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PROFESSOR: All right.

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So, lecture 13 was
about a lot of things.

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We had algorithms for the
carpenter's rule theorem.

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We had three of those.

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Let's see, there's
the CDR strain

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of pseudo-triangulations,
which we're

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going to talk about
a bunch today.

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There's the energy method, which
we'll talk about a little bit.

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Then there was locked trees.

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I'll give you-- there's one
new result in locked trees

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I'll tell you about.

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And then we had the last topic,
which was-- cheat-- oh, right.

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Last topic for today
is four dimensions,

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which wasn't covered in class.

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And what we did talk
about in class is 3D.

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Essentially no updates
there, although open problems

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are still open.

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Fairly challenging.

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But 4D we didn't get
to talk about at all.

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And that's not too hard,
how to unfold any 4D chain.

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So that'll be fun.

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Also 5D chains we'll get--
and 6D and 7D, but not eight.

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Any dimension.

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First question for
today is why do we

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care so much about
expansiveness?

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And this is mostly--
mostly, I want

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to reference next lecture,
which will use expansiveness.

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But, indeed, the original
point of expansiveness

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was just a convenient way
to ensure self-intersection.

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Essentially, before the
idea of expansiveness,

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we had this idea of a
really complicated chain.

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that's not very complicated
but for the carpenter's rule

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problem, we had this
idea that, well,

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how do we distinguish
between a linkage

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that is locked within Epsilon?

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However you draw it, if
it's not self-touching,

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it's going to
jiggle a little bit.

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Versus something they
can eventually unfold.

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And there was no good
way to distinguish

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between just jiggling a little
bit and unfolding all the way.

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Which was frustrating,
and expansiveness gave us

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a way to guarantee that.

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Not only did it avoid
self-intersection,

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which is nice.

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If you expand, you
don't self-intersect.

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But it also gave a way
that things would actually

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be rigid if it was
impossible to do.

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And so you couldn't move at all.

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That reduced it to a rigidity
theory problem or, really,

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a tensegrity theory problem.

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Which was kind of what
this problem needed.

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That's why it was
open for 25 years

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before we could prove
it with tensegrities.

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So, that's one answer.

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But, in fact, now that we
know expansive motions exist,

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they're really handy.

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They let us prove that the
energy method worked, for one.

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And next class we're going
to use them to add thickness

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to our edges, add polygons or
regions attached to every edge.

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And if you have an
expansive motion,

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it turns out under
certain conditions,

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this thing will still work.

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Whereas with non-expansive
motions like those

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produced by the energy method
it won't work in general.

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So initially it was
convenience, but it turns out

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to be handy mathematically.

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Two, we've seen one example
just proving the energy method

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That's actually our next topic.

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We will see another,
next lecture.

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But a few people asked about
the proof of the energy method,

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so I wanted to review it.

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How do we know we don't get
stuck in a local minimum?

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So this is an obvious
question to ask

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if you're familiar
with gradient descent.

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If not, here's a
kind of picture.

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Imagine you have some
energy landscape.

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In reality, this
is a surface drawn

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over n times d
dimensions, but I will

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assume n times d equals
1 because I can draw it.

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So here's configurations.

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And then on the x-axis, or in
general all the other axes.

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And then the y-axis, which
will always be one thing,

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is the energy function.

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So sometimes the energy is high.

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At some points it's going to
actually shoot off to infinity.

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That's where you'd
self-intersect.

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Remember, the energy function
was sum over all edges.

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The w sum over all vertices.

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U of 1 over the distance
between u and dw.

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So when a distance goes to 0,
the energy shoots to infinity.

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Those are the points
we want to avoid.

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So those are bad things.

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But if we start at some
non self-intersecting

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configuration, we'll
have some finite energy.

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And the idea is, well, if we
followed the negative gradient,

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that will decrease energy
as fast as possible.

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And from here it's going
to basically follow

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this trajectory.

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And it would get stuck
in this little well.

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This is called a local minimum.

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But this local minimum
might be lower.

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And so the worry is that
it depends where you start.

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If I start over here,
I'll fall down there.

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But if I start in
this kind of well,

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I'll end up at this kind
of accumulation point.

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And what I really want
is maybe some kind

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of global energy minimum.

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That's the usual issue
with gradient descent.

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That's why people don't like to
use gradient descent sometimes.

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Because it gets stuck.

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Hey, Jason.

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However, with this
particular setting

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we actually proved that
the energy method does not

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get stuck in a local minimum.

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And I'll just review
that proof because it

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went by quickly in lecture.

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So the idea in our setting is
we know the carpenter's rule

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theorem.

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Carpenter's rule
theorem tells us

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that any configuration
where we're not done--

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and I'll start just by thinking
about a single chain, maybe

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an open chain or a closed chain.

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Any chain that is not
already straight or convex,

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not only can be
unfolded, but has

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an expansive motion
to unfolding.

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Expansive motion
tells us that if we

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look at any distance between
a vertex and an edge,

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this increases.

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That's not how we
define expansive.

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We define it between
every pair of vertices.

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But it turns out
if you have-- this

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is actually something we needed
for non-self-intersection.

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If you have a vertex and an
edge, and you know that-- well,

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this edge stays fixed
length because it's a bar.

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We know that these
distances increase.

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And this distance
increases by expansiveness.

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Then, in fact, this point
must go away from this bar.

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You can draw an ellipse.

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If you held these fixed, it
would move along an ellipse.

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Or held the sum fixed.

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But you've got to go
outside that ellipse.

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This guy's got to get farther
away from-- every vertex,

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it has to get farther
away from every bar

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if the pairwise vertex-vertex
distances increase.

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So if you take this
expansive motion,

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you know that it is an
energy-decreasing motion.

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So what that tells you is
that any configuration that's

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not already straight
or convex has

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an energy-decreasing motion.

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So this tells you, you never
get stuck in a local minimum.

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Because what's a local
minimum in one dimension?

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It's a place where you
cannot go down from here.

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And those are the places
we're worried about.

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And the claim is the
only places where

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you don't have a downhill motion
are places where you don't have

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expansive motions, and
therefore the configuration

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must already be
straight or convex.

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And then you're done.

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So you don't have to
worry about getting

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stuck with the energy method.

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There is one tricky
part which is,

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if you have a linkage that
consists of multiple pieces--

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maybe you have some open
chains, some closed chains,

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you can have some
guy stuck in here.

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They won't necessarily
straighten out,

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but-- there's a couple kinds
of things you could do.

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You could unfold this chain,
you could unfold this chain,

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or you could just let them
fly away from each other.

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All of those things
will decrease energy.

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And so the worry here would
be that these two linkages

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fly apart and never
really unfold locally.

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That would be bad for us because
we're always decreasing energy.

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There's always a
decreasing flow,

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but we may not actually finish.

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We might not finally make this.

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Our goal in this case, remember,
is to make it outer convex.

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Meaning that all the
outermost chains are straight.

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All that outermost
polygons become convex.

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So we want to argue this.

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And I'm just going to
sketch the proof here.

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Essentially, when you
have multiple components,

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once they're very,
very, very far away,

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their contribution
to energy, just

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by all the pairwise distances
across between two components,

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will be very, very tiny.

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And the claim is actually
the energy method

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won't get you all the way to
an outer-convex configuration,

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but it will get you within
some Epsilon of outer convex.

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You get to choose
whatever Epsilon you want.

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And so when these guys
are far enough away,

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their contribution to energy
is much less than Epsilon.

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Once they're, like, n squared
divided by Epsilon away

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or something, the
contribution between them

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will be far less than Epsilon.

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So, what really
matters in energy

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is all the distances
inside the linkages, inside

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the components.

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And we're following the gradient
motion, which decreases energy

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as fast as possible.

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And so you can prove
that it's better--

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if you want to decrease
energy as fast as possible,

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it's better at some point
to unfold the pieces.

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Not to just let them
continue flying apart.

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So waving my hands
a little bit there.

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But if you know that you're
at least Epsilon away

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from being done,
then the energy will

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be fairly high within
the components.

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And therefore it's worth
unfolding the components.

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But it's especially
clear for one component.

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All right.

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Any questions about that.?

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That's why the
energy method works.

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The next set of
topics is related

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to pointed
pseudo-triangulations,

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which are indeed very cool.

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There's dozens of
papers about them.

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They've spawned their
whole little sub-field

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of computational geometry.

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We touched on them a
little bit in lecture,

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but there's a lot
more to say about it.

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In particular, why they even
work for the carpenter's rule

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problem.

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I didn't make clear,
but I would like to.

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So a couple things about them.

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They were actually
originally invented

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in 1994 or the first
used, I guess, in 1994.

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At this point, they were
called geodesic triangulations

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for a particular reason.

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And so for the carpenter's
rule problem, we had a polygon

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and then we
pseudo-triangulated the inside

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and the outside of the polygon.

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Here we're just
pseudo-triangulating

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the inside.

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In that setting, they're
call geodesic triangulations.

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And it's for a particular data
structures problem, actually.

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This is sometimes
covered in 6851,

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which is Advanced
Data Structures.

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You can check out online.

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The goal is to
decompose this polygon

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in a kind of balanced way.

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So first you choose
a pseudo-triangle

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that's kind of most in
the middle of the polygon.

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Then you recurse on the sides.

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Actually number one is this one.

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And if you do this,
it turns out if you

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want to walk from any
point in this polygon

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to any other point
in the polygon,

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00:11:54,500 --> 00:11:57,240
you only visit log
n pseudo-triangles.

249
00:11:57,240 --> 00:11:58,890
Just a nice, small number.

250
00:11:58,890 --> 00:12:02,330
And this is a particular problem
called ray shooting, where

251
00:12:02,330 --> 00:12:07,620
you imagine you have a laser
to shoot in some direction.

252
00:12:07,620 --> 00:12:10,360
The polygon-- you want to know
when it exits the polygon, when

253
00:12:10,360 --> 00:12:13,450
it hits the boundary
and where it hits.

254
00:12:13,450 --> 00:12:15,975
One way to do that is to
walk through the pieces here,

255
00:12:15,975 --> 00:12:16,850
the pseudo-triangles.

256
00:12:16,850 --> 00:12:19,095
And if you only have to
walk through log n of them,

257
00:12:19,095 --> 00:12:21,470
it turns out walking through
a single triangle only takes

258
00:12:21,470 --> 00:12:22,200
log n time.

259
00:12:22,200 --> 00:12:24,120
Total amount of time
is log squared n.

260
00:12:24,120 --> 00:12:28,720
So this is a pretty good way
to do ray shooting in polygons.

261
00:12:28,720 --> 00:12:31,890
Which is nice.

262
00:12:31,890 --> 00:12:33,490
So that's where
they originally come

263
00:12:33,490 --> 00:12:38,980
from but they weren't used
very much until later.

264
00:12:38,980 --> 00:12:43,420
So let me tell you some
nice properties about them.

265
00:12:43,420 --> 00:12:51,010
Pseudo-triangulation
has 2n minus 3 edges

266
00:12:51,010 --> 00:12:55,140
and n minus 2 pseudo-triangles.

267
00:12:57,955 --> 00:12:58,455
Always.

268
00:13:01,640 --> 00:13:04,110
I should say, if you draw a
pointed pseudo-triangulation,

269
00:13:04,110 --> 00:13:07,900
remember pointed
means at every vertex

270
00:13:07,900 --> 00:13:09,110
of the pseudo-triangulation.

271
00:13:09,110 --> 00:13:11,000
You have some angle
which is-- you

272
00:13:11,000 --> 00:13:13,410
have a reflex angle
bigger than 180.

273
00:13:13,410 --> 00:13:16,430
This is the pointed property.

274
00:13:16,430 --> 00:13:20,500
Pseudo-triangles-- all
the faces look like this.

275
00:13:20,500 --> 00:13:22,750
They only have three
convex vertices.

276
00:13:22,750 --> 00:13:24,640
All the other
vertices are reflex.

277
00:13:27,840 --> 00:13:30,190
So this is pseudo-triangle.

278
00:13:30,190 --> 00:13:33,690
Pointed pseudo-triangulation
has both of these properties.

279
00:13:33,690 --> 00:13:35,690
And it's pretty easy to
prove this by induction.

280
00:13:35,690 --> 00:13:37,690
I won't go through
the proof here.

281
00:13:37,690 --> 00:13:39,530
It's interesting
in the way that it

282
00:13:39,530 --> 00:13:41,580
is different from triangulation.

283
00:13:41,580 --> 00:13:44,280
So n here is the
number of points.

284
00:13:44,280 --> 00:13:47,780
And I'm thinking about
pointed pseudo-triangulations

285
00:13:47,780 --> 00:13:48,510
of the points.

286
00:13:48,510 --> 00:13:57,000
Meaning, I take some
points and I add edges

287
00:13:57,000 --> 00:13:59,450
until I can't anymore.

288
00:13:59,450 --> 00:14:01,580
Subject to the pointed property.

289
00:14:01,580 --> 00:14:03,820
So I add this edge,
I'm still pointed.

290
00:14:03,820 --> 00:14:05,510
I add this edge,
I'm still pointed.

291
00:14:05,510 --> 00:14:07,400
I add this edge,
I'm still pointed.

292
00:14:07,400 --> 00:14:08,820
Can't add this edge.

293
00:14:08,820 --> 00:14:10,750
Can't really add that edge.

294
00:14:10,750 --> 00:14:19,530
Let's see if I can and this
one and this one and this one.

295
00:14:19,530 --> 00:14:20,990
Anymore?

296
00:14:20,990 --> 00:14:23,614
This one and now I'm done.

297
00:14:23,614 --> 00:14:26,280
I can tell that I'm done because
at this point, all of the faces

298
00:14:26,280 --> 00:14:28,160
are pseudo-triangles.

299
00:14:28,160 --> 00:14:30,830
They each have three
convex vertices.

300
00:14:30,830 --> 00:14:34,140
And as soon as you have that all
the faces are pseudo-triangles,

301
00:14:34,140 --> 00:14:36,447
we can show that you
can't add any more

302
00:14:36,447 --> 00:14:37,780
edges while still being pointed.

303
00:14:37,780 --> 00:14:39,910
I'm still pointed
because I only added

304
00:14:39,910 --> 00:14:42,997
edges that preserve pointedness.

305
00:14:42,997 --> 00:14:44,580
And this is a sort
of greedy algorithm

306
00:14:44,580 --> 00:14:47,369
to construct pointed
pseudo-triangulation.

307
00:14:47,369 --> 00:14:48,035
It always works.

308
00:14:50,679 --> 00:14:52,220
And we can count
the number of edges.

309
00:14:52,220 --> 00:14:54,650
I won't bother, but it
will always be 2n minus 3.

310
00:14:57,950 --> 00:15:02,070
By contrast, if you look
at regular triangulations--

311
00:15:02,070 --> 00:15:05,450
so a triangulation would
go all the way to the point

312
00:15:05,450 --> 00:15:10,230
where every face is a
triangle, like this.

313
00:15:13,520 --> 00:15:16,340
And in that case they have
more edges, obviously.

314
00:15:16,340 --> 00:15:18,380
In general they
will have-- here is

315
00:15:18,380 --> 00:15:23,530
two more edges because there
are two interior vertices.

316
00:15:23,530 --> 00:15:27,360
So, in general, it would
be 2n minus 3 plus i edges

317
00:15:27,360 --> 00:15:31,390
for a triangulation, where i is
the number of interior points.

318
00:15:31,390 --> 00:15:38,715
And it will have n minus
2 plus i faces, triangles.

319
00:15:43,920 --> 00:15:47,350
Pseudo-triangulations are quite
minimal in the number of edges.

320
00:15:47,350 --> 00:15:51,650
In particular, they are
minimally generically rigid.

321
00:15:51,650 --> 00:15:53,130
This is the number
of edges I want

322
00:15:53,130 --> 00:15:54,660
for a nice rigid structure.

323
00:15:54,660 --> 00:15:57,380
2 n minus 3 is
what I should have.

324
00:15:57,380 --> 00:16:00,910
And indeed, pointed
pseudo-triangulations

325
00:16:00,910 --> 00:16:03,690
are minimally generically rigid.

326
00:16:03,690 --> 00:16:04,620
That's the next thing.

327
00:16:08,330 --> 00:16:15,190
Minimally generically rigid.

328
00:16:15,190 --> 00:16:18,760
This was the Laman condition.

329
00:16:18,760 --> 00:16:20,410
We saw a couple
characterizations,

330
00:16:20,410 --> 00:16:22,500
but in particular,
you can fairly easily

331
00:16:22,500 --> 00:16:24,890
prove that they satisfy
the Laman condition.

332
00:16:24,890 --> 00:16:25,426
Why?

333
00:16:25,426 --> 00:16:27,800
Well, first thing to check is
they have 2n minus 3 edges.

334
00:16:27,800 --> 00:16:29,334
That, I just claim.

335
00:16:29,334 --> 00:16:30,750
Second thing to
check is that if I

336
00:16:30,750 --> 00:16:33,420
look at any subset of the
vertices, k of the vertices,

337
00:16:33,420 --> 00:16:37,340
they should only have at most
2k minus 3 edges among them.

338
00:16:37,340 --> 00:16:39,870
It should induce at
most 2k minus 3 edges.

339
00:16:39,870 --> 00:16:44,655
Well, let's look
at any k vertices.

340
00:16:49,080 --> 00:16:52,670
What are the edges
that they induce?

341
00:16:52,670 --> 00:16:54,220
Well, the thing
I know about them

342
00:16:54,220 --> 00:16:59,010
is that they will induce
a pointed set of edges.

343
00:17:03,800 --> 00:17:06,950
Because this structure without
the blue edges is pointed.

344
00:17:06,950 --> 00:17:08,994
If I remove edges
from it, it will still

345
00:17:08,994 --> 00:17:10,160
be pointed, [CHUCKLE] right?

346
00:17:10,160 --> 00:17:11,930
If you have a big
angle, you'll continue

347
00:17:11,930 --> 00:17:14,069
to have a big angle if
you remove edges from it.

348
00:17:14,069 --> 00:17:17,240
It's a matter which subset of
edges I look at, in particular,

349
00:17:17,240 --> 00:17:18,510
in the deuce subset.

350
00:17:18,510 --> 00:17:20,290
It will still be pointed.

351
00:17:20,290 --> 00:17:25,839
If I have a pointed set of
edges on some k vertices,

352
00:17:25,839 --> 00:17:27,920
I can use that greedy
algorithm to finish

353
00:17:27,920 --> 00:17:28,961
the pseudo-triangulation.

354
00:17:32,660 --> 00:17:40,340
As long as you're
pointed, you can add edges

355
00:17:40,340 --> 00:17:45,370
until you can't
add edges, subject

356
00:17:45,370 --> 00:17:46,500
to the pointed constraint.

357
00:17:50,880 --> 00:17:53,239
At all times
preserving pointedness.

358
00:17:53,239 --> 00:17:55,280
This is what I did to
generate this triangulation

359
00:17:55,280 --> 00:17:55,946
the first place.

360
00:17:55,946 --> 00:17:57,780
If I take some
subset of k vertices,

361
00:17:57,780 --> 00:18:00,380
I can do it again
on those k vertices.

362
00:18:00,380 --> 00:18:04,170
When I'm finished, I will
have a pseudo-triangulation.

363
00:18:09,600 --> 00:18:14,210
And that has 2n minus 3 edges.

364
00:18:14,210 --> 00:18:15,875
In this case, 2k minus 3 edges.

365
00:18:19,930 --> 00:18:22,080
So I started with some
pointed set of edges.

366
00:18:22,080 --> 00:18:22,896
I added stuff.

367
00:18:22,896 --> 00:18:24,020
I ended up with 2k minus 3.

368
00:18:24,020 --> 00:18:27,425
That means I started with,
at most, 2k minus 3 edges.

369
00:18:32,110 --> 00:18:34,721
2k minus 3.

370
00:18:34,721 --> 00:18:36,720
And so that's how you
prove the Laman condition.

371
00:18:36,720 --> 00:18:38,790
It's kind of trivial,
because pointedness

372
00:18:38,790 --> 00:18:41,510
is an inherited property.

373
00:18:41,510 --> 00:18:43,930
So if you believe in
2n minus 3, then you

374
00:18:43,930 --> 00:18:47,230
believe minimally
generically rigid.

375
00:18:47,230 --> 00:18:49,430
Cool.

376
00:18:49,430 --> 00:18:53,400
Even cooler is that the
converse is roughly true.

377
00:18:53,400 --> 00:18:56,690
In some sense, all
Laman graphs can

378
00:18:56,690 --> 00:18:58,810
be drawn a pseudo-triangulation.

379
00:18:58,810 --> 00:19:02,420
There's one catch, which is
that you need a plane area.

380
00:19:04,970 --> 00:19:22,260
Any planar Laman graph
has a pseudo-triangulation

381
00:19:22,260 --> 00:19:23,090
as a realization.

382
00:19:27,550 --> 00:19:32,520
Realization was just a way
to draw the graph in 2D.

383
00:19:32,520 --> 00:19:34,240
Now, pseudo-triangulations
are planar,

384
00:19:34,240 --> 00:19:36,520
meaning none of the
edges cross each other.

385
00:19:36,520 --> 00:19:39,260
So you definitely need a graph
that can be drawn in the plane

386
00:19:39,260 --> 00:19:40,150
without crossings.

387
00:19:40,150 --> 00:19:41,584
That's the planar constraint.

388
00:19:41,584 --> 00:19:43,500
But if you have a planar
minimally generically

389
00:19:43,500 --> 00:19:47,640
rigid graph, you can always draw
it as a pseudo-triangulation.

390
00:19:47,640 --> 00:19:49,760
So it's kind of a
converse of this theorem.

391
00:19:49,760 --> 00:19:51,510
And a sense in which
pseudo-triangulations

392
00:19:51,510 --> 00:19:56,560
are universal for planar
minimal generic rigidity.

393
00:19:56,560 --> 00:19:59,050
Which is pretty neat.

394
00:19:59,050 --> 00:20:03,860
I will not prove that
theorem, but I will give you

395
00:20:03,860 --> 00:20:06,290
a visual sense of what
the proof looks like.

396
00:20:06,290 --> 00:20:08,496
It's based on the Henneberg
number construction.

397
00:20:08,496 --> 00:20:10,370
So we know minimally
generically rigid graphs

398
00:20:10,370 --> 00:20:13,040
can always be drawn
in a Henneberg way,

399
00:20:13,040 --> 00:20:16,580
either by adding new
degree-two vertices,

400
00:20:16,580 --> 00:20:18,850
or by adding new
degree-three vertices

401
00:20:18,850 --> 00:20:21,350
and removing one of
the existing edges.

402
00:20:21,350 --> 00:20:24,190
And the claim is, you take
any Henneberg construction.

403
00:20:24,190 --> 00:20:26,810
You can do it-- you
can implement it

404
00:20:26,810 --> 00:20:28,830
in a pseudo-triangulation.

405
00:20:28,830 --> 00:20:30,620
So you start up with
your single edge,

406
00:20:30,620 --> 00:20:31,870
that's a pseudo-triangulation.

407
00:20:31,870 --> 00:20:33,250
You keep adding things.

408
00:20:33,250 --> 00:20:35,640
Preserving the fact
that at all times

409
00:20:35,640 --> 00:20:37,710
is a pointed
pseudo-triangulation.

410
00:20:37,710 --> 00:20:39,950
It's tricky, but
for example, if you

411
00:20:39,950 --> 00:20:43,860
know that you want to add this
red vertex next to these two,

412
00:20:43,860 --> 00:20:46,990
you find a place
that will guarantee

413
00:20:46,990 --> 00:20:48,970
that this is a pseudo-triangle.

414
00:20:48,970 --> 00:20:50,316
This is a pseudo-triangle.

415
00:20:50,316 --> 00:20:52,190
It will be pointed for
free because it's just

416
00:20:52,190 --> 00:20:54,700
a degree-two vertex.

417
00:20:54,700 --> 00:20:57,580
So the challenge is
preserving pseudo-triangles.

418
00:20:57,580 --> 00:20:59,800
And same for adding the
degree-three vertex.

419
00:20:59,800 --> 00:21:02,050
This is harder,
of course, but you

420
00:21:02,050 --> 00:21:04,384
need to-- you find
a little patch here

421
00:21:04,384 --> 00:21:06,050
where, if you choose
the point in there,

422
00:21:06,050 --> 00:21:08,870
you get pseudo-triangles
on all three sides.

423
00:21:08,870 --> 00:21:10,490
And also, that
removing the edge still

424
00:21:10,490 --> 00:21:13,130
preserves pseudo-triangulation.

425
00:21:13,130 --> 00:21:15,765
It's definitely not trivial to
do this, but it can be done.

426
00:21:15,765 --> 00:21:17,140
And it's kind of
neat that it can

427
00:21:17,140 --> 00:21:18,765
be done for any
Henneberg construction.

428
00:21:24,470 --> 00:21:27,727
The last thing I wanted to tell
you about pseudo-triangulations

429
00:21:27,727 --> 00:21:29,810
is why they work for the
carpenter's rule theorem.

430
00:21:29,810 --> 00:21:32,520
Why do they give
expansive motions.

431
00:21:32,520 --> 00:21:33,770
Well, they're rigid of course.

432
00:21:33,770 --> 00:21:35,103
They don't give you any motions.

433
00:21:35,103 --> 00:21:39,600
But what we claimed is that if
you remove a convex hull edge--

434
00:21:39,600 --> 00:21:49,920
so if we take a pointed
pseudo-triangulation,

435
00:21:49,920 --> 00:22:02,530
and then we remove a convex
hull edge, we of course

436
00:22:02,530 --> 00:22:04,030
get something that's flexible.

437
00:22:04,030 --> 00:22:06,570
Because we were minimally
generically rigid,

438
00:22:06,570 --> 00:22:08,160
so if you remove an
edge, you will now

439
00:22:08,160 --> 00:22:09,480
be generically flexible.

440
00:22:09,480 --> 00:22:11,605
It turns out, you're not
only generically flexible,

441
00:22:11,605 --> 00:22:14,760
you are actually flexible for
at least a little bit of time.

442
00:22:14,760 --> 00:22:18,990
So pointed pseudo-triangulation
minus convex hull edge.

443
00:22:18,990 --> 00:22:26,480
It's not only as flexible,
but it flexes expansively.

444
00:22:29,850 --> 00:22:32,870
Meaning, if you look at
all pairwise distances--

445
00:22:32,870 --> 00:22:37,300
expansively-- look at
all pairwise distances,

446
00:22:37,300 --> 00:22:39,990
they either stay the same,
they're connected by a bar,

447
00:22:39,990 --> 00:22:42,290
or they will increase.

448
00:22:42,290 --> 00:22:44,490
And this is what we wanted
for carpenter's rule.

449
00:22:44,490 --> 00:22:46,330
You could do that motion
for a little while.

450
00:22:46,330 --> 00:22:48,163
Then you might have to
switch to a different

451
00:22:48,163 --> 00:22:50,170
pseudo-triangulation
according to a flip, which

452
00:22:50,170 --> 00:22:51,410
I won't talk about.

453
00:22:51,410 --> 00:22:53,810
We already did in lecture.

454
00:22:53,810 --> 00:22:55,480
But at all times
the claim is there

455
00:22:55,480 --> 00:22:58,460
is an expansive
pseudo-triangulation motion

456
00:22:58,460 --> 00:23:00,050
by removing single edge.

457
00:23:00,050 --> 00:23:01,540
So why does this hold?

458
00:23:01,540 --> 00:23:03,000
This, it turns
out, you can prove

459
00:23:03,000 --> 00:23:07,660
using things similar
to what we know

460
00:23:07,660 --> 00:23:11,890
from our proof of the
carpenter's rule theorem

461
00:23:11,890 --> 00:23:12,645
that we covered.

462
00:23:15,322 --> 00:23:17,780
So it's kind of neat to see
the connection between the two.

463
00:23:27,520 --> 00:23:33,900
So, why is this true?

464
00:23:33,900 --> 00:23:35,890
Well, we already
argued that the thing

465
00:23:35,890 --> 00:23:40,952
is flexible without the
expansiveness constraint.

466
00:23:40,952 --> 00:23:42,910
So we want to add the
expansiveness constraint.

467
00:23:42,910 --> 00:23:46,320
So just like before, we're going
to add all pairwise struts.

468
00:23:53,500 --> 00:23:55,110
So between all
pairs of vertices,

469
00:23:55,110 --> 00:23:57,634
let's say they're already
connected by a bar,

470
00:23:57,634 --> 00:23:58,800
we add a strut between them.

471
00:23:58,800 --> 00:24:00,879
Meaning, that could
only increase in length.

472
00:24:00,879 --> 00:24:01,920
Now we have a tensegrity.

473
00:24:01,920 --> 00:24:05,030
We want to argue that tensegrity
is infinitesimally flexible.

474
00:24:08,910 --> 00:24:18,980
So, we're going to do that
using the duality just

475
00:24:18,980 --> 00:24:22,000
like CDR, just like the
proof the carpenter's rule

476
00:24:22,000 --> 00:24:23,520
theorem we saw.

477
00:24:23,520 --> 00:24:27,190
We say, OK we want to prove this
thing is flexible by duality.

478
00:24:27,190 --> 00:24:29,290
That's equivalent
to proving something

479
00:24:29,290 --> 00:24:32,060
about the equilibrium stresses.

480
00:24:32,060 --> 00:24:35,825
Now in the case of CDR
in a single chain, what

481
00:24:35,825 --> 00:24:37,700
we needed to say was
the equilibrium stresses

482
00:24:37,700 --> 00:24:40,460
were all zero.

483
00:24:40,460 --> 00:24:42,292
And then we said,
OK, if in order

484
00:24:42,292 --> 00:24:44,250
to prove the equilibrium
stresses are all zero,

485
00:24:44,250 --> 00:24:45,541
we use Maxwell-Cremona theorem.

486
00:24:45,541 --> 00:24:48,285
That's equivalent to saying all
polyhedral lifting are flat.

487
00:24:48,285 --> 00:24:49,660
And then we prove
that by looking

488
00:24:49,660 --> 00:24:51,590
at the maximum z-coordinate.

489
00:24:51,590 --> 00:24:55,250
In this case, we can't argue
that all the stresses are zero.

490
00:24:55,250 --> 00:24:56,750
It's not true.

491
00:24:56,750 --> 00:24:59,460
But if you recall
the duality claim,

492
00:24:59,460 --> 00:25:01,900
I said if you have a
tensegrity, the tensegrity

493
00:25:01,900 --> 00:25:05,310
is rigid if and only if the
underlying linkage is rigid.

494
00:25:05,310 --> 00:25:09,180
Like when you replace
struts with bars.

495
00:25:09,180 --> 00:25:11,450
Which is going to be true here.

496
00:25:11,450 --> 00:25:15,310
And there is a stress,
an equilibrium stress,

497
00:25:15,310 --> 00:25:18,710
that is non-zero on every strut.

498
00:25:18,710 --> 00:25:20,710
Equilibrium stress being
non-zero in every strut

499
00:25:20,710 --> 00:25:22,168
means that all the
struts basically

500
00:25:22,168 --> 00:25:23,970
have to stay fixed length.

501
00:25:23,970 --> 00:25:27,540
They become bars and
then you're done.

502
00:25:27,540 --> 00:25:29,250
What I need to
prove the opposite,

503
00:25:29,250 --> 00:25:30,970
I want to prove its
flexible, I need

504
00:25:30,970 --> 00:25:34,281
to find a strut where
there's no stress.

505
00:25:34,281 --> 00:25:36,030
If I find a strut where
there's no stress,

506
00:25:36,030 --> 00:25:39,650
then basically that strut can
expand and there's a motion.

507
00:25:39,650 --> 00:25:42,405
That is roughly
true. [CHUCKLE] It's

508
00:25:42,405 --> 00:25:44,030
a little bit more
complicated than that

509
00:25:44,030 --> 00:25:46,900
but I'm going to just
claim suffice as to prove.

510
00:25:55,410 --> 00:26:10,770
There's a strut that has zero
in every equilibrium stress.

511
00:26:17,060 --> 00:26:19,050
Which strut am I
going to choose?

512
00:26:19,050 --> 00:26:21,010
I am going to
choose the one that

513
00:26:21,010 --> 00:26:24,950
comes from this
convex hull edge, e.

514
00:26:24,950 --> 00:26:27,440
So I removed a convex
hull edge as a bar.

515
00:26:27,440 --> 00:26:30,310
It's going to get replaced
when I add all pairwise struts,

516
00:26:30,310 --> 00:26:32,730
by a strut.

517
00:26:32,730 --> 00:26:38,270
So I've got the vertices
on the corner here.

518
00:26:38,270 --> 00:26:41,550
One of these, I
replaced with a strut.

519
00:26:41,550 --> 00:26:43,530
The other guys are
whatever they are.

520
00:26:43,530 --> 00:26:45,320
This is a pseudo-triangulation.

521
00:26:45,320 --> 00:26:48,170
Not a very interesting
triangulation in this case.

522
00:26:48,170 --> 00:26:49,900
I'm going to add a vertex here.

523
00:26:49,900 --> 00:26:52,660
Now it's a pseudo-triangulation.

524
00:26:52,660 --> 00:26:57,080
So this guy got replaced
with a strut, for example.

525
00:26:57,080 --> 00:27:01,580
I claim this guy is
zero in every stress.

526
00:27:01,580 --> 00:27:03,820
Why is it zero in every stress?

527
00:27:03,820 --> 00:27:19,480
Well, I claim, we've
essentially proved this,

528
00:27:19,480 --> 00:27:26,880
in any equilibrium stress
non-zero stresses must

529
00:27:26,880 --> 00:27:35,970
be on or interior to
convex polygons of bars.

530
00:27:44,130 --> 00:27:47,560
OK, first believe this claim.

531
00:27:47,560 --> 00:27:50,890
The only place I have non-zero
stress is when I'm interior

532
00:27:50,890 --> 00:27:55,790
or on the boundary of a
convex polygon of bars.

533
00:27:55,790 --> 00:27:59,160
And then, so there might be
some non-zero stresses in here.

534
00:27:59,160 --> 00:28:01,060
These edges might be
non-zero stressed.

535
00:28:01,060 --> 00:28:02,740
Intuitively, what's
happening here,

536
00:28:02,740 --> 00:28:05,260
these guys are going to serve
as mountains in the lifting.

537
00:28:05,260 --> 00:28:06,750
And there's a hole.

538
00:28:06,750 --> 00:28:10,150
So you go deep into
the hole there.

539
00:28:10,150 --> 00:28:13,120
Claim is that is the picture.

540
00:28:13,120 --> 00:28:14,640
So you've got some
convex polygons.

541
00:28:17,200 --> 00:28:20,370
There's stuff in here
that's-- could be bad.

542
00:28:20,370 --> 00:28:21,200
Could be stressed.

543
00:28:21,200 --> 00:28:23,990
But the stuff outside the convex
polygons, whatever they are,

544
00:28:23,990 --> 00:28:26,600
have to be flat.

545
00:28:26,600 --> 00:28:28,564
Meaning they have zero stress.

546
00:28:28,564 --> 00:28:30,480
I'm jumping back and
forth between the lifting

547
00:28:30,480 --> 00:28:31,980
and the stresses
by Maxwell-Cremona.

548
00:28:31,980 --> 00:28:34,400
Those are the same thing.

549
00:28:34,400 --> 00:28:35,910
Now look at edge e.

550
00:28:35,910 --> 00:28:38,020
Edge e is a strut.

551
00:28:38,020 --> 00:28:40,510
The other edges here that I've
drawn, those are the bars.

552
00:28:40,510 --> 00:28:42,280
I haven't drawn all
the other struts.

553
00:28:42,280 --> 00:28:45,707
I should really use
a color for this guy.

554
00:28:45,707 --> 00:28:47,790
So this red edge-- there's
lots of other red edges

555
00:28:47,790 --> 00:28:50,040
I'm not drawing between
all pairs of vertices,

556
00:28:50,040 --> 00:28:53,010
but the only bars
are the white edges.

557
00:28:53,010 --> 00:28:56,800
And those white
edges-- well there's

558
00:28:56,800 --> 00:28:58,930
some convex polygons,
here's a convex polygon,

559
00:28:58,930 --> 00:29:00,760
here's a convex polygon.

560
00:29:00,760 --> 00:29:04,640
But none of the convex
polygons enclose e.

561
00:29:04,640 --> 00:29:06,440
Because e's on the convex hull.

562
00:29:06,440 --> 00:29:08,540
E's on the outside here.

563
00:29:08,540 --> 00:29:13,110
So it can't be
stressed by this claim.

564
00:29:13,110 --> 00:29:14,700
And so if you
believe this claim,

565
00:29:14,700 --> 00:29:16,480
then you believe
e has zero stress

566
00:29:16,480 --> 00:29:18,920
because it's not interior
to any convex bar polygon.

567
00:29:18,920 --> 00:29:21,700
Because it's a strut, not a bar.

568
00:29:21,700 --> 00:29:22,437
OK.

569
00:29:22,437 --> 00:29:23,520
Now let's prove the claim.

570
00:29:31,900 --> 00:29:36,880
It follows from the argument
we gave in, was it lecture 12?

571
00:29:36,880 --> 00:29:37,590
Previous one?

572
00:29:37,590 --> 00:29:39,465
Where we proved the
carpenter's rule theorem.

573
00:29:42,180 --> 00:29:43,960
But it's not obvious
that it follows.

574
00:29:43,960 --> 00:29:45,630
Let me explain why.

575
00:29:59,380 --> 00:30:01,220
I want to look at the
maximum z-coordinate.

576
00:30:01,220 --> 00:30:05,960
I want to look at the region
of all the points in the plane

577
00:30:05,960 --> 00:30:10,780
that, in the polyhedral lifting,
given by Maxwell-Cremona's

578
00:30:10,780 --> 00:30:14,550
theorem, they are at maximum z.

579
00:30:17,770 --> 00:30:19,310
That is some region.

580
00:30:19,310 --> 00:30:20,800
It could be two-dimensional.

581
00:30:20,800 --> 00:30:22,659
It could be one-dimensional.

582
00:30:22,659 --> 00:30:24,200
Can't be three-dimensional.
[CHUCKLE]

583
00:30:24,200 --> 00:30:25,340
It's part of the plane.

584
00:30:25,340 --> 00:30:28,870
So this region may have
some one-dimensional parts.

585
00:30:28,870 --> 00:30:33,220
It may have some two-dimensional
parts, like this whole part.

586
00:30:33,220 --> 00:30:36,100
All of this stuff might
lift to maximum z.

587
00:30:36,100 --> 00:30:40,480
It could be-- have some
non-convex parts, like this.

588
00:30:40,480 --> 00:30:40,980
Whatever.

589
00:30:40,980 --> 00:30:43,430
This is kind of generically
what it might look like.

590
00:30:43,430 --> 00:30:46,620
I guess it could have cycles
with not-filled interior.

591
00:30:46,620 --> 00:30:50,990
All these things are plausible,
except we had this lemma.

592
00:30:50,990 --> 00:30:54,410
This is the key lemma, the heart
of the proof of the carpenter's

593
00:30:54,410 --> 00:30:55,590
rule theorem.

594
00:30:55,590 --> 00:31:01,540
If you look at m and
you look at a vertex, v,

595
00:31:01,540 --> 00:31:03,560
of the boundary of m.

596
00:31:03,560 --> 00:31:06,400
I'll use this Dell
notation to mean--

597
00:31:06,400 --> 00:31:08,180
ignore the points in here.

598
00:31:08,180 --> 00:31:10,077
I want points along
the boundary here.

599
00:31:10,077 --> 00:31:12,160
And in particular I want
to look at these vertices

600
00:31:12,160 --> 00:31:12,826
in this drawing.

601
00:31:15,840 --> 00:31:19,872
Look at such a vertex, v, and
suppose that v is pointed.

602
00:31:19,872 --> 00:31:21,580
We didn't use this
terminology because we

603
00:31:21,580 --> 00:31:23,300
didn't have it at the time.

604
00:31:23,300 --> 00:31:26,470
But maybe v looks like this.

605
00:31:26,470 --> 00:31:30,040
And there is an angle
here, this reflex.

606
00:31:33,020 --> 00:31:34,450
That's a point of vertex.

607
00:31:34,450 --> 00:31:38,370
Now I'm looking here just at
the bars, ignoring the struts.

608
00:31:38,370 --> 00:31:48,955
So suppose vertex v has a reflex
angle among just the bars.

609
00:31:54,120 --> 00:32:00,730
Then the claim is locally, this
entire region must be in m.

610
00:32:05,740 --> 00:32:09,650
Must be locally at the
maximum z-coordinate.

611
00:32:09,650 --> 00:32:11,200
How did we prove that?

612
00:32:11,200 --> 00:32:12,180
Well.

613
00:32:12,180 --> 00:32:16,140
the claim is this had to
be flat because there's

614
00:32:16,140 --> 00:32:18,320
no mountains here.

615
00:32:18,320 --> 00:32:23,660
So if you think of a
reflex region and v

616
00:32:23,660 --> 00:32:25,290
is at the maximum
z-coordinate, right?

617
00:32:25,290 --> 00:32:28,110
It's on the boundary of
the maximum z-coordinate

618
00:32:28,110 --> 00:32:30,530
so it has maximum possible z.

619
00:32:30,530 --> 00:32:32,340
So there's no way you can go up.

620
00:32:32,340 --> 00:32:34,520
These edges could be mountains.

621
00:32:34,520 --> 00:32:38,934
But all the other edges
which are not drawn here,

622
00:32:38,934 --> 00:32:40,600
these blue-- there
are some other edges.

623
00:32:40,600 --> 00:32:43,892
Those are the
struts coming to v.

624
00:32:43,892 --> 00:32:45,100
Those all have to be valleys.

625
00:32:45,100 --> 00:32:47,516
They all have to lift to valleys
because stresses can only

626
00:32:47,516 --> 00:32:49,130
carry positive stress.

627
00:32:49,130 --> 00:32:51,050
Positive stress
correspond to valleys.

628
00:32:51,050 --> 00:32:53,500
These guys are all valleys
and this is at maximum z.

629
00:32:53,500 --> 00:32:56,770
You really can't use those
valleys, right? [CHUCKLE]

630
00:32:56,770 --> 00:32:58,720
You're up here at maximum z.

631
00:32:58,720 --> 00:33:01,932
If you use valleys, you go to
higher z, which is not allowed.

632
00:33:01,932 --> 00:33:03,890
So these actually all
have to be flat out here.

633
00:33:03,890 --> 00:33:05,610
Which means locally,
all this stuff

634
00:33:05,610 --> 00:33:09,380
is at maximum z That was the
proof we had a couple lectures

635
00:33:09,380 --> 00:33:10,130
ago.

636
00:33:10,130 --> 00:33:13,350
Now, once you know that
that's at maximum z,

637
00:33:13,350 --> 00:33:15,820
you know that a lot of
these things are impossible.

638
00:33:15,820 --> 00:33:18,290
Because look, here's a
reflex angle among bars.

639
00:33:21,170 --> 00:33:24,450
There might be more
bars here, but we

640
00:33:24,450 --> 00:33:27,230
know that this is pointed.

641
00:33:27,230 --> 00:33:29,800
Whatever's missing here,
there's some reflex angle.

642
00:33:29,800 --> 00:33:32,050
We're assuming we have a
pointed pseudo-triangulation.

643
00:33:32,050 --> 00:33:33,175
And then we remove an edge.

644
00:33:33,175 --> 00:33:34,160
It's still pointed.

645
00:33:34,160 --> 00:33:36,314
So that means all of
this has to be in m.

646
00:33:36,314 --> 00:33:37,230
Is it in this picture?

647
00:33:37,230 --> 00:33:39,240
No, contradiction.

648
00:33:39,240 --> 00:33:41,029
So you cannot have
one-dimensional parts

649
00:33:41,029 --> 00:33:43,070
because wherever you have
a one-dimensional part,

650
00:33:43,070 --> 00:33:46,100
there's a reflex
angle that's not in m.

651
00:33:46,100 --> 00:33:46,767
So that's bad.

652
00:33:46,767 --> 00:33:48,850
In fact, they all have to
be two-dimensional parts

653
00:33:48,850 --> 00:33:49,870
like this.

654
00:33:49,870 --> 00:33:51,730
Can it be a two-dimensional
part like this?

655
00:33:51,730 --> 00:33:56,186
No, because here's a reflex
angle that is not in m.

656
00:33:56,186 --> 00:33:57,400
Contradiction.

657
00:33:57,400 --> 00:34:00,710
So the only situation
you can have is a convex

658
00:34:00,710 --> 00:34:04,830
polygon with the
interior all in m.

659
00:34:04,830 --> 00:34:06,710
Sorry , wrong.

660
00:34:06,710 --> 00:34:07,920
The reverse.

661
00:34:07,920 --> 00:34:09,840
The exterior should all be in m.

662
00:34:13,840 --> 00:34:16,580
Because, you have all
these reflex angles.

663
00:34:16,580 --> 00:34:18,570
That has to contain
the pointed part.

664
00:34:18,570 --> 00:34:22,335
And locally, by this
lemma, it has to be in m.

665
00:34:22,335 --> 00:34:24,230
So all this stuff has
to locally be in m.

666
00:34:24,230 --> 00:34:26,920
The only way is for
all of that to be in m.

667
00:34:26,920 --> 00:34:27,996
Inside, we don't know.

668
00:34:27,996 --> 00:34:29,370
Could be stresses
there, and this

669
00:34:29,370 --> 00:34:31,190
could be a hole
where you go deeper.

670
00:34:31,190 --> 00:34:33,889
All of this is at the
maximum z-coordinate.

671
00:34:33,889 --> 00:34:35,080
The boundary, here.

672
00:34:35,080 --> 00:34:36,250
The inside, who knows.

673
00:34:36,250 --> 00:34:38,340
Could go down.

674
00:34:38,340 --> 00:34:39,929
So you can have
stresses interior

675
00:34:39,929 --> 00:34:42,900
to convex polygons
in bars, but not

676
00:34:42,900 --> 00:34:47,040
exterior to the convex polygon.

677
00:34:47,040 --> 00:34:49,489
M is either everything,
and there's no boundary,

678
00:34:49,489 --> 00:34:50,761
and then there's no stress.

679
00:34:50,761 --> 00:34:52,219
Or it's not
everything, and then it

680
00:34:52,219 --> 00:34:55,440
will have to have stresses only
interior to convex polygons.

681
00:34:55,440 --> 00:34:57,950
And that's what we're claiming,
is that stresses can only

682
00:34:57,950 --> 00:35:00,080
be interior to
convex bar polygons.

683
00:35:00,080 --> 00:35:03,950
And in particular then,
e is not stressed.

684
00:35:03,950 --> 00:35:05,170
The end.

685
00:35:05,170 --> 00:35:08,020
This is definitely harder
than the case we talked about

686
00:35:08,020 --> 00:35:10,201
for carpenter's rule theorem.

687
00:35:10,201 --> 00:35:11,700
When you have a
single chain, you're

688
00:35:11,700 --> 00:35:13,741
just trying to prove
there's an expansive motion.

689
00:35:13,741 --> 00:35:17,320
It's a lot easier than this
because m is much simpler.

690
00:35:17,320 --> 00:35:20,740
It could basically only
be a path or cycle.

691
00:35:20,740 --> 00:35:24,940
You only have a single chain.

692
00:35:24,940 --> 00:35:27,546
Here m could be more
complicated, but in the end,

693
00:35:27,546 --> 00:35:28,920
it's actually
pretty darn simple.

694
00:35:31,590 --> 00:35:32,720
Any questions about that?

695
00:35:35,702 --> 00:35:37,690
Cool.

696
00:35:37,690 --> 00:35:40,960
Perfectly clear?
[CHUCKLE] Definitely

697
00:35:40,960 --> 00:35:44,605
a bit complicated, but nice.

698
00:35:44,605 --> 00:35:46,230
Now, it turns out
pseudo-triangulations

699
00:35:46,230 --> 00:35:49,124
are really at the core
of expansive motions.

700
00:35:49,124 --> 00:35:51,040
I'll just mention one
more theorem about them.

701
00:35:55,350 --> 00:35:58,250
So you may recall
I mentioned, if you

702
00:35:58,250 --> 00:36:01,100
look at the space of
infinitesimal motions

703
00:36:01,100 --> 00:36:06,290
of a linkage or a
tensegrity, they form a cone.

704
00:36:06,290 --> 00:36:09,070
It lives in some
high-dimensional space,

705
00:36:09,070 --> 00:36:12,210
but I'll try to
draw a cone here.

706
00:36:12,210 --> 00:36:15,330
These guys go off to infinity.

707
00:36:15,330 --> 00:36:17,800
I'm drawing it in three
dimensions, I guess.

708
00:36:17,800 --> 00:36:20,370
You have some place here--
this is the zero zero

709
00:36:20,370 --> 00:36:24,010
zero motion we're nothing moves.

710
00:36:24,010 --> 00:36:26,310
And these are potential
different motion--

711
00:36:26,310 --> 00:36:29,350
each of these points corresponds
to an assignment of velocity

712
00:36:29,350 --> 00:36:32,590
vectors where you move and you
preserve all the constraints.

713
00:36:32,590 --> 00:36:35,200
If I take such a motion,
I can scale it up.

714
00:36:35,200 --> 00:36:36,840
It's still a motion.

715
00:36:36,840 --> 00:36:38,690
Motions form a cone.

716
00:36:38,690 --> 00:36:40,780
It's called a convex cone.

717
00:36:40,780 --> 00:36:43,900
Every motion can be scaled up
or down all the way to zero.

718
00:36:43,900 --> 00:36:55,220
So if we look at the cone
of expansive motions,

719
00:36:55,220 --> 00:36:57,930
it's also a cone.

720
00:36:57,930 --> 00:37:00,940
It's a cone of motions of a
particular tensegrity where

721
00:37:00,940 --> 00:37:04,830
we add all pairwise struts.

722
00:37:04,830 --> 00:37:09,430
Then, the pseudo-triangulations
correspond

723
00:37:09,430 --> 00:37:13,950
to these edges of the cone.

724
00:37:13,950 --> 00:37:15,740
These edges,
they're really rays.

725
00:37:15,740 --> 00:37:17,090
They go off to infinity.

726
00:37:17,090 --> 00:37:19,330
These things are
called extreme rays.

727
00:37:23,590 --> 00:37:25,870
Meaning, they're extreme
in a particular direction.

728
00:37:25,870 --> 00:37:29,600
Like if I want the motion
that is the most this way,

729
00:37:29,600 --> 00:37:32,800
then it will be this ray.

730
00:37:32,800 --> 00:37:35,350
They're kind of the
corners of this polyhedron.

731
00:37:35,350 --> 00:37:37,320
The edges of the
polyhedron, if you will.

732
00:37:37,320 --> 00:37:40,430
Although in higher
dimensions, it is edges

733
00:37:40,430 --> 00:37:46,580
but they're called extreme
rays in the case of a cone.

734
00:37:46,580 --> 00:37:49,350
The extreme rays in the
cone of expansive motions

735
00:37:49,350 --> 00:37:53,570
equal pointed
pseudo-triangulations

736
00:37:53,570 --> 00:37:54,420
minus one edge.

737
00:38:10,500 --> 00:38:13,400
I think I have a slide about it.

738
00:38:13,400 --> 00:38:14,850
Yes.

739
00:38:14,850 --> 00:38:18,330
So this is a paper by [? Roto ?]
[? Santos ?] [? Enstrenu ?].

740
00:38:18,330 --> 00:38:22,339
This is just a particular
example for five points.

741
00:38:22,339 --> 00:38:24,380
These are all the different
pseudo-triangulations

742
00:38:24,380 --> 00:38:25,960
minus an edge.

743
00:38:25,960 --> 00:38:28,700
In some cases, like here,
you get two triangles.

744
00:38:28,700 --> 00:38:29,690
So that becomes rigid.

745
00:38:29,690 --> 00:38:31,780
So they fill in the
whole clique to indicate

746
00:38:31,780 --> 00:38:33,880
that's a rigid component.

747
00:38:33,880 --> 00:38:36,210
But the claim is
for five points.

748
00:38:36,210 --> 00:38:38,010
These are the edges
of the expansive cone.

749
00:38:38,010 --> 00:38:39,801
Each of these guys has
an expansive motion.

750
00:38:39,801 --> 00:38:40,940
Like this guy would rotate.

751
00:38:40,940 --> 00:38:42,040
Pretty simple.

752
00:38:42,040 --> 00:38:44,800
This one would open up a little.

753
00:38:44,800 --> 00:38:46,460
Like that.

754
00:38:46,460 --> 00:38:49,440
In general, all these guys
have an expansive motion.

755
00:38:49,440 --> 00:38:51,500
Those are
characterizing the cone.

756
00:38:51,500 --> 00:38:55,530
Those are kind of the
extreme rays in this fairly

757
00:38:55,530 --> 00:38:56,480
high-dimensional cone.

758
00:38:56,480 --> 00:39:00,850
It has something like 10
dimensions if it's 5 times 2.

759
00:39:00,850 --> 00:39:03,230
And you can prove that
if you're interested.

760
00:39:03,230 --> 00:39:04,990
Read the paper.

761
00:39:04,990 --> 00:39:10,290
So, in fact,
pseudo-triangulations

762
00:39:10,290 --> 00:39:12,229
are really core to
expansive motions.

763
00:39:12,229 --> 00:39:14,020
If you wrote down the
linear program, which

764
00:39:14,020 --> 00:39:17,562
is the expansive
motions, then you

765
00:39:17,562 --> 00:39:19,020
are essentially
characterizing this

766
00:39:19,020 --> 00:39:20,686
as the polyhedron in
the linear program.

767
00:39:20,686 --> 00:39:22,260
If you know linear programming.

768
00:39:22,260 --> 00:39:23,960
And you follow the
simplex method.

769
00:39:23,960 --> 00:39:26,740
Simplex method is all about
finding where the edges are.

770
00:39:26,740 --> 00:39:28,660
And here they are.

771
00:39:28,660 --> 00:39:31,360
So you would actually, if you
implemented the linear program,

772
00:39:31,360 --> 00:39:33,550
ran it on a simplex
solver, you would

773
00:39:33,550 --> 00:39:36,250
get pointed
pseudo-triangulations

774
00:39:36,250 --> 00:39:36,940
automatically.

775
00:39:36,940 --> 00:39:37,940
They would just pop out.

776
00:39:37,940 --> 00:39:40,730
That's actually originally
how they were discovered.

777
00:39:40,730 --> 00:39:43,210
But then this is why they
were discovered that way.

778
00:39:47,530 --> 00:39:50,410
End of pseudo-triangulations.

779
00:39:50,410 --> 00:39:52,130
All right.

780
00:39:52,130 --> 00:39:54,595
Next question is have any of
the open problems been solved?

781
00:39:54,595 --> 00:39:56,480
[CHUCKLE] I get this
almost every lecture

782
00:39:56,480 --> 00:39:57,832
and usually the answer is, no.

783
00:39:57,832 --> 00:39:59,040
So I don't tell you anything.

784
00:39:59,040 --> 00:40:01,580
But this time, there
is a nice result

785
00:40:01,580 --> 00:40:06,390
which is related
to this picture.

786
00:40:06,390 --> 00:40:08,840
These are slides
from the lecture.

787
00:40:08,840 --> 00:40:12,340
We had this linear
locked tree, which

788
00:40:12,340 --> 00:40:14,070
is minimal among all
linear locked trees.

789
00:40:14,070 --> 00:40:15,260
Minimum number of edges.

790
00:40:15,260 --> 00:40:18,060
We have this
equilateral locked tree.

791
00:40:18,060 --> 00:40:19,960
All the edges were
the same length.

792
00:40:19,960 --> 00:40:21,630
Was not strongly locked.

793
00:40:21,630 --> 00:40:23,390
There was some positive
distances here,

794
00:40:23,390 --> 00:40:25,430
but it is locked.

795
00:40:25,430 --> 00:40:27,930
I didn't mention this question,
but an obvious open question

796
00:40:27,930 --> 00:40:30,140
is, can you get both
at the same time?

797
00:40:30,140 --> 00:40:32,937
Is there a linear
equilateral locked tree?

798
00:40:32,937 --> 00:40:34,770
If there were, it'd
probably be a lot easier

799
00:40:34,770 --> 00:40:36,950
to prove locked that this mess.

800
00:40:36,950 --> 00:40:41,340
So it'd be nice to get both but,
in fact, you cannot get both.

801
00:40:41,340 --> 00:40:44,710
There's this paper by a bunch
of people in the open problem

802
00:40:44,710 --> 00:40:47,400
session from two years ago.

803
00:40:47,400 --> 00:40:51,030
It finally got
published last year.

804
00:40:51,030 --> 00:40:53,390
And it proves a bunch
of things, but one

805
00:40:53,390 --> 00:40:54,890
of the things it
proves that there's

806
00:40:54,890 --> 00:40:59,260
no equilateral locked tree
that is also linear, also

807
00:40:59,260 --> 00:41:02,210
lives in a single dimension.

808
00:41:02,210 --> 00:41:06,280
And the way it proves
that, essentially,

809
00:41:06,280 --> 00:41:11,380
so you take any linear
equilateral tree.

810
00:41:11,380 --> 00:41:12,360
What does it look like?

811
00:41:18,820 --> 00:41:22,750
Well, it lives
along a line here.

812
00:41:22,750 --> 00:41:26,310
It's segmented into
equal-length chunks.

813
00:41:26,310 --> 00:41:29,780
So from a high level,
it looks like this.

814
00:41:29,780 --> 00:41:31,590
Now in reality,
there could be lots

815
00:41:31,590 --> 00:41:34,430
of edges along each
of these segments.

816
00:41:34,430 --> 00:41:36,360
And there's interesting
things happening

817
00:41:36,360 --> 00:41:38,430
inside one of these vertices.

818
00:41:38,430 --> 00:41:40,500
Could be something like this.

819
00:41:40,500 --> 00:41:45,630
That's one way to do a
tree-like structure in there.

820
00:41:45,630 --> 00:41:47,880
It could have parts
like this that

821
00:41:47,880 --> 00:41:51,410
go from side to side
without touching these guys.

822
00:41:51,410 --> 00:41:53,660
Who knows what's happening
here, but from a high-level

823
00:41:53,660 --> 00:41:57,860
perspective, it's a bunch
of segments in a path.

824
00:41:57,860 --> 00:42:00,980
The idea is this is
basically just looking

825
00:42:00,980 --> 00:42:05,149
at five of the vertices
out of all n of them.

826
00:42:05,149 --> 00:42:06,940
But if we just look at
those five vertices,

827
00:42:06,940 --> 00:42:07,856
everything looks good.

828
00:42:07,856 --> 00:42:08,960
This is in canonical form.

829
00:42:08,960 --> 00:42:11,220
Canonical form for
a tree is where

830
00:42:11,220 --> 00:42:13,280
all the edges
point to the right.

831
00:42:13,280 --> 00:42:16,830
Something like this
is nice and canonical.

832
00:42:16,830 --> 00:42:18,800
So right now this is canonical.

833
00:42:18,800 --> 00:42:22,060
What we do is look
inside the path here.

834
00:42:22,060 --> 00:42:26,010
Look for a break point where
things are not canonical.

835
00:42:26,010 --> 00:42:27,370
And then we fix it.

836
00:42:27,370 --> 00:42:29,840
So here's just an
example of that.

837
00:42:29,840 --> 00:42:33,460
Here, this is what it looks
like currently at a high level.

838
00:42:33,460 --> 00:42:36,060
Here's what it looks
like in reality.

839
00:42:36,060 --> 00:42:39,370
So imagine-- so the tree
is doing this stuff locally

840
00:42:39,370 --> 00:42:40,980
at all these vertices.

841
00:42:40,980 --> 00:42:44,020
Right now, we're coalescing this
all into one big mega-vertex.

842
00:42:44,020 --> 00:42:45,670
And then it looks like this.

843
00:42:45,670 --> 00:42:48,390
But suppose now we realize,
oh actually there's

844
00:42:48,390 --> 00:42:49,694
kind of a split here.

845
00:42:49,694 --> 00:42:51,360
There's stuff over
to the right of this.

846
00:42:51,360 --> 00:42:52,390
There's stuff over
to right of this.

847
00:42:52,390 --> 00:42:54,070
They're not directly connected.

848
00:42:54,070 --> 00:42:56,040
Then what we'll do
is pull it apart.

849
00:42:56,040 --> 00:42:57,640
Treat these as two
separate vertices.

850
00:42:57,640 --> 00:43:00,680
Increase the number of vertices
in this picture by one.

851
00:43:00,680 --> 00:43:02,930
So we end up splitting
w here, according

852
00:43:02,930 --> 00:43:05,630
to whatever is actually
happening in the linkage.

853
00:43:05,630 --> 00:43:08,410
And then we say, oh gosh, this
is not really canonical, right?

854
00:43:08,410 --> 00:43:10,040
This edge is
pointing to the left.

855
00:43:10,040 --> 00:43:12,020
These guys are pointing
to the right, still.

856
00:43:12,020 --> 00:43:13,330
But this one's
pointing to the left.

857
00:43:13,330 --> 00:43:15,788
So it looks like this picture
got a rightward edge and then

858
00:43:15,788 --> 00:43:16,810
a leftward edge.

859
00:43:16,810 --> 00:43:21,370
We fix it by just rotating
this edge 180 degrees, keeping

860
00:43:21,370 --> 00:43:24,150
at all times this thing
pointing to the right.

861
00:43:24,150 --> 00:43:26,410
So this comes
along for the ride.

862
00:43:26,410 --> 00:43:28,160
And then we will end
up with this picture.

863
00:43:28,160 --> 00:43:31,080
Now we're canonical again
with one more vertex.

864
00:43:31,080 --> 00:43:31,660
We'll repeat.

865
00:43:31,660 --> 00:43:33,826
Eventually we'll be canonical
with all the vertices.

866
00:43:33,826 --> 00:43:35,230
And we're done.

867
00:43:35,230 --> 00:43:37,980
So if you take any
two configurations,

868
00:43:37,980 --> 00:43:39,910
you canonicalize both of them.

869
00:43:39,910 --> 00:43:42,540
You will end up with
essentially the same picture.

870
00:43:42,540 --> 00:43:44,570
It doesn't matter
which route you choose.

871
00:43:44,570 --> 00:43:46,890
But it's already
known that it doesn't

872
00:43:46,890 --> 00:43:48,140
matter which route you choose.

873
00:43:48,140 --> 00:43:51,900
You can change the route
just by flopping the tree

874
00:43:51,900 --> 00:43:56,000
with one motion or a very
small number of motions.

875
00:43:56,000 --> 00:43:57,500
So if you have two
configurations,

876
00:43:57,500 --> 00:43:58,770
you canonicalize
both of them then.

877
00:43:58,770 --> 00:44:00,186
Then there's a
motion between them

878
00:44:00,186 --> 00:44:05,010
and you get a motion from one
configuration to the other.

879
00:44:05,010 --> 00:44:06,680
Therefore, there
are no locked trees

880
00:44:06,680 --> 00:44:08,395
that are equilateral and linear.

881
00:44:11,440 --> 00:44:13,930
Kind of nice.

882
00:44:13,930 --> 00:44:17,290
The last thing I want
to talk about is 4D.

883
00:44:17,290 --> 00:44:18,940
A couple people asked about 4D.

884
00:44:18,940 --> 00:44:20,800
Why is it so different from 3D?

885
00:44:20,800 --> 00:44:24,530
The essential reason is we
have one-dimensional bars,

886
00:44:24,530 --> 00:44:26,070
but four dimensions of motion.

887
00:44:26,070 --> 00:44:28,080
That's a gap of
three dimensions.

888
00:44:28,080 --> 00:44:30,860
Three dimensions is a lot.

889
00:44:30,860 --> 00:44:32,760
When you have
one-dimensional chains in 3D,

890
00:44:32,760 --> 00:44:34,134
you have a gap of
two dimensions.

891
00:44:34,134 --> 00:44:36,130
Two dimensions turns
out to not be a lot.

892
00:44:36,130 --> 00:44:38,570
[CHUCKLE] So why is that?

893
00:44:38,570 --> 00:44:40,400
Let me prove it to you.

894
00:44:40,400 --> 00:44:43,060
Here is actually an animation
of the motion, just for fun.

895
00:44:43,060 --> 00:44:44,830
This is in the textbook.

896
00:44:44,830 --> 00:44:48,590
The top row is if you are zoomed
out from the very beginning.

897
00:44:48,590 --> 00:44:51,230
It's this mess and you
basically pull out a string.

898
00:44:51,230 --> 00:44:53,660
It actually wiggles
around a lot.

899
00:44:53,660 --> 00:44:55,160
Here's what it looks
like zoomed in.

900
00:44:55,160 --> 00:44:56,710
This is the original mess.

901
00:44:56,710 --> 00:44:58,910
And you end up flopping
this over here,

902
00:44:58,910 --> 00:45:00,200
then flopping it over there.

903
00:45:00,200 --> 00:45:01,260
And it's hard to
see because this

904
00:45:01,260 --> 00:45:03,760
is a two-dimensional projection
of a four-dimensional thing.

905
00:45:03,760 --> 00:45:06,247
But eventually you pull
open the whole thing.

906
00:45:06,247 --> 00:45:07,580
So how does this algorithm work?

907
00:45:07,580 --> 00:45:08,680
It's actually very simple.

908
00:45:13,030 --> 00:45:14,610
For open chains,
it's very simple.

909
00:45:14,610 --> 00:45:16,680
Closed chains, a little
more complicated.

910
00:45:16,680 --> 00:45:20,220
Trees, also very simple.

911
00:45:20,220 --> 00:45:30,860
So unfolding 4D open chains.

912
00:45:30,860 --> 00:45:34,792
This is by Roxanna Cocan
and Joseph O'Rourke.

913
00:45:34,792 --> 00:45:36,500
Back before the
carpenter's rule theorem.

914
00:45:39,100 --> 00:45:41,380
This publication was later
but the original version

915
00:45:41,380 --> 00:45:45,550
was in 1998, I think, or '99.

916
00:45:45,550 --> 00:45:47,340
So what do we do?

917
00:45:47,340 --> 00:45:50,740
Well, what I do is
look at the end bar.

918
00:45:50,740 --> 00:45:52,710
So let's say this is
the end of the chain.

919
00:45:52,710 --> 00:45:56,190
From here it goes in some
direction, and stuff.

920
00:45:56,190 --> 00:45:59,440
What I'd like to do
is rotate this bar

921
00:45:59,440 --> 00:46:02,530
so that it extends this bar.

922
00:46:02,530 --> 00:46:04,490
So I have an
180-degree angle here.

923
00:46:04,490 --> 00:46:06,350
That's my goal.

924
00:46:06,350 --> 00:46:08,790
If I can do that, I
can fuse that vertex,

925
00:46:08,790 --> 00:46:10,490
never rotate it again.

926
00:46:10,490 --> 00:46:12,460
Treat this as a longer edge.

927
00:46:12,460 --> 00:46:15,040
Therefore the remaining
thing has n minus 1 edges

928
00:46:15,040 --> 00:46:16,590
if I had n edges originally.

929
00:46:16,590 --> 00:46:17,840
So I can just apply induction.

930
00:46:17,840 --> 00:46:19,370
In other words,
just repeat that.

931
00:46:19,370 --> 00:46:21,690
Eventually the whole
thing will be straight.

932
00:46:21,690 --> 00:46:24,100
So the challenge is,
can I make a move--

933
00:46:24,100 --> 00:46:28,230
can I fold this last
bar to extend the next--

934
00:46:28,230 --> 00:46:30,930
the second bar?

935
00:46:30,930 --> 00:46:31,840
Two issues.

936
00:46:31,840 --> 00:46:34,540
Two problems that
could happen here.

937
00:46:34,540 --> 00:46:39,080
One is that there's no
continuous motion to get there.

938
00:46:39,080 --> 00:46:41,060
The other problem is that
even instantaneously,

939
00:46:41,060 --> 00:46:43,900
if I just picked it
up and dropped it

940
00:46:43,900 --> 00:46:45,727
into the right
location, it might

941
00:46:45,727 --> 00:46:47,060
intersect the rest of the chain.

942
00:46:47,060 --> 00:46:51,590
Maybe the chain here comes
and intersects that ray.

943
00:46:51,590 --> 00:46:54,940
If it intersects that ray, I
can just jiggle the whole chain

944
00:46:54,940 --> 00:46:56,842
a little bit, and it won't.

945
00:46:56,842 --> 00:46:59,510
[CHUCKLE] Is that clear?

946
00:46:59,510 --> 00:47:02,450
Let's see, you've got
these one-dimensional bars,

947
00:47:02,450 --> 00:47:05,080
and this is a
one-dimensional target ray.

948
00:47:05,080 --> 00:47:08,610
So if a bar happens
to be on this ray,

949
00:47:08,610 --> 00:47:10,010
just wiggle it a tiny bit.

950
00:47:10,010 --> 00:47:12,720
It's going to come
off of that place.

951
00:47:12,720 --> 00:47:14,980
This will actually even
work in three dimensions.

952
00:47:14,980 --> 00:47:15,550
This step.

953
00:47:18,090 --> 00:47:22,152
There's an even easier way to do
it, is just to rotate this bar.

954
00:47:22,152 --> 00:47:23,860
If you rotate this
bar, you can basically

955
00:47:23,860 --> 00:47:25,570
point the ray wherever you want.

956
00:47:25,570 --> 00:47:29,080
And if you look at where
that bar is pointing,

957
00:47:29,080 --> 00:47:31,160
it's kind of hard
to draw the picture

958
00:47:31,160 --> 00:47:35,090
but you just have to miss all
of these one-dimensional things.

959
00:47:35,090 --> 00:47:36,970
So it's very easy to miss.

960
00:47:36,970 --> 00:47:39,480
So let's just assume for now--
we'll kind of see this again

961
00:47:39,480 --> 00:47:40,212
in a moment.

962
00:47:40,212 --> 00:47:41,920
Assume that this ray
happens to be clear.

963
00:47:41,920 --> 00:47:45,230
There maybe things that
come very close to it.

964
00:47:45,230 --> 00:47:49,030
Maybe they go on the backside
and then come on the front.

965
00:47:49,030 --> 00:47:51,910
But they never-- don't
actually touch the ray.

966
00:47:51,910 --> 00:47:57,350
Now think about where this bar
can go relative to this vertex.

967
00:47:57,350 --> 00:48:01,724
If I draw a little
sphere here, I'm

968
00:48:01,724 --> 00:48:03,640
going to draw it initially
in three dimensions

969
00:48:03,640 --> 00:48:05,410
but really it's in
four dimensions.

970
00:48:05,410 --> 00:48:08,060
So I have a little sphere.

971
00:48:08,060 --> 00:48:09,800
Where this bar is
currently, corresponds

972
00:48:09,800 --> 00:48:11,840
to a single point on the sphere.

973
00:48:11,840 --> 00:48:13,420
I'm currently here.

974
00:48:13,420 --> 00:48:17,150
Where I want to be corresponds
to another point on the sphere.

975
00:48:17,150 --> 00:48:23,080
My goal is to find a path on
the sphere that gets to the x.

976
00:48:23,080 --> 00:48:26,610
Gets to the buried treasure.

977
00:48:26,610 --> 00:48:29,250
This is essentially treating
this as a whole ray.

978
00:48:29,250 --> 00:48:31,529
So I don't even care
that it's short.

979
00:48:31,529 --> 00:48:33,320
If I just imagine it
going off to infinity,

980
00:48:33,320 --> 00:48:35,810
can I go from
position a position b?

981
00:48:39,990 --> 00:48:43,637
Now in three dimensions, this
is a two-dimensional sphere.

982
00:48:43,637 --> 00:48:45,470
And if you look at the
obstacles, the things

983
00:48:45,470 --> 00:48:47,390
I have to avoid,
those correspond

984
00:48:47,390 --> 00:48:51,259
to one-dimensional bars here.

985
00:48:51,259 --> 00:48:53,050
Which, if you project
them onto the sphere,

986
00:48:53,050 --> 00:48:55,640
correspond to great
circular arcs.

987
00:48:55,640 --> 00:48:58,080
So the worry would
be-- let me draw them

988
00:48:58,080 --> 00:48:59,575
in red, the obstacles.

989
00:48:59,575 --> 00:49:02,520
The things that will
kill you if you hit them.

990
00:49:02,520 --> 00:49:04,080
The worry would be
you have what we

991
00:49:04,080 --> 00:49:09,050
call a cage,
something like this,

992
00:49:09,050 --> 00:49:12,010
of one-dimensional
obstacles on the sphere that

993
00:49:12,010 --> 00:49:14,380
block this x-point.

994
00:49:14,380 --> 00:49:18,180
Block the target
from the source.

995
00:49:18,180 --> 00:49:19,700
And in 3D this--
if you draw what

996
00:49:19,700 --> 00:49:22,710
happens in the knitting
needles, this is what happens.

997
00:49:22,710 --> 00:49:25,520
You can't move the last link
to extend the previous one

998
00:49:25,520 --> 00:49:27,975
because one-dimensional
barriers are

999
00:49:27,975 --> 00:49:30,380
a problem on a
two-dimensional sphere

1000
00:49:30,380 --> 00:49:33,360
because 2 minus 1 equals 1.

1001
00:49:33,360 --> 00:49:36,510
But on a
three-dimensional sphere,

1002
00:49:36,510 --> 00:49:38,190
if you draw
one-dimensional obstacles,

1003
00:49:38,190 --> 00:49:41,070
they have basically no
effect on the sphere.

1004
00:49:41,070 --> 00:49:42,690
They do not
disconnect the surface

1005
00:49:42,690 --> 00:49:45,120
of a three-dimensional sphere.

1006
00:49:45,120 --> 00:49:47,650
Can you see that?
[CHUCKLE] The analogy--

1007
00:49:47,650 --> 00:49:51,350
if you just keep the difference
in dimensions the same--

1008
00:49:51,350 --> 00:49:52,964
so we have a
three-dimensional sphere

1009
00:49:52,964 --> 00:49:54,380
versus a one-dimensional
obstacle.

1010
00:49:54,380 --> 00:49:55,865
Let's go down a dimension.

1011
00:49:55,865 --> 00:49:57,240
You have a
two-dimensional sphere

1012
00:49:57,240 --> 00:49:58,820
and a zero-dimensional obstacle.

1013
00:49:58,820 --> 00:50:00,900
What's a zero-dimensional
obstacle?

1014
00:50:00,900 --> 00:50:02,250
A point.

1015
00:50:02,250 --> 00:50:05,600
So imagine you have all
these red points which

1016
00:50:05,600 --> 00:50:06,790
you can't touch.

1017
00:50:06,790 --> 00:50:09,740
And then you have the initial
position and the target

1018
00:50:09,740 --> 00:50:10,750
position, b.

1019
00:50:10,750 --> 00:50:12,290
How do you get there?

1020
00:50:12,290 --> 00:50:13,330
No problem.

1021
00:50:13,330 --> 00:50:15,420
There's tons of paths.

1022
00:50:15,420 --> 00:50:19,130
It's basically the same in four
dimensions, just harder to see.

1023
00:50:19,130 --> 00:50:24,520
You have a three-sphere, which
is the boundary of a four-ball.

1024
00:50:24,520 --> 00:50:26,930
And you have these
one-dimensional arcs on it.

1025
00:50:26,930 --> 00:50:28,960
And they just don't
block your way at all

1026
00:50:28,960 --> 00:50:30,900
because they're so
low-dimensional.

1027
00:50:30,900 --> 00:50:34,080
So that's intuitively why 4D
is easy, 5D is also easy, 6D,

1028
00:50:34,080 --> 00:50:35,030
and so on.

1029
00:50:35,030 --> 00:50:37,330
Because the obstacles
are so tiny.

1030
00:50:37,330 --> 00:50:38,830
The obstacles remain
one-dimensional

1031
00:50:38,830 --> 00:50:40,290
on this high-dimensional sphere.

1032
00:50:40,290 --> 00:50:42,320
You can just fix
one link at a time.

1033
00:50:42,320 --> 00:50:44,876
Eventually the whole
thing will be straight.

1034
00:50:44,876 --> 00:50:47,230
Ta-da, 4D!

1035
00:50:47,230 --> 00:50:50,030
Any final questions?

1036
00:50:50,030 --> 00:50:50,900
Yes?

1037
00:50:50,900 --> 00:50:56,412
AUDIENCE: Is the situation
of folding a surface

1038
00:50:56,412 --> 00:50:58,336
in four dimensions
roughly analogous

1039
00:50:58,336 --> 00:50:59,507
with folding [INAUDIBLE]?

1040
00:50:59,507 --> 00:51:01,340
PROFESSOR: Folding a
two-dimensional surface

1041
00:51:01,340 --> 00:51:02,870
in four dimensions?

1042
00:51:02,870 --> 00:51:05,060
Definitely, this
argument breaks down.

1043
00:51:05,060 --> 00:51:08,882
And so you can get
locked things, there.

1044
00:51:08,882 --> 00:51:10,840
If you'd asked earlier
I could show one example

1045
00:51:10,840 --> 00:51:13,177
we have of a kind
of locked surface.

1046
00:51:13,177 --> 00:51:15,010
I don't know if there's
actually a locked 2D

1047
00:51:15,010 --> 00:51:17,370
surface in 4D known, though.

1048
00:51:17,370 --> 00:51:18,510
I suspect there is one.

1049
00:51:18,510 --> 00:51:20,630
That could be an interesting
question to work on.

1050
00:51:23,810 --> 00:51:25,410
Think about that.

1051
00:51:25,410 --> 00:51:27,820
All right, see you Thursday.