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PROFESSOR: All right,
let's get started.

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We are continuing our
theme of folding polygons

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into convex polyhedra.

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Let's do a quick
reminder, we're talking

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about gluing up the boundary
of a polygon to itself.

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And we were representing that
with gluing trees last time.

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So I want to do an actual
example of a gluing tree.

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So this is how you
make a cube out

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of a cross, my favorite example.

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And a valid gluing,
it's an Alexandroff

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gluing because it
has no crossing, so

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its topologically a sphere.

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Everything is glued
to everything.

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And it never glues more
than 360 degrees of material

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to any point.

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In fact, it's going to
be 270 at every point,

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because that's the curvature
of every vertex of a cube.

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And the gluing tree was, if we
sort of turned this inside out

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and think of the polygon as
being on the outside instead

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of the inside, it's really
hard to see here how to convert

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that into a gluing tree,
though you can do it.

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It's the way that
these arrows nest.

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But it's a little easier
to think of on the sphere

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or on the polyhedron, because
then turning inside out

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is just upside down.

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It's not a big deal.

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So if we look at where
the cut are on the cube,

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it looks something like this.

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That should be a cross.

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This is the top
of the cross, this

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is the bottom of the cross.

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It goes around the back side.

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So then the gluing tree is
essentially this doubled.

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If we walk around the
outside of the cuts,

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that is the gluing tree.

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So I can see, there's this
segment, then this segment,

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then that segment.

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Then we're back here,
then we go up one, two,

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and then we turn around.

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That's a leaf, and so it's
going to look like this weird h

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shape.

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And these are vertices.

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That's right, all of
these should be vertices.

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And locally you can see, ah,
this is a 90 degree angle.

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This is actually the
top of the cross,

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remember the polygon
is now out here.

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That's going to
be the top across.

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This is where it turns,
it gets a little confusing

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because of the
way I've drawn it.

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Here's the bottom of the cross.

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Those parts are easy to see.

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The rest is little awkward.

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But for algorithms,
this is a great way

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to think about things.

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And also for proving bounds
and how many different ways

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to fold, it's really useful
to think about gluing trees.

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Of course, all of these
views are equivalent,

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but a lot of things will
be easier to analyze here.

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So, clear?

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Let me first tell you what
we're going to prove today.

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We're going to do combinatorial
bounds and algorithms

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for all of this.

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And let's see.

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So I want general gluings,
general situation,

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edge-to-edge gluings, and a
particular kind of polygon

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called bounded sharpness.

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And then I want two columns.

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The first question is, how
many gluings are there?

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And whenever we get a bound
on the number of gluings,

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we're also going to get
an enumeration algorithm.

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Remember, there are three goals.

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The first goal was decision.

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Is there any folding?

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Second goal was enumeration,
give me all the foldings.

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And third goal was
counting, I think.

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How many different
solutions are there?

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For [INAUDIBLE], this is mainly
about decision and enumeration.

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And this number
of gluings, that's

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the combinatorial problem.

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So it turns out, for
everything the we'll do,

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these two have the same answer.

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So I'm going to put
it in one column.

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But sometimes for
a decision, you

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can do it a little faster
than enumerating them.

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In particular, because
the number of gluings

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can be exponential.

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We saw last time that it
could actually be infinite.

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I'm going to fix
that, don't worry.

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We'll make it finite.

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And then the right
bound is exponential.

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There's upper and lower
bounds of two to the theta n.

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And also here, bounded
sharpness is interesting

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because it's polynomial.

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And so this is also polynomial.

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The one result here that
is not tight is this one.

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It's an open question whether
for the general setup, the most

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interesting scenario,
just give you a polygon,

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is there any gluing?

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The best algorithm we
know is exponential,

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even to decide whether
there's a gluing.

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So I put a star
there to say that.

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It is not tight.

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The open problem is, can
you do it in polynomial?

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I suspect yes, but we worked on
it a long time ago and failed.

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For the special case of
edge-to-edge gluings, which

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is when you only glue
whole edges of the polygons

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to other whole edges, there is
a n to the order one algorithm.

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You can think of this as
an edge-to-edge gluing

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if you imagine this
as being two edges.

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If I actually draw
a vertex there,

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then this is an
edge-to-edge gluing.

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And then there's a polynomial
algorithm to tell you,

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is there any folding?

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If you want to list all
of the gluings though,

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you need exponential
time because there

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can be exponentially many.

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But everything except
this one result

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is the best you could hope for.

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Bounded sharpness
is a natural sense

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in which this exponential
is kind of cheating.

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Let's see, let me tell you
about bounded sharpness.

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I want to bound how sharp an
angle can be in a polygon.

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This is a really sharp angle.

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If every angle-- is that
actually the sharpness I want?

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Actually, no.

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Sorry, I'm getting
inside out already.

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Actually, what I care
about are reflex vertices.

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I don't want to have a
super big angles here.

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So if every angle
is at least 360

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minus some epsilon--
for a constant epsilon--

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then I call my polygon
bounded sharpness.

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If your polygon has
bounded sharpness,

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we get a polynomial bound
on the number of gluings

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and a polynomial algorithm
to list all the gluings.

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This is really the sense in
which this is all practical.

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Give me a polygon that's going
to have some sharpest angle,

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call that the bound.

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If epsilon gets really tiny, if
your reflex angles get really

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close to 360, then
this bound will go up.

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But as long as you're not
too extreme, you're all set.

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I mean less than or
equal to, yes, thank you.

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Yeah, that would
be a weird polygon

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if all the angles were so big.

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I think that doesn't exist.

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So for example, convex polygons
always have bounded sharpness.

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Epsilon is 180 in
that situation.

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So, that's good.

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I told you about
edge-to-edge gluings.

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Just glue whole
edges to whole edges.

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What I haven't really told you
is how we made this finite,

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so let me go to that.

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And when I say
gluings here, I really

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mean a combinatorial
type of gluing,

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because we had rolling belts.

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So there were infinitely
many actual gluings,

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but I'm going to
think of most of those

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as being the same thing and
just distinguish essentially

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what the gluing tree looks
like combinatorially.

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So let me define that.

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The first part is what I call
the abstract gluing tree.

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So this is just what the
picture looks like in general.

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So for example, the one
over there-- the cube--

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combinatorially the
tree looks like this.

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It has four leaves and a kind
of wishbone configuration.

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So I'm not trying to measure
the lengths or anything, just

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that's the abstract
picture, no metric.

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And then the second
part is I'm going

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to specify which polygon
vertices and edges are where.

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And I'm going to specify that
at things called junctions.

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So in particular, at
leaves I want to say,

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oh this is vertex one.

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At places like this, where
three edges come together,

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I want to specify for each
one, oh this is vertex two,

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this is edge eight, this
is vertex 10, whatever.

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I'm making up these numbers.

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They don't necessarily
correspond,

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but they should appear in order
because this is the polygon

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after all, just
warped inside out.

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Also this one.

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Also, I didn't really draw
this picture big enough,

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but if I have two parts
of the gluing tree

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where a vertex glues
against something--

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it could be another vertex
or it could be an edge--

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I also consider those junctions.

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So junctions are going
to be all the vertices--

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the only exception is when
there's an edge gluing

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to an edge, I don't
consider that a junction

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because there's going to
be fairly many of those.

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I want there to be finite.

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So everything of
degree one, everything

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of degree three or
more, for degree two

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there has to be at
least one vertex.

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So degree not equal to 2,
or at least one vertex.

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Those are junctions.

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For each I'm going to
mark where they are.

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That's part of my abstract
gluing tree, specifying where

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the junctures are, how
they're configured.

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And then for each
one, wherever there's

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part of the polygon
coming together,

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I specify which vertex
or edge is coming there.

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Now when I specify
an edge, I don't

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say what point on the edge,
because that could be infinite.

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There could be a whole range
of points, there could be.

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We'll worry about that later.

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I just specify that
edge eight is there.

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So this is what I call
combinatorial type of gluing.

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And in this case, there are
only finitely many gluings,

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finitely many types, but
it can be exponential.

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It's still big, just
not quite as big.

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So just to get
warmed up a little,

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let's prove this exponential
upper bound that is only 2

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to the order n.

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That will in
particular convince you

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that there's only
finitely many of these,

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although I think that's probably
obvious that it's finite.

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It's conceivable it could
be more than exponential.

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This is not a particularly
exciting bound,

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but we're going to use
a lot of the same ideas

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in the other proofs, in
particular this one which

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is more interesting.

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I want to look at
leaves of my tree.

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And we have to remember back
to the previous lecture, who

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can be at the leaves?

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Well, if could be a vertex
or it could be an edge.

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When it was an edge, we
called it a fold point.

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When we folded in the
middle of an edge--

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so here's an edge of
the polygon and I end up

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subdividing it and gluing
the edge onto itself-- right

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there is 180 degrees
of curvature.

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Total amount of
curvature is 720,

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therefore I only
have-- I can't count,

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I'm jetlagged-- four of them.

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At most four fold points.

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Everybody else is a vertex.

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So if I have n vertices, I
have at most n plus 4 leaves.

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That's already feeling good.

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I pull a rabbit out
of a hat and tell you

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that if I have only
n plus 4 leaves,

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there are 2 to the order n
trees on that many leaves.

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Actually, I should be
a little more careful.

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If I can bound the number
of nodes in the tree,

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then I'll get an exponential
bound on the number of trees.

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And that will be part one,
the abstract gluing tree.

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But right now I have a bound on
the number of leaves, not the

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number of nodes.

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And there's a difference
here, because if you

247
00:14:39,720 --> 00:14:43,300
have a tree which has
a whole bunch of degree

248
00:14:43,300 --> 00:14:52,770
2 vertices, then that would be
bad, because I only have two

249
00:14:52,770 --> 00:14:55,770
leaves I have
arbitrarily many nodes.

250
00:14:55,770 --> 00:15:00,190
But every time I have one
of these degree 2 nodes,

251
00:15:00,190 --> 00:15:03,160
we already threw away
the degree 2 nodes that

252
00:15:03,160 --> 00:15:05,579
had no vertices at them, because
there are infinitely many

253
00:15:05,579 --> 00:15:06,870
of those and those don't count.

254
00:15:06,870 --> 00:15:12,340
So once we throw that away,
the only degree two junctions

255
00:15:12,340 --> 00:15:15,117
are the ones that have a vertex,
which is why-- these dots are

256
00:15:15,117 --> 00:15:17,200
supposed to be the actual
vertices of the polygon.

257
00:15:17,200 --> 00:15:19,530
So it might be an edge here,
it might be an edge there.

258
00:15:19,530 --> 00:15:21,660
There's at least one
vertex at each of them.

259
00:15:21,660 --> 00:15:25,460
So the total number of nodes
here is indeed order n.

260
00:15:25,460 --> 00:15:28,740
I mean, it's probably actually
at most n plus 4 again.

261
00:15:28,740 --> 00:15:31,470
But if we're sloppy you say,
well the number of leaves

262
00:15:31,470 --> 00:15:32,450
is at most n plus 4.

263
00:15:32,450 --> 00:15:37,310
The number of degree
2 nodes is at most n.

264
00:15:37,310 --> 00:15:39,290
Then there could be
some degree three nodes.

265
00:15:39,290 --> 00:15:41,440
It's also at most
n plus 4 in total,

266
00:15:41,440 --> 00:15:44,470
or n is all I care about.

267
00:15:44,470 --> 00:15:46,390
You could be more
careful and figure out

268
00:15:46,390 --> 00:15:47,340
what the consonant is.

269
00:15:47,340 --> 00:15:51,690
But we get from this,
this is order n nodes.

270
00:15:54,530 --> 00:15:57,470
And once you have that the tree
has a linear number of nodes,

271
00:15:57,470 --> 00:16:01,220
I'll tell you this is one of
the [? Kadhalan ?] problems,

272
00:16:01,220 --> 00:16:03,470
that there's only 2
the order n such trees.

273
00:16:03,470 --> 00:16:06,900
But that's only part one
of the combinatorial type

274
00:16:06,900 --> 00:16:07,600
of the gluing.

275
00:16:07,600 --> 00:16:10,580
We also have to worry about
which vertices are where.

276
00:16:13,540 --> 00:16:15,980
I want to show that that
is only exponential.

277
00:16:24,150 --> 00:16:29,110
So imagine at this
point you've fixed

278
00:16:29,110 --> 00:16:30,910
the combinatorial
structure of the tree,

279
00:16:30,910 --> 00:16:33,720
but you have no
idea what's what.

280
00:16:33,720 --> 00:16:37,430
So in particular,
at these points--

281
00:16:37,430 --> 00:16:41,450
and maybe also you've
defined some degree 2

282
00:16:41,450 --> 00:16:43,340
junctions, which you
know are supposed

283
00:16:43,340 --> 00:16:48,410
to have at least one vertex--
the first thing I want

284
00:16:48,410 --> 00:16:51,150
to specify is, for each
of these little dots that

285
00:16:51,150 --> 00:16:56,910
come together-- these are the
junctions-- is it a vertex,

286
00:16:56,910 --> 00:16:59,510
or is it an edge?

287
00:16:59,510 --> 00:17:01,570
So that's just binary
information for each one.

288
00:17:01,570 --> 00:17:03,220
I'll fill it in, say
if it's a vertex,

289
00:17:03,220 --> 00:17:05,900
I'll leave it open
if it's an edge.

290
00:17:05,900 --> 00:17:08,990
In fact, I know that at
most one edge comes together

291
00:17:08,990 --> 00:17:09,550
at any point.

292
00:17:09,550 --> 00:17:13,140
Otherwise, you'd have too
much material glued there.

293
00:17:13,140 --> 00:17:15,975
So maybe it looks
something like this.

294
00:17:15,975 --> 00:17:18,590
It could be a fold point,
maybe another fold point.

295
00:17:22,640 --> 00:17:24,040
Something like that.

296
00:17:24,040 --> 00:17:26,755
So that coloring, black
and white, there's only 2

297
00:17:26,755 --> 00:17:30,030
to the order and such colorings.

298
00:17:30,030 --> 00:17:36,240
Call it a vertex edge
coloring, because again there's

299
00:17:36,240 --> 00:17:37,940
order n dots.

300
00:17:37,940 --> 00:17:41,870
Each one, there's two choices,
black or white, vertex or edge.

301
00:17:45,610 --> 00:17:47,720
And now what we
really care about

302
00:17:47,720 --> 00:17:50,360
is where the vertices,
because remember, this

303
00:17:50,360 --> 00:17:53,330
is the polygon in order.

304
00:17:53,330 --> 00:17:57,600
So if I could tell you, let's
say which vertex is vertex one?

305
00:17:57,600 --> 00:18:00,990
Maybe this vertex is vertex one.

306
00:18:00,990 --> 00:18:04,240
Then I know this is vertex two,
I know this is vertex three,

307
00:18:04,240 --> 00:18:06,310
I know this is edge
three or whichever

308
00:18:06,310 --> 00:18:09,600
edge connects V3 three and V4.

309
00:18:09,600 --> 00:18:11,690
This is also edge
three, and so on.

310
00:18:11,690 --> 00:18:15,210
I could just walk around
and label them in order.

311
00:18:15,210 --> 00:18:18,380
All I needed to do is
know where vertex one is.

312
00:18:18,380 --> 00:18:21,680
There's only n
possibilities for that.

313
00:18:28,040 --> 00:18:31,540
There's n filled circles, which
are where the vertices are.

314
00:18:31,540 --> 00:18:34,910
One of them is V1, so there's
only n choices for it.

315
00:18:34,910 --> 00:18:40,340
You multiply all of these
things together, n times 2

316
00:18:40,340 --> 00:18:42,390
to the order n, times
2 to the order n,

317
00:18:42,390 --> 00:18:44,670
that result is 2 to
the some other order n.

318
00:18:47,320 --> 00:18:50,380
That's total number of choices.

319
00:18:50,380 --> 00:18:55,059
So that's a rough
but fine upper bound.

320
00:18:55,059 --> 00:18:56,850
I'm not going to try
to tune this constant.

321
00:18:56,850 --> 00:18:57,970
I guess it would
be an open problem

322
00:18:57,970 --> 00:18:59,761
to get a really good
bound on the constant.

323
00:18:59,761 --> 00:19:01,840
I'm not sure if even
one has been worked out,

324
00:19:01,840 --> 00:19:04,660
but what I will show you is
that this is pretty much tight,

325
00:19:04,660 --> 00:19:07,140
so you can't hope to do
much better than this

326
00:19:07,140 --> 00:19:10,750
because there's a 2 to
the omega n lower bound

327
00:19:10,750 --> 00:19:14,455
on the number of gluings.

328
00:19:17,220 --> 00:19:21,170
And that is this crazy example.

329
00:19:21,170 --> 00:19:24,860
I'll draw it also on the board.

330
00:19:24,860 --> 00:19:26,410
It's a little tricky to draw.

331
00:19:26,410 --> 00:19:32,500
It's a very spiky star,
something like that.

332
00:19:32,500 --> 00:19:45,090
So I have e points, and n
spikes, n convex corners.

333
00:19:45,090 --> 00:19:47,420
The convex angle,
I think, is alpha.

334
00:19:47,420 --> 00:19:49,772
Let me match the notation.

335
00:19:49,772 --> 00:19:52,060
Yes, convex angle is alpha.

336
00:19:52,060 --> 00:19:56,060
The reflex angle here is
beta, but otherwise it's

337
00:19:56,060 --> 00:19:57,020
completely symmetric.

338
00:19:57,020 --> 00:19:59,530
So all the reflex
angles are beta,

339
00:19:59,530 --> 00:20:01,616
all the convex angles are alpha.

340
00:20:01,616 --> 00:20:03,630
I have n spikes.

341
00:20:03,630 --> 00:20:09,730
I want these points--
it's hard to draw,

342
00:20:09,730 --> 00:20:13,450
but I want them to be very,
very close to the center.

343
00:20:13,450 --> 00:20:14,920
Take the limit.

344
00:20:14,920 --> 00:20:18,680
In the limit, alpha is 0.

345
00:20:18,680 --> 00:20:23,835
So alpha is going to be some
very tiny amount epsilon.

346
00:20:23,835 --> 00:20:24,335
Beta.

347
00:20:27,280 --> 00:20:30,690
What's the limit of beta?

348
00:20:30,690 --> 00:20:32,920
180, 360?

349
00:20:32,920 --> 00:20:35,080
No, it's not 360.

350
00:20:35,080 --> 00:20:36,160
That's what's important.

351
00:20:39,210 --> 00:20:44,520
360 times 1 minus
1 over n, yeah.

352
00:20:44,520 --> 00:20:46,680
It's like 360 minus 360 over n.

353
00:20:51,050 --> 00:20:54,170
Believe it or not,
this is a big number,

354
00:20:54,170 --> 00:20:56,380
in that it doesn't
depend on epsilon.

355
00:20:56,380 --> 00:21:00,599
So it's going to be, what,
a little bit smaller--

356
00:21:00,599 --> 00:21:02,140
going to be some
minus epsilon prime.

357
00:21:02,140 --> 00:21:04,170
But I really don't care
about the epsilons.

358
00:21:04,170 --> 00:21:06,620
They're kind of irrelevant.

359
00:21:06,620 --> 00:21:10,140
Let me just cross them out,
but they're really there.

360
00:21:10,140 --> 00:21:17,180
So alpha is basically 0, beta
is a chunk less than 360.

361
00:21:17,180 --> 00:21:22,120
What this means is in the limit
I can take one of these alphas

362
00:21:22,120 --> 00:21:25,147
and glue it into
one of the betas,

363
00:21:25,147 --> 00:21:26,605
and it will still
be less than 360.

364
00:21:26,605 --> 00:21:29,270
In fact I could
glue n of the alpha.

365
00:21:29,270 --> 00:21:34,017
I could glue all of these
spikes into one of these gaps.

366
00:21:34,017 --> 00:21:35,975
Remember, the limiting
picture looks like this.

367
00:21:39,062 --> 00:21:40,270
That's maybe more convincing.

368
00:21:40,270 --> 00:21:42,520
It doesn't look that way here.

369
00:21:42,520 --> 00:21:43,940
This is nothing.

370
00:21:43,940 --> 00:21:47,380
Of course, I can fit arbitrarily
many of them into this gap

371
00:21:47,380 --> 00:21:50,750
and it will still
sum to less than 360.

372
00:21:50,750 --> 00:21:54,200
So this is kind of the key
to why this example works.

373
00:21:56,960 --> 00:22:00,740
I've got tons a room for alphas
in the complement of betas.

374
00:22:04,680 --> 00:22:09,390
Now, I would like to make
this not only an exponential

375
00:22:09,390 --> 00:22:11,700
lower bound on the
general case, but also

376
00:22:11,700 --> 00:22:13,960
on the edge-to-edge
case, because that's

377
00:22:13,960 --> 00:22:16,460
a little bit stronger to say
even edge-to-edge gluings there

378
00:22:16,460 --> 00:22:17,880
can be exponentially many.

379
00:22:17,880 --> 00:22:21,860
To do that, I'm going to
take the midpoint of one

380
00:22:21,860 --> 00:22:24,600
of the edges, call
that a vertex,

381
00:22:24,600 --> 00:22:29,590
and take the perimeter antipode,
which is like here hopefully,

382
00:22:29,590 --> 00:22:30,404
more or less.

383
00:22:30,404 --> 00:22:32,570
So I want the left parameter
and the right perimeter

384
00:22:32,570 --> 00:22:35,930
to be equal, also
call that a vertex.

385
00:22:35,930 --> 00:22:38,351
And now one thing we know how
to do is perimeter halving.

386
00:22:38,351 --> 00:22:40,850
Now, we were only supposed to
do perimeter halving on convex

387
00:22:40,850 --> 00:22:43,850
polygons, but it turns out it
will work on this polygon too.

388
00:22:43,850 --> 00:22:46,826
So I'm going to glue this
half edge to this half edge,

389
00:22:46,826 --> 00:22:48,950
then I'm going to glue this
edge to this edge, then

390
00:22:48,950 --> 00:22:51,770
this one to this one, and so on.

391
00:22:51,770 --> 00:22:54,550
So just do that gluing.

392
00:22:54,550 --> 00:22:58,660
And the only thing
to worry about

393
00:22:58,660 --> 00:23:00,350
is that you glue
two betas together.

394
00:23:00,350 --> 00:23:01,940
That would be more than 360.

395
00:23:01,940 --> 00:23:05,680
Anything else, beta to any
number of alpha's is OK.

396
00:23:05,680 --> 00:23:09,344
At this point it is really
helpful to draw a gluing tree,

397
00:23:09,344 --> 00:23:10,830
so let me do that.

398
00:23:25,410 --> 00:23:30,560
So gluing tree for
perimeter halving is path.

399
00:23:30,560 --> 00:23:35,224
And do I give these guys
names, probably x and y?

400
00:23:35,224 --> 00:23:37,110
Yeah, that's what
I would call them,

401
00:23:37,110 --> 00:23:40,580
so naturally that's
what I called them.

402
00:23:40,580 --> 00:23:44,350
We have x over here,
we have y over here.

403
00:23:44,350 --> 00:23:46,690
And then we have the
perimeter in between.

404
00:23:46,690 --> 00:23:48,420
So let's just
check what happens.

405
00:23:48,420 --> 00:23:51,090
Here we get an
alpha on one side--

406
00:23:51,090 --> 00:23:54,805
I'm going to say the right
side is the bottom, let's say.

407
00:23:54,805 --> 00:23:56,430
I'm not going to try
to think about it.

408
00:23:56,430 --> 00:24:00,695
So there's an alpha,
and that meets a beta.

409
00:24:00,695 --> 00:24:02,320
And then it just
alternates from there.

410
00:24:02,320 --> 00:24:05,850
There's an alpha, it meets a
beta, but on the other side.

411
00:24:05,850 --> 00:24:09,400
And a beta meets
an alpha and so on.

412
00:24:09,400 --> 00:24:14,940
At the end, it's probably the
reverse, like alpha, beta.

413
00:24:14,940 --> 00:24:19,200
I've drawn it up
here at the very top.

414
00:24:19,200 --> 00:24:23,560
The alpha's red, betas are black
so you can distinguish them.

415
00:24:23,560 --> 00:24:26,340
But that's clearly an OK gluing
because it's just one alpha

416
00:24:26,340 --> 00:24:27,180
gluing to one beta.

417
00:24:27,180 --> 00:24:31,040
We know we can glue tons
of alphas into one beta.

418
00:24:31,040 --> 00:24:32,097
That's one gluing.

419
00:24:32,097 --> 00:24:33,930
The fun part is that
there are exponentially

420
00:24:33,930 --> 00:24:38,500
many of these gluings which look
like this, where you take some

421
00:24:38,500 --> 00:24:43,080
of the betas and turn
them into leaves,

422
00:24:43,080 --> 00:24:45,610
kind of squeeze them out.

423
00:24:45,610 --> 00:24:48,430
And the result is that
you get two alphas gluing

424
00:24:48,430 --> 00:24:50,720
to a beta, which is fine.

425
00:24:50,720 --> 00:24:52,630
The beta glues to
itself, which is fine.

426
00:24:52,630 --> 00:24:54,470
I mean, it's a vertex.

427
00:24:54,470 --> 00:24:58,060
It's comforting that it
has an angle almost 360.

428
00:24:58,060 --> 00:25:00,800
Therefore, there's very little
curvature out at that leaf,

429
00:25:00,800 --> 00:25:02,670
but we don't really
need to check that.

430
00:25:02,670 --> 00:25:06,060
It is just necessary
for this to work.

431
00:25:06,060 --> 00:25:10,560
Of course, beta is less
than 360 by itself.

432
00:25:10,560 --> 00:25:12,720
Here's an example where I
squeezed two of the betas

433
00:25:12,720 --> 00:25:15,270
right next to each
other, then three alpha's

434
00:25:15,270 --> 00:25:18,130
come together and
glue to a beta.

435
00:25:18,130 --> 00:25:21,480
But what's nice is we don't
have to think about the polygon.

436
00:25:21,480 --> 00:25:23,020
We just think about
the gluing tree.

437
00:25:23,020 --> 00:25:28,450
And we say, well if
I pull this down-- so

438
00:25:28,450 --> 00:25:31,530
I end up with
that-- that's going

439
00:25:31,530 --> 00:25:34,550
to be just fine as long as
I also pull one on the top.

440
00:25:34,550 --> 00:25:36,900
I have to squeeze the
same number of betas

441
00:25:36,900 --> 00:25:38,400
on the top and the
bottom, otherwise

442
00:25:38,400 --> 00:25:39,820
these links won't match up.

443
00:25:39,820 --> 00:25:43,690
As long as I do that,
I get a valid gluing.

444
00:25:43,690 --> 00:25:49,990
So what I'm going to do,
I have n betas total.

445
00:25:49,990 --> 00:25:55,300
So I have n over
2 betas on the top

446
00:25:55,300 --> 00:25:59,650
and n over 2 betas
on the bottom.

447
00:25:59,650 --> 00:26:08,020
I'm going to squeeze n over
4 on the top and n over 4

448
00:26:08,020 --> 00:26:10,510
on the bottom, so half of them.

449
00:26:16,160 --> 00:26:18,520
That way the lengths
will match up,

450
00:26:18,520 --> 00:26:22,130
and I happen to know there's
a lot of ways to do this.

451
00:26:22,130 --> 00:26:27,620
The actual number is n over
2 choose n over 4 for the top

452
00:26:27,620 --> 00:26:28,950
and the same for the bottom.

453
00:26:28,950 --> 00:26:31,760
So it's this
squared, and I'm just

454
00:26:31,760 --> 00:26:33,960
going to tell you that
is 2 to the theta n.

455
00:26:37,140 --> 00:26:42,090
Actually, I think you
can be pretty precise.

456
00:26:42,090 --> 00:26:43,150
Can you?

457
00:26:43,150 --> 00:26:44,510
I've forgotten.

458
00:26:44,510 --> 00:26:50,410
Isn't n choose n over
2-- it's very close to 2

459
00:26:50,410 --> 00:26:53,104
to the n over 2, I believe.

460
00:26:53,104 --> 00:26:55,660
Is that right?

461
00:26:55,660 --> 00:26:58,010
But not exactly, yeah.

462
00:26:58,010 --> 00:27:00,181
All right, I'm
going to be sloppy.

463
00:27:00,181 --> 00:27:01,760
It's 2 to the theta n.

464
00:27:01,760 --> 00:27:03,860
You could figure out what
the constant is there.

465
00:27:03,860 --> 00:27:09,450
I think I have a bound
written down, maybe.

466
00:27:09,450 --> 00:27:11,980
Oh, it is the power of 2, yeah.

467
00:27:11,980 --> 00:27:14,675
It is equal, I believe,
according to my notes.

468
00:27:17,990 --> 00:27:21,470
So this is going
to be 2 to the n

469
00:27:21,470 --> 00:27:25,415
over 4 squared, which is
2 to the n over 2 exactly.

470
00:27:30,250 --> 00:27:30,844
Yeah?

471
00:27:30,844 --> 00:27:31,760
AUDIENCE: [INAUDIBLE]?

472
00:27:37,260 --> 00:27:39,930
PROFESSOR: I only am allowed
to squeeze betas, not alphas.

473
00:27:39,930 --> 00:27:42,650
But I can do whatever pattern I
want on the top and separately

474
00:27:42,650 --> 00:27:45,150
any pattern I want
on the bottom.

475
00:27:45,150 --> 00:27:47,100
Because what's key
is by squeezing

476
00:27:47,100 --> 00:27:51,350
betas I preserve the parity,
because when I squeeze away

477
00:27:51,350 --> 00:27:53,280
a beta I bring two
alphas together.

478
00:27:53,280 --> 00:27:56,089
So the parity and who's matching
whom on the top and the bottom

479
00:27:56,089 --> 00:27:56,880
is always the same.

480
00:27:56,880 --> 00:27:58,220
And the beta on
the top will always

481
00:27:58,220 --> 00:28:00,270
be matching some number
of alpha's on the bottom,

482
00:28:00,270 --> 00:28:01,887
and vice versa.

483
00:28:01,887 --> 00:28:03,970
So it doesn't matter what
pattern I do on the top,

484
00:28:03,970 --> 00:28:05,360
or what pattern I
do on the bottom.

485
00:28:05,360 --> 00:28:07,110
It just matters they
have the same length,

486
00:28:07,110 --> 00:28:09,987
so I arbitrarily chose
them to both be n over 4.

487
00:28:09,987 --> 00:28:11,570
Well, it wasn't
arbitrary, but I chose

488
00:28:11,570 --> 00:28:12,790
it to be that
because I knew there

489
00:28:12,790 --> 00:28:13,960
would be a lot of those choices.

490
00:28:13,960 --> 00:28:15,043
There are more, of course.

491
00:28:15,043 --> 00:28:18,680
Really I should sum this
over all choices of n over 4,

492
00:28:18,680 --> 00:28:21,200
but this is good enough.

493
00:28:21,200 --> 00:28:23,325
That summation would not
improve the bound by much.

494
00:28:30,870 --> 00:28:32,340
So that's sort of the bad news.

495
00:28:32,340 --> 00:28:33,910
These are really nasty polygons.

496
00:28:33,910 --> 00:28:35,960
They have exponentially
many gluings.

497
00:28:35,960 --> 00:28:43,640
For fun, we actually
did this for small n.

498
00:28:43,640 --> 00:28:45,230
It's been a while.

499
00:28:45,230 --> 00:28:48,440
This was some time ago
that we made all these.

500
00:28:48,440 --> 00:28:50,260
I think it actually
was before 2002,

501
00:28:50,260 --> 00:28:52,980
but that was when
the paper appeared.

502
00:28:52,980 --> 00:28:56,390
So here is a four
star, and these

503
00:28:56,390 --> 00:28:59,430
are I think all of
the possible gluings.

504
00:28:59,430 --> 00:29:01,650
Maybe just some of
them, it's been awhile,

505
00:29:01,650 --> 00:29:02,960
I could think about it.

506
00:29:05,815 --> 00:29:07,940
Not drawn as gluing trees
here, but drawn as actual

507
00:29:07,940 --> 00:29:08,898
gulings on the polygon.

508
00:29:08,898 --> 00:29:10,820
Here instead of
drawing the gluings

509
00:29:10,820 --> 00:29:13,280
on the outside, which
is how they happen

510
00:29:13,280 --> 00:29:15,404
in the gluing three, I've
drawn them on the inside,

511
00:29:15,404 --> 00:29:16,870
just because it's
easier to draw.

512
00:29:16,870 --> 00:29:18,530
And then we built
them in the same way

513
00:29:18,530 --> 00:29:20,988
that I built one like I showed
you last time, where we just

514
00:29:20,988 --> 00:29:24,030
took a polygon, started
taping edges together,

515
00:29:24,030 --> 00:29:27,740
and these are the
roughly taped versions.

516
00:29:27,740 --> 00:29:30,122
And then you guess where
the crease signs are,

517
00:29:30,122 --> 00:29:32,080
and then you can draw
the crease lines on here.

518
00:29:32,080 --> 00:29:34,142
And two of them
are drawn sort of

519
00:29:34,142 --> 00:29:36,350
worked out where exactly
the crease lines have to be,

520
00:29:36,350 --> 00:29:40,030
according to getting all
the edge links to match.

521
00:29:40,030 --> 00:29:41,600
And then you can
make your polygons.

522
00:29:41,600 --> 00:29:47,082
That is n equals 4,
then we did n equals 8.

523
00:29:47,082 --> 00:29:49,165
I guess we want n to be
even here for convenience.

524
00:29:51,750 --> 00:29:54,740
And there's more, although
still not a huge number.

525
00:29:54,740 --> 00:30:01,410
I think these are the top sides
and these are the bottom sides.

526
00:30:01,410 --> 00:30:03,700
And you can see, in some
cases it was a little tricky

527
00:30:03,700 --> 00:30:05,700
to tell where the creases
are, because they were

528
00:30:05,700 --> 00:30:08,130
almost flat or possibly
actually exactly flat.

529
00:30:08,130 --> 00:30:10,060
Like here they are flat.

530
00:30:10,060 --> 00:30:11,540
But it's a fun exercise.

531
00:30:11,540 --> 00:30:14,097
You can, whenever you
have these gluings,

532
00:30:14,097 --> 00:30:15,180
reconstruct the polyhedra.

533
00:30:15,180 --> 00:30:18,180
Of course now that we have
algorithms for Alexandroff's

534
00:30:18,180 --> 00:30:21,230
theorem, we could try plugging
these into the programs

535
00:30:21,230 --> 00:30:26,290
and they should give us exact 3D
polyhedra, from which we could

536
00:30:26,290 --> 00:30:29,252
figure out where the
crease lines are.

537
00:30:29,252 --> 00:30:32,640
It would be a fun, small,
mini-project, I guess,

538
00:30:32,640 --> 00:30:33,290
to do that.

539
00:30:33,290 --> 00:30:34,900
This is all from ages ago.

540
00:30:37,800 --> 00:30:44,140
At this point I want to move
on to this result, which

541
00:30:44,140 --> 00:30:44,910
is the good news.

542
00:30:44,910 --> 00:30:47,118
This is sort of bad news
that there are exponentially

543
00:30:47,118 --> 00:30:49,910
many different gluings.

544
00:30:49,910 --> 00:30:52,425
It's not so bad in that at
least we can enumerate them

545
00:30:52,425 --> 00:30:53,550
in the same amount of time.

546
00:30:53,550 --> 00:30:54,934
We'll get to that later.

547
00:30:54,934 --> 00:30:56,600
While we're in the
combinatorial streak,

548
00:30:56,600 --> 00:30:58,380
I want to prove
that there's only

549
00:30:58,380 --> 00:31:00,790
a polynomial number
of combinatorially

550
00:31:00,790 --> 00:31:04,250
distinct gluings for
bounded sharpness polygons.

551
00:31:04,250 --> 00:31:07,050
So of course, this
star polygon does not

552
00:31:07,050 --> 00:31:08,510
have bounded sharpness.

553
00:31:08,510 --> 00:31:14,160
We set beta to be very,
very close to 360.

554
00:31:14,160 --> 00:31:16,830
It may be worth talking
about this a little bit,

555
00:31:16,830 --> 00:31:21,020
because it wasn't exactly 360.

556
00:31:21,020 --> 00:31:24,870
It was bounded away from 360 by
this value of 360 divided by n.

557
00:31:24,870 --> 00:31:26,910
Now I said that
was a big number,

558
00:31:26,910 --> 00:31:30,680
because it was much,
much bigger than alpha.

559
00:31:30,680 --> 00:31:32,780
Bigger than n times
alpha even, because I

560
00:31:32,780 --> 00:31:35,410
can make alpha very close to 0.

561
00:31:35,410 --> 00:31:37,420
But it is not what we
call bounded sharpness.

562
00:31:37,420 --> 00:31:41,120
Bounded sharpness was 360
minus a constant-- I really

563
00:31:41,120 --> 00:31:44,070
shouldn't have called it
epsilon because it's confusing.

564
00:31:44,070 --> 00:31:47,540
Let me call it gamma just
got another Greek letter.

565
00:31:50,310 --> 00:31:51,810
Gamma is not arbitrarily small.

566
00:31:51,810 --> 00:31:56,780
It is a constant value, whereas
over here it's not constant.

567
00:31:56,780 --> 00:32:00,640
It's a constant divided
by n, that's sub-constant.

568
00:32:00,640 --> 00:32:03,740
So this is not bounded
sharpness, just to check.

569
00:32:03,740 --> 00:32:07,830
But if we have all the
reflex angles, at most 360

570
00:32:07,830 --> 00:32:10,410
minus an actual
constant bigger than 0,

571
00:32:10,410 --> 00:32:12,400
then we can get a
polynomial bound

572
00:32:12,400 --> 00:32:15,740
on the number of combinatorially
distinct gluings.

573
00:32:15,740 --> 00:32:17,167
So let me show that to you.

574
00:32:43,494 --> 00:32:45,410
Here's the cool thing
about bounded sharpness.

575
00:32:53,190 --> 00:32:57,100
How many leaves can
the gluing tree have?

576
00:32:57,100 --> 00:33:01,150
Each leaf is either an edge,
like it's a fold point.

577
00:33:01,150 --> 00:33:05,120
In that case, the
curvature is 180 degrees.

578
00:33:05,120 --> 00:33:07,310
Or it's a vertex.

579
00:33:07,310 --> 00:33:12,050
If it's a vertex, the
curvature at that point

580
00:33:12,050 --> 00:33:16,094
is 360 minus the
angle at that point.

581
00:33:16,094 --> 00:33:22,020
360 minus the angle,
let's call it alpha.

582
00:33:22,020 --> 00:33:24,980
Now, we know that every
angle alpha is at most 360

583
00:33:24,980 --> 00:33:27,040
minus gamma.

584
00:33:27,040 --> 00:33:35,470
So if this is going to be
at least 360 minus gamma,

585
00:33:35,470 --> 00:33:36,390
which is just gamma.

586
00:33:39,070 --> 00:33:43,540
So every vertex, every
leaf of the tree,

587
00:33:43,540 --> 00:33:46,370
has curvature at least gamma.

588
00:33:46,370 --> 00:33:48,100
Gamma is a constant.

589
00:33:48,100 --> 00:33:58,450
Total curvature is 720, exactly.

590
00:33:58,450 --> 00:34:05,150
Therefore, the number
of leaves is at most 720

591
00:34:05,150 --> 00:34:09,480
over gamma, which is a constant.

592
00:34:12,880 --> 00:34:14,780
That's going to help.

593
00:34:14,780 --> 00:34:18,625
We don't have to worry
about trees with n leaves.

594
00:34:18,625 --> 00:34:20,000
We only have to
worry about trees

595
00:34:20,000 --> 00:34:23,679
with 20 leaves, some
constant number of leaves.

596
00:34:23,679 --> 00:34:28,090
Again, as the bound on sharpness
goes down, the number of leaves

597
00:34:28,090 --> 00:34:31,560
will go up in this
inverse proportional way,

598
00:34:31,560 --> 00:34:33,690
but it's a constant.

599
00:34:33,690 --> 00:34:37,400
So at least for part
one, we're golden.

600
00:34:37,400 --> 00:34:39,370
I mean, the number
of gluing trees,

601
00:34:39,370 --> 00:34:43,060
I said over here-- where is it?

602
00:34:43,060 --> 00:34:45,440
If I have n nodes
in my tree, I'll

603
00:34:45,440 --> 00:34:48,900
only have 2 to
the order n trees.

604
00:34:48,900 --> 00:34:51,090
So if I can reduce the
number of nodes in my tree,

605
00:34:51,090 --> 00:34:53,173
I will reduce the number
of trees correspondingly.

606
00:34:56,940 --> 00:34:59,000
If there's only a
constant number of nodes,

607
00:34:59,000 --> 00:35:00,983
there will be a constant
number of trees.

608
00:35:00,983 --> 00:35:03,880
It doesn't matter that a
constant gets exponentiated,

609
00:35:03,880 --> 00:35:06,520
it'll still be constant.

610
00:35:06,520 --> 00:35:08,570
So that's good news,
except we bounded again

611
00:35:08,570 --> 00:35:11,410
the number of leaves,
not the number of nodes.

612
00:35:11,410 --> 00:35:14,460
So we have to be a
little bit careful,

613
00:35:14,460 --> 00:35:23,240
because in particular we can
have degree 2 junctions, which

614
00:35:23,240 --> 00:35:25,400
are real junctions--
they might look like this

615
00:35:25,400 --> 00:35:27,820
where a vertex comes to an edge.

616
00:35:27,820 --> 00:35:30,100
There's got to be
about n of those,

617
00:35:30,100 --> 00:35:32,250
because the vertices
have to be somewhere.

618
00:35:32,250 --> 00:35:36,150
We're saying there aren't
very many at the leaves.

619
00:35:36,150 --> 00:35:38,580
And therefore, at the degree
three and higher junctions,

620
00:35:38,580 --> 00:35:40,559
there can't be very many.

621
00:35:40,559 --> 00:35:42,100
Maybe I haven't said
this, but if you

622
00:35:42,100 --> 00:35:44,150
have a tree with
L leaves, you'll

623
00:35:44,150 --> 00:35:49,210
only have at most L internal
nodes, branching nodes.

624
00:35:49,210 --> 00:35:52,710
But degree two nodes, there
could be arbitrarily many.

625
00:35:52,710 --> 00:35:59,510
So what we come to is, if
I have L leaves in a tree,

626
00:35:59,510 --> 00:36:03,055
then I'll only get 2
to the order L gluings.

627
00:36:03,055 --> 00:36:04,305
This is what we need to prove.

628
00:36:08,622 --> 00:36:10,080
It will tell us
that in particular,

629
00:36:10,080 --> 00:36:11,770
if we have a constant
number of leaves,

630
00:36:11,770 --> 00:36:14,730
as comes from the
bounded sharpens case,

631
00:36:14,730 --> 00:36:17,900
I will get-- constant
number of gluings?

632
00:36:17,900 --> 00:36:20,480
That's not right.

633
00:36:20,480 --> 00:36:25,020
There's some polynomial
on n times 2 to the order

634
00:36:25,020 --> 00:36:31,280
L. Ah, that's why.

635
00:36:36,090 --> 00:36:38,060
I should look at my
notes occasionally.

636
00:36:38,060 --> 00:36:39,940
The right bound to
n to the order L.

637
00:36:39,940 --> 00:36:43,210
In fact, what I wrote
is an open problem.

638
00:36:43,210 --> 00:36:48,420
I think maybe it's n to the
order 1 times 2 to the order L.

639
00:36:48,420 --> 00:36:52,386
This is what you'd call a fixed
parameter tractable bound,

640
00:36:52,386 --> 00:36:53,760
but no such upper
bound is known.

641
00:36:53,760 --> 00:36:55,926
That might be fun to work
on in the problem session.

642
00:36:55,926 --> 00:36:58,250
It might not be difficult,
just at the time

643
00:36:58,250 --> 00:37:00,000
I didn't know those
bounds were important.

644
00:37:03,070 --> 00:37:05,720
So we still have
to prove this, n

645
00:37:05,720 --> 00:37:15,840
to the order L. This is the
part where my notes are wrong.

646
00:37:21,207 --> 00:37:23,290
Here's the idea, we have
a small number of leaves,

647
00:37:23,290 --> 00:37:26,200
we're worried about
these degree 2 junctions.

648
00:37:26,200 --> 00:37:30,420
But I claim, actually
there isn't a lot going on.

649
00:37:33,690 --> 00:37:35,550
Because one of the
great things we can do,

650
00:37:35,550 --> 00:37:38,890
over here we said for
every dot, was it a vertex

651
00:37:38,890 --> 00:37:39,715
or was it an edge?

652
00:37:42,640 --> 00:37:46,490
We can't afford that anymore,
because there's n dots still.

653
00:37:46,490 --> 00:37:48,880
We can't afford
2 to the order n.

654
00:37:48,880 --> 00:37:51,680
Where we can afford it is at
the leaves, because there's only

655
00:37:51,680 --> 00:37:55,220
L of those, and at
the branching nodes.

656
00:37:55,220 --> 00:37:59,320
So everything except
these degree 2 problems

657
00:37:59,320 --> 00:38:00,580
we could specify.

658
00:38:00,580 --> 00:38:03,180
And really we could
specify anything we want.

659
00:38:03,180 --> 00:38:07,040
So yeah, we could specify, is
it a vertex or is it an edge?

660
00:38:07,040 --> 00:38:12,644
But we could make
life even easier-- hm,

661
00:38:12,644 --> 00:38:14,800
I have an idea for
solving this open problem.

662
00:38:14,800 --> 00:38:17,530
Do I want to solve it right now?

663
00:38:17,530 --> 00:38:18,520
Yeah, let's do it.

664
00:38:22,610 --> 00:38:25,126
Let me think for a second.

665
00:38:25,126 --> 00:38:26,910
I think that's going to work.

666
00:38:26,910 --> 00:38:29,160
OK, cool, cool.

667
00:38:29,160 --> 00:38:31,240
I needed to rewrite
these notes anyway,

668
00:38:31,240 --> 00:38:34,560
so I might as well
prove a stronger result.

669
00:38:34,560 --> 00:38:36,600
All right, so let
me draw a picture.

670
00:38:42,287 --> 00:38:44,870
There's branching nodes, and I
surely draw one of these longer

671
00:38:44,870 --> 00:38:47,300
so I can say, oh there's some
degree two junctions here.

672
00:38:47,300 --> 00:38:49,640
Those are annoying,
ignore those.

673
00:38:49,640 --> 00:38:52,950
For everything else-- the
leaves and the branching

674
00:38:52,950 --> 00:38:56,530
nodes-- I'm going to specify
the same binary thing.

675
00:38:56,530 --> 00:38:58,990
Is it a vertex,
or is it an edge?

676
00:38:58,990 --> 00:39:02,160
But now there's only
order L of them.

677
00:39:02,160 --> 00:39:05,610
There's L leaves, so there's at
most L minus 1 internal nodes.

678
00:39:05,610 --> 00:39:09,710
Each of them has some
degree, but the total number

679
00:39:09,710 --> 00:39:14,510
of these dots will be order
L if I ignore the degree 2.

680
00:39:14,510 --> 00:39:23,320
So that's 2 to the order
L vertex edge colorings,

681
00:39:23,320 --> 00:39:30,570
just like before
except now with L,

682
00:39:30,570 --> 00:39:37,420
of leaves and branching nodes.

683
00:39:37,420 --> 00:39:39,023
Oh, I see the problem.

684
00:39:39,023 --> 00:39:39,523
Darn it.

685
00:39:44,730 --> 00:39:48,240
What I'd like to do is mimic
the same proof and say, well

686
00:39:48,240 --> 00:39:53,141
choose where V1 is and then
just label them around.

687
00:39:53,141 --> 00:39:54,640
There's a problem
with that, though.

688
00:39:54,640 --> 00:39:55,530
Say V1 is here.

689
00:39:58,460 --> 00:39:59,840
Where's V2?

690
00:39:59,840 --> 00:40:02,430
Well, maybe this is an
edge, maybe this is an edge.

691
00:40:02,430 --> 00:40:03,960
This, I have no idea.

692
00:40:03,960 --> 00:40:05,030
Is it an edge?

693
00:40:05,030 --> 00:40:06,310
Is it a vertex?

694
00:40:06,310 --> 00:40:07,860
I can't tell.

695
00:40:07,860 --> 00:40:11,230
At degree 2 junctions,
I have a problem.

696
00:40:11,230 --> 00:40:13,770
That's kind of annoying.

697
00:40:13,770 --> 00:40:18,670
To fix that-- and so this is
still going to be open, alas.

698
00:40:18,670 --> 00:40:23,870
To fix that I need to-- I
really want to know which dot

699
00:40:23,870 --> 00:40:28,090
is which vertex for
reasons to be determined.

700
00:40:28,090 --> 00:40:30,750
But in particular, that
is part of number two.

701
00:40:33,980 --> 00:40:41,460
So I'm just going to give
up and say, well there's

702
00:40:41,460 --> 00:40:45,805
2n different things that they
could be, n vertices and edges.

703
00:40:48,940 --> 00:40:52,230
Do I want to do it that way?

704
00:40:52,230 --> 00:40:57,830
Maybe n choose order L. I'm
just going to write down

705
00:40:57,830 --> 00:41:01,790
wherever I have a filled dot--
I guess this one is not filled.

706
00:41:01,790 --> 00:41:04,191
Maybe this one's filled, filled.

707
00:41:04,191 --> 00:41:06,690
I just going to write down,
what is the vertex number there?

708
00:41:06,690 --> 00:41:17,480
This is maybe V3, V5,
V20, V21, whatever,

709
00:41:17,480 --> 00:41:19,490
for each of the filled
dots among the vertices

710
00:41:19,490 --> 00:41:23,850
that I can see, ignoring again
the degree two junctions.

711
00:41:23,850 --> 00:41:26,965
There's order L of them.

712
00:41:26,965 --> 00:41:31,041
I just going to for each one
pick out one of the labels.

713
00:41:31,041 --> 00:41:33,000
Again, the order is
determined, so this

714
00:41:33,000 --> 00:41:34,560
is the right number of them.

715
00:41:34,560 --> 00:41:40,270
And this is n to
the order L at most.

716
00:41:40,270 --> 00:41:42,187
So this is where I'm
being a bit wasteful.

717
00:41:42,187 --> 00:41:44,270
It would be interesting
to try to get around that.

718
00:41:47,300 --> 00:41:49,455
This was actual
vertex labelings.

719
00:41:56,040 --> 00:41:59,200
So now, among the
vertices I can see--

720
00:41:59,200 --> 00:42:02,490
the degree three
junctions, and the leaves,

721
00:42:02,490 --> 00:42:05,069
and higher degree
junctions-- I know

722
00:42:05,069 --> 00:42:06,360
which vertices are glued there.

723
00:42:06,360 --> 00:42:10,325
I can also figure out
which edges are glued there

724
00:42:10,325 --> 00:42:14,716
by similar labeling, although
I don't think I actually care.

725
00:42:14,716 --> 00:42:19,610
No, I probably care, because
it's part of number 2.

726
00:42:19,610 --> 00:42:23,110
So I do have to also label
the open circles which

727
00:42:23,110 --> 00:42:25,150
edge is there.

728
00:42:25,150 --> 00:42:28,890
But a similar bound holds.

729
00:42:28,890 --> 00:42:29,405
What's left?

730
00:42:33,560 --> 00:42:35,676
What's left are the
degree 2 vertices.

731
00:42:35,676 --> 00:42:37,050
I don't know
anything about them.

732
00:42:40,690 --> 00:42:47,330
This is particularly tricky
when I have some edge here.

733
00:42:47,330 --> 00:42:49,190
But let's first
think about the case

734
00:42:49,190 --> 00:42:53,990
where I have a
vertex, add a leaf,

735
00:42:53,990 --> 00:42:58,140
and I have these
degree two junctions.

736
00:42:58,140 --> 00:43:01,392
And I don't know which edge
is here, which vertex is here.

737
00:43:01,392 --> 00:43:03,100
In this case, because
everything's small,

738
00:43:03,100 --> 00:43:04,350
I might be able to figure out.

739
00:43:04,350 --> 00:43:07,960
But here, where's the
V3 to V4 transition?

740
00:43:07,960 --> 00:43:10,900
Could be here, or
it could be here.

741
00:43:10,900 --> 00:43:13,040
It could be here.

742
00:43:13,040 --> 00:43:16,330
I can't afford to figure it out.

743
00:43:16,330 --> 00:43:20,720
Fortunately, I don't
have to figure it out.

744
00:43:20,720 --> 00:43:21,580
It's not so obvious.

745
00:43:21,580 --> 00:43:23,920
From this picture you
can't see what's going on,

746
00:43:23,920 --> 00:43:26,520
but this came from a polygon.

747
00:43:26,520 --> 00:43:30,290
This is an actual vertex,
this is an actual vertex.

748
00:43:30,290 --> 00:43:37,450
So this edge length is the
total perimeter from V3 to V5

749
00:43:37,450 --> 00:43:39,690
along the polygon.

750
00:43:39,690 --> 00:43:43,040
I know how long that is.

751
00:43:43,040 --> 00:43:48,760
So also if that was
going clockwise from V3

752
00:43:48,760 --> 00:43:50,322
around the polygon--
actually, that's

753
00:43:50,322 --> 00:43:52,030
probably counter-clockwise
in the polygon

754
00:43:52,030 --> 00:43:53,420
because everything's inside out.

755
00:43:53,420 --> 00:43:57,255
If I go the other way--
let's be relative-- from V3,

756
00:43:57,255 --> 00:44:00,560
I can measure out
the same length.

757
00:44:00,560 --> 00:44:02,520
So in fact, I have
some polygon-- maybe

758
00:44:02,520 --> 00:44:04,050
it looks like this, whatever.

759
00:44:04,050 --> 00:44:11,270
I go from V3 to V5, and I
measure out that length.

760
00:44:11,270 --> 00:44:14,470
And I go the other
way from V3, and I

761
00:44:14,470 --> 00:44:17,910
know this is getting
glued to that.

762
00:44:17,910 --> 00:44:23,770
So I know where the vertices
are from the lengths.

763
00:44:23,770 --> 00:44:26,270
I can figure out exactly
what this picture must

764
00:44:26,270 --> 00:44:28,314
be when I have-- in
this case, I happen

765
00:44:28,314 --> 00:44:30,230
to maybe get a vertex
going to another vertex.

766
00:44:30,230 --> 00:44:34,310
In general, probably get some
vertex gluing to some edge,

767
00:44:34,310 --> 00:44:35,992
and maybe vice versa.

768
00:44:35,992 --> 00:44:38,200
But if I can figure out
exactly what that pattern is,

769
00:44:38,200 --> 00:44:42,660
I can figure out the
label of these guys,

770
00:44:42,660 --> 00:44:46,015
because I know exactly
what's happening there.

771
00:44:46,015 --> 00:44:46,640
So that's good.

772
00:44:46,640 --> 00:44:54,920
If it's a vertex to vertex edge
in the gluing tree like this,

773
00:44:54,920 --> 00:44:58,460
I can figure out everything
between that pair.

774
00:45:08,486 --> 00:45:10,737
Is that enough?

775
00:45:10,737 --> 00:45:11,236
Almost.

776
00:45:17,800 --> 00:45:22,116
So if I have a leaf where
this is a vertex, which

777
00:45:22,116 --> 00:45:26,260
I was usually denoting
by a filled circle,

778
00:45:26,260 --> 00:45:27,700
I have various
degree 2 junctions

779
00:45:27,700 --> 00:45:31,500
which I ignore until I get
to a degree three or higher

780
00:45:31,500 --> 00:45:33,600
junction.

781
00:45:33,600 --> 00:45:37,420
Now at this place,
even just looking

782
00:45:37,420 --> 00:45:40,990
at this dot and this dot,
I know at least one of them

783
00:45:40,990 --> 00:45:41,830
must be a vertex.

784
00:45:41,830 --> 00:45:45,440
Because I can't glue two edges
together plus other stuff.

785
00:45:45,440 --> 00:45:46,610
I can glue at most one edge.

786
00:45:46,610 --> 00:45:48,810
So maybe this one's an
edge, like in the picture

787
00:45:48,810 --> 00:45:50,360
that we did before.

788
00:45:50,360 --> 00:45:51,830
But one of them
has to be a vertex.

789
00:45:51,830 --> 00:45:54,770
And then I can figure out
exactly what's happening there,

790
00:45:54,770 --> 00:45:56,720
because I have this
vertex to vertex thing.

791
00:45:56,720 --> 00:45:57,770
And I know the labels.

792
00:45:57,770 --> 00:46:01,884
I know this is V5 and
this is V13, whatever.

793
00:46:01,884 --> 00:46:04,050
I know exactly what's
happening on the right side, I

794
00:46:04,050 --> 00:46:05,560
measure backwards, I
know exactly what's

795
00:46:05,560 --> 00:46:07,770
happening on the left side
and who's going to what.

796
00:46:07,770 --> 00:46:10,350
So that determines everything.

797
00:46:10,350 --> 00:46:15,910
The problem is when
I have an edge here,

798
00:46:15,910 --> 00:46:18,370
because then I don't know
where it is along the edge.

799
00:46:18,370 --> 00:46:20,917
I can't measure lengths
anymore, somewhere

800
00:46:20,917 --> 00:46:22,000
in the middle of the edge.

801
00:46:24,890 --> 00:46:33,010
And indeed, if I have
something like this,

802
00:46:33,010 --> 00:46:35,690
where all of these angles--
I mean this for example

803
00:46:35,690 --> 00:46:41,030
is in a convex polygon, this
could be a perimeter halving--

804
00:46:41,030 --> 00:46:43,520
and it's not determined
what's happening here

805
00:46:43,520 --> 00:46:46,860
because this is a rolling belt.

806
00:46:46,860 --> 00:46:48,730
I'll show you how
undetermined it is.

807
00:46:48,730 --> 00:46:52,110
Suppose you have a bunch
of vertices like this,

808
00:46:52,110 --> 00:46:58,690
and then some
vertices like this.

809
00:46:58,690 --> 00:47:02,970
Depending on where I
bend around for E5,

810
00:47:02,970 --> 00:47:04,950
these guys could be here.

811
00:47:04,950 --> 00:47:07,980
They could be here,
or here, or here.

812
00:47:07,980 --> 00:47:12,720
They could be in the
middle straddling this.

813
00:47:12,720 --> 00:47:17,720
In fact, there are about
n squared different places

814
00:47:17,720 --> 00:47:20,780
where these guys could be,
relative to these guys.

815
00:47:20,780 --> 00:47:24,370
Because there's going to be n
transitions from-- if this is n

816
00:47:24,370 --> 00:47:28,500
and this is n-- n transitions
from this versus that guy,

817
00:47:28,500 --> 00:47:30,880
then n transitions from
these versus that guy,

818
00:47:30,880 --> 00:47:35,530
n transitions versus that
guy, total of about n squared.

819
00:47:35,530 --> 00:47:39,440
Actually if this is n and this
is n, it's exactly n squared.

820
00:47:39,440 --> 00:47:41,690
So we're not quite
done, but I claim

821
00:47:41,690 --> 00:47:44,220
that there's really only
order n squared left.

822
00:47:44,220 --> 00:47:46,020
It's always from
the rolling belts.

823
00:47:46,020 --> 00:47:51,370
We know from last time there's
at most three rolling belts,

824
00:47:51,370 --> 00:47:54,050
so at most quadratic per belt.

825
00:47:54,050 --> 00:47:59,660
So total outcome is
polynomial for what's

826
00:47:59,660 --> 00:48:02,980
left, what we haven't specified,
for how the degree 2 vertices

827
00:48:02,980 --> 00:48:03,520
behave.

828
00:48:03,520 --> 00:48:05,710
These are all
degree 2 junctions.

829
00:48:05,710 --> 00:48:09,676
It's here, or they're
here, we don't know.

830
00:48:09,676 --> 00:48:13,040
So if you want to get the full
specification of what's where,

831
00:48:13,040 --> 00:48:14,657
you do all this work.

832
00:48:14,657 --> 00:48:16,240
And then for the
degree two junctions,

833
00:48:16,240 --> 00:48:18,712
you have to deal with
the rolling belts.

834
00:48:18,712 --> 00:48:20,170
The way to prove
this is say, well,

835
00:48:20,170 --> 00:48:23,190
if I have a leaf that's a
vertex, I can get rid of it

836
00:48:23,190 --> 00:48:24,510
and induct.

837
00:48:24,510 --> 00:48:27,020
In the end, I'll have
at most four leaves

838
00:48:27,020 --> 00:48:29,440
because there's at
most four fold points.

839
00:48:29,440 --> 00:48:32,810
So I have a constant
number of leaves.

840
00:48:32,810 --> 00:48:34,400
You look at each
belt, each of them

841
00:48:34,400 --> 00:48:40,020
is quadratic and polynomial.

842
00:48:40,020 --> 00:48:41,870
If you work out the
bounds here-- we've

843
00:48:41,870 --> 00:48:43,670
tried to be a little
more precise--

844
00:48:43,670 --> 00:48:49,880
you could prove an overall
bound of n to the 2L minus 2

845
00:48:49,880 --> 00:48:50,580
for L leaves.

846
00:48:53,220 --> 00:48:56,330
And we have slightly better
bounds for L equals 4

847
00:48:56,330 --> 00:49:02,370
and L equals 3, but
that's pretty good.

848
00:49:02,370 --> 00:49:05,910
But the proof I gave is
at least n to the order L.

849
00:49:05,910 --> 00:49:09,770
We could n to the order L
count different things here,

850
00:49:09,770 --> 00:49:12,210
and then some N to the
constant at the end.

851
00:49:18,200 --> 00:49:20,330
This ends the
combinatorial part.

852
00:49:20,330 --> 00:49:22,820
Now we're going to
get to algorithms.

853
00:49:22,820 --> 00:49:25,420
So we've proved all
of these things.

854
00:49:25,420 --> 00:49:27,070
Exponential upper
and lower bounds

855
00:49:27,070 --> 00:49:29,290
for the general case
and edge-to-edge case,

856
00:49:29,290 --> 00:49:32,800
and polynomial for
bounded sharpness,

857
00:49:32,800 --> 00:49:35,750
or when you have a reasonable
number of leaves in your tree.

858
00:50:08,300 --> 00:50:14,940
I'm going to start with this
edge-to-edge gluing algorithm.

859
00:50:32,140 --> 00:50:39,350
This is going to use a technique
called dynamic programming,

860
00:50:39,350 --> 00:50:43,590
which some of you have seen
before, some of you haven't.

861
00:50:43,590 --> 00:50:45,190
It's a very simple idea.

862
00:50:45,190 --> 00:50:46,720
You take your
problem that you want

863
00:50:46,720 --> 00:50:48,900
to solve-- which is
I have a polygon--

864
00:50:48,900 --> 00:50:52,146
and you split it up
into sub-problems.

865
00:50:52,146 --> 00:50:54,830
You solve each of
the sub-problems,

866
00:50:54,830 --> 00:50:57,570
and one of the sub-problems
is actually the whole problem,

867
00:50:57,570 --> 00:50:59,500
and then you're done.

868
00:50:59,500 --> 00:51:02,080
In our case, a
sub-problem is going

869
00:51:02,080 --> 00:51:05,950
to be some sub-chain
from one vertex

870
00:51:05,950 --> 00:51:09,250
to another vertex
of the polygon.

871
00:51:15,680 --> 00:51:18,640
So I have some
polygon, and I just

872
00:51:18,640 --> 00:51:26,060
look at some interval of
the boundary from VI to VJ.

873
00:51:26,060 --> 00:51:31,730
And the idea is, well suppose
that VI is glued to VJ.

874
00:51:31,730 --> 00:51:34,070
How many different
ways, or how should I

875
00:51:34,070 --> 00:51:35,170
glue the part in between?

876
00:51:38,540 --> 00:51:40,549
How should I glue
this into itself?

877
00:51:40,549 --> 00:51:42,090
Because there's no
crossings, there's

878
00:51:42,090 --> 00:51:45,900
going to be no gluings
from inside to outside.

879
00:51:45,900 --> 00:51:47,510
So that's the sub-problem.

880
00:51:47,510 --> 00:51:53,700
And of course if I say from
the interval from V1 to V1,

881
00:51:53,700 --> 00:51:55,500
that is the entire problem.

882
00:51:55,500 --> 00:51:57,440
Of course V1 is glued to V1.

883
00:51:57,440 --> 00:52:02,980
So in particular, the
V1 to V1 sub-problem

884
00:52:02,980 --> 00:52:05,740
is what we want to solve.

885
00:52:05,740 --> 00:52:08,549
So this is our goal.

886
00:52:08,549 --> 00:52:10,090
So I'm going to tell
you how to solve

887
00:52:10,090 --> 00:52:11,756
all of these sub-problems,
and therefore

888
00:52:11,756 --> 00:52:13,960
how to solve what we
actually want to solve.

889
00:52:13,960 --> 00:52:17,530
That's dynamic programming.

890
00:52:17,530 --> 00:52:22,992
So let's try to solve one of
these sub-problems, VI to VJ.

891
00:52:22,992 --> 00:52:24,200
How are we going to solve it?

892
00:52:27,710 --> 00:52:31,900
So I'm going to draw a picture.

893
00:52:31,900 --> 00:52:33,789
It looks like this.

894
00:52:33,789 --> 00:52:35,580
This is now thinking
about the gluing tree,

895
00:52:35,580 --> 00:52:39,070
so the polygon's on
the outside here.

896
00:52:39,070 --> 00:52:41,960
That's one of the main reasons
gluing trees were developed,

897
00:52:41,960 --> 00:52:43,450
just to describe this algorithm.

898
00:52:43,450 --> 00:52:45,540
This algorthim's
actually quite old,

899
00:52:45,540 --> 00:52:49,030
1996 by [INAUDIBLE]
and [INAUDIBLE].

900
00:52:49,030 --> 00:52:51,970
The edge-to-edge case was
before I joined this group.

901
00:52:51,970 --> 00:52:55,104
And then later together
we did the general case,

902
00:52:55,104 --> 00:52:57,145
unbounded sharpness, which
is what we'll do next.

903
00:52:59,737 --> 00:53:01,320
So this is at the
heart of everything.

904
00:53:01,320 --> 00:53:04,170
So we have VO to
VJ glued together.

905
00:53:04,170 --> 00:53:07,620
We want to know how to
glue together the rest.

906
00:53:07,620 --> 00:53:11,290
Well, maybe some vertices
get glued in here in addition

907
00:53:11,290 --> 00:53:15,890
to VI and VJ, maybe not.

908
00:53:15,890 --> 00:53:18,610
Let's think about what
gets glued to this edge.

909
00:53:18,610 --> 00:53:21,629
There's the edge
from VI to VI plus 1.

910
00:53:21,629 --> 00:53:23,420
And this is an edge-to-edge
gluing, so what

911
00:53:23,420 --> 00:53:25,280
gets glued there
is an actual edge.

912
00:53:27,790 --> 00:53:28,820
What edge could it be?

913
00:53:28,820 --> 00:53:33,256
Some edge over here, I
don't know which one.

914
00:53:33,256 --> 00:53:34,950
The power of the
dynamic programming

915
00:53:34,950 --> 00:53:36,330
is you don't need to know.

916
00:53:36,330 --> 00:53:41,490
It's some edge, VK, VK
plus 1 gets glued there.

917
00:53:44,920 --> 00:53:47,710
There's only one of them,
it's out there somewhere.

918
00:53:47,710 --> 00:53:49,406
It has to have
matching edge length,

919
00:53:49,406 --> 00:53:51,280
but there could be many
choices, all of which

920
00:53:51,280 --> 00:53:52,363
have matching edge length.

921
00:53:52,363 --> 00:53:54,220
Maybe all the edge
lengths are the same,

922
00:53:54,220 --> 00:54:05,550
but there's at most n
choices for that edge.

923
00:54:05,550 --> 00:54:08,940
It's actually at most J
minus I choices plus 1.

924
00:54:08,940 --> 00:54:12,716
But there's at most
n edges in here,

925
00:54:12,716 --> 00:54:14,590
so there's at most n
different possibilities.

926
00:54:14,590 --> 00:54:16,820
Just try them all.

927
00:54:16,820 --> 00:54:18,640
Think about all of them.

928
00:54:18,640 --> 00:54:20,910
I haven't said here whether
I'm doing-- or I have?

929
00:54:20,910 --> 00:54:22,962
I guess I've been pointing here.

930
00:54:22,962 --> 00:54:25,170
I really want to do both of
these algorithms at once,

931
00:54:25,170 --> 00:54:28,210
but I'll start with this
one, which is just find

932
00:54:28,210 --> 00:54:30,130
all the gluings in
exponential time, which

933
00:54:30,130 --> 00:54:31,880
is optimal in the worst
case because there

934
00:54:31,880 --> 00:54:33,296
can be exponentially
many outputs.

935
00:54:37,330 --> 00:54:41,020
So try all n choices
for VK VK plus 1.

936
00:54:41,020 --> 00:54:43,410
What happens when
I do one of them?

937
00:54:43,410 --> 00:54:44,570
I might choose one of them.

938
00:54:49,570 --> 00:55:00,190
Well, the new picture will
be-- so the topological picture

939
00:55:00,190 --> 00:55:03,400
is I still have VI glued to VJ.

940
00:55:06,000 --> 00:55:13,020
Then I also have here VK and
VI plus 1 glued together.

941
00:55:13,020 --> 00:55:18,339
And I have VI also
glued to VK plus 1.

942
00:55:18,339 --> 00:55:20,130
Now it's possible
actually VK plus 1 and VJ

943
00:55:20,130 --> 00:55:22,864
are the same thing, in which
case there's no loop here.

944
00:55:22,864 --> 00:55:24,530
And that's the case
when no extra vertex

945
00:55:24,530 --> 00:55:25,710
got glued in there.

946
00:55:25,710 --> 00:55:29,266
But this is the
generic situation.

947
00:55:29,266 --> 00:55:33,140
If this is link 0, then
these guys are the same.

948
00:55:33,140 --> 00:55:36,530
So really what I have now are
two different sub-problems.

949
00:55:36,530 --> 00:55:41,520
There's this thing, which
is VI plus 1, dot, dot, VK.

950
00:55:41,520 --> 00:55:45,685
And there's this one which
is VK plus 1 dot, dot, VJ.

951
00:55:48,760 --> 00:55:50,370
The idea with the
sub-problems is

952
00:55:50,370 --> 00:55:52,120
to solve the
sub-problems in order

953
00:55:52,120 --> 00:55:54,540
by increasing chain length.

954
00:55:54,540 --> 00:55:57,020
So start with very short ones.

955
00:55:57,020 --> 00:55:59,860
And I saw longer and longer
ones using the previous results.

956
00:55:59,860 --> 00:56:02,160
If I've already
computed the answers

957
00:56:02,160 --> 00:56:04,976
to VK plus 1 to VJ
because that is smaller

958
00:56:04,976 --> 00:56:06,350
than the original--
look, there's

959
00:56:06,350 --> 00:56:08,480
two edges fewer at least.

960
00:56:08,480 --> 00:56:11,130
This thing is at least 2
smaller than the original chain

961
00:56:11,130 --> 00:56:12,720
because here's two edges.

962
00:56:12,720 --> 00:56:14,520
So I wanted to solve VI to VJ.

963
00:56:14,520 --> 00:56:17,690
I guessed what VK
to VK plus 1 was,

964
00:56:17,690 --> 00:56:20,174
meaning I tried all the
possibilities for each one.

965
00:56:20,174 --> 00:56:22,340
I say, well what are all
the possibilities for this?

966
00:56:22,340 --> 00:56:24,620
What are all the
possibilities for this?

967
00:56:24,620 --> 00:56:26,340
Take the cross-product.

968
00:56:26,340 --> 00:56:28,500
Just multiply those
sets together.

969
00:56:28,500 --> 00:56:30,750
Those are all the
possible gluings.

970
00:56:30,750 --> 00:56:32,050
And then I sum that.

971
00:56:32,050 --> 00:56:38,380
I take all the options for
all the different values of K.

972
00:56:38,380 --> 00:56:40,010
And that's it.

973
00:56:40,010 --> 00:56:44,024
That output is all
the possible gluings.

974
00:56:44,024 --> 00:56:45,690
So this should actually
give another way

975
00:56:45,690 --> 00:56:47,600
to prove that the number
of different gluings

976
00:56:47,600 --> 00:56:51,960
is only exponential, but this is
just for the edge-to-edge case.

977
00:56:55,710 --> 00:56:58,270
Maybe I should write that down.

978
00:56:58,270 --> 00:57:10,020
So we have for each
choice of K output--

979
00:57:10,020 --> 00:57:18,660
let's say, and for each
solution to the smaller

980
00:57:18,660 --> 00:57:21,280
sub-problem VK plus 1 to VJ.

981
00:57:21,280 --> 00:57:36,187
And then for each solution
to VK plus 1 to VK output.

982
00:57:36,187 --> 00:57:37,020
This is very simple.

983
00:57:37,020 --> 00:57:40,190
This is just a product
over various things.

984
00:57:40,190 --> 00:57:45,610
There's only order
n choices here.

985
00:57:45,610 --> 00:57:47,690
This thing, it depends
how many solutions there

986
00:57:47,690 --> 00:57:51,860
are to those sub-problems, but
it's the product of the two.

987
00:57:51,860 --> 00:57:56,970
And then that's
the total output.

988
00:57:56,970 --> 00:58:00,050
Believe me, this runs
in exponential time.

989
00:58:00,050 --> 00:58:01,726
That's not very hard to prove.

990
00:58:01,726 --> 00:58:03,100
What's interesting
is we can make

991
00:58:03,100 --> 00:58:06,830
the same algorithm with
a little bit of tweaking

992
00:58:06,830 --> 00:58:09,280
get a polynomial time
decision algorithm.

993
00:58:12,080 --> 00:58:14,010
That's cool.

994
00:58:14,010 --> 00:58:18,170
How do we make it a
polynomial decision algorithm?

995
00:58:18,170 --> 00:58:21,350
I can't afford to store every
solution for every sub-problem

996
00:58:21,350 --> 00:58:23,220
because there can be
exponentially many.

997
00:58:23,220 --> 00:58:26,555
So I only get to store one
solution for every sub-problem.

998
00:58:29,280 --> 00:58:30,470
So this is the great idea.

999
00:58:36,370 --> 00:58:55,830
Let's say for each sub-problem,
let's say VI to VK only store

1000
00:58:55,830 --> 00:59:01,990
the solution that
minimizes the amount

1001
00:59:01,990 --> 00:59:04,170
of stuff glued into VI VJ.

1002
00:59:10,260 --> 00:59:17,440
So the sum of the
angles glued at VI,

1003
00:59:17,440 --> 00:59:22,190
which is the same thing as
VJ because they are already

1004
00:59:22,190 --> 00:59:24,270
glued together.

1005
00:59:24,270 --> 00:59:25,810
That's what we were told.

1006
00:59:25,810 --> 00:59:27,840
So I had many choices here.

1007
00:59:27,840 --> 00:59:31,080
I had all the choices for
K. I had all the choices

1008
00:59:31,080 --> 00:59:32,054
for up here.

1009
00:59:32,054 --> 00:59:33,470
I had all the
choices for up here.

1010
00:59:33,470 --> 00:59:36,011
Well actually now I'm just going
to think about one choice up

1011
00:59:36,011 --> 00:59:38,190
here, which is what
is the choice that

1012
00:59:38,190 --> 00:59:40,600
minimizes the angle glued here?

1013
00:59:40,600 --> 00:59:42,260
And what is the
choice over here that

1014
00:59:42,260 --> 00:59:44,650
minimizes the angle glued there?

1015
00:59:44,650 --> 00:59:45,800
Why the minimum?

1016
00:59:45,800 --> 00:59:48,140
Because some of these are wrong.

1017
00:59:48,140 --> 00:59:52,270
I should actually say
output if it's Alexandroff.

1018
00:59:52,270 --> 00:59:54,490
We only want to output
Alexandroff gluings.

1019
00:59:54,490 --> 00:59:56,050
We only want to
output gluings where

1020
00:59:56,050 --> 00:59:58,530
the total sum of the angles
is less than or equal to 360.

1021
00:59:58,530 --> 01:00:03,070
I need to check
that for every one.

1022
01:00:03,070 --> 01:00:06,260
If I want to have the most
chance of it being Alexandroff,

1023
01:00:06,260 --> 01:00:09,020
if I got the least possible
angle glued here and glued

1024
01:00:09,020 --> 01:00:11,530
here, that's my best
hope of it working.

1025
01:00:11,530 --> 01:00:14,700
Now I still have flexibility,
which is I get to choose k.

1026
01:00:14,700 --> 01:00:17,580
There's still n choices
for k, but everything else

1027
01:00:17,580 --> 01:00:19,510
is determined at that point.

1028
01:00:19,510 --> 01:00:23,730
Once I fix K I see, well
does this work, yes or no?

1029
01:00:23,730 --> 01:00:26,560
If it works, that
gives me one candidate

1030
01:00:26,560 --> 01:00:28,719
for how much material
gets glued here.

1031
01:00:28,719 --> 01:00:30,260
It could be a lot,
could be a little.

1032
01:00:30,260 --> 01:00:33,480
I try for all K, I
take the smallest one.

1033
01:00:33,480 --> 01:00:37,010
So it's the same loop, except
there's no four loops here.

1034
01:00:37,010 --> 01:00:38,750
It's just for every
choice of K, I

1035
01:00:38,750 --> 01:00:42,490
look at the solution to VK plus
1 to VJ, the solution for VI

1036
01:00:42,490 --> 01:00:43,860
plus 1 to VK.

1037
01:00:43,860 --> 01:00:46,250
If it's Alexandroff,
I don't output it.

1038
01:00:46,250 --> 01:00:50,710
But I check, is it the best
solution so far in terms

1039
01:00:50,710 --> 01:00:55,150
of the total angle
glued of VI and VJ.

1040
01:00:55,150 --> 01:00:57,720
I keep looping until I
find the very best one,

1041
01:00:57,720 --> 01:00:59,600
and I output that one.

1042
01:00:59,600 --> 01:01:03,550
That would be the
solution that I store.

1043
01:01:03,550 --> 01:01:08,700
The result is, the running
time is actually polynomial,

1044
01:01:08,700 --> 01:01:12,930
because the number of
sub-problems-- a sub-problem

1045
01:01:12,930 --> 01:01:14,880
is determined by two vertices.

1046
01:01:14,880 --> 01:01:16,800
There's only n of
these, n of these.

1047
01:01:16,800 --> 01:01:21,040
So the number of
sub-problems is n squared.

1048
01:01:21,040 --> 01:01:28,540
For each sub-problem I need to
look at order n choices for K.

1049
01:01:28,540 --> 01:01:31,270
And so the total
running time is n

1050
01:01:31,270 --> 01:01:37,950
squared times n,
which is n cubed.

1051
01:01:37,950 --> 01:01:39,630
And that will tell
you if there's

1052
01:01:39,630 --> 01:01:42,120
any hope of anything
working, if there's

1053
01:01:42,120 --> 01:01:46,980
any gluing this will find
one, because it's always

1054
01:01:46,980 --> 01:01:53,842
trying to minimize the angles
that get glued together

1055
01:01:53,842 --> 01:01:54,800
That's is edge-to-edge.

1056
01:02:00,030 --> 01:02:01,520
Now we can generalize.

1057
01:02:01,520 --> 01:02:04,790
It's a little bit
messy, but we can

1058
01:02:04,790 --> 01:02:07,060
generalize to arbitrary gluings.

1059
01:02:12,770 --> 01:02:15,150
Let me sketch a little
bit how that goes.

1060
01:02:22,030 --> 01:02:27,450
So we've seen now this result
for algorithms and this result

1061
01:02:27,450 --> 01:02:28,800
for algorithms.

1062
01:02:28,800 --> 01:02:31,000
And now I'm going to tell
you about this result.

1063
01:02:31,000 --> 01:02:33,170
I wish there was a result
here to tell you about,

1064
01:02:33,170 --> 01:02:33,921
but there isn't.

1065
01:02:33,921 --> 01:02:36,420
So this last part that I did
where the decision algorithm is

1066
01:02:36,420 --> 01:02:39,160
very efficient, we don't
know how to generalize that

1067
01:02:39,160 --> 01:02:42,065
to the general case,
arbitrary gluings.

1068
01:02:42,065 --> 01:02:43,440
This only works
for edge-to-edge.

1069
01:02:43,440 --> 01:02:46,060
We assumed a whole
edge, VK VK plus 1

1070
01:02:46,060 --> 01:02:49,480
was glued to VI VI plus 1.

1071
01:02:49,480 --> 01:02:52,190
But the enumeration
part where we said,

1072
01:02:52,190 --> 01:02:53,810
oh just give me
all the solutions,

1073
01:02:53,810 --> 01:02:55,100
that we can get to work.

1074
01:02:55,100 --> 01:02:58,250
It's just slightly
more complicated.

1075
01:02:58,250 --> 01:03:04,770
Let me tell you how it
gets more complicated.

1076
01:03:04,770 --> 01:03:10,930
So one difference is that there
are more kinds of sub-problems,

1077
01:03:10,930 --> 01:03:12,930
this is the heart
of the problem.

1078
01:03:12,930 --> 01:03:18,510
We have the VI to VJ sub-problem
just like before, that's easy.

1079
01:03:18,510 --> 01:03:23,550
And then we have the
VI to EJ sub-problem.

1080
01:03:23,550 --> 01:03:28,830
So this one was VI was glued to
VJ, very clear what that means.

1081
01:03:28,830 --> 01:03:33,455
Here we have VI is
glued to an edge EJ.

1082
01:03:36,190 --> 01:03:39,880
The trouble with that is we
don't know where on the edge

1083
01:03:39,880 --> 01:03:41,200
it's glued.

1084
01:03:41,200 --> 01:03:45,220
It's somewhere on the edge.

1085
01:03:45,220 --> 01:03:48,200
This is where
things get trickier.

1086
01:03:48,200 --> 01:03:49,750
And there's also
the symmetric case,

1087
01:03:49,750 --> 01:03:51,805
of course, where it's EI to VJ.

1088
01:03:51,805 --> 01:03:53,430
But we don't really
need to worry about

1089
01:03:53,430 --> 01:03:54,305
that, it's symmetric.

1090
01:04:02,670 --> 01:04:05,080
So what does a
solution look like?

1091
01:04:09,310 --> 01:04:12,560
So before, a solution
was an entire gluing.

1092
01:04:12,560 --> 01:04:15,602
Now, because the
gluings can be infinite.

1093
01:04:15,602 --> 01:04:17,560
There can be infinitely
many different gluings.

1094
01:04:17,560 --> 01:04:20,050
I can't actually list
every one explicitly.

1095
01:04:20,050 --> 01:04:22,140
I've actually got to
explain ranges of gluings

1096
01:04:22,140 --> 01:04:24,880
if I have any hope of
capturing all of them.

1097
01:04:24,880 --> 01:04:28,050
So what a solution
is going to look like

1098
01:04:28,050 --> 01:04:40,226
is the combinatorial type of the
gluing, which was this stuff.

1099
01:04:40,226 --> 01:04:41,600
You have an abstract
gluing tree,

1100
01:04:41,600 --> 01:04:43,770
you get which vertices
glue to what edges.

1101
01:04:43,770 --> 01:04:45,930
So you have pretty
much the whole picture,

1102
01:04:45,930 --> 01:04:49,710
just not exactly where each
vertex is glued to which edge.

1103
01:04:49,710 --> 01:04:55,010
Then I also will tell
you a total angle

1104
01:04:55,010 --> 01:05:05,822
of material glued at VI so far.

1105
01:05:05,822 --> 01:05:07,280
In both these cases,
I want to know

1106
01:05:07,280 --> 01:05:11,320
how much stuff in
my gluing at VI

1107
01:05:11,320 --> 01:05:14,500
so I can tell whether
it's Alexandroff.

1108
01:05:14,500 --> 01:05:22,760
And then in the case where I
glue a vertex to an edge, what

1109
01:05:22,760 --> 01:05:24,760
I'm going to do is
give you an interval.

1110
01:05:28,535 --> 01:05:33,020
The Interval's always
going to be starting at 0,

1111
01:05:33,020 --> 01:05:38,260
and going to some
value x. x is less than

1112
01:05:38,260 --> 01:05:42,815
or equal to the length
of the edge, VJ.

1113
01:05:46,510 --> 01:05:50,420
And this is the interval along
the edge where VI can glue.

1114
01:05:53,510 --> 01:05:55,790
So I'm going to tell
you an actual interval

1115
01:05:55,790 --> 01:06:00,320
here from the edge EJ.

1116
01:06:00,320 --> 01:06:02,420
As long as you glue VI
to somewhere between

1117
01:06:02,420 --> 01:06:04,810
there and there,
this stuff works.

1118
01:06:04,810 --> 01:06:06,980
Whatever I specified
works over there.

1119
01:06:06,980 --> 01:06:08,330
You can't go beyond some point.

1120
01:06:15,897 --> 01:06:17,230
That will turn out to be enough.

1121
01:06:32,420 --> 01:06:39,750
So while there's the VI to VJ
case, it's less interesting.

1122
01:06:39,750 --> 01:06:43,110
Let's think about now if
I want to actually solve

1123
01:06:43,110 --> 01:06:50,075
the VI to EJ case.

1124
01:06:53,370 --> 01:06:55,130
What could happen?

1125
01:06:55,130 --> 01:06:57,620
Before there was
only one-- we looked

1126
01:06:57,620 --> 01:07:02,200
at what glues VI to VI
plus 1 and it was one edge.

1127
01:07:02,200 --> 01:07:05,149
It was a little
easier to think about.

1128
01:07:05,149 --> 01:07:06,690
Let's not think
about the whole edge,

1129
01:07:06,690 --> 01:07:11,410
but let's think about locally
what happens just after VI.

1130
01:07:11,410 --> 01:07:13,480
Something gets glued there.

1131
01:07:13,480 --> 01:07:15,999
Now it could be EJ
actually gets glued there.

1132
01:07:15,999 --> 01:07:17,290
It could be we continue gluing.

1133
01:07:17,290 --> 01:07:19,380
That's kind of like
zipping, although not

1134
01:07:19,380 --> 01:07:21,650
exactly what we were
calling zipping before.

1135
01:07:21,650 --> 01:07:27,200
Here I call it
zipping, good choice.

1136
01:07:27,200 --> 01:07:30,620
So one option is that you
zip for a little while.

1137
01:07:30,620 --> 01:07:32,440
Now, how long would you zip?

1138
01:07:32,440 --> 01:07:33,930
A couple options.

1139
01:07:33,930 --> 01:07:35,760
You zip until you
hit a vertex, you

1140
01:07:35,760 --> 01:07:38,240
can't stop in the
middle of two edges.

1141
01:07:38,240 --> 01:07:40,160
So maybe you hit a
vertex here first,

1142
01:07:40,160 --> 01:07:41,630
maybe you hit a
vertex here first,

1143
01:07:41,630 --> 01:07:44,520
and the next vertex
here is farther away.

1144
01:07:44,520 --> 01:07:46,450
So there's two pictures.

1145
01:07:46,450 --> 01:07:48,841
Let me draw the two pictures.

1146
01:07:48,841 --> 01:07:51,205
It could be hit you
hit VI plus 1 first.

1147
01:07:56,320 --> 01:07:58,690
So here's EJ.

1148
01:07:58,690 --> 01:08:04,020
It could be you hit
the vertex of EJ first.

1149
01:08:11,330 --> 01:08:13,380
In this case, you
get a sub-problem

1150
01:08:13,380 --> 01:08:16,550
which is VI plus 1 to EJ.

1151
01:08:16,550 --> 01:08:18,910
These are glued to each other.

1152
01:08:18,910 --> 01:08:21,560
In this case, you get
this vertex, which I think

1153
01:08:21,560 --> 01:08:28,830
is VJ gets glued to the
edge EI, which is just

1154
01:08:28,830 --> 01:08:33,460
the symmetric case I
was talking about here.

1155
01:08:33,460 --> 01:08:37,319
Edge EI is glued to Vertex VJ.

1156
01:08:37,319 --> 01:08:38,819
So those are smaller
sub-problems.

1157
01:08:38,819 --> 01:08:40,964
You solve those, and
then from that you

1158
01:08:40,964 --> 01:08:42,630
will see whether this
actually gives you

1159
01:08:42,630 --> 01:08:44,700
a solution to the
original sub-problem

1160
01:08:44,700 --> 01:08:47,250
you wanted to solve.

1161
01:08:47,250 --> 01:08:48,670
That's the zip case.

1162
01:08:48,670 --> 01:08:52,220
There's another case
which we call tug.

1163
01:08:55,689 --> 01:08:59,720
In the tug case-- so
that was supposing,

1164
01:08:59,720 --> 01:09:03,590
we'll maybe nothing
else gets glued in here.

1165
01:09:03,590 --> 01:09:06,439
But maybe another vertex gets
glued in there, that's the tug.

1166
01:09:06,439 --> 01:09:09,750
To pick a vertex, I pull it in.

1167
01:09:09,750 --> 01:09:17,149
So in that case we get
a picture like this.

1168
01:09:20,770 --> 01:09:24,880
So here's EJ, here's
VI, and now there's

1169
01:09:24,880 --> 01:09:29,060
some other vertex VK
that gets glued in there.

1170
01:09:29,060 --> 01:09:29,779
Which vertex?

1171
01:09:29,779 --> 01:09:32,252
I don't know, try them all.

1172
01:09:32,252 --> 01:09:33,960
Just like before,
before we were choosing

1173
01:09:33,960 --> 01:09:35,566
an edge that gets glued in.

1174
01:09:35,566 --> 01:09:37,149
Now it's not necessarily
a whole edge,

1175
01:09:37,149 --> 01:09:41,020
but if there's any
vertex that glues in,

1176
01:09:41,020 --> 01:09:42,720
pick who's the next one.

1177
01:09:42,720 --> 01:09:46,930
And now I get a sub-problem
here which is VI to VK.

1178
01:09:46,930 --> 01:09:53,210
I get a sub-problem
here which is VK to EJ.

1179
01:09:53,210 --> 01:09:56,610
I solve each of those, I
take the cross-product,

1180
01:09:56,610 --> 01:10:01,680
I do that for all values
of K just like I did here,

1181
01:10:01,680 --> 01:10:03,680
except the labels have
changed a little bit,

1182
01:10:03,680 --> 01:10:04,970
and I find all the outputs.

1183
01:10:04,970 --> 01:10:09,390
And again, you can show that
it's exponential time at most.

1184
01:10:09,390 --> 01:10:13,990
So that's an algorithm.

1185
01:10:13,990 --> 01:10:19,750
Now, it's exponential time,
but really it's not so bad

1186
01:10:19,750 --> 01:10:24,680
because it's at most
cubic like we had before.

1187
01:10:24,680 --> 01:10:26,360
But we're doing
extra work, which

1188
01:10:26,360 --> 01:10:31,900
is something like the
number of gluings dealt

1189
01:10:31,900 --> 01:10:37,590
with at each stage
per sub-problem.

1190
01:10:37,590 --> 01:10:41,310
Now, it's not just the
number of actual solutions,

1191
01:10:41,310 --> 01:10:44,150
because we generate
all possible solutions

1192
01:10:44,150 --> 01:10:46,370
and then check whether
they're Alexandroff.

1193
01:10:46,370 --> 01:10:50,940
So if you have some example that
happens to be unique in the way

1194
01:10:50,940 --> 01:10:54,240
that it glues up, this bound
will not actually be very good.

1195
01:10:54,240 --> 01:10:56,120
It's still going to
consider lots of options.

1196
01:10:56,120 --> 01:11:02,730
But in the case of
bounded sharpness,

1197
01:11:02,730 --> 01:11:05,275
where the number of gluings
is only polynomial--

1198
01:11:05,275 --> 01:11:07,290
and it's polynomial
for a strong reason

1199
01:11:07,290 --> 01:11:10,320
that the trees can't
get very complicated--

1200
01:11:10,320 --> 01:11:13,290
then we can actually turn this
into a polynomial algorithm.

1201
01:11:13,290 --> 01:11:15,880
This will be n cubed
times that bound.

1202
01:11:15,880 --> 01:11:19,064
As long as your polygon
is bounded sharpness,

1203
01:11:19,064 --> 01:11:21,730
you can show that at every stage
of course the number of gluings

1204
01:11:21,730 --> 01:11:24,107
will also be polynomial.

1205
01:11:24,107 --> 01:11:25,690
And so then even the
number of gluings

1206
01:11:25,690 --> 01:11:27,773
you have to consider and
check will be polynomial.

1207
01:11:27,773 --> 01:11:29,970
So the whole thing will
be polynomial time,

1208
01:11:29,970 --> 01:11:33,720
that is this result
and this result.

1209
01:11:36,810 --> 01:11:39,042
So if your polygon is
vaguely reasonable,

1210
01:11:39,042 --> 01:11:40,750
this is actually a
really good algorithm.

1211
01:11:40,750 --> 01:11:42,208
This algorithm has
been implemented

1212
01:11:42,208 --> 01:11:45,990
by two groups, one
[INAUDIBLE] in mathematica

1213
01:11:45,990 --> 01:11:48,952
and another [INAUDIBLE],
which sadly is not--

1214
01:11:48,952 --> 01:11:50,660
his implementation
used to be on the web,

1215
01:11:50,660 --> 01:11:53,400
I can't find it anymore.

1216
01:11:53,400 --> 01:11:56,640
But you get either
implementation,

1217
01:11:56,640 --> 01:11:58,160
plug-in your
polygon, and it just

1218
01:11:58,160 --> 01:12:00,850
lists all the possible gluings.

1219
01:12:00,850 --> 01:12:03,390
So that's all the
theory for today.

1220
01:12:03,390 --> 01:12:06,110
Now I want to show you a whole
bunch of gluings for fun.

1221
01:12:06,110 --> 01:12:08,810
Once you have all the stuff,
we have these programs,

1222
01:12:08,810 --> 01:12:10,670
you get to play.

1223
01:12:10,670 --> 01:12:16,890
So our first play is not that.

1224
01:12:16,890 --> 01:12:18,800
That was the last play.

1225
01:12:18,800 --> 01:12:20,330
First thing, we saw this before.

1226
01:12:20,330 --> 01:12:23,090
I'll just remind you
what it looks like

1227
01:12:23,090 --> 01:12:25,440
and tell you a little bit
about what's going on here.

1228
01:12:25,440 --> 01:12:28,660
So this was the cube.

1229
01:12:28,660 --> 01:12:31,130
And then it's eventually
going to fold into a cross.

1230
01:12:31,130 --> 01:12:33,360
The real study here is
for the Latin Cross--

1231
01:12:33,360 --> 01:12:35,710
which is the picture
I keep drawing

1232
01:12:35,710 --> 01:12:39,790
at the beginning of lecture and
which will now be unfolded--

1233
01:12:39,790 --> 01:12:44,280
what other convex polyhedra
can you glue that polygon into?

1234
01:12:44,280 --> 01:12:47,670
So it's like you're not
told what the creases are.

1235
01:12:47,670 --> 01:12:50,940
You just want to look at
gluings, the resulting creases.

1236
01:12:50,940 --> 01:12:53,774
This is one convex
polyhedron you can make.

1237
01:12:53,774 --> 01:12:56,190
This is something that can
come from Alexandroff's theorem

1238
01:12:56,190 --> 01:12:59,780
that you actually get a flat,
doubly-covered convex polygon.

1239
01:12:59,780 --> 01:13:03,320
It's technically a polyhedron,
according to Alexandroff.

1240
01:13:03,320 --> 01:13:07,800
These are all edge-to-edge
gluings if the length two edges

1241
01:13:07,800 --> 01:13:11,070
are actually subdivided
into two length one edges.

1242
01:13:11,070 --> 01:13:14,400
And this video only enumerates
the edge-to-edge gluings,

1243
01:13:14,400 --> 01:13:17,355
because at this point we only
had the edge-to-edge algorithm.

1244
01:13:17,355 --> 01:13:20,090
I forget, this video
is '98, I think.

1245
01:13:20,090 --> 01:13:23,690
This algorithm was in '96, so
they had just implemented that.

1246
01:13:23,690 --> 01:13:25,136
And then that's when I joined.

1247
01:13:25,136 --> 01:13:27,260
I was just starting out as
a grad student and said,

1248
01:13:27,260 --> 01:13:29,030
OK let's make a video.

1249
01:13:29,030 --> 01:13:30,140
That would be cool.

1250
01:13:30,140 --> 01:13:33,160
So I think there's one more.

1251
01:13:33,160 --> 01:13:35,410
And this is the only
one where we actually

1252
01:13:35,410 --> 01:13:37,380
have the animations.

1253
01:13:37,380 --> 01:13:39,505
This octahedron, this
was tricky to find,

1254
01:13:39,505 --> 01:13:40,880
beceause at this
point there were

1255
01:13:40,880 --> 01:13:42,920
no algorithms for
Alexandroff's theorem.

1256
01:13:42,920 --> 01:13:46,080
So we found it by taking
a piece of cardboard,

1257
01:13:46,080 --> 01:13:48,470
gluing it up,
getting a protractor,

1258
01:13:48,470 --> 01:13:50,350
and measuring all
the dihedral angles,

1259
01:13:50,350 --> 01:13:51,600
and then typing them in.

1260
01:13:51,600 --> 01:13:54,170
And Lo and behold,
in that animation

1261
01:13:54,170 --> 01:13:58,090
they don't perfectly close
up, but it's very close,

1262
01:13:58,090 --> 01:14:00,110
up to the measurement
error of the protractor.

1263
01:14:00,110 --> 01:14:01,975
So for edge-to-edge
gluings, there

1264
01:14:01,975 --> 01:14:03,642
are exactly five
convex polyhedra

1265
01:14:03,642 --> 01:14:04,850
you can make from that cross.

1266
01:14:08,729 --> 01:14:10,520
But if you allow
non-edge-to-edge gluings--

1267
01:14:10,520 --> 01:14:13,130
and so some years later when
we came up with that algorithm

1268
01:14:13,130 --> 01:14:16,710
and implemented it, we
discovered there are 85 gluings

1269
01:14:16,710 --> 01:14:17,870
of the cross.

1270
01:14:17,870 --> 01:14:20,300
This is only six of them.

1271
01:14:20,300 --> 01:14:22,260
We're going to do it
starting with the cube,

1272
01:14:22,260 --> 01:14:26,130
and then we're going to go
in increasing order of faces.

1273
01:14:26,130 --> 01:14:30,000
So there's two flat,
doubly-covered quadrilaterals.

1274
01:14:30,000 --> 01:14:32,730
They're different.

1275
01:14:32,730 --> 01:14:35,030
Then there is a bunch of
tetrahedra that you can make.

1276
01:14:35,030 --> 01:14:36,071
None of them are regular.

1277
01:14:36,071 --> 01:14:40,540
One of them was edge-to-edge,
I think it might be this one.

1278
01:14:40,540 --> 01:14:43,880
I think there's some more
tetrahedra at the top there.

1279
01:14:43,880 --> 01:14:46,390
The blue labels here
say what they are.

1280
01:14:46,390 --> 01:14:48,650
For each one of
these we taped up

1281
01:14:48,650 --> 01:14:50,470
the-- we knew what
the gluing was.

1282
01:14:50,470 --> 01:14:53,512
We taped it up, and
then we played with it

1283
01:14:53,512 --> 01:14:55,470
until we figured out,
this must be the creases.

1284
01:14:55,470 --> 01:14:57,344
Then we actually drew
in that create pattern,

1285
01:14:57,344 --> 01:15:00,192
folded it up, and verified
that is the right answer.

1286
01:15:00,192 --> 01:15:01,900
Because again, at this
point no algorithm

1287
01:15:01,900 --> 01:15:04,260
for Alexandroff's theorem.

1288
01:15:04,260 --> 01:15:05,200
More tetrahedra.

1289
01:15:05,200 --> 01:15:07,610
Here we get to the
five-sided polyhedra.

1290
01:15:07,610 --> 01:15:10,730
I think this one is edge-to-edge
and is in the video.

1291
01:15:10,730 --> 01:15:14,119
It's a little hard to
tell from the-- actually,

1292
01:15:14,119 --> 01:15:15,410
you can tell from the pictures.

1293
01:15:15,410 --> 01:15:17,240
Here there's two sevens.

1294
01:15:17,240 --> 01:15:18,580
That means seven glued to seven.

1295
01:15:18,580 --> 01:15:20,500
That means this is a
fold point, this edge

1296
01:15:20,500 --> 01:15:21,630
is glued to that edge.

1297
01:15:21,630 --> 01:15:24,050
Over here, every edge
has only one label,

1298
01:15:24,050 --> 01:15:26,000
where again this is
considered two edges.

1299
01:15:26,000 --> 01:15:27,500
So this was the
edge-to-edge gluing.

1300
01:15:32,490 --> 01:15:34,350
I think a couple more slides.

1301
01:15:34,350 --> 01:15:36,590
We get another
pentahedron and then

1302
01:15:36,590 --> 01:15:42,990
starting to get hexahedra other
than the cube, lots of them.

1303
01:15:42,990 --> 01:15:44,340
Then we start getting octahedra.

1304
01:15:44,340 --> 01:15:46,940
The last one in the
video was octahedron.

1305
01:15:46,940 --> 01:15:51,141
It's not this one, this
one has some fold points.

1306
01:15:51,141 --> 01:15:54,050
It must be-- these
are all octahedra,

1307
01:15:54,050 --> 01:15:56,272
so it's one of these.

1308
01:15:56,272 --> 01:15:58,640
This guy.

1309
01:15:58,640 --> 01:16:01,220
Of course, it's hard to
tell from a static image,

1310
01:16:01,220 --> 01:16:04,910
but yeah, obviously that was
the octahedron from the video.

1311
01:16:04,910 --> 01:16:05,975
And that is all.

1312
01:16:05,975 --> 01:16:08,740
So actually, there's only
23 distinct polyhedra

1313
01:16:08,740 --> 01:16:12,680
you can make, but you can
make them out of 85 gluings.

1314
01:16:12,680 --> 01:16:14,260
So 85 gluings.

1315
01:16:14,260 --> 01:16:18,330
If one of them is symmetric,
the cube is symmetric.

1316
01:16:18,330 --> 01:16:19,650
So that's one.

1317
01:16:19,650 --> 01:16:22,500
And then the other
84, there's everything

1318
01:16:22,500 --> 01:16:25,170
and its reflectional
inverse, because this polygon

1319
01:16:25,170 --> 01:16:26,660
is symmetric.

1320
01:16:26,660 --> 01:16:31,860
So it's 84 divided by 2,
which is 42, a good number.

1321
01:16:31,860 --> 01:16:37,680
42 actual honest
gluings plus the cube.

1322
01:16:37,680 --> 01:16:40,276
But those 42 gluings,
a lot of them

1323
01:16:40,276 --> 01:16:42,150
generate the same
polyhedron, because there's

1324
01:16:42,150 --> 01:16:44,710
only 23 distinct
polyhedra in the end.

1325
01:16:44,710 --> 01:16:46,490
I don't have it here,
it's on my web page.

1326
01:16:46,490 --> 01:16:48,290
You could look up
all 85, and it says

1327
01:16:48,290 --> 01:16:50,000
which ones are the
same as which ones.

1328
01:16:50,000 --> 01:16:51,500
And it was a real surprise
when we were building them.

1329
01:16:51,500 --> 01:16:53,130
It was like, wait
this looks identical

1330
01:16:53,130 --> 01:16:54,546
to this other
polyhedron, and it's

1331
01:16:54,546 --> 01:16:56,280
glued in a very different way.

1332
01:16:56,280 --> 01:16:58,550
Sort of an extrasymmetry
of that polyhedron,

1333
01:16:58,550 --> 01:16:59,960
as opposed to the cross itself.

1334
01:16:59,960 --> 01:17:00,880
It was pretty cool.

1335
01:17:03,510 --> 01:17:06,740
An interesting
thing to note here

1336
01:17:06,740 --> 01:17:13,810
is that we only bisect
the edges, right?

1337
01:17:13,810 --> 01:17:16,190
There's a lot of
edge-to-edge stuff,

1338
01:17:16,190 --> 01:17:18,890
and then some of the
edges get cut in half.

1339
01:17:18,890 --> 01:17:21,710
And in this case actually, they
look mostly like fold points,

1340
01:17:21,710 --> 01:17:24,170
but that's not always the case.

1341
01:17:24,170 --> 01:17:28,970
It turns out, if you have a
polygon were all the edge links

1342
01:17:28,970 --> 01:17:33,600
are the same like this and
you look at non-edge-to-edge

1343
01:17:33,600 --> 01:17:35,496
gluings, and there
are no rolling belts--

1344
01:17:35,496 --> 01:17:36,870
because if there's
rolling belts,

1345
01:17:36,870 --> 01:17:39,370
it's infinite because you're
not going to have anything like

1346
01:17:39,370 --> 01:17:40,030
this.

1347
01:17:40,030 --> 01:17:46,890
Then it suffices to
subdivide every edge in half.

1348
01:17:46,890 --> 01:17:49,070
And then every non-edge-to-edge
gluing will become

1349
01:17:49,070 --> 01:17:50,540
an edge-to-edge gluing.

1350
01:17:50,540 --> 01:17:55,740
This is called the [INAUDIBLE]
half-length theorem

1351
01:17:55,740 --> 01:17:56,687
in our book.

1352
01:17:56,687 --> 01:17:58,270
It was proved by
this guy [INAUDIBLE],

1353
01:17:58,270 --> 01:18:01,070
who was one of the
implementers of the algorithms

1354
01:18:01,070 --> 01:18:03,070
for enumerating
all these gluings.

1355
01:18:03,070 --> 01:18:04,230
And he proved it afterward.

1356
01:18:04,230 --> 01:18:06,100
The code doesn't actually
exploit that fact,

1357
01:18:06,100 --> 01:18:07,200
but it's kind of neat.

1358
01:18:07,200 --> 01:18:09,330
All you have to worry
about are half edges

1359
01:18:09,330 --> 01:18:12,900
in this case where all the
edge links are the same.

1360
01:18:12,900 --> 01:18:13,990
All right, that's a cross.

1361
01:18:13,990 --> 01:18:15,705
And it's kind of
maybe even surprising

1362
01:18:15,705 --> 01:18:18,810
that there's only finitely
many different gluings,

1363
01:18:18,810 --> 01:18:21,970
but there's no way to
get a rolling belt here.

1364
01:18:21,970 --> 01:18:24,795
Let's look at-- oh, here
are all the polyhedra

1365
01:18:24,795 --> 01:18:28,680
in a pretty 3D form.

1366
01:18:28,680 --> 01:18:31,240
Here is a polygon
which is convex.

1367
01:18:31,240 --> 01:18:33,270
It is the equilateral triangle.

1368
01:18:33,270 --> 01:18:35,900
And so, no surprise, there
a rolling belts here.

1369
01:18:35,900 --> 01:18:38,450
But here's the picture of
how all the rolling belts fit

1370
01:18:38,450 --> 01:18:39,040
together.

1371
01:18:39,040 --> 01:18:41,400
In principle, you can
extract this picture

1372
01:18:41,400 --> 01:18:44,160
from this algorithm.

1373
01:18:44,160 --> 01:18:47,070
Although combined with an
algorithm for Alexandroff's

1374
01:18:47,070 --> 01:18:48,609
theorem, that's the tricky part.

1375
01:18:48,609 --> 01:18:50,150
So you have some
very simple gluings,

1376
01:18:50,150 --> 01:18:51,490
like could could fold
the triangle in half,

1377
01:18:51,490 --> 01:18:53,140
get this doubly-covered
triangle,

1378
01:18:53,140 --> 01:18:54,680
you can make a
regular tetrahedron,

1379
01:18:54,680 --> 01:18:57,230
or you can fold the triangle
into a flat doubly-covered

1380
01:18:57,230 --> 01:18:59,152
rectangle like at the top.

1381
01:18:59,152 --> 01:19:00,985
And then there are
rolling belts in between.

1382
01:19:00,985 --> 01:19:03,170
So this is one
rolling belt, this

1383
01:19:03,170 --> 01:19:04,710
is another possible
rolling belt,

1384
01:19:04,710 --> 01:19:06,960
another possible rolling
belt each with a gluing tree.

1385
01:19:06,960 --> 01:19:08,730
They're not all drawn here.

1386
01:19:08,730 --> 01:19:11,290
And this is sort of
the topology of all

1387
01:19:11,290 --> 01:19:15,720
of the possible gluings you
could make, which is nice.

1388
01:19:15,720 --> 01:19:20,000
So a triangle is
relatively simple.

1389
01:19:20,000 --> 01:19:24,740
The square is a little messier.

1390
01:19:24,740 --> 01:19:28,400
There's again a bunch of
particularly nice gluings

1391
01:19:28,400 --> 01:19:34,390
like the [INAUDIBLE], the
folding in half one way,

1392
01:19:34,390 --> 01:19:38,820
folding in half the other
way, making a letter,

1393
01:19:38,820 --> 01:19:42,340
and some tetrahedra that
are just drawn here.

1394
01:19:42,340 --> 01:19:44,221
They're in the middle.

1395
01:19:44,221 --> 01:19:45,970
And then there's these
crazy rolling belts

1396
01:19:45,970 --> 01:19:47,120
for how they fit together.

1397
01:19:47,120 --> 01:19:51,100
So this is in unit square
all the different gluings.

1398
01:19:51,100 --> 01:19:55,050
And one of the fun things,
some of them are octahedra,

1399
01:19:55,050 --> 01:19:59,930
I have a list here.

1400
01:19:59,930 --> 01:20:03,020
But I don't know actually
which one of these belts

1401
01:20:03,020 --> 01:20:03,830
has the octahedra.

1402
01:20:03,830 --> 01:20:05,540
It might be a couple of them.

1403
01:20:05,540 --> 01:20:10,300
And it turns out, for octahedra
they actually figured out

1404
01:20:10,300 --> 01:20:13,470
an algorithm just for octahedra
to solve Alexandroff's theorem.

1405
01:20:13,470 --> 01:20:17,040
This is again before
Alexandroff theorem algorithms.

1406
01:20:17,040 --> 01:20:19,710
And they wanted to compute, well
if I take all these polyhdra,

1407
01:20:19,710 --> 01:20:21,660
which one has the
maximum volume?

1408
01:20:21,660 --> 01:20:23,630
So if I take a
square paper, what's

1409
01:20:23,630 --> 01:20:26,560
the maximum volume
shape you can make?

1410
01:20:26,560 --> 01:20:30,900
And there it is.

1411
01:20:30,900 --> 01:20:33,950
It's kind of fun.

1412
01:20:33,950 --> 01:20:35,820
One of those octahedra.

1413
01:20:35,820 --> 01:20:41,010
And this is a prettier picture
of-- so here they actually

1414
01:20:41,010 --> 01:20:43,080
computed all the polyhedra.

1415
01:20:43,080 --> 01:20:45,100
Obviously not all
infinitely many of them,

1416
01:20:45,100 --> 01:20:48,360
but various snapshots
along each rolling belt.

1417
01:20:48,360 --> 01:20:50,550
And each of these
rolling belts corresponds

1418
01:20:50,550 --> 01:20:52,960
to one of the circles over here.

1419
01:20:56,400 --> 01:20:57,790
And that's it.

1420
01:20:57,790 --> 01:21:00,370
Any questions?

1421
01:21:00,370 --> 01:21:03,250
So that's the end
of fun with gluings.

1422
01:21:03,250 --> 01:21:06,770
We'll look at other kinds of
gluing problems next class.