I'm always amazed by the types of questions Brady asks the subject matter expert. It really helps ground the topic, and dig a little deeper than the presenter would have normally delved.
Knowing the right questions to ask is the mark of a skilled educator. Brady is more than just a cinematographer, he's also here to help with the education. He asks the questions because we can't. And asking the right questions helps bring the knowledge out of the expert in a way that non-experts can understand. It takes real talent to do that translation.
Yes, because he asks the kind of questions that a lay mathematically curious lay person would naively ask, and not the rigorous questions a mathematician would ask.
Brady has a great talent for asking the question his viewers might be asking. Whenever I have a question I would have asked if I was in the room, Brady asked it a few seconds later, as if he can hear it, it makes the videos almost as great as hearing the conversation in person
Wish I knew this back in math class when I got caught doodling. “No teacher, I’m not doodling in the margins, I’m just practicing my polygon subdivisions.”
this happened to me the other way around: build a tetraeder out of boredom the whole class, and afterwards the teacher (chemistry) came to me and praised what a nice model of i had made. ^^ i'm damn sure he knew that i just did this out of boredom, but it was his way to reinvite me back to participate. that teacher was such a cool dude. :)
This is oddly satisfying. I may not fully grasp it, just love that people are out there doing these things, just for the pure love of mathematics, I don’t need applications.
"We connect two vertices if the corresponding faces are contained in each other." Even with the previous definition of faces that was glossed over, this is surprisingly imprecise and difficult to parse. I don't know how anyone could understand what's going on if they watched this video at full speed.
I've used these before in 3D modeling, very useful when you want to essentially increase the resolution of your model without changing its shape. Though I will point out that just dividing flat triangle surfaces will only ever give you a flat result, so this alone won't actually make your model smoother or something... there are other methods for that though.
Based on my zero knowledge of 3D modeling, I'm guessing this is the first step. You divide the triangles into smaller triangles and then adjust the positions of the new vertices?
@@lucromel 3D modeling is actually usually done with quads, those quads then get subdivided and usually smoothed. Triangles are simply not great to work with when 3D modeling.
You might also want to subdivide triangles for shading purposes. You can assign different colours or normals to the interior vertices to get effects that you couldn't get with a single triangle.
I wonder what a Parker subdivision would be. Maybe it would have a double-edge or double-vertex. Maybe it would have a point labeled (-1, 0, 1) or something else not allowed. Maybe it would have a quadrilateral hidden in there. I can't decide.
Fun fact : You have link between the subdivision of simplexe and distributed computing, you have to make a Schelgel Subdivision and then associate a combinatoric interpretation of it. Another way to said it is : their is a way to subdivide triangle to obtain every pre-order possible between d variables ( for pre-order with 3 (or d) variables you start with the simplexe of dim 2 (or d-1)). The key word is Chromatic Subdivision have fun !
Could you make a video about the practical applications of topology? What have we learned from the study of deeper geometrics, and topology in the past century that we can see in our daily lives?
I wish they did a non-equilateral triangle, which he said would work at the beginning, just to get a better visualization of how the congruence builds on the sub-division.
I'd be curious about any special properties of the subdivisions. Something reminiscent of the Dehn invariant would be an interesting discovery. I guess it was unclear to me from the video whether the subdivisions are always of equal area (maybe that's the goal definitionally), particularly for the barycentric technique. What about the lengths of the sides? For example, as the angles become more acute or more obtuse, does the ratio between sides remain the same? The average length? And if those values differ with further subdivision, why do they do so and by how much? Do they vary in a predictable way? Arithmetically?
Are my eyes playing tricks on me, or does the edge-wise subdivision break the symmetry of the tetrahedron (@14:00)? I'm not sure why, but that seems really weird to me. I guess it has something to do with the fact the the four coordinates are not treated the same, so it matters which vertex you put on which axis.
Here's a simple way to subdivide a triangle not mentioned in the video: pick any of the vertices as a starting point, and draw a new edge to the middle of the opposite side (or any number of equal subdivisions of the opposite side if you prefer), forming two (or more) new triangles. From the new vertices, repeat the process.
I think these are called the "altitudes" of the triangle, and that this is actually a variant on barycentric subdivision, where you do one random point at a time for each of the triangles instead of all of them
@@cheeseburgermonkey7104 , not quite. The altitude or height of the triangle is perpendicular to the opposite side. In obtuse triangles, this can lay outside the triangle, so would not be suitable for division. What I described above is the median. You could also use the angle bisector for a similar division.
I explored these kind of subdivisions, and focused on the symmetrical ones, which create beautiful patterns. Would appreciate if you could check them out. You can find a video on my channel, which has a link to the article I wrote. (unfortunately, I cannot directly share links, my comment will be hidden)
regarding tetrahedron - didn't check it myself just yet - will definitely do that in a while - so that math rule yields just one specific additional edge in the middle instead of two possible edges? how is so? why one over the other one? and why not both - which actually breaks the subdivision?
After converting the coordinates to their cumulative sums, which is the first step that you need to do before taking the differences, the order of the coordinates starts to play a role. There is then only one pair of edge coordinates that results in just -1's or +1's.
The thumbnail took me for a ride. See a new numberphile video showing how to best make a triangle look like an arcane demon summoning symbol. Go over the video and all I see in the video preview is a dimly lit polygonal cthulhu with red eyes waving at me.
I clicked on this video and watched a Doritos ad about how everybody's obsessed with triangles now. This vid came out 4 mins ago. Who else had the same thing happen to them?
The tetrahedron has several connections that were drawn that don't adhere to the rule of not having a -1 ánd a +1 in the difference between points, so that contradicts what was said. Also, there are three internal edges that fit the rule of having a difference of 1 (ignoring the above, otherwise there will be none), so how is the single edge chosen? All three are symmetric. If you cut on all of those (you'd have to add a vertex at the intersection) the result is 8 extra pyramids, all nicely symmetric along one of their axes. For subdivision into pyramids this is not necessary, as cutting along a single internal edge as was shown is already enough to result in 4 (albeit asymmetric) pyramids, but it feels like we're missing part of the algorithm here. EDIT: I was wrong! I forgot to first convert the coordinates to cumulative sums. If you do this, all differences only have -1 or +1 and only one of the internal edges fits the rule, amazing!
In the tetrahedron, after the initial step, we're left with an octahedron, and then it was said we only needed one edge to have it fall apart into four tetrahedrons, and the animation only highlighted one. Is this true? Does an octahedron divide into four by adding a single edge? And why this particular edge? Symmetry suggests there would be four edges which could do the same job, why this one? Which rule picks this one and not the other three?
It's possible I made a mistake, but I tried calculating the vectors for the tetrahedron, and it did work out for me using his rule that 2 of the 3 possible vertex joins (not 4 possible btw, only 3) through the interior of the tetrahedron don't meet the criteria, so only 1 does. And the asymmetry did seem peculiar to me at first, but I think it's explained by the fact that the number we're calculating is inherently asymmetric, because we're calculating it from left to right and therefore there's a preferred direction geometrically too.
@@Alex_Deam The matching pairs of coordinates that could be used for the white line in the tetrahedron are: (1 1 0 0) and (0 0 1 1) (1 0 1 0) and (0 1 0 1) (1 0 0 1) and (0 1 1 0) None of these follow the rule (when subtracted they all produce a mix of 1 and -1) and I don't see any compelling reason to choose one over the other. I'm not sure whether he said either way whether this extra line did or did not have to obey the rule. I think, though, that it was at least implied that putting in all the lines that *do* obey the rule leaves you with a central octahedron.
@@kevinmartin7760 You forgot something that I also forgot the first time I tried to do this mentally - once you get the coordinates, you have to do that thing where you take the sum of the first digit, sum of the first two digits, sum of the first three digits, sum of all 4 digits. This creates 1222 and 0012 (doesn't work because of difference of 2), 1122 and 0112 (this is the one that Does work), and 1112 and 0122 (doesn't work because both 1 and -1 present).
I also got hung up for a while on the octahedron splitting along that single edge bit. The thing is that you can't think of it as an existing octahedron that you're cutting into the tetrahedrons. You really have to imagine the octahedron not being there already (because it isn't really, there's no reason to wait to create the extra edge), and instead just consider what the lines including the "new" white one create. If you spend some time with the visualization at 13:56, it helps you realize why they are tetrahedrons.
It's not clear to me why going into 3-d to find the new points in the edge subdivision is necessary. Aren't the edges just split into r equal parts? And in the resulting new grid, aren't the new lines all parallel to one of the edges or the original triangle? Or is that only the case for equilateral triangles?
7:20 Or you just check whether the difference of (vector between) two vertices also occurs in the original simplex. That's very easy to do because the original simplex' vertices are just the unit vectors, and in a simplex any two vertices are joined by an edge.
Question 🙋 when starting out the edgewise subdivision why aren't the location of the equilateral triangle X1, Y1, Z1 ?? AND when S starting to explain the subdivision of a 3D triangle where do the point locations come from? Like where is the 4th digit coming from? Isn't the XYZ format already 3D
Ooooh! Okay! I forgot about the location of the point is written with three numbers 😅 I thought he was writing the unit location like each axis is a ruler "1 inch" in each of the three positive direction.... Like a number line, my bad. 😂
Is one of these subdivisions (more) applicable to a Triangular Bezier Surface Patch? ISTM the baricentric or even just a radial division (adding just the center point and connecting all points) because you'll need to re-parametrize along each edge.
Question - Which method of subdividing triangles and tetrahedrons, will lead to the maximum number of triangles or tetrahedrons, irregular or otherwise in shape?
I like the edge-deciding trick for the edgewise subdivision. That'll give me something to think about for a while. I've got some vague ideas about graph adjacency and Grey codes. I wonder if I can nut it out before I give up and go off reading?
What might be interesting, and probably has implications in probability, would be a random subdivision. Pick a point in the interior of the triangle at random. Now pick a random point on an edge and connect that to the opposite vertex. You have a subdivision of the original triangle into two parts (with probability 0 that you picked the point which makes the two parts of equal area). What are the odds that the interior point lies in the larger subtriangle? The smaller? (On the line seems likely to be a probability 0 again, but I can't prove that rigorously.)
@@hrysp you know, I wasn't thinking of that, but yeah, it would apply. You would need to define what "pick at random" means to get definite answers. I was more proposing a class of problems or an area for work. You could definitely vary how you pick at random, or change the distribution of your picks (say, "randomly" pick, but it's biased along a Bell curve).
So I went for uniform distributions all around. We'll call the interior point p, and the point on the edge x. So, the probability that p lies within a certain area is proportional to that area, and the probability that x lies on a certain interval is proportional to the length of that interval. Using such assumptions, the probability that the interior point lands in the smaller sub-triangle is, I believe, 1/4.
What do you mean by "perfect-symmetry"? If you mean like a regular polygon, then yes: if a polyhedron's faces are identical and regular, then it's a Platonic one, and the one with most faces is the regular icosahedron.
@@rosiefay7283 Yeah, sorry, english is not my native language, and yes, this is exactly what I was thinking. Take the icosahedron and inflate it so it becomes a ball/sphere. I've been thinking about this for a long time in context of (world) maps for video games. Can't wrap my head around the fact that there doesn't exist any geometrical solution that leads to more than 20 plots/areas. If you divided one of the 20 'triangles' into four triangles I find it fascinating that these four are not similar!
@soilnrock1975 If you're willing to use non-regular polygons, you can get up to 120 (dysdiakis triacontahedron). If you want all the sides the same length, you can get 30 (rhombic triacontahedron).
brady triangles could like be a subdivision of a simplex in dimension 2308 and have to subtract to 1 and zero in base 4261. i imagine a verry fun extra for numberphile 2
I was left confused about what the point of this was. Could there not have been more context supplied from the beginning about what we were looking at?
Maybe a subdivision that divides the triangle into smaller triangles of the same fractional area while minimizing the change in angles of the new triangles. ?: N new triangles, each with 1/Nth the area and all new triangles are similar to the original. For what triangles and N is this possible?
It was so each vertex would be along a different axis, 3 points-> 3 dimensions. (That's also why the tetrahedron was in 4d space) it's less about being in 3d Spaceᵗᵐ and more just needing 3 axes
Brady subdivision? Sounds good, I guess. Now, a Parker subdivision, for that one I would pay good money! (Obviously for the Parker subdivision, we end up with a bunch of triangles and a square somewhere, with no obvious way to fix it.)
![АЛАУДИНОВ у Скабеевой: эти их ВСУ нас НЕ ДОГОНЯТ 😁 [Пародия]](http://i.ytimg.com/vi/an0anqU4WGQ/mqdefault.jpg)
Subdivided Triangle from this week's video looks nice on a tee (or holding tea) - numberphile.creator-spring.com/listing/subdivided-tri-numberphile
Which method of subdivision leads to the maximum number of triangles and tetrahedrons? Without caring how regular or irregular they are?
I'm always amazed by the types of questions Brady asks the subject matter expert. It really helps ground the topic, and dig a little deeper than the presenter would have normally delved.
Knowing the right questions to ask is the mark of a skilled educator. Brady is more than just a cinematographer, he's also here to help with the education. He asks the questions because we can't. And asking the right questions helps bring the knowledge out of the expert in a way that non-experts can understand. It takes real talent to do that translation.
Yes, because he asks the kind of questions that a lay mathematically curious lay person would naively ask, and not the rigorous questions a mathematician would ask.
he's a modern day Socrates. Hope he doesn't end up the same!
@@charlytaylor1748 bruh i
Brady has a great talent for asking the question his viewers might be asking. Whenever I have a question I would have asked if I was in the room, Brady asked it a few seconds later, as if he can hear it, it makes the videos almost as great as hearing the conversation in person
Nothing like starting your morning with a big cup of triangle subdivision.
How do you know? I'm watching this before my school at 7 am in my region!!
@@Devils_bride It’s always morning somewhere haha
@@andrewgries9011true, lol
I‘m there right with you guys
Wish I knew this back in math class when I got caught doodling. “No teacher, I’m not doodling in the margins, I’m just practicing my polygon subdivisions.”
this happened to me the other way around: build a tetraeder out of boredom the whole class, and afterwards the teacher (chemistry) came to me and praised what a nice model of i had made. ^^ i'm damn sure he knew that i just did this out of boredom, but it was his way to reinvite me back to participate. that teacher was such a cool dude. :)
I love how active in the conversation Brady has gotten over the years. He's asking so much and helping us all learn so much.
If at first you don't succeed in inventing a new subdivision rule, tri, tri again.
It's weird seeing stuff I used to doodle in school actually be a mathematical concept
If you haven't seen them yet, I'd really recommend Vihart's "doodeling in math class" videos
Next time I'm playing a 4d video game I'll thank Dr. Welker for working out all the tetrahedral meshes
4D miner :)
Miegakure exists
Legend of Zelda: Breath of the 4th Dimension
@@FrankHarwald it doesn't, does it
4D golf soon
Brady doing the heavy lifting of the explanation here by asking all the right questions!
This is oddly satisfying. I may not fully grasp it, just love that people are out there doing these things, just for the pure love of mathematics, I don’t need applications.
"We connect two vertices if the corresponding faces are contained in each other."
Even with the previous definition of faces that was glossed over, this is surprisingly imprecise and difficult to parse. I don't know how anyone could understand what's going on if they watched this video at full speed.
I love how Brady always know what question will be asked by viewers
I've used these before in 3D modeling, very useful when you want to essentially increase the resolution of your model without changing its shape. Though I will point out that just dividing flat triangle surfaces will only ever give you a flat result, so this alone won't actually make your model smoother or something... there are other methods for that though.
Based on my zero knowledge of 3D modeling, I'm guessing this is the first step. You divide the triangles into smaller triangles and then adjust the positions of the new vertices?
@@lucromel Yeah, it's basically setup for some other operation after.
@@lucromel 3D modeling is actually usually done with quads, those quads then get subdivided and usually smoothed. Triangles are simply not great to work with when 3D modeling.
Catmull-Clark subdivision surfaces come to mind.
You might also want to subdivide triangles for shading purposes. You can assign different colours or normals to the interior vertices to get effects that you couldn't get with a single triangle.
Kudos to the 3D modeler and animator for helping us visualize the subdivisions. Thank you.
Y’all make a great team. It couldn’t have been achieved with only one of you.
OMG I had a course with him in Erasmus! Great guy!
I wonder what a Parker subdivision would be. Maybe it would have a double-edge or double-vertex. Maybe it would have a point labeled (-1, 0, 1) or something else not allowed. Maybe it would have a quadrilateral hidden in there. I can't decide.
Fun fact : You have link between the subdivision of simplexe and distributed computing, you have to make a Schelgel Subdivision and then associate a combinatoric interpretation of it. Another way to said it is : their is a way to subdivide triangle to obtain every pre-order possible between d variables ( for pre-order with 3 (or d) variables you start with the simplexe of dim 2 (or d-1)). The key word is Chromatic Subdivision have fun !
Cool computer graphics to go with the presentation of the theory
Could you make a video about the practical applications of topology? What have we learned from the study of deeper geometrics, and topology in the past century that we can see in our daily lives?
I wish they did a non-equilateral triangle, which he said would work at the beginning, just to get a better visualization of how the congruence builds on the sub-division.
I'd be curious about any special properties of the subdivisions. Something reminiscent of the Dehn invariant would be an interesting discovery. I guess it was unclear to me from the video whether the subdivisions are always of equal area (maybe that's the goal definitionally), particularly for the barycentric technique. What about the lengths of the sides? For example, as the angles become more acute or more obtuse, does the ratio between sides remain the same? The average length? And if those values differ with further subdivision, why do they do so and by how much? Do they vary in a predictable way? Arithmetically?
The barycentric subdivision indeed has the property that all of the small triangles have the same area.
What a cool name - Volkmar.
Volkmar - The Dark Lord of Triangles!
Are my eyes playing tricks on me, or does the edge-wise subdivision break the symmetry of the tetrahedron (@14:00)? I'm not sure why, but that seems really weird to me. I guess it has something to do with the fact the the four coordinates are not treated the same, so it matters which vertex you put on which axis.
Truly amazing video Numberphile, fantastic explanation, brilliant work, very informative, very proud of you Numberphile. (:
Here's a simple way to subdivide a triangle not mentioned in the video: pick any of the vertices as a starting point, and draw a new edge to the middle of the opposite side (or any number of equal subdivisions of the opposite side if you prefer), forming two (or more) new triangles. From the new vertices, repeat the process.
I think these are called the "altitudes" of the triangle, and that this is actually a variant on barycentric subdivision, where you do one random point at a time for each of the triangles instead of all of them
@@cheeseburgermonkey7104 , not quite. The altitude or height of the triangle is perpendicular to the opposite side. In obtuse triangles, this can lay outside the triangle, so would not be suitable for division. What I described above is the median. You could also use the angle bisector for a similar division.
I explored these kind of subdivisions, and focused on the symmetrical ones, which create beautiful patterns. Would appreciate if you could check them out. You can find a video on my channel, which has a link to the article I wrote.
(unfortunately, I cannot directly share links, my comment will be hidden)
@@josvromans The "3000 Tiles with Triangle Subdivision Patterns" video is especially cool! Nicely done!
I would love to hear someone talking on how to subdivide triangles on the surface of a sphere and other noneuclidian planes.
this plays a major role in Synergetics. so much more to be said about it...
regarding tetrahedron - didn't check it myself just yet - will definitely do that in a while - so that math rule yields just one specific additional edge in the middle instead of two possible edges? how is so? why one over the other one? and why not both - which actually breaks the subdivision?
After converting the coordinates to their cumulative sums, which is the first step that you need to do before taking the differences, the order of the coordinates starts to play a role. There is then only one pair of edge coordinates that results in just -1's or +1's.
The edgewise subdivision method gives me goursat’s lemma vibes
I'm sure Matt Parker can come up with a previously unknown subdivision
Does it contain a parker square in the middle?
@@Krekkertje It'd be a parker triangle
It would be a "Parker" subdivision. It almost works.
Aww. Thought he was gonna make a Triforce...
for dimension of 3 also interesting question how to "combine" edges into tetrahedrons.
The thumbnail took me for a ride.
See a new numberphile video showing how to best make a triangle look like an arcane demon summoning symbol.
Go over the video and all I see in the video preview is a dimly lit polygonal cthulhu with red eyes waving at me.
It'd probably be Daoloth.
I clicked on this video and watched a Doritos ad about how everybody's obsessed with triangles now.
This vid came out 4 mins ago. Who else had the same thing happen to them?
Woah lol
Ah, I love the smell of Barycentric Subdivisions in the morning...
I did this type of subdivision. I was trying to recreate Conway's polyhedron operations, but only got to subdividing and expanding.
A fresh Numberphile video on my life anniversary, that's pretty nice. :D
Happy Birthday 🎉
@@cbuchner1 Thanks.
watching this while high is very fun and good
sleep ❌️
18 minute numberphile video about triangles ✅️
The tetrahedron has several connections that were drawn that don't adhere to the rule of not having a -1 ánd a +1 in the difference between points, so that contradicts what was said.
Also, there are three internal edges that fit the rule of having a difference of 1 (ignoring the above, otherwise there will be none), so how is the single edge chosen? All three are symmetric. If you cut on all of those (you'd have to add a vertex at the intersection) the result is 8 extra pyramids, all nicely symmetric along one of their axes. For subdivision into pyramids this is not necessary, as cutting along a single internal edge as was shown is already enough to result in 4 (albeit asymmetric) pyramids, but it feels like we're missing part of the algorithm here.
EDIT: I was wrong! I forgot to first convert the coordinates to cumulative sums. If you do this, all differences only have -1 or +1 and only one of the internal edges fits the rule, amazing!
Hi Professor Welker! Hi Brady!
In the tetrahedron, after the initial step, we're left with an octahedron, and then it was said we only needed one edge to have it fall apart into four tetrahedrons, and the animation only highlighted one. Is this true? Does an octahedron divide into four by adding a single edge? And why this particular edge? Symmetry suggests there would be four edges which could do the same job, why this one? Which rule picks this one and not the other three?
It's possible I made a mistake, but I tried calculating the vectors for the tetrahedron, and it did work out for me using his rule that 2 of the 3 possible vertex joins (not 4 possible btw, only 3) through the interior of the tetrahedron don't meet the criteria, so only 1 does. And the asymmetry did seem peculiar to me at first, but I think it's explained by the fact that the number we're calculating is inherently asymmetric, because we're calculating it from left to right and therefore there's a preferred direction geometrically too.
@@Alex_Deam The matching pairs of coordinates that could be used for the white line in the tetrahedron are:
(1 1 0 0) and (0 0 1 1)
(1 0 1 0) and (0 1 0 1)
(1 0 0 1) and (0 1 1 0)
None of these follow the rule (when subtracted they all produce a mix of 1 and -1) and I don't see any compelling reason to choose one over the other.
I'm not sure whether he said either way whether this extra line did or did not have to obey the rule. I think, though, that it was at least implied that putting in all the lines that *do* obey the rule leaves you with a central octahedron.
@@kevinmartin7760 You forgot something that I also forgot the first time I tried to do this mentally - once you get the coordinates, you have to do that thing where you take the sum of the first digit, sum of the first two digits, sum of the first three digits, sum of all 4 digits. This creates 1222 and 0012 (doesn't work because of difference of 2), 1122 and 0112 (this is the one that Does work), and 1112 and 0122 (doesn't work because both 1 and -1 present).
I also got hung up for a while on the octahedron splitting along that single edge bit. The thing is that you can't think of it as an existing octahedron that you're cutting into the tetrahedrons. You really have to imagine the octahedron not being there already (because it isn't really, there's no reason to wait to create the extra edge), and instead just consider what the lines including the "new" white one create. If you spend some time with the visualization at 13:56, it helps you realize why they are tetrahedrons.
@@ceegers You're right, I forgot that part. It is the order of this summation that picks a particular diagonal for the octahedron.
Looking forward to a Bradycentric subdivision of triangles that reveals the hidden secrets of the universe!
A nice bath of congruent triangles
It's not clear to me why going into 3-d to find the new points in the edge subdivision is necessary. Aren't the edges just split into r equal parts? And in the resulting new grid, aren't the new lines all parallel to one of the edges or the original triangle? Or is that only the case for equilateral triangles?
going to integer coordinates instead of calculating distance is way faster for a computer
@wiseSYW huh. Cool 😎
7:20 Or you just check whether the difference of (vector between) two vertices also occurs in the original simplex. That's very easy to do because the original simplex' vertices are just the unit vectors, and in a simplex any two vertices are joined by an edge.
My guy sounds like Gru, I love it.
Question 🙋 when starting out the edgewise subdivision why aren't the location of the equilateral triangle X1, Y1, Z1 ?? AND when S
starting to explain the subdivision of a 3D triangle where do the point locations come from? Like where is the 4th digit coming from? Isn't the XYZ format already 3D
Ooooh! Okay! I forgot about the location of the point is written with three numbers 😅 I thought he was writing the unit location like each axis is a ruler "1 inch" in each of the three positive direction.... Like a number line, my bad. 😂
1:20 I didn't remotely understand what that meant.
One day there will be so many things named after Brady, future generations will think he was a great master of science/maths.
Is one of these subdivisions (more) applicable to a Triangular Bezier Surface Patch? ISTM the baricentric or even just a radial division (adding just the center point and connecting all points) because you'll need to re-parametrize along each edge.
Question - Which method of subdividing triangles and tetrahedrons, will lead to the maximum number of triangles or tetrahedrons, irregular or otherwise in shape?
only at 3:20 so far but scanning ahead on the scrub bar, no mention of Loop Subdivision?
In the high school halls, in the shopping malls, be cool or be cast out.
In the basement bars, in the backs of cars, conform or be cast out.
I like the edge-deciding trick for the edgewise subdivision. That'll give me something to think about for a while.
I've got some vague ideas about graph adjacency and Grey codes. I wonder if I can nut it out before I give up and go off reading?
What might be interesting, and probably has implications in probability, would be a random subdivision. Pick a point in the interior of the triangle at random. Now pick a random point on an edge and connect that to the opposite vertex. You have a subdivision of the original triangle into two parts (with probability 0 that you picked the point which makes the two parts of equal area). What are the odds that the interior point lies in the larger subtriangle? The smaller? (On the line seems likely to be a probability 0 again, but I can't prove that rigorously.)
Ah yes Bertrand's paradox
@@hrysp you know, I wasn't thinking of that, but yeah, it would apply. You would need to define what "pick at random" means to get definite answers. I was more proposing a class of problems or an area for work. You could definitely vary how you pick at random, or change the distribution of your picks (say, "randomly" pick, but it's biased along a Bell curve).
So I went for uniform distributions all around. We'll call the interior point p, and the point on the edge x. So, the probability that p lies within a certain area is proportional to that area, and the probability that x lies on a certain interval is proportional to the length of that interval. Using such assumptions, the probability that the interior point lands in the smaller sub-triangle is, I believe, 1/4.
For computer graphics, there’s Catmull-Clark subdivision.
this video kept reminding me of the computerphile video where professor brailsford explains error correction
Am I correct in thinking a sphere-surface can not be divided in more than 20 IDENTICAL perfect-symmetry curved "areas"?
What do you mean by "perfect-symmetry"? If you mean like a regular polygon, then yes: if a polyhedron's faces are identical and regular, then it's a Platonic one, and the one with most faces is the regular icosahedron.
@@rosiefay7283 Yeah, sorry, english is not my native language, and yes, this is exactly what I was thinking. Take the icosahedron and inflate it so it becomes a ball/sphere. I've been thinking about this for a long time in context of (world) maps for video games.
Can't wrap my head around the fact that there doesn't exist any geometrical solution that leads to more than 20 plots/areas. If you divided one of the 20 'triangles' into four triangles I find it fascinating that these four are not similar!
you'd have a lot of fun with Buckminster Fuller's Synergetics
@soilnrock1975 If you're willing to use non-regular polygons, you can get up to 120 (dysdiakis triacontahedron). If you want all the sides the same length, you can get 30 (rhombic triacontahedron).
@@HaileISela I love Buckminster-Fuller. And also M. C. Escher. And Roger Penrose :-)
is the edgewise subdivision the delaunay triangulation of the set of points that are used?
Very nice work on the animations
Dudes been messing with to many triangles, sliced his finger.
5:10 This point 1,1,1 in the middle doesn't seem to be in the same plane as the larger triangle. Is this correct?
Fielding logic! This was an ingress staple!
Superficially confirms why Blender uses quads for modeling which are subsequently converted to triangles.
brady triangles could like be a subdivision of a simplex in dimension 2308 and have to subtract to 1 and zero in base 4261. i imagine a verry fun extra for numberphile 2
Who did that monster visualization? That was slick!
Family of Zoidberg..
@@pissfilth It looked more like Cthulhu to me.
Well, Zoidberg is essentially Pink Cthulhu...
If we’re doing Brady subdivisions, we also need to do Parker subdivisions.
all these triangles make a cube - Mr.Popo - DBZ abridged.
Want to learn triangle subdivision? Play "Ingress".
yeah that game is based no doubt
I was left confused about what the point of this was. Could there not have been more context supplied from the beginning about what we were looking at?
Maybe a subdivision that divides the triangle into smaller triangles of the same fractional area while minimizing the change in angles of the new triangles.
?: N new triangles, each with 1/Nth the area and all new triangles are similar to the original. For what triangles and N is this possible?
That appen if you use the baricenter of the simplicial to build the smaller simplicials.
Anyone else hearing Rush's Subdivisions in their head throughout this? "In the high school halls..."
What was the point of raising the triangle into the third dimension? It seemed to only complicate things unnecessarily.
It was so each vertex would be along a different axis, 3 points-> 3 dimensions. (That's also why the tetrahedron was in 4d space) it's less about being in 3d Spaceᵗᵐ and more just needing 3 axes
Brady subdivision? Sounds good, I guess. Now, a Parker subdivision, for that one I would pay good money!
(Obviously for the Parker subdivision, we end up with a bunch of triangles and a square somewhere, with no obvious way to fix it.)
Brady, did you go to Germany to film this? Or was the professor in England (or the US)?
We can apply it another shapes as well as 3D shapes.
Expanding triangles rules.
Expanding angles rules.
Now do it on a dodecahedron.
At a time general constant don't work what is that time?
Expanding angles
Can you do a video on Maxwell Kinematic Coupling.
Nice ceiling lights.
2:57 - Fhtagn! 🐙
Good Video
Building the first ever glass telescope.
The volume is quite low
Tri-force subdivision!
Call me simplistic, but why not just take the midpoints and connect them?
That’s the barycentric algorithm
yes
how quad remeshing works?
And then someone's going to try to generalize it. This is very interesting.
Connect point and lines we get results.
Is the clock in the background at 10:10 on purpose?
The animation between 14:57 & 15:17 is brought to you by Cher.
I spent way too long assuming the octahedron was subdivided into regular tetrahedra. I’m a little mad
This must be a school in Denmark.... so many Louis Poulsen lamps
Try this with a dodecahedron