Today's video is about subdividing triangles and possibly indications
of how such subdivisions happen in higher dimensional analogues
of triangles. So we start with a triangle and we use a subdivision that is classic
in mathematics, the Barycentric subdivision. - (Brady: This could be any triangle, basically?)
This could be any triangle, but for the time being let's let's be as an equilateral triangle. We want
to subdivide the area of this triangle into smaller pieces. The barycentric subdivision
proceeds as follows: in order to subdivide it we need more vertices first. And the vertices
here we get by looking at the so-called faces of the object. So faces are the vertices, the edges,
and the whole triangle. And for each of those we take the barycentre; which is like the centre
point in a in a certain sense, the sum of the vertices divided by the number of vertices;
and for the actual vertices this is just the vertex itself; for an edge it's the midpoint
of this edge; and then we have the barycentre of the whole triangle somewhere here. And now
we need to know how to draw the edges of the new triangle, and we need a rule for that. And
the rule is that we connect two vertices of this new set of vertices if the corresponding faces
are contained in each other. So for example, we keep this edge from here to here - so this is
now a single edge - because this is the centre point of this edge which contains this vertex.
We also have an edge from here to here because this vertex is part of the whole triangle.
We have no edge from here to here because this vertex is not contained in this edge.
so we continue with that so this one is we have an edge here because this edge is contained in
the whole triangle and so on. So this is the barycentric subdivision of this triangle;
we've got six new triangles. You can continue, you can now uh start subdividing again to a
new refinement like- now you go to each of those triangles, you do the same thing, subdivide, you
can do this to each of these six. Again subdivide and you get a finer and finer set of triangles
which covers the original area of these triangles. - (You were telling me there might
be a problem with this.) - Well it depends on what purpose you are
doing that for. I mean one purpose you could do is an applied purpose, you have many triangles that,
say, try to approximate a shape - you model some monster in a movie. Your original set of triangles
is too coarse, there are too many edges in your monster and you want to have it more smooth,
and then you try to do this by having smaller and smaller triangles. Now if you do barycentric
subdivision one of the disadvantages of this subdivision is that the angles in these triangles,
they become more and more acute or obtuse. So here you have a very acute triangle and here
even more and so on. So the angles become uneven, the triangles tend to be kind of lengthy, and
that's not what you want usually if you want to model. For mathematical purposes the barycentric
subdivision is perfectly fine, there are very many applications in mathematics of barycentric
subdivisions of triangles, higher dimensional simplices, and so on. The alternative is the
so-called Edgewise Subdivision. So now, again, we need first to create a new set of vertices and the
way we create this new set of vertices here is a different one. So I will draw a little second
picture, which now is supposed to be in 3-space. These are the coordinate axes of 3-dimensional
space, and now I place exactly this triangle here in this position. So the coordinates now
of these three points are 1 in the x Direction, 0 in the y direction and 0 in the z direction;
and here I get 0 in the x direction, 1 in the y direction, 0 in this z direction and
then here have 001. And now I choose a number r and I dilate this triangle by this number r.
So let's, for example, take r equals to 3. This point now is 300, this point here is somewhere
here 030, this point here now being 003. On this new triangle there are no more points
with integer coordinates. On this line here we also have the point with coordinates 120
and the point with coordinates 210. And then there is a point somewhere here in the middle
which has coordinates 111. So this gives us our new set of points that in the original
barycentric subdivision we had obtained by taking barycentres. So let me draw these points now here;
so now for r equals to 3, so the number we have chosen, is a kind of degree of freedom for the
subdivision. And now this vertex becomes like 300, this one is 030, and this 003. So this is
these correspondents between the triangle right here and the triangle I've drawn here. And then
we draw the additional vertices: these are these integer points here from these lines copied,
and then the point in the middle 111. Okay, now again we need a rule how to connect these points
in order to form the new triangles. - (Why do we need a rule? how come
I- I) (can't just do it by eye? Like it's not- you know,
I couldn't just join all the dots?) - Yeah, actually in- if you if you want
to create a very even subdivision then prob- there's a good guess; and the good guess is
this one is a triangle, this one here is a triangle, this is a triangle and so on. That
will be the edgewise subdivision. You could as well take for example a triangle this way,
that's also a possibility. Or even a very strange triangle: like here an edge, here and here.
So you need also a mathematical rule for doing this; and the actual need for that rule is more
apparent if you go to higher dimensions, if you subdivide simplices for example,
a tetrahedra, and even higher dimensional objects. - (So as I look at this it all seems pretty
obvious but you're creating) (all this rigor because later on there'll be times when it's not
obvious what dots to join.) That's exactly true yes. So you change these coordinates slightly;
now instead of 300 you replace this by- you take the first coordinate which is 3, then you add the
first two coordinates is 0 plus 3 is 3 and then you add all coordinates which again gives 3.
You do this here too; you start with a 2, the next sum is 2 plus 1 is 3,
2 plus 1 plus 0 is 3. 1 is copied, 1 plus 2 is 3, 1 plus 2 plus 0 is 3, here the 0 is copied, 0 plus
3 is 303 and then another 3. And this one here you replace by 123; 1,1 plus 1, 1 plus 1 plus 1
and so on. And now the rule is that you take these new coordinates and you subtract them from each
other. For example you- you pick this vertex and this vertex, there you have 333 for this
vertex. You subtract 123 and the difference is - taking coordinate-wise - you take 3 minus 1
is 2, 3 minus 2 is 1, and 3 minus 3 is 0. So you get this 210. But you could also
take, for example, this vertex and this vertex; so you get 333 minus 233 which
then is 3 minus 2 is 1, 3 minus 3 is 0, 3 minus 3 is 0. The rule says do
not connect the two vertices that yield 210, but the ones that yield 100; and the reason is you
want the difference only uses 1, -1s and 0s. Indeed you want that the difference is either
a combination of 1s and 0s or a combination of -1s and 0s. So this is not okay, this is okay,
and if you change the role of the 2 then it would be -100 - and again that would be okay.
(So it's almost like using this alternative number, each pair of coordinates has like sort of a code, a)
(relationship between them. And if that relationship number - this number - is 1- contains only 1s, only 0s,)
(or -1s you join them.) - That's right. If you now apply this rule you will get the picture you
would have guessed from the beginning. This is the picture. So this yields nine triangles.
And you see they're all actually congruent, it's all the same triangles - if you had started from an
equilateral triangle. - (Yeah, and presumably this is just a perfect thing, it can't work out any other)
(way that you ever get 1s and 0s that shouldn't be joined and-) - That's right, it's an if and only if.
(And if you'd used a different number other than 3 and you had lots and lots of divisions)
(along these lines - same rule applies?) - The same rule applies, just the number of triangles increases.
So the number of triangles you get is always this number r squared. In- in general if you go
into higher dimensions, like you do this and apply this to say a tetrahedron - that's the next
dimension - so that would be a 3-dimensional object - you would get r cubed many new tetrahedra.
In general you get r to the d many new triangles or higher dimensional analogs of triangles. 1, 2, 3, 4, 5, 6, 7, 8, 9
(Works a treat. And what- and the advantage of this is we seem to get more friendly triangles do we?)
That's right, that's right. We get more friendly triangles, it's more evenly distributed, um you get-
uh you get in 2 dimensions you get only - if you start with an equilateral triangle - you only get
congruent triangles. In the next dimension you get a small number of congruent simplices, and that's
also so for higher dimensions, is you can control how many different incongruent objects you get and
you also can control the deviation of their shape. - (So you've shown me 2-dimensions triangles but)
(you've also alluded to the fact this works in higher dimensions, we could get like a simplex,)
(like a 3-dimensional object and divide that up into three- 3D pieces could we?) - That's right. I
mean the the 3D simplex, that's the tetrahedron so- - (We'll have four numbers for that will we?) - For the edgewise
subdivision we have four numbers. For simplicity let's choose here r equals to 2 otherwise we
end up with too many dots and it becomes hard to visualise. So that means that now we place - in order
to get this picture here - we place this tetrahedron in 4-space, so in a 4-dimensional space;
and with the vertices being the unit coordinate vectors and then dilate this by 2.
So we end up with this for example being 2000, this being 0200, 0020,
0002. And then we get these new vertices, like that one, for example that would
be a 1001. - (In the middle of each line) - In the middle of each line. And then because r is
small, r is 2, we don't get any additional point in the middle. Like here you get something in the
interior of this triangle; and here you get nothing in the interior of each face and also nothing in
the interior of the whole tetrahedron. If r is 3 we get some- an additional point in
the interior of every these triangles, and if r is 4 we would get a point in the middle.
But now let's apply this rule, it's already a bit more tricky. So here you do of course the same on
each face, so that's the easy rule, essentially we have seen in dimension 2; but now this is
not yet a subdivision into tetrahedra. I mean, you see here there is a tetrahedron sitting on top, a
tetrahedra sitting on the front, the tetrahedron sitting here - but then in the middle you have
this object which is actually an octahedron and you need to further subdivide it and for that
suddenly this edge becomes important, you need to connect those two vertices. And in order to get
this to subdivide into tetrahedra you draw another edge here in the middle. This new edge now serves as an
edge that subdivides this octahedron into four tetrahedron. - (When I'm doing a subdivision here)
(and I'm using your magic rule you showed me here with the- you know, transferring the coordinates)
(and then subtracting them; is it still the 1s- the 1s and the -1s I'm looking for to decide)
(when to make a join?) - Exactly so you you you make a transformation of these coordinates, the same
way we did before - you take the first coordinate then add the first two, and the first three, and the
first four - and then you subtract these new coordinates from each other. If the resulting
vector is a 0,1 vector or it's a 0, -1 vector then you connect them by an edge.
(Okay. Presumably you can have multiple 1s and, you know, it could be- have- it did it have-)
(Does it only ever have one 1 or can I have multiple 1s?) - It can have multiple 1s.
(Yeah, it's made of only 1s and-) - It's all made of only 1s but it's not permissible to have 1 and -1,
that that would not be permissible. It's just either 1 and 0 or -1 and 0.
(So for sort of a computer programmer or a modeler or an animator or someone, they would prefer to subdivide)
(their triangles using this method presumably?) - Yes, yes - (You're not one of those people, you're a pure mathematician)
That's right. - (What do you like to use? Do you prefer this barycentric? Or do you like these edgewise?)
I mean both are good and both serve different purposes. So in in mathematics, I mean the
baryentric subdivision is a is a classical object that is used in algebraic topology and serves
therefore many purposes. The edgewise subdivision also actually came from algebraic topology but has
also many applications in combinatorics, algebraic topology, and other fields. - (You've shown me two ways)
(to subdivide a triangle today, are there other ways?) - Yeah there are very many different ways. You can
just imagine like you can combine the edgewise and the barycentric subdivision; like you first
subdivide say barycentrically and then subdivide each triangle edgewise by some r. But you can also
do different things as I shown you before; I mean the the fact that were choosing this edge here
instead of this edge here, this was in order to make our triangles as even as possible, but
mathematically it's also okay to remove this edge and instead have this edge. So there are many ways;
you can do flips on these triangles on these edges to create new ways of subdividing your
original triangle. - (You were talking earlier about a practical application, you were talking about)
(rendering a monster in a movie or something, but mathematicians are doing this just purely)
(for abstract purposes, just for for research and to create new mathematics are they? There's like)
(there's just weird places you go with this.) - That's right, I mean we have many questions, many questions
are have been answered but there are also many questions that are open about these objects.
So uh the this the complete understanding of the structure of the number of faces that appear when
you do this iteratively for barycentric subdivision or edgewise subdivision has only been unravelled in the
last 20 years. And there are still open questions, even though the objects look pretty simple.
(Can there be new ways to subdivide? Or that are useful? Or do you think pretty much all the ways)
(to subdivide a triangle are known now or?) - I think you can always come up with a new way of
subdividing if you have a new question. If there can be new mathematical questions where those two
methods are not good for, and there are new new methods for subdivision needed to solve these questions.
(It's exciting, I'd love to come up with a new subdivision myself, you could call it the Brady subdivision.)
Okay I- we gladly can work on that. - (All right let's- let's go upstairs and sort it out.) - Okay, let's go
...they're superhero triangles. - These are our superhero triangles, absolutely.
I've now given you permission to put click bait thumbnails on this video. Get Captain
America over here, I can't believe it! Superhero triangles! - Which one is the
actual centre of the triangle? - There is no actual centre, it depends on your viewpoint