity way out Here, where these six parallel lines intersect, you get 16 points which have 37 lines, which is A, which is one above the maximum. Terrence Tao recently did a, a paper on what's called the Orchard problem and came up with a proof that there's only so many sporadic solutions. And this is one of the three sporadic solutions that was unknown. This is, this is a solution. Apparently I was the first one to find. And this curve that's on, on here, this is a, a cubic curve that goes through all 15 of these points. If you pull pick any nine of these points, you'll get the same cubic curve. It turns out that once you build up a set of points with a particular algebraic structure, it starts to force a whole bunch of other points which match that structure, which is something I have exploited quite a bit in my researches and algebraic solutions to recreational math questions. So, so this is, this is how I managed to come up with this particular one. There's a larger blog that I did about the Orchard problem that you can find with searching Orchard Problem and by name Ed Peg Jr. Some of the other sporadic ones are for 11, 16 and 19. So here's the one on, on 11 points. Here's the one on 16 points which I was just showing. And here's the 1 on 19 points. At this point, no one knows if there are any other sporadic solutions. The other solutions here are all sort of normally generated solutions for maximizing lines for the Orchard problem. And I have code that generates all of these in demonstrations. So some of these are quite nice. This is what's known as the Poppas configuration, with nine points coming up with 10 lines. And in each of these there are numbers. And if you add up the numbers, you'll always get the same value. I was never able to come up with a set of numbers that would work for 11 points and 16 lines. But I did come up with numbers for all other cases. Oh, I also didn't solve the 19.1. So if anybody can come up with numbers on that, that might be an interesting line. We can go higher. What can we do in. In 3D space? So here's 149 points that give 241 lines of five. I'm not sure what records are in this category, but this is. This would be a difficult one to beat. In the Wolfram Function Repository, we have an item called Find Extraordinary Lines where you can basically put in any set of points, and it'll return all the lines that have more than two points in them. An extraordinary line is when you're given a group of points. It's any collection of three or more points which makes a line. If we move to triangles, we can move, we can find all sorts of incredible things. For example, this is Martin's theorem. If we take three points on the complex plane and make a triangle, we can use those three points. Say they're z1, z2, z3. If we do x x minus c1 times x minus c2 times x minus c3, multiply them together and take the derivative. You'll get these two points, which are the foci of an ellipse, which is tangent to the midpoints of the triangle. So you take a, a cubic that has three roots. Those, you get three complex roots. And the, and the, the roots of the derivative are the foci of this ellipse. And that's, that's one of the really nice results in triangle geometry. Let's see if this works. We can continue doing this where you've got points of an object. And Lucas and Gauss both looked at this problem where they took complex polynomials and calculated the roots of the derivative. And they noticed that the points were always on the inside. And they managed to prove that that was the case. So here we have a, an octagon on the outside with a heptagon inside, and so on down to a single point. And they will always basically converge on some point that's on the inside. One unsolved problem is you have end points. What is the maximal number of unit lengths that can be between the points? So here's one of the ways of arranging eight points to get 18 unit lines. And this is an unsolved problem. We don't even know whether any of the, any of the results pass, say, 16, are actually true or not. There might be other solutions that we've missed. So anybody who would like to work on a problem. This is one that's unsolved at the moment. I mentioned the chromatic number of the plane earlier. Well, what about the chromatic number of space? What is the minimal number of points you need to color space so that there are no two points a unit distance apart? Well, this was recently improved by Aubrey De Gray, again with this particular graph here. This is a, a minimal known device for proving that six colors are needed for 3D space. Whether there's a smaller object that will, that will prove this result is unknown. So this is kind of the 3D version of the Moser spindle. We can also ask how can we put a set of points so that they are all an integer distance apart? So here's a Solution found in 2006 with seven points and it's. You should make a restriction that no, no three points are on a line. Otherwise it's easy. You just put your points on the line and they're integer distances apart. And also no four points on a circle. With those two restrictions, that becomes a very hard problem. And here is a solution for seven points where they all have integer distances. It's an unsolved problem at the moment whether there's eight points that are all at integer distances from each other. So if anybody would like to try to solve that one, that might be a good problem to look at. We can also look at what's known as the rational distance problem. You have a unit square and you want to find a point which is a rational distance from all four corners of the square. With this demonstration I did, I find a few thousand solutions for three distances that are rational and one irrational. For example, the point. Well, let's go to a simpler point here. Let's see if I can find a point that's actually. Here we go. This point here has three rational distances and one irrational distance. So far, this is an unsolved problem. What I did is there's a database of hero triangles and I just basically turned those hero triangle lists, there was over a million of them, and converted them into this problem. This to see whether any solutions popped out. And none did. But if any had, then that would have been solving one of the unsolved problems in mathematics. You'll notice that there's an ellipse on the outside here that seems to be particularly fruitful. And there's also some other curves. Like there's a thin green curve here. I haven't really looked too much at, at this set of points since, since I put the demonstration up. But there might be some sort of solution hanging in here somewhere. Maybe there's a solution on this ellipse that solves the problem. Someone doesn't seem to work. Now. Let me try re evaluating this. If I can grab it here. Oh, let's just skip this one. One moment. Ah, let's see. My system is frozen. There we go. Let's get the next slide. I'm in a. I'm in a slideshow and what's happening? There's a error messages panel that's come up and I can't access it while I'm In a slideshow. All right. The club surface is a way of putting 27 points on a cubic surface. And that's a maximal number of lines that you can put where in on the surface. Everything on that line is. Is on the surface. And this is the maximal that can be done. Here's a curious object. You have four or five points and all of them are at a distance which is a power of the plan plastic constant. The. The plastic constant is one of the. The strange routes like the golden ratio, but it has a. The plastic constant and the super golden ratio both have a. A lot of strange properties. And one of them is that you can put five points so that you hit powers the first. The. The 0th, 1, 2, 3, 4. You're. You're only missing. You only have one repeat with the thirds here. So this is a strange pentagon where all the distances and diagonals all work nicely with the plastic constant. And if you expand that to 3D space, you can get 19 points that are all powers of the. Of the plastic constant. And I call this the plastic pyramid. This is just something I found playing around. Whether it means anything, I'm not sure. But so far no one has found anything that even approaches us. This is one of these objects that just was begging to be found. All I need to do is let's play around a little bit with some points. And this, this 19 point objects have just sort of popped out. And there's a, A, a demonstration for the. This set of cliques in the plastic constant. If we expand the lines, we can try to find a maximal number of cylinders that all touch each other. And this was an unsolved problem until just a few years ago when these seven cylinders were all found that touch each other. And you can adjust the angle with these. All of these cylinders touch each other. And this, this wasn't found until just recently when somebody decided to take a serious look at the problem. And with a. With some calculations and on a computer, the solution just kind of popped out. It's currently unsolved whether there's a perfect solution with eight cylinders that can work. We can also look at points in lines otherwise like we can do it on a sphere like this. We can do various. We can do things like circum circles and altitudes and whatnot. We can move the. The vertices around. So this is a demonstration triangle circles on a sphere. We can play around with triangles on a sphere and find their various properties. We can also look at triangles on a Poincare disk. And these also have many different types of geometric properties, we can look at the incenter circumcenter. All these various triangle items have solutions that we can look at. And we also have many different items on regular triangles as well. Where for example, the circumsphere and in sphere are two things we can find. With regular triangles, we can also move right up to 3D space. Tetrahedron centers aren't as well studied as triangle centers, but they also have many nice properties. I'm hoping to eventually get to a expansion of this study in the Encyclopedia of Triangle Centers, which is built by Clark Kimberling. There's currently, I believe, 60,000 triangle centers that are known for tetrahedra. The number that is that are known is about 20. And I'm thinking that that number can be greatly expanded with a with a dedicated computer search. Oops, I wonder why that tanked. I'll have to take a look at that. Here's another unsolved problem. You have points on a sphere and you want to find what is the minimal charge you can have. So here's the best known solution for seven points. Well, it turns out for seven points this is considered an unsolved problem and it really shouldn't be. I'm thinking that five quite possible to do a computer attack on seven points and basically verify that this is the best solution. I believe eight solutions has been proven, but nine hasn't, ten hasn't, eleven hasn't. The solution for twelve is the icosahedron. That's been proven, but anything above twelve is considered unsolved problem. And I'm thinking that the Thompson problem is one good project that a high school student could probably do, especially for seven points. I think there's a way of outlining all the possibilities for seven points and basically brute force proving that the current best known solution for 7 is the best and we haven't missed anything. I think that's well within the range of what could be done this year. The Heilbronn problem, you want the. You have a number of points in a unit sphere, for example, or in a unit square. And you want to make the smallest triangle formed by the points to be as large as possible. So that's pretty easy for five points. For four points, five points, these have been proven. But once you get up to say 11 points, all these red triangles are the smallest. Anything that's not red is, is not a smallest triangle. Once you get up to say 13 points or more, the solutions that are known are not proven. So this would be. This is a nice unsolved problem that I've looked at, I've actually improved, I think, five of the best known solutions. But this is a nice comics combinatorious problem. One thing I would like to do is any combinatorial problems that are like this one that we don't have on demonstrations, it would be good to have a demonstration that shows what's known about the problem. Currently we're at I think 12,800 and so demonstrations. I would like to get 13,000 demonstrations before we go to version 14, whenever that is at least months and months away because we just released 13.2, I believe. So it'd be really nice if we got in about 200 more demonstrations before version 14 came out. So we can have 13,000 for version 13. Here's another graph. This is what's called the, the Danzer graph. The way it's generated, if I can bring up the code here, is to just take circle point seven and all the possible subsets of seven and then to just combine them. So for example, this point is for 7, 6, 5 is at. From the circle point 7, which is a heptagon, this is the sum of the points 5, 6 and 7. And it turns out when you add up the circle points in that way or the points on a heptagon in that way, you get this unit distant graph. That's very nice. And that was something found by Danzer years ago. We didn't know about it. We just recently added this to Math World when it was brought to our attention. If you know of any nice graphs or nice combinatorial items that we don't have on either demonstrations or on Math World, just free to send us a message about them and we'll try to get those added. Here's another unsolved problem you have. Well, here we have a 52 by 52 square and on every row and column we have two cells picked out. And no. 3 of these cells are on a line. Well, it turns out it's not just the rows and columns, but it's all the diagonals, all the skew diagonals. There's no line you can draw through this anywhere that goes through three of these. Of these points that have been chosen. This. There's. There's no larger solution known for this problem called no. 3 in a line. But this problem hasn't been improved since 1986. The computers we have today are about, oh, 100 trillion times more powerful than, than the last time this problem was looked at. So this is one problem that's. That's right. For a Sudden improvement. We can also put points up on a plane so that all the distances between points are unique. So for 26 points, this is believed to be the tightest packing of 26 points so that all the, all the distances are different. Here's just six points requires a seven by seven square, a seven by seven grid. And that's the tightest packing for, for six points beyond say 20. It's unsolved whether these are our tightest known packings. Here's a, a set of planes. You have 12 irregular Pentagon faces. And when I was, I was looking through the literature on this, this was an unsolved problem. And I'm seeing why is this unsolved? You just need the equation of the plane. Well, it took me about an hour to solve this unsolved problem in mathematics. And it was, it was a pretty simple formula. So I came up with the equation of the planes and did this demonstration. And suddenly this set of equations was, it was used by Dice Lab to make some of these dice on Wikipedia and other places. It just kind of, okay, the problem is solved now. But unsolved problems in mathematics don't need to be especially difficult. And this was probably solved before many times, but no one thought that the solution was worth anything, so they were just left unpublished or something, I don't know. But there is a published solution now that I'm probably unworthy to have my name on. For larger polyhedra. This is what's called the angle 38 space filler. So we have one object here at the center and it has 38 faces and the faces are sort of reflected here. It's a reflection in an unusual way. But basically here, the object in the center in opaque red is surrounded by 38 copies of itself. It's a convex object. And this is the largest known space filling polyhedron that's convex. However, There were no 3D pictures of it anywhere. I was became interested in this object about 20 years ago and just a few months ago somebody finally managed to figure out what the vertices of the object were. And so I made this community post about it and posted code for, for generating everything. I'm hoping to go back to other space filling polyhedra and come up with nice solutions and animations for all of those as well. But there's, there's about 4,000 of them. Many of them have never been seen outside of the realm of cryptography. And that's sort of one of these objects that, that should come out so that people can take a look at it. With a Heptagon. One thing you might wonder is what is the minimal number of braces you need for, for making a heptagon rigid? So all of these lengths are identical. And a number of years ago I found this, and it's been in draft data for about 10 years. And a few years ago I noticed that this might be a rigid object, in which case it would be a record for bracing the heptagon. And it turns out that it was a rigid object. And not only that, you could take any two connected points and just remove them and anything attached to them and it would still be rigid. This is a super rigid object. So this is kind of an accidental discovery that wound up solving an unsolved problem in mathematics. Back at the, at the start we had this object here. Well, you can get that from what's called the super golden ratio. Where you solve this super golden ratio is roots of this problem or psi plus one equals psi to the third. Well, if you look at the complex roots, this point here is that root to the third. This the same root to the squared, the actual root, the root to zero or just plain one and the root to the fourth. And here's negative the root squared. So these two points are opposite the zero point, the origin here. Well, you get this quadrilateral at the outside with power 0, 1, 2 and 3, and you can split that quadrilateral into two similar quadrilateral just by adding these two points. And this quadrilateral and that quadrilateral have a ratio of the super golden ratio. This one to that one has a ratio of the super golden ratio. And if you keep on repeating this kind of division here, let's see if I can get this door here. I'm not sure why that's not working. Let's try this out here. And we'll try for 10. And it doesn't seem to be working right now. Oh, Well, this. Use the resource function. Let's see if that works. There we go. So if we take this set of steps and repeat it 10 times, you get this object, we can repeat it more and hopefully that will come up with. You get this, this weird fractally shape the, with the edge on the side, side. And if you take a bunch of those, these are all the same shape. They're all similar. You can kind of make this nautilus out of the same shape. And that's what I had prepared. We have 10 minutes left. I'll be glad to answer any questions or go over any of these. Someone asked a question do you have any advice for someone who wants to make their own mathematical game? Well, just. Just try it out. Try to find the various boundaries and solutions that it has and see if anything like that has been found before. One way of doing that is to go to. Let me see if I can find something here. If you know the parameters of a problem, you can go to the online Encyclopedia of Integer Sequences and just try plugging in something that comes out from your problem. Like, you might notice that 2, 3, 5 and 8 came out. Well, that might be. You found the Fibonacci sequence. One of the first things I do when I got something I think might be new as I'll try to turn it into a sequence and then look it up on the oais.org let's see, a bit unrelated. You mentioned graph embedding. When will Mathematica do built in GNN functionality? And I'm afraid I don't know what GNN functionality is. If there's any other questions, I'll be glad to answer them. All right, I invite everybody to visit our demonstrations project and if you come up with any ideas for anything you'd like to see there, or if you come up with new code, you can just send it to me edpwolfram.com and I'll be glad to take a look at them. And if you have any ideas for anything you'd like to see, feel free to write to me. All right, Many thanks, Sa.