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PROFESSOR: So today's class we
talked about vertex unfolding,

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unfolding orthogonal
polyhedra, and something else

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which there weren't
any questions

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about-- got them already--
Cauchey's rigidity theorem.

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So we will go through
those in turn.

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A bunch of small
questions and two

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cool new things, new updates.

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So first question is
about vertex unfolding.

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Why do we fold things
in a chain and kind

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of a line, linear style?

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Would you get any
benefit out of a tree?

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The answer is chains are
easy to avoid intersection.

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That's the reason we use them
on these kinds of unfoldings.

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We could dedicate a different
slab for every piece

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and this is great.

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And incidentally-- I didn't
mention this in lecture,

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but it's obvious once you think
about it for five minutes.

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If you want general cuts-- if
you're allowed to cut anywhere

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on the surface and you want
a vertex unfolding-- then

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anything is possible because
you could take any surface

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and chainify it in the same
way we do with the [INAUDIBLE]

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sections.

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You subdivide.

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First, you triangulate.

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You subdivide each triangle
into three triangles.

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Then you can hinge it
together in a chain.

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And then you can just
lay it out like this.

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So general vertex unfolding
is trivial for this reason.

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When you're only allowed
to cut along edges,

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things get much harder.

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The only technique we have right
now is facet path unfolding.

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But of course, vertex
unfolding convex polyhedra

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is an easier version of edge
unfolding context polyhedra,

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but we have no idea
how to do that.

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So maybe it's possible.

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We already know it's not
possible with chains.

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So if it's possible, it's
somehow possible with trees,

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I guess.

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We know tree style
unfoldings can

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be helpful with
general cuttings,

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but this is not directly
relevant to vertex unfolding.

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But this is the source unfolding
and the star unfolding again.

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So yeah, who knows?

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But so far, chains are just
a useful proof technique.

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All right, next question.

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This is vertex
unfolding summarized

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in the lecture notes
on the right side.

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So we end up with a
nice Eulerian graph.

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In this case, every vertex
happens to have even degree.

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In general, two of them
might have odd degree

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and we just get a path.

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But the observation in
this question-- maybe

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I'll go to the other side
so you can read both--

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is that we visit the same
vertex multiple times,

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like we attach these two
triangles at this vertex.

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We also attach these two
triangles at that vertex.

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Is that OK?

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Yes, we define it to be OK.

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I think I briefly mentioned
it's defined to be OK.

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Why?

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I don't know.

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It's hard to imagine
little dots of paper there,

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but somehow you imagine that
they can stay connected.

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In particular, it's
the natural definition

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of a hinge dissection.

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You could add hinges
at those two places.

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But also it's
impossible otherwise.

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This was a fun little
puzzle to think about.

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But if you take-- I actually
have a slide for it.

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I don't have to draw.

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It's hard to draw
an icosahedron.

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If you take an icosahedron,
it has 20 faces.

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And the dual is a dodecahedron
so it has 12 vertices.

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So there aren't
enough to go around,

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at least if you wanted
a chain unfolding.

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Well, actually any
tree unfolding,

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if you have 20
triangles, you're going

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to need 19 vertices to
connect them together.

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And there's only 12
vertices so you can't do it.

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Is that literally true?

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I guess you could imagine,
in a general tree unfolding,

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you could use one point
to connect together

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many triangles.

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But I'm pretty sure
it's not possible

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if you're allowed
to reuse vertices.

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So that's why we allow it.

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We get a more
interesting result when

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we allow revisiting vertices.

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Could be still an
interesting question like,

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which polyhedron have
vertex unfolding if we only

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get to use the vertex once?

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But I don't know
any results on it.

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Next question is, any
progress on vertex unfolding?

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And there's one new result.

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Before I showed you, I think,
one example of a vertex

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ununfoldable polyhedron--
this guy, box on a box.

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Initially, we said oh,
this is edge ununfoldable,

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but it's also vertex
ununfoldable because this face

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here is a donut and the
five sides of that cube

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have to fit in here, and there's
not enough area for them.

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And that's true also if
you're vertex unfolding.

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No matter how this stays
attached to the donut face,

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it's got a line here,
that's not possible.

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But this is an
unsatisfying example

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because you have a face
that is not a disk.

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And in general, you're
not topologically convex.

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you can't just move the vertices
around and make this a convex

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polyhedron, not
strictly convex anyway.

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So here is a new
example with Zach Abel.

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He took this class
two years ago,

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or five years ago, I guess.

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This is a topologically
convex polyhedron,

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so you could move the
vertices to make this convex.

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That's not so obvious.

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And it is vertex ununfoldable,
which we will prove.

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This is one of the
two results for today.

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The example is a dark prism
here, a triangular prism,

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and a light triangular
prism just at right angles

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to each other.

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Then you take the
union of that solid,

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and then you take the boundary.

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So first, you can check
all the faces are disks.

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And for topological
convexity, you

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want to check that if
you look at any two faces

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like this light S1
and this dark one,

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they should only share
one edge at most.

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They should share nothing,
a vertex, or an edge.

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That's topological
convexity because that's

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how things work in convex
if the polygons were convex.

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So like here, they
share just that edge.

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So it is topologically convex.

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Let's prove that it is
vertex ununfoldable.

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So in some sense, this is a
strengthening of the witch's

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hats and stuff because this
is also edge ununfoldable.

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But it's even cooler.

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It's vertex ununfoldable.

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Let's just look
at the connections

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between the dark and the light.

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There are one, two, three,
four vertices on this side

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and four vertices
on the backside that

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connect dark to light.

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I should make some Star
Wars reference I suppose.

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All those vertices
are symmetric,

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so we can just look
at any of them.

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Somewhere, the light side has
to be attached to the dark side.

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Where could that be?

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Let's say vertex A because
it's got to be somewhere,

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and it's symmetric.

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So vertex A has three
angles incident to it.

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There's alpha-- that's
this angle-- which is big.

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It reflects.

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There's this beta and
there's this gamma.

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Beta is 90 degrees
here because this

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is supposed to be a rectangle.

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Gamma is well, let me state
what I want from these angles.

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There's some parameters
in the construction

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I haven't yet told you.

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So we want alpha to
be bigger than 270.

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Theta equals 92.

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And we want alpha plus
gamma equal to 360.

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So this is possible if you
angle these guys correctly,

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is the short version.

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I guess the most
interesting is this alpha

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plus gamma equals 360.

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Is that obvious?

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Not especially, but I'm claiming
it's true in this picture.

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We'll see it in an
unfolding, actually,

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on the very next slide.

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Yeah, let's go there.

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Still have the same picture.

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But when unfolded, it's
supposed to look like this.

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So what happens at A?

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If you want to connect the
light side which is alpha

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that's forced, you can
attach by a vertex unfolding

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to beta or gamma.

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Well, beta is impossible
because 90 plus greater than 270

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is greater than 360.

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So you would get overlap
locally at this vertex

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if you attach these even
if you're vertex unfolding.

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So you've got to
attach the gamma.

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Alpha plus gamma is
supposed to be exactly 360.

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And so vertex unfolding
doesn't buy you anything.

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In order to not
locally overlap here,

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they must fit right like that.

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Then the only other
condition we need

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is that these prisms
are long enough

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so that you get
overlap over here.

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So it's not a purely
local overlap,

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but it's very close by.

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It's the same two faces overlap.

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Here, they happen to
overlap here and here.

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If this were shorter, if
this were like this short--

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this is the length
of the prism here--

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then you'd avoid overlap.

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And that's this picture.

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So this is an unfolding of
one of the two prism halves.

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This is unfolding of the other.

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If you attach them, these
vertices as is necessary,

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you get overlap.

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Unless the prism
were short, then you

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would get this
kind of unfolding.

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So that's the proof.

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You can actually make the proof
even simpler, kind of nifty.

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Although it's less obvious
that this example exists.

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You can give the coordinates
and you can check it exactly

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00:10:05,080 --> 00:10:07,720
and it has all the
properties you want.

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00:10:07,720 --> 00:10:10,440
Each prism has been
bent a little bit.

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00:10:10,440 --> 00:10:12,650
More or less the
same example, but now

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00:10:12,650 --> 00:10:14,980
it has the feature that
alpha plus beta prime is

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greater than 360, and
alpha plus gamma prime

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00:10:17,730 --> 00:10:19,830
is greater than 360.

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00:10:19,830 --> 00:10:22,620
So you just can't
attach anything.

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00:10:22,620 --> 00:10:25,870
Even locally, you
get pure overlap.

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So easy to check if
you're a computer.

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00:10:27,960 --> 00:10:32,770
Easier to check and proof and
verify if you're a computer.

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00:10:32,770 --> 00:10:35,840
Less easy visually.

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So that's what's known.

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00:10:37,010 --> 00:10:39,910
Open problem-- which I
should probably write down--

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00:10:39,910 --> 00:10:43,810
is if your faces
are convex polygons,

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is there a vertex
ununfoldable polyhedron?

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00:10:46,247 --> 00:10:47,330
That, we still don't know.

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00:10:47,330 --> 00:10:50,460
So the witch's hat, all
the faces are triangles.

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There's no edge unfolding,
but it has a vertex unfolding.

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These guys are nice.

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00:10:55,200 --> 00:10:57,160
They're topologically
convex, but the faces

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are still very nonconvex.

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And we use that because we want
a giant angle, a reflex angle

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00:11:02,890 --> 00:11:05,140
together with something to
add up to more than 360.

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00:11:05,140 --> 00:11:07,015
If you're always adding
up two convex angles,

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you don't get that.

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00:11:08,600 --> 00:11:11,690
But maybe you could combine
it with a polyhedron that's

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not Hamiltonian, so you have
to have three things joining it

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00:11:15,400 --> 00:11:18,290
together and then you
could get more than 360?

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00:11:18,290 --> 00:11:18,960
I don't know.

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00:11:18,960 --> 00:11:22,660
It could be an interesting
problem to work on.

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00:11:22,660 --> 00:11:24,770
But so far, it is open.

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So that is vertex unfolding.

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00:11:29,910 --> 00:11:35,510
Next question is about
general unfolding

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of nonconvex polyhedra.

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00:11:36,920 --> 00:11:39,670
Still it's open whether
every polyhedron, say

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00:11:39,670 --> 00:11:42,270
without boundary, has
a general unfolding.

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00:11:42,270 --> 00:11:44,175
Still we know that all
orthogonal polyhedra

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of genus zero have an unfolding.

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00:11:46,870 --> 00:11:48,760
That remains the
state of the art.

245
00:11:48,760 --> 00:11:51,230
But we have a
better way to unfold

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00:11:51,230 --> 00:11:53,910
orthogonal polyhedra
of genus zero.

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00:11:53,910 --> 00:11:56,420
And this is a new paper.

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00:11:56,420 --> 00:11:58,390
It's not yet online
but it will be shortly.

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00:11:58,390 --> 00:12:01,140
The final version just
went to the journal.

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00:12:01,140 --> 00:12:05,860
"Unfolding Orthogonal Polyhedra
with Quadratic Refinement."

251
00:12:05,860 --> 00:12:09,040
So two of the same authors--
Mirela Damian and Robin

252
00:12:09,040 --> 00:12:11,440
Flatland from the
exponential unfolding.

253
00:12:14,750 --> 00:12:17,610
You may recall, we saw
a solution in lecture

254
00:12:17,610 --> 00:12:19,250
that has exponential refinement.

255
00:12:19,250 --> 00:12:22,120
The number of cuts in particular
can be exponential in n

256
00:12:22,120 --> 00:12:23,210
in the worst case.

257
00:12:23,210 --> 00:12:24,980
This requires you take the grid.

258
00:12:24,980 --> 00:12:28,260
You refine it, at most,
quadratic number of times

259
00:12:28,260 --> 00:12:30,840
in each dimension.

260
00:12:30,840 --> 00:12:33,520
So the number of
cuts is polynomial,

261
00:12:33,520 --> 00:12:37,420
I guess at most, and to
the seventh if you've got

262
00:12:37,420 --> 00:12:40,460
n squared by n squared
by n squared by n.

263
00:12:40,460 --> 00:12:42,210
I think it's actually
much less than that.

264
00:12:42,210 --> 00:12:46,120
Probably more like n squared,
but I need to be careful.

265
00:12:46,120 --> 00:12:49,070
So I want to cover
this solution.

266
00:12:49,070 --> 00:12:52,262
It's a simple variation
actually on the previous one,

267
00:12:52,262 --> 00:12:53,970
but also there's some
questions about how

268
00:12:53,970 --> 00:12:55,210
the exponential one works.

269
00:12:55,210 --> 00:12:57,159
I want to go through
that one more time.

270
00:12:57,159 --> 00:12:58,950
We have new figures
for it from this paper.

271
00:12:58,950 --> 00:13:02,370
So it looks a little different,
maybe a little bit clearer.

272
00:13:02,370 --> 00:13:04,130
That was our girl.

273
00:13:04,130 --> 00:13:05,320
I need my hands.

274
00:13:10,660 --> 00:13:14,155
Remember, the general idea
was to slice your polyhedron

275
00:13:14,155 --> 00:13:17,280
in one direction so you end
up with these bands, strips.

276
00:13:17,280 --> 00:13:21,260
And then you visit the
strips in a treelike fashion.

277
00:13:21,260 --> 00:13:23,380
And in particular,
at the leaves,

278
00:13:23,380 --> 00:13:27,160
like the closest to
you, you've got a band

279
00:13:27,160 --> 00:13:28,950
and then you've
got a front face.

280
00:13:28,950 --> 00:13:32,900
And so that would look
something like this little box.

281
00:13:32,900 --> 00:13:39,570
So you've got a band that
goes around-- which way is it?

282
00:13:39,570 --> 00:13:42,390
Band is this way it.

283
00:13:42,390 --> 00:13:45,180
Should be an x and z.

284
00:13:45,180 --> 00:13:46,840
And then there's
the front face which

285
00:13:46,840 --> 00:13:49,090
here is drawn on the back.

286
00:13:49,090 --> 00:13:53,350
Those are the five out
of six faces of the box

287
00:13:53,350 --> 00:13:54,660
that are actually present.

288
00:13:54,660 --> 00:13:57,450
This face that's closest
to us is actually

289
00:13:57,450 --> 00:13:59,790
attached to another band.

290
00:13:59,790 --> 00:14:03,480
You could have leaves on
this side or that side.

291
00:14:03,480 --> 00:14:04,990
So what do we do?

292
00:14:04,990 --> 00:14:09,290
We enter at S and
we exit at T. And we

293
00:14:09,290 --> 00:14:12,680
are told that the first
turn you make from S

294
00:14:12,680 --> 00:14:14,090
should be a right turn.

295
00:14:14,090 --> 00:14:18,210
And the last turn before
T should be a left turn.

296
00:14:18,210 --> 00:14:20,910
And you start in
the unfolding, you

297
00:14:20,910 --> 00:14:22,566
will start going
vertical on S, that's

298
00:14:22,566 --> 00:14:24,316
why you should turn
right because we never

299
00:14:24,316 --> 00:14:25,570
want to turn left.

300
00:14:25,570 --> 00:14:29,580
We never want to go left in
the unfolding, which is here.

301
00:14:29,580 --> 00:14:32,090
And we also want to end in
the vertical state for T

302
00:14:32,090 --> 00:14:35,020
so that we can chain
these together.

303
00:14:35,020 --> 00:14:40,270
So we started S. We're told to
turn right, which we do here.

304
00:14:40,270 --> 00:14:41,930
Then we're basically
going to zigzag.

305
00:14:41,930 --> 00:14:45,470
And if you check along this
surface, we turn right.

306
00:14:45,470 --> 00:14:47,800
We go here, up there.

307
00:14:47,800 --> 00:14:51,820
And then we turn
left, and then right.

308
00:14:51,820 --> 00:14:54,440
And then this is a
left because we're

309
00:14:54,440 --> 00:14:57,040
supposed to be
doing it from below.

310
00:14:57,040 --> 00:15:02,080
And then a right with that
red thing, then a left, right,

311
00:15:02,080 --> 00:15:06,460
down here, and a
left, and we're done.

312
00:15:06,460 --> 00:15:09,235
And the number balances
out, so we start vertical,

313
00:15:09,235 --> 00:15:10,610
we end vertical.

314
00:15:10,610 --> 00:15:16,054
And then when we're done with
all these things, we thicken.

315
00:15:16,054 --> 00:15:18,220
It's not drawn here, but
these strips get thickened.

316
00:15:18,220 --> 00:15:20,680
And you end up attaching
these fatter versions,

317
00:15:20,680 --> 00:15:22,680
but basically you're
following the same path.

318
00:15:22,680 --> 00:15:23,840
We don't thicken
it at the beginning

319
00:15:23,840 --> 00:15:25,256
because we're
actually going to do

320
00:15:25,256 --> 00:15:27,730
this many times
on the same cell.

321
00:15:27,730 --> 00:15:30,150
And then we abstract
that picture

322
00:15:30,150 --> 00:15:34,240
to this iconography which says--
this I didn't talk about much

323
00:15:34,240 --> 00:15:35,660
in lecture, which
says, you should

324
00:15:35,660 --> 00:15:38,790
start by turning right
in S, and then you

325
00:15:38,790 --> 00:15:42,170
should end by turning
left and then visiting T.

326
00:15:42,170 --> 00:15:44,640
So that's one picture.

327
00:15:44,640 --> 00:15:49,100
But it depends
where you're going.

328
00:15:49,100 --> 00:15:51,770
There are actually eight
different versions.

329
00:15:51,770 --> 00:15:53,780
I showed you this one.

330
00:15:53,780 --> 00:15:55,930
But all those combinations
have to happen.

331
00:15:55,930 --> 00:16:00,600
And the top ones are when
you're all the way forward.

332
00:16:00,600 --> 00:16:02,950
The bottom ones are where
you're all the way back

333
00:16:02,950 --> 00:16:07,420
in these slabs, so that's
slightly different.

334
00:16:07,420 --> 00:16:11,390
And then there's whether S is on
the left or S is on the right,

335
00:16:11,390 --> 00:16:13,840
relative to T. And then
there's whether you

336
00:16:13,840 --> 00:16:17,790
start by turning to
the right from S or end

337
00:16:17,790 --> 00:16:19,320
by turning to the left.

338
00:16:19,320 --> 00:16:21,170
T is always the opposite.

339
00:16:21,170 --> 00:16:24,010
But those are the
eight possibilities

340
00:16:24,010 --> 00:16:25,520
from three different choices.

341
00:16:25,520 --> 00:16:26,790
Are you far forward?

342
00:16:26,790 --> 00:16:28,380
Are you far back?

343
00:16:28,380 --> 00:16:31,550
Do you turn right from
S or do you turn left?

344
00:16:31,550 --> 00:16:33,720
And is S to the left
or is S to the right?

345
00:16:33,720 --> 00:16:35,780
That's why you have
eight combinations.

346
00:16:35,780 --> 00:16:38,280
They all look
basically the same,

347
00:16:38,280 --> 00:16:39,675
but those are your base cases.

348
00:16:42,190 --> 00:16:46,670
So this is still all
review of the old version.

349
00:16:46,670 --> 00:16:51,240
And then the last part of
the old method is non-leaves.

350
00:16:51,240 --> 00:16:53,960
So in general, you
have a band which

351
00:16:53,960 --> 00:16:55,980
is here drawn as
two separate pieces,

352
00:16:55,980 --> 00:16:58,162
but really this is one band.

353
00:16:58,162 --> 00:16:59,370
The bands are easy to unfold.

354
00:16:59,370 --> 00:17:01,450
They go straight.

355
00:17:01,450 --> 00:17:03,400
But then you have
all these children.

356
00:17:03,400 --> 00:17:05,500
So you have your parent
which is the S and T part.

357
00:17:05,500 --> 00:17:07,990
You have to start
here and end here.

358
00:17:07,990 --> 00:17:10,460
Recursively, you want to
visit this whole subtree.

359
00:17:10,460 --> 00:17:14,114
So you might have children that
are on the front side, children

360
00:17:14,114 --> 00:17:15,155
that are on the backside.

361
00:17:19,119 --> 00:17:21,990
And again, we're told we
should initially turn right.

362
00:17:21,990 --> 00:17:25,310
So if we turn right, we
visit the first thing.

363
00:17:25,310 --> 00:17:28,180
We tell this leaf or
this child, actually--

364
00:17:28,180 --> 00:17:29,850
this is not necessarily a leaf.

365
00:17:29,850 --> 00:17:32,050
We tell this child that the
next thing we want to do

366
00:17:32,050 --> 00:17:34,930
is turn to the left because
we just turned right.

367
00:17:34,930 --> 00:17:37,290
Actually, we just
turned right twice,

368
00:17:37,290 --> 00:17:40,412
so we've just gone right, right.

369
00:17:40,412 --> 00:17:42,120
We'll actually be
unfolding initially up,

370
00:17:42,120 --> 00:17:43,510
so we turn right twice.

371
00:17:43,510 --> 00:17:45,610
Now, we better
turn left and then

372
00:17:45,610 --> 00:17:51,730
we will end up pointing
I believe down,

373
00:17:51,730 --> 00:17:53,597
because initially this
was pointing down.

374
00:17:53,597 --> 00:17:56,180
After you do this construction,
you'll be pointing down again.

375
00:17:56,180 --> 00:17:58,210
And that's why you
want to turn left here.

376
00:17:58,210 --> 00:17:59,660
So you unroll this thing.

377
00:17:59,660 --> 00:18:03,670
You can verify you are
actually just zigzagging.

378
00:18:03,670 --> 00:18:05,950
And so you go back and
forth through these.

379
00:18:05,950 --> 00:18:08,110
Verify this always
proceeds to the right.

380
00:18:08,110 --> 00:18:11,060
But we're using different
iconography here.

381
00:18:11,060 --> 00:18:14,370
This is an S on the left
that initially turns left.

382
00:18:14,370 --> 00:18:17,420
This is an S on the right
that initially turns right.

383
00:18:17,420 --> 00:18:22,250
That's why we need the different
forms of the recursive call.

384
00:18:22,250 --> 00:18:24,430
This is a picture where
initially we turn right on S

385
00:18:24,430 --> 00:18:26,721
and S is to the right, but
there are different versions

386
00:18:26,721 --> 00:18:29,324
of this picture just
like in the base cases

387
00:18:29,324 --> 00:18:31,240
where S is to the left
and where you initially

388
00:18:31,240 --> 00:18:32,860
have to turn left.

389
00:18:32,860 --> 00:18:33,860
But they look the same.

390
00:18:33,860 --> 00:18:36,450
You just visit them back
and forth like this.

391
00:18:36,450 --> 00:18:39,180
At some point, you finish
the front children.

392
00:18:39,180 --> 00:18:40,512
Then you go up.

393
00:18:40,512 --> 00:18:41,970
And here, it's a
little bit shifted

394
00:18:41,970 --> 00:18:45,020
but these would
normally be aligned.

395
00:18:45,020 --> 00:18:46,840
Then you have to visit
the top children.

396
00:18:46,840 --> 00:18:50,910
You do that by visiting
and going back and forth.

397
00:18:50,910 --> 00:18:54,190
Then, of course, when
you're done here,

398
00:18:54,190 --> 00:18:57,370
you have to unwind
everything you did.

399
00:18:57,370 --> 00:18:58,800
Then we go over here.

400
00:18:58,800 --> 00:19:00,940
We visit that guy,
back and forth.

401
00:19:00,940 --> 00:19:03,390
Eventually, we will end
up on the inside here,

402
00:19:03,390 --> 00:19:07,400
and having doubled everything,
we come back out to T.

403
00:19:07,400 --> 00:19:10,940
And all the time, we're
unfolding to the right.

404
00:19:10,940 --> 00:19:13,480
Now, what can we say about this?

405
00:19:13,480 --> 00:19:19,410
Well, what sucks is all of these
children get visited twice.

406
00:19:19,410 --> 00:19:22,210
And that's why you're doubling
at every recursive level.

407
00:19:22,210 --> 00:19:33,114
And so if your recursion tree
looks like a long binary tree,

408
00:19:33,114 --> 00:19:34,155
then you get exponential.

409
00:19:40,380 --> 00:19:41,220
Not quite.

410
00:19:41,220 --> 00:19:43,440
We'll come back to that.

411
00:19:43,440 --> 00:19:45,252
Here's an example of
it being exponential

412
00:19:45,252 --> 00:19:46,210
with this kind of tree.

413
00:19:46,210 --> 00:19:48,310
It's a little hard to
draw too many levels deep.

414
00:19:48,310 --> 00:19:53,440
So we have one level deep
and then another level deep.

415
00:19:53,440 --> 00:19:56,700
But you'll notice
that it's not uniform.

416
00:19:56,700 --> 00:20:00,030
This leaf gets visited
exponentially many times,

417
00:20:00,030 --> 00:20:02,650
but this one still
only got visited once.

418
00:20:02,650 --> 00:20:04,552
Of course, it's shallower.

419
00:20:04,552 --> 00:20:06,510
But even these two leaves
are at the same level

420
00:20:06,510 --> 00:20:08,760
and they did not get visited
the same number of times.

421
00:20:08,760 --> 00:20:10,450
So there's some hope
of improving things

422
00:20:10,450 --> 00:20:11,325
as you might imagine.

423
00:20:14,270 --> 00:20:17,340
So this is where we go.

424
00:20:17,340 --> 00:20:22,050
There's two observations
here, two ideas.

425
00:20:22,050 --> 00:20:26,020
Maybe I'll tell you
where we started.

426
00:20:26,020 --> 00:20:30,750
I was I think-- I can
never remember years.

427
00:20:30,750 --> 00:20:34,890
I was teaching this class
either two or five years ago

428
00:20:34,890 --> 00:20:38,269
and I had idea number one.

429
00:20:38,269 --> 00:20:40,310
And we worked on it in
the open problems session,

430
00:20:40,310 --> 00:20:41,230
but it was not enough.

431
00:20:41,230 --> 00:20:44,210
Idea number one is
just part of the story.

432
00:20:44,210 --> 00:20:46,850
And it's an idea that comes
from the other graduate

433
00:20:46,850 --> 00:20:49,660
class I teach which is
heavy-light decomposition

434
00:20:49,660 --> 00:20:51,100
and advanced data structures.

435
00:20:51,100 --> 00:20:53,030
So this is a very
common technique

436
00:20:53,030 --> 00:20:54,960
in advanced data structures.

437
00:20:54,960 --> 00:20:57,562
It goes back to
Sleator and Tarjan,

438
00:20:57,562 --> 00:21:01,570
if you've ever heard of splay
trees or link-cut trees.

439
00:21:01,570 --> 00:21:06,830
It's that era which was
the '80s, I believe, '83.

440
00:21:06,830 --> 00:21:07,830
It's a very useful tool.

441
00:21:07,830 --> 00:21:10,630
Whenever you have
general tree structure

442
00:21:10,630 --> 00:21:12,439
and you want to
make it balanced,

443
00:21:12,439 --> 00:21:14,230
you should think
heavy-light decomposition.

444
00:21:14,230 --> 00:21:16,900
There are actually a few
ways to balance a tree,

445
00:21:16,900 --> 00:21:19,960
but heavy-light decomposition
is usually the most powerful.

446
00:21:19,960 --> 00:21:27,390
So essentially
what it does is say

447
00:21:27,390 --> 00:21:30,780
if I could just treat
this thing, this path

448
00:21:30,780 --> 00:21:35,320
as a single node, and then treat
these guys as separate nodes,

449
00:21:35,320 --> 00:21:36,740
then I'd kind of be balanced.

450
00:21:36,740 --> 00:21:39,950
It would be like one super node
and then a bunch of children

451
00:21:39,950 --> 00:21:41,550
hanging off of it.

452
00:21:41,550 --> 00:21:44,370
One, two, three, four.

453
00:21:44,370 --> 00:21:47,140
So that's the picture
I'd like to make.

454
00:21:47,140 --> 00:21:50,759
I'm going to contract paths,
only paths of my tree.

455
00:21:50,759 --> 00:21:52,300
And it turns out
there's always a way

456
00:21:52,300 --> 00:21:54,730
to take paths, contract
them into single nodes,

457
00:21:54,730 --> 00:22:01,330
so that the resulting thing
has height at most log n,

458
00:22:01,330 --> 00:22:03,060
no matter what tree
you start with.

459
00:22:03,060 --> 00:22:05,520
And heavy-light decomposition
is a way to do this.

460
00:22:05,520 --> 00:22:07,810
It's very simple to define.

461
00:22:07,810 --> 00:22:15,435
For every node, you define
the idea of a heavy child.

462
00:22:19,890 --> 00:22:31,150
This is a child with more
than half the weight.

463
00:22:31,150 --> 00:22:39,706
The weight of a subtree is the
number of descendant nodes,

464
00:22:39,706 --> 00:22:41,330
so the number of
nodes in that subtree.

465
00:22:44,490 --> 00:22:47,700
So if we look at this
picture, you've got this node.

466
00:22:47,700 --> 00:22:50,760
It has one, two, three, four,
five, six, seven, eight,

467
00:22:50,760 --> 00:22:53,490
nine nodes below it.

468
00:22:53,490 --> 00:22:57,400
This subtree only has one,
so this subtree has seven.

469
00:22:57,400 --> 00:23:00,400
Seven is bigger
than half of nine.

470
00:23:00,400 --> 00:23:02,640
Most of the weight
is to the left.

471
00:23:02,640 --> 00:23:06,120
So we call this edge heavy.

472
00:23:06,120 --> 00:23:10,720
Similarly, this edge is
heavy because this is five.

473
00:23:10,720 --> 00:23:12,770
Five is more than half of seven.

474
00:23:12,770 --> 00:23:16,100
This one's only
three nodes below it.

475
00:23:16,100 --> 00:23:21,230
That is still more than half
because half of this is 2.5.

476
00:23:21,230 --> 00:23:23,760
So this edge is heavy.

477
00:23:23,760 --> 00:23:25,400
This one is not heavy.

478
00:23:25,400 --> 00:23:30,110
So actually, in
the real picture,

479
00:23:30,110 --> 00:23:31,960
it would look like this.

480
00:23:31,960 --> 00:23:35,259
This is a heavy path and
this guy's all by itself,

481
00:23:35,259 --> 00:23:36,300
so there's one more node.

482
00:23:36,300 --> 00:23:40,660
But it has the same effect
of balancing the tree.

483
00:23:40,660 --> 00:23:43,602
So what's the point?

484
00:23:43,602 --> 00:23:44,810
In general, I have some tree.

485
00:23:44,810 --> 00:23:46,930
In general, I might
have many children.

486
00:23:46,930 --> 00:23:49,950
No idea how many
children are out there.

487
00:23:49,950 --> 00:23:52,580
So every node can
have many children

488
00:23:52,580 --> 00:23:55,350
and these correspond to
the various bands that

489
00:23:55,350 --> 00:23:58,555
are attached to my parent band.

490
00:24:02,140 --> 00:24:04,080
So that's heavy child.

491
00:24:04,080 --> 00:24:12,630
In general though, every node
has at most one heavy child

492
00:24:12,630 --> 00:24:16,110
because you can't have
more than two halves.

493
00:24:16,110 --> 00:24:18,520
If I have more than half
of the weight in one child,

494
00:24:18,520 --> 00:24:21,860
I can't have more than half of
the weight in another child.

495
00:24:21,860 --> 00:24:24,930
So every node has at
most one heavy child.

496
00:24:24,930 --> 00:24:28,600
These edges connecting nodes
and their heavy children

497
00:24:28,600 --> 00:24:30,040
we call heavy edges.

498
00:24:34,080 --> 00:24:35,950
And whenever we
have a node, there's

499
00:24:35,950 --> 00:24:38,640
at most one heavy edge
going down from it,

500
00:24:38,640 --> 00:24:41,420
so heavy edges form heavy paths.

501
00:24:41,420 --> 00:24:42,170
That's the point.

502
00:24:42,170 --> 00:24:45,915
So this red thing is
called a heavy path.

503
00:24:50,470 --> 00:24:51,730
Cool.

504
00:24:51,730 --> 00:24:55,460
So what do we know
about heavy paths?

505
00:24:55,460 --> 00:24:59,200
The other edges, I
should say, these guys,

506
00:24:59,200 --> 00:25:01,260
we call light edges.

507
00:25:01,260 --> 00:25:02,890
This is the heavy-light
decomposition,

508
00:25:02,890 --> 00:25:05,370
so every edge is
either heavy or light.

509
00:25:05,370 --> 00:25:07,370
It could be a node
only has light children

510
00:25:07,370 --> 00:25:08,620
because it's well-distributed.

511
00:25:08,620 --> 00:25:11,120
In the worst case, it
has one heavy child,

512
00:25:11,120 --> 00:25:13,400
one heavy edge coming out of it.

513
00:25:13,400 --> 00:25:16,880
So we get heavy paths.

514
00:25:16,880 --> 00:25:19,835
I guess I'll say heavy
edges form paths.

515
00:25:25,350 --> 00:25:27,300
And if we look at
what I normally

516
00:25:27,300 --> 00:25:35,730
call the light depth
of a node, this

517
00:25:35,730 --> 00:25:38,600
is if I look at a
node in this tree

518
00:25:38,600 --> 00:25:40,770
and I measure how
many light edges do

519
00:25:40,770 --> 00:25:43,890
I have to traverse in order to
get from the root of the tree

520
00:25:43,890 --> 00:25:46,455
to that node, here it's
only one light edge.

521
00:25:46,455 --> 00:25:52,990
The number of light edges follow
on any path is at most log n,

522
00:25:52,990 --> 00:25:58,120
log base 2 of n, because every
time I follow a light edge,

523
00:25:58,120 --> 00:26:00,700
I know that I started
with say end nodes.

524
00:26:00,700 --> 00:26:01,720
I follow a light edge.

525
00:26:01,720 --> 00:26:06,110
I know that I have at most half
n nodes because my weight goes

526
00:26:06,110 --> 00:26:07,210
down by half.

527
00:26:07,210 --> 00:26:11,160
If I start with n, I can only
reduce by a factor of 2 log n

528
00:26:11,160 --> 00:26:12,730
times before I run out of nodes.

529
00:26:12,730 --> 00:26:16,340
Then I'm in a leaf when I get
down to the weight of one.

530
00:26:16,340 --> 00:26:19,030
So light depth of any
node is at most log n

531
00:26:19,030 --> 00:26:30,020
because weight halves every
step, every light step.

532
00:26:33,980 --> 00:26:37,230
So this is why, if you
contract the heavy paths

533
00:26:37,230 --> 00:26:41,955
to single nodes, the depth
of the remaining tree

534
00:26:41,955 --> 00:26:43,560
is only log n.

535
00:26:43,560 --> 00:26:44,820
So what does this tell us?

536
00:26:44,820 --> 00:26:47,630
If I could somehow
deal with paths

537
00:26:47,630 --> 00:26:52,350
efficiently and not
feel the recursive pain

538
00:26:52,350 --> 00:26:54,790
that we're getting here
from doubling everything

539
00:26:54,790 --> 00:26:59,210
along a path, then I
can chain them together

540
00:26:59,210 --> 00:27:01,000
in the usual bad recursive way.

541
00:27:01,000 --> 00:27:03,960
I can afford to double
on a balance tree.

542
00:27:03,960 --> 00:27:06,890
So if I could somehow treat
these paths as single things,

543
00:27:06,890 --> 00:27:07,630
we'd be all set.

544
00:27:07,630 --> 00:27:10,540
That was idea number one
and that sat for years.

545
00:27:10,540 --> 00:27:13,080
I knew this should be possible.

546
00:27:13,080 --> 00:27:18,110
But what we were missing
was idea number two.

547
00:27:18,110 --> 00:27:19,240
It's funny.

548
00:27:19,240 --> 00:27:21,060
If you stare at this
picture long enough,

549
00:27:21,060 --> 00:27:22,830
it's actually kind of obvious.

550
00:27:22,830 --> 00:27:26,840
It requires more
work after the idea.

551
00:27:26,840 --> 00:27:29,830
So here, the same
picture as before.

552
00:27:29,830 --> 00:27:32,880
I spread it out a little bit.

553
00:27:32,880 --> 00:27:35,910
So we end up visiting
this guy last.

554
00:27:35,910 --> 00:27:38,210
And then we end up
recursively visiting this one

555
00:27:38,210 --> 00:27:41,150
a second time, then
this one a second time,

556
00:27:41,150 --> 00:27:42,500
then this one a second time.

557
00:27:42,500 --> 00:27:43,920
We go down here.

558
00:27:43,920 --> 00:27:46,410
We go around here.

559
00:27:46,410 --> 00:27:48,210
Visit this one a second time.

560
00:27:48,210 --> 00:27:49,930
Unwind.

561
00:27:49,930 --> 00:27:53,120
Visit this one a second
time, this one a second time,

562
00:27:53,120 --> 00:27:55,372
this one a second
time, and we're up.

563
00:27:55,372 --> 00:27:56,830
Did I visit them
all a second time?

564
00:28:03,680 --> 00:28:06,190
No.

565
00:28:06,190 --> 00:28:08,820
There's one that I didn't
visit a second time, which

566
00:28:08,820 --> 00:28:11,940
is the last child.

567
00:28:11,940 --> 00:28:14,480
I only visited that one once.

568
00:28:14,480 --> 00:28:19,400
And that's all we need is
one child because I only

569
00:28:19,400 --> 00:28:21,550
need to do a path efficiently.

570
00:28:21,550 --> 00:28:24,800
So I'm recursively visiting
most of the children here.

571
00:28:24,800 --> 00:28:26,830
Those are going to
be my light children.

572
00:28:26,830 --> 00:28:28,630
The one heavy
child, I want to aim

573
00:28:28,630 --> 00:28:32,480
to visit that one last because
that one's only visited once

574
00:28:32,480 --> 00:28:35,597
so it doesn't get recursively
doubled at this level.

575
00:28:35,597 --> 00:28:37,930
Of course, the higher up the
level, it will get doubled.

576
00:28:37,930 --> 00:28:40,310
But there's only log n levels
we're going to get doubled,

577
00:28:40,310 --> 00:28:43,590
so we'll only add a factor of n.

578
00:28:43,590 --> 00:28:55,080
Idea number two is that the
last child gets visited only one

579
00:28:55,080 --> 00:28:56,120
at this recursive level.

580
00:29:03,564 --> 00:29:05,230
And if you've taken
an algorithms class,

581
00:29:05,230 --> 00:29:06,729
you know recursively
visiting things

582
00:29:06,729 --> 00:29:09,390
once versus twice
makes a big difference.

583
00:29:09,390 --> 00:29:13,960
Here it's a matter of doubling
or just staying the same.

584
00:29:13,960 --> 00:29:17,360
If we can rearrange to visit
these children first, then do

585
00:29:17,360 --> 00:29:19,890
this one last, then visit
this child first, then do

586
00:29:19,890 --> 00:29:22,196
this one last, then visit
this child first, then do

587
00:29:22,196 --> 00:29:26,300
this one last, and then do all
the children in whatever order,

588
00:29:26,300 --> 00:29:29,890
then we'll guarantee that this
path doesn't get recursively

589
00:29:29,890 --> 00:29:31,870
refined at this
level so it's really

590
00:29:31,870 --> 00:29:34,400
being treated as a
single node basically.

591
00:29:34,400 --> 00:29:36,520
And that lets you tie
together these unfoldings

592
00:29:36,520 --> 00:29:38,090
in an efficient way.

593
00:29:38,090 --> 00:29:41,350
The only issue is I need to be
able-- this construction isn't

594
00:29:41,350 --> 00:29:46,220
enough because this construction
basically forced this one

595
00:29:46,220 --> 00:29:49,030
to be the last child
to get visited.

596
00:29:49,030 --> 00:29:52,210
I need to be able to
control which child is last

597
00:29:52,210 --> 00:29:54,380
visited because the
heavy property is

598
00:29:54,380 --> 00:29:55,600
a property of the tree.

599
00:29:55,600 --> 00:29:57,830
One of these guys,
maybe this one is heavy.

600
00:29:57,830 --> 00:29:59,020
Maybe that one's heavy.

601
00:29:59,020 --> 00:30:02,260
We've got to be
able to end on that.

602
00:30:02,260 --> 00:30:04,180
So there are two cases.

603
00:30:04,180 --> 00:30:08,290
First case is that we always
start from the bottom here.

604
00:30:08,290 --> 00:30:13,770
And first case is
that on the top side

605
00:30:13,770 --> 00:30:18,190
is where the heavy child is.

606
00:30:18,190 --> 00:30:21,190
So this is the easy case because
the bottom, we don't care.

607
00:30:21,190 --> 00:30:23,910
We just visit the children
in whatever order's natural.

608
00:30:23,910 --> 00:30:25,190
Then we come up.

609
00:30:25,190 --> 00:30:28,464
And basically, we have the
freedom to spin around here.

610
00:30:28,464 --> 00:30:29,880
We have to turn
left, but we could

611
00:30:29,880 --> 00:30:33,680
either choose to immediately
go up here or keep going

612
00:30:33,680 --> 00:30:35,830
and go up here, or keep
going and go up here.

613
00:30:35,830 --> 00:30:37,680
That would visit
that one too early.

614
00:30:37,680 --> 00:30:38,650
Or go here.

615
00:30:38,650 --> 00:30:41,000
Turns out this is
the right choice.

616
00:30:41,000 --> 00:30:43,320
But we had freedom to just
spin around and choose

617
00:30:43,320 --> 00:30:44,650
when to go up.

618
00:30:44,650 --> 00:30:45,490
Just try them all.

619
00:30:45,490 --> 00:30:48,544
One of them will
end up being last.

620
00:30:48,544 --> 00:30:49,960
Basically, you
want this guy to be

621
00:30:49,960 --> 00:30:52,540
in the middle between
where you start

622
00:30:52,540 --> 00:30:55,620
and the one just to the
left here because you're

623
00:30:55,620 --> 00:30:58,280
kind of rainbowing
back and forth,

624
00:30:58,280 --> 00:31:00,900
zeroing in on the last guy.

625
00:31:00,900 --> 00:31:05,760
And just by pigeonholing, I
suppose, one of these choices

626
00:31:05,760 --> 00:31:10,530
will work because there's
k choices you could make

627
00:31:10,530 --> 00:31:13,030
on when to turn up that each
result in a different guy being

628
00:31:13,030 --> 00:31:13,850
last.

629
00:31:13,850 --> 00:31:15,990
And so one of them corresponds
to the correct one.

630
00:31:15,990 --> 00:31:18,330
That's easy to figure
out which one it is.

631
00:31:18,330 --> 00:31:20,260
So if it's on the top
side, it's really easy.

632
00:31:20,260 --> 00:31:22,817
You just pick it.

633
00:31:22,817 --> 00:31:23,650
This would be great.

634
00:31:23,650 --> 00:31:25,524
This would actually give
us linear refinement

635
00:31:25,524 --> 00:31:27,400
if we only had this case.

636
00:31:27,400 --> 00:31:34,920
The other case is when the
guy we want to make last,

637
00:31:34,920 --> 00:31:38,820
the heavy child, is
on the bottom side.

638
00:31:38,820 --> 00:31:40,661
In this case, we
need to do more work.

639
00:31:40,661 --> 00:31:42,410
So this is basically
the old construction.

640
00:31:42,410 --> 00:31:44,900
Here we don't really
have freedom from S.

641
00:31:44,900 --> 00:31:47,890
We can't just bypass this
guy because it will actually

642
00:31:47,890 --> 00:31:49,490
be hard to get him later.

643
00:31:49,490 --> 00:31:53,070
So we've got to visit
him and back and forth.

644
00:31:53,070 --> 00:31:55,690
But the one thing we could
do is just omit a child,

645
00:31:55,690 --> 00:31:58,270
just forget about
it for a while.

646
00:31:58,270 --> 00:31:59,355
Don't visit it.

647
00:31:59,355 --> 00:32:00,980
Visit everything else
in whatever order

648
00:32:00,980 --> 00:32:01,950
feels natural to you.

649
00:32:04,500 --> 00:32:07,790
And then we're going
to splice this guy in.

650
00:32:07,790 --> 00:32:10,770
So we'll see where
does the path go.

651
00:32:10,770 --> 00:32:15,850
At some point, it will go
above this proposed last guy.

652
00:32:15,850 --> 00:32:20,440
And what we're going to do,
which is on the next slide,

653
00:32:20,440 --> 00:32:26,140
is on the return trip-- so
remember we have this picture.

654
00:32:26,140 --> 00:32:28,590
We come out here.

655
00:32:28,590 --> 00:32:30,540
Then we recursively
double everything.

656
00:32:30,540 --> 00:32:34,290
So at some point, we are
coming here, doubling.

657
00:32:34,290 --> 00:32:36,730
And at that point, we turn in.

658
00:32:39,570 --> 00:32:44,390
But if S and T are
aligned like this,

659
00:32:44,390 --> 00:32:45,640
we are pointing the wrong way.

660
00:32:45,640 --> 00:32:48,390
We want it to get here,
but it doesn't even

661
00:32:48,390 --> 00:32:49,810
look like we're
in the same side.

662
00:32:49,810 --> 00:32:51,290
But we are.

663
00:32:51,290 --> 00:32:56,040
We just have to recursively
double everything again.

664
00:32:56,040 --> 00:32:59,040
In the end, when you recursively
double everything again,

665
00:32:59,040 --> 00:33:01,670
you will be here.

666
00:33:01,670 --> 00:33:03,860
And we haven't
finished doubling here.

667
00:33:03,860 --> 00:33:07,600
But in the end, you will
come out here and exit at T.

668
00:33:07,600 --> 00:33:10,360
So overall, these guys which
haven't been visited yet,

669
00:33:10,360 --> 00:33:13,000
they'll end up being
doubled from this procedure.

670
00:33:13,000 --> 00:33:15,640
These guys will end
up being quadrupled.

671
00:33:15,640 --> 00:33:20,156
So we're doing four recursive
calls for most of the children.

672
00:33:20,156 --> 00:33:22,030
Some of them might get
lucky and only double.

673
00:33:22,030 --> 00:33:27,290
But all we care about is if
this guy only gets visited once.

674
00:33:27,290 --> 00:33:29,710
That's the key.

675
00:33:29,710 --> 00:33:32,440
Because this will
be recursively here.

676
00:33:32,440 --> 00:33:33,440
It's hard to unwind.

677
00:33:33,440 --> 00:33:35,890
But then you'll end up on
the inside of this channel.

678
00:33:35,890 --> 00:33:37,730
Then you'll end up
visiting this guy.

679
00:33:37,730 --> 00:33:39,480
Then you'll end up
visiting this guy.

680
00:33:39,480 --> 00:33:40,580
And then you get out at T.

681
00:33:40,580 --> 00:33:42,490
So you won't come back to here.

682
00:33:42,490 --> 00:33:44,590
That's the key.

683
00:33:44,590 --> 00:33:46,234
That's all we need.

684
00:33:46,234 --> 00:33:48,150
You can write down a
recurrence at this point.

685
00:33:55,280 --> 00:33:57,460
If you look at the
level of refinement

686
00:33:57,460 --> 00:34:01,380
for an n vertex polyhedron
or an n slab polyhedron,

687
00:34:01,380 --> 00:34:04,990
it would be-- n, we'll
say is the number of nodes

688
00:34:04,990 --> 00:34:07,720
in the tree of
slabs, this picture.

689
00:34:07,720 --> 00:34:09,980
In general, that is
proportional to the number

690
00:34:09,980 --> 00:34:12,960
of vertices in the polyhedron.

691
00:34:12,960 --> 00:34:20,000
In the worst case, it's going
to be the max of two things-- R

692
00:34:20,000 --> 00:34:25,199
n minus 1, and 4
times R of n over 2.

693
00:34:28,469 --> 00:34:33,920
Well, I should say
n sub i, max over i.

694
00:34:33,920 --> 00:34:41,250
So what I mean to say here--
this is also technically an sub

695
00:34:41,250 --> 00:34:42,889
i.

696
00:34:42,889 --> 00:34:46,469
This is an upper
bound certainly.

697
00:34:46,469 --> 00:34:51,860
So if we look at the recursion,
how much refinement we need,

698
00:34:51,860 --> 00:34:53,790
the worst refinement
over the whole tree

699
00:34:53,790 --> 00:34:56,050
is the max of all
of its subtrees.

700
00:34:56,050 --> 00:34:59,390
So there's a [INAUDIBLE] here, a
[INAUDIBLE] here, and then also

701
00:34:59,390 --> 00:35:02,720
of course, what we need
to do within the node.

702
00:35:02,720 --> 00:35:05,770
And on the one hand,
there's the heavy situation,

703
00:35:05,770 --> 00:35:07,240
the heavy edge.

704
00:35:07,240 --> 00:35:10,230
That only decreases the
size of the tree by 1.

705
00:35:10,230 --> 00:35:12,320
But it only gets
recursively visited

706
00:35:12,320 --> 00:35:15,870
one time, 1 times
R of n minus 1.

707
00:35:15,870 --> 00:35:17,470
There's the other
light children.

708
00:35:17,470 --> 00:35:20,710
These are all light children.

709
00:35:20,710 --> 00:35:23,370
n sub i represents the number
of nodes in that child.

710
00:35:23,370 --> 00:35:27,752
We know that this
is at most n over 2

711
00:35:27,752 --> 00:35:29,710
because we know that at
most half of the weight

712
00:35:29,710 --> 00:35:31,770
lives in the light child.

713
00:35:31,770 --> 00:35:34,940
Now these get recursively
visited at most four times,

714
00:35:34,940 --> 00:35:35,730
up to four times.

715
00:35:35,730 --> 00:35:37,730
Worst case, it
will be four times.

716
00:35:40,580 --> 00:35:43,680
In this recurrence, this
first term doesn't matter.

717
00:35:43,680 --> 00:35:45,590
R of n is at most
R of n minus 1.

718
00:35:45,590 --> 00:35:47,250
Yeah, big deal.

719
00:35:47,250 --> 00:35:49,850
That's just saying R
of n increases with n.

720
00:35:49,850 --> 00:35:54,330
So really, this is going to be--
and then the max disappears.

721
00:35:54,330 --> 00:35:59,720
All we need is this is at
most 4 times R of n over 2.

722
00:35:59,720 --> 00:36:00,830
And this is a recurrence.

723
00:36:00,830 --> 00:36:06,560
If you read CLRS, this
follows the master theorem

724
00:36:06,560 --> 00:36:11,424
and you know immediately
this is order n squared.

725
00:36:11,424 --> 00:36:13,590
It'd be theta n squared,
but this is an upper bound.

726
00:36:13,590 --> 00:36:15,881
In the worst case, it will
actually be theta n squared.

727
00:36:15,881 --> 00:36:20,060
So the amount of refinement
is at most n squared.

728
00:36:20,060 --> 00:36:21,960
If this is not obvious
to you, an easy way

729
00:36:21,960 --> 00:36:28,630
to imagine it is you start
with something of size n,

730
00:36:28,630 --> 00:36:34,590
you're visiting four times
something of size n over 2.

731
00:36:34,590 --> 00:36:37,480
This is called a recurrence
tree, recursion tree.

732
00:36:37,480 --> 00:36:39,890
Each of these visits four
times something of size

733
00:36:39,890 --> 00:36:45,480
n over 4 and so on.

734
00:36:45,480 --> 00:36:49,520
Obviously, log n levels here.

735
00:36:49,520 --> 00:36:52,770
And at the bottom, you're
going to have a bunch of ones.

736
00:36:52,770 --> 00:36:54,730
That's when you stop recursing.

737
00:36:54,730 --> 00:36:56,530
And if you add up
everything level

738
00:36:56,530 --> 00:37:03,120
by level, how much
work am I doing here

739
00:37:03,120 --> 00:37:05,830
and how much work
am I doing here?

740
00:37:05,830 --> 00:37:07,840
2n.

741
00:37:07,840 --> 00:37:09,600
How much work am I doing here?

742
00:37:09,600 --> 00:37:13,160
It gets harder to add,
but there are 4 squared,

743
00:37:13,160 --> 00:37:16,070
16 things, each n over 4.

744
00:37:16,070 --> 00:37:17,070
So it's 4n.

745
00:37:17,070 --> 00:37:20,080
In general, it goes up by
a factor of 2 every time

746
00:37:20,080 --> 00:37:25,270
which means you're
dominated by the last layer.

747
00:37:25,270 --> 00:37:30,030
And if you count, it's
going to be n squared

748
00:37:30,030 --> 00:37:35,590
because this is 2 to the
0, 2 to the 1, 2 to the 2.

749
00:37:35,590 --> 00:37:41,610
After you get to log n,
it's 2 to the log n times n,

750
00:37:41,610 --> 00:37:43,759
so that is n squared.

751
00:37:43,759 --> 00:37:45,300
And this is a
geometric series if you

752
00:37:45,300 --> 00:37:46,710
want to add all
these numbers up.

753
00:37:46,710 --> 00:37:50,510
And so it's 2 times n squared.

754
00:37:50,510 --> 00:37:53,650
So that's the
proof of this bound

755
00:37:53,650 --> 00:37:56,680
and that's how you get
quadratic refinement.

756
00:37:56,680 --> 00:38:00,470
So either idea is kind
of easy, but the two

757
00:38:00,470 --> 00:38:03,920
together, pretty powerful.

758
00:38:03,920 --> 00:38:05,110
Questions about unfolding?

759
00:38:05,110 --> 00:38:07,560
I think this algorithm
should be pretty clear.

760
00:38:07,560 --> 00:38:09,750
One of the questions
was, are they practical?

761
00:38:09,750 --> 00:38:11,890
I think the answer is no.

762
00:38:11,890 --> 00:38:14,485
It has the same issue of
strip folding from origami.

763
00:38:14,485 --> 00:38:16,360
But I think they would
make great animations.

764
00:38:16,360 --> 00:38:19,380
This is still a cool
project to see virtually.

765
00:38:19,380 --> 00:38:23,060
But to do it physically
is probably crazy.

766
00:38:23,060 --> 00:38:26,240
No one has tried
as far as I know.

767
00:38:26,240 --> 00:38:26,990
Another idea.

768
00:38:26,990 --> 00:38:29,040
This is an open problem.

769
00:38:29,040 --> 00:38:30,995
So orthogonal was great.

770
00:38:30,995 --> 00:38:32,370
It has all these
nice properties.

771
00:38:32,370 --> 00:38:34,670
Can you generalize it, not
to arbitrary polyhedra,

772
00:38:34,670 --> 00:38:36,320
but to some other
kind of orthogonal,

773
00:38:36,320 --> 00:38:41,240
such as this is a
hexagonal lattice in 3D?

774
00:38:41,240 --> 00:38:43,460
I mean, it's just
extruded in one dimension,

775
00:38:43,460 --> 00:38:45,720
but then you have hex
grid in another dimension.

776
00:38:45,720 --> 00:38:47,260
You could try to do that.

777
00:38:47,260 --> 00:38:48,970
And my guess is if
you set up the bands

778
00:38:48,970 --> 00:38:54,410
to be in hex dimensions, it
just works, but I haven't tried.

779
00:38:54,410 --> 00:38:57,760
There's potentially some
low-hanging fruit here.

780
00:38:57,760 --> 00:39:00,030
There should be other 3D
structures that might work.

781
00:39:00,030 --> 00:39:01,655
A typical one in
computational geometry

782
00:39:01,655 --> 00:39:04,890
is called c-oriented
polyhedra where here we

783
00:39:04,890 --> 00:39:06,270
have three oriented polyhedra.

784
00:39:06,270 --> 00:39:07,686
There's three
different directions

785
00:39:07,686 --> 00:39:11,280
that the faces can
be perpendicular to.

786
00:39:11,280 --> 00:39:13,870
Take 10 of them, can you
apply the same technique?

787
00:39:13,870 --> 00:39:15,750
That gets dicier.

788
00:39:15,750 --> 00:39:18,640
But something like this I
feel like may be possible.

789
00:39:18,640 --> 00:39:21,010
You should try it out.

790
00:39:21,010 --> 00:39:24,710
But as far as I know,
this is all open.

791
00:39:24,710 --> 00:39:27,800
Last question is
Cauchy's rigidity theorem

792
00:39:27,800 --> 00:39:30,620
seems intuitively obvious.

793
00:39:30,620 --> 00:39:33,550
Or as I wrote it in the notes,
is Cauchey's rigidity theorem

794
00:39:33,550 --> 00:39:34,050
obvious?

795
00:39:34,050 --> 00:39:37,970
And my answer is no,
not obvious to me.

796
00:39:37,970 --> 00:39:39,890
I have two reasons
why it's not obvious.

797
00:39:39,890 --> 00:39:44,820
One is it's not true
in two dimensions.

798
00:39:44,820 --> 00:39:46,190
That's kind of a weak statement.

799
00:39:46,190 --> 00:39:52,230
But if I take a convex
polyhedron in 2D--

800
00:39:52,230 --> 00:39:53,610
so these are rigid faces.

801
00:39:53,610 --> 00:39:56,450
This is the equivalent
of a triangle.

802
00:39:56,450 --> 00:39:58,030
This is not rigid, of course.

803
00:40:01,770 --> 00:40:02,930
So that's one statement.

804
00:40:02,930 --> 00:40:05,180
And then the other thing
is convexity is necessary.

805
00:40:05,180 --> 00:40:08,150
There's this thing,
Steffen's polyhedron.

806
00:40:08,150 --> 00:40:10,620
Actually, originally, there
was a Connelly polyhedron.

807
00:40:10,620 --> 00:40:12,810
We've seen some
results by Connelly.

808
00:40:12,810 --> 00:40:14,990
He's a rigidity expert.

809
00:40:14,990 --> 00:40:17,470
And then this was
simplified by this guy Klaus

810
00:40:17,470 --> 00:40:20,740
Steffen in the '70s.

811
00:40:20,740 --> 00:40:24,970
This is a drawing
of it flexing in 3D.

812
00:40:24,970 --> 00:40:27,429
Nonconvex polyhedra
can be flexible.

813
00:40:27,429 --> 00:40:29,512
They happen to preserve
their volume as they flex.

814
00:40:35,060 --> 00:40:37,120
It's hard to say why a
theorem is not obvious.

815
00:40:37,120 --> 00:40:38,300
To me, it's not obvious.

816
00:40:38,300 --> 00:40:40,174
Of course, if you play
with convex polyhedra,

817
00:40:40,174 --> 00:40:42,790
they are rigid, so maybe
it's obvious from that.

818
00:40:42,790 --> 00:40:47,090
But you've got to prove
experimental evidence

819
00:40:47,090 --> 00:40:48,959
mathematically.

820
00:40:48,959 --> 00:40:49,625
Other questions?

821
00:40:55,280 --> 00:40:57,630
Then I have some fun
videos to show you.

822
00:40:57,630 --> 00:41:01,174
The first one is the making
of this origami piece which

823
00:41:01,174 --> 00:41:02,840
I've probably shown
a picture of before.

824
00:41:02,840 --> 00:41:04,640
It's by Brian Chan.

825
00:41:04,640 --> 00:41:06,650
It's one square paper, no cuts.

826
00:41:06,650 --> 00:41:10,675
Mens et Manus with a little
crane instead of-- what's

827
00:41:10,675 --> 00:41:13,460
it normally?

828
00:41:13,460 --> 00:41:14,220
Who knows?

829
00:41:14,220 --> 00:41:15,940
And then he also
made this nice box.

830
00:41:15,940 --> 00:41:17,930
And then the glass is
etched by Peter Houk

831
00:41:17,930 --> 00:41:21,260
who runs the glass lab at MIT.