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
