Mathematical Games Hosted by Ed Pegg Jr. [Episode 15: Pi and Other Topics]

Sa. Hello. Everyone will be starting very soon, waiting for people to file in for the meeting. I'm Ed Peg Jr. And we should start within a minute. Feel free to post any questions that you'd like me to answer, and I can see if I can address them at the end of the talk or at any time. If you ever have any thing you'd like to ask, feel free to post it and I'll be glad to try to address it. And let's go. Welcome to mathematical games, episode 15 PI and other topics. I am Ed Peck Jr. And I am the head of the Wolfram Demonstrations Project and various other areas of Wolfram research. And my main interest is recreational mathematics. One of the leaders in that was Martin Gardner, who wrote a series of columns for Mathematical Games called for a series of columns for Scientific American called Mathematical Games. And I've been trying to make a a series of talks sort of in the spirit of Martin Gardner's columns. One of my favorite types of columns by Martin Gardner was when he did a hodgepodge of topics, usually seven or nine different topics, and those would usually cover a wide variety of previously unsolved questions in mathematics. And I always found those very interesting. And almost always his readers would manage to solve the problem within a few issues. Of course, there's a few exceptions, but I always found it interesting to see mathematics being solved as I was reading issue by issue of his columns. And of course, many famous mathematicians like Richard Guy, John Conway, Elwood Burlekamp, and many others helped him to solve these problems. Here's a few of the demonstrations involving unsolved problems, and I'll be hitting on some of these today. So I started out by saying the first topic would be PI. In celebration of PI Day, we did a few posts on PI day@community.wolfram.com and one thing to notice is that the area of a circle is PI R squared. So if you take the first five numbers and square them and then convert those into characters, you get 1, 4, 9, 16, and 25, which makes these letters a D, I, P and Y, which is an anagram of PI day. So if you square numbers, you get PI day. And I thought that was interesting. There's various weird ways of getting PI, and one of them is to start with the minger sponge, which is this object here. And you can take slices of it as shown in a few demonstrations. If you cut it in half on a diagonal, you get this odd object, but under that in two dimensions, there's a fractal known as the Sierpinski carpet. And that's sort of the face of a minger sponge. And here's two different ways you can code up getting that fractal. We could also perhaps look at one dimension, a Cantor set, where we start with a line segment, we remove the middle third, then we remove the middle thirds of what's left and the middle thirds of those and so on. And eventually you get what's called Cantor dust, which is a set of Area 0. It's basically all the ternary numbers that, but they don't form a connected set of numbers. So there's a one, two dimensional, three dimensional ways of getting this, this concept of taking one third. But we don't need to take just one third. We can do other fractions. With the Wallace thieve. If you take 8/9 times 24, 25ths times 48, 49 and so on, you get PI over 4 eventually. So if you do the same sort of Sierpinski carpet, but with increasingly smaller squares, instead you get this object which approaches the area of a circle and that's known as the Wallace sieve PI approximation. And it's sort of shown in this formula here. We can also do this in three dimensions with this more complicated formula here where we're taking 1/3, then 1/5, then 1/7 and 1/9 and so on. If you take the odd values as squares, you get this weird object which happens to have the same volume as a sphere. And that's a way of making a minger sponge have the same volume as a sphere. You just take smaller sections each time and you get the area of a sphere. There's other strange ways of getting PI. We can use the canonical tetragonal anti wedge hexahedron, which is this object here. Here, here. It's shown with its dual and by canonical, basically that means the, the object is, is aligned so that for a unit sphere, the edges are all perpendicular to the sphere or tangent to the sphere. And it's like, it's like this. The, the sphere is tangent to each of the edges of the, of the polyhedron. And it's possible to do this with every polyhedron. Every polyhedron has a canonical form. It just so happens that for this object with six faces, its volume is very close to PI to four decimal places. And as I said, you can do that with any polyhedron. Here's the demonstration canonical polyhedra where you can basically take any polyhedron up to 10 faces and it will automatically make the canonical form for that, for that polyhedron. So here's the 97th of the 257 octahedra. Let's see if we can get a different one here. Not sure these are actually changing. Probably freezing up in some way. So works online. So if you want it's called the canonical Polyhedra demonstration on demonstrations.welcome.com. We can do more on PI. For example, 7 to the 7th divided by 4 to the 9 is very close to PI within five decimal places. We can look at other constants such as the golden ratio which, which satisfies this relationship. Phi to the phi squared equals phi plus one. And here's the golden ratio here. It turns out that if you have 3 coins tangent to each other, you've got a unit disk and three golden ratio disks. They'll make a perfect tangency to a wall, as shown here. So this is one odd way of getting the golden ratio using circles. If you move that up, what you need is the Tribonacci constant which is seen in the snub cube and the snub dodecahedron here. Here are the those two objects. It turns out that all the complicated vertices seen in these two, if you look at them in terms of the Tribonacci constant for for the snub cube or the Fibonacci constant in connect in addition to the golden ratio for the snub dodecahedron that all of the complicated vertices simplify to integers. If you look at them at in the root space of the Tribonacci constant and the Tibonacci constant plus the golden ratio, everything is is is integers in those two spaces. And if you use the Tribonacci constant as shown here, this is the Tribonacci constant here. T to the third equals t squared plus t plus one. If you use it in this way, you get five tangencies. You have the unit circle at the center here and five coins based on the this expression of the Tribonacci constant. And you get this perfect tangency of five disks. In constrained geometry such as this one, there's a few constants that come up a lot. There's this one, the plastic constant here. This one, the super golden ratio. This one, the Tribonacci constant which we just use this one based on 7 and this one based on 9. Plastic constant, super golden ratio Turbonacci constant. The last two are. We can look at the discriminants. The one based on 7 has a number field discriminant of 7 squared. The one based on 9 has a number field discriminant of 9 squared. It turns out that this value related to 7 is related to heptagons. If you're in the. If you, if you're dealing with something and you look at the discriminant of a polynomial in the space you're working with, and it turns out to be 49, you're in the space of heptagons. And similar to that, if you check on something, you've got this weird value and the polynomial is as a discriminant of 81. That means you're in the space of nonagons. And I found this happens more often than I would ever expect. For example, you can try that just using circle points. If you do circle point seven, which makes a heptagon, you get a whole bunch of strange polynomials. But if you look at the discriminants for all those polynomials, they almost all have discriminant of 49. And similarly for the nonagon, they all almost have. They almost all have discriminants of 81. So that's sort of my secret for figuring out strange geometries as I look at their number field discriminant. If you look at the actual polynomials themselves, for example, here's for circle point seven, the polynomials tend to look somewhat nasty. You get, you get all these strange terms, but when you just look at the discriminants, everything simplifies nicely and you can figure out what's going on. And these five discriminants happen to come up a lot with constrained geometry. But there's other weird geometries you can go into. For example, here's a trick we can do with E PI and the square root of 19. We do this, we add 24 and we take the 24th root and we get this value. And then we look at X to the third equals X or two times x plus one, and that solves to be the same value, seemingly. Now, I deliberately picked 11 decimals of accuracy. If we use more decimals of accuracy, you'll see that they diverge. But that's a very close fit. This particular polynomial has a discriminant of -76, which is another frequent one that's seen. And with that particular geometry space, we can get this covering of a disk with 12 identical disks. And this is the best possible solution found by Melison. Here's a similar trick with E PI in this square root of 163, which I'm sure many of you have seen. If you add 24 and take the 24th root, you get this value. And if you take a Look at x to the third equals 2 times 3x squared minus 2x plus 1. You get the same value to 33 places. This is in the field of modular forms. This particular polynomial is highly related to this particular value in j space. And for an explanation you can see my column shattering the plane. Another related item is circles of Descartes which, which has many of the examples of these tangent circles doing weird things with the bend of the radius. As something I mentioned last week that the bigger my neighbor problem had been solved where you have a deck of cards and you're basically playing war with each other. And each time an ace, king, queen or jack comes out, there's a penalty of 4, 3, 2 or 1 cards. And the question has been for about 100 years whether this game ever runs into a cycle or does it always end. And just last month it had been. It's been shown by Braden Casella that the game can go on forever with this particular starting arrangement. So that problem has now been solved. Here's a item known as Heish tilings. If you take this shape at the center, you can surround it with a corona of copies of itself in darker gray. You can surround it again with the light gray snakes and you can surround it a third time with another set of dark gray space. But at that point you can't surround it anymore. This, this basically has a he number of three. And it turns out there's a. There's a infinite set of these. Let's see if I can make this work. And for some reason this doesn't want to work at the moment. Why don't you want to work? Let's see here. Oh, that's embarrassing. I probably have too many things in this notebook. However, if you go to the demonstration, you can see that it works. And each time you surround a copy of itself, you can get the. You can see that the snakes continue to work. Let me see if I can get this to work. If I exit out of here. With the current all and then reevaluate this, Hopefully this will work quickly. If anybody has any questions, it would be a good time to ask them. All right, that's come back now. Let's see if I can make this work. All right. Sorry about that. So here's the. The smallest snake. Here's the next level. Next level, Next level. It turns out that all of these snakes there's. And there's an infinite number of them. You can just keep growing them. All of them have heesh level three and that they can be surrounded with three Coronas, but no more. And this is an object I found while looking at Craig Kaplan's paper. He's numbers of unmarked polyforms. I looked through all his data and I happened to notice he had an infinite series in the he's threes. And so he said, well, these are now called the pig snakes. And this and other results about Heath numbers are at this link here, polynomial.org.uk and I'll be posting this a bit later, but here's the site and it's got data on thousands of polyhedra. And it was through this that. The hat was found. David Smith found this particular shape and he couldn't figure out what was going on with it, so he contacted Craig Kaplan to figure out the HE number. But it turned out you could make as many coronas as you wanted, but they weren't periodic in any way. And as such, a little bit over a year ago, the hat was finally published to the world. And we have various functions relating to the hat. For example, HAT hexagonal tiling. You can, you can make as many layers of the hat tiling as you might like. Here's one of the strange patterns caused by the related object, the turtle. You get these weird spiral patterns and we have. If you look at our community site, we've got a lot of different material written about the hat and I gave a talk about it almost a year ago, introducing some of the things that had been found. Here's David Smith and Robert Fouthauer, who runs tessellations.com holding a sphere covered with the hat and pentagons. Dune is in the Nude, is in the news, and the symbols used in the poster happen to be superset, union, intersection and subset. And one on quirk about that. If you turn everything clockwise, the direction of the opening spells out lose, which I thought was interesting. A rectangle based on A4 paper has the property that it can be divided into similar rectangles. And it happens that these make a perfect series of the squares and double squares. And this has recently been made as a puzzle by Rex Perez. Patrick Hamlin wanted to see what the best approximation you could get to a square using a four rectangles. And here's his best two results. It turns out that it's impossible to divide a square into a four rectangles, but you can get very, very close. These two approximations are within three decimal places and five decimal places. You can also divide a square into acute, acute similar triangles. There's only one acute triangle that works, and that is the 45, 60, 75 triangle that was proven by Lakovich in 1990. There's two other triangles that work that are obtuse. And you can also do it in many ways with right triangles. But if you use nine non right triangles, there's only three triangles that can divide a square. And it was long and unsolved. Question, what is the fewest number of triangles that the square could be cut into? And Tom Forgatus got it down to 32 triangles as shown here. And here are the vertices used. They're all based on the square root of 3. The the A4 paper is based on square root of 2. So this is kind of a tour of different geometries that do weird things. I introduced a few earlier plastic constant, super golden ratio, Tribonacci constant, and square root of 2 and square root of 3 and the golden ratio. Those are the big main ones in weird geometry. So switching over a bit, smooth numbers. Here's a 19 factorial. And it happens that the largest prime factor is 19. So here's two numbers, 123,200 factors into these factors, all 11 or less. If you add one, it's equal to three to the six times 13 squared. So this is everything has factors of 13 or less. And as such, this is called a 13 smooth number. And this is an 11 smooth number. These are the two largest neighboring numbers that are both 13 smooths. So we can make a quick. We can make shorter versions than this for factorization. Here's one sample with format factorization. But I wrote this bit of code to make it even, even more compact. So this number is equal to the six five squared, seven times 11. And this is three to the six times 13 squared. So here's a factorization table for the numbers up to 144. I came into smooth numbers in a weird way. This number caused an error in some code. The fourth root of 91 over 10. But you could also represent that as 9.1. So when I tried to figure out what was going on, I looked at the continued fraction and I noticed it was sort of weird. You got these high values, which usually doesn't happen in a simple continued fraction. Usually that indicates something is happening. And this value happened to be very close to this value within seven decimal places. 33:19 to the fourth is equal to 9.1 unless you go to more decimal places. So it seems that 91 over 10 is equal to almost 33 to the 4th over 19 to the 4th. Or if you multiply with cross diagonals, 91 times 19 to the 4th is almost equal to 10 over 33 to the 4th. When you multiply those out, you get these two numbers, which are both 19 smooths, 2 times 3 to the 4th times 5 times 11 to the 4th and 7 times 13 times 19 to the 4th are 11 million blah blah blah. And that same number plus one. And it turns out these are the two highest values for two consecutive 19 smooth numbers. And it's a famous example in the ABC conjecture and other branches of number theory. This is sort of a well known weird exception for some strange things in number theory. But it turns out that it sort of falls out from the square root of 9.1. And playing around with that more, I wondered, well, what are some other 19 smooth numbers that can be found? And it turns out that this number here, which is all sevens and nines, is a 19 smooth number. 17 times 19 to the 6 is this weird number. So in an overnight run I decided to take a look at what are the smoothest numbers made out of two digits. And it turns out that these numbers now in the online encyclopedia of integer sequences are the smoothest values. And here's their representations and factorizations. Let's see here. I wonder if this is going to work. Now, Here's the largest square numbers you can make with particular sets of digits. For example, with the, with the numbers 6, 0 and 9. The squares of these numbers are all represented with just those three digits. And so far this has been unsolved. Question, there's two triples that are unsolved, 0, 1, 3 and 6, 7, 8. All other numbers have strange square numbers built from those digits. And I might note that these are only for sporadic solutions. We leave out the trivial solutions. For example, there's numbers of this form which we don't listen to. But here are cases with 0, 1 and 2 which are stranger. So this is the squares containing three different digits. And this is an unsolved question what are the extents to these? And whether there's any more than are in this table. These are all the known values as of last year. The Brune Taurus is another item I took a look at recently. If you take a look in one dimension, this sequence here has all items of order 2 from the Alphabet 0, 1, 2 and 3. For example 01, 02, 03 and 00 are all in there. Also 1 1, 12 is 1, 2 and 13 and it's cyclic. 30 is the last item in this sequence. Here I'll order four items from 1, 1 and 0 are cyclically represented. So here's 00010000000010011 and so on. It turns out that the 16 different ways of grabbing numbers from here are all represented within this. It's also possible to do that in order. 2. For example, every single square that's possible with these three colors is within this sequence. And I'm just basically moving the pattern around. Notice that this shape here is in all of these. I'm just basically moving around so that a particular chosen sequence is at the top. In a quaternary, I'm doing the same thing. Any, any possible 2 by 2 square of these four colors can be moved to the top here. In this Debruin taurus. In binary, any 3 by 3 item is within this our story. Any 3 by 2 rectangle is within here. And it is basically moving this shape around. And in this larger rectangle, any 3 by 3 where is represented in there somewhere. So if you want to generate all possible patterns in a small place, you'll want to use a Debrun Taurus. Here's a problem I looked at about seven years ago. You want to divide a triangle so that no two edges are shared. So for example, here we're dividing this triangle up and these four triangles, none of the two internal triangles shares a full edge. And same with this here and so on. All of these have that property. And one thing we want to avoid is taking a solution and then using that, sticking it into one of the smaller triangles. So wouldn't be all that interesting to take this solution and divide up say this triangle here. Because that's sort of an obvious extension. Once you have these, these, these found, is there a way to find more of them? And that's a problem I posed. One solution I came up was to look at flagal diagrams for planar graphs and picking vertex sets which might be used to to go backwards to these triangle solutions. For example, this one here can be used to find this solution here. However, there's a whole bunch of planar graphs that pop out billions, trillions and so on. And I was unsuccessful in gleaning those out. But I did manage to doodle and find other solutions such as this one where all the triangles are the same area except the central triangle. Just recently, last month, a person on one of the mast sites called Ravenclaw Prefect managed to find a whole bunch of weird solutions. For example, this here, this here, and here's some other odd solutions he found. And I finally asked him, well, how did you find these? And his answer was I Just basically doodled them and I started finding them. So here's a very hard problem in constrained geometry where the best solution solving method seems to be doodling by hand. And I don't know of anything other than this that sort of compares with that. So it's unclear to me how this can be extended more. But he's managed to find this set of infinite solutions. For example, these are in a series and they're non trivial extensions of each other. And he also has this one at the bottom which is also a very strange setup. But he's basically shown that there's an infinite number of these weird triangles with no shared edges. And I mentioned Schlegel diagram. So here's a wall from demonstration which has all the graph planar graphs up to 10 vertices. And we'll make the Schlegel diagram. I'm not sure whether this will actually work. Yes it does. When I originally made this, the. The TUT representation for a planar graph hadn't been finalized. I should redo this one so that it uses the TUT diagrams of a planar graph which. Which basically gives the Schlegel diagram automatically. Because my code for these isn't all that great. The. The code for TUT diagrams of a graph. And if you just do graph. What is it called? The graph layout Cut diagram automatically does this for you. Last year it was shown that life is omniperiodic. Here's the last solution that was found and we can run this. This is a demonstration that has all of the life oscillators in the Conway's Game of Life up to period 100. So there's a period 19. Let's see, period 41 was one of the last ones found. Oops, there's period. I've gone too far. So here's period 43, Period42. And period 41. It was an unsolved problem until last year whether there was a oscillator for every possible period in the game of Life. But that has now been solved. And I believe that my solutions listed here are the smallest known solutions for each one. On the Game of Life. Recently released was a free book. Here's a picture of it. But if you go to Conway Life book you can get this. You can get a download of this book for free. And. And this basically covers the game of Life in great detail. Many of these things were not solved until just recently. And this book does a really good summary of everything about the game of Life as it is known. As of last year, A question that was recently looked at. Is it possible to make a polycube so that no. 3 cubes are in a row in any direction? And Georg Arndt took a look at this and he was able to grow the solution right up to this object here. With 26 cubes. This is the largest possible connected polycube so that no. 2 cubes are in a line in any direction. Not just diagonally or orthogonally, but weird Nights moves aren't anything else like or anything else. And any direction. There's no three cube centers in a straight line and that's as large as you can go. Then he decided to take a look at similar problems. Can we avoid no.4 in a row in either direction with with simple polyforms? Here's the record with 15 squares. And it turns out that Rodolfo Kerchon found this earlier. He also looked at no. 4 in a line with the restriction that no cube was at the midpoint of two others so that you couldn't have three cubes right next to each other. With. With that restriction the growth was reasonable enough. You could solve the problem. And that ended with this 75 cube object. No. No. 4 cubes are in a straight line in any direction. And also there's no. 3 cubes so that one of the cubes is at the midpoint of the other two. The solution for if you can have a midpoint cube is unsolved because it has sort of explosive growth at the start. And it's unclear how to resolve that before it starts decreasing in size. Foreign. Let's see. It's 2024. There's various things we could say about 2024. It's the sum of the first nine cubes, except for one. Next year we'll be able to add one at the start. It's also the Chevyshev number and it's 765 squared times 7 plus 1 the square root. It's also a dodecahedral number and a tetrahedral number as shown here. It turns out that all the decahedral dodecahedral numbers are also tetrahedral numbers in a strange way. In a earlier talk I talked about the Goldberg tile and I said to write it up more. It's in OEIS and it was developed by M. Goldberg in 1955. And here's what it looks like. It's basically an equilateral triangle with half of a 45, 45, 90. Eight, 48 tile solution for the dodecagon. And for some reason I see there's a spinny wheel over here. So. Let's see. So here's a way of dividing a dodecagon into 48 identical tiles with the Goldberg tile. Andrew Hudson recently found some new tilings as shown here. And I haven't yet added these to my catalog of substitution tilings and this sort of a note to myself to do that. So here's one of the two that he found which is based on a passant number that hadn't been used before in a substitution tiling. And here's a new substitution tiling based on the square root of three. Sort of note to myself, add these to the big catalog. Here's a solution for dividing a triangle into five similar triangles. And a person asked whether this was a known object. Well, it was known, but it wasn't published by anyone so far as I know. So I decided to write it up more. It's based on the super Golan ratio psi. Yes, that's the, that's psi. Psi to the third equals psi squared plus one. And that's the super golden ratio as shown here. Whenever I'm doing with these constraints I look at the discriminant as I mentioned earlier, and this one has discriminate minus 31. Here are some points for this dissection here. And if you look at the square root space in the super golden ratio, everything, everything is basically fractions of 12 in that geometric space. And with that we can make a diagram of the solution as shown here. These values here are powers of the super golden ratio. For example, this 0 here means this value is the super golden ratio to the power of zero are one. This value here is the super golden ratio. One and two means it's the super golden ratio squared. And it turns out that A, A012 triangle, if you use powers of the super golden ratio, so one super golden ratio, super golden ratio squared. You get a triangle that has 120 degree angle here. And with five of these, this triangle and that triangle are similar, are identical. This triangle and this triangle are identical. And this one is, is scaled down one level. With these five triangles you can get an equilateral triangle. And I believe that this will now be named after Gosin Carlson Berg because he was the first to actually publish it. Even though quite a few people, including myself knew about it. We just hadn't published it yet. I mentioned last time that the chance that five points will make an ellipse is still an unsolved question. So I figured I would mention that again since I'm doing a hodgepodge. What are the Chances of five points making an ellipse. And you can get, you can see more details about that. In my previous talk, I just showed the super golden ratio 012 dissection and that's using a power triangle. We can do the same sort of thing with the Kepler triangle, which is a power, which is a power triangle based on the square root of the golden ratio. So if you take what I just said, the 0th power, the first power and the second power, you get a right triangle. So this is one, the square root of the golden ratio and the golden ratio itself makes a right triangle. And exactly five years ago today it turns out I, I managed to find this tiling here as a subset of solutions in my demonstration wheels of powered triangles. And that led to the column I mentioned shattering the plane, which has a lot of new substitution tilings. But I wanted to look at other power triangles recently and I found some interesting dissections of rectangles of triangles and series. With the super golden ratio. I found this series where I use 023 triangles and if you multiply it by that value again, you're just basically adding one to each of the powers. So 023 is equal to 1, 3, 4 and so on. All of these triangles are similar. They are in series. Each one has an area this much larger than the previous one. And all seven of them assemble together to make a nice rectangle with sides and ratio of the super golden ratio. You can also do that with the plastic constant. Here are seven triangles in series that make a dissection based on the plastic constant. You can also do that with the golden ratio. Here's six triangles in series which combine together to make a rectangle. If you just use three pieces, you get what's called Brugner's three piece tangram and that's shown here. And there's another series called the chi ratio which I used in my, in my chi tiling. You can get these triangles in series that's shown here. I haven't figured out how to make a rectangle with this one yet. A weird property I mentioned, heptagons had weird properties. Here's one of them. Take a heptagon and then make these lines. And I show some of these lines here. It turns out that the intersections of those lines define all but one of the points of another heptagon which has twice the area of the original heptagon. I show that here. And all the code for this is either on community.wolfram.com right now or soon will be. This is One of the things I wrote up a few years ago. But I plan to combine all these into one big mega post sometime soon. And it turns out that if you take three, if you take three points so that it's, it's not so it's a scaling triangle. This point, this point and that point. If you reflect on each edge you get, you get three new triangles. So this triangle reflected over here makes this triangle. This triangle reflected on this edge makes this triangle. And this triangle reflected on this edge makes this triangle. That's three extra triangles. And if you look at these points, you get a new triangle which is similar to the original 1, 2, 4 heptagon triangle with double the area. And that's a weird property that only one other triangle has. It's strange that the heptagon has this strange property. Another weird heptagon geometry item led to this graph here. And it turns out that this is all done with unit edges. All these edges have length one. You've got a heptagon on the outside and one layer in. You've got a heptagonal star. And then on the, on the inside you've got the order two heptagonal star. But it turns out they, that you can twist them so that everything has unit distances away from each other. And this object happens to be rigid. It sets a record for bracing a heptagon. And it turns out to be so rigid that you can take one of these edges, remove the two connecting vertices and everything attached to those two vertices and the object is still rigid after you've ripped all that away. It's a super rigid object and it sets various strange records and bracing geometry. Since I, I talked about measures, I, I've got this graph here. You've got a measure called the Gutenberg. You can take five of them and make a Twip Twips. You can take another five of them and make a cosmological quantum point length. You can take four of those and make a, a Johnson point, and eight of those makes a button, three of those makes a barley corn, and three of those equals an inch. Three inches make a palm, and so on. I made this ridiculous graph of strange measures where everything has unit multiples of previous items. For example, there are four inches in a hand, but there's two inches in a stick. There's three hands and a foot, and so on. This is a, a weird thing on measuring. I, I saw a, a much smaller version and I figured with all the things we have in measure data for Mathematica, I could make something much Larger. There's recently a finding with the nauru graph which is shown here. You can get that by rolling an octahedron and getting all 24 orientations. For this rolling octahedron, if you basically roll it on one edge, you get this connected object. But if you keep rolling, you basically go straight up on this order three Taurus and you get back to this orientation and everything is connected in this way to get the equivalent of the nauru graph that somebody else noticed that all the half twists of an order 2 Rubik's Cube also make a nauru grass. That was something that was just found last year and I thought that was interesting. We've got various things with polyforms here and I'll just go through these are. This something I found. Here's some items that Patrick Hamlin found. It's possible to divide a equilateral triangle into five diamonds as shown here. This is the smallest possible solution. And here's another polyform where eight different sizes will make an equilateral triangle. And this is an object I found five years ago. Today you can make an infinite series of this particular fractal triangle to make a copy of itself. And it turns out that with some rather simple code for the complicated object, you get this object I call the fractal Cal Nautilus. All these fractal triangles here are similar to each other. And I used that as a plot point in a novel I just released on Amazon last week called Non Sequitur to the Equator. And in it I've got a whole history of science fiction novels. For example, this one from 1888, there's a quote in that novel. Are credit cards issued to the women just as to the men? And Dr. Leets replies, Certainly that's from 1888. It's the invention of credit cards. But it wasn't until 1974 that this was realized where women could get credit cards freely in the United States. So I thought this was an interesting bit of science fiction. I found lots of science fiction tidbits while researching this novel. And so I've packed all of those and and many other math facts and science faction to this hard science fiction novel, Non Secretary of the Equator. And that's what I had prepared. Many thanks for listening. Let me see if I can address some of these questions here. The got some questions on the ellipse. Yeah, the Conway Lifebook really is a beautiful book. I highly recommend it. And I don't see any questions here I can answer quickly. So I'll. I'll just thank everyone for attending my talk. Many thanks to everyone. And I'll be back next month with another talk. So many thanks.