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PROFESSOR: This class is talking
about two lectures, both about

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protein folding, fixed angled
chains, things like that.

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There's a few
questions about them.

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One is mostly about
these open problems,

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equilateral, equiangular,
obtuse 3D chains,

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fixed angle open problem
is, are they locked?

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So the question is about,
what about any subset

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of those combinations?

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So this originally comes
from an open problem

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I think posed in 1999, one of
the first 3D linkage papers,

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and it asked whether equilateral
universal joints can lock,

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and that's still open.

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For universal joints,
these two constraints

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don't make a lot of
sense because who

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cares if it's
initially equiangular.

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As soon as you move it, it
will no longer be equiangular.

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And obtuseness I don't
think matters too much,

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although it potentially could.

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So for that problem, equilateral
seems to be the core.

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For fixed angle,
though, we conjecture

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that fixed angle equilateral is
not enough from this example.

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It's still not
proved to be locked.

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I don't know if it's hard,
but it's probably tedious,

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so it hasn't been done.

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This was the crossed
legs example.

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All the edge lengths
are the same,

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and if you don't allow
the touching part,

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then all the angles
are also the same.

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So this is everything
except the obtuse property

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and it's probably locked, so
dropping obtuse is no good.

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The other things you could drop
are equilateral or equiangular.

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This is if you drop equiangular.

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Fixed angle equilateral is
not terribly constraining

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because you can
simulate a long bar

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by having a lot of
180 degree angles.

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This is obtuse and
it's equilateral

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but it's not
equiangular and it's

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locked for a trivial reason.

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So this is dropping equiangular.

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If I drop equilateral,
I can also

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just make a knitting
needles example.

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I take a really long link, and
this thing is basically string,

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so I don't really care.

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This connection can be done
with a lot of obtuse angles,

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and then have a
really long link.

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So dropping any of
the three constraints

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makes it easy to lock, so
for fixed angle chains,

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you need equilateral,
equiangular, and obtuse.

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All these things
together potentially mean

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you're not locked,
but we don't know.

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That's the open problem.

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Of course, we don't
necessarily need

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exactly equilateral
or equiangular.

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Hopefully, within some small min
to max ratio would be enough,

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but we don't know.

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Next question is, why did we
model the ribosome as a cone?

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That seems rather simple.

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Is this realistic?

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And partly, when I
taught the class in 2010,

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I didn't use any images I
couldn't get permission for.

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This year I'm more lax.

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We'll ask for forgiveness
instead of permission.

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This is a paper
from Science 2000.

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This is what the ribosome
actually looks like.

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There's many different
figures of it,

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but I particularly like this
one because it highlights

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a tunnel in the center
of the ribosome.

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And so the idea is this is a
machine for converting MRNA

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into your proteins,
and the idea is

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the protein comes through here.

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There's a little
bump in the tunnel.

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Some people conjecture this
is where the amino acid gets

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attached, and then
it feeds through here

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and starts spitting out.

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And as I said, there's
barely enough room

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here for an alpha helix, so
probably not too much folding

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happens inside and the folding
should just happen over here.

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And the observation is if
you have a reasonable size

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protein that's only going
to be about this big,

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then there's this big,
flat wall at the exit,

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so you have a plane
there and this half space

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is more or less a big obstacle.

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It's the alpha cone model
where alpha is 180, I think,

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if alpha is the half angle.

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So it's not really a cone.

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It's a cone that's been opened
up all the way to a plane,

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but that's one of the situations
that's handled by the theorems

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that we talked about.

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So that's why the cone model.

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We generalize to
cones just because it

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works for general
cones, but the real one

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is a sort of flat cone.

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That's where that comes from.

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I think actually we proved the
theorem before we knew this,

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but then we looked it
up and it was true.

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And the last question
is about the lecture 21,

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which some of you may not have
watched because it's optional,

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but there's this model
called the HP model

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for protein folding.

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It's a model of
protein energy and it

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says you have a chain
of H and P nodes,

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and basically, the H nodes
are attracted to each other

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and the P nodes don't care.

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And the model is that the H
nodes are hydrophobic so they

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want to be next to each other
so they're not next to water,

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which is surrounding
the whole molecule,

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and the P nodes don't care,
or they're hydrophilic.

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And it's known that if
you have an HP string

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and you want to find the optimal
folding into a 2D or a 3D

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structure, that's NP-hard,
which is kind of weird

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because somehow
nature does it, maybe

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because it found the easy
instances by evolution,

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or maybe there's
something we're missing.

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This model doesn't
capture reality.

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We don't know.

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But it captures part of reality,
as least as we observe it.

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Unfortunately, these
hardness proofs

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are a little too
complicated to cover here,

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but I can at least answer
what are they reducing from.

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There's two proofs.

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I guess they were basically the
same time, I think around 2001.

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The first one here
is in 3D, and these

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are by two MIT professors.

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There's this big construction
but the reduction

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is from bin packing.

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So you have a bunch
of fixed size bins

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and you have a bunch of
items of varying sizes

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and you just want
to fit them all

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in using the fewest
bins possible.

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And the rough idea of the
construction of this one, this

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is how an individual
number is represented.

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The rough idea is you
have this big cube.

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You fill the sides
with stuff to protect

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from the outside boundary.

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The insides are just
used for connections,

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and then the front face
here is this stuff.

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You construct
these bins, and you

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construct numbers which
have to fit inside the bins.

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That's the rough idea.

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The details are complicated.

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The 2D proof by
several people is

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from Hamiltonicity in
maximum degree-4 graphs.

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So you have a graph
for every vertex that

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has at most four incident edges.

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You want to find a
Hamiltonian cycle, I think.

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It's much harder to see
this picture, I would say.

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There's no one diagram
that summarizes it.

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This is roughly
the construction,

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which is quite complicated,
and I'll just leave it at that.

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That's what they're
reduced from.

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I think an interesting
open problem

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would be to find a simple or
cleaner proof of these results.

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Now that it's known
that they're hard,

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it's probably easy
or hardness proofs.

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There's also some open
questions from lecture,

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like is it APX hard?

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Is it approximable to a 1
plus epsilon factor or 1 minus

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epsilon factor for any
epsilon or is there

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some limit to approximability?

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We saw a nice approximation
4/3 whatever, 3/4 whatever,

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but can you do better?

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Still open.

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So that was the questions,
and then this is a question

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that you usually ask
in every lecture.

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You didn't actually
ask it on this one,

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but I copied and pasted.

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So there's some
interesting progress

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from this class two years ago,
and I think in the open problem

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session initially, probably
also a class project

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related to this part.

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So this is back in lecture 20.

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We had this proof,
which is kind of fun,

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that flattening a
fixed angle chain,

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deciding whether there
was a flat folded state,

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is weakly NP hard,
weakly meaning

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it depended a lot on what
these numbers look like

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and it was a reduction
from partitions.

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You had to split up the
numbers into two equal parts,

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and if you did, then this
key would fit in the slot

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and you're OK.

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If you didn't, the key would
collide with something here.

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So this means the
problem is hard

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if your edge lengths
are vastly different.

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They would have to differ
by a ratio of exponential

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in the number of edges for
this to really be hard.

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So a natural question is,
well, what if all the lengths

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are equal, equilateral chains?

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Turns out that is still hard,
and this is a paper just

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published last year with
Sarah Eisenstadt, who

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took the class then.

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And there's a bunch
of results here.

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All of these are NP hardness
results, strong NP hardness,

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so it doesn't depend
on the numbers.

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We've got flattening results.

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There's also min flat
span and max flat span.

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Once you have that
flattening is hard,

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these are pretty
easy to show hard,

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so I won't talk
about them so much.

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But basically, we consider
different ranges of angles

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and what you might allow.

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In some cases, we can get
perfect equilateral chains.

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In some cases, they have to
range between, say, 1 and 2

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in length.

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A natural question
is obtuse angles.

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We don't quite know that.

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The best we have is 60 degrees
minus epsilon and larger.

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This is nice for
orthogonal chains,

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but it's not for
chains, it's for trees.

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So still some open questions
here, but lots of hardness

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results.

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I thought I'd show you roughly
what the hardness proofs look

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like because
they're kind of fun.

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They all follow this
kind of structure.

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This is a gadget, and
it's a fixed angle chain.

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Here I'm going to show
everything with 90 degrees

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because it's easier
to think about.

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And it's kind of like--
I'm thinking plunger,

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but it's something like that.

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You can decide which
parts get pushed in

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and which parts get popped out,
and this L shape can basically

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shift left and right to
three different places

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if I did it right-- this
one, this one, and this one.

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So that's a useful construction,
as you might imagine,

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and the idea is you take that
and you add on these guys.

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So I've got another kind
of plungey thing like this,

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and this guy can move left
and right to various extents.

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That's not very intuitive.

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So then this connects
to these little guys,

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and these can just
flip up or down

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if it's a flat embedding
of a fixed angle chain.

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And so the idea is you have
three of these pokey elements

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and they can attach.

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They bump into these
different things.

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And if it's up,
then this can be up,

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but when they're down-- like
this guy is currently down,

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but it could be pushed
up-- then this guy

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must be flipped down like that.

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That's the rough idea of how
these parts fit together.

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And so then this is how you end
up building some kind of three

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sat problem.

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So you have some variables.

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They can either
be true or false.

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And the idea is that the
true guys are on the top,

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the bottom guys
are on the bottom,

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and there's some complicated
interaction between them

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to make sure that they can't
both be true and false.

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00:12:44,232 --> 00:12:45,940
Basically, you want
these things to point

248
00:12:45,940 --> 00:12:49,160
in whenever possible because
then these guys can go down.

249
00:12:49,160 --> 00:12:51,140
Otherwise, you would collide.

250
00:12:51,140 --> 00:12:53,500
And if this guy's in,
the corresponding guys

251
00:12:53,500 --> 00:12:55,770
in the bottom must
be down, and that's

252
00:12:55,770 --> 00:12:58,550
enforced by this long
thing, which is either

253
00:12:58,550 --> 00:13:00,254
completely down
or completely up.

254
00:13:00,254 --> 00:13:01,920
So when it's completely
down, these guys

255
00:13:01,920 --> 00:13:03,378
all have to be
down, which is going

256
00:13:03,378 --> 00:13:06,994
to be a problem down below
if things aren't satisfied.

257
00:13:06,994 --> 00:13:09,160
When it's up, all of these
guys would have to be up.

258
00:13:09,160 --> 00:13:11,182
Some of them could be
up if you feel like it,

259
00:13:11,182 --> 00:13:12,640
but really, you'd
probably put them

260
00:13:12,640 --> 00:13:14,590
all down whenever you
can because then you

261
00:13:14,590 --> 00:13:17,536
can take these guys
and stick them down.

262
00:13:17,536 --> 00:13:18,910
So this is how
you set variables,

263
00:13:18,910 --> 00:13:20,640
and then the other
things I showed

264
00:13:20,640 --> 00:13:23,340
were clauses, essentially.

265
00:13:23,340 --> 00:13:26,100
At least one of those
pins had to be pushed down

266
00:13:26,100 --> 00:13:30,470
and you don't know which one,
and that's the hard part.

267
00:13:30,470 --> 00:13:32,930
Rough sketch of how
this proof looks.

268
00:13:35,590 --> 00:13:40,740
One thing is here,
for this to work,

269
00:13:40,740 --> 00:13:44,360
these edges can't be spinnable.

270
00:13:44,360 --> 00:13:46,670
When I go here, I must
immediately go back down.

271
00:13:46,670 --> 00:13:50,939
I can't flip it.

272
00:13:50,939 --> 00:13:51,980
So these are rigid edges.

273
00:13:51,980 --> 00:13:53,900
That's not really
the original problem,

274
00:13:53,900 --> 00:13:55,330
but you can simulate
rigid edges,

275
00:13:55,330 --> 00:13:57,310
and this is where
the angles get small.

276
00:13:57,310 --> 00:14:00,160
You can simulate rigid edges
with a very sharp zigzag

277
00:14:00,160 --> 00:14:02,260
because if you ever
flipped one of these,

278
00:14:02,260 --> 00:14:04,450
it would collide with
the previous edge.

279
00:14:04,450 --> 00:14:07,680
So for sufficiently
sharp angles,

280
00:14:07,680 --> 00:14:12,390
you can force parts to stay
straight in this funny way,

281
00:14:12,390 --> 00:14:15,650
which is not great, but
it's one way to force it.

282
00:14:15,650 --> 00:14:19,160
We can also do it with trees
or various other techniques,

283
00:14:19,160 --> 00:14:22,370
but that's how the proof looks.

284
00:14:22,370 --> 00:14:28,750
So that is flattening fixed
angle chains, strongly NP hard.

285
00:14:28,750 --> 00:14:34,870
We go next to one
more topic for fun.

286
00:14:34,870 --> 00:14:37,335
This is the topic of flips.

287
00:14:41,130 --> 00:14:44,590
I guess I have an
image for this.

288
00:14:44,590 --> 00:14:46,500
This is a problem
posed by Paul Erdos,

289
00:14:46,500 --> 00:14:48,670
I think when he was a student.

290
00:14:48,670 --> 00:14:50,190
He was a very famous
mathematician.

291
00:14:50,190 --> 00:14:53,500
1935, he asked this question.

292
00:14:53,500 --> 00:14:56,950
This is the whole question
as originally written.

293
00:14:56,950 --> 00:15:01,049
You have a polygon and you take
the convex hull of the polygon,

294
00:15:01,049 --> 00:15:03,340
and you take these regions
which are in the convex hull

295
00:15:03,340 --> 00:15:04,339
but outside the polygon.

296
00:15:04,339 --> 00:15:05,720
We call those pockets.

297
00:15:05,720 --> 00:15:08,140
And you imagine
flipping those outside,

298
00:15:08,140 --> 00:15:11,430
reflecting through
that supporting line,

299
00:15:11,430 --> 00:15:14,520
that tangent on the outside,
so then you get this polygon.

300
00:15:14,520 --> 00:15:16,440
Here you get two pockets.

301
00:15:16,440 --> 00:15:20,430
Maybe you flip both of them, and
then you get a convex polygon.

302
00:15:20,430 --> 00:15:22,970
Now you have no pockets,
and so you're done.

303
00:15:22,970 --> 00:15:26,040
And the question is as posed.

304
00:15:26,040 --> 00:15:28,950
Prove that after a finite
number of such steps,

305
00:15:28,950 --> 00:15:31,770
the polygon would become convex.

306
00:15:31,770 --> 00:15:34,980
So you might call this
a conjecture by Erdos

307
00:15:34,980 --> 00:15:37,240
that it is finite.

308
00:15:37,240 --> 00:15:39,920
He'd never published a proof.

309
00:15:39,920 --> 00:15:42,120
There's one issue with
the problem as stated.

310
00:15:42,120 --> 00:15:43,860
You can't actually
flip both pockets

311
00:15:43,860 --> 00:15:47,940
at once if you want
to avoid collisions.

312
00:15:47,940 --> 00:15:49,860
The relation to
linkages is we imagine

313
00:15:49,860 --> 00:15:52,910
these flips could actually
happen by a rotation,

314
00:15:52,910 --> 00:15:54,410
but you could also
just imagine them

315
00:15:54,410 --> 00:15:57,350
as reflecting instantaneously.

316
00:15:57,350 --> 00:15:58,520
There's this issue.

317
00:15:58,520 --> 00:16:00,870
If you flip two
pockets at once, you

318
00:16:00,870 --> 00:16:03,730
might have collisions
afterwards.

319
00:16:03,730 --> 00:16:06,580
The way this problem has been
interpreted by most people

320
00:16:06,580 --> 00:16:08,810
is don't flip them all at once.

321
00:16:08,810 --> 00:16:10,400
Just flip them one
at a time, and then

322
00:16:10,400 --> 00:16:13,420
you guarantee, because
this is a supporting line,

323
00:16:13,420 --> 00:16:14,970
a tangent line, of
the convex hull,

324
00:16:14,970 --> 00:16:18,060
when you flip one pocket out,
it will go to the other side

325
00:16:18,060 --> 00:16:20,000
so it can't intersect
the rest of the polygon.

326
00:16:20,000 --> 00:16:25,940
So you would avoid collision
if you do one flip at a time.

327
00:16:25,940 --> 00:16:30,420
Now, this was observed, I
guess, a few years later

328
00:16:30,420 --> 00:16:32,020
by Bela de Sz.

329
00:16:32,020 --> 00:16:40,980
Nagy, and he also
published the first proof.

330
00:16:40,980 --> 00:16:45,390
Before we get there,
a weird property

331
00:16:45,390 --> 00:16:48,760
and why you might worry about
this being finite or infinite

332
00:16:48,760 --> 00:16:50,740
is if you take
this quadrilateral

333
00:16:50,740 --> 00:16:54,270
and make this very narrow,
this edge very small

334
00:16:54,270 --> 00:16:56,980
relative to the
horizontal edge, then

335
00:16:56,980 --> 00:17:01,410
you can require arbitrarily
many flips to convexify.

336
00:17:01,410 --> 00:17:05,479
Even for n equals 4, you
could require a million flips.

337
00:17:05,479 --> 00:17:07,020
So definitely there's
some dependence

338
00:17:07,020 --> 00:17:11,223
on the ratio between the longest
length and the smallest length,

339
00:17:11,223 --> 00:17:12,764
though in open
problem, what we still

340
00:17:12,764 --> 00:17:15,210
don't know is whether you
can bound the number of flips

341
00:17:15,210 --> 00:17:17,380
in terms of n and that ratio.

342
00:17:17,380 --> 00:17:18,380
Is it pseudo-polynomial?

343
00:17:18,380 --> 00:17:19,119
Who knows?

344
00:17:25,770 --> 00:17:27,670
So it turns out
it's always finite,

345
00:17:27,670 --> 00:17:31,020
and there have been many proofs
over the years of this result.

346
00:17:31,020 --> 00:17:34,700
It's kind of been
rediscovered many times.

347
00:17:34,700 --> 00:17:40,360
Nagy solved it originally
in 1939, and he cited Erdos.

348
00:17:40,360 --> 00:17:42,110
These two guys
didn't cite anyone,

349
00:17:42,110 --> 00:17:45,590
so they may have come up with
the problem independently.

350
00:17:45,590 --> 00:17:48,770
This paper, these are
two Russian proofs.

351
00:17:48,770 --> 00:17:53,175
Then this paper cited--
they knew about everything.

352
00:17:53,175 --> 00:17:55,000
They cited all of them.

353
00:17:55,000 --> 00:17:56,737
They cited Reshetnyak.

354
00:17:56,737 --> 00:17:58,820
I don't think they necessarily
knew about Yusupov.

355
00:18:01,620 --> 00:18:04,230
Independently, Kaluza
posed the problem in 1981

356
00:18:04,230 --> 00:18:08,720
and Wegner solved it, and
then to finally clean it up,

357
00:18:08,720 --> 00:18:13,740
Grunbaum, who we saw
from un-un-unfoldable,

358
00:18:13,740 --> 00:18:15,519
knew about everything.

359
00:18:15,519 --> 00:18:16,810
Presumably, this is everything.

360
00:18:16,810 --> 00:18:18,860
Of course, we might
have missed one.

361
00:18:18,860 --> 00:18:23,290
But he knew all the above,
came up with his own proof.

362
00:18:23,290 --> 00:18:26,250
And then Godfried
Touissant, father

363
00:18:26,250 --> 00:18:30,100
of computational geometry,
one of them, knew about these

364
00:18:30,100 --> 00:18:33,420
and came up with yet
another, simpler proof.

365
00:18:33,420 --> 00:18:35,900
So the story of
this is kind of fun.

366
00:18:35,900 --> 00:18:42,070
When I taught this class for
the very first time, 2003 or so,

367
00:18:42,070 --> 00:18:43,710
I thought, OK, cool.

368
00:18:43,710 --> 00:18:45,170
This is a classic theorem.

369
00:18:45,170 --> 00:18:47,140
Everyone should know it.

370
00:18:47,140 --> 00:18:50,680
So I thought I'd cover the
latest proof that's presumably

371
00:18:50,680 --> 00:18:55,490
the best, so I covered
Godfried's proof.

372
00:18:55,490 --> 00:18:58,655
I was a little unhappy with
it, but I finished writing down

373
00:18:58,655 --> 00:19:01,030
the proof, and then one of
the students raised their hand

374
00:19:01,030 --> 00:19:03,420
and asked, is that really right?

375
00:19:03,420 --> 00:19:05,120
Can you do that in step two?

376
00:19:05,120 --> 00:19:07,080
And the answer was
no, you can't do that.

377
00:19:07,080 --> 00:19:10,130
It basically skipped a step.

378
00:19:10,130 --> 00:19:11,200
And so I thought, oh gee.

379
00:19:11,200 --> 00:19:13,900
And so I corresponded
with O'Rourke and Godfried

380
00:19:13,900 --> 00:19:16,665
that weekend and was like,
is something missing here?

381
00:19:16,665 --> 00:19:18,790
Maybe we should go back to
some of the other proofs

382
00:19:18,790 --> 00:19:20,780
because we've got
lots to choose from.

383
00:19:20,780 --> 00:19:25,110
So we went to the original,
which is a very short proof.

384
00:19:25,110 --> 00:19:28,410
It's only one page long,
maybe one and a half pages.

385
00:19:28,410 --> 00:19:29,620
And so this is fun.

386
00:19:29,620 --> 00:19:33,160
This is 1939, the way
mathematics used to be done.

387
00:19:33,160 --> 00:19:37,009
And it's funny because
Grunbaum's proof was based

388
00:19:37,009 --> 00:19:38,550
on Nagy's proof and
Toussaint's proof

389
00:19:38,550 --> 00:19:40,250
was based on Grunbaum's
proof, so in the end,

390
00:19:40,250 --> 00:19:41,680
these proofs were very similar.

391
00:19:41,680 --> 00:19:45,370
In fact, they differed
exactly in the one step,

392
00:19:45,370 --> 00:19:46,690
which was kind of omitted here.

393
00:19:46,690 --> 00:19:47,990
So I thought, OK great.

394
00:19:47,990 --> 00:19:52,010
Here we have a fill-in
on how to do it.

395
00:19:52,010 --> 00:19:54,710
And so the next class, I
went up and I presented it.

396
00:19:54,710 --> 00:19:55,790
I was really happy.

397
00:19:55,790 --> 00:19:58,350
Isn't this cool,
1939 mathematics?

398
00:19:58,350 --> 00:20:00,650
It's really awesome,
and it's one line

399
00:20:00,650 --> 00:20:02,230
that filled in this step.

400
00:20:02,230 --> 00:20:05,450
The same student raises
his hand and is like,

401
00:20:05,450 --> 00:20:07,660
I don't think that's true.

402
00:20:07,660 --> 00:20:10,675
So now the step was filled
in but it was wrong.

403
00:20:10,675 --> 00:20:12,550
So it turns out actually
most of these proofs

404
00:20:12,550 --> 00:20:16,000
are wrong, but not all
of them, fortunately,

405
00:20:16,000 --> 00:20:17,830
so the theorem is still true.

406
00:20:17,830 --> 00:20:21,900
So that one was wrong, this one
was wrong, this one was wrong,

407
00:20:21,900 --> 00:20:23,620
this one skipped a
step that was key,

408
00:20:23,620 --> 00:20:27,310
this one essentially also
skipped the step that was key.

409
00:20:27,310 --> 00:20:31,130
So in the end, there's two
correct proofs, Reshetnyak

410
00:20:31,130 --> 00:20:37,260
and Bing and Kazarinoff,
which was surprising.

411
00:20:37,260 --> 00:20:43,860
So I thought I'd show you one
of the errors in the Nagy proof.

412
00:20:43,860 --> 00:20:45,940
There's one other
interesting feature here,

413
00:20:45,940 --> 00:20:48,710
which is, why wasn't this
discovered until our class?

414
00:20:48,710 --> 00:20:52,650
The student was Blaise Gassand,
and so then we wrote a paper

415
00:20:52,650 --> 00:20:56,510
about it and we have our
own proof, of course.

416
00:20:56,510 --> 00:20:59,280
Now, some people
may have realized

417
00:20:59,280 --> 00:21:01,400
there was an error in
some of the proofs.

418
00:21:01,400 --> 00:21:04,370
Bing and Kazarinoff, one of
the correct proofs, a very nice

419
00:21:04,370 --> 00:21:08,565
one, wrote-- here's the
original Russian sentence.

420
00:21:08,565 --> 00:21:10,340
The English translation
is, "the proof

421
00:21:10,340 --> 00:21:13,930
of this theorem given
by Nagy is incorrect,"

422
00:21:13,930 --> 00:21:17,280
which leaves something
to be desired.

423
00:21:17,280 --> 00:21:19,890
And Grunbaum, who
can read Russian--

424
00:21:19,890 --> 00:21:23,680
we had to get them
translated-- mentioned this.

425
00:21:23,680 --> 00:21:26,380
So he noticed that point. "They
remarked that Nagy's proof is

426
00:21:26,380 --> 00:21:29,680
invalid but there's no
basis for this claim."

427
00:21:29,680 --> 00:21:37,920
Then it remained undiscovered
until 2005 or something.

428
00:21:37,920 --> 00:21:39,690
Kind of funny.

429
00:21:39,690 --> 00:21:41,790
Good thing there's so many
proofs to choose from.

430
00:21:41,790 --> 00:21:44,000
So this is Nagy's
original proof.

431
00:21:44,000 --> 00:21:47,150
You can see the example
where flipping two pockets

432
00:21:47,150 --> 00:21:48,555
simultaneously
causes a crossing.

433
00:21:51,727 --> 00:21:53,310
Don't read the whole
thing, obviously,

434
00:21:53,310 --> 00:21:55,760
but there's one
sentence here which is,

435
00:21:55,760 --> 00:22:00,270
if you take a polygon, call
it p0, and you flip a pocket

436
00:22:00,270 --> 00:22:02,810
and you get p1, and you flip
a pocket and you get p2,

437
00:22:02,810 --> 00:22:06,010
and you take the convex
hull of p and you get c0,

438
00:22:06,010 --> 00:22:09,640
and you take the convex hull
of p1 and you get c1, p2,

439
00:22:09,640 --> 00:22:10,520
you get c2.

440
00:22:10,520 --> 00:22:13,215
And then you interleave
these polygons

441
00:22:13,215 --> 00:22:16,020
so that the polygon is a convex
hull, next polygon is a convex

442
00:22:16,020 --> 00:22:18,910
hull, because you're
always flipping out,

443
00:22:18,910 --> 00:22:24,680
each of these polygons obviously
contains the foregoing ones.

444
00:22:24,680 --> 00:22:26,825
And so that seemed really
nice and we thought,

445
00:22:26,825 --> 00:22:28,230
oh, this is so elegant.

446
00:22:28,230 --> 00:22:32,076
They used this to prove that
the limit of the p's is convex

447
00:22:32,076 --> 00:22:33,450
because the limit
of the p's then

448
00:22:33,450 --> 00:22:36,760
would be the limit of the c's
because it was interleaving.

449
00:22:36,760 --> 00:22:38,670
But it's not true
that these things

450
00:22:38,670 --> 00:22:42,490
contain the previous ones
because if you have two pockets

451
00:22:42,490 --> 00:22:44,640
and you flip one of
them, it's this one

452
00:22:44,640 --> 00:22:46,270
pocket versus multiple
pockets issue.

453
00:22:46,270 --> 00:22:48,036
If you flip one of
them, that will not

454
00:22:48,036 --> 00:22:49,660
contain the convex
hull of the original

455
00:22:49,660 --> 00:22:51,940
because you haven't
flipped them all.

456
00:22:51,940 --> 00:22:54,300
If you flipped all the
pockets, than you would contain

457
00:22:54,300 --> 00:22:56,560
the convex hull of
the original, but that

458
00:22:56,560 --> 00:22:58,030
would end up arguing
that some flip

459
00:22:58,030 --> 00:23:01,030
sequences work, not all of them.

460
00:23:01,030 --> 00:23:04,010
So this is annoying,
and I think this

461
00:23:04,010 --> 00:23:07,310
is where I run out of slides.

462
00:23:07,310 --> 00:23:15,520
But we have some time, so I can
give you a sketch of the proof.

463
00:23:15,520 --> 00:23:19,660
This is the Bing and Kazarinoff
proof or our version of it.

464
00:23:22,200 --> 00:23:25,890
Give you an idea
of how this works.

465
00:23:25,890 --> 00:23:28,730
It's an easy proof, it's just
easy also to get it wrong

466
00:23:28,730 --> 00:23:31,080
or to skip one of the steps.

467
00:23:31,080 --> 00:23:34,281
I will just do a proof
by picture, I think.

468
00:23:36,990 --> 00:23:41,580
So suppose you have a
polygon on the plane,

469
00:23:41,580 --> 00:23:45,980
and let's say you flip a pocket.

470
00:23:45,980 --> 00:23:59,210
So this would look
something like that.

471
00:23:59,210 --> 00:24:02,850
A little hard to
do a reflection.

472
00:24:02,850 --> 00:24:04,360
So this is the
reflective polygon,

473
00:24:04,360 --> 00:24:09,850
and if we look at each of
the vertices over here,

474
00:24:09,850 --> 00:24:15,570
these guys don't move,
these guys moved over here,

475
00:24:15,570 --> 00:24:17,590
reflecting through that line.

476
00:24:17,590 --> 00:24:19,000
Observation one.

477
00:24:19,000 --> 00:24:26,340
If I take some point x
interior to the polygon,

478
00:24:26,340 --> 00:24:31,470
then these points that move
get farther away from x.

479
00:24:31,470 --> 00:24:32,330
Why?

480
00:24:32,330 --> 00:24:37,980
Because if you look
at this line here,

481
00:24:37,980 --> 00:24:40,830
let's say you look at a
vertex and where it goes.

482
00:24:40,830 --> 00:24:43,444
This line is the Voronoi
diagram of those two points.

483
00:24:43,444 --> 00:24:45,610
This is the perpendicular
bisector of this and this.

484
00:24:45,610 --> 00:24:47,580
That's the meaning
of reflection.

485
00:24:47,580 --> 00:24:49,580
So that means everything
to the left of the line

486
00:24:49,580 --> 00:24:52,332
here is closer to this
point than that point.

487
00:24:52,332 --> 00:24:53,790
Everything to the
right of the line

488
00:24:53,790 --> 00:24:55,960
is closer to this
point than that point.

489
00:24:55,960 --> 00:24:57,856
Now x, which is
interior to the polygon,

490
00:24:57,856 --> 00:25:00,230
must be to the left of the
line because the whole polygon

491
00:25:00,230 --> 00:25:02,190
is to the left of the line.

492
00:25:02,190 --> 00:25:05,705
So distance from x to
this vertex increases.

493
00:25:09,750 --> 00:25:12,180
I mean, x will remain
inside because as you flip,

494
00:25:12,180 --> 00:25:13,900
you only get bigger.

495
00:25:13,900 --> 00:25:15,970
So if I take a point x
and I look at the distance

496
00:25:15,970 --> 00:25:18,340
to some vertex, it can only
monotonically increase.

497
00:25:18,340 --> 00:25:20,390
It also can't get
arbitrarily large

498
00:25:20,390 --> 00:25:23,790
because the maximum
it could possibly be

499
00:25:23,790 --> 00:25:26,632
is half the perimeter
of the polygon.

500
00:25:26,632 --> 00:25:28,340
The perimeter of the
polygon is preserved

501
00:25:28,340 --> 00:25:31,260
so it can only stretch so far.

502
00:25:31,260 --> 00:25:35,170
So if you look at this distance,
it's monotonically increasing

503
00:25:35,170 --> 00:25:37,170
and it's bounded because
it can never get bigger

504
00:25:37,170 --> 00:25:38,727
than the perimeter
of the polygon.

505
00:25:38,727 --> 00:25:39,810
Therefore, it has a limit.

506
00:25:39,810 --> 00:25:41,340
That distance has
a limit because

507
00:25:41,340 --> 00:25:47,070
monotone bounded sequences
always have a unique limit.

508
00:25:47,070 --> 00:25:47,680
So cool.

509
00:25:47,680 --> 00:25:51,120
Distance from x to some
vertex has a limit.

510
00:25:51,120 --> 00:25:54,100
Well, I'm going to do this for
three different points, x, that

511
00:25:54,100 --> 00:25:58,330
lie on some non-degenerate
triangle, so not all on a line.

512
00:25:58,330 --> 00:26:00,960
That means the three
distances from these points

513
00:26:00,960 --> 00:26:04,150
to this vertex all
converge to some limit.

514
00:26:04,150 --> 00:26:06,210
And therefore, that
point converges

515
00:26:06,210 --> 00:26:08,300
to a limit, namely the
intersection of those three

516
00:26:08,300 --> 00:26:11,290
circles centered
at those points.

517
00:26:11,290 --> 00:26:12,890
So this proves that
the polygon has

518
00:26:12,890 --> 00:26:16,800
a limit because every
vertex has a limiting point.

519
00:26:16,800 --> 00:26:17,880
Cool.

520
00:26:17,880 --> 00:26:21,270
Now the tricky part is to
argue that limit is convex,

521
00:26:21,270 --> 00:26:24,140
and this is where
everyone had an issue.

522
00:26:27,880 --> 00:26:32,770
The first thing we argue is that
the angles converge to a limit.

523
00:26:32,770 --> 00:26:35,230
This is kind of a
technicality because we

524
00:26:35,230 --> 00:26:36,730
know the points
converge to a limit,

525
00:26:36,730 --> 00:26:39,390
so surely the angle does.

526
00:26:39,390 --> 00:26:43,476
The only issue is, well,
if all the points converge

527
00:26:43,476 --> 00:26:45,850
to the same point, then the
angle would not have a limit,

528
00:26:45,850 --> 00:26:47,510
but that's easy to
argue can't happen

529
00:26:47,510 --> 00:26:49,670
because these edge
links are preserved.

530
00:26:49,670 --> 00:26:53,059
So I will just skip that one.

531
00:26:53,059 --> 00:26:54,850
It follows from some
of the things we said.

532
00:26:59,310 --> 00:27:00,860
Now we get to the fun part.

533
00:27:00,860 --> 00:27:02,060
This polygon has a limit.

534
00:27:02,060 --> 00:27:04,380
The angles have limits.

535
00:27:04,380 --> 00:27:07,010
So I want to look at
the limiting angles.

536
00:27:07,010 --> 00:27:09,070
I want to in particular
look at the vertices that

537
00:27:09,070 --> 00:27:13,900
move because if you flip an
infinite number of times, that

538
00:27:13,900 --> 00:27:17,280
means some vertex must move
an infinite number of times,

539
00:27:17,280 --> 00:27:19,330
because every time you
flip, somebody moves.

540
00:27:19,330 --> 00:27:20,830
These guys are
considered not moving

541
00:27:20,830 --> 00:27:23,070
even though they're kind
of involved in the flip.

542
00:27:23,070 --> 00:27:28,470
So if you look at the moved
guys, what is their angle?

543
00:27:28,470 --> 00:27:31,380
So before, there
was some angle here,

544
00:27:31,380 --> 00:27:36,170
and afterwards, the interior
angle is the reverse.

545
00:27:36,170 --> 00:27:39,040
If this was reflex
before, it's convex now.

546
00:27:39,040 --> 00:27:42,990
If there were a convex
angle over here like this,

547
00:27:42,990 --> 00:27:48,220
it would become
reflex over here.

548
00:27:48,220 --> 00:27:51,450
You alternate between
being less than 180

549
00:27:51,450 --> 00:27:55,620
and greater than 180
every time you move.

550
00:27:55,620 --> 00:28:01,230
So if you have a vertex that is
moving infinitely many times,

551
00:28:01,230 --> 00:28:05,830
its angle must alternate
between less than 180 convex

552
00:28:05,830 --> 00:28:08,960
and greater than 180 reflex
infinitely many times.

553
00:28:08,960 --> 00:28:11,760
If that happens and you
have a unique limit angle,

554
00:28:11,760 --> 00:28:15,570
your limit angle must be 180.

555
00:28:15,570 --> 00:28:17,550
That's interesting.

556
00:28:17,550 --> 00:28:26,110
If our vertex moves
infinitely many times,

557
00:28:26,110 --> 00:28:42,317
then its limit angle
equals 180, must be flat.

558
00:28:42,317 --> 00:28:44,150
Very close to a
contradiction at this point.

559
00:28:49,090 --> 00:28:52,930
These guys are going to
end up looking like this.

560
00:28:52,930 --> 00:28:56,690
Well, let's look
at the other guys.

561
00:28:56,690 --> 00:28:58,425
So we have some limit polygon.

562
00:28:58,425 --> 00:29:00,550
In the limit polygon, there
are some flat vertices,

563
00:29:00,550 --> 00:29:03,020
but there must also be
some non-flat vertices.

564
00:29:03,020 --> 00:29:05,720
You can't just go straight
and hope to close a cycle.

565
00:29:05,720 --> 00:29:07,920
So there's maybe some
convex ones like this.

566
00:29:07,920 --> 00:29:11,350
There may be some reflex ones.

567
00:29:11,350 --> 00:29:14,570
At some point,
these vertices must

568
00:29:14,570 --> 00:29:19,300
stop moving because everyone
who moves infinitely many times

569
00:29:19,300 --> 00:29:20,890
has a limit angle flat.

570
00:29:20,890 --> 00:29:23,154
So anybody who is--
we call it "pointed"

571
00:29:23,154 --> 00:29:25,570
here, although it's a little
different from pointed pseudo

572
00:29:25,570 --> 00:29:26,810
triangulations.

573
00:29:26,810 --> 00:29:29,200
Anyone who's not
a flat angle must

574
00:29:29,200 --> 00:29:31,040
stop moving after finite time.

575
00:29:31,040 --> 00:29:34,810
So let's go to that time
when all of these guys

576
00:29:34,810 --> 00:29:35,680
have stopped moving.

577
00:29:40,020 --> 00:29:41,590
So your limit
polygon is something.

578
00:29:41,590 --> 00:29:43,960
We don't know whether
it's convex or whatever.

579
00:29:43,960 --> 00:29:45,830
It could have many flat angles.

580
00:29:45,830 --> 00:29:47,750
It's got to have at
least one of them

581
00:29:47,750 --> 00:29:50,590
if we assume there's
something infinite here.

582
00:29:50,590 --> 00:29:53,180
This is the limit polygon.

583
00:29:53,180 --> 00:29:55,880
Now, at finite time,
we know that these guys

584
00:29:55,880 --> 00:30:00,320
have stopped moving, meaning
we're done with those guys.

585
00:30:03,040 --> 00:30:07,170
I've drawn the limit here, but
also the finite thing, which

586
00:30:07,170 --> 00:30:08,910
must be on the inside, right?

587
00:30:08,910 --> 00:30:14,490
This must be
something like this.

588
00:30:14,490 --> 00:30:18,810
Somehow it's going to flip and
reach the infinitely many times

589
00:30:18,810 --> 00:30:21,980
and reach this in the limit.

590
00:30:21,980 --> 00:30:23,500
Hm.

591
00:30:23,500 --> 00:30:27,270
That looks weird.

592
00:30:27,270 --> 00:30:30,770
So the way to argue
this in the clean way

593
00:30:30,770 --> 00:30:36,210
is if you look at the
convex hull of the limit,

594
00:30:36,210 --> 00:30:39,355
let's say, convex hull
of the limit is this.

595
00:30:42,320 --> 00:30:45,120
And look at the convex
hull of this finite time

596
00:30:45,120 --> 00:30:47,750
when the squared vertices
have stopped moving.

597
00:30:47,750 --> 00:30:49,950
Convex hull will
be-- well, it's got

598
00:30:49,950 --> 00:30:52,650
to be at least this because
these guys are already there.

599
00:30:52,650 --> 00:30:55,666
The convex hull is defined
by the square points.

600
00:30:55,666 --> 00:30:57,290
You don't care about
the flat vertices.

601
00:30:57,290 --> 00:30:58,930
That won't affect
the convex hull.

602
00:30:58,930 --> 00:31:02,000
So that means the convex
hull equals the limit convex

603
00:31:02,000 --> 00:31:04,850
hull at this finite time.

604
00:31:04,850 --> 00:31:06,887
Now you're about to
do another flip, which

605
00:31:06,887 --> 00:31:08,720
means you're going to
go outside that convex

606
00:31:08,720 --> 00:31:12,500
hull and contradiction.

607
00:31:12,500 --> 00:31:13,920
Is that clear?

608
00:31:13,920 --> 00:31:16,430
Maybe go through that
part one more time.

609
00:31:16,430 --> 00:31:19,750
After finite time, when all
of the non-flat vertices

610
00:31:19,750 --> 00:31:23,400
have stopped moving, we have
reached the final convex hull.

611
00:31:23,400 --> 00:31:26,120
That means you can't do any more
flips because every flip makes

612
00:31:26,120 --> 00:31:28,440
the convex hull
bigger, so you actually

613
00:31:28,440 --> 00:31:30,706
had to stop at that time.

614
00:31:30,706 --> 00:31:31,830
That's kind of a fun proof.

615
00:31:31,830 --> 00:31:34,300
The key is really this
part, that if a vertex flips

616
00:31:34,300 --> 00:31:37,377
infinitely many times, then
that limit angle must be flat,

617
00:31:37,377 --> 00:31:39,710
and so they really don't
participate in the convex hull.

618
00:31:39,710 --> 00:31:43,170
This is the Bing and
Kazarinoff key idea.

619
00:31:43,170 --> 00:31:44,170
AUDIENCE: Two questions.

620
00:31:44,170 --> 00:31:47,470
Does anyone do this for 3D?

621
00:31:47,470 --> 00:31:49,320
PROFESSOR: 3D is a good.

622
00:31:49,320 --> 00:31:52,040
People have tried to
define flips for 3D,

623
00:31:52,040 --> 00:31:56,380
and I think there's never really
been a successful definition.

624
00:31:56,380 --> 00:31:58,030
AUDIENCE: [INAUDIBLE]?

625
00:31:58,030 --> 00:32:00,024
PROFESSOR: Yeah, and
exactly how to flip it.

626
00:32:00,024 --> 00:32:01,940
I mean, you can define
pocket in the same way,

627
00:32:01,940 --> 00:32:04,960
but then the boundary
won't be a single plane.

628
00:32:04,960 --> 00:32:09,180
It'll be some convex
cap, and so flipping, you

629
00:32:09,180 --> 00:32:11,100
can't really just reflect.

630
00:32:11,100 --> 00:32:13,820
It's kind of annoying.

631
00:32:13,820 --> 00:32:16,380
AUDIENCE: The
other question was,

632
00:32:16,380 --> 00:32:18,145
you have a sequence
of simple operations

633
00:32:18,145 --> 00:32:21,120
that takes you from
non-convex to convex.

634
00:32:21,120 --> 00:32:23,489
Does anybody use that
in a proof to say,

635
00:32:23,489 --> 00:32:25,530
these things are preserved
across the operations?

636
00:32:25,530 --> 00:32:27,805
You said perimeter
was preserved.

637
00:32:27,805 --> 00:32:32,000
What about shortest paths?

638
00:32:32,000 --> 00:32:34,080
PROFESSOR: So is this
useful for something?

639
00:32:34,080 --> 00:32:36,650
Shortest paths are
certainly not preserved.

640
00:32:36,650 --> 00:32:38,020
Edge lengths are preserved.

641
00:32:38,020 --> 00:32:40,478
Of course, we know how to do
that with the carpenter's rule

642
00:32:40,478 --> 00:32:42,900
theorem, just staying in 2D,
so it's not that exciting.

643
00:32:42,900 --> 00:32:46,594
But the operations are
definitely a lot simpler.

644
00:32:46,594 --> 00:32:48,260
I think the easy
answer to your question

645
00:32:48,260 --> 00:32:52,760
is there are many
natural, simple moves,

646
00:32:52,760 --> 00:32:54,810
and this is the first one
that people considered,

647
00:32:54,810 --> 00:32:56,268
but actually, there
are a lot more,

648
00:32:56,268 --> 00:32:58,300
and I'm going to
talk about those.

649
00:32:58,300 --> 00:33:03,110
The main application I know
for this stuff is basically,

650
00:33:03,110 --> 00:33:07,900
people wanted to generate random
closed walks in 3D typically,

651
00:33:07,900 --> 00:33:10,200
and so they wanted to find
a small set of operations

652
00:33:10,200 --> 00:33:12,190
they could just perform
randomly and hope

653
00:33:12,190 --> 00:33:14,220
that that was a rapidly
mixing Markov chain,

654
00:33:14,220 --> 00:33:16,600
so eventually, you'd have
a kind of random thing.

655
00:33:16,600 --> 00:33:18,830
I don't think there
are any rapid mixing

656
00:33:18,830 --> 00:33:20,410
results, at least
that I'm aware of.

657
00:33:23,020 --> 00:33:25,500
In order to hope to
get the space randomly,

658
00:33:25,500 --> 00:33:27,850
you would at least have to
be able to make anything.

659
00:33:27,850 --> 00:33:32,044
So for that, the question is,
can you convexify anything,

660
00:33:32,044 --> 00:33:34,210
because if you can convexify
anything, then at least

661
00:33:34,210 --> 00:33:37,000
you can make anything,
more or less.

662
00:33:37,000 --> 00:33:39,680
So that's where these
questions come from.

663
00:33:39,680 --> 00:33:41,240
We focus more on
the universality,

664
00:33:41,240 --> 00:33:43,750
but I think what people care
about is this random generation

665
00:33:43,750 --> 00:33:44,250
business.

666
00:33:46,704 --> 00:33:48,620
To that end, of course,
for random generation,

667
00:33:48,620 --> 00:33:50,670
you don't really
care about crossings.

668
00:33:50,670 --> 00:33:52,510
And so another fun
extension which

669
00:33:52,510 --> 00:33:54,880
was in, I think, our
paper for the first time,

670
00:33:54,880 --> 00:33:58,860
although there's a weaker
version in Grunbaum's paper,

671
00:33:58,860 --> 00:34:01,570
if you start with a
non-crossing, this proof

672
00:34:01,570 --> 00:34:03,200
still works.

673
00:34:03,200 --> 00:34:06,640
You need to add a little
more to the argument.

674
00:34:06,640 --> 00:34:10,199
Either you decrease the number
of crossings, which can only

675
00:34:10,199 --> 00:34:15,901
happen a finite number of times,
or this kind of stuff works.

676
00:34:15,901 --> 00:34:18,359
You can make crossing polygons,
too, which is kind of cool.

677
00:34:22,389 --> 00:34:24,989
So that's the end
of basic flips.

678
00:34:24,989 --> 00:34:31,620
Then we have something called a
flip turn, which is kind of fun

679
00:34:31,620 --> 00:34:34,040
and in some ways better behaved,
and gets to your question

680
00:34:34,040 --> 00:34:35,560
of what else is preserved.

681
00:34:35,560 --> 00:34:39,590
So let's say you have
a pocket like this.

682
00:34:39,590 --> 00:34:42,460
So normally with a
flip, we would reflect.

683
00:34:42,460 --> 00:34:46,679
With a flip turn, you reflect
and then also flip this way,

684
00:34:46,679 --> 00:34:54,810
so it's the same as rotating
180 degrees about the center.

685
00:34:54,810 --> 00:34:57,139
In addition to
preserving perimeter,

686
00:34:57,139 --> 00:35:00,310
this preserves the
edge directions.

687
00:35:00,310 --> 00:35:03,040
This matches this, this matches
this, this matches this.

688
00:35:03,040 --> 00:35:05,050
It does not preserve
the edge order, however.

689
00:35:05,050 --> 00:35:06,630
Here, we preserve
the edge order.

690
00:35:06,630 --> 00:35:09,970
Here, we've reversed the
order of those three edges.

691
00:35:09,970 --> 00:35:12,190
So all this is really
doing is permuting

692
00:35:12,190 --> 00:35:14,270
the sequence of edges.

693
00:35:14,270 --> 00:35:17,150
The edge directions and the
edge lengths are all the same.

694
00:35:17,150 --> 00:35:21,900
So this means you could make
at most n factorial moves here.

695
00:35:21,900 --> 00:35:24,220
Immediately, it's different
from this situation

696
00:35:24,220 --> 00:35:26,124
where even for n
equals 4, you could

697
00:35:26,124 --> 00:35:27,290
have arbitrarily many moves.

698
00:35:27,290 --> 00:35:28,760
Here, it's at most n factorial.

699
00:35:28,760 --> 00:35:30,320
That was the original bound.

700
00:35:30,320 --> 00:35:33,040
It turns out every
polygon convexifies

701
00:35:33,040 --> 00:35:37,640
after order n
squared flip turns.

702
00:35:37,640 --> 00:35:40,630
It's a fun proof but I don't
have time to cover it here.

703
00:35:43,150 --> 00:35:46,040
That's flip turns.

704
00:35:46,040 --> 00:35:50,550
Then we go to deflations.

705
00:35:50,550 --> 00:35:52,470
This is the inverse of a flip.

706
00:35:52,470 --> 00:35:56,070
So suppose you take a polygon
and you do an operation

707
00:35:56,070 --> 00:35:59,310
that, if flipped, would result
in the original polygon,

708
00:35:59,310 --> 00:36:01,960
exactly the opposite of a flip.

709
00:36:01,960 --> 00:36:05,420
This was conjectured to also
finish after finite time,

710
00:36:05,420 --> 00:36:06,715
but in fact it doesn't.

711
00:36:06,715 --> 00:36:10,880
If you take any quadrilateral
satisfying Kawasaki,

712
00:36:10,880 --> 00:36:13,110
so if you add up the
opposite edge lengths

713
00:36:13,110 --> 00:36:18,170
and they're equal-- so 6
plus 3 is 9, 4 plus 5 is 9--

714
00:36:18,170 --> 00:36:22,070
then there it is at least
a flat limit possibly.

715
00:36:22,070 --> 00:36:24,410
That's the Kawasaki thing.

716
00:36:24,410 --> 00:36:27,710
And in fact, it will
converge to that flat limit

717
00:36:27,710 --> 00:36:30,334
and it will take infinite
time to get there.

718
00:36:30,334 --> 00:36:31,250
So that's kind of fun.

719
00:36:31,250 --> 00:36:33,600
It gets very hard to
draw the pictures.

720
00:36:33,600 --> 00:36:37,950
This is really the main example
of an infinitely deflating

721
00:36:37,950 --> 00:36:38,770
polygon.

722
00:36:38,770 --> 00:36:43,710
We have a paper called
"Deflating the Pentagon," which

723
00:36:43,710 --> 00:36:50,050
got some fun political
views at some point.

724
00:36:50,050 --> 00:36:53,780
It's about a
pentagon, five sides.

725
00:36:53,780 --> 00:36:56,560
And essentially, unless you have
a flat vertex, in which case

726
00:36:56,560 --> 00:36:59,365
you are quadrilateral, there
is no infinitely deflating

727
00:36:59,365 --> 00:37:01,740
pentagon, which means you can
deflate the pentagon always

728
00:37:01,740 --> 00:37:02,620
in finite time.

729
00:37:05,130 --> 00:37:06,920
It is open for
hexagons and higher

730
00:37:06,920 --> 00:37:08,820
whether there is another
example different

731
00:37:08,820 --> 00:37:11,060
from the quadrilateral.

732
00:37:11,060 --> 00:37:12,950
Then there's the idea of a pop.

733
00:37:12,950 --> 00:37:15,500
This is an even simpler
operation than a flip.

734
00:37:15,500 --> 00:37:18,350
You just take two
edges, like here we're

735
00:37:18,350 --> 00:37:24,680
taking these two edges,
and, ignoring crossings,

736
00:37:24,680 --> 00:37:26,890
you just flip as if
that were a pocket lid,

737
00:37:26,890 --> 00:37:28,260
so if this were a pocket.

738
00:37:28,260 --> 00:37:29,380
Then you get this polygon.

739
00:37:29,380 --> 00:37:31,150
Now here, you can be
forced to get crossings.

740
00:37:31,150 --> 00:37:33,220
No matter how you flip,
you might get a crossing.

741
00:37:33,220 --> 00:37:36,320
Still, we're wondering--
who cares about crossings--

742
00:37:36,320 --> 00:37:39,840
are pops enough to make
anything, or to convexify?

743
00:37:39,840 --> 00:37:41,350
And the answer is no.

744
00:37:41,350 --> 00:37:43,750
There's this set of polygons
called alternating polygons

745
00:37:43,750 --> 00:37:46,040
where the vertices
alternate between the x-axis

746
00:37:46,040 --> 00:37:47,370
and the y-axis.

747
00:37:47,370 --> 00:37:50,320
And you can prove that no
matter what pop you do,

748
00:37:50,320 --> 00:37:53,270
you are still an
alternating polygon,

749
00:37:53,270 --> 00:37:55,270
and alternating polygons,
you can also prove,

750
00:37:55,270 --> 00:37:57,900
are never convex,
so you're stuck.

751
00:37:57,900 --> 00:38:03,280
This was open for many years but
finally solved a few years ago.

752
00:38:03,280 --> 00:38:07,690
There's also pop turns, which
is where you take two edges

753
00:38:07,690 --> 00:38:11,320
and you do a 180 degree
rotation like this.

754
00:38:11,320 --> 00:38:13,910
And there, we can prove
if you allow crossings,

755
00:38:13,910 --> 00:38:15,870
you can convexify any polygon.

756
00:38:15,870 --> 00:38:18,420
I don't have a figure of that
because it's just an algorithm.

757
00:38:18,420 --> 00:38:21,840
We haven't actually run
it on a nice example.

758
00:38:21,840 --> 00:38:24,560
If you avoid crossings, we can
characterize when it's possible

759
00:38:24,560 --> 00:38:26,160
and when it's impossible.

760
00:38:26,160 --> 00:38:28,557
And those are pretty much
all the simple operations

761
00:38:28,557 --> 00:38:29,890
that at least have been studied.

762
00:38:29,890 --> 00:38:34,330
There's probably more to
think about, but that's it,

763
00:38:34,330 --> 00:38:37,530
and that's the end of
my part of the class.